Let the equations of two adjacent sides of a parallelogram $$\\mathrm{ABCD}$$ be $$2 x-3 y=-23$$ and $$5 x+4 y=23$$. If the equation of its one diagonal $$\\mathrm{AC}$$ is $$3 x+7 y=23$$ and the distance of A from the other diagonal is $$\\mathrm{d}$$, then $$50 \\mathrm{~d}^{2}$$ is equal to ____________.
Let equation of $A B$ be $2 x-3 y=-23 \\ldots$ (i)\n
and equation of $B C$ be $5 x+4 y=23\\ldots$ (ii)\n
Equation of $A C$ is $3 x+7 y=23 \\ldots$ (iii)\n
Solving Eqs. (i) and (ii), we get\n
$x=-1$, and $y=7$\n
$\\therefore$ Co-ordinate of $B$ is $(-1,7)$\n
On solving Eqs. (ii) and (iii), we get\n
$$\nx=3, y=2\n$$\n
$\\therefore$ Co-ordinate of $C$ is $(3,2)$\n
On solving Eqs. (i) and (iii), we get $x=-4$ and $y=5$\n
$\\therefore$ Co-ordinate of $A$ is $(-4,5)$.\n
Let $E$ be the intersection point of diagonal co-ordinate of \n
$E$ is $\\left(\\frac{-4+3}{2}, \\frac{5+2}{2}\\right)$ or $\\left(-\\frac{1}{2}, \\frac{7}{2}\\right)$ \n
$\\because E$ is mid-point of $A C$\n
$$\n\\begin{aligned}\n& \\text { Equation of } B D \\text { is } y-7=\\left(\\frac{7-\\frac{7}{2}}{-1+\\frac{1}{2}}\\right)(x+1) \\\\\\\\\n& \\Rightarrow 7 x+y=0\n\\end{aligned}\n$$\n
Distance of $A$ from diagonal $B D=\\frac{|7 \\times(-4)+5|}{\\sqrt{7^2+1^2}}$\n
If the sum of squares of all real values of $$\\alpha$$, for which the lines $$2 x-y+3=0,6 x+3 y+1=0$$ and $$\\alpha x+2 y-2=0$$ do not form a triangle is $$p$$, then the greatest integer less than or equal to $$p$$ is _________.
Let $$A(a, b), B(3,4)$$ and $$C(-6,-8)$$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $$P(2 a+3,7 b+5)$$ from the line $$2 x+3 y-4=0$$ measured parallel to the line $$x-2 y-1=0$$ is
Let $$\\mathrm{A}$$ be the point of intersection of the lines $$3 x+2 y=14,5 x-y=6$$ and $$\\mathrm{B}$$ be the point of intersection of the lines $$4 x+3 y=8,6 x+y=5$$. The distance of the point $$P(5,-2)$$ from the line $$\\mathrm{AB}$$ is
Solving lines $$\\mathrm{L}_1(3 \\mathrm{x}+2 \\mathrm{y}=14)$$ and $$\\mathrm{L}_2(5 \\mathrm{x}-\\mathrm{y}=6)$$ to get $$\\mathrm{A}(2,4)$$ and solving lines $$\\mathrm{L}_3(4 \\mathrm{x}+3 \\mathrm{y}=8)$$ and $$\\mathrm{L}_4(6 \\mathrm{x}+\\mathrm{y}=5)$$ to get $$\\mathrm{B}\\left(\\frac{1}{2}, 2\\right)$$.
\n
Finding Eqn. of $$\\mathrm{AB}: 4 \\mathrm{x}-3 \\mathrm{y}+4=0$$
If $$x^2-y^2+2 h x y+2 g x+2 f y+c=0$$ is the locus of a point, which moves such that it is always equidistant from the lines $$x+2 y+7=0$$ and $$2 x-y+8=0$$, then the value of $$g+c+h-f$$ equals
Cocus of point $$\\mathrm{P}(\\mathrm{x}, \\mathrm{y})$$ whose distance from Gives\n$$X+2 y+7=0$$ & $$2 x-y+8=0$$ are equal is $$\\frac{x+2 y+7}{\\sqrt{5}}= \\pm \\frac{2 x-y+8}{\\sqrt{5}}$$
\n
$$(x+2 y+7)^2-(2 x-y+8)^2=0$$
\n
Combined equation of lines
\n
$$\\begin{aligned}\n& (x-3 y+1)(3 x+y+15)=0 \\\\\n& 3 x^2-3 y^2-8 x y+18 x-44 y+15=0 \\\\\n& x^2-y^2-\\frac{8}{3} x y+6 x-\\frac{44}{3} y+5=0 \\\\\n& x^2-y^2+2 h x y+2 g x 2+2 f y+c=0 \\\\\n& h=\\frac{4}{3}, g=3, f=-\\frac{22}{3}, c=5 \\\\\n& g+c+h-f=3+5-\\frac{4}{3}+\\frac{22}{3}=8+6=14\n\\end{aligned}$$
The vertices of a triangle are $$\\mathrm{A}(-1,3), \\mathrm{B}(-2,2)$$ and $$\\mathrm{C}(3,-1)$$. A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7508,
"subject": "General Science",
"question": "Consider the set of all lines px + qy + r = 0 such that 3p + 2q + 4r = 0. Which one of the following statements\nis true?\n",
"options": [
{
"text": "The lines are not concurrent"
},
{
"text": "The lines are concurrent at the point $$\\left( {{3 \\over 4},{1 \\over 2}} \\right)$$"
},
{
"text": "The lines are all parallel "
},
{
"text": "Each line passes through the origin"
}
],
"answer": "The lines are concurrent at the point $$\\left( {{3 \\over 4},{1 \\over 2}} \\right)$$",
"solution": "**Answer:** The lines are concurrent at the point $$\\left( {{3 \\over 4},{1 \\over 2}} \\right)$$\n\nEquation of lines;\n
x = $${3 \\over 4}$$ and y = $${2 \\over 4}$$ = $${1 \\over 2}$$\n
For any value of p,q and r, the equation of set of lines will pan through $$\\left( {{3 \\over 4},{1 \\over 2}} \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7509,
"subject": "General Science",
"question": "Locus of mid point of the portion between the axes of \n
$$ \\Rightarrow a\\sin t - b\\cos t = 3y$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7511,
"subject": "General Science",
"question": "If the equation of the locus of a point equidistant from the point $$\\left( {{a_{1,}}{b_1}} \\right)$$ and $$\\left( {{a_{2,}}{b_2}} \\right)$$ is \n $$\\left( {{a_1} - {a_2}} \\right)x + \\left( {{b_1} - {b_2}} \\right)y + c = 0$$ , then the value of $$'c'$$ is :",
"options": [
{
"text": "$$\\sqrt {{a_1}^2 + {b_1}^2 - {a_2}^2 - {b_2}^2} $$ "
},
{
"text": "$${1 \\over 2}\\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \\right)$$ "
},
{
"text": "$${{a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2}$$ "
},
{
"text": "$${1 \\over 2}\\left( {{a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2} \\right)$$."
}
],
"answer": "$${1 \\over 2}\\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \\right)$$ ",
"solution": "**Answer:** $${1 \\over 2}\\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \\right)$$ \n\nSince, the points $\\left(a_1, b_1\\right)$ and $\\left(a_2, b_2\\right)$ satisfy the equation. So, that\n
$$\n\\begin{aligned}\n& a_1\\left(a_1-a_2\\right)+b_1\\left(b_1-b_2\\right)+c=0 ~~........(1) \\\\\\\\\n& \\text { and } a_2\\left(a_1-a_2\\right)+b_2\\left(b_1-b_2\\right)+c=0 ~~.........(2) \\\\\\\\\n& \\text { On adding Eqs. (i) and (ii), we get } \\\\\\\\\n& \\left(a_1+a_2\\right)\\left(a_1-a_2\\right)+\\left(b_1+b_2\\right)\\left(b_1-b_2\\right)+2 c=0 \\\\\\\\\n& \\Rightarrow 2 c=-\\left(a_1^2-a_2^2+b_1^2-b_2^2\\right) \\\\\\\\\n& \\Rightarrow c=\\frac{1}{2}\\left(a_2^2+b_2^2-a_1^2-b_1^2\\right)\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7512,
"subject": "General Science",
"question": "Let $$A\\left( {2, - 3} \\right)$$ and $$B\\left( {-2, 1} \\right)$$ be vertices of a triangle $$ABC$$. If the centroid of this triangle moves on the line $$2x + 3y = 1$$, then the locus of the vertex $$C$$ is the line :",
"options": [
{
"text": "$$3x - 2y = 3$$"
},
{
"text": "$$2x - 3y = 7$$"
},
{
"text": "$$3x + 2y = 5$$"
},
{
"text": "$$2x + 3y = 9$$"
}
],
"answer": "$$2x + 3y = 9$$",
"solution": "**Answer:** $$2x + 3y = 9$$\n\nLet the vertex $$C$$ be $$(h,k),$$ then the \n
$$ \\Rightarrow $$ 5x = 7y",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7517,
"subject": "General Science",
"question": "A square ABCD has all its vertices on the curve x2y2 = 1. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is _________.",
"options": [],
"answer": "80",
"solution": "**Answer:** 80\n\nx2y2 = 1\n
Area $$ = 4\\left( {{1 \\over 2}} \\right)(OA)(OB) = 2{(OB)^2} = 4\\sqrt 5 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7518,
"subject": "General Science",
"question": "Let A be a fixed point (0, 6) and B be a moving point (2t, 0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is : ",
"options": [
{
"text": "3x2 $$-$$ 2y $$-$$ 6 = 0"
},
{
"text": "3x2 + 2y $$-$$ 6 = 0"
},
{
"text": "2x2 + 3y $$-$$ 9 = 0"
},
{
"text": "2x2 $$-$$ 3y + 9 = 0"
}
],
"answer": "2x2 + 3y $$-$$ 9 = 0",
"solution": "**Answer:** 2x2 + 3y $$-$$ 9 = 0\n\nA(0, 6) and B(2t, 0)
Perpendicular bisector of AB is
$$(y - 3) = {t \\over 3}(x - t)$$
So, $$C = \\left( {0,3 - {{{t^2}} \\over 3}} \\right)$$
A point $$P$$ moves so that the sum of squares of its distances from the points $$(1,2)$$ and $$(-2,1)$$ is 14. Let $$f(x, y)=0$$ be the locus of $$\\mathrm{P}$$, which intersects the $$x$$-axis at the points $$\\mathrm{A}$$, $$\\mathrm{B}$$ and the $$y$$-axis at the points C, D. Then the area of the quadrilateral ACBD is equal to :
If the point $$\\left(\\alpha, \\frac{7 \\sqrt{3}}{3}\\right)$$ lies on the curve traced by the mid-points of the line segments of the lines $$x \\cos \\theta+y \\sin \\theta=7, \\theta \\in\\left(0, \\frac{\\pi}{2}\\right)$$ between the co-ordinates axes, then $$\\alpha$$ is equal to :
Let $$\\mathrm{A}(-1,1)$$ and $$\\mathrm{B}(2,3)$$ be two points and $$\\mathrm{P}$$ be a variable point above the line $$\\mathrm{AB}$$ such that the area of $$\\triangle \\mathrm{PAB}$$ is 10. If the locus of $$\\mathrm{P}$$ is $$\\mathrm{a} x+\\mathrm{by}=15$$, then $$5 \\mathrm{a}+2 \\mathrm{~b}$$ is :
If the locus of the point, whose distances from the point $$(2,1)$$ and $$(1,3)$$ are in the ratio $$5: 4$$, is $$a x^2+b y^2+c x y+d x+e y+170=0$$, then the value of $$a^2+2 b+3 c+4 d+e$$ is equal to :
are perpendicular to each other for :",
"options": [
{
"text": "two values of $$a$$"
},
{
"text": "$$\\forall \\,a$$ "
},
{
"text": "for one value of $$a$$ "
},
{
"text": "for no values of $$a$$ "
}
],
"answer": "two values of $$a$$",
"solution": "**Answer:** two values of $$a$$\n\n$$3a + {a^2} - 2 = 0 \\Rightarrow {a^2} + 3a - 2 = 0;$$\n
$$ \\Rightarrow a = {{ - 3 \\pm \\sqrt {9 + 8} } \\over 2} = {{ - 3 \\pm \\sqrt {17} } \\over 2}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7525,
"subject": "General Science",
"question": "If the pair of straight lines $${x^2} - 2pxy - {y^2} = 0$$ and $${x^2} - 2qxy - {y^2} = 0$$ be such that each pair bisects the angle between the other pair, then :",
"options": [
{
"text": "$$pq = -1$$ "
},
{
"text": "$$p = q$$ "
},
{
"text": "$$p = -q$$ "
},
{
"text": "$$pq = 1$$."
}
],
"answer": "$$pq = -1$$ ",
"solution": "**Answer:** $$pq = -1$$ \n\nEquation of bisectors of second pair of straight lines is, \n
$$ \\Rightarrow pq = - 1.$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7526,
"subject": "General Science",
"question": "If the sum of the slopes of the lines given by $${x^2} - 2cxy - 7{y^2} = 0$$ is four times their product $$c$$ has the value :",
"options": [
{
"text": "$$-2$$ "
},
{
"text": "$$-1$$ "
},
{
"text": "$$2$$ "
},
{
"text": "$$1$$"
}
],
"answer": "$$2$$ ",
"solution": "**Answer:** $$2$$ \n\nLet the lines be $$y = {m_1}x$$ and $$y = {m_2}x$$ then\n
So, combined equation of angle bisector is $$ {x^2} + 3xy - {y^2} = 0$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7530,
"subject": "General Science",
"question": "The lines $\\mathrm{L}_1, \\mathrm{~L}_2, \\ldots, \\mathrm{L}_{20}$ are distinct. For $\\mathrm{n}=1,2,3, \\ldots, 10$ all the lines $\\mathrm{L}_{2 \\mathrm{n}-1}$ are parallel to each other and all the lines $L_{2 n}$ pass through a given point $P$. The maximum number of points of intersection of pairs of lines from the set $\\left\\{\\mathrm{L}_1, \\mathrm{~L}_2, \\ldots, \\mathrm{L}_{20}\\right\\}$ is equal to ___________.",
"options": [],
"answer": "101",
"solution": "**Answer:** 101\n\n
To find the maximum number of points of intersection of pairs of lines from the given set, we need to consider how the lines are arranged based on the given conditions.
Firstly, there are 10 lines (${L}_1, {L}_3, ..., {L}_{19}$) that are parallel to each other. Since parallel lines do not intersect with each other, these 10 lines will not contribute to the number of intersection points among themselves.
Secondly, there are 10 lines (${L}_2, {L}_4, ..., {L}_{20}$) that all pass through a given point $P$. Although these lines intersect at $P$, they only contribute one unique point of intersection to the total count.
To calculate the maximum number of intersection points, we need to consider the total number of ways to pick pairs of lines from the 20 lines available without restrictions, then subtract the combinations that do not result in intersections, which includes the combinations of parallel lines among themselves and the concurrent lines through point $P$.
This calculation is represented as:
$$Total = ^{20}C_2 - ^{10}C_2 - ^{10}C_2 + 1$$
Here, $^{20}C_2$ calculates the total number of ways to pick any two lines out of 20, which includes intersecting and non-intersecting lines. $^{10}C_2$ is subtracted twice: once for the set of parallel lines (${L}_1, {L}_3, ..., {L}_{19}$) that don't intersect among themselves and once more for the set of concurrent lines (${L}_2, {L}_4, ..., {L}_{20}$) intersecting only at point $P$. Since all the concurrent lines intersect at the same point, we add 1 back to include this intersection point.
Carrying out this calculation gives us the total number of distinct intersection points as $101$.
Let $$\\mathrm{R}$$ be the interior region between the lines $$3 x-y+1=0$$ and $$x+2 y-5=0$$ containing the origin. The set of all values of $$a$$, for which the points $$\\left(a^2, a+1\\right)$$ lie in $$R$$, is :
$$\\therefore\\,\\,\\,$$ P1 : P2 = 2 : $${1 \\over 2}$$ = 4 : 1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7540,
"subject": "General Science",
"question": "Two vertical poles AB = 15 m and CD = 10 m are standing apart on a horizontal ground with points A\nand C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m)\nabove the line AC is :",
"options": [
{
"text": "10/3"
},
{
"text": "5"
},
{
"text": "20/3"
},
{
"text": "6"
}
],
"answer": "6",
"solution": "**Answer:** 6\n\n\n Equation of AD : $$y = {{10} \\over a}x$$
Equation of BC : $${x \\over a} + {y \\over {15}} = 1$$
$$ \\therefore $$ y coordinate of point P = 6 = height of point P above the line AC.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7541,
"subject": "General Science",
"question": "
A ray of light passing through the point P(2, 3) reflects on the x-axis at point A and the reflected ray passes through the point Q(5, 4). Let R be the point that divides the line segment AQ internally into the ratio 2 : 1. Let the co-ordinates of the foot of the perpendicular M from R on the bisector of the angle PAQ be ($$\\alpha$$, $$\\beta$$). Then, the value of 7$$\\alpha$$ + 3$$\\beta$$ is equal to ____________.
The equations of two sides $$\\mathrm{AB}$$ and $$\\mathrm{AC}$$ of a triangle $$\\mathrm{ABC}$$ are $$4 x+y=14$$ and $$3 x-2 y=5$$, respectively. The point $$\\left(2,-\\frac{4}{3}\\right)$$ divides the third side $$\\mathrm{BC}$$ internally in the ratio $$2: 1$$, the equation of the side $$\\mathrm{BC}$$ is
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7543,
"subject": "General Science",
"question": "If $${x_1},{x_2},{x_3}$$ and $${y_1},{y_2},{y_3}$$ are both in G.P. with the same common ratio, then the points $$\\left( {{x_1},{y_1}} \\right),\\left( {{x_2},{y_2}} \\right)$$ and $$\\left( {{x_3},{y_3}} \\right)$$ :",
"options": [
{
"text": "are vertices of a triangle"
},
{
"text": "lie on a straight line "
},
{
"text": "lie on an ellipse "
},
{
"text": "lie on a circle "
}
],
"answer": "lie on a straight line ",
"solution": "**Answer:** lie on a straight line \n\nTaking co-ordinates as\n
Eliminating $$b$$ from $$(2)$$ and $$(3),$$ we get \n
$${a^2} - 4 = 0 \\Rightarrow a = \\pm 2 \\Rightarrow b = - 3$$, $$1$$\n
$$\\therefore$$ Equation of straight lines are\n
$${x \\over 2} + {y \\over { - 3}} = 1$$ \n
or $${x \\over { - 2}} + {y \\over 1} = 1$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7546,
"subject": "General Science",
"question": "The line parallel to the $$x$$ - axis and passing through the intersection of the lines $$ax + 2by + 3b = 0$$ and $$bx - 2ay - 3a = 0,$$ where $$(a, b)$$ $$ \\ne $$ $$(0, 0)$$ is :",
"options": [
{
"text": "below the $$x$$ - axis at a distance of $${3 \\over 2}$$ from it "
},
{
"text": "below the $$x$$ - axis at a distance of $${2 \\over 3}$$ from it "
},
{
"text": "above the $$x$$ - axis at a distance of $${3 \\over 2}$$ from it "
},
{
"text": "above the $$x$$ - axis at a distance of $${2 \\over 3}$$ from it "
}
],
"answer": "below the $$x$$ - axis at a distance of $${3 \\over 2}$$ from it ",
"solution": "**Answer:** below the $$x$$ - axis at a distance of $${3 \\over 2}$$ from it \n\nThe line passing through the intersection of lines \n
c $$-$$ 3 = $$ \\pm 2\\sqrt 6 $$ c = 3 $$ \\pm $$ 2$$\\sqrt 6 $$\n
Line can be x + y = 3 $$ \\pm $$ 2$$\\sqrt 6 $$\n
x + y = 3 $$-$$ 2$$\\sqrt 6 $$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7552,
"subject": "General Science",
"question": "A square, of each side 2, lies above the x-axis and has one vertex at the origin. If\none of the sides passing through the origin makes an angle 30o with the positive direction of the x-axis, then the sum of the x-coordinates of the vertices of the square is :\n",
"options": [
{
"text": "$$2\\sqrt 3 - 1$$ "
},
{
"text": "$$2\\sqrt 3 - 2$$"
},
{
"text": "$$\\sqrt 3 - 2$$"
},
{
"text": "$$\\sqrt 3 - 1$$"
}
],
"answer": "$$2\\sqrt 3 - 2$$",
"solution": "**Answer:** $$2\\sqrt 3 - 2$$\n\n\n
$$\\therefore\\,\\,\\,$$ Sum of the x - coordinate of the vertices \n
= 0 + $$\\sqrt 3 $$ + $$\\sqrt 3 $$ $$-$$ 1 + ($$-$$ 1) = 2$$\\sqrt 3 $$ $$-$$ 2",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 7553,
"subject": "General Science",
"question": "The sides of a rhombus ABCD are parallel to the lines, x $$-$$ y + 2 = 0 and 7x $$-$$ y + 3 = 0. If the diagonals of the rhombus intersect P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the coordinate of A is :",
"options": [
{
"text": "$${5 \\over 2}$$"
},
{
"text": "$${7 \\over 4}$$"
},
{
"text": "2"
},
{
"text": "$${7 \\over 2}$$"
}
],
"answer": "$${5 \\over 2}$$",
"solution": "**Answer:** $${5 \\over 2}$$\n\nLet the coordinate A be (0, c)\n
Equations of the given lines are \n
x $$-$$ y + 2 = 0 and 7x $$-$$ y + 3 = 0\n
We know that the diagonals of the rhombus will be parallel to the angle bisectors of the two given lines; y = x + 2 and y = 7x + 3\n
$$\\therefore\\,\\,\\,$$ equation of angle bisectors is given as : \n
$${{x - y + 2} \\over {\\sqrt 2 }} = \\pm {{7x - y + 3} \\over {5\\sqrt 2 }}$$\n
5x $$-$$ 5y + 10 = $$ \\pm $$ (7x $$-$$ y + 3)\n
$$\\therefore\\,\\,\\,$$ Parallel equations of the diagonals are 2x + 4y $$-$$ 7 = 0\n
and 12x $$-$$ 6y + 13 = 0\n
$$\\therefore\\,\\,\\,$$ slopes of diagonals are $${{ - 1} \\over 2}$$ and 2.\n
Now, slope of the diagonal from A(0, c) and passing through P(1, 2) is (2 $$-$$ c)\n
$$\\therefore\\,\\,\\,$$ 2 $$-$$ c = 2 $$ \\Rightarrow $$ c = 0 (not possible)\n
$$ \\therefore $$$$\\,\\,\\,$$ 2 $$-$$ c = $${{ - 1} \\over 2}$$ $$ \\Rightarrow $$ c = $${5 \\over 2}$$\n
$$\\therefore\\,\\,\\,$$ Coordinate of A is $${5 \\over 2}$$. ",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 7554,
"subject": "General Science",
"question": "A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and\nthe perpendicular from the origin to this line makes an angle of 60o with the line x + y = 0. Then an equation\nof the line L is :",
"options": [
{
"text": "x + $$\\sqrt 3 $$y = 8"
},
{
"text": "$$\\sqrt 3 $$x + y = 8"
},
{
"text": "( $$\\sqrt 3 $$ + 1)x + ( $$\\sqrt 3 $$ – 1)y = 8 $$\\sqrt 2 $$"
},
{
"text": "( $$\\sqrt 3 $$ - 1)x + ( $$\\sqrt 3 $$ + 1)y = 8 $$\\sqrt 2 $$"
}
],
"answer": "( $$\\sqrt 3 $$ - 1)x + ( $$\\sqrt 3 $$ + 1)y = 8 $$\\sqrt 2 $$",
"solution": "**Answer:** ( $$\\sqrt 3 $$ - 1)x + ( $$\\sqrt 3 $$ + 1)y = 8 $$\\sqrt 2 $$\n\n
\nThe equation of line is \nx cos $$\\theta $$ + y sin $$\\theta $$ = p
\n$$ \\Rightarrow $$ x cos (75o) + y sin (75o) = 4
equation of line is 4x $$-$$ 3y + 24 = 0",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7559,
"subject": "General Science",
"question": "If in a parallelogram ABDC, the coordinates of A, B and C are respectively (1, 2), (3, 4) and (2, 5), then the\nequation of the diagonal AD is :",
"options": [
{
"text": "5x + 3y – 11 = 0"
},
{
"text": "5x – 3y + 1 = 0"
},
{
"text": "3x – 5y + 7 = 0"
},
{
"text": "3x + 5y – 13 = 0"
}
],
"answer": "5x – 3y + 1 = 0",
"solution": "**Answer:** 5x – 3y + 1 = 0\n\nco-ordinates of point D are (4, 7)\n
$$ \\Rightarrow $$ line AD is 5x $$-$$ 3y + 1 = 0",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7560,
"subject": "General Science",
"question": "Two sides of a parallelogram are along the lines, x + y = 3 & x – y + 3 = 0. If its diagonals intersect at (2, 4), then one of its vertex is :",
"options": [
{
"text": "(2, 1)"
},
{
"text": "(2, 6)"
},
{
"text": "(3, 5)"
},
{
"text": "(3, 6)"
}
],
"answer": "(3, 6)",
"solution": "**Answer:** (3, 6)\n\n\n Solving \n
and $${{{x_1} + 0} \\over 2} = 2;\\,\\,{x_i} = 4$$\n
similarly y1 = 5\n
C $$ \\Rightarrow $$ (4, 5)\n
Now equation of BC is x $$-$$ y = $$-$$ 1\n
and equation of CD is x + y = 9\n
Solving x + y = 9 and x $$-$$ y = $$-$$ 3\n
Point D is (3, 6)",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 7561,
"subject": "General Science",
"question": "A point on the straight line, 3x + 5y = 15 which is equidistant from the coordinate axes will lie only\nin :",
"options": [
{
"text": "1st and 2nd qudratants"
},
{
"text": "4th qudratant"
},
{
"text": "1st and 2nd and 4th qudratants"
},
{
"text": "1st qudratant"
}
],
"answer": "1st and 2nd qudratants",
"solution": "**Answer:** 1st and 2nd qudratants\n\n\n Let the point is P(x, y).\n
According to the question, the point P(x, y) is equidistance from both x and y axis.\n
$$ \\therefore $$ |x| = |y|\n
$$ \\Rightarrow $$ x = $$ \\pm $$ y\n
So the point P lies on the either x = y or x = - y line. And point P(x, y) also lies on the straight line 3x + 5y = 15.\n
Form the graph, you can see the point P can either be on 1st qudratant or 2nd qudratant.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7562,
"subject": "General Science",
"question": "If the perpendicular bisector of the line segment joining the points P(1 ,4) and Q(k, 3) has y-intercept equal to –4, then a value of k is :",
"options": [
{
"text": "$$\\sqrt {14} $$"
},
{
"text": "-4"
},
{
"text": "–2 "
},
{
"text": "$$\\sqrt {15} $$"
}
],
"answer": "-4",
"solution": "**Answer:** -4\n\n$${m_{PQ}} = {{4 - 3} \\over {1 - k}} $$\n
$$ \\therefore $$ Slope of perpendicular bisector of PQ, $$ {m_ \\bot } = k - 1$$
mid point of PQ = $$\\left( {{{k + 1} \\over 2},{7 \\over 2}} \\right)$$
$${{{k^2} - 1} \\over 2} = {{15} \\over 2} \\Rightarrow k = \\pm 4$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7563,
"subject": "General Science",
"question": "A ray of light coming from the point (2, $$2\\sqrt 3 $$) is incident at an angle 30o on the line x = 1 at the\npoint A. The ray gets reflected on the line x = 1 and meets x-axis at the point B. Then, the line AB\npasses through the point :",
"options": [
{
"text": "(3, -$$\\sqrt 3 $$)"
},
{
"text": "(4, -$$\\sqrt 3 $$)"
},
{
"text": "$$\\left( {4, - {{\\sqrt 3 } \\over 2}} \\right)$$"
},
{
"text": "$$\\left( {3, - {1 \\over {\\sqrt 3 }}} \\right)$$"
}
],
"answer": "(3, -$$\\sqrt 3 $$)",
"solution": "**Answer:** (3, -$$\\sqrt 3 $$)\n\n\n
(3, -$$\\sqrt 3 $$) satisfy the line.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7564,
"subject": "General Science",
"question": "A man is walking on a straight line. The arithmetic mean\nof the reciprocals of the intercepts of this line on the\ncoordinate axes is $${1 \\over 4}$$. Three stones A, B and C are placed at the points\n(1, 1), (2, 2) and (4, 4) respectively. Then, which of these stones is / are on the path of the man?",
"options": [
{
"text": "A only"
},
{
"text": "All the three"
},
{
"text": "C only"
},
{
"text": "B only"
}
],
"answer": "B only",
"solution": "**Answer:** B only\n\nGiven, position of A = (1, 1)
Position of B = (2, 2)
Position of C = (4, 4)
Let x-intercept be a and y-intercept be b.
Equation of line traced is
$${x \\over a} + {y \\over b} = 1$$
This is the equation of path, let a point (h, k) lie on this path.
Then, $${h \\over a} + {k \\over b} = 1$$
Also, AM of reciprocal of a and b = $${1 \\over 4}$$
On comparing Eqs. (i) and (ii), we get (h, k) = (2, 2)
Hence, the required stone is B(2, 2).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7565,
"subject": "General Science",
"question": "The intersection of three lines x $$-$$ y = 0, x + 2y = 3 and 2x + y = 6 is a :",
"options": [
{
"text": "Right angled triangle"
},
{
"text": "Equilateral triangle"
},
{
"text": "None of the above"
},
{
"text": "Isosceles triangle"
}
],
"answer": "Isosceles triangle",
"solution": "**Answer:** Isosceles triangle\n\nThe given three lines are x $$-$$ y = 0, x + 2y = 3 and 2x + y = 6 then point of intersection,
lines x $$-$$ y = 0 and x + 2y = 3 is (1, 1)
lines x $$-$$ y = 0 and 2x + y = 6 is (2, 2)
and lines x + 2y = 3 and 2x + y = 0 is (3, 0)
The triangle ABC has vertices A(1, 1), B(2, 2) and C(3, 0)
$$ \\therefore $$ AB = $$\\sqrt 2 $$, BC = $$\\sqrt 5 $$ and AC = $$\\sqrt 5 $$
$$ \\therefore $$ $$\\Delta$$ABC is isosceles",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7566,
"subject": "General Science",
"question": "The maximum value of z in the following equation z = 6xy + y2, where 3x + 4y $$ \\le $$ 100 and 4x + 3y $$ \\le $$ 75 for x $$ \\ge $$ 0 and y $$ \\ge $$ 0 is __________.",
"options": [],
"answer": "904",
"solution": "**Answer:** 904\n\n\n\n 3x + 4y $$ \\le $$ 100
4x + 3y $$ \\le $$ 75
x $$ \\ge $$ 0, y $$ \\ge $$ 0
Feasible region is shown in the graph
Let maximum value of 6xy + y2 = c
For a solution with feasible region,
6xy + y2 = c and 4x + 3y = 75 must have at least one positive solution.
$$\\Rightarrow c \\le {{{{225}^2}} \\over {56}} \\approx 904$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7567,
"subject": "General Science",
"question": "The number of integral values of m so that the abscissa of point of intersection of lines 3x + 4y = 9 and y = mx + 1 is also an integer, is :",
"options": [
{
"text": "1"
},
{
"text": "2"
},
{
"text": "3"
},
{
"text": "0"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\n3x + 4(mx + 1) = 9
$$ \\therefore $$ Two integral value of m.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7568,
"subject": "General Science",
"question": "The point P (a, b) undergoes the following three transformations successively :
(a) reflection about the line y = x.
(b) translation through 2 units along the positive direction of x-axis.
(c) rotation through angle $${\\pi \\over 4}$$ about the origin in the anti-clockwise direction.
If the co-ordinates of the final position of the point P are $$\\left( { - {1 \\over {\\sqrt 2 }},{7 \\over {\\sqrt 2 }}} \\right)$$, then the value of 2a + b is equal to :",
"options": [
{
"text": "13"
},
{
"text": "9"
},
{
"text": "5"
},
{
"text": "7"
}
],
"answer": "9",
"solution": "**Answer:** 9\n\nImage of A(a, b) along y = x is B(b, a). Translating it 2 units it becomes C(b + 2, a).
$$\\Rightarrow$$ 2a + b = 9",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7569,
"subject": "General Science",
"question": "Two sides of a parallelogram are along the lines 4x + 5y = 0 and 7x + 2y = 0. If the equation of one of the diagonals of the parallelogram is 11x + 7y = 9, then other diagonal passes through the point :",
"options": [
{
"text": "(1, 2)"
},
{
"text": "(2, 2)"
},
{
"text": "(2, 1)"
},
{
"text": "(1, 3)"
}
],
"answer": "(2, 2)",
"solution": "**Answer:** (2, 2)\n\nBoth the lines pass through origin.
point D is equal to intersection of 4x + 5y = 0 & 11x + 7y = 9
So, coordinates of point $$D = \\left( {{5 \\over 3}, - {4 \\over 3}} \\right)$$
Also, point B is point of intersection of 7x + 2y = 0 and 11x + 7y = 9
So, coordinates of point $$B = \\left( { - {2 \\over 3},{7 \\over 3}} \\right)$$
diagonals of parallelogram intersect at middle let middle point of B, D
diagonal AC passes through (2, 2)",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7570,
"subject": "General Science",
"question": "
A line, with the slope greater than one, passes through the point $$A(4,3)$$ and intersects the line $$x-y-2=0$$ at the point B. If the length of the line segment $$A B$$ is $$\\frac{\\sqrt{29}}{3}$$, then $$B$$ also lies on the line :\n
Let $$m_{1}, m_{2}$$ be the slopes of two adjacent sides of a square of side a such that $$a^{2}+11 a+3\\left(m_{1}^{2}+m_{2}^{2}\\right)=220$$. If one vertex of the square is $$(10(\\cos \\alpha-\\sin \\alpha), 10(\\sin \\alpha+\\cos \\alpha))$$, where $$\\alpha \\in\\left(0, \\frac{\\pi}{2}\\right)$$ and the equation of one diagonal is $$(\\cos \\alpha-\\sin \\alpha) x+(\\sin \\alpha+\\cos \\alpha) y=10$$, then $$72\\left(\\sin ^{4} \\alpha+\\cos ^{4} \\alpha\\right)+a^{2}-3 a+13$$ is equal to :
A light ray emits from the origin making an angle 30$$^\\circ$$ with the positive $$x$$-axis. After getting reflected by the line $$x+y=1$$, if this ray intersects $$x$$-axis at Q, then the abscissa of Q is :
The straight lines $$\\mathrm{l_{1}}$$ and $$\\mathrm{l_{2}}$$ pass through the origin and trisect the line segment of the line L : $$9 x+5 y=45$$ between the axes. If $$\\mathrm{m}_{1}$$ and $$\\mathrm{m}_{2}$$ are the slopes of the lines $$\\mathrm{l_{1}}$$ and $$\\mathrm{l_{2}}$$, then the point of intersection of the line $$\\mathrm{y=\\left(m_{1}+m_{2}\\right)}x$$ with L lies on :
\n Also given that $m_1$ and $m_2$ are the slopes of the lines $l_1$ and $l_2$, respectively.\n
Let $A$ be the point of intersection of line $l_1$ and $L$, then
co-ordinates of $A$ are $\\left(\\frac{2 \\times 5+1 \\times 0}{2+1}, \\frac{2 \\times 0+1 \\times 9}{2+1}\\right)=\\left(\\frac{10}{3}, 3\\right)$\n
And let $B$ be the point of intersection of line $l_2$ and $L$, then
co-ordinates of $B$ are $$\n\\left(\\frac{1 \\times 5+2 \\times 0}{1+2}, \\frac{1 \\times 0+2 \\times 9}{1+2}\\right)=\\left(\\frac{5}{3}, 6\\right)\n$$\n
Now, slope of line $l_1,\\left(m_1\\right)=\\frac{3-0}{\\frac{10}{3}-0}=\\frac{9}{10}$\n
and slope of lines $l_2,\\left(m_2\\right)=\\frac{6-0}{\\frac{5}{3}-0}=\\frac{18}{5}$\n
$$\n\\begin{aligned}\n\\therefore \\text { line } y & =\\left(m_1+m_2\\right) x \\\\\\\\\n& =\\left(\\frac{9}{10}+\\frac{18}{5}\\right) x=\\frac{45}{10} x=\\frac{9}{2} x .......(ii)\n\\end{aligned}\n$$\n
Let $$A(-2,-1), B(1,0), C(\\alpha, \\beta)$$ and $$D(\\gamma, \\delta)$$ be the vertices of a parallelogram $$A B C D$$. If the point $$C$$ lies on $$2 x-y=5$$ and the point $$D$$ lies on $$3 x-2 y=6$$, then the value of $$|\\alpha+\\beta+\\gamma+\\delta|$$ is equal to ___________.
Let $$\\alpha, \\beta, \\gamma, \\delta \\in \\mathbb{Z}$$ and let $$A(\\alpha, \\beta), B(1,0), C(\\gamma, \\delta)$$ and $$D(1,2)$$ be the vertices of a parallelogram $$\\mathrm{ABCD}$$. If $$A B=\\sqrt{10}$$ and the points $$\\mathrm{A}$$ and $$\\mathrm{C}$$ lie on the line $$3 y=2 x+1$$, then $$2(\\alpha+\\beta+\\gamma+\\delta)$$ is equal to
A line passing through the point $$\\mathrm{A}(9,0)$$ makes an angle of $$30^{\\circ}$$ with the positive direction of $$x$$-axis. If this line is rotated about A through an angle of $$15^{\\circ}$$ in the clockwise direction, then its equation in the new position is :
A ray of light coming from the point $$\\mathrm{P}(1,2)$$ gets reflected from the point $$\\mathrm{Q}$$ on the $$x$$-axis and then passes through the point $$R(4,3)$$. If the point $$S(h, k)$$ is such that $$P Q R S$$ is a parallelogram, then $$hk^2$$ is equal to:
Let a variable line of slope $$m>0$$ passing through the point $$(4,-9)$$ intersect the coordinate axes at the points $$A$$ and $$B$$. The minimum value of the sum of the distances of $$A$$ and $$B$$ from the origin is
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7579,
"subject": "General Science",
"question": "If 0 $$ \\le $$ x < $${\\pi \\over 2}$$, then the number of values of x for which sin x $$-$$ sin 2x + sin 3x = 0, is :",
"options": [
{
"text": "3"
},
{
"text": "1"
},
{
"text": "4"
},
{
"text": "2"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\nsin x $$-$$ sin 2x + sin 3x = 0 $$x \\in \\left[ {0,{\\pi \\over 2}} \\right)$$\n
two solutions",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7580,
"subject": "General Science",
"question": "If $$\\sqrt 3 ({\\cos ^2}x) = (\\sqrt 3 - 1)\\cos x + 1$$, the number of solutions of the given equation when $$x \\in \\left[ {0,{\\pi \\over 2}} \\right]$$ is __________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n$$\\sqrt 3 ({\\cos ^2}x) = (\\sqrt 3 - 1)\\cos x + 1$$\n
$$ \\Rightarrow $$ $$\\sqrt 3 {\\cos ^2}x - \\sqrt 3 \\cos x + \\cos x - 1 = 0$$
$$ \\Rightarrow \\sqrt 3 \\cos x(\\cos x - 1) + (\\cos x - 1) = 0$$
$$ \\Rightarrow (\\cos x - 1)(\\sqrt 3 \\cos x + 1) = 0$$
$$\\cos x = 1$$
$$ \\Rightarrow x = 0$$ $$ [as x \\in \\left[ {0,{\\pi \\over 2}} \\right]$$]
and $$\\cos x = - {1 \\over {\\sqrt 3 }}$$ (not possible in $$x \\in \\left[ {0,{\\pi \\over 2}} \\right]$$]\n
$$ \\therefore $$ Number of solution = 1",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7581,
"subject": "General Science",
"question": "The number of solutions of the equation $4 \\sin ^2 x-4 \\cos ^3 x+9-4 \\cos x=0 ; x \\in[-2 \\pi, 2 \\pi]$ is :",
"options": [
{
"text": "0"
},
{
"text": "3"
},
{
"text": "1"
},
{
"text": "2"
}
],
"answer": "0",
"solution": "**Answer:** 0\n\n
We start by recognizing that $\n\\sin^2 x + \n\\cos^2 x = 1$. Substituting $\n\\sin^2 x = 1 - \n\\cos^2 x$ into the original equation gives:
\n
$$\n4(1 - \n\\cos^2 x) - 4\n\\cos^3 x + 9 - 4\n\\cos x = 0\n$$
\n
Rearranging and simplifying this equation, we have:
\n
$$\n4 - 4\n\\cos^2 x - 4\n\\cos^3 x + 9 - 4\n\\cos x = 0$$
\n
$$\n4\n\\cos^3 x + 4\n\\cos^2 x + 4\n\\cos x - 13 = 0$$
\n\n
$4 \\cos ^3 x+4 \\cos ^2 x+4 \\cos x=13$
\n
Observing the bounds given, $x \\in [-2\\pi, 2\\pi]$, we want to find how many solutions satisfy this cubic equation in terms of $\n\\cos x$. However, we note that the left-hand side (LHS) of the equation, representing a combination of cosines, could at most approach a maximum sum when $\n\\cos x = 1$, that being $4(1) + 4(1) + 4(1) = 12$. Yet, we have the equation set to equal 13, which is impossible given the maximum sum of the LHS can only be 12.
\n
The above reasoning indicates that, within the domain specified, there is no value of $x$ for which the equation holds true. Therefore, the number of solutions to the equation is zero.
If $$2 \\tan ^2 \\theta-5 \\sec \\theta=1$$ has exactly 7 solutions in the interval $$\\left[0, \\frac{n \\pi}{2}\\right]$$, for the least value of $$n \\in \\mathbf{N}$$, then $$\\sum_\\limits{k=1}^n \\frac{k}{2^k}$$ is equal to:
If $$\\alpha,-\\frac{\\pi}{2}<\\alpha<\\frac{\\pi}{2}$$ is the solution of $$4 \\cos \\theta+5 \\sin \\theta=1$$, then the value of $$\\tan \\alpha$$ is
If $$2 \\sin ^3 x+\\sin 2 x \\cos x+4 \\sin x-4=0$$ has exactly 3 solutions in the interval $$\\left[0, \\frac{\\mathrm{n} \\pi}{2}\\right], \\mathrm{n} \\in \\mathrm{N}$$, then the roots of the equation $$x^2+\\mathrm{n} x+(\\mathrm{n}-3)=0$$ belong to :
Let $$|\\cos \\theta \\cos (60-\\theta) \\cos (60+\\theta)| \\leq \\frac{1}{8}, \\theta \\epsilon[0,2 \\pi]$$. Then, the sum of all $$\\theta \\in[0,2 \\pi]$$, where $$\\cos 3 \\theta$$ attains its maximum value, is :\n
Let $$S=\\left\\{\\sin ^2 2 \\theta:\\left(\\sin ^4 \\theta+\\cos ^4 \\theta\\right) x^2+(\\sin 2 \\theta) x+\\left(\\sin ^6 \\theta+\\cos ^6 \\theta\\right)=0\\right.$$ has real roots $$\\}$$. If $$\\alpha$$ and $$\\beta$$ be the smallest and largest elements of the set $$S$$, respectively, then $$3\\left((\\alpha-2)^2+(\\beta-1)^2\\right)$$ equals __________.
If cot$$\\alpha$$ = 1 and sec$$\\beta$$ = $$ - {5 \\over 3}$$, where $$\\pi < \\alpha < {{3\\pi } \\over 2}$$ and $${\\pi \\over 2} < \\beta < \\pi $$, then the value of $$\\tan (\\alpha + \\beta )$$ and the quadrant in which $$\\alpha$$ + $$\\beta$$ lies, respectively are :
",
"options": [
{
"text": "$$ - {1 \\over 7}$$ and IVth quadrant"
},
{
"text": "7 and Ist quadrant"
},
{
"text": "$$-$$7 and IVth quadrant"
},
{
"text": "$$ {1 \\over 7}$$ and Ist quadrant"
}
],
"answer": "$$ - {1 \\over 7}$$ and IVth quadrant",
"solution": "**Answer:** $$ - {1 \\over 7}$$ and IVth quadrant\n\n$\\because \\cot \\alpha=1, \\quad \\alpha \\in\\left(\\pi, \\frac{3 \\pi}{2}\\right)$
then $\\tan \\alpha=1$
and $\\sec \\beta=-\\frac{5}{3}, \\quad \\beta \\in\\left(\\frac{\\pi}{2}, \\pi\\right)$
For $$\\alpha, \\beta \\in(0, \\pi / 2)$$, let $$3 \\sin (\\alpha+\\beta)=2 \\sin (\\alpha-\\beta)$$ and a real number $$k$$ be such that $$\\tan \\alpha=k \\tan \\beta$$. Then, the value of $$k$$ is equal to
= $${3 \\over 4}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7608,
"subject": "General Science",
"question": "The equation y = sinx sin (x + 2) – sin2\n (x + 1) represents a straight line lying in :\n",
"options": [
{
"text": "first, second and fourth quadrants"
},
{
"text": "first, third and fourth quadrants"
},
{
"text": "second and third quadrants only"
},
{
"text": "third and fourth quadrants only "
}
],
"answer": "third and fourth quadrants only ",
"solution": "**Answer:** third and fourth quadrants only \n\ny = sinx.sin(x+2) - sin2(x+1)
$$ \\Rightarrow 1 \\ge A \\ge {3 \\over 4}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7618,
"subject": "General Science",
"question": "If m and M are the minimum and the maximum values of\n
4 + $${1 \\over 2}$$ sin2 2x $$-$$ 2cos4 x, x $$ \\in $$ R, then M $$-$$ m is equal to :",
"options": [
{
"text": "$${{15} \\over 4}$$ "
},
{
"text": "$${{9} \\over 4}$$"
},
{
"text": "$${{7} \\over 4}$$"
},
{
"text": "$${{1} \\over 4}$$"
}
],
"answer": "$${{9} \\over 4}$$",
"solution": "**Answer:** $${{9} \\over 4}$$\n\nGiven,\n
ymax = $$\\sqrt {{{75} \\over 4} + {1 \\over 4}} $$ = $$\\sqrt {19} $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7620,
"subject": "General Science",
"question": "The number of integral values of 'k' for which the equation $$3\\sin x + 4\\cos x = k + 1$$ has a solution, k$$\\in$$R is ___________.",
"options": [],
"answer": "11",
"solution": "**Answer:** 11\n\nWe know,\n
$$ - \\sqrt {{a^2} + {b^2}} \\le a\\cos x + b\\sin x \\le \\sqrt {{a^2} + {b^2}} $$
$$ \\therefore $$ Set of integers = $$ - 6, - 5, - 4, - 3, - 2, - 1,0,1,2,3,4$$ = Total 11 intergers.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7621,
"subject": "General Science",
"question": "
The set of all values of $$\\lambda$$ for which the equation $${\\cos ^2}2x - 2{\\sin ^4}x - 2{\\cos ^2}x = \\lambda $$ has a real solution $$x$$, is :
$$\\cos ^2 2x - 2 \\sin ^4 x - 2 \\cos ^2 x = \\lambda$$\n\n
Using the trigonometric identities $$\\cos^2x = 1 - \\sin^2x$$ and $$\\cos^22x = 1 - 2\\sin^2x$$, we can rewrite the equation in terms of $$\\cos^2x$$:\n\n
$$\\lambda = 4 \\cos ^4 x - 4 \\cos ^2 x + 1 - 2(1 - 2 \\cos ^2 x + \\cos ^4 x) - 2 \\cos ^2 x$$\n
$$\\lambda = 2 \\cos ^4 x - 2 \\cos ^2 x - 1$$\n\n
We can factor out a 2 and rewrite this as :\n\n
$$\\lambda = 2(\\cos ^4 x - \\cos ^2 x - \\frac{1}{2})$$\n
$$\\lambda = 2[(\\cos ^2 x - \\frac{1}{2})^2 - \\frac{3}{4}]$$\n\n
So $\\lambda_{\\max }=2\\left[\\frac{1}{4}-\\frac{3}{4}\\right]=2 \\times\\left(\\frac{-2}{4}\\right)=-1$ (maximum value) and \n
$\\lambda_{\\min }=2\\left[0-\\frac{3}{4}\\right]=-\\frac{3}{2}($ minimum value $)$ \n
So range of the value of $x$ is $\\left[\\frac{-3}{2},-1\\right]$.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7622,
"subject": "General Science",
"question": "The value of
The value of $$36\\left(4 \\cos ^{2} 9^{\\circ}-1\\right)\\left(4 \\cos ^{2} 27^{\\circ}-1\\right)\\left(4 \\cos ^{2} 81^{\\circ}-1\\right)\\left(4 \\cos ^{2} 243^{\\circ}-1\\right)$$ is :
If the value of $$\\frac{3 \\cos 36^{\\circ}+5 \\sin 18^{\\circ}}{5 \\cos 36^{\\circ}-3 \\sin 18^{\\circ}}$$ is $$\\frac{a \\sqrt{5}-b}{c}$$, where $$a, b, c$$ are natural numbers and $$\\operatorname{gcd}(a, c)=1$$, then $$a+b+c$$ is equal to :
To find the value of $$\\frac{3 \\cos 36^{\\circ}+5 \\sin 18^{\\circ}}{5 \\cos 36^{\\circ}-3 \\sin 18^{\\circ}}$$ in the form $$\\frac{a \\sqrt{5}-b}{c}$$, we need to simplify the given expression. Let's start by using some fundamental trigonometric identities.
\n\n
We know that:
\n\n
$$\\cos 36^{\\circ} = \\frac{\\sqrt{5} + 1}{4}$$
\n\n
$$\\sin 18^{\\circ} = \\frac{\\sqrt{5} - 1}{4}$$
\n\n
First, substitute these values into the expression:
or $$\\,\\,\\,\\left| {\\overrightarrow {AD} } \\right| = \\sqrt {16 + 16 + 1} = \\sqrt {33} $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7643,
"subject": "General Science",
"question": "Let $$\\overrightarrow a ,\\overrightarrow b $$ and $$\\overrightarrow c $$ be three non-zero vectors such that no two of these are collinear. If the vector $$\\overrightarrow a + 2\\overrightarrow b $$ is collinear with $$\\overrightarrow c $$ and $$\\overrightarrow b + 3\\overrightarrow c $$ is collinear with $$\\overrightarrow a $$ ($$\\lambda $$ being some non-zero scalar) then $$\\overrightarrow a + 2\\overrightarrow b + 6\\overrightarrow c $$ equals to :",
"options": [
{
"text": "$\\overrightarrow{0}$"
},
{
"text": "$$\\lambda \\overrightarrow b $$ "
},
{
"text": "$$\\lambda \\overrightarrow c $$"
},
{
"text": "$$\\lambda \\overrightarrow a $$"
}
],
"answer": "$\\overrightarrow{0}$",
"solution": "**Answer:** $\\overrightarrow{0}$\n\nIf $\\overrightarrow{\\mathbf{a}}+2 \\overrightarrow{\\mathbf{b}}$ is collinear with $\\overrightarrow{\\mathbf{c}}$, then\n
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7645,
"subject": "General Science",
"question": "Let $$a, b$$ and $$c$$ be distinct non-negative numbers. If the vectors $$a\\widehat i + a\\widehat j + c\\widehat k,\\,\\,\\widehat i + \\widehat k$$ and $$c\\widehat i + c\\widehat j + b\\widehat k$$ lie in a plane, then $$c$$ is :",
"options": [
{
"text": "the Geometric Mean of $$a$$ and $$b$$"
},
{
"text": "the Arithmetic Mean of $$a$$ and $$b$$"
},
{
"text": "equal to zero "
},
{
"text": "the Harmonic Mean of $$a$$ and $$b$$"
}
],
"answer": "the Geometric Mean of $$a$$ and $$b$$",
"solution": "**Answer:** the Geometric Mean of $$a$$ and $$b$$\n\nVector $$a\\overrightarrow i + a\\overrightarrow j + c\\overrightarrow k ,\\,\\,\\overrightarrow i + \\overrightarrow k $$ \n
and $$c\\overrightarrow i + c\\overrightarrow j + b\\overrightarrow k $$ are coplanar\n
$$\\left| {\\matrix{\n a & a & c \\cr \n 1 & 0 & 1 \\cr \n c & c & b \\cr \n\n } } \\right| = 0 \\Rightarrow {c^2} = ab$$ \n
$$ \\Rightarrow c = \\sqrt {ab} $$\n
$$\\therefore$$ $$c$$ is $$G.M.$$ of $$a$$ and $$b.$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7646,
"subject": "General Science",
"question": "The non-zero vectors are $${\\overrightarrow a ,\\overrightarrow b }$$ and $${\\overrightarrow c }$$ are related by $${\\overrightarrow a = 8\\overrightarrow b }$$ and $${\\overrightarrow c = - 7\\overrightarrow b \\,\\,.}$$ Then the angle between $${\\overrightarrow a }$$ and $${\\overrightarrow c }$$ is :",
"options": [
{
"text": "$$0$$ "
},
{
"text": "$${\\pi \\over 4}$$"
},
{
"text": "$${\\pi \\over 2}$$ "
},
{
"text": "$$\\pi $$ "
}
],
"answer": "$$\\pi $$ ",
"solution": "**Answer:** $$\\pi $$ \n\nClearly $$\\overrightarrow a = - {8 \\over 7}\\overrightarrow c $$\n
$$ \\Rightarrow \\overrightarrow a ||\\overrightarrow c $$ and are opposite in direction \n
$$\\therefore$$ Angle between $$\\overrightarrow a $$ and $$\\overrightarrow c $$ is $$\\pi .$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7647,
"subject": "General Science",
"question": "Let $$\\overrightarrow a $$, $$\\overrightarrow b $$, $$\\overrightarrow c $$ be three non-zero vectors which are pairwise non-collinear. If $\\overrightarrow a+3 \\overrightarrow b$ is collinear with $\\overrightarrow c$ and $\\overrightarrow b+2 \\overrightarrow c$ is collinear with $\\overrightarrow a$, then $\\overrightarrow a+\\overrightarrow b+6 \\overrightarrow c$ is :",
"options": [
{
"text": "$\\overrightarrow a+\\overrightarrow c$"
},
{
"text": "$\\overrightarrow c$"
},
{
"text": "$\\overrightarrow a$"
},
{
"text": "$\\overrightarrow 0$"
}
],
"answer": "$\\overrightarrow 0$",
"solution": "**Answer:** $\\overrightarrow 0$\n\n
We are given that $\\overrightarrow a + 3 \\overrightarrow b$ is collinear with $\\overrightarrow c$, and $\\overrightarrow b + 2 \\overrightarrow c$ is collinear with $\\overrightarrow a$. This means we can write:
\n\n
$\\overrightarrow a + 3 \\overrightarrow b = \\lambda \\overrightarrow c \\quad ...(i)$
\n
$\\overrightarrow b + 2 \\overrightarrow c = \\mu \\overrightarrow a \\quad ...(ii)$
\n\n
for some scalars $\\lambda$ and $\\mu$.
\n
We are trying to find $\\overrightarrow a + \\overrightarrow b + 6\\overrightarrow c$ in terms of $\\overrightarrow a$, $\\overrightarrow b$, and $\\overrightarrow c$. We can also express this as :
\n
$\\overrightarrow a + 3 \\overrightarrow b + 6\\overrightarrow c = (\\lambda + 6) \\overrightarrow c \\quad ...(iii)$
\n
by adding $6\\overrightarrow c$ to both sides of equation (i).
\n
Now, from equation (ii), multiplying by 3 gives us :
\n
$3\\overrightarrow b + 6 \\overrightarrow c = 3\\mu \\overrightarrow a \\quad ...(iv)$
\n
Adding $\\overrightarrow a$ to both sides of equation (iv) gives :
\n
$\\overrightarrow a + 3 \\overrightarrow b + 6\\overrightarrow c = (1 + 3\\mu) \\overrightarrow a \\quad ...(v)$
\n
Now, we have two expressions for $\\overrightarrow a + 3 \\overrightarrow b + 6\\overrightarrow c$, one in terms of $\\overrightarrow c$ (from equation iii) and one in terms of $\\overrightarrow a$ (from equation v). Setting these equal to each other gives :
\n
$(\\lambda + 6) \\overrightarrow c = (1 + 3\\mu) \\overrightarrow a \\quad ...(vi)$
\n
Since $\\overrightarrow a$ and $\\overrightarrow c$ are not collinear, this equation can only hold if the coefficients on both sides are zero, hence :
\n
$\\lambda + 6 = 0$ and $1 + 3\\mu = 0$
\n
This gives $\\lambda = -6$ and $\\mu = -\\frac{1}{3}$.
\n
Finally, substituting $\\lambda = -6$ into equation (iii) gives :
\n
$\\overrightarrow a + 3 \\overrightarrow b + 6\\overrightarrow c = 0$
\n
So, $\\overrightarrow a + \\overrightarrow b + 6\\overrightarrow c = \\overrightarrow 0$.
\n
Therefore, the correct answer is Option D : $\\overrightarrow 0$.
\n",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7648,
"subject": "General Science",
"question": "If the vectors $$\\overrightarrow {AB} = 3\\widehat i + 4\\widehat k$$ and $$\\overrightarrow {AC} = 5\\widehat i - 2\\widehat j + 4\\widehat k$$ are the sides of a triangle $$ABC,$$ then the length of the median through $$A$$ is :",
"options": [
{
"text": "$$\\sqrt {18} $$ "
},
{
"text": "$$\\sqrt {72} $$"
},
{
"text": "$$\\sqrt {33} $$"
},
{
"text": "$$\\sqrt {45} $$ "
}
],
"answer": "$$\\sqrt {33} $$",
"solution": "**Answer:** $$\\sqrt {33} $$\n\nAs $$M$$ is mid point of $$BC$$ \n
$$ = 4\\overrightarrow i + \\overrightarrow j + 4\\overrightarrow k $$\n
Length of median $$AM$$ \n
$$ = \\sqrt {16 + 1 + 16} = \\sqrt {33} $$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7649,
"subject": "General Science",
"question": "Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are $$\\overrightarrow a ,\\overrightarrow b ,\\overrightarrow c $$ and $${{\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\over 4}$$ respectively, then the position vector of the orthocentre of this triangle, is :",
"options": [
{
"text": "$${\\overrightarrow a + \\overrightarrow b + \\overrightarrow c }$$ "
},
{
"text": "$$ - \\left( {{{\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\over 2}} \\right)$$"
},
{
"text": "$$\\overrightarrow 0 $$ "
},
{
"text": "$$\\left( {{{\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\over 2}} \\right)$$ "
}
],
"answer": "$$\\left( {{{\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\over 2}} \\right)$$ ",
"solution": "**Answer:** $$\\left( {{{\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\over 2}} \\right)$$ \n\n\n
Given,\n
Position vector of circumcentre, $$\\overrightarrow C = {{\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\over 4}$$\n
We know, position vector of centroid, $$\\overrightarrow G = {{\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\over 3}$$\n
Now, let $$\\overrightarrow R $$ be the orthocentre of the triangle.\n
We know, $$\\overrightarrow G $$ $$ = {{2\\overrightarrow C + \\overrightarrow R } \\over 3}$$\n
$$ \\Rightarrow $$ 3$$\\overrightarrow G $$ $$ = 2\\overrightarrow C + \\overrightarrow R $$\n
$$ \\Rightarrow $$ $$\\overrightarrow R = 3\\overrightarrow G - 2\\overrightarrow C $$\n
= $$\\left( {\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\right) - 2\\left( {{{\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\over 4}} \\right)$$\n
= $${{\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\over 2}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7650,
"subject": "General Science",
"question": "If the position vectors of the vertices A, B and C of a $$\\Delta $$ ABC are respectively $$4\\widehat i + 7\\widehat j + 8\\widehat k,$$ $$2\\widehat i + 3\\widehat j + 4\\widehat k,$$ and $$2\\widehat i + 5\\widehat j + 7\\widehat k,$$ then the position vectors of the point, where the bisector of $$\\angle $$A meets BC is :",
"options": [
{
"text": "$${1 \\over 2}\\left( {4\\widehat i + 8\\widehat j + 11\\widehat k} \\right)$$"
},
{
"text": "$${1 \\over 3}\\left( {6\\widehat i + 11\\widehat j + 15\\widehat k} \\right)$$"
},
{
"text": "$${1 \\over 3}\\left( {6\\widehat i + 13\\widehat j + 18\\widehat k} \\right)$$"
},
{
"text": "$${1 \\over 4}\\left( {8\\widehat i + 14\\widehat j + 19\\widehat k} \\right)$$"
}
],
"answer": "$${1 \\over 3}\\left( {6\\widehat i + 13\\widehat j + 18\\widehat k} \\right)$$",
"solution": "**Answer:** $${1 \\over 3}\\left( {6\\widehat i + 13\\widehat j + 18\\widehat k} \\right)$$\n\nSuppose angular bisector of A meets BC at D(x, , z) \n
Therefore, position vector of point p = $${1 \\over 3}$$ (6i + 13j + 18k)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7651,
"subject": "General Science",
"question": "If $$\\overrightarrow \\alpha $$ = $$\\left( {\\lambda - 2} \\right)\\overrightarrow a + \\overrightarrow b $$ and $$\\overrightarrow \\beta = \\left( {4\\lambda - 2} \\right)\\overrightarrow a + 3\\overrightarrow b $$ be two given vectors $$\\overrightarrow a $$ and $$\\overrightarrow b $$ are non-collinear. The value of $$\\lambda $$ for which vectors $$\\overrightarrow \\alpha $$ and $$\\overrightarrow \\beta $$ are collinear, is -",
"options": [
{
"text": "4"
},
{
"text": "3"
},
{
"text": "$$-$$3"
},
{
"text": "$$-$$4"
}
],
"answer": "$$-$$4",
"solution": "**Answer:** $$-$$4\n\n$$\\overrightarrow \\alpha = \\left( {\\lambda - 2} \\right)\\overrightarrow \\alpha + \\overrightarrow b $$\n
$$\\lambda = - 4$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7652,
"subject": "General Science",
"question": "The lines\n $$\\overrightarrow r = \\left( {\\widehat i - \\widehat j} \\right) + l\\left( {2\\widehat i + \\widehat k} \\right)$$ and \n $$\\overrightarrow r = \\left( {2\\widehat i - \\widehat j} \\right) + m\\left( {\\widehat i + \\widehat j + \\widehat k} \\right)$$",
"options": [
{
"text": "do not intersect for any values of $$l$$ and m"
},
{
"text": "intersect for all values of $$l$$ and m"
},
{
"text": "intersect when $$l$$ = 2 and m = $${1 \\over 2}$$"
},
{
"text": "intersect when $$l$$ = 1 and m = 2"
}
],
"answer": "do not intersect for any values of $$l$$ and m",
"solution": "**Answer:** do not intersect for any values of $$l$$ and m\n\nL1 = $$\\overrightarrow r = \\left( {\\widehat i - \\widehat j} \\right) + l\\left( {2\\widehat i + \\widehat k} \\right)$$\n
Equating coefficient of $$\\widehat i$$, $$\\widehat j$$ and $$\\widehat k$$ of L1\n and L2\n
2l + 1 = m + 2 ... (1)\n
–1 = –1 + m ...(2)\n
l = –m ...(3)\n
from (ii) m = 0\n
from (iii) $$l$$ = 0\n
These values of m and $$l$$ do not satisfy equation (1).\n
Hence the two lines do not intersect for any values of $$l$$ and m.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7653,
"subject": "General Science",
"question": "If $$\\overrightarrow a $$\nand $$\\overrightarrow b $$\nare unit vectors, then the greatest value of\n
$$\\sqrt 3 \\left| {\\overrightarrow a + \\overrightarrow b } \\right| + \\left| {\\overrightarrow a - \\overrightarrow b } \\right|$$ is_____.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nLet angle between $$\\overrightarrow a $$ and $$\\overrightarrow b $$\n be $$\\theta $$.\n
$$\\sqrt 3 \\left| {\\overrightarrow a + \\overrightarrow b } \\right| + \\left| {\\overrightarrow a - \\overrightarrow b } \\right|$$\n
Let $${\\lambda ^2} = 1$$, possible unit vector $$ = {{\\widehat i - \\widehat j + \\widehat k} \\over {\\sqrt 3 }}$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7655,
"subject": "General Science",
"question": "Let a vector $$\\alpha \\widehat i + \\beta \\widehat j$$ be obtained by rotating the vector $$\\sqrt 3 \\widehat i + \\widehat j$$ by an angle 45$$^\\circ$$ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices ($$\\alpha$$, $$\\beta$$), (0, $$\\beta$$) and (0, 0) is equal to :",
"options": [
{
"text": "$${1 \\over {\\sqrt 2 }}$$"
},
{
"text": "$${1 \\over 2}$$"
},
{
"text": "1"
},
{
"text": "2$${\\sqrt 2 }$$"
}
],
"answer": "$${1 \\over 2}$$",
"solution": "**Answer:** $${1 \\over 2}$$\n\n\n ($$\\alpha$$, $$\\beta$$) $$ \\equiv $$ (2 cos 75$$^\\circ$$, 2 sin 75$$^\\circ$$)
Area = $${1 \\over 2}$$ (2 cos 75$$^\\circ$$) (2 sin 75$$^\\circ$$)
= sin(150$$^\\circ$$) = $${1 \\over 2}$$ square unit",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7656,
"subject": "General Science",
"question": "A vector $$\\overrightarrow a $$ has components 3p and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to new system, $$\\overrightarrow a $$ has components p + 1 and $$\\sqrt {10} $$, then the value of p is equal to :",
"options": [
{
"text": "1"
},
{
"text": "$$ - {5 \\over 4}$$"
},
{
"text": "$${4 \\over 5}$$"
},
{
"text": "$$-$$1"
}
],
"answer": "$$-$$1",
"solution": "**Answer:** $$-$$1\n\n$${\\left| {\\overrightarrow a } \\right|_{old}} = {\\left| {\\overrightarrow a } \\right|_{new}}$$
$$ \\Rightarrow $$ p = $${5 \\over 4}$$, $$-$$ 1",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7657,
"subject": "General Science",
"question": "Let a, b and c be distinct positive numbers. If the vectors $$a\\widehat i + a\\widehat j + c\\widehat k,\\widehat i+\\widehat k$$ and $$c\\widehat i + c\\widehat j + b\\widehat k$$ are co-planar, then c is equal to :",
"options": [
{
"text": "$${2 \\over {{1 \\over a} + {1 \\over b}}}$$"
},
{
"text": "$${{a + b} \\over 2}$$"
},
{
"text": "$${1 \\over a} + {1 \\over b}$$"
},
{
"text": "$$\\sqrt {ab} $$"
}
],
"answer": "$$\\sqrt {ab} $$",
"solution": "**Answer:** $$\\sqrt {ab} $$\n\nBecause vectors are coplanar
Hence, $$\\left| {\\matrix{\n a & a & c \\cr \n 1 & 0 & 1 \\cr \n c & c & b \\cr \n\n } } \\right| = 0$$
$$ \\Rightarrow {c^2} = ab \\Rightarrow c = \\sqrt {ab} $$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7658,
"subject": "General Science",
"question": "Let A, B, C be three points whose position vectors respectively are\n
$$\\overrightarrow a = \\widehat i + 4\\widehat j + 3\\widehat k$$
\n
$$\\overrightarrow b = 2\\widehat i + \\alpha \\widehat j + 4\\widehat k,\\,\\alpha \\in R$$
\n
$$\\overrightarrow c = 3\\widehat i - 2\\widehat j + 5\\widehat k$$
\n
If $$\\alpha$$ is the smallest positive integer for which $$\\overrightarrow a ,\\,\\overrightarrow b ,\\,\\overrightarrow c $$ are noncollinear, then the length of the median, in $$\\Delta$$ABC, through A is :
Let the vectors $$\\vec{a}=(1+t) \\hat{i}+(1-t) \\hat{j}+\\hat{k}, \\vec{b}=(1-t) \\hat{i}+(1+t) \\hat{j}+2 \\hat{k}$$ and $$\\vec{c}=t \\hat{i}-t \\hat{j}+\\hat{k}, t \\in \\mathbf{R}$$ be such that for $$\\alpha, \\beta, \\gamma \\in \\mathbf{R}, \\alpha \\vec{a}+\\beta \\vec{b}+\\gamma \\vec{c}=\\overrightarrow{0} \\Rightarrow \\alpha=\\beta=\\gamma=0$$. Then, the set of all values of $$t$$ is :
",
"options": [
{
"text": "a non-empty finite set"
},
{
"text": "equal to $$\\mathbf{N}$$"
},
{
"text": "equal to $$\\mathbf{R}-\\{0\\}$$"
},
{
"text": "equal to $$\\mathbf{R}$$"
}
],
"answer": "equal to $$\\mathbf{R}-\\{0\\}$$",
"solution": "**Answer:** equal to $$\\mathbf{R}-\\{0\\}$$\n\n
Clearly $$\\overrightarrow a $$, $$\\overrightarrow b $$, $$\\overrightarrow c $$ are non-coplanar
For any vector $$\\vec{a}=a_{1} \\hat{i}+a_{2} \\hat{j}+a_{3} \\hat{k}$$, with $$10\\left|a_{i}\\right|<1, i=1,2,3$$, consider the following statements :
",
"options": [
{
"text": "Only (B) is true"
},
{
"text": "Only (A) is true"
},
{
"text": "Neither (A) nor (B) is true"
},
{
"text": "Both (A) and (B) are true"
}
],
"answer": "Both (A) and (B) are true",
"solution": "**Answer:** Both (A) and (B) are true\n\nWe have,\n
Hence, (B) is also true.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7663,
"subject": "General Science",
"question": "
If the points $$\\mathrm{P}$$ and $$\\mathrm{Q}$$ are respectively the circumcenter and the orthocentre of a $$\\triangle \\mathrm{ABC}$$, then $$\\overrightarrow{\\mathrm{PA}}+\\overrightarrow{\\mathrm{PB}}+\\overrightarrow{\\mathrm{PC}}$$ is\nequal to :
The circumcenter of a triangle is equidistant from the vertices of the triangle. It is the center of the circumcircle, the circle that passes through all three vertices of the triangle.\n\n
2. **Orthocenter $ Q $**:\n
The orthocenter of a triangle is the point of intersection of its three altitudes. An altitude of a triangle is a perpendicular line segment drawn from a vertex to its opposite side (or its extension).\n\n
3. **Centroid $ G $**:\n
The centroid of a triangle is the point of intersection of its medians. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. The centroid always divides each median in a 2:1 ratio, with the larger segment being closer to the vertex.\n\n
With the above definitions, it's known that the centroid divides the line segment joining the circumcenter and orthocenter in the ratio 2 : 1, meaning :\n
An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If $$\\overrightarrow {OP} = \\overrightarrow u ,\\overrightarrow {OR} = \\overrightarrow v $$, and $$\\overrightarrow {OQ} = \\alpha \\overrightarrow u + \\beta \\overrightarrow v $$, then $$\\alpha ,{\\beta ^2}$$ are the roots of the equation :
",
"options": [
{
"text": "$${x^2} + x - 2 = 0$$"
},
{
"text": "$$3{x^2} + 2x - 1 = 0$$"
},
{
"text": "$$3{x^2} - 2x - 1 = 0$$"
},
{
"text": "$${x^2} - x - 2 = 0$$"
}
],
"answer": "$${x^2} - x - 2 = 0$$",
"solution": "**Answer:** $${x^2} - x - 2 = 0$$\n\nAn arc $P Q$ of a circle subtends a right angle at its centre $O$. The mid-point of an $\\operatorname{arc} P Q$ is $R$. So, $P R=R Q$ \n
Also given that, $$\\overrightarrow {OP} = \\overrightarrow u ,\\overrightarrow {OR} = \\overrightarrow v $$, and $$\\overrightarrow {OQ} = \\alpha \\overrightarrow u + \\beta \\overrightarrow v $$\n
\n
Let $$\\overrightarrow {OP} = \\overrightarrow u$$\n
Now, $\\alpha$ and $\\beta^2$ are roots of the quadratic equation is $x^2-x-2=0$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7665,
"subject": "General Science",
"question": "
If the points with position vectors $$\\alpha \\hat{i}+10 \\hat{j}+13 \\hat{k}, 6 \\hat{i}+11 \\hat{j}+11 \\hat{k}, \\frac{9}{2} \\hat{i}+\\beta \\hat{j}-8 \\hat{k}$$ are collinear, then $$(19 \\alpha-6 \\beta)^{2}$$ is equal to :
The position vectors of the vertices $$\\mathrm{A}, \\mathrm{B}$$ and $$\\mathrm{C}$$ of a triangle are $$2 \\hat{i}-3 \\hat{j}+3 \\hat{k}, 2 \\hat{i}+2 \\hat{j}+3 \\hat{k}$$ and $$-\\hat{i}+\\hat{j}+3 \\hat{k}$$ respectively. Let $$l$$ denotes the length of the angle bisector $$\\mathrm{AD}$$ of $$\\angle \\mathrm{BAC}$$ where $$\\mathrm{D}$$ is on the line segment $$\\mathrm{BC}$$, then $$2 l^2$$ equals :
Let $$\\vec{a}, \\vec{b}$$ and $$\\vec{c}$$ be three non-zero vectors such that $$\\vec{b}$$ and $$\\vec{c}$$ are non-collinear. If $$\\vec{a}+5 \\vec{b}$$ is collinear with $$\\vec{c}, \\vec{b}+6 \\vec{c}$$ is collinear with $$\\vec{a}$$ and $$\\vec{a}+\\alpha \\vec{b}+\\beta \\vec{c}=\\overrightarrow{0}$$, then $$\\alpha+\\beta$$ is equal to
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7668,
"subject": "General Science",
"question": "If $$\\overrightarrow a \\,\\,,\\,\\,\\overrightarrow b \\,\\,,\\,\\,\\overrightarrow c $$ are vectors such that $$\\left[ {\\overrightarrow a \\,\\overrightarrow b \\,\\overrightarrow c } \\right] = 4$$ then $$\\left[ {\\overrightarrow a \\, \\times \\overrightarrow b \\,\\,\\overrightarrow b \\times \\,\\overrightarrow c \\,\\,\\overrightarrow c \\, \\times \\overrightarrow a } \\right] = $$ ",
"options": [
{
"text": "$$16$$ "
},
{
"text": "$$64$$ "
},
{
"text": "$$4$$ "
},
{
"text": "$$8$$ "
}
],
"answer": "$$16$$ ",
"solution": "**Answer:** $$16$$ \n\nWe have, $$\\left[ {\\overrightarrow a \\times \\overrightarrow b \\,\\,\\overrightarrow b \\times \\overrightarrow c \\,\\,\\overrightarrow c \\times \\overrightarrow a } \\right]$$\n
$$ = \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right).\\,\\,\\left\\{ {\\left( {\\overrightarrow b \\times \\overrightarrow c } \\right) \\times \\left( {\\overrightarrow c \\times \\overrightarrow a } \\right)} \\right\\}$$\n
$$ = \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right).\\,\\,\\left\\{ {\\left( {\\overrightarrow m \\,.\\,\\overrightarrow a } \\right)\\overrightarrow c - \\left( {\\overrightarrow m \\,.\\,\\overrightarrow c } \\right)\\overrightarrow a } \\right\\}$$\n
( where $$\\overrightarrow m = \\overrightarrow b \\times \\overrightarrow c $$ )\n
$$ = \\left\\{ {\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right).\\overrightarrow c } \\right\\}.\\,\\left\\{ {\\overrightarrow a \\,.\\,\\left( {\\overrightarrow b \\times \\overrightarrow c } \\right)} \\right\\}$$\n
$$ = {\\left[ {\\overrightarrow a \\,\\overrightarrow b \\,\\overrightarrow c } \\right]^2} = {4^2} = 16.$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7669,
"subject": "General Science",
"question": "If $$\\overrightarrow u \\,,\\overrightarrow v $$ and $$\\overrightarrow w $$ are three non-coplanar vectors, then $$\\,\\left( {\\overrightarrow u + \\overrightarrow v - \\overrightarrow w } \\right).\\left( {\\overrightarrow u - \\overrightarrow v } \\right) \\times \\left( {\\overrightarrow v - \\overrightarrow w} \\right)$$ equals :",
"options": [
{
"text": "$$3\\overrightarrow u .\\overrightarrow v \\times \\overrightarrow w $$ "
},
{
"text": "$$0$$"
},
{
"text": "$$\\overrightarrow u .\\overrightarrow v \\times \\overrightarrow w $$ "
},
{
"text": "$$\\overrightarrow u .\\overrightarrow w \\times \\overrightarrow v $$ "
}
],
"answer": "$$\\overrightarrow u .\\overrightarrow v \\times \\overrightarrow w $$ ",
"solution": "**Answer:** $$\\overrightarrow u .\\overrightarrow v \\times \\overrightarrow w $$ \n\n$$\\left( {\\overrightarrow u + \\overrightarrow v - \\overrightarrow w } \\right).\\left( {\\overrightarrow u \\times \\overrightarrow v - \\overrightarrow u \\times \\overrightarrow w - \\overrightarrow v \\times \\overrightarrow v + \\overrightarrow v \\times \\overrightarrow w } \\right)$$\n
$$ = \\left( {\\overrightarrow u + \\overrightarrow v - \\overrightarrow w } \\right).\\left( {\\overrightarrow u \\times \\overrightarrow v - \\overrightarrow u \\times \\overrightarrow w + \\overrightarrow v \\times \\overrightarrow w } \\right) = \\overrightarrow u .\\left( {\\overrightarrow u \\times \\overrightarrow v } \\right)$$\n
$$ - \\overrightarrow u .\\left( {\\overrightarrow u \\times \\overrightarrow w } \\right) + \\overrightarrow u .\\left( {\\overrightarrow v \\times \\overrightarrow w } \\right) + \\overrightarrow v .\\left( {\\overrightarrow u \\times \\overrightarrow v } \\right) - \\overrightarrow v .\\left( {\\overrightarrow u \\times \\overrightarrow w } \\right)$$\n
$$ + \\overrightarrow v .\\left( {\\overrightarrow v \\times \\overrightarrow w } \\right) - \\overrightarrow w .\\left( {\\overrightarrow u \\times \\overrightarrow v } \\right) + \\overrightarrow w .\\left( {\\overrightarrow u \\times \\overrightarrow w } \\right) - \\overrightarrow w .\\left( {\\overrightarrow u \\times \\overrightarrow w } \\right)$$\n
$$ = \\overrightarrow u .\\left( {\\overrightarrow v \\times \\overrightarrow w } \\right) - \\overrightarrow v .\\left( {\\overrightarrow u \\times \\overrightarrow w } \\right) - \\overrightarrow w .\\left( {\\overrightarrow u \\times \\overrightarrow v } \\right)$$\n
$$ = \\left[ {\\left. {\\overrightarrow u \\overrightarrow v \\overrightarrow w } \\right)} \\right] + \\left[ {\\left. {\\overrightarrow v \\overrightarrow w \\overrightarrow u } \\right)} \\right] - \\left[ {\\overrightarrow w \\overrightarrow u \\overrightarrow v } \\right]$$\n
$$ = \\overrightarrow u .\\left( {\\overrightarrow v \\times \\overrightarrow w } \\right)$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7670,
"subject": "General Science",
"question": "If $${\\overrightarrow a ,\\overrightarrow b ,\\overrightarrow c }$$ are non-coplanar vectors and $$\\lambda $$ is a real number, then the vectors $${\\overrightarrow a + 2\\overrightarrow b + 3\\overrightarrow c ,\\,\\,\\lambda \\overrightarrow b + 4\\overrightarrow c }$$ and $$\\left( {2\\lambda - 1} \\right)\\overrightarrow c $$ are non coplanar for :",
"options": [
{
"text": "no value of $$\\lambda $$ "
},
{
"text": "all except one value of $$\\lambda $$ "
},
{
"text": "all except two values of $$\\lambda $$ "
},
{
"text": "all values of $$\\lambda $$ "
}
],
"answer": "all except two values of $$\\lambda $$ ",
"solution": "**Answer:** all except two values of $$\\lambda $$ \n\nVectors $$\\overrightarrow a + 2\\overrightarrow b + 3\\overrightarrow c ,\\lambda \\overrightarrow b + 4\\overrightarrow c ,\\,\\,\\,$$
and $$\\left( {2\\lambda - 1} \\right)\\overrightarrow c $$ are \n
$$\\therefore$$ Forces are noncoplanar for all $$\\lambda ,$$ \n
except $$\\lambda = 0,{1 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7671,
"subject": "General Science",
"question": "Let $$\\overrightarrow a \\,\\, = \\,\\,\\widehat i - \\widehat k,\\,\\,\\,\\,\\,\\overrightarrow b \\,\\,\\, = \\,\\,\\,x\\widehat i + \\widehat j\\,\\,\\, + \\,\\,\\,\\left( {1 - x} \\right)\\widehat k$$ and $$\\overrightarrow c \\,\\, = \\,\\,y\\widehat i + x\\widehat j + \\left( {1 + x - y} \\right)\\widehat k.$$ Then $$\\left[ {\\overrightarrow a ,\\overrightarrow b ,\\overrightarrow c } \\right]$$ depends on :",
"options": [
{
"text": "only $$y$$ "
},
{
"text": "only $$x$$ "
},
{
"text": "both $$x$$ and $$y$$ "
},
{
"text": "neither $$x$$ nor $$y$$ "
}
],
"answer": "neither $$x$$ nor $$y$$ ",
"solution": "**Answer:** neither $$x$$ nor $$y$$ \n\n$$\\overrightarrow a = \\widehat j - \\widehat k,\\overrightarrow b = x\\widehat i + \\overrightarrow j + \\left( {1 - x} \\right)\\widehat k$$ \n
and $$\\overrightarrow c = y\\widehat i + x\\widehat j + \\left( {1 + x - y} \\right)\\widehat k$$\n
$$\\left[ {\\overrightarrow a \\,\\overrightarrow b \\,\\overrightarrow c } \\right] = \\overrightarrow a .\\overrightarrow b \\times \\overrightarrow c = \\left| {\\matrix{\n 1 & 0 & { - 1} \\cr \n x & 1 & {1 - x} \\cr \n y & x & {1 + x - y} \\cr \n\n } } \\right|$$\n
$$ = 1\\left[ {1 + x - y - x + {x^2}} \\right] - \\left[ { - {x^2} - y} \\right]$$\n
$$ = 1 - y + {x^2} - {x^2} + y = 1$$\n
Hence $$\\left[ {\\overrightarrow a \\,\\overrightarrow b \\,\\overrightarrow c } \\right]$$ is independent of $$x$$ and $$y$$ both.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7672,
"subject": "General Science",
"question": "If $$\\overrightarrow a ,\\overrightarrow b ,\\overrightarrow c $$ are non coplanar vectors and $$\\lambda $$ is a real number then
$$\\left[ {\\lambda \\left( {\\overrightarrow a + \\overrightarrow b } \\right)\\,\\,\\,\\,\\,\\,\\,\\,{\\lambda ^2}\\overrightarrow b \\,\\,\\,\\,\\,\\,\\,\\,\\lambda \\overrightarrow c } \\right] = \\left[ {\\overrightarrow a \\,\\,\\,\\,\\,\\,\\,\\,\\overrightarrow b + \\overrightarrow c \\,\\,\\,\\,\\,\\,\\,\\,\\overrightarrow b } \\right]$$ for :",
"options": [
{
"text": "exactly one value of $$\\lambda $$ "
},
{
"text": "no value of $$\\lambda $$"
},
{
"text": "exactly three values of $$\\lambda $$"
},
{
"text": "exactly two values of $$\\lambda $$"
}
],
"answer": "no value of $$\\lambda $$",
"solution": "**Answer:** no value of $$\\lambda $$\n\n$$\\left[ {\\lambda \\left( {\\overrightarrow a + \\overrightarrow b } \\right){\\lambda ^2}\\overrightarrow b \\,\\,\\,\\lambda \\overrightarrow c } \\right] = \\left[ {\\overrightarrow a \\,\\,\\overrightarrow b + \\overrightarrow c \\,\\,\\overrightarrow b } \\right]$$\n
$$ \\Rightarrow {\\lambda ^4}\\left[ {\\overrightarrow a + \\overrightarrow b \\,\\,\\overrightarrow b \\overrightarrow c } \\right] = \\left[ {\\overrightarrow a \\,\\,\\overrightarrow b + \\overrightarrow c \\,\\,\\overrightarrow b } \\right]$$\n
$$ \\Rightarrow {\\lambda ^4}\\left\\{ {\\left[ {\\overrightarrow a \\,\\overrightarrow b \\,\\overrightarrow c } \\right] + \\left[ {\\overrightarrow b \\,\\overrightarrow b \\,\\overrightarrow c } \\right]} \\right\\} = \\left[ {\\overrightarrow a \\,\\overrightarrow b \\,\\overrightarrow b } \\right] + \\left[ {\\overrightarrow a \\,\\overrightarrow c \\,\\overrightarrow b } \\right]$$\n
$$ \\Rightarrow {\\lambda ^4}\\left[ {\\overrightarrow a \\,\\overrightarrow b \\,\\overrightarrow c } \\right] = - \\left[ {\\overrightarrow a \\,\\overrightarrow b \\,\\overrightarrow c } \\right]$$\n
$$ \\Rightarrow {\\lambda ^4} = - 1$$\n
$$ \\Rightarrow \\lambda \\,\\,$$ has no real values. ",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7673,
"subject": "General Science",
"question": "If $$\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) \\times \\overrightarrow c = \\overrightarrow a \\times \\left( {\\overrightarrow b \\times \\overrightarrow c } \\right)$$ where $${\\overrightarrow a ,\\overrightarrow b }$$ and $${\\overrightarrow c }$$ are any three vectors such that $$\\overrightarrow a .\\overrightarrow b \\ne 0,\\,\\,\\overrightarrow b .\\overrightarrow c \\ne 0$$ then $${\\overrightarrow a }$$ and $${\\overrightarrow c }$$ are :",
"options": [
{
"text": "inclined at an angle of $${\\pi \\over 3}$$ between them "
},
{
"text": "inclined at an angle of $${\\pi \\over 6}$$ between them "
},
{
"text": "perpendicular "
},
{
"text": "parallel "
}
],
"answer": "parallel ",
"solution": "**Answer:** parallel \n\n$$\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) \\times \\overrightarrow c $$\n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\overrightarrow a \\times \\left( {\\overrightarrow b \\times \\overrightarrow c } \\right),\\overrightarrow a .\\overrightarrow b \\ne 0,\\,\\,\\overrightarrow b .\\overrightarrow c \\ne 0$$\n
$$ \\Rightarrow \\left( {\\overrightarrow a .\\overrightarrow c } \\right).\\overrightarrow b - \\left( {\\overrightarrow b .\\overrightarrow c } \\right)\\overrightarrow a $$\n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = \\left( {\\overrightarrow a .\\overrightarrow c } \\right).\\overrightarrow b - \\left( {\\overrightarrow a .\\overrightarrow b } \\right).\\overrightarrow c $$\n
$$ \\Rightarrow \\left( {\\overrightarrow a .\\overrightarrow b } \\right).\\overrightarrow c = \\left( {\\overrightarrow b .\\overrightarrow c } \\right)\\overrightarrow a $$\n
$$ \\Rightarrow \\overrightarrow a ||\\overrightarrow c .$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7674,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = \\widehat i + \\widehat j + \\widehat k,\\overrightarrow b = \\widehat i - \\widehat j + 2\\widehat k$$ and $$\\overrightarrow c = x\\widehat i + \\left( {x - 2} \\right)\\widehat j - \\widehat k\\,\\,.$$ If the vectors $$\\overrightarrow c $$ lies in the plane of $$\\overrightarrow a $$ and $$\\overrightarrow b $$, then $$x$$ equals :",
"options": [
{
"text": "$$-4$$ "
},
{
"text": "$$-2$$"
},
{
"text": "$$0$$ "
},
{
"text": "$$1.$$"
}
],
"answer": "$$-2$$",
"solution": "**Answer:** $$-2$$\n\nGiven $$\\overrightarrow a = \\widehat i + \\widehat j + \\widehat k,\\overrightarrow b = \\widehat i - \\widehat j + 2\\widehat k$$ \n
and $$\\overrightarrow c = x\\widehat i + \\left( {x - 2} \\right)\\widehat j - \\widehat k$$\n
If $$\\overrightarrow c $$ lies in the plane of $$\\overrightarrow a $$ and $$\\overrightarrow b ,$$ \n
then $$\\left[ {\\overrightarrow a \\,\\overrightarrow {b\\,} \\overrightarrow c } \\right] = 0$$\n
$$ \\Rightarrow 2x = - 4\\,\\,\\,\\,\\,\\,\\,\\,\\, \\Rightarrow x = - 2$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7675,
"subject": "General Science",
"question": "The vector $$\\overrightarrow a = \\alpha \\widehat i + 2\\widehat j + \\beta \\widehat k$$ lies in the plane of the vectors \n $$\\overrightarrow b = \\widehat i + \\widehat j$$ and $$\\overrightarrow c = \\widehat j + \\widehat k$$ and bisects the angle between $$\\overrightarrow b $$ and $$\\overrightarrow c $$.Then which one of the following gives possible values of $$\\alpha $$ and $$\\beta $$ ? ",
"options": [
{
"text": "$$\\alpha = 2,\\,\\,\\beta = 2$$ "
},
{
"text": "$$\\alpha = 1,\\,\\,\\beta = 2$$"
},
{
"text": "$$\\alpha = 2,\\,\\,\\beta = 1$$"
},
{
"text": "$$\\alpha = 1,\\,\\,\\beta = 1$$"
}
],
"answer": "$$\\alpha = 1,\\,\\,\\beta = 1$$",
"solution": "**Answer:** $$\\alpha = 1,\\,\\,\\beta = 1$$\n\nAs $$\\overrightarrow a $$ lies in the plane of $$\\overrightarrow b $$ and $$\\overrightarrow c $$\n
$$\\therefore$$ $$\\overrightarrow a = \\overrightarrow b + \\lambda \\overrightarrow c $$\n
$$ \\Rightarrow \\alpha \\widehat i + 2\\widehat j + \\beta \\widehat k = \\widehat i + \\widehat j + \\lambda \\left( {\\widehat j + \\widehat k} \\right)$$\n
$$ \\Rightarrow \\alpha = 1,2 = 1 + \\lambda ,\\,\\beta = \\lambda \\Rightarrow \\alpha = 1,\\beta = 1$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7676,
"subject": "General Science",
"question": "If $$\\overrightarrow u ,\\overrightarrow v ,\\overrightarrow w $$ are non-coplanar vectors and $$p,q$$ are real numbers, then the equality $$\\left[ {3\\overrightarrow u \\,\\,p\\overrightarrow v \\,\\,p\\overrightarrow w } \\right] - \\left[ {p\\overrightarrow v \\,\\,\\overrightarrow w \\,\\,q\\overrightarrow u } \\right] - \\left[ {2\\overrightarrow w \\,\\,q\\overrightarrow v \\,\\,q\\overrightarrow u } \\right] = 0$$ holds for : ",
"options": [
{
"text": "exactly two values of $$(p,q)$$"
},
{
"text": "more than two but not all values of $$(p,q)$$ "
},
{
"text": "all values of $$(p,q)$$ "
},
{
"text": "exactly one value of $$(p,q)$$"
}
],
"answer": "exactly one value of $$(p,q)$$",
"solution": "**Answer:** exactly one value of $$(p,q)$$\n\n$$\\left[ {3\\overrightarrow u \\,\\,p\\overrightarrow v \\,\\,p\\overrightarrow \\omega } \\right] - \\left[ {p\\overrightarrow v \\,\\,\\overrightarrow \\omega \\,\\,q\\overrightarrow u } \\right] - \\left[ {2\\overrightarrow \\omega \\,\\,q\\overrightarrow v \\,\\,q\\overrightarrow u } \\right] = 0$$\n
$$ \\Rightarrow \\left( {3{p^2} - pq + 2{q^2}} \\right)\\left[ {\\overrightarrow u \\,\\,\\overrightarrow v \\,\\,\\overrightarrow \\omega } \\right] = 0$$\n
$$ \\Rightarrow 3{p^2} - pq + 2{q^2} = 0\\,\\,$$ $$\\,\\,\\,\\,\\left( \\, \\right.$$ As $$\\,\\,\\,\\,\\left[ {\\overrightarrow u \\,\\,\\overrightarrow v \\,\\,\\overrightarrow \\omega } \\right] \\ne 0$$ $$\\left. {} \\right)$$\n
$$ \\Rightarrow p = 0,q = 0,p = {q \\over 2}$$ $$ \\Rightarrow p = 0,q = 0$$\n
$$\\therefore$$ Exactly one value of $$\\left( {p,q} \\right)$$ ",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7677,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = \\widehat j - \\widehat k$$ and $$\\overrightarrow c = \\widehat i - \\widehat j - \\widehat k.$$ Then the vector $$\\overrightarrow b $$ satisfying $$\\overrightarrow a \\times \\overrightarrow b + \\overrightarrow c = \\overrightarrow 0 $$ and $$\\overrightarrow a .\\overrightarrow b = 3$$ :",
"options": [
{
"text": "$$2\\widehat i - \\widehat j + 2\\widehat k$$ "
},
{
"text": "$$\\widehat i - \\widehat j - 2\\widehat k$$"
},
{
"text": "$$\\widehat i + \\widehat j - 2\\widehat k$$"
},
{
"text": "$$-\\widehat i +\\widehat j - 2\\widehat k$$"
}
],
"answer": "$$-\\widehat i +\\widehat j - 2\\widehat k$$",
"solution": "**Answer:** $$-\\widehat i +\\widehat j - 2\\widehat k$$\n\n$$\\overrightarrow c = \\overrightarrow b \\times \\overrightarrow a $$\n
$$ \\Rightarrow \\overrightarrow b .\\overrightarrow c = \\overrightarrow b .\\left( {\\overrightarrow b \\times \\overrightarrow a } \\right) \\Rightarrow \\overrightarrow b .\\overrightarrow c = 0$$\n
$$\\overrightarrow b = \\left( {3 + 2{b_3}} \\right)\\widehat i + \\left( {3 + {b_3}} \\right)\\widehat j + {b_3}\\widehat k$$ \n
From the option given, it is clear that $${b_3}$$ equal to either $$2$$ or $$-2.$$ \n
$${b_3} = 2$$ \n
then $$\\overrightarrow b = 7\\widehat i + 5\\widehat j + 2\\widehat k$$ which is not possible\n
If $${b_3} = - 2,$$ then $$\\overrightarrow b = - \\widehat i + \\widehat j - 2\\widehat k$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7678,
"subject": "General Science",
"question": "If $$\\overrightarrow a = {1 \\over {\\sqrt {10} }}\\left( {3\\widehat i + \\widehat k} \\right)$$ and $$\\overrightarrow b = {1 \\over 7}\\left( {2\\widehat i + 3\\widehat j - 6\\widehat k} \\right),$$ then the value \n
of $$\\left( {2\\overrightarrow a - \\overrightarrow b } \\right)\\left[ {\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) \\times \\left( {\\overrightarrow a + 2\\overrightarrow b } \\right)} \\right]$$ is :",
"options": [
{
"text": "$$-3$$ "
},
{
"text": "$$5$$ "
},
{
"text": "$$3$$ "
},
{
"text": "$$-5$$ "
}
],
"answer": "$$-5$$ ",
"solution": "**Answer:** $$-5$$ \n\nWe have $$\\overrightarrow a .\\overrightarrow b = 0,\\,\\,\\overrightarrow a .\\overrightarrow a = 1,\\,\\,\\overrightarrow b .\\overrightarrow b = 1$$\n
$$\\left( {2\\overrightarrow a - \\overrightarrow b } \\right).\\left[ {\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) \\times \\left( {\\overrightarrow a + 2\\overrightarrow b } \\right)} \\right]$$\n
$$ = \\left( {2\\overrightarrow a - \\overrightarrow b } \\right).\\left[ {\\left\\{ {\\overrightarrow a .\\left( {\\overrightarrow a + 2\\overrightarrow b } \\right)} \\right\\}\\overrightarrow b - \\left\\{ {\\overrightarrow b .\\left( {\\overrightarrow a + 2\\overrightarrow b } \\right)\\overrightarrow a } \\right\\}} \\right]$$\n
$$ = \\left( {2\\overrightarrow a - \\overrightarrow b } \\right).\\left[ {\\left( {\\overrightarrow a .\\overrightarrow a + 2\\overrightarrow a .\\overrightarrow b } \\right)\\overrightarrow b - \\left( {\\overrightarrow a .\\overrightarrow b + 2\\overrightarrow b .\\overrightarrow b } \\right)\\overrightarrow a } \\right]$$\n
$$ = \\left( {2\\overrightarrow a - \\overrightarrow b } \\right).\\left[ {\\overrightarrow b - 2\\overrightarrow a } \\right]$$\n
$$ = 4\\overrightarrow a .\\overrightarrow b - \\overrightarrow b .\\overrightarrow b - 4\\overrightarrow a .\\overrightarrow a $$\n
$$ = - 5$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7679,
"subject": "General Science",
"question": "If $$\\left[ {\\overrightarrow a \\times \\overrightarrow b \\,\\,\\,\\,\\overrightarrow b \\times \\overrightarrow c \\,\\,\\,\\,\\overrightarrow c \\times \\overrightarrow a } \\right] = \\lambda {\\left[ {\\overrightarrow a\\,\\,\\,\\,\\,\\,\\,\\, \\overrightarrow b \\,\\,\\,\\,\\,\\,\\,\\,\\overrightarrow c } \\right]^2}$$ then $$\\lambda $$ is equal to :",
"options": [
{
"text": "$$0$$ "
},
{
"text": "$$1$$"
},
{
"text": "$$2$$"
},
{
"text": "$$3$$"
}
],
"answer": "$$1$$",
"solution": "**Answer:** $$1$$\n\n$$L.H.S$$ $$ = \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right).\\left[ {\\left( {\\overrightarrow b \\times \\overrightarrow c } \\right) \\times \\left( {\\overrightarrow c \\times \\overrightarrow a } \\right)} \\right]$$\n
$$ = \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right).\\left[ {\\left( {\\overrightarrow b \\times \\overrightarrow c .\\overrightarrow a } \\right)} \\right]\\overrightarrow c - \\left( {\\overrightarrow b \\times \\overrightarrow c .\\overrightarrow c } \\right)\\left. {\\overrightarrow a } \\right]$$\n
$$ = \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right).\\left[ {\\left[ {\\overrightarrow b \\,\\overrightarrow c \\,\\overrightarrow a } \\right]\\overrightarrow c } \\right]$$ $$\\,\\,\\,\\,\\,\\,\\left[ \\, \\right.$$As $$\\overrightarrow b \\times \\overrightarrow c .\\overrightarrow c = 0$$ $$\\left. \\, \\right]$$\n
$$ = \\left[ {\\overrightarrow a \\,\\overrightarrow b \\,\\overrightarrow c } \\right].\\left( {\\overrightarrow a \\times \\overrightarrow b .\\overrightarrow c } \\right) = {\\left[ {\\overrightarrow a \\,\\,\\,\\,\\,\\overrightarrow b \\,\\,\\,\\,\\,\\overrightarrow c } \\right]^2}$$\n
$$\\left[ {\\overrightarrow a \\times \\overrightarrow b \\,\\,\\,\\overrightarrow b \\times \\overrightarrow c \\,\\,\\,\\overrightarrow c \\times \\overrightarrow a } \\right] = {\\left[ {\\overrightarrow a \\,\\,\\,\\,\\,\\overrightarrow b \\,\\,\\,\\,\\,\\overrightarrow c } \\right]^2}$$\n
So $$\\,\\,\\,\\,\\,\\lambda = 1$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7680,
"subject": "General Science",
"question": "Let $$\\overrightarrow a ,\\overrightarrow b $$ and $$\\overrightarrow c $$ be three non-zero vectors such that no two of them are collinear and
$$\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) \\times \\overrightarrow c = {1 \\over 3}\\left| {\\overrightarrow b } \\right|\\left| {\\overrightarrow c } \\right|\\overrightarrow a .$$ If $$\\theta $$ is the angle between vectors $$\\overrightarrow b $$ and $${\\overrightarrow c }$$ , then a value of sin $$\\theta $$ is :",
"options": [
{
"text": "$${2 \\over 3}$$"
},
{
"text": "$${{ - 2\\sqrt 3 } \\over 3}$$ "
},
{
"text": "$${{ 2\\sqrt 2 } \\over 3}$$"
},
{
"text": "$${{ - \\sqrt 2 } \\over 3}$$ "
}
],
"answer": "$${{ 2\\sqrt 2 } \\over 3}$$",
"solution": "**Answer:** $${{ 2\\sqrt 2 } \\over 3}$$\n\n$$\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) \\times \\overrightarrow c = {1 \\over 3}\\left| {\\overrightarrow b } \\right|\\left| {\\overrightarrow c } \\right|\\overrightarrow a $$\n
$$ \\Rightarrow - \\overrightarrow c \\times \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) = {1 \\over 3}\\left| {\\overrightarrow b } \\right|\\left| {\\overrightarrow c } \\right|\\overrightarrow a $$\n
$$ \\Rightarrow - \\left( {\\overrightarrow c .\\overrightarrow b } \\right)\\overrightarrow a + \\left( {\\overrightarrow c .\\overrightarrow a } \\right)\\overrightarrow b = {1 \\over 3}\\left| {\\overrightarrow b } \\right|\\left| {\\overrightarrow c } \\right|\\overrightarrow a $$\n
$$ \\Rightarrow - \\left| {\\overrightarrow b } \\right|\\left| {\\overrightarrow c } \\right|\\cos \\theta \\overrightarrow a + \\left( {\\overrightarrow c .\\overrightarrow a } \\right)\\overrightarrow b = {1 \\over 3}\\left| {\\overrightarrow b } \\right|\\left| {\\overrightarrow c } \\right|\\overrightarrow a $$ \n
$$\\therefore$$ $$\\,\\,\\,\\overrightarrow a ,\\,\\overrightarrow b ,\\,\\overrightarrow c $$ are non collinear, the above equation is possible only when\n
$$ - \\cos \\theta = {1 \\over 3}$$ and $$\\overrightarrow c .\\overrightarrow a = 0$$\n
$$ \\Rightarrow \\sin \\theta = {{2\\sqrt 2 } \\over 3};\\theta \\in \\,{\\rm I}{\\rm I}$$ quad",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7681,
"subject": "General Science",
"question": "Let $$\\overrightarrow a ,\\overrightarrow b $$ and $$\\overrightarrow c $$ be three unit vectors such that $$\\overrightarrow a \\times \\left( {\\overrightarrow b \\times \\overrightarrow c } \\right) = {{\\sqrt 3 } \\over 2}\\left( {\\overrightarrow b + \\overrightarrow c } \\right).$$ If $${\\overrightarrow b }$$ is not parallel to $${\\overrightarrow c },$$ then the angle between $${\\overrightarrow a }$$ and $${\\overrightarrow b }$$ is:",
"options": [
{
"text": "$${{2\\pi } \\over 3}$$ "
},
{
"text": "$${{5\\pi } \\over 6}$$"
},
{
"text": "$${{3\\pi } \\over 4}$$"
},
{
"text": "$${{\\pi } \\over 2}$$"
}
],
"answer": "$${{5\\pi } \\over 6}$$",
"solution": "**Answer:** $${{5\\pi } \\over 6}$$\n\n$$\\overrightarrow a \\times \\left( {\\overrightarrow b \\times \\overrightarrow c } \\right) = {{\\sqrt 3 } \\over 2}\\left( {\\overrightarrow b + \\overrightarrow c } \\right)$$\n
$$ \\Rightarrow \\left( {\\overrightarrow a .\\overrightarrow c } \\right)\\overrightarrow b - \\left( {\\overrightarrow a .\\overrightarrow b } \\right)\\overrightarrow c = {{\\sqrt 3 } \\over 2}\\overrightarrow b + {{\\sqrt 3 } \\over 2}\\overrightarrow c $$\n
$$\\left[ \\, \\right.$$ As $$\\overrightarrow a $$ and $$\\overrightarrow b $$ are unit vectors $$\\left. \\, \\right]$$\n
where $$\\theta $$ is the angle between $$\\overrightarrow a $$ and $$\\overrightarrow b $$ \n
$$\\theta = {{5\\pi } \\over 6}$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7682,
"subject": "General Science",
"question": "Let $$\\overrightarrow a $$, $$\\overrightarrow b $$ and $$\\overrightarrow c $$ be three unit vectors, out of which vectors $$\\overrightarrow b $$ and $$\\overrightarrow c $$ are non-parallel. If $$\\alpha $$ and $$\\beta $$ are the angles which vector $$\\overrightarrow a $$ makes with vectors $$\\overrightarrow b $$ and $$\\overrightarrow c $$ respectively and $$\\overrightarrow a $$ $$ \\times $$ ($$\\overrightarrow b $$ $$ \\times $$ $$\\overrightarrow c $$) = $${1 \\over 2}\\overrightarrow b $$, then $$\\left| {\\alpha - \\beta } \\right|$$ is equal to : ",
"options": [
{
"text": "90o"
},
{
"text": "30o"
},
{
"text": "45o"
},
{
"text": "60o"
}
],
"answer": "30o",
"solution": "**Answer:** 30o\n\n$$\\left( {\\overrightarrow a .\\overrightarrow c } \\right)\\overrightarrow b - \\left( {\\overrightarrow a .\\overrightarrow b } \\right).\\overrightarrow c = {1 \\over 2}\\overrightarrow b $$\n
$$ \\because $$ $$\\overrightarrow b \\,\\,$$ & $$\\overrightarrow c \\,\\,$$ are linearly independent\n
$$ \\therefore $$ $$\\overrightarrow a \\,$$.$$\\overrightarrow c \\,$$ = $${1 \\over 2}$$ & $$\\overrightarrow a .\\overrightarrow b $$ = 0\n
(All given vectors are unit vectors)\n
$$ \\therefore $$ $$\\overrightarrow a $$^$$\\overrightarrow c $$ = 60o & $$\\overrightarrow a $$^$$\\overrightarrow b $$ = 90o\n
$$ \\therefore $$ $$\\left| {\\alpha - \\beta } \\right| = {30^o}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7683,
"subject": "General Science",
"question": "Let $$\\alpha $$ $$ \\in $$ R and the three vectors
$$\\overrightarrow a = \\alpha \\widehat i + \\widehat j + 3\\widehat k$$, $$\\overrightarrow b = 2\\widehat i + \\widehat j - \\alpha \\widehat k$$
and $$\\overrightarrow c = \\alpha \\widehat i - 2\\widehat j + 3\\widehat k$$.
Then the set\nS = {$$\\alpha $$ :\n$$\\overrightarrow a $$ ,\n$$\\overrightarrow b $$ and\n$$\\overrightarrow c $$ are coplanar} :",
"options": [
{
"text": "contains exactly two numbers only one of which is positive"
},
{
"text": "is singleton"
},
{
"text": "contains exactly two positive numbers"
},
{
"text": "is empty"
}
],
"answer": "is empty",
"solution": "**Answer:** is empty\n\nSince these vectors are coplanar then,
sum of distinct solutions $$=$$ $$-$$ 1",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7686,
"subject": "General Science",
"question": "Let $$\\overrightarrow a $$ = $$\\widehat i - \\widehat j$$, $$\\overrightarrow b $$ = $$\\widehat i + \\widehat j + \\widehat k$$ and $$\\overrightarrow c $$ \n
be a vector such that $$\\overrightarrow a $$ × $$\\overrightarrow c $$ + $$\\overrightarrow b $$ = $$\\overrightarrow 0 $$
and $$\\overrightarrow a $$ . $$\\overrightarrow c $$ = 4, then |$$\\overrightarrow c $$|2 is equal to :",
"options": [
{
"text": "8"
},
{
"text": "$$19 \\over 2$$"
},
{
"text": "9"
},
{
"text": "$$17 \\over 2$$"
}
],
"answer": "$$19 \\over 2$$",
"solution": "**Answer:** $$19 \\over 2$$\n\nGiven that, \n
$$\\overrightarrow a \\times \\overrightarrow c + \\overrightarrow b = \\overrightarrow 0 $$\n
$$ \\Rightarrow $$ $$\\overrightarrow a \\times \\left( {\\overrightarrow a \\times \\overrightarrow c } \\right) + \\overrightarrow a \\times \\overrightarrow b = \\overrightarrow 0 $$\n
$$ \\Rightarrow $$ $$\\left( {\\overrightarrow a \\cdot \\overrightarrow c } \\right)\\overrightarrow a - \\left( {\\overrightarrow a \\cdot \\overrightarrow a } \\right)\\overrightarrow c + \\overrightarrow a \\times \\overrightarrow b = \\overrightarrow 0 $$\n
given that\n
$$\\overrightarrow a \\cdot \\overrightarrow c = 4$$\n
and $$\\overrightarrow a \\cdot \\overrightarrow a = {\\left| {\\overrightarrow a } \\right|^2} = {\\left( {\\sqrt 2 } \\right)^2} = 2$$\n
$$ \\Rightarrow $$ $$4\\overrightarrow a - 2\\overrightarrow c + \\overrightarrow a \\times \\overrightarrow b = 0$$\n
Now $$\\overrightarrow a \\times \\overrightarrow b $$\n
$$\\overrightarrow u = \\widehat i + \\widehat j + 2 \\widehat k$$ or \n $$\\overrightarrow u = \\widehat i + \\widehat j + 4 \\widehat k$$\n
$$ \\therefore $$ cos $$\\theta $$ = $${{{\\overrightarrow u .\\overrightarrow w } \\over {\\left| {\\overrightarrow u } \\right|\\left| {\\overrightarrow w } \\right|}}}$$\n
$$ \\Rightarrow $$ $$\\lambda $$ = 1",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7689,
"subject": "General Science",
"question": "Let x0 be the point of Local maxima of $$f(x) = \\overrightarrow a .\\left( {\\overrightarrow b \\times \\overrightarrow c } \\right)$$, where $$\\overrightarrow a = x\\widehat i - 2\\widehat j + 3\\widehat k$$, $$\\overrightarrow b = - 2\\widehat i + x\\widehat j - \\widehat k$$, $$\\overrightarrow c = 7\\widehat i - 2\\widehat j + x\\widehat k$$. Then the value of $$\\overrightarrow a .\\overrightarrow b + \\overrightarrow b .\\overrightarrow c + \\overrightarrow c .\\overrightarrow a $$ at x = x0 is :",
"options": [
{
"text": "14"
},
{
"text": "-30"
},
{
"text": "-4"
},
{
"text": "-22"
}
],
"answer": "-22",
"solution": "**Answer:** -22\n\n$$f(x) = \\overrightarrow a \\,.\\,(\\overrightarrow b \\times \\overrightarrow c )$$
$$ = \\left[ {\\overrightarrow a \\,\\overrightarrow b \\,\\overrightarrow c } \\right]$$
$$ \\therefore $$ $$\\overrightarrow a = - 3\\widehat i - 2\\widehat j + 3\\widehat k$$
$$\\overrightarrow b = - 2\\widehat i - 3\\widehat j - \\widehat k$$
$$\\overrightarrow c = 7\\widehat i - 2\\widehat j - 3\\widehat k$$
Now, $$\\overrightarrow a \\,.\\,\\overrightarrow b \\, + \\,\\overrightarrow b \\,.\\,\\overrightarrow c \\, + \\,\\overrightarrow c \\,.\\,\\overrightarrow a $$
$$ = - 22$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7690,
"subject": "General Science",
"question": "If the volume of a parallelopiped, whose coterminus edges are given by the vectors $$\\overrightarrow a = \\widehat i + \\widehat j + n\\widehat k$$, $$\\overrightarrow b = 2\\widehat i + 4\\widehat j - n\\widehat k$$ and $$\\overrightarrow c = \\widehat i + n\\widehat j + 3\\widehat k$$ ($$n \\ge 0$$), is 158 cu. units, then :",
"options": [
{
"text": "n = 7"
},
{
"text": "$$\\overrightarrow b .\\overrightarrow c = 10$$"
},
{
"text": "$$\\overrightarrow a .\\overrightarrow c = 17$$"
},
{
"text": "n = 9"
}
],
"answer": "$$\\overrightarrow b .\\overrightarrow c = 10$$",
"solution": "**Answer:** $$\\overrightarrow b .\\overrightarrow c = 10$$\n\nWe know, Volume(V) = $$\\left[ {\\overrightarrow a \\overrightarrow b \\overrightarrow c } \\right]$$\n
$$ \\therefore $$ $$\\overrightarrow a = \\widehat i + \\widehat j + 8\\widehat k$$,
$$\\overrightarrow b = 2\\widehat i + 4\\widehat j - 8\\widehat k$$ and
$$\\overrightarrow c = \\widehat i + 8\\widehat j + 3\\widehat k$$\n
Now $${\\overrightarrow a .\\overrightarrow c }$$ = 1 + 8 + 24 = 33\n
$${\\overrightarrow b .\\overrightarrow c }$$ = 2 + 32 - 24 = 10",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 7691,
"subject": "General Science",
"question": "Let three vectors $$\\overrightarrow a ,\\overrightarrow b $$ and $$\\overrightarrow c $$ be such that $$\\overrightarrow c $$ is coplanar with $$\\overrightarrow a $$ and $$\\overrightarrow b $$, \n $$\\overrightarrow a .\\overrightarrow c $$ = 7 and\n$$\\overrightarrow b $$ is perpendicular to $$\\overrightarrow c $$, where $$\\overrightarrow a = - \\widehat i + \\widehat j + \\widehat k$$ and $$\\overrightarrow b = 2\\widehat i + \\widehat k$$ , then the value of $$2{\\left| {\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\right|^2}$$ is _____.",
"options": [],
"answer": "75",
"solution": "**Answer:** 75\n\n$$\\overrightarrow c = \\lambda (\\overrightarrow b \\times (\\overrightarrow a \\times \\overrightarrow b ))$$
$$ = \\lambda ((\\overrightarrow b \\,.\\,\\overrightarrow b )\\overrightarrow a - (\\overrightarrow b \\,.\\,\\overrightarrow a )\\overrightarrow b )$$
$$ = \\lambda (5( - \\widehat i + \\widehat j + \\widehat k) + 2\\widehat i + \\widehat k)$$
$$ = 2\\left( {{1 \\over 4} + {{49} \\over 4} + 25} \\right) = 25 + 50 = 75$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7692,
"subject": "General Science",
"question": "If $$\\overrightarrow a $$ and $$\\overrightarrow b $$ are perpendicular, then $$\\overrightarrow a \\times \\left( {\\overrightarrow a \\times \\left( {\\overrightarrow a \\times \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)} \\right)} \\right)$$ is equal to :",
"options": [
{
"text": "$${1 \\over 2}|\\overrightarrow a {|^4}\\overrightarrow b $$"
},
{
"text": "$$\\overrightarrow 0 $$"
},
{
"text": "$$\\overrightarrow a \\times \\overrightarrow b $$"
},
{
"text": "$$|\\overrightarrow a {|^4}\\overrightarrow b $$"
}
],
"answer": "$$|\\overrightarrow a {|^4}\\overrightarrow b $$",
"solution": "**Answer:** $$|\\overrightarrow a {|^4}\\overrightarrow b $$\n\n$$\\overrightarrow a \\,.\\,\\overrightarrow b = 0$$
$$\\overrightarrow a \\times (\\overrightarrow a \\times \\overrightarrow b ) = (\\overrightarrow a \\,.\\,\\overrightarrow b )\\overrightarrow a - (\\overrightarrow a \\,.\\,\\overrightarrow a )\\overrightarrow b = - |\\overrightarrow a {|^2}\\overrightarrow b $$
Now, $$\\overrightarrow a \\times (\\overrightarrow a \\times ( - |\\overrightarrow a {|^2}\\overrightarrow b ))$$
$$ = - |\\overrightarrow a {|^2}(\\overrightarrow a \\times (\\overrightarrow a \\times \\overrightarrow b ))$$
$$ = - |\\overrightarrow a {|^2}( - |\\overrightarrow a {|^2}\\overrightarrow b ) = |\\overrightarrow a {|^4}\\overrightarrow b $$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7693,
"subject": "General Science",
"question": "Let $$\\overrightarrow c $$ be a vector perpendicular to the vectors, $$\\overrightarrow a $$ = $$\\widehat i$$ + $$\\widehat j$$ $$-$$ $$\\widehat k$$ and $$\\overrightarrow b $$ = $$\\widehat i$$ + 2$$\\widehat j$$ + $$\\widehat k$$. If $$\\overrightarrow c \\,.\\,\\left( {\\widehat i + \\widehat j + 3\\widehat k} \\right)$$ = 8 then the value of $$\\overrightarrow c $$ . $$\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)$$ is equal to __________.",
"options": [],
"answer": "28",
"solution": "**Answer:** 28\n\n$$\\overrightarrow a \\times \\overrightarrow b = \\left| {\\matrix{\n {\\widehat i} & {\\widehat j} & {\\widehat k} \\cr \n 1 & 1 & { - 1} \\cr \n 1 & 2 & 1 \\cr \n\n } } \\right| = (3, - 2,1)$$
$$\\overrightarrow c \\bot \\overrightarrow a ,\\overrightarrow c \\bot \\overrightarrow b \\Rightarrow C||\\overrightarrow a \\times \\overrightarrow b $$
$$\\overrightarrow c = \\lambda (\\overrightarrow a \\times \\overrightarrow b )$$
$$ \\Rightarrow \\overrightarrow c = \\lambda (3\\widehat i - 2\\widehat j + \\widehat k)$$
Given, $$\\overrightarrow c .(\\widehat i + \\widehat j + 3\\widehat k) = 8$$
$$\\overrightarrow b = - \\beta \\widehat i - \\alpha \\widehat j - \\widehat k$$ and
$$\\overrightarrow c = \\widehat i - 2\\widehat j - \\widehat k$$
such that $$\\overrightarrow a \\,.\\,\\overrightarrow b = 1$$ and $$\\overrightarrow b \\,.\\,\\overrightarrow c = - 3$$, then $${1 \\over 3}\\left( {\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)\\,.\\,\\overrightarrow c } \\right)$$ is equal to _____________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n$$\\overrightarrow a .\\overrightarrow b = 1 \\Rightarrow - \\alpha \\beta - \\alpha \\beta - 3 = 1$$
$$ \\Rightarrow \\alpha \\beta = - 2$$ .... (i)
$$\\overrightarrow b .\\overrightarrow c = - 3 \\Rightarrow - \\beta + 2\\alpha + 1 = - 3$$
$$2\\alpha - \\beta = - 4$$ ..... (ii)
Solving (i) & (ii) $$\\alpha$$ = $$-$$1, $$\\beta$$ = 2,
$${1 \\over 3}((\\overrightarrow a \\, \\times \\overrightarrow b )\\,.\\,\\overrightarrow c ) = {1 \\over 3}\\left| {\\matrix{\n { - 1} & 2 & 3 \\cr \n { - 2} & 1 & { - 1} \\cr \n 1 & { - 2} & { - 1} \\cr \n\n } } \\right| = 2$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7696,
"subject": "General Science",
"question": "Let O be the origin. Let $$\\overrightarrow {OP} = x\\widehat i + y\\widehat j - \\widehat k$$ and $$\\overrightarrow {OQ} = - \\widehat i + 2\\widehat j + 3x\\widehat k$$, x, y$$\\in$$R, x > 0, be such that $$\\left| {\\overrightarrow {PQ} } \\right| = \\sqrt {20} $$ and the vector $$\\overrightarrow {OP} $$ is perpendicular $$\\overrightarrow {OQ} $$. If $$\\overrightarrow {OR} $$ = $$3\\widehat i + z\\widehat j - 7\\widehat k$$, z$$\\in$$R, is coplanar with $$\\overrightarrow {OP} $$ and $$\\overrightarrow {OQ} $$, then the value of x2 + y2 + z2 is equal to :",
"options": [
{
"text": "2"
},
{
"text": "9"
},
{
"text": "7"
},
{
"text": "1"
}
],
"answer": "9",
"solution": "**Answer:** 9\n\n$$\\overrightarrow {OP} = x\\widehat i + y\\widehat j - \\widehat k\\,$$\n
$$ \\therefore $$ x2 + y2 + z2 = 9",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7697,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = 2\\widehat i + \\widehat j - 2\\widehat k$$ and $$\\overrightarrow b = \\widehat i + \\widehat j$$. If $$\\overrightarrow c $$ is a vector such that $$\\overrightarrow a .\\,\\overrightarrow c = \\left| {\\overrightarrow c } \\right|,\\left| {\\overrightarrow c - \\overrightarrow a } \\right| = 2\\sqrt 2 $$ and the angle between $$(\\overrightarrow a \\times \\overrightarrow b )$$ and $$\\overrightarrow c $$ is $${\\pi \\over 6}$$, then the value of $$\\left| {\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) \\times \\overrightarrow c } \\right|$$ is :",
"options": [
{
"text": "$${2 \\over 3}$$"
},
{
"text": "4"
},
{
"text": "3"
},
{
"text": "$${3 \\over 2}$$"
}
],
"answer": "$${3 \\over 2}$$",
"solution": "**Answer:** $${3 \\over 2}$$\n\n$$\\left| {\\overrightarrow a } \\right| = 3 = a;\\overrightarrow a \\,.\\,\\overrightarrow c = c$$
Now, $$\\left| {\\overrightarrow c - \\overrightarrow a } \\right| = 2\\sqrt 2 $$
$$ \\Rightarrow {c^2} + {a^2} - 2\\overrightarrow c \\,.\\,\\overrightarrow a = 8$$
$$ \\Rightarrow {c^2} + 9 - 2(c) = 8$$
$$ \\Rightarrow {c^2} - 2c + 1 = 0 \\Rightarrow c = 1 = \\left| {\\overrightarrow c } \\right|$$
Also, $$\\overrightarrow a \\times \\overrightarrow b = 2\\widehat i - 2\\widehat j + \\widehat k$$
Given, $$(\\overrightarrow a \\times \\overrightarrow b ) = \\left| {\\overrightarrow a \\times \\overrightarrow b } \\right|\\left| {\\overrightarrow c } \\right|\\sin {\\pi \\over 6}$$
$$ = (3)(1)(1/2)$$
$$ = 3/2$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7698,
"subject": "General Science",
"question": "Let a vector $${\\overrightarrow a }$$ be coplanar with vectors $$\\overrightarrow b = 2\\widehat i + \\widehat j + \\widehat k$$ and $$\\overrightarrow c = \\widehat i - \\widehat j + \\widehat k$$. If $${\\overrightarrow a}$$ is perpendicular to $$\\overrightarrow d = 3\\widehat i + 2\\widehat j + 6\\widehat k$$, and $$\\left| {\\overrightarrow a } \\right| = \\sqrt {10} $$. Then a possible value of $$[\\matrix{\n {\\overrightarrow a } & {\\overrightarrow b } & {\\overrightarrow c } \\cr \n\n } ] + [\\matrix{\n {\\overrightarrow a } & {\\overrightarrow b } & {\\overrightarrow d } \\cr \n\n } ] + [\\matrix{\n {\\overrightarrow a } & {\\overrightarrow c } & {\\overrightarrow d } \\cr \n\n } ]$$ is equal to :",
"options": [
{
"text": "$$-$$42"
},
{
"text": "$$-$$40"
},
{
"text": "$$-$$29"
},
{
"text": "$$-$$38"
}
],
"answer": "$$-$$42",
"solution": "**Answer:** $$-$$42\n\n$$\\overrightarrow a = \\lambda \\overrightarrow b + \\mu \\overrightarrow c = \\widehat i(2\\lambda + \\mu ) + \\widehat j(\\lambda - \\mu ) + \\widehat k(\\lambda + \\mu )$$
$$[\\overrightarrow a \\overrightarrow b \\overrightarrow c ]$$ = 0
$$[\\overrightarrow a \\overrightarrow b \\overrightarrow c ] + [\\overrightarrow a \\overrightarrow b \\overrightarrow d ] + [\\overrightarrow a \\overrightarrow c \\overrightarrow d ] = [\\overrightarrow a \\overrightarrow b + \\overrightarrow c \\overrightarrow d ]$$
$$ = 3\\lambda (12) + \\lambda (6) = 42\\lambda = - 42$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7699,
"subject": "General Science",
"question": "Let three vectors $$\\overrightarrow a $$, $$\\overrightarrow b $$ and $$\\overrightarrow c $$ be such that $$\\overrightarrow a \\times \\overrightarrow b = \\overrightarrow c $$, $$\\overrightarrow b \\times \\overrightarrow c = \\overrightarrow a $$ and $$\\left| {\\overrightarrow a } \\right| = 2$$. Then which one of the following is not true?",
"options": [
{
"text": "$$\\overrightarrow a \\times \\left( {(\\overrightarrow b + \\overrightarrow c ) \\times (\\overrightarrow b \\times \\overrightarrow c )} \\right) = \\overrightarrow 0 $$"
},
{
"text": "Projection of $$\\overrightarrow a $$ on $$(\\overrightarrow b \\times \\overrightarrow c )$$ is 2"
},
{
"text": "$$\\left[ {\\matrix{\n {\\overrightarrow a } & {\\overrightarrow b } & {\\overrightarrow c } \\cr \n\n } } \\right] + \\left[ {\\matrix{\n {\\overrightarrow c } & {\\overrightarrow a } & {\\overrightarrow b } \\cr \n\n } } \\right] = 8$$"
},
{
"text": "$${\\left| {3\\overrightarrow a + \\overrightarrow b - 2\\overrightarrow c } \\right|^2} = 51$$"
}
],
"answer": "$${\\left| {3\\overrightarrow a + \\overrightarrow b - 2\\overrightarrow c } \\right|^2} = 51$$",
"solution": "**Answer:** $${\\left| {3\\overrightarrow a + \\overrightarrow b - 2\\overrightarrow c } \\right|^2} = 51$$\n\n(1) $$\\overrightarrow a \\times \\left( {(\\overrightarrow b + \\overrightarrow c ) \\times (\\overrightarrow b \\times \\overrightarrow c )} \\right)$$
$$ = \\overrightarrow a ( - \\overrightarrow b \\times \\overrightarrow c + \\overrightarrow c \\times \\overrightarrow b ) = - 2\\left( {\\overrightarrow a \\times (\\overrightarrow b \\times \\overrightarrow c )} \\right)$$
$$ = - 2(\\overrightarrow a \\times \\overrightarrow a ) = \\overrightarrow 0 $$
(2) Projection of $$\\overrightarrow a $$ on $$\\overrightarrow b \\times \\overrightarrow c $$
$$ = {{\\overrightarrow a \\,.\\,(\\overrightarrow b \\times \\overrightarrow c )} \\over {\\left| {\\overrightarrow b \\times \\overrightarrow c } \\right|}} = {{\\overrightarrow a .\\overrightarrow a } \\over {\\left| {\\overrightarrow a } \\right|}} = \\left| {\\overrightarrow a } \\right| = 2$$
(3) $$\\left[ {\\overrightarrow a \\overrightarrow b \\overrightarrow c } \\right] + \\left[ {\\overrightarrow c \\overrightarrow a \\overrightarrow b } \\right] = 2\\left[ {\\overrightarrow a \\overrightarrow b \\overrightarrow c } \\right] = 2\\overrightarrow a .(\\overrightarrow b \\times \\overrightarrow c )$$
$$ = 2\\overrightarrow a .\\overrightarrow a = 2{\\left| {\\overrightarrow a } \\right|^2} = 8$$
(4) $$\\overrightarrow a \\times \\overrightarrow b = \\overrightarrow c $$ and $$\\overrightarrow b \\times \\overrightarrow c = \\overrightarrow a $$
$$ \\Rightarrow \\overrightarrow a ,\\overrightarrow b ,\\overrightarrow c $$ are mutually $$ \\bot $$ vectors.
$$\\therefore$$ $$\\left| {\\overrightarrow a \\times \\overrightarrow b } \\right| = \\left| {\\overrightarrow c } \\right| \\Rightarrow \\left| {\\overrightarrow a } \\right|\\left| {\\overrightarrow b } \\right| = \\left| {\\overrightarrow c } \\right| \\Rightarrow \\left| {\\overrightarrow b } \\right| = {{\\left| {\\overrightarrow c } \\right|} \\over 2}$$
Also, $$\\left| {\\overrightarrow b \\times \\overrightarrow c } \\right| = \\left| {\\overrightarrow a } \\right| \\Rightarrow \\left| {\\overrightarrow b } \\right|\\left| {\\overrightarrow c } \\right| = 2 \\Rightarrow \\left| {\\overrightarrow c } \\right| = 2$$ & $$\\left| {\\overrightarrow b } \\right| = 1$$
$${\\left| {3\\overrightarrow a + \\overrightarrow b - 2\\overrightarrow c } \\right|^2} = (3\\overrightarrow a + \\overrightarrow b - 2\\overrightarrow c ).(3\\overrightarrow a + \\overrightarrow b - 2\\overrightarrow c )$$
$$ = 9{\\left| {\\overrightarrow a } \\right|^2} + {\\left| {\\overrightarrow b } \\right|^2} + 4{\\left| {\\overrightarrow c } \\right|^2}$$
$$(2 + a + b)\\widehat i + (a + 2b + c)\\widehat j - (b + c)\\widehat k,(1 + b)\\widehat i + 2b\\widehat j - b\\widehat k$$ and $$(2 + b)\\widehat i + 2b\\widehat j + (1 - b)\\widehat k$$, $$a,b,c, \\in R$$
be co-planar. Then which of the following is true?",
"options": [
{
"text": "2b = a + c"
},
{
"text": "3c = a + b"
},
{
"text": "a = b + 2c"
},
{
"text": "2a = b + c"
}
],
"answer": "2b = a + c",
"solution": "**Answer:** 2b = a + c\n\nIf the vectors are co-planar,
$$ = 2b - a - c = 0$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7701,
"subject": "General Science",
"question": "Let $$\\overrightarrow a $$, $$\\overrightarrow b $$ and $$\\overrightarrow c $$ be three vectors such that $$\\overrightarrow a $$ = $$\\overrightarrow b $$ $$\\times$$ ($$\\overrightarrow b $$ $$\\times$$ $$\\overrightarrow c $$). If magnitudes of the vectors $$\\overrightarrow a $$, $$\\overrightarrow b $$ and $$\\overrightarrow c $$ are $$\\sqrt 2 $$, 1 and 2 respectively and the angle between $$\\overrightarrow b $$ and $$\\overrightarrow c $$ is $$\\theta \\left( {0 < \\theta < {\\pi \\over 2}} \\right)$$, then the value of 1 + tan$$\\theta$$ is equal to :",
"options": [
{
"text": "$$\\sqrt 3 + 1$$"
},
{
"text": "2"
},
{
"text": "1"
},
{
"text": "$${{\\sqrt 3 + 1} \\over {\\sqrt 3 }}$$"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\n$$\\overrightarrow a = \\left( {\\overrightarrow b .\\,\\overrightarrow c } \\right)\\overrightarrow b - \\left( {\\overrightarrow b \\,.\\,\\overrightarrow b } \\right)\\overrightarrow c $$
$$ = 1.2\\cos \\theta \\overrightarrow b - \\overrightarrow c $$
$$ \\Rightarrow \\overrightarrow a = 2\\cos \\theta \\overrightarrow b - \\overrightarrow c $$
$${\\left| {\\overrightarrow a } \\right|^2} = {(2\\cos \\theta )^2} + {2^2} - 2.2\\cos \\theta \\overrightarrow b \\,.\\,\\overrightarrow c $$
$$ \\therefore $$ $$1 + \\tan \\theta = 2$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7702,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = \\widehat i - \\alpha \\widehat j + \\beta \\widehat k$$, $$\\overrightarrow b = 3\\widehat i + \\beta \\widehat j - \\alpha \\widehat k$$ and $$\\overrightarrow c = -\\alpha \\widehat i - 2\\widehat j + \\widehat k$$, where $$\\alpha$$ and $$\\beta$$ are integers. If $$\\overrightarrow a \\,.\\,\\overrightarrow b = - 1$$ and $$\\overrightarrow b \\,.\\,\\overrightarrow c = 10$$, then $$\\left( {\\overrightarrow a \\, \\times \\overrightarrow b } \\right).\\,\\overrightarrow c $$ is equal to ___________.",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\n$$\\overrightarrow a = (1, - \\alpha ,\\beta )$$
$$ = 3 + 2 + 4 = 9$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7703,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = \\widehat i + \\widehat j + \\widehat k$$ and $$\\overrightarrow b = \\widehat j - \\widehat k$$. If $$\\overrightarrow c $$ is a vector such that $$\\overrightarrow a \\times \\overrightarrow c = \\overrightarrow b $$ and $$\\overrightarrow a .\\overrightarrow c = 3$$, then $$\\overrightarrow a .(\\overrightarrow b \\times \\overrightarrow c )$$ is equal to :",
"options": [
{
"text": "$$-$$2"
},
{
"text": "$$-$$6"
},
{
"text": "6"
},
{
"text": "2"
}
],
"answer": "$$-$$2",
"solution": "**Answer:** $$-$$2\n\n$$\\left| {\\overrightarrow a } \\right| = \\sqrt 3 $$; $$\\overrightarrow a .\\overrightarrow c = 3$$; $$\\overrightarrow a \\times \\overrightarrow b = - 2\\widehat i + \\widehat j + \\widehat k$$, $$\\overrightarrow a \\times \\overrightarrow c = \\overrightarrow b $$
Cross with $$\\overrightarrow a $$,
$$\\overrightarrow a \\times (\\overrightarrow a \\times \\overrightarrow c ) = \\overrightarrow a \\times \\overrightarrow b $$
$$ \\Rightarrow (\\overrightarrow a .\\overrightarrow c )\\overrightarrow a - {a^2}\\overrightarrow c = \\overrightarrow a \\times \\overrightarrow b $$
$$ \\Rightarrow 3\\overrightarrow a - 3\\overrightarrow c = - 2\\widehat i + \\widehat j + \\widehat k$$
$$ \\Rightarrow 3\\widehat i + 3\\widehat j + 3\\widehat k - 3\\overrightarrow c = - 2\\widehat i + \\widehat j + \\widehat k$$
$$\\therefore$$ $$\\overrightarrow a .(\\overrightarrow b \\times \\overrightarrow c ) = (\\overrightarrow a \\times \\overrightarrow b ).\\overrightarrow c = {{ - 10} \\over 3} + {2 \\over 3} + {2 \\over 3} = - 2$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7704,
"subject": "General Science",
"question": "Let $$\\overrightarrow a ,\\overrightarrow b ,\\overrightarrow c $$ three vectors mutually perpendicular to each other and have same magnitude. If a vector $${ \\overrightarrow r } $$ satisfies.\n
$$\\overrightarrow a \\times \\{ (\\overrightarrow r - \\overrightarrow b ) \\times \\overrightarrow a \\} + \\overrightarrow b \\times \\{ (\\overrightarrow r - \\overrightarrow c ) \\times \\overrightarrow b \\} + \\overrightarrow c \\times \\{ (\\overrightarrow r - \\overrightarrow a ) \\times \\overrightarrow c \\} = \\overrightarrow 0 $$, then $$\\overrightarrow r $$ is equal to :",
"options": [
{
"text": "$${1 \\over 3}(\\overrightarrow a + \\overrightarrow b + \\overrightarrow c )$$"
},
{
"text": "$${1 \\over 3}(2\\overrightarrow a + \\overrightarrow b - \\overrightarrow c )$$"
},
{
"text": "$${1 \\over 2}(\\overrightarrow a + \\overrightarrow b + \\overrightarrow c )$$"
},
{
"text": "$${1 \\over 2}(\\overrightarrow a + \\overrightarrow b + 2\\overrightarrow c )$$"
}
],
"answer": "$${1 \\over 2}(\\overrightarrow a + \\overrightarrow b + \\overrightarrow c )$$",
"solution": "**Answer:** $${1 \\over 2}(\\overrightarrow a + \\overrightarrow b + \\overrightarrow c )$$\n\nSuppose $$\\overrightarrow r = x\\overrightarrow a + y\\overrightarrow b + 2\\overrightarrow c $$
and $$\\left| {\\overrightarrow a } \\right| = \\left| {\\overrightarrow b } \\right| = \\left| {\\overrightarrow c } \\right| = k$$
$$\\overrightarrow a \\times \\{ (\\overrightarrow r - \\overrightarrow b ) \\times \\overrightarrow a \\} + \\overrightarrow b \\times \\{ (\\overrightarrow r - \\overrightarrow c ) \\times \\overrightarrow b \\} + \\overrightarrow c \\times \\{ (\\overrightarrow r - \\overrightarrow a ) \\times \\overrightarrow c \\} = \\overrightarrow 0 $$
$$ \\Rightarrow {k^2}(\\overrightarrow r - \\overrightarrow b ) - {k^2}x\\overrightarrow a + {k^2}(\\overrightarrow r - \\overrightarrow c ) - {k^2}y\\overrightarrow b + {k^2}(\\overrightarrow r - \\overrightarrow a ) - {k^2}z\\overrightarrow c = \\overrightarrow 0 $$
$$ \\Rightarrow 3\\overrightarrow r -(\\overrightarrow a + \\overrightarrow b + \\overrightarrow c ) - \\overrightarrow r = \\overrightarrow 0 $$
$$ \\Rightarrow \\overrightarrow r = {{\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\over 2}$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7705,
"subject": "General Science",
"question": "
If $$\\overrightarrow a \\,.\\,\\overrightarrow b = 1,\\,\\overrightarrow b \\,.\\,\\overrightarrow c = 2$$ and $$\\overrightarrow c \\,.\\,\\overrightarrow a = 3$$, then the value of $$\\left[ {\\overrightarrow a \\times \\left( {\\overrightarrow b \\times \\overrightarrow c } \\right),\\,\\overrightarrow b \\times \\left( {\\overrightarrow c \\times \\overrightarrow a } \\right),\\,\\overrightarrow c \\times \\left( {\\overrightarrow b \\times \\overrightarrow a } \\right)} \\right]$$ is :
",
"options": [
{
"text": "0"
},
{
"text": "$$ - 6\\overrightarrow a \\,.\\,\\left( {\\overrightarrow b \\times \\overrightarrow c } \\right)$$"
},
{
"text": "$$ - 12\\overrightarrow c \\,.\\,\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)$$"
},
{
"text": "$$ - 12\\overrightarrow b \\,.\\,\\left( {\\overrightarrow c \\times \\overrightarrow a } \\right)$$"
}
],
"answer": "0",
"solution": "**Answer:** 0\n\n
$$\\because$$ $$\\overrightarrow a \\times \\left( {\\overrightarrow b \\times \\overrightarrow c } \\right) = 3\\overrightarrow b - \\overrightarrow c = \\overrightarrow u $$
\n
$$\\overrightarrow b \\times \\left( {\\overrightarrow c \\times \\overrightarrow a } \\right) = \\overrightarrow c - 2\\overrightarrow a = \\overrightarrow v $$
\n
$$\\overrightarrow c \\times \\left( {\\overrightarrow b \\times \\overrightarrow a } \\right) = 3\\overrightarrow b - 2\\overrightarrow a = \\overrightarrow w $$
\n
$$\\therefore$$ $$\\overrightarrow u + \\overrightarrow v = \\overrightarrow w $$
\n
So, vectors $$\\overrightarrow u $$, $$\\overrightarrow v $$ and $$\\overrightarrow w $$ are coplanar, hence their Scalar triple product will be zero.
Let a vector $$\\overrightarrow c $$ be coplanar with the vectors $$\\overrightarrow a = - \\widehat i + \\widehat j + \\widehat k$$ and $$\\overrightarrow b = 2\\widehat i + \\widehat j - \\widehat k$$. If the vector $$\\overrightarrow c $$ also satisfies the conditions $$\\overrightarrow c \\,.\\,\\left[ {\\left( {\\overrightarrow a + \\overrightarrow b } \\right) \\times \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)} \\right] = - 42$$ and $$\\left( {\\overrightarrow c \\times \\left( {\\overrightarrow a - \\overrightarrow b } \\right)} \\right)\\,.\\,\\widehat k = 3$$, then the value of $$|\\overrightarrow c {|^2}$$ is equal to :
Given, $$\\overrightarrow c \\,.\\,\\left[ {\\left( {\\overrightarrow a + \\overrightarrow b } \\right) \\times \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)} \\right] = - 42$$
Given, $$\\overrightarrow a = 2\\widehat i - \\widehat j + 5\\widehat k$$ and $$\\overrightarrow b = \\alpha \\widehat i + \\beta \\widehat j + 2\\widehat k$$
\n
Also, $$\\left( {\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) \\times i} \\right)\\,.\\,\\widehat k = {{23} \\over 2}$$
\n
$$ \\Rightarrow \\left( {\\left( {\\overrightarrow a \\,.\\,\\widehat i} \\right)\\overrightarrow b - \\left( {\\overrightarrow b \\,.\\,\\widehat i} \\right)\\,.\\,\\overline a } \\right)\\,.\\,\\widehat k = {{23} \\over 2}$$
\n
$$ \\Rightarrow \\left( {2\\,.\\,\\overrightarrow b - \\alpha \\,.\\,\\overrightarrow a } \\right)\\,.\\,\\widehat k = {{23} \\over 2}$$
Let $$\\overrightarrow{\\mathrm{a}}=3 \\hat{i}+\\hat{j}$$ and $$\\overrightarrow{\\mathrm{b}}=\\hat{i}+2 \\hat{j}+\\hat{k}$$. Let $$\\overrightarrow{\\mathrm{c}}$$ be a vector satisfying $$\\overrightarrow{\\mathrm{a}} \\times(\\overrightarrow{\\mathrm{b}} \\times \\overrightarrow{\\mathrm{c}})=\\overrightarrow{\\mathrm{b}}+\\lambda \\overrightarrow{\\mathrm{c}}$$. If $$\\overrightarrow{\\mathrm{b}}$$ and $$\\overrightarrow{\\mathrm{c}}$$ are non-parallel, then the value of $$\\lambda$$ is :
$$\\overrightarrow a = 3\\widehat i + \\widehat j$$ & $$\\overrightarrow b = \\widehat i + 2\\widehat j + \\widehat k$$
\n
$$\\overrightarrow a \\times (\\overrightarrow b \\times \\overrightarrow c ) = (\\overrightarrow a \\,.\\,\\overrightarrow c )\\overrightarrow b - (\\overrightarrow a \\,.\\,\\overrightarrow b )\\overrightarrow c = \\overrightarrow b + \\lambda \\overrightarrow c $$
\n
If $$\\overrightarrow b $$ & $$\\overrightarrow c $$ are non-parallel
\n
then $$\\overrightarrow a \\,.\\,\\overrightarrow c = 1$$ & $$\\overrightarrow a \\,.\\,\\overrightarrow b = - \\lambda $$
\n
but $$\\overrightarrow a \\,.\\,\\overrightarrow b = 5 \\Rightarrow \\lambda = - 5$$
Let $$\\vec{v}=\\alpha \\hat{i}+2 \\hat{j}-3 \\hat{k}, \\vec{w}=2 \\alpha \\hat{i}+\\hat{j}-\\hat{k}$$ and $$\\vec{u}$$ be a vector such that $$|\\vec{u}|=\\alpha>0$$. If the minimum value of the scalar triple product $$\\left[ {\\matrix{\n {\\overrightarrow u } & {\\overrightarrow v } & {\\overrightarrow w } \\cr \n\n } } \\right]$$ is $$-\\alpha \\sqrt{3401}$$, and $$|\\vec{u} \\cdot \\hat{i}|^{2}=\\frac{m}{n}$$ where $$m$$ and $$n$$ are coprime natural numbers, then $$m+n$$ is equal to ____________.
$$\\left[ {\\matrix{\n {\\overrightarrow u } & {\\overrightarrow v } & {\\overrightarrow w } \\cr \n\n } } \\right] = \\overrightarrow u .\\left( {\\overrightarrow v \\times \\overrightarrow w } \\right)$$\n\n
$$\\left( {\\overrightarrow a \\times \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) + \\overrightarrow b \\times \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)} \\right) \\times \\left( {\\overrightarrow a - \\overrightarrow b } \\right)$$
\n
$$ = \\left( {\\overrightarrow a \\left( {\\overrightarrow a .\\,\\overrightarrow b } \\right) - \\overrightarrow b \\left( {\\overrightarrow a .\\,\\overrightarrow a } \\right) + \\overrightarrow a \\left( {\\overrightarrow b .\\,\\overrightarrow b } \\right) - \\overrightarrow b \\left( {\\overrightarrow a .\\,\\overrightarrow b } \\right)} \\right) \\times \\left( {\\overrightarrow a - \\overrightarrow b } \\right)$$
\n
$$ = \\left( {\\overrightarrow a .\\,\\overrightarrow b } \\right)\\left( {\\overrightarrow a \\times \\overrightarrow a - \\overrightarrow a \\times \\overrightarrow b } \\right) - \\left( {\\overrightarrow a .\\,\\overrightarrow a } \\right)\\left( {\\overrightarrow b \\times \\overrightarrow a - \\overrightarrow b \\times \\overrightarrow b } \\right) + \\left( {\\overrightarrow b .\\,\\overrightarrow b } \\right)\\left( {\\overrightarrow a \\times \\overrightarrow a - \\overrightarrow a \\times \\overrightarrow b } \\right) - \\left( {\\overrightarrow a .\\,\\overrightarrow b } \\right)\\left( {\\overrightarrow b \\times \\overrightarrow a - \\overrightarrow b \\times \\overrightarrow b } \\right)$$
\n
$$ = \\left( {\\overrightarrow a .\\,\\overrightarrow b } \\right)\\left( {\\overrightarrow b \\times \\overrightarrow a } \\right) - \\left( {\\overrightarrow a .\\,\\overrightarrow a } \\right)\\left( {\\overrightarrow b \\times \\overrightarrow a } \\right) + \\left( {\\overrightarrow b .\\,\\overrightarrow b } \\right)\\left( {\\overrightarrow b \\times \\overrightarrow a } \\right) - \\left( {\\overrightarrow a .\\,\\overrightarrow b } \\right)\\left( {\\overrightarrow b \\times \\overrightarrow a } \\right)$$
\n
$$ = \\left( {\\overrightarrow b \\times \\overrightarrow a } \\right)\\left( {\\overrightarrow b .\\,\\overrightarrow b - \\overrightarrow a .\\,\\overrightarrow a } \\right)$$
\n
$$ = \\left( {5\\overrightarrow b \\times \\overrightarrow a } \\right)\\left( {5 + {\\lambda ^2} - 13 - {\\lambda ^2}} \\right)$$
\n
$$ = 8\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)$$
$${\\left| {\\overrightarrow a \\times \\left( {\\overrightarrow a - \\overrightarrow b } \\right) + \\overrightarrow b \\times \\left( {\\overrightarrow a - \\overrightarrow b } \\right)} \\right|^2}$$
\n
$$ = {\\left| {2\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)} \\right|^2} = 4\\,.\\,35 = 140$$
If $$\\overrightarrow a ,\\overrightarrow b ,\\overrightarrow c $$ are three non-zero vectors and $$\\widehat n$$ is a unit vector perpendicular to $$\\overrightarrow c $$ such that $$\\overrightarrow a = \\alpha \\overrightarrow b - \\widehat n,(\\alpha \\ne 0)$$ and $$\\overrightarrow b \\,.\\overrightarrow c = 12$$, then $$\\left| {\\overrightarrow c \\times (\\overrightarrow a \\times \\overrightarrow b )} \\right|$$ is equal to :
$$\\widehat n = \\alpha \\overrightarrow b - \\overrightarrow a $$
\n
$$\\overrightarrow c \\times \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) = \\left( {\\overrightarrow c \\,.\\,\\overrightarrow b } \\right)\\overrightarrow a - \\left( {\\overrightarrow c \\,.\\,\\overrightarrow a } \\right)\\overrightarrow b $$
\n
$$ = 12\\overrightarrow a - \\left( {\\overrightarrow c \\,.\\,\\left( {\\alpha \\overrightarrow b - \\widehat n} \\right)} \\right)\\overrightarrow b $$
\n
$$ = 12\\overrightarrow a - (12\\alpha - 0)\\overrightarrow b $$
\n
$$ = 12\\left( {\\overrightarrow a - \\alpha \\overrightarrow b } \\right)$$
\n
$$\\therefore$$ $$\\left| {\\overrightarrow c \\times \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)} \\right| = 12$$
Let $$\\overrightarrow a $$, $$\\overrightarrow b $$ and $$\\overrightarrow c $$ be three non-zero non-coplanar vectors. Let the position vectors of four points $$A,B,C$$ and $$D$$ be $$\\overrightarrow a - \\overrightarrow b + \\overrightarrow c ,\\lambda \\overrightarrow a - 3\\overrightarrow b + 4\\overrightarrow c , - \\overrightarrow a + 2\\overrightarrow b - 3\\overrightarrow c $$ and $$2\\overrightarrow a - 4\\overrightarrow b + 6\\overrightarrow c $$ respectively. If $$\\overrightarrow {AB} ,\\overrightarrow {AC} $$ and $$\\overrightarrow {AD} $$ are coplanar, then $$\\lambda$$ is equal to __________.
Let $$\\overrightarrow a = - \\widehat i - \\widehat j + \\widehat k,\\overrightarrow a \\,.\\,\\overrightarrow b = 1$$ and $$\\overrightarrow a \\times \\overrightarrow b = \\widehat i - \\widehat j$$. Then $$\\overrightarrow a - 6\\overrightarrow b $$ is equal to :
If the four points, whose position vectors are $$3\\widehat i - 4\\widehat j + 2\\widehat k,\\widehat i + 2\\widehat j - \\widehat k, - 2\\widehat i - \\widehat j + 3\\widehat k$$ and $$5\\widehat i - 2\\alpha \\widehat j + 4\\widehat k$$ are coplanar, then $$\\alpha$$ is equal to :
Let $$\\overrightarrow a $$, $$\\overrightarrow b $$ and $$\\overrightarrow c $$ be three non zero vectors such that $$\\overrightarrow b $$ . $$\\overrightarrow c $$ = 0 and $$\\overrightarrow a \\times (\\overrightarrow b \\times \\overrightarrow c ) = {{\\overrightarrow b - \\overrightarrow c } \\over 2}$$. If $$\\overrightarrow d $$ be a vector such that $$\\overrightarrow b \\,.\\,\\overrightarrow d = \\overrightarrow a \\,.\\,\\overrightarrow b $$, then $$(\\overrightarrow a \\times \\overrightarrow b )\\,.\\,(\\overrightarrow c \\times \\overrightarrow d )$$ is equal to
Let $$\\overrightarrow u = \\widehat i - \\widehat j - 2\\widehat k,\\overrightarrow v = 2\\widehat i + \\widehat j - \\widehat k,\\overrightarrow v .\\,\\overrightarrow w = 2$$ and $$\\overrightarrow v \\times \\overrightarrow w = \\overrightarrow u + \\lambda \\overrightarrow v $$. Then $$\\overrightarrow u .\\,\\overrightarrow w $$ is equal to :
For three vectors to be coplanar, their scalar triple product must be zero. We have the vectors A, B, and C, and we know the given relation between λ and μ:\n\n
So, the correct answer is 2290 (Option D).",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7718,
"subject": "General Science",
"question": "
Let $$a, b, c$$ be three distinct real numbers, none equal to one. If the vectors $$a \\hat{i}+\\hat{\\mathrm{j}}+\\hat{\\mathrm{k}}, \\hat{\\mathrm{i}}+b \\hat{j}+\\hat{\\mathrm{k}}$$ and $$\\hat{\\mathrm{i}}+\\hat{\\mathrm{j}}+c \\hat{\\mathrm{k}}$$ are coplanar, then $$\\frac{1}{1-a}+\\frac{1}{1-b}+\\frac{1}{1-c}$$ is equal to :
Let $$\\lambda \\in \\mathbb{Z}, \\vec{a}=\\lambda \\hat{i}+\\hat{j}-\\hat{k}$$ and $$\\vec{b}=3 \\hat{i}-\\hat{j}+2 \\hat{k}$$. Let $$\\vec{c}$$ be a vector such that $$(\\vec{a}+\\vec{b}+\\vec{c}) \\times \\vec{c}=\\overrightarrow{0}, \\vec{a} \\cdot \\vec{c}=-17$$ and $$\\vec{b} \\cdot \\vec{c}=-20$$. Then $$|\\vec{c} \\times(\\lambda \\hat{i}+\\hat{j}+\\hat{k})|^{2}$$ is equal to :
We are given that $(\\vec{a} + \\vec{b} + \\vec{c}) \\times \\vec{c} = 0$ which implies $(\\vec{a} + \\vec{b}) \\times \\vec{c} = 0$. So, $\\vec{c}$ is in the direction of $\\vec{a} + \\vec{b}$.\n\n
If four distinct points with position vectors $$\\vec{a}, \\vec{b}, \\vec{c}$$ and $$\\vec{d}$$ are coplanar, then $$[\\vec{a} \\,\\,\\vec{b} \\,\\,\\vec{c}]$$ is equal to :
Let the vectors $$\\vec{u}_{1}=\\hat{i}+\\hat{j}+a \\hat{k}, \\vec{u}_{2}=\\hat{i}+b \\hat{j}+\\hat{k}$$ and $$\\vec{u}_{3}=c \\hat{i}+\\hat{j}+\\hat{k}$$ be coplanar. If the vectors $$\\vec{v}_{1}=(a+b) \\hat{i}+c \\hat{j}+c \\hat{k}, \\vec{v}_{2}=a \\hat{i}+(b+c) \\hat{j}+a \\hat{k}$$ and $$\\vec{v}_{3}=b \\hat{i}+b \\hat{j}+(c+a) \\hat{k}$$ are also coplanar, then $$6(\\mathrm{a}+\\mathrm{b}+\\mathrm{c})$$ is equal to :
$$\n\\begin{aligned}\n& \\Rightarrow\\left|\\begin{array}{ccc}\na+b & c & c \\\\\na & b+c & a \\\\\nb & b & c+a\n\\end{array}\\right|=0 \\\\\n& \\Rightarrow(a+b)\\left[b c+b a+c^2+c a-a b\\right]-c\\left[a c+a^2-a b\\right] \\\\\\\\\n& \\quad+c\\left[a b-b^2-b c\\right]=0\n\\end{aligned}\n$$\n
$$\n\\begin{aligned}\n& \\Rightarrow a b c+a c^2+a^2 c+b^2 c+b c^2+a b c-a c^2-a^2 c \\\\\\\\\n& +a b c+a b c-b^2 c-b c^2=0 \\\\\\\\\n& \\Rightarrow 4 a b c=0 \\Rightarrow a b c=0 \\\\\\\\\n& \\begin{array}{lr}\n\\text { So, } a+b+c-2=0 [from (i)]\\\\\\\\\n\\Rightarrow a+b+c=2\n\\end{array} \\\\\\\\\n& \\Rightarrow 6(a+b+c)=12\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7722,
"subject": "General Science",
"question": "
Let the position vectors of the points A, B, C and D be\n$$5 \\hat{i}+5 \\hat{j}+2 \\lambda \\hat{k}, \\hat{i}+2 \\hat{j}+3 \\hat{k},-2 \\hat{i}+\\lambda \\hat{j}+4 \\hat{k}$$ and $$-\\hat{i}+5 \\hat{j}+6 \\hat{k}$$. Let the set $$S=\\{\\lambda \\in \\mathbb{R}$$ :\nthe points A, B, C and D are coplanar $$\\}$$. \n
Then $$\\sum_\\limits{\\lambda \\in S}(\\lambda+2)^{2}$$ is equal to :
",
"options": [
{
"text": "$$\\frac{37}{2}$$"
},
{
"text": "25"
},
{
"text": "13"
},
{
"text": "41"
}
],
"answer": "41",
"solution": "**Answer:** 41\n\nGiven, position vectors of the points $A, B, C$ and $D$ be \n
Let the vectors $$\\vec{a}, \\vec{b}, \\vec{c}$$ represent three coterminous edges of a parallelopiped of volume V. Then the volume of the parallelopiped, whose coterminous edges are represented by $$\\vec{a}, \\vec{b}+\\vec{c}$$ and $$\\vec{a}+2 \\vec{b}+3 \\vec{c}$$ is equal to :
",
"options": [
{
"text": "3 V"
},
{
"text": "2 V"
},
{
"text": "6 V"
},
{
"text": "V"
}
],
"answer": "V",
"solution": "**Answer:** V\n\nGiven that the volume $V$ of the parallelepiped formed by the vectors $\\vec{a}$, $\\vec{b}$, and $\\vec{c}$ is represented by the scalar triple product $[\\vec{a},\\vec{b},\\vec{c}]$, which is the determinant of the 3 x 3 matrix with vectors $\\vec{a}$, $\\vec{b}$, and $\\vec{c}$ as its rows (or columns).\n\n
When the vectors representing the coterminous edges of the parallelepiped are $\\vec{a}$, $\\vec{b} + \\vec{c}$, and $\\vec{a} + 2\\vec{b} + 3\\vec{c}$, the volume $V$ of the parallelepiped is represented by :\n\n
The sum of all values of $$\\alpha$$, for which the points whose position vectors are $$\\hat{i}-2 \\hat{j}+3 \\hat{k}, 2 \\hat{i}-3 \\hat{j}+4 \\hat{k},(\\alpha+1) \\hat{i}+2 \\hat{k}$$ and $$9 \\hat{i}+(\\alpha-8) \\hat{j}+6 \\hat{k}$$ are coplanar, is equal to :
Let $$\\vec{a}=\\hat{i}+\\hat{j}+\\hat{k}, \\vec{b}=2 \\hat{i}+4 \\hat{j}-5 \\hat{k}$$ and $$\\vec{c}=x \\hat{i}+2 \\hat{j}+3 \\hat{k}, x \\in \\mathbb{R}$$.\nIf $$\\vec{d}$$ is the unit vector in the direction of $$\\vec{b}+\\vec{c}$$ such that $$\\vec{a} \\cdot \\vec{d}=1$$, then $$(\\vec{a} \\times \\vec{b}) \\cdot \\vec{c}$$ is equal to
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7727,
"subject": "General Science",
"question": "If $$\\left| {\\overrightarrow a } \\right| = 5,\\left| {\\overrightarrow b } \\right| = 4,\\left| {\\overrightarrow c } \\right| = 3$$ thus what will be the value of $$\\left| {\\overrightarrow a .\\overrightarrow b + \\overrightarrow b .\\overrightarrow c + \\overrightarrow c .\\overrightarrow a } \\right|,$$ given that $$\\overrightarrow a + \\overrightarrow b + \\overrightarrow c = 0$$ :",
"options": [
{
"text": "$$25$$"
},
{
"text": "$$50$$ "
},
{
"text": "$$-25$$"
},
{
"text": "$$-50$$"
}
],
"answer": "$$25$$",
"solution": "**Answer:** $$25$$\n\nWe have, $$\\overrightarrow a + \\overrightarrow b + \\overrightarrow c = \\overrightarrow 0 $$\n
$$ \\Rightarrow {\\left( {\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\right)^2} = 0$$\n
$$ \\Rightarrow {\\left| {\\overrightarrow a } \\right|^2} + {\\left| {\\overrightarrow b } \\right|^2} + {\\left| {\\overrightarrow c } \\right|^2}$$\n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, + 2\\left( {\\overrightarrow a \\,.\\,\\overrightarrow b + \\overrightarrow b \\,.\\,\\overrightarrow c + \\overrightarrow c \\,.\\,\\overrightarrow a } \\right) = 0$$\n
$$ \\Rightarrow 25 + 16 + 9 + 2\\left( {\\overrightarrow a \\,.\\,\\overrightarrow b + \\overrightarrow b \\,.\\,\\overrightarrow c + \\overrightarrow c \\,.\\,\\overrightarrow a } \\right) = 0$$\n
$$ \\Rightarrow \\left( {\\overrightarrow a \\,.\\,\\overrightarrow b + \\overrightarrow b \\,.\\,\\overrightarrow c + \\overrightarrow c \\,.\\,\\overrightarrow a } \\right) = - 25.$$\n
$$\\therefore$$ $$\\left| {\\overrightarrow a \\,.\\,\\overrightarrow b + \\overrightarrow b \\,.\\,\\overrightarrow c + \\overrightarrow c \\,.\\,\\overrightarrow a } \\right| = 25.$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7728,
"subject": "General Science",
"question": "$$\\overrightarrow a \\,,\\overrightarrow b \\,,\\overrightarrow c $$ are $$3$$ vectors, such that
$$\\overrightarrow a + \\overrightarrow b + \\overrightarrow c = 0$$ , $$\\left| {\\overrightarrow a } \\right| = 1\\,\\,\\,\\left| {\\overrightarrow b } \\right| = 2,\\,\\,\\,\\left| {\\overrightarrow c } \\right| = 3,$$,\n
then $${\\overrightarrow a .\\overrightarrow b + \\overrightarrow b .\\overrightarrow c + \\overrightarrow c .\\overrightarrow a }$$ is equal to :",
"options": [
{
"text": "$$1$$"
},
{
"text": "$$0$$"
},
{
"text": "$$-7$$ "
},
{
"text": "$$7$$"
}
],
"answer": "$$-7$$ ",
"solution": "**Answer:** $$-7$$ \n\n$$\\overrightarrow a + \\overrightarrow b + \\overrightarrow c = 0$$ \n
$$ \\Rightarrow \\left( {\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\right).\\left( {\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\right) = 0$$\n
$${\\left| {\\overrightarrow a } \\right|^2} + {\\left| {\\overrightarrow b } \\right|^2} + {\\left| {\\overrightarrow c } \\right|^2} + 2\\left( {\\overrightarrow a .\\overrightarrow b + \\overrightarrow b .\\overrightarrow c + \\overrightarrow c .\\overrightarrow a } \\right) = 0$$\n
$$\\overrightarrow a .\\overrightarrow b + \\overrightarrow b .\\overrightarrow c + \\overrightarrow c .\\overrightarrow a = {{ - 1 - 4 - 9} \\over 2}$$\n
$$ = - 7$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7729,
"subject": "General Science",
"question": "A particle acted on by constant forces $$4\\widehat i + \\widehat j - 3\\widehat k$$ and $$3\\widehat i + \\widehat j - \\widehat k$$ is displaced from the point $$\\widehat i + 2\\widehat j + 3\\widehat k$$ to the point $$\\,5\\widehat i + 4\\widehat j + \\widehat k.$$ The total work done by the forces is : ",
"options": [
{
"text": "$$50$$ units "
},
{
"text": "$$20$$ units "
},
{
"text": "$$30$$ units "
},
{
"text": "$$40$$ units "
}
],
"answer": "$$40$$ units ",
"solution": "**Answer:** $$40$$ units \n\nThe work done by a force on a particle is given by the dot product of the force and the displacement vector of the particle. The displacement vector can be found by subtracting the initial position from the final position:\n
\nThe total work done by the two forces is equal to the sum of the work done by each force. The work done by each force can be calculated as the dot product of the force and the displacement:\n
\nTherefore, the total work done by the forces is 40 J (joules) or 40 units.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7730,
"subject": "General Science",
"question": "Let $$\\overrightarrow u ,\\overrightarrow v ,\\overrightarrow w $$ be such that $$\\left| {\\overrightarrow u } \\right| = 1,\\,\\,\\,\\left| {\\overrightarrow v } \\right|2,\\,\\,\\,\\left| {\\overrightarrow w } \\right|3.$$ If the projection $${\\overrightarrow v }$$ along $${\\overrightarrow u }$$ is equal to that of $${\\overrightarrow w }$$ along $${\\overrightarrow u }$$ and $${\\overrightarrow v },$$ $${\\overrightarrow w }$$ are perpendicular to each other then $$\\left| {\\overrightarrow u - \\overrightarrow v + \\overrightarrow w } \\right|$$ equals :",
"options": [
{
"text": "$$14$$ "
},
{
"text": "$${\\sqrt {7} }$$"
},
{
"text": "$${\\sqrt {14} }$$ "
},
{
"text": "$$2$$"
}
],
"answer": "$${\\sqrt {14} }$$ ",
"solution": "**Answer:** $${\\sqrt {14} }$$ \n\nProjection of $$\\overrightarrow v $$ along $$\\overrightarrow u = {{\\overrightarrow v .\\overrightarrow u } \\over {\\left| {\\overrightarrow u } \\right|}} = {{\\overrightarrow v .\\overrightarrow u } \\over 2}$$\n
projection of $$\\overrightarrow w $$ along $$\\overrightarrow u = {{\\overrightarrow w .\\overrightarrow u } \\over {\\left| {\\overrightarrow u } \\right|}} = {{\\overrightarrow w .\\overrightarrow u } \\over 2}$$\n
Given $${{\\overrightarrow v .\\overrightarrow u } \\over 2} = {{\\overrightarrow w .\\overrightarrow u } \\over 2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( 1 \\right)$$\n
Also, $$\\overrightarrow v .\\overrightarrow w = 0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( 2 \\right)$$\n
Now $${\\left| {\\overrightarrow u - \\overrightarrow v + \\overrightarrow w } \\right|^2}$$ \n
$$ = {\\left| {\\overrightarrow u } \\right|^2} + {\\left| {\\overrightarrow v } \\right|^2} + {\\left| {\\overrightarrow w } \\right|^2} - $$\n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\overrightarrow u .\\overrightarrow v - 2\\overrightarrow v .\\overrightarrow w + 2\\overrightarrow u .\\overrightarrow w $$\n
$$ = 1 + 4 + 9 + 0$$ [ From $$(1)$$ and $$(2)$$ ] $$=14$$ \n
$$\\therefore$$ $$\\left| {\\overrightarrow u - \\overrightarrow v + \\overrightarrow w } \\right| = \\sqrt {14} $$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7731,
"subject": "General Science",
"question": "The values of a, for which the points $$A, B, C$$ with position vectors $$2\\widehat i - \\widehat j + \\widehat k,\\,\\,\\widehat i - 3\\widehat j - 5\\widehat k$$ and $$a\\widehat i - 3\\widehat j + \\widehat k$$ respectively are the vertices of a right angled triangle with $$C = {\\pi \\over 2}$$ are :",
"options": [
{
"text": "$$2$$ and $$1$$ "
},
{
"text": "$$-2$$ and $$-1$$ "
},
{
"text": "$$-2$$ and $$1$$ "
},
{
"text": "$$2$$ and $$-1$$ "
}
],
"answer": "$$2$$ and $$1$$ ",
"solution": "**Answer:** $$2$$ and $$1$$ \n\n$$\\overrightarrow {CA} = \\left( {2 - a} \\right)\\widehat i + 2\\widehat j;$$\n
On solving $$(i)$$ and $$(ii)$$, we get $$\\lambda = - 3,\\mu = 2$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7733,
"subject": "General Science",
"question": "Let $$\\overrightarrow a $$ and $$\\overrightarrow b $$ be two unit vectors. If the vectors $$\\,\\overrightarrow c = \\widehat a + 2\\widehat b$$ and $$\\overrightarrow d = 5\\widehat a - 4\\widehat b$$ are perpendicular to each other, then the angle between $$\\overrightarrow a $$ and $$\\overrightarrow b $$ is :",
"options": [
{
"text": "$${\\pi \\over 6}$$ "
},
{
"text": "$${\\pi \\over 2}$$"
},
{
"text": "$${\\pi \\over 3}$$"
},
{
"text": "$${\\pi \\over 4}$$"
}
],
"answer": "$${\\pi \\over 3}$$",
"solution": "**Answer:** $${\\pi \\over 3}$$\n\nLet $$\\overrightarrow c = \\widehat a + 2\\widehat b$$ and $$\\overrightarrow d = 5\\widehat a - 4\\widehat b$$\n
Since $$\\overrightarrow c $$ and $$\\overrightarrow d $$ are perpendicular to each other \n
$$\\therefore$$ $$\\overrightarrow c .\\overrightarrow d = 0 \\Rightarrow \\left( {\\widehat a + 2\\widehat b} \\right).\\left( {5\\widehat a - 4\\widehat b} \\right) = 0$$\n
$$ \\Rightarrow 5 + 6\\widehat a.\\widehat b - 8 = 0$$ $$\\,\\,\\,\\,\\,\\,$$ (as $$\\widehat a.\\widehat a = 1$$)\n
$$ \\Rightarrow \\widehat a.\\widehat b = {1 \\over 2} \\Rightarrow \\theta = {\\pi \\over 3}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7734,
"subject": "General Science",
"question": "Let $$ABCD$$ be a parallelogram such that $$\\overrightarrow {AB} = \\overrightarrow q ,\\overrightarrow {AD} = \\overrightarrow p $$ and $$\\angle BAD$$ be an acute angle. If $$\\overrightarrow r $$ is the vector that coincide with the altitude directed from the vertex $$B$$ to the side $$AD,$$ then $$\\overrightarrow r $$ is given by :",
"options": [
{
"text": "$$\\overrightarrow r = 3\\overrightarrow q - {{3\\left( {\\overrightarrow p .\\overrightarrow q } \\right)} \\over {\\left( {\\overrightarrow p .\\overrightarrow p } \\right)}}\\overrightarrow p $$ "
},
{
"text": "$$\\overrightarrow r = - \\overrightarrow q + {{\\left( {\\overrightarrow p .\\overrightarrow q } \\right)} \\over {\\left( {\\overrightarrow p .\\overrightarrow p } \\right)}}\\overrightarrow p $$ "
},
{
"text": "$$\\vec r = \\vec q - {{\\left( {\\vec p.\\vec q} \\right)} \\over {\\left( {\\vec p.\\vec p} \\right)}}\\vec p$$ "
},
{
"text": "$$\\overrightarrow r = - 3\\overrightarrow q - {{3\\left( {\\overrightarrow p .\\overrightarrow q } \\right)} \\over {\\left( {\\overrightarrow p .\\overrightarrow p } \\right)}}$$ "
}
],
"answer": "$$\\overrightarrow r = - \\overrightarrow q + {{\\left( {\\overrightarrow p .\\overrightarrow q } \\right)} \\over {\\left( {\\overrightarrow p .\\overrightarrow p } \\right)}}\\overrightarrow p $$ ",
"solution": "**Answer:** $$\\overrightarrow r = - \\overrightarrow q + {{\\left( {\\overrightarrow p .\\overrightarrow q } \\right)} \\over {\\left( {\\overrightarrow p .\\overrightarrow p } \\right)}}\\overrightarrow p $$ \n\nLet $$ABCD$$ be a parallelogram such that\n
$$\\overrightarrow {AB} = \\overrightarrow q ,\\overrightarrow {AD} = \\overrightarrow p $$ and $$\\angle BAD$$ be an acute angle.\n
We have\n
\n
$$\\overrightarrow {AX} = \\left( {{{\\overrightarrow p .\\overrightarrow q } \\over {\\left| {\\overrightarrow p } \\right|}}} \\right)\\left( {{{\\overrightarrow p } \\over {\\left| {\\overrightarrow p } \\right|}}} \\right) = {{\\overrightarrow p .\\overrightarrow q } \\over {{{\\left| {\\overrightarrow p } \\right|}^2}}}\\overrightarrow p $$\n
Let $$\\overrightarrow r = \\overrightarrow {BX} = \\overrightarrow {BA} + \\overrightarrow {AX} $$ \n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, = - \\overrightarrow q + {{\\overrightarrow p .\\overrightarrow q } \\over {{{\\left| {\\overrightarrow p } \\right|}^2}}}\\overrightarrow p $$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 7735,
"subject": "General Science",
"question": "In a triangle ABC, right angled at the vertex A, if the position vectors of A, B and C are respectively 3$$\\widehat i$$ + $$\\widehat j$$ $$-$$ $$\\widehat k$$, $$-$$$$\\widehat i$$ + 3$$\\widehat j$$ + p$$\\widehat k$$ and 5$$\\widehat i$$ + q$$\\widehat j$$ $$-$$ 4$$\\widehat k$$, then the point (p, q) lies\non a line :",
"options": [
{
"text": "parallel to x-axis. "
},
{
"text": "parallel to y-axis."
},
{
"text": "making an acute angle with the positive direction of x-axis."
},
{
"text": "making an obtuse angle with the positive direction of x-axis. "
}
],
"answer": "making an acute angle with the positive direction of x-axis.",
"solution": "**Answer:** making an acute angle with the positive direction of x-axis.\n\nGiven, \n
$$\\overrightarrow A = 3\\widehat i + \\widehat j - \\widehat k$$\n
$$\\overrightarrow B = - \\widehat i + 3\\widehat j - p\\widehat k$$\n
$$\\overrightarrow C = 5\\widehat i + 9\\widehat j - 4\\widehat k$$\n
$$ \\therefore $$ line makes an angle less than 90o or acute angle with the positive direction of x-axis.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 7736,
"subject": "General Science",
"question": "Let $$\\overrightarrow u $$ be a vector coplanar with the vectors $$\\overrightarrow a = 2\\widehat i + 3\\widehat j - \\widehat k$$ and $$\\overrightarrow b = \\widehat j + \\widehat k$$. If $$\\overrightarrow u $$ is perpendicular to $$\\overrightarrow a $$ and $$\\overrightarrow u .\\overrightarrow b = 24$$, then $${\\left| {\\overrightarrow u } \\right|^2}$$ is equal to",
"options": [
{
"text": "336"
},
{
"text": "315"
},
{
"text": "256"
},
{
"text": "84"
}
],
"answer": "336",
"solution": "**Answer:** 336\n\nYou should know that, when $$\\overrightarrow u $$ is coplanar with $$\\overrightarrow a $$ and $$\\overrightarrow b $$ then we can write $$\\overrightarrow u = x\\overrightarrow a + y\\overrightarrow b $$ \n
Here, $$\\overrightarrow u $$ is perpendicular with $$\\overrightarrow a $$ then, \n
$$\\overrightarrow u .\\overrightarrow a = 0$$\n
$$ \\Rightarrow \\,\\,\\,\\,\\left( {x\\,\\overrightarrow a + y\\overrightarrow b } \\right)\\,.\\overrightarrow a = 0$$\n
$$ \\Rightarrow \\,\\,\\,\\,x\\,.\\overrightarrow {\\,a} \\,.\\,\\overrightarrow a \\, + \\,y\\,.\\,\\overrightarrow a \\,.\\,\\overrightarrow b = 0$$\n
$$ \\Rightarrow \\,\\,\\,\\,x\\,.\\,{\\left| {\\overrightarrow a } \\right|^2} + \\,y\\overrightarrow a \\,.\\,\\overrightarrow b = 0$$\n
$$ \\Rightarrow \\,\\,\\,\\,x + y = 12.......\\left( 2 \\right)$$\n
By solvig (1) and (2) we get, \n
x = - 2 and y = 14\n
Now, $${\\left| {\\overrightarrow u } \\right|^2} = \\overrightarrow u .\\overrightarrow u $$\n
$$ = \\,\\,\\,\\,\\left( {x\\overrightarrow a + y\\overrightarrow b } \\right).\\overrightarrow u $$\n
$$ = \\,\\,\\,\\,x\\overrightarrow a .\\overrightarrow u + y\\overrightarrow b .\\overrightarrow u $$\n
$$ = \\,\\,\\,\\,x\\overrightarrow a .\\overrightarrow u + y\\overrightarrow b .\\overrightarrow u $$\n
$$ = \\,\\,\\,\\,0 + 14 \\times 24$$ [as $$\\overrightarrow a .\\overrightarrow u = 0$$ and $$\\overrightarrow u .\\overrightarrow b = 24]$$\n
$$=\\,\\,\\,\\,$$ 336",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7737,
"subject": "General Science",
"question": "Let $$\\sqrt 3 \\widehat i + \\widehat j,$$ $$\\widehat i + \\sqrt 3 \\widehat j$$ and $$\\beta \\widehat i + \\left( {1 - \\beta } \\right)\\widehat j$$ respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is $${3 \\over {\\sqrt 2 }}$$, then the sum of all possible values of $$\\beta $$ is : ",
"options": [
{
"text": "4"
},
{
"text": "1"
},
{
"text": "2"
},
{
"text": "3"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\nAngle bisector is x $$-$$ y = 0\n
$$ \\Rightarrow $$ $$\\beta $$ = 2 or $$-$$ 1",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7738,
"subject": "General Science",
"question": "Let A (3, 0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G\ndivides BM in the ratio, 2 : 1, then cos ($$\\angle $$GOA) (O being the origin) is equal to :",
"options": [
{
"text": "$${1 \\over {\\sqrt {15} }}$$"
},
{
"text": "$${1 \\over {6\\sqrt {10} }}$$"
},
{
"text": "$${1 \\over {\\sqrt {30} }}$$"
},
{
"text": "$${1 \\over {2\\sqrt {15} }}$$"
}
],
"answer": "$${1 \\over {\\sqrt {15} }}$$",
"solution": "**Answer:** $${1 \\over {\\sqrt {15} }}$$\n\nG is the centroid of $$\\Delta $$ABC
and $${\\lambda _3}$$ = $$-1 - 2{\\lambda _1}$$ = - 1 + 1 = 0",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7741,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = \\widehat i + \\widehat j + \\sqrt 2 \\widehat k,$$ $$\\overrightarrow b = {b_1}\\widehat i + {b_2}\\widehat j + \\sqrt 2 \\widehat k$$, $$\\overrightarrow c = 5\\widehat i + \\widehat j + \\sqrt 2 \\widehat k$$ be three vectors such that the projection vector of $$\\overrightarrow b $$ on $$\\overrightarrow a $$ is $$\\overrightarrow a $$. \n If $$\\overrightarrow a + \\overrightarrow b $$ is perpendicular to $$\\overrightarrow c $$ , then $$\\left| {\\overrightarrow b } \\right|$$ is equal to : ",
"options": [
{
"text": "$$\\sqrt {32} $$"
},
{
"text": "6"
},
{
"text": "$$\\sqrt {22} $$"
},
{
"text": "4"
}
],
"answer": "6",
"solution": "**Answer:** 6\n\nProjection of $$\\overrightarrow b $$ on $$\\overrightarrow a $$ is $$\\overrightarrow a $$\n
$$ \\therefore $$ $${{\\overrightarrow b \\cdot \\overrightarrow a } \\over {\\left| {\\overrightarrow a } \\right|}} = \\left| {\\overrightarrow a } \\right|$$\n
$$ \\therefore $$ $$\\overrightarrow a = \\left( {4\\widehat i + 2\\widehat j + 4\\widehat k} \\right)$$\n
Then $$\\overrightarrow a .\\widehat k - 4 $$\n
= 4 - 4 = 0",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7743,
"subject": "General Science",
"question": "Let $$\\overrightarrow a $$, $$\\overrightarrow b $$ and $$\\overrightarrow c $$ be three unit vectors such that\n $${\\left| {\\overrightarrow a - \\overrightarrow b } \\right|^2}$$ + $${\\left| {\\overrightarrow a - \\overrightarrow c } \\right|^2}$$ = 8.\n
Then $${\\left| {\\overrightarrow a + 2\\overrightarrow b } \\right|^2}$$ + $${\\left| {\\overrightarrow a + 2\\overrightarrow c } \\right|^2}$$ is equal to ______.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nGiven, $$\\left| {\\overrightarrow a } \\right| = \\left| {\\overrightarrow b } \\right| = \\left| {\\overrightarrow c } \\right| = 1$$\n
$${\\left| {\\overrightarrow a - \\overrightarrow b } \\right|^2}$$ + $${\\left| {\\overrightarrow a - \\overrightarrow c } \\right|^2}$$ = 8\n
$$ \\Rightarrow $$ $${\\left| {\\overrightarrow a } \\right|^2} + {\\left| {\\overrightarrow b } \\right|^2} - 2\\overrightarrow a .\\overrightarrow b + {\\left| {\\overrightarrow a } \\right|^2} + {\\left| {\\overrightarrow c } \\right|^2} - 2\\overrightarrow a .\\overrightarrow c $$ = 8\n
$$ \\Rightarrow $$ $$\\overrightarrow a .\\overrightarrow b + \\overrightarrow a .\\overrightarrow c $$ = -2\n
Now, $${\\left| {\\overrightarrow a + 2\\overrightarrow b } \\right|^2}$$ + $${\\left| {\\overrightarrow a + 2\\overrightarrow c } \\right|^2}$$\n
= $${\\left| {\\overrightarrow a } \\right|^2} + 4{\\left| {\\overrightarrow b } \\right|^2} + 4\\overrightarrow a .\\overrightarrow b + {\\left| {\\overrightarrow a } \\right|^2} + 4{\\left| {\\overrightarrow c } \\right|^2} + 4\\overrightarrow a .\\overrightarrow c $$\n
= 10 + 4$$\\left( {\\overrightarrow a .\\overrightarrow b + \\overrightarrow a .\\overrightarrow c } \\right)$$\n
= 10 + 4(-2)\n
= 2",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7744,
"subject": "General Science",
"question": "Let a, b c $$ \\in $$ R be such that a2\n + b2\n + c2\n = 1. If $$a\\cos \\theta = b\\cos \\left( {\\theta + {{2\\pi } \\over 3}} \\right) = c\\cos \\left( {\\theta + {{4\\pi } \\over 3}} \\right)$$,\n where\n$${\\theta = {\\pi \\over 9}}$$, then the angle between the vectors\n$$a\\widehat i + b\\widehat j + c\\widehat k$$ and $$b\\widehat i + c\\widehat j + a\\widehat k$$ is :",
"options": [
{
"text": "0"
},
{
"text": "$${{\\pi \\over 9}}$$"
},
{
"text": "$${{{2\\pi } \\over 3}}$$"
},
{
"text": "$${{\\pi \\over 2}}$$"
}
],
"answer": "$${{\\pi \\over 2}}$$",
"solution": "**Answer:** $${{\\pi \\over 2}}$$\n\nLet, $$\\overrightarrow {{a_1}} = a\\widehat i + b\\widehat j + c\\widehat k$$
and $$\\overrightarrow {{a_2}} = b\\widehat i + c\\widehat j + a\\widehat k$$
$$ \\Rightarrow $$ $$\\alpha = {\\pi \\over 2}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7745,
"subject": "General Science",
"question": "Let the vectors $$\\overrightarrow a $$, $$\\overrightarrow b $$, $$\\overrightarrow c $$\n be such that\n $$\\left| {\\overrightarrow a } \\right| = 2$$, $$\\left| {\\overrightarrow b } \\right| = 4$$\n and $$\\left| {\\overrightarrow c } \\right| = 4$$. If the projection of\n $$\\overrightarrow b $$\n on $$\\overrightarrow a $$\n is equal to the projection of $$\\overrightarrow c $$\n on $$\\overrightarrow a $$\n and $$\\overrightarrow b $$\n is perpendicular to $$\\overrightarrow c $$,\n then the value of\n $$\\left| {\\overrightarrow a + \\vec b - \\overrightarrow c } \\right|$$\n is ___________.",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\nProjection of $$\\overrightarrow b $$\n on $$\\overrightarrow a $$\n= Projection of $$\\overrightarrow c $$\n on $$\\overrightarrow a $$\n
$$ \\Rightarrow $$ $${{\\overrightarrow b .\\overrightarrow a } \\over {\\left| {\\overrightarrow a } \\right|}} = {{\\overrightarrow c .\\overrightarrow a } \\over {\\left| {\\overrightarrow a } \\right|}}$$\n
$$ \\Rightarrow $$ $$\\overrightarrow b .\\overrightarrow a = \\overrightarrow c .\\overrightarrow a $$\n
$$ \\because $$ $$\\overrightarrow b $$\n is perpendicular to $$\\overrightarrow c $$\n
$$ \\therefore $$ $$\\overrightarrow b .\\overrightarrow c = 0$$\n
Let $$\\left| {\\overrightarrow a + \\vec b - \\overrightarrow c } \\right|$$ = k\n
Square both sides\n
k2 = $${{{\\left( {\\overrightarrow a } \\right)}^2}}$$ + $${{{\\left( {\\overrightarrow b } \\right)}^2}}$$ + $${{{\\left( {\\overrightarrow c } \\right)}^2}}$$ + $$2\\overrightarrow a .\\overrightarrow b $$ - $$2\\overrightarrow b .\\overrightarrow c $$ - $$2\\overrightarrow a .\\overrightarrow c $$\n
$$ \\Rightarrow $$ k2 = $${{{\\left( {\\overrightarrow a } \\right)}^2}}$$ + $${{{\\left( {\\overrightarrow b } \\right)}^2}}$$ + $${{{\\left( {\\overrightarrow c } \\right)}^2}}$$\n
$$ \\Rightarrow $$ k2 = 22 + 42 + 42 = 36\n
$$ \\Rightarrow $$ k = 6",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7746,
"subject": "General Science",
"question": "If $$\\overrightarrow x $$ and $$\\overrightarrow y $$ be two non-zero vectors such that\n$$\\left| {\\overrightarrow x + \\overrightarrow y } \\right| = \\left| {\\overrightarrow x } \\right|$$ and $${2\\overrightarrow x + \\lambda \\overrightarrow y }$$ is perpendicular to $${\\overrightarrow y }$$,\nthen the value of $$\\lambda $$ is _________ .\n",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n$$\\left| {\\overrightarrow x + \\overrightarrow y } \\right| = \\left| {\\overrightarrow x } \\right|$$\n Squaring both sides we get\n
$${\\left| {\\overrightarrow x } \\right|^2} + 2\\overrightarrow x .\\overrightarrow y + {\\left| {\\overrightarrow y } \\right|^2} = {\\left| {\\overrightarrow x } \\right|^2}$$\n
$$ \\Rightarrow $$ $$2\\overrightarrow x .\\overrightarrow y + \\overrightarrow y .\\overrightarrow y $$ = 0 ....(1)\n
Given $${2\\overrightarrow x + \\lambda \\overrightarrow y }$$ is perpendicular to $${\\overrightarrow y }$$\n
$$ \\therefore $$ $$\\left( {2\\overrightarrow x + \\lambda \\overrightarrow y } \\right).\\overrightarrow y $$ = 0\n
$$ \\Rightarrow $$ $$2\\overrightarrow x .\\overrightarrow y + \\lambda \\overrightarrow y .\\overrightarrow y $$ = 0 ....(2)\n
Comparing (1) & (2) we get, $$\\lambda $$ = 1",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7747,
"subject": "General Science",
"question": "Let $$\\overrightarrow x $$ be a vector in the plane containing vectors $$\\overrightarrow a = 2\\widehat i - \\widehat j + \\widehat k$$ and $$\\overrightarrow b = \\widehat i + 2\\widehat j - \\widehat k$$. If the vector $$\\overrightarrow x $$ is perpendicular to $$\\left( {3\\widehat i + 2\\widehat j - \\widehat k} \\right)$$ and its projection on $$\\overrightarrow a $$ is $${{17\\sqrt 6 } \\over 2}$$, then the value of $$|\\overrightarrow x {|^2}$$ is equal to __________.",
"options": [],
"answer": "486",
"solution": "**Answer:** 486\n\nLet, $$\\overrightarrow x = k(\\overrightarrow a + \\lambda \\overrightarrow b )$$
$$\\overrightarrow x$$ is perpendicular to $$3\\widehat i + 2\\widehat j - \\widehat k$$
$$ = {{85} \\over {14}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7749,
"subject": "General Science",
"question": "Let $$\\overrightarrow a $$, $$\\overrightarrow b $$, $$\\overrightarrow c $$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $$\\theta$$, with the vector $$\\overrightarrow a $$ + $$\\overrightarrow b $$ + $$\\overrightarrow c $$. Then 36cos22$$\\theta$$ is equal to ___________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n$${\\left| {\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\right|^2} = {\\left| {\\overrightarrow a } \\right|^2} + {\\left| {\\overrightarrow b } \\right|^2} + {\\left| {\\overrightarrow c } \\right|^2} + 2(\\overrightarrow a \\,.\\,\\overrightarrow b + \\overrightarrow a \\,.\\,\\overrightarrow c + \\overrightarrow b \\,.\\,\\overrightarrow c ) = 3$$
$$ \\Rightarrow \\left| {\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\right| = \\sqrt 3 \\overrightarrow a .(\\overrightarrow a + \\overrightarrow b + \\overrightarrow c ) = \\left| {\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\right|\\cos \\theta $$
$$ \\Rightarrow 1 = \\sqrt 3 \\cos \\theta $$
$$ \\Rightarrow \\cos 2\\theta = - {1 \\over 3}$$
$$ \\Rightarrow 36{\\cos ^2}2\\theta = 4$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7750,
"subject": "General Science",
"question": "In a triangle ABC, if $$\\left| {\\overrightarrow {BC} } \\right| = 3$$, $$\\left| {\\overrightarrow {CA} } \\right| = 5$$ and $$\\left| {\\overrightarrow {BA} } \\right| = 7$$, then the projection of the vector $$\\overrightarrow {BA} $$ on $$\\overrightarrow {BC} $$ is equal to :",
"options": [
{
"text": "$${{19} \\over 2}$$"
},
{
"text": "$${{13} \\over 2}$$"
},
{
"text": "$${{11} \\over 2}$$"
},
{
"text": "$${{15} \\over 2}$$"
}
],
"answer": "$${{11} \\over 2}$$",
"solution": "**Answer:** $${{11} \\over 2}$$\n\n
$$ \\Rightarrow \\alpha = 6$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7752,
"subject": "General Science",
"question": "If $$\\left( {\\overrightarrow a + 3\\overrightarrow b } \\right)$$ is perpendicular to $$\\left( {7\\overrightarrow a - 5\\overrightarrow b } \\right)$$ and $$\\left( {\\overrightarrow a - 4\\overrightarrow b } \\right)$$ is perpendicular to $$\\left( {7\\overrightarrow a - 2\\overrightarrow b } \\right)$$, then the angle between $$\\overrightarrow a $$ and $$\\overrightarrow b $$ (in degrees) is _______________.",
"options": [],
"answer": "60",
"solution": "**Answer:** 60\n\n$$\\left( {\\overrightarrow a + 3\\overrightarrow b } \\right) \\bot \\left( {7\\overrightarrow a - 5\\overrightarrow b } \\right)$$
$$ \\therefore $$ $$\\left( {\\overrightarrow a + 3\\overrightarrow b } \\right)\\,.\\,\\left( {7\\overrightarrow a - 5\\overrightarrow b } \\right) = 0$$
$$ \\Rightarrow $$ $$7{\\left| {\\overrightarrow a } \\right|^2} - 15{\\left| {\\overrightarrow b } \\right|^2} + 16\\overrightarrow a \\,.\\,\\overrightarrow b = 0$$ ....(1)
Also, $$\\left( {\\overrightarrow a - 4\\overrightarrow b } \\right)\\,.\\,\\left( {7\\overrightarrow a - 2\\overrightarrow b } \\right) = 0$$
$$ \\Rightarrow $$ $$7{\\left| {\\overrightarrow a } \\right|^2} + 8{\\left| {\\overrightarrow b } \\right|^2} - 30\\overrightarrow a \\,.\\,\\overrightarrow b = 0$$ .....(2)\n
Equation (1) × 30\n
$$210{\\left| {\\overrightarrow a } \\right|^2} - 450{\\left| {\\overrightarrow b } \\right|^2} + 480\\overrightarrow a \\,.\\,\\overrightarrow b = 0$$ ....(3)\n
Equation (2) × 16\n
$$112{\\left| {\\overrightarrow a } \\right|^2} + 128{\\left| {\\overrightarrow b } \\right|^2} - 480\\overrightarrow a \\,.\\,\\overrightarrow b = 0$$ .....(4)\n
from (3) & (4)
$$322{\\left| {\\overrightarrow a } \\right|^2} = 322{\\left| {\\overrightarrow b } \\right|^2}$$\n
$$ \\Rightarrow $$ $${\\left| {\\overrightarrow a } \\right|^2} = {\\left| {\\overrightarrow b } \\right|^2}$$\n
$$ \\Rightarrow $$ $$\\left| {\\overrightarrow a } \\right| = \\left| {\\overrightarrow b } \\right|$$\n
From equation (2),\n
$$15\\left| {\\overrightarrow a } \\right| = 30\\overrightarrow a .\\overrightarrow b $$\n
$$ \\Rightarrow $$ $$15{\\left| {\\overrightarrow a } \\right|^2} = 30\\left| {\\overrightarrow a } \\right|.\\left| {\\overrightarrow b } \\right|\\cos \\theta $$\n
$$\\therefore$$ $$\\theta = 60^\\circ $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7753,
"subject": "General Science",
"question": "If the projection of the vector $$\\widehat i + 2\\widehat j + \\widehat k$$ on the sum of the two vectors $$2\\widehat i + 4\\widehat j - 5\\widehat k$$ and $$ - \\lambda \\widehat i + 2\\widehat j + 3\\widehat k$$ is 1, then $$\\lambda$$ is equal to __________.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n$$\\overrightarrow a = \\widehat i + 2\\widehat j + \\widehat k$$
$$\\overrightarrow b = (2 - \\lambda )\\widehat i + 6\\widehat j - 2\\widehat k$$
$${{\\overrightarrow a \\,.\\,\\overrightarrow b } \\over {|\\overrightarrow b |}} = 1,\\overrightarrow a \\,.\\,\\overrightarrow b = 12 - \\lambda $$
$$\\left( {\\overrightarrow a \\,.\\,\\overrightarrow b } \\right) = |\\overrightarrow b {|^2}$$
20$$\\lambda$$ = 100 $$\\Rightarrow$$ $$\\lambda$$ = 5",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7754,
"subject": "General Science",
"question": "Let $$\\overrightarrow a $$ and $$\\overrightarrow b $$ be two vectors such that $$\\left| {2\\overrightarrow a + 3\\overrightarrow b } \\right| = \\left| {3\\overrightarrow a + \\overrightarrow b } \\right|$$ and the angle between $$\\overrightarrow a $$ and $$\\overrightarrow b $$ is 60$$^\\circ$$. If $${1 \\over 8}\\overrightarrow a $$ is a unit vector, then $$\\left| {\\overrightarrow b } \\right|$$ is equal to :",
"options": [
{
"text": "4"
},
{
"text": "6"
},
{
"text": "5"
},
{
"text": "8"
}
],
"answer": "5",
"solution": "**Answer:** 5\n\n$${\\left| {3\\overrightarrow a + \\overrightarrow b } \\right|^2} = {\\left| {2\\overrightarrow a + 3\\overrightarrow b } \\right|^2}$$
$$\\left( {3\\overrightarrow a + \\overrightarrow b } \\right).\\left( {3\\overrightarrow a + \\overrightarrow b } \\right) = \\left( {2\\overrightarrow a + 3\\overrightarrow b } \\right).\\left( {2\\overrightarrow a + 3\\overrightarrow b } \\right)$$
$$9\\overrightarrow a .\\,\\overrightarrow a + 6\\overrightarrow a \\,.\\,\\overrightarrow b + \\overrightarrow b \\,.\\,\\overrightarrow b = 4\\overrightarrow a \\,.\\,\\overrightarrow a + 12\\overrightarrow a \\,.\\,\\overrightarrow b + 9\\overrightarrow b \\,.\\,\\overrightarrow b $$
$$5{\\left| {\\overrightarrow a } \\right|^2} - 6\\overrightarrow a \\,.\\,\\overrightarrow b = 8{\\left| {\\overrightarrow b } \\right|^2}$$
$$5{(8)^2} - 6.8\\,.\\,\\left| {\\overrightarrow b } \\right|\\cos 60^\\circ = 8{\\left| {\\overrightarrow b } \\right|^2}$$ $$\\because$$ $$\\left( \\matrix{\n {1 \\over 8}\\left| {\\overrightarrow a } \\right| = 1 \\hfill \\cr \n \\Rightarrow \\left| {\\overrightarrow a } \\right| = 8 \\hfill \\cr} \\right)$$
$$40 - 3\\left| {\\overrightarrow b } \\right| = {\\left| {\\overrightarrow b } \\right|^2}$$
$$ \\Rightarrow {\\left| {\\overrightarrow b } \\right|^2} + 3\\left| {\\overrightarrow b } \\right| - 40 = 0$$
$$\\left| {\\overrightarrow b } \\right| = - 8$$, $$\\left| {\\overrightarrow b } \\right| = 5$$
Let $$\\overrightarrow a = {a_1}\\widehat i + {a_2}\\widehat j + {a_3}\\widehat k$$ $${a_i} > 0$$, $$i = 1,2,3$$ be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of $$\\overrightarrow a $$ on the vector $$3\\widehat i + 4\\widehat j$$ be 7. Let $$\\overrightarrow b $$ be a vector obtained by rotating $$\\overrightarrow a $$ with 90$$^\\circ$$. If $$\\overrightarrow a $$, $$\\overrightarrow b $$ and x-axis are coplanar, then projection of a vector $$\\overrightarrow b $$ on $$3\\widehat i + 4\\widehat j$$ is equal to:
$$\\overrightarrow b = - 2r\\widehat i + r\\widehat j + r\\widehat k$$
\n
$$\\overrightarrow b = r( - 2\\widehat i + \\widehat j + \\widehat k)$$
\n
Now $$\\left| {\\overrightarrow a } \\right| = \\left| {\\overrightarrow b } \\right|$$
\n
$$5\\sqrt 3 = \\left| r \\right|\\sqrt b \\Rightarrow \\left| r \\right| = {5 \\over {\\sqrt 2 }}$$
\n
$$\\Rightarrow$$ Projection of $$\\overrightarrow b $$ on $$3\\widehat i + 4\\widehat j = \\left| {{{\\overrightarrow b \\,.\\,\\left( {3\\widehat i + 4\\widehat j} \\right)} \\over {\\sqrt {{3^2} + {4^2}} }}} \\right|$$
Let S be the set of all a $$\\in R$$ for which the angle between the vectors $$\n\\vec{u}=a\\left(\\log _{e} b\\right) \\hat{i}-6 \\hat{j}+3 \\hat{k}$$ and $$\\vec{v}=\\left(\\log _{e} b\\right) \\hat{i}+2 \\hat{j}+2 a\\left(\\log _{e} b\\right) \\hat{k}$$, $$(b>1)$$ is acute. Then S is equal to :
Let $$\\vec{a}=5 \\hat{i}-\\hat{j}-3 \\hat{k}$$ and $$\\vec{b}=\\hat{i}+3 \\hat{j}+5 \\hat{k}$$ be two vectors. Then which one of the following statements is TRUE ?
",
"options": [
{
"text": "Projection of $$\\vec{a}$$ on $$\\vec{b}$$ is $$\\frac{-13}{\\sqrt{35}}$$ and the direction of the projection vector is opposite to the direction \nof $$\\vec{b}$$."
},
{
"text": "Projection of $$\\vec{a}$$ on $$\\vec{b}$$ is $$\\frac{13}{\\sqrt{35}}$$ and the direction of the projection vector is opposite to the direction \nof $$\\vec{b}$$."
},
{
"text": "Projection of $$\\vec{a}$$ on $$\\vec{b}$$ is $$\\frac{13}{\\sqrt{35}}$$ and the direction of the projection vector is same as of $$\\vec{b}$$."
},
{
"text": "Projection of $$\\vec{a}$$ on $$\\vec{b}$$ is $$\\frac{-13}{\\sqrt{35}}$$ and the direction of the projection vector is same as of $$\\vec{b}$$."
}
],
"answer": "Projection of $$\\vec{a}$$ on $$\\vec{b}$$ is $$\\frac{-13}{\\sqrt{35}}$$ and the direction of the projection vector is opposite to the direction \nof $$\\vec{b}$$.",
"solution": "**Answer:** Projection of $$\\vec{a}$$ on $$\\vec{b}$$ is $$\\frac{-13}{\\sqrt{35}}$$ and the direction of the projection vector is opposite to the direction \nof $$\\vec{b}$$.\n\n$\\begin{aligned} & \\text { Projection of }\\vec{a} \\text { on } \\vec{b} =\\frac{\\vec{a} \\cdot \\vec{b}}{|\\vec{b}|} \\\\\\\\ & = \\frac{(5 \\hat{i}-\\hat{j}-3 \\hat{k}) \\cdot(\\hat{i}+3 \\hat{j}+5 \\hat{k})}{\\sqrt{1^2+3^2+5^2}}=\\frac{5-3-15}{\\sqrt{35}} \\\\\\\\ & = \\frac{-13}{\\sqrt{35}}\\end{aligned}$\n
Negative sign indicates that direction of the projection vector is opposite to the direction \nof $$\\vec{b}$$.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7758,
"subject": "General Science",
"question": "
Let $$\\vec{a}=2 \\hat{i}-7 \\hat{j}+5 \\hat{k}, \\vec{b}=\\hat{i}+\\hat{k}$$ and $$\\vec{c}=\\hat{i}+2 \\hat{j}-3 \\hat{k}$$ be three given vectors. If $$\\overrightarrow{\\mathrm{r}}$$ is a vector such that $$\\vec{r} \\times \\vec{a}=\\vec{c} \\times \\vec{a}$$ and $$\\vec{r} \\cdot \\vec{b}=0$$, then $$|\\vec{r}|$$ is equal to :
Let $$\\vec{a}=2 \\hat{i}+\\hat{j}+\\hat{k}$$, and $$\\vec{b}$$ and $$\\vec{c}$$ be two nonzero vectors such that $$|\\vec{a}+\\vec{b}+\\vec{c}|=|\\vec{a}+\\vec{b}-\\vec{c}|$$ and $$\\vec{b} \\cdot \\vec{c}=0$$. Consider the following two statements:
\n
(A) $$|\\vec{a}+\\lambda \\vec{c}| \\geq|\\vec{a}|$$ for all $$\\lambda \\in \\mathbb{R}$$.
\n
(B) $$\\vec{a}$$ and $$\\vec{c}$$ are always parallel.
\n
Then,
",
"options": [
{
"text": "only (B) is correct"
},
{
"text": "both (A) and (B) are correct"
},
{
"text": "only (A) is correct"
},
{
"text": "neither (A) nor (B) is correct"
}
],
"answer": "only (A) is correct",
"solution": "**Answer:** only (A) is correct\n\n$|\\vec{a}+\\vec{b}+\\vec{c}|=|\\vec{a}+\\vec{b}-\\vec{c}|$\n
If the vectors $$\\overrightarrow a = \\lambda \\widehat i + \\mu \\widehat j + 4\\widehat k$$, $$\\overrightarrow b = - 2\\widehat i + 4\\widehat j - 2\\widehat k$$ and $$\\overrightarrow c = 2\\widehat i + 3\\widehat j + \\widehat k$$ are coplanar and the projection of $$\\overrightarrow a $$ on the vector $$\\overrightarrow b $$ is $$\\sqrt {54} $$ units, then the sum of all possible values of $$\\lambda + \\mu $$ is equal to :
The vector $$\\overrightarrow a = - \\widehat i + 2\\widehat j + \\widehat k$$ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is $$\\overrightarrow b $$. Then the projection of $$3\\overrightarrow a + \\sqrt 2 \\overrightarrow b $$ on $$\\overrightarrow c = 5\\widehat i + 4\\widehat j + 3\\widehat k$$ is :
First, we write $\\overrightarrow{b}$ as a linear combination of $\\overrightarrow{a}$ and $\\overrightarrow{j}$ since $\\overrightarrow{b}$ is a rotation of $\\overrightarrow{a}$ about the y-axis.
The magnitude of $\\overrightarrow{b}$ is the same as the magnitude of $\\overrightarrow{a}$ because a rotation doesn't change the magnitude of a vector. This gives us :
Let $$\\overrightarrow \\alpha = 4\\widehat i + 3\\widehat j + 5\\widehat k$$ and $$\\overrightarrow \\beta = \\widehat i + 2\\widehat j - 4\\widehat k$$. Let $${\\overrightarrow \\beta _1}$$ be parallel to $$\\overrightarrow \\alpha $$ and $${\\overrightarrow \\beta _2}$$ be perpendicular to $$\\overrightarrow \\alpha $$. If $$\\overrightarrow \\beta = {\\overrightarrow \\beta _1} + {\\overrightarrow \\beta _2}$$, then the value of $$5{\\overrightarrow \\beta _2}\\,.\\left( {\\widehat i + \\widehat j + \\widehat k} \\right)$$ is :
Let a unit vector $$\\hat{u}=x \\hat{i}+y \\hat{j}+z \\hat{k}$$ make angles $$\\frac{\\pi}{2}, \\frac{\\pi}{3}$$ and $$\\frac{2 \\pi}{3}$$ with the vectors $$\\frac{1}{\\sqrt{2}} \\hat{i}+\\frac{1}{\\sqrt{2}} \\hat{k}, \\frac{1}{\\sqrt{2}} \\hat{j}+\\frac{1}{\\sqrt{2}} \\hat{k}$$ and $$\\frac{1}{\\sqrt{2}} \\hat{i}+\\frac{1}{\\sqrt{2}} \\hat{j}$$ respectively. If $$\\vec{v}=\\frac{1}{\\sqrt{2}} \\hat{i}+\\frac{1}{\\sqrt{2}} \\hat{j}+\\frac{1}{\\sqrt{2}} \\hat{k}$$ then $$|\\hat{u}-\\vec{v}|^2$$ is equal to
Let a unit vector which makes an angle of $$60^{\\circ}$$ with $$2 \\hat{i}+2 \\hat{j}-\\hat{k}$$ and an angle of $$45^{\\circ}$$ with $$\\hat{i}-\\hat{k}$$ be $$\\vec{C}$$. Then $$\\vec{C}+\\left(-\\frac{1}{2} \\hat{i}+\\frac{1}{3 \\sqrt{2}} \\hat{j}-\\frac{\\sqrt{2}}{3} \\hat{k}\\right)$$ is:
For $$\\lambda>0$$, let $$\\theta$$ be the angle between the vectors $$\\vec{a}=\\hat{i}+\\lambda \\hat{j}-3 \\hat{k}$$ and $$\\vec{b}=3 \\hat{i}-\\hat{j}+2 \\hat{k}$$. If the vectors $$\\vec{a}+\\vec{b}$$ and $$\\vec{a}-\\vec{b}$$ are mutually perpendicular, then the value of (14 cos $$\\theta)^2$$ is equal to
Let $$\\overrightarrow{\\mathrm{a}}=\\hat{i}+2 \\hat{j}+3 \\hat{k}, \\overrightarrow{\\mathrm{b}}=2 \\hat{i}+3 \\hat{j}-5 \\hat{k}$$ and $$\\overrightarrow{\\mathrm{c}}=3 \\hat{i}-\\hat{j}+\\lambda \\hat{k}$$ be three vectors. Let $$\\overrightarrow{\\mathrm{r}}$$ be a unit vector along $$\\vec{b}+\\vec{c}$$. If $$\\vec{r} \\cdot \\vec{a}=3$$, then $$3 \\lambda$$ is equal to:
The set of all $$\\alpha$$, for which the vectors $$\\vec{a}=\\alpha t \\hat{i}+6 \\hat{j}-3 \\hat{k}$$ and $$\\vec{b}=t \\hat{i}-2 \\hat{j}-2 \\alpha t \\hat{k}$$ are inclined at an obtuse angle for all $$t \\in \\mathbb{R}$$, is
$$\n\\begin{array}{ll}\n\\Rightarrow & (\\vec{a}+\\vec{b}) \\times \\vec{c}=-\\vec{c} \\times \\vec{c} \\\\\\\\\n\\Rightarrow & \\vec{a} \\times \\vec{c}+\\vec{b} \\times \\vec{c}=\\overrightarrow{0} \\\\\\\\\n\\Rightarrow & \\vec{b} \\times \\vec{c}=\\vec{c} \\times \\vec{a} \\\\\\\\\n\\text { Similarly, } & \\vec{a} \\times \\vec{b}=\\vec{b} \\times \\vec{c} \\\\\\\\\n\\text { Hence, } & \\vec{a} \\times \\vec{b}=\\vec{b} \\times \\vec{c}=\\vec{c} \\times \\vec{a}\n\\end{array}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7770,
"subject": "General Science",
"question": "If $$\\left| {\\overrightarrow a } \\right| = 4,\\left| {\\overrightarrow b } \\right| = 2$$ and the angle between $${\\overrightarrow a }$$ and $${\\overrightarrow b }$$ is $$\\pi /6$$ then $${\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)^2}$$ is equal to :",
"options": [
{
"text": "$$48$$ "
},
{
"text": "$$16$$"
},
{
"text": "$$\\overrightarrow a $$ "
},
{
"text": "none of these "
}
],
"answer": "$$16$$",
"solution": "**Answer:** $$16$$\n\n$${\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)^2} = {\\left| {\\overrightarrow a } \\right|^2}{\\left| {\\overrightarrow b } \\right|^2}\\,\\,{\\sin ^2}{\\pi \\over 6}$$ \n
$$ = 16 \\times 4 \\times {1 \\over 4} = 16$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7771,
"subject": "General Science",
"question": "If the vectors $$\\overrightarrow c ,\\overrightarrow a = x\\widehat i + y\\widehat j + z\\widehat k$$ and $$\\widehat b = \\widehat j$$ are such that $$\\overrightarrow a ,\\overrightarrow c $$ and $$\\overrightarrow b $$ form a right handed system then $${\\overrightarrow c }$$ is : ",
"options": [
{
"text": "$$z\\widehat i - x\\widehat k$$ "
},
{
"text": "$$\\overrightarrow 0 $$ "
},
{
"text": "$$y\\widehat j$$ "
},
{
"text": "$$ - z\\widehat i + x\\widehat k$$ "
}
],
"answer": "$$z\\widehat i - x\\widehat k$$ ",
"solution": "**Answer:** $$z\\widehat i - x\\widehat k$$ \n\nSince $$\\overrightarrow a ,\\overrightarrow c ,\\overrightarrow b $$ form a right handed system,\n
$$\\therefore$$ $$\\overrightarrow c = \\overrightarrow b \\times \\overrightarrow a = \\left| {\\matrix{\n {\\widehat i} & {\\widehat j} & {\\widehat k} \\cr \n 0 & 1 & 0 \\cr \n x & y & z \\cr \n\n } } \\right|$$\n
$$ = z\\widehat i - x\\widehat k$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7772,
"subject": "General Science",
"question": "$$\\overrightarrow a = 3\\widehat i - 5\\widehat j$$ and $$\\overrightarrow b = 6\\widehat i + 3\\widehat j$$ are two vectors and $$\\overrightarrow c $$ is a vector such that $$\\overrightarrow c = \\overrightarrow a \\times \\overrightarrow b $$ then $$\\left| {\\overrightarrow a } \\right|:\\left| {\\overrightarrow b } \\right|:\\left| {\\overrightarrow c } \\right|$$ =",
"options": [
{
"text": "$$\\sqrt {34} :\\sqrt {45} :\\sqrt {39} $$ "
},
{
"text": "$$\\sqrt {34} :\\sqrt {45} :39$$ "
},
{
"text": "$$34:39:45$$ "
},
{
"text": "$$\\,39:35:34$$ "
}
],
"answer": "$$\\sqrt {34} :\\sqrt {45} :39$$ ",
"solution": "**Answer:** $$\\sqrt {34} :\\sqrt {45} :39$$ \n\n
To solve this problem, let's take it step by step, beginning with calculating each of the vector magnitudes (or norms) and then finding the magnitude of the cross product vector $\\overrightarrow{c}$.
\n\n
Given vectors $\\overrightarrow{a}$ and $\\overrightarrow{b}$ are:
For our provided vectors, the $k$ component of both vectors is $0$, thus their cross product will be purely in the $k$ direction. The determinant simplifies to:
Finally, the ratio of the magnitudes of vectors $\\overrightarrow{a}$, $\\overrightarrow{b}$, and $\\overrightarrow{c}$ is thus:
\n\n
$\\sqrt{34}:\\sqrt{45}:39.$
\n\n
Therefore, the correct answer is:
\n\n
Option B) $\\sqrt{34}:\\sqrt{45}:39.$
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7773,
"subject": "General Science",
"question": "If $$\\overrightarrow a \\times \\overrightarrow b = \\overrightarrow b \\times \\overrightarrow c = \\overrightarrow c \\times \\overrightarrow a $$ then $$\\overrightarrow a + \\overrightarrow b + \\overrightarrow c = $$ ",
"options": [
{
"text": "$$abc$$ "
},
{
"text": "$$-1$$"
},
{
"text": "$$0$$"
},
{
"text": "$$2$$"
}
],
"answer": "$$0$$",
"solution": "**Answer:** $$0$$\n\nLet $$\\overrightarrow a + \\overrightarrow b + \\overrightarrow c = \\overrightarrow r .$$ Then\n
$$\\overrightarrow a \\times \\left( {\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\right) = \\overrightarrow a \\times \\overrightarrow r $$\n
$$ \\Rightarrow 0 + \\overrightarrow a \\times \\overrightarrow b + \\overrightarrow a \\times \\overrightarrow c = \\overrightarrow a \\times \\overrightarrow r $$\n
$$ \\Rightarrow \\overrightarrow a \\times \\overrightarrow b - \\overrightarrow c \\times \\overrightarrow a = \\overrightarrow a \\times \\overrightarrow r $$\n
$$ \\Rightarrow \\overrightarrow a \\times \\overrightarrow r = \\overrightarrow 0 $$\n
Similarly $$\\overrightarrow b \\times \\overrightarrow r = \\overrightarrow 0 \\,\\,\\,\\& \\,\\,\\,\\overrightarrow c \\times \\overrightarrow r = \\overrightarrow 0 $$\n
Above three conditions will be satisfied for non-zero vectors if and only if $$\\overrightarrow r = \\overrightarrow 0 $$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7774,
"subject": "General Science",
"question": "A tetrahedron has vertices at $$O(0,0,0), A(1,2,1) B(2,1,3)$$ and $$C(-1,1,2).$$ Then the angle between the faces $$OAB$$ and $$ABC$$ will be :",
"options": [
{
"text": "$${90^ \\circ }$$ "
},
{
"text": "$${\\cos ^{ - 1}}\\left( {{{19} \\over {35}}} \\right)$$ "
},
{
"text": "$${\\cos ^{ - 1}}\\left( {{{17} \\over {31}}} \\right)$$"
},
{
"text": "$${30^ \\circ }$$"
}
],
"answer": "$${\\cos ^{ - 1}}\\left( {{{19} \\over {35}}} \\right)$$ ",
"solution": "**Answer:** $${\\cos ^{ - 1}}\\left( {{{19} \\over {35}}} \\right)$$ \n\nVector perpendicular to the face $$OAB$$\n
$$\n\\begin{aligned}\n& \\text { Now, } \\overrightarrow{\\mathbf{w}} \\cdot(\\overrightarrow{\\mathbf{u}} \\times \\overrightarrow{\\mathbf{v}})=\\left|\\begin{array}{rrr}\n1 & 2 & 3 \\\\\n1 & 1 & 0 \\\\\n1 & -1 & 0\n\\end{array}\\right|=-6 \\\\\\\\\n& \\begin{array}{ll}\n\\Rightarrow |\\overrightarrow{\\mathbf{w}} \\cdot(\\overrightarrow{\\mathbf{u}} \\times \\overrightarrow{\\mathbf{v}})|=6 \\\\\\\\\n\\text { and } \\overrightarrow{\\mathbf{u}} \\times \\overrightarrow{\\mathbf{v}}=(\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}) \\times(\\hat{\\mathbf{i}}-\\hat{\\mathbf{j}})-2 \\hat{\\mathbf{k}} \\\\\\\\\n\\Rightarrow |\\overrightarrow{\\mathbf{u}} \\times \\overrightarrow{\\mathbf{v}}|=2 \\\\\\\\\n\\therefore |\\overrightarrow{\\mathbf{w}} \\cdot \\hat{\\mathbf{n}}|=\\frac{6}{2}=3\n\\end{array}\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7776,
"subject": "General Science",
"question": "Let $$\\overrightarrow a ,\\overrightarrow b $$ and $$\\overrightarrow c $$ be non-zero vectors such that $$\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) \\times \\overrightarrow c = {1 \\over 3}\\left| {\\overrightarrow b } \\right|\\left| {\\overrightarrow c } \\right|\\overrightarrow a \\,\\,.$$ If $$\\theta $$ is the acute angle between the vectors $${\\overrightarrow b }$$ and $${\\overrightarrow c },$$ then $$sin\\theta $$ equals :",
"options": [
{
"text": "$${{2\\sqrt 2 } \\over 3}$$ "
},
{
"text": "$${{\\sqrt 2 } \\over 3}$$"
},
{
"text": "$${2 \\over 3}$$ "
},
{
"text": "$${1 \\over 3}$$"
}
],
"answer": "$${{2\\sqrt 2 } \\over 3}$$ ",
"solution": "**Answer:** $${{2\\sqrt 2 } \\over 3}$$ \n\nGiven $$\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) \\times \\overrightarrow c = {1 \\over 3}\\left| {\\overrightarrow b } \\right|\\left| {\\overrightarrow c } \\right|\\overrightarrow a $$\n
Clearly $$\\overrightarrow a $$ and $$\\overrightarrow b $$ are noncollinear\n
$$ \\Rightarrow \\left( {\\overrightarrow a .\\overrightarrow c } \\right)\\overrightarrow b - \\left( {\\overrightarrow b .\\overrightarrow c } \\right)\\overrightarrow a = {1 \\over 3}\\left| {\\overrightarrow b } \\right|\\left| {\\overrightarrow c } \\right|\\overrightarrow a $$\n
$$\\therefore$$ $$\\overrightarrow a .\\overrightarrow c = 0$$ \n
and $$ - \\overrightarrow b .\\overrightarrow c = {1 \\over 3}\\left| {\\overrightarrow b } \\right|\\left| {\\overrightarrow c } \\right| \\Rightarrow \\cos \\theta = {{ - 1} \\over 3}$$\n
$$\\,\\,\\,\\,\\,\\,\\,\\left[ {} \\right.$$ $$\\theta $$ is acute angle between $$\\overrightarrow b $$ and $$\\overrightarrow c $$ $$\\left. {} \\right]$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7777,
"subject": "General Science",
"question": "For any vector $${\\overrightarrow a }$$ , the value of $${\\left( {\\overrightarrow a \\times \\widehat i} \\right)^2} + {\\left( {\\overrightarrow a \\times \\widehat j} \\right)^2} + {\\left( {\\overrightarrow a \\times \\widehat k} \\right)^2}$$ is equal to :",
"options": [
{
"text": "$$3{\\overrightarrow a ^2}$$ "
},
{
"text": "$${\\overrightarrow a ^2}$$"
},
{
"text": "$$2{\\overrightarrow a ^2}$$"
},
{
"text": "$$4{\\overrightarrow a ^2}$$"
}
],
"answer": "$$2{\\overrightarrow a ^2}$$",
"solution": "**Answer:** $$2{\\overrightarrow a ^2}$$\n\nLet $$\\overrightarrow a = x\\overrightarrow i + y\\overrightarrow j + z\\overrightarrow k $$\n
$$\\overrightarrow a \\times \\overrightarrow i = z\\overrightarrow j - y\\overrightarrow k $$\n
$$ \\Rightarrow {\\left( {\\overrightarrow a \\times \\overrightarrow i } \\right)^2} = {y^2} + {z^2}$$\n
and $${\\left( {\\overrightarrow a \\times \\overrightarrow k } \\right)^2} = {x^2} + {y^2}$$\n
$$ \\Rightarrow {\\left( {\\overrightarrow a \\times \\overrightarrow i } \\right)^2} + {\\left( {\\overrightarrow a \\times \\overrightarrow j } \\right)^2} + {\\left( {\\overrightarrow a \\times \\overrightarrow k } \\right)^2}$$\n
$$ = 2\\left( {{x^2} + {y^2} + {z^2}} \\right) = 2\\overrightarrow a 2$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7778,
"subject": "General Science",
"question": "If $$\\widehat u$$ and $$\\widehat v$$ are unit vectors and $$\\theta $$ is the acute angle between them, then $$2\\widehat u \\times 3\\widehat v$$ is a unit vector for :",
"options": [
{
"text": "no value of $$\\theta $$ "
},
{
"text": "exactly one value of $$\\theta $$ "
},
{
"text": "exactly two values of $$\\theta $$ "
},
{
"text": "more than two values of $$\\theta $$ "
}
],
"answer": "exactly one value of $$\\theta $$ ",
"solution": "**Answer:** exactly one value of $$\\theta $$ \n\nGiven $$\\left| {2\\widehat u \\times 3\\widehat v} \\right| = 1$$ \n
and $$\\theta $$ is acute angle between $$\\widehat u$$ \n
Hence, there is exactly one value of $$\\theta $$ \n
for which $$2\\widehat u \\times 3\\widehat v$$ is a unit vector.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7779,
"subject": "General Science",
"question": "The vectors $$\\overrightarrow a $$ and $$\\overrightarrow b $$ are not perpendicular and $$\\overrightarrow c $$ and $$\\overrightarrow d $$ are two vectors satisfying $$\\overrightarrow b \\times \\overrightarrow c = \\overrightarrow b \\times \\overrightarrow d $$ and $$\\overrightarrow a .\\overrightarrow d = 0\\,\\,.$$ Then the vector $$\\overrightarrow d $$ is equal to :",
"options": [
{
"text": "$$\\overrightarrow c + \\left( {{{\\overrightarrow a .\\overrightarrow c } \\over {\\overrightarrow a .\\overrightarrow b }}} \\right)\\overrightarrow b $$ "
},
{
"text": "$$\\overrightarrow b + \\left( {{{\\overrightarrow b .\\overrightarrow c } \\over {\\overrightarrow a .\\overrightarrow b }}} \\right)\\overrightarrow c $$ "
},
{
"text": "$$\\overrightarrow c - \\left( {{{\\overrightarrow a .\\overrightarrow c } \\over {\\overrightarrow a .\\overrightarrow b }}} \\right)\\overrightarrow b $$ "
},
{
"text": "$$\\overrightarrow b - \\left( {{{\\overrightarrow b .\\overrightarrow c } \\over {\\overrightarrow a .\\overrightarrow b }}} \\right)\\overrightarrow c $$ "
}
],
"answer": "$$\\overrightarrow c - \\left( {{{\\overrightarrow a .\\overrightarrow c } \\over {\\overrightarrow a .\\overrightarrow b }}} \\right)\\overrightarrow b $$ ",
"solution": "**Answer:** $$\\overrightarrow c - \\left( {{{\\overrightarrow a .\\overrightarrow c } \\over {\\overrightarrow a .\\overrightarrow b }}} \\right)\\overrightarrow b $$ \n\n$$\\overrightarrow a .\\overrightarrow b \\ne 0,\\overrightarrow a .\\overrightarrow d = 0$$\n
Now, $$\\overrightarrow b \\times \\overrightarrow c = \\overrightarrow b \\times \\overrightarrow d $$\n
$$ \\Rightarrow \\overrightarrow a \\times \\left( {\\overrightarrow b \\times \\overrightarrow c } \\right) = \\overrightarrow a \\times \\left( {\\overrightarrow b \\times \\overrightarrow d } \\right)$$\n
$$ \\Rightarrow \\left( {\\overrightarrow a .\\overrightarrow c } \\right)\\overrightarrow b - \\left( {\\overrightarrow a .\\overrightarrow b } \\right)\\overrightarrow c = \\left( {\\overrightarrow a .\\overrightarrow d } \\right)\\overrightarrow b - \\left( {\\overrightarrow a .\\overrightarrow b } \\right)\\overrightarrow d $$\n
$$ \\Rightarrow \\left( {\\overrightarrow a .\\overrightarrow b } \\right)\\overrightarrow d = - \\left( {\\overrightarrow a .\\overrightarrow c } \\right)\\overrightarrow b + \\left( {\\overrightarrow a .\\overrightarrow b } \\right)\\overrightarrow c $$\n
$$\\overrightarrow d = \\overrightarrow c - \\left( {{{\\overrightarrow a .\\overrightarrow c } \\over {\\overrightarrow a .\\overrightarrow b }}} \\right)\\overrightarrow b $$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7780,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = 2\\widehat i + \\widehat j -2 \\widehat k$$ and $$\\overrightarrow b = \\widehat i + \\widehat j$$.\n
Let $$\\overrightarrow c $$ be a vector such that $$\\left| {\\overrightarrow c - \\overrightarrow a } \\right| = 3$$,\n
$$\\left| {\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) \\times \\overrightarrow c } \\right| = 3$$ and the angle between $$\\overrightarrow c $$ and $\\overrightarrow a \\times \\overrightarrow b$ is $$30^\\circ $$.\n
Then $$\\overrightarrow a .\\overrightarrow c $$ is equal to :",
"options": [
{
"text": "2"
},
{
"text": "5"
},
{
"text": "$${1 \\over 8}$$"
},
{
"text": "$${{25} \\over 8}$$"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\nGiven:\n
$$\\overrightarrow a = 2\\widehat i + \\widehat j - 2\\widehat k,\\,\\,\\overrightarrow b = \\widehat i + \\widehat j$$\n
$$ \\Rightarrow $$ $$\\left| {\\overrightarrow a } \\right| = 3$$\n
$$ \\therefore $$ $$\\overrightarrow a \\times \\overrightarrow b = 2\\widehat i - 2\\widehat j + \\widehat k$$\n
$$\\left| {\\overrightarrow a \\times \\overrightarrow b } \\right| = \\sqrt {{2^2} + {2^2} + {1^2}} = 3$$\n
We have $$\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) \\times \\overrightarrow c = \\left| {\\overrightarrow a \\times \\overrightarrow b } \\right|\\left| {\\overrightarrow c } \\right|\\sin 30^\\circ$$\n
$$ \\Rightarrow $$ $$\\left| {\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) \\times \\overrightarrow c } \\right| = 3\\left| {\\overrightarrow c } \\right|.{1 \\over 2}$$\n
$$ \\therefore $$ $$\\left| {\\overrightarrow c } \\right| = 2$$\n
Now $$\\left| {\\overrightarrow c - \\overrightarrow a } \\right| = 3$$\n
On squaring, we get\n
$$ \\Rightarrow $$ $${c^2} + {a^2} - 2 - \\overrightarrow c .\\overrightarrow a = 9$$\n
$$ \\Rightarrow $$ $$4 + 9 - 2 - \\overrightarrow a .\\overrightarrow c = 9$$\n
$$ \\Rightarrow $$ $$\\overrightarrow a .\\overrightarrow c = 2$$ [$$ \\because $$ $$\\overrightarrow c .\\overrightarrow a \\,\\, = \\,\\,\\overrightarrow a .\\overrightarrow c $$]",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7781,
"subject": "General Science",
"question": "The area (in sq. units) of the parallelogram whose diagonals are along the vectors $$8\\widehat i - 6\\widehat j$$ and $$3\\widehat i + 4\\widehat j - 12\\widehat k,$$ is : ",
"options": [
{
"text": "26"
},
{
"text": "65"
},
{
"text": "20"
},
{
"text": "52"
}
],
"answer": "65",
"solution": "**Answer:** 65\n\nWhen diagonal $${\\overrightarrow {{d_1}} }$$ and $${\\overrightarrow {{d_2}} }$$ are given of a parallelogram then the area of parallelogram = $${1 \\over 2}\\left| {\\overrightarrow {{d_1}} \\times \\overrightarrow {{d_2}} } \\right|$$\n
$$ \\Rightarrow $$ $$6\\widehat i - 6\\widehat j + {9 \\over 2}\\widehat k$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7783,
"subject": "General Science",
"question": "If $$\\overrightarrow a ,\\,\\,\\overrightarrow b ,$$ and $$\\overrightarrow C $$ are unit vectors such that $$\\overrightarrow a + 2\\overrightarrow b + 2\\overrightarrow c = \\overrightarrow 0 ,$$ then $$\\left| {\\overrightarrow a \\times \\overrightarrow c } \\right|$$ is equal to :",
"options": [
{
"text": "$${{\\sqrt {15} } \\over 4}$$ "
},
{
"text": "$${{1} \\over {4}}$$"
},
{
"text": "$${{15} \\over {16}}$$"
},
{
"text": "$${{\\sqrt {15} } \\over 16}$$"
}
],
"answer": "$${{\\sqrt {15} } \\over 4}$$ ",
"solution": "**Answer:** $${{\\sqrt {15} } \\over 4}$$ \n\nGiven,
\n$$\\overrightarrow a + 2\\overrightarrow b + 2\\overrightarrow c = \\overrightarrow 0 $$
\n$$ \\Rightarrow $$ $$\\overrightarrow a + 2\\overrightarrow c = - 2\\overrightarrow b $$ \n \nSquaring both sides,
\n$${\\left| {\\overrightarrow a } \\right|^2} + 4\\overrightarrow a .\\overrightarrow c + 4{\\left| {\\overrightarrow c } \\right|^2} = 4{\\left| {\\overrightarrow b } \\right|^2}$$
\n$$ \\Rightarrow $$ 1 + $$4\\overrightarrow a .\\overrightarrow c $$ + 4 = 4 [as $$\\left| {\\overrightarrow a } \\right|^2$$ = $$\\left| {\\overrightarrow b } \\right|^2$$ = $$\\left| {\\overrightarrow c } \\right|^2$$ = 1]
\n$$ \\Rightarrow $$ $$\\overrightarrow a .\\overrightarrow c = - {1 \\over 4}$$
\n$$ \\Rightarrow $$ $$\\left| {\\overrightarrow a } \\right|\\left| {\\overrightarrow c } \\right|\\cos \\theta = - {1 \\over 4}$$
$$ = - 10\\widehat i + 5\\widehat j$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7790,
"subject": "General Science",
"question": "Let $$\\overrightarrow a $$\n, $$\\overrightarrow b $$\n and $$\\overrightarrow c $$\n be three unit vectors such that\n $$\\overrightarrow a + \\vec b + \\overrightarrow c = \\overrightarrow 0 $$. If $$\\lambda = \\overrightarrow a .\\vec b + \\vec b.\\overrightarrow c + \\overrightarrow c .\\overrightarrow a $$ and\n $$\\overrightarrow d = \\overrightarrow a \\times \\vec b + \\vec b \\times \\overrightarrow c + \\overrightarrow c \\times \\overrightarrow a $$, then the ordered pair, $$\\left( {\\lambda ,\\overrightarrow d } \\right)$$ is equal to :",
"options": [
{
"text": "$$\\left( {{3 \\over 2},3\\overrightarrow a \\times \\overrightarrow c } \\right)$$"
},
{
"text": "$$\\left( { - {3 \\over 2},3\\overrightarrow c \\times \\overrightarrow b } \\right)$$"
},
{
"text": "$$\\left( { - {3 \\over 2},3\\overrightarrow a \\times \\overrightarrow b } \\right)$$"
},
{
"text": "$$\\left( {{3 \\over 2},3\\overrightarrow b \\times \\overrightarrow c } \\right)$$"
}
],
"answer": "$$\\left( { - {3 \\over 2},3\\overrightarrow a \\times \\overrightarrow b } \\right)$$",
"solution": "**Answer:** $$\\left( { - {3 \\over 2},3\\overrightarrow a \\times \\overrightarrow b } \\right)$$\n\n$$\\overrightarrow a + \\vec b + \\overrightarrow c = \\overrightarrow 0 $$\n
$$ \\Rightarrow $$ $${\\left| {\\overrightarrow a + \\overrightarrow b + \\overrightarrow c } \\right|^2}$$ = 0\n
$$ \\Rightarrow $$ $${{{\\left| {\\overrightarrow a } \\right|}^2}}$$ + $${{{\\left| {\\overrightarrow b } \\right|}^2}}$$ + $${{{\\left| {\\overrightarrow c } \\right|}^2}}$$ + $${2\\left( {\\overrightarrow a .\\overrightarrow b } \\right)}$$ + $${2\\left( {\\overrightarrow b .\\overrightarrow c } \\right)}$$ + $${2\\left( {\\overrightarrow c .\\overrightarrow {a} } \\right)}$$ = 0\n
$$ \\Rightarrow $$ 3 + $${2\\left( {\\overrightarrow a .\\overrightarrow b + \\overrightarrow b .\\overrightarrow c + \\overrightarrow c .\\overrightarrow a } \\right)}$$ = 0\n
$$\\overrightarrow d = \\overrightarrow a \\times \\vec b + \\vec b \\times \\overrightarrow c + \\overrightarrow c \\times \\overrightarrow a $$\n
= $${\\overrightarrow a \\times \\overrightarrow b + \\overrightarrow b \\times \\left( { - \\overrightarrow a - \\overrightarrow b } \\right) + \\left( { - \\overrightarrow a - \\overrightarrow b } \\right) \\times \\overrightarrow a }$$\n
= $${\\overrightarrow a \\times \\overrightarrow b - \\overrightarrow b \\times \\overrightarrow a + }$$ $${\\overrightarrow b \\times \\overrightarrow b - \\overrightarrow a \\times \\overrightarrow a - \\overrightarrow b \\times \\overrightarrow a }$$\n
= $${\\overrightarrow a \\times \\overrightarrow b + \\overrightarrow a \\times \\overrightarrow b + \\overrightarrow a \\times \\overrightarrow b }$$\n
= $${3\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)}$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7791,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = \\widehat i - 2\\widehat j + \\widehat k$$ and $$\\overrightarrow b = \\widehat i - \\widehat j + \\widehat k$$ be two\nvectors. If $$\\overrightarrow c $$ is a vector such that $$\\overrightarrow b \\times \\overrightarrow c = \\overrightarrow b \\times \\overrightarrow a $$ and $$\\overrightarrow c .\\overrightarrow a = 0$$, then $$\\overrightarrow c .\\overrightarrow b $$ is equal to",
"options": [
{
"text": "$$ - {1 \\over 2}$$"
},
{
"text": "$$ - {3 \\over 2}$$"
},
{
"text": "$${1 \\over 2}$$"
},
{
"text": "-1"
}
],
"answer": "$$ - {1 \\over 2}$$",
"solution": "**Answer:** $$ - {1 \\over 2}$$\n\n$$\\overrightarrow a = \\widehat i - 2\\widehat j + \\widehat k$$\n
$$\\overrightarrow b = \\widehat i - \\widehat j + \\widehat k$$\n
$$\\left| {\\overrightarrow a } \\right|$$ = $$\\sqrt 6 $$, $$\\left| {\\overrightarrow b } \\right|$$ = $$\\sqrt 3 $$\n
and $${\\overrightarrow a .\\overrightarrow b }$$ = 4\n
Given $$\\overrightarrow b \\times \\overrightarrow c = \\overrightarrow b \\times \\overrightarrow a $$\n
$$ \\Rightarrow $$ $${\\left( {\\overrightarrow b \\times \\overrightarrow c } \\right)}$$ - $${\\left( {\\overrightarrow b \\times \\overrightarrow a } \\right)}$$ = 0\n
$$ \\Rightarrow $$ $${\\overrightarrow b \\times \\left( {\\overrightarrow c - \\overrightarrow a } \\right)}$$ = 0\n
$$ \\therefore $$ $${\\overrightarrow b \\parallel \\left( {\\overrightarrow c - \\overrightarrow a } \\right)}$$\n
$$ \\Rightarrow $$ $${\\left( {\\overrightarrow c - \\overrightarrow a } \\right) = \\lambda \\overrightarrow b }$$\n
$$ \\Rightarrow $$ $${\\overrightarrow c = \\overrightarrow a + \\lambda \\overrightarrow b }$$\n
$$ \\Rightarrow $$ $${\\overrightarrow c .\\overrightarrow a = \\overrightarrow a .\\overrightarrow a + \\lambda \\overrightarrow a .\\overrightarrow b }$$\n
$$ \\Rightarrow $$ 0 = $${{{\\left| {\\overrightarrow a } \\right|}^2} + \\lambda \\left( {\\overrightarrow a .\\overrightarrow b } \\right)}$$\n
$$ \\Rightarrow $$ $$\\lambda $$ = $${{{ - {{\\left| {\\overrightarrow a } \\right|}^2}} \\over {\\overrightarrow a .\\overrightarrow b }}}$$ = $${{ - 6} \\over 4} = - {3 \\over 2}$$\n
$$ \\therefore $$ $$\\overrightarrow c $$ = $${\\overrightarrow a - {3 \\over 2}\\overrightarrow b }$$\n
$$ \\Rightarrow $$ $$\\overrightarrow c $$ = ($$\\widehat i - 2\\widehat j + \\widehat k$$) - $${3 \\over 2}$$($$\\widehat i - \\widehat j + \\widehat k$$)\n
$$ \\therefore $$ $$\\overrightarrow c .\\overrightarrow b $$ = $$ - {1 \\over 2}\\left( {\\widehat i + \\widehat j + \\widehat k} \\right)$$($$\\widehat i - \\widehat j + \\widehat k$$)\n
$$ \\therefore $$ $$\\overrightarrow c .\\overrightarrow b $$ = $$ - {1 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7792,
"subject": "General Science",
"question": "Let $$\\overrightarrow a $$, $$\\overrightarrow b $$ and $$\\overrightarrow c $$ be three vectors such that $$\\left| {\\overrightarrow a } \\right| = \\sqrt 3 $$,\n$$\\left| {\\overrightarrow b } \\right| = 5,\\overrightarrow b .\\overrightarrow c = 10$$ and the angle between $$\\overrightarrow b $$ and $$\\overrightarrow c $$\nis $${\\pi \\over 3}$$. If $${\\overrightarrow a }$$ is perpendicular to the vector $$\\overrightarrow b \\times \\overrightarrow c $$ , then $$\\left| {\\overrightarrow a \\times \\left( {\\overrightarrow b \\times \\overrightarrow c } \\right)} \\right|$$ is equal to _____.",
"options": [],
"answer": "30",
"solution": "**Answer:** 30\n\nGiven $$\\left| {\\overrightarrow a } \\right| = \\sqrt 3 $$,\n$$\\left| {\\overrightarrow b } \\right| = 5$$\n
Given $$\\overrightarrow b .\\overrightarrow c = 10$$\n
And the angle between $$\\overrightarrow b $$ and $$\\overrightarrow c $$\nis $${\\pi \\over 3}$$\n
$${\\overrightarrow a }$$ is perpendicular to the vector $$\\overrightarrow b \\times \\overrightarrow c $$\n
$$ \\therefore $$ $$\\overrightarrow a .\\left( {\\overrightarrow b \\times \\overrightarrow c } \\right)$$ = 0 and angle between them is $${\\pi \\over 2}$$\n
Now $$\\left| {\\overrightarrow a \\times \\left( {\\overrightarrow b \\times \\overrightarrow c } \\right)} \\right|$$\n
= $$\\left| {\\overrightarrow a } \\right|\\left| {\\overrightarrow b \\times \\overrightarrow c } \\right|\\sin {\\pi \\over 2}$$\n
= $$\\left| {\\overrightarrow a } \\right|$$.$${\\left| {\\overrightarrow b } \\right|.\\left| {\\overrightarrow c } \\right|}$$$$\\sin {\\pi \\over 3}$$.1\n
$$ \\therefore $$ $$\\lambda $$ = $${8 \\over {10}}$$ = 0.8\n",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7794,
"subject": "General Science",
"question": "If $$\\overrightarrow a = 2\\widehat i + \\widehat j + 2\\widehat k$$, then the value of
\n$${\\left| {\\widehat i \\times \\left( {\\overrightarrow a \\times \\widehat i} \\right)} \\right|^2} + {\\left| {\\widehat j \\times \\left( {\\overrightarrow a \\times \\widehat j} \\right)} \\right|^2} + {\\left| {\\widehat k \\times \\left( {\\overrightarrow a \\times \\widehat k} \\right)} \\right|^2}$$ is equal to____\n",
"options": [],
"answer": "18",
"solution": "**Answer:** 18\n\nLet $$\\overrightarrow a = x\\widehat i + y\\widehat j + z\\widehat k$$
Now $$\\widehat i \\times \\left( {\\overrightarrow a \\times \\widehat i} \\right) = \\left( {\\widehat i.\\widehat i} \\right)\\overrightarrow a - \\left( {\\widehat i.\\overrightarrow a } \\right)\\widehat i$$
= $$y\\widehat j + z\\widehat k$$
Similarly $$\\widehat j \\times \\left( {\\overrightarrow a \\times \\widehat j} \\right) = x\\widehat i + z\\widehat k$$
$$\\widehat k \\times \\left( {\\overrightarrow a \\times \\widehat k} \\right) = x\\widehat i + y\\widehat j$$
Now $${\\left| {y\\widehat j + z\\widehat k} \\right|^2} + {\\left| {x\\widehat i + z\\widehat k} \\right|^2} + {\\left| {x\\widehat i + y\\widehat j} \\right|^2}$$
= $$2({x^2} + {y^2} + {z^2}) $$\n
Given $$\\overrightarrow a = 2\\widehat i + \\widehat j + 2\\widehat k$$\n
$$ \\therefore $$ x = 2, y = 1, z = 2\n
= 2(4 + 1 + 4) = 18",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7795,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = \\widehat i + 2\\widehat j - \\widehat k$$, $$\\overrightarrow b = \\widehat i - \\widehat j$$ and $$\\overrightarrow c = \\widehat i - \\widehat j - \\widehat k$$ be three given vectors. If $$\\overrightarrow r $$ is a vector such that $$\\overrightarrow r \\times \\overrightarrow a = \\overrightarrow c \\times \\overrightarrow a $$ and $$\\overrightarrow r .\\,\\overrightarrow b = 0$$, then $$\\overrightarrow r .\\,\\overrightarrow a $$ is equal to __________.",
"options": [],
"answer": "12",
"solution": "**Answer:** 12\n\n
Given, $$\\overrightarrow a = \\widehat i + 2\\widehat j - \\widehat k$$,
\n
$$\\overrightarrow b = \\widehat i - \\widehat j$$,
\n
$$\\overrightarrow c = \\widehat i - \\widehat j - \\widehat k$$
\n
$$\\overrightarrow r \\times \\overrightarrow a = \\overrightarrow c \\times \\overrightarrow a $$
\n
$$ \\Rightarrow \\overrightarrow r \\times \\overrightarrow a - \\overrightarrow c \\times \\overrightarrow a = 0$$
\n
$$ \\Rightarrow (\\overrightarrow r - \\overrightarrow c ) \\times \\overrightarrow a = 0$$
\n
$$\\therefore$$ $$\\overrightarrow r - \\overrightarrow c = \\lambda \\overrightarrow a $$
\n
$$ \\Rightarrow \\overrightarrow r = \\lambda \\overrightarrow a + \\overrightarrow c $$
\n
$$ \\Rightarrow \\overrightarrow r \\,.\\,\\overrightarrow b = \\lambda \\overrightarrow a \\,.\\,\\overrightarrow b + \\overrightarrow c \\,.\\,\\overrightarrow b $$ (taking dot with $$\\overrightarrow b $$)
\n
$$ \\Rightarrow 0 = \\lambda \\overrightarrow a \\,.\\,\\overrightarrow b + \\overrightarrow c \\,.\\,\\overrightarrow b $$ [$$\\because$$ $$\\overrightarrow r \\,.\\,\\overrightarrow b = 0$$]
\n
$$ \\Rightarrow \\lambda (\\widehat i + 2\\widehat j - \\widehat k)\\,.\\,(\\widehat i - \\widehat j) + (\\widehat i - \\widehat j - \\widehat k)\\,.\\,(\\widehat i - \\widehat j) = 0$$
\n
$$ \\Rightarrow \\lambda (1 - 2) + 2 = 0$$
\n
$$ \\Rightarrow \\lambda = 2$$
\n
$$\\therefore$$ $$\\overrightarrow r = 2\\overrightarrow a + \\overrightarrow c $$
\n
$$ \\Rightarrow \\overrightarrow r \\,.\\,\\overrightarrow a = 2\\overrightarrow a \\,.\\,\\overrightarrow a + \\overrightarrow c \\,.\\,\\overrightarrow a $$ [taking dot with $${\\overrightarrow a }$$]
\n
$$ = 2{\\left| {\\overrightarrow a } \\right|^2} + \\overrightarrow a \\,.\\,\\overrightarrow c $$
\n
$$ = 2(1 + 4 + 1) + (1 - 2 + 1)$$
\n
$$ \\Rightarrow \\overrightarrow r \\,.\\,\\overrightarrow a = 12$$
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7796,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = \\widehat i + \\alpha \\widehat j + 3\\widehat k$$ and $$\\overrightarrow b = 3\\widehat i - \\alpha \\widehat j + \\widehat k$$. If the area of the parallelogram whose adjacent sides are represented by the vectors $$\\overrightarrow a $$ and $$\\overrightarrow b $$ is $$8\\sqrt 3 $$ square units, then $$\\overrightarrow a $$ . $$\\overrightarrow b $$ is equal to __________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n$$\\overrightarrow a = \\widehat i + \\alpha \\widehat j + 3\\widehat k$$
$$\\overrightarrow b = 3\\widehat i - \\alpha \\widehat j + \\widehat k$$
Area of parallelogram = $$\\left| {\\overrightarrow a \\times \\overrightarrow b } \\right|$$
$$\\overrightarrow r = 3\\widehat i - \\widehat j + 2\\widehat k$$
$$\\alpha + |\\overrightarrow r {|^2} = 1 + 14 = 15$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7798,
"subject": "General Science",
"question": "Let $$\\overrightarrow a $$ and $$\\overrightarrow b $$ be two non-zero vectors perpendicular to each other and $$|\\overrightarrow a | = |\\overrightarrow b |$$. If $$|\\overrightarrow a \\times \\overrightarrow b | = |\\overrightarrow a |$$, then the angle between the vectors $$\\left( {\\overrightarrow a + \\overrightarrow b + \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)} \\right)$$ and $${\\overrightarrow a }$$ is equal to :",
"options": [
{
"text": "$${\\sin ^{ - 1}}\\left( {{1 \\over {\\sqrt 6 }}} \\right)$$"
},
{
"text": "$${\\cos ^{ - 1}}\\left( {{1 \\over {\\sqrt 2 }}} \\right)$$"
},
{
"text": "$${\\sin ^{ - 1}}\\left( {{1 \\over {\\sqrt 3 }}} \\right)$$"
},
{
"text": "$${\\cos ^{ - 1}}\\left( {{1 \\over {\\sqrt 3 }}} \\right)$$"
}
],
"answer": "$${\\cos ^{ - 1}}\\left( {{1 \\over {\\sqrt 3 }}} \\right)$$",
"solution": "**Answer:** $${\\cos ^{ - 1}}\\left( {{1 \\over {\\sqrt 3 }}} \\right)$$\n\n$$\\overrightarrow a $$ is perpendicular to $$\\overrightarrow b $$
$$ \\therefore $$ $$\\overrightarrow a $$ . $$\\overrightarrow b $$ = 0
Given, | $$\\overrightarrow a $$ $$\\times$$ $$\\overrightarrow b $$ | = | $$\\overrightarrow a $$ |
and | $$\\overrightarrow a $$ | = | $$\\overrightarrow b $$ |
$$ \\therefore $$ | $$\\overrightarrow a $$ $$\\times$$ $$\\overrightarrow b $$ | = | $$\\overrightarrow a $$ | = | $$\\overrightarrow b $$ | = k(assume)
Now, angle between $$\\overrightarrow a $$ and ($$\\overrightarrow a $$ + $$\\overrightarrow b $$ + ($$\\overrightarrow a $$ $$\\times$$ $$\\overrightarrow b $$))
$$\\cos \\theta = {{\\overrightarrow a ((\\overrightarrow a + \\overrightarrow b + (\\overrightarrow a \\times \\overrightarrow b ))} \\over {|\\overrightarrow a |.|\\overrightarrow a + \\overrightarrow b + (\\overrightarrow a \\times \\overrightarrow b )|}}$$
$$ = {{|\\overrightarrow a {|^2} + \\,\\overrightarrow a .\\,\\overrightarrow b + \\overrightarrow a (\\overrightarrow a \\times \\overrightarrow b )} \\over {|\\overrightarrow a |.|\\overrightarrow a + \\overrightarrow b + (\\overrightarrow a \\times \\overrightarrow b )|}}$$
[Note $$\\overrightarrow a .(\\overrightarrow a \\times \\overrightarrow b ) = [\\overrightarrow a \\overrightarrow a \\overrightarrow b ] = 0$$]
$$ = {{|\\overrightarrow a {|^2} + 0 + 0} \\over {|\\overrightarrow a |.|\\overrightarrow a + \\overrightarrow b + (\\overrightarrow a \\times \\overrightarrow b )|}}$$
Now, $$|\\overrightarrow a + \\overrightarrow b + (\\overrightarrow a \\times \\overrightarrow b ){|^2}$$
$$ = |\\overrightarrow a {|^2} + |\\overrightarrow a {|^2} + |\\overrightarrow a \\times \\overrightarrow b {|^2} + $$
$$2\\overrightarrow a .\\overrightarrow b + 2\\overrightarrow b .(\\overrightarrow a \\times \\overrightarrow b ) + 2\\overrightarrow a .(\\overrightarrow a \\times \\overrightarrow b )$$
$$ = {k^2} + {k^2} + {k^2}$$
$$ \\therefore $$ $$|\\overrightarrow a + \\overrightarrow b + (\\overrightarrow a \\times \\overrightarrow b ){|^2} = 3{k^2}$$
$$ \\Rightarrow |\\overrightarrow a + \\overrightarrow b + (\\overrightarrow a \\times \\overrightarrow b )| = \\sqrt 3 k$$
$$ \\Rightarrow \\theta = {\\cos ^{ - 1}}\\left( {{1 \\over {\\sqrt 3 }}} \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7799,
"subject": "General Science",
"question": "If the shortest distance between the lines $$\\overrightarrow {{r_1}} = \\alpha \\widehat i + 2\\widehat j + 2\\widehat k + \\lambda (\\widehat i - 2\\widehat j + 2\\widehat k)$$, $$\\lambda$$ $$\\in$$ R, $$\\alpha$$ > 0 and $$\\overrightarrow {{r_2}} = - 4\\widehat i - \\widehat k + \\mu (3\\widehat i - 2\\widehat j - 2\\widehat k)$$, $$\\mu$$ $$\\in$$ R is 9, then $$\\alpha$$ is equal to ____________.",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\nIf $$\\overrightarrow r = \\overrightarrow a + \\lambda \\overrightarrow b $$ and $$\\overrightarrow r = \\overrightarrow c + \\lambda \\overrightarrow d $$ then shortest distance between two lines is
$$L = {{(\\overrightarrow a - \\overrightarrow c ).(\\overrightarrow b \\times \\overrightarrow d )} \\over {|b \\times d|}}$$
$$\\therefore$$ $$\\overrightarrow a - \\overrightarrow c = ((\\alpha + 4)\\widehat i + 2\\widehat j + 3\\widehat k)$$
$${{\\overrightarrow b \\times \\overrightarrow d } \\over {|b \\times d|}} = {{(2\\widehat i + 2\\widehat j + \\widehat k)} \\over 3}$$
or $$\\alpha$$ = 6",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7800,
"subject": "General Science",
"question": "Let $$\\overrightarrow p = 2\\widehat i + 3\\widehat j + \\widehat k$$ and $$\\overrightarrow q = \\widehat i + 2\\widehat j + \\widehat k$$ be two vectors. If a vector $$\\overrightarrow r = (\\alpha \\widehat i + \\beta \\widehat j + \\gamma \\widehat k)$$ is perpendicular to each of the vectors ($$(\\overrightarrow p + \\overrightarrow q )$$ and $$(\\overrightarrow p - \\overrightarrow q )$$, and $$\\left| {\\overrightarrow r } \\right| = \\sqrt 3 $$, then $$\\left| \\alpha \\right| + \\left| \\beta \\right| + \\left| \\gamma \\right|$$ is equal to _______________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n$$\\overrightarrow p = 2\\widehat i + 3\\widehat j + \\widehat k$$ (Given )
$$\\overrightarrow r = \\pm \\left( { - \\widehat i - \\widehat j - \\widehat k} \\right)$$
According to question
$$\\overrightarrow r = \\alpha \\widehat i + \\beta \\widehat j + \\gamma \\widehat k$$
So, |$$\\alpha$$| = 1, |$$\\beta$$| = 1, |$$\\gamma$$| = 1
$$\\Rightarrow$$ $$\\left| \\alpha \\right| + \\left| \\beta \\right| + \\left| \\gamma \\right|$$ = 3",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7801,
"subject": "General Science",
"question": "If $$\\left| {\\overrightarrow a } \\right| = 2,\\left| {\\overrightarrow b } \\right| = 5$$ and $$\\left| {\\overrightarrow a \\times \\overrightarrow b } \\right|$$ = 8, then $$\\left| {\\overrightarrow a .\\,\\overrightarrow b } \\right|$$ is equal to :",
"options": [
{
"text": "6"
},
{
"text": "4"
},
{
"text": "3"
},
{
"text": "5"
}
],
"answer": "6",
"solution": "**Answer:** 6\n\n$$\\left| {\\overrightarrow a } \\right| = 2,\\left| {\\overrightarrow b } \\right| = 5$$
$$\\left| {\\overrightarrow a \\times \\overrightarrow b } \\right| = \\left| {\\overrightarrow a } \\right|\\left| {\\overrightarrow b } \\right|\\sin \\theta = \\pm 8$$
$$\\sin \\theta = \\pm \\,{4 \\over 5}$$
$$\\therefore$$ $$\\overrightarrow a .\\,\\overrightarrow b = \\left| {\\overrightarrow a } \\right|\\left| {\\overrightarrow b } \\right|\\cos \\theta $$
$$\\left| {\\overrightarrow a .\\,\\overrightarrow b } \\right| = 6$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7802,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = \\widehat i + \\widehat j + 2\\widehat k$$ and $$\\overrightarrow b = - \\widehat i + 2\\widehat j + 3\\widehat k$$. Then the vector product $$\\left( {\\overrightarrow a + \\overrightarrow b } \\right) \\times \\left( {\\left( {\\overrightarrow a \\times \\left( {\\left( {\\overrightarrow a - \\overrightarrow b } \\right) \\times \\overrightarrow b } \\right)} \\right) \\times \\overrightarrow b } \\right)$$ is equal to :",
"options": [
{
"text": "$$5(34\\widehat i - 5\\widehat j + 3\\widehat k)$$"
},
{
"text": "$$7(34\\widehat i - 5\\widehat j + 3\\widehat k)$$"
},
{
"text": "$$7(30\\widehat i - 5\\widehat j + 7\\widehat k)$$"
},
{
"text": "$$5(30\\widehat i - 5\\widehat j + 7\\widehat k)$$"
}
],
"answer": "$$7(34\\widehat i - 5\\widehat j + 3\\widehat k)$$",
"solution": "**Answer:** $$7(34\\widehat i - 5\\widehat j + 3\\widehat k)$$\n\n$$\\overrightarrow a = \\widehat i + \\widehat j + 2\\widehat k$$
$$\\overrightarrow b = - \\widehat i + 2\\widehat j + 3\\widehat k$$
$$\\overrightarrow a + \\overrightarrow b = 3\\widehat j + 5\\widehat k;\\overrightarrow a.\\overrightarrow b = - 1 + 2 + 6 = 7$$
$$\\left( {\\left( {\\overrightarrow a \\times \\left( {\\left( {\\overrightarrow a - \\overrightarrow b } \\right) \\times \\overrightarrow b } \\right)} \\right) \\times \\overrightarrow b } \\right)$$
$$\\left( {\\left( {\\overrightarrow a \\times \\left( {\\overrightarrow a \\times \\overrightarrow b - \\overrightarrow b \\times \\overrightarrow b } \\right)} \\right) \\times \\overrightarrow b } \\right)$$
$$\\left( {\\overrightarrow a \\times \\left( {\\overrightarrow a \\times \\overrightarrow b - 0} \\right)} \\right) \\times \\overrightarrow b $$
$$\\left( {\\overrightarrow a \\times \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)} \\right) \\times \\overrightarrow b $$
$$\\left( {\\left( {\\overrightarrow a .\\overrightarrow b } \\right)\\overrightarrow a - \\left( {\\overrightarrow a .\\overrightarrow a } \\right)\\overrightarrow b } \\right) \\times \\overrightarrow b $$
$$\\left( {\\overrightarrow a .\\overrightarrow b } \\right)\\overrightarrow a \\times \\overrightarrow b - \\left( {\\overrightarrow a .\\overrightarrow a } \\right)\\left( {\\overrightarrow b \\times \\overrightarrow b } \\right)$$
$$\\left( {\\overrightarrow a .\\overrightarrow b } \\right)\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)$$
$$\\overrightarrow a \\times \\overrightarrow b = \\left| {\\matrix{\n i & j & k \\cr \n 1 & 1 & 2 \\cr \n { - 1} & 2 & 3 \\cr \n\n } } \\right| = - \\widehat i - 5\\widehat j + 3\\widehat k$$
= $$7(34\\widehat i - 5\\widehat j + 3\\widehat k)$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7803,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = \\widehat i + \\widehat j + \\widehat k,\\overrightarrow b $$ and $$\\overrightarrow c = \\widehat j - \\widehat k$$ be three vectors such that $$\\overrightarrow a \\times \\overrightarrow b = \\overrightarrow c $$ and $$\\overrightarrow a \\,.\\,\\overrightarrow b = 1$$. If the length of projection vector of the vector $$\\overrightarrow b $$ on the vector $$\\overrightarrow a \\times \\overrightarrow c $$ is l, then the value of 3l2 is equal to _____________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n$$\\overrightarrow a \\times \\overrightarrow b = \\overrightarrow c $$
Take Dot with $$\\overrightarrow c $$
$$\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right).\\,\\overrightarrow c = {\\left| {\\overrightarrow c } \\right|^2} = 2$$
Projection of $$\\overrightarrow b $$ or $$\\overrightarrow a \\times \\overrightarrow c = l$$
$${{\\left| {\\overrightarrow b \\,.\\,(\\overrightarrow a \\times \\overrightarrow c )} \\right|} \\over {|\\overrightarrow a \\times \\overrightarrow c |}} = l$$
$$3{l^2} = 2$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7804,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = \\widehat i + 5\\widehat j + \\alpha \\widehat k$$, $$\\overrightarrow b = \\widehat i + 3\\widehat j + \\beta \\widehat k$$ and $$\\overrightarrow c = - \\widehat i + 2\\widehat j - 3\\widehat k$$ be three vectors such that, $$\\left| {\\overrightarrow b \\times \\overrightarrow c } \\right| = 5\\sqrt 3 $$ and $${\\overrightarrow a }$$ is perpendicular to $${\\overrightarrow b }$$. Then the greatest amongst the values of $${\\left| {\\overrightarrow a } \\right|^2}$$ is _____________.",
"options": [],
"answer": "90",
"solution": "**Answer:** 90\n\nSince, $$\\overrightarrow a .\\,\\overrightarrow b = 0$$
$$ \\Rightarrow \\left| {\\overrightarrow a } \\right|_{\\max }^2 = {(26 + {\\alpha ^2})_{\\max }} = 90$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7805,
"subject": "General Science",
"question": "Let $$\\overrightarrow a = 2\\widehat i - \\widehat j + 2\\widehat k$$ and $$\\overrightarrow b = \\widehat i + 2\\widehat j - \\widehat k$$. Let a vector $$\\overrightarrow v $$ be in the plane containing $$\\overrightarrow a $$ and $$\\overrightarrow b $$. If $$\\overrightarrow v $$ is perpendicular to the vector $$3\\widehat i + 2\\widehat j - \\widehat k$$ and its projection on $$\\overrightarrow a $$ is 19 units, then $${\\left| {2\\overrightarrow v } \\right|^2}$$ is equal to _____________.",
"options": [],
"answer": "1494",
"solution": "**Answer:** 1494\n\n$$\\overrightarrow a = 2\\widehat i - \\widehat j + 2\\widehat k$$
$$\\overrightarrow b = \\widehat i + 2\\widehat j - \\widehat k$$
$$\\overrightarrow c = 3\\widehat i + 2\\widehat j - \\widehat k$$
$$\\overrightarrow v = x\\overrightarrow a + y\\overrightarrow b $$
$$\\overrightarrow v \\left( {3\\widehat i + 2\\widehat j - \\widehat k} \\right) = 0$$
$$\\overrightarrow v .\\widehat a = 19$$
$$\\overrightarrow v = \\lambda \\overrightarrow c \\times \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)$$
$$\\overrightarrow v = \\lambda \\left[ {\\left( {\\overrightarrow c .\\overrightarrow b } \\right)\\overrightarrow a - \\left( {\\overrightarrow c .\\overrightarrow a } \\right)\\overrightarrow b } \\right]$$
Let $$\\overrightarrow a = \\alpha \\widehat i + 3\\widehat j - \\widehat k$$, $$\\overrightarrow b = 3\\widehat i - \\beta \\widehat j + 4\\widehat k$$ and $$\\overrightarrow c = \\widehat i + 2\\widehat j - 2\\widehat k$$ where $$\\alpha ,\\,\\beta \\in R$$, be three vectors. If the projection of $$\\overrightarrow a $$ on $$\\overrightarrow c $$ is $${{10} \\over 3}$$ and $$\\overrightarrow b \\times \\overrightarrow c = - 6\\widehat i + 10\\widehat j + 7\\widehat k$$, then the value of $$\\alpha + \\beta $$ is equal to :
Let $$\\overrightarrow a = \\widehat i - 2\\widehat j + 3\\widehat k$$, $$\\overrightarrow b = \\widehat i + \\widehat j + \\widehat k$$ and $$\\overrightarrow c $$ be a vector such that $$\\overrightarrow a + \\left( {\\overrightarrow b \\times \\overrightarrow c } \\right) = \\overrightarrow 0 $$ and $$\\overrightarrow b \\,.\\,\\overrightarrow c = 5$$. Then the value of $$3\\left( {\\overrightarrow c \\,.\\,\\overrightarrow a } \\right)$$ is equal to _________.
Equation (i) and equation (ii) are contradicting.",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7808,
"subject": "General Science",
"question": "
Let $$\\overrightarrow a = \\alpha \\widehat i + 2\\widehat j - \\widehat k$$ and $$\\overrightarrow b = - 2\\widehat i + \\alpha \\widehat j + \\widehat k$$, where $$\\alpha \\in R$$. If the area of the parallelogram whose adjacent sides are represented by the vectors $$\\overrightarrow a $$ and $$\\overrightarrow b $$ is $$\\sqrt {15({\\alpha ^2} + 4)} $$, then the value of $$2{\\left| {\\overrightarrow a } \\right|^2} + \\left( {\\overrightarrow a \\,.\\,\\overrightarrow b } \\right){\\left| {\\overrightarrow b } \\right|^2}$$ is equal to :
Let $$\\overrightarrow a $$ be a vector which is perpendicular to the vector $$3\\widehat i + {1 \\over 2}\\widehat j + 2\\widehat k$$. If $$\\overrightarrow a \\times \\left( {2\\widehat i + \\widehat k} \\right) = 2\\widehat i - 13\\widehat j - 4\\widehat k$$, then the projection of the vector $$\\overrightarrow a $$ on the vector $$2\\widehat i + 2\\widehat j + \\widehat k$$ is :
If $$\\overrightarrow a = 2\\widehat i + \\widehat j + 3\\widehat k$$, $$\\overrightarrow b = 3\\widehat i + 3\\widehat j + \\widehat k$$ and $$\\overrightarrow c = {c_1}\\widehat i + {c_2}\\widehat j + {c_3}\\widehat k$$ are coplanar vectors and $$\\overrightarrow a \\,.\\,\\overrightarrow c = 5$$, $$\\overrightarrow b \\bot \\overrightarrow c $$, then $$122({c_1} + {c_2} + {c_3})$$ is equal to ___________.
Let $$\\overrightarrow a $$ and $$\\overrightarrow b $$ be the vectors along the diagonals of a parallelogram having area $$2\\sqrt 2 $$. Let the angle between $$\\overrightarrow a $$ and $$\\overrightarrow b $$ be acute, $$|\\overrightarrow a | = 1$$, and $$|\\overrightarrow a \\,.\\,\\overrightarrow b | = |\\overrightarrow a \\times \\overrightarrow b |$$. If $$\\overrightarrow c = 2\\sqrt 2 \\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) - 2\\overrightarrow b $$, then an angle between $$\\overrightarrow b $$ and $$\\overrightarrow c $$ is :
Let $$\\overrightarrow a = \\widehat i + \\widehat j - \\widehat k$$ and $$\\overrightarrow c = 2\\widehat i - 3\\widehat j + 2\\widehat k$$. Then the number of vectors $$\\overrightarrow b $$ such that $$\\overrightarrow b \\times \\overrightarrow c = \\overrightarrow a $$ and $$|\\overrightarrow b | \\in $$ {1, 2, ........, 10} is :
Let $$\\overrightarrow b = \\widehat i + \\widehat j + \\lambda \\widehat k$$, $$\\lambda$$ $$\\in$$ R. If $$\\overrightarrow a $$ is a vector such that $$\\overrightarrow a \\times \\overrightarrow b = 13\\widehat i - \\widehat j - 4\\widehat k$$ and $$\\overrightarrow a \\,.\\,\\overrightarrow b + 21 = 0$$, then $$\\left( {\\overrightarrow b - \\overrightarrow a } \\right).\\,\\left( {\\widehat k - \\widehat j} \\right) + \\left( {\\overrightarrow b + \\overrightarrow a } \\right).\\,\\left( {\\widehat i - \\widehat k} \\right)$$ is equal to _____________.
Let $$\\theta$$ be the angle between the vectors $$\\overrightarrow a $$ and $$\\overrightarrow b $$, where $$|\\overrightarrow a | = 4,$$ $$|\\overrightarrow b | = 3$$ and $$\\theta \\in \\left( {{\\pi \\over 4},{\\pi \\over 3}} \\right)$$. Then $${\\left| {\\left( {\\overrightarrow a - \\overrightarrow b } \\right) \\times \\left( {\\overrightarrow a + \\overrightarrow b } \\right)} \\right|^2} + 4{\\left( {\\overrightarrow a \\,.\\,\\overrightarrow b } \\right)^2}$$ is equal to __________.
$${\\left| {\\left( {\\overrightarrow a - \\overrightarrow b } \\right) \\times \\left( {\\overrightarrow a + \\overrightarrow b } \\right)} \\right|^2} + 4{\\left( {\\overrightarrow a \\,.\\,\\overrightarrow b } \\right)^2}$$
\n
$$ \\Rightarrow {\\left| {\\overrightarrow a \\times \\overrightarrow a + \\overrightarrow a \\times \\overrightarrow b - \\overrightarrow b \\times \\overrightarrow a - \\overrightarrow b \\times \\overrightarrow b } \\right|^2} + 4{\\left( {\\overrightarrow a \\,.\\,\\overrightarrow b } \\right)^2}$$
\n
$$ \\Rightarrow {\\left| {2\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)} \\right|^2} + 4{\\left( {\\overrightarrow a \\,.\\,\\overrightarrow b } \\right)^2}$$
\n
$$ \\Rightarrow 4{\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right)^2} + {\\left( {\\overrightarrow a \\,.\\,\\overrightarrow b } \\right)^2}$$
\n
$$ \\Rightarrow 4{\\left| {\\overrightarrow a } \\right|^2}{\\left| {\\overrightarrow b } \\right|^2} = 4\\,.\\,16\\,.\\,9 = 576$$
Let $$\\widehat a$$ and $$\\widehat b$$ be two unit vectors such that $$|(\\widehat a + \\widehat b) + 2(\\widehat a \\times \\widehat b)| = 2$$. If $$\\theta$$ $$\\in$$ (0, $$\\pi$$) is the angle between $$\\widehat a$$ and $$\\widehat b$$, then among the statements :
\n
(S1) : $$2|\\widehat a \\times \\widehat b| = |\\widehat a - \\widehat b|$$
\n
(S2) : The projection of $$\\widehat a$$ on ($$\\widehat a$$ + $$\\widehat b$$) is $${1 \\over 2}$$
",
"options": [
{
"text": "Only (S1) is true."
},
{
"text": "Only (S2) is true."
},
{
"text": "Both (S1) and (S2) are true."
},
{
"text": "Both (S1) and (S2) are false."
}
],
"answer": "Both (S1) and (S2) are true.",
"solution": "**Answer:** Both (S1) and (S2) are true.\n\n
$$\\left| {\\widehat a + \\widehat b + 2(\\widehat a \\times \\widehat b)} \\right| = 2,\\,\\theta \\in (0,\\,\\pi )$$
\n
$$ \\Rightarrow {\\left| {\\widehat a + \\widehat b + 2(\\widehat a \\times \\widehat b)} \\right|^2} = 4$$
where $$\\theta$$ is angle between $$\\widehat a$$ and $$\\widehat b$$.
\n
$$\\therefore$$ $$2\\left| {\\widehat a \\times \\widehat b} \\right| = \\sqrt 3 = \\left| {\\widehat a - \\widehat b} \\right|$$
\n
(S1) is correct.
\n
And projection of $$\\widehat a$$ on $$(\\widehat a + \\widehat b) = \\left| {{{\\widehat a\\,.\\,(\\widehat a + \\widehat b)} \\over {\\left| {\\widehat a + \\widehat b} \\right|}}} \\right| = {1 \\over 2}$$
Let $$\\widehat a$$, $$\\widehat b$$ be unit vectors. If $$\\overrightarrow c $$ be a vector such that the angle between $$\\widehat a$$ and $$\\overrightarrow c $$ is $${\\pi \\over {12}}$$, and $$\\widehat b = \\overrightarrow c + 2\\left( {\\overrightarrow c \\times \\widehat a} \\right)$$, then $${\\left| {6\\overrightarrow c } \\right|^2}$$ is equal to :
Let $$\\mathrm{ABC}$$ be a triangle such that $$\\overrightarrow{\\mathrm{BC}}=\\overrightarrow{\\mathrm{a}}, \\overrightarrow{\\mathrm{CA}}=\\overrightarrow{\\mathrm{b}}, \\overrightarrow{\\mathrm{AB}}=\\overrightarrow{\\mathrm{c}},|\\overrightarrow{\\mathrm{a}}|=6 \\sqrt{2},|\\overrightarrow{\\mathrm{b}}|=2 \\sqrt{3}$$ and $$\\vec{b} \\cdot \\vec{c}=12$$. Consider the statements :\n
\n$\\Rightarrow 12+12=-6 \\sqrt{2} \\cdot 2 \\sqrt{3} \\cos (\\pi-\\angle A C B)$\n
\n$\\therefore \\cos (\\angle A C B)=\\sqrt{\\frac{2}{3}}$ \n
\n$$\n\\begin{aligned}\n&\\therefore \\angle A C B=\\cos ^{-1} \\sqrt{\\frac{2}{3}} \\\\\\\\\n&\\therefore S(2) \\text { is correct. }\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7818,
"subject": "General Science",
"question": "
Let $$\\vec{a}=\\hat{i}-\\hat{j}+2 \\hat{k}$$ and let $$\\vec{b}$$ be a vector such that $$\\vec{a} \\times \\vec{b}=2 \\hat{i}-\\hat{k}$$ and $$\\vec{a} \\cdot \\vec{b}=3$$. Then the projection of $$\\vec{b}$$ on the vector $$\\vec{a}-\\vec{b}$$ is :
$$\\overrightarrow a = \\widehat i - \\widehat j + 2\\widehat k$$
\n
$$\\overrightarrow a \\times \\overrightarrow b = 2\\widehat i - \\widehat k$$
\n
$$\\overrightarrow a \\,.\\,\\overrightarrow b = 3$$
\n
$$|\\overrightarrow a \\times \\overrightarrow b {|^2} + |\\overrightarrow a \\,.\\,\\overrightarrow b {|^2} = |\\overrightarrow a {|^2}\\,.\\,|\\overrightarrow b {|^2}$$
\n
$$ \\Rightarrow 5 + 9 = 6|\\overrightarrow b {|^2}$$
\n
$$ \\Rightarrow |b {|^2} = {7 \\over 3}$$
\n
$$|\\overrightarrow a - \\overrightarrow b | = \\sqrt {|\\overrightarrow a {|^2} + |\\overrightarrow b {|^2} - 2\\overrightarrow a \\,.\\,\\overrightarrow b } = \\sqrt {{7 \\over 3}} $$
\n
projection of $$\\overrightarrow b $$ on $$\\overrightarrow a - \\overrightarrow b = {{\\overrightarrow b \\,.\\,(\\overrightarrow a - \\overrightarrow b )} \\over {|\\overrightarrow a - \\overrightarrow b |}}$$
\n
$$ = {{\\overrightarrow b \\,.\\,\\overrightarrow a - |\\overrightarrow b {|^2}} \\over {|\\overrightarrow a - \\overrightarrow b |}} = {{3 - {7 \\over 3}} \\over {\\sqrt {{7 \\over 3}} }}$$
Let $$\\overrightarrow{\\mathrm{a}}=\\alpha \\hat{i}+\\hat{j}-\\hat{k}$$ and $$\\overrightarrow{\\mathrm{b}}=2 \\hat{i}+\\hat{j}-\\alpha \\hat{k}, \\alpha>0$$. If the projection of $$\\overrightarrow{\\mathrm{a}} \\times \\overrightarrow{\\mathrm{b}}$$ on the vector $$-\\hat{i}+2 \\hat{j}-2 \\hat{k}$$ is 30, then $$\\alpha$$ is equal to :
Let $$\\vec{a}=\\alpha \\hat{i}+\\hat{j}+\\beta \\hat{k}$$ and $$\\vec{b}=3 \\hat{i}-5 \\hat{j}+4 \\hat{k}$$ be two vectors, such that $$\\vec{a} \\times \\vec{b}=-\\hat{i}+9 \\hat{j}+12 \\hat{k}$$. Then the projection of $$\\vec{b}-2 \\vec{a}$$ on $$\\vec{b}+\\vec{a}$$ is equal to :
$$\\overrightarrow b - 2\\overrightarrow a = 3\\widehat i - 5\\widehat j + 4\\widehat k - 2\\left( { - 3\\widehat i + \\widehat j - \\widehat k} \\right)$$
\n
$$\\overrightarrow b - 2\\overrightarrow a = 9\\widehat i - 7\\widehat j + 6\\widehat k$$
\n
$$\\overrightarrow b + \\overrightarrow a = \\left( {3\\widehat i - 5\\widehat j + 4\\widehat k} \\right) + \\left( { - 3\\widehat i + \\widehat j - \\widehat k} \\right)$$
\n
$$\\overrightarrow b + \\overrightarrow a = - 4\\widehat j + 3\\widehat k$$
\n
Projection of $$\\overrightarrow b - 2\\overrightarrow a $$ on $$\\overrightarrow b + \\overrightarrow a $$ is $$ = {{\\left( {\\overrightarrow b - 2\\overrightarrow a } \\right)\\,.\\,\\left( {\\overrightarrow b + \\overrightarrow a } \\right)} \\over {\\left| {\\overrightarrow b + \\overrightarrow a } \\right|}}$$
Let $$\\overrightarrow a $$, $$\\overrightarrow b $$, $$\\overrightarrow c $$ be three non-coplanar vectors such that $$\\overrightarrow a $$ $$\\times$$ $$\\overrightarrow b $$ = 4$$\\overrightarrow c $$, $$\\overrightarrow b $$ $$\\times$$ $$\\overrightarrow c $$ = 9$$\\overrightarrow a $$ and $$\\overrightarrow c $$ $$\\times$$ $$\\overrightarrow a $$ = $$\\alpha$$$$\\overrightarrow b $$, $$\\alpha$$ > 0. If $$\\left| {\\overrightarrow a } \\right| + \\left| {\\overrightarrow b } \\right| + \\left| {\\overrightarrow c } \\right| = {1 \\over {36}}$$, then $$\\alpha$$ is equal to __________.
Let a vector $$\\vec{a}$$ has magnitude 9. Let a vector $$\\vec{b}$$ be such that for every $$(x, y) \\in \\mathbf{R} \\times \\mathbf{R}-\\{(0,0)\\}$$, the vector $$(x \\vec{a}+y \\vec{b})$$ is perpendicular to the vector $$(6 y \\vec{a}-18 x \\vec{b})$$. Then the value of $$|\\vec{a} \\times \\vec{b}|$$ is equal to :
Let $$\\hat{a}$$ and $$\\hat{b}$$ be two unit vectors such that the angle between them is $$\\frac{\\pi}{4}$$. If $$\\theta$$ is the angle between the vectors $$(\\hat{a}+\\hat{b})$$ and $$(\\hat{a}+2 \\hat{b}+2(\\hat{a} \\times \\hat{b}))$$, then the value of $$164 \\,\\cos ^{2} \\theta$$ is equal to :
Let $$\\vec{a}, \\vec{b}, \\vec{c}$$ be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and\n$$(\\vec{a} \\times \\vec{b}) \\cdot(\\vec{b} \\times \\vec{c})+(\\vec{b} \\times \\vec{c}) \\cdot(\\vec{c} \\times \\vec{a})+(\\vec{c} \\times \\vec{a}) \\cdot(\\vec{a} \\times \\vec{b})=168$$, then $$|\\vec{a}|+|\\vec{b}|+|\\vec{c}|$$ is equal to :
Let $$\\vec{a}$$ and $$\\vec{b}$$ be two vectors such that $$|\\vec{a}+\\vec{b}|^{2}=|\\vec{a}|^{2}+2|\\vec{b}|^{2}, \\vec{a} \\cdot \\vec{b}=3$$ and $$|\\vec{a} \\times \\vec{b}|^{2}=75$$. Then $$|\\vec{a}|^{2}$$ is equal to __________.
$$ \\Rightarrow $$ $25|\\overrightarrow{\\mathrm{r}}|^{2}=339$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7827,
"subject": "General Science",
"question": "Let $\\vec{a}, \\vec{b}, \\vec{c}$ be three vectors such that\n
$|\\vec{a}|=\\sqrt{31}, 4|\\vec{b}|=|\\vec{c}|=2$ and $2(\\vec{a} \\times \\vec{b})=3(\\vec{c} \\times \\vec{a})$.\n
If the angle between $\\vec{b}$ and $\\vec{c}$ is $\\frac{2 \\pi}{3}$, then $\\left(\\frac{\\vec{a} \\times \\vec{c}}{\\vec{a} \\cdot \\vec{b}}\\right)^{2}$ is equal to __________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n$2(\\vec{a} \\times \\vec{b})=3(\\vec{c} \\times \\vec{a})$\n
$$A(2,6,2), B(-4,0, \\lambda), C(2,3,-1)$$ and $$D(4,5,0),|\\lambda| \\leq 5$$ are the vertices of a quadrilateral $$A B C D$$. If its area is 18 square units, then $$5-6 \\lambda$$ is equal to __________.
Let $$\\vec{a}$$ and $$\\vec{b}$$ be two vectors such that $$|\\vec{a}|=\\sqrt{14},|\\vec{b}|=\\sqrt{6}$$ and $$|\\vec{a} \\times \\vec{b}|=\\sqrt{48}$$. Then $$(\\vec{a} \\cdot \\vec{b})^{2}$$ is equal to ___________.
Let a unit vector $$\\widehat{O P}$$ make angles $$\\alpha, \\beta, \\gamma$$ with the positive directions of the co-ordinate axes $$\\mathrm{OX}$$, $$\\mathrm{OY}, \\mathrm{OZ}$$ respectively, where $$\\beta \\in\\left(0, \\frac{\\pi}{2}\\right)$$. If $$\\widehat{\\mathrm{OP}}$$ is perpendicular to the plane through points $$(1,2,3),(2,3,4)$$ and $$(1,5,7)$$, then which one of the following is true?
If $$\\overrightarrow a = \\widehat i + 2\\widehat k,\\overrightarrow b = \\widehat i + \\widehat j + \\widehat k,\\overrightarrow c = 7\\widehat i - 3\\widehat j + 4\\widehat k,\\overrightarrow r \\times \\overrightarrow b + \\overrightarrow b \\times \\overrightarrow c = \\overrightarrow 0 $$ and $$\\overrightarrow r \\,.\\,\\overrightarrow a = 0$$. Then $$\\overrightarrow r \\,.\\,\\overrightarrow c $$ is equal to :
Let $$\\overrightarrow a = 4\\widehat i + 3\\widehat j$$ and $$\\overrightarrow b = 3\\widehat i - 4\\widehat j + 5\\widehat k$$. If $$\\overrightarrow c $$ is a vector such that $$\\overrightarrow c .\\left( {\\overrightarrow a \\times \\overrightarrow b } \\right) + 25 = 0,\\overrightarrow c \\,.(\\widehat i + \\widehat j + \\widehat k) = 4$$, and projection of $$\\overrightarrow c $$ on $$\\overrightarrow a $$ is 1, then the projection of $$\\overrightarrow c $$ on $$\\overrightarrow b $$ equals :
Let $$\\overrightarrow a = \\widehat i + 2\\widehat j + \\lambda \\widehat k,\\overrightarrow b = 3\\widehat i - 5\\widehat j - \\lambda \\widehat k,\\overrightarrow a \\,.\\,\\overrightarrow c = 7,2\\overrightarrow b \\,.\\,\\overrightarrow c + 43 = 0,\\overrightarrow a \\times \\overrightarrow c = \\overrightarrow b \\times \\overrightarrow c $$. Then $$\\left| {\\overrightarrow a \\,.\\,\\overrightarrow b } \\right|$$ is equal to :
Let $$|\\vec{a}|=2,|\\vec{b}|=3$$ and the angle between the vectors $$\\vec{a}$$ and $$\\vec{b}$$ be $$\\frac{\\pi}{4}$$. Then $$|(\\vec{a}+2 \\vec{b}) \\times(2 \\vec{a}-3 \\vec{b})|^{2}$$ is equal to :
If $$\\delta > 0$$ and the area of the triangle $$\\mathrm{ABC}$$ is $$5 \\sqrt{6}$$, then $$\\overrightarrow{C B} \\cdot \\overrightarrow{C A}$$ is equal to
Let $$\\vec{a}=\\hat{i}+4 \\hat{j}+2 \\hat{k}, \\vec{b}=3 \\hat{i}-2 \\hat{j}+7 \\hat{k}$$ and $$\\vec{c}=2 \\hat{i}-\\hat{j}+4 \\hat{k}$$. If a vector $$\\vec{d}$$ satisfies $$\\vec{d} \\times \\vec{b}=\\vec{c} \\times \\vec{b}$$ and $$\\vec{d} \\cdot \\vec{a}=24$$, then $$|\\vec{d}|^{2}$$ is equal to :
",
"options": [
{
"text": "313"
},
{
"text": "413"
},
{
"text": "423"
},
{
"text": "323"
}
],
"answer": "413",
"solution": "**Answer:** 413\n\nGiven that $$\\vec{d} \\times \\vec{b} = \\vec{c} \\times \\vec{b}$$, we can rewrite this as:\n\n
Let $$\\vec{a}=3 \\hat{i}+\\hat{j}-\\hat{k}$$ and $$\\vec{c}=2 \\hat{i}-3 \\hat{j}+3 \\hat{k}$$. If $$\\vec{b}$$ is a vector such that $$\\vec{a}=\\vec{b} \\times \\vec{c}$$ and $$|\\vec{b}|^{2}=50$$, then $$|72-| \\vec{b}+\\left.\\vec{c}\\right|^{2} \\mid$$ is equal to __________.
We know that the magnitude of the cross product of two vectors is equal to the product of the magnitudes of the vectors and the sine of the angle between them :
We have the dot product $$\\vec{b} \\cdot \\vec{c} = |\\vec{b}||\\vec{c}|\\cos\\theta$$, and we can use the relationship between sine and cosine: $$\\cos\\theta = \\sqrt{1 - \\sin^2\\theta} = \\frac{\\sqrt{99}}{10}$$.
Let $$\\vec{a}=\\hat{i}+2 \\hat{j}+3 \\hat{k}$$ and $$\\vec{b}=\\hat{i}+\\hat{j}-\\hat{k}$$. If $$\\vec{c}$$ is a vector such that $$\\vec{a} \\cdot \\vec{c}=11,\n\\vec{b} \\cdot(\\vec{a} \\times \\vec{c})=27$$ and $$\\vec{b} \\cdot \\vec{c}=-\\sqrt{3}|\\vec{b}|$$, then $$|\\vec{a} \\times \\vec{c}|^{2}$$ is equal to _________.
Let $$\\vec{a}$$ be a non-zero vector parallel to the line of intersection of the two planes described by $$\\hat{i}+\\hat{j}, \\hat{i}+\\hat{k}$$ and $$\\hat{i}-\\hat{j}, \\hat{j}-\\hat{k}$$. If $$\\theta$$ is the angle between the vector $$\\vec{a}$$ and the vector $$\\vec{b}=2 \\hat{i}-2 \\hat{j}+\\hat{k}$$ and $$\\vec{a} \\cdot \\vec{b}=6$$, then the ordered pair $$(\\theta,|\\vec{a} \\times \\vec{b}|)$$ is equal to :
",
"options": [
{
"text": "$$\\left(\\frac{\\pi}{3}, 3 \\sqrt{6}\\right)$$"
},
{
"text": "$$\\left(\\frac{\\pi}{3}, 6\\right)$$"
},
{
"text": "$$\\left(\\frac{\\pi}{4}, 3 \\sqrt{6}\\right)$$"
},
{
"text": "$$\\left(\\frac{\\pi}{4}, 6\\right)$$"
}
],
"answer": "$$\\left(\\frac{\\pi}{4}, 6\\right)$$",
"solution": "**Answer:** $$\\left(\\frac{\\pi}{4}, 6\\right)$$\n\nWe have, $$\\vec{a}$$ is non-zero vector parallel to the line of intersection of the two planes described by $\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}, \\hat{\\mathbf{i}}+\\hat{\\mathbf{k}}$ and $\\hat{\\mathbf{i}}-\\hat{\\mathbf{j}}, \\hat{\\mathbf{j}}-\\hat{\\mathbf{k}}$.\n
Let $\\mathbf{n}_1$ and $\\mathbf{n}_2$ are the normal vector to the plane $\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}, \\hat{\\mathbf{i}}+\\hat{\\mathbf{k}}$ and $\\hat{\\mathbf{i}}-\\hat{\\mathbf{j}}, \\hat{\\mathbf{j}}-\\hat{\\mathbf{k}}$, respectively.\n
Let $$\\vec{a}=2 \\hat{i}+7 \\hat{j}-\\hat{k}, \\vec{b}=3 \\hat{i}+5 \\hat{k}$$ and $$\\vec{c}=\\hat{i}-\\hat{j}+2 \\hat{k}$$. Let $$\\vec{d}$$ be a vector which is perpendicular to both $$\\vec{a}$$ and $$\\vec{b}$$, and $$\\vec{c} \\cdot \\vec{d}=12$$. Then $$(-\\hat{i}+\\hat{j}-\\hat{k}) \\cdot(\\vec{c} \\times \\vec{d})$$ is equal to :
",
"options": [
{
"text": "24"
},
{
"text": "42"
},
{
"text": "44"
},
{
"text": "48"
}
],
"answer": "44",
"solution": "**Answer:** 44\n\nIf $\\vec{d}$ is $\\perp$ to both $\\vec{a}$ and $\\vec{b}$ then\n
Let O be the origin and the position vector of the point P be $$ - \\widehat i - 2\\widehat j + 3\\widehat k$$. If the position vectors of the points A, B and C are $$ - 2\\widehat i + \\widehat j - 3\\widehat k,2\\widehat i + 4\\widehat j - 2\\widehat k$$ and $$ - 4\\widehat i + 2\\widehat j - \\widehat k$$ respectively, then the projection of the vector $$\\overrightarrow {OP} $$ on a vector perpendicular to the vectors $$\\overrightarrow {AB} $$ and $$\\overrightarrow {AC} $$ is :
",
"options": [
{
"text": "$$\\frac{7}{3}$$"
},
{
"text": "3"
},
{
"text": "$$\\frac{10}{3}$$"
},
{
"text": "$$\\frac{8}{3}$$"
}
],
"answer": "3",
"solution": "**Answer:** 3\n\nGiven, the position vector of point P is :\n$ \\overrightarrow{OP} = -\\widehat{i} - 2\\widehat{j} + 3\\widehat{k} $\n\n
To find the projection of $ \\overrightarrow{OP} $ onto $ \\overrightarrow{AB} \\times \\overrightarrow{AC} $, we need to find the dot product between $ \\overrightarrow{OP} $ and the normalized vector $ \\overrightarrow{AB} \\times \\overrightarrow{AC} $.\n\n
First, find the magnitude of $ \\overrightarrow{AB} \\times \\overrightarrow{AC} $ :\n\n
The projection of vector $ \\overrightarrow{OP} $ onto a vector perpendicular to both $ \\overrightarrow{AB} $ and $ \\overrightarrow{AC} $ (which is $ \\overrightarrow{AB} \\times \\overrightarrow{AC}$) is given by :\n\n
The area of the quadrilateral $$\\mathrm{ABCD}$$ with vertices $$\\mathrm{A}(2,1,1), \\mathrm{B}(1,2,5), \\mathrm{C}(-2,-3,5)$$ and $$\\mathrm{D}(1,-6,-7)$$ is equal to :
Let $$\\vec{a}=6 \\hat{i}+9 \\hat{j}+12 \\hat{k}, \\vec{b}=\\alpha \\hat{i}+11 \\hat{j}-2 \\hat{k}$$ and $$\\vec{c}$$ be vectors such that $$\\vec{a} \\times \\vec{c}=\\vec{a} \\times \\vec{b}$$. If
$$\\vec{a} \\cdot \\vec{c}=-12, \\vec{c} \\cdot(\\hat{i}-2 \\hat{j}+\\hat{k})=5$$, then $$\\vec{c} \\cdot(\\hat{i}+\\hat{j}+\\hat{k})$$ is equal to _______________.
Let $$\\vec{a}=2 \\hat{i}+3 \\hat{j}+4 \\hat{k}, \\vec{b}=\\hat{i}-2 \\hat{j}-2 \\hat{k}$$ and $$\\vec{c}=-\\hat{i}+4 \\hat{j}+3 \\hat{k}$$.\n\nIf $$\\vec{d}$$ is a vector perpendicular to both $$\\vec{b}$$ and $$\\vec{c}$$, and $$\\vec{a} \\cdot \\vec{d}=18$$, then $$|\\vec{a} \\times \\vec{d}|^{2}$$ is equal to :
\n",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7846,
"subject": "General Science",
"question": "Let $\\overrightarrow{\\mathrm{a}}=\\hat{i}+\\hat{j}+\\hat{k}, \\overrightarrow{\\mathrm{b}}=-\\hat{i}-8 \\hat{j}+2 \\hat{k}$ and $\\overrightarrow{\\mathrm{c}}=4 \\hat{i}+\\mathrm{c}_2 \\hat{j}+\\mathrm{c}_3 \\hat{k}$ be three vectors such that $\\overrightarrow{\\mathrm{b}} \\times \\overrightarrow{\\mathrm{a}}=\\overrightarrow{\\mathrm{c}} \\times \\overrightarrow{\\mathrm{a}}$. If the angle between the vector $\\overrightarrow{\\mathrm{c}}$ and the vector $3 \\hat{i}+4 \\hat{j}+\\hat{k}$ is $\\theta$, then the greatest integer less than or equal to $\\tan ^2 \\theta$ is _______________.",
"options": [],
"answer": "38",
"solution": "**Answer:** 38\n\n$\\begin{aligned} & \\vec{a}=\\hat{i}+\\hat{j}+k \\\\\\\\ & \\vec{b}=\\hat{i}+8 \\hat{j}+2 k \\\\\\\\ & \\vec{c}=4 \\hat{i}+c_2 \\hat{j}+c_3 k \\\\\\\\ & \\vec{b} \\times \\vec{a}=\\vec{c} \\times \\vec{a} \\\\\\\\ & (\\vec{b}-\\vec{c}) \\times \\vec{a}=0 \\\\\\\\ & \\vec{b}-\\vec{c}=\\lambda \\vec{\\alpha} \\\\\\\\ & \\vec{b}=\\vec{c}+\\lambda \\vec{\\alpha}\\end{aligned}$\n
Let the position vectors of the vertices $$\\mathrm{A}, \\mathrm{B}$$ and $$\\mathrm{C}$$ of a triangle be $$2 \\hat{i}+2 \\hat{j}+\\hat{k}, \\hat{i}+2 \\hat{j}+2 \\hat{k}$$ and $$2 \\hat{i}+\\hat{j}+2 \\hat{k}$$ respectively. Let $$l_1, l_2$$ and $$l_3$$ be the lengths of perpendiculars drawn from the ortho center of the triangle on the sides $$\\mathrm{AB}, \\mathrm{BC}$$ and $$\\mathrm{CA}$$ respectively, then $$l_1^2+l_2^2+l_3^2$$ equals:
Let $$\\vec{a}=3 \\hat{i}+2 \\hat{j}+\\hat{k}, \\vec{b}=2 \\hat{i}-\\hat{j}+3 \\hat{k}$$ and $$\\vec{c}$$ be a vector such that $$(\\vec{a}+\\vec{b}) \\times \\vec{c}=2(\\vec{a} \\times \\vec{b})+24 \\hat{j}-6 \\hat{k}$$ and $$(\\vec{a}-\\vec{b}+\\hat{i}) \\cdot \\vec{c}=-3$$. Then $$|\\vec{c}|^2$$ is equal to ________.
Let $$\\vec{a}=3 \\hat{i}+\\hat{j}-2 \\hat{k}, \\vec{b}=4 \\hat{i}+\\hat{j}+7 \\hat{k}$$ and $$\\vec{c}=\\hat{i}-3 \\hat{j}+4 \\hat{k}$$ be three vectors. If a vectors $$\\vec{p}$$ satisfies $$\\vec{p} \\times \\vec{b}=\\vec{c} \\times \\vec{b}$$ and $$\\vec{p} \\cdot \\vec{a}=0$$, then $$\\vec{p} \\cdot(\\hat{i}-\\hat{j}-\\hat{k})$$ is equal to
The distance of the point $$Q(0,2,-2)$$ form the line passing through the point $$P(5,-4, 3)$$ and perpendicular to the lines $$\\vec{r}=(-3 \\hat{i}+2 \\hat{k})+\\lambda(2 \\hat{i}+3 \\hat{j}+5 \\hat{k}), \\lambda \\in \\mathbb{R}$$ and $$\\vec{r}=(\\hat{i}-2 \\hat{j}+\\hat{k})+\\mu(-\\hat{i}+3 \\hat{j}+2 \\hat{k}), \\mu \\in \\mathbb{R}$$ is :
Let $$\\vec{a}$$ and $$\\vec{b}$$ be two vectors such that $$|\\vec{a}|=1,|\\vec{b}|=4$$, and $$\\vec{a} \\cdot \\vec{b}=2$$. If $$\\vec{c}=(2 \\vec{a} \\times \\vec{b})-3 \\vec{b}$$ and the angle between $$\\vec{b}$$ and $$\\vec{c}$$ is $$\\alpha$$, then $$192 \\sin ^2 \\alpha$$ is equal to ________.
Let $$\\overrightarrow{O A}=\\vec{a}, \\overrightarrow{O B}=12 \\vec{a}+4 \\vec{b} \\text { and } \\overrightarrow{O C}=\\vec{b}$$, where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then $$\\mathrm{{{area\\,of\\,the\\,quadrilateral\\,OA\\,BC} \\over {area\\,of\\,S}}}$$ is equal to _________.
Let $$\\vec{a}=\\hat{i}+\\alpha \\hat{j}+\\beta \\hat{k}, \\alpha, \\beta \\in \\mathbb{R}$$. Let a vector $$\\vec{b}$$ be such that the angle between $$\\vec{a}$$ and $$\\vec{b}$$ is $$\\frac{\\pi}{4}$$ and $$|\\vec{b}|^2=6$$. If $$\\vec{a} \\cdot \\vec{b}=3 \\sqrt{2}$$, then the value of $$\\left(\\alpha^2+\\beta^2\\right)|\\vec{a} \\times \\vec{b}|^2$$ is equal to
Let $$\\vec{a}$$ and $$\\vec{b}$$ be two vectors such that $$|\\vec{b}|=1$$ and $$|\\vec{b} \\times \\vec{a}|=2$$. Then $$|(\\vec{b} \\times \\vec{a})-\\vec{b}|^2$$ is equal to
To find the value of $$|(\\vec{b} \\times \\vec{a})-\\vec{b}|^2$$, we can use properties of vector operations and magnitudes. Given $$|\\vec{b}| = 1$$ and $$|\\vec{b} \\times \\vec{a}| = 2$$, let's break down the calculation step by step:
\n
Firstly, we observe that the cross product of two vectors $$\\vec{b} \\times \\vec{a}$$ is orthogonal (perpendicular) to both $$\\vec{b}$$ and $$\\vec{a}$$. This means that when we take the dot product of $$\\vec{b} \\times \\vec{a}$$ with either $$\\vec{b}$$ or $$\\vec{a}$$, the result will be zero due to the orthogonal property. Specifically,
\n
$$(\\vec{b} \\times \\vec{a}) \\cdot \\vec{b}= 0 $$(because $$\\vec{b} \\times \\vec{a}$$ is perpendicular to both $$\\vec{b}$$ and $$\\vec{a}$$)
\n
Next, to find the magnitude squared of the vector $$(\\vec{b} \\times \\vec{a}) - \\vec{b}$$, we apply the formula for the magnitude squared of a vector subtraction, which can be expressed as:
Let $$\\overrightarrow{\\mathrm{a}}=\\mathrm{a}_1 \\hat{i}+\\mathrm{a}_2 \\hat{j}+\\mathrm{a}_3 \\hat{k}$$ and $$\\overrightarrow{\\mathrm{b}}=\\mathrm{b}_1 \\hat{i}+\\mathrm{b}_2 \\hat{j}+\\mathrm{b}_3 \\hat{k}$$ be two vectors such that $$|\\overrightarrow{\\mathrm{a}}|=1, \\vec{a} \\cdot \\vec{b}=2$$ and $$|\\vec{b}|=4$$. If $$\\vec{c}=2(\\vec{a} \\times \\vec{b})-3 \\vec{b}$$, then the angle between $$\\vec{b}$$ and $$\\vec{c}$$ is equal to:
Statement I : Let $$\\vec{a}=\\hat{i}+2 \\hat{j}-3 \\hat{k}$$ and $$\\vec{b}=2 \\hat{i}+\\hat{j}-\\hat{k}$$. Then the vector $$\\vec{r}$$ satisfying $$\\vec{a} \\times \\vec{r}=\\vec{a} \\times \\vec{b}$$ and $$\\vec{a} \\cdot \\vec{r}=0$$ is of magnitude $$\\sqrt{10}$$.
\n
Statement II : In a triangle $$A B C, \\cos 2 A+\\cos 2 B+\\cos 2 C \\geq-\\frac{3}{2}$$.
",
"options": [
{
"text": "Both Statement I and Statement II are correct.\n"
},
{
"text": "Both Statement I and Statement II are incorrect.\n"
},
{
"text": "Statement I is correct but Statement II is incorrect.\n"
},
{
"text": "Statement I is incorrect but Statement II is correct."
}
],
"answer": "Statement I is incorrect but Statement II is correct.",
"solution": "**Answer:** Statement I is incorrect but Statement II is correct.\n\n
Let $$\\vec{a}=2 \\hat{i}+\\alpha \\hat{j}+\\hat{k}, \\vec{b}=-\\hat{i}+\\hat{k}, \\vec{c}=\\beta \\hat{j}-\\hat{k}$$, where $$\\alpha$$ and $$\\beta$$ are integers and $$\\alpha \\beta=-6$$. Let the values of the ordered pair $$(\\alpha, \\beta)$$, for which the area of the parallelogram of diagonals $$\\vec{a}+\\vec{b}$$ and $$\\vec{b}+\\vec{c}$$ is $$\\frac{\\sqrt{21}}{2}$$, be $$\\left(\\alpha_1, \\beta_1\\right)$$ and $$\\left(\\alpha_2, \\beta_2\\right)$$. Then $$\\alpha_1^2+\\beta_1^2-\\alpha_2 \\beta_2$$ is equal to
Let three vectors ,$$\\overrightarrow{\\mathrm{a}}=\\alpha \\hat{i}+4 \\hat{j}+2 \\hat{k}, \\overrightarrow{\\mathrm{b}}=5 \\hat{i}+3 \\hat{j}+4 \\hat{k}, \\overrightarrow{\\mathrm{c}}=x \\hat{i}+y \\hat{j}+z \\hat{k}$$ form a triangle such that $$\\vec{c}=\\vec{a}-\\vec{b}$$ and the area of the triangle is $$5 \\sqrt{6}$$. If $$\\alpha$$ is a positive real number, then $$|\\vec{c}|^2$$ is equal to:
The area of the triangle formed by vectors $$\\vec{a}$$ and $$\\vec{b}$$ is given by the magnitude of the cross product of $$\\vec{a}$$ and $$\\vec{b}$$, divided by 2:
Let $$\\overrightarrow{O A}=2 \\vec{a}, \\overrightarrow{O B}=6 \\vec{a}+5 \\vec{b}$$ and $$\\overrightarrow{O C}=3 \\vec{b}$$, where $$O$$ is the origin. If the area of the parallelogram with adjacent sides $$\\overrightarrow{O A}$$ and $$\\overrightarrow{O C}$$ is 15 sq. units, then the area (in sq. units) of the quadrilateral $$O A B C$$ is equal to:
Let $$\\mathrm{ABC}$$ be a triangle of area $$15 \\sqrt{2}$$ and the vectors $$\\overrightarrow{\\mathrm{AB}}=\\hat{i}+2 \\hat{j}-7 \\hat{k}, \\overrightarrow{\\mathrm{BC}}=\\mathrm{a} \\hat{i}+\\mathrm{b} \\hat{j}+\\mathrm{c} \\hat{k}$$ and $$\\overrightarrow{\\mathrm{AC}}=6 \\hat{i}+\\mathrm{d} \\hat{j}-2 \\hat{k}, \\mathrm{~d}>0$$. Then the square of the length of the largest side of the triangle $$\\mathrm{ABC}$$ is _________.
Let $$\\overrightarrow{\\mathrm{a}}=4 \\hat{i}-\\hat{j}+\\hat{k}, \\overrightarrow{\\mathrm{b}}=11 \\hat{i}-\\hat{j}+\\hat{k}$$ and $$\\overrightarrow{\\mathrm{c}}$$ be a vector such that $$(\\overrightarrow{\\mathrm{a}}+\\overrightarrow{\\mathrm{b}}) \\times \\overrightarrow{\\mathrm{c}}=\\overrightarrow{\\mathrm{c}} \\times(-2 \\overrightarrow{\\mathrm{a}}+3 \\overrightarrow{\\mathrm{b}})$$.\nIf $$(2 \\vec{a}+3 \\vec{b}) \\cdot \\vec{c}=1670$$, then $$|\\vec{c}|^2$$ is equal to:
Let $$\\vec{a}=9 \\hat{i}-13 \\hat{j}+25 \\hat{k}, \\vec{b}=3 \\hat{i}+7 \\hat{j}-13 \\hat{k}$$ and $$\\vec{c}=17 \\hat{i}-2 \\hat{j}+\\hat{k}$$ be three given vectors. If $$\\vec{r}$$ is a vector such that $$\\vec{r} \\times \\vec{a}=(\\vec{b}+\\vec{c}) \\times \\vec{a}$$ and $$\\vec{r} \\cdot(\\vec{b}-\\vec{c})=0$$, then $$\\frac{|593 \\vec{r}+67 \\vec{a}|^2}{(593)^2}$$ is equal to __________.
If $$\\mathrm{A}(1,-1,2), \\mathrm{B}(5,7,-6), \\mathrm{C}(3,4,-10)$$ and $$\\mathrm{D}(-1,-4,-2)$$ are the vertices of a quadrilateral ABCD, then its area is :
Let $$\\overrightarrow{\\mathrm{a}}=\\hat{i}-3 \\hat{j}+7 \\hat{k}, \\overrightarrow{\\mathrm{b}}=2 \\hat{i}-\\hat{j}+\\hat{k}$$ and $$\\overrightarrow{\\mathrm{c}}$$ be a vector such that $$(\\overrightarrow{\\mathrm{a}}+2 \\overrightarrow{\\mathrm{b}}) \\times \\overrightarrow{\\mathrm{c}}=3(\\overrightarrow{\\mathrm{c}} \\times \\overrightarrow{\\mathrm{a}})$$.\nIf $$\\vec{a} \\cdot \\vec{c}=130$$, then $$\\vec{b} \\cdot \\vec{c}$$ is equal to __________.
Let $$\\vec{a}=2 \\hat{i}+5 \\hat{j}-\\hat{k}, \\vec{b}=2 \\hat{i}-2 \\hat{j}+2 \\hat{k}$$ and $$\\vec{c}$$ be three vectors such that $$(\\vec{c}+\\hat{i}) \\times(\\vec{a}+\\vec{b}+\\hat{i})=\\vec{a} \\times(\\vec{c}+\\hat{i})$$. If $$\\vec{a} \\cdot \\vec{c}=-29$$, then $$\\vec{c} \\cdot(-2 \\hat{i}+\\hat{j}+\\hat{k})$$ is equal to:
Consider three vectors $$\\vec{a}, \\vec{b}, \\vec{c}$$. Let $$|\\vec{a}|=2,|\\vec{b}|=3$$ and $$\\vec{a}=\\vec{b} \\times \\vec{c}$$. If $$\\alpha \\in\\left[0, \\frac{\\pi}{3}\\right]$$ is the angle between the vectors $$\\vec{b}$$ and $$\\vec{c}$$, then the minimum value of $$27|\\vec{c}-\\vec{a}|^2$$ is equal to:
Let $$\\vec{a}=2 \\hat{i}+\\hat{j}-\\hat{k}, \\vec{b}=((\\vec{a} \\times(\\hat{i}+\\hat{j})) \\times \\hat{i}) \\times \\hat{i}$$. Then the square of the projection of $$\\vec{a}$$ on $$\\vec{b}$$ is:
Let $$\\overrightarrow{\\mathrm{a}}=6 \\hat{i}+\\hat{j}-\\hat{k}$$ and $$\\overrightarrow{\\mathrm{b}}=\\hat{i}+\\hat{j}$$. If $$\\overrightarrow{\\mathrm{c}}$$ is a is vector such that $$|\\overrightarrow{\\mathrm{c}}| \\geq 6, \\overrightarrow{\\mathrm{a}} \\cdot \\overrightarrow{\\mathrm{c}}=6|\\overrightarrow{\\mathrm{c}}|,|\\overrightarrow{\\mathrm{c}}-\\overrightarrow{\\mathrm{a}}|=2 \\sqrt{2}$$ and the angle between $$\\vec{a} \\times \\vec{b}$$ and $$\\vec{c}$$ is $$60^{\\circ}$$, then $$|(\\vec{a} \\times \\vec{b}) \\times \\vec{c}|$$ is equal to:
Let $$\\vec{a}=2 \\hat{i}-3 \\hat{j}+4 \\hat{k}, \\vec{b}=3 \\hat{i}+4 \\hat{j}-5 \\hat{k}$$ and a vector $$\\vec{c}$$ be such that $$\\vec{a} \\times(\\vec{b}+\\vec{c})+\\vec{b} \\times \\vec{c}=\\hat{i}+8 \\hat{j}+13 \\hat{k}$$. If $$\\vec{a} \\cdot \\vec{c}=13$$, then $$(24-\\vec{b} \\cdot \\vec{c})$$ is equal to _______.
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7871,
"subject": "Physics",
"question": "The power factor of $$AC$$ circuit having resistance $$(R)$$ and inductance $$(L)$$ connected in series and an angular velocity $$\\omega $$ is ",
"options": [
{
"text": "$$R/\\omega L$$ "
},
{
"text": "$$R/{\\left( {{R^2} + {\\omega ^2}{L^2}} \\right)^{1/2}}$$ "
},
{
"text": "$$\\omega L/R$$ "
},
{
"text": "$$R/{\\left( {{R^2} - {\\omega ^2}{L^2}} \\right)^{1/2}}$$ "
}
],
"answer": "$$R/{\\left( {{R^2} + {\\omega ^2}{L^2}} \\right)^{1/2}}$$ ",
"solution": "**Answer:** $$R/{\\left( {{R^2} + {\\omega ^2}{L^2}} \\right)^{1/2}}$$ \n\nThe impedance triangle for resistance $$\\left( R \\right)$$ and inductor $$(L)$$ connected in series is shown in the figure.\n
\n
Power factor $$\\cos \\phi = {R \\over {\\sqrt {{R^2} + {\\omega ^2}{L^2}} }}$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 7872,
"subject": "Physics",
"question": "In an $$LCR$$ series $$a.c.$$ circuit, the voltage across each of the components, $$L,C$$ and $$R$$ is $$50V$$. The voltage across the $$L.C$$ combination will be :",
"options": [
{
"text": "$$100V$$ "
},
{
"text": "$$50\\sqrt 2 $$ "
},
{
"text": "$$50$$ $$V$$ "
},
{
"text": "$$0$$ $$V$$ (zero) "
}
],
"answer": "$$0$$ $$V$$ (zero) ",
"solution": "**Answer:** $$0$$ $$V$$ (zero) \n\nSince the phase difference between $$L$$ & $$C$$ is $$\\pi ,$$\n
$$\\therefore$$ net voltage difference across $$LC=50-50=0$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 7873,
"subject": "Physics",
"question": "In a $$LCR$$ circuit capacitance is changed from $$C$$ to $$2$$ $$C$$. For the resonant frequency to remain unchaged, the inductance should be changed from $$L$$ to ",
"options": [
{
"text": "$$L/2$$ "
},
{
"text": "$$2L$$ "
},
{
"text": "$$4L$$ "
},
{
"text": "$$L/4$$"
}
],
"answer": "$$L/2$$ ",
"solution": "**Answer:** $$L/2$$ \n\nFor resonant frequency to remain same $$LC$$ should be const. $$LC=$$ const\n
$$ \\Rightarrow LC = L' \\times 2C \\Rightarrow L' = {L \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7874,
"subject": "Physics",
"question": "Alternating current can not be measured by $$D.C.$$ ammeter because ",
"options": [
{
"text": "Average value of current for complete cycle is zero "
},
{
"text": "$$A.C.$$ Changes direction "
},
{
"text": "$$A.C.$$ can not pass through $$D.C.$$ Ammeter "
},
{
"text": "$$D.C.$$ Ammeter will get damaged. "
}
],
"answer": "Average value of current for complete cycle is zero ",
"solution": "**Answer:** Average value of current for complete cycle is zero \n\n$$D.C.$$ ammeter measure average current in $$AC$$ current, average current is zero for complete cycle. Hence reading will be zero.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7875,
"subject": "Physics",
"question": "The self inductance of the motor of an electric fan is $$10$$ $$H$$. In order to impart maximum power at $$50$$ $$Hz$$, it should be connected to a capacitance of ",
"options": [
{
"text": "$$8\\mu F$$ "
},
{
"text": "$$4\\mu F$$"
},
{
"text": "$$2\\mu F$$"
},
{
"text": "$$1\\mu F$$"
}
],
"answer": "$$1\\mu F$$",
"solution": "**Answer:** $$1\\mu F$$\n\nFor maximum power, $${X_L} = X{}_C,$$ which yields \n
$$\\therefore$$ $$C = 0.1 \\times {10^{ - 5}}F = 1\\mu F$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7876,
"subject": "Physics",
"question": "A circuit has a resistance of $$12$$ $$ohm$$ and an impedance of $$15$$ $$ohm$$. The power factor of the circuit will be ",
"options": [
{
"text": "$$0.4$$ "
},
{
"text": "$$0.8$$ "
},
{
"text": "$$0.125$$ "
},
{
"text": "$$1.25$$ "
}
],
"answer": "$$0.8$$ ",
"solution": "**Answer:** $$0.8$$ \n\nPower factor $$ = \\cos \\phi = {R \\over Z} = {{12} \\over {15}} = {4 \\over 5} = 0.8$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 7877,
"subject": "Physics",
"question": "The phase difference between the alternating current and $$emf$$ is $${\\pi \\over 2}.$$ Which of the following cannot be the constituent of the circuit? ",
"options": [
{
"text": "$$R,L$$ "
},
{
"text": "$$C$$ alone "
},
{
"text": "$$L$$ alone "
},
{
"text": "$$L, C$$ "
}
],
"answer": "$$R,L$$ ",
"solution": "**Answer:** $$R,L$$ \n\n
The phase difference between the alternating current and emf in an AC circuit depends on the components in the circuit:
\n
\n
In a purely resistive ($R$) circuit, the current and emf are in phase, meaning the phase difference is $0$.
\n
In a purely inductive ($L$) circuit, the current lags behind the emf by $\\frac{\\pi}{2}$, meaning the phase difference is $\\frac{\\pi}{2}$.
\n
In a purely capacitive ($C$) circuit, the current leads the emf by $\\frac{\\pi}{2}$, again meaning the phase difference is $\\frac{\\pi}{2}$.
\n
In an $L$-$R$ or $L$-$C$ circuit, the phase difference depends on the relative values of $L$, $R$, and $C$ and can be anywhere between $0$ and $\\frac{\\pi}{2}$.
\n
\n
Therefore, if the phase difference between the alternating current and emf is $\\frac{\\pi}{2}$, then the circuit cannot contain only a resistor ($R$) since that would give a phase difference of $0$. So, the answer is Option A: $R,L$. The phase difference would not be $\\frac{\\pi}{2}$ if the circuit contains both a resistor and an inductor.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7878,
"subject": "Physics",
"question": "In a series resonant $$LCR$$ circuit, the voltage across $$R$$ is $$100$$ volts and $$R = 1\\,k\\Omega $$ with $$C = 2\\mu F.$$ The resonant frequency $$\\omega $$ is $$200$$ $$rad/s$$. At resonance the voltage across $$L$$ is ",
"options": [
{
"text": "$$2.5 \\times {10^{ - 2}}V$$ "
},
{
"text": "$$40$$ $$V$$ "
},
{
"text": "$$250$$ $$V$$ "
},
{
"text": "$$4 \\times {10^{ - 3}}V$$ "
}
],
"answer": "$$250$$ $$V$$ ",
"solution": "**Answer:** $$250$$ $$V$$ \n\nAcross resistor, $$I = {V \\over R} = {{100} \\over {1000}} = 0.1A$$\n
$$I{X_L} = 0.1 \\times 2500 = 250V$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7879,
"subject": "Physics",
"question": "In an $$a.c.$$ circuit the voltage applied is $$E = {E_0}\\,\\sin \\,\\omega t.$$ The resulting current in the circuit is $$I = {I_0}\\sin \\left( {\\omega t - {\\pi \\over 2}} \\right).$$ The power consumption in the circuit is given by ",
"options": [
{
"text": "$$P = \\sqrt 2 {E_0}{I_0}$$ "
},
{
"text": "$$P = {{{E_0}{I_0}} \\over {\\sqrt 2 }}$$ "
},
{
"text": "$$P=zero$$ "
},
{
"text": "$$P = {{{E_0}{I_0}} \\over 2}$$ "
}
],
"answer": "$$P=zero$$ ",
"solution": "**Answer:** $$P=zero$$ \n\nKEY CONCEPT : We know that power consumed in a.c. circuit is given by, \n
$$\\left( {\\,\\,} \\right.$$ as $$\\left. {\\,\\,\\cos {\\pi \\over 2} = 0\\,\\,} \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7880,
"subject": "Physics",
"question": "In a series $$LCR$$ circuit $$R = 200\\Omega $$ and the voltage and the frequency of the main supply is $$220V$$ and $$50$$ $$Hz$$ respectively. On taking out the capacitance from the circuit the current lags behind the voltage by $${30^ \\circ }.$$ On taking out the inductor from the circuit the current leads the voltage by $${30^ \\circ }.$$ The power dissipated in the $$LCR$$ circuit is ",
"options": [
{
"text": "$$305$$ $$W$$ "
},
{
"text": "$$210$$ $$W$$ "
},
{
"text": "$$zero$$ $$W$$ "
},
{
"text": "$$242$$ $$W$$ "
}
],
"answer": "$$242$$ $$W$$ ",
"solution": "**Answer:** $$242$$ $$W$$ \n\nWhen capacitance is taken out, the circular is $$LR.$$ \n
$$ = {{220 \\times 220} \\over {200}} = 242\\,W$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7881,
"subject": "Physics",
"question": "A fully charged capacitor $$C$$ with initial charge $${q_0}$$ is connected to a coil of self inductance $$L$$ at $$t=0.$$ The time at which the energy is stored equally between the electric and the magnetic fields is : ",
"options": [
{
"text": "$${\\pi \\over 4}\\sqrt {LC} $$ "
},
{
"text": "$$2\\pi \\sqrt {LC} $$ "
},
{
"text": "$$\\sqrt {LC} $$ "
},
{
"text": "$$\\pi \\sqrt {LC} $$ "
}
],
"answer": "$${\\pi \\over 4}\\sqrt {LC} $$ ",
"solution": "**Answer:** $${\\pi \\over 4}\\sqrt {LC} $$ \n\nEnergy stored in magnetic field $$ = {1 \\over 2}L{i^2}$$\n
Energy stored in electric field $$ = {1 \\over 2}{{{q^2}} \\over C}$$\n
Also $$q = {q_0}\\,\\cos \\,\\omega t$$ and $$\\omega = {1 \\over {\\sqrt {LC} }}$$\n
On solving $$t = {\\pi \\over 4}\\sqrt {LC} $$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 7882,
"subject": "Physics",
"question": "An arc lamp requires a direct current of\n10 A at 80 V to function. If it is connected\nto a 220 V (rms), 50 Hz AC supply, the\nseries inductor needed for it to work is\nclose to :",
"options": [
{
"text": "0.044 H"
},
{
"text": "0.065 H"
},
{
"text": "80 H"
},
{
"text": "0.08 H"
}
],
"answer": "0.065 H",
"solution": "**Answer:** 0.065 H\n\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7883,
"subject": "Physics",
"question": "A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5 $$\\Omega $$, L=25 mH and C=1000 $$\\mu $$F. The total impedance, and phase difference between the voltage across the source and the current will respectively be :",
"options": [
{
"text": "10 $$\\Omega $$ and tan$$-$$1 $$\\left( {{5 \\over 3}} \\right)$$"
},
{
"text": "$$7\\,\\Omega $$ and 45o "
},
{
"text": "$$10\\,\\Omega $$ and tan$$-$$1$$\\left( {{8 \\over 3}} \\right)$$"
},
{
"text": "$$7\\,\\Omega $$ and tan$$-$$1$$\\left( {{5 \\over 3}} \\right)$$"
}
],
"answer": "$$7\\,\\Omega $$ and 45o ",
"solution": "**Answer:** $$7\\,\\Omega $$ and 45o \n\n
It is given that e0 = 283 V; $$\\omega$$ = 320.
\n
The inductor reactance is XL = 320 $$\\times$$ 25 $$\\times$$ 10$$-$$3 = 8 $$\\Omega$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7884,
"subject": "Physics",
"question": "In an a.c. circuit, the instantaneous e.m.f. and current are given by \ne = 100 sin 30 t \ni = 20 sin $$\\left( {30t - {\\pi \\over 4}} \\right)$$ \nIn one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively ",
"options": [
{
"text": "50, 0 "
},
{
"text": "50, 10 "
},
{
"text": "$${{1000} \\over {\\sqrt 2 }},10$$ "
},
{
"text": "$${{50} \\over {\\sqrt 2 }}$$ "
}
],
"answer": "$${{1000} \\over {\\sqrt 2 }},10$$ ",
"solution": "**Answer:** $${{1000} \\over {\\sqrt 2 }},10$$ \n\nWattless current,\n
= $${{1000} \\over {\\sqrt 2 }}$$ watt. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7885,
"subject": "Physics",
"question": "An ideal capacitor of capacitance $$0.2\\,\\mu F$$ is charged to a potential difference of $$10$$ $$V.$$ The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance $$0.5$$ $$mH.$$ The current at a time when the potential difference across the capacitor is $$5$$ $$V,$$ is : ",
"options": [
{
"text": "$$0.34\\,\\,A$$ "
},
{
"text": "$$0.25\\,\\,A$$"
},
{
"text": "$$0.17\\,\\,A$$"
},
{
"text": "$$0.15\\,\\,A$$"
}
],
"answer": "$$0.17\\,\\,A$$",
"solution": "**Answer:** $$0.17\\,\\,A$$\n\nCapacitance, C = 0.2 $$\\mu $$F = 0.2 $$ \\times $$ 10$$-$$6 F\n
Inductance, L = 0.5 m H = 0.5 $$ \\times $$ 10$$-$$3 H\n
= 0.17 A. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7886,
"subject": "Physics",
"question": "A circuit connected to an ac source of emf\ne = e0sin(100t) with t in seconds, gives a phase\ndifference of $$\\pi $$/4 between the emf e and\ncurrent i. Which of the following circuits will\nexhibit this ?",
"options": [
{
"text": "RC circuit with R = 1 k$$\\Omega $$ and C = 1μF"
},
{
"text": "RL circuit with R = 1k$$\\Omega $$ and L = 1mH"
},
{
"text": "RC circuit with R = 1k$$\\Omega $$ and C = 10 μF"
},
{
"text": "RL circuit with R = 1 k$$\\Omega $$ and L = 10 mH"
}
],
"answer": "RC circuit with R = 1k$$\\Omega $$ and C = 10 μF",
"solution": "**Answer:** RC circuit with R = 1k$$\\Omega $$ and C = 10 μF\n\nGiven phase difference = $${\\pi \\over 4}$$ and $$\\omega $$ = 100 rad/s
L $$ \\leftrightarrow $$ m, C $$ \\leftrightarrow $$ $${1 \\over k}$$, R $$ \\leftrightarrow $$ b",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 7890,
"subject": "Physics",
"question": "In LC circuit the inductance L = 40 mH and\n capacitance C = 100 $$\\mu $$F. If a voltage\n V(t) = 10sin(314t) is applied to the circuit, the\n current in the circuit is given as :",
"options": [
{
"text": "0.52 cos 314 t"
},
{
"text": "5.2 cos 314 t"
},
{
"text": "0.52 sin 314 t"
},
{
"text": "10 cos 314 t"
}
],
"answer": "0.52 cos 314 t",
"solution": "**Answer:** 0.52 cos 314 t\n\n\n
= 1.1 $$ \\times $$ 10–2 H",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7892,
"subject": "Physics",
"question": "A 750 Hz, 20 V (rms) source is connected to a\nresistance of 100 $$\\Omega $$, an inductance of 0.1803 H\nand a capacitance of 10 $$\\mu $$F all in series. The\ntime in which the resistance (heat capacity\n2 J/oC) will get heated by 10oC. (assume no loss\nof heat to the surroudnings) is close to :",
"options": [
{
"text": "348 s"
},
{
"text": "418 s"
},
{
"text": "245 s"
},
{
"text": "365 s"
}
],
"answer": "348 s",
"solution": "**Answer:** 348 s\n\nf = 750 Hz, Vrms = 20 V,\n
$$ \\Rightarrow $$ t = $${{2\\left( {10} \\right)} \\over {0.0575}}$$ = 348 sec",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7893,
"subject": "Physics",
"question": "In a series LR circuit, power of 400W is dissipated from a source of 250 V, 50 Hz. The power factor\nof the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value C is added in\nseries to the L and R. Taking the value of C as $$\\left( {{n \\over {3\\pi }}} \\right)$$ $$\\mu $$F, then value of n is __________.",
"options": [],
"answer": "400",
"solution": "**Answer:** 400\n\nGiven, power factor of LR circuit,\n
$$ \\Rightarrow $$ n = 400",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 7894,
"subject": "Physics",
"question": "A series L-C-R circuit is designed to resonate at an angular frequency $$\\omega$$0 = 105 rad/s. The circuit draws 16W power from 120V source at resonance. The value of resistance 'R' in the circuit is _________ $$\\Omega$$.",
"options": [],
"answer": "900",
"solution": "**Answer:** 900\n\nGiven, angular frequency at resonance, $$\\omega$$0 = 105 rads$$-$$1
$$ = 30 \\times 30 = 900\\,\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7895,
"subject": "Physics",
"question": "A transmitting station releases waves of wavelength 960 m. A capacitor of 2.56 $$\\mu$$F is used in the resonant circuit. The self inductance of coil necessary for resonance is __________ $$\\times$$ 10$$-$$8 H.",
"options": [],
"answer": "10",
"solution": "**Answer:** 10\n\n$$\\lambda$$ = 960 m
C = 2.56 $$\\mu$$F = 2.56 $$\\times$$ 10$$-$$6 F
c = 3 $$\\times$$ 108 m/s
L = ?
Now at resonance, $${\\omega _0} = {1 \\over {\\sqrt {LC} }}$$
[Resonant frequency]
$$2\\pi {f_0} = {1 \\over {\\sqrt {LC} }}$$
On substituting $${f_0} = {c \\over \\lambda }$$, we have $$2\\pi {c \\over \\lambda } = {1 \\over {\\sqrt {LC} }}$$
$$ \\Rightarrow L = 10 \\times {10^{ - 8}}$$ H",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7896,
"subject": "Physics",
"question": "An LCR circuit contains resistance of 110$$\\Omega$$ and a supply of 220 V at 300 rad/s angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by 45$$^\\circ$$. If on the other hand, only inductor is removed the current leads by 45$$^\\circ$$ with the applied voltage. The rms current flowing in the circuit will be :",
"options": [
{
"text": "1 A"
},
{
"text": "2.5 A"
},
{
"text": "2 A"
},
{
"text": "1.5 A"
}
],
"answer": "2 A",
"solution": "**Answer:** 2 A\n\nSince $$\\phi $$ remain same, circuit is in resonance
So, $${I_{rms}} = {{\\sqrt {I_1^2 + I_2^2} } \\over {\\sqrt 2 }}$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 7899,
"subject": "Physics",
"question": "In a series LCR resonant circuit, the quality factor is measured as 100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be __________.",
"options": [],
"answer": "283",
"solution": "**Answer:** 283\n\nQuality factor = $${{{X_L}} \\over R} = {{\\omega L} \\over R}$$
$$Q = {1 \\over {\\sqrt {LC} }}{L \\over R}$$
$$Q = \\left( {{1 \\over {\\sqrt C }}} \\right){{\\sqrt L } \\over R}$$
Q' = 282.84",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 7900,
"subject": "Physics",
"question": "A sinusoidal voltage of peak value 250 V is applied to a series LCR circuit, in which R = 8$$\\Omega$$, L = 24 mH and C = 60 $$\\mu$$F. The value of power dissipated at resonant condition is 'x' kW. The value of x to the nearest integer is ____________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nAt resonance power (P)
Choose the most appropriate answer from the options given below :",
"options": [
{
"text": "(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)"
},
{
"text": "(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)"
},
{
"text": "(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)"
},
{
"text": "(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)"
}
],
"answer": "(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)",
"solution": "**Answer:** (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)\n\n\n (a) phase difference b/w current & voltage in a purely resistive AC circuit is zero
(b) phase difference b/w current & voltage in a pure inductive AC circuit is $${\\pi \\over 2}$$; current lags voltage.
(c) phase difference b/w current & voltage in a pure capacitive AC circuit is $${\\pi \\over 2}$$; current lead voltage.
(d) phase difference b/w current & voltage in an LCR series circuit is = $${\\tan ^{ - 1}}\\left( {{{{X_C} - {X_L}} \\over R}} \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7903,
"subject": "Physics",
"question": "What happens to the inductive reactance and the current in a purely inductive circuit if the frequency is halved?",
"options": [
{
"text": "Both, inducting reactance and current will be doubled."
},
{
"text": "Inductive reactance will be doubled and current will be halved."
},
{
"text": "Both, inductive reactance and current will be halved."
},
{
"text": "Inductive reactance will be halved and current will be doubled."
}
],
"answer": "Inductive reactance will be halved and current will be doubled.",
"solution": "**Answer:** Inductive reactance will be halved and current will be doubled.\n\n$${X_L} = \\omega L$$
& $$I' = {{2V} \\over {{X_L}}} = 2I$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7904,
"subject": "Physics",
"question": "In a series LCR resonance circuit, if we change the resistance only, from a lower to higher value :",
"options": [
{
"text": "The bandwidth of resonance circuit will increase."
},
{
"text": "The resonance frequency will increase."
},
{
"text": "The quality factor will increase."
},
{
"text": "The quality factor and the resonance frequency will remain constant."
}
],
"answer": "The bandwidth of resonance circuit will increase.",
"solution": "**Answer:** The bandwidth of resonance circuit will increase.\n\n$${\\omega } = {1 \\over {\\sqrt {LC} }}$$\n
So if R increase then $$\\Delta \\beta$$ will increase too.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7905,
"subject": "Physics",
"question": "An AC source rated 220 V, 50 Hz is connected to a resistor. The time taken by the current to change from its maximum to the rms value is :",
"options": [
{
"text": "2.5 ms"
},
{
"text": "25 ms"
},
{
"text": "2.5 s"
},
{
"text": "0.25 ms"
}
],
"answer": "2.5 ms",
"solution": "**Answer:** 2.5 ms\n\n$$I = {I_0}\\sin \\omega t $$\n
$$ = {6 \\over {6\\sqrt 2 }} = {1 \\over {\\sqrt 2 }}$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 7907,
"subject": "Physics",
"question": "AC voltage V(t) = 20 sin$$\\omega$$t of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m2. The amplitude of the oscillating displacement current for the applied AC voltage is _________. [Take $$\\varepsilon $$0 = 8.85 $$\\times$$ 10$$-$$12 F/m]",
"options": [
{
"text": "55.58 $$\\mu$$A"
},
{
"text": "21.14 $$\\mu$$A"
},
{
"text": "27.79 $$\\mu$$A"
},
{
"text": "83.37 $$\\mu$$A"
}
],
"answer": "27.79 $$\\mu$$A",
"solution": "**Answer:** 27.79 $$\\mu$$A\n\nGiven,
AC voltage, V(t) = 20 sin $$\\omega$$t volt.
Frequency, f = 50Hz
Separation between the plates, d = 2 mm = 2 $$\\times$$ 10$$-$$3 m
Area, A = 1 m2
As, $$C = {{{\\varepsilon _0}A} \\over d}$$
where, $${{\\varepsilon _0}}$$ = absolute electrical permittivity of free space = 8.854 $$\\times$$ 10$$-$$12 N$$-$$1 kg2m$$-$$2
$$\\therefore$$ The amplitude of the oscillating displacement current for applied AC voltage will be approximately 27.79 $$\\mu$$A.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7908,
"subject": "Physics",
"question": "In an LCR series circuit, an inductor 30 mH and a resistor 1 $$\\Omega$$ are connected to an AC source of angular frequency 300 rad/s. The value of capacitance for which, the current leads the voltage by 45$$^\\circ$$ is $${1 \\over x} \\times {10^{ - 3}}$$ F. Then the value of x is ____________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nGiven,
Inductance, L = 30 mH
Resistance, R = 1 $$\\Omega$$
Angular frequency, $$\\omega$$ = 300 rad/s
We know that in L-C-R circuit, $$\\tan \\phi = {{{X_C} - {X_L}} \\over R}$$
According to question, the value of capacitance is $${1 \\over x} \\times {10^{ - 3}}F$$. So, on comparing it with Eq. (i), we can say x = 3.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7909,
"subject": "Physics",
"question": "For a series LCR circuit with R = 100 $$\\Omega$$, L = 0.5 mH and C = 0.1 pF connected across 220V$$-$$50 Hz AC supply, the phase angle between current and supplied voltage and the nature of the circuit is :",
"options": [
{
"text": "0$$^\\circ$$, resistive circuit"
},
{
"text": "$$ \\approx $$ 90$$^\\circ$$, predominantly inductive circuit"
},
{
"text": "0$$^\\circ$$, resonance circuit"
},
{
"text": "$$ \\approx $$ 90$$^\\circ$$, predominantly capacitive circuit"
}
],
"answer": "$$ \\approx $$ 90$$^\\circ$$, predominantly capacitive circuit",
"solution": "**Answer:** $$ \\approx $$ 90$$^\\circ$$, predominantly capacitive circuit\n\nR = 100$$\\Omega$$
$${X_L} = \\omega L = 50\\pi \\times {10^{ - 3}}$$
& $$\\left| {{X_C} - {X_L}} \\right| > > R$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7910,
"subject": "Physics",
"question": "A series LCR circuit of R = 5$$\\Omega$$, L = 20 mH and C = 0.5 $$\\mu$$F is connected across an AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is ______________ $$\\times$$ 102 W.",
"options": [],
"answer": "125",
"solution": "**Answer:** 125\n\nXL = XC (due to resonance)
Z = R so $${i_{rms}} = {V \\over Z} = {V \\over R}$$
$${{{V^2}} \\over R} = {{250 \\times 250} \\over 5} = 125 \\times {10^2}W$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7911,
"subject": "Physics",
"question": "Match List - I with List - II
At resonance, current have maximum value.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7912,
"subject": "Physics",
"question": "A 10 $$\\Omega$$ resistance is connected across 220V $$-$$ 50 Hz AC supply. The time taken by the current to change from its maximum value to the rms value is :",
"options": [
{
"text": "2.5 ms"
},
{
"text": "1.5 ms"
},
{
"text": "3.0 ms"
},
{
"text": "4.5 ms"
}
],
"answer": "2.5 ms",
"solution": "**Answer:** 2.5 ms\n\n
= 2.5 ms",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7913,
"subject": "Physics",
"question": "A 0.07 H inductor and a 12$$\\Omega$$ resistor are connected in series to a 220V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take $$\\pi$$ as $${{22} \\over 7}$$]",
"options": [
{
"text": "8.8 A and $${\\tan ^{ - 1}}\\left( {{{11} \\over 6}} \\right)$$"
},
{
"text": "88 A and $${\\tan ^{ - 1}}\\left( {{{11} \\over 6}} \\right)$$"
},
{
"text": "0.88 A and $${\\tan ^{ - 1}}\\left( {{{11} \\over 6}} \\right)$$"
},
{
"text": "8.8 A and $${\\tan ^{ - 1}}\\left( {{{6} \\over 11}} \\right)$$"
}
],
"answer": "8.8 A and $${\\tan ^{ - 1}}\\left( {{{11} \\over 6}} \\right)$$",
"solution": "**Answer:** 8.8 A and $${\\tan ^{ - 1}}\\left( {{{11} \\over 6}} \\right)$$\n\n$$\\phi = {\\tan ^{ - 1}}\\left( {{{{X_L}} \\over R}} \\right)$$
$$I = {V \\over Z} = {{220} \\over {25.059}} = 8.77A$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 7914,
"subject": "Physics",
"question": "A 100$$\\Omega$$ resistance, a 0.1 $$\\mu$$F capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz.",
"options": [
{
"text": "0.70 H"
},
{
"text": "70.3 mH"
},
{
"text": "7.03 $$\\times$$ 10$$-$$5 H"
},
{
"text": "70.3 H"
}
],
"answer": "70.3 H",
"solution": "**Answer:** 70.3 H\n\nC = 0.1 $$\\mu$$F = 10$$-$$7 F
by putting values $$L \\simeq 70.3$$ Hz.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7915,
"subject": "Physics",
"question": "A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance R = 3 k$$\\Omega$$, an inductor of inductive reactance XL = 250 $$\\pi$$$$\\Omega$$ and an unknown capacitor. The value of capacitance to maximize the average power should be : (Take $$\\pi$$2 = 10)",
"options": [
{
"text": "4 $$\\mu$$F"
},
{
"text": "25 $$\\mu$$F"
},
{
"text": "400 $$\\mu$$F"
},
{
"text": "40 $$\\mu$$F"
}
],
"answer": "4 $$\\mu$$F",
"solution": "**Answer:** 4 $$\\mu$$F\n\nFrom maximum average power
XL = XC
250$$\\pi$$ = $${1 \\over {2\\pi (50)C}}$$
$$ \\Rightarrow $$ C = 4 $$\\times$$ 10$$-$$6",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7916,
"subject": "Physics",
"question": "The alternating current is given by $$i = \\left\\{ {\\sqrt {42} \\sin \\left( {{{2\\pi } \\over T}t} \\right) + 10} \\right\\}A$$
The r.m.s. value of of this current is ................. A.",
"options": [],
"answer": "11",
"solution": "**Answer:** 11\n\n$$f_{rms}^2 = f_{1\\,rms}^2 + f_{2\\,rms}^2$$
$$ = 121 \\Rightarrow {f_{rms}}$$ = 11 A",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7917,
"subject": "Physics",
"question": "An ac circuit has an inductor and a resistor resistance R in series, such that XL = 3R. Now, a capacitor is added in series such that XC = 2R. The ratio of new power factor with the old power factor of the circuit is $$\\sqrt 5 :x$$. The value of x is ___________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n $$\\cos \\phi = {R \\over {\\sqrt {{R^2} + 3{R^2}} }}$$
An inductor of 0.5 mH, a capacitor of 200 $$\\mu$$F and a resistor of 2 $$\\Omega$$ are connected in series with a 220 V ac source. If the current is in phase with the emf, the frequency of ac source will be ____________ $$\\times$$ 102 Hz.
A telegraph line of length 100 km has a capacity of 0.01 $$\\mu$$F/km and it carries an alternating current at 0.5 kilo cycle per second. If minimum impedance is required, then the value of the inductance that needs to be introduced in series is _____________ mH. (if $$\\pi$$ = $$\\sqrt{10}$$)
A sinusoidal voltage V(t) = 210 sin 3000 t volt is applied to a series LCR circuit in which L = 10 mH, C = 25 $$\\mu$$F and R = 100 $$\\Omega$$. The phase difference ($$\\Phi $$) between the applied voltage and resultant current will be :
Statement I : The reactance of an ac circuit is zero. It is possible that the circuit contains a capacitor and an inductor.
\n
Statement II : In ac circuit, the average power delivered by the source never becomes zero.
\n
In the light of the above statements, choose the correct answer from the options given below.
",
"options": [
{
"text": "Both Statement I and Statement II are true."
},
{
"text": "Both Statement I and Statement II are false."
},
{
"text": "Statement I is true but Statement II is false."
},
{
"text": "Statement I is false but Statement II is true."
}
],
"answer": "Statement I is true but Statement II is false.",
"solution": "**Answer:** Statement I is true but Statement II is false.\n\n
A resistance of 40 $$\\Omega$$ is connected to a source of alternating current rated 220 V, 50 Hz. Find the time taken by the current to change from its maximum value to the rms value :
A series LCR circuit with $$R = {{250} \\over {11}}\\,\\Omega $$ and $${X_L} = {{70} \\over {11}}\\,\\Omega $$ is connected across a 220 V, 50 Hz supply. The value of capacitance needed to maximize the average power of the circuit will be _________ $$\\mu$$F. (Take : $$\\pi = {{22} \\over 7}$$)
",
"options": [],
"answer": "500",
"solution": "**Answer:** 500\n\nFor maximum power\n\n
To increase the resonant frequency in series LCR circuit,
",
"options": [
{
"text": "source frequency should be increased."
},
{
"text": "another resistance should be added in series with the first resistance."
},
{
"text": " another capacitor should be added in series with the first capacitor."
},
{
"text": " the source frequency should be decreased."
}
],
"answer": " another capacitor should be added in series with the first capacitor.",
"solution": "**Answer:** another capacitor should be added in series with the first capacitor.\n\n
When you walk through a metal detector carrying a metal object in your pocket, it raises an alarm. This phenomenon works on :
",
"options": [
{
"text": "Electromagnetic induction"
},
{
"text": "Resonance in ac circuits"
},
{
"text": "Mutual induction in ac circuits"
},
{
"text": "Interference of electromagnetic waves"
}
],
"answer": "Resonance in ac circuits",
"solution": "**Answer:** Resonance in ac circuits\n\n
Metal detector works on the principle of resonance in ac circuits.
In a series $$L R$$ circuit $$X_{L}=R$$ and power factor of the circuit is $$P_{1}$$. When capacitor with capacitance $$C$$ such that $$X_{L}=X_{C}$$ is put in series, the power factor becomes $$P_{2}$$. The ratio $$\\frac{P_{1}}{P_{2}}$$ is:
A direct current of $$4 \\mathrm{~A}$$ and an alternating current of peak value $$4 \\mathrm{~A}$$ flow through resistance of $$3\\, \\Omega$$ and $$2\\,\\Omega$$ respectively. The ratio of heat produced in the two resistances in same interval of time will be :
To light, a $$50 \\mathrm{~W}, 100 \\mathrm{~V}$$ lamp is connected, in series with a capacitor of capacitance $$\\frac{50}{\\pi \\sqrt{x}} \\mu F$$, with $$200 \\mathrm{~V}, 50 \\mathrm{~Hz} \\,\\mathrm{AC}$$ source. The value of $$x$$ will be ___________.
A series LCR circuit has $$\\mathrm{L}=0.01\\, \\mathrm{H}, \\mathrm{R}=10\\, \\Omega$$ and $$\\mathrm{C}=1 \\mu \\mathrm{F}$$ and it is connected to ac voltage of amplitude $$\\left(\\mathrm{V}_{\\mathrm{m}}\\right) 50 \\mathrm{~V}$$. At frequency $$60 \\%$$ lower than resonant frequency, the amplitude of current will be approximately :
The equation of current in a purely inductive circuit is $$5 \\sin \\left(49\\, \\pi t-30^{\\circ}\\right)$$. If the inductance is $$30 \\,\\mathrm{mH}$$ then the equation for the voltage across the inductor, will be :
\n
$$\\left\\{\\right.$$ Let $$\\left.\\pi=\\frac{22}{7}\\right\\}$$\n
The frequencies at which the current amplitude in an LCR series circuit becomes $$\\frac{1}{\\sqrt{2}}$$ times its maximum value, are $$212\\,\\mathrm{rad} \\,\\mathrm{s}^{-1}$$ and $$232 \\,\\mathrm{rad} \\,\\mathrm{s}^{-1}$$. The value of resistance in the circuit is $$R=5 \\,\\Omega$$. The self inductance in the circuit is __________ $$\\mathrm{mH}$$.
An alternating emf $$\\mathrm{E}=440 \\sin 100 \\pi \\mathrm{t}$$ is applied to a circuit containing an inductance of $$\\frac{\\sqrt{2}}{\\pi} \\mathrm{H}$$. If an a.c. ammeter is connected in the circuit, its reading will be :
A circuit element $$\\mathrm{X}$$ when connected to an a.c. supply of peak voltage $$100 \\mathrm{~V}$$ gives a peak current of $$5 \\mathrm{~A}$$ which is in phase with the voltage. A second element $$\\mathrm{Y}$$ when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by $$\\frac{\\pi}{2}$$. If $$\\mathrm{X}$$ and $$\\mathrm{Y}$$ are connected in series to the same supply, what will be the rms value of the current in ampere?
Element Y should be inductive with, $$\nX_{L}=\\frac{100}{5}=20 \\Omega\n$$\n\n
When X and Y are connector in series,\n\n
$$\n\\begin{aligned}\n&Z=\\sqrt{20^{2}+20^{2}}=20 \\sqrt{2} \\Omega \\\\\\\\\n&I=\\frac{100}{Z}=\\frac{100}{20 \\sqrt{2}}=\\frac{5}{\\sqrt{2}} \\\\\\\\\n&i_{\\mathrm{rms}}=\\frac{1}{\\sqrt{2}} I \\\\\\\\\n&=\\frac{5}{2}\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7937,
"subject": "Physics",
"question": "An alternating voltage source $\\mathrm{V}=260 \\sin (628 \\mathrm{t}$ ) is connected across a pure inductor of $5 \\mathrm{mH}$ Inductive reactance in the circuit is :",
"options": [
{
"text": "$6.28 \\Omega$"
},
{
"text": "$0.318 \\Omega$"
},
{
"text": "$0.5 \\Omega$"
},
{
"text": "$3.14 \\Omega$"
}
],
"answer": "$3.14 \\Omega$",
"solution": "**Answer:** $3.14 \\Omega$\n\n$\\omega$ = 628 rad/s\n
$X_{L}=L \\omega$\n\n
$$\n\\begin{aligned}\n& =5 \\mathrm{mH} \\times 628 \\\\\\\\\n& =3.14 \\Omega\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7938,
"subject": "Physics",
"question": "A series $\\mathrm{LCR}$ circuit consists of $\\mathrm{R}=80 \\Omega, \\mathrm{X}_{\\mathrm{L}}=100 \\Omega$, and $\\mathrm{X}_{\\mathrm{C}}=40 \\Omega$. The input
voltage is 2500 $\\cos (100 \\pi \\mathrm{t}) \\mathrm{V}$. The amplitude of current, in the circuit, is _________ A.",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\n$\\omega=100 \\pi$\n\n
A series LCR circuit is connected to an ac source of $$220 \\mathrm{~V}, 50 \\mathrm{~Hz}$$. The circuit contain a resistance $$\\mathrm{R}=100 ~\\Omega$$ and an inductor of inductive reactance $$\\mathrm{X}_{\\mathrm{L}}=79.6 ~\\Omega$$. The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be _________ $$\\mu \\mathrm{F}$$.
",
"options": [],
"answer": "40",
"solution": "**Answer:** 40\n\nTo maximize the average rate at which energy \nsupplied i.e. power will be maximum. \n
So in LCR circuit power will be maximum at the\ncondition of resonance and in resonance condition\n
If $$\\mathrm{R}, \\mathrm{X}_{\\mathrm{L}}$$, and $$\\mathrm{X}_{\\mathrm{C}}$$ represent resistance, inductive reactance and capacitive reactance. Then which of the following is dimensionless :
So, $\\frac{R}{\\sqrt{X_{L} X_{C}}}$ is dimensionless.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7941,
"subject": "Physics",
"question": "
An inductor of $$0.5 ~\\mathrm{mH}$$, a capacitor of $$20 ~\\mu \\mathrm{F}$$ and resistance of $$20 ~\\Omega$$ are connected in series with a $$220 \\mathrm{~V}$$ ac source. If the current is in phase with the emf, the amplitude of current of the circuit is $$\\sqrt{x}$$ A. The value of $$x$$ is ___________
In a series LR circuit with $$\\mathrm{X_L=R}$$, power factor P1. If a capacitor of capacitance C with $$\\mathrm{X_C=X_L}$$ is added to the circuit the power factor becomes P2. The ratio of P1 to P2 will be :
An inductor of inductance 2 $$\\mathrm{\\mu H}$$ is connected in series with a resistance, a variable capacitor and an AC source of frequency 7 kHz. The value of capacitance for which maximum current is drawn into the circuit $$\\frac{1}{x}\\mathrm{F}$$, where the value of $$x$$ is ___________.
A series LCR circuit is connected to an AC source of 220 V, 50 Hz. The circuit contains a resistance R = 80$$\\Omega$$, an inductor of inductive reactance $$\\mathrm{X_L=70\\Omega}$$, and a capacitor of capacitive reactance $$\\mathrm{X_C=130\\Omega}$$. The power factor of circuit is $$\\frac{x}{10}$$. The value of $$x$$ is :
In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes $$x$$ times its initial resonant frequency $$\\omega_0$$. The value of $$x$$ is :
An LCR series circuit of capacitance 62.5 nF and resistance of 50 $$\\Omega$$, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is __________ mH.
\n
(Take $$\\pi^2=10$$)
",
"options": [],
"answer": "100",
"solution": "**Answer:** 100\n\n $\\because$ For maximum amplitude of current, circuit should be at resonance.\n
Statement I : An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.
\n
Statement II : An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor.
\n
In the light of above statements, choose the correct answer form the options given below:
",
"options": [
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Statement I is true but statement II is false"
},
{
"text": "Statement I is false but statement II is true"
},
{
"text": "Both Statement I and Statement II are true"
}
],
"answer": "Both Statement I and Statement II are false",
"solution": "**Answer:** Both Statement I and Statement II are false\n\n
Statement I: An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.
\n
This statement is incorrect. Electrical resonance occurs in an AC circuit when the capacitive reactance and inductive reactance are equal, causing the impedance of the circuit to be minimum. This typically happens in a series RLC circuit or a parallel RLC circuit. If the circuit contains only a capacitor or an inductor, it cannot undergo electrical resonance as there is no counterpart reactance to balance the impedance.
\n\n
Statement II: An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor.
\n
This statement is also incorrect. An AC circuit containing a pure capacitor or a pure inductor will have a power factor of 0, not a non-zero power factor. The power factor of a capacitor is -1, and the power factor of an inductor is +1, but when only considering the reactive components, the power factor is 0. In such a circuit, no real power is consumed, and the circuit only has reactive power. The energy is alternately stored and released by the capacitor and inductor, but no energy is dissipated as heat or used to perform work.
\n\n
As both statements are incorrect, the correct answer would be an option that states both statements are false.
Statement I : When the frequency of an a.c source in a series LCR circuit increases, the current in the circuit first increases, attains a maximum value and then decreases.
\n
Statement II : In a series LCR circuit, the value of power factor at resonance is one.
\n
In the light of given statements, choose the most appropriate answer from the options given below.
",
"options": [
{
"text": "Both Statement I and Statement II are False."
},
{
"text": "Statement I is incorrect but Statement II is true."
},
{
"text": "Both Statement I and Statement II are true."
},
{
"text": "Statement I is correct but Statement II is false."
}
],
"answer": "Both Statement I and Statement II are true.",
"solution": "**Answer:** Both Statement I and Statement II are true.\n\n
Statement I is true because in a series LCR circuit, the current first increases as the frequency increases, reaching a maximum value when the circuit is at resonance. At resonance, the inductive reactance (XL) and capacitive reactance (XC) cancel each other out, resulting in the lowest impedance (Z) and the highest current. As the frequency continues to increase beyond resonance, the current in the circuit decreases.
\n
Statement II is also true because, at resonance in a series LCR circuit, the inductive reactance (XL) and capacitive reactance (XC) are equal and cancel each other out. This results in the impedance (Z) being purely resistive. The power factor at resonance is given by the cosine of the phase angle (θ), and since the phase angle is 0° at resonance, the power factor is 1.
\n
Thus, both statements are true, and the correct answer is Both Statement I and Statement II are true.
Statement I : Maximum power is dissipated in a circuit containing an inductor, a capacitor and a resistor connected in series with an AC source, when resonance occurs
\n
Statement II : Maximum power is dissipated in a circuit containing pure resistor due to zero phase difference between current and voltage.
\n
In the light of the above statements, choose the correct answer from the options given below:
",
"options": [
{
"text": "Statement I is false but Statement II is true"
},
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Statement I is true but Statement II is false"
},
{
"text": "Both Statement I and Statement II are true"
}
],
"answer": "Both Statement I and Statement II are true",
"solution": "**Answer:** Both Statement I and Statement II are true\n\n
In a series LCR circuit connected to an AC source, resonance occurs at a particular frequency at which the inductive reactance is equal to the capacitive reactance, resulting in the minimum impedance of the circuit. At this frequency, the circuit draws maximum current from the source, and thus, the maximum power is dissipated in the circuit. Therefore, Statement I is true.
\n
In a circuit containing only a resistor, the power dissipated is given by P = VI = I$^2$R, where V is the voltage across the resistor, I is the current flowing through the resistor, and R is the resistance of the resistor. The voltage and current are in phase in a purely resistive circuit, which means that the power is maximized. Therefore, Statement II is also true.
A capacitor of capacitance $$150.0 ~\\mu \\mathrm{F}$$ is connected to an alternating source of emf given by $$\\mathrm{E}=36 \\sin (120 \\pi \\mathrm{t}) \\mathrm{V}$$. The maximum value of current in the circuit is approximately equal to :
where $$E_\\text{max}$$ is the maximum voltage, $$\\omega$$ is the angular frequency, and $$C$$ is the capacitance.
\n
Given the emf equation: $$E = 36 \\sin(120\\pi t) \\, \\text{V}$$, we can determine that $$E_\\text{max} = 36\\, \\text{V}$$ and $$\\omega = 120\\pi \\, \\text{rad/s}$$.
\n
The capacitance is given as $$150.0\\, \\mu\\text{F} = 150.0 \\times 10^{-6}\\, \\text{F}$$.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7951,
"subject": "Physics",
"question": "A parallel plate capacitor has a capacitance $\\mathrm{C}=200~ \\mathrm{pF}$. It is connected to $230 \\mathrm{~V}$ ac supply with an angular frequency $300~ \\mathrm{rad} / \\mathrm{s}$. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are :",
"options": [
{
"text": "$14.3 ~\\mu \\mathrm{A}$ and $143 ~\\mu \\mathrm{A}$"
},
{
"text": "$13.8 ~\\mu \\mathrm{A}$ and $13.8 ~\\mu \\mathrm{A}$"
},
{
"text": "$13.8 ~\\mu \\mathrm{A}$ and $138 ~\\mu \\mathrm{A}$"
},
{
"text": "$1.38 ~\\mu \\mathrm{A}$ and $1.38 ~\\mu \\mathrm{A}$"
}
],
"answer": "$13.8 ~\\mu \\mathrm{A}$ and $13.8 ~\\mu \\mathrm{A}$",
"solution": "**Answer:** $13.8 ~\\mu \\mathrm{A}$ and $13.8 ~\\mu \\mathrm{A}$\n\n
To solve this problem, we need to understand how to calculate the rms current in an AC circuit containing a capacitor, as well as the concept of displacement current.
\n\n
First, the rms (root mean square) value of the current ($I_{\\text{rms}}$) in a capacitor when connected to an AC supply is given by:
Therefore, the rms value of the conduction current is $13.8\\, \\mu \\text{A}$.
\n\n
Now, for the displacement current (which is essentially the current that would \"flow\" through the dielectric of the capacitor as a result of the changing electric field). In an AC circuit, the displacement current and the conduction current are the same. That's because the displacement current is necessary to sustain the changing electric field between the plates of the capacitor, and this changing field is what causes the conduction current in the leads and the rest of the circuit.
\n\n
Hence, the rms value of the displacement current is the same as the conduction current, which is $13.8\\, \\mu \\text{A}$.
\n\n
So, the correct option is:
\n\n
Option B
\n\n
$13.8\\, \\mu \\text{A}$ and $13.8\\, \\mu \\text{A}$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7952,
"subject": "Physics",
"question": "In series LCR circuit, the capacitance is changed from $C$ to $4 C$. To keep the resonance frequency unchanged, the new inductance should be:",
"options": [
{
"text": "increased by $2 \\mathrm{~L}$"
},
{
"text": "reduced by $\\frac{1}{4} \\mathrm{~L}$"
},
{
"text": "reduced by $\\frac{3}{4} \\mathrm{~L}$"
},
{
"text": "increased to $4 \\mathrm{~L}$"
}
],
"answer": "reduced by $\\frac{3}{4} \\mathrm{~L}$",
"solution": "**Answer:** reduced by $\\frac{3}{4} \\mathrm{~L}$\n\n
The resonance frequency $$ f_0 $$ of an LCR circuit is given by:
\n\n$$ f_0 = \\frac{1}{2\\pi\\sqrt{LC}} $$\n\n
where:
\n
\n
$$ L $$ is the inductance.
\n
$$ C $$ is the capacitance.
\n
\n\n
To keep the resonance frequency unchanged when the capacitance is changed from $$ C $$ to $$ 4C $$, we must adjust the inductance $$ L $$ to a new value $$ L' $$ such that:
Thus, the new inductance $$ L' $$ is one fourth of the original inductance $$ L $$. Therefore, to achieve the same resonance frequency with the capacitance increased to $$ 4C $$, the inductance should be reduced to a quarter of its initial value:
\n\n$$ L' = \\frac{L}{4} $$\n\n
This means we have reduced the inductance by $$ \\frac{3}{4}L $$, so the correct answer is:
A series LCR circuit with $$\\mathrm{L}=\\frac{100}{\\pi} \\mathrm{mH}, \\mathrm{C}=\\frac{10^{-3}}{\\pi} \\mathrm{F}$$ and $$\\mathrm{R}=10 \\Omega$$, is connected across an ac source of $$220 \\mathrm{~V}, 50 \\mathrm{~Hz}$$ supply. The power factor of the circuit would be ________.
An AC voltage $$V=20 \\sin 200 \\pi t$$ is applied to a series LCR circuit which drives a current $$I=10 \\sin \\left(200 \\pi t+\\frac{\\pi}{3}\\right)$$. The average power dissipated is:
An alternating voltage $$V(t)=220 \\sin 100 \\pi t$$ volt is applied to a purely resistive load of $$50 \\Omega$$. The time taken for the current to rise from half of the peak value to the peak value is:
A series L.R circuit connected with an ac source $$E=(25 \\sin 1000 t) V$$ has a power factor of $$\\frac{1}{\\sqrt{2}}$$. If the source of emf is changed to $$\\mathrm{E}=(20 \\sin 2000 \\mathrm{t}) \\mathrm{V}$$, the new power factor of the circuit will be :
A capacitor of reactance $$4 \\sqrt{3} \\Omega$$ and a resistor of resistance $$4 \\Omega$$ are connected in series with an ac source of peak value $$8 \\sqrt{2} \\mathrm{~V}$$. The power dissipation in the circuit is __________ W.
To calculate the power dissipation in the circuit, we follow a systematic approach. We're provided with the reactance of the capacitor ($$X_C = 4 \\sqrt{3} \\Omega$$), the resistance ($$R = 4 \\Omega$$), and the peak value of the AC voltage source ($$V_{peak} = 8 \\sqrt{2} V$$). The power dissipated in an AC circuit is primarily through the resistive component, as inductors and capacitors store and release energy but do not dissipate it as heat.
\n\n
First, we need to determine the effective impedance of the series circuit, which combines the resistance (R) and the capacitive reactance (X_C) in a series configuration. We calculate the impedance (Z) using the formula:
\n\n
$$Z = \\sqrt{R^2 + X_C^2}$$
\n\n
Plugging in the given values:
\n\n
$$Z = \\sqrt{(4)^2 + (4 \\sqrt{3})^2}$$
\n\n
$$Z = \\sqrt{16 + 48} = \\sqrt{64} = 8 \\Omega$$
\n\n
Next, we convert the peak voltage to RMS (root mean square) voltage because power calculations in AC circuits are performed using RMS values. The formula to convert peak voltage ($$V_{peak}$$) to RMS voltage ($$V_{RMS}$$) is:
Now, to find the RMS current ($$I_{RMS}$$) in the circuit, we use Ohm's law as applied to AC circuits, which is $$I_{RMS} = \\frac{V_{RMS}}{Z}$$:
\n\n
$$I_{RMS} = \\frac{8}{8} = 1 \\, \\mathrm{A}$$
\n\n
Finally, the power dissipated in the circuit is calculated using the formula for power in resistive components of an AC circuit, which is $$P = I_{RMS}^2 \\times R$$:
\n\n
$$P = (1)^2 \\times 4 = 4 \\, \\mathrm{W}$$
\n\n
Therefore, the power dissipation in the circuit is 4 W.
A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb :
To understand the impact of placing a dielectric between the plates of the capacitor on the glow of the bulb, we need to consider the properties and behavior of capacitors in an AC circuit.
\n\n
When a capacitor is connected in series with a bulb in an AC circuit, the impedance of the capacitor plays a significant role in determining the current through the circuit. The impedance $ Z_C $ of a capacitor in an AC circuit is given by:
\n\n
$$ Z_C = \\frac{1}{\\omega C} $$
\n\n
where:
\n\n
\n\n
$ Z_C $ is the capacitive reactance (impedance of the capacitor).
\n\n
$ \\omega $ is the angular frequency of the AC supply ( $ \\omega = 2 \\pi f $ , where $ f $ is the frequency).
\n\n
$ C $ is the capacitance of the capacitor.
\n\n
\n\n
When we place a dielectric between the plates of the capacitor, the capacitance $ C $ increases. The capacitance with a dielectric can be described as:
\n\n
$$ C' = \\kappa C $$
\n\n
where:
\n\n
\n\n
$ C' $ is the new capacitance with the dielectric present.
\n\n
$ \\kappa $ is the dielectric constant ( $ \\kappa > 1 $ ).
\n\n
\n\n
Since $ C' > C $, the new capacitive reactance $ Z_C' $ can be given as:
Since $\\kappa > 1$, $ Z_C' < Z_C $. This means the impedance of the capacitor decreases when a dielectric is placed between its plates. In a series circuit, the overall impedance decreases when the impedance of one component decreases, leading to an increase in the current through the circuit.
\n\n
Thus, with an increase in current, the bulb will glow brighter. Therefore, the correct option is:
When a coil is connected across a $$20 \\mathrm{~V}$$ dc supply, it draws a current of $$5 \\mathrm{~A}$$. When it is connected across $$20 \\mathrm{~V}, 50 \\mathrm{~Hz}$$ ac supply, it draws a current of $$4 \\mathrm{~A}$$. The self inductance of the coil is __________ $$\\mathrm{mH}$$. (Take $$\\pi=3$$)
Let's first determine the resistance of the coil when connected to a DC supply. The current drawn by the coil in a DC circuit can be used to calculate its resistance using Ohm's law:
\n\n
$$R = \\frac{V}{I}$$
\n\n
Given the DC supply voltage $$V_{DC} = 20 \\, \\mathrm{V}$$ and the current $$I_{DC} = 5 \\, \\mathrm{A}$$, the resistance $$R$$ is:
Next, let's use the information given for the AC supply. When connected to an AC supply, the coil's impedance $$Z$$ can be determined using the given current. The total voltage and current in an AC circuit are related to the impedance by the formula:
\n\n
$$Z = \\frac{V}{I}$$
\n\n
Given the AC supply voltage $$V_{AC} = 20 \\, \\mathrm{V}$$ and the current $$I_{AC} = 4 \\, \\mathrm{A}$$, the impedance $$Z$$ is:
In an ac circuit, the instantaneous current is zero, when the instantaneous voltage is maximum. In this case, the source may be connected to :
\n
A. pure inductor.
\n
B. pure capacitor.
\n
C. pure resistor.
\n
D. combination of an inductor and capacitor.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "A, B and C only\n"
},
{
"text": "A and B only\n"
},
{
"text": "A, B and D only\n"
},
{
"text": "B, C and D only"
}
],
"answer": "A, B and D only\n",
"solution": "**Answer:** A, B and D only\n\n\n
To understand which answer is correct, we must consider the relationship between voltage and current in various types of circuits, namely circuits with pure inductors, pure capacitors, pure resistors, and a combination of inductors and capacitors (LC circuits).
\n\n
In a purely resistive circuit, the voltage and current are in phase, meaning when the voltage is maximum, the current is also maximum. Therefore, option C (pure resistor) cannot result in the instantaneous current being zero when the instantaneous voltage is maximum.
\n\n
In a purely inductive circuit, the current lags the voltage by $90^\\circ$, or in other words, when the voltage is at its maximum, the current is zero. This is because the inductor opposes changes in current, leading to this phase difference.
\n\n
In a purely capacitive circuit, the current leads the voltage by $90^\\circ$. This means when the voltage is at its maximum value, the current through the capacitor is zero because the current reaches its maximum or minimum before the voltage does.
\n\n
In an LC circuit (inductor-capacitor), under certain conditions like at its resonant frequency, the circuit behaves as if it is purely resistive in nature where voltage and current are in phase. However, it's more nuanced because the question likely implies a condition not at resonance but rather at a general characteristic of LC circuits. In LC circuits, aside from the resonance condition, the presence of both inductive and capacitive components can lead to scenarios where the inductive and capacitive reactances cancel each other, particularly in the case where oscillations are involved, and indeed, there can be moments when the voltage is maximum and the current is minimum (zero in the ideal case) due to the energy being alternately stored in the magnetic field of the inductor and the electric field of the capacitor.
\n\n
Thus, reflecting on the scenarios:
\n\n
\n\n
Pure inductor - Fits the description: instantaneous current is zero when the instantaneous voltage is maximum.
\n\n
Pure capacitor - Also fits the description for the reason described, taking into account ideal conditions.
\n\n
Pure resistor - Does not fit as voltage and current are in phase.
\n\n
Combination of an inductor and capacitor (LC circuit) - Can fit the description under certain non-resonance conditions where the behavior of energy exchange between L and C at a specific instant can result in the instant current being zero when the voltage is maximum, but this is more complex and dependent on specific conditions unlike the straightforward lag or lead in purely inductive or capacitive circuits.
\n\n
\n\n
Therefore, the correct answer is Option C: A, B, and D only.
The given alternating current (AC) can be represented as $$i=6+\\sqrt{56} \\sin (100 \\pi t+\\pi / 3)$$ A, where $$6$$ is the DC component and $$\\sqrt{56} \\sin (100 \\pi t+\\pi / 3)$$ is the AC component of the current. The RMS (Root Mean Square) value of an alternating current is a measure of the equivalent direct current (DC) that will produce the same power in a resistor. The RMS value is mostly relevant for the AC component of the current, as the DC component's effective value is just its magnitude itself.
\n\n
The RMS value of the total current is not straightforward because the presence of the DC component affects how we calculate the RMS value. However, when calculating RMS values for a signal consisting of a superposition of AC and DC components, one notable property is that the RMS value of the combined signal is the square root of the sum of the squares of the RMS values of the separate AC and DC components.
\n\n
First, let's acknowledge the components separately:\n\n
\n
The DC component is: $$6$$ A
\n
The AC component is: $$\\sqrt{56} \\sin (100 \\pi t+\\pi / 3)$$ A
\n
\n
For the DC component, the RMS value is simply its magnitude:\n\n
$$I_{RMS, DC} = 6$$ A
\n\n
For the AC component, the RMS value is calculated using the formula for the RMS value of a sinusoidal function, which is $$I_{RMS} = \\frac{I_{max}}{\\sqrt{2}}$$, where $$I_{max}$$ is the peak value of the current. In this case, $$I_{max} = \\sqrt{56}$$.
\n\n
Therefore, the RMS value of the AC component is:\n\n
A coil of negligible resistance is connected in series with $$90 \\Omega$$ resistor across $$120 \\mathrm{~V}, 60 \\mathrm{~Hz}$$ supply. A voltmeter reads $$36 \\mathrm{~V}$$ across resistance. Inductance of the coil is :
To find the inductance of the coil, we need to analyze the given circuit and use the information provided. The circuit consists of a resistor and an inductor in series, connected to an AC supply. Here are the given values:
4. Voltage across the resistor, $$V_R = 36 \\mathrm{~V}$$
\n\n
First, we calculate the current through the resistor (which is the same as the current through the inductor, since they are in series) using Ohm's law:
An alternating emf $$\\mathrm{E}=110 \\sqrt{2} \\sin 100 \\mathrm{t}$$ volt is applied to a capacitor of $$2 \\mu \\mathrm{F}$$, the rms value of current in the circuit is ________ $$\\mathrm{mA}$$.
We are given a capacitor with a capacitance $$C = 2 \\mu \\mathrm{F} = 2 \\times 10^{-6} \\, \\text{F}$$ and we need to determine the RMS current. The capacitive reactance $$\\mathrm{X_C}$$ is given by:
\n\n
\n$$\\mathrm{X_C} = \\frac{1}{\\omega C}$$\n
\n\n
where $$\\omega$$ is the angular frequency. From the given formula for emf, we see that:
A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:
To solve this problem, we need to understand the relationship between the current amplitude in a series LCR circuit at resonance and the resistance $ R $. At resonance, the impedance $ Z $ of the series LCR circuit is equal to the resistance $ R $, and thus:
\n\n
$$ Z = R $$
\n\n
The amplitude of the current $ I_0 $ at resonance is given by Ohm's law:
\n\n
$$ I_0 = \\frac{V_0}{R} $$
\n\n
where $ V_0 $ is the amplitude of the voltage supplied.
\n\n
Now, if the resistance $ R $ is halved while keeping the voltage amplitude $ V_0 $, capacitance $ C $, and inductance $ L $ the same, the new resistance becomes:
\n\n
$$ R_{new} = \\frac{R}{2} $$
\n\n
The new current amplitude $ I_0' $ at resonance is given by the modified Ohm's law:
An alternating voltage of amplitude $$40 \\mathrm{~V}$$ and frequency $$4 \\mathrm{~kHz}$$ is applied directly across the capacitor of $$12 \\mu \\mathrm{F}$$. The maximum displacement current between the plates of the capacitor is nearly :
Now, the maximum current ($$I_{max}$$) in the circuit can be calculated using Ohm's law, considering the maximum voltage across the capacitor and its capacitive reactance:
A series LCR circuit is subjected to an ac signal of $$200 \\mathrm{~V}, 50 \\mathrm{~Hz}$$. If the voltage across the inductor $$(\\mathrm{L}=10 \\mathrm{~mH})$$ is $$31.4 \\mathrm{~V}$$, then the current in this circuit is _______.
For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is $$2.5 \\mathrm{~nF}$$. If resistance of $$200 \\Omega$$ and $$100 \\mathrm{~mH}$$ inductor is being used in the given circuit. The frequency of ac source is _________ $$\\times 10^3 \\mathrm{~Hz}$$ (given $$\\mathrm{a}^2=10$$)
To solve this problem, we need to use the concept of resonance in an LCR (inductor-capacitor-resistor) circuit. At resonance, the inductive reactance and capacitive reactance cancel each other out. The condition for resonance in an LCR circuit is given by:
\n\n
$$\\omega L = \\frac{1}{\\omega C}$$
\n\n
Where:\n\n
\n\n
$$\\omega$$ is the angular frequency
\n\n
$$L$$ is the inductance
\n\n
$$C$$ is the capacitance
\n\n
\n\n\n\n
Rewriting for angular frequency:
\n\n
$$\\omega^2 = \\frac{1}{LC}$$
\n\n
The angular frequency $$\\omega$$ is related to the frequency $ f $ by:
\n\n
$$\\omega = 2 \\pi f$$
\n\n
Substituting this into the equation for angular frequency gives:
Statement I : In an LCR series circuit, current is maximum at resonance.
\n
Statement II : Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to same voltage source.
\n
In the light of the above statements, choose the correct from the options given below :
",
"options": [
{
"text": "Statement I is true but Statement II is false\n"
},
{
"text": "Statement I is false but Statement II is true\n"
},
{
"text": "Both Statement I and Statement II are true\n"
},
{
"text": "Both Statement I and Statement II are false"
}
],
"answer": "Both Statement I and Statement II are true\n",
"solution": "**Answer:** Both Statement I and Statement II are true\n\n\n
Statement I : True
\n
Statement II : True
\n
Current in purely resistive circuit is equal to current in LCR circuit with same resistance at resonance, otherwise more.
When a $$d c$$ voltage of $$100 \\mathrm{~V}$$ is applied to an inductor, a $$d c$$ current of $$5 \\mathrm{~A}$$ flows through it. When an ac voltage of $$200 \\mathrm{~V}$$ peak value is connected to inductor, its inductive reactance is found to be $$20 \\sqrt{3} \\Omega$$. The power dissipated in the circuit is _________ W.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7971,
"subject": "Physics",
"question": "In a transformer, number of turns in the primary coil are $$140$$ and that in the secondary coil are $$280.$$ If current in primary coil is $$4A,$$ then that in the secondary coil is ",
"options": [
{
"text": "$$4A$$ "
},
{
"text": "$$2A$$ "
},
{
"text": "$$6A$$ "
},
{
"text": "$$10A$$ "
}
],
"answer": "$$2A$$ ",
"solution": "**Answer:** $$2A$$ \n\n$${N_p} = 140,\\,\\,{N_s} = 280,\\,\\,{I_p} = 4A,\\,\\,{I_s} = ?$$ \n
For a transformer $${{{I_s}} \\over {{I_p}}} = {{{N_p}} \\over {{N_s}}}$$\n
$$ \\Rightarrow {{{I_s}} \\over 4} = {{140} \\over {280}} \\Rightarrow {I_s} = 2A$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7972,
"subject": "Physics",
"question": "The core of any transformer is laminated so as to",
"options": [
{
"text": "reduce the energy loss due to eddy currents"
},
{
"text": "make it light weight"
},
{
"text": "make it robust and strong"
},
{
"text": "increase the secondary voltage "
}
],
"answer": "reduce the energy loss due to eddy currents",
"solution": "**Answer:** reduce the energy loss due to eddy currents\n\nLaminated core provide less area of cross-section for the current to flow. Because of this, resistance of the core increases and current decreases thereby decreasing the eddy current losses.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7973,
"subject": "Physics",
"question": "In an $$AC$$ generator, a coil with $$N$$ turns, all of the same area $$A$$ and total resistance $$R,$$ rotates with frequency $$\\omega $$ in a magnetic field $$B.$$ The maximum value of $$emf$$ generated in the coil is ",
"options": [
{
"text": "$$N.A.B.R.$$$$\\omega $$ "
},
{
"text": "$$N.A.B$$"
},
{
"text": "$$N.A.B.R.$$ "
},
{
"text": "$$N.A.B.$$$$\\omega $$ "
}
],
"answer": "$$N.A.B.$$$$\\omega $$ ",
"solution": "**Answer:** $$N.A.B.$$$$\\omega $$ \n\n$$e = - {{d\\phi } \\over {dt}} = - {{d\\left( {N\\overrightarrow B .\\overrightarrow A } \\right)} \\over {dt}}$$\n
$$ \\Rightarrow {e_{\\max }} = NBA\\omega $$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 7974,
"subject": "Physics",
"question": "A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns, giving the output power at 230 V. If the current in the primary of the transformer is 5 A, and its efficiency is 90%, the output current would be ; ",
"options": [
{
"text": "50 A"
},
{
"text": "45 A"
},
{
"text": "25 A"
},
{
"text": "20 A"
}
],
"answer": "45 A",
"solution": "**Answer:** 45 A\n\nEfficiency n = 0.9 = $${{{P_s}} \\over {{P_p}}}$$\n
= $$\\,\\,\\,\\,$$ 45 A",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7975,
"subject": "Physics",
"question": "A power transmission line feeds input power at 2300 V to a srep down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be : ",
"options": [
{
"text": "50 A"
},
{
"text": "45 A"
},
{
"text": "35 A"
},
{
"text": "25 A"
}
],
"answer": "45 A",
"solution": "**Answer:** 45 A\n\nGiven,\n
$$ \\Rightarrow $$ I = 45 A",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7976,
"subject": "Physics",
"question": "A transformer consisting of 300 turns in the\nprimary and 150 turns in the secondary gives\noutput power of 2.2 kW. If the current in the\nsecondary coil is 10A, then the input voltage\nand current in the primary coil are :",
"options": [
{
"text": "220 V and 20 A"
},
{
"text": "220 V and 10A"
},
{
"text": "440 V and 5A"
},
{
"text": "440 V and 20 A"
}
],
"answer": "440 V and 5A",
"solution": "**Answer:** 440 V and 5A\n\nGiven NP = 300, Ns = 150, P0 = 2200W
\nAlso, P0 = ViIi \n$$ \\Rightarrow {I_i} = {{2200} \\over {440}} = 5A$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7977,
"subject": "Physics",
"question": "A common transistor radio set requires 12 V (D.C.) for its operation. The D.C. source is constructed by using a transformer and a rectifier circuit, which are operated at 220 V (A.C.) on standard domestic A.C. supply. The number of turns of secondary coil are 24, then the number of turns of primary are ___________.",
"options": [],
"answer": "440",
"solution": "**Answer:** 440\n\nIn a transformer,
where, Np = number of turns in primary circuit, Ns = number of turns in secondary circuit = 24, Vp = potential of primary circuit = 220 V and Vs = potential of secondary circuit = 12 V
A transformer operating at primary voltage $$8 \\,\\mathrm{kV}$$ and secondary voltage $$160 \\mathrm{~V}$$ serves a load of $$80 \\mathrm{~kW}$$. Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be
",
"options": [
{
"text": "$$800 \\,\\Omega$$ and $$1.06 \\,\\Omega$$"
},
{
"text": "$$10 \\,\\Omega$$ and $$500 \\,\\Omega$$"
},
{
"text": "$$800 \\,\\Omega$$ and $$0.32 \\,\\Omega$$"
},
{
"text": "$$1.06 \\,\\Omega$$ and $$500 \\,\\Omega$$"
}
],
"answer": "$$800 \\,\\Omega$$ and $$0.32 \\,\\Omega$$",
"solution": "**Answer:** $$800 \\,\\Omega$$ and $$0.32 \\,\\Omega$$\n\n
A square shaped coil of area $$70 \\mathrm{~cm}^{2}$$ having 600 turns rotates in a magnetic field of $$0.4 ~\\mathrm{wbm}^{-2}$$, about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field. If the coil completes 500 revolution in a minute, the instantaneous emf when the plane of the coil is inclined at $$60^{\\circ}$$ with the field, will be ____________ V. (Take $$\\pi=\\frac{22}{7}$$)
Resonance occurs when both $L$ and $C$ are present (C-Z) and \n
quality factor determines sharpness of resonance (D-III)",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 7982,
"subject": "Physics",
"question": "In an ac generator, a rectangular coil of 100 turns each having area $14 \\times 10^{-2} \\mathrm{~m}^{2}$ is rotated at $360 ~\\mathrm{rev} / \\mathrm{min}$ about an axis perpendicular to a uniform magnetic field of magnitude $3.0 \\mathrm{~T}$. The maximum value of the emf produced will be ________ $V$.\n
\n$\\left(\\right.$ Take $\\left.\\pi=\\frac{22}{7}\\right)$",
"options": [],
"answer": "1584",
"solution": "**Answer:** 1584\n\n
An ideal transformer with purely resistive load operates at $$12 ~\\mathrm{kV}$$ on the primary side. It supplies electrical energy to a number of nearby houses at $$120 \\mathrm{~V}$$. The average rate of energy consumption in the houses served by the transformer is 60 $$\\mathrm{kW}$$. The value of resistive load $$(\\mathrm{Rs})$$ required in the secondary circuit will be ___________ $$\\mathrm{m} \\Omega$$.
The power delivered to the houses is given as 60 kW. This power is supplied at a voltage of 120 V. The power consumed in a resistive load can be found using the formula $P = V^2/R$, where P is the power, V is the voltage, and R is the resistance.
\n
We can rearrange this formula to solve for the resistance:
Primary side of a transformer is connected to $$230 \\mathrm{~V}, 50 \\mathrm{~Hz}$$ supply. Turns ratio of primary to secondary winding is $$10: 1$$. Load resistance connected to secondary side is $$46 \\Omega$$. The power consumed in it is :
A power transmission line feeds input power at $$2.3 \\mathrm{~kV}$$ to a step down transformer with its primary winding having 3000 turns. The output power is delivered at $$230 \\mathrm{~V}$$ by the transformer. The current in the primary of the transformer is $$5 \\mathrm{~A}$$ and its efficiency is $$90 \\%$$. The winding of transformer is made of copper. The output current of transformer is _________ $$A$$.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7986,
"subject": "Physics",
"question": "In an oscillating $$LC$$ circuit the maximum charge on the capacitor is $$Q$$. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is ",
"options": [
{
"text": "$${Q \\over 2}$$ "
},
{
"text": "$${Q \\over {\\sqrt 3 }}$$ "
},
{
"text": "$${Q \\over {\\sqrt 2 }}$$ "
},
{
"text": "$$Q$$"
}
],
"answer": "$${Q \\over {\\sqrt 2 }}$$ ",
"solution": "**Answer:** $${Q \\over {\\sqrt 2 }}$$ \n\nWhen the capacitor is completely charged, the total energy in the $$L.C$$ circuit is with the capacitor and that energy is $$E = {1 \\over 2}{{{Q^2}} \\over C}$$\n
When half energy is with the capacitor in the form of electric field between the plates of the capacitor we get \n$${E \\over 2} = {1 \\over 2}{{Q{'^2}} \\over C}$$ where $$Q'$$ is the charge on plate of the capacitor\n
$$ \\Rightarrow t = RC\\,in\\,\\left( {2.5} \\right)$$\n
$$ \\Rightarrow R = 2.71 \\times {10^6}\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7988,
"subject": "Physics",
"question": "A coil of self inductance 10 mH and resistance 0.1 $$\\Omega $$ is connected through a switch to a battery of internal\nresistance 0.9 $$\\Omega $$. After the switch is closed, the time taken for the current to attain 80% of the saturation\nvalue is: [take ln 5 = 1.6] \n",
"options": [
{
"text": "0.324 s"
},
{
"text": "0.002 s"
},
{
"text": "0.103 s"
},
{
"text": "0.016 s"
}
],
"answer": "0.016 s",
"solution": "**Answer:** 0.016 s\n\nL = 10 × 10–3 H, r1 = 0.1 $$\\Omega $$
\net/L = 5 \nt = L ln 5 = 10 × 10–3 × 1.6 = 16 × 10–3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7989,
"subject": "Physics",
"question": "A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t = 0, then\nthe time at which the energy stored in the inductor reaches $$\\left( {{1 \\over n}} \\right)$$ times of its maximum value, is :",
"options": [
{
"text": "$${L \\over R}\\ln \\left( {{{\\sqrt n } \\over {\\sqrt n + 1}}} \\right)$$"
},
{
"text": "$${L \\over R}\\ln \\left( {{{\\sqrt n } \\over {\\sqrt n - 1}}} \\right)$$"
},
{
"text": "$${L \\over R}\\ln \\left( {{{\\sqrt n + 1} \\over {\\sqrt n - 1}}} \\right)$$"
},
{
"text": "$${L \\over R}\\ln \\left( {{{\\sqrt n - 1} \\over {\\sqrt n }}} \\right)$$"
}
],
"answer": "$${L \\over R}\\ln \\left( {{{\\sqrt n } \\over {\\sqrt n - 1}}} \\right)$$",
"solution": "**Answer:** $${L \\over R}\\ln \\left( {{{\\sqrt n } \\over {\\sqrt n - 1}}} \\right)$$\n\nP.E. in inductor, $$U = {1 \\over 2}L{I^2}$$
$$t = {L \\over R}\\ln \\left( {{{\\sqrt n } \\over {\\sqrt n - 1}}} \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7990,
"subject": "Physics",
"question": "An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 mH inductor and 5 $$\\Omega $$\nresistor. The ratio of the currents at time t = $$\\infty $$ and at t = 40 s is close to : (Take e2 = 7.389)",
"options": [
{
"text": "1.06"
},
{
"text": "0.84"
},
{
"text": "1.15"
},
{
"text": "1.46"
}
],
"answer": "1.06",
"solution": "**Answer:** 1.06\n\ni = i0(1 - $${e^{ - {{Rt} \\over L}}}$$)\n
Then most appropriate option is 1.06",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7991,
"subject": "Physics",
"question": "An inductor of 10 mH is connected to a 20V battery through a resistor of 10 k$$\\Omega$$ and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after 1 $$\\mu$$s is $${x \\over {100}}$$ mA. Then x is equal to ___________. (Take e$$-$$1 = 0.37)",
"options": [],
"answer": "74",
"solution": "**Answer:** 74\n\n$${I_{\\max }} = {V \\over R} = {{20V} \\over {10K\\Omega }} = 2$$ mA
A coil of inductance 1 H and resistance $$100 \\,\\Omega$$ is connected to a battery of 6 V. Determine approximately :
\n
(a) The time elapsed before the current acquires half of its steady - state value.
\n
(b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on. (Given $$\\ln 2=0.693, \\mathrm{e}^{-3 / 2}=0.25$$)
",
"options": [
{
"text": "t = 10 ms; U = 2 mJ"
},
{
"text": "t = 10 ms; U = 1 mJ"
},
{
"text": "t = 7 ms; U = 1 mJ"
},
{
"text": "t = 7 ms; U = 2 mJ"
}
],
"answer": "t = 7 ms; U = 1 mJ",
"solution": "**Answer:** t = 7 ms; U = 1 mJ\n\n
A capacitor of capacitance 500 $$\\mu$$F is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. The maximum current in LC circuit will be _______ A.
A coil has an inductance of $$2 \\mathrm{H}$$ and resistance of $$4 ~\\Omega$$. A $$10 \\mathrm{~V}$$ is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be ___________ $$\\times 10^{-2} \\mathrm{~J}$$.
To find the energy stored in the magnetic field after the current has built up to its equilibrium value, we first need to find the steady-state current in the coil.
\n
When the current reaches its equilibrium value, the coil behaves like a resistor because the back-emf induced by the changing magnetic field is zero. Ohm's law can be applied:
\n
$$I = \\frac{V}{R}$$
\n
where
\n
\n
$$I$$ is the current
\n
$$V$$ is the voltage across the coil (10 V)
\n
$$R$$ is the resistance of the coil (4 Ω)
\n
\n
Plugging in the values:
\n
$$I = \\frac{10}{4}$$\n$$I = 2.5 \\mathrm{~A}$$
\n
Now that we have the steady-state current, we can find the energy stored in the magnetic field using the formula:
\n
$$W = \\frac{1}{2}LI^2$$
\n
where
\n
\n
$$W$$ is the energy stored in the magnetic field
\n
$$L$$ is the inductance of the coil (2 H)
\n
$$I$$ is the steady-state current (2.5 A)
\n
\n
Plugging in the values:
\n
$$W = \\frac{1}{2}(2)(2.5)^2$$
\n$$W = 1(6.25)$$
\n$$W = 6.25 \\mathrm{~J}$$
\n
To express this in terms of $$10^{-2} \\mathrm{~J}$$, divide by $$10^{-2}$$:
\n
$$6.25 \\div 10^{-2} = 625$$
\n
Therefore, the energy stored in the magnetic field after the current has built up to its equilibrium value is 625$$\\times 10^{-2} \\mathrm{~J}$$.
An oscillating LC circuit consists of a $$75 ~\\mathrm{mH}$$ inductor and a $$1.2 ~\\mu \\mathrm{F}$$ capacitor. If the maximum charge to the capacitor is $$2.7 ~\\mu \\mathrm{C}$$. The maximum current in the circuit will be ___________ $$\\mathrm{mA}$$
A capacitor of capacitance $$100 \\mu \\mathrm{F}$$ is charged to a potential of $$12 \\mathrm{~V}$$ and connected to a $$6.4 \\mathrm{~mH}$$ inductor to produce oscillations. The maximum current in the circuit would be :
A series combination of resistor of resistance $$100 ~\\Omega$$, inductor of inductance $$1 ~\\mathrm{H}$$ and capacitor of capacitance $$6.25 ~\\mu \\mathrm{F}$$ is connected to an ac source. The quality factor of the circuit will be __________
$$ \\Rightarrow r \\propto {1 \\over {{v^2}}}$$ and $$r \\propto {1 \\over m}.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8005,
"subject": "Physics",
"question": "It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its\nenergy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The values of pd and pc are respectively : ",
"options": [
{
"text": "(0, 1)"
},
{
"text": "(0.89, 0.28)"
},
{
"text": "(0.28, 0.89)"
},
{
"text": "(0, 0)"
}
],
"answer": "(0.89, 0.28)",
"solution": "**Answer:** (0.89, 0.28)\n\n\n
= 0.28 ",
"topic": "Algebra",
"subtopic": "Prealgebra / Basic Algebra"
},
{
"id": 8006,
"subject": "Physics",
"question": "Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential\ndifference. The ratio of final speeds of hydrogen and helium ions is close to :",
"options": [
{
"text": "2 : 1"
},
{
"text": "1 : 2"
},
{
"text": "5 : 7"
},
{
"text": "10 : 7"
}
],
"answer": "2 : 1",
"solution": "**Answer:** 2 : 1\n\nWe know, kinetic energy K = qV\n
$$\\sqrt {{d_1}} $$ and $$\\sqrt {{d_2}} $$ are the impact parameters corresponding to scattering angles 60$$^\\circ$$ and 90$$^\\circ$$ respectively, when an $$\\alpha$$ particle is approaching a gold nucleus. For d1 = x d2, the value of x will be ____________.
Choose the correct option from the following options given below :
",
"options": [
{
"text": "In the ground state of Rutherford's model electrons are in stable equilibrium. While in Thomson's model electrons always experience a net-force."
},
{
"text": "An atom has a nearly continuous mass distribution in a Rutherford's model but has a highly non-uniform mass distribution in Thomson's model."
},
{
"text": "A classical atom based on Rutherford's model is doomed to collapse."
},
{
"text": "The positively charged part of the atom possesses most of the mass in Rutherford's model but not in Thomson's model."
}
],
"answer": "A classical atom based on Rutherford's model is doomed to collapse.",
"solution": "**Answer:** A classical atom based on Rutherford's model is doomed to collapse.\n\n
An atom based on classical theory of Rutherford's model should collapse as the electrons in continuous circular motion that is a continuously accelerated charge should emit EM waves and so should lose energy. These electrons losing energy should soon fall into heavy nucleus collapsing the whole atom.
The energy of $$\\mathrm{He}^{+}$$ ion in its first excited state is, (The ground state energy for the Hydrogen atom is $$-13.6 ~\\mathrm{eV})$$ :
",
"options": [
{
"text": "$$-13.6 ~\\mathrm{eV}$$"
},
{
"text": "$$-27.2 ~\\mathrm{eV}$$"
},
{
"text": "$$-3.4 ~\\mathrm{eV}$$"
},
{
"text": "$$-54.4 ~\\mathrm{eV}$$"
}
],
"answer": "$$-13.6 ~\\mathrm{eV}$$",
"solution": "**Answer:** $$-13.6 ~\\mathrm{eV}$$\n\nThe energy levels of a one-electron ion can be described by the formula:\n
\n$$\nE_n = -\\frac{Z^2}{n^2} \\times E_0\n$$\n
\nwhere $E_n$ is the energy of the nth level, Z is the atomic number (number of protons), n is the principal quantum number, and $E_0$ is the ground state energy of the hydrogen atom (-13.6 eV).\n
\nFor the $$\\mathrm{He}^{+}$$ ion, the atomic number Z is 2 (since helium has 2 protons). We are looking for the energy of the first excited state, which corresponds to n = 2. Plugging these values into the formula, we get:\n
\nSo, the energy of the $$\\mathrm{He}^{+}$$ ion in its first excited state is $$-13.6 ~\\mathrm{eV}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8010,
"subject": "Physics",
"question": "
If 917 $$\\mathop A\\limits^o $$ be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be ___________ $$\\mathop A\\limits^o $$.
The energy difference formula for transitions between energy levels in a hydrogen atom, which is given by
\n
$\n\\Delta E = -13.6 \\, \\text{eV} \\times \\left(\\frac{1}{n_1^2} - \\frac{1}{n_2^2}\\right)\n$
\n
where ($n_1$) and ($n_2$) are the initial and final energy levels, respectively. For the Lyman series, the electron transitions to the ground state (($n_1$ = 1)), and for the Balmer series, the electron transitions to the first excited state (($n_1$ = 2)).
\n
From the given information, we have the lowest wavelength in the Lyman series, ($\\lambda_1 = 917 \\, \\text{Å}$). Therefore, the energy difference ($\\Delta E$) for the Lyman series is
\n
$\n\\Delta E = \\frac{hc}{\\lambda_1}\n$
\n
where (h) is Planck's constant and (c) is the speed of light.
\n
Similarly, for the Balmer series, the energy difference ($\\Delta E$) is
When a metal target is bombarded with high-energy electrons, the phenomenon known as X-ray emission occurs. This is due to the excitation of the inner-shell electrons in the metal atoms by the high-energy electrons. When these inner-shell electrons drop back to their original energy levels, they emit energy in the form of X-ray photons.
The ratio of wavelength of spectral lines $$\\mathrm{H}_{\\alpha}$$ and $$\\mathrm{H}_{\\beta}$$ in the Balmer series is $$\\frac{x}{20}$$. The value of $$x$$ is _________.
The Balmer series corresponds to electronic transitions in a hydrogen atom that terminate in the second (n=2) energy level. The spectral lines in the Balmer series are often labeled according to a Greek letter scheme, with H$_\\alpha$ corresponding to the n=3 to n=2 transition, H$_\\beta$ corresponding to the n=4 to n=2 transition, and so on.
\n
The wavelength of a spectral line in the Balmer series can be calculated using the Rydberg formula:
where $R_H$ is the Rydberg constant for hydrogen, $n$ is the principal quantum number corresponding to the initial energy level, and $\\lambda$ is the wavelength of the spectral line.
\n
Using this formula, the wavelength of the H$_\\alpha$ line is:
Statement I : Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford's model.
\n
Statement II : An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford's model.
\n
In the light of the above statements, choose the most appropriate from the options given below
",
"options": [
{
"text": "Both Statement I and Statement II are true\n"
},
{
"text": "Statement I is true but Statement II is false\n"
},
{
"text": "Statement I is false but Statement II is true\n"
},
{
"text": "Both statement I and statement II are false"
}
],
"answer": "Statement I is true but Statement II is false\n",
"solution": "**Answer:** Statement I is true but Statement II is false\n\n\n
According to Rutherford atomic model, most of mass of atom and all its positive charge is concentrated in tiny nucleus & electron revolve around it.
\n
According to Thomson atomic model, atom is spherical cloud of positive charge with electron embedded in it.
\n
Hence, Statement I is true but statement II false.
In an alpha particle scattering experiment distance of closest approach for the $$\\alpha$$ particle is $$4.5 \\times 10^{-14} \\mathrm{~m}$$. If target nucleus has atomic number 80 , then maximum velocity of $$\\alpha$$-particle is __________ $$\\times 10^5 \\mathrm{~m} / \\mathrm{s}$$ approximately.
\n
($$\\frac{1}{4 \\pi \\epsilon_0}=9 \\times 10^9 \\mathrm{SI}$$ unit, mass of $$\\alpha$$ particle $$=6.72 \\times 10^{-27} \\mathrm{~kg}$$)
Therefore the energy required to remove electron from \n
$$n = 2$$ is $$+3.4eV.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8016,
"subject": "Physics",
"question": "The wavelengths involved in the spectrum of deuterium $$\\left( {{}_1^2\\,D} \\right)$$ are slightly different from that of hydrogen spectrum, because ",
"options": [
{
"text": "the size of the two nuclei are different "
},
{
"text": "the nuclear forces are different in the two cases "
},
{
"text": "the masses of the two nuclei are different "
},
{
"text": "the attraction between the electron and the nucleus is different in the two cases "
}
],
"answer": "the masses of the two nuclei are different ",
"solution": "**Answer:** the masses of the two nuclei are different \n\nThe wavelength of spectrum is given by\n
For different $$M,R$$ is different and therefore $$\\lambda $$ is different ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8017,
"subject": "Physics",
"question": "Which of the following atoms has the lowest ionization potential ? ",
"options": [
{
"text": "$${}_7^{14}N$$ "
},
{
"text": "$${}_{55}^{133}\\,Cs$$ "
},
{
"text": "$${}_{18}^{40}\\,Ar$$ "
},
{
"text": "$${}_8^{16}\\,O$$ "
}
],
"answer": "$${}_{55}^{133}\\,Cs$$ ",
"solution": "**Answer:** $${}_{55}^{133}\\,Cs$$ \n\nThe ionisation potential increases from left to right in a period and decreases from top to bottom in a group. Therefore ceasium will have the lowest ionisation potential. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8018,
"subject": "Physics",
"question": "If the binding energy of the electron in a hydrogen atom is $$13.6eV,$$ the energy required to remove the electron from the first excited state of $$L{i^{ + + }}$$ is ",
"options": [
{
"text": "$$30.6$$ $$eV$$ "
},
{
"text": "$$13.6$$ $$eV$$ "
},
{
"text": "$$3.4$$ $$eV$$ "
},
{
"text": "$$122.4$$ $$eV$$ "
}
],
"answer": "$$30.6$$ $$eV$$ ",
"solution": "**Answer:** $$30.6$$ $$eV$$ \n\n$${E_n} = - {{13.6} \\over {{n^2}}}{Z^2}eV/$$atom\n
For lithium ion $$Z=3;$$ $$n=2$$ (for first excited state)\n
$${E_n} = - {{13.6} \\over {{2^2}}} \\times {3^2} = - 30.6eV$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8019,
"subject": "Physics",
"question": "Which of the following transitions in hydrogen atoms emit photons of highest frequency ?",
"options": [
{
"text": "$$n = 1$$ to $$n=2$$ "
},
{
"text": "$$n = 2$$ to $$n=6$$ "
},
{
"text": "$$n = 6$$ to $$n=2$$ "
},
{
"text": "$$n = 2$$ to $$n=1$$ "
}
],
"answer": "$$n = 2$$ to $$n=1$$ ",
"solution": "**Answer:** $$n = 2$$ to $$n=1$$ \n\nWe have no find the frequency of emitted photons. For emission of photons the transition must take place from a higher energy level to a lower energy level which are given only in options $$(c)$$ and $$(d)$$.\n
$$\\therefore$$ $${v_1} > {v_2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8020,
"subject": "Physics",
"question": "Suppose an electron is attracted towards the origin by a force $${k \\over r}$$ where $$'k'$$ is a constant and $$'r'$$ is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the $${n^{th}}$$ orbital of the electron is found to be $$'{r_n}'$$ and the kinetic energy of the electron to be $$'{T_n}'.$$ \n
$$ \\Rightarrow m{v^2} = $$ constant $$ \\Rightarrow $$ kinetic energy is constant\n
$$ \\Rightarrow T$$ is independent of $$n.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8021,
"subject": "Physics",
"question": "The transition from the state $$n=4$$ to $$n=3$$ in a hydrogen like atom result in ultra violet radiation. Infrared radiation will be obtained in the transition from : ",
"options": [
{
"text": "$$3 \\to 2$$ "
},
{
"text": "$$4 \\to 2$$"
},
{
"text": "$$5 \\to 4$$"
},
{
"text": "$$2 \\to 1$$"
}
],
"answer": "$$5 \\to 4$$",
"solution": "**Answer:** $$5 \\to 4$$\n\nIt is given that transition from the state $$n=4$$ to $$n=3$$ in a hydrogen like atom result in ultraviolet radiation. For infrared radiation the energy gap should be less. The only option is $$5$$ $$ \\to 4.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8022,
"subject": "Physics",
"question": "Energy required for the electron excitation in $$L{i^{ + + }}$$ from the first to the third Bohr orbit is : ",
"options": [
{
"text": "$$36.3$$ $$eV$$ "
},
{
"text": "$$108.8$$ $$eV$$"
},
{
"text": "$$122.4$$ $$eV$$ "
},
{
"text": "$$12.1$$ $$eV$$ "
}
],
"answer": "$$108.8$$ $$eV$$",
"solution": "**Answer:** $$108.8$$ $$eV$$\n\nEnergy of excitation, \n
$$ \\Rightarrow \\Delta E = 13.6{\\left( 3 \\right)^2}\\left( {{1 \\over {{1^2}}} - {1 \\over {{3^2}}}} \\right) = 108.8\\,eV$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8023,
"subject": "Physics",
"question": "Hydrogen atom is excited from ground state to another state with principal quantum number equal to $$4.$$ Then the number of spectral lines in the emission spectra will be : ",
"options": [
{
"text": "$$2$$ "
},
{
"text": "$$3$$ "
},
{
"text": "$$5$$ "
},
{
"text": "$$6$$ "
}
],
"answer": "$$6$$ ",
"solution": "**Answer:** $$6$$ \n\nThe possible number of the spectral lines is given\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8024,
"subject": "Physics",
"question": "A diatomic molecule is made of two masses $${m_1}$$ and $${m_2}$$ which are separated by a distance $$r.$$ If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by: ($$n$$ is an integer) ",
"options": [
{
"text": "$${{{{\\left( {{m_1} + {m_2}} \\right)}^2}{n^2}{h^2}} \\over {2m_1^2m_2^2{r^2}}}$$ "
},
{
"text": "$${{{n^2}{h^2}} \\over {2\\left( {{m_1} + {m_2}} \\right){r^2}}}$$ "
},
{
"text": "$${{2{n^2}{h^2}} \\over {\\left( {{m_1} + {m_2}} \\right){r^2}}}$$ "
},
{
"text": "$${{\\left( {{m_1} + {m_2}} \\right){n^2}{h^2}} \\over {2{m_1}{m_2}{r^2}}}$$ "
}
],
"answer": "$${{\\left( {{m_1} + {m_2}} \\right){n^2}{h^2}} \\over {2{m_1}{m_2}{r^2}}}$$ ",
"solution": "**Answer:** $${{\\left( {{m_1} + {m_2}} \\right){n^2}{h^2}} \\over {2{m_1}{m_2}{r^2}}}$$ \n\nThe energy of the system of two atoms of diatomic\n
[ As $$n > > 1,$$ so we can assume n - 1 = n and 2n - 1 = 2n ]\n
$$ \\Rightarrow $$ $$\\upsilon $$ $$ \\propto $$ $$\\left[ {{{2n} \\over {{n^2}{{\\left( {n } \\right)}^2}}}} \\right]$$ $$ \\propto $$ $$\\left[ {{{2n} \\over {{n^4}}}} \\right]$$ $$ \\propto $$ $$\\left[ {{2 \\over {{n^3}}}} \\right]$$\n\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8026,
"subject": "Physics",
"question": "The radiation corresponding to $$3 \\to 2$$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field $$3 \\times {10^{ - 4}}\\,T.$$ If the radius of the larger circular path followed by these electrons is $$10.0$$ $$mm$$, the work function of the metal is close to: ",
"options": [
{
"text": "$$1.8$$ $$eV$$ "
},
{
"text": "$$1.1$$ $$eV$$ "
},
{
"text": "$$0.8$$ $$eV$$ "
},
{
"text": "$$1.6$$ $$eV$$ "
}
],
"answer": "$$1.1$$ $$eV$$ ",
"solution": "**Answer:** $$1.1$$ $$eV$$ \n\nRadius of circular path followed by electron is given by, \n
Then, $${\\lambda _1} = {\\lambda _2} = 4{\\lambda _3} = 9{\\lambda _4}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8028,
"subject": "Physics",
"question": "As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom/ion : ",
"options": [
{
"text": "kinetic energy decreases, potential energy increases but total energy remains same "
},
{
"text": "kinetic energy and total energy decrease but potential energy increases "
},
{
"text": "its kinetic energy increases but potential energy and total energy decrease"
},
{
"text": "kinetic energy, potential energy and total energy decrease "
}
],
"answer": "its kinetic energy increases but potential energy and total energy decrease",
"solution": "**Answer:** its kinetic energy increases but potential energy and total energy decrease\n\n$$U = - K{{z{e^2}} \\over r};\\,\\,T.E = {k \\over 2}{{ze^2} \\over r}$$\n
$$K.E = {k \\over 2}{{z{e^2}} \\over r}.$$ Here $$r$$ decreases",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8029,
"subject": "Physics",
"question": "A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :",
"options": [
{
"text": "2"
},
{
"text": "3"
},
{
"text": "4"
},
{
"text": "5"
}
],
"answer": "4",
"solution": "**Answer:** 4\n\nEnergy released when hydrogen atom makes transition from n = 2 to n = 1 is, \n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8031,
"subject": "Physics",
"question": "According to Bohr’s theory, the time averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the nth orbit is proportional to : (n = principal quantum number)\n",
"options": [
{
"text": "$${n^{ - 4}}$$ "
},
{
"text": "$${n^{ - 5}}$$ "
},
{
"text": "n$$-$$3 "
},
{
"text": "n$$-$$2"
}
],
"answer": "$${n^{ - 5}}$$ ",
"solution": "**Answer:** $${n^{ - 5}}$$ \n\nMagnetic field at the center of neucleus of H-atom \n
$$ \\Rightarrow $$$$\\,\\,\\,$$ B $$ \\propto $$ n $$-$$5",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8032,
"subject": "Physics",
"question": "Muon ($$\\mu $$$$-$$) is a negatively charged (|q| = |e|) particle with a mass m$$\\mu $$ = 200 me, where me is the mass of the electron and e is the electronic charge. If $$\\mu $$$$-$$ is bond to a proton to form a hydrogen like atom, identify the correct statements.\n (A) Radis of the muonic orbit is 200 times smaller than that of the electron. \n (B) The speed of the $$\\mu $$$$-$$ in the nth orbit is $${1 \\over {200}}$$ times that of the electron in the nth orbit.\n (C) The ionization energy of muonic atom is 200 timesmore than of an hydroen atom. \n (D) The momentum of the muon in the nth orbit is 200 times more than that of the electron. ",
"options": [
{
"text": "(A), (B), (D)"
},
{
"text": "(A), (C), (D)"
},
{
"text": "(B), (D)"
},
{
"text": "(C), (D)"
}
],
"answer": "(A), (C), (D)",
"solution": "**Answer:** (A), (C), (D)\n\n
Only, m$$\\mu$$ = 200 me rest are same. So, statement (C) is correct: The ionisation energy of muonic atom is 200 times more than that of an hydrogen atom.
\n
(iv) Since Momentum $$\\propto$$ Energy. So, statement (D) is correct: The momentum of the muon in he nth orbit is 200 times more than that of the electron.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8033,
"subject": "Physics",
"question": "The energy required to remove the electron from a singly ionized Helium atom is $$2.2$$ times the energies required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is :",
"options": [
{
"text": "$$20$$ $$eV$$ "
},
{
"text": "$$34$$ $$eV$$ "
},
{
"text": "$$79$$ $$eV$$ "
},
{
"text": "$$109$$ $$eV$$ "
}
],
"answer": "$$79$$ $$eV$$ ",
"solution": "**Answer:** $$79$$ $$eV$$ \n\nEnergy required to remove e$$-$$ from singly ionized Helium atom\n
$$\\therefore\\,\\,\\,\\,$$ Total energy required to ionize Helium atom completely \n
= (54.4 + 24.72) eV\n
= 79.12 eV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8034,
"subject": "Physics",
"question": "If the series limit frequency of the Lyman series is $${\\nu _L}$$, then the series limit frequency of the Pfund series is: ",
"options": [
{
"text": "$${\\nu _L}/25$$ "
},
{
"text": "$$25{\\nu _L}$$"
},
{
"text": "$$16{\\nu _L}$$ "
},
{
"text": "$${\\nu _L}/16$$ "
}
],
"answer": "$${\\nu _L}/25$$ ",
"solution": "**Answer:** $${\\nu _L}/25$$ \n\nNote :\n
(1) In Lyman Series, transition happens in n = 1 state \n from n = 2, 3, . . . . . $$ \\propto $$\n
(2) In Balmer Series, transition happens in n = 2 state \n from n = 3, 4, . . . . . $$ \\propto $$\n
(3) In Paschen Series, transition happens in n = 3 state\n from n = 4, 5, . . . . . $$ \\propto $$\n
(4) In Bracktt Series, transition happens in n = 4 state\n from n = 5, 6 . . . . . . $$ \\propto $$\n
(5) In Pfund Series, transition happens in n = 5 state\n from n = 6, 7, . . . . $$ \\propto $$\n
$$\\therefore\\,\\,\\,$$ $$\\gamma $$P = $${{cRZ{}^2} \\over {25}}$$ = $${{{\\gamma _L}} \\over {25}}$$ [as $$\\gamma $$L = cRZ2]",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8035,
"subject": "Physics",
"question": "An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let\n$${\\lambda _n}$$, $${\\lambda _g}$$ be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let\n$${\\Lambda _n}$$ be the wavelength of the emitted photon in the transition from the nth state to the ground state. For large n, (A, B are constants)",
"options": [
{
"text": "$${\\Lambda _n} \\approx A + {B \\over {\\lambda _n^2}}$$"
},
{
"text": "$${\\Lambda _n} \\approx A + B{\\lambda _n}$$ "
},
{
"text": "$$\\Lambda _n^2 \\approx A + B\\lambda _n^2$$ "
},
{
"text": "$$\\Lambda _n^2 \\approx \\lambda$$"
}
],
"answer": "$${\\Lambda _n} \\approx A + {B \\over {\\lambda _n^2}}$$",
"solution": "**Answer:** $${\\Lambda _n} \\approx A + {B \\over {\\lambda _n^2}}$$\n\nWe know, \n
Wavelength of emitted photon from n2 state to n1 state is \n
Let A = $${1 \\over {R{Z^2}}}$$ and B = $${1 \\over {{K^2}R{Z^2}}}$$\n
$$ \\Rightarrow $$$$\\,\\,\\,$$ $$\\Lambda $$n = A + $${B \\over {\\lambda _n^2}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8036,
"subject": "Physics",
"question": "The electron in a hydrogen atom first jumps from the third excited state to the second excited state and\nsubsequently to the first excited state. The ratio of the respective wavelengths, $${{{\\lambda _1}} \\over {{\\lambda _2}}}$$, of the photons emitted\nin this process is : ",
"options": [
{
"text": "$${{22} \\over 5}$$"
},
{
"text": "$${7 \\over 5}$$"
},
{
"text": "$${9 \\over 7}$$"
},
{
"text": "$${{20} \\over 7}$$"
}
],
"answer": "$${{20} \\over 7}$$",
"solution": "**Answer:** $${{20} \\over 7}$$\n\nn = 1 (Ground state)\n n = 2 (First excitate state)\n n = 3 (Second excitate state)\n n = 4 (Third excitate state)\n
\n$$ \\therefore $$ m = 2, and now $$13.6 \\times 4\\left( {{1 \\over 4} - {1 \\over {{n^2}}}} \\right) = {{hc} \\over {108.5 \\times {{10}^{ - 9}}}}$$ \nn = 5",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8039,
"subject": "Physics",
"question": "In Li+ +, electron in first Bohr orbit is excited to a level by a radiation of wavelength $$\\lambda $$. When the ion gets\ndeexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral\nlines are observed. What is the value of $$\\lambda $$?\n (Given : H = 6.63 × 10–34 Js; c = 3 × 108\n ms\n–1)",
"options": [
{
"text": "10.8 nm"
},
{
"text": "12.3 nm"
},
{
"text": "9.4 nm"
},
{
"text": "11.4 nm"
}
],
"answer": "10.8 nm",
"solution": "**Answer:** 10.8 nm\n\n\n$${{hc} \\over \\lambda } = 13.6\\,ev(g)\\left\\{ {1 - {1 \\over {16}}} \\right\\}$$
\n$${\\lambda _2} = {{5 \\times 660 \\times 64} \\over {36 \\times 12}} = 489\\,nm$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8041,
"subject": "Physics",
"question": "Radiation coming from transitions\nn = 2 to n = 1 of hydrogen atoms fall on He+\nions in n = 1 and n = 2 states. The possible\ntransition of helium ions as they absorb energy\nfrom the radiation is :",
"options": [
{
"text": "n = 1 $$ \\to $$ n = 4"
},
{
"text": "n = 2 $$ \\to $$ n = 5"
},
{
"text": "n = 2 $$ \\to $$ n = 4"
},
{
"text": "n = 2 $$ \\to $$ n = 3"
}
],
"answer": "n = 2 $$ \\to $$ n = 4",
"solution": "**Answer:** n = 2 $$ \\to $$ n = 4\n\nEnergy released for tension n = 2 to n = 1 of hydrogen atom\n
\n(A) n = 1 to n = 4 \n $$E = 13.6 \\times {2^2} \\times \\left( {{1 \\over {{1^2}}} - {1 \\over {{4^2}}}} \\right) = 13.6 \\times {{15} \\over 4}eV$$
\n(B) n = 2 to n = 4 \n $$E = 13.6 \\times {2^2} \\times \\left( {{1 \\over {{2^2}}} - {1 \\over {{4^2}}}} \\right) = 13.6 \\times {{3} \\over 4}eV$$
\n(C) n = 2 to n = 5 \n $$E = 13.6 \\times {2^2} \\times \\left( {{1 \\over {{2^2}}} - {1 \\over {{5^2}}}} \\right) = 13.6 \\times {{21} \\over 25}eV$$
\n(D) n = 2 to n = 3 \n $$E = 13.6 \\times {2^2} \\times \\left( {{1 \\over {{2^2}}} - {1 \\over {{3^2}}}} \\right) = 13.6 \\times {{5} \\over 9}eV$$
\nSo, possible transition is n = 2 $$\\to$$ n = 4",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8042,
"subject": "Physics",
"question": "A particle of mass m moves in a circular orbit in a central potential field U(r) = $${1 \\over 2}$$ kr2. If Bohr 's\nquantization conditions are applied, radii of possible orbitls and energy levels vary with quantum number n as : \n",
"options": [
{
"text": "rn $$ \\propto $$ $$\\sqrt n $$, En $$ \\propto $$ n"
},
{
"text": "rn $$ \\propto $$ $$\\sqrt n $$, En $$ \\propto $$ $${1 \\over n}$$"
},
{
"text": "rn $$ \\propto $$ n, En $$ \\propto $$ n"
},
{
"text": "rn $$ \\propto $$ n2, En $$ \\propto $$ $${1 \\over {{n^2}}}$$"
}
],
"answer": "rn $$ \\propto $$ $$\\sqrt n $$, En $$ \\propto $$ n",
"solution": "**Answer:** rn $$ \\propto $$ $$\\sqrt n $$, En $$ \\propto $$ n\n\nForce due to this field, F = $$ - {{\\partial U} \\over {\\partial r}}$$\n
For circular orbit, $${{m{v^2}} \\over r}$$ = -kr\n
$$ \\Rightarrow $$ v $$ \\propto $$ r ..... (1)\n
Fron Bohr’s quantization condition\n
mvr = $${{nh} \\over {2\\pi }}$$ .....(2)\n
From (1) and (2),\n
$${r_n}$$ $$ \\propto $$ $${n^{{1 \\over 2}}}$$\n
Given, U(r) = $${1 \\over 2}$$kr2\n
$$ \\Rightarrow $$ En = $$ - {1 \\over 2}U\\left( r \\right)$$ = $$ - {1 \\over 4}k{r^2}$$\n
$$ \\Rightarrow $$ En $$ \\propto $$ n",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8043,
"subject": "Physics",
"question": "In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $$\\lambda $$. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be: ",
"options": [
{
"text": "$${{25} \\over {16}}$$ $$\\lambda $$"
},
{
"text": "$${{27} \\over {20}}$$ $$\\lambda $$"
},
{
"text": "$${{16} \\over {25}}$$ $$\\lambda $$"
},
{
"text": "$${{20} \\over {27}}$$ $$\\lambda $$"
}
],
"answer": "$${{20} \\over {27}}$$ $$\\lambda $$",
"solution": "**Answer:** $${{20} \\over {27}}$$ $$\\lambda $$\n\nFor M $$ \\to $$ L steel\n
$$\\lambda ' = {{20} \\over {27}}\\lambda $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8044,
"subject": "Physics",
"question": "A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980$$\\mathop A\\limits^ \\circ $$. The radius of the atom in the excited state, in terms of Bohr radius a0 will be : (hc = 12500 eV$$\\mathop A\\limits^ \\circ $$)",
"options": [
{
"text": "4a0"
},
{
"text": "9a0"
},
{
"text": "25a0"
},
{
"text": "16a0"
}
],
"answer": "16a0",
"solution": "**Answer:** 16a0\n\nEnergy of photon $$ = {{12500} \\over {980}} = 12.75eV$$\n
$$ \\therefore $$ Electron will excite to n = 4\n
Since 'R' $$ \\propto $$ n2\n
$$ \\therefore $$ Radius of atom will be 16a0",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8045,
"subject": "Physics",
"question": "A He+ ion is in its first excited state. Its\nionization energy is :-",
"options": [
{
"text": "13.60 eV"
},
{
"text": "6.04 eV"
},
{
"text": "48.36 eV"
},
{
"text": "54.40 eV"
}
],
"answer": "13.60 eV",
"solution": "**Answer:** 13.60 eV\n\n$$T.E. = - \\left( {13.6} \\right)\\left( {{{{Z^2}} \\over {{n^2}}}} \\right)eV$$
\nz = n = 2
\n$$ \\Rightarrow $$ Ionisation energy = –T.E.\n= 13.6 eV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8046,
"subject": "Physics",
"question": "The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 $$ \\times $$ 10-16 s. The frequency of revolution of the electron in its first excited state (in s-1) is :\n",
"options": [
{
"text": "5.6 $$ \\times $$ 1012"
},
{
"text": "1.6 $$ \\times $$ 1014"
},
{
"text": "7.8 $$ \\times $$ 1014"
},
{
"text": "6.2 $$ \\times $$ 1015"
}
],
"answer": "7.8 $$ \\times $$ 1014",
"solution": "**Answer:** 7.8 $$ \\times $$ 1014\n\nTime period of revolution of electron in nth orbit\n
= $${{20} \\over {27}}$$ $$ \\times $$ 6561 = 4860 $$\\mathop A\\limits^o $$ = 486 nm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8048,
"subject": "Physics",
"question": "The energy required to ionise a hydrogen like\nion in its ground state is 9 Rydbergs. What is\nthe wavelength of the radiation emitted when\nthe electron in this ion jumps from the second\nexcited state to the ground state ?",
"options": [
{
"text": "35.8 nm"
},
{
"text": "11.4 nm"
},
{
"text": "8.6 nm"
},
{
"text": "24.2 nm"
}
],
"answer": "11.4 nm",
"solution": "**Answer:** 11.4 nm\n\nSo, ionisation energy = (13.6 Z2) eV\n
$$ \\Rightarrow $$ $$\\nu $$ $$ \\propto $$ $${1 \\over {{n^3}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8050,
"subject": "Physics",
"question": "In the line spectra of hydrogen atoms, difference between the largest and the shortest wavelengths of the Lyman series is 304 $$\\mathop A\\limits^0 $$. The corresponding difference for the Paschan series in $$\\mathop A\\limits^0 $$ is :\n___________.",
"options": [],
"answer": "10553",
"solution": "**Answer:** 10553\n\nFor Lyman series :\n
= $$\\left( {{{144} \\over 7} - 9} \\right) \\times 912$$ = 10553 $$\\mathop A\\limits^0 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8051,
"subject": "Physics",
"question": "A particle of mass 200 MeV/c2 collides with a\nhydrogen atom at rest. Soon after the collision\nthe particle comes to rest, and the atom\nrecoils and goes to its first excited state. The\ninitial kinetic energy of the particle (in eV) is\n $${N \\over 4}$$. The value of N is :\n (Given the mass of the hydrogen atom to be 1\nGeV/c2) ______ .",
"options": [],
"answer": "51",
"solution": "**Answer:** 51\n\nGiven, mparticle = 200 MeV/c2 = m(Assume)\n
This lost energy is used by the hydrogen atom to move from ground state to the first excited state. We know the the energy required by the hydrogen atom to move from ground state to first excited state is 10.2 eV.\n
$$ \\therefore $$ N = 51",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8052,
"subject": "Physics",
"question": "According to Bohr atomic model, in which of the following transitions will the frequency be maximum?",
"options": [
{
"text": "n = 2 to n = 1"
},
{
"text": "n = 3 to n = 2"
},
{
"text": "n = 4 to n = 3"
},
{
"text": "n = 5 to n = 4"
}
],
"answer": "n = 2 to n = 1",
"solution": "**Answer:** n = 2 to n = 1\n\nLet, nf, ni be the final and initial orbit.
$$\\therefore$$ Frequency will be maximum for transition n = 2 to n = 1.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8053,
"subject": "Physics",
"question": "The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is :",
"options": [
{
"text": "194.8 nm"
},
{
"text": "490.7 nm"
},
{
"text": "913.3 nm"
},
{
"text": "121.8 nm"
}
],
"answer": "121.8 nm",
"solution": "**Answer:** 121.8 nm\n\n$$\\Delta$$E = 10.2 eV
$$ \\simeq $$ 121.8 nm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8054,
"subject": "Physics",
"question": "If $$\\lambda$$1 and $$\\lambda$$2 are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of $$\\lambda$$1 : $$\\lambda$$2 is :",
"options": [
{
"text": "7 : 135"
},
{
"text": "7 : 108"
},
{
"text": "1 : 9"
},
{
"text": "1 : 3"
}
],
"answer": "7 : 135",
"solution": "**Answer:** 7 : 135\n\nFor Lyman series\n
$$ \\Rightarrow $$ $${{{\\lambda _1}} \\over {{\\lambda _2}}} = {7 \\over {135}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8055,
"subject": "Physics",
"question": "The first three spectral lines of H-atom in the Balmer series are given $$\\lambda$$1, $$\\lambda$$2, $$\\lambda$$3 considering the Bohr atomic model, the wave lengths of first and third spectral lines $$\\left( \\frac{\\lambda_{1} }{\\lambda_{3} } \\right) $$ are related by a factor of approximately 'x' $$\\times$$ 10$$-$$1.
The value of x, to the nearest integer, is _________.",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\nFor 1st line
$$x \\approx 15$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 8056,
"subject": "Physics",
"question": "If an electron is moving in the nth orbit of the hydrogen atom, then its velocity (vn) for the nth orbit is given as :",
"options": [
{
"text": "$${v_n} \\propto {1 \\over n}$$"
},
{
"text": "vn $$ \\propto $$ n2"
},
{
"text": "vn $$ \\propto $$ n"
},
{
"text": "$${v_n} \\propto {1 \\over {{n^2}}}$$"
}
],
"answer": "$${v_n} \\propto {1 \\over n}$$",
"solution": "**Answer:** $${v_n} \\propto {1 \\over n}$$\n\nWe know velocity of electron in nth shell of hydrogen atom is given by
$$v = {{2\\pi kZ{e^2}} \\over {nh}}$$
$$ \\therefore $$ $$v \\propto {1 \\over n}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8057,
"subject": "Physics",
"question": "Which level of the single ionized carbon has the same energy as the ground state energy of hydrogen atom?",
"options": [
{
"text": "8"
},
{
"text": "6"
},
{
"text": "1"
},
{
"text": "4"
}
],
"answer": "6",
"solution": "**Answer:** 6\n\n$${E_n} = - 13.6{{{Z^2}} \\over {{n^2}}}$$
$$ \\Rightarrow $$ n = 6",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8058,
"subject": "Physics",
"question": "The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra is lying in the visible region?",
"options": [
{
"text": "Brackett series"
},
{
"text": "Balmer series"
},
{
"text": "Paschen series"
},
{
"text": "Lyman series"
}
],
"answer": "Balmer series",
"solution": "**Answer:** Balmer series\n\nBalmer series of hydrogen atomic spectrum is lying in the visible region, when electron jumps from a higher energy level to n = 2 orbit.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8059,
"subject": "Physics",
"question": "A particle of mass m moves in a circular orbit in a central potential field U(r) = U0r4. If Bohr's quantization conditions are applied, radii of possible orbitals rn vary with $${n^{{1 \\over \\alpha }}}$$, where $$\\alpha$$ is ____________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n$$\\overrightarrow F = - {{d\\overrightarrow u } \\over {dr}}$$
So the value of $$\\alpha = 3$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8060,
"subject": "Physics",
"question": "Imagine that the electron in a hydrogen atom is replaced by a muon ($$\\mu$$). The mass of muon particle is 207 times that of an electron and charge is equal to the charge of an electron. The ionization potential of this hydrogen atom will be :",
"options": [
{
"text": "13.6 eV"
},
{
"text": "2815.2 eV"
},
{
"text": "331.2 eV"
},
{
"text": "27.2 eV"
}
],
"answer": "2815.2 eV",
"solution": "**Answer:** 2815.2 eV\n\nmm = 207 me\n
In hydrogen atom one electron present. Now that the electron in hydrogen atom is replaced by a muon ($$\\mu$$).\n
We know, Energy(E) = $$ - {{{e^4}m} \\over {8\\varepsilon _0^2{n^2}{h^2}}}$$\n
Em = $$ - {{{e^4}\\left( {207} \\right){m_e}} \\over {8\\varepsilon _0^2{n^2}{h^2}}}$$\n
= 13.6 $$ \\times $$ 207 = 2815.2 eV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8061,
"subject": "Physics",
"question": "In Bohr's atomic model, the electron is assumed to revolve in a circular orbit of radius 0.5 $$\\mathop A\\limits^o $$. If the speed of electron is 2.2 $$\\times$$ 166 m/s, then the current associated with the electron will be _____________ $$\\times$$ 10$$-$$2 mA. [Take $$\\pi$$ as $${{22} \\over 7}$$]",
"options": [],
"answer": "112",
"solution": "**Answer:** 112\n\n$$I = {e \\over T} = {{e\\omega } \\over {2\\pi }} = {{eV} \\over {2\\pi r}}$$
= 112 $$\\times$$ 10$$-$$2 mA",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8062,
"subject": "Physics",
"question": "The K$$\\alpha$$ X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atoms with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be __________ keV. (Round off to the nearest integer)
= 10 keV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8063,
"subject": "Physics",
"question": "A particular hydrogen like ion emits radiation of frequency 2.92 $$\\times$$ 1015 Hz when it makes transition from n = 3 to n = 1. The frequency in Hz of radiation emitted in transition from n = 2 to n = 1 will be :",
"options": [
{
"text": "0.44 $$\\times$$ 1015"
},
{
"text": "6.57 $$\\times$$ 1015"
},
{
"text": "4.38 $$\\times$$ 1015"
},
{
"text": "2.46 $$\\times$$ 1015"
}
],
"answer": "2.46 $$\\times$$ 1015",
"solution": "**Answer:** 2.46 $$\\times$$ 1015\n\n$$n{f_1} = k\\left( {{1 \\over 1} - {1 \\over {{3^2}}}} \\right)$$
$${{{f_1}} \\over {{f_2}}} = {{8/9} \\over {3/4}} \\Rightarrow {f_2} = 2.46 \\times {10^{15}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8064,
"subject": "Physics",
"question": "X different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are exited to states with principal quantum number n = 6 ? The value of X is ______________.",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\nNo. of different wavelengths = $${{n(n - 1)} \\over 2}$$
Statement I : In hydrogen atom, the frequency of radiation emitted when an electron jumps from lower energy orbit (E1) to higher energy orbit (E2), is given as hf = E1 $$-$$ E2
\n
Statement II : The jumping of electron from higher energy orbit (E2) to lower energy orbit (E1) is associated with frequency of radiation given as f = (E2 $$-$$ E1)/h
\n
This condition is Bohr's frequency condition.
\n
In the light of the above statements, choose the correct answer from the options given below :
",
"options": [
{
"text": "Both Statement I and Statement II are true."
},
{
"text": "Both Statement I and Statement II are false."
},
{
"text": "Statement I is correct but Statement II is false."
},
{
"text": "Statement I is incorrect but Statement II is true."
}
],
"answer": "Statement I is incorrect but Statement II is true.",
"solution": "**Answer:** Statement I is incorrect but Statement II is true.\n\n
Radiation is not emitted but absorbed when an electron jumps from low energy to high energy.
A beam of monochromatic light is used to excite the electron in Li+ + from the first orbit to the third orbit. The wavelength of monochromatic light is found to be x $$\\times$$ 10$$-$$10 m. The value of x is ___________.
In Bohr's atomic model of hydrogen, let K, P and E are the kinetic energy, potential energy and total energy of the electron respectively. Choose the correct option when the electron undergoes transitions to a higher level :
",
"options": [
{
"text": "All K, P and E increase."
},
{
"text": "K decreases, P and E increase."
},
{
"text": "P decreases, K and E increase."
},
{
"text": "K increases, P and E decrease."
}
],
"answer": "K decreases, P and E increase.",
"solution": "**Answer:** K decreases, P and E increase.\n\n
A hydrogen atom in its first excited state absorbs a photon of energy x $$\\times$$ 10$$-$$2 eV and excited to a higher energy state where the potential energy of electron is $$-$$1.08 eV. The value of x is ______________.
Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength $$\\lambda$$. The value of principal quantum number '$$n$$' of the excited state will be : ($$\\mathrm{R}:$$ Rydberg constant)
In the hydrogen spectrum, $$\\lambda$$ be the wavelength of first transition line of Lyman series. The wavelength difference will be \"a$$\\lambda$$'' between the wavelength of $$3^{\\text {rd }}$$ transition line of Paschen series and that of $$2^{\\text {nd }}$$ transition line of Balmer series where $$\\mathrm{a}=$$ ___________.
Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.
An electron of a hydrogen like atom, having $$Z=4$$, jumps from $$4^{\\text {th }}$$ energy state to $$2^{\\text {nd }}$$ energy state. The energy released in this process, will be :
\n
(Given Rch = $$13.6~\\mathrm{eV}$$)
\n
Where R = Rydberg constant
\n
c = Speed of light in vacuum
\n
h = Planck's constant
",
"options": [
{
"text": "$$10.5 ~\\mathrm{eV}$$"
},
{
"text": "$$40.8 ~\\mathrm{eV}$$"
},
{
"text": "$$13.6 ~\\mathrm{eV}$$"
},
{
"text": "$$3.4 ~\\mathrm{eV}$$"
}
],
"answer": "$$40.8 ~\\mathrm{eV}$$",
"solution": "**Answer:** $$40.8 ~\\mathrm{eV}$$\n\nThe energy difference between the 4th and 2nd energy states of a hydrogen-like atom can be calculated using the formula for the energy levels of a hydrogen-like atom:\n\n
$\\begin{aligned} \\Delta \\mathrm{E} & =13.6 \\mathrm{Z}^2\\left(\\frac{1}{\\mathrm{n}_1^2}-\\frac{1}{\\mathrm{n}_2^2}\\right) \\\\\\\\ \\mathrm{Z} & =4 \\text { (hydrogen like atom) } \\\\ \\mathrm{n}_1 & =2, \\mathrm{n}_2=4 \\\\\\\\ \\Delta \\mathrm{E} & =13.6(4)^2\\left(\\frac{1}{4}-\\frac{1}{16}\\right) \\\\\\\\ & =13.6 \\times\\left(\\frac{16-4}{64}\\right) \\times 16 \\\\\\\\ \\Delta \\mathrm{E} & =13.6 \\times \\frac{12}{64} \\times 16 \\\\\\\\ \\Delta \\mathrm{E} & =40.8 \\mathrm { eV}\\end{aligned}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8079,
"subject": "Physics",
"question": "The radius of electron's second stationary orbit in Bohr's atom is R. The radius of 3rd orbit will be",
"options": [
{
"text": "2.25R"
},
{
"text": "$3 \\mathrm{R}$"
},
{
"text": "$\\frac{\\mathrm{R}}{3}$"
},
{
"text": "$9 \\mathrm{R}$"
}
],
"answer": "2.25R",
"solution": "**Answer:** 2.25R\n\n$r \\propto \\frac{n^{2}}{Z}$\n\n
$$\n\\begin{aligned}\n& \\frac{r_{2 { nd }}}{r_{3 \\mathrm{rd}}}=\\left(\\frac{n_{2}}{n_{3}}\\right)^{2} \\\\\\\\\n& \\Rightarrow \\frac{R}{r_{3 r d}}=\\left(\\frac{2}{3}\\right)^{2} \\\\\\\\\n& \\Rightarrow r_{3 \\mathrm{rd}}=\\frac{9}{4} R \\\\\\\\\n& =2.25 R\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8080,
"subject": "Physics",
"question": "If the binding energy of ground state electron in a hydrogen atom is $13.6\\, \\mathrm{eV}$, then, the energy required to remove the electron from the second excited state of $\\mathrm{Li}^{2+}$ will be : $x \\times 10^{-1} \\mathrm{eV}$. The value of $x$ is ________.
A light of energy $$12.75 ~\\mathrm{eV}$$ is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is $$\\frac{x}{\\pi} \\times 10^{-17} ~\\mathrm{eVs}$$. The value of $$x$$ is ___________ (use $$h=4.14 \\times 10^{-15} ~\\mathrm{eVs}, c=3 \\times 10^{8} \\mathrm{~ms}^{-1}$$ ).
",
"options": [],
"answer": "828",
"solution": "**Answer:** 828\n\nLet the electron jumps to $n^{\\text {th }}$ orbit so\n\n
Speed of an electron in Bohr's $$7^{\\text {th }}$$ orbit for Hydrogen atom is $$3.6 \\times 10^{6} \\mathrm{~m} / \\mathrm{s}$$. The corresponding speed of the electron in $$3^{\\text {rd }}$$ orbit, in $$\\mathrm{m} / \\mathrm{s}$$ is :
The wavelength of the radiation emitted is $$\\lambda_0$$ when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will $$\\frac{20}{x}\\lambda_0$$. The value of $$x$$ is _____________.
",
"options": [],
"answer": "27",
"solution": "**Answer:** 27\n\n Second excited state $$ \\to $$ first excited state , $n=3$ to $n=2$\n
A photon is emitted in transition from n = 4 to n = 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h = 4 $$\\times$$ 10$$^{-15}$$ eVs) :
The radius of $$2^{\\text {nd }}$$ orbit of $$\\mathrm{He}^{+}$$ of Bohr's model is $$r_{1}$$ and that of fourth orbit of $$\\mathrm{Be}^{3+}$$ is represented as $$r_{2}$$. Now the ratio $$\\frac{r_{2}}{r_{1}}$$ is $$x: 1$$. The value of $$x$$ is ___________.
To find the value of $$x$$, we need to first determine the expressions for the radii of the specified orbits for $$\\mathrm{He}^{+}$$ and $$\\mathrm{Be}^{3+}$$ according to Bohr's model. The radius of an orbit in a hydrogen-like atom (an atom with only one electron) is given by:
$$n$$ is the principal quantum number (orbit number)
\n
$$h$$ is the Planck's constant
\n
$$\\epsilon_0$$ is the vacuum permittivity
\n
$$Z$$ is the atomic number (number of protons in the nucleus)
\n
$$e$$ is the elementary charge
\n
$$m_e$$ is the mass of the electron
\n
$$\\pi$$ is the mathematical constant pi
\n
\n\n
In this problem, we are looking at the 2nd orbit of $$\\mathrm{He}^{+}$$ (which has an atomic number $$Z = 2$$) and the 4th orbit of $$\\mathrm{Be}^{3+}$$ (which has an atomic number $$Z = 4$$). Let's calculate the radii for these orbits:
\n\n
For the 2nd orbit of $$\\mathrm{He}^{+}$$ ($$n_1 = 2$$ and $$Z_1 = 2$$):
Therefore, the value of $$x$$ in the ratio $$\\frac{r_{2}}{r_{1}} = x: 1$$ is $$\\boxed{2}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8086,
"subject": "Physics",
"question": "
A $$12.5 \\mathrm{~eV}$$ electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:
",
"options": [
{
"text": "2"
},
{
"text": "4"
},
{
"text": "3"
},
{
"text": "1"
}
],
"answer": "3",
"solution": "**Answer:** 3\n\nThe energy of an electron in an excited state of hydrogen is given by the equation:\n
\nCode snippet
\n$$E = -13.6 \\frac{1}{n^2} \\mathrm{~eV}$$
\nwhere n is the principal quantum number of the state.\n
\nThe energy of the electron beam is 12.5 eV, which is enough to excite the electron to the n = 3 state.
The possible transitions from n = 3 to lower energy states are:\n
\nn = 3 to n = 2, with a wavelength of 656.33 nm (H-alpha)
\nn = 3 to n = 1, with a wavelength of 102.57 nm (Lyman-alpha)
\nn = 2 to n = 1, with a wavelength of 121.57 nm (Lyman-beta)
\nTherefore, there are 3 possible spectral lines that can be emitted.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8087,
"subject": "Physics",
"question": "
The angular momentum for the electron in Bohr's orbit is L. If the electron is assumed to revolve in second orbit of hydrogen atom, then the change in angular momentum will be
According to Bohr's model of the hydrogen atom, the angular momentum of an electron in an orbit is an integral multiple of Planck's constant divided by $2\\pi$ (or $h/2\\pi$, where $h$ is the Planck's constant). This can be expressed as:
\n
$$ L = n \\frac{h}{2\\pi} $$
\n
where $n$ is the principal quantum number or the orbit number.
\n
So, for the first orbit ($n=1$), the angular momentum $L_1$ is:
Since $\\frac{h}{2\\pi}$ is equal to the initial angular momentum $L_1$, the change in angular momentum when the electron moves to the second orbit is $L$.
The formula to calculate the radius of an orbit for a hydrogen-like atom/ion is:
\n
$$\nr_n = r_0 \\frac{n^2}{Z}\n$$
\n
where:
\n
\n
$r_n$ is the radius of the nth orbit,
\n
$n$ is the principal quantum number (the orbit number),
\n
$r_0$ is the Bohr radius (radius of the first Bohr orbit in the hydrogen atom), and
\n
$Z$ is the atomic number (the number of protons in the nucleus).
\n
\n
We're dealing with a Li²⁺ ion and we're interested in the fifth orbit ($n = 5$), and given that $r_0$ is 0.51 Å and $Z$ for Li is 3, we can substitute these values into the formula:
A small particle of mass $$m$$ moves in such a way that its potential energy $$U=\\frac{1}{2} m ~\\omega^{2} r^{2}$$ where $$\\omega$$ is constant and $$r$$ is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of $$n^{\\text {th }}$$ orbit will be proportional to,
According to Bohr's quantization of angular momentum, the angular momentum $$L$$ of a particle in a circular orbit is given by:
\n
$$L = n\\hbar$$
\n
Where $$n$$ is an integer and $$\\hbar$$ is the reduced Planck's constant. The angular momentum $$L$$ can also be expressed as:
\n
$$L = mvr$$
\n
Where $$m$$ is the mass of the particle, $$v$$ is its linear velocity, and $$r$$ is the radius of the orbit.
\n
Now, we are given the potential energy $$U = \\frac{1}{2} m\\omega^2r^2$$. Since the particle is in a circular orbit, its centripetal force is provided by the gradient of the potential energy:
Experimentally it is found that $$12.8 ~\\mathrm{eV}$$ energy is required to separate a hydrogen atom into a proton and an electron. So the orbital radius of the electron in a hydrogen atom is $$\\frac{9}{x} \\times 10^{-10} \\mathrm{~m}$$. The value of the $$x$$ is __________.
The binding energy of an electron in a hydrogen atom is given by the formula:
\n
$$ E = \\frac{k e^2}{2 r} $$
\n
where:
\n
\n
$E$ is the energy of the electron,
\n
$k$ is Coulomb's constant ($9 \\times 10^9 \\, \\text{Nm}^2/\\text{C}^2$),
\n
$e$ is the charge of the electron ($1.6 \\times 10^{-19} \\, \\text{C}$), and
\n
$r$ is the radius of the orbit.
\n
\n
In this scenario, the energy $E$ required to separate a hydrogen atom into a proton and an electron is given as $12.8 \\, \\text{eV}$, which needs to be converted into joules using the conversion factor $1 \\, \\text{eV} = 1.6 \\times 10^{-19} \\, \\text{J}$. So,
Let's analyze each of the given statements to determine which ones are correct:
\n\n
(A) The angular momentum of an electron in $n^{\\text{th}}$ orbit is an integral multiple of $\\hbar$.
\n
This statement reflects Bohr's quantization rule for angular momentum in the Bohr model of the hydrogen atom. According to this rule, the angular momentum of an electron in a stationary orbit is quantized and given by:
\n$$ L = n\\hbar $$\n
where $n$ is a principal quantum number (which can be any positive integer), and $\\hbar$ is the reduced Planck's constant. Therefore, this statement (A) is correct.
\n\n
(B) Nuclear forces do not obey inverse square law.
\n
Nuclear forces, specifically strong nuclear forces, do act over short ranges within the nucleus but do not obey the inverse square law, which is characteristic of the electromagnetic and gravitational forces. Thus, statement (B) is correct.
\n\n
(C) Nuclear forces are spin dependent.
\n
The strength of the nuclear force can depend on the spin alignment of the nucleons. This is why some isotopes are more stable than others depending on spin-related factors. Thus, statement (C) is correct.
\n\n
(D) Nuclear forces are central and charge independent.
\n
The strong nuclear force is indeed charge independent, meaning it is the same regardless of the types of nucleons involved (neutrons or protons). However, nuclear forces are not always central; they can be tensor forces too, which involve more complex interactions that are not purely central. Therefore, the entirety of statement (D) is not correct; the statement should specify that nuclear forces are charge independent but may not always be central.
\n\n
(E) Stability of nucleus is inversely proportional to the value of packing fraction.
\n
This statement (E) is correct.
\n\n
Based on the above explanations, the correct statements are (A), (B), (C) and (E). Hence, the correct answer is:
\n$$ \\text{Option C : (A), (B), (C), (D) only} $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8092,
"subject": "Physics",
"question": "A particular hydrogen-like ion emits the radiation of frequency $3 \\times 10^{15} \\mathrm{~Hz}$ when it makes transition from $n=2$ to $n=1$. The frequency of radiation emitted in transition from $n=3$ to $n=1$ is $\\frac{x}{9} \\times 10^{15} \\mathrm{~Hz}$, when $x=$ ________ .",
"options": [],
"answer": "32",
"solution": "**Answer:** 32\n\n
The emission frequency of radiation from a hydrogen-like ion during electron transitions can be understood using the formula derived from Rydberg's equation for hydrogen-like atoms, which is given as:
$n_1$ and $n_2$ are the principal quantum numbers for the initial and final energy levels, respectively, with $n_2 < n_1$.
Given a transition from $n=2$ to $n=1$ emits radiation with a frequency of $3 \\times 10^{15} \\mathrm{~Hz}$, we can write:
$$\\nu_1 = 3 \\times 10^{15} \\mathrm{~Hz}$$
For the transition from $n=3$ to $n=1$, we can use the same principle to find the frequency of the emitted radiation, $\\nu_2$, which will depend on the difference in energy levels involved in the transition. The frequency is directly proportional to this energy difference, so we can compare the two transitions using their respective frequencies:
Hence, in the given formula $\\frac{x}{9} \\times 10^{15} \\mathrm{~Hz}$ for the frequency, $x$ is equal to $32$.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8093,
"subject": "Physics",
"question": "The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly :",
"options": [
{
"text": "$13.6 \\mathrm{eV}$"
},
{
"text": "$1.5 \\mathrm{eV}$"
},
{
"text": "$12.1 \\mathrm{eV}$"
},
{
"text": "$1.9 \\mathrm{eV}$"
}
],
"answer": "$12.1 \\mathrm{eV}$",
"solution": "**Answer:** $12.1 \\mathrm{eV}$\n\n
The energy $E_n$ of an electron in the $n$th energy level of a hydrogen atom is given by the formula:
\n\n
$ E_n = -\\frac{13.6 \\, \\text{eV}}{n^2} $
\n\n
where:
\n\n
\n
$E_n$ is the energy in electron volts (eV),
\n
$13.6 \\, \\text{eV}$ is the ionization energy of hydrogen (the energy required to remove the electron from the ground state, $n=1$),
\n
$n$ is the principal quantum number (an integer starting from 1).
\n
\n
The Balmer Series
\n\n
The Balmer series is a specific set of spectral lines of hydrogen that are visible to the naked eye. It results from the electron making transitions between energy levels, specifically from levels with $n \\geq 3$ (i.e., from higher energy states) down to $n = 2$. These transitions release energy in the form of electromagnetic radiation, which we observe as the Balmer series.
\n\n\n
Ground State Energy ($E_1$): The energy of the electron in the ground state ($n=1$) of hydrogen is:
Minimum Energy for Balmer Series Transition: The minimum energy transition within the Balmer series is from $n=3$ to $n=2$, but to understand the total energy required for an electron to emit radiation in the Balmer series, starting from the ground state, we consider the energy needed to first excite the electron from $n=1$ to $n=3$.
\n
Energy Difference ($\\Delta E$): The energy emitted as radiation for the electron to transition from the ground state to a state where it can participate in the Balmer series is the difference between the ground state energy and the energy of the $n=3$ state:
This calculation shows that the minimum energy required by a hydrogen atom in the ground state to emit radiation in the Balmer series is approximately 12.1 eV. The energy difference essentially represents the energy that must be supplied to an electron to excite it from the ground state ($n=1$) to an excited state ($n=3$) from which it can then transition to $n=2$, emitting radiation observable as part of the Balmer series.
The radius of the nth stationary orbit in the Bohr model of an atom is directly proportional to the square of its principal quantum number n and inversely proportional to the atomic number Z. This relationship is represented by the formula $$r_n = \\frac{n^2h^2}{4\\pi^2kme^2Z},$$ where :
\n\n
\n\n
\n\n
$n$ is the principal quantum number,
\n\n
\n\n
\n\n
$h$ is Planck's constant,
\n\n
\n\n
\n\n
$k$ is the Coulomb constant,
\n\n
\n\n
\n\n
$m$ is the mass of the electron,
\n\n
\n\n
\n\n
$e$ is the charge of the electron, and
\n\n
\n\n
\n\n
$Z$ is the atomic number of the nucleus (for hydrogen, Z=1).
\n\n
\n\n
\n\n
To find the radius of the fourth orbit ($r_4$), in comparison to the third orbit ($r_3$), we apply the formula with $n=4$ for the fourth orbit and $n=3$ for the third orbit, and simplify as follows:
Thus, the radius of the fourth stationary orbit is $\\frac{16}{9}$ times the radius of the third stationary orbit, $R$. Therefore, $$r_4 = \\frac{16}{9} R$$.
When a hydrogen atom going from $$n=2$$ to $$n=1$$ emits a photon, its recoil speed is $$\\frac{x}{5} \\mathrm{~m} / \\mathrm{s}$$. Where $$x=$$ ________. (Use, mass of hydrogen atom $$=1.6 \\times 10^{-27} \\mathrm{~kg}$$)
Hydrogen atom is bombarded with electrons accelerated through a potential difference of $$\\mathrm{V}$$, which causes excitation of hydrogen atoms. If the experiment is being performed at $$\\mathrm{T}=0 \\mathrm{~K}$$, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be $$\\frac{\\alpha}{10} \\mathrm{~V}$$, where $$\\alpha=$$ __________.
For minimum potential difference electron has to make transition from $$n=3$$ to $$n=2$$ state but first electron has to reach to $$\\mathrm{n}=3$$ state from ground state. So, energy of bombarding electron should be equal to energy difference of $$\\mathrm{n=3}$$ and $$\\mathrm{n=1}$$ state.
To determine the relationship between the magnetic moment and the principal quantum number $ n $, we need to understand the formula for the magnetic moment of an electron in a Bohr orbit.
\n\n
The magnetic moment ($ \\mu_n $) of an electron in the nth Bohr orbit is given by:
$ Z $ is the atomic number (for a hydrogen atom, $ Z = 1 $)
\n\n
$ h $ is Planck's constant
\n\n
\n\n
Since the Bohr's magneton ($ \\mu_B $) is defined as:
\n\n
$$ \\mu_B = \\frac{e \\hbar}{2m} $$
\n\n
The magnetic moment for the nth orbit can be written as:
\n\n
$$ \\mu_n = n \\cdot \\mu_B $$
\n\n
From the above formula, it is clear that the magnetic moment $ \\mu_n $ is directly proportional to the principal quantum number $ n $. Mathematically, this relationship can be expressed as:
A electron of hydrogen atom on an excited state is having energy $$\\mathrm{E}_{\\mathrm{n}}=-0.85 \\mathrm{~eV}$$. The maximum number of allowed transitions to lower energy level is _________.
To determine how many spectral lines will be emitted due to transitions of electrons in a hydrogen atom when it is given an energy of $10.2 \\, \\text{eV}$, we first need to ascertain which energy level the electron will reach with this energy and then count the possible transitions (spectral lines) as it returns to the ground state.
\n\n
Energy Levels of Hydrogen Atom :
\n\n
The energy levels $ E_n $ of a hydrogen atom can be calculated using the formula:
\n\n
$ E_n = -\\frac{13.6 \\, \\text{eV}}{n^2} $
\n\n
where $ n $ is the principal quantum number.
\n\n
Ground State Energy :
\n\n
The ground state (n=1) energy is $ E_1 = -13.6 \\, \\text{eV} $.
\n\n
Determine the Excited State :
\n\n
If the ground state electron is given $10.2 \\, \\text{eV}$, its total energy becomes:
Now, we find the principal quantum number $ n $ for which the energy is closest to $-3.4 \\, \\text{eV}$:
\n\n
\n
For $ n = 2 $: $ E_2 = -\\frac{13.6}{2^2} = -3.4 \\, \\text{eV} $
\n
For $ n = 3 $: $ E_3 = -\\frac{13.6}{3^2} = -1.51 \\, \\text{eV} $
\n
\n
Since $-3.4 \\, \\text{eV}$ matches exactly with $ E_2 $, the electron reaches the second energy level ($ n = 2 $).
\n\n
Counting Spectral Lines :
\n\n
When the electron falls back to the ground state from $ n = 2 $, it can do so in a single transition:
\n\n
\n
$ n = 2 $ to $ n = 1 $
\n
\n
Thus, only one spectral line will be emitted during this transition.
\n\n
Conclusion :
\n\n
The number of spectral lines emitted when a hydrogen atom in the ground state is given $10.2 \\, \\text{eV}$ and the electron transitions back to the ground state from $ n = 2 $ is just one.
A hydrogen atom changes its state from $$n=3$$ to $$n=2$$. Due to recoil, the percentage change in the wave length of emitted light is approximately $$1 \\times 10^{-n}$$. The value of $$n$$ is _______.
[Given Rhc $$=13.6 \\mathrm{~eV}, \\mathrm{hc}=1242 \\mathrm{~eV} \\mathrm{~nm}, \\mathrm{h}=6.6 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s}$$ mass of the hydrogenatom $$=1.6 \\times 10^{-27} \\mathrm{~kg}$$]
According to Bohr's theory, one of the postulates specifies that the angular momentum of an electron in orbit around a nucleus is quantized. This quantization can be expressed by the formula:
\n\n
$$ L = n\\frac{h}{2\\pi} $$
\n\n
Where:
\n\n
\n\n
$L$ is the angular momentum of the electron,
\n\n
$n$ is the principal quantum number (or the orbit number in simpler terms), which can take positive integer values (1, 2, 3, ...),
\n\n
$h$ is Planck's constant ($6.62607015 \\times 10^{-34} m^2 kg / s$), and
\n\n
$\\frac{h}{2\\pi}$ is often denoted as $\\hbar$ (h-bar), known as the reduced Planck's constant.
\n\n
\n\n
For an electron in the 4th orbit ($n = 4$) of a hydrogen atom, we substitute $n = 4$ into the equation:
\n\n
$$ L = 4\\frac{h}{2\\pi} $$
\n\n
Therefore, the moment of momentum (or angular momentum) of an electron in the $4^{\\text{th}}$ orbit of a hydrogen atom is:
\n\n
$$ L = 4\\frac{h}{2\\pi} = 2\\frac{2h}{2\\pi} = 2\\frac{h}{\\pi} $$
An electron rotates in a circle around a nucleus having positive charge $$\\mathrm{Ze}$$. Correct relation between total energy (E) of electron to its potential energy (U) is :
In the context of an electron orbiting around a nucleus with a positive charge of $$\\mathrm{Ze}$$, we are dealing with classical physics approximations and the electrostatic force between the electron and the nucleus. In such a setup, the electron's potential energy (U) is due to electrostatic interaction, and it is given by Coulomb's law:
\n\n
$$U = -\\frac{kZe^2}{r}$$
\n\n
Where:
\n\n
\n\n
$U$ is the potential energy of the electron,
\n\n
$k$ is Coulomb's constant,
\n\n
$Z$ is the atomic number (number of protons in the nucleus),
\n\n
$e$ is the charge of an electron, and
\n\n
$r$ is the radius of the orbit of the electron around the nucleus.
\n\n
\n\n
The negative sign indicates that the potential energy is negative because the electron and nucleus attract each other.
\n\n
The total energy (E) of the electron in orbit is the sum of its kinetic energy (K) and its potential energy (U). Since the electron is in a stable orbit, its kinetic energy can be shown to be exactly half the magnitude of its potential energy but positive:
\n\n
$$K = -\\frac{1}{2}U$$
\n\n
Therefore,
\n\n
$$E = K + U = -\\frac{1}{2}U + U = \\frac{1}{2}U$$
\n\n
To find a relation between total energy (E) and potential energy (U), we rearrange the equation as follows:
\n\n
$$2E = U$$
\n\n
This is to say, the total energy (E) is half the magnitude of potential energy (U) but negative, and the correct relationship between them, when looking for a positive proportionality, yields to $2E = U$. Hence, the correct option is:
The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is $$915\\mathop A\\limits^o$$. The longest wavelength of spectral lines in the Balmer series will be _______ $$\\mathop A\\limits^o$$.
To determine the longest wavelength associated with the Paschen series, we need to understand what the Paschen series is and how its wavelength can be calculated. The Paschen series pertains to the spectral line emissions of the hydrogen atom as an electron transitions from higher energy levels (n > 3) down to n = 3. The formula used to calculate the wavelength ($\\lambda$) of the emitted photon during such a transition in the hydrogen atom is given by the Rydberg formula:
$R_{H}$ is the Rydberg constant for hydrogen, $R_{H} = 1.097 \\times 10^7 \\, \\text{m}^{-1}$.
\n\n
$n$ is the principal quantum number of the higher energy level (For the longest wavelength, we use the smallest possible value of $n$ that is larger than 3, which is $n=4$, since for longer wavelength transitions, the difference in energy levels should be the smallest).
\n\n
\n\n
To find the longest wavelength, we plug in $n=4$ into the Rydberg equation, as this corresponds to the smallest energy transition within the Paschen series ($n=4$ to $n=3$):
The wavelength of light emitted when an electron transitions between energy levels in a hydrogen atom is given by the Rydberg formula:
\n\n
$$\\frac{1}{\\lambda} = R \\left( \\frac{1}{n_1^2} - \\frac{1}{n_2^2} \\right)$$
\n\n
where:
\n\n
\n\n
$$\\lambda$$ is the wavelength of the emitted light,
\n\n
$$R$$ is the Rydberg constant,
\n\n
$$n_1$$ is the lower energy level, and
\n\n
$$n_2$$ is the higher energy level.
\n\n
\n\n
The Balmer series corresponds to electron transitions where the lower energy level, $$n_1$$, is 2, and the Lyman series corresponds to transitions where $$n_1$$ is 1. The shortest wavelength in each series occurs for transitions from the highest possible energy level ($$n_2 = \\infty$$) to the specified lower level ($$n_1$$).
\n\n
For the Balmer series (shortest wavelength):
\n\n
$$n_1 = 2, n_2 = \\infty$$,
\n\n
Putting these values into the Rydberg formula gives:
\n\n
$$\\frac{1}{\\lambda_{B}} = R \\left( \\frac{1}{2^2} - \\frac{1}{\\infty^2} \\right)$$
\n\n
$$\\frac{1}{\\lambda_{B}} = R \\left( \\frac{1}{4} \\right)$$
Radius of a certain orbit of hydrogen atom is 8.48 $$\\mathop A\\limits^o$$. If energy of electron in this orbit is $$E / x$$. then $$x=$$ ________ (Given $$\\mathrm{a}_0=0.529$$ $$\\mathop A\\limits^o$$, $$E=$$ energy of electron in ground state).
Let's approach this problem by understanding the basics and applying the Bohr model to find the energy levels of a hydrogen atom.
\n\n
The energy of an electron in a hydrogen atom for any given orbit can be defined using the formula:
\n\n
$$E_n = \\frac{E}{n^2}$$
\n\n
where:
\n\n
\n\n
$E$ is the energy of the electron in the ground state ($n=1$), and
\n\n
$n$ is the principal quantum number (orbit number).
\n\n
\n\n
The radius of an orbit in the hydrogen atom, according to the Bohr model, is given by:
\n\n
$$r_n = n^2 a_0$$
\n\n
where:
\n\n
\n\n
$r_n$ is the radius of the nth orbit,
\n\n
$a_0$ is the Bohr radius ($0.529$ angstroms or $\\mathop A\\limits^o$), and
\n\n
$n$ is the principal quantum number (orbit number).
\n\n
\n\n
Given that:
\n\n
\n\n
The radius of a certain orbit of the hydrogen atom is $8.48 \\mathop A\\limits^o$, and
\n\n
The energy of an electron in this orbit is $E/x$,
\n\n
We need to find $x$.
\n\n
\n\n
First, let's find $n$, the principal quantum number for the orbit with radius $8.48$ angstroms:
\n\n
$$8.48 = n^2 \\times 0.529$$
\n\n
Solving for $n^2$:
\n\n
$$n^2 = \\frac{8.48}{0.529}$$
\n\n
$$n^2 \\approx 16.03$$
\n\n
For simplicity and practicality in the quantum model, $n^2$ approximately equal to 16 would imply $n = 4$, considering $n$ must be a whole number and $16.03$ is close to $16$, which is a perfect square of $4$.
\n\n
Now, to find $x$, we'll use the energy relationship. Since the energy levels of the hydrogen atom are inversely proportional to the square of the principal quantum number $n$,
According to the given information, $E_{orbit} = \\frac{E}{x}$, which means:
\n\n
$$\\frac{E}{x} = \\frac{E}{16}$$
\n\n
Hence,
\n\n
$$x = 16$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8111,
"subject": "Physics",
"question": "In the nuclear fusion reaction \n$$${}_1^2H + {}_1^3H \\to {}_2^4He + n$$$\n given that the repulsive potential energy between the two nuclei is $$ \\sim 7.7 \\times {10^{ - 14}}J$$, the temperature at which the gases must be heated to initiate the reaction is nearly \n [ Boltzmann's Constant $$k = 1.38 \\times {10^{ - 23}}\\,J/K$$ ]",
"options": [
{
"text": "$${10^7}\\,\\,K$$ "
},
{
"text": "$${10^5}\\,\\,K$$ "
},
{
"text": "$${10^3}\\,\\,K$$ "
},
{
"text": "$${10^9}\\,\\,K$$ "
}
],
"answer": "$${10^9}\\,\\,K$$ ",
"solution": "**Answer:** $${10^9}\\,\\,K$$ \n\nThe average kinetic energy per molecule $$ = {3 \\over 2}kT$$\n
This kinetic energy should be able to provide the repulsive potential energy\n
$$ = \\left[ {8{M_p} + 9{M_N} - {M_O}} \\right]{c^2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8117,
"subject": "Physics",
"question": "This question contains Statement- 1 and Statement- 2. Of the four choices given after the statements, choose the one that best describes the two statements: \n Statement- 1:\n Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion and \n
Statement- 2:\n For heavy nuclei, binding energy per nucleon increases with increasing $$Z$$ while for light nuclei it decreases with increasing $$Z.$$
",
"options": [
{
"text": "Statement - $$1$$ is false, Statement - $$2$$ is true "
},
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is a correct explanation for Statement - $$1$$ "
},
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is not a correct explanation for Statement - $$1$$ "
},
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is false"
}
],
"answer": "Statement - $$1$$ is true, Statement - $$2$$ is false",
"solution": "**Answer:** Statement - $$1$$ is true, Statement - $$2$$ is false\n\nWe know that energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. Therefore statement $$(1)$$ is correct.\n
The second statement is false because for heavy nuclei the binding energy per nucleon decreases with increasing $$Z$$ and for light nuclei, B.E/nucleon increases with increasing $$Z$$ and for light nuclei, $$B.E/$$nucleon increases with increasing $$Z.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8118,
"subject": "Physics",
"question": "A nucleus of mass $$M+$$$$\\Delta m$$ is at rest and decays into two daughter nuclei of equal mass $${M \\over 2}$$ each. Speed of light is $$c.$$ \n
The binding energy per nucleon for the parent nucleus is $${E_1}$$ and that for the daughter nuclei is $${E_2}.$$ Then
",
"options": [
{
"text": "$${E_2} = 2{E_1}$$ "
},
{
"text": "$${E_1} > {E_2}$$ "
},
{
"text": "$${E_2} > {E_1}$$ "
},
{
"text": "$${E_1} = 2{E_2}$$ "
}
],
"answer": "$${E_2} > {E_1}$$ ",
"solution": "**Answer:** $${E_2} > {E_1}$$ \n\nIn nuclear fission, the binding energy per nucleon of daughter nuclei is greater than the parent nucleon of daughter nuclei is greater than the parent nucleus. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8119,
"subject": "Physics",
"question": "A nucleus of mass $$M+$$$$\\Delta m$$ is at rest and decays into two daughter nuclei of equal mass $${M \\over 2}$$ each. Speed of light is $$c.$$ \n
$$E = {{81 \\times {{10}^{ - 15}}} \\over {1.6 \\times {{10}^{ - 19}}}} = 0.511MeV$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8121,
"subject": "Physics",
"question": "Imagine that a reactor converts all given mass into energy and that it operates at a power level of 109 watt. The mass of the fuel consumed per hour in the reactor will be : (velocity of light, c is \n3×108 m/s)\n",
"options": [
{
"text": "0.96 gm"
},
{
"text": "0.8 gm "
},
{
"text": "4 $$ \\times $$ 10$$-$$2 gm"
},
{
"text": "6.6 $$ \\times $$ 10$$-$$5 gm"
}
],
"answer": "4 $$ \\times $$ 10$$-$$2 gm",
"solution": "**Answer:** 4 $$ \\times $$ 10$$-$$2 gm\n\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8122,
"subject": "Physics",
"question": "Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV)\n",
"options": [
{
"text": "30.2 MeV"
},
{
"text": "32.4 MeV"
},
{
"text": "23.6 MeV"
},
{
"text": "25.8 MeV"
}
],
"answer": "23.6 MeV",
"solution": "**Answer:** 23.6 MeV\n\n1H2 + 1H2 $$ \\to $$ 2He4\n
No. of proton in one dueteron = 2 \n
$$\\therefore\\,\\,\\,$$ Total protons in two dueterons = 2 $$ \\times $$ 2 = 4\n
$$\\therefore\\,\\,\\,$$ Binding energy of two dueteron\n
= 1.1 $$ \\times $$ 4 = 4 : 4 MeV\n
In (2He4) no of protons = 4\n
$$\\therefore\\,\\,\\,$$ Binding energy of (2He4) nuclei = 4 $$ \\times $$ 7 = 28 Mev\n
Energy released in this process = 28 $$-$$ 4.4 = 23.6 MeV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8123,
"subject": "Physics",
"question": "Consider the nuclear fission\n
Ne20 $$ \\to $$ 2He4 + C12\n
Given that the binding energy/ nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement - \n",
"options": [
{
"text": "8.3 MeV energy will be released"
},
{
"text": "energy of 11.9 MeV has to be supplied"
},
{
"text": "energy of 12.4 MeV will be supplied"
},
{
"text": "energy of 3.6 MeV will be released"
}
],
"answer": "energy of 11.9 MeV has to be supplied",
"solution": "**Answer:** energy of 11.9 MeV has to be supplied\n\nNe20$$~ \\to $$ 2He4 + C12\n
Q – value, EB = (BE)react $$-$$ (BE)product\n
= (20 × 8.03) – ((2 × 7.07 × 4) + 7.86 × 12)\n
= 9.72 MeV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8124,
"subject": "Physics",
"question": "In a reactor, 2 kg of 92U235 fuel is fully used up\nin 30 days. The energy released per fission is\n200 MeV. Given that the Avogadro number,\nN = 6.023 $$ \\times $$ 1026 per kilo mole and 1 eV =\n1.6 × 10–19 J. The power output of the reactor is\nclose to",
"options": [
{
"text": "125 MW"
},
{
"text": "60 MW"
},
{
"text": "54 MW"
},
{
"text": "35 MW"
}
],
"answer": "60 MW",
"solution": "**Answer:** 60 MW\n\nNumber of uranium atoms in 2 kg\n
= 1.33 $$ \\times $$ 106 kWh",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8127,
"subject": "Physics",
"question": "A nucleus of mass M emits $$\\gamma$$ -ray photon of frequency 'v'. The loss of internal energy by the nucleus is :
= 5.38 MeV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8129,
"subject": "Physics",
"question": "From the given data, the amount of energy required to break the nucleus of aluminium $$_{13}^{27}$$Al is __________ x $$\\times$$ 10$$-$$3 J.
Mass of neutron = 1.00866 u
Mass of proton = 1.00726 u
Mass of Aluminium nucleus = 27.18846 u
(Assume 1 u corresponds to x J of energy)
(Round off to the nearest integer)",
"options": [],
"answer": "27",
"solution": "**Answer:** 27\n\n$$\\Delta$$m = (Zmp + (A $$-$$ Z)mn) $$-$$ MAl
Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments 'B' and 'C' of mass numbers 105 and 115. The binding energy of nucleons in 'B' and 'C' is 6.4 MeV per nucleon. The energy Q released per fission will be :
A nucleus of mass $$M$$ at rest splits into two parts having masses $$\\frac{M^{\\prime}}{3}$$ and $${{2M'} \\over 3}(M' < M)$$. The ratio of de Broglie wavelength of two parts will be :
Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below :
\n
$${ }_{1}^{2} X+{ }_{1}^{2} X={ }_{2}^{4} Y$$
\n
The binding energies per nucleon for $$\\frac{2}{1} X$$ and $${ }_{2}^{4} Y$$ are $$1.1 \\,\\mathrm{MeV}$$ and $$7.6 \\,\\mathrm{MeV}$$ respectively. The energy released in this process is _______________ $$\\mathrm{MeV}$$.
The energy released per fission of nucleus of $$^{240}$$X is 200 MeV. The energy released if all the atoms in 120g of pure $$^{240}$$X undergo fission is ____________ $$\\times$$ 10$$^{25}$$ MeV.
\n
(Given $$\\mathrm{N_A=6\\times10^{23}}$$)
",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\n$120 \\mathrm{~g}$ of $^{240}X$ will have $\\frac{1}{2}$ mole of $X$\n
\nNumber of atom of $X=\\frac{1}{2} \\times N_{A}=3 \\times 10^{23}$ atom\n
\nEnergy released $=3 \\times 10^{23} \\times 200 ~ \\mathrm{MeV}$\n
In the given nuclear reaction, the approximate amount of energy released will be:
\n
[Given, mass of $${ }_{92}^{238} \\mathrm{~A}=238.05079 \\times 931.5 ~\\mathrm{MeV} / \\mathrm{c}^{2},$$
\n
mass of $${ }_{90}^{234} B=234 \\cdot 04363 \\times 931 \\cdot 5 ~\\mathrm{MeV} / \\mathrm{c}^{2},$$
\n
mass of $$\\left.{ }_{2}^{4} D=4 \\cdot 00260 \\times 931 \\cdot 5 ~\\mathrm{MeV} / \\mathrm{c}^{2}\\right]$$
",
"options": [
{
"text": "2.12 MeV"
},
{
"text": "4.25 MeV"
},
{
"text": "3.82 MeV"
},
{
"text": "5.9 MeV"
}
],
"answer": "4.25 MeV",
"solution": "**Answer:** 4.25 MeV\n\nThe energy released in a nuclear reaction can be determined by the mass difference between the reactants and the products, multiplied by the speed of light squared, as per Einstein's mass-energy equivalence relation, $$E=mc^2$$.\n
\nIn the given nuclear reaction, the energy released is:\n
\nSo, the approximate amount of energy released in the given nuclear reaction is 4.25 MeV.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8138,
"subject": "Physics",
"question": "
A common example of alpha decay is $${ }_{92}^{238} \\mathrm{U} \\longrightarrow{ }_{90}^{234} \\mathrm{Th}+{ }_{2} \\mathrm{He}^{4}+\\mathrm{Q}$$
The energy released $$(Q)$$ during the alpha decay of $${ }_{92}^{238} \\mathrm{U}$$ is __________ MeV
",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nTo find the energy released during the alpha decay, we first need to calculate the mass difference between the reactants and the products.\n
\nMass difference = Mass of Uranium-238 - (Mass of Thorium-234 + Mass of Helium-4)\n
\nThe energy released (Q) during the alpha decay of $$^{238}U$$ is approximately 4.1 MeV.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8139,
"subject": "Physics",
"question": "
A nucleus disintegrates into two nuclear parts, in such a way that ratio of their nuclear sizes is $$1: 2^{1 / 3}$$. Their respective speed have a ratio of $$n: 1$$. The value of $n$ is __________.
",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nLet the masses of the two nuclear parts be $$m_1$$ and $$m_2$$, and their respective speeds be $$v_1$$ and $$v_2$$. According to the problem, the ratio of their nuclear sizes is $$1 : 2^{1/3}$$. Since the nuclear size is proportional to the cube root of the mass, we can write:\n
\nThus, the value of $$n$$ is 2.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8140,
"subject": "Physics",
"question": "
For a nucleus $${ }_{\\mathrm{A}}^{\\mathrm{A}} \\mathrm{X}$$ having mass number $$\\mathrm{A}$$ and atomic number $$\\mathrm{Z}$$
\n
A. The surface energy per nucleon $$\\left(b_{\\mathrm{s}}\\right)=-a_{1} A^{2 / 3}$$.
\n
B. The Coulomb contribution to the binding energy $$\\mathrm{b}_{\\mathrm{c}}=-a_{2} \\frac{Z(Z-1)}{A^{4 / 3}}$$
\n
C. The volume energy $$\\mathrm{b}_{\\mathrm{v}}=a_{3} A$$
\n
D. Decrease in the binding energy is proportional to surface area.
\n
E. While estimating the surface energy, it is assumed that each nucleon interacts with 12 nucleons. ( $$a_{1}, a_{2}$$ and $$a_{3}$$ are constants)
\n
Choose the most appropriate answer from the options given below:
",
"options": [
{
"text": "C, D only"
},
{
"text": "B, C, E only"
},
{
"text": "B, C only"
},
{
"text": "A, B, C, D only"
}
],
"answer": "C, D only",
"solution": "**Answer:** C, D only\n\n
In the semi-empirical mass formula, the terms have the following forms:
\n
\n
The volume term (binding energy) is proportional to the volume of the nucleus, and therefore to the number of nucleons $A$.
\n
The surface term takes into account that the nucleons at the surface of the nucleus have fewer nearest neighbors than those in the interior, and so the binding energy is less for surface nucleons. This term is usually proportional to $A^{2/3}$.
\n
The Coulomb term arises because the protons in the nucleus repel each other. This term is usually proportional to $\\frac{Z(Z-1)}{A^{1/3}}$.
\n
The symmetry term arises because of the Pauli exclusion principle, but it's not mentioned here.
\n
The pairing term arises because of the tendency of protons and neutrons to pair up, but it's not mentioned here.
\n
\n
Now, let's check the statements:
\n
\n
A: The surface energy per nucleon $b_{s}=-a_{1} A^{2 / 3}$ is incorrect. The correct form should be proportional to $A^{2/3}$, but it should not be per nucleon.
\n
B: The Coulomb contribution to the binding energy $b_{c}=-a_{2} \\frac{Z(Z-1)}{A^{4 / 3}}$ is incorrect. The correct form should be proportional to $\\frac{Z(Z-1)}{A^{1/3}}$.
\n
C: The volume energy $b_{v}=a_{3} A$ is correct.
\n
D: Decrease in the binding energy is proportional to surface area is correct. This represents the surface term in the semi-empirical mass formula.
\n
E: While estimating the surface energy, it is assumed that each nucleon interacts with 12 nucleons is incorrect. While the exact number can vary depending on the model used, it's generally agreed that each nucleon interacts with a few nearest neighbors, not necessarily 12.
A nucleus with mass number 242 and binding energy per nucleon as $$7.6~ \\mathrm{MeV}$$ breaks into two fragment each with mass number 121. If each fragment nucleus has binding energy per nucleon as $$8.1 ~\\mathrm{MeV}$$, the total gain in binding energy is _________ $$\\mathrm{MeV}$$.
The total binding energy of a nucleus is the binding energy per nucleon multiplied by the number of nucleons (protons and neutrons), which is the mass number.
\n
The initial total binding energy of the nucleus is $242 \\times 7.6 \\, \\text{MeV}$.
\n
After the break, each fragment has a total binding energy of $121 \\times 8.1 \\, \\text{MeV}$.
Since there are two such fragments, the final total binding energy is $2 \\times 121 \\times 8.1 \\, \\text{MeV}$.
\n
The gain in binding energy is the final total binding energy minus the initial total binding energy. Therefore, the gain in binding energy is:
In a nuclear fission process, a high mass nuclide $$(A \\approx 236)$$ with binding energy $$7.6 \\mathrm{~MeV} /$$ Nucleon dissociated into middle mass nuclides $$(\\mathrm{A} \\approx 118)$$, having binding energy of $$8.6 \\mathrm{~MeV} / \\mathrm{Nucleon}$$. The energy released in the process would be ______ $$\\mathrm{MeV}$$.
To determine the energy released in a nuclear fission process, we use the difference in binding energy (BE) before and after the fission. The formula for energy released ($$Q$$ value) in the process is given by:
$$Q = (\\text{Total BE of products}) - (\\text{Total BE of reactants})$$
In this case, the reactant is a high mass nuclide with atomic mass $A \\approx 236$ and a binding energy of $$7.6 \\mathrm{MeV}/\\mathrm{nucleon}$$. Each of the two middle mass nuclides formed as products has atomic mass $A \\approx 118$ and a binding energy of $$8.6 \\mathrm{MeV}/\\mathrm{nucleon}$$.
Therefore, we calculate the total binding energy of reactant and products as follows:
This calculation demonstrates how the difference in binding energy per nucleon before and after fission leads to the release of energy, consistent with the mass-energy equivalence principle.
The atomic mass of $${ }_6 \\mathrm{C}^{12}$$ is $$12.000000 \\mathrm{~u}$$ and that of $${ }_6 \\mathrm{C}^{13}$$ is $$13.003354 \\mathrm{~u}$$. The required energy to remove a neutron from $${ }_6 \\mathrm{C}^{13}$$, if mass of neutron is $$1.008665 \\mathrm{~u}$$, will be :
The mass defect in a particular reaction is $$0.4 \\mathrm{~g}$$. The amount of energy liberated is $$n \\times 10^7 \\mathrm{~kWh}$$, where $$n=$$ __________. (speed of light $$\\left.=3 \\times 10^8 \\mathrm{~m} / \\mathrm{s}\\right)$$
The explosive in a Hydrogen bomb is a mixture of $${ }_1 \\mathrm{H}^2,{ }_1 \\mathrm{H}^3$$ and $${ }_3 \\mathrm{Li}^6$$ in some condensed form. The chain reaction is given by
In a nuclear fission reaction of an isotope of mass $$M$$, three similar daughter nuclei of same mass are formed. The speed of a daughter nuclei in terms of mass defect $$\\Delta M$$ will be :
A nucleus at rest disintegrates into two smaller nuclei with their masses in the ratio of $$2: 1$$. After disintegration they will move :
",
"options": [
{
"text": "in opposite directions with speed in the ratio of $$1: 2$$ respectively.\n"
},
{
"text": "in the same direction with same speed.\n"
},
{
"text": "in opposite directions with speed in the ratio of $$2: 1$$ respectively.\n"
},
{
"text": "in opposite directions with the same speed."
}
],
"answer": "in opposite directions with speed in the ratio of $$1: 2$$ respectively.\n",
"solution": "**Answer:** in opposite directions with speed in the ratio of $$1: 2$$ respectively.\n\n\n
In nuclear disintegration, the conservation of momentum plays a crucial role in determining the motion of the resulting fragments. Since the original nucleus is at rest, its total initial momentum is zero. After the disintegration, the total momentum of the system must still be zero to conserve momentum.
\n\n
Given the mass ratio of the resulting two smaller nuclei is $2:1$, let's denote the masses of the two nuclei as $2m$ and $m$, respectively.
\n\n
Applying Conservation of Momentum
\n\n
For the nucleus with mass $2m$ and velocity $v_1$ and for the nucleus with mass $m$ and velocity $v_2$, the conservation of momentum equation is:
\n\n
$ 2m \\cdot v_1 + m \\cdot v_2 = 0 $
\n\n
Given that momentum is a vector quantity, and the total initial momentum was zero, the nuclei must move in opposite directions for their momenta to cancel each other out. Therefore, we rearrange the equation:
\n\n
$ 2m \\cdot v_1 = -m \\cdot v_2 $
\n\n
$ 2 \\cdot v_1 = -v_2 $
\n\n
$ v_2 = -2 \\cdot v_1 $
\n\n
The negative sign indicates that $v_1$ and $v_2$ are in opposite directions. Taking magnitudes and considering the ratio:
This equation shows that the velocity of the lighter fragment (mass $m$) is twice the velocity of the heavier fragment (mass $2m$). Since they must move in opposite directions for momentum conservation, this confirms:
\n\n
Correct Answer:
\n\n
Option A - in opposite directions with speed in the ratio of $1:2$ respectively.
The energy released in the fusion of $$2 \\mathrm{~kg}$$ of hydrogen deep in the sun is $$E_H$$ and the energy released in the fission of $$2 \\mathrm{~kg}$$ of $${ }^{235} \\mathrm{U}$$ is $$E_U$$. The ratio $$\\frac{E_H}{E_U}$$ is approximately:\n(Consider the fusion reaction as $$4_1^1H+2 \\mathrm{e}^{-} \\rightarrow{ }_2^4 \\mathrm{He}+2 v+6 \\gamma+26.7 \\mathrm{~MeV}$$, energy released in the fission reaction of $${ }^{235} \\mathrm{U}$$ is $$200 \\mathrm{~MeV}$$ per fission nucleus and $$\\mathrm{N}_{\\mathrm{A}}= 6.023 \\times 10^{23})$$
To determine the ratio $$\\frac{E_H}{E_U}$$, let's first consider the energy released in the fusion of hydrogen and the energy released in the fission of $$^{235}U$$.
Next, to convert this energy into electron volts ($$\\mathrm{eV}$$), we use the conversion factor: $$1 \\mathrm{~J} = 6.242 \\times 10^{12} \\mathrm{~MeV}$$.
A star has $$100 \\%$$ helium composition. It starts to convert three $${ }^4 \\mathrm{He}$$ into one $${ }^{12} \\mathrm{C}$$ via triple alpha process as $${ }^4 \\mathrm{He}+{ }^4 \\mathrm{He}+{ }^4 \\mathrm{He} \\rightarrow{ }^{12} \\mathrm{C}+\\mathrm{Q}$$. The mass of the star is $$2.0 \\times 10^{32} \\mathrm{~kg}$$ and it generates energy at the rate of $$5.808 \\times 10^{30} \\mathrm{~W}$$. The rate of converting these $${ }^4 \\mathrm{He}$$ to $${ }^{12} \\mathrm{C}$$ is $$\\mathrm{n} \\times 10^{42} \\mathrm{~s}^{-1}$$, where $$\\mathrm{n}$$ is _________. [ Take, mass of $${ }^4 \\mathrm{He}=4.0026 \\mathrm{u}$$, mass of $${ }^{12} \\mathrm{C}=12 \\mathrm{u}$$]
To determine the rate of converting $${ }^4 \\mathrm{He}$$ to $${ }^{12} \\mathrm{C}$$, we need to calculate the energy released per reaction and use the given power production of the star to find the rate of reactions. The relevant nuclear reaction is:
We use Einstein's mass-energy equivalence principle, $$E = mc^2$$, to find the energy released per reaction. The conversion factor between atomic mass units and energy is $$1 \\, \\mathrm{u} = 931.5 \\, \\mathrm{MeV}$$.
\n\n
\n\n
Energy released per reaction: $$0.0078 \\, \\mathrm{u} \\times 931.5 \\, \\mathrm{MeV/u} = 7.2627 \\, \\mathrm{MeV}$$
\n\n\n\n
We convert this energy into joules. $$1 \\, \\mathrm{MeV} = 1.60218 \\times 10^{-13} \\, \\mathrm{J}$$:
\n\n
\n\n
Energy per reaction: $$7.2627 \\, \\mathrm{MeV} \\times 1.60218 \\times 10^{-13} \\, \\mathrm{J/MeV} = 1.163 \\times 10^{-12} \\, \\mathrm{J}$$
\n\n\n\n
The power generated by the star is given as $$5.808 \\times 10^{30} \\, \\mathrm{W}$$. The rate of the reaction is the power divided by the energy per reaction:
\n\n
\n\n
$$\\text{Rate of reactions} = \\frac{\\text{Power}}{\\text{Energy per reaction}}$$
Which of the following nuclear fragments corresponding to nuclear fission between neutron $$\\left({ }_0^1 \\mathrm{n}\\right)$$ and uranium isotope $$\\left({ }_{92}^{235} \\mathrm{U}\\right)$$ is correct :
In order to identify the correct nuclear fragments resulting from the fission of uranium-235 by a neutron, $$\\left({ }_0^1 \\mathrm{n}\\right) + \\left({ }_{92}^{235} \\mathrm{U}\\right)$$, we need to apply the conservation of mass number and atomic number. These conservation laws tell us that the sum of mass numbers (top numbers, representing the total count of protons and neutrons) and the sum of atomic numbers (bottom numbers, representing the total count of protons) before and after the fission must be equal. Let's apply these rules to each option:
\n\n
For the uranium-235 fission reaction, before the reaction, the total mass number is $$235 + 1 = 236$$ and the total atomic number is $$92$$ (since a neutron doesn't contribute to the atomic number).
Total mass number after = $$144 + 89 + 3(1) = 233 + 3 = 236$$
\n\n
Total atomic number after = $$56 + 36 = 92$$
\n\n
\n\n\n\n
By examining the conservation of mass numbers and atomic numbers, Option D is identified as the correct option because both the total mass number and atomic number after the reaction match exactly with the total mass number and atomic number before the reaction. Therefore, the nuclear fission fragments and the neutron count for the reaction are accurately represented by $${ }_{56}^{144} \\mathrm{Ba}+{ }_{36}^{89} \\mathrm{Kr}+3{ }_0^1 \\mathrm{n}$$.
The disintegration energy $$Q$$ for the nuclear fission of $${ }^{235} \\mathrm{U} \\rightarrow{ }^{140} \\mathrm{Ce}+{ }^{94} \\mathrm{Zr}+n$$ is\n_______ $$\\mathrm{MeV}$$.
\n
Given atomic masses of $${ }^{235} \\mathrm{U}: 235.0439 u ;{ }^{140} \\mathrm{Ce}: 139.9054 u, { }^{94} \\mathrm{Zr}: 93.9063 u ; n: 1.0086 u$$,\nValue of $$c^2=931 \\mathrm{~MeV} / \\mathrm{u}$$.
To determine the nuclear binding energy of the isotope $$_5^{12}B$$, we need to consider the mass defect concept. The mass defect is the difference between the sum of the individual masses of nucleons (protons and neutrons) and the actual mass of the nucleus.
\n\n
Let's calculate the mass defect first. The isotope $$_5^{12}B$$ has 5 protons and 7 neutrons (since the total number of nucleons is 12). Therefore, the mass defect can be written as:
\n\n
$$\\Delta M = (5 M_p + 7 M_n) - M_0$$
\n\n
Once we have the mass defect, the binding energy can be found using Einstein's mass-energy equivalence principle, which is given by:
\n\n
$$E = \\Delta M \\cdot c^2$$
\n\n
Substituting the mass defect into this equation, we get:
$$^{236}_{92} X \\rightarrow \\, ^{141}_{56} Y + \\, ^{92}_{36} Z + 3 R$$
\n\n
We need to determine the identity of particles denoted by $ R $. Let's use the conservation of charge and mass number (nucleon number) to identify $ R $.
\n\n
First, for the conservation of nucleon number (mass number), we have:
\n\n
$$236 = 141 + 92 + 3 \\times A_R$$
\n\n
Where $ A_R $ is the mass number of $ R $. This simplifies to:
\n\n
$$236 = 233 + 3A_R$$
\n\n
$$3A_R = 236 - 233$$
\n\n
$$3A_R = 3$$
\n\n
$$A_R = 1$$
\n\n
Next, we use the conservation of charge (atomic number), we have:
\n\n
$$92 = 56 + 36 + 3Z_R$$
\n\n
Where $ Z_R $ is the atomic number of $ R $. This simplifies to:
\n\n
$$92 = 92 + 3Z_R$$
\n\n
$$3Z_R = 92 - 92$$
\n\n
$$3Z_R = 0$$
\n\n
$$Z_R = 0$$
\n\n
Since the particle $ R $ has a mass number of 1 and atomic number of 0, it must be a neutron.
\n\n
So, the identity of the emitted particles $ R $ is:
Binding energy of a certain nucleus is $$18 \\times 10^8 \\mathrm{~J}$$. How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus:
To determine the difference between the total mass of all the nucleons and the nuclear mass of the given nucleus, we need to use the concept of mass-energy equivalence provided by Einstein's famous equation:
\n\n
$$ E= \\Delta m \\cdot c^2 $$
\n\n
where:\n\n
\n\n
E is the binding energy.
\n\n
$$ \\Delta m $$ is the mass defect (difference in mass).
\n\n
$$ c $$ is the speed of light in vacuum, which is approximately $$3 \\times 10^8 \\, \\text{m/s}$$.
If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is\n________ $$\\times 10^{-2} \\mathrm{~MeV}$$. (Given $$1 \\mathrm{u}=931 \\mathrm{~MeV} / \\mathrm{c}^2$$, atomic mass of helium $$=4.002603 \\mathrm{u}$$)
To find the energy released when three helium nuclei combine to form a carbon nucleus, we first need to understand that this process is essentially nuclear fusion, forming a heavier nucleus from lighter ones. The mass defect in this fusion process is the key to calculating the energy released, according to Einstein's equation $$E = \\Delta mc^2$$, where $$E$$ is the energy released, $$\\Delta m$$ is the mass defect, and $$c$$ is the speed of light.
\n\n
The atomic mass of a helium nucleus (also known as an alpha particle) is $$4.002603 \\, \\text{u}$$.
\n\n
1. Calculate the total initial mass of three helium nuclei:
2. The atomic mass of a carbon nucleus formed by the fusion of three helium nuclei is not directly given, but we can infer it's approximately $$12 \\, \\text{u}$$, based on knowledge of isotopes and considering that the question appears to simplify the carbon nucleus to a mass number of 12 (common carbon-12 isotope).
\n\n
3. Calculate the mass defect ($$\\Delta m$$):
\n
$$\\Delta m = \\text{Total initial mass} - \\text{Final mass}$$
4. Convert the mass defect to energy. Given $$1 \\, \\text{u} = 931 \\, \\text{MeV/c}^2$$, the energy released is calculated using the formula $$E = \\Delta mc^2$$:
Therefore, the energy released in this reaction is approximately $$727.1839 \\times 10^{-2} \\, \\text{MeV}$$. The exact value might differ slightly depending on how the atomic mass of the carbon nucleus is considered or rounded in specific scenarios, but based on the information provided, this is a suitable approximation.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8157,
"subject": "Physics",
"question": "If $${N_0}$$ is the original mass of the substance of half-life period $${t_{1/2}} = 5$$ years, then the amount of substance left after $$15$$ years is ",
"options": [
{
"text": "$${N_0}/8$$ "
},
{
"text": "$${N_0}/16$$"
},
{
"text": "$${N_0}/2$$"
},
{
"text": "$${N_0}/4$$"
}
],
"answer": "$${N_0}/8$$ ",
"solution": "**Answer:** $${N_0}/8$$ \n\nAfter every half-life, the mass of the substance reduces to half its initial value.\n
$${N_0}\\mathop \\to \\limits^{5\\,years} \\,\\,{{{N_0}} \\over 2}\\mathop \\to \\limits^{5\\,years} {{{N_0}} \\over {{2^2}}}\\mathop \\to \\limits^{5\\,years} {{{N_0}} \\over 8}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8158,
"subject": "Physics",
"question": "At a specific instant emission of radioactive compound is deflected in a magnetic field. The compound can emit \n $$\\eqalign{\n & \\left( i \\right)\\,\\,\\,\\,\\,\\,\\,electrons\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( {ii} \\right)\\,\\,\\,\\,\\,\\,\\,protons \\cr \n & \\left( {iii} \\right)\\,\\,\\,H{e^{2 + }}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( {iv} \\right)\\,\\,\\,\\,\\,\\,\\,neutrons \\cr} $$\n
The emission at instant can be
",
"options": [
{
"text": "$$i, ii, iii$$ "
},
{
"text": "$$i, ii, iii, iv$$ "
},
{
"text": "$$iv$$ "
},
{
"text": "$$ii, iii$$ "
}
],
"answer": "$$i, ii, iii$$ ",
"solution": "**Answer:** $$i, ii, iii$$ \n\nCharged particles are deflected in magnetic field.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8159,
"subject": "Physics",
"question": "When a $${U^{238}}$$ nucleus originally at rest, decays by emitting an alpha particle having a speed $$'u',$$ the recoil speed of the residual nucleus is ",
"options": [
{
"text": "$${{4\\mu } \\over {238}}$$ "
},
{
"text": "$$ - {{4\\mu } \\over {234}}$$ "
},
{
"text": "$$ {{4\\mu } \\over {234}}$$"
},
{
"text": "$$ - {{4\\mu } \\over {238}}$$"
}
],
"answer": "$$ {{4\\mu } \\over {234}}$$",
"solution": "**Answer:** $$ {{4\\mu } \\over {234}}$$\n\nHere, conservation of linear momentum can be applied\n
\n
$$238 \\times 0 = 4u + 234v $$\n
$$\\therefore$$ $$v = - {4 \\over {234}}u$$\n
$$\\therefore$$ speed $$ = |\\overrightarrow v | = {4 \\over {234}}u$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8160,
"subject": "Physics",
"question": "A radioactive sample at any instant has its disintegration rate $$5000$$ disintegrations per minute. After $$5$$ minutes, the rate is $$1250$$ disintegrations per minute. Then, the decay constant (per minute) is ",
"options": [
{
"text": "$$0.4$$ $$ln2$$ "
},
{
"text": "$$0.2$$ $$ln2$$ "
},
{
"text": "$$0.1$$ $$ln2$$ "
},
{
"text": "$$0.8$$ $$ln2$$ "
}
],
"answer": "$$0.4$$ $$ln2$$ ",
"solution": "**Answer:** $$0.4$$ $$ln2$$ \n\n$$\\lambda = {1 \\over t}{\\log _e}{{{A_0}} \\over A}$$\n
Therefore the atomic number should decrease by $$16$$\n
The number of $${\\beta ^ - }$$ - particles released $$=4$$\n
Therefore the atomic number should increase by $$4.$$\n
Also the number of $${\\beta ^ + }$$ particles released is $$2,$$ which \n
should decrease the atomic number by $$2.$$\n
Therefore the final atomic number is \n
$$92-16+4-2=78$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8162,
"subject": "Physics",
"question": "Which of the following cannot be emitted by radioactive substances during their decay ? ",
"options": [
{
"text": "Protons "
},
{
"text": "Neutrinoes "
},
{
"text": "Helium nuclei "
},
{
"text": "Electrons "
}
],
"answer": "Protons ",
"solution": "**Answer:** Protons \n\nThe radioactive substances emit $$\\alpha $$ -particles (Helium nucleus), $$\\beta $$ -particles (electrons) and neutrinoes. ",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8163,
"subject": "Physics",
"question": "Which of the following radiations has the least wavelength ? ",
"options": [
{
"text": "$$\\gamma $$ - rays "
},
{
"text": "$$\\beta $$ - rays "
},
{
"text": "$$\\alpha $$ - rays"
},
{
"text": "$$X$$ - rays "
}
],
"answer": "$$\\gamma $$ - rays ",
"solution": "**Answer:** $$\\gamma $$ - rays \n\nThe electromagnetic spectrum is as follows\n
\n
$$\\therefore$$ $$\\gamma $$-rays has least wavelength.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8164,
"subject": "Physics",
"question": "A nucleus disintegrated into two nuclear parts which have their velocities in the ratio of $$2:1.$$ The ratio of their nuclear sizes will be",
"options": [
{
"text": "$${3^{{\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}}:1$$ "
},
{
"text": "$$1:{2^{1/3}}$$ "
},
{
"text": "$${2^{1/3}}:1$$ "
},
{
"text": "$$1:{3^{{\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}}$$ "
}
],
"answer": "$$1:{2^{1/3}}$$ ",
"solution": "**Answer:** $$1:{2^{1/3}}$$ \n\nFrom conservation of momentum $${m_1}{v_1} = {m_2}{v_2}$$\n
$$ \\Rightarrow T = $$ half life period $$ = {{15} \\over 3} = 5\\,\\,$$ minutes",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8167,
"subject": "Physics",
"question": "The intensity of gamma radiation from a given source is $$L$$. On passing through $$36$$ $$mm$$ of lead, it is reduced to $${{\\rm I} \\over 8}.$$ The thickness of lead which will reduce the intensity to $${{\\rm I} \\over 2}$$ will be ",
"options": [
{
"text": "$$9mm$$ "
},
{
"text": "$$6mm$$ "
},
{
"text": "$$12mm$$ "
},
{
"text": "$$18mm$$ "
}
],
"answer": "$$12mm$$ ",
"solution": "**Answer:** $$12mm$$ \n\nKEY CONCEPT : Intensity $$I = {I_0}.{e^{ - \\mu d}},$$\n
or $$\\,\\,\\,d = {{36} \\over 3} = 12\\,mm$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8168,
"subject": "Physics",
"question": "The $$'rad'$$ is the correct unit used to report the measurement of ",
"options": [
{
"text": "the ability of a beam of gamma ray photons to produce ions in a target "
},
{
"text": "the energy delivered by radiation to a target "
},
{
"text": "the biological effect of radiation "
},
{
"text": "the rate of decay of radioactive source "
}
],
"answer": "the biological effect of radiation ",
"solution": "**Answer:** the biological effect of radiation \n\nThe risk posed to a human being by any radiation exposure depends partly upon the absorbed dose, the amount of energy absorbed per gram of tissue. Absorbed dose is expressed in rad. A rad is equal to $$100$$ $$ergs$$ of energy absorbed by $$1$$ gram of tissue. The more modern, internationally adopted unit is the gray (named after the English medical physicist $$L.$$ $$H.$$ Gray); one gray equals $$100$$ rad.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8169,
"subject": "Physics",
"question": "In gamma ray emission from a nucleus",
"options": [
{
"text": "only the proton number changes "
},
{
"text": "both the neutron number and the proton number change "
},
{
"text": "there is no change in the proton number and the neutron number "
},
{
"text": "only the neutron number changes "
}
],
"answer": "there is no change in the proton number and the neutron number ",
"solution": "**Answer:** there is no change in the proton number and the neutron number \n\nThere is no change in the proton number and the neutron number as the $$\\gamma $$ - emission takes place as a result of excitation or de-excitation of nuclei. $$\\gamma $$-rays have no charge or mass.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8170,
"subject": "Physics",
"question": "The half-life period of a ratio-active element $$X$$ is same as the mean life time of another ratio-active element $$Y.$$ Initially they have the same number of atoms. Then ",
"options": [
{
"text": "$$X$$ and $$Y$$ decay at same rate always "
},
{
"text": "$$X$$ will decay faster than $$Y$$ "
},
{
"text": "$$Y$$ will decay faster than $$X$$ "
},
{
"text": "$$X$$ and $$Y$$ have same decay rate initially "
}
],
"answer": "$$Y$$ will decay faster than $$X$$ ",
"solution": "**Answer:** $$Y$$ will decay faster than $$X$$ \n\nAccording to question, \n
Half life of $$X,\\,{T_{1/2}} = {\\tau _{av}},\\,\\,\\,$$ average life of $$Y$$\n
$$\\therefore$$ $$Y$$ will decay faster than $$X.$$ [ as $$N$$ is same ]",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8171,
"subject": "Physics",
"question": "A radioactive nucleus (initial mass number $$A$$ and atomic number $$Z$$ emits $$3\\,\\alpha $$- particles and $$2$$ positrons. The ratio of number of neutrons to that of protons in the final nucleus will be ",
"options": [
{
"text": "$${{A - Z - 8} \\over {Z - 4}}$$ "
},
{
"text": "$${{A - Z - 4} \\over {Z - 8}}$$ "
},
{
"text": "$${{A - Z - 12} \\over {Z - 4}}$$ "
},
{
"text": "$${{A - Z - 4} \\over {Z - 2}}$$ "
}
],
"answer": "$${{A - Z - 4} \\over {Z - 8}}$$ ",
"solution": "**Answer:** $${{A - Z - 4} \\over {Z - 8}}$$ \n\n$${}_Z^AX\\mathop \\to \\limits^{A - 12} {}_{Z - 8}Y + 3{}_2^4{X_e} + {}_t^0e$$\n
Number of protons, Np = Z - 8\n
Number of neutrons, Nn = $$A - 12 - \\left( {Z - 8} \\right)$$\n
$$\\therefore$$ Required ratio $$ = {{A - Z - 4} \\over {Z - 8}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8172,
"subject": "Physics",
"question": "The half life of a radioactive substance is $$20$$ minutes. The approximate time interval $$\\left( {{t_2} - {t_1}} \\right)$$ between the time $${{t_2}}$$ when $${2 \\over 3}$$ of it had decayed and time $${{t_1}}$$ when $${1 \\over 3}$$ of it had decayed is : ",
"options": [
{
"text": "$$14$$ min "
},
{
"text": "$$20$$ min "
},
{
"text": "$$28$$ min "
},
{
"text": "$$7$$ min "
}
],
"answer": "$$20$$ min ",
"solution": "**Answer:** $$20$$ min \n\nNumber of undecayed atom after time $${t_2};$$\n
$${{{N_0}} \\over 3} = {N_0}{e^{ - \\lambda {t_2}}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( i \\right)$$\n
$${t_2} - {t_1} = 20\\,$$ min",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8173,
"subject": "Physics",
"question": "Half-lives of two radioactive elements $$A$$ and $$B$$ are $$20$$ minutes and $$40$$ minutes, respectively. Initially, the samples have equal number of nuclei. After $$80$$ minutes, the ratio of decayed number of $$A$$ and $$B$$ nuclei will be: ",
"options": [
{
"text": "$$1:4$$ "
},
{
"text": "$$5:4$$ "
},
{
"text": "$$1:16$$ "
},
{
"text": "$$4:1$$ "
}
],
"answer": "$$5:4$$ ",
"solution": "**Answer:** $$5:4$$ \n\nFor $${A_{t{\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}} = 20\\,\\,$$ min, $$t=80$$ min, number of half lifes $$n=4$$\n
For $${B_{t{\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}} = 40\\,\\,$$ min, $$t=80$$ min, number of half lifes $$n=4$$\n
$$ = {{15} \\over {16}} \\times {4 \\over 3} = {5 \\over 4}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8174,
"subject": "Physics",
"question": "A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At\nsometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by : ",
"options": [
{
"text": "$$t = {T \\over {\\log (1.3)}}$$ "
},
{
"text": "$$t = T\\log (1.3)$$ "
},
{
"text": "$$t = {T \\over 2}{{\\log 2} \\over {\\log 1.3}}$$ "
},
{
"text": "$$t = T{{\\log 1.3} \\over {\\log 2}}$$ "
}
],
"answer": "$$t = T{{\\log 1.3} \\over {\\log 2}}$$ ",
"solution": "**Answer:** $$t = T{{\\log 1.3} \\over {\\log 2}}$$ \n\nLet initially there are total N0 number of nuclei.\n
At time t $${{{N_B}} \\over {{N_A}}}$$ = 0.3 (given)\n
$$ \\Rightarrow $$ $${{N_B}}$$ = 0.3$${{N_A}}$$\n
N0 = NA + NB = NA + 0.3NA\n
$$ \\therefore $$ NA = $${{{N_0}} \\over {1.3}}$$\n
If T is half-life, then $$\\lambda $$ = $${{{{\\ln} } 2} \\over T}$$\n
$$ \\Rightarrow $$ t = $${{\\ln \\left( {1.3} \\right)} \\over {{{\\ln 2} \\over T}}}$$ = $${{\\ln \\left( {1.3} \\right)} \\over {\\ln 2}}T$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8175,
"subject": "Physics",
"question": "A solution containing active cobalt $${^{60}_{27}}Co$$ having activity of $$0.8$$ $$\\mu Ci$$ and decay constant $$\\lambda $$ is injected in an animal's body. If $$1\\,c{m^3}$$ of blood is drawn from the animal's body after $$10$$ hrs of injection, the activity found was $$300$$ decays per minute What is the volume of blood that is flowing in the body ? $$\\left( {\\,\\,Ci = 3.7 \\times {{10}^{10}}\\,} \\right.$$ decays per second and at $$t=10$$ hrs $$\\left. {{e^{ - \\lambda t}} = 0.84} \\right)$$",
"options": [
{
"text": "$$6$$ liters"
},
{
"text": "$$7$$ liters"
},
{
"text": "$$4$$ liters"
},
{
"text": "$$5$$ liters"
}
],
"answer": "$$5$$ liters",
"solution": "**Answer:** $$5$$ liters\n\nInitial activity, No = 0.8 $$\\mu $$Ci\n
Activity at time t, N = N0 e$$-$$$$\\lambda $$t\n
Activity in 1 cm3 blood after 10 hr, n = 300 decays per minute \n
= $${{300} \\over {60}}\\,dps$$\n
= 5 dps.\n
Activity in whole blood after to hr = N0 e$$-$$$$\\lambda $$$$ \\times $$10\n
$$\\therefore\\,\\,\\,\\,$$ Volume of the total blood = $${{{N_0}{e^{ - 10\\lambda }}} \\over n}$$\n
$$ \\simeq $$ 5 liters.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8176,
"subject": "Physics",
"question": "An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of 8 : 27. The ratio of the radii of the nuclei (assumed to be spherical) is : ",
"options": [
{
"text": "8 : 27"
},
{
"text": "4 : 9"
},
{
"text": "3 : 2"
},
{
"text": "2 : 3"
}
],
"answer": "3 : 2",
"solution": "**Answer:** 3 : 2\n\n
The two nuclei have velocity in ratio 8 : 27. By conservation of momentum, we have
Thus, ratio of radii of nuclei $${r_1}:{r_2} = 3:2$$.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8177,
"subject": "Physics",
"question": "At some instant, a radioactive sample S1 having an activity 5 $$\\mu $$Ci has twice the number of nuclei as another sample S2 which has an activity of 10 $$\\mu $$Ci. The half lives of S1 and S2 are : ",
"options": [
{
"text": "20 years and 5 years, respectively "
},
{
"text": "20 years and 10years, respectively "
},
{
"text": "5 years and 20 years, respectively "
},
{
"text": "10 years and 20 years, respectively"
}
],
"answer": "20 years and 5 years, respectively ",
"solution": "**Answer:** 20 years and 5 years, respectively \n\nAs per question, N1\n = 2N2\n
So half life of sample S1 should be four timrs than sample S2.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8178,
"subject": "Physics",
"question": "Half lives of two radioactive nuclei A and B are 10 minutes and 20 minutes, respectively, If initially a sample\nhas equal number of nuclei, then after 60 minutes, the ratio of decayed numbers of nuclei A and B will be :\n",
"options": [
{
"text": "9 : 8"
},
{
"text": "1 : 8"
},
{
"text": "8 : 1"
},
{
"text": "3 : 8"
}
],
"answer": "9 : 8",
"solution": "**Answer:** 9 : 8\n\nNA = N0$${\\left( {{1 \\over 2}} \\right)^{{t \\over {{T_{{1 \\over 2}}}}}}}$$\n = N0$${\\left( {{1 \\over 2}} \\right)^{{{60} \\over {10}}}}$$\n = N0$${\\left( {{1 \\over 2}} \\right)^6}$$\n
\n$$ \\Rightarrow t = {2 \\over {4\\lambda }} = \\left( {{1 \\over {2\\lambda }}} \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8180,
"subject": "Physics",
"question": "Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to -\n",
"options": [
{
"text": "200"
},
{
"text": "150"
},
{
"text": "400"
},
{
"text": "360"
}
],
"answer": "200",
"solution": "**Answer:** 200\n\nat t = 0, A0 = $${{dN} \\over {dt}}$$ = 1600C/s\n
at t = 8s, A = 100 C/s\n
$${A \\over {{A_0}}}$$ = $${1 \\over {16}}$$ in 8 sec\n
Therefor half life is t1/2 = 2 sec\n
$$ \\therefore $$ Activity at t = 6 will be 1600$${\\left( {{1 \\over 2}} \\right)^3}$$ \n
= 200 C/s\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8181,
"subject": "Physics",
"question": "Two radioactive materials A and B have decay\nconstants 10$$\\lambda $$ and $$\\lambda $$, respectively. It initially\nthey have the same number of nuclei, then the\nratio of the number of nuclei of A to that of B\nwill be 1/e after a time :",
"options": [
{
"text": "1/9$$\\lambda $$"
},
{
"text": "11/10$$\\lambda $$"
},
{
"text": "1/10$$\\lambda $$"
},
{
"text": "1/11$$\\lambda $$"
}
],
"answer": "1/9$$\\lambda $$",
"solution": "**Answer:** 1/9$$\\lambda $$\n\nN1 = N0e–10$$\\lambda $$t\n ; N2 = N0e–$$\\lambda $$t
\n$$ \\Rightarrow t = {1 \\over {9\\lambda }}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8182,
"subject": "Physics",
"question": "In a radioactive decay chain, the initial nucleus is $${}_{90}^{232}$$Th. At the end there are 6 $$\\alpha $$-particles and 4 $$\\beta $$-particles which are emitted. If the end nucleus is $${}_Z^A$$X, A and Z are given by : ",
"options": [
{
"text": "A = 208; Z = 80"
},
{
"text": "A = 208; Z = 82"
},
{
"text": "A = 200; Z = 81 "
},
{
"text": "A = 202; Z = 80"
}
],
"answer": "A = 208; Z = 82",
"solution": "**Answer:** A = 208; Z = 82\n\n$${}_{90}^{232}$$Th $$\\buildrel \\, \\over\n \\longrightarrow $$ $${}_{78}^{208}$$Y + $${}_2^4$$He\n
$${}_{78}^{208}$$Y $$\\buildrel \\, \\over\n \\longrightarrow $$ $${}_{82}^{208}$$ X + 4$$\\beta $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8183,
"subject": "Physics",
"question": "At a given instant, say t = 0, two radioactive substance A and B have equal activities. the ratio $${{{R_B}} \\over {{R_A}}}$$ of their activities after time t itself decays with time t as e$$-$$3t. If the half-life of A is ln2, the half-life of B is : ",
"options": [
{
"text": "4ln2"
},
{
"text": "$${{\\ln 2} \\over 2}$$"
},
{
"text": "$${{\\ln 2} \\over 4}$$"
},
{
"text": "2ln2"
}
],
"answer": "$${{\\ln 2} \\over 4}$$",
"solution": "**Answer:** $${{\\ln 2} \\over 4}$$\n\nWe know, \n
$$ \\therefore $$ Half life of B is $$\\left( {{t_{{1 \\over 2}}}} \\right) = {{\\ln 2} \\over {{\\lambda _B}}} = {{\\ln 2} \\over 4}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8184,
"subject": "Physics",
"question": "A sample of radioactive material A, that has an activity of 10 mCi(1 Ci = 3.7 $$ \\times $$ 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive materail B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively : ",
"options": [
{
"text": "5 days and 10 days"
},
{
"text": "10 days and 40 days"
},
{
"text": "20 days and 5 days"
},
{
"text": "20 days and 10 days"
}
],
"answer": "20 days and 5 days",
"solution": "**Answer:** 20 days and 5 days\n\nLet number of nuclei present in material A is NA and in meterial B is NB According to question, \n
$$ \\therefore $$ By checking options you can see possible value of $${\\left( {{T_{{1 \\over 2}}}} \\right)_A}$$ = 20 days and $${\\left( {{T_{{1 \\over 2}}}} \\right)_B}$$ = 5 days",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8185,
"subject": "Physics",
"question": "The ratio of mass densities of nuclei of 40Ca\nand 16O is close to :-",
"options": [
{
"text": "1"
},
{
"text": "5"
},
{
"text": "0.1"
},
{
"text": "2"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\nDensities of nucleus happens to be constant,\nirrespective of mass number.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8186,
"subject": "Physics",
"question": "The activity of a radioactive sample falls from 700 s–1 to 500 s–1 in 30 minutes. Its half life is close\nto:\n",
"options": [
{
"text": "62 min"
},
{
"text": "66 min"
},
{
"text": "72 min"
},
{
"text": "52 min"
}
],
"answer": "62 min",
"solution": "**Answer:** 62 min\n\nA = A0 e-$$\\lambda $$t\n
$$ \\Rightarrow $$ $$N\\left( {t/2} \\right) = {3 \\over 4}{N_0}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8188,
"subject": "Physics",
"question": "The radius R of a nucleus of mass number A can be estimated by the formula R = (1.3 $$ \\times $$ 10–15)A1/3 m.\n It follows that the mass density of a nucleus is of the order of :\n
(Mprot. $$ \\cong $$ Mneut $$ \\simeq $$ 1.67 $$ \\times $$ 10–27 kg)",
"options": [
{
"text": "1024 kg m–3"
},
{
"text": "1010 kg m–3"
},
{
"text": "1017 kg m–3"
},
{
"text": "103 kg m–3"
}
],
"answer": "1017 kg m–3",
"solution": "**Answer:** 1017 kg m–3\n\n$$R = (1.3 \\times {10^{ - 15}}){A^{{1 \\over 3}}}$$
$$p \\approx {10^{17}}kg/m$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8189,
"subject": "Physics",
"question": "A radioactive nucleus decays by two different\nprocesses. The half life for the first process is\n10 s and that for the second is 100 s. The\neffective half life of the nucleus is close to :",
"options": [
{
"text": "12 sec"
},
{
"text": "9 sec"
},
{
"text": "55 sec"
},
{
"text": "6 sec"
}
],
"answer": "9 sec",
"solution": "**Answer:** 9 sec\n\nT1 = 10 sec\n
Correct is e+ + e- $$ \\to $$ 2$$\\gamma $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8191,
"subject": "Physics",
"question": "Two radioactive substances X and Y originally have N1 and N2 nuclei respectively. Half life of X is half of the half life of Y. After three half lives of Y, number of nuclei of both are equal. The ratio $${{{N_1}} \\over {{N_2}}}$$ will be equal to :",
"options": [
{
"text": "$${1 \\over 8}$$"
},
{
"text": "$${3 \\over 1}$$"
},
{
"text": "$${1 \\over 3}$$"
},
{
"text": "$${8 \\over 1}$$"
}
],
"answer": "$${8 \\over 1}$$",
"solution": "**Answer:** $${8 \\over 1}$$\n\nLet Half life of x = t
then half life of y = 2t
when 3 half life of y is completed then 6 half life of x is completed.
$$ \\therefore $$ Now x have = $${{{N_1}} \\over {{2^6}}}$$ nuclei
$${T_{eff}} = {{{T_1}{T_2}} \\over {{T_1} + {T_2}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8196,
"subject": "Physics",
"question": "The decay of a proton to neutron is :",
"options": [
{
"text": "always possible as it is associated only with $$\\beta$$+ decay"
},
{
"text": "possible only inside the nucleus"
},
{
"text": "not possible as proton mass is less than the neutron mass"
},
{
"text": "not possible but neutron to proton conversation is possible"
}
],
"answer": "possible only inside the nucleus",
"solution": "**Answer:** possible only inside the nucleus\n\nPositron emission or Beta plus decay is a subtype of radioactive decay called Beta decay, in which a proton inside a nucleus is converted into a neutron while releasing a positron and an electron neutrino.\n
So, decay of a proton to neutron is possible only inside the nucleus. Free proton cannot decay to neutron as mass of proton is less compared to neutron so to decay into higher mass, proton need extra energy which free proton can’t get, only proton inside nucleus can get.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8197,
"subject": "Physics",
"question": "A radioactive material decays by simultaneous emissions of two particles with half lives of 1400 years and 700 years respectively. What will be the time after which one third of the material remains ? (Take ln 3 = 1.1)",
"options": [
{
"text": "740 years"
},
{
"text": "1110 years"
},
{
"text": "700 years"
},
{
"text": "340 years"
}
],
"answer": "740 years",
"solution": "**Answer:** 740 years\n\nThe given situation can be shown as
Here, radioactive material X is decayed into two particles Y and Z with their respective decay constant, $$\\lambda$$a and $$\\lambda$$b. It means that
$$\\Rightarrow$$ t $$\\approx$$ 740 yr",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8198,
"subject": "Physics",
"question": "A radioactive substance decays to $${\\left( {{1 \\over {16}}} \\right)^{th}}$$ of its initial activity in 80 days. The half life of the radioactive substance expressed in days is ____________.",
"options": [],
"answer": "20",
"solution": "**Answer:** 20\n\n$${N_0}\\buildrel {{{{t_1}} \\over 2}} \\over\n \\longrightarrow {{{N_0}} \\over 2}\\buildrel {{{{t_1}} \\over 2}} \\over\n \\longrightarrow {{{N_0}} \\over 4}\\buildrel {{{{t_1}} \\over 2}} \\over\n \\longrightarrow {{{N_0}} \\over 8}\\buildrel {{{{t_1}} \\over 2}} \\over\n \\longrightarrow {{{N_0}} \\over {16}}$$
$$4 \\times {t_{1/2}} = 80$$
$${t_{1/2}} = 20$$ days",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8199,
"subject": "Physics",
"question": "Some nuclei of a radioactive material are undergoing radioactive decay. The time gap between the instances when a quarter of the nuclei have decayed and when half of the nuclei have decayed is given as :
$$A = \\lambda N = 1.618 \\times {10^{13}} = 16.18 \\times {10^{12}}$$ dps",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8201,
"subject": "Physics",
"question": "The nuclear activity of a radioactive element becomes $${\\left( {{1 \\over 8}} \\right)^{th}}$$ of its initial value in 30 years. The half-life of radioactive element is _____________ years.",
"options": [],
"answer": "10",
"solution": "**Answer:** 10\n\nWe know, $$A = {A_0}{e^{ - \\lambda t}}$$\n
$$ \\Rightarrow $$ $${t_{1/2}}$$ = 10 years",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8202,
"subject": "Physics",
"question": "If 'f' denotes the ratio of the number of nuclei decayed (Nd) to the number of nuclei at t = 0 (N0) then for a collection of radioactive nuclei, the rate of change of 'f' with respect to time is given as :
$$ \\Rightarrow $$ $${{df} \\over {dt}} = \\lambda {e^{ - \\lambda t}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8203,
"subject": "Physics",
"question": "A radioactive sample has an average life of 30 ms and is decaying. A capacitor of capacitance 200 $$\\mu$$F is first charged and later connected with resistor 'R'. If the ratio of charge on capacitor to the activity of radioactive sample is fixed with respect to time then the value of 'R' should be _____________ $$\\Omega$$.",
"options": [],
"answer": "150",
"solution": "**Answer:** 150\n\nTm =30 ms
A. Atoms of each element emit characteristics spectrum.
B. According to Bohr's Postulate, an electron in a hydrogen atom, revolves in a certain stationary orbit.
C. The density of nuclear matter depends on the size of the nucleus.
D. A free neutron is stable but a free proton decay is possible.
E. Radioactivity is an indication of the instability of nuclei.
Choose the correct answer from the options given below :",
"options": [
{
"text": "A, B, C, D and E"
},
{
"text": "A, B and E only"
},
{
"text": "B and D only"
},
{
"text": "A, C and E only"
}
],
"answer": "A, B and E only",
"solution": "**Answer:** A, B and E only\n\n(A) True, atom of each element emits characteristic spectrum.
(B) True, according to Bohr's postulates $$mvr = {{nh} \\over {2\\pi }}$$ and hence electron resides into orbits of specific radius called stationary orbits.
(C) False, density of nucleus is constant.
(D) False, A free neutron is unstable decays into proton and electron and antineutrino.
(E) True, unstable nucleus show radioactivity.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8205,
"subject": "Physics",
"question": "There are 1010 radioactive nuclei in a given radioactive element, its half-life time is 1 minute. How many nuclei will remain after 30 seconds? $$\\left( {\\sqrt 2 = 1.414} \\right)$$",
"options": [
{
"text": "2 $$\\times$$ 1010"
},
{
"text": "7 $$\\times$$ 109"
},
{
"text": "105"
},
{
"text": "4 $$\\times$$ 1010"
}
],
"answer": "7 $$\\times$$ 109",
"solution": "**Answer:** 7 $$\\times$$ 109\n\n$${N \\over {{N_0}}} = {\\left( {{1 \\over 2}} \\right)^{{t \\over {{t^{1/2}}}}}}$$
$$ \\Rightarrow N = {10^{10}} \\times {\\left( {{1 \\over 2}} \\right)^{{1 \\over 2}}} = {{{{10}^{10}}} \\over {\\sqrt 2 }} \\approx 7 \\times {10^9}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8206,
"subject": "Physics",
"question": "The half life period of radioactive element x is same as the mean life time of another radioactive element y. Initially they have the same number of atoms. Then :",
"options": [
{
"text": "x-will decay faster than y."
},
{
"text": "y-will decay faster than x."
},
{
"text": "x and y have same decay rate initially and later on different decay rate."
},
{
"text": "x and y decay at the same rate always."
}
],
"answer": "y-will decay faster than x.",
"solution": "**Answer:** y-will decay faster than x.\n\nGiven, ($$\\tau$$1/2)x = ($$\\tau$$)y
Here, $$\\tau$$1/2 = half-life period of radioactive element and $$\\tau$$ = mean life period of radioactive element.
Therefore, y will decay faster than x.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8207,
"subject": "Physics",
"question": "
The activity of a radioactive material is 2.56 $$\\times$$ 10$$-$$3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2 $$\\times$$ 10$$-$$5 Ci ?
The half life of a radioactive substance is 5 years. After x years a given sample of the radioactive substance gets reduced to 6.25% of its initial value. The value of x is ____________.
A radioactive nucleus can decay by two different processes. Half-life for the first process is 3.0 hours while it is 4.5 hours for the second process. The effective half-life of the nucleus will be:
A sample contains 10$$-$$2 kg each of two substances A and B with half lives 4 s and 8 s respectively. The ratio of their atomic weights is 1 : 2. The ratio of the amounts of A and B after 16 s is $${x \\over {100}}$$. The value of x is ___________.
The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minute. 10 minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is :
The activity of a radioactive material is $$6.4 \\times 10^{-4}$$ curie. Its half life is 5 days. The activity will become $$5 \\times 10^{-6}$$ curie after :
A freshly prepared radioactive source of half life 2 hours 30 minutes emits radiation which is 64 times the permissible safe level. The minimum time, after which it would be possible to work safely with source, will be _________ hours.
Two radioactive materials A and B have decay constants $$25 \\lambda$$ and $$16 \\lambda$$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of B to that of A will be \"e\" after a time $$\\frac{1}{a \\lambda}$$. The value of a is _________.
A free neutron decays into a proton but a free proton does not decay into neutron. This is because
",
"options": [
{
"text": "neutron is an uncharged particle"
},
{
"text": "neutron has larger rest mass than proton"
},
{
"text": "neutron is a composite particle made of a proton and an electron"
},
{
"text": "proton is a charged particle"
}
],
"answer": "neutron has larger rest mass than proton",
"solution": "**Answer:** neutron has larger rest mass than proton\n\nAs neutron has more rest mass than proton it will require energy to decay proton into neutron.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8224,
"subject": "Physics",
"question": "Given below are two statements: one is labelled as Assertion $\\mathbf{A}$ and the other is labelled as Reason $\\mathbf{R}$\n
\nAssertion A: The nuclear density of nuclides ${ }_{5}^{10} \\mathrm{~B},{ }_{3}^{6} \\mathrm{Li},{ }_{26}^{56} \\mathrm{Fe},{ }_{10}^{20} \\mathrm{Ne}$ and ${ }_{83}^{209} \\mathrm{Bi}$ can be arranged as $\\rho_{\\mathrm{Bi}}^{\\mathrm{N}}>\\rho_{\\mathrm{Fe}}^{\\mathrm{N}}>\\rho_{\\mathrm{Ne}}^{\\mathrm{N}}>\\rho_{\\mathrm{B}}^{\\mathrm{N}}>\\rho_{\\mathrm{Li}}^{\\mathrm{N}}$\n
\nReason R: The radius $R$ of nucleus is related to its mass number $A$ as $R=R_{0} A^{1 / 3}$, where $R_{0}$ is a constant.\n
\nIn the light of the above statements, choose the correct answer from the options given below",
"options": [
{
"text": "${Both ~\\mathbf{A}}$ and $\\mathbf{R}$ are true and $\\mathbf{R}$ is the correct explanation of $\\mathbf{A}$"
},
{
"text": "Both $\\mathbf{A}$ and $\\mathbf{R}$ are true but $\\mathbf{R}$ is NOT the correct explanation of $\\mathbf{A}$"
},
{
"text": "$\\mathbf{A}$ is false but $\\mathbf{R}$ is true"
},
{
"text": "$\\mathbf{A}$ is true but $\\mathbf{R}$ is false"
}
],
"answer": "$\\mathbf{A}$ is false but $\\mathbf{R}$ is true",
"solution": "**Answer:** $\\mathbf{A}$ is false but $\\mathbf{R}$ is true\n\n
A radioactive nucleus decays by two different process. The half life of the first process is 5 minutes and that of the second process is $30 \\mathrm{~s}$. The effective half-life of the nucleus is calculated to be $\\frac{\\alpha}{11} \\mathrm{~s}$. The value of $\\alpha$ is __________.
Substance A has atomic mass number 16 and half life of 1 day. Another substance B has atomic mass number 32 and half life of $$\\frac{1}{2}$$ day. If both A and B simultaneously start undergo radio activity at the same time with initial mass 320 g each, how many total atoms of A and B combined would be left after 2 days.
If a radioactive element having half-life of $$30 \\mathrm{~min}$$ is undergoing beta decay, the fraction of radioactive element remains undecayed after $$90 \\mathrm{~min}$$. will be
\n$$\n\\begin{aligned}\n\\text { Number of remaining } & =\\left(\\frac{N_{0}}{2^{3}}\\right) \\\\\\\\\n& =\\frac{N_{0}}{8}\n\\end{aligned}\n$$\n
\n$\\therefore \\quad$ Fraction will be $\\frac{1}{8}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8228,
"subject": "Physics",
"question": "
A radioactive element $$_{92}^{242}$$X emits two $$\\alpha$$-particles, one electron and two positrons. The product nucleus is represented by $$_{\\mathrm{P}}^{234}$$Y. The value of P is __________.
A nucleus disintegrates into two smaller parts, which have their velocities in the ratio 3 : 2. The ratio of their nuclear sizes will be $${\\left( {{x \\over 3}} \\right)^{{1 \\over 3}}}$$. The value of '$$x$$' is :-
Assume that protons and neutrons have equal masses. Mass of a nucleon is $$1.6\\times10^{-27}$$ kg and radius of nucleus is $$1.5\\times10^{-15}~\\mathrm{A^{1/3}}$$ m. The approximate ratio of the nuclear density and water density is $$n\\times10^{13}$$. The value of $$n$$ is __________.
\nDensity of water $=1000 \\mathrm{~kg} / \\mathrm{m}^{3}$\n
\n$\\frac{\\text { Density of nucleus }}{\\text { Density of water }}=\\frac{\\frac{32}{9 \\pi} \\times 10^{17}}{1000}$\n
\n$=\\frac{320}{9 \\pi} \\times 10^{13}$\n
\n$=11.32 \\times 10^{13}$\n
\nvalue of $n=11$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8233,
"subject": "Physics",
"question": "The half-life of a radioactive nucleus is 5 years. The fraction of the original sample that would decay in 15 years is:",
"options": [
{
"text": "$\\frac{1}{8}$"
},
{
"text": "$\\frac{3}{4}$"
},
{
"text": "$\\frac{7}{8}$"
},
{
"text": "$\\frac{1}{4}$"
}
],
"answer": "$\\frac{7}{8}$",
"solution": "**Answer:** $\\frac{7}{8}$\n\nThe decay of a radioactive nucleus is an exponential process, and the fraction of the original sample that remains after time $t$ is given by:\n
\n$N(t) = N_0 e^{-\\lambda t}$\n
\nwhere $N_0$ is the initial number of nuclei, $N(t)$ is the number of nuclei remaining after time $t$, and $\\lambda$ is the decay constant, which is related to the half-life $T_{1/2}$ by the equation:\n
\n$\\lambda = \\frac{\\ln(2)}{T_{1/2}}$\n
\nIn this problem, the half-life of the nucleus is given as 5 years, so we have:\n
\nTo find the fraction that has decayed, we subtract this expression from 1, since the fraction that remains plus the fraction that has decayed must add up to 1:\n
\nWe know that the half-life of the nucleus is 5 years, which means that the fraction of nuclei remaining after one half-life is 1/2. Therefore, after 3 half-lives (which is equivalent to 15 years), the fraction of nuclei remaining is:\n
\nTherefore, the fraction of the original sample that would decay in 15 years is $\\frac{7}{8}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8234,
"subject": "Physics",
"question": "
Given below are two statements: one is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$
\n
Assertion A : The binding energy per nucleon is practically independent of the atomic number for nuclei of mass number in the range 30 to 170 .
\n
Reason R : Nuclear force is short ranged.
\n
In the light of the above statements, choose the correct answer from the options given below
",
"options": [
{
"text": "$$\\mathrm{A}$$ is false but $$\\mathbf{R}$$ is true"
},
{
"text": "$$\\mathrm{A}$$ is true but $$\\mathbf{R}$$ is false"
},
{
"text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$"
},
{
"text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true but $$\\mathbf{R}$$ is NOT the correct explanation of $$\\mathbf{A}$$"
}
],
"answer": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$",
"solution": "**Answer:** Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$\n\nThe statement about the binding energy per nucleon is true, and is known as the semi-empirical mass formula. According to this formula, the binding energy per nucleon for nuclei in the range of mass numbers 30 to 170 is nearly constant, with a maximum value around mass number 60.\n
\nThe statement about nuclear force being short ranged is also true. The strong nuclear force that binds nucleons together is a short-range force that acts only over distances of a few femtometers.\n
\nTherefore, both Assertion A and Reason R are true, and Reason R provides a valid explanation for Assertion A. The correct answer is:\n
\nSo, Both Assertion A and Reason R are true, and Reason R is the correct explanation for Assertion A.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8235,
"subject": "Physics",
"question": "
Two radioactive elements A and B initially have same number of atoms. The half life of A is same as the average life of B. If $$\\lambda_{A}$$ and $$\\lambda_{B}$$ are decay constants of A and B respectively, then choose the correct relation from the given options.
We are given that the half-life of A is the same as the average life of B. The relationship between half-life ($T_{1/2}$) and the decay constant ($\\lambda$) is:
\n
$$T_{1/2} = \\frac{\\ln 2}{\\lambda}$$
\n
For the average life ($\\tau$), the relationship with the decay constant is:
\n
$$\\tau = \\frac{1}{\\lambda}$$
\n
According to the given information, the half-life of A is equal to the average life of B:
\n
$$T_{1/2(A)} = \\tau_{B}$$
\n
Now, we can substitute the relationships for half-life and average life:
The decay constant for a radioactive nuclide is 1.5 $$\\times$$ 10$$^{-5}$$ s$$^{-1}$$. Atomic weight of the substance is 60 g mole$$^{-1}$$, ($$N_A=6\\times10^{23}$$). The activity of 1.0 $$\\mu$$g of the substance is ___________ $$\\times$$ 10$$^{10}$$ Bq.
The activity of a radioactive substance is defined as the rate of decay or disintegration of the substance. It is given by the following formula:
\n
$$A = \\lambda N$$
\n
where $A$ is the activity, $\\lambda$ is the decay constant, and $N$ is the number of radioactive atoms present.
\n
We can use this formula to find the activity of 1.0 $\\mu$g (or $10^{-6}$ g) of the substance. First, we need to find the number of radioactive atoms present in 1.0 $\\mu$g of the substance. We can use the following formula to do this:
\n
$$N = \\frac{m}{M} N_A$$
\n
where $m$ is the mass of the substance, $M$ is its molar mass, and $N_A$ is Avogadro's number.
A radio active material is reduced to $$1 / 8$$ of its original amount in 3 days. If $$8 \\times 10^{-3} \\mathrm{~kg}$$ of the material is left after 5 days the initial amount of the material is
The decay of a radioactive material follows an exponential decay law, which can be expressed as:
\n
$$ N = N_0 \\cdot \\left(\\frac{1}{2}\\right)^{\\frac{t}{T}} $$
\n
where:
\n
\n
$$N$$ is the final amount of the material,
\n
$$N_0$$ is the initial amount of the material,
\n
$$t$$ is the elapsed time,
\n
$$T$$ is the half-life of the material.
\n
\n
From the problem, we know that the material is reduced to $$1/8$$ of its original amount in 3 days. Therefore, the half-life of the material can be calculated as follows:
This simplifies to $$2^{-3} = 2^{-\\frac{3}{T}} $$, which gives $$ T = 1 \\, \\text{day} $$.
\n
Knowing the half-life, we can now find the initial amount of the material. We know that after 5 days, $$8 \\times 10^{-3} \\, \\text{kg}$$ of the material is left. Therefore, we can write:
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8239,
"subject": "Physics",
"question": "The radius of a nucleus of mass number 64 is 4.8 fermi. Then the mass number of another nucleus having radius of 4 fermi is $\\frac{1000}{x}$, where $x$ is _______.",
"options": [],
"answer": "27",
"solution": "**Answer:** 27\n\n
According to the empirical formula relating the radius of a nucleus ($ R $) with its mass number ($ A $), we know that the radius of a nucleus is proportional to the cube root of its mass number. This relationship is given as:
\n\n
$$ R = R_0 A^{1/3} $$
\n\n
where $ R_0 $ is a constant with an approximate value of 1.2 fermis.
\n\n
Given that for a nucleus with a mass number 64 has a radius of 4.8 fermis, we can write:
\n\n
$$ 4.8 \\text{ fermi} = R_0 \\times 64^{1/3} $$
\n\n
Now another nucleus has a radius of 4 fermis:
\n\n
$$ 4 \\text{ fermi} = R_0 \\times A'^{1/3} $$
\n\n
Where $ A' $ is the mass number of the other nucleus.
\n\n
Let's solve for $ R_0 $ from the first equation:
\n\n
$$ R_0 = \\frac{4.8 \\text{ fermi}}{64^{1/3}} $$
\n\n
Now we're going to find the mass number $ A' $ using the second equation and substituting $ R_0 $ from the above:
A nucleus has mass number $$A_1$$ and volume $$V_1$$. Another nucleus has mass number $$A_2$$ and Volume $$V_2$$. If relation between mass number is $$A_2=4 A_1$$, then $$\\frac{V_2}{V_1}=$$ __________.
$$ = 1.1 \\times {10^{ - 10}}F$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8243,
"subject": "Physics",
"question": "A 10 $$\\mu $$F capacitor is fully charged to a potential\ndifference of 50 V. After removing the source\nvoltage it is connected to an uncharged\ncapacitor in parallel. Now the potential\ndifference across them becomes 20 V. The\ncapacitance of the second capacitor is :",
"options": [
{
"text": "20 $$\\mu $$F"
},
{
"text": "15 $$\\mu $$F"
},
{
"text": "10 $$\\mu $$F"
},
{
"text": "30 $$\\mu $$F"
}
],
"answer": "15 $$\\mu $$F",
"solution": "**Answer:** 15 $$\\mu $$F\n\nInitially,\n
Charge on capacitor 10 μF\n
Q = CV = (10 μF) (50V)\n
Q = 500 μC\n
Final Charge on 10 μF capacitor\n
Q = CV = (10 μF) (20V)\n
Q = 200 μC\n
From charge conservation,\n
Charge on unknown capacitor\n
Q = 500 μC – 200 μC = 300 μC\n
$$ \\Rightarrow $$ Capacitance (C) = $${Q \\over V}$$ = $${{300} \\over {20}}$$ = 15 $$\\mu $$F",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8244,
"subject": "Physics",
"question": "Match List I with List II.
Capacitance of an isolated conducting sphere of radius R1 becomes n times when it is enclosed by a concentric conducting sphere of radius R2 connected to earth. The ratio of their radii $$\\left( {{{{R_2}} \\over {{R_1}}}} \\right)$$ is :
Given below are two statements: One is labeled as Assertion A and the other is labeled as Reason R.
\n
Assertion A : Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one.
\n
Reason R : Capacitance of metallic spheres depend on the radii of spheres
\n
In light of the above statements, choose the correct answer from the options given below.
",
"options": [
{
"text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true but $$\\mathbf{R}$$ is not the correct explanation of $$\\mathbf{A}$$"
},
{
"text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are true and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$"
},
{
"text": "$$\\mathbf{A}$$ is false but $$\\mathbf{R}$$ is true"
},
{
"text": "$$\\mathbf{A}$$ is true but $$\\mathbf{R}$$ is false"
}
],
"answer": "$$\\mathbf{A}$$ is false but $$\\mathbf{R}$$ is true",
"solution": "**Answer:** $$\\mathbf{A}$$ is false but $$\\mathbf{R}$$ is true\n\nThe amount of charge on each sphere will be the same if they are charged to the same potential. Whether the sphere is solid or hollow doesn't affect the amount of charge stored on the sphere as long as they have the same radii and are charged to the same potential.\n
So, capacitance does not depend on its charge, it depends only on the radius of the conductor (R).\n
Therefore, Reason $\\mathrm{R}$ is true.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8247,
"subject": "Physics",
"question": "A capacitance of 2 $$\\mu $$F is required in an electrical circuit across a potential difference of 1.0 kV. A large\nnumber of 1 $$\\mu $$F capacitors are available which can withstand a potential difference of not more than 300 V.\nThe minimum number of capacitors required to achieve this is: ",
"options": [
{
"text": "2"
},
{
"text": "16"
},
{
"text": "32"
},
{
"text": "24"
}
],
"answer": "32",
"solution": "**Answer:** 32\n\nTo get a capacitance of 2 μF arrangement of capacitors of capacitance 1μF as shown in figure\n8 capacitors of 1μF in parallel with four such branches in series i.e., 32 such capacitors are\nrequired.\n\n
$$ \\Rightarrow $$ $${{C_{eq}} = 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8248,
"subject": "Physics",
"question": "Two capacitors of capacities 2C and C are joined in parallel and charged up to potential V. The battery is removed and the capacitor of capacity C is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be :",
"options": [
{
"text": "$${V \\over {K + 2}}$$"
},
{
"text": "$${V \\over K}$$"
},
{
"text": "$${{3V} \\over {K + 2}}$$"
},
{
"text": "$${{3V} \\over K}$$"
}
],
"answer": "$${{3V} \\over {K + 2}}$$",
"solution": "**Answer:** $${{3V} \\over {K + 2}}$$\n\n
Two parallel plate capacitors of capacity C and 3C are connected in parallel combination and charged to a potential difference 18 V. The battery is then disconnected and the space between the plates of the capacitor of capacity C is completely filled with a material of dielectric constant 9. The final potential difference across the combination of capacitors will be ___________ V.
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 8250,
"subject": "Physics",
"question": "A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor ",
"options": [
{
"text": "decreases "
},
{
"text": "remains unchanged "
},
{
"text": "becomes infinite "
},
{
"text": "increases "
}
],
"answer": "remains unchanged ",
"solution": "**Answer:** remains unchanged \n\nThe capacitancce of parallel plate capacitor in which a metal plate of thickness $$t$$ is inserted is given by\n
Here $$t \\to 0\\,\\,\\,\\,\\,\\,$$ $$\\therefore$$ $$C = {{{\\varepsilon _0}A} \\over d}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8251,
"subject": "Physics",
"question": "A parallel plate capacitor with air between the plates has capacitance of $$9$$ $$pF.$$ The separation between its plates is $$'d'.$$ The space between the plates has dielectric constant $${k_1}$$ $$=3$$ and thickness $${d \\over 3}$$ while the other one has dielectric constant $${k_2} = 6$$ and thickness $${{2d} \\over 3}$$. Capacitance of the capacitor is now ",
"options": [
{
"text": "$$1.8$$ $$pF$$ "
},
{
"text": "$$45$$ $$pF$$ "
},
{
"text": "$$40.5$$ $$pF$$ "
},
{
"text": "$$20.25$$ $$pF$$ "
}
],
"answer": "$$40.5$$ $$pF$$ ",
"solution": "**Answer:** $$40.5$$ $$pF$$ \n\n\n
The given capacitance is equal to two capacitances connected in series where\n
$$\\therefore$$ $${C_{eq}} = {9 \\over 2}{{{\\varepsilon _0}A} \\over d} = {9 \\over 2} \\times 9pF = 40.5pF$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8252,
"subject": "Physics",
"question": "A parallel plate capacitor is made of two circular plates separated by a distance $$5$$ $$mm$$ and with a dielectric of dielectric constant $$2.2$$ between them. When the electric field in the dielectric is $$3 \\times {10^4}\\,V/m$$ the charge density of the positive plate will be close to: ",
"options": [
{
"text": "$$6 \\times {10^{ - 7}}\\,\\,C/{m^2}$$ "
},
{
"text": "$$3 \\times {10^{ - 7}}\\,\\,C/{m^2}$$ "
},
{
"text": "$$3 \\times {10^4}\\,\\,C/{m^2}$$ "
},
{
"text": "$$6 \\times {10^4}\\,\\,C/{m^2}$$ "
}
],
"answer": "$$6 \\times {10^{ - 7}}\\,\\,C/{m^2}$$ ",
"solution": "**Answer:** $$6 \\times {10^{ - 7}}\\,\\,C/{m^2}$$ \n\nElectric field in presence of dielectric between the two plates of a parallel plate capacitor is given by,\n
$$E = {\\sigma \\over {K{\\varepsilon _0}}}$$\n
Then, charge density\n
$$\\sigma = K{\\varepsilon _0}E$$\n
$$ = 2.2 \\times 8.85 \\times {10^{ - 12}} \\times 3 \\times {10^4} \\approx 6 \\times {10^{ - 7}}\\,\\,C/{m^2}$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8253,
"subject": "Physics",
"question": "A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material\nof dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be :",
"options": [
{
"text": "0.9 n C"
},
{
"text": "1.2 n C"
},
{
"text": "0.3 n C"
},
{
"text": "2.4 n C"
}
],
"answer": "1.2 n C",
"solution": "**Answer:** 1.2 n C\n\nCharge on Capacitor initially, \n
Qi = CV\n
After inserting dielectric of dielectric constant = K,\n
new capacitance, Qf = (KC) $$ \\vee $$\n
$$\\therefore\\,\\,\\,$$ Induced charges on dielectric \n
= 1.2 nC",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8254,
"subject": "Physics",
"question": "The parallel combination of two air filled\nparallel plate capacitors of capacitance C and\nnC is connected to a battery of voltage, V. When\nthe capacitors are fully charged, the battery is\nremoved and after that a dielectric material of\ndielectric constant K is placed between the two\nplates of the first capacitor. The new potential\ndifference of the combined system is :-",
"options": [
{
"text": "V"
},
{
"text": "$${V \\over {K + n}}$$"
},
{
"text": "$${{(n+1)V} \\over {K + n}}$$"
},
{
"text": "$${{nV} \\over {K + n}}$$"
}
],
"answer": "$${{(n+1)V} \\over {K + n}}$$",
"solution": "**Answer:** $${{(n+1)V} \\over {K + n}}$$\n\n\nInitially Q = CV(1 + n)
\n$$ \\therefore $$ Ceq = (K + n)C
\n$$ \\therefore $$ $$V = {{CV\\left( {1 + n} \\right)} \\over {\\left( {K + n} \\right)C}} = {{V\\left( {1 + n} \\right)} \\over {\\left( {K + n} \\right)}}$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 8255,
"subject": "Physics",
"question": "Voltage rating of a parallel plate capacitor is\n500V. Its dielectric can withstand a maximum\nelectric field of 106 V/m. The plate area is\n10–4 m2. What is the dielectric constant is the\ncapacitance is 15 pF?\n(given $$\\varepsilon $$0 = 8.86 × 10–12 C2/Nm2)",
"options": [
{
"text": "8.5"
},
{
"text": "4.5"
},
{
"text": "3.8"
},
{
"text": "6.2"
}
],
"answer": "8.5",
"solution": "**Answer:** 8.5\n\nA = 10–4 m2
\nk $$ \\approx $$ 8.5\n",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8256,
"subject": "Physics",
"question": "A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is :",
"options": [
{
"text": "508 pJ"
},
{
"text": "692 pJ"
},
{
"text": "560 pJ"
},
{
"text": "600 pJ"
}
],
"answer": "508 pJ",
"solution": "**Answer:** 508 pJ\n\nInitial energy of capacitor\n
W = 508 J",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8257,
"subject": "Physics",
"question": "A parallel plate capacitor has plate of length\n'l', width ‘w’ and separation of plates is ‘d’. It\nis connected to a battery of emf V. A dielectric\nslab of the same thickness ‘d’ and of dielectric\nconstant k = 4 is being inserted between the\nplates of the capacitor. At what length of the\nslab inside plates, will the energy stored in the\ncapacitor be two times the initial energy\nstored?",
"options": [
{
"text": "$${l \\over 4}$$"
},
{
"text": "$${l \\over 2}$$"
},
{
"text": "$${{2l} \\over 3}$$"
},
{
"text": "$${l \\over 3}$$"
}
],
"answer": "$${l \\over 3}$$",
"solution": "**Answer:** $${l \\over 3}$$\n\n\n
$$ \\Rightarrow $$ x = $${l \\over 3}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8258,
"subject": "Physics",
"question": "For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $${3 \\over 4}$$d, where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0) is given by the following relation :",
"options": [
{
"text": "$$C' = {{3 + K} \\over {4K}}{C_0}$$"
},
{
"text": "$$C' = {{4 + K} \\over {3}}{C_0}$$"
},
{
"text": "$$C' = {{4K} \\over {K + 3}}{C_0}$$"
},
{
"text": "$$C' = {{4} \\over {3 + K}}{C_0}$$"
}
],
"answer": "$$C' = {{4K} \\over {K + 3}}{C_0}$$",
"solution": "**Answer:** $$C' = {{4K} \\over {K + 3}}{C_0}$$\n\n\n\n $${C_0} = {{{ \\in _0}A} \\over d}$$
$$ \\therefore $$ $$C' = {{4K} \\over {(K + 3)}}{C_0}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8259,
"subject": "Physics",
"question": "A parallel plate capacitor has plate area 100 m2 and plate separation of 10 m. The space between the plates is filled up to a thickness 5 m with a material of dielectric constant of 10. The resultant capacitance of the system is 'x' pF.
The value of $$\\varepsilon $$0 = 8.85 $$\\times$$ 10$$-$$12 F.m$$-$$1.
The value of 'x' to the nearest integer is _____________.",
"options": [],
"answer": "161",
"solution": "**Answer:** 161\n\nArea = 100 m2\n
$$ \\Rightarrow $$ $${C_{eq}} = 161$$ PF",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8260,
"subject": "Physics",
"question": "A parallel plate capacitor with plate area 'A' and distance of separation 'd' is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :
$$\\varepsilon (x) = {\\varepsilon _0} + kx$$, for $$\\left( {0 < x \\le {d \\over 2}} \\right)$$
$${C_{eq}} = {C \\over 2} = {{kA} \\over {2\\ln \\left( {{{2{\\varepsilon _0} + kd} \\over {2{\\varepsilon _0}}}} \\right)}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8261,
"subject": "Physics",
"question": "If qf is the free charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge qb an be expressed as :",
"options": [
{
"text": "$${q_b} = {q_f}\\left( {1 - {1 \\over {\\sqrt k }}} \\right)$$"
},
{
"text": "$${q_b} = {q_f}\\left( {1 - {1 \\over k}} \\right)$$"
},
{
"text": "$${q_b} = {q_f}\\left( {1 + {1 \\over {\\sqrt k }}} \\right)$$"
},
{
"text": "$${q_b} = {q_f}\\left( {1 + {1 \\over k}} \\right)$$"
}
],
"answer": "$${q_b} = {q_f}\\left( {1 - {1 \\over k}} \\right)$$",
"solution": "**Answer:** $${q_b} = {q_f}\\left( {1 - {1 \\over k}} \\right)$$\n\n
When a dielectric is inserted in a capacitor
Due to free charge $$\\overrightarrow E = {\\overrightarrow E _0}$$ only
After dielectric $$E' = {{{E_0}} \\over k}$$
$${q_B} = {q_f}\\left( {1 - {1 \\over k}} \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8262,
"subject": "Physics",
"question": "A parallel plate capacitor of capacitance 200 $$\\mu$$F is connected to a battery of 200 V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be ____________J.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n$$\\Delta U = {1 \\over 2}(\\Delta C){V^2}$$
A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :
",
"options": [
{
"text": "increase by 50%"
},
{
"text": "decrease by 15%"
},
{
"text": "increase by 25%"
},
{
"text": "increase by 33%"
}
],
"answer": "increase by 50%",
"solution": "**Answer:** increase by 50%\n\n
Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness $${d \\over 2}$$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor?
A parallel plate capacitor with width $$4 \\mathrm{~cm}$$, length $$8 \\mathrm{~cm}$$ and separation between the plates of $$4 \\mathrm{~mm}$$ is connected to a battery of $$20 \\mathrm{~V}$$. A dielectric slab of dielectric constant 5 having length $$1 \\mathrm{~cm}$$, width $$4 \\mathrm{~cm}$$ and thickness $$4 \\mathrm{~mm}$$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ____________ $$\\epsilon_{0}$$ J. (Where $$\\epsilon_{0}$$ is the permittivity of free space)
Two capacitors, each having capacitance $$40 \\,\\mu \\mathrm{F}$$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $$\\mathrm{K}$$ such that the equivalence capacitance of the system became $$24 \\,\\mu \\mathrm{F}$$. The value of $$\\mathrm{K}$$ will be :
A slab of dielectric constant $$\\mathrm{K}$$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $$\\frac{3}{4} \\mathrm{~d}$$, where $$\\mathrm{d}$$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be :
\n
(Given $$\\mathrm{C}_{0}$$ = capacitance of capacitor with air as medium between plates.)
",
"options": [
{
"text": "$$\\frac{4 K C_{0}}{3+K}$$"
},
{
"text": "$$\\frac{3 K C_{0}}{3+K}$$"
},
{
"text": "$$\\frac{3+K}{4 K C_{0}}$$"
},
{
"text": "$$\\frac{K}{4+K}$$"
}
],
"answer": "$$\\frac{4 K C_{0}}{3+K}$$",
"solution": "**Answer:** $$\\frac{4 K C_{0}}{3+K}$$\n\n
A capacitor has capacitance 5$$\\mu$$F when it's parallel plates are separated by air medium of thickness d. A slab of material of dielectric constant 1.5 having area equal to that of plates but thickness $$\\frac{d}{2}$$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be __________ $$\\mu$$F.
A parallel plate capacitor has plate area 40 cm$$^2$$ and plates separation 2 mm. The space between the plates is filled with a dielectric medium of a thickness 1 mm and dielectric constant 5. The capacitance of the system is :
A parallel plate capacitor with air between the plate has a capacitance of 15pF. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes $$\\frac{x}{4}$$ pF. The value of $$x$$ is ____________.
The distance between two plates of a capacitor is $$\\mathrm{d}$$ and its capacitance is $$\\mathrm{C}_{1}$$, when air is the medium between the plates. If a metal sheet of thickness $$\\frac{2 d}{3}$$ and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes $$\\mathrm{C}_{2}$$. The ratio $$\\frac{\\mathrm{C}_{2}}{\\mathrm{C}_{1}}$$ is
When a metal sheet of thickness ($\\frac{2d}{3}$) is introduced between the plates of a capacitor, it divides the capacitor into two separate capacitors. The metal sheet acts as a new plate in each capacitor, and because the metal is a conductor, it is at the same potential as the plates on either side.
\n
The capacitance of the original capacitor with air between the plates is given by:
where ($\\epsilon_0$) is the permittivity of free space, ($\\mathrm{A}$) is the area of one of the plates, and ($\\mathrm{d}$) is the distance between the plates.
\n
When the metal sheet of thickness ($\\frac{2d}{3}$) is introduced, the distance between the plates is reduced by ($\\frac{2d}{3}$), so the capacitance of each of the new capacitors is:
A parallel plate capacitor with plate separation $$5 \\mathrm{~mm}$$ is charged up by a battery. It is found that on introducing a dielectric sheet of thickness $$2 \\mathrm{~mm}$$, while keeping the battery connections intact, the capacitor draws $$25 \\%$$ more charge from the battery than before. The dielectric constant of the sheet is _________.
A parallel plate capacitor of capacitance $$12.5 \\mathrm{~pF}$$ is charged by a battery connected between its plates to potential difference of $$12.0 \\mathrm{~V}$$. The battery is now disconnected and a dielectric slab $$(\\epsilon_{\\mathrm{r}}=6)$$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ________ $$\\times10^{-12} \\mathrm{~J}$$.
A capacitor has air as dielectric medium and two conducting plates of area $$12 \\mathrm{~cm}^2$$ and they are $$0.6 \\mathrm{~cm}$$ apart. When a slab of dielectric having area $$12 \\mathrm{~cm}^2$$ and $$0.6 \\mathrm{~cm}$$ thickness is inserted between the plates, one of the conducting plates has to be moved by $$0.2 \\mathrm{~cm}$$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $$\\epsilon_0=8.834 \\times 10^{-12} \\mathrm{~F} / \\mathrm{m}$$)
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8275,
"subject": "Physics",
"question": "A fully charged capacitor has a capacitance $$'C'$$. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity $$'s'$$ and mass $$'m'.$$ If the temperature of the block is raised by $$'\\Delta T',$$ the potential difference $$'v'$$ across the capacitance is ",
"options": [
{
"text": "$${{mCAT} \\over s}$$ "
},
{
"text": "$$\\sqrt {{{2mCAT} \\over s}} $$ "
},
{
"text": "$$\\sqrt {{{2msAT} \\over C}} $$ "
},
{
"text": "$${{ms\\Delta T} \\over C}$$ "
}
],
"answer": "$$\\sqrt {{{2msAT} \\over C}} $$ ",
"solution": "**Answer:** $$\\sqrt {{{2msAT} \\over C}} $$ \n\nApplying conservation of energy,\n
$${1 \\over 2}C{V^2} = m.s\\Delta T;\\,\\,\\,$$\n
$$V = \\sqrt {{{2m.s.\\Delta T} \\over C}} $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8276,
"subject": "Physics",
"question": "A parallel plate capacitor is made by stacking $$n$$ equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is $$'C'$$ then the resultant capacitance is ",
"options": [
{
"text": "$$\\left( {n + 1} \\right)C$$ "
},
{
"text": "$$\\left( {n - 1} \\right)C$$ "
},
{
"text": "$$nC$$ "
},
{
"text": "$$C$$ "
}
],
"answer": "$$\\left( {n - 1} \\right)C$$ ",
"solution": "**Answer:** $$\\left( {n - 1} \\right)C$$ \n\nAs $$n$$ plates are joined, it means $$(n-1)$$ capacitor joined in parallel.\n
$$\\therefore$$ resultant capacitance $$=(n-1)C$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8277,
"subject": "Physics",
"question": "Two capacitors $${C_1}$$ and $${C_2}$$ are charged to $$120$$ $$V$$ and $$200$$ $$V$$ respectively. It is found that connecting them together the potential on each one can be made zero. Then ",
"options": [
{
"text": "$$5{C_1} = 3{C_2}$$ "
},
{
"text": "$$3{C_1} = 5{C_2}$$ "
},
{
"text": "$$3{C_1} + 5{C_2} = 0$$ "
},
{
"text": "$$9{C_1} = 4{C_2}$$ "
}
],
"answer": "$$3{C_1} = 5{C_2}$$ ",
"solution": "**Answer:** $$3{C_1} = 5{C_2}$$ \n\n\n
$$ \\Rightarrow 3{C_1} = 5{C_2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8278,
"subject": "Physics",
"question": "Three capacitors each of 4 $$\\mu $$F are to be connected in such a way that the effective capacitance is 6 $$\\mu $$F. This can be done by connecting them :",
"options": [
{
"text": "all in series"
},
{
"text": "two in series and one in parallel\n"
},
{
"text": "all in parallel"
},
{
"text": "two in parallel and one in series"
}
],
"answer": "two in series and one in parallel\n",
"solution": "**Answer:** two in series and one in parallel\n\n\n(a) $${1 \\over {{C_{eq}}}} = {1 \\over 4} + {1 \\over 4} + {1 \\over 4} = {3 \\over 4}$$\n
(d) $${C_{eq}} = {{\\left( {4 + 4} \\right) \\times 4} \\over {\\left( {4 + 4} \\right) + 4}} = {8 \\over 3}\\mu F$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8279,
"subject": "Physics",
"question": "A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and\nis conneced to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost\nin this process by the time the charge is redistributed between them is (in nJ) _____",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\n\n Ui = $${1 \\over 2}CV_0^2$$\n
Ceq = $${{{C_1}{C_2}} \\over {{C_1} + {C_2}}}$$ = $${{8 \\times 2} \\over {8 + 2}}$$ = 1.6 $$\\mu $$F",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 8281,
"subject": "Physics",
"question": "Two capacitors of capacitances C and 2C are\ncharged to potential differences V and 2V,\nrespectively. These are then connected in\nparallel in such a manner that the positive\nterminal of one is connected to the negative\nterminal of the other. The final energy of this\nconfiguration is :",
"options": [
{
"text": "Zero"
},
{
"text": "$${3 \\over 2}C{V^2}$$"
},
{
"text": "$${9 \\over 2}C{V^2}$$"
},
{
"text": "$${{25} \\over 6}C{V^2}$$"
}
],
"answer": "$${3 \\over 2}C{V^2}$$",
"solution": "**Answer:** $${3 \\over 2}C{V^2}$$\n\n\n
$${U_f} = {3 \\over 2}C{V^2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8282,
"subject": "Physics",
"question": "Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be :",
"options": [
{
"text": "4 : 1"
},
{
"text": "1 : 2"
},
{
"text": "2 : 1"
},
{
"text": "1 : 4"
}
],
"answer": "1 : 4",
"solution": "**Answer:** 1 : 4\n\nGiven, C1 = C2 = C
When both capacitors are connected in series, their equivalent capacitance will be
$$\\therefore$$ Cs : Cp = 1 : 4",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8283,
"subject": "Physics",
"question": "Consider the combination of 2 capacitors C1 and C2 with C2 > C1, when connected in parallel, the equivalent capacitance is $${{15} \\over 4}$$ times the equivalent capacitance of the same connected in series. Calculate the ratio of capacitors, $${{{C_2}} \\over {{C_1}}}$$.",
"options": [
{
"text": "$${{15} \\over {11}}$$"
},
{
"text": "No Solutions"
},
{
"text": "$${{29} \\over {15}}$$"
},
{
"text": "$${{15} \\over {4}}$$"
}
],
"answer": "No Solutions",
"solution": "**Answer:** No Solutions\n\nWhen connected in parallel
The total charge on the system of capacitors $$C_{1}=1 \\mu \\mathrm{F}, C_{2}=2 \\mu \\mathrm{F}, \\mathrm{C}_{3}=4 \\mu \\mathrm{F}$$ and $$\\mathrm{C}_{4}=3 \\mu \\mathrm{F}$$ connected in parallel is :
\n
(Assume a battery of $$20 \\mathrm{~V}$$ is connected to the combination)
A capacitor of capacitance $$\\mathrm{C}$$ and potential $$\\mathrm{V}$$ has energy $$\\mathrm{E}$$. It is connected to another capacitor of capacitance $$2 \\mathrm{C}$$ and potential $$2 \\mathrm{~V}$$. Then the loss of energy is $$\\frac{x}{3} \\mathrm{E}$$, where $$x$$ is _______.
$$\\begin{aligned}\n& \\text { Energy loss }=\\frac{1}{2} \\frac{C_1 C_2}{C_1+C_2}\\left(V_1-V_2\\right)^2 \\\\\n& =\\frac{2}{3} \\cdot E \\\\\n& \\therefore x=2\n\\end{aligned}$$
Three capacitors of capacitances $$25 \\mu \\mathrm{F}, 30 \\mu \\mathrm{F}$$ and $$45 \\mu \\mathrm{F}$$ are connected in parallel to a supply of $$100 \\mathrm{~V}$$. Energy stored in the above combination is E. When these capacitors are connected in series to the same supply, the stored energy is $$\\frac{9}{x} \\mathrm{E}$$. The value of $$x$$ is _________.
$$ = 32 \\times {10^{ - 32}}J$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8289,
"subject": "Physics",
"question": "A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be ",
"options": [
{
"text": "$$1/2$$ "
},
{
"text": "$$1$$ "
},
{
"text": "$$2$$ "
},
{
"text": "$$1/4$$ "
}
],
"answer": "$$1/2$$ ",
"solution": "**Answer:** $$1/2$$ \n\nRequired ratio\n
$$\\left( \\, \\right.$$ as $$V=e$$ $$\\left. \\, \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8290,
"subject": "Physics",
"question": "The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains 4 $$\\mu $$C charge, its radius will be :\n [ Take : $${1 \\over {4\\,\\pi { \\in _0}}} = $$ 9 $$ \\times $$ 109 N $$-$$ m2/C2 ]",
"options": [
{
"text": "20 mm"
},
{
"text": "32 mm"
},
{
"text": "28 mm"
},
{
"text": "16 mm"
}
],
"answer": "16 mm",
"solution": "**Answer:** 16 mm\n\nEnergy of sphere = $${{{Q^2}} \\over {2C}}$$\n
\n$$ = {{15} \\over 4} \\times {10^{ - 6}} = 3.75{\\rm{ }} \\times {\\rm{ }}{10^{-6}}{\\rm{ }}J$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8292,
"subject": "Physics",
"question": "A 5 $$\\mu $$F capacitor is charged fully by a 220 V\nsupply. It is then disconnected from the supply\nand is connected in series to another\nuncharged 2.5 $$\\mu $$F capacitor. If the energy\nchange during the charge redistribution is\n$${X \\over {100}}J$$ then value of X to the nearest integer is\n_____.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nui = $$\\frac{1}{2} $$ $$ \\times $$ 5 $$ \\times $$ 10-6$$ \\times $$220\n
Value of X is approximate 4",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8293,
"subject": "Physics",
"question": "A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connected\nin parallel with another uncharged capacitor of capacitance $${C \\over 2}$$. The energy loss in the process\nafter the charge is distributed between the two capacitors is :",
"options": [
{
"text": "$${1 \\over 2}CV_0^2$$"
},
{
"text": "$${1 \\over 4}CV_0^2$$"
},
{
"text": "$${1 \\over 3}CV_0^2$$"
},
{
"text": "$${1 \\over 6}CV_0^2$$"
}
],
"answer": "$${1 \\over 6}CV_0^2$$",
"solution": "**Answer:** $${1 \\over 6}CV_0^2$$\n\nHeat loss = $${1 \\over 2}\\left( {{{{C_1}{C_2}} \\over {{C_1} + {C_2}}}} \\right)V_0^2$$\n
= $${1 \\over 6}CV_0^2$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8294,
"subject": "Physics",
"question": "A parallel plate capacitor whose capacitance C is 14 pF is charged by a battery to a potential difference V = 12 V between its plates. The charging battery is now disconnected and a porcelin plate with k = 7 is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of _____________ pJ. (Assume no friction)",
"options": [],
"answer": "864",
"solution": "**Answer:** 864\n\n$${U_i} = {1 \\over 2}c{v^2}$$
A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is ___________ nJ.
A capacitor of capacitance $$900 \\mu \\mathrm{F}$$ is charged by a $$100 \\mathrm{~V}$$ battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as $$x \\times 10^{-} { }^{2} \\mathrm{~J}$$. The value of $$x$$ is _____________.
A parallel plate capacitor of capacitance $$2 \\mathrm{~F}$$ is charged to a potential $$\\mathrm{V}$$, The energy stored in the capacitor is $$E_{1}$$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $$\\mathrm{E}_{2}$$. The ratio $$\\mathrm{E}_{2} / \\mathrm{E}_{1}$$ is :
Step 2: Determine the energy stored when two capacitors are connected in parallel
\n\n
When the charged capacitor (capacitor 1) is connected to an identical uncharged capacitor (capacitor 2), the charge will redistribute between the two capacitors. The total capacitance of the parallel combination is:
After connection, this charge will be shared equally by the two capacitors because they are identical. Therefore, the voltage across each capacitor in the parallel combination will be:
A $$600 ~\\mathrm{pF}$$ capacitor is charged by $$200 \\mathrm{~V}$$ supply. It is then disconnected from the supply and is connected to another uncharged $$600 ~\\mathrm{pF}$$ capacitor. Electrostatic energy lost in the process is ____________ $$\\mu \\mathrm{J}$$
When the charged capacitor is connected to the uncharged capacitor, the charge will distribute equally between them because they have the same capacitance. Therefore, the final voltage across each capacitor is half of the initial voltage, i.e., 100 V.
\n
The energy in each capacitor after the redistribution is:
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8300,
"subject": "Physics",
"question": "Two identical capacitors have same capacitance $C$. One of them is charged to the potential $V$ and other to the potential $2 \\mathrm{~V}$. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :",
"options": [
{
"text": "$\\frac{1}{4} \\mathrm{CV}^2$"
},
{
"text": "$\\frac{3}{4} \\mathrm{CV}^2$"
},
{
"text": "$\\frac{1}{2} \\mathrm{CV}^2$"
},
{
"text": "$2 \\mathrm{CV}^2$"
}
],
"answer": "$\\frac{1}{4} \\mathrm{CV}^2$",
"solution": "**Answer:** $\\frac{1}{4} \\mathrm{CV}^2$\n\n
To solve the problem, let's start by considering the energy stored in each capacitor before they are connected together.\n\n
The energy stored in a capacitor is given by the formula:
\n\n
$$E = \\frac{1}{2} C V^2$$
\n\n
Where $E$ is the energy, $C$ is the capacitance, and $V$ is the potential.\n\n
For the first capacitor charged to the potential $V$, the energy stored is:
\n\n
$$E_1 = \\frac{1}{2} C V^2$$
\n\n
For the second capacitor charged to the potential $2V$, the energy stored is:
\n\n
$$E_2 = \\frac{1}{2} C (2V)^2 = \\frac{1}{2} C \\cdot 4V^2 = 2 CV^2$$
\n\n
The total initial energy stored in the system is the sum of $E_1$ and $E_2$:
When the two capacitors are connected together, their potentials will become equal because they are identical capacitors. Let's denote this final potential as $V_f$. The total charge before and after the connection remains constant because charge is conserved. Therefore, we can write:
\n\n
$$C \\cdot V + C \\cdot 2V = 2C \\cdot V_f$$
\n\n
Simplifying this equation gives us the final potential:
\n\n
$$V + 2V = 2 V_f$$
\n\n
$$3V = 2 V_f$$
\n\n
$$V_f = \\frac{3}{2} V$$
\n\n
The final energy stored in the system when the capacitors are connected is now the total energy stored across both capacitors at the final potential $V_f$:
\n\n
$$E_{\\text{final}} = 2 \\cdot \\frac{1}{2} C V_f^2 = 2 \\cdot \\frac{1}{2} C \\left(\\frac{3}{2} V\\right)^2 = C \\cdot \\frac{9}{4} V^2$$
\n\n
The decrease in energy $ \\Delta E $ of the combined system is the initial energy minus the final energy:
\n\n
$$\\Delta E = E_{\\text{initial}} - E_{\\text{final}}$$
\n\n
$$\\Delta E = \\frac{5}{2} CV^2 - \\frac{9}{4} CV^2$$
\n\n
$$\\Delta E = \\frac{10}{4} CV^2 - \\frac{9}{4} CV^2$$
A $$16 \\Omega$$ wire is bend to form a square loop. A $$9 \\mathrm{~V}$$ battery with internal resistance $$1 \\Omega$$ is connected across one of its sides. If a $$4 \\mu F$$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $$\\frac{x}{2} \\mu J$$, where $$x=$$ _________
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8302,
"subject": "Physics",
"question": "A parallel plate condenser with a dielectric of dielectric constant $$K$$ between the plates has a capacity $$C$$ and is charged to a potential $$V$$ volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is ",
"options": [
{
"text": "zero "
},
{
"text": "$${1 \\over 2}\\,\\left( {K - 1} \\right)\\,C{V^2}$$ "
},
{
"text": "$${{C{V^2}\\left( {K - 1} \\right)} \\over K}$$ "
},
{
"text": "$$\\left( {K - 1} \\right)\\,C{V^2}$$ "
}
],
"answer": "zero ",
"solution": "**Answer:** zero \n\n
First, let's consider the capacitor with the dielectric between its plates. The charge on the capacitor is Q, its capacitance is C (which includes the effect of the dielectric), and the potential difference (voltage) across its plates is V. According to the formula for the energy stored in a capacitor :
\n
$$U = \\frac{1}{2} CV^2$$
\n
we find that the energy is equal to half of the product of the capacitance and the square of the voltage.
\n
Now, consider the process where the dielectric slab is removed from the capacitor. When the dielectric is removed, the capacitance of the capacitor decreases. However, because the capacitor is not connected to anything that can supply or absorb charge, the charge Q on the capacitor stays the same. Since the charge stays the same but the capacitance decreases, the voltage V across the capacitor must increase to keep Q = CV true.
\n
The energy of the capacitor without the dielectric is still given by :
\n
$$U = \\frac{1}{2} CV^2$$
\n
but now C is smaller and V is larger. However, because both C and V2 change in such a way that their product remains constant, the energy of the capacitor doesn't change when the dielectric is removed.
\n
Therefore, the energy of the capacitor before the dielectric is removed is the same as the energy of the capacitor after the dielectric is removed. When the dielectric is reinserted, the process is just reversed, so again no net work is done.
\n
So the total work done by the system in the process of removing the dielectric and then reinserting it is zero. This is because work is the transfer of energy, and in this case, the energy of the system (the charged capacitor) does not change. Therefore, no energy is transferred, so no work is done.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8303,
"subject": "Physics",
"question": "A parallel plate capacitor with area 200 cm2 and separation between the plates 1.5 cm, is connected across a battery of emf V. If the force of attraction between the plates is $$25 \\times {10^{ - 6}}N,$$ the value of V is approximately : $$\\left( {{ \\in _o} = 8.85 \\times {{10}^{ - 12}}{{{C^2}} \\over {N.{m^2}}}} \\right)$$",
"options": [
{
"text": "250 V"
},
{
"text": "100 V"
},
{
"text": "300 V"
},
{
"text": "150 V"
}
],
"answer": "250 V",
"solution": "**Answer:** 250 V\n\nGiven area of Parallel plate capacitor, A = 200 cm2\n
Separation between the plates, d = 1.5 cm\n
Force of attraction between the plates, F = 25 × 10–6 N\n
F = QE\n
$$ \\Rightarrow $$ F = $${{{Q^2}} \\over {2A{ \\in _0}}}$$\n
[ As E due to parallel plate = $${\\sigma \\over {2{ \\in _0}}} = {Q \\over {2A{ \\in _0}}}$$]\n
Also we know, Q = CV = $${{{ \\in _0}AV} \\over d}$$\n
= $$1.5 \\times {10^{ - 2}}\\sqrt {{{25} \\over {8.85}}} $$ = 250 V",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8304,
"subject": "Physics",
"question": "A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is : \n
A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be.
E between two plates is $${\\sigma \\over {{\\varepsilon _0}}}$$ and due to one plate is $${\\sigma \\over {2{\\varepsilon _0}}}$$ so the force will be halved
A parallel plate capacitor is formed by two plates each of area 30$$\\pi$$ cm2 separated by 1 mm. A material of dielectric strength 3.6 $$\\times$$ 107 Vm$$-$$1 is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7 $$\\times$$ 10$$-$$6C, the value of dielectric constant of the material is :
Two identical thin metal plates has charge $$q_{1}$$ and $$q_{2}$$ respectively such that $$q_{1}>q_{2}$$. The plates were brought close to each other to form a parallel plate capacitor of capacitance C. The potential difference between them is :
$$ \\Rightarrow V = {{{q_1} - {q_2}} \\over {2C}}$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8310,
"subject": "Physics",
"question": "Two parallel plate capacitors $C_{1}$ and $C_{2}$ each having capacitance of $10 \\mu \\mathrm{F}$ are individually charged by a 100 V D.C. source. Capacitor $C_{1}$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor $\\mathrm{C}_{2}$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_{1}$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be ________ V.\n
A capacitor of capacitance $$\\mathrm{C}$$ is charged to a potential V. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is :
The electric field inside a parallel plate capacitor is uniform and given by $\\mathbf{E}=\\frac{V}{d} \\hat{\\mathbf{j}}$ where $d$ is the separation between the plates.
The electric flux through a closed surface enclosing only the positive plate of the capacitor is given by $\\Phi_E = \\oint_S \\mathbf{E}\\cdot d\\mathbf{A}$.
Since the electric field is perpendicular to the surface of the plate, the flux through the surface will be constant and given by $\\Phi_E = E A$, where $A$ is the area of the plate.
\n
The area of the positive plate is $A = \\frac{Q}{\\varepsilon_0 V}$, where $Q = C V$ is the charge on the positive plate. Therefore, the electric flux through the surface is given by:
Thus, the answer is $\\frac{C V}{\\varepsilon_0}$.
\n",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8312,
"subject": "Physics",
"question": "Let $$C$$ be the capacitance of a capacitor discharging through a resistor $$R.$$ Suppose $${t_1}$$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $${t_2}$$ is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio $${t_1}/{t_2}$$ will be ",
"options": [
{
"text": "$$1$$ "
},
{
"text": "$${1 \\over 2}$$ "
},
{
"text": "$${1 \\over 4}$$ "
},
{
"text": "$$2$$ "
}
],
"answer": "$${1 \\over 4}$$ ",
"solution": "**Answer:** $${1 \\over 4}$$ \n\nInitial energy of capacitor, $${E_1} = {{q_1^2} \\over {2C}}$$\n
Final energy of capacitor, $${E_2} = {1 \\over 2}{E_1} = {{q_1^2} \\over {4C}} = {\\left( {{{{{{q_1}} \\over {\\sqrt 2 }}} \\over {2C}}} \\right)^2}$$\n
$$\\therefore$$ $${t_1}=$$ time for the charge to reduce to $${1 \\over {\\sqrt 2 }}$$ of its initial value \n
and $${t_2} = $$ time for the charge to reduce to $${1 \\over 4}$$ of its initial value\n
$$ = {1 \\over 2}{{\\ln \\left( {{1 \\over 2}} \\right)} \\over {2\\ln \\left( {{1 \\over 2}} \\right)}} = {1 \\over 4}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8313,
"subject": "Physics",
"question": "The material filled between the plates of a parallel plate capacitor has resistivity 200 $$\\Omega$$m. The value of capacitance of the capacitor is 2 pF. If a potential difference of 40 V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is : (given the value of relative permittivity of material is 50) ",
"options": [
{
"text": "9.0 $$\\mu$$A"
},
{
"text": "9.0 mA"
},
{
"text": "0.9 mA"
},
{
"text": "0.9 $$\\mu$$A"
}
],
"answer": "0.9 mA",
"solution": "**Answer:** 0.9 mA\n\n$$\\rho$$ = 200 $$\\Omega$$m
$$ = {{80} \\over {{{10}^4} \\times 8.85}}$$ = 903 $$\\mu$$A = 0.9 mA",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8314,
"subject": "Physics",
"question": "A capacitor is connected to a 20 V battery through a resistance of 10$$\\Omega$$. It is found that the potential difference across the capacitor rises to 2 V in 1 $$\\mu$$s. The capacitance of the capacitor is __________ $$\\mu$$F. Given : $$\\ln \\left( {{{10} \\over 9}} \\right) = 0.105$$",
"options": [
{
"text": "9.52"
},
{
"text": "0.95"
},
{
"text": "0.105"
},
{
"text": "1.85"
}
],
"answer": "0.95",
"solution": "**Answer:** 0.95\n\nGiven, the peak voltage of the battery, V0 = 20V
The voltage of the battery, V = 2V
Time, t = 1 $$\\mu$$s = 1 $$\\times$$ 10$$-$$6s
Resistance of the capacitor, R = 10 $$\\Omega$$
As we know that,
V = V0(1 $$-$$ e$$-$$t/RC)
Substituting the values in the above equation, we get
$$\\therefore$$ The capacitance of the capacitor is 0.95 $$\\mu$$F.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8315,
"subject": "Physics",
"question": "
A capacitor is discharging through a resistor R. Consider in time t1, the energy stored in the capacitor reduces to half of its initial value and in time t2, the charge stored reduces to one eighth of its initial value. The ratio t1/t2 will be
The electric field between the two parallel plates of a capacitor of $$1.5 \\mu \\mathrm{F}$$ capacitance drops to one third of its initial value in $$6.6 \\mu \\mathrm{s}$$ when the plates are connected by a thin wire. The resistance of this wire is ________ $$\\Omega$$. (Given, $$\\log 3=1.1$$)
To find the resistance of the wire connecting the two plates of the capacitor, we need to apply the formula that relates the time constant $$\\tau$$ (in seconds) of a capacitor-resistor (CR) circuit to the capacitance C (in Farads) and resistance R (in Ohms). The time constant $$\\tau$$ is given by:
\n\n
$$\\tau = R \\times C$$
\n\n
The time constant also defines the time it takes for the voltage across the capacitor (and therefore the electric field between the plates, since they are directly related) to drop to approximately $$\\frac{1}{e}$$ (where $$e$$ is the base of natural logarithms, approximately equal to 2.718) of its initial value. However, the question states that the electric field drops to one-third of its initial value. Using the natural logarithm properties, we can relate this decay process to the concept of the time constant.
\n\n
The formula for the voltage (or electric field) across a discharging capacitor as a function of time $$t$$ is:
\n\n
$$V(t) = V_0 \\times e^{-\\frac{t}{RC}}$$
\n\n
where:
\n\n
\n\n
$$V(t)$$ is the voltage across the capacitor at time $$t$$,
\n\n
$$V_0$$ is the initial voltage,
\n\n
$$R$$ is the resistance,
\n\n
$$C$$ is the capacitance, and
\n\n
$$t$$ is the time.
\n\n
\n\n
Given that the electric field drops to one-third of its initial value in $$6.6 \\mu \\mathrm{s}$$, we can set $$V(t)$$ to $$\\frac{1}{3}V_0$$ and solve for $$R$$:
Given that $$\\ln\\left(\\frac{1}{3}\\right) = \\ln(3^{-1}) = -\\ln(3) = -1.1$$ (since $$\\log 3 = 1.1$$ and using the natural log instead of common log), we have:
\n\n
$$-1.1 = -\\frac{6.6}{R \\times 1.5}$$
\n\n
Solving for $$R$$:
\n\n
$$R = \\frac{6.6}{1.5 \\times 1.1}$$
\n\n
Now, calculate the value of $$R$$:
\n\n
$$R = \\frac{6.6}{1.65} = 4\\, \\Omega$$
\n\n
Therefore, the resistance of the wire connecting the two plates of the capacitor is $$4 \\, \\Omega$$.
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8317,
"subject": "Physics",
"question": "A circular disc of radius $$R$$ is removed from a bigger circular disc of radius $$2R$$ such that the circumferences of the discs coincide. The center of mass of the new disc is $$\\alpha R$$ form the center of the bigger disc. The value of $$\\alpha $$ is ",
"options": [
{
"text": "$$1/4$$ "
},
{
"text": "$$1/3$$ "
},
{
"text": "$$1/2$$ "
},
{
"text": "$$1/6$$ "
}
],
"answer": "$$1/3$$ ",
"solution": "**Answer:** $$1/3$$ \n\n \n Let the mass per unit area be $$\\sigma .$$ \n Then the mass of the complete disc\n $$ = \\sigma \\left[ {\\pi {{\\left( {2R} \\right)}^2}} \\right] = 4\\pi \\sigma {R^2}$$\n
The mass of the removed disc $$ = \\sigma \\left( {\\pi {R^2}} \\right) = \\pi \\sigma {R^2}$$ \n
So mass of the remaining disc = $$4\\pi \\sigma {R^2}$$ - $$\\pi \\sigma {R^2}$$ = $$3\\pi \\sigma {R^2}$$\n
Let center of mass of $$3\\pi \\sigma {R^2}$$ mass is at x distance from origin O.\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8318,
"subject": "Physics",
"question": "A body $$A$$ of mass $$M$$ while falling vertically downloads under gravity breaks into two-parts; a body $$B$$ of mass $${1 \\over 3}$$ $$M$$ and a body $$C$$ of mass $${2 \\over 3}$$ $$M.$$ The center of mass of bodies $$B$$ and $$C$$ taken together shifts compared to that of bodies $$B$$ and $$C$$ taken together shifts compared to that of body $$A$$ towards ",
"options": [
{
"text": "does not shift "
},
{
"text": "depends on height of breaking "
},
{
"text": "body $$B$$ "
},
{
"text": "body $$C$$ "
}
],
"answer": "does not shift ",
"solution": "**Answer:** does not shift \n\nThe center of mass does not shift as no external force is applied horizontally. So the center of mass of the system continues its original path. It is only the internal forces which comes into play while breaking. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8319,
"subject": "Physics",
"question": "Distance of the center of mass of a solid uniform cone from its vertex is $$z{}_0$$. If the radius of its base is $$R$$ and its height is $$h$$ then $$z{}_0$$ is equal to : ",
"options": [
{
"text": "$${{5h} \\over 8}$$ "
},
{
"text": "$${{3{h^2}} \\over {8R}}$$ "
},
{
"text": "$${{{h^2}} \\over {4R}}$$ "
},
{
"text": "$${{3h} \\over 4}$$"
}
],
"answer": "$${{3h} \\over 4}$$",
"solution": "**Answer:** $${{3h} \\over 4}$$\n\n \nLet the density of solid cone $$\\rho $$.\n
= $${3 \\over 4}\\left( {{{2a + b} \\over {3a + b}}} \\right)L$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8321,
"subject": "Physics",
"question": "The centre of mass of solid hemisphere of radius 8 cm is x from the centre of the flat surface. Then\nvalue of x is __________.\n",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nAs we know c.o.m. or hemisphere = $${{3R} \\over 8}$$\n
$$ \\therefore $$ x = $${{3R} \\over 8}$$ = $${{3 \\times 8} \\over 8}$$ = 3 cm",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8322,
"subject": "Physics",
"question": "The position of the centre of mass of a uniform semi-circular wire of radius 'R' placed in x-y plane with its centre at the origin and the line joining its ends as x-axis is given by $$\\left( {0,{{xR} \\over \\pi }} \\right)$$. Then, the value of | x | is ______________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nCentre of mass of half ring is located at a distance $${{2R} \\over \\pi }$$ from centre of the ring on its axis of symmetry so position of centre of mass in the given question will be $$\\left( {0,{{xR} \\over \\pi }} \\right)$$
The distance of centre of mass from end A of a one dimensional rod (AB) having mass density $$\\rho=\\rho_{0}\\left(1-\\frac{x^{2}}{L^{2}}\\right) \\mathrm{kg} / \\mathrm{m}$$ and length L (in meter) is $$\\frac{3 L}{\\alpha} \\mathrm{m}$$. The value of $$\\alpha$$ is ___________. (where x is the distance from end A)
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8324,
"subject": "Physics",
"question": "Consider a two particle system with particles having masses $${m_1}$$ and $${m_2}$$. If the first particle is pushed towards the center of mass through a distance $$d,$$ by what distance should the second particle is moved, so as to keep the center of mass at the same position? ",
"options": [
{
"text": "$${{{m_2}} \\over {{m_1}}}\\,\\,d$$ "
},
{
"text": "$${{{m_1}} \\over {{m_1} + {m_2}}}d$$ "
},
{
"text": "$${{{m_1}} \\over {{m_2}}}d$$ "
},
{
"text": "$$d$$ "
}
],
"answer": "$${{{m_1}} \\over {{m_2}}}d$$ ",
"solution": "**Answer:** $${{{m_1}} \\over {{m_2}}}d$$ \n\nInitially, \n \n $$0 = {{{m_1}\\left( { - {x_1}} \\right) + {m_2}{x_2}} \\over {{m_1} + {m_2}}} \\Rightarrow {m_1}{x_1} = {m_2}{x_2}$$\n
Finally,\n
mass m1 moved towards the center a distance d so the distance of mass m1 from the origin is x1 - d, and now let mass m2 need to move d' to keep the center at the origin.\n \n $$\\therefore$$ $$0 = {{{m_1}\\left( {d - {x_1}} \\right) + {m_2}\\left( {{x_2} - d'} \\right)} \\over {{m_1} + {m_2}}}$$ \n
Two blocks of masses 10 kg and 30 kg are placed on the same straight line with coordinates (0, 0) cm and (x, 0) cm respectively. The block of 10 kg is moved on the same line through a distance of 6 cm towards the other block. The distance through which the block of 30 kg must be moved to keep the position of centre of mass of the system unchanged is :
",
"options": [
{
"text": "4 cm towards the 10 kg block"
},
{
"text": "2 cm away from the 10 kg block"
},
{
"text": "2 cm towards the 10 kg block"
},
{
"text": "4 cm away from the 10 kg block"
}
],
"answer": "2 cm towards the 10 kg block",
"solution": "**Answer:** 2 cm towards the 10 kg block\n\n
Three identical spheres each of mass M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3 m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be $$\\sqrt x$$ m. The value of x is ____________.
Two bodies of mass $$1 \\mathrm{~kg}$$ and $$3 \\mathrm{~kg}$$ have position vectors $$\\hat{i}+2 \\hat{j}+\\hat{k}$$ and $$-3 \\hat{i}-2 \\hat{j}+\\hat{k}$$ respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector :
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 8328,
"subject": "Physics",
"question": "Three identical spheres each of mass $2 \\mathrm{M}$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $4 \\mathrm{~m}$ each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is $\\frac{4 \\sqrt{2}}{x}$, where the value of $x$ is ___________ .",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n1. Center of Mass Coordinates:\n\n
The center of mass (COM) of a system of particles is calculated as:
In a system two particles of masses $$m_1=3 \\mathrm{~kg}$$ and $$m_2=2 \\mathrm{~kg}$$ are placed at certain distance from each other. The particle of mass $$m_1$$ is moved towards the center of mass of the system through a distance $$2 \\mathrm{~cm}$$. In order to keep the center of mass of the system at the original position, the particle of mass $$m_2$$ should move towards the center of mass by the distance _________ $$\\mathrm{cm}$$.
To solve this problem, we can make use of the concept of center of mass. The center of mass (CM) of a system remains unchanged if the internal forces act within the system without any external force. When one mass moves toward the CM, to keep the CM at the same position, the other mass must move in a way that the product of each mass with its displacement relative to the CM remains constant.
\n\n
The formula to ensure that the center of mass remains unchanged can be derived from the principle of conservation of momentum or simply by understanding that the weighted average position (considering masses as weights) does not change.
\n\n
Let's denote:\n- $$x_1$$ as the distance moved by $$m_1$$ towards the CM,\n- $$x_2$$ as the distance $$m_2$$ needs to move towards the CM,\n- The total mass of the system as $$M = m_1 + m_2$$.
\n\n
Since $$m_1$$ moves towards the CM by 2 cm, we apply the principle that the weighted sum of displacements (taking mass into account) remains 0 to maintain the center of mass at its original position:
\n\n
$$m_1 \\cdot x_1 + m_2 \\cdot x_2 = 0$$
\n\n
Given that $$m_1 = 3 \\, \\text{kg}$$, $$m_2 = 2 \\, \\text{kg}$$, and $$x_1 = 2 \\, \\text{cm}$$, we substitute these values into the equation:
\n\n
$$3 \\cdot 2 + 2 \\cdot x_2 = 0$$
\n\n
Solving for $$x_2$$ gives:
\n\n
$$6 + 2x_2 = 0$$
\n\n
$$2x_2 = -6$$
\n\n
$$x_2 = -3 \\, \\text{cm}$$
\n\n
This means the mass $$m_2$$ should move $$3 \\, \\mathrm{cm}$$ towards the center of mass to keep the center of mass of the system at the original position. The negative sign indicates the direction is towards the center of mass, similar to $$m_1$$'s movement direction in relation to keeping the CM stationary.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8330,
"subject": "Physics",
"question": "A block of mass $$0.50$$ $$kg$$ is moving with a speed of $$2.00$$ $$m{s^{ - 1}}$$ on a smooth surface. It strike another mass of $$1.0$$ $$kg$$ and then they move together as a single body. The energy loss during the collision is :",
"options": [
{
"text": "$$0.16J$$ "
},
{
"text": "$$1.00J$$ "
},
{
"text": "$$0.67J$$ "
},
{
"text": "$$0.34$$ $$J$$ "
}
],
"answer": "$$0.67J$$ ",
"solution": "**Answer:** $$0.67J$$ \n\nLet $$m$$ = 0.50 kg and $$M$$ = 1.0 kg\n
Initial kinetic energy of the system when 1 kg mass is at rest,\n
$$ = \\left( {1 - {1 \\over 3}} \\right)J = 0.67J$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8331,
"subject": "Physics",
"question": "Statement - 1 : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. \n
Statement - 2 : Principle of conservation of momentum holds true for all kinds of collisions.",
"options": [
{
"text": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is the correct explanation of Statement - 1 "
},
{
"text": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is not the correct explanation of Statement - 1 "
},
{
"text": "Statement - 1 is false, Statement - 2 is true "
},
{
"text": "Statement - 1 is true, Statement - 2 is false"
}
],
"answer": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is the correct explanation of Statement - 1 ",
"solution": "**Answer:** Statement - 1 is true, Statement - 2 is true; Statement - 2 is the correct explanation of Statement - 1 \n\nIn completely inelastic collision, \n
$${m_1}{v_1} + {m_2}{v_2} = {m_1}v + {m_2}v$$ \n
after collision both particle have common velocity $$v$$, so all energy is not lost \n
$$\\therefore$$ Statement - $$1$$ is true\n
The principle of conservation of momentum applicable for all kinds of collisions.\n
So statement - $$2$$ is also true. \n
Statement - $$2$$ explains statement - $$1$$ correctly because applying the principle of conservation of momentum, we can get the common velocity and hence the kinetic energy of the combined body. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8332,
"subject": "Physics",
"question": "This question has statement $${\\rm I}$$ and statement $${\\rm I}$$$${\\rm I}$$. Of the four choices given after the statements, choose the one that best describes the two statements.\n
Statement - $${\\rm I}$$: A point particle of mass $$m$$ moving with speed $$\\upsilon $$ collides with stationary point particle of mass $$M.$$ If the maximum energy loss possible is given as $$f\\left( {{1 \\over 2}m{v^2}} \\right)$$, then $$f = \\left( {{m \\over {M + m}}} \\right).$$\n
Statement - $${\\rm II}$$: Maximum energy loss occurs when the particles get stuck together as a result of the collision.
",
"options": [
{
"text": "Statement - $${\\rm I}$$ is true, Statement - $${\\rm II}$$ is true; Statement - $${\\rm II}$$ is the correct explanation of Statement - $${\\rm I}$$."
},
{
"text": "Statement - $${\\rm I}$$ is true, Statement - $${\\rm II}$$ is true; Statement - $${\\rm II}$$ is not the correct explanation of Statement - $${\\rm I}$$."
},
{
"text": "Statement - $${\\rm I}$$ is true, Statement - $${\\rm II}$$ is false "
},
{
"text": "Statement - $${\\rm I}$$ is false, Statement - $${\\rm II}$$ true."
}
],
"answer": "Statement - $${\\rm I}$$ is false, Statement - $${\\rm II}$$ true.",
"solution": "**Answer:** Statement - $${\\rm I}$$ is false, Statement - $${\\rm II}$$ true.\n\nInitial energy = $${{{P^2}} \\over {2m}}$$, where $$P$$ is the momentum and m is the mass of the moving particle.\n
Loss of energy is maximum when collision is inelastic means when the particles get stuck together as a result of the collision.\n
So after collision energy = $${{{P^2}} \\over {2\\left( {m + M} \\right)}}$$\n
$$\\therefore$$ Maximum energy loss $$ = {{{P^2}} \\over {2m}} - {{{P^2}} \\over {2\\left( {m + M} \\right)}}$$.\n \n
Statement $${\\rm I}{\\rm I}$$ says \"Maximum energy loss occurs when the particles get stuck together as a result of the collision.\" This is a case of perfectly inelastic collision.\n
Hence statement $${\\rm I}$$$${\\rm I}$$ is correct.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8333,
"subject": "Physics",
"question": "A particle of mass $$m$$ moving in the $$x$$ direction with speed $$2v$$ is hit by another particle of mass $$2m$$ moving in the $$y$$ direction with speed $$v.$$ If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to: ",
"options": [
{
"text": "$$56\\% $$ "
},
{
"text": "$$62\\% $$"
},
{
"text": "$$44\\% $$"
},
{
"text": "$$50\\% $$"
}
],
"answer": "$$56\\% $$ ",
"solution": "**Answer:** $$56\\% $$ \n\n\n Applying conservation of linear momentum in x direction\n
$$2mv = \\left( {2m + m} \\right){V_x}$$\n
$$ \\Rightarrow {V_x} = {2 \\over 3}v$$\n
Applying conservation of linear momentum in y direction\n
$$2mv = \\left( {2m + m} \\right){V_y}$$\n
$$ \\Rightarrow {V_y} = {2 \\over 3}v$$\n
Final speed of 3m mass, $${V_f}$$ = $$\\sqrt {V_x^2 + V_y^2} $$\n
$$ \\simeq 56\\% $$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8334,
"subject": "Physics",
"question": "A neutron moving with a speed ‘v’ makes a head on collision with a stationary\nhydrogen atom in ground state. The minimum kinetic energy of the neutron for\nwhich inelastic collision will take place is :",
"options": [
{
"text": "10.2 eV"
},
{
"text": "16.8 eV"
},
{
"text": "12.1 eV"
},
{
"text": "20.4 eV"
}
],
"answer": "20.4 eV",
"solution": "**Answer:** 20.4 eV\n\nLet, velocity offer collision = v1\n
$$ \\therefore $$ From conservation of momentum, \n
$$ \\Rightarrow $$ $${1 \\over 2}$$mv2 = 20.4 eV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8335,
"subject": "Physics",
"question": "The mass of a hydrogen molecule is 3.32 $$\\times$$ 10-27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of 45o to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly: ",
"options": [
{
"text": "2.35 $$\\times$$ 103 N m-2"
},
{
"text": "4.70 $$\\times$$ 103 N m-2"
},
{
"text": "2.35 $$\\times$$ 102 N m-2"
},
{
"text": "4.70 $$\\times$$ 102 N m-2"
}
],
"answer": "2.35 $$\\times$$ 103 N m-2",
"solution": "**Answer:** 2.35 $$\\times$$ 103 N m-2\n\nConsidering one hydrogen molecule : \n
$$\\therefore\\,\\,\\,$$ As direction of $$\\Delta $$P is towards negative i so force on the molecule will also be towards negative i direction. From Newton's third law, the reaction force will be on the wall in positive i direction with same magnitude.\n
$$\\therefore\\,\\,\\,$$ Force on the wall = $${{{2mv} \\over {\\sqrt 2 }}}$$ $$ \\times $$ n\n
$$\\therefore\\,\\,\\,$$ Pressure on the wall, P = $${F \\over A}$$\n
= 2.35 $$ \\times $$ 103 N m$$-$$2",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8336,
"subject": "Physics",
"question": "A proton of mass m collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90o with respect to each other. The mass of unknown particle is : ",
"options": [
{
"text": "$${m \\over 2}$$"
},
{
"text": "m"
},
{
"text": "$${m \\over {\\sqrt 3 }}$$"
},
{
"text": "2 m"
}
],
"answer": "m",
"solution": "**Answer:** m\n\n
In figure (i) before collision, m' is mass of unknown particle; m is mass of proton; v1 is initial velocity.
\n\n
(i) Before collision :
\n
\n\n
Now, in figure (ii), v1 is final velocity of unknown particle and v2 is final velocity of proton.
\n\n
(ii) After collision :
\n
\n
By conservation of momentum, we have
\n
Momentum before collision = Momentum after collision
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8337,
"subject": "Physics",
"question": "In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If\nthe final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative\nvelocity between the two particles, after collision, is :",
"options": [
{
"text": "$${{{v_0}} \\over {\\sqrt 2 }}$$ "
},
{
"text": "$${{v_0}} \\over 4$$ "
},
{
"text": "$$\\sqrt 2 {v_0}$$"
},
{
"text": "$${{v_0}} \\over 2$$ "
}
],
"answer": "$$\\sqrt 2 {v_0}$$",
"solution": "**Answer:** $$\\sqrt 2 {v_0}$$\n\nFrom conservation of linear momentum,\n
$$\\therefore$$ $${{v_1} - {v_2}}$$ = $$\\sqrt 2 {v_0}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8338,
"subject": "Physics",
"question": "If 1022 gas molecules each of mass 10–26 kg\ncollide with a surface (perpendicular to it)\nelastically per second over an area 1 m2 with\na speed 104 m/s, the pressure exerted by the gas\nmolecules will be of the order of :",
"options": [
{
"text": "108 N/m2"
},
{
"text": "1016 N/m2"
},
{
"text": "104 N/m2"
},
{
"text": "2 N/m2"
}
],
"answer": "2 N/m2",
"solution": "**Answer:** 2 N/m2\n\n
Momentum imparted to the surface in one collision,
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8339,
"subject": "Physics",
"question": "A particle of mass 'm' is moving with speed '2v'\nand collides with a mass '2m' moving with\nspeed 'v' in the same direction. After collision,\nthe first mass is stopped completely while the\nsecond one splits into two particles each of\nmass 'm', which move at angle 45° with respect\nto the origianl direction.\nThe speed of each of the moving particle will\nbe :-",
"options": [
{
"text": "2 $$\\sqrt2$$v"
},
{
"text": "v / (2 $$\\sqrt2$$ )"
},
{
"text": "v / $$\\sqrt2$$"
},
{
"text": "$$\\sqrt2$$v"
}
],
"answer": "2 $$\\sqrt2$$v",
"solution": "**Answer:** 2 $$\\sqrt2$$v\n\nInitial momentum · Pi\n = 2mv + 2mv = 4 mv \nLet v' be the speed of $$l$$ particle\n\n$$ \\therefore 2{{mv} \\over {\\sqrt 2 }} = 4mv$$ \n$$ \\Rightarrow $$ v' = $${2\\sqrt 2 }$$ v",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8340,
"subject": "Physics",
"question": "A body of mass 2 kg makes an eleastic collision\nwith a second body at rest and continues to move\nin the original direction but with one fourth of its\noriginal speed. What is the mass of the second\nbody ?",
"options": [
{
"text": "1.2 kg"
},
{
"text": "1.0 kg"
},
{
"text": "1.8 kg"
},
{
"text": "1.5 kg"
}
],
"answer": "1.2 kg",
"solution": "**Answer:** 1.2 kg\n\nBy conservation of linear momentum:
\n$$ \\Rightarrow v - {{{v_0}} \\over 4} = {v_0} \\Rightarrow m = {6 \\over 5} = 1.2\\,kg$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8341,
"subject": "Physics",
"question": "A body of mass m1 moving with an unknown\nvelocity of $${v_1}\\mathop i\\limits^ \\wedge $$, undergoes a collinear collision\nwith a body of mass m2 moving with a velocity\n$${v_2}\\mathop i\\limits^ \\wedge $$ . After collision, m1 and m2 move with\nvelocities of $${v_3}\\mathop i\\limits^ \\wedge $$ and $${v_4}\\mathop i\\limits^ \\wedge $$ , respectively.\nIf m2 = 0.5 m1 and v3 = 0.5 v1, then v1 is :-",
"options": [
{
"text": "$${v_4} - {{{v_2}} \\over 2}$$"
},
{
"text": "$${v_4} - {{{v_2}} \\over 4}$$"
},
{
"text": "$${v_4} - {v_2}$$"
},
{
"text": "$${v_4} + {v_2}$$"
}
],
"answer": "$${v_4} - {v_2}$$",
"solution": "**Answer:** $${v_4} - {v_2}$$\n\nApplying linear momentum conservation
\n$${m_1}{v_1}\\widehat i + {m_2}{v_2}\\widehat i = {m_1}{v_3}\\widehat i + {m_2}{v_4}\\widehat i$$
\nm1v1 + 0.5 m1v2 = m1(0.5 v1) + 0.5 m1v4
\n0.5 m1v1 = 0.5 m1(v4 - v2)
\nv1 = v4 - v2",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 8342,
"subject": "Physics",
"question": "An alpha-particle of mass m suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is ",
"options": [
{
"text": "2m"
},
{
"text": "4m"
},
{
"text": "1.5m"
},
{
"text": "3.5m"
}
],
"answer": "4m",
"solution": "**Answer:** 4m\n\nWe have following collision, where mass of $\\alpha$ particle $=m$ and mass of nucleus $=M$\n
\n Let $\\alpha$ particle rebounds with velocity $v_1$, then Given :\n
final energy of $\\alpha=36 \\%$ of initial energy\n
$$\n\\begin{aligned}\n\\Rightarrow \\frac{1}{2} m v_1^2 =0.36 \\times \\frac{1}{2} m v^2 \\\\\\\\\n\\Rightarrow v_1 = 0.6 v .......(i)\n\\end{aligned}\n$$\n
As unknown nucleus gained $64 \\%$ of energy of $\\alpha$, we have\n
$$\n\\begin{aligned}\n& \\frac{1}{2} M v_2^2=0.64 \\times \\frac{1}{2} m v^2 \\\\\n& \\Rightarrow v_2=\\sqrt{\\frac{m}{M}} \\times 0.8 v .........(ii)\n\\end{aligned}\n$$\n
From momentum conservation, we have\n
$$\nm v=M v_2-m v_1\n$$\n
Substituting values of $v_1$ and $v_2$ from Eqs. (i) and (ii), we have\n
$$\n\\begin{array}{rlrl}\nm v =M \\sqrt{\\frac{m}{M}} \\times 0.8 v-m \\times 0.6 v \\\\\\\\\n\\Rightarrow 16 m v =\\sqrt{m M} \\times 0.8 v \\\\\\\\\n\\Rightarrow 2 m =\\sqrt{m M} \\\\\\\\\n\\Rightarrow 4 m^2 =m M \\Rightarrow M=4 m\n\\end{array}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8343,
"subject": "Physics",
"question": "A simple pendulum, made of a string of length $$\\ell $$ and a bob of mass m, is released from a small angle $${{\\theta _0}}$$. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It\nbounces back and goes up to an angle $${{\\theta _1}}$$. Then M is given by : ",
"options": [
{
"text": "$${m \\over 2}\\left( {{{{\\theta _0} + {\\theta _1}} \\over {{\\theta _0} - {\\theta _1}}}} \\right)$$"
},
{
"text": "$${m \\over 2}\\left( {{{{\\theta _0} - {\\theta _1}} \\over {{\\theta _0} + {\\theta _1}}}} \\right)$$"
},
{
"text": "$$m\\left( {{{{\\theta _0} + {\\theta _1}} \\over {{\\theta _0} - {\\theta _1}}}} \\right)$$"
},
{
"text": "$$m\\left( {{{{\\theta _0} - {\\theta _1}} \\over {{\\theta _0} + {\\theta _1}}}} \\right)$$"
}
],
"answer": "$$m\\left( {{{{\\theta _0} + {\\theta _1}} \\over {{\\theta _0} - {\\theta _1}}}} \\right)$$",
"solution": "**Answer:** $$m\\left( {{{{\\theta _0} + {\\theta _1}} \\over {{\\theta _0} - {\\theta _1}}}} \\right)$$\n\n\n
$$ \\Rightarrow $$ $${M \\over m} = {{{\\theta _0} - {\\theta _1}} \\over {{\\theta _0} + {\\theta _1}}} \\Rightarrow M = {{{\\theta _0} - \\theta 1} \\over {{\\theta _0} + {\\theta _1}}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8344,
"subject": "Physics",
"question": "A body of mass 1 kg falls freely from a height of 100 m, on a platform mass 3 kg which is mounted on a spring having spring constant k = 1.25 $$ \\times $$ 106 N/m. The body sticks to the platform and the spring's maximum compression is found to be x. Given that g = 10 ms–2\n, the value of x will be close to : ",
"options": [
{
"text": "8 cm"
},
{
"text": "4 cm"
},
{
"text": "40 cm"
},
{
"text": "80 cm"
}
],
"answer": "4 cm",
"solution": "**Answer:** 4 cm\n\nvelocity of 1 kg block just before it collides with 3kg block \n
= $$\\sqrt {2gh} = \\sqrt {2000} $$ m/s\n
Applying momentum conversation just before and just after collision. \n
solving x $$ \\approx $$ 4 cm",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8345,
"subject": "Physics",
"question": "A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms–1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is - (g = 10 ms–2)",
"options": [
{
"text": "30 m "
},
{
"text": "40 m"
},
{
"text": "20 m"
},
{
"text": "10 m"
}
],
"answer": "40 m",
"solution": "**Answer:** 40 m\n\n\n Time taken for the particles to collide, \n
$$ \\therefore $$ Height above tower = 40 m",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8346,
"subject": "Physics",
"question": "Two bodies of the same mass are moving with the same speed, but in different directions in a\nplane. They have a completely inelastic collision and move together thereafter with a final speed\nwhich is half of their initial speed. The angle between the initial velocities of the two bodies (in\ndegree) is ________.",
"options": [],
"answer": "120",
"solution": "**Answer:** 120\n\n\n
$$ \\Rightarrow $$ cos $$\\theta $$ = $${1 \\over 2}$$\n
$$ \\Rightarrow $$ $$\\theta $$ = 60o\n
$$ \\therefore $$ 2$$\\theta $$ = 120o",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8347,
"subject": "Physics",
"question": "A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg\ncollides with the block and sticks to it. If the velocity of the bullet is 20 m/s in the horizontal\ndirection just before the collision then the kinetic energy just before the combined system strikes\nthe floor, is [Take g = 10 m/s2\n . Assume there is no rotational motion and loss of energy after the\ncollision is negligable.]",
"options": [
{
"text": "23 J"
},
{
"text": "21 J"
},
{
"text": "20 J"
},
{
"text": "19 J"
}
],
"answer": "21 J",
"solution": "**Answer:** 21 J\n\n\n
= $$ - {1 \\over 8}m{u^2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8350,
"subject": "Physics",
"question": "A particle of mass m is dropped from a height\nh above the ground. At the same time another\nparticle of the same mass is thrown vertically\nupwards from the ground with a speed of $$\\sqrt {2gh} $$. If they collide head-on completely\ninelastically, the time taken for the combined\nmass to reach the ground, in units of $$\\sqrt {{h \\over g}} $$ is :",
"options": [
{
"text": "$$\\sqrt {{1 \\over 2}} $$"
},
{
"text": "$${1 \\over 2}$$"
},
{
"text": "$$\\sqrt {{3 \\over 2}} $$"
},
{
"text": "$$\\sqrt {{3 \\over 4}} $$"
}
],
"answer": "$$\\sqrt {{3 \\over 2}} $$",
"solution": "**Answer:** $$\\sqrt {{3 \\over 2}} $$\n\n\n Particles will collide after time t = $${h \\over {\\sqrt {2gh} }}$$ = $$\\sqrt {{h \\over {2g}}} $$\n
Velocity of (A) just before collision, VA = 0 + gt = $$g\\sqrt {{h \\over {2g}}} $$ = $$\\sqrt {{{gh} \\over 2}} $$\n
Velocity of (B) just before collision, \n VB = $${\\sqrt {2gh} }$$ - $$g\\sqrt {{h \\over {2g}}} $$ \n
height from ground where collision takes place = h1 = h - $${1 \\over 2}g{t^2}$$ = h - $${1 \\over 2}g.{h \\over {2g}}$$ = $${{3h} \\over 4}$$\n
Time taken by combined mass to reach the\n ground = $$\\sqrt {{{2 \\times {{3h} \\over 4}} \\over {2g}}} $$ = $$\\sqrt {{{3h} \\over {2g}}} $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8351,
"subject": "Physics",
"question": "A body A, of mass m = 0.1 kg has an initial\nvelocity of 3$$\\widehat i$$ ms-1 . It collides elastically with\nanother body, B of the same mass which has\nan initial velocity of 5$$\\widehat j$$ ms-1. After collision,\nA moves with a velocity $$\\overrightarrow v = 4\\left( {\\widehat i + \\widehat j} \\right)$$. The\nenergy of B after collision is written as $${x \\over {10}}$$. The value of x is ___________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nBy conservation of linear momentum :\n
$$ \\Rightarrow $$ $$v = {u \\over {\\sqrt 2 }}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8353,
"subject": "Physics",
"question": "A ball with a speed of 9 m/s collides with another identical ball at rest. After the collision, the direction of each ball makes an angle of 30$$^\\circ$$ with the original direction. The ratio of velocities of the balls after collision is x : y, where x is __________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nThe situation is shown below
Using conservation of linear momentum in y-direction,
pi = pf
As, pi = 0
and pf = mv1 sin30$$^\\circ$$ $$-$$ mv2 sin30$$^\\circ$$
$$\\Rightarrow$$ 0 = m $$\\times$$ $${1 \\over 2}$$v1 $$-$$ m $$\\times$$ $${1 \\over 2}$$v2
$$\\Rightarrow$$ v1 = v2 or v1 : v2 = 1 : 1
Since, v1 : v2 = x : y (given)
$$\\therefore$$ x = 1",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8354,
"subject": "Physics",
"question": "Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Body 'P' having mass M moving with speed 'u' has head-on collision elastically with another body 'Q' having mass 'm' initially at rest. If m << M, body 'Q' will have a maximum speed equal to '2u' after collision.
Reason R : During elastic collision, the momentum and kinetic energy are both conserved.
In the light of the above statements, choose the most appropriate answer from the options given below :",
"options": [
{
"text": "A is correct but R is not correct."
},
{
"text": "A is not correct but R is correct."
},
{
"text": "Both A and R are correct and R is the correct explanation of A."
},
{
"text": "Both A and R are correct but R is NOT the correct explanation of A."
}
],
"answer": "Both A and R are correct and R is the correct explanation of A.",
"solution": "**Answer:** Both A and R are correct and R is the correct explanation of A.\n\n\n
m < < M
e = $${{{v_2} - {v_1}} \\over {{u_1} - {u_2}}}$$
For elastic collision $$ \\to $$ e = 1
1 = $${{{v_2} - u} \\over {u - 0}}$$
u = v2 $$-$$ u
v2 = 2u
In elastic collision kinetic energy & momentum are conserved.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8355,
"subject": "Physics",
"question": "An object of mass m1 collides with another object of mass m2, which is at rest. After the collision the objects move with equal speeds in opposite direction. The ratio of the masses m2 : m1 is :",
"options": [
{
"text": "1 : 1"
},
{
"text": "3 : 1"
},
{
"text": "2 : 1"
},
{
"text": "1 : 2"
}
],
"answer": "3 : 1",
"solution": "**Answer:** 3 : 1\n\nBefore collision
$$ \\Rightarrow $$ 3m1 = m2 $$ \\Rightarrow $$ $${{{m_2}} \\over {{m_1}}} = {3 \\over 1}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8356,
"subject": "Physics",
"question": "A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass 'm' travelling along the surface hits at one end of the rod with a velocity 'u' in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses $$\\left( {{m \\over M}} \\right)$$ is $${1 \\over x}$$. The value of 'x' will be ____________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nThe given situation can be shown as
Before collision,
As the collision is perfectly elastic, therefore momentum is conserved, i.e.
pinitially = pfinally
$$\\Rightarrow$$ mu = Mv ..... (i)
Angular momentum will also be conserved about point O.
ratio of masses $$\\left( {{m \\over M}} \\right) = {1 \\over x}$$.
Comparing it with Eq. (iv), we get x = 4.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8357,
"subject": "Physics",
"question": "A body of mass 2 kg moving with a speed of 4 m/s. makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial peed. The speed of the two body centre of mass is $${x \\over {10}}$$ m/s. Then the value of x is ___________.",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\npi = pf
x = 25",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8358,
"subject": "Physics",
"question": "A body of mass M moving at speed V0 collides elastically with a mass 'm' at rest. After the collision, the two masses move at angles $$\\theta$$1 and $$\\theta$$2 with respect to the initial direction of motion of the body of mass M. The largest possible value of the ratio M/m, for which the angles $$\\theta$$1 and $$\\theta$$2 will be equal, is :",
"options": [
{
"text": "4"
},
{
"text": "1"
},
{
"text": "3"
},
{
"text": "2"
}
],
"answer": "3",
"solution": "**Answer:** 3\n\n Given $$\\theta$$1 = $$\\theta$$2 = $$\\theta$$
from momentum conservation
in x-direction MV0 = MV1 cos$$\\theta$$ + mV2 cos$$\\theta$$
in y-direction 0 = MV1 sin$$\\theta$$ $$-$$ mV2 sin$$\\theta$$
A body of mass M at rest explodes into three pieces, in the ratio of masses 1 : 1 : 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms$$-$$1 and 40 ms$$-$$1 respectively. The velocity of the third piece will be :
",
"options": [
{
"text": "15 ms$$-$$1"
},
{
"text": "25 ms$$-$$1"
},
{
"text": "35 ms$$-$$1"
},
{
"text": "50 ms$$-$$1"
}
],
"answer": "25 ms$$-$$1",
"solution": "**Answer:** 25 ms$$-$$1\n\nGiven problem a body of mass $M$ explodes into three pieces of mass ratio $1: 1: 2$\n
$$ \\therefore $$ Mass of fragments will be $x, x, 2 x$ \n
Hence, $M=x+x+2 x=4 x \\mathrm{~kg}$\n
As in the process of explosion no external forces are involved and explosion occurs due to internal forces. Thus, momentum of the system will be conserved.\n
What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of 5 times its mass?
\n
(Assume the collision to be head-on elastic collision)
Two billiard balls of mass 0.05 kg each moving in opposite directions with 10 ms$$-$$1 collide and rebound with the same speed. If the time duration of contact is t = 0.005 s, then what is the force exerted on the ball due to each other?
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8362,
"subject": "Physics",
"question": "A ball is dropped from a height of $20 \\mathrm{~m}$. If the coefficient of restitution for the collision between ball and floor is $0.5$, after hitting the floor, the ball rebounds to a height of ________ $\\mathrm{m}$.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nWe know, $h^{\\prime}=e^{2} h$\n\n
A body of mass 1 kg collides head on elastically with a stationary body of mass 3 kg. After collision, the smaller body reverses its direction of motion and moves with a speed of 2 m/s. The initial speed of the smaller body before collision is ___________ ms$$^{-1}$$.
A particle of mass m moving with velocity v collides with a stationary particle of mass 2m. After collision, they stick together and continue to move together with velocity
When two particles collide, the total momentum before the collision is equal to the total momentum after the collision, according to the law of conservation of momentum. Therefore, we can write:
\n
$$mv = (m+2m) v_f$$
\n
where $v_f$ is the final velocity of the combined particles.
A body starts falling freely from height $$H$$ hits an inclined plane in its path at height $$h$$. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of $$\\frac{H}{h}$$ for which the body will take the maximum time to reach the ground is __________.
A stationary particle breaks into two parts of masses $$m_A$$ and $$m_B$$ which move with velocities $$v_A$$ and $$v_B$$ respectively. The ratio of their kinetic energies $$\\left(K_B: K_A\\right)$$ is :
A stationary particle breaks into two parts with masses $$m_A$$ and $$m_B$$, which then move with velocities $$v_A$$ and $$v_B$$, respectively. We need to determine the ratio of their kinetic energies $$K_A$$ and $$K_B$$.
\n\n
Since the initial momentum of the particle is zero, the momentum of the two parts must be equal and opposite to conserve momentum:
However, using the momentum relationship, we can substitute $$m_A v_A = m_B v_B$$ into the kinetic energy ratio, leading us to simplify:
\n\n
$$ \\frac{K_A}{K_B} = \\frac{v_A}{v_B} $$
\n\n
Therefore, the ratio of their kinetic energies $$\\left(K_B: K_A\\right)$$ is:
\n\n
$$ \\frac{K_B}{K_A} = \\frac{v_B}{v_A} $$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8367,
"subject": "Physics",
"question": "Consider the following two statements : \n
$$A.$$ Linear momentum of a system of particles is zero\n
$$B.$$ Kinetic energy of a system of particles is zero.\n
then ",
"options": [
{
"text": "$$A$$ does not imply $$B$$ and $$B$$ does not imply $$A$$"
},
{
"text": "$$A$$ implies $$B$$ but $$B$$ does not imply $$A$$ "
},
{
"text": "$$A$$ does not imply $$B$$ but $$B$$ implies $$A$$ "
},
{
"text": "$$A$$ implies $$B$$ and $$B$$ implies $$A$$ "
}
],
"answer": "$$A$$ does not imply $$B$$ and $$B$$ does not imply $$A$$",
"solution": "**Answer:** $$A$$ does not imply $$B$$ and $$B$$ does not imply $$A$$\n\n
The correct answer is Option A : $$A$$ does not imply $$B$$ and $$B$$ does not imply $$A$$.
\n
Here's why :
\n
Statement $$A:$$ The linear momentum of a system of particles being zero does not mean that the kinetic energy is also zero. For example, consider two equal mass particles moving with the same speed but in opposite directions. The linear momentum of the system will be zero because momentum is a vector quantity and the two momenta will cancel out. However, kinetic energy is a scalar quantity and does not cancel out in this way. Each particle has kinetic energy due to its motion, so the total kinetic energy of the system is not zero.
\n
Statement $$B:$$ The kinetic energy of a system of particles being zero also does not imply that the linear momentum is zero. If the kinetic energy is zero, it means that all the particles are at rest (since kinetic energy is associated with motion). However, the linear momentum will also be zero in this case because momentum depends on both mass and velocity, and the velocity of each particle is zero.
\n
But remember that zero linear momentum doesn't exclusively mean all particles are at rest. As explained above, it could be the scenario where particles have equal and opposite momenta, thereby cancelling each other. Hence, $$B$$ doesn't imply $$A$$.
\n
So, neither statement implies the other. Hence, Option A is correct.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8368,
"subject": "Physics",
"question": "A machine gun fires a bullet of mass $$40$$ $$g$$ with a velocity $$1200m{s^{ - 1}}.$$ The man holding it can exert a maximum force of $$144$$ $$N$$ on the gun. How many bullets can he fire per second at the most? ",
"options": [
{
"text": "Two "
},
{
"text": "Four "
},
{
"text": "One "
},
{
"text": "Three "
}
],
"answer": "Three ",
"solution": "**Answer:** Three \n\nAssume the man can fire $$n$$ bullets in one second.\n
$$\\therefore$$ change in momentum per second $$ = n \\times mv = F$$\n
[ $$m=$$ mass of bullet, $$v=$$ velocity, $$F$$ = force) ] \n
$$\\therefore$$ $$n = {F \\over {mv}} = {{144 \\times 1000} \\over {40 \\times 1200}} = 3$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8369,
"subject": "Physics",
"question": "A player caught a cricket ball of mass $$150$$ $$g$$ moving at a rate of $$20$$ $$m/s.$$ If the catching process is completed in $$0.1s,$$ the force of the blow exerted by the ball on the hand of the player is equal to ",
"options": [
{
"text": "$$150$$ $$N$$ "
},
{
"text": "$$3$$ $$N$$ "
},
{
"text": "$$30$$ $$N$$ "
},
{
"text": "$$300$$ $$N$$ "
}
],
"answer": "$$30$$ $$N$$ ",
"solution": "**Answer:** $$30$$ $$N$$ \n\nWe know, Force$$ \\times $$ time = Impulse = Change in momentum\n
$$\\therefore$$ $$F \\times t = m\\left( {v - u} \\right)$$\n
$$ \\Rightarrow $$ $$F = {{m\\left( {v - u} \\right)} \\over t} = {{0.15\\left( {0 - 20} \\right)} \\over {0.1}} = 30N$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8370,
"subject": "Physics",
"question": "A bomb of mass $$16kg$$ at rest explodes into two pieces of masses $$4$$ $$kg$$ and $$12$$ $$kg.$$ The velocity of the $$12$$ $$kg$$ mass is $$4\\,\\,m{s^{ - 1}}.$$ The kinetic energy of the other mass is ",
"options": [
{
"text": "$$144$$ $$J$$ "
},
{
"text": "$$288$$ $$J$$ "
},
{
"text": "$$192$$ $$J$$ "
},
{
"text": "$$96$$ $$J$$ "
}
],
"answer": "$$288$$ $$J$$ ",
"solution": "**Answer:** $$288$$ $$J$$ \n\nHere linear momentum is conserved as no external force is acting on the bomb.\n
Let the velocity and mass of $$4$$ $$kg$$ piece be $${v_1}$$ and $${m_1}$$ and that of $$12$$ $$kg$$ piece be $${v_2}$$ and $${m_2}$$.\n \n Applying conservation of linear momentum\n
$$\\therefore$$ $$K.E{_1} = {1 \\over 2}{m_1}v_1^2 = {1 \\over 2} \\times 4 \\times 144 = 288\\,J$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8371,
"subject": "Physics",
"question": "A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is : ",
"options": [
{
"text": "$$2\\sqrt {{k \\over p}} $$"
},
{
"text": "$$2\\sqrt {{p \\over k}} $$"
},
{
"text": "$$\\sqrt {{{2p} \\over 2}} $$"
},
{
"text": "$$\\sqrt {{{2k} \\over p}} $$"
}
],
"answer": "$$2\\sqrt {{p \\over k}} $$",
"solution": "**Answer:** $$2\\sqrt {{p \\over k}} $$\n\n$${{dp} \\over {dt}} = F = kt$$\n
$$\\int_P^{3P} {dP} = \\int_0^T {kt\\,dt} $$\n
$$2p = {{K{T^2}} \\over 2}$$\n
$$T = 2\\sqrt {{P \\over K}} $$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8372,
"subject": "Physics",
"question": "A wedge of mass M = 4m lies on a frictionless\nplane. A particle of mass m approaches the\nwedge with speed v. There is no friction\nbetween the particle and the plane or between\nthe particle and the wedge. The maximum\nheight climbed by the particle on the wedge is\ngiven by :-",
"options": [
{
"text": "$${{{v^2}} \\over {g}}$$"
},
{
"text": "$${{2{v^2}} \\over {7g}}$$"
},
{
"text": "$${{{v^2}} \\over {2g}}$$"
},
{
"text": "$${{2{v^2}} \\over {5g}}$$"
}
],
"answer": "$${{2{v^2}} \\over {5g}}$$",
"solution": "**Answer:** $${{2{v^2}} \\over {5g}}$$\n\nInitial condition can be shown in the figure\nbelow\n
\n As mass $m$ collides with wedge, let both wedge and mass move with speed $v^{\\prime}$. Then,\n\n\n\n
\n
Given:\n
Mass of the wedge (M) = 4m\n
Mass of the particle (m)\n
Initial speed of the particle (v)\n
There is no friction.
\n
Step 1 : Conservation of Linear Momentum\n
Before the collision, the momentum of the system is just the momentum of the particle because the wedge is at rest. After the collision, both the particle and the wedge will be moving. Let's denote the final common velocity as $v'$.
\n
We can write the conservation of momentum as :
\n
$$mv = (m + 4m)v'$$
\n
which simplifies to
\n
$$v' = \\frac{v}{5} \\tag{1}$$
\n
Step 2 : Conservation of Mechanical Energy\n
We apply the conservation of energy before and after the collision. Before the collision, only the particle has kinetic energy. After the collision, both the particle and the wedge have kinetic energy and the particle has potential energy due to its height h on the wedge.
So, the maximum height climbed by the particle on the wedge is given by $\\frac{2v^2}{5g}$.
\n
Therefore, the correct answer is Option D :
\n
$$\\frac{2v^2}{5g}$$
\n",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 8373,
"subject": "Physics",
"question": "A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The\nman pushes his son so that he starts moving at a speed of 0.70 ms–1 with respect to the man. The speed of the\nman with respect to the surface is :",
"options": [
{
"text": "0.28 ms–1"
},
{
"text": "0.47 ms–1"
},
{
"text": "0.20 ms–1"
},
{
"text": "0.14 ms–1"
}
],
"answer": "0.20 ms–1",
"solution": "**Answer:** 0.20 ms–1\n\n50 V1 = 20 V2
\nV1 + V2 = 0.70
\nV1 = 0.20",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8374,
"subject": "Physics",
"question": "Particle A of mass m1\n moving with velocity $$\\left( {\\sqrt3\\widehat i + \\widehat j} \\right)m{s^{ - 1}}$$ collides with another particle B of mass m2\nwhich is at rest initially. Let $$\\overrightarrow {{V_1}} $$\n and $$\\overrightarrow {{V_2}} $$\n be the velocities of particles A and B after collision\nrespectively. If m1\n = 2m2\n and after collision $$\\overrightarrow {{V_1}} = $$$$\\left( {\\widehat i + \\sqrt 3 \\widehat j} \\right)$$\n, the angle between $$\\overrightarrow {{V_1}} $$\n and $$\\overrightarrow {{V_2}} $$\n is :",
"options": [
{
"text": "105o"
},
{
"text": "15o"
},
{
"text": "-45o"
},
{
"text": "60o"
}
],
"answer": "105o",
"solution": "**Answer:** 105o\n\nGiven m1\n = 2m2\n So let, m2 = m and m1 = 2m\n
From momentum conservation\n
$${\\overrightarrow p _i}$$ = $${\\overrightarrow p _f}$$\n
$$ \\Rightarrow $$ (2m)$$\\left( {\\sqrt 3 \\widehat i + \\widehat j} \\right)$$ + 0 = 2m$$\\left( {\\widehat i + \\sqrt 3 \\widehat j} \\right)$$ + m$${\\overrightarrow V _2}$$\n
$$ \\Rightarrow $$ $${\\overrightarrow V _2}$$ = 2$$\\left( {\\sqrt 3 \\widehat i + \\widehat j} \\right)$$ - 2$$\\left( {\\widehat i + \\sqrt 3 \\widehat j} \\right)$$\n
Also given after collision $$\\overrightarrow {{V_1}} = \\left( {\\widehat i + \\sqrt3\\widehat j} \\right)m{s^{ - 1}}$$\n\n
For angle between $${\\overrightarrow V _1}$$ & $${\\overrightarrow V _2}$$,\n
cos $$\\theta $$ = $${{{{\\overrightarrow V }_1}.{{\\overrightarrow V }_2}} \\over {\\left| {{{\\overrightarrow V }_1}} \\right|\\left| {{{\\overrightarrow V }_2}} \\right|}}$$\n
$$ \\Rightarrow $$ $$\\theta $$ = 105o",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8375,
"subject": "Physics",
"question": "Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies (K.E.)A : (K.E.)B will be $${{A \\over 1}}$$, so the value of A will be ________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nGiven that, $${{{M_1}} \\over {{M_2}}} = {1 \\over 2}$$
$$ \\Rightarrow $$ $$ \\therefore $$ A = 2",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8376,
"subject": "Physics",
"question": "Two particles having masses 4 g and 16 g respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is n : 2. The value of n will be ___________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n$$\\because$$ relation b/w kinetic energy & momentum is
$$ \\Rightarrow $$ $$n = 1$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8377,
"subject": "Physics",
"question": "If the Kinetic energy of a moving body becomes four times its initial Kinetic energy, then the percentage change in its momentum will be :",
"options": [
{
"text": "100%"
},
{
"text": "200%"
},
{
"text": "300%"
},
{
"text": "400%"
}
],
"answer": "100%",
"solution": "**Answer:** 100%\n\nK2 = 4K1
% change = $${{\\Delta P} \\over {{P_1}}} \\times 100 = {{2m{v_1} - m{v_1}} \\over {m{v_1}}} \\times 100 = 100\\% $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8378,
"subject": "Physics",
"question": "A bullet of '4 g' mass is fired from a gun of mass 4 kg. If the bullet moves with the muzzle speed of 50 ms$$-$$1, the impulse imparted to the gun and velocity of recoil of gun are :",
"options": [
{
"text": "0.2 kg ms$$-$$1, 0.1 ms$$-$$1"
},
{
"text": "0.4 kg ms$$-$$1, 0.05 ms$$-$$1"
},
{
"text": "0.2 kg ms$$-$$1, 0.05 ms$$-$$1"
},
{
"text": "0.4 kg ms$$-$$1, 0.1 ms$$-$$1"
}
],
"answer": "0.2 kg ms$$-$$1, 0.05 ms$$-$$1",
"solution": "**Answer:** 0.2 kg ms$$-$$1, 0.05 ms$$-$$1\n\nmBullet = 4g, MGun = 4 kg
A man of 60 kg is running on the road and suddenly jumps into a stationary trolly car of mass 120 kg. Then, the trolly car starts moving with velocity 2 ms$$-$$1. The velocity of the running man was ___________ ms$$-$$1, when he jumps into the car.
",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\nTotal momentum of (man + trolley) system is always conserved\n
Initially man was moving with velocity v1 and trolley was at rest, finally both were moving with velocity 2 ms$$-$$1 after man jumps on the trolley.\n
A batsman hits back a ball of mass 0.4 kg straight in the direction of the bowler without changing its initial speed of 15 ms$$-$$1. The impulse imparted to the ball is ___________ Ns.
Two bodies A and B of masses 5 kg and 8 kg are moving such that the momentum of body B is twice that of the body A. The ratio of their kinetic energies will be :
A ball of mass $$0.15 \\mathrm{~kg}$$ hits the wall with its initial speed of $$12 \\mathrm{~ms}^{-1}$$ and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is $$100 \\mathrm{~N}$$, calculate the time duration of the contact of ball with the wall.
A body of mass $$8 \\mathrm{~kg}$$ and another of mass $$2 \\mathrm{~kg}$$ are moving with equal kinetic energy. The ratio of their respective momentum will be :
A body of mass $$10 \\mathrm{~kg}$$ is projected at an angle of $$45^{\\circ}$$ with the horizontal. The trajectory of the body is observed to pass through a point $$(20,10)$$. If $$\\mathrm{T}$$ is the time of flight, then its momentum vector, at time $$\\mathrm{t}=\\frac{\\mathrm{T}}{\\sqrt{2}}$$, is _____________.
In two different experiments, an object of mass $$5 \\mathrm{~kg}$$ moving with a speed of $$25 \\mathrm{~ms}^{-1}$$ hits two different walls and comes to rest within (i) 3 second, (ii) 5 seconds, respectively. Choose the correct option out of the following :
",
"options": [
{
"text": "Impulse and average force acting on the object will be same for both the cases."
},
{
"text": "Impulse will be same for both the cases but the average force will be different."
},
{
"text": "Average force will be same for both the cases but the impulse will be different."
},
{
"text": "Average force and impulse will be different for both the cases."
}
],
"answer": "Impulse will be same for both the cases but the average force will be different.",
"solution": "**Answer:** Impulse will be same for both the cases but the average force will be different.\n\n
$$\\Delta$$P = impulse = same since acceleration is different force acting will be different.
100 balls each of mass $$\\mathrm{m}$$ moving with speed $$v$$ simultaneously strike a wall normally and reflected back with same speed, in time $$\\mathrm{t ~s}$$. The total force exerted by the balls on the wall is
",
"options": [
{
"text": "$$\\frac{200 m v}{t}$$"
},
{
"text": "$$\\frac{100 m v}{t}$$"
},
{
"text": "$$\\frac{m v}{100 t}$$"
},
{
"text": "$$200 m v t$$"
}
],
"answer": "$$\\frac{200 m v}{t}$$",
"solution": "**Answer:** $$\\frac{200 m v}{t}$$\n\nWhen the balls strike the wall, the change in momentum of each ball is given by:\n
$$\\Delta p = mv - (-mv) = 2mv$$\n
Since there are 100 balls, the total change in momentum of all the balls is $$\\Delta P = 2m(100v) = 200mv.$$ The time taken for all the balls to strike the wall is $\\mathrm{t}$ seconds. Therefore, the average force exerted on the wall is given by:\n\n
Therefore, the total force exerted by the balls on the wall is $$\\boxed{F = \\frac{200mv}{\\mathrm{t}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8389,
"subject": "Physics",
"question": "A machine gun of mass $10 \\mathrm{~kg}$ fires $20 \\mathrm{~g}$ bullets at the rate of 180 bullets per minute with a speed of $100 \\mathrm{~m} \\mathrm{~s}^{-1}$ each. The recoil velocity of the gun is",
"options": [
{
"text": "$ 0.02 \\mathrm{~m} / \\mathrm{s}$"
},
{
"text": "$1.5 \\mathrm{~m} / \\mathrm{s}$"
},
{
"text": "$2.5 \\mathrm{~m} / \\mathrm{s}$"
},
{
"text": "$0.6 \\mathrm{~m} / \\mathrm{s}$"
}
],
"answer": "$0.6 \\mathrm{~m} / \\mathrm{s}$",
"solution": "**Answer:** $0.6 \\mathrm{~m} / \\mathrm{s}$\n\n
A ball of mass $$200 \\mathrm{~g}$$ rests on a vertical post of height $$20 \\mathrm{~m}$$. A bullet of mass $$10 \\mathrm{~g}$$, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance $$30 \\mathrm{~m}$$ and the bullet at a distance of $$120 \\mathrm{~m}$$ from the foot of the post. The value of initial velocity of the bullet will be (if $$g=10 \\mathrm{~m} / \\mathrm{s}^{2}$$) :
A bullet of $$10 \\mathrm{~g}$$ leaves the barrel of gun with a velocity of $$600 \\mathrm{~m} / \\mathrm{s}$$. If the barrel of gun is $$50 \\mathrm{~cm}$$ long and mass of gun is $$3 \\mathrm{~kg}$$, then value of impulse supplied to the gun will be :
",
"options": [
{
"text": "12 Ns"
},
{
"text": "3 Ns"
},
{
"text": "6 Ns"
},
{
"text": "36 Ns"
}
],
"answer": "6 Ns",
"solution": "**Answer:** 6 Ns\n\nFirst, we need to find the velocity of the gun after the bullet is fired. We can use conservation of momentum to do this. The total momentum of the system of the gun and bullet is conserved before and after the bullet is fired. Therefore, we can write\n
\n$$m_g u_g + m_b u_b = m_g v_g + m_b v_b$$\n
\nwhere $$u_g = 0$$ is the initial velocity of the gun, $$u_b = 600 \\mathrm{~m/s}$$ is the initial velocity of the bullet, $$m_g = 3 \\mathrm{~kg}$$ is the mass of the gun, $$m_b = 0.01 \\mathrm{~kg}$$ is the mass of the bullet, $$v_b = 0$$ is the final velocity of the bullet (since it has left the gun), and $$v_g$$ is the final velocity of the gun.\n
\nTherefore, the velocity of the gun after the bullet is fired is $$v_g = 2 \\mathrm{~m/s}$$.\n
\nNext, we need to find the impulse on the gun. The impulse-momentum theorem states that the impulse on an object is equal to the change in momentum of that object. Therefore, the impulse on the gun is given by\n
\n$$I = \\Delta p = m_g \\Delta v$$\n
\nwhere $$\\Delta v = v_f - u_g$$ is the change in velocity of the gun. Since the initial velocity of the gun is zero, we can write $$\\Delta v = v_g$$.\n
\nTherefore, the impulse on the gun is $$6 \\mathrm{~Ns}$$, which is the correct answer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8392,
"subject": "Physics",
"question": "
An average force of $$125 \\mathrm{~N}$$ is applied on a machine gun firing bullets each of mass $$10 \\mathrm{~g}$$ at the speed of $$250 \\mathrm{~m} / \\mathrm{s}$$ to keep it in position. The number of bullets fired per second by the machine gun is :
To find the number of bullets fired per second, we can use the concept of momentum. When the machine gun fires bullets, it experiences a backward force due to the conservation of momentum. The force applied on the machine gun is used to balance this backward force.
Two bodies of mass $$4 \\mathrm{~g}$$ and $$25 \\mathrm{~g}$$ are moving with equal kinetic energies. The ratio of magnitude of their linear momentum is :
A body of mass $$1000 \\mathrm{~kg}$$ is moving horizontally with a velocity $$6 \\mathrm{~m} / \\mathrm{s}$$. If $$200 \\mathrm{~kg}$$ extra mass is added, the final velocity (in $$\\mathrm{m} / \\mathrm{s}$$) is:
An artillery piece of mass $$M_1$$ fires a shell of mass $$M_2$$ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is:
A spherical body of mass $$100 \\mathrm{~g}$$ is dropped from a height of $$10 \\mathrm{~m}$$ from the ground. After hitting the ground, the body rebounds to a height of $$5 \\mathrm{~m}$$. The impulse of force imparted by the ground to the body is given by : (given, $$\\mathrm{g}=9.8 \\mathrm{~m} / \\mathrm{s}^2$$)
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8398,
"subject": "Physics",
"question": "Two identical particles move towards each other with velocity $$2v$$ and $$v$$ respectively. The velocity of center of mass is ",
"options": [
{
"text": "$$v$$ "
},
{
"text": "$$v/3$$ "
},
{
"text": "$$v/2$$ "
},
{
"text": "zero"
}
],
"answer": "$$v/2$$ ",
"solution": "**Answer:** $$v/2$$ \n\n \nThe velocity of center of mass of two particle system is\n
A solid circular disc of mass $$50 \\mathrm{~kg}$$ rolls along a horizontal floor so that its center of mass has a speed of $$0.4 \\mathrm{~m} / \\mathrm{s}$$. The absolute value of work done on the disc to stop it is ________ J.
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8400,
"subject": "Physics",
"question": "A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ________ m/s. (Take g = 10 m/s2)",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n\n
Let the speed of bob at lowest position be v1 and at the highest position be v2.
Maximum tension is at lowest position and minimum tension is at the highest position. Now, using, conservation of mechanical energy,
$$ \\Rightarrow {v_2}^2 = 25 \\Rightarrow {v_2} = 5$$ m/s
Thus, velocity of bob at highest position 5 m/s.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8401,
"subject": "Physics",
"question": "
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k2rt2, where k is a constant. The power delivered to the particle by the force acting on it is given as
A stone tide to a spring of length L is whirled in a vertical circle with the other end of the spring at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is $$\\sqrt {x({u^2} - gL)} $$. The value of x is -
A pendulum of length 2 m consists of a wooden bob of mass 50 g. A bullet of mass 75 g is fired towards the stationary bob with a speed v. The bullet emerges out of the bob with a speed $${v \\over 3}$$ and the bob just completes the vertical circle. The value of v is ___________ ms$$-$$1. (if g = 10 m/s2).
",
"options": [],
"answer": "10",
"solution": "**Answer:** 10\n\nLet minimum velocity at the lowest point for the bob to compleate the circular path = v'\n
At the point $3$, both the tension $T_{3}$ and the weight $m g$ of the body act towards the centre of the circle. So $T_{3}+m g$ provides the centripetal force necessary for the rotation of the body.\n
At the point $1$, the tension $T_{1}$ acts vertically upwards i.e., towards the centre of the circle and the weight $m g$ of the body acts vertically downwards i.e., in an opposite direction. So $T_{1}-m g$ provides the necessary centripetal force here.\n\n
If the velocity of the body at the highest point $3$ be less than $\\sqrt{g r}$, the string will slack and the body will drop down instead of rotating in the circular path. So this minimum velocity of the body at the highest point is called the critical velocity.\n
Minimum velocity at the lowest point for maintaining the critical velocity :\n
Now, as the body goes from $3$ to $1$, its height increases by $2 r$. So its potential energy increases by $m g \\times 2 r$. From the principle of conservation of mechanical energy, we have\n\n
The K.E. of the body at $3$ - Its K.E. at $1=$ Increase in P.E. \n
or, $1 / 2 m v_{3}^{2}-1 / 2 m v_{1}^{2}=2 m g r$\n\n
or, $v_{3}^{2}=v_{1}^{2}+4 g r$\n\n
When $v_{3}=\\sqrt{g r}, v_{1}$ is minimurn\n\n
$\\therefore\\left(v_{1}\\right)_{\\min }=\\sqrt{g r+4 g r}=\\sqrt{5 g r}$\n\n
Minimum tension :\n
When the body moves with the critical velocity at the highest point, the tension in the string becomes zero; then $m g=\\frac{m v_{3}^{2}}{r}$ When this condition is satisfied, the tension in the string at the lowest point $1$ becomes minimum.\n\n
A car is moving on a circular path of radius 600 m such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of 54 km/hr is $$t(1-e^{-\\pi/2})s$$. The value of t is ____________.
A particle is moving in a circle of radius $$50 \\mathrm{~cm}$$ in such a way that at any instant the normal and tangential components of it's acceleration are equal. If its speed at $$\\mathrm{t}=0$$ is $$4 \\mathrm{~m} / \\mathrm{s}$$, the time taken to complete the first revolution will be $$\\frac{1}{\\alpha}\\left[1-e^{-2 \\pi}\\right] \\mathrm{s}$$, where $$\\alpha=$$ _________.
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8406,
"subject": "Physics",
"question": "The minimum velocity (in $$m{s^{ - 1}}$$) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction $$0.6$$ to avoid skidding is ",
"options": [
{
"text": "$$60$$ "
},
{
"text": "$$30$$ "
},
{
"text": "$$15$$ "
},
{
"text": "$$25$$ "
}
],
"answer": "$$30$$ ",
"solution": "**Answer:** $$30$$ \n\nFor no skidding along curved track,\n
The maximum velocity possible $${v_{\\max }} = \\sqrt {\\mu rg} $$\n
Here $$\\mu = 0.6,\\,r = 150m,\\,g = 9.8$$\n
$$\\therefore$$ $${v_{\\max }} = \\sqrt {0.6 \\times 150 \\times 9.8} \\simeq 30m/s$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8407,
"subject": "Physics",
"question": "Which of the following statements is FALSE for a particle moving in a circle with a constant\nangular speed? ",
"options": [
{
"text": "The velocity vector is tangent to the circle."
},
{
"text": "The acceleration vector is tangent to the circle."
},
{
"text": "The acceleration vector points to the centre of the circle."
},
{
"text": "The velocity and acceleration vectors are perpendicular to each other."
}
],
"answer": "The acceleration vector is tangent to the circle.",
"solution": "**Answer:** The acceleration vector is tangent to the circle.\n\nOnly option $$(b)$$ is false since acceleration vector acts along the radius of the circle or towards center of the circle for uniform circular motion and velocity vector always acts along the tangent of the circle.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8408,
"subject": "Physics",
"question": "For a particle in uniform circular motion the acceleration $$\\overrightarrow a $$ at a point P(R, θ) on the circle of radius R is (here θ is measured from the x–axis)",
"options": [
{
"text": "$$ - {{{v^2}} \\over R}\\cos \\theta \\widehat i + {{{v^2}} \\over R}\\sin \\theta \\widehat j$$"
},
{
"text": "$$ - {{{v^2}} \\over R}\\sin \\theta \\widehat i + {{{v^2}} \\over R}\\cos \\theta \\widehat j$$"
},
{
"text": "$$ - {{{v^2}} \\over R}\\cos \\theta \\widehat i - {{{v^2}} \\over R}\\sin \\theta \\widehat j$$"
},
{
"text": "$${{{v^2}} \\over R}\\widehat i + {{{v^2}} \\over R}\\widehat j$$"
}
],
"answer": "$$ - {{{v^2}} \\over R}\\cos \\theta \\widehat i - {{{v^2}} \\over R}\\sin \\theta \\widehat j$$",
"solution": "**Answer:** $$ - {{{v^2}} \\over R}\\cos \\theta \\widehat i - {{{v^2}} \\over R}\\sin \\theta \\widehat j$$\n\nFor a particle in uniform circular motion,\n
$${a_c} = {{{v^2}} \\over R}$$ towards the center of the circle\n
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8409,
"subject": "Physics",
"question": "Two cars of masses m1 and m2 are moving in circles of radii r1 and r2, respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal acceleration is",
"options": [
{
"text": "m1r1 : m2r2"
},
{
"text": "m1 : m2"
},
{
"text": "r1 : r2"
},
{
"text": "1 : 1"
}
],
"answer": "r1 : r2",
"solution": "**Answer:** r1 : r2\n\nWe know, $$a = r\\,{w^2} = r \\times {\\left( {{{2\\pi } \\over T}} \\right)^2}$$\n
$$\\therefore$$ $${{{a_1}} \\over {{a_2}}} = {{{r_1}} \\over {{r_2}}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8410,
"subject": "Physics",
"question": "A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of $$3.5$$ revolutions per second. A coin placed at a distnce of 1.25 cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is : (g = 10 m/s2)",
"options": [
{
"text": "0.5"
},
{
"text": "0.3"
},
{
"text": "0.7"
},
{
"text": "0.6"
}
],
"answer": "0.6",
"solution": "**Answer:** 0.6\n\n
We have
\n
$$m{\\omega ^2}r = \\mu mg$$ ...... (1)
\n
Given : rate of rotation = 3.5 rev/s
\n
$$\\Rightarrow$$ 1 revolution = 2$$\\pi$$ rad
\n
\n
That is, 3.5 revolutions = 3.5 $$\\times$$ 2$$\\pi$$ rad
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8411,
"subject": "Physics",
"question": "A particle is moving along a circular path with a constant speed of 10 ms–1. What is the magnitude of the change in velocity of the particle, when it moves through an angle of 60o around the centre of the circle? ",
"options": [
{
"text": "zero "
},
{
"text": "10 m/s"
},
{
"text": "$$10\\sqrt 2 m/s$$"
},
{
"text": "$$10\\sqrt 3 m/s$$"
}
],
"answer": "10 m/s",
"solution": "**Answer:** 10 m/s\n\n\n $$\\left| {\\Delta \\overrightarrow v } \\right| = \\sqrt {v_1^2 + v_2^2 + 2{v_1}{v_2}\\cos \\left( {\\pi - \\theta } \\right)} $$\n
$$ = 2v\\sin {\\theta \\over 2}$$ since $$\\left[ {\\left| {\\overline v {}_1} \\right| = \\left| {{{\\overline v }_2}} \\right|} \\right]$$\n
= 10 m/s",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8412,
"subject": "Physics",
"question": "A body is projected at t = 0 with a velocity 10 ms–1\n at an angle of 60o with the horizontal. The radius of curvature of its trajectory at t = 1s is R. neglecting air resistance and taking acceleration due to gravity g = 10 ms–2, the value of R is : ",
"options": [
{
"text": "2.8 m"
},
{
"text": "5.1 m"
},
{
"text": "2.5 m"
},
{
"text": "10.3 m"
}
],
"answer": "2.8 m",
"solution": "**Answer:** 2.8 m\n\n\n vx = 10cos60o = 5 m/s\nvy = 10cos30o = $$5\\sqrt 3 $$ m/s\n
R = $${{100\\left( {2 - \\sqrt 3 } \\right)} \\over {10\\cos 15}} = 2.8m$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8413,
"subject": "Physics",
"question": "A particle of mass m is fixed to one end of a\nlight spring having force constant k and\nunstretched length $$\\ell $$. The other end is fixed. The\nsystem is given an angular speed $$\\omega $$ about the\nfixed end of the spring such that it rotates in a\ncircle in gravity free space. Then the stretch in\nthe spring is :",
"options": [
{
"text": "$${{m\\ell {\\omega ^2}} \\over {k - m{\\omega ^2}}}$$"
},
{
"text": "$${{m\\ell {\\omega ^2}} \\over {k - m{\\omega}}}$$"
},
{
"text": "$${{m\\ell {\\omega ^2}} \\over {k + m{\\omega ^2}}}$$"
},
{
"text": "$${{m\\ell {\\omega ^2}} \\over {k + m{\\omega}}}$$"
}
],
"answer": "$${{m\\ell {\\omega ^2}} \\over {k - m{\\omega ^2}}}$$",
"solution": "**Answer:** $${{m\\ell {\\omega ^2}} \\over {k - m{\\omega ^2}}}$$\n\n\n At elongated position (x),\n
$$ \\Rightarrow $$ x = $${{m\\ell {\\omega ^2}} \\over {k - m{\\omega ^2}}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8414,
"subject": "Physics",
"question": "A clock has a continuously moving second's hand of 0.1 m length. The average acceleration of the\ntip of the hand (in units of ms–2) is of the order of :",
"options": [
{
"text": "10-3"
},
{
"text": "10-1"
},
{
"text": "10-2"
},
{
"text": "10-4"
}
],
"answer": "10-3",
"solution": "**Answer:** 10-3\n\nR = 0.1 m\n
Average acceleration is of the order of 10–3.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8415,
"subject": "Physics",
"question": "A particle is moving with uniform speed along the circumference of a circle of radius R under the action of a central fictitious force F which is inversely proportional to R3. Its time period of revolution will be given by :",
"options": [
{
"text": "$$T \\propto {R^{{4 \\over 3}}}$$"
},
{
"text": "$$T \\propto {R^{{5 \\over 2}}}$$"
},
{
"text": "$$T \\propto {R^{{3 \\over 2}}}$$"
},
{
"text": "$$T \\propto {R^2}$$"
}
],
"answer": "$$T \\propto {R^2}$$",
"solution": "**Answer:** $$T \\propto {R^2}$$\n\n$$F \\propto {1 \\over {{R^3}}}$$
$$T \\propto {R^2}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8416,
"subject": "Physics",
"question": "A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius 20 cm. If the block takes 40 s to complete one round, the normal force by the side walls of the groove is :",
"options": [
{
"text": "9.859 $$\\times$$ 10$$-$$2 N"
},
{
"text": "0.0314 N"
},
{
"text": "9.859 $$\\times$$ 10$$-$$4 N"
},
{
"text": "6.28 $$\\times$$ 10$$-$$3 N"
}
],
"answer": "9.859 $$\\times$$ 10$$-$$4 N",
"solution": "**Answer:** 9.859 $$\\times$$ 10$$-$$4 N\n\nNormal force will provide the necessary centripetal force.
$$ \\Rightarrow $$ N = m$$\\omega$$2R
Also, $$\\omega$$ = $${{2\\pi } \\over t}$$
N = (0.2)$$\\left( {{{4{\\pi ^2}} \\over {{T^2}}}} \\right)$$(0.2)
$$ \\therefore $$ N = 9.859 $$\\times$$ 10$$-$$4 N",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8417,
"subject": "Physics",
"question": "Statement I : A cyclist is moving on an unbanked road with a speed of 7 kmh$$-$$1 and takes a sharp circular turn along a path of radius of 2m without reducing the speed. The static friction coefficient is 0.2. The cyclist will not slip and pass the curve. (g = 9.8 m/s2)
Statement II : If the road is banked at an angle of 45$$^\\circ$$, cyclist can cross the curve of 2m radius with the speed of 18.5 kmh$$-$$1 without slipping.
In the light of the above statements, choose the correct answer from the options given below.",
"options": [
{
"text": "Statement I is incorrect and statement II is correct"
},
{
"text": "Both statement I and statement II are true"
},
{
"text": "Statement I is correct and statement II is incorrect"
},
{
"text": "Both statement I and statement II are false"
}
],
"answer": "Both statement I and statement II are true",
"solution": "**Answer:** Both statement I and statement II are true\n\nOn a horizontal ground,
Therefore, $$N = 10.2 \\times {10^3}$$ kg-m/s2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8419,
"subject": "Physics",
"question": "A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r such that $$r = {L \\over {\\sqrt 2 }}$$. The speed of particle will be :",
"options": [
{
"text": "$${\\sqrt {rg} }$$"
},
{
"text": "$${\\sqrt {2rg} }$$"
},
{
"text": "$${2\\sqrt {rg} }$$"
},
{
"text": "$${\\sqrt {{{rg} \\over 2}} }$$"
}
],
"answer": "$${\\sqrt {rg} }$$",
"solution": "**Answer:** $${\\sqrt {rg} }$$\n\n
$$\\tan \\theta = {{{v^2}} \\over {rg}} \\Rightarrow v = \\sqrt {rg} $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8420,
"subject": "Physics",
"question": "A huge circular arc of length 4.4 ly subtends an angle '4s' at the centre of the circle. How long it would take for a body to complete 4 revolution if its speed is 8 AU per second?
we get time = 4.5 $$\\times$$ 1010 sec.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8421,
"subject": "Physics",
"question": "
A disc with a flat small bottom beaker placed on it at a distance R from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity $$\\omega$$. The coefficient of static friction between the bottom of the beaker and the surface of the disc is $$\\mu$$. The beaker will revolve with the disc if :
The force that prevents the beaker from sliding is the static friction force. For an object in uniform circular motion, the net force acting on the object (which is the friction force in this case) is equal to the centripetal force.
\n
The static friction force is given by the normal force (which is equal to the weight of the object) multiplied by the coefficient of static friction, which is :\n
$$F_{\\text{friction}} = \\mu mg.$$
\n
The centripetal force needed to keep an object moving in a circle of radius R at angular velocity ω is given by :\n
$$F_{\\text{centripetal}} = mR\\omega^2.$$
\n
For the beaker not to slide off, the static friction force must be at least as large as the centripetal force. Therefore, we have :\n
$$\\mu mg \\geq mR\\omega^2.$$
\n
After canceling the mass m from both sides, we get :\n
For a particle in uniform circular motion, the acceleration $$\\overrightarrow a $$ at any point P(R, $$\\theta$$) on the circular path of radius R is (when $$\\theta$$ is measured from the positive x-axis and v is uniform speed) :
As the particle in uniform circular motion experiences only centripetal acceleration of magnitude $$\\omega$$2R or $${{{v^2}} \\over R}$$ directed towards centre so from diagram,
A curved in a level road has a radius 75 m. The maximum speed of a car turning this curved road can be 30 m/s without skidding. If radius of curved road is changed to 48 m and the coefficient of friction between the tyres and the road remains same, then maximum allowed speed would be ___________ m/s.
A stone of mass m, tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is
",
"options": [
{
"text": "the same throughout the motion."
},
{
"text": "minimum at the highest position of the circular path."
},
{
"text": "minimum at the lowest position of the circular path."
},
{
"text": "minimum when the rope is in the horizontal position. "
}
],
"answer": "minimum at the highest position of the circular path.",
"solution": "**Answer:** minimum at the highest position of the circular path.\n\n
\n
At any $$\\theta :T - mg\\cos \\theta = {{m{v^2}} \\over R}$$
\n
$$ \\Rightarrow T = mg\\cos \\theta + {{m{v^2}} \\over R}$$
\n
Since v is constant,
\n
$$\\Rightarrow$$ T will be minimum when cos$$\\theta$$ is minimum.
\n
$$\\Rightarrow$$ $$\\theta$$ = 180$$^\\circ$$ corresponds to Tminimum.
A fly wheel is accelerated uniformly from rest and rotates through 5 rad in the first second. The angle rotated by the fly wheel in the next second, will be :
A boy ties a stone of mass 100 g to the end of a 2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80 N. If the maximum speed with which the stone can revolve is $${K \\over \\pi }$$ rev./min. The value of K is :
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8427,
"subject": "Physics",
"question": "A stone of mass $1 \\mathrm{~kg}$ is tied to end of a massless string of length $1 \\mathrm{~m}$. If the breaking tension of the string is $400 \\mathrm{~N}$, then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is :",
"options": [
{
"text": "$20 \\mathrm{~ms}^{-1}$"
},
{
"text": "$40 \\mathrm{~ms}^{-1}$"
},
{
"text": "$400 \\mathrm{~ms}^{-1}$"
},
{
"text": "$10 \\mathrm{~ms}^{-1}$"
}
],
"answer": "$20 \\mathrm{~ms}^{-1}$",
"solution": "**Answer:** $20 \\mathrm{~ms}^{-1}$\n\n\n
$$ \\Rightarrow $$ $v=20 \\mathrm{~m} / \\mathrm{s}$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8428,
"subject": "Physics",
"question": "A body is moving with constant speed, in a circle of radius $10 \\mathrm{~m}$. The body completes one revolution in $4 \\mathrm{~s}$. At the end of 3rd second, the displacement of body (in $\\mathrm{m}$ ) from its starting point is :",
"options": [
{
"text": "$15 \\pi$"
},
{
"text": "30"
},
{
"text": "$10 \\sqrt{2}$"
},
{
"text": "$5 \\pi$"
}
],
"answer": "$10 \\sqrt{2}$",
"solution": "**Answer:** $10 \\sqrt{2}$\n\n$$\n\\begin{aligned}\n& \\mathrm{\\omega}=\\frac{2 \\pi}{\\mathrm{T}}=\\frac{2 \\pi}{4}=\\frac{\\pi}{2} \\mathrm{rad} / \\mathrm{s} \\\\\\\\\n& \\theta=\\mathrm{\\omega t} \\\\\\\\\n& \\theta=\\frac{\\pi}{2} \\times 3 \\\\\\\\\n& \\theta=\\frac{3 \\pi}{2} \\mathrm{rad}\n\\end{aligned}\n$$\n
\n
$$\n\\begin{aligned}\n& r=10 \\mathrm{~m} \\\\\\\\\n& T=4 \\mathrm{sec} \\\\\\\\\n& d=\\sqrt{2}(10) \\mathrm{m}\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8429,
"subject": "Physics",
"question": "A stone tied to $180 \\mathrm{~cm}$ long string at its end is making 28 revolutions in horizontal circle in every minute. The magnitude of acceleration of stone is $\\frac{1936}{x} ms^{-2}$. The value of $x$ ________. (Take $\\pi=\\frac{22}{7}$ )",
"options": [],
"answer": "125",
"solution": "**Answer:** 125\n\nAcceleration of stone $a=\\frac{v^{2}}{r}=\\omega^{2} R$\n\n
An object moves at a constant speed along a circular path in a horizontal plane with center at the origin. When the object is at $$x=+2~\\mathrm{m}$$, its velocity is $$\\mathrm{ - 4\\widehat j}$$ m/s. The object's velocity (v) and acceleration (a) at $$x=-2~\\mathrm{m}$$ will be
A car is moving on a horizontal curved road with radius 50 m. The approximate maximum speed of car will be, if friction between tyres and road is 0.34. [take g = 10 ms$$^{-2}$$]
A car is moving with a constant speed of 20 m/s in a circular horizontal track of radius 40 m. A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be : (Take g = 10 m/s$$^2$$)
A body of mass 200g is tied to a spring of spring constant 12.5 N/m, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s. Then the ratio of extension in the spring to its natural length will be :
A vehicle of mass $$200 \\mathrm{~kg}$$ is moving along a levelled curved road of radius $$70 \\mathrm{~m}$$ with angular velocity of $$0.2 ~\\mathrm{rad} / \\mathrm{s}$$. The centripetal force acting on the vehicle is:
",
"options": [
{
"text": "$$560 \\mathrm{~N}$$"
},
{
"text": "$$14 \\mathrm{~N}$$"
},
{
"text": "$$2800 \\mathrm{~N}$$"
},
{
"text": "$$2240 \\mathrm{~N}$$"
}
],
"answer": "$$560 \\mathrm{~N}$$",
"solution": "**Answer:** $$560 \\mathrm{~N}$$\n\nThe centripetal force acting on an object moving along a circular path of radius $$r$$ and angular velocity $$\\omega$$ is given by:\n
\n$$F_c=mr\\omega^2$$\n
\nwhere $$m$$ is the mass of the object.\n
\nIn this problem, the vehicle of mass $$m=200 \\mathrm{~kg}$$ is moving along a circular path of radius $$r=70 \\mathrm{~m}$$ with angular velocity $$\\omega=0.2 ~\\mathrm{rad}/\\mathrm{s}$$.
Substituting the given values into the formula, we get:\n
\nTherefore, the centripetal force acting on the vehicle is $$560~\\mathrm{N}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8435,
"subject": "Physics",
"question": "
A coin placed on a rotating table just slips when it is placed at a distance of $$1 \\mathrm{~cm}$$ from the center. If the angular velocity of the table in halved, it will just slip when placed at a distance of _________ from the centre :
",
"options": [
{
"text": "1 cm"
},
{
"text": "8 cm"
},
{
"text": "4 cm"
},
{
"text": "2 cm"
}
],
"answer": "4 cm",
"solution": "**Answer:** 4 cm\n\nWhen a coin is placed on a rotating table and is just about to slip, the centrifugal force acting on the coin equals the maximum static friction force. Let's denote the mass of the coin as $$m$$, the initial angular velocity as $$\\omega_1$$, and the final angular velocity as $$\\omega_2$$.\n
\nInitially, when the coin is placed at a distance of 1 cm from the center, the centrifugal force acting on the coin is:\n
\n$$F_1 = m r_1 \\omega_1^2$$\n
\nwhere $$r_1 = 1 \\mathrm{~cm}$$.\n
\nWhen the angular velocity is halved ($$\\omega_2 = \\frac{1}{2}\\omega_1$$), the centrifugal force acting on the coin when it just slips is:\n
\n$$F_2 = m r_2 \\omega_2^2 = m r_2 \\left(\\frac{1}{2}\\omega_1\\right)^2$$\n
\nSince the coin is just about to slip in both cases, the maximum static friction force remains the same. Therefore, we can equate the centrifugal forces:\n
\n$$m r_1 \\omega_1^2 = m r_2 \\left(\\frac{1}{2}\\omega_1\\right)^2$$\n
\nCanceling the mass $$m$$ and the initial angular velocity $$\\omega_1^2$$ from both sides, we get:\n
\nSo, the coin will just slip when placed at a distance of 4 cm from the center when the angular velocity is halved.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8436,
"subject": "Physics",
"question": "
A particle is moving with constant speed in a circular path. When the particle turns by an angle $$90^{\\circ}$$, the ratio of instantaneous velocity to its average velocity is $$\\pi: x \\sqrt{2}$$. The value of $$x$$ will be -
A small block of mass $$100 \\mathrm{~g}$$ is tied to a spring of spring constant $$7.5 \\mathrm{~N} / \\mathrm{m}$$ and length $$20 \\mathrm{~cm}$$. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity $$5 ~\\mathrm{rad} / \\mathrm{s}$$ about point $$\\mathrm{A}$$, then tension in the spring is -
In this problem, the spring is stretched due to the circular motion of the block, so the effective radius of the circular motion becomes the natural length of the spring plus the extension in the spring $(r=0.2+x)$.
\n
The spring force, which is also the centripetal force, is given by $F_c=Kx=m\\omega^2 r$, where $K$ is the spring constant, $x$ is the extension of the spring, $m$ is the mass of the block, $\\omega$ is the angular velocity, and $r$ is the radius of the circular path.
\n
Plugging in the values and solving for $x$ and $Kx$ gives the extension in the spring as $x = 0.1$ m and the tension in the spring (which is the spring force) as $Kx=7.5 \\times 0.1 = 0.75$ N.
\nSubstituting the given values:\n
\n$7.5x = 0.1 \\cdot (5^2) \\cdot (0.2 + x),$\n
\nwhich simplifies to:\n
\n$7.5x = 5 \\cdot (x + 0.2)$
\nSolving for $x$ gives $x = 0.1$ m. The tension in the spring is then $kx = 7.5 \\cdot 0.1 = 0.75$ N.\n \n
A child of mass $$5 \\mathrm{~kg}$$ is going round a merry-go-round that makes 1 rotation in $$3.14 \\mathrm{~s}$$. The radius of the merry-go-round is $$2 \\mathrm{~m}$$. The centrifugal force on the child will be
To calculate the centrifugal force acting on the child, we need to find the angular velocity of the merry-go-round and then apply the formula for centrifugal force.
\n
The merry-go-round makes 1 rotation in 3.14 seconds, so its angular velocity ($$\\omega$$) can be calculated as:
\n
$$\\omega = \\frac{2\\pi}{T}$$, where $$T$$ is the time period for one rotation.
Now, the formula for centrifugal force ($$F$$) is:
\n
$$F = m \\cdot r \\cdot \\omega^2$$, where $$m$$ is the mass of the child, $$r$$ is the radius of the merry-go-round, and $$\\omega$$ is the angular velocity.
Therefore, the centrifugal force on the child is $$40\\,\\text{N}$$.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8439,
"subject": "Physics",
"question": "A ball of mass $0.5 \\mathrm{~kg}$ is attached to a string of length $50 \\mathrm{~cm}$. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is $400 \\mathrm{~N}$. The maximum possible value of angular velocity of the ball in $\\mathrm{rad} / \\mathrm{s}$ is, :",
"options": [
{
"text": "1600"
},
{
"text": "20"
},
{
"text": "40"
},
{
"text": "1000"
}
],
"answer": "40",
"solution": "**Answer:** 40\n\n
To find the maximum possible angular velocity (ω) of the ball, we need to consider the maximum tension the string can bear without breaking. This tension provides the centripetal force needed to keep the ball in a circular path.
\n\n
The centripetal force (Fc) required for circular motion is given by the formula:
\n\n
$ F_{c} = m r \\omega^2 $
\n\n
Where:
\n\n
m = mass of the ball (0.5 kg)
\n\n
r = radius of the circle (0.5 m, since 50 cm = 0.5 m)
\n\n
ω = angular velocity in rad/s
\n\n
The maximum tension the string can bear is also the maximum centripetal force (Fc,max) that can be provided by the string, which is 400 N.
\n\n
Now we can set up the equation with the given values:
Therefore, the maximum possible angular velocity of the ball is 40 rad/s, which corresponds to Option C.
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8440,
"subject": "Physics",
"question": "A particle moving in a circle of radius $\\mathrm{R}$ with uniform speed takes time $\\mathrm{T}$ to complete one revolution.
\nIf this particle is projected with the same speed at an angle $\\theta$ to the horizontal, the maximum height attained by it is equal to $4 R$. The angle of projection $\\theta$ is then given by :",
"options": [
{
"text": "$\\sin ^{-1}\\left[\\frac{2 \\mathrm{gT}^2}{\\pi^2 \\mathrm{R}}\\right]^{\\frac{1}{2}}$"
},
{
"text": "$\\sin ^{-1}\\left[\\frac{\\pi^2 \\mathrm{R}}{2 \\mathrm{gT}^2}\\right]^{\\frac{1}{2}}$"
},
{
"text": "$\\cos ^{-1}\\left[\\frac{\\pi \\mathrm{R}}{2 \\mathrm{gT}^2}\\right]^{\\frac{1}{2}}$"
},
{
"text": "$\\cos ^{-1}\\left[\\frac{2 \\mathrm{gT}^2}{\\pi^2 \\mathrm{R}}\\right]^{\\frac{1}{2}}$"
}
],
"answer": "$\\sin ^{-1}\\left[\\frac{2 \\mathrm{gT}^2}{\\pi^2 \\mathrm{R}}\\right]^{\\frac{1}{2}}$",
"solution": "**Answer:** $\\sin ^{-1}\\left[\\frac{2 \\mathrm{gT}^2}{\\pi^2 \\mathrm{R}}\\right]^{\\frac{1}{2}}$\n\n
To solve for the angle of projection $\\theta$, we will first establish the relationship between the variables given and then derive the formula using kinematics.
\n\n
The time $T$ for one revolution at speed $v$ in a circle of radius $R$ is related to the circumference of the circle by the formula:
\n\n
$$ v = \\frac{2\\pi R}{T} $$
\n\n
When the particle is projected with the same speed $v$ at an angle $\\theta$ to the horizontal, its vertical component of velocity is given by $v_y=v\\sin\\theta$.
\n\n
The maximum height $H$ reached by the projectile can be found from the kinematic equation:
\n\n
$$ H = \\frac{v_y^2}{2g} = \\frac{(v\\sin\\theta)^2}{2g} $$
\n\n
We are given that the maximum height attained $H$ is equal to $4R$, so:
\n\n
$$ 4R = \\frac{(v\\sin\\theta)^2}{2g} $$
\n\n
Substitute $v$ from the first equation into the second one:
A train is moving with a speed of $$12 \\mathrm{~m} / \\mathrm{s}$$ on rails which are $$1.5 \\mathrm{~m}$$ apart. To negotiate a curve radius $$400 \\mathrm{~m}$$, the height by which the outer rail should be raised with respect to the inner rail is (Given, $$g=10 \\mathrm{~m} / \\mathrm{s}^2)$$ :
A coin is placed on a disc. The coefficient of friction between the coin and the disc is $$\\mu$$. If the distance of the coin from the center of the disc is $$r$$, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is :
When the coin is on the disc and the disc starts rotating, centrifugal force acts on the coin, trying to push it away from the center. The friction between the coin and the disc opposes this motion. For the coin to not slip, the frictional force must be equal to the centrifugal force acting on the coin.
The normal force (N) acting on the coin is equal to the weight of the coin, which can be represented as:
$$N = mg$$
where $$m$$ is the mass of the coin, and $$g$$ is the acceleration due to gravity.
The centrifugal force ($$f$$) that acts on the coin due to the rotation of the disc is given by:
$$f = m\\omega^2r$$
where $$\\omega$$ is the angular velocity, and $$r$$ is the distance of the coin from the center of the disc.
Friction force (f) is also given by the formula:
$$f = \\mu N$$
where $$\\mu$$ is the coefficient of static friction between the coin and the disc.
For the coin to not slip, the centrifugal force must be equal to the frictional force, hence:
$$\\mu mg = m\\omega^2r$$
Dividing both sides by $$mr$$, we get:
$$\\omega = \\sqrt{\\frac{\\mu g}{r}}$$
This equation shows that the maximum angular velocity ($$\\omega$$) that can be given to the disc to prevent the coin from slipping off depends on the coefficient of friction ($$\\mu$$), the acceleration due to gravity ($$g$$), and the distance of the coin from the center ($$r$$).
If the radius of curvature of the path of two particles of same mass are in the ratio $$3: 4$$, then in order to have constant centripetal force, their velocities will be in the ratio of :
A stone of mass $$900 \\mathrm{~g}$$ is tied to a string and moved in a vertical circle of radius $$1 \\mathrm{~m}$$ making $$10 \\mathrm{~rpm}$$. The tension in the string, when the stone is at the lowest point is (if $$\\pi^2=9.8$$ and $$g=9.8 \\mathrm{~m} / \\mathrm{s}^2$$) :
A clock has $$75 \\mathrm{~cm}, 60 \\mathrm{~cm}$$ long second hand and minute hand respectively. In 30 minutes duration the tip of second hand will travel $$x$$ distance more than the tip of minute hand. The value of $$x$$ in meter is nearly (Take $$\\pi=3.14$$) :
To determine the distance traveled by the tips of the second hand and the minute hand, we need to calculate the circumference of the circles they make. Let's start by calculating the distances traveled by both hands over a period of 30 minutes.
\n\n
First, the circumference formula is given by:
\n\n
$$C = 2\\pi r$$
\n\n
For the second hand:
\n\n
The length of the second hand is $$75 \\space \\text{cm}$$. The second hand completes one full revolution every 60 seconds, so in 1 minute, the second hand travels:
$$ \\text{Total distance traveled by second hand} = 30 \\times 2\\pi \\times 75 \\text{ cm} $$
\n\n
Now, let's calculate it with $$\\pi = 3.14$$:
\n\n
$$ \\text{Total distance traveled by second hand} = 30 \\times 2 \\times 3.14 \\times 75 \\text{ cm} $$
\n\n
$$ \\text{Total distance traveled by second hand} = 30 \\times 471 \\text{ cm} = 14130 \\text{ cm} $$
\n\n
For the minute hand:
\n\n
The length of the minute hand is $$60 \\text{ cm}$$. The minute hand completes one full revolution every 60 minutes, so in 30 minutes, the minute hand travels:
A man carrying a monkey on his shoulder does cycling smoothly on a circular track of radius $$9 \\mathrm{~m}$$ and completes 120 resolutions in 3 minutes. The magnitude of centripetal acceleration of monkey is (in $$\\mathrm{m} / \\mathrm{s}^2$$ ) :
First, let's calculate the centripetal acceleration experienced by the monkey while the man does cycling smoothly on a circular track. The formula for centripetal acceleration ($a_c$) is given by:
\n\n
$a_c = \\frac{v^2}{r}$
\n\n
where\n\n
\n
$v$ is the velocity of the monkey and the man cycling,
\n
$r$ is the radius of the circular track.
\n
\n
To find $v$, we first need to find the circumference of the circle, which is given by:
\n\n
$C = 2\\pi r$
\n\n
Given the radius $r = 9 \\, \\mathrm{m}$, the circumference $C$ is:
\n\n
$C = 2\\pi \\times 9 = 18\\pi \\, \\mathrm{m}$
\n\n
Then, we determine the total distance travelled by calculating how many times they complete the circle in the given time. With 120 revolutions in 3 minutes (180 seconds), the total distance $D$ travelled is:
A car of $$800 \\mathrm{~kg}$$ is taking turn on a banked road of radius $$300 \\mathrm{~m}$$ and angle of banking $$30^{\\circ}$$. If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely: $$(\\mathrm{g}=10 \\mathrm{~m} / \\mathrm{s}^2, \\sqrt{3}=1.73)$$
When a car takes a turn on a banked road, the forces involved are the gravitational force, the normal force from the surface, and frictional force (if any). The net force provides the necessary centripetal force for the circular motion. The angle of banking and static friction contribute to the maximum speed the car can achieve without slipping.
\n\n
The forces acting on the car are as follows:
\n\n
1. The normal force ($$N$$) acts perpendicular to the surface of the road.
\n\n
2. Gravitational force ($$mg$$) acts downward.
\n\n
3. Frictional force ($$f$$), which can provide additional centripetal force if needed. It acts parallel to the surface of the road, towards the center of the circle.
\n\n
The normal force and the gravitational force components can be resolved into two directions: perpendicular and parallel to the road surface. The maximum speed is achieved when all available forces (normal, frictional) are utilized to provide the necessary centripetal force ($$F_c$$) without slipping.
\n\n
The centripetal force required for circular motion is given by:
\n\n
$F_c = \\frac{mv^2}{r}$
\n\n
Where:
\n\n
\n\n
$$m$$ = mass of the car = $$800 kg$$
\n\n
$$v$$ = speed of the car
\n\n
$$r$$ = radius of the turn = $$300 m$$
\n\n
\n\n
On a banked curve, the maximum velocity can be calculated using the formula:
Hence, the closest option to the calculated maximum speed without slipping is:
\n\n
Option A: 51.4 m/s
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8448,
"subject": "Physics",
"question": "Consider telecommunication through optical fibres. Which of the following statements is not true? ",
"options": [
{
"text": "Optical fibres can be of graded refractive index"
},
{
"text": "Optical fibres are subject to electromagnetic interference from outside "
},
{
"text": "Optical fibres have extremely low transmission loss "
},
{
"text": "Optical fibres may have homogeneous core with a suitable cladding. "
}
],
"answer": "Optical fibres are subject to electromagnetic interference from outside ",
"solution": "**Answer:** Optical fibres are subject to electromagnetic interference from outside \n\nOptical fibres form a dielectric wave guide and are free from electromagnetic interference or radio frequency interference.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8449,
"subject": "Physics",
"question": "This question has Statement - $$1$$ and Statement - $$2$$. Of the four choices given after the statements, choose the one that best describes the two statements.\n Statement -$$1$$ : Sky wave signals are used for long distance radio communication. These signals are in general, less stable then ground wave signals.\n Statement -$$2$$ : The state of ionosphere varies from hour to hour, day to day and season to season. ",
"options": [
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is true, Statement - $$2$$ is the correct explanation of Statement - $$1$$."
},
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is true, Statement - $$2$$ is not the correct explanation of Statement - $$1$$."
},
{
"text": "Statement - $$1$$ is false, Statement - $$2$$ is true "
},
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is false."
}
],
"answer": "Statement - $$1$$ is true, Statement - $$2$$ is true, Statement - $$2$$ is not the correct explanation of Statement - $$1$$.",
"solution": "**Answer:** Statement - $$1$$ is true, Statement - $$2$$ is true, Statement - $$2$$ is not the correct explanation of Statement - $$1$$.\n\nFor long distance communication, sky wave signals are used.\n
Also, the state of ionosphere varies every time. So, both statements are correct.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8450,
"subject": "Physics",
"question": "A radar has a power of $$1kW$$ and is operating at a frequency of $$10$$ $$GHz.$$ It is located on a mountain top of height $$500$$ $$m.$$ The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth $$ = 6.4 \\times {10^6}m$$) is : ",
"options": [
{
"text": "$$80$$ $$km$$ "
},
{
"text": "$$16$$ $$km$$ "
},
{
"text": "$$40$$ $$km$$ "
},
{
"text": "$$64$$ $$km$$ "
}
],
"answer": "$$80$$ $$km$$ ",
"solution": "**Answer:** $$80$$ $$km$$ \n\n\n
Let $$d$$ is the maximum distance, upto it the objects \n
$$d = \\sqrt {{{500}^2} + 2 \\times 6.4 \\times {{10}^6}} = 80km$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8451,
"subject": "Physics",
"question": "A signal is to be transmitted through a wave of wavelength $$\\lambda $$, using a linear antenna. The length l of the antenna and effective power radiated Peff will be given respectively as :\n
We know that for a linear antenna of length l, (i) it's length of antenna is comparable to the wavelength of the signal and (ii) the power radiated is proportional to 1/$$\\lambda$$2. That is, the power radiated for the antenna increases with decreasing $$\\lambda$$ thereby increasing the frequency.
\n
Power of the antenna $$P = \\mu {\\left( {{1 \\over l}} \\right)^2}$$
\n
Here, $$\\mu$$ = K; hence, the correct option is (D).
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8452,
"subject": "Physics",
"question": "A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized\nfor transmission. How many telephonic channels can be transmitted simultaneously if each channel\nrequires a bandwidth of 5 kHz ?",
"options": [
{
"text": "2 $$\\times$$ 106"
},
{
"text": "2 $$\\times$$ 103"
},
{
"text": "2 $$\\times$$ 105"
},
{
"text": "2 $$\\times$$ 104"
}
],
"answer": "2 $$\\times$$ 105",
"solution": "**Answer:** 2 $$\\times$$ 105\n\n
The carrier frequency is 10 GHz, and only 10% of it is utilized for transmission. Therefore, the bandwidth used for transmission is:
We should convert this to kHz because the channel bandwidth is given in kHz:
\n
$1 \\text{ GHz} = 1,000,000 \\text{ kHz}$
\n
Now, if each telephonic channel requires a bandwidth of 5 kHz, the number of channels that can be transmitted simultaneously is:
\n
$\\text{Number of channels} = \\frac{\\text{Total bandwidth}}{\\text{Bandwidth per channel}}$
\n
$\\text{Number of channels} = \\frac{1,000,000 \\text{ kHz}}{5 \\text{ kHz}}$
\n
$\\text{Number of channels} = 200,000$
\n
So, 200,000 telephonic channels can be transmitted simultaneously.
\n
Among the given options, the correct answer is Option C: $2 \\times 10^5$.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8453,
"subject": "Physics",
"question": "In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwith. The number of channels accomodated for transmitting TV signals of band width 6 MHz are (Take velocity of light c = 3 $$ \\times $$ 108m/s, h = 6.6 $$ \\times $$ 10$$-$$34 J-s)",
"options": [
{
"text": "3.75 $$ \\times $$ 106"
},
{
"text": "3.86 $$ \\times $$ 106"
},
{
"text": "6.25 $$ \\times $$ 105"
},
{
"text": "4.87 $$ \\times $$ 105"
}
],
"answer": "6.25 $$ \\times $$ 105",
"solution": "**Answer:** 6.25 $$ \\times $$ 105\n\nGiven, \n
= 6.25 $$ \\times $$ 105",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8454,
"subject": "Physics",
"question": "A TV transmission tower has a height of 140 m and the height of the receiving antenna is 40 m. What is the maximum distance upto which signals can be broadcasted from this tower is LOS (Line of Sight) mode ? (Given : radius of earth = 6.4 × 106 m). ",
"options": [
{
"text": "40 km"
},
{
"text": "65 km"
},
{
"text": "48 km"
},
{
"text": "80 km"
}
],
"answer": "65 km",
"solution": "**Answer:** 65 km\n\nMaximum distance upto which signal can be broadcasted is \n
dmax = $$\\sqrt {2R{h_T}} + \\sqrt {2R{h_R}} $$\n
where hT and hR are heights of transmitter tower and height of reserver respectively. Putting all values -\n
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8456,
"subject": "Physics",
"question": "The wavelength of the carrier waves in a\nmodern optical fiber communication network\nis close to :",
"options": [
{
"text": "1500 nm"
},
{
"text": "2400 nm"
},
{
"text": "600 nm"
},
{
"text": "900 nm"
}
],
"answer": "1500 nm",
"solution": "**Answer:** 1500 nm\n\nWavelength of carrier waves in modern optical\nfiber communication is most widely used near\nabout 1500 nm.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8457,
"subject": "Physics",
"question": "In a line of sight radio communication, a\ndistance of about 50 km is kept between the\ntransmitting and receiving antennas. If the\nheight of the receiving antenna is 70m, then the\nminimum height of the transmitting antenna\nshould be : \n(Radius of the Earth = 6.4 × 106 m).",
"options": [
{
"text": "40 m"
},
{
"text": "20 m"
},
{
"text": "51 m"
},
{
"text": "32 m"
}
],
"answer": "32 m",
"solution": "**Answer:** 32 m\n\nRange = $$\\sqrt {2R{h_T}} + \\sqrt {2R{h_R}} $$
\nBy solving hT = 32 m.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8458,
"subject": "Physics",
"question": "The physical sizes of the transmitter and\nreceiver antenna in a communication system\nare :-",
"options": [
{
"text": "proportional to carrier frequency"
},
{
"text": "inversely proportional to carrier frequency"
},
{
"text": "inversely proportional to modulation\nfrequency"
},
{
"text": "independent of both carrier and modulation\nfrequency"
}
],
"answer": "inversely proportional to carrier frequency",
"solution": "**Answer:** inversely proportional to carrier frequency\n\nThe physical size of antenna of receiver and transmitter both are inversely proportional\nto carrier frequency. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8459,
"subject": "Physics",
"question": "Given below in the the left column are different\nmodes of communication using the kinds of\nwaves given the right column.\n\n
\nMobile Phones – Microwaves",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8460,
"subject": "Physics",
"question": "A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is $$-$$5 dB per km and cable length is 20 km. The power received at receiver is 10$$-$$x W. The value of x is ________.
[Gain in $$dB = 10{\\log _{10}}\\left( {{{{P_o}} \\over {{P_i}}}} \\right)$$]",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nGiven, power of transmitted signal, Pi = 0.1 kW = 0.1 $$\\times$$ 103W = 102 W
Rate of attenuation, R = $$-$$5 dB/km
Length of cable, l = 20 km
Power received at receiver, Px = 10$$-$$x W
Total loss, $$\\beta$$ = R $$\\times$$ l = $$-$$5 $$\\times$$ 20 = $$-$$ 100 dB
$$\\because$$ Gain ($$\\beta$$) = $$10{\\log _{10}}{{{P_0}} \\over {{P_i}}}$$
Hence, x = 8",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8461,
"subject": "Physics",
"question": "A 25 m long antenna is mounted on an antenna tower. The height of the antenna tower is 75 m. The wavelength (in meter) of the signal transmitted by this antenna would be :",
"options": [
{
"text": "200"
},
{
"text": "400"
},
{
"text": "100"
},
{
"text": "300"
}
],
"answer": "100",
"solution": "**Answer:** 100\n\nGiven that, height of peak of antenna : H = 25 m.
As, we know that
$$\\lambda$$ = 4H
$$ \\therefore $$ $$\\lambda$$ = 4 $$\\times$$ 25
$$ \\Rightarrow $$ $$\\lambda$$ = 100 m",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8462,
"subject": "Physics",
"question": "Two identical antennas mounted on identical towers are separated from each other by a distance of 45 km. What should nearly be the minimum height of receiving antenna to receive the signals in line of sight? (Assume radius of earth is 6400 km)",
"options": [
{
"text": "158.2 m"
},
{
"text": "79.1 m"
},
{
"text": "19.77 m"
},
{
"text": "39.55 m"
}
],
"answer": "39.55 m",
"solution": "**Answer:** 39.55 m\n\n
$$ = {{45 \\times 45} \\over {8 \\times 6400}}$$ Km
$$ = 0.03955$$ Km
$$ = 39.55$$ m",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8463,
"subject": "Physics",
"question": "For VHF signal broadcasting, ___________ km2 of maximum service area will be covered by an antenna tower of height 30 m, if the receiving antenna is placed at ground. Let radius of the earth be 6400 km. (Round off to the Nearest Integer) (Take $$\\pi$$ as 3.14)",
"options": [],
"answer": "1206",
"solution": "**Answer:** 1206\n\n$$d = \\sqrt {2hR} $$
$$ \\approx $$ 1206 km2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8464,
"subject": "Physics",
"question": "Match List - I with List - II.
Stratosphere :\n The stratosphere starts just above the troposphere and extends to 50 kilometers (31 miles)\nhigh.\n
Mesosphere :\n The mesosphere starts just above the stratosphere and extends to 85 kilometers (53 miles)\nhigh.\n
Thermosphere :\n The Thermosphere starts just above the mesosphere and extends to 600 kilometers (372 miles)\nhigh.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8465,
"subject": "Physics",
"question": "A TV transmission tower antenna is at a height of 20 m. Suppose that the receiving antenna is at.
(i) ground level
(ii) a height of 5 m.
The increase in antenna range in case (ii) relative to case (i) is n%.
The value of n, to the nearest integer, is ___________.",
"options": [],
"answer": "50",
"solution": "**Answer:** 50\n\nFor calculation of Range from tower.
$$ = {8 \\over {16}} \\times 100$$ = 50%",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8466,
"subject": "Physics",
"question": "What should be the height of transmitting antenna and the population covered if the television telecast is to cover a radius of 150 km? The average population density around the tower is 2000/km2 and the value of Re = 6.5 $$\\times$$ 106 m.",
"options": [
{
"text": "Height = 1241 m
Population covered = 7 $$\\times$$ 105"
},
{
"text": "Height = 1731 m
Population covered = 1413 $$\\times$$ 105"
},
{
"text": "Height = 1800 m
Population covered = 1413 $$\\times$$ 108"
},
{
"text": "Height = 1600 m
Population covered = 2 $$\\times$$ 105"
}
],
"answer": "Height = 1731 m
Population covered = 1413 $$\\times$$ 105",
"solution": "**Answer:** Height = 1731 m
Population covered = 1413 $$\\times$$ 105\n\n$$d = \\sqrt {2{R_e}h} $$
Area covered = $$\\pi$$d2
$$h = {{{d^2}} \\over {2{R_e}}}$$ = 1731 m
Population covered = 2000 $$\\times$$ $$\\pi$$(d2)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8467,
"subject": "Physics",
"question": "A transmitting antenna at top of a tower has a height of 50 m and the height of receiving antenna is 80 m. What is range of communication for Line of Sight (LoS) mode?
$$ = 57.28$$ km",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8468,
"subject": "Physics",
"question": "A transmitting antenna has a height of 320 m and that of receiving antenna is 2000 m. The maximum distance between them for satisfactory communication in line of sight mode is 'd'. The value of 'd' is ................. km.",
"options": [],
"answer": "224",
"solution": "**Answer:** 224\n\n$${d_m} = \\sqrt {2R{h_T}} + \\sqrt {2R{h_R}} $$
dm = 224 km",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 8469,
"subject": "Physics",
"question": "An antenna is mounted on a 400 m tall building. What will be the wavelength of signal that can be radiated effectively by the transmission tower upto a range of 44 km?",
"options": [
{
"text": "37.8 m"
},
{
"text": "605 m"
},
{
"text": "75.6 m"
},
{
"text": "302 m"
}
],
"answer": "605 m",
"solution": "**Answer:** 605 m\n\nh : height of antenna
$$\\lambda$$ : wavelength of signal
h < $$\\lambda$$
$$\\lambda$$ > h
$$\\lambda$$ > 400 m",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8470,
"subject": "Physics",
"question": "If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at 160 m, then the maximum range of LOS communication is ___________ km. (Take radius of Earth = 6400 km)",
"options": [],
"answer": "64",
"solution": "**Answer:** 64\n\nhT = hR = 160 ........ (i)
The TV transmission tower at a particular station has a height of 125 m. For doubling the coverage of its range, the height of the tower should be increased by
We do not transmit low frequency signal to long distances because-
\n
(a) The size of the antenna should be comparable to signal wavelength which is unreal solution for a signal of longer wavelength.
\n
(b) Effective power radiated by a long wavelength baseband signal would be high.
\n
(c) We want to avoid mixing up signals transmitted by different transmitter simultaneously.
\n
(d) Low frequency signal can be sent to long distances by superimposing with a high frequency wave as well.
\n
Therefore, the most suitable option will be :
",
"options": [
{
"text": "All statements are true"
},
{
"text": "(a), (b) and (c) are true only"
},
{
"text": "(a), (c) and (d) are true only"
},
{
"text": "(b), (c) and (d) are true only"
}
],
"answer": "(a), (c) and (d) are true only",
"solution": "**Answer:** (a), (c) and (d) are true only\n\n
For longer wavelength, size of antenna would increase. Also, mixing of signals needs to be avoided.
\n
Also, we can use modulation to send low frequency signal by superimposing them with high frequency signals.
The height of a transmitting antenna at the top of a tower is 25 m and that of receiving antenna is, 49 m. The maximum distance between them, for satisfactory communication in LOS (Line-Of-Sight) is K$$\\sqrt5$$ $$\\times$$ 102 m. The value of K is ___________.
\n
(Assume radius of Earth is 64 $$\\times$$ 10+5 m) [Calculate upto nearest integer value]
An antenna is placed in a dielectric medium of dielectric constant 6.25. If the maximum size of that antenna is 5.0 mm, it can radiate a signal of minimum frequency of __________ GHz.
The required height of a TV tower which can cover the population of $$6.03$$ lakh is $$h$$. If the average population density is 100 per square $$\\mathrm{km}$$ and the radius of earth is $$6400 \\mathrm{~km}$$, then the value of $$h$$ will be _____________ $$m$$.
\nSalellite communication $\\rightarrow 3.7-4.2 \\mathrm{GHz}$
\n$\\therefore$ A-II, B-I, C-IV, D-III",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8483,
"subject": "Physics",
"question": "The height of transmitting antenna is $180 \\mathrm{~m}$ and the height of the receiving antenna is $245 \\mathrm{~m}$. The maximum distance between them for satisfactory communication in line of sight will be :
\nwhere $d_{\\max}$ is the maximum distance between the antennas, $R$ is the radius of the earth, $h_t$ is the height of the transmitting antenna above the earth's surface, and $h_r$ is the height of the receiving antenna above the earth's surface.\n
\nTherefore, the maximum distance between the transmitting and receiving antennas for line of sight communication is $\\boxed{104\\,\\mathrm{km}}$.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 8484,
"subject": "Physics",
"question": "
To radiate EM signal of wavelength $$\\lambda$$ with high efficiency, the antennas should have a minimum size equal to:
",
"options": [
{
"text": "$$\\frac{\\lambda}{2}$$"
},
{
"text": "$$\\lambda$$"
},
{
"text": "$$\\frac{\\lambda}{4}$$"
},
{
"text": "$$2 \\lambda$$"
}
],
"answer": "$$\\frac{\\lambda}{4}$$",
"solution": "**Answer:** $$\\frac{\\lambda}{4}$$\n\nThe minimum length of an antenna to efficiently radiate an electromagnetic wave is one quarter of the wavelength of the signal, which is commonly known as a quarter-wave antenna. \n
\nThis is because when the length of the antenna is equal to a quarter of the wavelength of the signal, the voltage and current at the feed point of the antenna are in phase, resulting in efficient radiation of the signal. An antenna that is shorter than a quarter wavelength will not efficiently couple with the signal, while an antenna that is longer than a quarter wavelength may have complex radiation patterns.\n
\nTherefore, the minimum length of an antenna to efficiently radiate an electromagnetic wave is $$\\frac{\\lambda}{4}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8485,
"subject": "Physics",
"question": "
Match List - I with List - II
\n
\n
\n
\n
\n
\n
\n
\n
\n\n
\n
\n
List - I (Layer of atmosphere)
\n
\n
List - II (Approximate height over earth's surface)
\n
\n\n\n
\n
(A)
\n
$$\\mathrm{F_1}$$ - Layer
\n
(I)
\n
10 km
\n
\n
\n
(B)
\n
$$\\mathrm{D}$$ - Layer
\n
(II)
\n
170 - 190 km
\n
\n
\n
(C)
\n
Troposphere
\n
(III)
\n
100 km
\n
\n
\n
(D)
\n
$$\\mathrm{E}$$ - Layer
\n
(IV)
\n
65 - 75 km
\n
\n\n
\n
Choose the correct answer from the options given below:
",
"options": [
{
"text": "A - II, B - IV, C - III, D - I"
},
{
"text": "A - II, B-I, C - IV, D - III"
},
{
"text": "A - II, B - IV, C - I, D - III"
},
{
"text": "A - III, B - IV, C - I, D - II"
}
],
"answer": "A - II, B - IV, C - I, D - III",
"solution": "**Answer:** A - II, B - IV, C - I, D - III\n\nLet's match the layers of the atmosphere (List - I) with their approximate heights over the Earth's surface (List - II):\n
\n(A) $$\\mathrm{F_1}$$ - Layer: This layer is part of the ionosphere, which is located within the thermosphere. The $$\\mathrm{F_1}$$ layer is at an approximate height of 170 - 190 km. So, A matches with II.\n
\n(B) $$\\mathrm{D}$$ - Layer: This layer is also part of the ionosphere and is the lowest layer of the ionosphere. The $$\\mathrm{D}$$ layer is at an approximate height of 65 - 75 km. So, B matches with IV.\n
\n(C) Troposphere: This is the lowest layer of the Earth's atmosphere, where weather occurs, and it extends up to approximately 10 km over the Earth's surface. So, C matches with I.\n
\n(D) $$\\mathrm{E}$$ - Layer: This layer is another part of the ionosphere, which is above the $$\\mathrm{D}$$ layer and below the $$\\mathrm{F_1}$$ layer. The $$\\mathrm{E}$$ layer is at an approximate height of 100 km. So, D matches with III.\n
\nThus, the correct matching is:\n
\nA - II, B - IV, C - I, D - III",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8486,
"subject": "Physics",
"question": "
In satellite communication, the uplink frequency band used is :
",
"options": [
{
"text": "$$3.7-4.2 \\mathrm{~GHz}$$"
},
{
"text": "$$5.925-6.425 \\mathrm{~GHz}$$"
},
{
"text": "$$76-88 \\mathrm{~MHz}$$"
},
{
"text": "$$420-890 \\mathrm{~MHz}$$"
}
],
"answer": "$$5.925-6.425 \\mathrm{~GHz}$$",
"solution": "**Answer:** $$5.925-6.425 \\mathrm{~GHz}$$\n\nIn satellite communication, the uplink frequency band refers to the range of frequencies used to transmit signals from the Earth to the satellite. The specific frequency bands used for uplink communication can vary depending on the type of satellite and the service it provides. However, one of the most common frequency bands used for satellite uplinks in geostationary satellite communication is the C-band.\n
\n$$5.925-6.425 \\mathrm{~GHz}$$, is the typical frequency range used for uplink communication in the C-band.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8487,
"subject": "Physics",
"question": "
A transmitting antenna is kept on the surface of the earth. The minimum height of receiving antenna required to receive the signal in line of sight at $$4 \\mathrm{~km}$$ distance from it is $$x \\times 10^{-2} \\mathrm{~m}$$. The value of $$x$$ is __________.
\n
(Let, radius of earth $$\\mathrm{R}=6400 \\mathrm{~km}$$ )
To determine the minimum height of the receiving antenna required to receive the signal in line of sight at a distance of 4 km from the transmitting antenna, we need to consider the curvature of the Earth. Let's denote the height of the receiving antenna as $$h$$ and the distance between the antennas as $$d = 4 \\mathrm{~km}$$.
\n
We can use the Pythagorean theorem to relate the Earth's radius $$R$$, the height of the receiving antenna $$h$$, and the distance between the antennas $$d$$:
\n
$$(R + h)^2 = R^2 + d^2$$
\n
Now, we can plug in the given values for $$R$$ and $$d$$:
\n
$$(6400 + h)^2 = 6400^2 + 4^2$$
\n
Expanding the equation and subtracting $$6400^2$$ from both sides gives:
\n
$$2 \\times 6400 \\times h + h^2 = 16$$
\n
Since $$h$$ is much smaller than the Earth's radius, we can ignore the $$h^2$$ term:
The power radiated from a linear antenna of length $$l$$ is proportional to
\n
(Given, $$\\lambda=$$ Wavelength of wave):
",
"options": [
{
"text": "$$\\left(\\frac{l}{\\lambda}\\right)^{2}$$"
},
{
"text": "$$\\frac{l}{\\lambda^{2}}$$"
},
{
"text": "$$\\frac{l^{2}}{\\lambda}$$"
},
{
"text": "$$\\frac{l}{\\lambda}$$"
}
],
"answer": "$$\\left(\\frac{l}{\\lambda}\\right)^{2}$$",
"solution": "**Answer:** $$\\left(\\frac{l}{\\lambda}\\right)^{2}$$\n\nThe power radiated by a linear antenna is typically determined by its physical length relative to the wavelength of the signal it is transmitting or receiving. The power radiated from a linear antenna of length 'l' is proportional to the square of the ratio of its length to the wavelength, represented as $(\\frac{l}{\\lambda})^{2}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8489,
"subject": "Physics",
"question": "
A TV transmitting antenna is $$98 \\mathrm{~m}$$ high and the receiving antenna is at the ground level. If the radius of the earth is $$6400 \\mathrm{~km}$$, the surface area covered by the transmitting antenna is approximately:
The range (or distance to the horizon) for line-of-sight communication is given by the formula:
\n
$d = \\sqrt{2hR}$,
\n
where:
\n
\n
$d$ is the range,
\n
$h$ is the height of the antenna, and
\n
$R$ is the radius of the Earth.
\n
\n
Given that $h = 98 \\, \\text{m} = 98 \\times 10^{-3} \\, \\text{km}$ and $R = 6400 \\, \\text{km}$, we can substitute these values into the formula to find $d$:
The surface area $A$ covered by the transmitting antenna is approximately a circle with radius equal to the range $d$. The area of a circle is given by the formula $A = \\pi r^2$, where $r$ is the radius. Therefore, we can substitute $d$ into this formula to find $A$:
The range of transmission (R) of a TV tower (or any radio tower) depends on the height of the tower (h). This relationship is given by the formula:
\n
$$\nR = \\sqrt{2hR_E}\n$$
\n
where $R_E$ is the radius of the Earth. This formula comes from the geometry of the situation, considering the curvature of the Earth.
\n
Now, if the height of the tower is increased, this will increase the range of transmission. Specifically, because the height appears under the square root in the formula, the range of transmission is proportional to the square root of the height. So, if the height is multiplied by some factor, the range will be multiplied by the square root of that factor.
\n
In this case, the height of the tower is increased by 21%, which means the new height is $h_2 = 1.21h_1$. If we substitute this into the formula for the range, we get:
This shows that the new range is 1.1 times the original range, or in other words, the range has increased by 10%.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8491,
"subject": "Physics",
"question": "A signal of $$5$$ $$kHz$$ frequency is amplitude modulated on a carrier wave of frequency $$2$$ $$MHz.$$ The frequencies of the resultant signal is/are : ",
"options": [
{
"text": "$$2005$$ $$kHz, 2000$$ $$kHz$$ and $$1995$$ $$kHz$$"
},
{
"text": "$$2000$$ $$kHz$$ and $$1995$$ $$kHz$$ "
},
{
"text": "$$2$$ $$MHz$$ only "
},
{
"text": "$$2005$$ $$kHz$$ and $$1995$$ $$kHz$$"
}
],
"answer": "$$2005$$ $$kHz, 2000$$ $$kHz$$ and $$1995$$ $$kHz$$",
"solution": "**Answer:** $$2005$$ $$kHz, 2000$$ $$kHz$$ and $$1995$$ $$kHz$$\n\nAmplitude modulated wave consists of three frequencies are \n
i.e. $$2005$$ $$KHz,$$ $$2000KHz,$$ $$1995$$ $$KHz$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8492,
"subject": "Physics",
"question": "Choose the correct statement : ",
"options": [
{
"text": "In amplitude modulation the amplitude of the high frequency carrier wave is make to vary in proportion to the amplitude of the audio signal. "
},
{
"text": "In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal."
},
{
"text": "In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal. "
},
{
"text": "In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal."
}
],
"answer": "In amplitude modulation the amplitude of the high frequency carrier wave is make to vary in proportion to the amplitude of the audio signal. ",
"solution": "**Answer:** In amplitude modulation the amplitude of the high frequency carrier wave is make to vary in proportion to the amplitude of the audio signal. \n\nIn amplitude modulation, the amplitude of the high frequency carrier wave made to vary in proportional to the amplitude of audio signal.\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8493,
"subject": "Physics",
"question": "A modulated signal Cm(t) has the form Cm(t) = 30 sin 300 $$\\pi $$t + 10 (cos 200 $$\\pi $$t −cos 400 $$\\pi $$t). The carrier frequency c, the modulating frequency (message frequency) f$$\\omega $$, and the modulation index $$\\mu $$ are respectively given by :",
"options": [
{
"text": "fc = 200 Hz; f$$\\omega $$ = 50 Hz; $$\\mu $$ = $${1 \\over 2}$$"
},
{
"text": "fc = 150 Hz; f$$\\omega $$ = 50 Hz; $$\\mu $$ = $${2 \\over 3}$$ "
},
{
"text": "fc = 150 Hz; f$$\\omega $$ = 30 Hz; $$\\mu $$ = $${1 \\over 3}$$"
},
{
"text": "fc = 200 Hz; f$$\\omega $$ = 30 Hz; $$\\mu $$ = $${1 \\over 2}$$"
}
],
"answer": "fc = 150 Hz; f$$\\omega $$ = 50 Hz; $$\\mu $$ = $${2 \\over 3}$$ ",
"solution": "**Answer:** fc = 150 Hz; f$$\\omega $$ = 50 Hz; $$\\mu $$ = $${2 \\over 3}$$ \n\nGiven, \n
Cm(t) = 30 sin 300$$\\pi $$t + 10 (cos 200 $$\\pi $$t $$-$$ cos 400 $$\\pi $$t)\n
Standard equation of amplitude modulated wave, \n
Cm(t) = Ac sin ($$\\omega $$ct) $$-$$ $${{\\mu {A_c}} \\over 2}$$ cos $$\\left( {{\\omega _c} + {\\omega _m}} \\right)$$ t + $${{\\mu {A_c}} \\over 2}$$ cos $$\\left( {{\\omega _c} - {\\omega _m}} \\right)$$ t\n
$$ \\Rightarrow $$ $$\\mu $$ = $${{10} \\over {15}}$$ = $${2 \\over 3}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8494,
"subject": "Physics",
"question": "An audio signal consists of two distinct sounds : one a human speech signal in the frequency band of 200 Hz to 2700 Hz, while the other is a high frequency music signal in the frequency band of 10200 Hz to 15200 Hz. The ratio of the AM signal bandwidth required to send both the signals together to the AM signal bandwidth required to send just the human speech is :",
"options": [
{
"text": "3"
},
{
"text": "5"
},
{
"text": "6"
},
{
"text": "2"
}
],
"answer": "6",
"solution": "**Answer:** 6\n\n
To calculate the ratio of the AM signal bandwidth required to send both the signals together to the AM signal bandwidth required to send just the human speech, we need to consider the frequency range covered by both signals.
\n
For the human speech signal, the frequency band is from 200 Hz to 2700 Hz. Therefore, the bandwidth required to send just the human speech is :
\n
Bandwidth for speech = 2700 Hz - 200 Hz = 2500 Hz
\n
For the high-frequency music signal, the frequency band is from 10200 Hz to 15200 Hz. So, the bandwidth required to send the music signal is :
\n
Bandwidth for music = 15200 Hz - 10200 Hz = 5000 Hz
\n
To send both signals together, we need to consider the combined frequency range. The combined frequency range covers from 200 Hz to 15200 Hz. Hence, the bandwidth required to send both signals together is :
\n
Total bandwidth = 15200 Hz - 200 Hz = 15000 Hz
\n
Now, let's calculate the ratio :
\n
Ratio = Total bandwidth / Bandwidth for speech\n
$$ \\Rightarrow $$ Ratio = 15000 Hz / 2500 Hz\n
$$ \\Rightarrow $$ Ratio = 6
\n
Therefore, the correct answer is Option C : 6.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8495,
"subject": "Physics",
"question": "A signal of frequency 20 kHz and peak voltage of 5 Volt is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 Volts. Choose the correct statement.",
"options": [
{
"text": "Modulation index = 5, side frequency bands are at 1400 kHz and 1000 kHz"
},
{
"text": "Modulation index = 5, side frequency bands are at 21.2 kHz and 18.8 kHz\n"
},
{
"text": "Modulation index = 0.8, side frequency bands are at 1180 kHz and 1220 kHz"
},
{
"text": "Modulation index = 0.2, side frequency bands are at 1220 kHz and 1180 kHz\n"
}
],
"answer": "Modulation index = 0.2, side frequency bands are at 1220 kHz and 1180 kHz\n",
"solution": "**Answer:** Modulation index = 0.2, side frequency bands are at 1220 kHz and 1180 kHz\n\n\nModulation index (m) = $${{{V_m}} \\over {{V_0}}} = {5 \\over {25}} = 0.2$$\n
Given, carrier wave, \n
Fc = 1.2 $$ \\times $$ 106 Hz = 1200 kHz\n
and frequency of modulate wave,\n
Fm = 20 kHz\n
$$\\therefore\\,\\,\\,$$ Side frequency bands are\n
F1 = 1200 + 20 = 1220 kHz\n
F2 = 1200 $$-$$ 80 = 1180 kHz",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8496,
"subject": "Physics",
"question": "In amplitude modulation, sinusoidal carrier frequency used is denoted by $${\\omega _c}$$ and the signal frequency is\ndenoted by $${\\omega _m}$$. The bandwidth ($$\\Delta {\\omega _m}$$) of the signal is such that $$\\Delta {\\omega _m}$$ < < $$\\omega _c$$. Which of the following frequencies is not contained in the modulated wave?",
"options": [
{
"text": "$${\\omega _m}$$"
},
{
"text": "$${\\omega _c}$$ "
},
{
"text": "$${\\omega _m}$$ + $${\\omega _c}$$"
},
{
"text": "$${\\omega _c}$$ - $${\\omega _m}$$"
}
],
"answer": "$${\\omega _m}$$",
"solution": "**Answer:** $${\\omega _m}$$\n\nModulated carrier wave contains frequency $${\\omega _c}$$ and\n$${\\omega _c}$$ ± $${\\omega _m}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8497,
"subject": "Physics",
"question": "A carrier wave of peak voltage 14 V is used for transmitting a message signal. The peak voltage of modulating signal given to achieve a modulation index of 80% will be : ",
"options": [
{
"text": "7 V"
},
{
"text": "28 V"
},
{
"text": "11.2 V"
},
{
"text": "22.4 V"
}
],
"answer": "11.2 V",
"solution": "**Answer:** 11.2 V\n\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8498,
"subject": "Physics",
"question": "The number of amplitude modulted broadcast stations that can be accomodated in a $$300$$ $$kHz$$ band width for the highest modulating frequency $$15$$ $$kHz$$ will be : ",
"options": [
{
"text": "$$20$$"
},
{
"text": "$$15$$ "
},
{
"text": "$$10$$ "
},
{
"text": "$$8$$ "
}
],
"answer": "$$10$$ ",
"solution": "**Answer:** $$10$$ \n\nGiven, modulating frequency = 15 kHz \n
$$\\therefore\\,\\,\\,\\,$$ Bandwidth of one channel = 2 $$ \\times $$ 15 = 30 kHz\n
$$\\therefore\\,\\,\\,\\,$$ No of stations = $${{300} \\over {30}}$$ = 10",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8499,
"subject": "Physics",
"question": "The carrier frequency of a transmitter is provided by a tank circuit of a coil of inductance 49 $$\\mu $$H and a capacitance of 2.5 nF. It is modulated by an audio signal of $$12$$ $$kHz.$$ The frequency range occupied by the side bands is : ",
"options": [
{
"text": "13482 kHz $$-$$ 13494 kHz"
},
{
"text": "442 kHz $$-$$ 466 kHz"
},
{
"text": "63 kHz $$-$$ 75 kHz"
},
{
"text": "18 kHz $$-$$ 30 kHz"
}
],
"answer": "442 kHz $$-$$ 466 kHz",
"solution": "**Answer:** 442 kHz $$-$$ 466 kHz\n\nThe tank circuit of an inductance and a capacitance is\n
Given carrier frequency is modulated by audio signal of 12 kHz. Therefore,
\n
f = 454.5 kHz $$\\pm$$ 12 kHz
$$\\Rightarrow$$ f = 442 kHz $$-$$ 466 kHz
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8500,
"subject": "Physics",
"question": "The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency will you allot ?",
"options": [
{
"text": "2900 kHz"
},
{
"text": "2750 kHz"
},
{
"text": "2250 kHz"
},
{
"text": "2000 kz"
}
],
"answer": "2000 kz",
"solution": "**Answer:** 2000 kz\n\nfcarrier = $${{250} \\over {0.1}}$$ = 2500 KHZ\n
$$ \\therefore $$ Range of signal = 2250 Hz to 2750 Hz \n
Now check all options : for 2000 KHZ\n
fmod = 200 Hz\n
$$ \\therefore $$ Range = 1800 KHZ to 2200 KHZ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8501,
"subject": "Physics",
"question": "An amplitude modulated signal is given by V(t) = 10[1 + 0.3cos(2.2 $$ \\times $$ 104\nt)] sin(5.5 $$ \\times $$ 105\nt). Here t is in\nseconds. The sideband frequencies (in kHz) are, [Given $$\\pi $$ = 22/7]\n",
"options": [
{
"text": "892.5 and 857.5"
},
{
"text": "89.25 and 85.75"
},
{
"text": "1785 and 1715"
},
{
"text": "178.5 and 171.5"
}
],
"answer": "89.25 and 85.75",
"solution": "**Answer:** 89.25 and 85.75\n\nV(t) = 10 + $${3 \\over 2}$$ [2cos A sinB]\n
\n$$ \\Rightarrow {v_0}\\sin \\left( {{\\omega _o}t} \\right) + {A \\over 2}\\left[ {\\sin \\left( {{\\omega _o} - \\omega } \\right)t + \\sin \\left( {{\\omega _o} + \\omega } \\right)t} \\right]$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8504,
"subject": "Physics",
"question": "A message signal of frequency 100 MHz and\npeak voltage 100 V is used to execute\namplitude modulation on a carrier wave of\nfrequency 300 GHz and peak voltage 400 V.\nThe modulation index and difference between\nthe two side band frequencies are :",
"options": [
{
"text": "0.25; 1 × 108 Hz"
},
{
"text": "4; 2 × 108 Hz"
},
{
"text": "4; 1 × 108 Hz"
},
{
"text": "0.25; 2 × 108 Hz"
}
],
"answer": "0.25; 2 × 108 Hz",
"solution": "**Answer:** 0.25; 2 × 108 Hz\n\nfm = 100 MHz = 108 Hz, (Vm)0 = 100 V
\nfc = 300 GHz, (Vc)0 = 400 V
\n$$ \\therefore $$ UBF – LBF = 2fm = 2 × 108 Hz",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8505,
"subject": "Physics",
"question": "In an amplitude modulator circuit, the carrier wave is given by, C(t) = 4 sin(20000 $$\\pi $$t) while modulating signal\nis given by, m(t) = 2 sin (2000 $$\\pi $$t). The values of modulation index and lower side band frequency are :",
"options": [
{
"text": "0.3 and 9 kHz"
},
{
"text": "0.5 and 10 kHz"
},
{
"text": "0.4 and 10 kHz"
},
{
"text": "0.5 and 9 kHz"
}
],
"answer": "0.5 and 9 kHz",
"solution": "**Answer:** 0.5 and 9 kHz\n\nC(t) = 4sin(20000 $$\\pi $$t), Ac = 4\n
$$ \\Rightarrow $$ vm = (5 + 3 cos 6280t)\nsin(211 × 104 t)\n
maximum Amp. = 5 + 3 = 8 V\n
minimum Amp. = 5 – 3 = 2 V\n
From the given option nearest value of minimum Amplitude = $${5 \\over 2}$$ V",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 8507,
"subject": "Physics",
"question": "An audio signal vm = 20 sin 2$$\\pi$$(1500t) amplitude modulates a carrier vc = 80 sin 2$$\\pi$$(100,000t).
The value of percent modulation is _________.",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\nGiven, audio signal,
Vm = 20 sin 2$$\\pi$$(1500t) .... (i)
Carrier signal, Vc = 80 sin 2$$\\pi$$(100000t) .... (ii)
Statement I : A speech signal of 2 kHz is used to modulate a carrier signal of 1 MHz. The bandwidth requirement for the signal is 4 kHz.
Statement II : The side band frequencies are 1002 kHz and 998 kHz.
In the light of the above statements, choose the correct answer from the options given below :",
"options": [
{
"text": "Statement I is false but Statement II is true"
},
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Statement I is true but Statement II is false"
},
{
"text": "Both Statement I and Statement II are true"
}
],
"answer": "Both Statement I and Statement II are true",
"solution": "**Answer:** Both Statement I and Statement II are true\n\nSide band = (fc $$-$$ fm) to (fc + fm)
= (1000 $$-$$ 2) KHz to (1000 + 2) Khz
= 998 KHz to 1002 KHz
Band width = 2fm
= 2 $$\\times$$ 2 KHz
= 4 KHz
Both statements are true.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8509,
"subject": "Physics",
"question": "If a message signal of frequency 'fm' is amplitude modulated with a carrier signal of frequency 'fc' and radiated through an antenna, the wavelength of the corresponding signal in air is :",
"options": [
{
"text": "$${c \\over {{f_c} + {f_m}}}$$"
},
{
"text": "$${c \\over {{f_c}}}$$"
},
{
"text": "$${c \\over {{f_c} - {f_m}}}$$"
},
{
"text": "$${c \\over {{f_m}}}$$"
}
],
"answer": "$${c \\over {{f_c}}}$$",
"solution": "**Answer:** $${c \\over {{f_c}}}$$\n\nGiven frequency of massage signal = fm
Frequency of carrier signal = fc
The wavelength of the corresponding signal in air is $$ \\Rightarrow $$ $$\\lambda = {c \\over {{f_c}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8510,
"subject": "Physics",
"question": "The maximum and minimum amplitude of an amplitude modulated wave is 16V and 8V respectively. The modulation index for this amplitude modulated wave is x $$\\times$$ 10$$-$$2. The value of x is __________.",
"options": [],
"answer": "33",
"solution": "**Answer:** 33\n\n$${A_m} = {{{A_{\\max }} - {A_{\\min }}} \\over 2}$$
$$ \\therefore $$ x = 33",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8511,
"subject": "Physics",
"question": "If the highest frequency modulating a carrier is 5 kHz, then the number of AM broadcast stations accommodated in a 90 kHz bandwidth are _________.",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\nB. W. (Bandwidth) = 2 $$\\times$$ maximum frequency at modulating signal
= 2 $$\\times$$ 5kHz
= 10 kHz
$$ \\therefore $$ No of stations accommodate
= $${{90} \\over {10}}$$ = 9",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8512,
"subject": "Physics",
"question": "A carrier signal C(t) = 25 sin(2.512 $$\\times$$ 1010t) is amplitude modulated by a message signal m(t) = 5 sin(1.57 $$\\times$$ 108t) and transmitted through an antenna. What will be the bandwidth of the modulated signal?",
"options": [
{
"text": "50 MHz"
},
{
"text": "8 GHz"
},
{
"text": "1987.5 MHz"
},
{
"text": "2.01 GHz"
}
],
"answer": "50 MHz",
"solution": "**Answer:** 50 MHz\n\n$$\\beta = 2{f_{m(t)}}$$
$$ \\Rightarrow $$ $$\\beta = 50$$ MHz",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8513,
"subject": "Physics",
"question": "A carrier wave Vc(t) = 160 sin(2$$\\pi$$ $$\\times$$ 106t) volts is made to vary between Vmax = 200 V and Vmin = 120 V by a message signal Vm(t) = Am sin(2$$\\pi$$ $$\\times$$ 103t) volts. The peak voltage Am of the modulating signal is ___________.",
"options": [],
"answer": "40",
"solution": "**Answer:** 40\n\nGiven, VC(t) = 160 sin(2$$\\pi$$ $$\\times$$ 106t) V
Vmax = 200 V, Vmin = 120 V and Vm(t) = Am sin(2$$\\pi$$ $$\\times$$ 103t) V.
Since, maximum amplitude,
Amax = Am + Ac
$$\\Rightarrow$$ Vmax = Vm + Vc $$\\Rightarrow$$ 200 = Vm + 160
$$\\Rightarrow$$ Vm = 200 $$-$$ 160 $$\\Rightarrow$$ Vm = 40
$$\\therefore$$ Peak voltage, Am = 40",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8514,
"subject": "Physics",
"question": "In amplitude modulation, the message signal
Vm(t) = 10 sin (2$$\\pi$$ $$\\times$$ 105t) volts and
Carrier signal
VC(t) = 20 sin(2$$\\pi$$ $$\\times$$ 107 t) volts
The modulated signal now contains the message signal with lower side band and upper side band frequency, therefore the bandwidth of modulated signal is $$\\alpha$$ kHz. The value of $$\\alpha$$ is :",
"options": [
{
"text": "200 kHz"
},
{
"text": "50 kHz"
},
{
"text": "100 kHz "
},
{
"text": "0"
}
],
"answer": "200 kHz",
"solution": "**Answer:** 200 kHz\n\nBandwidth = 2 $$\\times$$ fm
= 2 $$\\times$$ 105 HZ = 200 KHZ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8515,
"subject": "Physics",
"question": "A message signal of frequency 20 kHz and peak voltage of 20 volt is used to modulate a carrier wave of frequency 1 MHz and peak voltage of 20 volt. The modulation index will be :",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nModulation index
$$\\mu = {{{A_m}} \\over {{A_c}}} = {{20} \\over {20}} = 1$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8516,
"subject": "Physics",
"question": "The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency 11.21 MHz, peak voltage 15 V is amplitude modulated by a 7.7 kHz sine wave of 5V amplitude are $${a \\over {10}}V$$ and $${b \\over {10}}V$$ respectively. Then the value of $${a \\over b}$$ is ____________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n
$$ \\Rightarrow {a \\over b} = 1$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8517,
"subject": "Physics",
"question": "The maximum amplitude for an amplitude modulated wave is found to be 12V while the minimum amplitude is found to be 3V. The modulation index is 0.6x where x is ____________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nAmax = Ac + Am = 12
Amin = Ac $$-$$ Am = 3
$$\\Rightarrow$$ Ac = $${{15} \\over 2}$$ & Am = $${9 \\over 2}$$
$$\\Rightarrow$$ x = 1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8518,
"subject": "Physics",
"question": "An amplitude modulated wave is represented by Cm(t) = 10(1 + 0.2 cos 12560t) sin(111 $$\\times$$ 104t) volts. The modulating frequency in kHz will be ................. .",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nWm = 12560 = 2$$\\pi$$fm
$${f_m} = {{12560} \\over {2\\pi }}$$
= 2000 Hz",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8519,
"subject": "Physics",
"question": "A bandwidth of 6 MHz is available for A.M. transmission. If the maximum audio signal frequency used for modulating the carrier wave is not to exceed 6 kHz. The number of stations that can be broadcasted within this band simultaneously without interfering with each other will be __________.",
"options": [],
"answer": "500",
"solution": "**Answer:** 500\n\nSignal bandwidth = 2 fm = 12 kHz
$$\\therefore$$ N = $${{6MHZ} \\over {12kHZ}} = {{6 \\times {{10}^6}} \\over {12 \\times {{10}^3}}} = 500$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8520,
"subject": "Physics",
"question": "A carrier wave with amplitude of 250 V is amplitude modulated by a sinusoidal base band signal of amplitude 150V. The ratio of minimum amplitude to maximum amplitude for the amplitude modulated wave is 50 : x, then value of x is ____________.",
"options": [],
"answer": "200",
"solution": "**Answer:** 200\n\nGiven, the amplitude of the carrier wave, Ac = 250 V
The amplitude of the message wave, Am = 150 V
We know that,
The maximum amplitude, Amax = Ac + Am
Substituting the values in the above equation, we get
Amax = 250 + 150 = 400 V
We know that,
The minimum amplitude, Amin = Ac $$-$$ Am
Substituting the values in the above equation, we get
Amin = 250 $$-$$ 150 = 100 V
Thus, the ratio of the minimum amplitude to the maximum amplitude of the modulated wave is
The value of the x = 4.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8521,
"subject": "Physics",
"question": "
Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an audio signal requires a bandwidth of 8 kHz, how many channels can be accommodated for transmission :
Amplitude modulated wave is represented by $${V_{AM}} = 10\\,[1 + 0.4\\cos (2\\pi \\times {10^4}t)]\\cos (2\\pi \\times {10^7}t)$$. The total bandwidth of the amplitude modulated wave is :
Choose the correct statement for amplitude modulation :
",
"options": [
{
"text": "Amplitude of modulating signal is varied in accordance with the information signal."
},
{
"text": "Amplitude of modulated signal is varied in accordance with the information signal."
},
{
"text": "Amplitude of carrier signal is varied in accordance with the information signal."
},
{
"text": "Amplitude of modulated signal is varied in accordance with the modulating signal."
}
],
"answer": "Amplitude of carrier signal is varied in accordance with the information signal.",
"solution": "**Answer:** Amplitude of carrier signal is varied in accordance with the information signal.\n\nIn amplitude modulation, amplitude of carrier signal\nis varied according to the message signal.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8524,
"subject": "Physics",
"question": "
A sinusoidal wave y(t) = 40sin(10 $$\\times$$ 106 $$\\pi$$t) is amplitude modulated by another sinusoidal wave x(t) = 20sin (1000 $$\\pi$$t). The amplitude of minimum frequency component of modulated signal is :
A baseband signal of 3.5 MHz frequency is modulated with a carrier signal of 3.5 GHz frequency using amplitude modulation method. What should be the minimum size of antenna required to transmit the modulated signal?
A speech signal given by 11 sin(2200 $$\\pi$$t) V is used for amplitude modulation with a carrier signal given by 44 sin(6600 $$\\pi$$t) V. The minimum amplitude of modulated wave will be :
In AM modulation, a signal is modulated on a carrier wave such that maximum and minimum amplitudes are found to be 6 V and 2 V respectively. The modulation index is :
The maximum and minimum voltage of an amplitude modulated signal are $$60 \\mathrm{~V}$$ and $$20 \\mathrm{~V}$$ respectively. The percentage modulation index will be :
In an Amplitude Modulated (AM) signal, the amplitude of the carrier signal is varied in accordance with the information being sent. The extent of this variation is given by the modulation index, often denoted by $\\mu$.
\n
In the problem statement, we are given a peak-to-peak variation in the signal of 8V. This is essentially twice the amplitude of the modulating signal, because the peak-to-peak value represents the total variation from the minimum to the maximum. Thus, we find the amplitude of the modulating signal ($A_m$) as follows:
\n
$A_m = \\frac{8V}{2} = 4V$
\n
The maximum amplitude of the AM wave is 9V. This includes the amplitude of the carrier signal and the amplitude of the modulating signal. We therefore find the amplitude of the carrier signal ($A_c$) as follows:
\n
$A_c = 9V - A_m = 9V - 4V = 5V$
\n
Finally, the modulation index $\\mu$ is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier signal:
\n
$\\mu = \\frac{A_m}{A_c} = \\frac{4V}{5V} = 0.8$
\n
Thus, the modulation index of the AM signal is 0.8, meaning that the amplitude of the carrier signal is varied by 80% of its original amplitude to encode the information being sent. The correct answer among the provided options is therefore Option A: 0.8.
A modulating signal $$2 \\sin \\left(6.28 \\times 10^{6}\\right) t$$ is added to the carrier signal $$4 \\sin \\left(12.56 \\times 10^{9}\\right) t$$ for amplitude modulation. The combined signal is passed through a non-linear square law device. The output is then passed through a band pass filter. The bandwidth of the output signal of band pass filter will be _________ $$\\mathrm{MHz}$$.
In an amplitude modulation, a modulating signal having amplitude of X Volt is superimposed with a carrier signal of amplitude Y Volt in first case. Then, in second case, the same modulating signal is superimposed with different carrier signal of amplitude 2Y Volt. The ratio of modulation index in the two cases respectively will be :
",
"options": [
{
"text": "2 : 1"
},
{
"text": "1 : 1"
},
{
"text": "4 : 1"
},
{
"text": "1 : 2"
}
],
"answer": "2 : 1",
"solution": "**Answer:** 2 : 1\n\nThe modulation index is a measure of the degree of modulation and is given by the ratio of the amplitude of the modulating signal to the amplitude of the carrier signal.\n\n
In the first case, the modulation index is $${X \\over Y}$$.\n\n
In the second case, the modulation index is $${X \\over {2Y}}$$.\n\n
The ratio of the modulation index in the two cases is $${{{X \\over Y}} \\over {{X \\over {2Y}}}} = {2 \\over 1}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8536,
"subject": "Physics",
"question": "
Given below are two statements :
\n
Statement I : For transmitting a signal, size of antenna ( $l$ ) should be comparable to wavelength of signal (at least $l=\\frac{\\lambda}{4}$ in dimension)
\n
Statement II: In amplitude modulation, amplitude of carrier wave remains constant (unchanged).
\n
In the light of the above statements, choose the most appropriate answer from the options given below
",
"options": [
{
"text": "Statement I is incorrect but Statement II is correct"
},
{
"text": "Both Statement I and Statement II are correct"
},
{
"text": "Both Statement I and Statement II are incorrect"
},
{
"text": " Statement I is correct but Statement II is incorrect"
}
],
"answer": " Statement I is correct but Statement II is incorrect",
"solution": "**Answer:** Statement I is correct but Statement II is incorrect\n\n
For transmitting a signal, antenna should have a\nsize comparable to the wavelength of the signal i.e., at least\n$${1 \\over 4}$$ in dimension, so that the antenna properly senses the\ntime variation of the signal.
\n
The amplitude of the carrier wave is varied in accordance\nwith the message signal is termed as amplitude modulation.
\n
So, statement I is correct but statement II is incorrect.
The amplitude of $$15 \\sin (1000 \\pi \\mathrm{t})$$ is modulated by $$10 \\sin (4 \\pi \\mathrm{t})$$ signal. The amplitude modulated signal contains frequency (ies) of
\n
A. $$500 \\mathrm{~Hz}$$
\n
B. $$2 \\mathrm{~Hz}$$
\n
C. $$250 \\mathrm{~Hz}$$
\n
D. $$498 \\mathrm{~Hz}$$
\n
E. $$502 \\mathrm{~Hz}$$
\n
Choose the correct answer from the options given below:
",
"options": [
{
"text": "B only"
},
{
"text": "A only"
},
{
"text": "A, D and E only"
},
{
"text": "A and B only"
}
],
"answer": "A, D and E only",
"solution": "**Answer:** A, D and E only\n\nCarrier wave frequency\n\n
A sinusoidal carrier voltage is amplitude modulated. The resultant amplitude modulated wave has maximum and minimum amplitude of $$120 \\mathrm{~V}$$ and $$80 \\mathrm{~V}$$ respectively. The amplitude of each sideband is :
The modulation index (m) of an amplitude-modulated wave is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier wave. It is a measure of the degree of modulation of the carrier wave and is given by the following equation:
\n\n$$ m = \\frac{V_{max} - V_{min}}{V_{max} + V_{min}} $$\n
\nWhere $V_{max}$ is the maximum peak-to-peak voltage and $V_{min}$ is the minimum peak-to-peak voltage.\n
\nPlugging in the values from the question, we have:\n
\nSo, the modulation index is 0.4, which means the degree of modulation is 0.4.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8541,
"subject": "Physics",
"question": "
A message signal of frequency 5 kHz is used to modulate a carrier signal of frequency 2 MHz. The bandwidth for amplitude modulation is :
The amplitude of $$15 \\sin (1000 \\pi \\mathrm{t})$$ is modulated by $$10 \\sin (4 \\pi \\mathrm{t})$$ signal. The amplitude modulated signal contains frequencies of
\n
A. $$500 \\mathrm{~Hz}$$
\n
B. $$2 \\mathrm{~Hz}$$
\n
C. $$250 \\mathrm{~Hz}$$
\n
D. $$498 \\mathrm{~Hz}$$
\n
E. $$502 \\mathrm{~Hz}$$
\n
Choose the correct answer from the options given below:
",
"options": [
{
"text": "A and C only"
},
{
"text": "A and B only"
},
{
"text": "A and D only"
},
{
"text": "A, D and E only"
}
],
"answer": "A, D and E only",
"solution": "**Answer:** A, D and E only\n\nWhen an amplitude modulated signal is created, the frequency components of the resulting signal consist of the carrier frequency, the sum, and the difference of the carrier and modulating frequencies.\n
\nIn this case, the carrier signal is $$15 \\sin (1000 \\pi t)$$, which has a frequency of
\nSo, the amplitude modulated signal contains frequencies of 500 Hz, 2 Hz, 498 Hz, and 502 Hz.
The correct answer is (A, D, and E only).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8543,
"subject": "Physics",
"question": "
A message signal of frequency $$3 ~\\mathrm{kHz}$$ is used to modulate a carrier signal of frequency $$1.5 ~\\mathrm{MHz}$$. The bandwidth of the amplitude modulated wave is
The bandwidth of an amplitude-modulated wave is given by $$2B$$, where $$B$$ is the bandwidth of the modulating signal.
In this case, the modulating signal is a message signal of frequency $$3~\\mathrm{kHz}$$, so the bandwidth of the amplitude modulated wave is $$2(3~\\mathrm{kHz})=6~\\mathrm{kHz}$$.
\n
Therefore, the correct answer is $$6~\\mathrm{kHz}$$.
A carrier wave of amplitude 15 V is modulated by a sinusoidal base band signal of magnitude 3 V. The ratio of maximum amplitude to minimum amplitude in an amplitude modulated wave is
",
"options": [
{
"text": "1"
},
{
"text": "5"
},
{
"text": "2"
},
{
"text": "$$\\frac{3}{2}$$"
}
],
"answer": "$$\\frac{3}{2}$$",
"solution": "**Answer:** $$\\frac{3}{2}$$\n\nThe maximum and minimum amplitude of a modulated wave are given by:\n
\nMaximum amplitude = carrier wave amplitude + base band signal magnitude = 15V + 3V = 18V\n
\nMinimum amplitude = carrier wave amplitude - base band signal magnitude = 15V - 3V = 12V\n
\nTherefore, the ratio of maximum amplitude to minimum amplitude in an amplitude modulated wave is 18/12 = 1.5 or $$\\frac{3}{2}$$. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8545,
"subject": "Physics",
"question": "
For an amplitude modulated wave the minimum amplitude is $$3 \\mathrm{~V}$$, while the modulation index is $$60 \\%$$. The maximum amplitude of the modulated wave is:
$A_{\\max}$ is the maximum amplitude of the modulated wave,
\n
$A_{\\min}$ is the minimum amplitude of the modulated wave.
\n
\n
Given that the modulation index is 60% or 0.6 and the minimum amplitude $A_{\\min}$ is 3 V, we can rearrange this formula to solve for the maximum amplitude $A_{\\max}$:
\n
$$ 0.6 = \\frac{A_{\\max} - 3}{A_{\\max} + 3} $$
\n
Solving this equation gives:
\n
$$ A_{\\max} = 12 \\, \\text{V} $$
\n
So, the correct answer is 12 V.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8546,
"subject": "Physics",
"question": "A 200 $$\\Omega $$ resistor has a certain color code. If one\nreplaces the red color by green in the code, the\nnew resistance will be :",
"options": [
{
"text": "500 $$\\Omega $$"
},
{
"text": "100 $$\\Omega $$"
},
{
"text": "200 $$\\Omega $$"
},
{
"text": "300 $$\\Omega $$"
}
],
"answer": "500 $$\\Omega $$",
"solution": "**Answer:** 500 $$\\Omega $$\n\n200 $$\\Omega $$ = Red + Black + Brown
\nGreen = 5
\nSo, Green + Black + Brown = 500 $$\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8547,
"subject": "Physics",
"question": "The resistance of the series combination of two resistances is $$S.$$ When they are jointed in parallel the total resistance is $$P.$$ If $$S = nP$$ then the Minimum possible value of $$n$$ is ",
"options": [
{
"text": "$$2$$ "
},
{
"text": "$$3$$ "
},
{
"text": "$$4$$ "
},
{
"text": "$$1$$ "
}
],
"answer": "$$4$$ ",
"solution": "**Answer:** $$4$$ \n\n\n
Minimum value of $$n$$ is $$4$$ ",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 8548,
"subject": "Physics",
"question": "A uniform metallic wire has a resistance of 18 $$\\Omega $$ and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is -\n",
"options": [
{
"text": "12 $$\\Omega $$"
},
{
"text": "2 $$\\Omega $$"
},
{
"text": "4 $$\\Omega $$"
},
{
"text": "8 $$\\Omega $$"
}
],
"answer": "4 $$\\Omega $$",
"solution": "**Answer:** 4 $$\\Omega $$\n\n\n
Req berween any two vertex will be\n
$${1 \\over {{{\\mathop{\\rm R}\\nolimits} _{eq}}}} = {1 \\over {12}} + {1 \\over 6} \\Rightarrow {{\\mathop{\\rm R}\\nolimits} _{eq.}} = 4\\Omega $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8549,
"subject": "Physics",
"question": "The equivalent resistance of series combination of two resistors is 's'. When they are connected in parallel, the equivalent resistance is 'p'. If s = np, then the minimum value for n is ____________. (Round off to the Nearest Integer)",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n$$s = np$$
So, minimum value for n = 4",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 8550,
"subject": "Physics",
"question": "Two wires of same length and thickness having specific resistances 6$$\\Omega$$ cm and 3$$\\Omega$$ cm respectively are connected in parallel. The effective resistivity is $$\\rho$$$$\\Omega$$ cm. The value of $$\\rho$$, to the nearest integer, is ____________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nLet length of each wire is l and area A. When they are connected in parallel then their effective area 2A.
$$ \\Rightarrow \\rho = 4$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8551,
"subject": "Physics",
"question": "What equal length of an iron wire and a copper-nickel alloy wire, each of 2 mm diameter connected parallel to give an equivalent resistance of 3$$\\Omega$$ ?
(Given resistivities of iron and copper-nickel alloy wire are 12 $$\\mu$$$$\\Omega$$ and 51 $$\\mu$$$$\\Omega$$ cm respectively)",
"options": [
{
"text": "82 m"
},
{
"text": "97 m"
},
{
"text": "110 m"
},
{
"text": "90 m"
}
],
"answer": "97 m",
"solution": "**Answer:** 97 m\n\n$${{{R_1}{R_2}} \\over {{R_1} + {R_2}}} = 3$$
$$\\Rightarrow$$ l = 97 m",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8552,
"subject": "Physics",
"question": "If you are provided a set of resistances 2$$\\Omega$$, 4$$\\Omega$$, 6$$\\Omega$$ and 8$$\\Omega$$. Connect these resistances so as to obtain an equivalent resistance of $${{46} \\over 3}$$$$\\Omega$$.",
"options": [
{
"text": "4$$\\Omega$$ and 6$$\\Omega$$ are in parallel with 2$$\\Omega$$ and 8$$\\Omega$$ in series"
},
{
"text": "6$$\\Omega$$ and 8$$\\Omega$$ are in parallel with 2$$\\Omega$$ and 4$$\\Omega$$ in series"
},
{
"text": "2$$\\Omega$$ and 6$$\\Omega$$ are in parallel with 4$$\\Omega$$ and 8$$\\Omega$$ in series"
},
{
"text": "2$$\\Omega$$ and 4$$\\Omega$$ are in parallel with 6$$\\Omega$$ and 8$$\\Omega$$ in series"
}
],
"answer": "2$$\\Omega$$ and 4$$\\Omega$$ are in parallel with 6$$\\Omega$$ and 8$$\\Omega$$ in series",
"solution": "**Answer:** 2$$\\Omega$$ and 4$$\\Omega$$ are in parallel with 6$$\\Omega$$ and 8$$\\Omega$$ in series\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8553,
"subject": "Physics",
"question": "Five identical cells each of internal resistance 1$$\\Omega$$ and emf 5V are connected in series and in parallel with an external resistance 'R'. For what value of 'R', current in series and parallel combination will remain the same?",
"options": [
{
"text": "1 $$\\Omega$$"
},
{
"text": "25 $$\\Omega$$ "
},
{
"text": "5 $$\\Omega$$"
},
{
"text": "10 $$\\Omega$$ "
}
],
"answer": "1 $$\\Omega$$",
"solution": "**Answer:** 1 $$\\Omega$$\n\n$${i_1} = {{25} \\over {5 + R}}$$
R = 1$$\\Omega$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 8554,
"subject": "Physics",
"question": "First, a set of n equal resistors of 10 $$\\Omega$$ each are connected in series to a battery of emf 20V and internal resistance 10$$\\Omega$$. A current I is observed to flow. Then, the n resistors are connected in parallel to the same battery. It is observed that the current is increased 20 times, then the value of n is ............... .",
"options": [],
"answer": "20",
"solution": "**Answer:** 20\n\nIn series
$$n = 20$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8555,
"subject": "Physics",
"question": "A square shaped wire with resistance of each side 3$$\\Omega$$ is bent to form a complete circle. The resistance between two diametrically opposite points of the circle in unit of $$\\Omega$$ will be ___________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n Req = 3$$\\Omega$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8556,
"subject": "Physics",
"question": "Two resistors R1 = (4 $$\\pm$$ 0.8) $$\\Omega$$ and R2 = (4 $$\\pm$$ 0.4) $$\\Omega$$ are connected in parallel. The equivalent resistance of their parallel combination will be :",
"options": [
{
"text": "(4 $$\\pm$$ 0.4) $$\\Omega$$"
},
{
"text": "(2 $$\\pm$$ 0.4) $$\\Omega$$"
},
{
"text": "(2 $$\\pm$$ 0.3) $$\\Omega$$"
},
{
"text": "(4 $$\\pm$$ 0.3) $$\\Omega$$"
}
],
"answer": "(2 $$\\pm$$ 0.3) $$\\Omega$$",
"solution": "**Answer:** (2 $$\\pm$$ 0.3) $$\\Omega$$\n\nGiven,
R1 = (4 $$\\pm$$ 0.8) $$\\Omega$$
R2 = (4 $$\\pm$$ 0.4) $$\\Omega$$
Equivalent resistance when the resistors are connected in parallel is given by
$$\\therefore$$ The equivalent resistance in parallel combination is $${R_{eq}} = (2 \\pm 0.3)\\Omega $$.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 8557,
"subject": "Physics",
"question": "
What will be the most suitable combination of three resistors A = 2$$\\Omega$$, B = 4$$\\Omega$$, C = 6$$\\Omega$$ so that $$\\left( {{{22} \\over 3}} \\right)$$$$\\Omega$$ is equivalent resistance of combination?
",
"options": [
{
"text": "Parallel combination of A and C connected in series with B."
},
{
"text": "Parallel combination of A and B connected in series with C."
},
{
"text": "Series combination of A and C connected in parallel with B."
},
{
"text": "Series combination of B and C connected in parallel with A."
}
],
"answer": "Parallel combination of A and B connected in series with C.",
"solution": "**Answer:** Parallel combination of A and B connected in series with C.\n\n
An electric cable of copper has just one wire of radius 9 mm. Its resistance is 14 $$\\Omega$$. If this single copper wire of the cable is replaced by seven identical well insulated copper wires each of radius 3 mm connected in parallel, then the new resistance of the combination will be :
Eight copper wire of length $$l$$ and diameter $$d$$ are joined in parallel to form a single composite conductor of resistance $$R$$. If a single copper wire of length $$2 l$$ have the same resistance $$(R)$$ then its diameter will be ____________ d.
\nStatement I : The equivalent resistance of resistors in a series combination is smaller than least resistance used in the combination.\n
\nStatement II : The resistivity of the material is independent of temperature.\n
\nIn the light of the above statements, choose the correct answer from the options given below :",
"options": [
{
"text": "Statement I is true but Statement II is false"
},
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Statement I is false but Statement II is true"
}
],
"answer": "Both Statement I and Statement II are false",
"solution": "**Answer:** Both Statement I and Statement II are false\n\nStatement I is incorrect, as the equivalent resistance of resistors in a series combination is always greater than the largest individual resistance used in the combination. This is because the total voltage drop across the resistors in series is equal to the sum of the voltage drops across each resistor, and the current through each resistor is the same. Therefore, the resistance of the entire series combination is the sum of the individual resistances.\n
\nStatement II is also incorrect, as the resistivity of most materials changes with temperature. This is because temperature affects the mobility of charge carriers in a material, which in turn affects its resistivity. For example, the resistivity of metals generally increases with temperature, while the resistivity of semiconductors generally decreases with temperature.\n
\nTherefore, the correct answer is Both Statement I and Statement II are false",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 8562,
"subject": "Physics",
"question": "
10 resistors each of resistance 10 $$\\Omega$$ can be connected in such as to get maximum and minimum equivalent resistance. The ratio of maximum and minimum equivalent resistance will be ___________.
When resistors are connected in series, the equivalent resistance is the sum of individual resistances. Therefore, the maximum equivalent resistance will be when all 10 resistors are connected in series, giving a total resistance of 100 $$\\Omega$$.
\n
When resistors are connected in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances. Therefore, the minimum equivalent resistance will be when all 10 resistors are connected in parallel, giving a total resistance of 1 $$\\Omega$$.
\n
The ratio of maximum to minimum equivalent resistance is thus $$\\frac{100}{1}=100$$.
A wire of resistance $$\\mathrm{R}$$ and length $$\\mathrm{L}$$ is cut into 5 equal parts. If these parts are joined parallely, then resultant resistance will be :
A wire of resistance $$20 \\Omega$$ is divided into 10 equal parts, resulting pairs. A combination of two parts are connected in parallel and so on. Now resulting pairs of parallel combination are connected in series. The equivalent resistance of final combination is _________ $$\\Omega$$.
Let's start by understanding the process of dividing the wire and recombining its parts to form the final configuration. Initially, we have a wire with a resistance of $$20 \\Omega$$. This wire is divided into 10 equal parts, each part then has a resistance of:\n\n
$$\\frac{20 \\Omega}{10} = 2 \\Omega$$
\n\n
Since each part has the same length and presumably the same material and cross-sectional area, then each part will have the same resistance of $$2 \\Omega$$.
\n\n
When two parts are connected in parallel, the equivalent resistance, $$R_{\\text{parallel}}$$, of this configuration can be calculated using the formula for two resistors in parallel:
Now, since the original wire was divided into 10 equal parts, and pairs of these parts are connected in parallel, this results in $$\\frac{10}{2} = 5$$ pairs. Each of these pairs has an equivalent resistance of $$1 \\Omega$$.
\n\n
Finally, these pairs are all connected in series. The total resistance of resistors in series is simply the sum of their individual resistances. Therefore, the equivalent resistance of the final configuration, $$R_{\\text{series}}$$, is:
So, the equivalent resistance of the final combination is $$5 \\Omega$$.
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 8565,
"subject": "Physics",
"question": "The mass of product liberated on anode in an electrochemical cell depends on (where $$t$$ is the time period for which the current is passed). ",
"options": [
{
"text": "$${\\left( {It} \\right)^{1/2}}$$"
},
{
"text": "$$It$$ "
},
{
"text": "$$I/t$$ "
},
{
"text": "$${I^2}t$$ "
}
],
"answer": "$$It$$ ",
"solution": "**Answer:** $$It$$ \n\nAccording to Faraday's first law of electrolysis \n
$$m = ZIt \\Rightarrow m \\propto It$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8566,
"subject": "Physics",
"question": "The nagative $$Zn$$ pole of a Daniell cell, sending a constant current through a circuit, decreases in mass by $$0.13g$$ in $$30$$ minutes. If the electrochemical equivalent of $$Zn$$ and $$Cu$$ are $$32.5$$ and $$31.5$$ respectively, the increase in the mass of the positive $$Cu$$ pole in this time is ",
"options": [
{
"text": "$$0.180$$ $$g$$ "
},
{
"text": "$$0.141$$ $$g$$ "
},
{
"text": "$$0.126$$ $$g$$ "
},
{
"text": "$$0.242$$ $$g$$ "
}
],
"answer": "$$0.126$$ $$g$$ ",
"solution": "**Answer:** $$0.126$$ $$g$$ \n\nAccording to Faraday's first law of electrolysis $$m = z \\times q$$\n
For same $$q,$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$m \\propto Z$$\n
$$ = {{31.5} \\over {32.5}} \\times 0.13 = 0.126\\,g$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8567,
"subject": "Physics",
"question": "The series combination of two batteries, both\nof the same emf 10 V, but different internal\nresistance of 20$$\\Omega $$ and 5$$\\Omega $$, is connected to the\nparallel combination of two resistors 30$$\\Omega $$ and\nR $$\\Omega $$. The voltage difference across the battery\nof internal resistance 20$$\\Omega $$ is zero, the value of\nR (in $$\\Omega $$) is : _______",
"options": [],
"answer": "30",
"solution": "**Answer:** 30\n\n\n
$$ \\Rightarrow $$ R = 30 $$\\Omega $$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 8568,
"subject": "Physics",
"question": "In an electrical circuit, a battery is connected to pass 20C of charge through it in a certain given time. The potential difference between two plates of the battery is maintained at 15V. The workdone by the battery is __________J.",
"options": [],
"answer": "300",
"solution": "**Answer:** 300\n\n
The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of 2$$\\Omega$$. The value of internal resistance of each cell is
Two cells of same emf but different internal resistances r1 and r2 are connected in series with a resistance R. The value of resistance R, for which the potential difference across second cell is zero, is :
Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are $$r_{1}$$ and $$r_{2}$$ $$\\left(r_{1}>r_{2}\\right)$$. If the potential difference across the source of internal resistance $$r_{1}$$ is zero, then the value of R will be :
Two identical cells, when connected either in parallel or in series gives same current in an external resistance $$5 ~\\Omega$$. The internal resistance of each cell will be ___________ $$\\Omega$$.
$\\Rightarrow r=5 \\Omega$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8573,
"subject": "Physics",
"question": "A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold, then, an electric current will ",
"options": [
{
"text": "flow from Antimony to Bismuth at the hot junction "
},
{
"text": "flow from Bismuth to Antimony at the cold junction "
},
{
"text": "now flow through the thermocouple "
},
{
"text": "flow from Antimony to Bismuth at the cold junction"
}
],
"answer": "flow from Antimony to Bismuth at the cold junction",
"solution": "**Answer:** flow from Antimony to Bismuth at the cold junction\n\nAt cold junction, current flows from Antimony to Bismuth (because current flows from metal occurring later in the series to metal occurring earlier in the thermoelectric series). ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8574,
"subject": "Physics",
"question": "A copper rod of cross-sectional area A carries a uniform current I through it. At temperature T, if the volume charge density of the rod is $$\\rho $$, how long will the changes take to travel a distance d ? ",
"options": [
{
"text": "$${{2\\rho \\,d\\,A} \\over {\\rm I}}$$"
},
{
"text": "$${{2\\rho \\,d\\,A} \\over {{\\rm I}\\,T}}$$"
},
{
"text": "$${{\\rho \\,d\\,A} \\over {{\\rm I}\\,}}$$"
},
{
"text": "$${{\\rho \\,d\\,A} \\over {{\\rm I}\\,T}}$$"
}
],
"answer": "$${{\\rho \\,d\\,A} \\over {{\\rm I}\\,}}$$",
"solution": "**Answer:** $${{\\rho \\,d\\,A} \\over {{\\rm I}\\,}}$$\n\n
Given : Volume charge density of rod = $$\\rho$$.
\n
We know that current $$I = neA{v_d}$$; where n is number of electrons, e is electronic charge, A is area, vd is drift velocity
\n
\n
Since volume charge density is $$\\rho$$ = ne. Therefore,
Thus, time required to travel distance $$d = {{\\rho dA} \\over I}$$
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8575,
"subject": "Physics",
"question": "Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm2, is $$\\upsilon $$. If the electron density in copper is 9 $$ \\times $$ 1028/m3 the value of $$\\upsilon $$. in mm/s is close to (Take charge of electron to be = 1.6 $$ \\times $$ 10$$-$$19C)",
"options": [
{
"text": "0.02"
},
{
"text": "3"
},
{
"text": "2"
},
{
"text": "0.2"
}
],
"answer": "0.02",
"solution": "**Answer:** 0.02\n\nWe know, \n
= 0.02 m/s",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8576,
"subject": "Physics",
"question": "A current through a wire depends on time as
i = $$\\alpha$$0t + $$\\beta$$t2
where $$\\alpha$$0 = 20 A/s and $$\\beta$$ = 8 As$$-$$2. Find the charge crossed through a section of the wire in 15 s.",
"options": [
{
"text": "2250 C"
},
{
"text": "2100 C"
},
{
"text": "260 C"
},
{
"text": "11250 C"
}
],
"answer": "11250 C",
"solution": "**Answer:** 11250 C\n\nGiven, $$i = {\\alpha _0}t + \\beta {t^2}$$
$$\\therefore$$ q = 11250 C",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8577,
"subject": "Physics",
"question": "A cylindrical wire of radius 0.5 mm and conductivity 5 $$\\times$$ 107 S/m is subjected to an electric field of 10 mV/m. The expected value of current in the wire will be x3$$\\pi$$ mA. The value of x is _________.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nWe know that current density,
$$ \\therefore $$ x = 5",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8578,
"subject": "Physics",
"question": "A current of 10A exists in a wire of cross-sectional area of 5 mm2 with a drift velocity of 2 $$\\times$$ 10$$-$$3 ms$$-$$1. The number of free electrons in each cubic meter of the wire is ___________.",
"options": [
{
"text": "625 $$\\times$$ 1025"
},
{
"text": "1 $$\\times$$ 1023"
},
{
"text": "2 $$\\times$$ 1025"
},
{
"text": "2 $$\\times$$ 106"
}
],
"answer": "625 $$\\times$$ 1025",
"solution": "**Answer:** 625 $$\\times$$ 1025\n\n$$I = neA{V_d}$$
$$ = {{{{10}^{25}}} \\over {16}} = 6.25 \\times {10^{27}} = 625 \\times {10^{25}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8579,
"subject": "Physics",
"question": "A current of 5 A is passing through a non-linear magnesium wire of cross-section 0.04 m2. At every point the direction of current density is at an angle of 60$$^\\circ$$ with the unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is :
The current density in a cylindrical wire of radius r = 4.0 mm is 1.0 $$\\times$$ 106 A/m2. The current through the outer portion of the wire between radial distances $${r \\over 2}$$ and r is x$$\\pi$$ A; where x is __________.
The current density in a cylindrical wire of radius 4 mm is 4 $$\\times$$ 106 Am$$-$$2. The current through the outer portion of the wire between radial distances $${R \\over 2}$$ and R is ____________ $$\\pi$$ A.
A $$1 \\mathrm{~m}$$ long copper wire carries a current of $$1 \\mathrm{~A}$$. If the cross section of the wire is $$2.0 \\mathrm{~mm}^{2}$$ and the resistivity of copper is $$1.7 \\times 10^{-8}\\, \\Omega \\mathrm{m}$$, the force experienced by moving electron in the wire is ____________ $$\\times 10^{-23} \\mathrm{~N}$$.
\n
(charge on electorn $$=1.6 \\times 10^{-19} \\,\\mathrm{C}$$)
(A) The drift velocity of electrons decreases with the increase in the temperature of conductor.
\n
(B) The drift velocity is inversely proportional to the area of cross-section of given conductor.
\n
(C) The drift velocity does not depend on the applied potential difference to the conductor.
\n
(D) The drift velocity of electron is inversely proportional to the length of the conductor.
\n
(E) The drift velocity increases with the increase in the temperature of conductor.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "(A) and (B) only"
},
{
"text": "(A) and (D) only"
},
{
"text": "(B) and (E) only"
},
{
"text": "(B) and (C) only"
}
],
"answer": "(A) and (D) only",
"solution": "**Answer:** (A) and (D) only\n\n
We know, Resistivity $\\rho=\\frac{m}{n e^2 \\tau}$\n
Where $\\tau$ is relaxation time As temperature $\\uparrow, \\tau \\downarrow, \\rho \\uparrow, \\mathrm{R} \\uparrow, \\mathrm{i} \\downarrow, \\mathrm{v}_{\\mathrm{d}} \\downarrow$ as $i=$ ne $\\mathrm{A} v_{\\mathrm{d}}$\n\n
Statement (A) is correct and hence statement (E) is incorrect.\n
$$\n\\mathrm{R}=\\rho \\frac{l}{A}, I=\\frac{V}{R}=\\frac{V A}{\\rho l}=n e A v_d \\rightarrow v_d=\\text { constant }\n$$\n
For a given $\\mathrm{V}, v_{\\mathrm{d}}$ is independent of $\\mathrm{A}$. Hence, statement (B) is incorrect\n\n
Form above $v_d \\propto$ V statement (C) is also incorrect. Since $v_d \\propto \\frac{1}{\\ell}$ as seen from above statement, so statement (D) is correct.
The drift velocity of electrons for a conductor connected in an electrical circuit is $$\\mathrm{V}_{\\mathrm{d}}$$. The conductor in now replaced by another conductor with same material and same length but double the area of cross section. The applied voltage remains same. The new drift velocity of electrons will be
As drift velocity does not depends upon area, so it will remain same, even after changing area of conductor.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8586,
"subject": "Physics",
"question": "
The charge flowing in a conductor changes with time as $$\\mathrm{Q}(\\mathrm{t})=\\alpha \\mathrm{t}-\\beta \\mathrm{t}^{2}+\\gamma \\mathrm{t}^{3}$$. Where $$\\alpha, \\beta$$ and $$\\gamma$$ are constants. Minimum value of current is :
The number density of free electrons in copper is nearly $$8 \\times 10^{28} \\mathrm{~m}^{-3}$$. A copper wire has its area of cross section $$=2 \\times 10^{-6} \\mathrm{~m}^{2}$$ and is carrying a current of $$3.2 \\mathrm{~A}$$. The drift speed of the electrons is ___________ $$\\times 10^{-6} \\mathrm{ms}^{-1}$$
where $I$ is the current, $n$ is the number density of free electrons, $e$ is the charge of an electron, $A$ is the cross-sectional area of the wire, and $v$ is the drift speed of the electrons.
A current of $$2 \\mathrm{~A}$$ flows through a wire of cross-sectional area $$25.0 \\mathrm{~mm}^{2}$$. The number of free electrons in a cubic meter are $$2.0 \\times 10^{28}$$. The drift velocity of the electrons is __________ $$\\times 10^{-6} \\mathrm{~ms}^{-1}$$ (given, charge on electron $$=1.6 \\times 10^{-19} \\mathrm{C}$$ ).
The drift velocity $v_d$ can be found using the formula for current $I$ in a conductor:
\n
$I = nqAv_d$,
\n
where:
\n
\n
$n$ is the number density of free electrons (number of free electrons per unit volume),
\n
$q$ is the charge of an electron,
\n
$A$ is the cross-sectional area of the conductor, and
\n
$v_d$ is the drift velocity of the electrons.
\n
\n
We can rearrange the above formula to solve for $v_d$:
\n
$v_d = \\frac{I}{nqA}$.
\n
Given that $I = 2 \\, \\text{A}$, $n = 2.0 \\times 10^{28} \\, \\text{m}^{-3}$, $q = 1.6 \\times 10^{-19} \\, \\text{C}$, and $A = 25.0 \\, \\text{mm}^{2} = 25.0 \\times 10^{-6} \\, \\text{m}^{2}$, we can substitute these values into the formula to find $v_d$:
Therefore, the drift velocity of the electrons is $25 \\times 10^{-6} \\, \\text{ms}^{-1}$.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8589,
"subject": "Physics",
"question": "The current in a conductor is expressed as $I=3 t^2+4 t^3$, where $I$ is in Ampere and $t$ is in second. The amount of electric charge that flows through a section of the conductor during $t=1 \\mathrm{~s}$ to $t=2 \\mathrm{~s}$ is __________ C.",
"options": [],
"answer": "22",
"solution": "**Answer:** 22\n\n
To find the amount of electric charge that flows through a section of the conductor, we have to integrate the current over the given time interval. The current $I(t)$ as a function of time $t$ is given by:
\n\n$$ I=3t^2+4t^3 $$\n\n
The electric charge $Q$ that flows through the conductor from time $t = 1$ s to $t = 2$ s is calculated by integrating the current $I(t)$ with respect to time over this interval:
\n\n$$ Q = \\int_{t_1}^{t_2} I(t) \\, dt $$\n\n
Substituting the given limits ($t_1=1$ and $t_2=2$) and the expression for $I(t)$, we get:
The electric current through a wire varies with time as $$I=I_0+\\beta t$$, where $$I_0=20 \\mathrm{~A}$$ and $$\\beta=3 \\mathrm{~A} / \\mathrm{s}$$. The amount of electric charge crossed through a section of the wire in $$20 \\mathrm{~s}$$ is :
To calculate the amount of electric charge $Q$ that crosses through a section of the wire over a period of $20$ seconds, given the current varies with time as $I = I_0 + \\beta t$, where $I_0 = 20$ A (initial current) and $\\beta = 3$ A/s (rate of change of current with time), we use the concept of integration from calculus because the current is not constant but changes linearly with time.
\n\n
The electric charge $Q$ is the integral of current $I$ over the time interval from $0$ to $20$ seconds. The formula for $Q$ is given by:
\n\n
$Q = \\int_{0}^{T} I(t) \\, dt$
\n\n
Where $I(t) = I_0 + \\beta t$ and $T = 20$ s. Substituting the given values:
\n\n
$Q = \\int_{0}^{20} (20 + 3t) \\, dt$
\n\n
This integral can be solved in two parts:
\n\n
1. The integral of the constant term $20$:
\n\n
$\\int 20 \\, dt = 20t$
\n\n
2. The integral of the linear term $3t$:
\n\n
$\\int 3t \\, dt = \\frac{3}{2}t^2$
\n\n
Thus, combining these and evaluating from $0$ to $20$ seconds:
$$\\therefore$$ $${P_2} = {{{V^2}} \\over {R/4}} = 4\\left( {{{{V^2}} \\over R}} \\right) = 4{P_1}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8593,
"subject": "Physics",
"question": "A $$220$$ volt, $$1000$$ watt bulb is connected across a $$110$$ $$volt$$ mains supply. The power consumed will be ",
"options": [
{
"text": "$$750$$ watt "
},
{
"text": "$$500$$ watt "
},
{
"text": "$$250$$ watt "
},
{
"text": "$$1000$$ watt "
}
],
"answer": "$$250$$ watt ",
"solution": "**Answer:** $$250$$ watt \n\nWe know that $$R = {{V_{rated}^2} \\over {{P_{rated}}}} = {{{{\\left( {220} \\right)}^2}} \\over {1000}}$$\n
When this bulb is connected to $$110$$ volt mains supply we get\n
$$P = {{{V^2}} \\over R} = {{{{\\left( {110} \\right)}^2} \\times 1000} \\over {{{\\left( {220} \\right)}^2}}} = {{1000} \\over 4} = 250W$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8594,
"subject": "Physics",
"question": "The electrochemical equivalent of a metal is $${3.35109^{ - 7}}$$ $$kg$$ per Coulomb. The mass of the metal liberated at the cathode when a $$3A$$ current is passed for $$2$$ seconds will be ",
"options": [
{
"text": "$$6.6 \\times {10^{57}}/kg$$ "
},
{
"text": "$$9.9 \\times {10^{ - 7}}\\,kg$$ "
},
{
"text": "$$19.8 \\times {10^{ - 7}}\\,kg$$ "
},
{
"text": "$$1.1 \\times {10^{ - 7}}\\,kg$$ "
}
],
"answer": "$$19.8 \\times {10^{ - 7}}\\,kg$$ ",
"solution": "**Answer:** $$19.8 \\times {10^{ - 7}}\\,kg$$ \n\nThe mass liberated $$m,$$ electrochemical equivalent of a metal $$Z,$$ are related as $$m = Zit$$\n
$$ = 19.8 \\times {10^{ - 7}}\\,kg$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8595,
"subject": "Physics",
"question": "The thermo $$emf$$ of a thermocouple varies with temperature $$\\theta $$ of the hot junction as $$E = a\\theta + b{\\theta ^2}$$ in volts where the ratio $$a/b$$ is $${700^ \\circ }C.$$ If the cold junction is kept at $${0^ \\circ }C,$$ then the neutral temperature is ",
"options": [
{
"text": "$${1400^ \\circ }C$$ "
},
{
"text": "$${350^ \\circ }C$$"
},
{
"text": "$${700^ \\circ }C$$"
},
{
"text": "No neutral temperature is possible for this termocouple."
}
],
"answer": "No neutral temperature is possible for this termocouple.",
"solution": "**Answer:** No neutral temperature is possible for this termocouple.\n\nNeutral temperature is the temperature of a hot junction at which $$E$$ is maximum.\n
Neutral temperature can never be negative hence no $$\\theta $$ is possible.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8596,
"subject": "Physics",
"question": "A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be ",
"options": [
{
"text": "four times "
},
{
"text": "doubled "
},
{
"text": "halved "
},
{
"text": "one fourth "
}
],
"answer": "doubled ",
"solution": "**Answer:** doubled \n\n$$H = {{{V^2}t} \\over R}$$\n
Resistance of half the coil $$ = {R \\over 2}$$\n
$$\\therefore$$ As $$R$$ reduces to half, $$'H'$$ will be doubled.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8597,
"subject": "Physics",
"question": "The resistance of hot tungsten filament is about $$10$$ times the cold resistance. What will be resistance of $$100$$ $$W$$ and $$200$$ $$V$$ lamp when not in use ? ",
"options": [
{
"text": "$$20\\Omega $$ "
},
{
"text": "$$40\\Omega $$"
},
{
"text": "$$200\\Omega $$"
},
{
"text": "$$400\\Omega $$"
}
],
"answer": "$$40\\Omega $$",
"solution": "**Answer:** $$40\\Omega $$\n\n$$P = Vi = {{{V_2}} \\over R}$$\n
NOTE : We may use this expression as an approximation because the difference in the answers is appreciable. For accurate results one should use $$R = {R_0}{e^{\\alpha \\Delta T}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8600,
"subject": "Physics",
"question": "The resistance of a wire is $$5$$ ohm at $${50^ \\circ }C$$ and $$6$$ ohm at $${100^ \\circ }C.$$ The resistance of the wire at $${0^ \\circ }C$$ will be ",
"options": [
{
"text": "$$3$$ ohm"
},
{
"text": "$$2$$ ohm "
},
{
"text": "$$1$$ ohm "
},
{
"text": "$$4$$ ohm"
}
],
"answer": "$$4$$ ohm",
"solution": "**Answer:** $$4$$ ohm\n\nKEY CONCEPT : We know that\n
Thus the bulb marked $$25W$$-$$220$$ will fuse.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8603,
"subject": "Physics",
"question": "The supply voltage to room is $$120V.$$ The resistance of the lead wires is $$6\\Omega $$. A $$60$$ $$W$$ bulb is already switched on. What is the decrease of voltage across the bulb, when a $$240$$ $$W$$ heater is switched on in parallel to the bulb? ",
"options": [
{
"text": "zero "
},
{
"text": "$$2.9$$ Volt"
},
{
"text": "$$13.3$$ Volt"
},
{
"text": "$$10.04$$ Volt "
}
],
"answer": "$$10.04$$ Volt ",
"solution": "**Answer:** $$10.04$$ Volt \n\n\n
Power of bulb $$=60W$$ $$\\left( {given} \\right)$$\n
$${V_1} - {V_2} = 117.073 - 106.66 = 10.04$$ Volt (approximately)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8604,
"subject": "Physics",
"question": "In a large building, three are $$15$$ bulbs of $$40$$ $$W$$, $$5$$ bulbs of $$100$$ $$W$$, $$5$$ fans of $$80$$ $$W$$ and $$1$$ heater of $$1$$ $$kW.$$ The voltage of electric mains is $$220$$ $$V.$$ The minimum capacity of the main fuse of the building will be: ",
"options": [
{
"text": "$$8$$ $$A$$ "
},
{
"text": "$$10$$ $$A$$ "
},
{
"text": "$$12$$ $$A$$ "
},
{
"text": "$$14$$ $$A$$ "
}
],
"answer": "$$12$$ $$A$$ ",
"solution": "**Answer:** $$12$$ $$A$$ \n\nTotal power consumed by electrical appliances in the building, $${P_{total}} = 2500W$$\n
(Minimum capacity of main fuse)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8605,
"subject": "Physics",
"question": "The resistance of an electrical toaster has a temperature dependence given by R(T) = R0 [1 + $$\\alpha $$(T − T0)] in its range of operation. At T0 = 300 K, R = 100 $$\\Omega $$ and at T = 500 K, R = 120 $$\\Omega $$. The toaster is connected to a voltage source at 200 V and its temperature is raised at a constant rate\nfrom 300 to 500 K in 30 s. The total work done in raising the temperature is :",
"options": [
{
"text": "400 $$\\ln \\,{{1.5} \\over {1.3}}\\,J$$"
},
{
"text": "200 $$\\ln \\,{{2} \\over {3}}\\,J$$"
},
{
"text": "60000 $$\\ln \\,{{6} \\over {5}}\\,J$$"
},
{
"text": "300 J"
}
],
"answer": "60000 $$\\ln \\,{{6} \\over {5}}\\,J$$",
"solution": "**Answer:** 60000 $$\\ln \\,{{6} \\over {5}}\\,J$$\n\nGiven, \n
$$ = 60000\\ln \\left( {{6 \\over 5}} \\right)\\,J$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8606,
"subject": "Physics",
"question": "A constant voltages is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be : ",
"options": [
{
"text": "Doubled "
},
{
"text": "Halved"
},
{
"text": "Unchanged"
},
{
"text": "Increased 8 times "
}
],
"answer": "Increased 8 times ",
"solution": "**Answer:** Increased 8 times \n\n
Since rate of heat $$ = {{{V^2}} \\over R} \\Rightarrow $$ rate of heat $$ \\propto {1 \\over R}$$, where $$R = {{\\rho L} \\over A} = {{\\rho L} \\over {\\pi {r^2}}}$$; here r is radius of wire and L is length of wire. Therefore,
\n
$$R \\propto {L \\over {{r^2}}}$$.
\n
Thus, rate of heat $$ \\propto {{{r^2}} \\over L}$$.
\n
When length in halved and radius is doubled then, rate of heat $$ \\propto {{{{(2r)}^2}} \\over {L/2}} = {{8{r^2}} \\over L}$$
\n
Therefore, rate of heat developed in wire will be increased 8 times.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8607,
"subject": "Physics",
"question": "A heating element has a resistance of 100 $$\\Omega $$ at room temperature. When it is connected to a supply of 220 V, a steady current of 2 A passes in it and temperature is 500oC more than room temperature. what is the temperature coefficient of resistance of the heating element ? ",
"options": [
{
"text": "0.5 $$ \\times $$ 10$$-$$4oC$$-$$1"
},
{
"text": "5 $$ \\times $$ 10$$-$$4oC$$-$$1"
},
{
"text": "1 $$ \\times $$ 10$$-$$4oC$$-$$1"
},
{
"text": "2 $$ \\times $$ 10$$-$$4oC$$-$$1"
}
],
"answer": "2 $$ \\times $$ 10$$-$$4oC$$-$$1",
"solution": "**Answer:** 2 $$ \\times $$ 10$$-$$4oC$$-$$1\n\nWhen temperature increased by 500oC then, nrw registance \n
$$ \\Rightarrow $$ $$ \\propto $$ = $${{10} \\over {100 \\times 500}} = 2 \\times {10^{ - 4}}$$ oC$$-$$1 ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8608,
"subject": "Physics",
"question": "A 2 W carbon resistor is color coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is -",
"options": [
{
"text": "0.4 mA"
},
{
"text": "20 mA"
},
{
"text": "63 mA"
},
{
"text": "100 mA"
}
],
"answer": "20 mA",
"solution": "**Answer:** 20 mA\n\nP = i2R.\n
$$ \\therefore $$ for imax, R must be minimum\n
from color coding R = 50 $$ \\times $$ 102$$\\Omega $$\n
$$ \\therefore $$ imax = 20mA",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8609,
"subject": "Physics",
"question": "A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is -",
"options": [
{
"text": "11 $$ \\times $$ 10–5 W"
},
{
"text": "11 $$ \\times $$ 10–3 W"
},
{
"text": "11 $$ \\times $$ 105 W"
},
{
"text": "11 $$ \\times $$ 10–4 W"
}
],
"answer": "11 $$ \\times $$ 10–5 W",
"solution": "**Answer:** 11 $$ \\times $$ 10–5 W\n\nP = I2R\n
4.4 = 4 $$ \\times $$ 10$$-$$6 R\n
R = 1.1 $$ \\times $$ 106 $$\\Omega $$\n
P' = $${{{{11}^2}} \\over R}$$ = $${{{{11}^2}} \\over {1.1}}$$ $$ \\times $$ 10$$-$$6 = 11 $$ \\times $$ 10$$-$$5 W",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8610,
"subject": "Physics",
"question": "Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be : \n",
"options": [
{
"text": "240 W"
},
{
"text": "60 W"
},
{
"text": "30 W"
},
{
"text": "120 W"
}
],
"answer": "240 W",
"solution": "**Answer:** 240 W\n\nIn series condition, equivalent resistance is 2R \n
thus power consumed is 60W = $${{{\\varepsilon ^2}} \\over {2R}}$$\n
In parallel condition, equivalent resistance is R/2 thus new power is \n
P' = $${{{\\varepsilon ^2}} \\over {\\left( {R/2} \\right)}}$$\n
or P' = 4P = 240W",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 8611,
"subject": "Physics",
"question": "Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P1 and P2 respectively, then : \n",
"options": [
{
"text": "P1 = 4W, P2 = 16 W"
},
{
"text": "P1 = 16W, P2 = 4 W"
},
{
"text": "P1 = 9W, P2 = 16 W"
},
{
"text": "P1 = 16W, P2 = 9 W"
}
],
"answer": "P1 = 16W, P2 = 4 W",
"solution": "**Answer:** P1 = 16W, P2 = 4 W\n\n$${R_1} = {{{{220}^2}} \\over {25}}$$\n
$$ \\therefore $$ efficiency $$ = {{1000} \\over {1000 + {i^2}R}} \\times 100 = 96\\% $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8614,
"subject": "Physics",
"question": "In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters\nof 1 kW. The voltage of electric main is 220 V. The minimum fuse capacity (rated value) of the\nbuilding will be :",
"options": [
{
"text": "15 A"
},
{
"text": "20 A"
},
{
"text": "25 A"
},
{
"text": "10 A"
}
],
"answer": "20 A",
"solution": "**Answer:** 20 A\n\nTotal power is
$$ \\therefore $$ Current = $${{4325} \\over {220}}$$ = 19.66 A $$ \\simeq $$ 20 A",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8615,
"subject": "Physics",
"question": "A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is :",
"options": [
{
"text": "0.50 W"
},
{
"text": "0.072 W"
},
{
"text": "0.10 W"
},
{
"text": "0.125 W"
}
],
"answer": "0.10 W",
"solution": "**Answer:** 0.10 W\n\nPR = 0.5 W\n
$$ \\Rightarrow $$ i2R = 0.5 W\n
iR = 2.5\n
$$ \\Rightarrow $$ i = 0.2 A & R = 12.5 $$\\Omega $$\n
Also, V = E – ir\n
$$ \\Rightarrow $$ 2.5 = 3 – (0.2)r\n
$$ \\Rightarrow $$ r = 2.5 $$\\Omega $$\n
Power dissipated in internal resistance\n
= i2r = (0.2)2(2.5) = 0.1 W",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8616,
"subject": "Physics",
"question": "A resistor develops 500 J of thermal energy in 20 s when a current of 1.5A is passed through it. If the current is increased from 1.5A to 3A, what will be the energy developed in 20 s.",
"options": [
{
"text": "1000 J"
},
{
"text": "2000 J"
},
{
"text": "1500 J"
},
{
"text": "500 J"
}
],
"answer": "2000 J",
"solution": "**Answer:** 2000 J\n\n$${H_1} = i_1^2R\\Delta t$$
$$ \\Rightarrow {H_2} = 2000J$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8617,
"subject": "Physics",
"question": "The energy dissipated by a resistor is 10 mJ in 1 s when an electric current of 2 mA flows through it. The resistance is ___________$$\\Omega$$. (Round off to the Nearest Integer)",
"options": [],
"answer": "2500",
"solution": "**Answer:** 2500\n\n
Given, energy dissipated by a resistor, H = 10 mJ = 10 $$\\times$$ 10$$-$$3 J
\n
Time, t = 1 s
\n
Electric current, I = 2 mA = 2 $$\\times$$ 10$$-$$3 A
\n
Resistance, R = ?
\n
According to Joule's law of heating,
\n
H = I2Rt
\n
$$ \\Rightarrow R = {H \\over {{I^2}T}}$$ ....... (i)
$$ \\Rightarrow R = {{{{10}^{ - 2}}} \\over {4 \\times {{10}^{ - 6}}}} \\Rightarrow R = 0.25 \\times {10^4}$$
\n
$$ \\Rightarrow R = 2500\\,\\Omega $$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8618,
"subject": "Physics",
"question": "An electric bulb rated as 200 W at 100 V is used in a circuit having 200 V supply. The resistance 'R' that must be put in series with the bulb so that the bulb delivers the same power is _____________ $$\\Omega$$.",
"options": [],
"answer": "50",
"solution": "**Answer:** 50\n\nPower, $$P = {{{V^2}} \\over {{R_B}}}$$
To produce same power, same voltage (i.e. 100 V) should be across the bulb.
Hence, R = RB
R = 50 $$\\Omega$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8619,
"subject": "Physics",
"question": "The resistance of a conductor at 15$$^\\circ$$C is 16$$\\Omega$$ and at 100$$^\\circ$$C is 20$$\\Omega$$. What will be the temperature coefficient of resistance of the conductor?",
"options": [
{
"text": "0.010$$^\\circ$$C$$-$$1"
},
{
"text": "0.033$$^\\circ$$C$$-$$1"
},
{
"text": "0.003$$^\\circ$$C$$-$$1"
},
{
"text": "0.042$$^\\circ$$C$$-$$1"
}
],
"answer": "0.003$$^\\circ$$C$$-$$1",
"solution": "**Answer:** 0.003$$^\\circ$$C$$-$$1\n\n16 = R0 [1 + $$\\alpha$$ (15 $$-$$ T0)]
20 = R0 [1 + $$\\alpha$$ (100 $$-$$ T0)]
Assuming T0 = 0$$^\\circ$$C, as a general convention.
$$\\Rightarrow$$ $$\\alpha$$ = 0.003$$^\\circ$$C$$-$$1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8620,
"subject": "Physics",
"question": "An electric bulb of 500 watt at 100 volt is used in a circuit having a 200 V supply. Calculate the resistance R to be connected in series with the bulb so that the power delivered by the bulb is 500 W.",
"options": [
{
"text": "20 $$\\Omega$$"
},
{
"text": "30 $$\\Omega$$"
},
{
"text": "5 $$\\Omega$$"
},
{
"text": "10 $$\\Omega$$"
}
],
"answer": "20 $$\\Omega$$",
"solution": "**Answer:** 20 $$\\Omega$$\n\n500 watt at 100 v
P = Vi
500 = Vi
i = 5 Amp
V = i $$\\times$$ R
R = 20",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8621,
"subject": "Physics",
"question": "A resistor dissipates 192 J of energy in 1s when a current of 4A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5s in _________ J.",
"options": [],
"answer": "3840",
"solution": "**Answer:** 3840\n\nE = i2Rt
192 = 16 (R) (1)
R = 12$$\\Omega$$
E1 = (8)2 (12) (5)
= 3840 J",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8622,
"subject": "Physics",
"question": "Due to cold weather a 1 m water pipe of cross-sectional area 1 cm2 is filled with ice at $$-$$10$$^\\circ$$C. Resistive heating is used to melt the ice. Current of 0.5A is passed through 4 k$$\\Omega$$ resistance. Assuming that all the heat produced is used for melting, what is the minimum time required? (Given latent heat of fusion for water/ice = 3.33 $$\\times$$ 105 J kg$$-$$1, specific heat of ice = 2 $$\\times$$ 103 J kg$$-$$1 and density of ice = 103 kg/m3",
"options": [
{
"text": "0.353 s"
},
{
"text": "35.3 s"
},
{
"text": "3.53 s"
},
{
"text": "70.6 s"
}
],
"answer": "35.3 s",
"solution": "**Answer:** 35.3 s\n\n\nGiven, the length of the water pipe, L = 1 m
The cross-sectional area of the water pipe, A = 1 cm2 = 10$$-$$4 m2
The temperature of the ice = $$-$$ 10$$^\\circ$$C
Current passing in the conductor, I = 0.5 A
Resistance of the conductor, R = 4 k$$\\Omega$$
The latent heat of fusion for ice, Lf = 3.33 $$\\times$$ 105 J/kg
The density of the ice, d = 1000 kg/m3
The specific heat of the ice, cp, ice = 2 $$\\times$$ 103 J/kg
Heat required to melt the ice at 10$$^\\circ$$C to 0$$^\\circ$$C
Thus, the minimum time required to melt the ice is 35.3 s.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8623,
"subject": "Physics",
"question": "A uniform heating wire of resistance 36$$\\Omega$$ is connected across a potential difference of 240 V. The wire is then cut into half and potential difference of 240V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1 : x, where x is ____________",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nFor Case I,
The potential difference of the uniform wire, V = 240 V
The resistance of the uniform wire, R1 = 36 $$\\Omega$$
The value of the x = 4.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8624,
"subject": "Physics",
"question": "
Resistance of the wire is measured as 2 $$\\Omega$$ and 3 $$\\Omega$$ at 10$$^\\circ$$C and 30$$^\\circ$$C respectively. Temperature co-efficient of resistance of the material of the wire is :
A resistor develops 300 J of thermal energy in 15 s, when a current of 2 A is passed through it. If the current increases to 3 A, the energy developed in 10 s is ____________ J.
Statement I : A uniform wire of resistance $$80 \\,\\Omega$$ is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be $$5 \\,\\Omega$$.
\n
Statement II: Two resistances 2R and 3R are connected in parallel in a electric circuit. The value of thermal energy developed in 3R and 2R will be in the ratio $$3: 2$$.
\n
In the light of the above statements, choose the most appropriate answer from the option given below
",
"options": [
{
"text": "Both statement I and statement II are correct"
},
{
"text": "Both statement I and statement II are incorrect"
},
{
"text": "Statement I is correct but statement II is incorrect"
},
{
"text": "Statement I is incorrect but statement II is correct"
}
],
"answer": "Statement I is correct but statement II is incorrect",
"solution": "**Answer:** Statement I is correct but statement II is incorrect\n\n
An electrical bulb rated 220 V, 100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is 220 V, the power consumed by the 100 W bulb will be about _______ W.
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 8628,
"subject": "Physics",
"question": "The $\\mathrm{H}$ amount of thermal energy is developed by a resistor in $10 \\mathrm{~s}$ when a current of $4 \\mathrm{~A}$ is passed through it. If the current is increased to $16 \\mathrm{~A}$, the thermal energy developed by the resistor in $10 \\mathrm{~s}$ will be :",
"options": [
{
"text": "$\\frac{\\mathrm{H}}{4}$"
},
{
"text": "$16 \\mathrm{H}$"
},
{
"text": "H"
},
{
"text": "$4 \\mathrm{H}$"
}
],
"answer": "$16 \\mathrm{H}$",
"solution": "**Answer:** $16 \\mathrm{H}$\n\n$H \\propto i^{2}$ for $t=$ constant\n\n
A potential $$\\mathrm{V}_{0}$$ is applied across a uniform wire of resistance $$R$$. The power dissipation is $$P_{1}$$. The wire is then cut into two equal halves and a potential of $$V_{0}$$ is applied across the length of each half. The total power dissipation across two wires is $$P_{2}$$. The ratio $$P_{2}: \\mathrm{P}_{1}$$ is $$\\sqrt{x}: 1$$. The value of $$x$$ is ___________.
",
"options": [],
"answer": "16",
"solution": "**Answer:** 16\n\nLet's analyze the initial situation where the potential $$V_0$$ is applied across the entire length of the wire with resistance $$R$$.\n
\nThe power dissipation, $$P_1$$, can be calculated using the formula:\n
\n$$P_1 = \\frac{V_0^2}{R}$$\n
\nNow, let's consider the case where the wire is cut into two equal halves. Each half will have half the original resistance, $$\\frac{R}{2}$$. The potential $$V_0$$ is applied across the length of each half.\n
\nFor each half of the wire, the power dissipation, $$P'$$, can be calculated using the formula:\n
\nComparing this to the given ratio $$\\sqrt{x} : 1$$, we have:\n
\n$$\\frac{P_2}{P_1} = \\sqrt{x}$$\n
\nSo, $$\\sqrt{x} = 4$$.\n
\nSquaring both sides, we get:\n
\n$$x = 16$$\n
\nThe value of $$x$$ is 16.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8631,
"subject": "Physics",
"question": "
The current flowing through a conductor connected across a source is $$2 \\mathrm{~A}$$ and 1.2 $$\\mathrm{A}$$ at $$0^{\\circ} \\mathrm{C}$$ and $$100^{\\circ} \\mathrm{C}$$ respectively. The current flowing through the conductor at $$50^{\\circ} \\mathrm{C}$$ will be ___________ $$\\times 10^{2} \\mathrm{~mA}$$.
",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\nOur approach to this problem uses the fact that the voltage across the conductor remains constant as it is connected to the same source. By analyzing the relation between the currents and resistances at different temperatures, you can find the current flowing through the conductor at $$50^{\\circ} \\mathrm{C}$$.\n
\nFirst, you establish a relationship between the currents and resistances at $$0^{\\circ} \\mathrm{C}$$ and $$100^{\\circ} \\mathrm{C}$$:\n
\n$$i_0 R_0 = i_{100} R_{100}$$\n
\nPlugging in the given values for $$i_0$$ and $$i_{100}$$:\n
\nNow, you need to find the current $$i_{50}$$ at $$50^{\\circ} \\mathrm{C}$$. To do this, you calculate the resistance $$R_{50}$$ using the found value of $$\\alpha$$:\n
\nThus, the current flowing through the conductor at $$50^{\\circ} \\mathrm{C}$$ is $$15 \\times 10^2 \\mathrm{~mA}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8632,
"subject": "Physics",
"question": "
Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be:
",
"options": [
{
"text": "4 : 1"
},
{
"text": "1 : 4"
},
{
"text": "2 : 1"
},
{
"text": "1 : 2"
}
],
"answer": "4 : 1",
"solution": "**Answer:** 4 : 1\n\nLet's consider the power $$P$$ dissipated by a single heater filament with resistance $$R$$ when connected to a voltage $$V$$.\n
\nThe power dissipated by the heater filament is given by:\n
\n$$P = \\frac{V^2}{R}$$\n
\nWhen the two identical heater filaments are connected in parallel, the equivalent resistance $$R_P$$ is given by:\n
When a potential difference $$V$$ is applied across a wire of resistance $$R$$, it dissipates energy at a rate $$W$$. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the energy dissipation rate will become:
An electric toaster has resistance of $$60 \\Omega$$ at room temperature $$\\left(27^{\\circ} \\mathrm{C}\\right)$$. The toaster is connected to a $$220 \\mathrm{~V}$$ supply. If the current flowing through it reaches $$2.75 \\mathrm{~A}$$, the temperature attained by toaster is around : ( if $$\\alpha=2 \\times 10^{-4}$$/$$^\\circ \\mathrm{C}$$)
To find the power dissipation of the bulb when it's connected to a $$100 \\mathrm{V}$$ supply instead of its rated $$200 \\mathrm{V}$$ supply, we can use the relation between power (P), voltage (V), and resistance (R), which is given by $$P = \\frac{V^2}{R}$$. The resistance of the bulb can be considered constant in this case, allowing us to calculate the change in power dissipation due to the change in voltage.
\n\n
First, let's find the resistance of the bulb based on its rated conditions:
\n\n
$$P = \\frac{V^2}{R} \\Rightarrow R = \\frac{V^2}{P}$$
Water boils in an electric kettle in 20 minutes after being switched on. Using the same main supply, the length of the heating element should be _________ to __________ times of its initial length if the water is to be boiled in 15 minutes.
When an electric kettle is used to heat water, the time taken to boil the water depends on the power of the heating element. The power supplied to the heating element is inversely proportional to the heating time. If we want to reduce the boiling time, the power needs to be increased. The power of the heating element is given by:
\n\n
$$P = \\frac{V^2}{R}$$
\n\n
where $$P$$ is the power, $$V$$ is the voltage, and $$R$$ is the resistance of the heating element. The resistance $$R$$ of the heating element is proportional to its length $$L$$ while the material and cross-sectional area remain constant.
\n\n
So, we can write:
\n\n
$$R \\propto L$$
\n\n
To achieve boiling in 15 minutes instead of 20 minutes, the power needs to increase, which implies the resistance must decrease. Let the initial length of the heating element be $$L_0$$, and let the new length needed be $$L$$. The time taken to heat is inversely proportional to the power:
\n\n
$$\\frac{T_1}{T_2} = \\frac{P_2}{P_1}$$
\n\n
Given that:
\n\n
$$T_1 = 20 \\text{ minutes}$$
\n\n
$$T_2 = 15 \\text{ minutes}$$
\n\n
We need to find the ratio:
\n\n
$$\\frac{20}{15} = \\frac{P_2}{P_1}$$
\n\n
Simplifying:
\n\n
$$\\frac{4}{3} = \\frac{P_2}{P_1}$$
\n\n
The power is inversely proportional to the resistance:
\n\n
$$\\frac{P_2}{P_1} = \\frac{R_1}{R_2}$$
\n\n
Thus:
\n\n
$$\\frac{4}{3} = \\frac{R_1}{R_2}$$
\n\n
Since resistance is proportional to length:
\n\n
$$\\frac{R_1}{R_2} = \\frac{L_1}{L_2}$$
\n\n
Hence:
\n\n
$$\\frac{4}{3} = \\frac{L_1}{L_2}$$
\n\n
Simplifying, we find:
\n\n
$$L_2 = \\frac{3}{4} L_1$$
\n\n
This means the length of the heating element should be decreased to $$\\frac{3}{4}$$ of its initial length.
The number of electrons flowing per second in the filament of a $$110 \\mathrm{~W}$$ bulb operating at $$220 \\mathrm{~V}$$ is : (Given $$\\mathrm{e}=1.6 \\times 10^{-19} \\mathrm{C}$$)
$$\\begin{aligned}\n& P=v \\times i \\Rightarrow i=\\frac{110}{220}=\\frac{1}{2} \\mathrm{~A} \\\\\n& i=n e \\Rightarrow n=\\frac{1}{2 \\times 1.6 \\times 10^{-19}}=31.25 \\times 10^{17}\n\\end{aligned}$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8639,
"subject": "Physics",
"question": "If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter a ",
"options": [
{
"text": "low resistance in parallel "
},
{
"text": "high resistance in parallel "
},
{
"text": "high resistance in series "
},
{
"text": "low resistance in series "
}
],
"answer": "high resistance in series ",
"solution": "**Answer:** high resistance in series \n\nKEY CONCEPT : To convert a galvanometer into a voltmeter we connect a high resistance in series with the galvanometer.\n
The same procedure needs to be done if ammeter is to be used as a voltmeter.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8640,
"subject": "Physics",
"question": "The thermo $$e.m.f.$$ of a thermo -couple is $$25$$ $$\\mu V/{}^ \\circ C$$ at room temperature. A galvanometer of $$40$$ $$ohm$$ resistance, capable of detecting current as low as $${10^{ - 5}}\\,A,$$ is connected with the thermo couple. The smallest temperature difference that can be detected by this system is ",
"options": [
{
"text": "$${16^0}C$$ "
},
{
"text": "$${12^0}C$$"
},
{
"text": "$${8^0}C$$"
},
{
"text": "$${20^0}C$$"
}
],
"answer": "$${16^0}C$$ ",
"solution": "**Answer:** $${16^0}C$$ \n\nLet $$\\theta $$ be the smallest temperature difference that can be detected by the thermocouple, then\n
$$\\therefore$$ $$\\theta = {16^ \\circ }C.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8641,
"subject": "Physics",
"question": "An ammeter reads upto $$1$$ ampere. Its internal resistance is $$0.81$$ $$ohm$$. To increase the range to $$10$$ $$A$$ the value of the required shunt is ",
"options": [
{
"text": "$$0.03\\,\\Omega $$ "
},
{
"text": "$$0.3\\,\\Omega $$ "
},
{
"text": "$$0.9\\,\\Omega $$"
},
{
"text": "$$0.09\\,\\Omega $$ "
}
],
"answer": "$$0.09\\,\\Omega $$ ",
"solution": "**Answer:** $$0.09\\,\\Omega $$ \n\n$${i_g} \\times G = \\left( {i - {i_g}} \\right)S$$\n
$$\\therefore$$ $$S = {{{i_g} \\times G} \\over {i - {i_g}}} = {{1 \\times 0.81} \\over {10 - 1}} = 0.09\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8642,
"subject": "Physics",
"question": "A moving coil galvanometer has $$150$$ equal divisions. Its current sensitivity is $$10$$- divisions per milliampere and voltage sensitivity is $$2$$ divisions per millivolt. In order that each division reads $$1$$ volt, the resistance in $$ohms$$ needed to be connected in series with the coil will be - ",
"options": [
{
"text": "$${10^5}$$ "
},
{
"text": "$${10^3}$$"
},
{
"text": "$$9995$$ "
},
{
"text": "$$99995$$ "
}
],
"answer": "$$9995$$ ",
"solution": "**Answer:** $$9995$$ \n\nKEY CONCEPT : Resistance of Galvanometer,\n
From equations $$(i)$$ and $$(iii),$$ $${q_2} = {q \\over {1 + {{{Z_2}} \\over {{z_1}}}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8644,
"subject": "Physics",
"question": "This questions has Statement - $${\\rm I}$$ and Statement - $${\\rm I}$$$${\\rm I}$$. Of the four choices given after the Statements, choose the one that best describes into two Statements.\n
Statement - $${\\rm I}$$ : Higher the range, greater is the resistance of ammeter. \n Statement - $${\\rm I}$$$${\\rm I}$$ : To increase the range of ammeter, additional shunt needs to be used across it.
",
"options": [
{
"text": "Statement - $${\\rm I}$$ is true, Statement - $${\\rm II}$$ is true, Statement - $${\\rm II}$$ is the correct explanation of statement - $${\\rm I}$$."
},
{
"text": "Statement - $${\\rm I}$$ is true, Statement - $${\\rm II}$$ is true, Statement - $${\\rm II}$$ is not the correct explanation of statement - $${\\rm I}$$."
},
{
"text": "Statement - $${\\rm I}$$ is true, Statement - $${\\rm II}$$ is false"
},
{
"text": "Statement - $${\\rm I}$$ is false, Statement - $${\\rm II}$$ is true"
}
],
"answer": "Statement - $${\\rm I}$$ is false, Statement - $${\\rm II}$$ is true",
"solution": "**Answer:** Statement - $${\\rm I}$$ is false, Statement - $${\\rm II}$$ is true\n\nStatements $${\\rm I}$$ is false and Statement $${\\rm I}$$$${\\rm I}$$ is true\n
Therefore for $${\\rm I}$$ to increase, $$S$$ should decrease, So additional $$S$$ can be connected across it.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8645,
"subject": "Physics",
"question": "A 50 $$\\Omega $$ resistance is connected to a battery of 5 V. A galvanometer of resistance 100 $$\\Omega $$ is to be used as an ammeter to measure current through the resistance, for this a resistance rs is connected to the galvanometer. Which of the following connections should be employed if the measured current is within 1% of thecurrent without the ammeter in the circuit ?",
"options": [
{
"text": "rs = 0.5 $$\\Omega $$ in parallel with the galvanometer"
},
{
"text": "rs = 0.5 $$\\Omega $$ in series with the galvanometer"
},
{
"text": "rs = 1 $$\\Omega $$ in series with galvanometer"
},
{
"text": "rs =1 $$\\Omega $$ in parallel with galvanometer"
}
],
"answer": "rs = 0.5 $$\\Omega $$ in parallel with the galvanometer",
"solution": "**Answer:** rs = 0.5 $$\\Omega $$ in parallel with the galvanometer\n\n\n
Current in the circuit without ammeter \n
I $$=$$ $${5 \\over {50}} = 0.1$$ A\n
$$ \\therefore $$ With ammeter current $$=$$ 0.1 $$ \\times $$ $${{99} \\over {100}}$$ = 0.099 A\n
$$ \\Rightarrow $$ rs $$=$$ $${{50} \\over {99.5}} = 0.5\\,\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8646,
"subject": "Physics",
"question": "A galvanometer having a coil resistance of $$100\\,\\Omega $$ gives a full scale deflection, when a currect of $$1$$ $$mA$$ is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of $$10$$ $$A,$$ is :",
"options": [
{
"text": "$$0.1\\,\\Omega $$ "
},
{
"text": "$$3\\,\\Omega $$ "
},
{
"text": "$$0.01\\,\\Omega $$"
},
{
"text": "$$2\\,\\Omega $$"
}
],
"answer": "$$0.01\\,\\Omega $$",
"solution": "**Answer:** $$0.01\\,\\Omega $$\n\n$${\\rm I}gG = \\left( {{\\rm I} - {\\rm I}g} \\right)s$$\n
$$\\therefore$$ $$S \\approx 0.01\\,\\,\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8647,
"subject": "Physics",
"question": "When a current of 5 mA is passed through a galvanometer having a coil of resistance 15$$\\Omega $$, it shows full\nscale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a\nvoltmeter of range 0 – 10V is:",
"options": [
{
"text": "4.005 × 103 $$\\Omega $$"
},
{
"text": "1.985 × 103 $$\\Omega $$"
},
{
"text": "2.535 × 103 $$\\Omega $$"
},
{
"text": "2.045 × 103 $$\\Omega $$"
}
],
"answer": "1.985 × 103 $$\\Omega $$",
"solution": "**Answer:** 1.985 × 103 $$\\Omega $$\n\nGiven : Current through the galvanometer,\n
ig = 5 × 10–3 A\n
Galvanometer resistance, G = 15 $$\\Omega $$\n
Let resistance R to be put in series with the galvanometer to\nconvert it into a voltmeter.\n
V = ig (R + G)\n
10 = 5 × 10–3 (R + 15)\n
$$ \\therefore $$ R = 2000 – 15 = 1985\n
= 1.985 × 103 $$\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8648,
"subject": "Physics",
"question": "In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 k$$\\Omega $$ are used. The figure of merit of the galvanometer is 60 $$\\mu A/$$division. In the absence of shunt resistance, the galvanometer produces a deflection of $$\\theta $$ = 9 divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of $$\\theta /2,$$ is closest to :",
"options": [
{
"text": "500 $$\\Omega $$"
},
{
"text": "220 $$\\Omega $$"
},
{
"text": "55 $$\\Omega $$"
},
{
"text": "110 $$\\Omega $$"
}
],
"answer": "110 $$\\Omega $$",
"solution": "**Answer:** 110 $$\\Omega $$\n\n
Current required by unit deflection is 60 $$\\mu$$A.
\n
For, $$\\theta$$ = 9 current is I = 9 $$\\times$$ 60 $$\\mu$$A
\n
$$\\Rightarrow$$ I = 540 $$\\mu$$A = 540 $$\\times$$ 10$$-$$6 A
100 S = 11000 $$\\Rightarrow$$ S = 110 $$\\Omega$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8649,
"subject": "Physics",
"question": "A galvanometer with its coil resistance 25 $$\\Omega $$ requires a current of 1 mA for its full deflection. In order to construct an ammeter to read upto a current of 2 A, the approximate value of the shunt resistance should be : ",
"options": [
{
"text": "$$2.5 \\times {10^{ - 3}}\\,\\Omega $$ "
},
{
"text": "$$1.25 \\times {10^{ - 2}}\\Omega $$"
},
{
"text": "$$1.25 \\times {10^{ - 3}}\\Omega $$"
},
{
"text": "$$2.5 \\times {10^{ - 2}}\\Omega $$"
}
],
"answer": "$$1.25 \\times {10^{ - 2}}\\Omega $$",
"solution": "**Answer:** $$1.25 \\times {10^{ - 2}}\\Omega $$\n\n\n
$$ \\Rightarrow $$ $$\\,\\,\\,$$ S = 1.25 $$ \\times $$ 10$$-$$2 $$\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8650,
"subject": "Physics",
"question": "A moving coil galvanometer has resistance 50$$\\Omega $$\nand it indicates full deflection at 4mA current.\nA voltmeter is made using this galvanometer\nand a 5 k$$\\Omega $$ resistance. The maximum voltage,\nthat can be measured using this voltmeter, will\nbe close to :",
"options": [
{
"text": "15 V"
},
{
"text": "10 V"
},
{
"text": "40 V"
},
{
"text": "20 V"
}
],
"answer": "20 V",
"solution": "**Answer:** 20 V\n\nG = 50 $$\\Omega $$
\nS = 5000 $$\\Omega $$
\nIg = 4 × 10–3
\nV = ig (G + S)
\nV = 4 × 10–3 (50 + 5000)
\n= 4 × 10–3 (5050) = 20.2 volt",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8651,
"subject": "Physics",
"question": "A moving coil galvanometer, having a resistance G, produces full scale deflection when a current Ig flows\nthrough it. This galvanometer can be converted into (i) an ammeter of range 0 to I0(I0 > Ig) by connecting a\nshunt resistance RA to it and (ii) into a voltmeter of range 0 to V (V = GI0) by connecting a series resistance\nRV to it. Then,",
"options": [
{
"text": "$${R_A}{R_V} = {G^2}$$ and $${{{R_A}} \\over {{R_V}}} = {{{I_g}} \\over {\\left( {{I_0} - {I_g}} \\right)}}$$"
},
{
"text": "$${R_A}{R_V} = {G^2}\\left( {{{{I_g}} \\over {{I_0} - {I_g}}}} \\right)$$ and $${{{R_A}} \\over {{R_V}}} = {\\left( {{{{I_0} - {I_g}} \\over {{I_g}}}} \\right)^2}$$"
},
{
"text": "$${R_A}{R_V} = {G^2}\\left( {{{{I_0} - {I_g}} \\over {{I_g}}}} \\right)$$ and $${{{R_A}} \\over {{R_V}}} = {\\left( {{{{I_g}} \\over {{I_0} - {I_g}}}} \\right)^2}$$"
},
{
"text": "$${R_A}{R_V} = {G^2}$$ and $${{{R_A}} \\over {{R_V}}} = {\\left( {{{{I_g}} \\over {{I_0} - {I_g}}}} \\right)^2}$$"
}
],
"answer": "$${R_A}{R_V} = {G^2}$$ and $${{{R_A}} \\over {{R_V}}} = {\\left( {{{{I_g}} \\over {{I_0} - {I_g}}}} \\right)^2}$$",
"solution": "**Answer:** $${R_A}{R_V} = {G^2}$$ and $${{{R_A}} \\over {{R_V}}} = {\\left( {{{{I_g}} \\over {{I_0} - {I_g}}}} \\right)^2}$$\n\nGalvanometer is converted into ammeter of range 0 to I0 :\n\n (Io - Ig)RA = IgG\n $$ \\Rightarrow $$ $${R_A} = {{{I_g}G} \\over {\\left( {{I_0} - {I_g}} \\right)}}$$............(1)\n
Galvanometer is converted into voltmeter of range 0 to V :\n\n V = Ig(RV + G)\n Also given, V = GI0\n $$ \\therefore $$ GI0 = Ig(RV + G)\n $$ \\Rightarrow $$ $${R_V} = {{G\\left( {{I_0} - {I_g}} \\right)} \\over {{I_g}}}$$ .........(2)\n
From equation (1) and (2)\n $${R_A}{R_V} = {G^2}$$\n $${{{R_A}} \\over {{R_V}}} = {\\left( {{{{I_g}} \\over {{I_0} - {I_g}}}} \\right)^2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8652,
"subject": "Physics",
"question": "A moving coil galvanometer allows a full scale\ncurrent of 10–4 A. A series resistance of 2 M$$\\Omega $$\nis required to convert the above galvanometer\ninto a voltmeter of range 0-5 V. Therefore the\nvalue of shunt resistance required to convert the\nabove galvanometer into an ammeter of range\n0.10 mA is :",
"options": [
{
"text": "200 $$\\Omega $$"
},
{
"text": "500 $$\\Omega $$"
},
{
"text": "100 $$\\Omega $$"
},
{
"text": "None of the options are correct"
}
],
"answer": "None of the options are correct",
"solution": "**Answer:** None of the options are correct\n\nGiven data,\n
$\\therefore$ No option is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8653,
"subject": "Physics",
"question": "The resistance of a galvanometer is 50 ohm and\nthe maximum current which can be passed\nthrough it is 0.002 A. What resistance must be\nconnected to it in order to convert it into an\nammeter of range 0 – 0.5 A ?",
"options": [
{
"text": "0.02 ohm"
},
{
"text": "0.2 ohm"
},
{
"text": "0.002 ohm"
},
{
"text": "0.5 ohm"
}
],
"answer": "0.2 ohm",
"solution": "**Answer:** 0.2 ohm\n\n
\nWe have \nIg Rg = (0.5 – Ig) S
\n$$ \\Rightarrow $$ (0.002) (50) = (0.5 – Ig) S
\n$$ \\Rightarrow S \\simeq {{0.002 \\times 50} \\over {0.5}} = 0.2\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8654,
"subject": "Physics",
"question": "A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 $$ \\times $$ 10–4 A passes through it, its needle ( pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of : ",
"options": [
{
"text": "200 ohm"
},
{
"text": "250 ohm"
},
{
"text": "6200 ohm"
},
{
"text": "6250 ohm"
}
],
"answer": "200 ohm",
"solution": "**Answer:** 200 ohm\n\nIg = 4 $$ \\times $$ 10$$-$$4 $$ \\times $$ 25 = 10$$-$$2 A\n
\n
2.5 = (50 + R) 10$$-$$2\n
$$ \\therefore $$ R = 200 $$\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8655,
"subject": "Physics",
"question": "A galvanometer having a resistance of 20 $$\\Omega $$ and 30 divisions on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt, is:",
"options": [
{
"text": "120 $$\\Omega $$"
},
{
"text": "125 $$\\Omega $$"
},
{
"text": "80 $$\\Omega $$"
},
{
"text": "100 $$\\Omega $$"
}
],
"answer": "80 $$\\Omega $$",
"solution": "**Answer:** 80 $$\\Omega $$\n\nRg = 20$$\\Omega $$\n
R = 80 ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8656,
"subject": "Physics",
"question": "A circuit to verify Ohm's law uses ammeter and voltmeter in series or parallel connected correctly\nto the resistor. In the circuit :",
"options": [
{
"text": "Ammeter is always connected in series and voltmeter in parallel"
},
{
"text": "Both ammeter and voltmeter must be connected in series"
},
{
"text": "Both ammeter and voltmeter must be connected in parallel"
},
{
"text": "Ammeter is always used in parallel and voltmeter is series"
}
],
"answer": "Ammeter is always connected in series and voltmeter in parallel",
"solution": "**Answer:** Ammeter is always connected in series and voltmeter in parallel\n\nAmmeter is always connected in series and\nvoltmeter is connected in parallel.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8657,
"subject": "Physics",
"question": "A galvanometer is used in laboratory for\ndetecting the null point in electrical\nexperiments. If, on passing a current of 6 mA it\nproduces a deflection of 2o, its figure of merit\nis close to :",
"options": [
{
"text": "6 $$ \\times $$ 10–3 A/div"
},
{
"text": "666o A/div"
},
{
"text": "3 $$ \\times $$ 10–3 A/div"
},
{
"text": "333o A/div"
}
],
"answer": "3 $$ \\times $$ 10–3 A/div",
"solution": "**Answer:** 3 $$ \\times $$ 10–3 A/div\n\nfigure of merit = $${I \\over \\theta }$$\n
= $${{6 \\times {{10}^{ - 3}}} \\over 2}$$\n
= 3 $$ \\times $$ 10–3 A/div",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8658,
"subject": "Physics",
"question": "Two resistors 400$$\\Omega $$ and 800$$\\Omega $$ are connected in series across a 6 V battery. The potential difference measured by a voltmeter of 10 k$$\\Omega $$ across 400 $$\\Omega $$ resistor is close to :\n",
"options": [
{
"text": "2.05 V"
},
{
"text": "1.95 V"
},
{
"text": "2 V"
},
{
"text": "1.8 V"
}
],
"answer": "1.95 V",
"solution": "**Answer:** 1.95 V\n\n\n
= 1.95 V",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8659,
"subject": "Physics",
"question": "A galvanometer of resistance G is converted\ninto a voltmeter of range 0 – 1 V by connecting\na resistance R1 in series with it. The additional\nresistance that should be connected in series\nwith R1 to increase the range of the voltmeter\nto 0 – 2 V will be :",
"options": [
{
"text": "G"
},
{
"text": "R1"
},
{
"text": "R1 + G"
},
{
"text": "R1 - G"
}
],
"answer": "R1 + G",
"solution": "**Answer:** R1 + G\n\n1 = ig(G + R1) ....(1)\n
$$ \\Rightarrow $$ R2\n = R1\n + G",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8660,
"subject": "Physics",
"question": "A galvanometer having a coil resistance\n100 $$\\Omega $$ gives a full scale deflection when a\ncurrent of 1 mA is passed through it. What is\nthe value of the resistance which can convert\nthis galvanometer into a voltmeter giving full\nscale deflection for a potential difference of\n10 V?",
"options": [
{
"text": "8.9 k$$\\Omega $$"
},
{
"text": "10 k$$\\Omega $$"
},
{
"text": "9.9 k$$\\Omega $$"
},
{
"text": "7.9 k$$\\Omega $$"
}
],
"answer": "9.9 k$$\\Omega $$",
"solution": "**Answer:** 9.9 k$$\\Omega $$\n\nig = 1 mA , Rg = 100 $$\\Omega $$\n
V = ig(R + Rg)\n
$$ \\Rightarrow $$10 = 1 × 10–3 (R + 100)\n
$$ \\Rightarrow $$R = 9.9 k$$\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8661,
"subject": "Physics",
"question": "For full scale deflection of total 50 divisions, 50 mV voltage is required in galvanometer. The resistance of galvanometer if its current sensitivity is 2 div/mA will be :",
"options": [
{
"text": "1$$\\Omega$$"
},
{
"text": "5$$\\Omega$$"
},
{
"text": "4$$\\Omega$$"
},
{
"text": "2$$\\Omega$$"
}
],
"answer": "2$$\\Omega$$",
"solution": "**Answer:** 2$$\\Omega$$\n\n$${I_{\\max }} = {{50} \\over 2} = 25$$ mA
$$R = {V \\over I} = {{50mV} \\over {25mA}} = 2\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8662,
"subject": "Physics",
"question": "Consider a galvanometer shunted with 5$$\\Omega$$ resistance and 2% of current passes through it. What is the resistance of the given galvanometer ?",
"options": [
{
"text": "300 $$\\Omega$$"
},
{
"text": "344 $$\\Omega$$"
},
{
"text": "245 $$\\Omega$$"
},
{
"text": "226 $$\\Omega$$"
}
],
"answer": "245 $$\\Omega$$",
"solution": "**Answer:** 245 $$\\Omega$$\n\n 0.02i Rg = 0.98i $$\\times$$ 5
A 72 $$\\Omega$$ galvanometer is shunted by a resistance of 8 $$\\Omega$$. The percentage of the total current which passes through the galvanometer is :
If n represents the actual number of deflections in a converted galvanometer of resistance G and shunt resistance S. Then the total current I when its figure of merit is K will be:
A teacher in his physics laboratory allotted an experiment to determine the resistance (G) of a galvanometer. Students took the observations for $${1 \\over 3}$$ deflection in the galvanometer. Which of the below is true for measuring value of G?
",
"options": [
{
"text": "$${1 \\over 3}$$ deflection method cannot be used for determining the resistance of the galvanometer."
},
{
"text": "$${1 \\over 3}$$ deflection method can be used and in this case the G equals to twice the value of shunt resistances."
},
{
"text": "$${1 \\over 3}$$ deflection method can be used and in this case, the G equals to three times the value of shunt resistances."
},
{
"text": "$${1 \\over 3}$$ deflection method can be used and in this case the G value equals to the shunt resistances."
}
],
"answer": "$${1 \\over 3}$$ deflection method can be used and in this case the G equals to twice the value of shunt resistances.",
"solution": "**Answer:** $${1 \\over 3}$$ deflection method can be used and in this case the G equals to twice the value of shunt resistances.\n\n
The circuit for the given situation is:
\n
\n
Since G and S are in parallel,
\n
$$\\Rightarrow$$ $${i \\over 3}$$ $$\\times$$ G = $${2i \\over 3}$$ $$\\times$$ S
\n
$$\\Rightarrow$$ G = 2S
\n
$$\\Rightarrow$$ G equals twice the value of shunt resistance.
Two identical cells each of emf 1.5 V are connected in parallel across a parallel combination of two resistors each of resistance 20 $$\\Omega$$. A voltmeter connected in the circuit measures 1.2 V. The internal resistance of each cell is :
Given below are two statements : One is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$.
\n
Assertion A : For measuring the potential difference across a resistance of $$600 \\Omega$$, the voltmeter with resistance $$1000 \\Omega$$ will be preferred over voltmeter with resistance $$4000 \\Omega$$.
\n
Reason R : Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.
\n
In the light of the above statements, choose the most appropriate answer from the options given below.
",
"options": [
{
"text": "$$\\mathbf{A}$$ is not correct but $$\\mathbf{R}$$ is correct"
},
{
"text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$"
},
{
"text": "$$\\mathbf{A}$$ is correct but $$\\mathbf{R}$$ is not correct"
},
{
"text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct but $$\\mathbf{R}$$ is not the correct explanation of $$\\mathbf{A}$$"
}
],
"answer": "$$\\mathbf{A}$$ is not correct but $$\\mathbf{R}$$ is correct",
"solution": "**Answer:** $$\\mathbf{A}$$ is not correct but $$\\mathbf{R}$$ is correct\n\n
Assertion A is incorrect because the preferred voltmeter for measuring the potential difference across a resistance of 600 ohm is actually the voltmeter with a higher resistance, not a lower resistance. The reason for this is that when a voltmeter is connected in parallel with the resistance being measured, it will draw current away from the resistance, reducing the potential difference across it. A higher resistance voltmeter will draw less current and therefore have a smaller effect on the potential difference being measured.
\n\n
In this case, the voltmeter with resistance 4000 ohm would be preferred because it would draw less current than the voltmeter with resistance 1000 ohm, leading to a more accurate measurement of the potential difference across the resistance of 600 ohm.
\n\n
Therefore, assertion A is incorrect because it states that the voltmeter with resistance 1000 ohm is preferred, when in reality the voltmeter with resistance 4000 ohm is preferred.
\n\n\n
Reason R states that a voltmeter with higher resistance will draw smaller current than a voltmeter with lower resistance. This is correct because Ohm's law states that the current through a resistor is proportional to the voltage across it and inversely proportional to the resistance. Thus, a voltmeter with higher resistance will draw less current, making it a better choice for measuring the potential difference across another resistance.
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8668,
"subject": "Physics",
"question": "The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by $50 \\%$. The percentage change in voltage sensitivity of the galvanometer will be :",
"options": [
{
"text": "$0 \\%$"
},
{
"text": "$75 \\%$"
},
{
"text": "$100 \\%$"
},
{
"text": "$50 \\%$"
}
],
"answer": "$0 \\%$",
"solution": "**Answer:** $0 \\%$\n\nCurrent sensitivity $=$ Voltage sensitivity $\\times R$\n\n
Current sensitivity is made $1.5$ times.\n\n
$R$ also increase $1.5$ times.\n\n
Hence voltage sensitivity $=\\frac{1.5 \\times \\text { current sensitivity }}{1.5 \\times R}$\n\n
A cell of emf 90 V is connected across series combination of two resistors each of 100$$\\Omega$$ resistance. A voltmeter of resistance 400$$\\Omega$$ is used to measure the potential difference across each resistor. The reading of the voltmeter will be :
When a resistance of $$5 ~\\Omega$$ is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of $$250 \\mathrm{~mA}$$, however when $$1050 ~\\Omega$$ resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is ____________ $$\\Omega$$.
\nSolving for the resistance of the galvanometer, $$R_G$$:\n
\n$$95 R_G = 4750$$\n
\n$$R_G = 50 \\ \\Omega$$\n
\nSo, the resistance of the galvanometer is $$50 \\ \\Omega$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8671,
"subject": "Physics",
"question": "
Two identical cells each of emf $$1.5 \\mathrm{~V}$$ are connected in series across a $$10 ~\\Omega$$ resistance. An ideal voltmeter connected across $$10 ~\\Omega$$ resistance reads $$1.5 \\mathrm{~V}$$. The internal resistance of each cell is __________ $$\\Omega$$.
",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nLet the internal resistance of each cell be $$r$$.\n\n
Since the two cells are connected in series, their internal resistances add up, and the total internal resistance of the series combination is $$2r$$. The total EMF of the series combination of cells is $$1.5\\,\\text{V} + 1.5\\,\\text{V} = 3\\,\\text{V}$$. \n\n
Let's use Kirchhoff's Voltage Law (KVL) for the closed loop in the circuit:\n
$$\\text{EMF}_{total} - I(R + 2r) = 0$$\n\n
We are given that the voltage across the $$10\\,\\Omega$$ resistor, as measured by the ideal voltmeter, is $$1.5\\,\\text{V}$$. According to Ohm's law, the current in the circuit can be determined as:\n
So, the internal resistance of each cell is $$5\\,\\Omega$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8672,
"subject": "Physics",
"question": "
The current sensitivity of moving coil galvanometer is increased by $$25 \\%$$. This increase is achieved only by changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be:
",
"options": [
{
"text": "+25%"
},
{
"text": "$$-$$50%"
},
{
"text": "$$-$$25%"
},
{
"text": "Zero"
}
],
"answer": "+25%",
"solution": "**Answer:** +25%\n\nThe current sensitivity ($I_s$) of a moving coil galvanometer is given by:\n
\n$$I_s = \\frac{nBA}{C}$$\n
\nwhere $$n$$ is the number of turns in the coil, $$B$$ is the magnetic field, $$A$$ is the area of the coil, and $$C$$ is the torsional constant of the suspension wire.\n
\nWe are told that the current sensitivity is increased by 25% by changing only the number of turns and the area of the cross section while keeping the resistance of the galvanometer coil constant. Let the new number of turns be $$n'$$, and the new area be $$A'$$.\n
\nSince the current sensitivity is increased by 25%, we have:\n
\n$$I_s' = 1.25I_s = \\frac{n' B A'}{C}$$\n
\nThe resistance ($R$) of the galvanometer coil is given by:\n
\n$$R = \\rho \\frac{l}{A}$$\n
\nwhere $$\\rho$$ is the resistivity of the wire and $$l$$ is the length of the wire. Since the resistance is kept constant, we can write:\n
\n$$R' = \\rho \\frac{l'}{A'} = R$$\n
\nNow, the voltage sensitivity ($V_s$) of the galvanometer is given by:\n
\n$$V_s = I_s \\times R$$\n
\nWe want to find the percentage change in the voltage sensitivity. Let the new voltage sensitivity be $$V_s'$$:\n
\n$$V_s' = I_s' \\times R'$$\n
\nSince $$R = R'$$, we can write:\n
\n$$V_s' = 1.25I_s \\times R = 1.25V_s$$\n
\nThe percentage change in the voltage sensitivity is:\n
\n$$\\frac{V_s' - V_s}{V_s} \\times 100\\% = \\frac{1.25V_s - V_s}{V_s} \\times 100\\% = 25\\%$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8673,
"subject": "Physics",
"question": "In an ammeter, $5 \\%$ of the main current passes through the galvanometer. If resistance of the galvanometer is $\\mathrm{G}$, the resistance of ammeter will be :",
"options": [
{
"text": "$199 \\mathrm{~G}$"
},
{
"text": "$200 \\mathrm{~G}$"
},
{
"text": "$\\frac{G}{20}$"
},
{
"text": "$\\frac{\\mathrm{G}}{199}$"
}
],
"answer": "$\\frac{G}{20}$",
"solution": "**Answer:** $\\frac{G}{20}$\n\n
When an ammeter is designed, a shunt resistor ($R_s$) is placed in parallel with the galvanometer to ensure that only a small fraction of the total current passes through the galvanometer itself. This is done because the galvanometer is usually a sensitive instrument designed to measure small currents, and passing a large current through it could damage it.
\n\n
In this scenario, we are told that $5\\%$ of the main current passes through the galvanometer. This means that the remaining $95\\%$ of the current must pass through the shunt. Let's denote the total current as $I$, the current through the galvanometer as $I_g$, and the current through the shunt as $I_s$. Therefore, we have:
\n\n
$$ I_g = \\frac{5}{100} I $$
\n
$$ I_s = I - I_g = I - \\frac{5}{100} I = \\frac{95}{100} I $$
\n\n
Since the galvanometer and shunt are in parallel, the voltage across each must be the same:
\n\n
$$ V_g = V_s $$
\n\n
According to Ohm's law, $V = IR$, where $V$ is the voltage, $I$ is the current, and $R$ is the resistance. Hence, for the galvanometer and the shunt:
\n\n
$$ I_g G = I_s R_s $$
\n\n
By substituting $I_g$ and $I_s$ from the above proportionality, we get:
\n\n
$$ \\left(\\frac{5}{100} I\\right) G = \\left(\\frac{95}{100} I\\right) R_s $$
\n\n
Let's solve for $R_s$:
\n\n
$$ R_s = \\frac{5}{95} G $$\n
\n
Further simplifying this:
\n\n
$$ R_s = \\frac{G}{19} $$\n
\n
The total resistance of the ammeter $R_a$ can be found using the parallel resistance formula:
Hence, the correct answer is Option C: $\\frac{G}{20}$.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8674,
"subject": "Physics",
"question": "A galvanometer has a resistance of $50 ~\\Omega$ and it allows maximum current of $5 \\mathrm{~mA}$. It can be converted into voltmeter to measure upto $100 \\mathrm{~V}$ by connecting in series a resistor of resistance :",
"options": [
{
"text": "$19500 \\Omega$"
},
{
"text": "$5975 \\Omega$"
},
{
"text": "$20050 \\Omega$"
},
{
"text": "$19950 \\Omega$"
}
],
"answer": "$19950 \\Omega$",
"solution": "**Answer:** $19950 \\Omega$\n\n
To convert a galvanometer into a voltmeter to measure higher voltages, you need to add a series resistance to it. Let's figure out the required resistance value.
\n\n
First, let's find the maximum voltage that can be directly measured by the galvanometer without any additional resistance. We know the maximum current, $I$, that the galvanometer can safely measure is $5 \\text{ mA}$, and the resistance of the galvanometer, $R_g$, is $50 \\Omega$.
\n\n
Using Ohm's law $ V = I \\times R $, the maximum voltage $V_g$ the galvanometer can measure is:
Next, to measure up to $100 \\text{ V}$, we need to add a series resistor $R_s$ so that the total voltage drop when the maximum current is flowing is $100 \\text{ V}$. The voltage drop across the additional resistor $R_s$ when the maximum current flows would be the total voltage minus the voltage across the galvanometer:
\n\n
$ V_s = V_{\\text{total}} - V_g $
\n\n
$ V_s = 100 \\text{ V} - 0.25 \\text{ V} $
\n\n
$ V_s = 99.75 \\text{ V} $
\n\n
Applying Ohm's law to the series resistor to find its resistance value:
Therefore, the resistance of the series resistor required to convert the galvanometer into a voltmeter that can measure up to $100 \\text{ V}$ is $19950 \\Omega$.
A current of $$200 \\mu \\mathrm{A}$$ deflects the coil of a moving coil galvanometer through $$60^{\\circ}$$. The current to cause deflection through $$\\frac{\\pi}{10}$$ radian is :
The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of $$24 \\Omega$$ is applied. The resistance of galvanometer coil will be :
A galvanometer having coil resistance $$10 \\Omega$$ shows a full scale deflection for a current of $$3 \\mathrm{~mA}$$. For it to measure a current of $$8 \\mathrm{~A}$$, the value of the shunt should be:
To determine the value of the shunt resistor required to convert the galvanometer into an ammeter capable of measuring a current of $$8 \\mathrm{~A}$$, we need to use the concept of shunting where the additional resistor (shunt) is placed in parallel with the galvanometer.
\n\n
The formula for calculating the value of the shunt resistor $ R_s $ is given by:
\n\n
$$ R_s = \\frac{R_g \\cdot I_g}{I - I_g} $$
\n\n
Where:
\n\n
$ R_g $ = resistance of the galvanometer = $$10 \\Omega$$ \n\n
$ I_g $ = full-scale deflection current of the galvanometer = $$3 \\mathrm{~mA}$$ = $$0.003 \\mathrm{~A}$$
\n\n
$ I $ = total current to be measured = $$8 \\mathrm{~A}$$
Two resistance of $$100 \\Omega$$ and $$200 \\Omega$$ are connected in series with a battery of $$4 \\mathrm{~V}$$ and negligible internal resistance. A voltmeter is used to measure voltage across $$100 \\Omega$$ resistance, which gives reading as $$1 \\mathrm{~V}$$. The resistance of voltmeter must be _______ $$\\Omega$$.
A galvanometer has a coil of resistance $$200 \\Omega$$ with a full scale deflection at $$20 \\mu \\mathrm{A}$$. The value of resistance to be added to use it as an ammeter of range $$(0-20) \\mathrm{mA}$$ is :
To convert a galvanometer into an ammeter, we need to add a shunt resistance in parallel with the galvanometer's coil. The purpose of the shunt resistance is to bypass the majority of the current while allowing only a small fraction of it to pass through the galvanometer, thereby preventing it from being damaged by high currents.
\n\n
Given parameters:\n\n
\n
Resistance of the galvanometer's coil, $$R_g = 200 \\Omega$$
\n
Full-scale deflection current of the galvanometer, $$I_g = 20 \\mu \\mathrm{A} = 20 \\times 10^{-6} \\mathrm{A}$$
Therefore, the value of the resistance to be added to use the galvanometer as an ammeter of range $$(0-20) \\mathrm{mA}$$ is $$0.20 \\Omega$$. Thus, the correct answer is:
A galvanometer of resistance $$100 \\Omega$$ when connected in series with $$400 \\Omega$$ measures a voltage of upto $$10 \\mathrm{~V}$$. The value of resistance required to convert the galvanometer into ammeter to read upto $$10 \\mathrm{~A}$$ is $$x \\times 10^{-2} \\Omega$$. The value of $$x$$ is :
To convert a galvanometer into an ammeter to measure larger currents, a low resistance known as the shunt resistance ($$R_{\\text{sh}}$$) is connected in parallel with the galvanometer. The value of this shunt resistance can be calculated using the principles of parallel circuits and the desired maximum current the ammeter should read.
\n\n
The original configuration of the galvanometer allows it to measure up to $$10\\,\\text{V}$$, and it has a resistance of $$100\\,\\Omega$$. When connected in series with a $$400\\,\\Omega$$ resistor, the total resistance in the circuit is $$100\\,\\Omega + 400\\,\\Omega = 500\\,\\Omega$$. Given this configuration measures up to $$10\\,\\text{V}$$, we can calculate the maximum current it is designed to measure using Ohm's law:
Now, to recalibrate the device to measure up to $$10\\,\\text{A}$$, we require the calculation of the shunt resistor $$R_{\\text{sh}}$$ that needs to be connected in parallel with the galvanometer. The total current $$I$$ will now be $$10\\,\\text{A}$$, and the part of this current flowing through the galvanometer ($$I_g$$) remains $$0.02\\,\\text{A}$$ (as before, to ensure we do not exceed the device's original maximum measuring capability), leaving the rest to flow through the shunt. Thus, $$I - I_g$$ flows through the shunt.
\n\n
Since the voltage drop across both the shunt and the galvanometer must be the same for parallel components, we use Ohm’s Law $$V = IR$$ for both and set up an equation to calculate $$R_{\\text{sh}}$$:
\n\n
$I_g R_g = (I - I_g) R_{\\text{sh}}$
\n\n
Substituting known values ($$R_g = 100\\,\\Omega$$, $$I = 10\\,\\text{A}$$, and $$I_g = 0.02\\,\\text{A}$$):
Expressing this in terms of $$\\times 10^{-2} \\Omega$$ gives $$R_{\\text{sh}} \\approx 20.04 \\times 10^{-2} \\Omega$$. Therefore, the value of $$x$$ is approximately $$20.04$$, and rounding it according to the provided options leads to the closest value:
\n\n
Option B: 20
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8681,
"subject": "Physics",
"question": "An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii are in the ratio of $${4 \\over 3}$$ and $${2 \\over 3}$$, then the ratio of the current passing through the wires will be ",
"options": [
{
"text": "$$8/9$$ "
},
{
"text": "$$1/3$$ "
},
{
"text": "$$3$$ "
},
{
"text": "$$2$$ "
}
],
"answer": "$$1/3$$ ",
"solution": "**Answer:** $$1/3$$ \n\n\n
$$ = {3 \\over 4} \\times {4 \\over 9} = {1 \\over 3}\\,\\,$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8682,
"subject": "Physics",
"question": "The Kirchhoff's first law $$\\left( {\\sum i = 0} \\right)$$ and second law $$\\left( {\\sum iR = \\sum E} \\right),$$ where the symbols have their usual meanings, are respectively based on ",
"options": [
{
"text": "conservation of charge, conservation of momentum"
},
{
"text": "conservation of energy, conservation of charge "
},
{
"text": "conservation of momentum, conservation of charge "
},
{
"text": "conservation of charge, conservation of energy"
}
],
"answer": "conservation of charge, conservation of energy",
"solution": "**Answer:** conservation of charge, conservation of energy\n\nNOTE : Kirchhoff's first law is based on conservation of charge and Kirchhoffs second law is based on conservation of energy.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8683,
"subject": "Physics",
"question": "Two sources of equal $$emf$$ are connected to an external resistance $$R.$$ The internal resistance of the two sources are $${R_1}$$ and $${R_2}\\left( {{R_1} > {R_1}} \\right).$$ If the potential difference across the source having internal resistance $${R_2}$$ is zero, then ",
"options": [
{
"text": "$$R = {R_2} - {R_1}$$ "
},
{
"text": "$$R = {R_2} \\times \\left( {{R_1} + {R_2}} \\right)/\\left( {{R_2} - {R_1}} \\right)$$ "
},
{
"text": "$$R = {R_1}{R_2}/\\left( {{R_2} - {R_1}} \\right)$$ "
},
{
"text": "$$R = {R_1}{R_2}/\\left( {{R_1} - {R_2}} \\right)$$ "
}
],
"answer": "$$R = {R_2} - {R_1}$$ ",
"solution": "**Answer:** $$R = {R_2} - {R_1}$$ \n\n\n
$$\\therefore$$ $$R = {R_2} - {R_1}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8684,
"subject": "Physics",
"question": "Which of the following statements is false?",
"options": [
{
"text": "Kirchhoff’s second law represents energy conservation."
},
{
"text": "Wheatstone bridge is the most sensitive when all the four resistances are of the same order of\nmagnitude."
},
{
"text": "In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is\ndisturbed."
},
{
"text": "A rheostat can be used as a potential divider."
}
],
"answer": "In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is\ndisturbed.",
"solution": "**Answer:** In a balanced wheatstone bridge if the cell and the galvanometer are exchanged, the null point is\ndisturbed.\n\nThere is no change in null point, if the cell and the\ngalvanometer are exchanged in a balanced wheatstone\nbridge.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8685,
"subject": "Physics",
"question": "Two batteries with e.m.f 12 V and 13 V are connected in parallel across a load resistor of 10 $$\\Omega $$. The\ninternal resistances of the two batteries are 1 $$\\Omega $$ and 2 $$\\Omega $$ respectively. The voltage across the load lies between :",
"options": [
{
"text": "11.7 V and 11.8 V"
},
{
"text": "11.6 V and 11.7 V"
},
{
"text": "11.5 V and 11.6 V"
},
{
"text": "11.4 V and 11.5 V"
}
],
"answer": "11.5 V and 11.6 V",
"solution": "**Answer:** 11.5 V and 11.6 V\n\n\n
Let potential at S, T, R = V and potential at P, Q, U, = 0\n
$$ \\Rightarrow $$$$\\,\\,\\,$$ V = 11.56 Volt.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8686,
"subject": "Physics",
"question": "In an electric circuit, a cell of certain emf provides a potential difference of 1.25 V across a load resistance of 5$$\\Omega$$. However, it provides a potential difference of 1 V across a load resistance of 2$$\\Omega$$. The emf of the cell is given by $${x \\over {10}}V$$. Then the value of x is ______________.",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\n
In case (a) $$\\varepsilon = {{1.25} \\over 5}(5 + r)$$
$$ \\Rightarrow 4\\varepsilon = 5 + r$$ ..... (1)
In case (b), $$\\varepsilon = {1 \\over 2}(2 + r)$$
or x = 15",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8687,
"subject": "Physics",
"question": "A 16 $$\\Omega$$ wire is bend to form a square loop. A 9V supply having internal resistance of 1$$\\Omega$$ is connected across one of its sides. The potential drop across the diagonals of the square loop is _______________ $$\\times$$ 10$$-$$1 V",
"options": [],
"answer": "45",
"solution": "**Answer:** 45\n\nHere assume current as
By KVL in outer loop
9 $$-$$ 12i $$-$$ 4i = 0
16i = 9
8i = $${9 \\over 2}$$ = 4.5
= 45 $$\\times$$ 10-1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8688,
"subject": "Physics",
"question": "In a meter bridge experiment null point is obtained at $$20$$ $$cm$$, from one end of the wire when resistance $$X$$ is balanced against another resistance $$Y.$$ If $$X < Y$$, then where will be the new position of the null point from the same end, if one decides to balance a resistance of $$4$$ $$X$$ against $$Y$$ ",
"options": [
{
"text": "$$40$$ $$cm$$ "
},
{
"text": "$$80$$ $$cm$$ "
},
{
"text": "$$50$$ $$cm$$ "
},
{
"text": "$$70$$ $$cm$$ "
}
],
"answer": "$$50$$ $$cm$$ ",
"solution": "**Answer:** $$50$$ $$cm$$ \n\nIn the first case $${X \\over Y} = {{20} \\over {80}} = {1 \\over 4}$$\n
In the second case $${{4X} \\over Y} = {\\ell \\over {100 - \\ell }} \\Rightarrow \\ell = 50$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8689,
"subject": "Physics",
"question": "On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The\nresistance of their series combination is 1 k$$\\Omega $$. How much was the resistance on the left slot before\ninterchanging the resistances?",
"options": [
{
"text": "910 $$\\Omega $$"
},
{
"text": "990 $$\\Omega $$"
},
{
"text": "505 $$\\Omega $$"
},
{
"text": "550 $$\\Omega $$"
}
],
"answer": "550 $$\\Omega $$",
"solution": "**Answer:** 550 $$\\Omega $$\n\n\n
When two resistance $$\\mathrm{R_1}$$ and $$\\mathrm{R_2}$$ connected in series and introduced into the left gap of a meter bridge and a resistance of 10 $$\\Omega$$ is introduced into the right gap, a null point is found at 60 cm from left side. When $$\\mathrm{R_1}$$ and $$\\mathrm{R_2}$$ are connected in parallel and introduced into the left gap, a resistance of 3 $$\\Omega$$ is introduced into the right gap to get null point at 40 cm from left end. The product of $$\\mathrm{R_1}$$ $$\\mathrm{R_2}$$ is ____________$$\\Omega^2$$
In a metre bridge experiment the balance point is obtained if the gaps are closed by 2$$\\Omega$$ and 3$$\\Omega$$. A shunt of X $$\\Omega$$ is added to 3$$\\Omega$$ resistor to shift the balancing point by 22.5 cm. The value of X is ___________.
\nas $3 \\Omega$ is shunted the balance point will shift towards $3 \\Omega$. So, new length $l^{\\prime}=22.5+I=62.5$\n
\nSo, $\\frac{62.5}{37.5}=\\frac{2}{3 x}(3+x)$\n
\n$\\Rightarrow x=2 \\Omega$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8692,
"subject": "Physics",
"question": "In a metre-bridge when a resistance in the left gap is $2 \\Omega$ and unknown resistance in the right gap, the balance length is found to be $40 \\mathrm{~cm}$. On shunting the unknown resistance with $2 \\Omega$, the balance length changes by :",
"options": [
{
"text": "$62.5 $ "
},
{
"text": "$22.5 \\mathrm{~cm}$"
},
{
"text": "$20 \\mathrm{~cm}$"
},
{
"text": "$65 \\mathrm{~cm}$"
}
],
"answer": "$22.5 \\mathrm{~cm}$",
"solution": "**Answer:** $22.5 \\mathrm{~cm}$\n\n
To solve this problem, let's first understand that a meter bridge setup is based on the principle of a Wheatstone bridge, in which two unknown resistances are in such a configuration that if the bridge is balanced, the ratio of the resistances on one side is equal to the ratio of resistances on the other side. In a balanced condition, no current flows through the galvanometer that is connected diagonally across the bridge.
\n
We can express the condition of the balance as follows:
\n
$$\n\\frac{R_1}{R_2} = \\frac{L_1}{L_2}\n$$
\n
where $R_1$ is the known resistance $2 Ω$, $R_2$ is the unknown resistance, $L_1$ is the balance length $40 cm$ and $L_2$ is the length of the remaining wire on the meter bridge (100 cm - 40 cm = 60 cm).
\n
Let's calculate the initial unknown resistance ($R_2$) using the balance condition:
Now, when the unknown resistance $R_2$ is shunted with a 2 Ω resistor, the new combined resistance ($R'_2$) can be calculated using the parallel resistance formula:
The resistance per centimeter of a meter bridge wire is $$r$$, with $$X \\Omega$$ resistance in left gap. Balancing length from left end is at $$40 \\mathrm{~cm}$$ with $$25 \\Omega$$ resistance in right gap. Now the wire is replaced by another wire of $$2 r$$ resistance per centimeter. The new balancing length for same settings will be at
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8694,
"subject": "Physics",
"question": "An energy source will supply a constant current into the load if its internal resistance is ",
"options": [
{
"text": "very large as compared to the load resistance "
},
{
"text": "equal to the resistance of the load "
},
{
"text": "non-zero but less than the resistance of the load"
},
{
"text": "zero "
}
],
"answer": "zero ",
"solution": "**Answer:** zero \n\n$$I = {E \\over {R + r}},\\,$$ Internal resistance $$\\left( r \\right)$$ is \n
zero, $$I = {E \\over R} = $$ constant.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8695,
"subject": "Physics",
"question": "When $$5V$$ potential difference is applied across a wire of length $$0.1$$ $$m,$$ the drift speed of electrons is $$2.5 \\times {10^{ - 4}}\\,\\,m{s^{ - 1}}.$$ If the electron density in the wire is $$8 \\times {10^{28}}\\,\\,{m^{ - 3}},$$ the resistivity of the material is close to :",
"options": [
{
"text": "$$1.6 \\times {10^{ - 6}}\\Omega m$$ "
},
{
"text": "$$1.6 \\times {10^{ - 5}}\\Omega m$$ "
},
{
"text": "$$1.6 \\times {10^{ - 8}}\\Omega m$$ "
},
{
"text": "$$1.6 \\times {10^{ - 7}}\\Omega m$$ "
}
],
"answer": "$$1.6 \\times {10^{ - 5}}\\Omega m$$ ",
"solution": "**Answer:** $$1.6 \\times {10^{ - 5}}\\Omega m$$ \n\n$$V = IR = \\left( {neA{v_d}} \\right)\\rho {\\ell \\over A}$$\n
$$ = 1.6 \\times {10^{ - 5}}\\Omega m$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8696,
"subject": "Physics",
"question": "A conducting wire of length 'l', area of cross-section A and electric resistivity $$\\rho$$ is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current.
If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be :",
"options": [
{
"text": "$$4{{VA} \\over {\\rho l}}$$"
},
{
"text": "$${3 \\over 4}{{VA} \\over {\\rho l}}$$"
},
{
"text": "$${1 \\over 4}{{VA} \\over {\\rho l}}$$"
},
{
"text": "$${1 \\over 4}{{\\rho l} \\over {VA}}$$"
}
],
"answer": "$${1 \\over 4}{{VA} \\over {\\rho l}}$$",
"solution": "**Answer:** $${1 \\over 4}{{VA} \\over {\\rho l}}$$\n\nWe know that
$$ \\Rightarrow $$ $$I = {1 \\over 4}{{VA} \\over {\\rho l}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8697,
"subject": "Physics",
"question": "In the experiment of Ohm's law, a potential difference of 5.0 V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00 mm. The measured current in the conductor is 2.00 A. The maximum permissible percentage error in the resistivity of the conductor is :",
"options": [
{
"text": "3.9"
},
{
"text": "8.4"
},
{
"text": "7.5"
},
{
"text": "3.0"
}
],
"answer": "3.9",
"solution": "**Answer:** 3.9\n\n$$V = I \\times \\rho {l \\over A}$$
A uniform metallic wire carries a current 2 A, when 3.4 V battery is connected across it. The mass of uniform metallic wire is 8.92 $$\\times$$ 10$$^{-3}$$ kg, density is 8.92 $$\\times$$ 10$$^{3}$$ kg/m$$^3$$ and resistivity is 1.7 $$\\times$$ 10$$^{-8}~\\Omega$$-$$\\mathrm{m}$$. The length of wire is :
\n$l=10 \\mathrm{~m}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8699,
"subject": "Physics",
"question": "The length of a wire of a potentiometer is $$100$$ $$cm$$, and the $$e.$$ $$m.$$ $$f.$$ of its standard cell is $$E$$ volt. It is employed to measure the $$e.m.f.$$ of a battery whose internal resistance in $$0.5\\Omega .$$ If the balance point is obtained at $$1=30$$ $$cm$$ from the positive end, the $$e.m.f.$$ of the battery is \n
where $$i$$ is the current in the potentiometer wire.
",
"options": [
{
"text": "$${{30E} \\over {100.5}}$$ "
},
{
"text": "$${{30E} \\over {\\left( {100 - 0.5} \\right)}}$$ "
},
{
"text": "$${{30\\left( {E - 0.5i} \\right)} \\over {100}}$$ "
},
{
"text": "$${{30E} \\over {100}} - 0.5i$$, where i is the current in the potentiometer\nwire"
}
],
"answer": "$${{30E} \\over {100}} - 0.5i$$, where i is the current in the potentiometer\nwire",
"solution": "**Answer:** $${{30E} \\over {100}} - 0.5i$$, where i is the current in the potentiometer\nwire\n\nPotential gradient along wire, K = $${E \\over {100}}$$ volt/cm\n
For battery V = E' – ir, where E' is emf of battery.\n
or K × 30 = E' – ir, where current i is drawn from battery\n
or $${{E \\times 30} \\over {100}}$$ = E' + 0.5i\n
or E' = $${{30E} \\over {100}} - 0.5i$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8700,
"subject": "Physics",
"question": "In a potentiometer experiment the balancing with a cell is at length $$240$$ $$cm.$$ On shunting the cell with a resistance of $$2\\Omega ,$$ the balancing length becomes $$120$$ $$cm$$. The internal resistance of the cell is ",
"options": [
{
"text": "$$0.5\\Omega $$ "
},
{
"text": "$$1\\Omega $$"
},
{
"text": "$$2\\Omega $$"
},
{
"text": "$$4\\Omega $$"
}
],
"answer": "$$2\\Omega $$",
"solution": "**Answer:** $$2\\Omega $$\n\nThe internal resistance of the cell, \n
$$ = {{240 - 120} \\over {120}} \\times 2 = 2\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8701,
"subject": "Physics",
"question": "In a potentiometer experiment, it is found that no current passes through the galvanometer when the\nterminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a\nresistance of 5 $$\\Omega$$, a balance is found when the cell is connected across 40 cm of the wire. Find the internal\nresistance of the cell.",
"options": [
{
"text": "2.5 $$\\Omega$$"
},
{
"text": "1 $$\\Omega$$"
},
{
"text": "1.5 $$\\Omega$$"
},
{
"text": "2 $$\\Omega$$"
}
],
"answer": "1.5 $$\\Omega$$",
"solution": "**Answer:** 1.5 $$\\Omega$$\n\nInternal resistance of potentiometer, \n
= 1.5 $$\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8702,
"subject": "Physics",
"question": "An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and esistance 5 $$\\Omega $$. The value of R, to give a potential difference of 5 mV across 10 cm of potentiometer wire, is : ",
"options": [
{
"text": "480 $$\\Omega $$"
},
{
"text": "495 $$\\Omega $$"
},
{
"text": "490 $$\\Omega $$"
},
{
"text": "395 $$\\Omega $$"
}
],
"answer": "395 $$\\Omega $$",
"solution": "**Answer:** 395 $$\\Omega $$\n\n\n Let current flowing in the wire is i.\n
$$ \\therefore $$ R = 395 $$\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8703,
"subject": "Physics",
"question": "The balancing length for a cell is 560 cm in a potentiometer experiment. When an external\nresistance of 10 $$\\Omega $$ is connected in parallel to the cell, the balancing length changes by 60 cm. If\nthe internal resistance of the ceil is $${N \\over {10}}$$\n$$\\Omega $$ , where N is an integer then value of N is .............",
"options": [],
"answer": "12",
"solution": "**Answer:** 12\n\nLet the emf of cell is $$\\varepsilon $$ internal resistance is 'r' and potential gradient is x.\n
$$ \\Rightarrow $$ N = 12",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8704,
"subject": "Physics",
"question": "The length of a potentiometer wire is 1200 cm\nand it carries a current of 60 mA. For a cell of\nemf 5V and internal resistance of 20$$\\Omega $$, the null\npoint on it is found to be a 1000 cm. The\nresistance of whole wire is :",
"options": [
{
"text": "80$$\\Omega $$"
},
{
"text": "60$$\\Omega $$"
},
{
"text": "120$$\\Omega $$"
},
{
"text": "100$$\\Omega $$"
}
],
"answer": "100$$\\Omega $$",
"solution": "**Answer:** 100$$\\Omega $$\n\nLet Resistance per unit length of potentiometer wire = $$\\lambda $$\n
A cell, shunted by a 8 $$\\Omega$$ resistance, is balanced across a potentiometer wire of length 3 m. The balancing length is 2 m when the cell is shunted by 4 $$\\Omega$$ resistance. The value of internal resistance of the cell will be ____________ $$\\Omega$$.
A potentiometer wire of length 10 m and resistance 20 $$\\Omega$$ is connected in series with a 25 V battery and an external resistance 30 $$\\Omega$$. A cell of emf E in secondary circuit is balanced by 250 cm long potentiometer wire. The value of E (in volt) is $${x \\over {10}}$$. The value of x is __________.
In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is 3 : 2, the difference in the balancing length of the potentiometer wire in above two cases will be ___________ cm.
In a potentiometer arrangement, a cell of emf 1.20 V gives a balance point at 36 cm length of wire. This cell is now replaced by another cell of emf 1.80 V. The difference in balancing length of potentiometer wire in above conditions will be ___________ cm.
A potentiometer wire of length $$300 \\mathrm{~cm}$$ is connected in series with a resistance 780 $$\\Omega$$ and a standard cell of emf $$4 \\mathrm{V}$$. A constant current flows through potentiometer wire. The length of the null point for cell of emf $$20\\, \\mathrm{mV}$$ is found to be $$60 \\mathrm{~cm}$$. The resistance of the potentiometer wire is ____________ $$\\Omega$$.
In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf $$1.5 \\mathrm{~V}$$ is found to be $$60 \\mathrm{~cm}$$. If this cell is replaced by another cell of emf E, the length-of null point increases by $$40 \\mathrm{~cm}$$. The value of $$E$$ is $$\\frac{x}{10} V$$. The value of $$x$$ is ____________.
",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\nE1 = 1.5 V, l1 = 60 cm, l2 = 40 cm + 60 cm = 100 cm \n
With the help of potentiometer, we can determine the value of emf of a given cell. The sensitivity of the potentiometer is
\n
(A) directly proportional to the length of the potentiometer wire
\n
(B) directly proportional to the potential gradient of the wire
\n
(C) inversely proportional to the potential gradient of the wire
\n
(D) inversely proportional to the length of the potentiometer wire
\n
Choose the correct option for the above statements :
",
"options": [
{
"text": "A only"
},
{
"text": "B and D only"
},
{
"text": "A and C only"
},
{
"text": "inversely C only"
}
],
"answer": "A and C only",
"solution": "**Answer:** A and C only\n\n
A potentiometer is an electrical instrument used to measure the electromotive force (EMF) of a cell.
\n
The sensitivity of a potentiometer is defined as the change in potential difference per unit length of the wire.
\n
It is directly proportional to the length of the potentiometer wire (Option A), because a longer wire has a larger potential difference that can be measured.
\n
It is inversely proportional to the potential gradient of the wire (Option C), because a smaller potential gradient results in a smaller change in potential difference per unit length.
\n
Therefore, the correct answer is (C) A and C only.
A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5$$\\Omega$$. When a resistance of 15$$\\Omega$$ is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ___________$$\\Omega$$.
To measure the internal resistance of a battery, potentiometer is used. For $$R=10 \\Omega$$, the balance point is observed at $$l=500 \\mathrm{~cm}$$ and for $$\\mathrm{R}=1 \\Omega$$ the balance point is observed at $$l=400 \\mathrm{~cm}$$. The internal resistance of the battery is approximately :
To measure the internal resistance of a battery using a potentiometer, we need to understand the principle behind it. The potentiometer is used to measure the voltage across the battery (emf) under different conditions. The balance length corresponds to the emf of the battery when no current is drawn, whereas under load, it corresponds to the terminal voltage.
\n\n
Let's denote the emf of the battery by $$E$$ and the internal resistance by $$r$$. According to Ohm's law, when a resistance $$R$$ is connected across the battery, the terminal voltage $$V$$ is given by:
\n\n
$$ V = E - Ir $$
\n\n
where $$I$$ is the current through the circuit.
\n\n
From the problem, the balance length $$l$$ is proportional to the voltage across the potentiometer wire. Thus, we can write:
\n\n
$$ \\frac{V_1}{V_2} = \\frac{l_1}{l_2} $$
\n\n
In the first condition, when $$R = 10 \\Omega$$ and the balance length $$l_1 = 500 \\, \\text{cm}$$:
\n\n
$$ V_1 = E - I_1 r $$
\n\n
For the second condition, when $$R = 1 \\Omega$$ and the balance length $$l_2 = 400 \\, \\text{cm}$$:
Since this value is closest to $$0.3 \\Omega$$, the correct option is:
\n\n
Option B: $$0.3 \\Omega$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8714,
"subject": "Physics",
"question": "The length of a given cylindrical wire is increased by $$100\\% $$. Due to the consequent decrease in diameter the change in the resistance of the wire will be",
"options": [
{
"text": "$$200\\% $$"
},
{
"text": "$$100\\% $$ "
},
{
"text": "$$50\\% $$"
},
{
"text": "$$300\\% $$"
}
],
"answer": "$$300\\% $$",
"solution": "**Answer:** $$300\\% $$\n\n$${R_f} = {n^2}{R_1}$$\n
Here $$n=2$$ (length becomes twice)\n
$$\\therefore$$ $${R_f} = 4{R_i}$$\n
New resistance $$=400$$ of $${R_i}$$\n
$$\\therefore$$ Increase $$ = 300\\% $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8715,
"subject": "Physics",
"question": "Thermistors are usually made of ",
"options": [
{
"text": "metal oxides with high temperature coefficient of resistivity "
},
{
"text": "metals with high temperature coefficient of resistivity "
},
{
"text": "metals with low temperature coefficient of resistivity "
},
{
"text": "semiconducting materials having low temperature "
}
],
"answer": "metal oxides with high temperature coefficient of resistivity ",
"solution": "**Answer:** metal oxides with high temperature coefficient of resistivity \n\nThermistors are usually made of metal-oxides with high temperature coefficient of resistivity.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8716,
"subject": "Physics",
"question": "A material $$'B'$$ has twice the specific resistance of $$'A'.$$ A circular wire made of $$'B'$$ has twice the diameter of a wire made of $$'A'$$. Then for the two wires to have the same resistance, the ratio $${l \\over B}/{l \\over A}$$ of their respective lengths must be",
"options": [
{
"text": "$$1$$ "
},
{
"text": "$${l \\over 2}$$ "
},
{
"text": "$${l \\over 4}$$ "
},
{
"text": "$$2$$ "
}
],
"answer": "$$2$$ ",
"solution": "**Answer:** $$2$$ \n\n$${\\rho _B} = 2{\\rho _A}$$\n
$$\\therefore$$ Resistance will increase by $$0.2\\% $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8718,
"subject": "Physics",
"question": "A uniform wire of length 1 and radius r has a resistance of 100 $$\\Omega $$. It is recast into a wire of radius $${r \\over 2}.$$ The resistance of new wire will be : ",
"options": [
{
"text": "1600 $$\\Omega $$ "
},
{
"text": "400 $$\\Omega $$"
},
{
"text": "200 $$\\Omega $$"
},
{
"text": "100 $$\\Omega $$"
}
],
"answer": "1600 $$\\Omega $$ ",
"solution": "**Answer:** 1600 $$\\Omega $$ \n\nResistance of a wire of length l and radius r is given\nby\n
The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be ____________ %.
A wire of resistance R1 is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is :
",
"options": [
{
"text": "9 : 1"
},
{
"text": "1 : 9"
},
{
"text": "4 : 1"
},
{
"text": "3 : 1"
}
],
"answer": "9 : 1",
"solution": "**Answer:** 9 : 1\n\nLength is increased by twice of its original length.
So, if original length is $l_1$ then final length is $l_2$= $l_1$ + 2$l_1$ = 3$l_1$.\n
Two metallic wires of identical dimensions are connected in series. If $$\\sigma_{1}$$ and $$\\sigma_{2}$$ are the conductivities of the these wires respectively, the effective conductivity of the combination is :
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
\n
Assertion A: Alloys such as constantan and manganin are used in making standard resistance coils.
\n
Reason R: Constantan and manganin have very small value of temperature coefficient of resistance.
\n
In the light of the above statements, choose the correct answer from the options given below.
",
"options": [
{
"text": "Both A and R are true and R is the correct explanation of A."
},
{
"text": "Both A and R are true but R is NOT the correct explanation of A."
},
{
"text": "A is true but R is false."
},
{
"text": "A is false but R is true."
}
],
"answer": "Both A and R are true and R is the correct explanation of A.",
"solution": "**Answer:** Both A and R are true and R is the correct explanation of A.\n\n
Since they have low temperature coefficient of\nresistance, their resistance remains almost\nconstant.
A $$1 \\mathrm{~m}$$ long wire is broken into two unequal parts $$\\mathrm{X}$$ and $$\\mathrm{Y}$$. The $$\\mathrm{X}$$ part of the wire is streched into another wire W. Length of $$W$$ is twice the length of $$X$$ and the resistance of $$\\mathrm{W}$$ is twice that of $$\\mathrm{Y}$$. Find the ratio of length of $$\\mathrm{X}$$ and $$\\mathrm{Y}$$.
A hollow cylindrical conductor has length of 3.14 m, while its inner and outer diameters are 4 mm and 8 mm respectively. The resistance of the conductor is $$n\\times10^{-3}\\Omega$$. If the resistivity of the material is $$\\mathrm{2.4\\times10^{-8}\\Omega m}$$. The value of $$n$$ is ___________.
",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nResistance of the hollow cylindrical conductor is given by, \n
A wire of resistance $$160 ~\\Omega$$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be
",
"options": [
{
"text": "$$640 ~\\Omega$$"
},
{
"text": "$$40 ~\\Omega$$"
},
{
"text": "$$16 ~\\Omega$$"
},
{
"text": "$$10 ~\\Omega$$"
}
],
"answer": "$$10 ~\\Omega$$",
"solution": "**Answer:** $$10 ~\\Omega$$\n\nLet the original length of the wire be L and its cross-sectional area be A. Then, its resistance R is given by:
\n$$R = \\frac{\\rho L}{A}$$
\nwhere $$\\rho$$ is the resistivity of the material of the wire.\n
\nWhen the wire is melted and drawn into a wire of one-fourth of its length, its new length is L/4 and its new cross-sectional area is 4A (since the same amount of material is now spread over a longer length). Therefore, its new resistance R' is given by:
\nTherefore, the new resistance of the wire is $$10 ~\\Omega$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8735,
"subject": "Physics",
"question": "
A rectangular parallelopiped is measured as $$1 \\mathrm{~cm} \\times 1 \\mathrm{~cm} \\times 100 \\mathrm{~cm}$$. If its specific resistance is $$3 \\times 10^{-7} ~\\Omega \\mathrm{m}$$, then the resistance between its two opposite rectangular faces will be ___________ $$\\times 10^{-7} ~\\Omega$$.
The resistance of a material can be calculated using the formula:
\n
$ R = \\rho \\frac{L}{A} $
\n
where
\n
\n
$R$ is the resistance,
\n
$\\rho$ (rho) is the resistivity or specific resistance of the material,
\n
$L$ is the length (or distance over which the resistance is being measured), and
\n
$A$ is the cross-sectional area through which the current flows.
\n
\n
In the context of a rectangular parallelepiped, the "length" and "cross-sectional area" can vary depending on which faces of the shape are considered.
\n
In this particular calculation, the resistance is being measured between the two smaller faces of the parallelepiped, which are squares of side length 1 cm:
The length of a metallic wire is increased by $$20 \\%$$ and its area of cross section is reduced by $$4 \\%$$. The percentage change in resistance of the metallic wire is __________.
Wheatstone bridge principle is used to measure the specific resistance $$\\left(S_1\\right)$$ of given wire, having length $$L$$, radius $$r$$. If $$X$$ is the resistance of wire, then specific resistance is ; $$S_1=X\\left(\\frac{\\pi r^2}{L}\\right)$$. If the length of the wire gets doubled then the value of specific resistance will be :
The specific resistance (or resistivity) of a material is a fundamental property that describes how much the material resists the flow of electric current. The resistivity is typically denoted by the symbol $$\\rho$$ (rho), and it can be calculated by using the resistance $$X$$ of a uniform specimen of the material, along with its physical dimensions. In the case of a wire, the resistivity formula in terms of its resistance $$X$$, length $$L$$, and cross-sectional area $$A=\\pi r^2$$ is given by:
\n\n$$\n\\rho = \\frac{X \\cdot A}{L}\n$$\n\n
This formula is a reinterpretation of Ohm's law, and it states that the specific resistance is proportional to the area of the cross-section of the wire and inversely proportional to its length.
\n\n
Given that the specific resistance of the wire $$S_1$$ is determined using the formula:
\n\n$$\nS_1 = X \\left(\\frac{\\pi r^2}{L}\\right)\n$$\n\n
Now, let's see what happens to the specific resistance if the length of the wire is doubled. If we denote the new length as $$2L$$, the resistance of the wire with the new length will change because resistance is directly proportional to the length of the wire. However, resistivity (specific resistance) is an intrinsic property of the material and does not depend on its length or shape, only on its temperature. Therefore, even if we change the length of the wire, the specific resistance should remain the same.
\n\n
If we calculate the new resistance $$X'$$ with the doubled length, it would be:
\n\n$$\nX' = X \\left(\\frac{2L}{L}\\right) = 2X\n$$\n\n
Thus, we would use the new resistance $$X'$$ and the new length $$2L$$ to calculate the specific resistance again:
Since this formula is essentially the same as our original formula for $$S_1$$, we can conclude that:
\n\n$$\nS_1' = S_1\n$$\n\n
Therefore, the value of the specific resistance $$S_1$$ will remain the same even if the length of the wire gets doubled. The correct answer to the question is:
Two conductors have the same resistances at $$0^{\\circ} \\mathrm{C}$$ but their temperature coefficients of resistance are $$\\alpha_1$$ and $$\\alpha_2$$. The respective temperature coefficients for their series and parallel combinations are :
At room temperature $$(27^{\\circ} \\mathrm{C})$$, the resistance of a heating element is $$50 \\Omega$$. The temperature coefficient of the material is $$2.4 \\times 10^{-4}{ }^{\\circ} \\mathrm{C}^{-1}$$. The temperature of the element, when its resistance is $$62 \\Omega$$, is __________$${ }^{\\circ} \\mathrm{C}$$.
We can start solving this problem by first understanding that the resistance of a material changes with temperature, and this change can be quantified using the temperature coefficient of resistance $ \\alpha $. The relationship between the resistance of a material at any temperature $ T $ and its resistance at a reference temperature $ T_0 $ is given by the formula:
\n\n
$ R = R_0(1 + \\alpha(T - T_0)) $
\n\n
Where:
\n\n
\n\n
$ R $ is the resistance at temperature $ T $ (in this case, $ 62 \\Omega $).
\n\n
$ R_0 $ is the resistance at reference temperature $ T_0 $ (in this case, $ 50 \\Omega $).
\n\n
$ \\alpha $ is the temperature coefficient of resistance (in this case, $ 2.4 \\times 10^{-4} \\, ^\\circ\\mathrm{C}^{-1} $).
\n\n
$ T $ is the unknown temperature we need to find.
\n\n
$ T_0 $ is the reference temperature, given as $ 27^\\circ \\mathrm{C} $.
\n\n
\n\n
By substituting the given values into the formula, we get:
Two wires $$A$$ and $$B$$ are made up of the same material and have the same mass. Wire $$A$$ has radius of $$2.0 \\mathrm{~mm}$$ and wire $$B$$ has radius of $$4.0 \\mathrm{~mm}$$. The resistance of wire $$B$$ is $$2 \\Omega$$. The resistance of wire $$A$$ is ________ $$\\Omega$$.
Resistance of a wire at $$0^{\\circ} \\mathrm{C}, 100^{\\circ} \\mathrm{C}$$ and $$t^{\\circ} \\mathrm{C}$$ is found to be $$10 \\Omega, 10.2 \\Omega$$ and $$10.95 \\Omega$$ respectively. The temperature $$t$$ in Kelvin scale is _________.
To determine the temperature $$t$$ in the Kelvin scale, we need to use the relationship between the resistance of a wire and temperature. The general formula for the resistance $R$ of a wire as a function of temperature is:
\n\n
$$ R_t = R_0 (1 + \\alpha t) $$
\n\n
where:
\n\n
\n\n
$$R_t$$ is the resistance at temperature $$t$$
\n\n
$$R_0$$ is the resistance at the reference temperature (usually $$0^{\\circ} \\mathrm{C}$$)
\n\n
$$\\alpha$$ is the temperature coefficient of resistance
\n\n
$$t$$ is the temperature
\n\n
\n\n
We are given the following resistances:
\n\n
\n\n
Resistance at $$0^{\\circ} \\mathrm{C}$$: $$R_0 = 10 \\Omega$$
\n\n
Resistance at $$100^{\\circ} \\mathrm{C}$$: $$R_{100} = 10.2 \\Omega$$
\n\n
Resistance at $$t^{\\circ} \\mathrm{C}$$: $$R_t = 10.95 \\Omega$$
\n\n
\n\n
First, we need to find the temperature coefficient of resistance $$\\alpha$$. Using the resistance at $$100^{\\circ} \\mathrm{C}$$:
A wire of resistance $$R$$ and radius $$r$$ is stretched till its radius became $$r / 2$$. If new resistance of the stretched wire is $$x ~R$$, then value of $$x$$ is ________.
The resistance $$R$$ of a wire is given by the formula:
\n$$\nR = \\rho \\frac{l}{A},\n$$\n
where:
\n
\n
$$\\rho$$ is the resistivity of the material,
\n
$$l$$ is the length of the wire,
\n
$$A$$ is the cross-sectional area of the wire.
\n
\n\n
If we have a cylindrical wire, the cross-sectional area can be expressed as $$A = \\pi r^2$$, where $$r$$ is the radius of the cylinder. Therefore, the resistance of the original wire can be written as:
\n$$\nR = \\rho \\frac{l}{\\pi r^2}.\n$$\n\n
When the wire is stretched such that its radius becomes $$r / 2$$, its volume would remain constant, given that the volume of a cylinder is $$V = A \\cdot l = \\pi r^2 \\cdot l$$. Assuming the volume before and after the stretching is the same, and since the area is now a quarter of the original (because when the radius is halved, the area, which is proportional to the square of the radius, is reduced to a quarter), the length must have increased to four times the original to preserve the volume. That is,
A wire of length $$10 \\mathrm{~cm}$$ and radius $$\\sqrt{7} \\times 10^{-4} \\mathrm{~m}$$ connected across the right gap of a meter bridge. When a resistance of $$4.5 \\Omega$$ is connected on the left gap by using a resistance box, the balance length is found to be at $$60 \\mathrm{~cm}$$ from the left end. If the resistivity of the wire is $$\\mathrm{R} \\times 10^{-7} \\Omega \\mathrm{m}$$, then value of $$\\mathrm{R}$$ is :
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8746,
"subject": "Physics",
"question": "This question has Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements.\n
Statement 1 : Davisson - Germer experiment established the wave nature of electrons.
Statement 2 : If electrons have wave nature, they can interfere and show diffraction. ",
"options": [
{
"text": "Statement 1 is true, Statement 2 is false"
},
{
"text": "Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1. "
},
{
"text": "Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1."
},
{
"text": "Statement 1 is false, Statement 2 is true. "
}
],
"answer": "Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1. ",
"solution": "**Answer:** Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1. \n\nDavisson-Germer experiment showed that electron beams can undergo diffraction when passed through atomic crystals. This shows the wave nature of electrons as waves can exhibit interference and diffraction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8747,
"subject": "Physics",
"question": "A beam of electrons of energy E scatters from\na target having atomic spacing of 1 $$\\mathop A\\limits^o $$. The first\nmaximum intensity occurs at $$\\theta $$ = 60o. Then E (in\neV) is ______.\n (Planck constant h = 6.64 × 10–34 Js, 1 eV =\n1.6 × 10–19 J, electron mass m = 9.1 × 10–31 kg)",
"options": [],
"answer": "50.47",
"solution": "**Answer:** 50.47\n\nGiven d = 1 $$\\mathop A\\limits^o $$\n
Statement I : Davisson-Germer experiment establishes the wave nature of electrons.
\n
Statement II : If electrons have wave nature, they can interfere and show diffraction.
\n
In the light of the above statements choose the correct answer from the option given below :
",
"options": [
{
"text": "Both Statement I and Statement II are true."
},
{
"text": "Both Statement I and Statement II are false."
},
{
"text": "Statement I is true but Statement II is false."
},
{
"text": "Statement I is false but Statement II is true."
}
],
"answer": "Both Statement I and Statement II are true.",
"solution": "**Answer:** Both Statement I and Statement II are true.\n\n
Davisson-Germer experiment is done and establishes the wave nature of electrons. Interference and diffraction establishes wave nature.
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R
\n
Assertion A : The beam of electrons show wave nature and exhibit interference and diffraction.
\n
Reason R : Davisson Germer Experimentally verified the wave nature of electrons.
\n
In the light of the above statements, choose the most appropriate answer from the options given below :
",
"options": [
{
"text": "A is not correct but R is correct."
},
{
"text": "Both A and R are correct and R is the correct explanation of A"
},
{
"text": "Both A and R are correct but R is Not the correct explanation of A"
},
{
"text": "A is correct but R is not correct"
}
],
"answer": "Both A and R are correct and R is the correct explanation of A",
"solution": "**Answer:** Both A and R are correct and R is the correct explanation of A\n\n
The assertion A and the reason R are both correct statements, and the reason R is the correct explanation of the assertion A.
\n\n
Explanation :
\n\n
The assertion A states that the beam of electrons exhibit wave nature and show interference and diffraction. This statement is correct because electrons exhibit both particle-like and wave-like behavior. When electrons are accelerated to high speeds, they have a wavelength associated with them, and this wavelength can interfere and diffract just like any other wave.
\n\n
The reason R states that the wave nature of electrons was experimentally verified by Davisson Germer. This statement is also correct because Davisson and Germer performed an experiment in 1927 where they observed diffraction patterns in a beam of electrons that were scattered off a nickel crystal. This observation provided strong evidence for the wave nature of electrons.\n
\n
Therefore, both assertion A and reason R are correct statements, and reason R is the correct explanation of assertion A.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8751,
"subject": "Physics",
"question": "Formation of covalent bonds in compounds exhibits ",
"options": [
{
"text": "wave nature of electron "
},
{
"text": "particle nature of electron "
},
{
"text": "both wave and particle nature of electron "
},
{
"text": "none of these "
}
],
"answer": "wave nature of electron ",
"solution": "**Answer:** wave nature of electron \n\nFormation of covalent bond is best explained by molecular orbital theory.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8752,
"subject": "Physics",
"question": "If the kinetic energy of a free electron doubles, it's deBroglie wavelength changes by the factor ",
"options": [
{
"text": "$$2$$ "
},
{
"text": "$${1 \\over 2}$$ "
},
{
"text": "$${\\sqrt 2 }$$ "
},
{
"text": "$${1 \\over {\\sqrt 2 }}$$ "
}
],
"answer": "$${1 \\over {\\sqrt 2 }}$$ ",
"solution": "**Answer:** $${1 \\over {\\sqrt 2 }}$$ \n\nde-Broglie wavelength, \n
If $$K.E$$ is doubled, wavelength becomes $${\\lambda \\over {\\sqrt 2 }}$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 8753,
"subject": "Physics",
"question": "A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The\ncollision is head on, and elastic. The ratio of the de-Broglie wavelengths $${\\lambda _A}$$ to $${\\lambda _B}$$ after the collision is:",
"options": [
{
"text": "$${{{\\lambda _A}} \\over {{\\lambda _B}}} = {1 \\over 3}$$ "
},
{
"text": "$${{{\\lambda _A}} \\over {{\\lambda _B}}} = 2$$"
},
{
"text": "$${{{\\lambda _A}} \\over {{\\lambda _B}}} = {2 \\over 3}$$"
},
{
"text": "$${{{\\lambda _A}} \\over {{\\lambda _B}}} = {1 \\over 2}$$"
}
],
"answer": "$${{{\\lambda _A}} \\over {{\\lambda _B}}} = 2$$",
"solution": "**Answer:** $${{{\\lambda _A}} \\over {{\\lambda _B}}} = 2$$\n\nFrom question, mA = M; mB = $${m \\over 2}$$\n
uA = V and uB = 0\n
Let after collision velocity of A = V1 and\n
velocity of B = V2\n
Applying law of conservation of momentum,\n
mu = mv1 + $$\\left( {{m \\over 2}} \\right){v_2}$$\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8757,
"subject": "Physics",
"question": "Both the nucleus and the atom of some element arein their respective first excited states. They get de-excted by emitting photons of wavelengths $$\\lambda $$N, $$\\lambda $$A respectively. The ratio $${{{}^\\lambda N} \\over {{}^\\lambda A}}$$is closest to : ",
"options": [
{
"text": "10$$-$$6"
},
{
"text": "10"
},
{
"text": "10$$-$$10"
},
{
"text": "10$$-$$1"
}
],
"answer": "10$$-$$6",
"solution": "**Answer:** 10$$-$$6\n\n
We know that $$E = {{hc} \\over \\lambda }$$
\n
So, for atom $${E_A} = {{hc} \\over {{\\lambda _A}}}$$
\n
And for neutron $${E_N} = {{hc} \\over {{\\lambda _N}}}$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8758,
"subject": "Physics",
"question": "In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10–12 m, the minimum electron energy required is close to -",
"options": [
{
"text": "25 keV"
},
{
"text": "500 keV"
},
{
"text": "100 keV"
},
{
"text": "1 keV"
}
],
"answer": "25 keV",
"solution": "**Answer:** 25 keV\n\n$$\\lambda $$ = $${h \\over p}$$ {$$\\lambda $$ = 7.5 $$ \\times $$ 10$$-$$12}\n
v $$ = 1.45 \\times {10^6}$$ m/s",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8760,
"subject": "Physics",
"question": "A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V. Another particle B of mass ' 4 m' and charge 'q' is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths $${{{\\lambda _A}} \\over {{\\lambda _B}}}$$ is close to :",
"options": [
{
"text": "4.47"
},
{
"text": "10.00"
},
{
"text": "14.14"
},
{
"text": "0.07"
}
],
"answer": "14.14",
"solution": "**Answer:** 14.14\n\nK.E. acquired by charge = K = qV\n
$$ \\Rightarrow $$ $${1 \\over {{\\lambda ^2}}} = {1 \\over {\\lambda _1^2}} + {1 \\over {\\lambda _2^2}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8762,
"subject": "Physics",
"question": "A nucleus A, with a finite de-broglie\nwavelength $$\\lambda $$A, undergoes spontaneous fission\ninto two nuclei B and C of equal mass. B flies\nin the same direction as that of A, while C flies\nin the opposite direction with a velocity equal\nto half of that of B. The de-Broglie wavelengths\n$$\\lambda $$B and $$\\lambda $$C of B and C are respectively :",
"options": [
{
"text": "$$\\lambda $$A, 2$$\\lambda $$A"
},
{
"text": "2$$\\lambda $$A, $$\\lambda $$A"
},
{
"text": "$$\\lambda $$A, $$\\lambda $$A/2"
},
{
"text": "$$\\lambda $$A/2, $$\\lambda $$A"
}
],
"answer": "$$\\lambda $$A/2, $$\\lambda $$A",
"solution": "**Answer:** $$\\lambda $$A/2, $$\\lambda $$A\n\nLet mass of B and C is m each. By momentum conservation
\n$$ \\Rightarrow \\lambda = {{{\\lambda _x}{\\lambda _y}} \\over {\\left| {{\\lambda _x} - {\\lambda _y}} \\right|}}$$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8764,
"subject": "Physics",
"question": "An electron of mass m and magnitude of charge\n|e| initially at rest gets accelerated by a constant\nelectric field E. The rate of change of de-Broglie\nwavelength of this electron at time t ignoring\nrelativistic effects is :",
"options": [
{
"text": "$${{ - h} \\over {\\left| e \\right|Et}}$$"
},
{
"text": "$${{ - h} \\over {\\left| e \\right|E\\sqrt t }}$$"
},
{
"text": "$${{ - h} \\over {\\left| e \\right|E{t^2}}}$$"
},
{
"text": "$${{\\left| e \\right|Et} \\over h}$$"
}
],
"answer": "$${{ - h} \\over {\\left| e \\right|E{t^2}}}$$",
"solution": "**Answer:** $${{ - h} \\over {\\left| e \\right|E{t^2}}}$$\n\nF = |e| E\n
$$ \\therefore $$ $$\\lambda $$e > $$\\lambda $$p > $$\\lambda $$He++",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8767,
"subject": "Physics",
"question": "A particle is moving 5 times as fast as an\nelectron. The ratio of the de-Broglie wavelength\nof the particle to that of the electron is 1.878 $$ \\times $$\n10–4. The mass of the particle is close to",
"options": [
{
"text": "1.2 $$ \\times $$ 10–28 kg"
},
{
"text": "9.1 $$ \\times $$ 10–31 kg"
},
{
"text": "4.8 $$ \\times $$ 10–27 kg"
},
{
"text": "9.7 $$ \\times $$ 10–28 kg"
}
],
"answer": "9.7 $$ \\times $$ 10–28 kg",
"solution": "**Answer:** 9.7 $$ \\times $$ 10–28 kg\n\nLet mass of particle = m\n
$$ = 2\\sqrt 2 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8775,
"subject": "Physics",
"question": "An electron of mass me and a proton of mass mp = 1836 me are moving with the same speed. The ratio of their de Broglie wavelength $${{{}^\\lambda electron} \\over {{}^\\lambda proton}}$$ will be :",
"options": [
{
"text": "1"
},
{
"text": "1836"
},
{
"text": "$${1 \\over {1836}}$$"
},
{
"text": "918"
}
],
"answer": "1836",
"solution": "**Answer:** 1836\n\nGiven mass of electron = me
$${{{\\lambda _e}} \\over {{\\lambda _p}}} = 1836$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8776,
"subject": "Physics",
"question": "Given below are two statements : one is labeled as Assertion A and the other is labelled as Reason R.
Assertion A : An electron microscope can achieve better resolving power than an optical microscope.
Reason R : The de Broglie's wavelength of the electrons emitted from an electron gun is much less than wavelength of visible light.
In the light of the above statements, choose the correct answer from the options given below :",
"options": [
{
"text": "A is false but R is true."
},
{
"text": "Both A and R are true but R is NOT the correct explanation of A."
},
{
"text": "Both A and R are true and R is the correct explanation of A."
},
{
"text": "A is true but R is false."
}
],
"answer": "Both A and R are true and R is the correct explanation of A.",
"solution": "**Answer:** Both A and R are true and R is the correct explanation of A.\n\nResolution limit ($$\\Delta$$$$\\theta$$) = $${{1.22\\lambda } \\over d}$$
Resolution power = $${1 \\over {{\\mathop{\\rm Re}\\nolimits} solution\\,\\lim it}}$$
Since, wavelength of electron is much less than visible light, its resolving power will be much more.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8777,
"subject": "Physics",
"question": "The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths? (mp = 1.00727u me = 0.00055u)",
"options": [
{
"text": "41.4 : 1"
},
{
"text": "(1860)2 : 1"
},
{
"text": "1860 : 1"
},
{
"text": "43 : 1"
}
],
"answer": "43 : 1",
"solution": "**Answer:** 43 : 1\n\n$${\\lambda _e} = {{12.27} \\over {\\sqrt V }}\\mathop A\\limits^o $$
$${\\lambda _p} = {{0.286} \\over {\\sqrt V }}\\mathop A\\limits^o $$
$${{{\\lambda _e}} \\over {{\\lambda _p}}} = {{12.27} \\over {0.286}} = 43$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8778,
"subject": "Physics",
"question": "An electron of mass m and a photon have same energy E. The ratio of wavelength of electron to that of photon is : (c being the velocity of light)",
"options": [
{
"text": "$${1 \\over c}{\\left( {{{2m} \\over E}} \\right)^{1/2}}$$"
},
{
"text": "$${1 \\over c}{\\left( {{E \\over {2m}}} \\right)^{1/2}}$$"
},
{
"text": "$${\\left( {{E \\over {2m}}} \\right)^{1/2}}$$"
},
{
"text": "$$c{(2mE)^{1/2}}$$"
}
],
"answer": "$${1 \\over c}{\\left( {{E \\over {2m}}} \\right)^{1/2}}$$",
"solution": "**Answer:** $${1 \\over c}{\\left( {{E \\over {2m}}} \\right)^{1/2}}$$\n\nFor photon, E = $${{hc} \\over \\lambda }$$
$${\\lambda _p} = {{hc} \\over E}$$ .... (i)
For electron, $${\\lambda _e} = {{hc} \\over {\\sqrt {2mE} }}$$ .... (ii)
$${{{\\lambda _e}} \\over {{\\lambda _p}}} = {{{{hc} \\over {\\sqrt {2mE} }}} \\over {{{hc} \\over E}}} = \\sqrt {{E \\over {2m{c^2}}}} = {1 \\over c}{\\left( {{E \\over {2m}}} \\right)^{1/2}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8779,
"subject": "Physics",
"question": "A particle is travelling 4 time as fast as an electron. Assuming the ratio of de-Broglie wavelength of a particle to that of electron is 2 : 1, the mass of the particle is :",
"options": [
{
"text": "$${1 \\over {16}}$$ times the mass of e$$-$$"
},
{
"text": "8 times the mass of e$$-$$"
},
{
"text": "16 times the mass of e$$-$$"
},
{
"text": "$${1 \\over {8}}$$ times the mass of e$$-$$"
}
],
"answer": "$${1 \\over {8}}$$ times the mass of e$$-$$",
"solution": "**Answer:** $${1 \\over {8}}$$ times the mass of e$$-$$\n\n$$\\lambda = {h \\over p}$$
$${\\lambda _c} = {{2mc{\\lambda ^2}} \\over h}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8781,
"subject": "Physics",
"question": "An electron of mass me and a proton of mass mp are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is ",
"options": [
{
"text": "$${{{m_e}} \\over {{m_p}}}$$"
},
{
"text": "1"
},
{
"text": "$${{{m_p}} \\over {{m_e}}}$$"
},
{
"text": "$$\\sqrt {{{{m_p}} \\over {{m_e}}}} $$"
}
],
"answer": "$$\\sqrt {{{{m_p}} \\over {{m_e}}}} $$",
"solution": "**Answer:** $$\\sqrt {{{{m_p}} \\over {{m_e}}}} $$\n\n$$\\lambda = {h \\over p} = {h \\over {\\sqrt {2km} }}$$
$${{{\\lambda _e}} \\over {{\\lambda _p}}} = \\sqrt {{{{k_p}{m_p}} \\over {{k_e}{m_e}}}} = \\sqrt {{{{m_p}} \\over {{m_e}}}} $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8782,
"subject": "Physics",
"question": "What should be the order of arrangement of de-Broglie wavelength of electron ($$\\lambda$$e), an $$\\alpha$$-particle ($$\\lambda$$a) and proton ($$\\lambda$$p) given that all have the same kinetic energy?",
"options": [
{
"text": "$$\\lambda$$e = $$\\lambda$$p = $$\\lambda$$$$\\alpha$$"
},
{
"text": "$$\\lambda$$e < $$\\lambda$$p < $$\\lambda$$$$\\alpha$$"
},
{
"text": "$$\\lambda$$e > $$\\lambda$$p > $$\\lambda$$$$\\alpha$$"
},
{
"text": "$$\\lambda$$e = $$\\lambda$$p > $$\\lambda$$$$\\alpha$$"
}
],
"answer": "$$\\lambda$$e > $$\\lambda$$p > $$\\lambda$$$$\\alpha$$",
"solution": "**Answer:** $$\\lambda$$e > $$\\lambda$$p > $$\\lambda$$$$\\alpha$$\n\n$$\\lambda = {h \\over p} = {h \\over {\\sqrt {2mE} }} \\propto {1 \\over {\\sqrt m }}$$
m$$\\alpha$$ > mp > me
So, $$\\lambda$$e > $$\\lambda$$p > $$\\lambda$$$$\\alpha$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8783,
"subject": "Physics",
"question": "A particle of mass 4M at rest disintegrates into two particles of mass M and 3M respectively having non zero velocities. The ratio of de-Broglie wavelength of particle of mass M to that of mass 3M will be :",
"options": [
{
"text": "1 : 3"
},
{
"text": "3 : 1"
},
{
"text": "1 : $$\\sqrt 3 $$"
},
{
"text": "1 : 1"
}
],
"answer": "1 : 1",
"solution": "**Answer:** 1 : 1\n\n$$\\lambda = {h \\over p}$$
both the particles will move with momentum same in magnitude & opposite in direction.
So De-Broglie wavelength of both will be same i.e. ratio 1 : 1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8784,
"subject": "Physics",
"question": "An electron moving with speed v and a photon moving with speed c, have same D-Broglie wavelength. The ratio of kinetic energy of electron to that of photon is :",
"options": [
{
"text": "$${{3c} \\over v}$$"
},
{
"text": "$${v \\over {3c}}$$"
},
{
"text": "$${v \\over {2c}}$$"
},
{
"text": "$${{2c} \\over v}$$"
}
],
"answer": "$${v \\over {2c}}$$",
"solution": "**Answer:** $${v \\over {2c}}$$\n\n$${\\lambda _e} = {\\lambda _{Ph}}$$
$$ = {v \\over {2c}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8785,
"subject": "Physics",
"question": "A particle of mass 9.1 $$\\times$$ 10$$-$$31 kg travels in a medium with a speed of 106 m/s and a photon of a radiation of linear momentum 10$$-$$27 kg m/s travels in vacuum. The wavelength of photon is __________ times the wavelength of the particle.",
"options": [],
"answer": "910",
"solution": "**Answer:** 910\n\nFor photon $${\\lambda _1} = {h \\over P} = {{6.6 \\times {{10}^{ - 34}}} \\over {{{10}^{ - 27}}}}$$
$$\\therefore$$ $${{{\\lambda _1}} \\over {{\\lambda _2}}} = 910$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8786,
"subject": "Physics",
"question": "The de-Broglie wavelength of a particle having kinetic energy E is $$\\lambda$$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?",
"options": [
{
"text": "$${1 \\over 9}$$E"
},
{
"text": "$${7 \\over 9}$$E"
},
{
"text": "E"
},
{
"text": "$${16 \\over 9}$$E"
}
],
"answer": "$${7 \\over 9}$$E",
"solution": "**Answer:** $${7 \\over 9}$$E\n\n$$\\lambda = {h \\over {mv}} = {h \\over {\\sqrt {2mE} }}$$, $$mv = \\sqrt {2mE} $$
Extra energy given = $${{16} \\over 9}E - E = {7 \\over 9}E$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8787,
"subject": "Physics",
"question": "A moving proton and electron have the same de-Broglie wavelength. If K and P denote the K.E. and momentum respectively. Then choose the correct option :",
"options": [
{
"text": "Kp < Ke and Pp = Pe"
},
{
"text": "Kp = Ke and Pp = Pe"
},
{
"text": "Kp < Ke an Pp < Pe"
},
{
"text": "Kp > Ke and Pp = Pe"
}
],
"answer": "Kp < Ke and Pp = Pe",
"solution": "**Answer:** Kp < Ke and Pp = Pe\n\n$${\\lambda _p} = {h \\over {{P_p}}}$$
$${\\lambda _e} = {h \\over {{P_e}}}$$
$$\\because$$ $${\\lambda _p} = {\\lambda _e}$$
$$ \\Rightarrow {P_p} = {P_e}$$
$${(K)_p} = {{P_p^2} \\over {2{m_p}}}$$
$${(K)_e} = {{P_e^2} \\over {2{m_e}}}$$
Kp < Ke as mp > me
Option (a)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8788,
"subject": "Physics",
"question": "Consider two separate ideal gases of electrons and protons having same number of particles. The temperature of both the gases are same. The ratio of the uncertainty in determining the position of an electron to that of a proton is proportional to :-",
"options": [
{
"text": "$${\\left( {{{{m_p}} \\over {{m_e}}}} \\right)^{3/2}}$$"
},
{
"text": "$$\\sqrt {{{{m_e}} \\over {{m_p}}}} $$"
},
{
"text": "$$\\sqrt {{{{m_p}} \\over {{m_e}}}} $$"
},
{
"text": "$${{{m_p}} \\over {{m_e}}}$$"
}
],
"answer": "$$\\sqrt {{{{m_p}} \\over {{m_e}}}} $$",
"solution": "**Answer:** $$\\sqrt {{{{m_p}} \\over {{m_e}}}} $$\n\n$$\\Delta x\\,.\\,\\Delta p \\ge {h \\over {4\\pi }}$$
$$\\Delta x = {h \\over {4\\pi m\\Delta v}}$$
$$v = \\sqrt {{{3KT} \\over m}} $$
$${{\\Delta {x_e}} \\over {\\Delta {x_p}}} = \\sqrt {{{{m_p}} \\over {{m_e}}}} $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8789,
"subject": "Physics",
"question": "The temperature of an ideal gas in 3-dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is :
$$\\therefore$$ The corresponding de-Broglie wavelength of an electron approximately at 300 K is 6.26 nm.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8790,
"subject": "Physics",
"question": "
The de Broglie wavelengths for an electron and a photon are $$\\lambda$$e and $$\\lambda$$p respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two ?
An $$\\alpha$$ particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths $$({\\lambda _\\alpha }:{\\lambda _{C12}})$$ is :
An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are Ee and pe and that of photon are Eph and pph respectively. Which of the following is correct?
A proton, a neutron, an electron and an $$\\alpha$$ particle have same energy. If $$\\lambda$$p, $$\\lambda$$n, $$\\lambda$$e and $$\\lambda$$a are the de Broglie's wavelengths of proton, neutron, electron and $$\\alpha$$ particle respectively, then choose the correct relation from the following :
An electron (mass $$\\mathrm{m}$$) with an initial velocity $$\\vec{v}=v_{0} \\hat{i}\\left(v_{0}>0\\right)$$ is moving in an electric field $$\\vec{E}=-E_{0} \\hat{i}\\left(E_{0}>0\\right)$$ where $$E_{0}$$ is constant. If at $$\\mathrm{t}=0$$ de Broglie wavelength is $$\\lambda_{0}=\\frac{h}{m v_{0}}$$, then its de Broglie wavelength after time t is given by
The equation $$\\lambda=\\frac{1.227}{x} \\mathrm{~nm}$$ can be used to find the de-Brogli wavelength of an electron. In this equation $$x$$ stands for :
\n
Where
\n
$$\\mathrm{m}=$$ mass of electron
\n
$$\\mathrm{P}=$$ momentum of electron
\n
$$\\mathrm{K}=$$ Kinetic energy of electron
\n
$$\\mathrm{V}=$$ Accelerating potential in volts for electron
An $$\\alpha$$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:
A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of $$\\lambda$$. An alpha particle having certain kinetic energy has the same de-Brogle wavelength $$\\lambda$$. The ratio of kinetic energy of proton and that of alpha particle is:
$\\Rightarrow \\frac{k_{P}}{k_{\\alpha}}=\\left(\\frac{m_{\\alpha}}{m_{P}}\\right)=\\frac{4}{1}=4: 1$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8799,
"subject": "Physics",
"question": "An electron accelerated through a potential difference $V_{1}$ has a de-Broglie wavelength of $\\lambda$. When the potential is changed to $V_{2}$, its de-Broglie wavelength increases by $50 \\%$. The value of $\\left(\\frac{V_{1}}{V_{2}}\\right)$ is equal to",
"options": [
{
"text": "$\\frac{3}{2}$"
},
{
"text": "4"
},
{
"text": "3"
},
{
"text": "$\\frac{9}{4}$"
}
],
"answer": "$\\frac{9}{4}$",
"solution": "**Answer:** $\\frac{9}{4}$\n\n
The ratio of de-Broglie wavelength of an $$\\alpha$$ particle and a proton accelerated from rest by the same potential is $$\\frac{1}{\\sqrt m}$$, the value of m is -
Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de-Broglie wavelength of $$\\lambda_0$$. IF the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be :
",
"options": [
{
"text": "3 $$\\lambda_0$$"
},
{
"text": "9 $$\\lambda_0$$"
},
{
"text": "$$\\frac{\\lambda_0}{\\sqrt2}$$"
},
{
"text": "$$\\frac{\\lambda_0}{2}$$"
}
],
"answer": "$$\\frac{\\lambda_0}{\\sqrt2}$$",
"solution": "**Answer:** $$\\frac{\\lambda_0}{\\sqrt2}$$\n\nWhen electron is accelerated through potential difference $V$, then
\n$$\n\\begin{aligned}\n& \\text { K.E. }=\\mathrm{eV} \\\\\\\\\n& \\Rightarrow \\lambda=\\frac{\\mathrm{h}}{\\sqrt{2 \\mathrm{~m}(\\mathrm{KE})}}=\\frac{\\mathrm{h}}{\\sqrt{2 \\mathrm{meV}}} \\\\\\\\\n& \\therefore \\lambda \\alpha \\frac{1}{\\sqrt{\\mathrm{V}}} \\\\\\\\\n& \\therefore \\frac{\\lambda}{\\lambda_0}=\\sqrt{\\frac{20}{40}} \\\\\\\\\n& \\therefore \\lambda=\\frac{\\lambda_0}{\\sqrt{2}}\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8802,
"subject": "Physics",
"question": "An $$\\alpha$$-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:",
"options": [
{
"text": "$${\\lambda _\\alpha } > {\\lambda _p} < {\\lambda _e}$$"
},
{
"text": "$${\\lambda _\\alpha } > {\\lambda _p} > {\\lambda _e}$$"
},
{
"text": "$${\\lambda _\\alpha } = {\\lambda _p} = {\\lambda _e}$$"
},
{
"text": "$${\\lambda _\\alpha } < {\\lambda _p} < {\\lambda _e}$$"
}
],
"answer": "$${\\lambda _\\alpha } < {\\lambda _p} < {\\lambda _e}$$",
"solution": "**Answer:** $${\\lambda _\\alpha } < {\\lambda _p} < {\\lambda _e}$$\n\n$\\lambda=\\frac{h}{m v}=\\frac{h}{\\sqrt{2 m k}}$\n
\nSo, $\\lambda \\propto \\frac{1}{\\sqrt{m}}$\n
\nSo, $\\lambda_{e}>\\lambda_{p}>\\lambda_{\\alpha}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8803,
"subject": "Physics",
"question": "The de Broglie wavelength of an electron having kinetic energy $\\mathrm{E}$ is $\\lambda$. If the kinetic energy of electron becomes $\\frac{E}{4}$, then its de-Broglie wavelength will be :",
"options": [
{
"text": "$\\sqrt{2} \\lambda$"
},
{
"text": "$2 \\lambda$"
},
{
"text": "$\\frac{\\lambda}{2}$"
},
{
"text": "$\\frac{\\lambda}{\\sqrt{2}}$"
}
],
"answer": "$2 \\lambda$",
"solution": "**Answer:** $2 \\lambda$\n\n\n$$\n\\lambda = \\frac{h}{\\sqrt{2mE}}\n$$\n
\nwhere $h$ is Planck's constant, $m$ is the mass of the particle, and $E$ is its kinetic energy.\n
\nWe are given that the de Broglie wavelength of an electron with kinetic energy $E$ is $\\lambda$, and we want to find the de Broglie wavelength of the same electron when its kinetic energy becomes $\\frac{E}{4}$.\n
\nTo do this, we can use the formula for the de Broglie wavelength again, but with the new kinetic energy $\\frac{E}{4}$:\n
\nwhere we have used the fact that $\\sqrt{\\frac{1}{4}} = \\frac{1}{\\sqrt{4}} = \\frac{1}{2}$ to simplify the expression.\n
\nTherefore, the de Broglie wavelength of the electron when its kinetic energy becomes $\\frac{E}{4}$ is twice its original value, or $\\boxed{2\\lambda}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8804,
"subject": "Physics",
"question": "
A proton and an $$\\alpha$$-particle are accelerated from rest by $$2 \\mathrm{~V}$$ and $$4 \\mathrm{~V}$$ potentials, respectively. The ratio of their de-Broglie wavelength is :
\nTherefore, the ratio of their de-Broglie wavelengths is 4 : 1.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8805,
"subject": "Physics",
"question": "
The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy :
\n
(Assume $$m_{p}=m_{e} \\times 1849$$ )
",
"options": [
{
"text": "1:43"
},
{
"text": "1:62"
},
{
"text": "2:43"
},
{
"text": "1:30"
}
],
"answer": "1:43",
"solution": "**Answer:** 1:43\n\nThe de Broglie wavelength (λ) of a particle can be found using the formula:\n
\n$$\n\\lambda = \\frac{h}{p}\n$$\n
\nwhere h is the Planck constant and p is the momentum of the particle. The momentum of a particle can be expressed in terms of its kinetic energy (K) and mass (m) as follows:\n
\n$$\np = \\sqrt{2mK}\n$$\n
\nCombining these two equations, we get:\n
\n$$\n\\lambda = \\frac{h}{\\sqrt{2mK}}\n$$\n
\nNow, we are given that the kinetic energy of the proton and electron is the same. Let's denote the masses of the proton and electron as $$m_p$$ and $$m_e$$, respectively. We are given the relationship between the two masses:\n
\n$$\nm_p = 1849 \\times m_e\n$$\n
\nLet's find the ratio of the de Broglie wavelengths of the proton ($λ_p$) and the electron ($λ_e$):\n
\nThus, the ratio of the de Broglie wavelengths of the proton and electron having the same kinetic energy is 1:43.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8806,
"subject": "Physics",
"question": "
The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is $$\\lambda_1$$. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes
The de Broglie wavelength of a particle is given by:
\n
$$\\lambda = \\frac{h}{p}$$
\n
where h is Planck's constant and p is the momentum of the particle. The momentum of a gas molecule can be related to its kinetic energy (which is related to the temperature of the gas) by:
\n
$$p = \\sqrt{2mK}$$
\n
where m is the mass of the molecule and K is the kinetic energy of the molecule.
\n
At a given temperature T, the average kinetic energy of a molecule in a gas is given by:
\n
$$K = \\frac{3}{2} kT$$
\n
where k is Boltzmann's constant.
\n
Therefore, the de Broglie wavelength of a molecule in a gas is given by:
Proton $$(\\mathrm{P})$$ and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, $$\\mathrm{m}_{\\mathrm{p}}=1849 \\mathrm{~m}_{\\mathrm{e}}$$ ):
The kinetic energy of an electron, $$\\alpha$$-particle and a proton are given as $$4 \\mathrm{~K}, 2 \\mathrm{~K}$$ and $$\\mathrm{K}$$ respectively. The de-Broglie wavelength associated with electron $$(\\lambda \\mathrm{e}), \\alpha$$-particle $$((\\lambda \\alpha)$$ and the proton $$(\\lambda p)$$ are as follows:
This equation shows that the de Broglie wavelength of a particle is inversely proportional to the square root of its mass and its kinetic energy. This means that the particle with the smallest mass and kinetic energy will have the largest de Broglie wavelength.
\n
Given that the kinetic energies of the electron, alpha particle, and proton are $4K$, $2K$, and $K$, respectively, and knowing that the mass of the electron is less than the mass of the proton and the mass of the proton is less than the mass of the alpha particle, we can infer that the de Broglie wavelength of the electron is greater than the de Broglie wavelength of the proton, which in turn is greater than the de Broglie wavelength of the alpha particle.
\n
Therefore, the correct answer is $\\lambda_{\\alpha} < \\lambda_{p} < \\lambda_{e}$.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8809,
"subject": "Physics",
"question": "The de Broglie wavelengths of a proton and an $\\alpha$ particle are $\\lambda$ and $2 \\lambda$ respectively. The ratio of the velocities of proton and $\\alpha$ particle will be :",
"options": [
{
"text": "$8: 1$"
},
{
"text": "$1: 2$"
},
{
"text": "$1: 8$"
},
{
"text": "$4: 1$"
}
],
"answer": "$8: 1$",
"solution": "**Answer:** $8: 1$\n\n
To find the ratio of velocities of two particles based on their de Broglie wavelengths, we can use the de Broglie wavelength formula, which relates the momentum of a particle to its wavelength. The de Broglie's wavelength formula is given by:
\n\n
$$ \\lambda = \\frac{h}{p} $$
\n\n
where: \n\n
$\\lambda$ is the de Broglie wavelength,
\n\n
$h$ is the Planck constant, and
\n\n
$p$ is the momentum of the particle.
\n\n
The momentum $p$ of a particle can also be expressed as the product of its mass $m$ and velocity $v$:
\n\n
$$ p = mv $$
\n\n
So, we can rewrite the de Broglie wavelength equation in terms of mass and velocity:
\n\n
$$ \\lambda = \\frac{h}{mv} $$
\n\n
For the proton (let's use subscript $p$ for proton), the wavelength is $\\lambda$:
\n\n
$$ \\lambda_p = \\frac{h}{m_p v_p}$$
\n\n
For the $\\alpha$ particle (let's use subscript $\\alpha$ for alpha), the wavelength is $2\\lambda$:
\n\n
$$ 2\\lambda = \\frac{h}{m_\\alpha v_\\alpha}$$
\n\n
We are interested in finding the ratio of the velocities $\\frac{v_p}{v_\\alpha}$. Using the given data about wavelengths:
\n\n
$$ \\lambda_p = \\lambda $$
\n\n
$$ \\lambda_\\alpha = 2\\lambda $$
\n\n
Using the de Broglie equation for both particles:
\n\n
$$ \\frac{h}{m_p v_p} = \\lambda $$
\n\n
$$ \\frac{h}{m_\\alpha v_\\alpha} = 2\\lambda $$
\n\n
Dividing the second equation by the first equation gives us:
Since we know an $\\alpha$ particle consists of 2 protons and 2 neutrons (essentially four nucleons), the mass of an $\\alpha$ particle is roughly four times the mass of a proton ($m_\\alpha \\approx 4m_p$).
\n\n
Substituting $m_\\alpha$ with $4m_p$ in the equation:
The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is $$25 \\%$$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:
Since $$p_{electron} = p_{photon}$$ (from the de-Broglie relation), we can replace $$p_{photon}$$ with $$p_{electron}$$ which is $$m_{e}v_{electron}$$:
To determine the relationship between the de-Broglie wavelengths of a proton, an electron, and an alpha particle with the same energies, we need to use the de-Broglie wavelength formula:
\n\n
\n\n
$$\\lambda = \\frac{h}{p}$$
\n\n\n\n
\n\n
where $ h $ is Planck's constant and $ p $ is the momentum of the particle.
\n\n\n\n
For particles with the same kinetic energy $ E $, we have:
\n\n
\n\n
$$E = \\frac{p^2}{2m}$$
\n\n\n\n
\n\n
Solving for $ p $:
\n\n\n\n
\n\n
$$p = \\sqrt{2mE}$$
\n\n\n\n
\n\n
Substituting this into the de-Broglie equation, we get:
\n\n\n\n
\n\n
$$\\lambda = \\frac{h}{\\sqrt{2mE}}$$
\n\n\n\n
\n\n
Since all three particles have the same energy $ E $, the de-Broglie wavelength is inversely proportional to the square root of the mass $ m $:
\n\n\n\n
\n\n
$$\\lambda \\propto \\frac{1}{\\sqrt{m}}$$
\n\n\n\n
\n\n
The masses of the particles are as follows:
\n\n\n\n
\n\n
Mass of proton $ m_p $ is approximately $ 1.67 \\times 10^{-27} $ kg
\n\n
Mass of electron $ m_e $ is approximately $ 9.11 \\times 10^{-31} $ kg
\n\n
Mass of alpha particle $ m_{\\alpha} $ is approximately $ 4 \\times 1.67 \\times 10^{-27} $ kg (since it has 2 protons and 2 neutrons)
\n\n
\n\n
Comparatively:
\n\n
\n\n
Mass of alpha particle $ m_{\\alpha} $ is the largest.
\n\n
Mass of proton $ m_p $ is intermediate.
\n\n
Mass of electron $ m_e $ is the smallest.
\n\n
\n\n
Therefore, the de-Broglie wavelength will be:
\n\n
\n\n
Largest for the electron ($ \\lambda_{\\mathrm{e}} $)
\n\n
Intermediate for the proton ($ \\lambda_{\\mathrm{p}} $)
\n\n
Smallest for the alpha particle ($ \\lambda_{\\alpha} $)
A proton and an electron have the same de Broglie wavelength. If $$\\mathrm{K}_{\\mathrm{p}}$$ and $$\\mathrm{K}_{\\mathrm{e}}$$ be the kinetic energies of proton and electron respectively, then choose the correct relation :
To determine the correct relation between the kinetic energies of a proton ($$\\mathrm{K}_{\\mathrm{p}}$$) and an electron ($$\\mathrm{K}_{\\mathrm{e}}$$) when they have the same de Broglie wavelength, we need to use the de Broglie wavelength formula:
\n\n
\n\n
$$\\lambda = \\frac{h}{p}$$
\n\n\n\n
where $$\\lambda$$ is the de Broglie wavelength, $$h$$ is Planck's constant, and $$p$$ is the momentum of the particle.
\n\n
The momentum $$p$$ of a particle is given by:
\n\n
\n\n
$$p = \\sqrt{2mK}$$
\n\n\n\n
where $$m$$ is the mass of the particle and $$K$$ is its kinetic energy. For a proton and an electron with the same de Broglie wavelength:
To solve for the ratio of the kinetic energies of a proton and an electron with the same de-Broglie wavelength, let us first recall the relationship between kinetic energy, momentum, and the de-Broglie wavelength.
\n\n
The de-Broglie wavelength $$\\lambda$$ is given by:
\n\n
$$\\lambda = \\frac{h}{p}$$
\n\n
where $$h$$ is Planck’s constant and $$p$$ is the momentum.
\n\n
Since the proton and the electron have the same de-Broglie wavelength, their momenta must be equal:
Given that the mass of the proton $$m_{\\text{p}}$$ is 1836 times the mass of the electron $$m_{\\text{e}}$$, we have:
\n\n
$$\\frac{m_{\\text{p}}}{m_{\\text{e}}} = 1836$$
\n\n
Thus, the ratio of the kinetic energies is:
\n\n
$$\\frac{K_{\\text{e}}}{K_{\\text{p}}} = 1836$$
\n\n
Therefore, the correct answer is:
\n\n
Option C
\n\n
$$1: 1836$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8814,
"subject": "Physics",
"question": "Photon of frequency $$v$$ has a momentum associated with it. If $$c$$ is the velocity of light, the momentum is ",
"options": [
{
"text": "$$hv/c$$ "
},
{
"text": "$$v/c$$ "
},
{
"text": "$$h$$ $$v$$ $$c$$ "
},
{
"text": "$$hv/{c^2}$$ "
}
],
"answer": "$$hv/c$$ ",
"solution": "**Answer:** $$hv/c$$ \n\nEnergy of a photon of frequency $$v$$ is given by $$E = hv.$$\n
Also, $$E = m{c^2},\\,\\,m{c^2} = hv$$\n
$$ \\Rightarrow mc = {{hv} \\over C} \\Rightarrow p = {{hv} \\over c}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8815,
"subject": "Physics",
"question": "If a source of power $$4kW$$ produces $${10^{20}}$$ photons/second, the radiation belongs to a part of the spectrum called ",
"options": [
{
"text": "$$X$$ -rays "
},
{
"text": "ultraviolet rays "
},
{
"text": "microwaves "
},
{
"text": "$$\\gamma $$ - rays "
}
],
"answer": "$$X$$ -rays ",
"solution": "**Answer:** $$X$$ -rays \n\nPower, $$P = {{nhv} \\over t}$$\n
$$ \\Rightarrow v = {{P \\times t} \\over {nh}}$$\n
$$ = {{4 \\times {{10}^3} \\times 1} \\over {{{10}^{20}} \\times 6.63 \\times {{10}^{ - 34}}}} = 6 \\times {10^{16}}Hz$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8816,
"subject": "Physics",
"question": "A Laser light of wavelength 660 nm is used to weld Retina detachment. If a Laser pulse of width 60 ms and power 0.5 kW is used the approximate number of photons in the pulse are :\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8817,
"subject": "Physics",
"question": "In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to : ",
"options": [
{
"text": "2020 nm"
},
{
"text": "250 nm"
},
{
"text": "1700 nm"
},
{
"text": "220 nm"
}
],
"answer": "250 nm",
"solution": "**Answer:** 250 nm\n\nThe minimum wavelength of emitted photons is\n
$$\\lambda $$ = $${{1240} \\over {5.6 - 0.7}}nm$$ = 250 nm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8818,
"subject": "Physics",
"question": "A 2 mW laser operates at wavelength of 500 nm. The number of photons that will be emitted per second is :\n[Given Planck's constant h = 6.6 × 10–34 Js, speed of light c = 3.0 × 108\n m/s]",
"options": [
{
"text": "5 × 1015"
},
{
"text": "1.5 × 1016"
},
{
"text": "1 × 1016"
},
{
"text": "2 × 1016"
}
],
"answer": "5 × 1015",
"solution": "**Answer:** 5 × 1015\n\n$$2 \\times {10^{ - 3}} = {{hc} \\over \\lambda }{{dn} \\over {dt}}$$
\n$$ = {{1000} \\over {6.6 \\times 3}} \\times 10 = 5 \\times {10^{15}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8819,
"subject": "Physics",
"question": "A beam of electromagnetic radiation of intensity 6.4 × 10–5 W/cm2 is comprised of wavelength,\n$$\\lambda $$ = 310 nm. It falls normally on a metal (work function $$\\phi $$ = 2eV) of surface area of 1 cm2. If one\nin 103 photons ejects an elctron, total number of electrons ejected in 1 s is 10x. (hc = 1240 eVnm,\n1eV = 1.6 × 10–19 J), then x is _____.",
"options": [],
"answer": "11",
"solution": "**Answer:** 11\n\nEnergy of photon = $${{1240} \\over {310}}$$ = 4 eV\n
Energy is greater than work function so photoelectric effect will take place.\n
Total number of electrons = $${{{{10}^{14}}} \\over {{{10}^3}}}$$ = 1011",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8820,
"subject": "Physics",
"question": "Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of\nphotons of X-rays to the number density of photons of the visible light of the given wavelengths is :",
"options": [
{
"text": "$${1 \\over {500}}$$"
},
{
"text": "500"
},
{
"text": "250"
},
{
"text": "$${1 \\over {250}}$$"
}
],
"answer": "$${1 \\over {500}}$$",
"solution": "**Answer:** $${1 \\over {500}}$$\n\nGiven, wavelength of x ray ($$\\lambda $$1) = 1 nm\n
And wavelength of visible light ($$\\lambda $$2) = 500 nm\n
we know, Power$$(P) = {{nhc} \\over \\lambda }$$\n
Statement I : Two photons having equal linear momenta have equal wavelengths.
Statement II : If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease.
In the light of the above statements, choose the correct answer from the options given below.",
"options": [
{
"text": "Statement I is false but Statement II is true"
},
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Statement I is true but Statement II is false"
}
],
"answer": "Statement I is true but Statement II is false",
"solution": "**Answer:** Statement I is true but Statement II is false\n\nAs we know, $$\\lambda = {h \\over p} = {h \\over {\\sqrt {2mK} }}$$
If linear momenta of two photons are equal, then their wavelengths is also equal.
Also, if the wavelength is decreased, then the momentum and energy of photon will increase.
Hence, option (d) is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8822,
"subject": "Physics",
"question": "The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be :",
"options": [
{
"text": "4.34 m/s"
},
{
"text": "2.19 m/s"
},
{
"text": "3.25 m/s"
},
{
"text": "4.17 m/s"
}
],
"answer": "4.17 m/s",
"solution": "**Answer:** 4.17 m/s\n\n($$\\Delta$$E) Releases when photon going from n = 5 to n = 1
A source of monochromatic light liberates 9 $$\\times$$ 1020 photon per second with wavelength 600 nm when operated at 400 W. The number of photons emitted per second with wavelength of 800 nm by the source of monochromatic light operating at same power will be :
A parallel beam of light of wavelength $$900 \\mathrm{~nm}$$ and intensity $$100 \\,\\mathrm{Wm}^{-2}$$ is incident on a surface perpendicular to the beam. The number of photons crossing $$1 \\mathrm{~cm}^{2}$$ area perpendicular to the beam in one second is :
If a source of electromagnetic radiation having power $$15 \\mathrm{~kW}$$ produces $$10^{16}$$ photons per second, the radiation belongs to a part of spectrum is.
A small object at rest, absorbs a light pulse of power $$20 \\mathrm{~mW}$$ and duration $$300 \\mathrm{~ns}$$. Assuming speed of light as $$3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}$$, the momentum of the object becomes equal to :
A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is $$x \\times 10^{15} \\mathrm{~Hz}$$. The value of $$x$$ is ____________.
\n
(Given h $$=4.25 \\times 10^{-15} ~\\mathrm{eVs}$$ )
When a monochromatic light is incident on hydrogen atoms in the ground state (n = 1), the hydrogen atoms can absorb energy and transition to higher energy levels. When the atoms return to lower energy levels, they emit radiation of different wavelengths corresponding to the energy differences between the energy levels.
\n
The energy levels of the hydrogen atom are given by the formula:
\n
$$E_n = -\\frac{13.6 \\mathrm{~eV}}{n^2}$$
\n
where $$E_n$$ is the energy of the nth level and $$n$$ is the principal quantum number.
\n
Since the hydrogen atoms emit radiation of six different wavelengths, there must be six different transitions from the excited states back to lower energy levels.
\n
The six transitions correspond to the following energy level changes:
\n\n
From n = 2 to n = 1
\n
From n = 3 to n = 1
\n
From n = 3 to n = 2
\n
From n = 4 to n = 1
\n
From n = 4 to n = 2
\n
From n = 4 to n = 3
\n\n
The highest energy level involved is n = 4. Therefore, the incident light must have a frequency high enough to excite the hydrogen atoms from the ground state (n = 1) to n = 4.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8829,
"subject": "Physics",
"question": "Monochromatic light of frequency $6 \\times 10^{14} \\mathrm{~Hz}$ is produced by a laser. The power emitted is $2 \\times 10^{-3} \\mathrm{~W}$.
How many photons per second on an average, are emitted by the source ?
To find out the number of photons emitted per second by the laser, we can use the relationship between the energy of a single photon, the total energy emitted per second (power), and the number of photons emitted per second. The energy $E$ of a single photon is given by Planck's equation:
\n\n
$$ E = hf $$
\n\n
where:
\n
\n
$ h $ is Planck's constant ($ 6.63 \\times 10^{-34} \\mathrm{Js} $), and
\n
$ f $ is the frequency of the light ($ 6 \\times 10^{14} \\mathrm{Hz} $).
$$ N = \\frac{2 \\times 10^{-3}}{3.978 \\times 10^{-19}} $$
\n
$$ N = 5.03 \\times 10^{15} \\text{ photons per second} $$
\n\n
The number of photons emitted per second is approximately $5 \\times 10^{15}$. Therefore, the correct answer, rounded to one significant figure, is:
\n\n
Option A: $5 \\times 10^{15}$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8830,
"subject": "Physics",
"question": "Conductivity of a photodiode starts changing only if the wavelength of incident light is less than $660 \\mathrm{~nm}$. The band gap of photodiode is found to be $\\left(\\frac{\\mathrm{X}}{8}\\right) \\mathrm{eV}$. The value of $\\mathrm{X}$ is :
To find the value of $$ X $$ in the band gap $$ \\left(\\frac{\\mathrm{X}}{8}\\right) \\mathrm{eV} $$, we need to understand the relationship between the wavelength of light that can result in changes in the conductivity of a photodiode and the photodiode's band gap energy.
\n
The band gap energy ($$ E_{g} $$) of a material is the minimum energy required for an electron to transition from the valence band to the conduction band, thus creating a hole-electron pair and allowing conductivity to occur. When light with a certain wavelength ($$ \\lambda $$) is incident upon the photodiode, if the energy of the photons is greater than or equal to the band gap energy of the photodiode, then the photons can excite electrons and change the conductivity of the photodiode.
\n
The energy of a photon ($$ E_{photon} $$) is given by the equation:
\n
$$ E_{photon} = \\frac{hc}{\\lambda} $$
\n
where:
\n
\n
$$ h $$ is Planck's constant, which is given as $$ 6.6 \\times 10^{-34} $$ J·s.
\n
$$ c $$ is the speed of light in a vacuum ($$ \\approx 3 \\times 10^{8} $$ m/s).
\n
$$ \\lambda $$ is the wavelength of the incident light.
\n
\n
Then the band gap energy in terms of electron volts is found by converting the energy from joules to electron volts (eV) using the charge of an electron $$ e $$ ($$ 1.6 \\times 10^{-19} $$ C):
\n
$$ E_{g} = \\frac{hc}{\\lambda e} $$
\n
Given that the photodiode starts conducting when the wavelength of light is less than $$ 660 \\mathrm{~nm} $$, we should use this wavelength as the threshold wavelength $$ \\lambda_{threshold} $$:
Two sources of light emit with a power of $$200 \\mathrm{~W}$$. The ratio of number of photons of visible light emitted by each source having wavelengths $$300 \\mathrm{~nm}$$ and $$500 \\mathrm{~nm}$$ respectively, will be :
If the total energy transferred to a surface in time $$\\mathrm{t}$$ is $$6.48 \\times 10^5 \\mathrm{~J}$$, then the magnitude of the total momentum delivered to this surface for complete absorption will be:
In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at $$10.2 \\mathrm{~V}$$. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is ________ nm. (Given hc $$=1245 \\mathrm{~eV} \\mathrm{~nm}, \\mathrm{e}=1.6 \\times 10^{-19} \\mathrm{C}$$).
The Franck-Hertz experiment provides evidence for quantized energy levels within atoms. When atoms are excited by electrons with a specific kinetic energy, they can jump to higher energy levels. Upon returning to lower levels, they emit photons whose energies correspond to the difference between these levels. The first dip in the current-voltage graph for hydrogen, observed at $$10.2 \\mathrm{~V}$$, corresponds to the energy required to excite a hydrogen atom to its first excitation level. The wavelength of the light emitted when the atom returns to its ground state can be calculated using the energy of the photon emitted.
\n\n
To find the wavelength ($$\\lambda$$) of light emitted, we use the relationship between energy ($$E$$), Planck's constant ($$h$$), the speed of light ($$c$$), and wavelength ($$\\lambda$$), given in the equation form as $$E = \\frac{hc}{\\lambda}$$.
\n\n
However, we are given $$hc$$ in electron volts per nanometer ($$1245 \\mathrm{~eV} \\cdot \\mathrm{nm}$$), and the energy is also given in terms of voltage ($$10.2 \\mathrm{~V}$$). First, we convert the energy into electron volts (eV) using the formula: $$E = eV$$, where $$e$$ is the charge of an electron ($$1.6 \\times 10^{-19} \\mathrm{C}$$).
\n\n
The energy in electron volts can be directly calculated as:\n\n
Reflection, diffraction, interference, and polarization are all phenomena that can be explained by the wave nature of light. These phenomena are evidence that light behaves as a wave, evident through various experimental observations:
\n\n
\n\n
Reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. The laws of reflection can be explained by treating light as waves.
\n\n
Diffraction is the bending of light around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle. This phenomenon is observed when a wave encounters an obstacle or a slit that is comparable in size to its wavelength, which is best explained by wave theory.
\n\n
Interference occurs when two or more waves superimpose to form a resultant wave of greater, lower, or the same amplitude. This phenomenon illustrates the wave nature of light through constructive and destructive interference patterns, such as those observed in the double-slit experiment.
\n\n
Polarization is a property applying to transverse waves that specifies the geometrical orientation of the oscillations. For light waves, this can be seen as the direction in which the electric field oscillates. Polarization can only occur with waves that have transverse components, further supporting the wave nature of light.
\n\n
\n\n
On the other hand, the photoelectric effect cannot be explained solely by the wave nature of light. It is the emission of electrons or other free carriers when light shines on a material. Electrons emitted in this manner can be called photoelectrons. The phenomenon is best explained by Albert Einstein's quantum theory of light, where light is considered as quanta of energy called photons. This effect demonstrates the particle aspect of light, wherein each photon has a discrete packet of energy equal to $$hf$$, where $$h$$ is Planck's constant and $$f$$ is the frequency of the light.
\n\n
Thus, the photoelectric effect is the correct answer because it specifically requires the particle theory of light for its explanation, unlike reflection, diffraction, interference, and polarization, which are well-explained by the wave nature of light.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8835,
"subject": "Physics",
"question": "Sodium and copper have work functions $$2.3$$ $$eV$$ and $$4.5$$ $$eV$$ respectively. Then the ratio of the wavelengths is nearest to ",
"options": [
{
"text": "$$1:2$$ "
},
{
"text": "$$4:1$$ "
},
{
"text": "$$2:1$$ "
},
{
"text": "$$1:4$$ "
}
],
"answer": "$$2:1$$ ",
"solution": "**Answer:** $$2:1$$ \n\nWe know that work function is the energy required and energy $$E = h\\upsilon $$\n
$$ \\Rightarrow \\lambda = {{6.63 \\times {{10}^{ - 34}} \\times 3 \\times {{10}^8}} \\over {40 \\times 1.6 \\times {{10}^{ - 16}}}} = 310nm$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8838,
"subject": "Physics",
"question": "According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal $$Vs$$ the frequency, of the incident radiation gives as straight the whose slope ",
"options": [
{
"text": "depends both on the intensity of the radiation and the metal used"
},
{
"text": "depends on the intensity of the radiation "
},
{
"text": "depends on the nature of the metal used "
},
{
"text": "is the same for the all metals and independent of the intensity of the radiation "
}
],
"answer": "is the same for the all metals and independent of the intensity of the radiation ",
"solution": "**Answer:** is the same for the all metals and independent of the intensity of the radiation \n\n
Einstein's photoelectric equation is given by
\n
$$K_{\\max} = h\\nu - \\phi,$$
\n
where $K_{\\max}$ is the maximum kinetic energy of the emitted photoelectrons, $h$ is Planck's constant, $\\nu$ is the frequency of the incident radiation, and $\\phi$ is the work function of the metal (the minimum energy needed to eject an electron).
\n
If you plot $K_{\\max}$ against $\\nu$, you will get a straight line with slope $h$ (Planck's constant) and y-intercept $-\\phi$ (the negative of the work function). The slope of the line (which is $h$) does not depend on the intensity of the radiation or the type of metal used. Instead, it is a universal constant.
\n
Therefore, the correct answer is:
\n
Option D: The slope is the same for all metals and independent of the intensity of the radiation.
\n",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 8839,
"subject": "Physics",
"question": "A photocell is illuminated by a small bright source placed $$1$$ $$m$$ away. When the same source of light is placed $${1 \\over 2}$$ $$m$$ away, the number of electrons emitted by photo-cathode would ",
"options": [
{
"text": "increases by a factor of $$4$$ "
},
{
"text": "decreases by a factor of $$4$$ "
},
{
"text": "increases by a factor of $$2$$ "
},
{
"text": "decreases by a factor of $$2$$ "
}
],
"answer": "increases by a factor of $$4$$ ",
"solution": "**Answer:** increases by a factor of $$4$$ \n\n$$I \\propto {1 \\over {{r^2}}};{{{I_1}} \\over {{I_2}}} = {\\left( {{{{r_2}} \\over {{r_1}}}} \\right)^2} = {1 \\over 4}$$\n
$${I_2} \\to 4\\,\\,$$ times $${I_1}$$\n
When intensity becomes 4 times, no. of photoelectrons emitted would increase by $$4$$ times, since number of electrons emitted per second is directly proportional to intensity.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8840,
"subject": "Physics",
"question": "The time taken by a photoelectron to come out after the photon strikes is approximately ",
"options": [
{
"text": "$${10^{ - 4}}\\,s$$ "
},
{
"text": "$${10^{ - 10}}\\,s$$ "
},
{
"text": "$${10^{ - 16}}\\,s$$ "
},
{
"text": "$${10^{ - 1}}\\,s$$ "
}
],
"answer": "$${10^{ - 10}}\\,s$$ ",
"solution": "**Answer:** $${10^{ - 10}}\\,s$$ \n\nEmission of photo-electron starts from the surface\nafter incidence of photons in about $${10^{ - 10}}s.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8841,
"subject": "Physics",
"question": "The threshold frequency for a metallic surface corresponds to an energy of $$6.2$$ $$eV$$ and the stopping potential for a radiation incident on this surface is $$5V.$$ The incident radiation lies in ",
"options": [
{
"text": "ultra-violet region "
},
{
"text": "infra-red region "
},
{
"text": "visible region "
},
{
"text": "$$x$$-ray region"
}
],
"answer": "ultra-violet region ",
"solution": "**Answer:** ultra-violet region \n\n
The energy of the incident radiation that causes photoelectrons to be emitted can be calculated using the stopping potential, V, via the equation
\n
$$E = eV,$$
\n
where e is the elementary charge.
\n
The elementary charge, e, is approximately equal to $$1.602 \\times 10^{-19}$$ C. So the energy of the incident radiation is
This is less than the threshold energy of $$6.2 \\, \\text{eV},$$ which means that the incident radiation does not have enough energy to overcome the work function of the metal and thus cannot cause photoelectrons to be emitted.
\n
The regions of the electromagnetic spectrum are generally classified by energy as follows:
\n
\n
Infrared: Less than about $$1.24 \\, \\text{eV}$$
\n
Visible: Between about $$1.24 \\, \\text{eV}$$ and $$3.1 \\, \\text{eV}$$
\n
Ultraviolet: Between about $$3.1 \\, \\text{eV}$$ and $$124 \\, \\text{eV}$$
\n
X-rays: Greater than about $$124 \\, \\text{eV}$$
\n
\n
Therefore, the incident radiation falls in the ultraviolet region.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8842,
"subject": "Physics",
"question": "The surface of a metal is illuminated with the light of $$400$$ $$nm.$$ The kinetic energy of the ejected photoelectrons was found to be $$1.68$$ $$eV.$$ The work function of the metal is : $$\\left( {hc = 1240eV.nm} \\right)$$ ",
"options": [
{
"text": "$$1.41$$ $$eV$$ "
},
{
"text": "$$1.51$$ $$eV$$ "
},
{
"text": "$$1.68$$ $$eV$$ "
},
{
"text": "$$3.09$$ $$eV$$ "
}
],
"answer": "$$1.41$$ $$eV$$ ",
"solution": "**Answer:** $$1.41$$ $$eV$$ \n\n
The photoelectric effect equation, which relates the energy of the incident light to the kinetic energy of the ejected photoelectrons and the work function of the metal, is given by:
\n
$E = K_{\\max} + W$,
\n
where
\n
$E$ is the energy of the incident light,
\n
$K_{\\max}$ is the maximum kinetic energy of the photoelectrons, and
\n
$W$ is the work function of the metal.
\n
The energy of the incident light can be calculated using the formula $E = \\frac{hc}{\\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\\lambda$ is the wavelength of the light. However, given that $hc = 1240$ eV⋅nm, we can simplify this to $E = \\frac{1240}{\\lambda}$.
\n
Substituting the given values into this equation, we have:
\n
$E = \\frac{1240}{400} = 3.1$ eV.
\n
We can then substitute these values into the photoelectric effect equation:
Rounding to two decimal places, the work function of the metal is therefore approximately $1.42$ eV.
\n
Thus, Option A: $1.41$ eV is the closest to the correct answer.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8843,
"subject": "Physics",
"question": "Statement - $$1$$ : When ultraviolet light is incident on a photocell, its stopping potential is $${V_0}$$ and the maximum kinetic energy of the photoelectrons is $${K_{\\max }}$$. When the ultraviolet light is replaced by $$X$$-rays, both $${V_0}$$ and $${K_{\\max }}$$ increase.\n
Statement - $$2$$ : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.
",
"options": [
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is the correct explanation of Statement - $$1$$ "
},
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is not the correct explanation of Statement - $$1$$ "
},
{
"text": "Statement - $$1$$ is is false, Statement - $$2$$ is true "
},
{
"text": "Statement - $$1$$ is is true, Statement - $$2$$ is false"
}
],
"answer": "Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is not the correct explanation of Statement - $$1$$ ",
"solution": "**Answer:** Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is not the correct explanation of Statement - $$1$$ \n\nStatement 1 is true. The energy of an incident photon (from the ultraviolet light or X-rays) on a photocell is given by Planck's equation, $E = h\\nu$, where $h$ is Planck's constant and $\\nu$ is the frequency of the light. X-rays have a higher frequency than ultraviolet light, so they deliver more energy to the photoelectrons. This results in a higher stopping potential ($V_0$) and maximum kinetic energy ($K_{\\max}$) for the photoelectrons.\n
\nStatement 2 is also true. However, while the speeds (and hence kinetic energies) of photoelectrons do vary, this variation is not because of a range of frequencies in the incident light. Rather, it's due to the interaction of the incident photons with electrons at different energy levels in the metal. A single frequency of light can produce photoelectrons with a range of speeds because the electrons they encounter can have a variety of binding energies.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8844,
"subject": "Physics",
"question": "This question has Statement - $$1$$ and Statement - $$2$$. Of the four choices given after the statements, choose the one that best describes the two statements.\n
Statement - $$1$$ : A metallic surface is irradiated by a monochromatic light of frequency $$v > {v_0}$$ (the threshold frequency). The maximum kinetic energy and the stopping potential are $${K_{\\max }}$$ and $${V_0}$$ respectively. If the frequency incident on the surface is doubled, both the $${K_{\\max }}$$ anmd $${V_0}$$ are also doubled.\n Statement - $$2$$ : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.
",
"options": [
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is true, Statement - $$2$$ is the correct explanation of Statement - $$1$$."
},
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is true, Statement - $$2$$ is not the correct explanation of Statement - $$1$$."
},
{
"text": "Statement - $$1$$ is false, Statement - $$2$$ is true."
},
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is false."
}
],
"answer": "Statement - $$1$$ is false, Statement - $$2$$ is true.",
"solution": "**Answer:** Statement - $$1$$ is false, Statement - $$2$$ is true.\n\nBy Einstein photoelectric equation, \n
$${K_{\\max }} = e{V_0} = hv - h{v_0}$$\n
When $$v$$ is doubled, $${K_{\\max }}$$ and $${V_0}$$ become more than double. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8845,
"subject": "Physics",
"question": "Radiation of wavelength $$\\lambda ,$$ is incident on a photocell. The fastest emitted electron has speed $$v.$$ If the wavelength is changed to $${{3\\lambda } \\over 4},$$ the speed of the fastest emitted electron will be:",
"options": [
{
"text": "$$ = v{\\left( {{4 \\over 3}} \\right)^{{1 \\over 2}}}$$ "
},
{
"text": "$$ = v{\\left( {{3 \\over 4}} \\right)^{{1 \\over 2}}}$$ "
},
{
"text": "$$ > v{\\left( {{4 \\over 3}} \\right)^{{1 \\over 2}}}$$ "
},
{
"text": "$$ < v{\\left( {{4 \\over 3}} \\right)^{{1 \\over 2}}}$$ "
}
],
"answer": "$$ > v{\\left( {{4 \\over 3}} \\right)^{{1 \\over 2}}}$$ ",
"solution": "**Answer:** $$ > v{\\left( {{4 \\over 3}} \\right)^{{1 \\over 2}}}$$ \n\n$$h{v_0}^2 - h{v_0} = {1 \\over 2}m{v^2}$$\n
$$ \\Rightarrow $$ V1 $$ = {{hc} \\over e}$$ $$\\left[ {{1 \\over {{\\lambda _3}}} - {3 \\over {2{\\lambda _1}}} + {1 \\over {2{\\lambda _2}}}} \\right]$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8847,
"subject": "Physics",
"question": "A photoelectric surface is illuminated successively by monochromatic light of wavelengths $$\\lambda $$ and $${\\lambda \\over 2}.$$ If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is :",
"options": [
{
"text": "$${{hc} \\over {3\\lambda }}$$ "
},
{
"text": "$${{hc} \\over {2\\lambda }}$$"
},
{
"text": "$${{hc} \\over {\\lambda }}$$"
},
{
"text": "$${3\\,{hc} \\over {\\lambda }}$$"
}
],
"answer": "$${{hc} \\over {2\\lambda }}$$",
"solution": "**Answer:** $${{hc} \\over {2\\lambda }}$$\n\nWe know,\n
$$ \\Rightarrow $$ $$\\phi $$ = $${{hc} \\over {2\\lambda }}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8848,
"subject": "Physics",
"question": "The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to 3n, the maximum velocity of the ejected photoelectrons will be :",
"options": [
{
"text": "less than $$\\sqrt 3 $$ v"
},
{
"text": "v"
},
{
"text": "more than $$\\sqrt 3 \\,v$$ "
},
{
"text": "equal to $$\\sqrt 3 \\,v$$"
}
],
"answer": "more than $$\\sqrt 3 \\,v$$ ",
"solution": "**Answer:** more than $$\\sqrt 3 \\,v$$ \n\n
The given maximum velocity is v and frequency is n. We know that the kinetic energy is given by
\n
$$KE = {1 \\over 2}m{v^2} = hn - \\phi $$
\n
where h is Planck's constant and $$\\phi$$ is the work function. Therefore, the kinetic energy of the incident light is
\n
$${E_1} = hn - \\phi $$ ..... (1)
\n
When the frequency of the incident light is increased to 3n, then the kinetic energy is given by
\n
$${1 \\over 2}mv_1^2 = 3hn - \\phi $$
\n
$$ \\Rightarrow {E_2} = 3hn - \\phi $$ ..... (2)
\n
Substituting $$hn = {E_1} + \\phi $$ [from Eq. (1)] in Eq. (2), we get
\n$$ \\Rightarrow {K_{\\max }} = \\left( {1237} \\right)\\left( {{{380 - 260} \\over {380 \\times 260}}} \\right)$$\n= 1.5 eV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8851,
"subject": "Physics",
"question": "When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping\npotential for the current is –V0/2. When the surface is illuminated by monochromatic light of frequency v/2, the stopping potential is – V0. The threshold frequency for photoelectric emission is : ",
"options": [
{
"text": "2$$v$$"
},
{
"text": "$${4 \\over 3}v$$"
},
{
"text": "$${{3v} \\over 2}$$"
},
{
"text": "$${{5v} \\over 3}$$"
}
],
"answer": "$${{3v} \\over 2}$$",
"solution": "**Answer:** $${{3v} \\over 2}$$\n\nEinstein’s photoelectric equation in the two cases\nis given by\n
$$ \\Rightarrow $$ $${\\upsilon _0} = {3 \\over 2}\\upsilon $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8852,
"subject": "Physics",
"question": "A metal plate of area 1 $$ \\times $$ 10–4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons.\nThe number of emitted photoelectrons per second and their maximum energy, respectively, will be ",
"options": [
{
"text": "1014 and 10 eV"
},
{
"text": "1012 and 5 eV"
},
{
"text": "1011 and 5 eV"
},
{
"text": "1010 and 5 eV"
}
],
"answer": "1011 and 5 eV",
"solution": "**Answer:** 1011 and 5 eV\n\n
Given that the area of the metal plate $$A = 1 \\times 10^{-4} m^2$$, and the intensity of the radiation
$$I = 16 mW/m^2 = 16 \\times 10^{-3} W/m^2$$.
\n
The power $$P$$ falling on the metal plate is given by the product of the intensity of the radiation and the area:
The work function of the metal is given as 5 eV, which is the energy required to remove an electron from the metal. Therefore, the maximum energy of the photoelectrons $$E_{\\text{max}}$$ is given by the energy of the incident photons minus the work function:
\n
$$E_{\\text{max}} = E_{\\text{photon}} - \\text{Work function} = 10 eV - 5 eV = 5 eV$$.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8853,
"subject": "Physics",
"question": "The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 $$ \\times $$ 107)ct + sin(6.28 $$ \\times $$ 107)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ? \n (Take c = 3 $$ \\times $$ 108 ms$$-$$1, h = 6.6 $$ \\times $$ 10$$-$$34J-s)",
"options": [
{
"text": "6.82 eV"
},
{
"text": "12.5 eV"
},
{
"text": "8.52 eV"
},
{
"text": "7.72 eV"
}
],
"answer": "7.72 eV",
"solution": "**Answer:** 7.72 eV\n\nGiven that,\n
$$ \\simeq $$ 7.7 $$eV$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8854,
"subject": "Physics",
"question": "Surface of certain metal is first illuminated with light of wavelength $$\\lambda $$1 = 350 nm and then, by light of wavelength $$\\lambda $$2 = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to : \n
(Energy of photon n = $${{1240} \\over {\\lambda (in\\,mm)}}$$eV)",
"options": [
{
"text": "1.8"
},
{
"text": "2.5"
},
{
"text": "5.6"
},
{
"text": "1.4"
}
],
"answer": "1.8",
"solution": "**Answer:** 1.8\n\nLet speed of photon electron in first case is 2v\n
= 1V",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8856,
"subject": "Physics",
"question": "The surface of a metal is illuminated alternately\nwith photons of energies E1 = 4 eV and E2 = 2.5 eV\nrespectively. The ratio of maximum speeds of the\nphotoelectrons emitted in the two cases is 2. The\nwork function of the metal in (eV) is _____.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nGiven,\n
$$ \\therefore $$ Work function($$\\phi $$) of the metal = 2 eV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8857,
"subject": "Physics",
"question": "When the wavelength of radiation falling on a metal is changed from 500 nm to 200 nm, the\nmaximum kinetic energy of the photoelectrons becomes three times larger. The work function of\nthe metal is close to :",
"options": [
{
"text": "1.02 eV"
},
{
"text": "0.81 eV"
},
{
"text": "0.61 eV"
},
{
"text": "0.52 eV"
}
],
"answer": "0.61 eV",
"solution": "**Answer:** 0.61 eV\n\nK1 = $${{hc} \\over {500}} - {\\phi _0}$$\n
$$ \\Rightarrow $$ $$\\phi $$0 = 0.61 eV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8858,
"subject": "Physics",
"question": "When radiation of wavelength $$\\lambda $$ is used to\nilluminate a metallic surface, the stopping\npotential is V. When the same surface is\nilluminated with radiation of wavelength 3$$\\lambda $$,\nthe stopping potential is\n$${V \\over 4}$$. If the threshold\nwavelength for the metallic surface is n$$\\lambda $$ then\nvalue of n will be __________.",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\n$${{hc} \\over \\lambda } - \\phi $$ = eV .....(i)\n
So, n = 9",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8859,
"subject": "Physics",
"question": "Radiation, with wavelength 6561 $$\\mathop A\\limits^o $$ falls on a\nmetal surface to produce photoelectrons. The\nelectrons are made to enter a uniform magnetic\nfield of 3 × 10–4 T. If the radius of the largest\ncircular path followed by the electrons is\n10 mm, the work function of the metal is\nclose to :",
"options": [
{
"text": "1.8eV"
},
{
"text": "0.8eV"
},
{
"text": "1.1eV"
},
{
"text": "1.6eV"
}
],
"answer": "1.1eV",
"solution": "**Answer:** 1.1eV\n\nLet the work function be $$\\phi $$.\n
$$ \\therefore $$ $$\\phi $$ = 1.9 - 0.8 = 1.1 eV",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8860,
"subject": "Physics",
"question": "When photon of energy 4.0 eV strikes the\nsurface of a metal A, the ejected photoelectrons\nhave maximum kinetic energy TA eV end\nde-Broglie wavelength $$\\lambda _A$$. The maximum\nkinetic energy of photoelectrons liberated from\nanother metal B by photon of energy 4.50 eV\nis TB = (TA – 1.5) eV. If the de-Broglie\nwavelength of these photoelectrons $$\\lambda _B$$ = 2$$\\lambda _A$$,\nthen the work function of metal B is :",
"options": [
{
"text": "1.5eV"
},
{
"text": "4eV"
},
{
"text": "2eV"
},
{
"text": "3eV"
}
],
"answer": "4eV",
"solution": "**Answer:** 4eV\n\nWe know, de-Broglie wavelength\n
$$ \\Rightarrow $$ $$\\phi $$B = 4 eV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8861,
"subject": "Physics",
"question": "The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is :",
"options": [
{
"text": "400 nm"
},
{
"text": "329 nm"
},
{
"text": "309 nm"
},
{
"text": "382 nm"
}
],
"answer": "382 nm",
"solution": "**Answer:** 382 nm\n\nFrom the photoelectric effect equation
$${\\lambda _2} \\simeq 382$$ nm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8862,
"subject": "Physics",
"question": "Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x : y. The value of x is ___________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n$$K{E_{\\max }} = hv - \\phi $$
$$v_1^2 - v_2^2 = {{2h} \\over m}({f_1} - {f_2})$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8865,
"subject": "Physics",
"question": "The radiation corresponding to 3 $$\\to$$ 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The electrons are passed through a magnetic field of 5 $$\\times$$ 10$$-$$4 T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is : (Mass of electron = 9.1 $$\\times$$ 10$$-$$31 kg)",
"options": [
{
"text": "1.36 eV"
},
{
"text": "1.88 eV"
},
{
"text": "0.82 eV"
},
{
"text": "0.16 eV"
}
],
"answer": "0.82 eV",
"solution": "**Answer:** 0.82 eV\n\nEnergy of photon can be given as
$$\\phi$$ = (1.89 $$-$$ 1.075) eV = 0.82 eV",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8866,
"subject": "Physics",
"question": "A certain metallic surface is illuminated by monochromatic radiation of wavelength $$\\lambda$$. The stopping potential for photoelectric current for this radiation is 3V0. If the same surface is illuminated with a radiation of wavelength 2$$\\lambda$$, the stopping potential is V0. The threshold wavelength of this surface for photoelectric effect is ____________ $$\\lambda$$.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n$$KE = {{hc} \\over \\lambda } - \\phi hc$$
$${\\lambda _t} = 4{\\lambda _0}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8867,
"subject": "Physics",
"question": "When radiation of wavelength $$\\lambda$$ is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is :",
"options": [
{
"text": "2$$\\lambda$$"
},
{
"text": "4$$\\lambda$$"
},
{
"text": "8$$\\lambda$$"
},
{
"text": "6$$\\lambda$$"
}
],
"answer": "4$$\\lambda$$",
"solution": "**Answer:** 4$$\\lambda$$\n\n$${V_s} = hv - \\phi $$
$$ = \\left( {{{hc} \\over {6.4}}} \\right) \\times 4 = 4\\lambda $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8868,
"subject": "Physics",
"question": "A light beam of wavelength 500 nm is incident on a metal having work function of 1.25 eV, placed in a magnetic field of intensity B. The electrons emitted perpendicular to the magnetic field B, with maximum kinetic energy are bent into circular are of radius 30 cm. The value of B is ___________ $$\\times$$ 10$$-$$7 T.
Given hc = 20 $$\\times$$ 10$$-$$26 J-m, mass of electron = 9 $$\\times$$ 10$$-$$31 kg",
"options": [],
"answer": "125",
"solution": "**Answer:** 125\n\nBy photoelectric equation
$$ = 125 \\times {10^{ - 7}}T$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8869,
"subject": "Physics",
"question": "An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000$$\\mathop A\\limits^o $$. What is the maximum kinetic energy of the emitted photoelectron? ",
"options": [
{
"text": "7.61 eV"
},
{
"text": "1.41 eV"
},
{
"text": "3.3 eV"
},
{
"text": "No photoelectron would be emitted"
}
],
"answer": "1.41 eV",
"solution": "**Answer:** 1.41 eV\n\ninitially, energy of electron = + 3eV
finally, in 2nd excited state,
energy of electron = $$ - {{(13.6eV)} \\over {{3^2}}}$$
$$ = - 1.51eV$$
Loss in energy is emitted as photon,
So, photon energy $${{hc} \\over \\lambda } = 4.51eV$$
$$\\Delta$$V = $${V_{{s_1}}} - {V_{{s_2}}}$$ = 1.93 $$-$$ 0.6 = 1.33 V",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8871,
"subject": "Physics",
"question": "In a photoelectric experiment, increasing the intensity of incident light :",
"options": [
{
"text": "increases the number of photons incident and also increases the K.E. of the ejected electrons"
},
{
"text": "increases the frequency of photons incident and increases the K.E. of the ejected electrons"
},
{
"text": "increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged"
},
{
"text": "increases the number of photons incident and the K.E. of the ejected electrons remains unchanged"
}
],
"answer": "increases the number of photons incident and the K.E. of the ejected electrons remains unchanged",
"solution": "**Answer:** increases the number of photons incident and the K.E. of the ejected electrons remains unchanged\n\n$$\\to$$ Increasing intensity means number of incident photons are increased.
$$\\to$$ Kinetic energy of ejected electrons depend on the frequency of incident photons, not the intensity.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8872,
"subject": "Physics",
"question": "A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm?",
"options": [
{
"text": "0.96 V"
},
{
"text": "1.25 V"
},
{
"text": "0.24 V"
},
{
"text": "1.5 V"
}
],
"answer": "1.25 V",
"solution": "**Answer:** 1.25 V\n\n$$k{E_{\\max }} = {{hc} \\over {{\\lambda _i}}} + \\phi $$
or $$e{V_o} = {{hc} \\over {{\\lambda _i}}} + \\phi $$
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :
\n
Assertion A : The photoelectric effect does not takes place, if the energy of the incident radiation is less than the work function of a metal.
\n
Reason R : Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal.
\n
In the light of the above statements, choose the most appropriate answer from the options given below.
",
"options": [
{
"text": "Both A and R are correct and R is the correct explanation of A."
},
{
"text": "Both A and R are correct but R is not the correct explanation of A."
},
{
"text": "A is correct but R is not correct."
},
{
"text": "A is not correct but R is correct."
}
],
"answer": "Both A and R are correct but R is not the correct explanation of A.",
"solution": "**Answer:** Both A and R are correct but R is not the correct explanation of A.\n\n
When energy of incident radiation is equal to the work function of the metal, then the KE of photoelectrons would be zero. But this statement does not comment on the situation when energy is less than the work function.
Let K1 and K2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $$\\lambda$$1 and $$\\lambda$$2, respectively are incident on a metallic surface. If $$\\lambda$$1 = 3$$\\lambda$$2 then :
A metal surface is illuminated by a radiation of wavelength 4500 $$\\mathop A\\limits^o $$. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90$$^\\circ$$ with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately :
The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 $$\\mathop A\\limits^o $$ is 0.42 V. If the threshold frequency is x $$\\times$$ 1013 /s, where x is _________ (nearest integer).
The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :
When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v1. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v2. If v2 = x v1, the value of x will be __________.
A metal exposed to light of wavelength $$800 \\mathrm{~nm}$$ and and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength $$500 \\mathrm{~nm}$$ is used. The workfunction of the metal is : (Take hc $$=1230 \\,\\mathrm{eV}-\\mathrm{nm}$$ ).
Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be
The kinetic energy of emitted electron is E when the light incident on the metal has wavelength $$\\lambda$$. To double the kinetic energy, the incident light must have wavelength:
The threshold frequency of a metal is $$f_{0}$$. When the light of frequency $$2 f_{0}$$ is incident on the metal plate, the maximum velocity of photoelectrons is $$v_{1}$$. When the frequency of incident radiation is increased to $$5 \\mathrm{f}_{0}$$, the maximum velocity of photoelectrons emitted is $$v_{2}$$. The ratio of $$v_{1}$$ to $$v_{2}$$ is :
",
"options": [
{
"text": "$$\\frac{v_{1}}{v_{2}}=\\frac{1}{2}$$"
},
{
"text": "$$\\frac{v_{1}}{v_{2}}=\\frac{1}{16}$$"
},
{
"text": "$$\\frac{v_{1}}{v_{2}}=\\frac{1}{4}$$"
},
{
"text": "$$\\frac{v_{1}}{v_{2}}=\\frac{1}{8}$$"
}
],
"answer": "$$\\frac{v_{1}}{v_{2}}=\\frac{1}{2}$$",
"solution": "**Answer:** $$\\frac{v_{1}}{v_{2}}=\\frac{1}{2}$$\n\n$$\n\\begin{aligned}\n& \\frac{1}{2} m v^2=h f-h f_0 \\\\\\\\\n& \\Rightarrow \\frac{1}{2} m v_1^2=2 h f_0-h f_0=h f_0 \\\\\\\\\n& \\text { also, } \\frac{1}{2} m v_2^2=5 h f_0-h f_0=4 h f_0\n\\end{aligned}\n$$\n
taking ratio,\n
$$\n\\frac{v_1^2}{v_2^2}=\\frac{1}{4} \\Rightarrow \\frac{v_1}{v_2}=\\frac{1}{2}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8885,
"subject": "Physics",
"question": "If the two metals $\\mathrm{A}$ and $\\mathrm{B}$ are exposed to radiation of wavelength $350 \\mathrm{~nm}$. The work functions of metals $\\mathrm{A}$ and $\\mathrm{B}$ are $4.8 \\mathrm{eV}$ and $2.2 \\mathrm{eV}$. Then choose the correct option.",
"options": [
{
"text": "Metal B will not emit photo-electrons"
},
{
"text": "Both metals $\\mathrm{A}$ and $\\mathrm{B}$ will not emit photo-electrons"
},
{
"text": "Metal A will not emit photo-electrons\n"
},
{
"text": "Both metals A and B will emit photo-electrons"
}
],
"answer": "Metal A will not emit photo-electrons\n",
"solution": "**Answer:** Metal A will not emit photo-electrons\n\n\n$$\n\\phi=\\frac{h c}{\\lambda}=\\frac{1240}{350} \\mathrm{eV}=3.54 \\mathrm{eV}\n$$\n
$\\therefore$ Only metal B will emit photoelectron.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8886,
"subject": "Physics",
"question": "
The threshold wavelength for photoelectric emission from a material is 5500 $$\\mathop A\\limits^o $$. Photoelectrons will be emitted, when this material is illuminated with monochromatic radiation from a
\n
A. 75 W infra-red lamp
\n
B. 10 W infra-red lamp
\n
C. 75 W ultra-violet lamp
\n
D. 10 W ultra-violet lamp
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "C only"
},
{
"text": "A and D only"
},
{
"text": "C and D only"
},
{
"text": "B and C only"
}
],
"answer": "C and D only",
"solution": "**Answer:** C and D only\n\nWavelength of infra-red $=700 \\mathrm{~nm}$ (minimum)\n
\nWavelength of UV $=100-400 \\mathrm{~nm}$\n
\nSince we need $\\lambda<5000$ Å\n
\n$\\Rightarrow$ Only UV would be able to emit photoelectrons.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8887,
"subject": "Physics",
"question": "
Given below are two statements :
\n
Statement I : Stopping potential in photoelectric effect does not depend on the power of the light source.
\n
Statement II : For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light.
\n
In the light of above statements, choose the most appropriate answer from the options given below
",
"options": [
{
"text": "Both Statement I and Statement II are incorrect"
},
{
"text": "Statement I is correct but Statement II is incorrect"
},
{
"text": "Both Statement I and Statement II are correct"
},
{
"text": "Statement I is incorrect but Statement II is correct"
}
],
"answer": "Both Statement I and Statement II are correct",
"solution": "**Answer:** Both Statement I and Statement II are correct\n\nStatement I is correct as stopping potential is\nindependent of power of light used.
\nStatement II is correct as maximum kinetic energy\nof photoelectron depends on wavelength of light.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8888,
"subject": "Physics",
"question": "
From the photoelectric effect experiment, following observations are made. Identify which of these are correct.
\n
A. The stopping potential depends only on the work function of the metal.
\n
B. The saturation current increases as the intensity of incident light increases.
\n
C. The maximum kinetic energy of a photo electron depends on the intensity of the incident light.
\n
D. Photoelectric effect can be explained using wave theory of light.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "A, B, D only"
},
{
"text": "A, C, D only"
},
{
"text": "B, C only"
},
{
"text": "B only"
}
],
"answer": "B only",
"solution": "**Answer:** B only\n\n(A) From Einstein's equation\n
\n$$\nK_{\\max }=e V_{s}=h v-\\phi\n$$\n
\nForm the stopping potential $\\left(\\mathrm{V}_{\\mathrm{s}}\\right)$ depends on $\\phi$ $\\& \\,v$.\n
\n(B) Saturation current is proportional to intensity, i.e., number of incident photons.\n
\n(C) $K_{\\max }$ only depends on nature of photon and $\\phi$.\n
\n(D) Einstein used particle behaviour of photon to explain photon electric effect.\n
\nOnly B is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8889,
"subject": "Physics",
"question": "
Given below are two statements:
\n
Statement I : Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface.
\n
Statement II : Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light.
\n
In the light of above statements, choose the correct answer form the options given below
",
"options": [
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Statement I is true but statement II is false"
},
{
"text": "Statement I is false but statement II is true"
},
{
"text": "Both Statement I and Statement II are false"
}
],
"answer": "Statement I is true but statement II is false",
"solution": "**Answer:** Statement I is true but statement II is false\n\nNow let's analyze both statements:\n
\nStatement I is correct. According to the photoelectric effect, the ability to emit electrons from a metallic surface depends on the energy of the incident light. The energy of a photon is given by $$E = h\\nu$$, where $$h$$ is Planck's constant and $$\\nu$$ is the frequency of the light. Ultraviolet rays have higher frequencies and thus higher energies compared to microwaves and infrared rays, making them more effective for the emission of electrons from a metallic surface.\n
\nStatement II is incorrect. The maximum kinetic energy of photoelectrons is given by $$K_{max} = h\\nu - h\\nu_0$$, where $$\\nu_0$$ is the threshold frequency. This equation shows that the maximum kinetic energy of photoelectrons is directly proportional to the frequency of the incident light (above the threshold frequency), not inversely proportional.\n
\nBased on the analysis, the correct answer is:\n
\nStatement I is true, and Statement II is false.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8890,
"subject": "Physics",
"question": "
An atom absorbs a photon of wavelength $$500 \\mathrm{~nm}$$ and emits another photon of wavelength $$600 \\mathrm{~nm}$$. The net energy absorbed by the atom in this process is $$n \\times 10^{-4} ~\\mathrm{eV}$$. The value of n is __________. [Assume the atom to be stationary during the absorption and emission process] (Take $$\\mathrm{h}=6.6 \\times 10^{-34} ~\\mathrm{Js}$$ and $$\\mathrm{c}=3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}$$ )
",
"options": [],
"answer": "4125",
"solution": "**Answer:** 4125\n\nThe energy $$E$$ of a photon is related to its wavelength $$\\lambda$$ by the formula:\n
\n$$E=\\frac{hc}{\\lambda}$$\n
\nwhere $$h$$ is Planck's constant and $$c$$ is the speed of light. \n\nIn this problem, we are given that an atom absorbs a photon of wavelength $$\\lambda_1=500~\\mathrm{nm}$$ and emits another photon of wavelength $$\\lambda_2=600~\\mathrm{nm}$$. We can use the formula above to calculate the energy absorbed by the atom:\n
\nWe need to express this energy in electron volts (eV), which is a more convenient unit for atomic and molecular energies. To do this, we can divide the energy in joules by the charge of an electron:\n
\nTherefore, the value of $$n$$ is $$\\boxed{4125}$$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8891,
"subject": "Physics",
"question": "
The difference between threshold wavelengths for two metal surfaces $$\\mathrm{A}$$ and $$\\mathrm{B}$$ having work function $$\\phi_{A}=9 ~\\mathrm{eV}$$ and $$\\phi_{B}=4 \\cdot 5 ~\\mathrm{eV}$$ in $$\\mathrm{nm}$$ is:
\n
$$\\{$$ Given, hc $$=1242 ~\\mathrm{eV} \\mathrm{nm}\\}$$
Threshold wavelength ($$\\lambda_{threshold}$$) is the maximum wavelength of light required to remove an electron from a metal surface, i.e., to overcome the work function ($$\\phi$$). The relationship between work function and threshold wavelength is given by:
$$\\phi$$ is the work function in electron-volts (eV)
\n
$$h$$ is the Planck's constant
\n
$$c$$ is the speed of light
\n
$$\\lambda_{threshold}$$ is the threshold wavelength
\n
\n\n
In this problem, we are given the product $$hc = 1242 ~\\mathrm{eV}\\,\\mathrm{nm}$$, and the work functions $$\\phi_{A} = 9 ~\\mathrm{eV}$$ and $$\\phi_{B} = 4.5 ~\\mathrm{eV}$$. Let's calculate the threshold wavelengths for metal surfaces $$\\mathrm{A}$$ and $$\\mathrm{B}$$:
So, the difference between the threshold wavelengths for the two metal surfaces is $$\\boxed{138}\\,\\mathrm{nm}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8892,
"subject": "Physics",
"question": "
A metallic surface is illuminated with radiation of wavelength $$\\lambda$$, the stopping potential is $$V_{0}$$. If the same surface is illuminated with radiation of wavelength $$2 \\lambda$$. the stopping potential becomes $$\\frac{V_{o}}{4}$$. The threshold wavelength for this metallic surface will be
The photoelectric effect occurs when light (or more generally, electromagnetic radiation) incident on a metallic surface causes the ejection of electrons from the surface. The energy of the incident photons must be greater than the work function of the metal (denoted by $$\\phi_0$$) for the electrons to be ejected.
\n
The energy of the ejected electrons can be expressed as the difference between the energy of the incident photons and the work function of the metal. The maximum kinetic energy of the ejected electrons is given by:
\n
$$K_{max} = h\\nu - \\phi_0$$
\n
where $$h$$ is the Planck's constant, $$\\nu$$ is the frequency of the incident light, and $$\\phi_0$$ is the work function of the metal.
\n
We can also write the maximum kinetic energy in terms of the stopping potential ($$V_0$$) and the elementary charge ($$e$$) of an electron:
\n
$$K_{max} = eV_0$$
\n
Equating these two expressions for the maximum kinetic energy, we get:
\n
$$eV_0 = h\\nu - \\phi_0$$
\n
We can express the frequency $$\\nu$$ in terms of the speed of light $$c$$ and the wavelength $$\\lambda$$:
\n
$$\\nu = \\frac{c}{\\lambda}$$
\n
Substituting this expression for frequency in the equation, we get:
\n
$$eV_0 = h\\frac{c}{\\lambda} - \\phi_0$$
\n
Now, we have two cases:
\n\n
The stopping potential is $$V_0$$, and the wavelength is $$\\lambda$$.
\n
The stopping potential is $$\\frac{V_0}{4}$$, and the wavelength is $$2\\lambda$$.
\n\n
For the first case, we use the photoelectric effect equation as is:
\n
$$eV_0 = \\frac{hc}{\\lambda} - \\phi_0$$
\n
For the second case, we replace $$V_0$$ with $$\\frac{V_0}{4}$$ and $$\\lambda$$ with $$2\\lambda$$:
A. The photocurrent is proportional to the intensity of the incident radiation
\n
B. Maximum Kinetic energy with which photoelectrons are emitted depends on the intensity of incident light.
\n
C. Max. K.E with which photoelectrons are emitted depends on the frequency of incident light.
\n
D. The emission of photoelectrons require a minimum threshold intensity of incident radiation.
\n
E. Max. K.E of the photoelectrons is independent of the frequency of the incident light.
\n
Choose the correct answer from the options given below:
",
"options": [
{
"text": "A and E only"
},
{
"text": "A and B only"
},
{
"text": "B and C only"
},
{
"text": "A and C only"
}
],
"answer": "A and C only",
"solution": "**Answer:** A and C only\n\n
The photoelectric effect is the phenomenon of emission of electrons (or photoelectrons) from the surface of a metal when it is illuminated by light of sufficient energy. The observations from the photoelectric effect led to the development of quantum theory.
\n
According to the principles of the photoelectric effect:
\n
A. The photocurrent (number of photoelectrons ejected per unit time) is indeed proportional to the intensity of the incident radiation. More intense light means more photons hitting the surface and thus more electrons being ejected.
\n
B. The maximum kinetic energy of the photoelectrons does not depend on the intensity of the incident light but rather on its frequency. Increasing the intensity of light increases the number of photoelectrons (current) but does not increase their maximum kinetic energy.
\n
C. The maximum kinetic energy of the photoelectrons does indeed depend on the frequency of the incident light. If the frequency of the incident light is below a certain threshold frequency specific to the metal, no photoelectrons are emitted regardless of the intensity of the light. Above this threshold, the maximum kinetic energy of the photoelectrons increases linearly with the frequency of the light.
\n
D. The emission of photoelectrons does not require a minimum threshold intensity of incident radiation, but rather a minimum threshold frequency.
\n
E. The maximum kinetic energy of the photoelectrons is not independent of the frequency of the incident light, but rather depends on it.
The work functions of Aluminium and Gold are $$4.1 ~\\mathrm{eV}$$ and and $$5.1 ~\\mathrm{eV}$$ respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is
We are given the work functions of Aluminium and Gold as $$4.1 ~\\mathrm{eV}$$ and $$5.1 ~\\mathrm{eV}$$, respectively.
\n
The stopping potential ($$V_s$$) is related to the frequency ($$f$$) of the incident light by the equation:
\n
$$eV_s = h(f - f_0)$$
\n
Where $$e$$ is the charge of an electron, $$h$$ is Planck's constant, and $$f_0$$ is the threshold frequency. The threshold frequency is related to the work function ($$\\phi$$) by:
\n
$$\\phi = hf_0$$
\n
So, we can write the equation for stopping potential as:
\n
$$V_s = \\frac{h}{e}(f - \\frac{\\phi}{h})$$
\n
This equation represents a straight line with slope $$\\frac{h}{e}$$. Therefore, the slope of the stopping potential versus frequency plot is the same for both Aluminium and Gold. The ratio of the slopes is:
\n
$$\\frac{\\text{Slope for Gold}}{\\text{Slope for Aluminium}} = \\frac{\\frac{h}{e}}{\\frac{h}{e}} = 1$$
\n
So , the ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is 1.
The threshold frequency, $$ \\nu_0 $$, corresponds to the minimum frequency of light required to eject electrons from the surface of a metal, a phenomenon known as the photoelectric effect. The work function, represented by $$ \\phi $$, is the minimum energy needed to remove an electron from the surface of the metal.
\n\n
The energy of a photon is given by the equation $$ E = h\\nu $$, where:
\n
\n
$$ E $$ is the energy of the photon,
\n
$$ h $$ is Planck's constant ($$ h = 6.626 \\times 10^{-34} \\, \\text{J} \\cdot \\text{s} $$), and
\n
$$ \\nu $$ is the frequency of the photon.
\n
\n\n
To find the threshold frequency for a metal with a work function of $$ 6.63 \\, \\text{eV} $$, we must first express the work function in joules (since Planck's constant is in joules per second). To convert electron volts to joules, use the conversion factor $$ 1 \\, \\text{eV} = 1.602 \\times 10^{-19} \\, \\text{J} $$:
In a photoelectric effect experiment a light of frequency 1.5 times the threshold frequency is made to fall on the surface of photosensitive material. Now if the frequency is halved and intensity is doubled, the number of photo electrons emitted will be:
When a metal surface is illuminated by light of wavelength $$\\lambda$$, the stopping potential is $$8 \\mathrm{~V}$$. When the same surface is illuminated by light of wavelength $$3 \\lambda$$, stopping potential is $$2 \\mathrm{~V}$$. The threshold wavelength for this surface is:
The work function of a substance is $$3.0 \\mathrm{~eV}$$. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately;
UV light of $$4.13 \\mathrm{~eV}$$ is incident on a photosensitive metal surface having work function $$3.13 \\mathrm{~eV}$$. The maximum kinetic energy of ejected photoelectrons will be:
To find the maximum kinetic energy of the ejected photoelectrons, we'll use the photoelectric effect equation:
\n\n$$KE_{\\text{max}} = h\\nu - \\phi$$\n\n
where $$KE_{\\text{max}}$$ is the maximum kinetic energy of the ejected electrons, $$h$$ is Planck's constant, $$\\nu$$ is the frequency of the incident light, and $$\\phi$$ is the work function of the metal.
\n\n
However, in this problem, we are given the energy of the UV light in electronvolts (eV) directly, which simplifies the problem. The energy of the UV light in electronvolts also represents the energy of the photons ($$h\\nu$$) hitting the metal surface. Thus, we can calculate the maximum kinetic energy of the ejected photoelectrons using the given energies directly:
Given below are two statements: one is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason R.
\n
Assertion A: Number of photons increases with increase in frequency of light.
\n
Reason R: Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.
\n
In the light of the above statements, choose the most appropriate answer from the options given below:
",
"options": [
{
"text": "$$\\mathbf{A}$$ is not correct but $$\\mathbf{R}$$ is correct.\n"
},
{
"text": "$$\\mathbf{A}$$ is correct but $$\\mathbf{R}$$ is not correct.\n"
},
{
"text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$.\n"
},
{
"text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct and $$\\mathbf{R}$$ is NOT the correct explanation of $$\\mathbf{A}$$."
}
],
"answer": "$$\\mathbf{A}$$ is not correct but $$\\mathbf{R}$$ is correct.\n",
"solution": "**Answer:** $$\\mathbf{A}$$ is not correct but $$\\mathbf{R}$$ is correct.\n\n\n
In order to determine the most appropriate answer to the question, let's analyze the given Assertion A and Reason R in detail.
\n\n
Assertion A: Number of photons increases with increase in frequency of light.
\n\n
This statement is not correct. The number of photons is determined by the intensity (or power) of the light and is given by the formula:
\n\n
$$N = \\frac{P}{hf}$$
\n\n
where $$N$$ is the number of photons per second, $$P$$ is the power (intensity) of the light, $$h$$ is Planck's constant, and $$f$$ is the frequency of the light. As the frequency increases, the energy per photon increases, but it does not necessarily mean that the number of photons increases unless the power of the light also increases proportionally.
\n\n
Reason R: Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.
\n\n
This statement is correct. According to the photoelectric effect, the maximum kinetic energy of emitted electrons is given by:
\n\n
$$K.E_{\\text{max}} = hf - \\phi$$
\n\n
where $$K.E_{\\text{max}}$$ is the maximum kinetic energy of the emitted electrons, $$h$$ is Planck's constant, $$f$$ is the frequency of the incident radiation, and $$\\phi$$ is the work function of the material. As the frequency of the incident light increases, the kinetic energy of the emitted electrons increases.
\n\n
Now, let's match the statements with the options given:
\n\n
Option A: $$\\mathbf{A}$$ is not correct but $$\\mathbf{R}$$ is correct.
\n\n
This option is correct because Assertion A is incorrect while Reason R is correct.
\n\n
Option B: $$\\mathbf{A}$$ is correct but $$\\mathbf{R}$$ is not correct.
\n\n
This option is incorrect because Assertion A is not correct.
\n\n
Option C: Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$.
\n\n
This option is incorrect because Assertion A is not correct.
\n\n
Option D: Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct and $$\\mathbf{R}$$ is NOT the correct explanation of $$\\mathbf{A}$$.
\n\n
This option is incorrect because Assertion A is not correct.
\n\n
Therefore, the most appropriate answer is:
\n\n
Option A: $$\\mathbf{A}$$ is not correct but $$\\mathbf{R}$$ is correct.
Which of the following statement is not true about stopping potential $$(\\mathrm{V}_0)$$ ?
",
"options": [
{
"text": "It depends upon frequency of the incident light.\n"
},
{
"text": "It is $$1 / \\mathrm{e}$$ times the maximum kinetic energy of electrons emitted.\n"
},
{
"text": "It increases with increase in intensity of the incident light.\n"
},
{
"text": "It depends on the nature of emitter material."
}
],
"answer": "It increases with increase in intensity of the incident light.\n",
"solution": "**Answer:** It increases with increase in intensity of the incident light.\n\n\n
Stopping potential is independent of intensity of \nlight. It depends on frequency of light.
When UV light of wavelength $$300 \\mathrm{~nm}$$ is incident on the metal surface having work function $$2.13 \\mathrm{~eV}$$, electron emission takes place. The stopping potential is :
To find the stopping potential ($V_s$) when UV light of wavelength $300 $ nm is incident on a metal surface with a work function of $2.13$ eV, we can use the photoelectric equation which relates the energy of the incident photons, the work function of the metal, and the kinetic energy of the emitted electrons.
\n\n
The energy (E) of the photons can be calculated using the equation:
\n\n
$E = \\frac{hc}{\\lambda}$
\n\n
where $h$ is the Planck constant, $c$ is the speed of light, and $\\lambda$ is the wavelength of the incident light. Given that $hc = 1240 $ eV nm, we can calculate the energy of the UV light photons directly.
\n\n
Substituting $hc = 1240 $ eV nm and $\\lambda = 300 $ nm into the equation gives:
Next, we can calculate the maximum kinetic energy of the emitted electrons using the photoelectric effect equation:
\n\n
$K_{max} = E - \\phi$
\n\n
where $K_{max}$ is the maximum kinetic energy of the emitted electrons, $E$ is the energy of the incident photons, and $\\phi$ is the work function of the metal.
\n\n
Given $E = 4.13$ eV and the work function $\\phi = 2.13$ eV, we have:
The stopping potential ($V_s$) is related to the maximum kinetic energy of the emitted electrons by the equation:
\n\n
$K_{max} = eV_s$
\n\n
where $e$ is the elementary charge (the charge of an electron), and $V_s$ is the stopping potential. Since $e = 1$ when using energy in eV and potential in volts, the stopping potential $V_s$ can be directly equated to the kinetic energy in eV:
\n\n
$V_s = K_{max} = 2 \\, \\text{V}$
\n\n
Therefore, the stopping potential is $2$ V, which corresponds to Option B.
In photoelectric experiment energy of $$2.48 \\mathrm{~eV}$$ irradiates a photo sensitive material. The stopping potential was measured to be $$0.5 \\mathrm{~V}$$. Work function of the photo sensitive material is :
To find the work function of the photo sensitive material, we can use the photoelectric equation which relates the kinetic energy of the ejected electrons to the photon energy and the work function ($\\phi$) of the material:
\n\n
$KE_{\\text{max}} = h\\nu - \\phi$
\n\n
Where $KE_{\\text{max}}$ is the maximum kinetic energy of the ejected electrons, $h\\nu$ is the energy of the incoming photon, and $\\phi$ is the work function of the material.
\n\n
However, the kinetic energy of the ejected electrons can also be related to the stopping potential ($V_s$) by the equation:
\n\n
$KE_{\\text{max}} = e \\cdot V_s$
\n\n
Substituting this into the first equation gives:
\n\n
$e \\cdot V_s = h\\nu - \\phi$
\n\n
Here, $e$ is the charge of an electron ($1.6 \\times 10^{-19} \\, \\text{C}$), but since we are dealing with energies in electronvolts (eV), and $1 \\, \\text{eV} = 1.6 \\times 10^{-19} \\, \\text{J}$, we can directly use the values given without converting the units:
\n\n
$\\phi = h\\nu - e \\cdot V_s$
\n\n
Where $h\\nu = 2.48 \\, \\text{eV}$ is the energy of the irradiating photons, and $V_s = 0.5 \\, \\text{V}$.
Therefore, the work function ($\\phi$) of the photo sensitive material is $1.98 \\, \\text{eV}$, which corresponds to Option A.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8904,
"subject": "Physics",
"question": "When the current changes from $$ + 2A$$ to $$-2A$$ in $$0.05$$ second, an $$e.m.f.$$ of $$8$$ $$V$$ is inducted in a coil. The coefficient of self- induction of the coil is ",
"options": [
{
"text": "$$0.2H$$ "
},
{
"text": "$$0.4H$$ "
},
{
"text": "$$0.8$$ $$H$$ "
},
{
"text": "$$0.1$$ $$H$$ "
}
],
"answer": "$$0.1$$ $$H$$ ",
"solution": "**Answer:** $$0.1$$ $$H$$ \n\n$$e = - {{\\Delta \\phi } \\over {\\Delta t}} = {{ - \\Delta \\left( {LI} \\right)} \\over {\\Delta t}} = - L{{\\Delta I} \\over {\\Delta t}}$$\n
$$\\therefore$$ $$\\left| e \\right| = L{{\\Delta I} \\over {\\Delta t}} \\Rightarrow 8 = L \\times {4 \\over {0.05}}$$\n
$$ \\Rightarrow L = {{8 \\times 0.05} \\over 4} = 0.1H$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8905,
"subject": "Physics",
"question": "Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon",
"options": [
{
"text": "the rates at which currents are changing in the two coils "
},
{
"text": "relative position and orientation of the two coils "
},
{
"text": "the currents in the two coils "
},
{
"text": "the materials of the wires of the coils "
}
],
"answer": "relative position and orientation of the two coils ",
"solution": "**Answer:** relative position and orientation of the two coils \n\nMutual conductance depends on the relative position and orientation of the two coils. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8906,
"subject": "Physics",
"question": "A coil of inductance $$300$$ $$mH$$ and resistance $$2\\,\\Omega $$ is connected to a source of voltage $$2$$ $$V$$. The current reaches half of its steady state value in ",
"options": [
{
"text": "$$0.1$$ $$s$$ "
},
{
"text": "$$0.05$$ $$s$$ "
},
{
"text": "$$0.3$$ $$s$$ "
},
{
"text": "$$0.15$$ $$s$$ "
}
],
"answer": "$$0.1$$ $$s$$ ",
"solution": "**Answer:** $$0.1$$ $$s$$ \n\nKEY CONCEPT : The charging of inductance given\n
$$ = 2.4\\pi \\times {10^{ - 4}}H$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8910,
"subject": "Physics",
"question": "Two coaxial solenoids of different radius carry current $$I$$ in the same direction. $$\\overrightarrow {{F_1}} $$ be the magnetic force on the inner solenoid due to the outer one and $$\\overrightarrow {{F_2}} $$ be the magnetic force on the outer solenoid due to the inner one. Then : ",
"options": [
{
"text": "$$\\overrightarrow {{F_1}} $$ is radially in wards and $$\\overrightarrow {{F_2}} = 0$$ "
},
{
"text": "$$\\overrightarrow {{F_1}} $$ is radially outwards and $$\\overrightarrow {{F_2}} = 0$$ "
},
{
"text": "$$\\overrightarrow {{F_1}} = \\overrightarrow {{F_2}} = 0$$ "
},
{
"text": "$$\\overrightarrow {{F_1}} $$ is radially inwards and $$\\overrightarrow {{F_2}} $$ is radially outards "
}
],
"answer": "$$\\overrightarrow {{F_1}} = \\overrightarrow {{F_2}} = 0$$ ",
"solution": "**Answer:** $$\\overrightarrow {{F_1}} = \\overrightarrow {{F_2}} = 0$$ \n\n$$\\mathop {F{}_1}\\limits^ \\to = \\mathop {F{}_2}\\limits^ \\to = 0$$ \n
because of action and reaction pair",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8911,
"subject": "Physics",
"question": "A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current I = Io cos ($$\\omega $$t). The emf induced in the smaller inner loop is nearly : ",
"options": [
{
"text": "$${{\\pi {\\mu _o}{I_o}} \\over 2}.{{{a^2}} \\over b}\\,\\omega \\sin \\left( {\\omega t} \\right)$$"
},
{
"text": "$${{\\pi {\\mu _o}{I_o}} \\over 2}.{{{a^2}} \\over b}\\,\\omega \\cos \\left( {\\omega t} \\right)$$ "
},
{
"text": "$$\\pi {\\mu _o}{I_o}\\,{{{a^2}} \\over b}\\omega \\sin \\left( {\\omega t} \\right)$$ "
},
{
"text": "$${{\\pi {\\mu _o}{I_o}\\,{b^2}} \\over a}\\omega \\cos \\left( {\\omega t} \\right)$$ "
}
],
"answer": "$${{\\pi {\\mu _o}{I_o}} \\over 2}.{{{a^2}} \\over b}\\,\\omega \\sin \\left( {\\omega t} \\right)$$",
"solution": "**Answer:** $${{\\pi {\\mu _o}{I_o}} \\over 2}.{{{a^2}} \\over b}\\,\\omega \\sin \\left( {\\omega t} \\right)$$\n\nMutual inductance, \n
M = $${{{\\mu _0}\\pi {N_1}{N_2}\\,{a^2}} \\over {2b}}$$\n
here $${{N_1}}$$ = N2 = 1\n
$$\\therefore\\,\\,\\,$$ M = $${{{\\mu _0}\\pi {a^2}} \\over {2b}}$$ \n
= $${{\\pi {\\mu _0}{I_0}} \\over 2}$$ . $${{{a^2}} \\over b}$$ $$\\omega $$ sin $$\\omega $$ t ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8912,
"subject": "Physics",
"question": "There are two long co-axial solenoids of same length $$l$$. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2, respectively. The ratio of mutual inductance to the self - inductance of the inner-coil is :",
"options": [
{
"text": "$${{{n_2}} \\over {{n_1}}}.{{{r_2}^2} \\over {{r_1}^2}}$$"
},
{
"text": "$${{{n_2}} \\over {{n_1}}}$$"
},
{
"text": "$${{{n_1}} \\over {{n_2}}}$$"
},
{
"text": "$${{{n_2}} \\over {{n_1}}}.{{{r_1}} \\over {{r_2}}}$$"
}
],
"answer": "$${{{n_2}} \\over {{n_1}}}$$",
"solution": "**Answer:** $${{{n_2}} \\over {{n_1}}}$$\n\n$$M = {\\mu _0}\\,{n_1}\\,{n_2}\\,\\pi r_1^2$$\n
$$L = {\\mu _0}\\,n_1^2\\,\\pi r_1^2$$\n
$$ \\Rightarrow \\,\\,{M \\over L} = {{{n_2}} \\over {{n_1}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8913,
"subject": "Physics",
"question": "Two coils 'P' and 'Q' are separated by some\ndistance. When a current of 3 A flows through\ncoil 'P', a magnetic flux of 10–3 Wb passes\nthrough 'Q'. No current is passed through 'Q'.\nWhen no current passes through 'P' and a\ncurrent of 2 A passes through 'Q', the flux\nthrough 'P' is :-",
"options": [
{
"text": "3.67 × 10–4 Wb"
},
{
"text": "3.67 × 10–3 Wb"
},
{
"text": "6.67 × 10–4 Wb"
},
{
"text": "6.67 × 10–3 Wb"
}
],
"answer": "6.67 × 10–4 Wb",
"solution": "**Answer:** 6.67 × 10–4 Wb\n\nMutual induction
\n$$ \\Rightarrow {\\phi _p} = 6.67{\\rm{ }} \\times {\\rm{ }}{10^{-4}}{\\rm{ }}Wb$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8914,
"subject": "Physics",
"question": "The total number of turns and cross-section area\nin a solenoid is fixed. However, its length L is varied\nby adjusting the separation between windings. The\ninductance of solenoid will be proportional to :",
"options": [
{
"text": "1/L2"
},
{
"text": "1/L"
},
{
"text": "L"
},
{
"text": "L2"
}
],
"answer": "1/L",
"solution": "**Answer:** 1/L\n\n$$\\phi $$ = NBA = LI
\nN $$\\mu $$0 nI$$\\pi $$R2 = LI
\n$$N{\\mu _0}{N \\over l}l\\pi {R^2} = LI$$
\nN and R constant
\nSelf inductance (L) $$ \\propto {1 \\over l} \\propto {1 \\over {length}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8915,
"subject": "Physics",
"question": "A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear\ndimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per\nunit length of the frame the same, then the self inductance of the coil: \n",
"options": [
{
"text": "decreases by a factor of $$9\\sqrt 3 $$"
},
{
"text": "increases by a factor of 27"
},
{
"text": "decreases by a factor of 9"
},
{
"text": "increases by a factor of 3"
}
],
"answer": "increases by a factor of 3",
"solution": "**Answer:** increases by a factor of 3\n\nTotal length L will remain constant\n
So self inductance will become 3 times",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 8916,
"subject": "Physics",
"question": "The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is -\n",
"options": [
{
"text": "740 J"
},
{
"text": "637.5 J"
},
{
"text": "540 J"
},
{
"text": "437.5 J"
}
],
"answer": "437.5 J",
"solution": "**Answer:** 437.5 J\n\n$$L{{di} \\over {dt}} = 25$$\n
$$ = {5 \\over 6} \\times 525 = 437.5$$ J",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8917,
"subject": "Physics",
"question": "In a fluorescent lamp choke (a small\ntransformer) 100 V of reverse voltage is\nproduced when the choke current changes\nuniformly from 0.25 A to 0 in a duration of\n0.025 ms. The self-inductance of the choke\n(in mH) is estimated to be ________.",
"options": [],
"answer": "10",
"solution": "**Answer:** 10\n\nV = $$\\left| {L{{di} \\over {dt}}} \\right|$$\n
$$ \\Rightarrow $$ L = $${V \\over {\\left| {{{di} \\over {dt}}} \\right|}}$$ = $${{100} \\over {{{0.25} \\over {0.025 \\times {{10}^{ - 3}}}}}}$$ = 10 mH",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8918,
"subject": "Physics",
"question": "A coil of inductance 2 H having negligible resistance is connected to a source of supply whose voltage is given by V = 3t volt. (where t is in second). If the voltage is applied when t = 0, then the energy stored in the coil after 4 s is _______J.",
"options": [],
"answer": "144",
"solution": "**Answer:** 144\n\n$$L{{di} \\over {dt}} = \\varepsilon $$
$$ = {9 \\over 8} \\times {{{{(4)}^4}} \\over {4 \\times 2}} = 144$$ J",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8919,
"subject": "Physics",
"question": "The time taken for the magnetic energy to reach 25% of its maximum value, when a solenoid of resistance R, inductance L is connected to a battery, is :",
"options": [
{
"text": "infinite"
},
{
"text": "$${L \\over R}$$ ln10"
},
{
"text": "$${L \\over R}$$ ln2"
},
{
"text": "$${L \\over R}$$ ln5"
}
],
"answer": "$${L \\over R}$$ ln2",
"solution": "**Answer:** $${L \\over R}$$ ln2\n\nMagnetic energy, U = $${1 \\over 2}$$LI$$_0^2$$
$$ \\Rightarrow t = {L \\over R}\\ln 2$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8920,
"subject": "Physics",
"question": "An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds :",
"options": [
{
"text": "0.4"
},
{
"text": "0.8"
},
{
"text": "0.125"
},
{
"text": "0.2"
}
],
"answer": "0.2",
"solution": "**Answer:** 0.2\n\n$$U = {1 \\over 2}L{i^2} = 64 \\Rightarrow L = 2$$
$${i^2}R = 640$$
$$R = {{640} \\over {{{(8)}^2}}} = 10$$
$$\\tau = {L \\over R} = {1 \\over 5} = 0.2$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8921,
"subject": "Physics",
"question": "A small square loop of side 'a' and one turn is placed inside a larger square loop of side b and one turn (b >> a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side 'b', then the coefficient of mutual inductance between the two loops is :",
"options": [
{
"text": "$${{{\\mu _0}} \\over {4\\pi }}8\\sqrt 2 {{{a^2}} \\over b}$$"
},
{
"text": "$${{{\\mu _0}} \\over {4\\pi }}{{8\\sqrt 2 } \\over a}$$"
},
{
"text": "$${{{\\mu _0}} \\over {4\\pi }}8\\sqrt 2 {{{b^2}} \\over a}$$"
},
{
"text": "$${{{\\mu _0}} \\over {4\\pi }}{{8\\sqrt 2 } \\over b}$$"
}
],
"answer": "$${{{\\mu _0}} \\over {4\\pi }}8\\sqrt 2 {{{a^2}} \\over b}$$",
"solution": "**Answer:** $${{{\\mu _0}} \\over {4\\pi }}8\\sqrt 2 {{{a^2}} \\over b}$$\n\n $$B = \\left[ {{{{\\mu _0}} \\over {4\\pi }}{I \\over {b/2}} \\times 2\\sin 45} \\right] \\times 4$$
A 10 $$\\Omega$$, 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 $$\\mu$$s. The average e.m.f. induced in the coil is ____________ V.
The current in a coil of self inductance 2.0 H is increasing according to I = 2 sin(t2) A. The amount of energy spent during the period when current changes from 0 to 2 A is ____________ J.
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8924,
"subject": "Physics",
"question": "$12 \\mathrm{~V}$ battery connected to a coil of resistance $6 \\Omega$ through a switch, drives a constant current in the circuit. The switch is opened in $1 \\mathrm{~ms}$. The emf induced across the coil is $20 \\mathrm{~V}$. The inductance of the coil is :",
"options": [
{
"text": "$5 ~ \\mathrm{mH}$"
},
{
"text": "$8 ~\\mathrm{mH}$"
},
{
"text": "$10~ \\mathrm{mH}$"
},
{
"text": "$12 ~\\mathrm{mH}$"
}
],
"answer": "$10~ \\mathrm{mH}$",
"solution": "**Answer:** $10~ \\mathrm{mH}$\n\nWhen the switch is closed, the circuit is a simple DC circuit and the current in the circuit is given by Ohm's law:
\nWhen the switch is opened, the current in the circuit drops to zero instantaneously.
However, the magnetic field generated by the current in the coil does not disappear immediately, and it continues to produce a back EMF that opposes the change in current.
This back EMF induces a voltage across the coil that can be calculated using Faraday's law of induction: $\\mathcal{E} = -L\\frac{\\Delta I}{\\Delta t}$, where $\\mathcal{E}$ is the induced voltage, $L$ is the inductance of the coil, and $\\Delta I/\\Delta t$ is the rate of change of current in the coil.\n
\nIn this case, we know that the induced voltage is $20\\text{V}$ and the rate of change of current is
$\\Delta I/\\Delta t = -2\\text{A}/(1\\text{ms}) = -2\\times 10^3\\text{A/s}$.
Substituting these values into the equation above, we get: $20\\text{V} = -L\\times(-2\\times 10^3\\text{A/s})$.\n
\nSolving for $L$, we get: $L = \\frac{20\\text{V}}{2\\times 10^3\\text{A/s}} = 0.01\\text{H}$.\n
\nTherefore, the inductance of the coil is $0.01\\text{H}$, or 10 mH.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8925,
"subject": "Physics",
"question": "
Two concentric circular coils with radii $$1 \\mathrm{~cm}$$ and $$1000 \\mathrm{~cm}$$, and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be ___________ $$\\times 10^{-8} \\mathrm{H}$$. (Take, $$\\pi^{2}=10$$ )
So, the mutual inductance between the two concentric coils is $$4 \\times 10^{-8} \\mathrm{H}$$.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8926,
"subject": "Physics",
"question": "A transformer has an efficiency of $80 \\%$ and works at $10 \\mathrm{~V}$ and $4 \\mathrm{~kW}$. If the secondary voltage is $240 \\mathrm{~V}$, then the current in the secondary coil is :",
"options": [
{
"text": "$1.33 \\mathrm{~A}$"
},
{
"text": "$13.33 \\mathrm{~A}$"
},
{
"text": "$1.59 \\mathrm{~A}$"
},
{
"text": "$15.1 \\mathrm{~A}$"
}
],
"answer": "$13.33 \\mathrm{~A}$",
"solution": "**Answer:** $13.33 \\mathrm{~A}$\n\n
To find the current in the secondary coil of the transformer, we first need to calculate the output power, taking into account the efficiency. The efficiency ($$ \\eta $$) of the transformer is given by the ratio of the output power ($$ P_{\\text{out}} $$) to the input power ($$ P_{\\text{in}} $$) times 100%.
\n
The given efficiency is $$ \\eta = 80\\% $$ or $$ \\eta = 0.8 $$ in decimal form. The input power is also given as $$ P_{\\text{in}} = 4 \\text{kW} $$ or $$ P_{\\text{in}} = 4000 \\text{W} $$.
\n
Let's calculate $$ P_{\\text{out}} $$ using the efficiency formula:
Two coils have mutual inductance $$0.002 \\mathrm{~H}$$. The current changes in the first coil according to the relation $$\\mathrm{i}=\\mathrm{i}_0 \\sin \\omega \\mathrm{t}$$, where $$\\mathrm{i}_0=5 \\mathrm{~A}$$ and $$\\omega=50 \\pi$$ rad/s. The maximum value of emf in the second coil is $$\\frac{\\pi}{\\alpha} \\mathrm{~V}$$. The value of $$\\alpha$$ is _______.
A small square loop of wire of side $$l$$ is placed inside a large square loop of wire of side $$L\\left(L=l^2\\right)$$. The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is $$\\sqrt{x} \\times 10^{-7} \\mathrm{H}$$, where $$x=$$ _________.
The current in an inductor is given by $$\\mathrm{I}=(3 \\mathrm{t}+8)$$ where $$\\mathrm{t}$$ is in second. The magnitude of induced emf produced in the inductor is $$12 \\mathrm{~mV}$$. The self-inductance of the inductor _________ $$\\mathrm{mH}$$.
The induced emf ($\\varepsilon$) in an inductor is given by Faraday's law of electromagnetic induction, which in its differential form for an inductor can be expressed as:
\n\n
$$\\varepsilon = L \\frac{dI}{dt}$$
\n\n
where:
\n\n
\n\n
$\\varepsilon$ is the induced emf in the inductor,
\n\n
$L$ is the inductance of the inductor,
\n\n
$\\frac{dI}{dt}$ is the rate of change of current through the inductor.
\n\n
\n\n
Given that the current $I = (3t + 8)$, where $t$ is in seconds, we can find the rate of change of current by differentiating $I$ with respect to $t$.
\n\n
$$\\frac{dI}{dt} = \\frac{d}{dt}(3t + 8) = 3$$
\n\n
The given magnitude of induced emf is $12 \\, \\text{mV} = 12 \\times 10^{-3} \\, \\text{V}$ (since $1\\,\\text{mV} = 10^{-3} \\, \\text{V}$).
\n\n
Now, plug these values into the formula to find $L$:
In a coil, the current changes from $$-2 \\mathrm{~A}$$ to $$+2 \\mathrm{~A}$$ in $$0.2 \\mathrm{~s}$$ and induces an emf of $$0.1 \\mathrm{~V}$$. The self inductance of the coil is :
To find the self-inductance of the coil, we can use the formula for induced electromotive force (emf), which is given by Faraday's law of electromagnetic induction as it applies to self-induction:
\n\n
$$ \\text{emf} = - L \\frac{\\Delta I}{\\Delta t} $$
\n\n
Where:
\n\n
\n\n
$ \\text{emf} $ = induced voltage in volts (V)
\n\n
$ L $ = self-inductance of the coil in henries (H)
\n\n
$ \\Delta I $ = change in current in amperes (A)
\n\n
$ \\Delta t $ = time interval in seconds (s) over which the current change occurs
Substituting the given values into the formula, we get:
\n\n
$$ 0.1 = -L \\frac{4}{0.2} $$
\n\n
Solving for $ L $, the equation becomes:
\n\n
$$ 0.1 = -L \\cdot 20 $$
\n\n
Therefore:
\n\n
$$ L = - \\frac{0.1}{20} $$
\n\n
Calculating the value of $ L $:
\n\n
$$ L = -0.005 \\, \\text{H} $$
\n\n
Or, expressing $ L $ in millihenries (mH):
\n\n
$$ L = -5 \\, \\text{mH} $$
\n\n
However, considering the absolute value (since inductance is a magnitude and cannot be negative in this context):
\n\n
$$ L = 5 \\, \\text{mH} $$
\n\n
Thus, the self-inductance of the coil is 5 mH, which corresponds to Option D.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8931,
"subject": "Physics",
"question": "In a uniform magnetic field of induction $$B$$ a wire in the form of a semicircle of radius $$r$$ rotates about the diameter of the circle with an angular frequency $$\\omega .$$ The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $$R,$$ the mean power generated per period of rotation is ",
"options": [
{
"text": "$${{{{\\left( {B\\pi r\\omega } \\right)}^2}} \\over {2R}}$$ "
},
{
"text": "$${{{{\\left( {B\\pi {r^2}\\omega } \\right)}^2}} \\over {8R}}$$ "
},
{
"text": "$${{B\\pi {r^2}\\omega } \\over {2R}}$$ "
},
{
"text": "$${{{{\\left( {B\\pi r{\\omega ^2}} \\right)}^2}} \\over {8R}}$$ "
}
],
"answer": "$${{{{\\left( {B\\pi {r^2}\\omega } \\right)}^2}} \\over {8R}}$$ ",
"solution": "**Answer:** $${{{{\\left( {B\\pi {r^2}\\omega } \\right)}^2}} \\over {8R}}$$ \n\n$$\\phi = \\overrightarrow B .\\overrightarrow A ;\\phi = BA\\cos \\,\\omega t$$\n
$${e_{t = 3}} = - 10\\,V$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8934,
"subject": "Physics",
"question": "A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to : ",
"options": [
{
"text": "development of air current when the plate is placed "
},
{
"text": "induction of electrical charge on the plate "
},
{
"text": "shielding of magnetic lines of force as aluminium is a para-magnetic material."
},
{
"text": "electromagnetic induction in the aluminium plate giving rise to electromagnetic damping."
}
],
"answer": "electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.",
"solution": "**Answer:** electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.\n\nBecause of the Lenz's law of conservation of energy.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8935,
"subject": "Physics",
"question": "A circular loop of radius $$0.3$$ $$cm$$ lies center of the small loop is on the axis of the bigger loop. The distance between their centers is $$15$$ $$cm.$$ If a current of $$2.0$$ $$A$$ flows through the smaller loop, than the flux linked with bigger loop is ",
"options": [
{
"text": "$$9.1 \\times {10^{ - 11}}\\,$$ weber "
},
{
"text": "$$6 \\times {10^{ - 11}}\\,$$ weber "
},
{
"text": "$$3.3 \\times {10^{ - 11}}\\,$$ weber "
},
{
"text": "$$6.6 \\times {10^{ - 9}}\\,$$ weber "
}
],
"answer": "$$9.1 \\times {10^{ - 11}}\\,$$ weber ",
"solution": "**Answer:** $$9.1 \\times {10^{ - 11}}\\,$$ weber \n\nAs we know, Magnetic flux, $$\\phi = B.A$$\n
$$ = 9.216 \\times {10^{ - 11}} = 9.2 \\times {10^{ - 11}}\\,\\,$$ weber",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8936,
"subject": "Physics",
"question": "A conducting metal circular-wire-loop of radius r is placed perpendicular to a\nmagnetic field which varies with time as \n B = B0e$${^{{{ - t} \\over r}}}$$ , where B0 and $$\\tau $$ are constants, at time t = 0. If the resistance of the loop is R then the heat generated in the loop after a long time (t $$ \\to $$ $$\\infty $$) is :",
"options": [
{
"text": "$${{{\\pi ^2}{r^4}B_0^4} \\over {2\\tau R}}$$ "
},
{
"text": "$${{{\\pi ^2}{r^4}B_0^2} \\over {2\\tau R}}$$ "
},
{
"text": "$${{{\\pi ^2}{r^4}B_0^2R} \\over \\tau }$$ "
},
{
"text": "$${{{\\pi ^2}{r^4}B_0^2} \\over {\\tau R}}$$ "
}
],
"answer": "$${{{\\pi ^2}{r^4}B_0^2} \\over {2\\tau R}}$$ ",
"solution": "**Answer:** $${{{\\pi ^2}{r^4}B_0^2} \\over {2\\tau R}}$$ \n\nGiven, \n
= $${{{\\pi ^2}{r^4}B_0^2} \\over {2\\tau R}}$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8937,
"subject": "Physics",
"question": "At the center of a fixed large circular coil of radius R, a much smaller circular coil of radius r is placed. The two coils are concentric and are in the same plane. The larger coil carries a current I. The smaller coil is set to rotate with a constant angular velocity $$\\omega $$ about an axis along their common diameter. Calculate the emf induced in their smaller coil after a time t of its start of rotation. ",
"options": [
{
"text": "$${{{\\mu _o}{\\rm I}} \\over {2\\,R}}$$ $$\\omega $$ $$\\pi $$ r2 sin$$\\omega $$ t"
},
{
"text": "$${{{\\mu _o}{\\rm I}} \\over {4\\,R}}$$ $$\\omega $$ $$\\pi $$ r2 sin$$\\omega $$ t"
},
{
"text": "$${{{\\mu _o}{\\rm I}} \\over {4\\,R}}$$ $$\\omega $$ r2 sin$$\\omega $$ t"
},
{
"text": "$${{{\\mu _o}{\\rm I}} \\over {2\\,R}}$$ $$\\omega $$ r2 sin$$\\omega $$ t"
}
],
"answer": "$${{{\\mu _o}{\\rm I}} \\over {2\\,R}}$$ $$\\omega $$ $$\\pi $$ r2 sin$$\\omega $$ t",
"solution": "**Answer:** $${{{\\mu _o}{\\rm I}} \\over {2\\,R}}$$ $$\\omega $$ $$\\pi $$ r2 sin$$\\omega $$ t\n\n
We know that electric flux $$\\phi = \\overrightarrow B \\,.\\,\\overrightarrow A $$
\n
$$ \\Rightarrow \\phi = BA\\cos \\omega t$$
\n
Now, $$B = {{{\\mu _0}} \\over 2}{I \\over R}$$ is magnetic field due to circular coil of radius R and $$A = \\pi {r^2}$$ is area of circular coil of radius r. Therefore,
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8938,
"subject": "Physics",
"question": "A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity $$\\omega ,$$ the maxium e.m.f. induced in the coil will be: ",
"options": [
{
"text": "3 nBA$$\\omega $$"
},
{
"text": "$${3 \\over 2}$$ nBA$$\\omega $$"
},
{
"text": "nBA$$\\omega $$"
},
{
"text": "$${1 \\over 2}$$ nBA$$\\omega $$"
}
],
"answer": "nBA$$\\omega $$",
"solution": "**Answer:** nBA$$\\omega $$\n\nFlux in the coil, $$\\phi $$ = nBA sin($$\\omega $$t)\n
When n = no. of turns\n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ A = Area of coil \n
$$\\left| e \\right| = {{d\\phi } \\over {dt}}$$\n
= nBA$$\\omega $$ cos$$\\omega $$t\n
$$\\therefore\\,\\,\\,$$ emax = nBA$$\\omega $$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8939,
"subject": "Physics",
"question": "A conducting circular loop made of a thin wire, has area 3.5 $$ \\times $$ 10$$-$$3 m2 and resistance 10 $$\\Omega $$. It is placed perpendicular to a time dependent magnetic field B(t) = (0.4T)sin(50$$\\pi $$t). The field is uniform in space. Then the net charge flowing through the loop during t = 0 s and t = 10 ms is close to : ",
"options": [
{
"text": "0.14 mC"
},
{
"text": "0.7 mC"
},
{
"text": "0.21 mC"
},
{
"text": "0.6 mC"
}
],
"answer": "0.14 mC",
"solution": "**Answer:** 0.14 mC\n\nAt t = 0 s\n
B(0) = 0.4 sin (0) = 0\n
and at t = 10 ms\n
B(10) = 0.4 sin (50$$\\pi $$$$ \\times $$10$$ \\times $$10-3)\n
= 0.4 sin $$\\left( {{\\pi \\over 2}} \\right)$$\n
= 0.14 mC",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8940,
"subject": "Physics",
"question": "A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is - ",
"options": [
{
"text": "2mV"
},
{
"text": "12 mV"
},
{
"text": "6 mV"
},
{
"text": "1 mV"
}
],
"answer": "12 mV",
"solution": "**Answer:** 12 mV\n\n
We can apply Faraday's law of electromagnetic induction to solve this problem. Faraday's law states that the induced electromotive force (emf) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.
\n
The cube is moving through a magnetic field, so it's behaving like a conductor moving through a magnetic field. The induced emf or voltage can be calculated by using the formula:
\n
emf = B $$ \\times $$ v $$ \\times $$ d
\n
Where:\n
B is the magnetic field strength,\n
v is the velocity of the conductor, and\n
d is the length of the conductor perpendicular to the direction of motion and magnetic field.
\n
In this case, the cube is moving in the y-direction and the magnetic field is in the z-direction. So, the faces perpendicular to the x-axis are involved. The length of the conductor (d) perpendicular to the motion and magnetic field is the edge length of the cube, which is 2 cm or 0.02 m.
\n
So, plugging the given values into the formula:
\n
emf = 0.1 T $$ \\times $$ 6 m/s $$ \\times $$ 0.02 m = 0.012 V = 12 mV
\n
So, the potential difference between the two faces of the cube perpendicular to the x-axis is 12 mV.
Therefore, Option B is correct.
\n",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 8941,
"subject": "Physics",
"question": "Two concentric circular coils, C1 and C2 are\n placed in the XY plane. C1 has 500 turns, and\n a radius of 1 cm. C2 has 200 turns and radius\n of 20 cm. C2 carries a time dependent current\n I(t) = (5t2 – 2t + 3) A where t is in s. The emf\n induced in C1 (in mV), at the instant t = 1 s is\n $${4 \\over x}$$. The value of x is ___ .",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n$${B_2} = {{{\\mu _0}{I_2}{N_2}} \\over {2{R_2}}}$$
$$ \\therefore $$ $$x = 5$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8942,
"subject": "Physics",
"question": "A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of\na metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field\nchanges with time at a steady rate $${{dB} \\over {dt}}$$ = 0.032 Ts–1. The induced current in the loop is close to\n(Resistivity of the metal wire is 1.23 $$ \\times $$ 10–8 $$\\Omega $$m)\n",
"options": [
{
"text": "0.53 A"
},
{
"text": "0.43 A"
},
{
"text": "0.34 A"
},
{
"text": "0.61 A"
}
],
"answer": "0.61 A",
"solution": "**Answer:** 0.61 A\n\nWe know, $$\\phi = BA$$
$$ \\Rightarrow $$ $$i = 612.68 \\times {10^{ - 3}} = 0.61A$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8943,
"subject": "Physics",
"question": "A circular coil of radius 10 cm is placed in a\nuniform magnetic field of 3.0 $$ \\times $$ 10–5 T with its\nplane perpendicular to the field initially. It is\nrotated at constant angular speed about an\naxis along the diameter of coil and\nperpendicular to magnetic field so that it\nundergoes half of rotation in 0.2 s. The\nmaximum value of EMF induced (in $$\\mu $$V) in the\ncoil will be close to the integer _______.",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\nAt any time\nflux $$\\phi $$ = BA cos $$\\omega t$$\n
= 15 $$\\mu $$V",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8944,
"subject": "Physics",
"question": "A planar loop of wire rotates in a uniform magnetic field. Initially at t = 0, the plane of the loop\nis perpendicular to the magnetic field. If it rotates with a period of 10 s about an axis in its plane\nthen the magnitude of induced emf will be maximum and minimum, respectively at :",
"options": [
{
"text": "2.5 s and 7.5 s"
},
{
"text": "5.0 s and 10.0 s"
},
{
"text": "5.0 s and 7.5 s"
},
{
"text": "2.5 s and 5.0 s"
}
],
"answer": "2.5 s and 5.0 s",
"solution": "**Answer:** 2.5 s and 5.0 s\n\nFlux $$\\phi $$ = $$\\overrightarrow B .\\overrightarrow A $$ = BAcos$$\\omega $$t\n
$$ \\Rightarrow $$ t = $${T \\over 4}$$ or $${3T \\over 4}$$ i.e. 2.5 s or 7.5 s.\n
For induced emf to be minimum i.e zero.\n
$${{2\\pi t} \\over T}$$ = n$$\\pi $$\n
$$ \\Rightarrow $$ t = $$n{\\pi \\over 2}$$\n
$$ \\Rightarrow $$ Induced emf is zero at t = 5 s, 10 s",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8945,
"subject": "Physics",
"question": "A loop ABCDEFA of straight edges has six corner points A(0, 0, 0), B(5, 0, 0), C(5, 5, 0), D (0, 5,\n0), E(0, 5, 5) and F(0, 0, 5). The magnetic field in this region is $$\\overrightarrow B = \\left( {3\\widehat i + 4\\widehat k} \\right)T$$\n. The quantity of\nflux through the loop ABCDEFA (in Wb) is _______.",
"options": [],
"answer": "175",
"solution": "**Answer:** 175\n\n$$\\phi $$ = $$\\overrightarrow B .\\overrightarrow A $$ = $$\\left( {3\\widehat i + 4\\widehat k} \\right).\\left( {25\\widehat i + 25\\widehat k} \\right)$$\n
$$ \\Rightarrow $$ $$\\phi $$ = (3 $$ \\times $$ 25) + (4 $$ \\times $$ 25) = 175 weber",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 8946,
"subject": "Physics",
"question": "A long solenoid of radius R carries a time (t) - dependent current I(t)=I0t(1 - t). A ring of radius 2R is placed coaxially near its middle. During the time interval 0 $$ \\le $$ t $$ \\le $$ 1, the induced current (IR) and the induced EMF(VR) in the ring change as :",
"options": [
{
"text": "Direction of IR remains unchanged and VR is zero at t = 0.25"
},
{
"text": "Direction of IR remains unchanged and VR is maximum at t = 0.5"
},
{
"text": "At t = 0.25 direction of IR reverses and VR is maximum"
},
{
"text": "At t = 0.5 direction of IR reverses and VR is zero"
}
],
"answer": "At t = 0.5 direction of IR reverses and VR is zero",
"solution": "**Answer:** At t = 0.5 direction of IR reverses and VR is zero\n\nI(t) = I0t(1 - t)\n
after t = 0.5, IR reverses its direction.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8947,
"subject": "Physics",
"question": "Consider a circular coil of wire carrying constant current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by $$\\phi $$i. The magnetic flux through the area of the circular coil area is given by $$\\phi $$0. Which of the following option is correct?\n",
"options": [
{
"text": "$$\\phi $$i = $$\\phi $$0"
},
{
"text": "$$\\phi $$i < $$\\phi $$0"
},
{
"text": "$$\\phi $$i $$>$$ $$\\phi $$0"
},
{
"text": "$$\\phi $$i = - $$\\phi $$0"
}
],
"answer": "$$\\phi $$i = - $$\\phi $$0",
"solution": "**Answer:** $$\\phi $$i = - $$\\phi $$0\n\nAs, magnetic field lines forms a closed loop, hence each line from circular area will pass through outer area\nin opposite direction hence $$\\phi $$i = - $$\\phi $$0.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8948,
"subject": "Physics",
"question": "If the maximum value of accelerating potential provided by a ratio frequency oscillator is 12 kV. The number of revolution made by a proton in a cyclotron to achieve one sixth of the speed of light is ...............
A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be :
The electrical power dissipated due to the current induced in a coil placed in a time-varying magnetic field can be determined by considering Faraday's Law of Induction and the resistance of the coil.
\n
Faraday's Law of Induction
\n
The induced EMF ($\\mathcal{E}$) in a coil with $N$ turns experiencing a time-varying magnetic flux ($\\Phi_B$) is given by:
\n
$$ \\mathcal{E} = -N \\frac{d\\Phi_B}{dt} $$
\n
Resistance of the Coil
\n
The resistance ($R$) of a wire depends on its length ($L$) and cross-sectional area ($A$) as well as the resistivity ($\\rho$) of the material:
\n
$$ R = \\frac{\\rho L}{A} $$
\n
For a coil of radius $r$ and with wire of radius $a$, assuming the wire is wound tightly with a length approximated by the circumference of the coil multiplied by the number of turns, we have:
\n
$$ L \\approx 2\\pi r N $$
\n
And the cross-sectional area of the wire is given by:
\n
$$ A = \\pi a^2 $$
\n
Thus, the resistance becomes:
\n
$$ R \\approx \\frac{\\rho \\cdot 2\\pi r N}{\\pi a^2} = \\frac{2\\rho r N}{a^2} $$
\n
Power Dissipated
\n
The electrical power ($P$) dissipated in the coil is related to the induced current ($I$) and the resistance ($R$):
\n
$$ P = I^2 R $$
\n
Using Ohm's Law, the induced current $I$ can be expressed as:
\n
$$ I = \\frac{\\mathcal{E}}{R} $$
\n
Substituting the expressions for $\\mathcal{E}$ and $R$:
\n
$$ I = \\frac{N \\frac{d\\Phi_B}{dt}}{\\frac{2\\rho r N}{a^2}} = \\frac{a^2}{2\\rho r} \\cdot \\frac{d\\Phi_B}{dt} $$
\n
Hence, the power dissipated:
\n
$$ P = \\left(\\frac{a^2}{2\\rho r} \\cdot \\frac{d\\Phi_B}{dt}\\right)^2 \\cdot \\frac{2\\rho r N}{a^2} $$
\n
$$ P = \\frac{a^4}{4\\rho^2 r^2} \\left( \\frac{d\\Phi_B}{dt} \\right)^2 \\cdot \\frac{2\\rho r N}{a^2} $$
\n
$$ P = \\frac{a^2}{2\\rho r} \\cdot N \\left( \\frac{d\\Phi_B}{dt} \\right)^2 $$
\n
Case when the number of turns is halved and the radius of the wire is doubled
A metallic rod of length 20 cm is placed in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal component of the Earth's magnetic field at that place is 4 $$\\times$$ 10$$-$$3 T and the angle of dip is 45$$^\\circ$$. The emf induced in the rod is ___________ mV.
The magnetic flux through a coil perpendicular to its plane is varying according to the relation $$\\phi = (5{t^3} + 4{t^2} + 2t - 5)$$ Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at t = 2 s will be,
The electric current in a circular coil of 2 turns produces a magnetic induction B1 at its centre. The coil is unwound and in rewound into a circular coil of 5 tuns and the same current produces a magnetic induction B2 at its centre. The ratio of $${{{B_2}} \\over {{B_1}}}$$ is
Magnetic flux (in weber) in a closed circuit of resistance 20 $$\\Omega$$ varies with time t(s) at $$\\phi$$ = 8t2 $$-$$ 9t + 5. The magnitude of the induced current at t = 0.25 s will be ____________ mA.
In a coil of resistance $$8 \\,\\Omega$$, the magnetic flux due to an external magnetic field varies with time as $$\\phi=\\frac{2}{3}\\left(9-t^{2}\\right)$$. The value of total heat produced in the coil, till the flux becomes zero, will be _____________ $$J$$.
A conducting circular loop is placed in $$X-Y$$ plane in presence of magnetic field $$\\overrightarrow{\\mathrm{B}}=\\left(3 \\mathrm{t}^{3} \\,\\hat{j}+3 \\mathrm{t}^{2}\\, \\hat{k}\\right)$$ in SI unit. If the radius of the loop is $$1 \\mathrm{~m}$$, the induced emf in the loop, at time, $$\\mathrm{t}=2 \\mathrm{~s}$$ is $$\\mathrm{n} \\pi \\,\\mathrm{V}$$. The value of $$\\mathrm{n}$$ is ___________.
A coil is placed in magnetic field such that plane of coil is perpendicular to the direction of magnetic field. The magnetic flux through a coil can be changed :
\n
A. By changing the magnitude of the magnetic field within the coil.
\n
B. By changing the area of coil within the magnetic field.
\n
C. By changing the angle between the direction of magnetic field and the plane of the coil.
\n
D. By reversing the magnetic field direction abruptly without changing its magnitude.
\n
Choose the most appropriate answer from the options given below :
",
"options": [
{
"text": "A, B and D only"
},
{
"text": "A, B and C only"
},
{
"text": "A and B only"
},
{
"text": "A and C only"
}
],
"answer": "A, B and C only",
"solution": "**Answer:** A, B and C only\n\nThe magnitude of magnetic flux is given by :\n\n
$$\\Phi = BA \\cos \\theta$$\n\n
where $$B$$ is the magnitude of the magnetic field, $$A$$ is the area of the coil, and $$\\theta$$ is the angle between the normal to the area and the direction of the magnetic field.\n
The correct option is A, B, and C only.\n\n
A. The magnetic flux through a coil can be changed by changing the magnitude of the magnetic field within the coil. A stronger magnetic field will increase the magnetic flux, while a weaker magnetic field will decrease the magnetic flux.\n\n
B. The magnetic flux through a coil can also be changed by changing the area of the coil within the magnetic field. A larger area of the coil will result in a greater magnetic flux, while a smaller area will result in a smaller magnetic flux.\n\n
C. The magnetic flux through a coil can also be changed by changing the angle between the direction of the magnetic field and the plane of the coil. When the angle is perpendicular to the plane of the coil, the magnetic flux is at its maximum. When the angle is parallel to the plane of the coil, the magnetic flux is zero.\n\n
D. Reversing the magnetic field direction abruptly without changing its magnitude will not change the magnetic flux through the coil. Magnetic flux is proportional to the dot product of the magnetic field and the area vector of the coil. If the magnitude of the magnetic field remains the same and the direction is reversed, the dot product remains the same and the magnetic flux remains unchanged.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8957,
"subject": "Physics",
"question": "
A square loop of area 25 cm$$^2$$ has a resistance of 10 $$\\Omega$$. The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec, will be
A conducting circular loop of radius $$\\frac{10}{\\sqrt\\pi}$$ cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is :
An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional area $$24 \\mathrm{~cm}^{2}$$. The two ends of the wire are connected to a resistor. The total resistance in the circuit is $$12 ~\\Omega$$. If an externally applied uniform magnetic field in the core along its axis changes from $$1.5 \\mathrm{~T}$$ in one direction to $$1.5 ~\\mathrm{T}$$ in the opposite direction, the charge flowing through a point in the circuit during the change of magnetic field will be ___________ $$\\mathrm{mC}$$.
",
"options": [],
"answer": "60",
"solution": "**Answer:** 60\n\nThe induced emf in the circuit is given by Faraday's law of electromagnetic induction, which is $\\mathcal{E}=-d\\phi/dt$, where $\\phi$ is the magnetic flux through the circuit.\n
\nThe magnetic flux through the circuit is proportional to the magnetic field through the core, so we can write $\\phi=NBA$, where $N$ is the number of turns in the loop, $B$ is the magnetic field through the core, and $A$ is the cross-sectional area of the core.\n
\nAs the magnetic field changes from $1.5\\mathrm{~T}$ in one direction to $-1.5\\mathrm{~T}$ in the opposite direction, the change in magnetic flux is $\\Delta\\phi=2NBA$.\n
\nThe induced emf drives a current $I$ through the resistor in the circuit, and the current and the resistance are related by Ohm's law, which is $I=\\mathcal{E}/R$. Substituting the expression for $\\mathcal{E}$ into this equation, we get $I=-d\\phi/dtR$.\n
\nThe charge $Q$ that flows through the circuit during the change in magnetic field is given by $Q=\\int Idt$. Substituting the expression for $I$ into this equation and integrating with respect to time, we get $Q=-\\Delta\\phi/R$, where $\\Delta\\phi$ is the change in magnetic flux and $R$ is the resistance of the circuit.\n
\nSubstituting the given values into this expression, we get:\n
\nTherefore, the charge flowing through a point in the circuit during the change of magnetic field is $60\\mathrm{~mC}$, which is the same as the provided answer.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8960,
"subject": "Physics",
"question": "
Given below are two statements:
\n
Statement I : If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double.
\n
Statement II : Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio
\n
In the light of the above statements, choose the correct answer from the options given below :
",
"options": [
{
"text": "Statement I is true but Statement II is false"
},
{
"text": "Statement I is false but Statement II is true"
},
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Both Statement I and Statement II are true"
}
],
"answer": "Statement I is true but Statement II is false",
"solution": "**Answer:** Statement I is true but Statement II is false\n\n
Statement I: If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double.
\n
This statement is true. The formula for current sensitivity ($I_s$) of a moving coil galvanometer is given by:
\n
$I_s = \\frac{NAB}{k}$
\n
where:
\n
\n
$N$ is the number of turns in the coil,
\n
$A$ is the area of the coil,
\n
$B$ is the magnetic field strength, and
\n
$k$ is the spring constant of the coil.
\n
\n
From this formula, you can see that the current sensitivity is directly proportional to the number of turns (N). If $N$ is doubled, then the current sensitivity will also double.
\n
Statement II: Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio.
\n
This statement is false. The formula for voltage sensitivity ($V_s$) of a moving coil galvanometer is given by:
\n
$V_s = I_s R = \\frac{NAB}{k} R$
\n
where:
\n
\n
$R$ is the resistance of the coil.
\n
\n
From this formula, you can see that the voltage sensitivity is proportional to the number of turns ($N$) but also inversely proportional to the coil resistance ($R$). If you double the number of turns ($N$), you also double the length of the wire making up the coil, and thus, you double the resistance ($R$) of the coil. The doubling of $N$ is offset by the doubling of $R$, so the overall voltage sensitivity remains the same.
\n
Therefore, increasing the current sensitivity by only increasing the number of turns in the coil will not increase the voltage sensitivity in the same ratio.
A. moving the coil with uniform speed inside uniform magnetic field
\n
B. moving the coil with non uniform speed inside uniform magnetic field
\n
C. rotating the coil inside the uniform magnetic field
\n
D. changing the area of the coil inside the uniform magnetic field
\n
Choose the correct answer from the options given below:
",
"options": [
{
"text": "A and C only"
},
{
"text": "C and D only"
},
{
"text": "B and D only"
},
{
"text": "B and C only"
}
],
"answer": "C and D only",
"solution": "**Answer:** C and D only\n\nIf the coil is simply moved at uniform or non-uniform speed in a uniform magnetic field without changing the orientation of the coil or the area of the coil enclosed by the magnetic field, the magnetic flux through the coil does not change, and no emf is induced according to Faraday's Law of electromagnetic induction.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8962,
"subject": "Physics",
"question": "
A rectangular loop of length $$2.5 \\mathrm{~m}$$ and width $$2 \\mathrm{~m}$$ is placed at $$60^{\\circ}$$ to a magnetic field of $$4 \\mathrm{~T}$$. The loop is removed from the field in $$10 \\mathrm{~sec}$$. The average emf induced in the loop during this time is
According to Faraday's Law of Electromagnetic Induction, the induced emf in a closed circuit is equal to the negative rate of change of magnetic flux through the circuit. Mathematically, it is expressed as:
\n\n
$$ \\varepsilon = -\\frac{d\\Phi}{dt} $$
\n\n
Where $ \\varepsilon $ is the induced emf, and $ \\Phi $ is the magnetic flux.
\n\n
To find the magnetic flux $ \\Phi $ through the rectangular loop, we use the formula:
\n\n
$$ \\Phi = B \\cdot A \\cdot \\cos(\\theta) $$
\n\n
Where:
\n\n
\n\n
$ B $ is the magnetic field strength,
\n\n
$ A $ is the area of the loop, and
\n\n
$ \\theta $ is the angle between the magnetic field lines and the normal (perpendicular) to the plane of the loop.
When the loop is removed from the magnetic field, the final magnetic flux $ \\Phi_{final} $ is zero, because the loop is no longer within the magnetic field. Thus, the change in magnetic flux $ \\Delta\\Phi $ is:
The magnetic flux $$\\phi$$ (in weber) linked with a closed circuit of resistance $$8 \\Omega$$ varies with time (in seconds) as $$\\phi=5 t^2-36 t+1$$. The induced current in the circuit at $$t=2 \\mathrm{~s}$$ is __________ A.
A coil is places perpendicular to a magnetic field of $$5000 \\mathrm{~T}$$. When the field is changed to $$3000 \\mathrm{~T}$$ in $$2 \\mathrm{~s}$$, an induced emf of $$22 \\mathrm{~V}$$ is produced in the coil. If the diameter of the coil is $$0.02 \\mathrm{~m}$$, then the number of turns in the coil is:
Faraday law $$\\rightarrow \\oint \\vec{E} \\cdot \\overrightarrow{d l}=\\frac{d \\phi_{\\mathrm{B}}}{d t}$$
\n
Gauss' law for electricity $$\\rightarrow \\oint \\overrightarrow{\\mathrm{E}} \\cdot \\overrightarrow{\\mathrm{dA}}=\\frac{\\mathrm{Q}}{\\varepsilon_0}$$
\n
Gauss ' law for magnetism $$\\rightarrow \\oint \\vec{B} \\cdot \\overrightarrow{d A}=0$$
A square loop of side $$10 \\mathrm{~cm}$$ and resistance $$0.7 \\Omega$$ is placed vertically in east-west plane. A uniform magnetic field of $$0.20 T$$ is set up across the plane in north east direction. The magnetic field is decreased to zero in $$1 \\mathrm{~s}$$ at a steady rate. Then, magnitude of induced emf is $$\\sqrt{x} \\times 10^{-3} \\mathrm{~V}$$. The value of $$x$$ is __________.
Let's identify each law listed in List I and match it with the corresponding mathematical expression listed in List II.
\n\nGauss's law of magnetostatics states that the total magnetic flux through a closed surface is zero, as magnetic monopoles do not exist. This is given by the formula:\n\n
$$ \\oint \\vec{B} \\cdot \\vec{da} = 0 $$
\n\n
So, (A) matches with (II).
\n\nFaraday's law of electromagnetic induction states that the induced electromotive force (emf) in any closed loop is equal to the negative of the time rate of change of the magnetic flux through the loop. It is given by:\n\n
\n\nAmpere's law relates the integrated magnetic field around a closed loop to the electric current passing through the loop. The mathematical expression given in the context of magnetostatics (without the displacement current) is:\n\n
$$ \\oint \\vec{B} \\cdot \\vec{dl} = \\mu_0 I $$
\n\n
Consequently, (C) matches with (IV).
\n\n
Lastly, Gauss's law of electrostatics states that the total electric flux out of a closed surface is proportional to the charge enclosed within the surface:
Based on these matches, the correct answer must link A-II, B-III, C-IV, and D-I:
\n\n
The correct option is:
\n\n
Option C
\n\n
A-II, B-III, C-IV, D-I
\n\n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 8968,
"subject": "Physics",
"question": "A metal conductor of length $$1$$ $$m$$ rotates vertically about one of its ends at angular velocity $$5$$ radians per second. If the horizontal component of earth's magnetic field is $$0.2 \\times {10^{ - 4}}T,$$ then the $$e.m.f.$$ developed between the two ends of the conductor is",
"options": [
{
"text": "$$5mV$$ "
},
{
"text": "$$50\\mu V$$ "
},
{
"text": "$$5\\mu V$$"
},
{
"text": "$$50mV$$ "
}
],
"answer": "$$50\\mu V$$ ",
"solution": "**Answer:** $$50\\mu V$$ \n\n$$\\ell = 1m,\\,\\,\\omega = 5\\,rad/s,\\,\\,B = 0.2 \\times {10^{ - 4}}T$$\n
$$\\varepsilon = {{B\\omega {\\ell ^2}} \\over 2} = {{0.2 \\times {{10}^{ - 4}} \\times 5 \\times 1} \\over 2} = 50\\mu V$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8969,
"subject": "Physics",
"question": "A boat is moving due east in a region where the earth's magnetic fields is $$5.0 \\times {10^{ - 5}}$$ $$N{A^{ - 1}}\\,{m^{ - 1}}$$ due north and horizontal. The best carries a vertical aerial $$2$$ $$m$$ long. If the speed of the boat is $$1.50\\,m{s^{ - 1}},$$ the magnitude of the induced $$emf$$ in the wire of aerial is : ",
"options": [
{
"text": "$$0.75$$ $$mV$$ "
},
{
"text": "$$0.50$$ $$mV$$ "
},
{
"text": "$$0.15$$ $$mV$$ "
},
{
"text": "$$1$$ $$mV$$ "
}
],
"answer": "$$0.15$$ $$mV$$ ",
"solution": "**Answer:** $$0.15$$ $$mV$$ \n\nInduced $$emf$$ $$ = v{B_H}l = 1.5 \\times 5 \\times {10^{ - 5}} \\times 2$$ \n
$$ = 15 \\times {10^{ - 5}} = 0.15\\,mV$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8970,
"subject": "Physics",
"question": "A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 ms–1, at right angles to the horizontal component of the earth's magnetic field of 0.3 $$ \\times $$ 10–4 Wb/m2. The value of the induced emf in wire is : ",
"options": [
{
"text": "0.3 $$ \\times $$ 10–3 V"
},
{
"text": "2.5 $$ \\times $$ 10–3 V"
},
{
"text": "1.5 $$ \\times $$ 10–3 V"
},
{
"text": "1.1 $$ \\times $$ 10–3 V"
}
],
"answer": "1.1 $$ \\times $$ 10–3 V",
"solution": "**Answer:** 1.1 $$ \\times $$ 10–3 V\n\nInduied emf = Bv$$\\ell $$ sin 45o\n
= 1.1 $$ \\times $$ 10$$-$$3 V",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8971,
"subject": "Physics",
"question": "An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5 $$\\times$$ 10$$-$$4 Wb/m2 and the angle of dip is 60$$^\\circ$$. The emf induced between the tips of the plane wings will be __________.",
"options": [
{
"text": "88.37 mV"
},
{
"text": "62.50 mV"
},
{
"text": "54.125 mV"
},
{
"text": "108.25 mV"
}
],
"answer": "108.25 mV",
"solution": "**Answer:** 108.25 mV\n\n$$\\varepsilon $$ind = (Bv) LV and Bv = BTotal sin60o
= 108.25 mV ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8972,
"subject": "Physics",
"question": "A circular conducting coil of radius 1 m is being heated by the change of magnetic field $$\\overrightarrow B $$ passing perpendicular to the plane in which the coil is laid. The resistance of the coil is 2 $$\\mu$$$$\\Omega$$. The magnetic field is slowly switched off such that its magnitude changes in time as
The energy dissipated by the coil before the magnetic field is switched off completely is E = ___________ mJ.",
"options": [],
"answer": "80",
"solution": "**Answer:** 80\n\n$$\\phi = \\overrightarrow B .\\overrightarrow S $$
Heat = 80 mJ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8973,
"subject": "Physics",
"question": "A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s$$-$$1 in a uniform horizontal magnetic field of 3.0 $$\\times$$ 10$$-$$2 T. The maximum emf induced the coil will be ................. $$\\times$$ 10$$-$$2 volt (rounded off to the nearest integer)",
"options": [],
"answer": "60",
"solution": "**Answer:** 60\n\nMaximum emf $$\\varepsilon = N\\,\\omega AB$$
N = 20, $$\\omega$$ = 50, B = 3 $$\\times$$ 10$$-$$2 T
Rounded off to nearest integer = 60",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8974,
"subject": "Physics",
"question": "
A metallic conductor of length 1 m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5 rad s$$-$$1. If the horizontal component of earth's magnetic field is 0.2 $$\\times$$ 10$$-$$4 T, then emf induced between the two ends of the conductor is :
A circular coil of 1000 turns each with area 1m2 is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of 0.07T. The maximum voltage generation will be ___________ V.
Spherical insulating ball and a spherical metallic ball of same size and mass are dropped from the same height. Choose the correct statement out of the following
\n
{Assume negligible air friction}
",
"options": [
{
"text": "Metal ball will reach the earth's surface earlier than the insulating ball"
},
{
"text": "Both will reach the earth's surface simultaneously."
},
{
"text": "Insulating ball will reach the earth's surface earlier than the metal ball"
},
{
"text": "Time taken by them to reach the earth's surface will be independent of the properties of their materials"
}
],
"answer": "Insulating ball will reach the earth's surface earlier than the metal ball",
"solution": "**Answer:** Insulating ball will reach the earth's surface earlier than the metal ball\n\nThe correct answer is option C: the insulating ball will reach the earth's surface earlier than the metal ball.\n\n
When the two balls are dropped from the same height, they will experience the same gravitational force, which will cause them to accelerate downwards. However, According to Faraday’s law of electromagnetic induction motion of metal is opposed by earth magnetic field. Which will slightly reduce its acceleration.\n\n
On the other hand, the insulating ball will not experience any electromagnetic induction. Therefore, it will accelerate downwards at a slightly faster rate than the metal ball. As a result, the insulating ball will reach the earth's surface earlier than the metal ball.\n\n
So, option C is the correct statement.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8977,
"subject": "Physics",
"question": "
A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B = 0.8 T. When released the radius of the loop starts shrinking at a constant rate of 2 cms$$^{-1}$$. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be __________ mV.
A wire of length 1m moving with velocity 8 m/s at right angles to a magnetic field of 2T. The magnitude of induced emf, between the ends of wire will be __________.
\n = 2 × 8 × 1 = 16 V",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8979,
"subject": "Physics",
"question": "A $20 \\mathrm{~cm}$ long metallic rod is rotated with $210~ \\mathrm{rpm}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field $0.2 \\mathrm{~T}$ parallel to the axis exists everywhere. The emf developed between the centre and the ring is ____________ $\\mathrm{mV}$.\n
\nTake $\\pi=\\frac{22}{7}$\n",
"options": [],
"answer": "88",
"solution": "**Answer:** 88\n\nGiven that the rod is rotating at 210 rpm, we first convert this to radians per second:\n
\nSo the emf developed between the center and the ring is 88 mV.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8980,
"subject": "Physics",
"question": "
A conducting circular loop is placed in a uniform magnetic field of $$0.4 \\mathrm{~T}$$ with its plane perpendicular to the field. Somehow, the radius of the loop starts expanding at a constant rate of $$1 \\mathrm{~mm} / \\mathrm{s}$$. The magnitude of induced emf in the loop at an instant when the radius of the loop is $$2 \\mathrm{~cm}$$ will be ___________ $$\\mu \\mathrm{V}$$.
The problem involves a conducting circular loop placed in a uniform magnetic field with its plane perpendicular to the field. The radius of the loop is expanding at a constant rate, and we are asked to find the magnitude of the induced emf in the loop at an instant when the radius of the loop is $$2 \\mathrm{~cm}$$.
\n
The magnetic flux through a circular loop of radius $$r$$ and area $$A = \\pi r^2$$ placed in a uniform magnetic field $$B$$ perpendicular to the plane of the loop is given by:
\n$$\\Phi_B = B A = B \\pi r^2$$
\n
The induced emf in the loop is given by Faraday's law of electromagnetic induction:
\n$$\\mathcal{E} = -\\frac{d\\Phi_B}{dt}$$
\n
In this case, the radius of the loop is expanding at a constant rate of $$10^{-3} \\mathrm{~m/s}$$, which means that the rate of change of the area of the loop is:
Therefore, the magnitude of the induced emf in the loop at an instant when the radius of the loop is $$2 \\mathrm{~cm}$$ is $$50.24 $$ $$ \\simeq $$ 50 $\\mu \\mathrm{V}$.
Given below are two statements: one is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$
\n
Assertion A: A bar magnet dropped through a metallic cylindrical pipe takes more time to come down compared to a non-magnetic bar with same geometry and mass.
\n
Reason R: For the magnetic bar, Eddy currents are produced in the metallic pipe which oppose the motion of the magnetic bar.
\n
In the light of the above statements, choose the correct answer from the options given below
",
"options": [
{
"text": "A is true but R is false"
},
{
"text": "A is false but R is true"
},
{
"text": "Both A and R are true but R is NOT the correct explanation of A"
},
{
"text": "Both A and R are true and R is the correct explanation of A"
}
],
"answer": "Both A and R are true and R is the correct explanation of A",
"solution": "**Answer:** Both A and R are true and R is the correct explanation of A\n\n
Assertion A is true because a bar magnet dropped through a metallic cylindrical pipe takes more time to come down compared to a non-magnetic bar with the same geometry and mass. This is due to the effect of the magnet's magnetic field on the metallic pipe.
\n
Reason R is also true because when the magnetic bar moves through the metallic pipe, it induces a changing magnetic field in the pipe. This changing magnetic field, in turn, induces Eddy currents in the pipe. According to Lenz's law, these Eddy currents produce their own magnetic field, which opposes the motion of the magnetic bar, causing it to fall more slowly through the pipe.
\n
Since both Assertion A and Reason R are true and R provides the correct explanation for A, the correct answer is Both A and R are true and R is the correct explanation of A.
A square loop of side $$2.0 \\mathrm{~cm}$$ is placed inside a long solenoid that has 50 turns per centimetre and carries a sinusoidally varying current of amplitude $$2.5 \\mathrm{~A}$$ and angular frequency $$700 ~\\mathrm{rad} ~\\mathrm{s}^{-1}$$. The central axes of the loop and solenoid coincide. The amplitude of the emf induced in the loop is $$x \\times 10^{-4} \\mathrm{~V}$$. The value of $$x$$ is __________.
In this problem, a square loop is inside a long solenoid, and there's a varying current flowing through the solenoid. Because the current is changing, it induces a changing magnetic field inside the solenoid.
\n
According to Faraday's law of electromagnetic induction, a changing magnetic field will induce an electromotive force (emf) in a loop placed in that field. In this case, the loop is the square loop inside the solenoid.
\n
The formula used here is based on Faraday's law, which states that the induced emf in a loop is equal to the rate of change of magnetic flux through the loop. This is given by:
\n
$ \\text{emf} = -\\frac{d \\Phi}{dt} $
\n
where $\\Phi$ is the magnetic flux.
\n
The magnetic field inside a solenoid is given by $B = \\mu_0 n I$, where $\\mu_0$ is the permeability of free space, $n$ is the number of turns per unit length in the solenoid, and $I$ is the current through the solenoid.
\n
The magnetic flux through the square loop is then given by $\\Phi = B \\cdot A = \\mu_0 n I A$, where $A$ is the area of the loop.
\n
When the current is sinusoidal, i.e., $I(t) = I_0 \\sin(\\omega t)$, its derivative with respect to time is $dI/dt = I_0 \\omega \\cos(\\omega t)$, where $\\omega$ is the angular frequency.
\n
Hence, the rate of change of flux becomes:
\n
$ \\frac{d \\Phi}{dt} = \\mu_0 n A \\frac{dI}{dt} = \\mu_0 n A I_0 \\omega \\cos(\\omega t) $
\n
The emf, which is equal to the negative of the rate of change of flux, will have a maximum value (the amplitude) when $\\cos(\\omega t) = 1$, giving:
\n
$\n\\text{Emf amplitude} = \\mu_0 n A I_0 \\omega$
An emf of $$0.08 \\mathrm{~V}$$ is induced in a metal rod of length $$10 \\mathrm{~cm}$$ held normal to a uniform magnetic field of $$0.4 \\mathrm{~T}$$, when moves with a velocity of:
The emf induced in a rod moving through a magnetic field is given by Faraday's law of electromagnetic induction, specifically, in the form of motional emf, which states that:
\n
$ \\text{emf} = B \\cdot L \\cdot v $
\n
where:
\n
\n
(B) is the magnetic field strength,
\n
(L) is the length of the rod, and
\n
(v) is the velocity of the rod.
\n
\n
In this case, we are given the emf, (B), and (L), and we need to solve for (v). Rearranging the equation gives:
Certain galvanometers have a fixed core made of non magnetic metallic material. The function of this metallic material is
",
"options": [
{
"text": "to bring the coil to rest quickly"
},
{
"text": "to produce large deflecting torque on the coil"
},
{
"text": "to oscillate the coil in magnetic field for longer period of time"
},
{
"text": "to make the magnetic field radial"
}
],
"answer": "to bring the coil to rest quickly",
"solution": "**Answer:** to bring the coil to rest quickly\n\n
The function of the non-magnetic metallic core in a galvanometer is to provide a path for the induced current generated by the moving coil in the magnetic field. This induced current opposes the motion of the coil and brings it to rest quickly due to the effect known as eddy current damping.
\n
When the coil swings past its equilibrium position, a change in magnetic flux occurs. According to Faraday's law of electromagnetic induction, this change in magnetic flux generates an induced current, known as an eddy current. The eddy current creates its own magnetic field which opposes the original change in flux, causing the coil to quickly come to rest.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8985,
"subject": "Physics",
"question": "A coil of 200 turns and area $0.20 \\mathrm{~m}^2$ is rotated at half a revolution per second and is placed in uniform magnetic field of $0.01 \\mathrm{~T}$ perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is $\\frac{2 \\pi}{\\beta}$ volt. The value of $\\beta$ is _______.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n$\\begin{aligned} & \\phi=\\mathrm{NAB} \\cos (\\omega \\mathrm{t}) \\\\\\\\ & \\varepsilon=-\\frac{\\mathrm{d} \\phi}{\\mathrm{dt}}=\\mathrm{NAB} \\omega \\sin (\\omega \\mathrm{t}) \\\\\\\\ & \\varepsilon_{\\max }=\\mathrm{NAB} \\omega \\\\\\\\ & =200 \\times 0.2 \\times 0.01 \\times \\pi \\\\\\\\ & =\\frac{4 \\pi}{10}=\\frac{2 \\pi}{5} \\mathrm{volt}\\end{aligned}$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8986,
"subject": "Physics",
"question": "A rectangular loop of sides $12 \\mathrm{~cm}$ and $5 \\mathrm{~cm}$, with its sides parallel to the $x$-axis and $y$-axis respectively, moves with a velocity of $5 \\mathrm{~cm} / \\mathrm{s}$ in the positive $x$ axis direction, in a space containing a variable magnetic field in the positive $z$ direction. The field has a gradient of $10^{-3} \\mathrm{~T} / \\mathrm{cm}$ along the negative $x$ direction and it is decreasing with time at the rate of $10^{-3} \\mathrm{~T} / \\mathrm{s}$. If the resistance of the loop is $6 \\mathrm{~m} \\Omega$, the power dissipated by the loop as heat is __________ $\\times 10^{-9} \\mathrm{~W}$.",
"options": [],
"answer": "216",
"solution": "**Answer:** 216\n\n\n\n
$\\mathrm{B}_0$ is the magnetic field at origin\n
$$\n\\begin{aligned}\n& \\frac{d B}{d x}=-\\frac{10^{-3}}{10^{-2}} \\\\\\\\\n& \\int_{B_0}^B d B=-\\int_0^x 10^{-1} d x \\\\\\\\\n& B-B_0=-10^{-1} x \\\\\\\\\n& B=\\left(B_0-\\frac{x}{10}\\right)\n\\end{aligned}\n$$\n\n
Motional emf in $\\mathrm{AB}=0$\n\n
Motional emf in $\\mathrm{CD}=0$\n
Motional emf in $\\mathrm{AD}=\\varepsilon_1=\\mathrm{B}_0 \\ell \\mathrm{v}$\n
A horizontal straight wire $$5 \\mathrm{~m}$$ long extending from east to west falling freely at right angle to horizontal component of earths magnetic field $$0.60 \\times 10^{-4} \\mathrm{~Wbm}^{-2}$$. The instantaneous value of emf induced in the wire when its velocity is $$10 \\mathrm{~ms}^{-1}$$ is _________ $$\\times 10^{-3} \\mathrm{~V}$$.
A ceiling fan having 3 blades of length $$80 \\mathrm{~cm}$$ each is rotating with an angular velocity of 1200 $$\\mathrm{rpm}$$. The magnetic field of earth in that region is $$0.5 \\mathrm{G}$$ and angle of dip is $$30^{\\circ}$$. The emf induced across the blades is $$\\mathrm{N} \\pi \\times 10^{-5} \\mathrm{~V}$$. The value of $$\\mathrm{N}$$ is _________.
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8989,
"subject": "Physics",
"question": "Electromagnetic waves are transverse in nature is evident by ",
"options": [
{
"text": "polarization "
},
{
"text": "interference "
},
{
"text": "reflection "
},
{
"text": "diffraction "
}
],
"answer": "polarization ",
"solution": "**Answer:** polarization \n\nThe phenomenon of polarisation is shown only by transverse waves.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8990,
"subject": "Physics",
"question": "An electromagnetic wave of frequency $$v=3.0$$ $$MHz$$ passes from vacuum into a dielectric medium with permittivity $$ \\in = 4.0.$$ Then ",
"options": [
{
"text": "wave length is halved and frequency remains unchanged "
},
{
"text": "wave length is doubled and the frequency becomes half "
},
{
"text": "wave length is doubled and the frequency remains unchanged "
},
{
"text": "wave length and frequency both remain unchanged."
}
],
"answer": "wave length is halved and frequency remains unchanged ",
"solution": "**Answer:** wave length is halved and frequency remains unchanged \n\nFrequency remains constant during refraction\n
$$\\therefore$$ wavelength is halved and frequency remains unchanged",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8991,
"subject": "Physics",
"question": "A radiation of energy $$E$$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is ",
"options": [
{
"text": "$$Ec$$ "
},
{
"text": "$$2E/c$$ "
},
{
"text": "$$E/c$$ "
},
{
"text": "$$E/{c^2}$$ "
}
],
"answer": "$$2E/c$$ ",
"solution": "**Answer:** $$2E/c$$ \n\nMomentum of photon $$ = {E \\over c}$$\n
Change in momentum $$ = {{2E} \\over c}$$\n
$$=$$ momentum transferred to the surface\n
(the photon will reflect with same magnitude of momentum in opposite direction)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8992,
"subject": "Physics",
"question": "The $$rms$$ value of the electric field of the light coming from the Sun is $$720$$ $$N/C.$$ The average total energy density of the electromagnetic wave is ",
"options": [
{
"text": "$$4.58 \\times {10^{ - 6}}\\,J/{m^3}$$ "
},
{
"text": "$$6.37 \\times {10^{ - 9}}\\,J/{m^3}$$ "
},
{
"text": "$$81.35 \\times {10^{ - 12}}\\,J/{m^3}$$ "
},
{
"text": "$$3.3 \\times {10^{ - 3}}\\,J/{m^3}$$ "
}
],
"answer": "$$4.58 \\times {10^{ - 6}}\\,J/{m^3}$$ ",
"solution": "**Answer:** $$4.58 \\times {10^{ - 6}}\\,J/{m^3}$$ \n\n$${E_{rms}} = 720$$\n
$$ = 4.58 \\times {10^{ - 6}}\\,J/{m^3}$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8993,
"subject": "Physics",
"question": "An electromagnetic wave in vacuum has the electric and magnetic field $$\\mathop E\\limits^ \\to $$and $$\\mathop B\\limits^ \\to $$, which are always perpendicular to each other. The direction of polarization is given by $$\\mathop X\\limits^ \\to $$ and that of wave propagation by $$\\mathop k\\limits^ \\to $$. Then ",
"options": [
{
"text": "$$\\mathop X\\limits^ \\to ||\\mathop B\\limits^ \\to $$ and $$\\mathop X\\limits^ \\to ||\\mathop B\\limits^ \\to \\times \\mathop E\\limits^ \\to $$ "
},
{
"text": "$$\\mathop X\\limits^ \\to ||\\mathop E\\limits^ \\to $$ and $$\\mathop k\\limits^ \\to ||\\mathop E\\limits^ \\to \\times \\mathop B\\limits^ \\to $$ "
},
{
"text": "$$\\mathop X\\limits^ \\to ||\\mathop B\\limits^ \\to $$ and $$\\mathop k\\limits^ \\to ||\\mathop E\\limits^ \\to \\times \\mathop B\\limits^ \\to $$ "
},
{
"text": "$$\\mathop X\\limits^ \\to ||\\mathop E\\limits^ \\to $$ and $$\\mathop k\\limits^ \\to ||\\mathop B\\limits^ \\to \\times \\mathop E\\limits^ \\to $$ "
}
],
"answer": "$$\\mathop X\\limits^ \\to ||\\mathop E\\limits^ \\to $$ and $$\\mathop k\\limits^ \\to ||\\mathop E\\limits^ \\to \\times \\mathop B\\limits^ \\to $$ ",
"solution": "**Answer:** $$\\mathop X\\limits^ \\to ||\\mathop E\\limits^ \\to $$ and $$\\mathop k\\limits^ \\to ||\\mathop E\\limits^ \\to \\times \\mathop B\\limits^ \\to $$ \n\nas The $$E.M.$$ wave are transverse in nature i.e.,\n
$$ = {{\\overrightarrow k \\times \\overrightarrow E } \\over {\\mu \\omega }} = \\overrightarrow H \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( i \\right)$$\n
where $$\\overrightarrow H = {{\\overrightarrow B } \\over \\mu }$$\n
and $${{\\overrightarrow k \\times \\overrightarrow H } \\over {\\omega \\varepsilon }} = - \\overrightarrow E \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( {ii} \\right)$$\n
$$\\overrightarrow k $$ is $$ \\bot \\,\\,\\overrightarrow H $$ and $$\\overrightarrow k $$ is also $$ \\bot $$ to $$\\overrightarrow E $$\n
or In other words $$\\overrightarrow X ||\\overrightarrow E $$ and $$\\overrightarrow k ||\\overrightarrow E \\times \\overrightarrow B $$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8994,
"subject": "Physics",
"question": "The magnetic field in a travelling electromagnetic wave has a peak value of $$20$$ $$n$$$$T$$. The peak value of electric field strength is : ",
"options": [
{
"text": "$$3V/m$$ "
},
{
"text": "$$6V/m$$ "
},
{
"text": "$$9V/m$$ "
},
{
"text": "$$12V/m$$ "
}
],
"answer": "$$6V/m$$ ",
"solution": "**Answer:** $$6V/m$$ \n\nFrom question, \n
$${B_0} = 20nT = 20 \\times {10^{ - 9}}T$$\n
( as velocity of light in vacuum $$C = 3 \\times {10^8}\\,\\,m{s^{ - 1}}$$ )\n
$${\\overrightarrow E _0} = {\\overrightarrow B _0} \\times \\overrightarrow C $$\n
$$\\left| {{{\\overrightarrow E }_0}} \\right| = \\left| {\\overrightarrow B } \\right|.\\left| {\\overrightarrow C } \\right| = 20 \\times {10^{ - 9}} \\times 3 \\times {10^8}$$\n
$$ = 6\\,\\,V/m.$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8995,
"subject": "Physics",
"question": "During the propagation of electromagnetic waves in a medium : ",
"options": [
{
"text": "Electric energy density is double of the magnetic energy density. "
},
{
"text": "Electric energy density is half of the magnetic energy density. "
},
{
"text": "Electric energy density is equal to the magnetic energy density."
},
{
"text": "Both electric and magnetic energy of densities are zero. "
}
],
"answer": "Electric energy density is equal to the magnetic energy density.",
"solution": "**Answer:** Electric energy density is equal to the magnetic energy density.\n\n$${E_0} = C{B_0}$$ and $$C = {1 \\over {\\sqrt {{\\mu _0}{\\varepsilon _0}} }}$$\n
Electric energy density $$ = {1 \\over 2}{\\varepsilon _0}{E_0}^2 = {\\mu _E}$$\n
Magnetic energy density $$ = {1 \\over 2}{{{B_0}^2} \\over {{\\mu _0}}} = {\\mu _B}$$\n
Thus, $${\\mu _E} = {\\mu _B}$$\n
Energy is equally divided between electric and magnetic field",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8996,
"subject": "Physics",
"question": "A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100 W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 5 m from the lamp will be nearly : ",
"options": [
{
"text": "1.34 V/m"
},
{
"text": "2.68 V/m"
},
{
"text": "4.02 V/m"
},
{
"text": "5.36 V/m"
}
],
"answer": "2.68 V/m",
"solution": "**Answer:** 2.68 V/m\n\nSince $\\quad U_E \\times c=I=$ Intensity\n
$$\n\\begin{aligned}\n& \\frac{1}{2} \\varepsilon_0 E^2 \\times c=I=\\frac{P}{4 \\pi R^2}=\\frac{3 / 100 \\times P}{4 \\pi R^2} \\\\\\\\\nE & =\\sqrt{\\frac{6 P}{\\varepsilon_0 \\times c \\times 100 \\times 4 \\pi R^2}} \\\\\\\\\n& =\\sqrt{\\frac{6 \\times 100}{8.85 \\times 10^{-12} \\times 3 \\times 10^8 \\times 100 \\times 4 \\pi \\times(5)^2}} \\\\\\\\\n& =2.68 \\mathrm{~V} / \\mathrm{m}\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8997,
"subject": "Physics",
"question": "An electromagnetic wave of frequency 1 $$ \\times $$ 1014 hertz is propagating along z - axis. The amplitude of electric field is 4 V/m. If $$ \\in $$0 = 8.8 $$ \\times $$ 10$$-$$12 C2/N-m2, then average energy density of electric field will be : ",
"options": [
{
"text": "35.2 $$ \\times $$ 10$$-$$10 J/m3"
},
{
"text": "35.2 $$ \\times $$ 10$$-$$11 J/m3"
},
{
"text": "35.2 $$ \\times $$ 10$$-$$12 J/m3"
},
{
"text": "35.2 $$ \\times $$ 10$$-$$13 J/m3"
}
],
"answer": "35.2 $$ \\times $$ 10$$-$$12 J/m3",
"solution": "**Answer:** 35.2 $$ \\times $$ 10$$-$$12 J/m3\n\nWe have\n
$$\nE=E_0 \\sin (\\omega t-k x)\n$$\n
Since, energy density is $\\mu_{\\mathrm{E}}=\\frac{1}{2} \\varepsilon_0 E^2$\n
$$\n\\begin{aligned}\n\\Rightarrow \\bar{\\mu}_{\\mathrm{E}} & =\\frac{1}{2} \\varepsilon_0 \\bar{E}^2=\\frac{1}{2} \\varepsilon_0 E_{\\mathrm{rms}}^2 \\\\\\\\\n& =\\frac{1}{2} \\varepsilon_0 \\frac{E_0^2}{2}=\\frac{1}{4} \\varepsilon_0 E_0^2 \\\\\\\\\n& =\\frac{1}{4} \\times 8.8 \\times 10^{-12} \\times 4^2=35.2 \\times 10^{-12} \\mathrm{~J} / \\mathrm{m}^3\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 8998,
"subject": "Physics",
"question": "For plane electromagnetic waves propagating in the z direction, which one of the following combination gives the correct possible direction for $$\\overrightarrow E $$ and $$\\overrightarrow B $$ field respectively ? ",
"options": [
{
"text": "$$\\left( {\\widehat i + 2\\widehat j} \\right)\\,\\,$$ and $$\\left( {2\\widehat i - \\widehat j} \\right)$$"
},
{
"text": "$$\\left(-\\, {2\\widehat i - 3\\widehat j} \\right)$$ and $$\\left( {3\\widehat i - 2\\widehat j} \\right)$$"
},
{
"text": "$$\\left( {2\\widehat i + 3\\widehat j} \\right)$$ and $$\\left( {\\widehat i + 2\\widehat j} \\right)$$"
},
{
"text": "$$\\left( {3\\widehat i + 4\\widehat j} \\right)$$ and $$\\left( {4\\widehat i - 3\\widehat j} \\right)$$"
}
],
"answer": "$$\\left(-\\, {2\\widehat i - 3\\widehat j} \\right)$$ and $$\\left( {3\\widehat i - 2\\widehat j} \\right)$$",
"solution": "**Answer:** $$\\left(-\\, {2\\widehat i - 3\\widehat j} \\right)$$ and $$\\left( {3\\widehat i - 2\\widehat j} \\right)$$\n\nSince $\\vec{E}$ and $\\vec{B}$ are mutually perpendicular. Therefore,\n
$$\n\\begin{gathered}\n\\vec{E} \\times \\vec{B}=\\vec{c}=c \\hat{k} \\\\\\\\\n\\Rightarrow(-2 \\hat{i}-3 \\hat{j}) \\cdot(3 \\hat{i}-2 \\hat{j})=-6+6=0 \\\\\\\\\n\\Rightarrow(-2 \\hat{i}-3 \\hat{j}) \\times(3 \\hat{i}-2 \\hat{j})=(6+9) \\hat{k}=15 \\hat{k}\n\\end{gathered}\n$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 8999,
"subject": "Physics",
"question": "An electromagnetic wave travelling in the x-direction has frequency of 23 $$ \\times $$ 1014 Hz and electric field amplitude of 27 Vm$$-$$1. From the options given below, which one describes the magnetic field for this wave ?",
"options": [
{
"text": "$$\\overrightarrow B $$ (x, t) = (3 $$ \\times $$ 10$$-$$8T) $$\\widehat j$$\n
$$\n\\begin{aligned}\nB & =B_o \\sin (K x-\\omega t)=B_o \\sin 2 \\pi\\left(\\frac{x}{\\lambda}-f t\\right) \\\\\\\\\n& =9 \\times 10^{-8} \\sin 2 \\pi\\left(\\frac{x}{1.5 \\times 10^6}-2 \\times 10^{14} t\\right)\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9000,
"subject": "Physics",
"question": "Consider an electromagnetic wave propagating in vacuum. Choose the correct\nstatement :",
"options": [
{
"text": "For an electromagnetic wave propagating in +x direction the electric field is $$\\vec E = {1 \\over {\\sqrt 2 }}{E_{yz}}{\\mkern 1mu} \\left( {x,t} \\right)\\left( {\\hat y - \\hat z} \\right)$$ \n
and the magnetic field is $$\\vec B = {1 \\over {\\sqrt 2 }}{B_{yz}}{\\mkern 1mu} \\left( {x,t} \\right)\\left( {\\hat y + \\hat z} \\right)$$"
},
{
"text": "For an electromagnetic wave propagating in +x direction the electric field is\n$$\\vec E = {1 \\over {\\sqrt 2 }}{E_{yz{\\mkern 1mu} }}\\left( {y,z,t} \\right)\\left( {\\hat y + \\hat z} \\right)$$ \n
and the magnetic field is $$\\vec B = {1 \\over {\\sqrt 2 }}{B_{yz{\\mkern 1mu} }}\\left( {y,z,t} \\right)\\left( {\\hat y + \\hat z} \\right)$$"
},
{
"text": "For an electromagnetic wave propagating in + y direction the electric field is\n$$\\overrightarrow E = {1 \\over {\\sqrt 2 }}{E_{yz{\\mkern 1mu} }}\\left( {x,t} \\right)\\widehat y$$ \n and the magnetic field is $$\\vec B = {1 \\over {\\sqrt 2 }}{B_{yz{\\mkern 1mu} }}\\left( {x,t} \\right)\\widehat z$$"
},
{
"text": "For an electromagnetic wave propagating in + y direction the electric field is \n$$\\overrightarrow E = {1 \\over {\\sqrt 2 }}{E_{yz{\\mkern 1mu} }}\\left( {x,t} \\right)\\widehat z$$ \n and the magnetic field is $$\\overrightarrow B = {1 \\over {\\sqrt 2 }}{B_{z{\\mkern 1mu} }}\\left( {x,t} \\right)\\widehat y$$"
}
],
"answer": "For an electromagnetic wave propagating in +x direction the electric field is $$\\vec E = {1 \\over {\\sqrt 2 }}{E_{yz}}{\\mkern 1mu} \\left( {x,t} \\right)\\left( {\\hat y - \\hat z} \\right)$$ \n
and the magnetic field is $$\\vec B = {1 \\over {\\sqrt 2 }}{B_{yz}}{\\mkern 1mu} \\left( {x,t} \\right)\\left( {\\hat y + \\hat z} \\right)$$",
"solution": "**Answer:** For an electromagnetic wave propagating in +x direction the electric field is $$\\vec E = {1 \\over {\\sqrt 2 }}{E_{yz}}{\\mkern 1mu} \\left( {x,t} \\right)\\left( {\\hat y - \\hat z} \\right)$$ \n
and the magnetic field is $$\\vec B = {1 \\over {\\sqrt 2 }}{B_{yz}}{\\mkern 1mu} \\left( {x,t} \\right)\\left( {\\hat y + \\hat z} \\right)$$\n\nAs wave is propagating in + x direction, then $$\\overrightarrow E $$ and $$\\overrightarrow B $$ should be function of $$\\left( {x,t} \\right)$$ and must be in y $$-$$ z plane.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9001,
"subject": "Physics",
"question": "The electric field component of a monochromatic radiation is given by\n
$$\\overrightarrow E $$ = 2 E0 $$\\widehat i$$ cos kz cos $$\\omega $$t\n
Its magnetic field $$\\overrightarrow B $$ is then given by : ",
"options": [
{
"text": "$${{2{E_0}} \\over c}$$ $$\\widehat j$$ sin kz cos $$\\omega $$t"
},
{
"text": "$$-$$ $${{2{E_0}} \\over c}$$ $$\\widehat j$$ sin kz sin $$\\omega $$t"
},
{
"text": "$${{2{E_0}} \\over c}$$ $$\\widehat j$$ sin kz sin $$\\omega $$t"
},
{
"text": "$${{2{E_0}} \\over c}$$ $$\\widehat j$$ cos kz cos $$\\omega $$t"
}
],
"answer": "$${{2{E_0}} \\over c}$$ $$\\widehat j$$ sin kz sin $$\\omega $$t",
"solution": "**Answer:** $${{2{E_0}} \\over c}$$ $$\\widehat j$$ sin kz sin $$\\omega $$t\n\n
Therefore, its magnetic field $$\\overrightarrow B $$ is given by $$B = {{2{E_0}} \\over c}\\widehat j\\sin kz\\sin \\omega t$$
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9002,
"subject": "Physics",
"question": "Magnetic field in a plane electromagnetic wave is given by\n
$$\\overrightarrow B $$ = B0 sin (k x + $$\\omega $$t) $$\\widehat j\\,T$$ \n
Expression for corresponding electric field will be : \n Where c is speed of light. ",
"options": [
{
"text": "$$\\overrightarrow E $$ = B0 c sin (k x + $$\\omega $$t) $$\\widehat k$$ V/m"
},
{
"text": "$$\\overrightarrow E $$ = $${{{B_0}} \\over c}$$ sin (k x + $$\\omega $$t) $$\\widehat k$$ V/m"
},
{
"text": "$$\\overrightarrow E $$ = $$-$$ B0 c sin (kx +$$\\omega $$t) $$\\widehat k$$ V/m"
},
{
"text": "$$\\overrightarrow E $$ = B0 c sin (kx $$-$$$$\\omega $$t) $$\\widehat k$$ V/m"
}
],
"answer": "$$\\overrightarrow E $$ = B0 c sin (k x + $$\\omega $$t) $$\\widehat k$$ V/m",
"solution": "**Answer:** $$\\overrightarrow E $$ = B0 c sin (k x + $$\\omega $$t) $$\\widehat k$$ V/m\n\nThe relation between electric and magnetic field is , \n
C = $${{\\overrightarrow E } \\over {\\overrightarrow B }}$$\n
$$ \\Rightarrow $$$$\\,\\,\\,$$ $$\\overrightarrow E $$ = C $$\\overrightarrow B $$\n
Electric field component is perpendicular to the direction of magnetic field. Given magnetic field is along y $$-$$ axis, \n
So, electric field along z $$-$$ axis will be \n
$$\\overrightarrow E $$ = B0 C sin (kx + $$\\omega $$t) $$\\,\\widehat k$$ v/m",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9003,
"subject": "Physics",
"question": "A plane polarized monochromatic EM wave is traveling in vacuum along z direction such that at t = t1 it is found that the electric field is zero at a spatial point z1. The next zero that occurs in its neighbourhood is at z2. The frequency of the electroagnetic wave is :",
"options": [
{
"text": "$${{3 \\times {{10}^8}} \\over {\\left| {{z_2} - {z_1}} \\right|}}$$"
},
{
"text": "$${{1.5 \\times {{10}^8}} \\over {\\left| {{z_2} - {z_1}} \\right|}}$$"
},
{
"text": "$${{6 \\times {{10}^8}} \\over {\\left| {{z_2} - {z_1}} \\right|}}$$"
},
{
"text": "$${1 \\over {{t_1} + {{\\left| {{z_2} - {z_1}} \\right|} \\over {3 \\times {{10}^8}}}}}$$"
}
],
"answer": "$${{1.5 \\times {{10}^8}} \\over {\\left| {{z_2} - {z_1}} \\right|}}$$",
"solution": "**Answer:** $${{1.5 \\times {{10}^8}} \\over {\\left| {{z_2} - {z_1}} \\right|}}$$\n\n
Since $$c = f\\lambda \\Rightarrow f = {c \\over \\lambda }$$. Here, $$\\lambda = 2|{z_2} - {z_1}|$$ and $$c = 3 \\times {10^8}$$. Therefore,
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9004,
"subject": "Physics",
"question": "An EM wave from air enters a medium. The electric fields are
\n$$\\overrightarrow {{E_1}} $$ = $${E_{01}}\\widehat x\\cos \\left[ {2\\pi v\\left( {{z \\over c} - t} \\right)} \\right]$$ in air and \n
where the wave number k and frequency $$\\nu $$ refer to their values\nin air. The medium is non-magnetic. If $${\\varepsilon _{{r_1}}}$$ and $${\\varepsilon _{{r_2}}}$$ refer to relative permittivities of air and medium\nrespectively, which of the following options is correct ? \n",
"options": [
{
"text": "$${{{\\varepsilon _{{r_1}}}} \\over {{\\varepsilon _{{r_2}}}}} = 4$$ "
},
{
"text": "$${{{\\varepsilon _{{r_1}}}} \\over {{\\varepsilon _{{r_2}}}}} = 2$$"
},
{
"text": "$${{{\\varepsilon _{{r_1}}}} \\over {{\\varepsilon _{{r_2}}}}} = {1 \\over 4}$$"
},
{
"text": "$${{{\\varepsilon _{{r_1}}}} \\over {{\\varepsilon _{{r_2}}}}} = {1 \\over 2}$$"
}
],
"answer": "$${{{\\varepsilon _{{r_1}}}} \\over {{\\varepsilon _{{r_2}}}}} = {1 \\over 4}$$",
"solution": "**Answer:** $${{{\\varepsilon _{{r_1}}}} \\over {{\\varepsilon _{{r_2}}}}} = {1 \\over 4}$$\n\nElectric field in air, \n
and k = $${\\omega \\over c}$$ = $${{1.5 \\times {{10}^{11}}} \\over {3 \\times {{10}^8}}}$$ = 0.5 × 103\n
Note : When wave is propagating in positive z-direction then sign of kz and $${\\omega t}$$ should be opposite.\n From option you can see only option D can be correct.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9007,
"subject": "Physics",
"question": "An electromagnetic wave is represented by the electric field $$\\overrightarrow E = {E_0}\\widehat n\\sin \\left[ {\\omega t + \\left( {6y - 8z} \\right)} \\right]$$\n. Taking unit\nvectors in x, y and z directions to be $$\\widehat i,\\widehat j,\\widehat k$$\n, the direction of propagation $$\\widehat s$$, is :",
"options": [
{
"text": "$$\\widehat s = {{3\\widehat i - 4\\widehat j} \\over 5}$$"
},
{
"text": "$$\\widehat s = {{ - 4\\widehat k + 3\\widehat j} \\over 5}$$"
},
{
"text": "$$\\widehat s = \\left( {{{ - 3\\widehat j + 4\\widehat k} \\over 5}} \\right)$$"
},
{
"text": "$$\\widehat s = {{4\\widehat j - 3\\widehat k} \\over 5}$$"
}
],
"answer": "$$\\widehat s = \\left( {{{ - 3\\widehat j + 4\\widehat k} \\over 5}} \\right)$$",
"solution": "**Answer:** $$\\widehat s = \\left( {{{ - 3\\widehat j + 4\\widehat k} \\over 5}} \\right)$$\n\n$$\\overrightarrow E = {E_0}\\widehat n\\sin \\left( {\\omega t + \\left( {6y - 8z} \\right)} \\right)$$
\n$$ = {E_0}\\widehat n\\sin \\left( {\\omega t + \\overrightarrow k .\\overrightarrow r } \\right)$$
\nwhere $$\\overrightarrow r = x\\widehat i + y\\widehat j + z\\widehat k$$ and $$\\overrightarrow k .\\overrightarrow r = 6y - 8z$$
\n$$ \\Rightarrow \\overrightarrow k = 6\\widehat j - 8\\widehat k$$
\ndirection of propagation
$$\\widehat s = - \\widehat k = \\left( {{{ - 3\\widehat j + 4\\widehat k} \\over 5}} \\right)$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 9008,
"subject": "Physics",
"question": "The electric field of a plane electromagnetic\nwave is given by \n$$\\overrightarrow E = {E_0}\\widehat i\\cos (kz)cos(\\omega t)$$ The corresponding magnetic field $$\\overrightarrow B $$ is then given by",
"options": [
{
"text": "$$\\overrightarrow B = {{{E_0}} \\over C}\\widehat j\\sin (kz)\\sin (\\omega t)$$"
},
{
"text": "$$\\overrightarrow B = {{{E_0}} \\over C}\\widehat j\\sin (kz)\\cos (\\omega t)$$"
},
{
"text": "$$\\overrightarrow B = {{{E_0}} \\over C}\\widehat j\\cos (kz)\\sin (\\omega t)$$"
},
{
"text": "$$\\overrightarrow B = {{{E_0}} \\over C}\\widehat k\\sin (kz)\\cos (\\omega t)$$"
}
],
"answer": "$$\\overrightarrow B = {{{E_0}} \\over C}\\widehat j\\sin (kz)\\sin (\\omega t)$$",
"solution": "**Answer:** $$\\overrightarrow B = {{{E_0}} \\over C}\\widehat j\\sin (kz)\\sin (\\omega t)$$\n\n$$ \\therefore \\overrightarrow E \\times \\overrightarrow B \\parallel \\overrightarrow v $$
\nGiven that wave is propagating along positive z-axis and $$\\overrightarrow E $$ along positive x-axis. Hence $$\\overrightarrow B $$ along y-axis.
\nFrom Maxwell equation\n \n$$\\overrightarrow V \\times \\overrightarrow E = - {{\\partial B} \\over {\\partial t}}$$
\ni.e. $${{\\partial E} \\over {\\partial Z}} = - {{\\partial B} \\over {dt}}$$ and $${B_0} = {{{E_0}} \\over C}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9009,
"subject": "Physics",
"question": "50 W/m2 energy density of sunlight is normally\nincident on the surface of a solar panel. Some\npart of incident energy (25%) is reflected from\nthe surface and the rest is absorbed. The force\nexerted on 1m2 surface area will be close to\n(c = 3 × 108 m/s) :-",
"options": [
{
"text": "20 × 10–8 N"
},
{
"text": "35 × 10–8 N"
},
{
"text": "10 × 10–8 N"
},
{
"text": "15 × 10–8 N"
}
],
"answer": "20 × 10–8 N",
"solution": "**Answer:** 20 × 10–8 N\n\nRadiation pressure for 100% reflection = $${{2I} \\over C}$$
\nRadiation pressure for 0% reflection = $${I \\over C}$$
\n$$ \\therefore $$ Force = P × (Area) = 20.83 × 10–8 N",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9010,
"subject": "Physics",
"question": "The magnetic field of a plane electromagnetic\nwave is given by : \n$$$\\overline B = {B_0}\\widehat i\\left[ {\\cos (kz - \\omega t)} \\right] + {B_i}\\widehat j\\cos (kz + \\omega t)$$$\nB0 = 3 × 10–5 T and B1 = 2 × 10–6 T.\n The rms\nvalue of the force experienced by a stationary\ncharge Q = 10–4 C at z = 0 is closest to :",
"options": [
{
"text": "0.6 N"
},
{
"text": "0.9 N"
},
{
"text": "3 × 10–2 N"
},
{
"text": "0.1 N"
}
],
"answer": "0.6 N",
"solution": "**Answer:** 0.6 N\n\nMaximum electric field E = (B) (C)
\n$${F_{rms}} = {{{F_0}} \\over {\\sqrt 2 }} = 0.6\\,N$$ (approx)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9011,
"subject": "Physics",
"question": "The magnetic field of an electromagnetic wave\nis given by :-
\n$$\\overrightarrow E $$ and $$\\overrightarrow B $$ are perpendicular to each other
\n$$ \\Rightarrow \\overrightarrow E .\\overrightarrow B = 0$$
\n$$ \\Rightarrow $$ Either direction of $$\\overrightarrow E $$ $$\\widehat i - 2\\widehat j$$ or $$-\n \\widehat i + 2\\widehat j$$ from given option
Also wave propagation direction is parallel to $$\\overrightarrow E \\times \\overrightarrow B $$ which is $$- \\widehat k$$
\n$$ \\Rightarrow $$ $$\\overrightarrow E $$ is along ($$-\n \\widehat i + 2\\widehat j$$)",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 9012,
"subject": "Physics",
"question": "A plane electromagnetic wave travels in free\nspace along the x-direction. The electric field\ncomponent of the wave at a particular point of\nspace and time is E = 6 V m–1 along y-direction.\nIts corresponding magnetic field component,\nB would be :",
"options": [
{
"text": "2 × 10–8 T along y-direction"
},
{
"text": "6 × 10–8 T along z-direction"
},
{
"text": "2 × 10–8 T along z-direction"
},
{
"text": "6 × 10–8 T along x-direction"
}
],
"answer": "2 × 10–8 T along z-direction",
"solution": "**Answer:** 2 × 10–8 T along z-direction\n\nThe direction of propagation of an EM wave is direction of $$\\overrightarrow E \\times \\overrightarrow B $$
\n$$\\widehat i = \\widehat j \\times \\widehat B$$
\nB = 2 × 10–8 T along z direction.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9013,
"subject": "Physics",
"question": "A light wave is incident normally on a glass slab of refractive index 1.5. If 4 % of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be : ",
"options": [
{
"text": "6 V/m"
},
{
"text": "10 V/m"
},
{
"text": "30 V/m"
},
{
"text": "24 V/m"
}
],
"answer": "24 V/m",
"solution": "**Answer:** 24 V/m\n\nPrefracted = $${{96} \\over {100}}Pi$$\n
A1$$\\sqrt {{{64} \\over {100}} \\times {{\\left( {30} \\right)}^2}} = 24$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9014,
"subject": "Physics",
"question": "A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by :\n [Given permittivity of space $$ \\in $$0 = 9 $$ \\times $$ 10–12 SI units, Speed of light c = 3 $$ \\times $$ 108 m/s]",
"options": [
{
"text": "2 kV/m"
},
{
"text": "1 kV/m"
},
{
"text": "1.4 kV/m"
},
{
"text": "0.7 kV/m"
}
],
"answer": "1.4 kV/m",
"solution": "**Answer:** 1.4 kV/m\n\nIntensity of EM wave is given by\n
$$ = 1.4$$ kv/m",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9015,
"subject": "Physics",
"question": "An electromagnetic wave of intensity 50 Wm–2 enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by: ",
"options": [
{
"text": "$$\\left( {{1 \\over {\\sqrt n }},{1 \\over {\\sqrt n }}} \\right)$$"
},
{
"text": "$$\\left( {\\sqrt n ,\\sqrt n } \\right)$$"
},
{
"text": "$$\\left( {\\sqrt n ,{1 \\over {\\sqrt n }}} \\right)$$"
},
{
"text": "$$\\left( {{1 \\over {\\sqrt n }},\\sqrt n } \\right)$$"
}
],
"answer": "$$\\left( {\\sqrt n ,{1 \\over {\\sqrt n }}} \\right)$$",
"solution": "**Answer:** $$\\left( {\\sqrt n ,{1 \\over {\\sqrt n }}} \\right)$$\n\nC $$ = {1 \\over {\\sqrt {{\\mu _0}{ \\in _0}} }}$$\n
$$ \\therefore $$ correct answer is 3 $$ \\times $$ 104 N/C",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9018,
"subject": "Physics",
"question": "The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then : ",
"options": [
{
"text": "$${U_E} = {{{U_B}} \\over 2}$$"
},
{
"text": "$${U_E} > {U_B}$$"
},
{
"text": "$${U_E} < {U_B}$$"
},
{
"text": "$${U_E} = {U_B}$$"
}
],
"answer": "$${U_E} = {U_B}$$",
"solution": "**Answer:** $${U_E} = {U_B}$$\n\nEnergy density of magnetic field (UB) = $${{{B^2}} \\over {2{\\mu _0}}}$$\n
$$ \\therefore $$ $$\\left| {\\overrightarrow B } \\right| = {{\\left| {\\overrightarrow E } \\right|} \\over C}$$\n
$$ = {{6.3} \\over {3 \\times {{10}^8}}}$$\n
$$ = 2.1 \\times {10^{ - 8}}\\,$$ T\n
Now we have to find the direction of $$\\overrightarrow B $$.\n
We know, $$\\widehat E \\times \\widehat B = \\widehat C$$ and given $$\\overrightarrow E $$ is in y-direction and wave moving in positive x-direction.\n
$$ \\Rightarrow $$ $$\\widehat B = \\widehat K$$\n
$$ \\therefore $$ $$\\overrightarrow B = \\left| {\\overrightarrow B } \\right|\\widehat B$$\n
$$ = 2.1 \\times {10^{ - 8}}\\,\\widehat K\\,T$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9020,
"subject": "Physics",
"question": "The mean intensity of radiation on the surface of the Sun is about 108 W/m2 . The rms value of the corresponding magnetic field is closet to :",
"options": [
{
"text": "102 T"
},
{
"text": "10$$-$$2 T"
},
{
"text": "10$$-$$4 T"
},
{
"text": "1 T"
}
],
"answer": "10$$-$$4 T",
"solution": "**Answer:** 10$$-$$4 T\n\nI = $${\\varepsilon _0}\\,C\\,E_{rms}^2$$\n
Brms $$ \\approx $$ 10$$-$$4",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9021,
"subject": "Physics",
"question": "The electric field of a plane electromagnetic wave is given by\n $$\\overrightarrow E = {E_0}\\left( {\\widehat x + \\widehat y} \\right)\\sin \\left( {kz - \\omega t} \\right)$$\n Its magnetic field will be given by :",
"options": [
{
"text": "$${{{E_0}} \\over c}\\left( {\\widehat x + \\widehat y} \\right)\\sin \\left( {kz - \\omega t} \\right)$$"
},
{
"text": "$${{{E_0}} \\over c}\\left( {\\widehat x - \\widehat y} \\right)\\sin \\left( {kz - \\omega t} \\right)$$"
},
{
"text": "$${{{E_0}} \\over c}\\left( {\\widehat x - \\widehat y} \\right)\\cos \\left( {kz - \\omega t} \\right)$$"
},
{
"text": "$${{{E_0}} \\over c}\\left( { - \\widehat x + \\widehat y} \\right)\\sin \\left( {kz - \\omega t} \\right)$$"
}
],
"answer": "$${{{E_0}} \\over c}\\left( { - \\widehat x + \\widehat y} \\right)\\sin \\left( {kz - \\omega t} \\right)$$",
"solution": "**Answer:** $${{{E_0}} \\over c}\\left( { - \\widehat x + \\widehat y} \\right)\\sin \\left( {kz - \\omega t} \\right)$$\n\nGiven, $$\\overrightarrow E = {E_0}\\left( {\\widehat x + \\widehat y} \\right)\\sin \\left( {kz - \\omega t} \\right)$$\n
We know, direction of propagation, $$\\overrightarrow C = \\overrightarrow E \\times \\overrightarrow B $$\n
Here direction of propagation = $$\\widehat k$$\n
$$ \\therefore $$ $$\\widehat k$$ = $$\\overrightarrow E \\times \\overrightarrow B $$\n
and $$\\widehat E = {{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}$$\n
$$ \\therefore $$ $$\\widehat k = \\left( {{{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}} \\right) \\times \\overrightarrow B $$\n
$$ \\Rightarrow $$ $$\\widehat B = {{ - \\widehat i + \\widehat j} \\over {\\sqrt 2 }}$$\n
$$ \\therefore $$ $$\\widehat B$$ = $${{{E_0}} \\over c}\\left( { - \\widehat x + \\widehat y} \\right)\\sin \\left( {kz - \\omega t} \\right)$$\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9022,
"subject": "Physics",
"question": "For a plane electromagnetic wave, the magnetic field at a point x and time t is\n
= 36 sin (0.5$$ \\times $$103x+1.5$$ \\times $$1011t) V/m\n
Also, $$\\overrightarrow s = {{\\overrightarrow E \\times \\overrightarrow B } \\over {{\\mu _0}}}$$\n
$$ \\Rightarrow $$ $$ - \\widehat i = {{\\overrightarrow E \\times \\widehat k} \\over {{\\mu _0}}}$$\n
$$ \\therefore $$ Direction of $${\\overrightarrow E = - \\widehat j}$$\n
$$ \\therefore $$ Instantaneous electric field,\n
$$\\overrightarrow E \\left( {x,t} \\right) = \\left[ { - 36\\sin \\left( {0.5 \\times {{10}^3}x + 1.5 \\times {{10}^{11}}t} \\right)\\widehat j} \\right]{V \\over m}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9023,
"subject": "Physics",
"question": "Suppose that intensity of a laser is $${{315} \\over \\pi }$$ W/m2. The rms electric field, in units of V/m associated\nwith this source is close to the nearest integer is __________.\n
= $${E_0}q\\left( {0.8\\widehat i + \\widehat j + 0.2\\widehat k} \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9030,
"subject": "Physics",
"question": "A plane electromagnetic wave of frequency\n25 GHz is propagating in vacuum along the\nz-direction. At a particular point in space and\ntime, the magnetic field is given by $$\\overrightarrow B = 5 \\times {10^{ - 8}}\\widehat jT$$. The corresponding electric field $$\\overrightarrow E $$ is (speed of light c = 3 × 108 ms–1)",
"options": [
{
"text": "15 $$\\widehat i$$V / m"
},
{
"text": "-15 $$\\widehat i$$V / m"
},
{
"text": "1.66 × 10–16 $$\\widehat i$$V / m"
},
{
"text": "-1.66 × 10–16 $$\\widehat i$$V / m"
}
],
"answer": "15 $$\\widehat i$$V / m",
"solution": "**Answer:** 15 $$\\widehat i$$V / m\n\n$$\\overrightarrow E = \\overrightarrow B \\times \\overrightarrow V $$\n
= $${15\\,\\widehat i}$$ V/m",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9031,
"subject": "Physics",
"question": "The electric field of a plane electromagnetic wave is given by\n $$\\overrightarrow E = {E_0}{{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}\\cos \\left( {kz + \\omega t} \\right)$$\n
At t = 0, a positively charged particle is at the point (x, y, z) = $$\\left( {0,0,{\\pi \\over k}} \\right)$$.\n If its instantaneous velocity at\n(t = 0) is $${v_0}\\widehat k$$\n, the force acting on it due to the wave is :",
"options": [
{
"text": "parallel to $$\\widehat k$$"
},
{
"text": "parallel to $${{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}$$"
},
{
"text": "antiparallel to $${{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}$$"
},
{
"text": "zero"
}
],
"answer": "antiparallel to $${{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}$$",
"solution": "**Answer:** antiparallel to $${{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}$$\n\n$$\\overrightarrow E = {E_0}{{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}\\cos \\left( {kz + \\omega t} \\right)$$\n
$$\\overrightarrow E $$ at t = 0 at z = $${\\pi \\over k}$$ is given by\n
As $$\\overrightarrow E \\times \\overrightarrow B = \\overrightarrow c $$\n
Force due to magnetic field is in direction $$q\\left( {\\overrightarrow v \\times \\overrightarrow B } \\right)$$ and given $${\\overrightarrow v }$$ parallel to $$\\widehat k$$.\n
$$\\overrightarrow F = $$ $$q\\left( {\\overrightarrow E + \\overrightarrow v \\times \\overrightarrow B } \\right)$$\n
$$ \\therefore $$ $$\\overrightarrow F $$ is in direction of $${\\overrightarrow E }$$\n
which is antiparallel to $${{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9032,
"subject": "Physics",
"question": "If the magnetic field in a plane electromagnetic wave is given by\n $$\\overrightarrow B $$ = 3 $$ \\times $$ 10-8 sin(1.6 $$ \\times $$ 103x + 48 $$ \\times $$ 1010t)$$\\widehat j$$ T, then what will be expression for electric field ?\n",
"options": [
{
"text": "$$\\overrightarrow E $$ = (9sin(1.6 $$ \\times $$ 103x + 48 $$ \\times $$ 1010t)$$\\widehat k$$ V/m)"
},
{
"text": "$$\\overrightarrow E $$ = (60sin(1.6 $$ \\times $$ 103x + 48 $$ \\times $$ 1010t)$$\\widehat k$$ V/m)"
},
{
"text": "$$\\overrightarrow E $$ = (3 $$ \\times $$ 10-8 sin(1.6 $$ \\times $$ 103x + 48 $$ \\times $$ 1010t)$$\\widehat i$$ V/m)"
},
{
"text": "$$\\overrightarrow E $$ = (3 $$ \\times $$ 10-8 sin(1.6 $$ \\times $$ 103x + 48 $$ \\times $$ 1010t)$$\\widehat j$$ V/m)"
}
],
"answer": "$$\\overrightarrow E $$ = (9sin(1.6 $$ \\times $$ 103x + 48 $$ \\times $$ 1010t)$$\\widehat k$$ V/m)",
"solution": "**Answer:** $$\\overrightarrow E $$ = (9sin(1.6 $$ \\times $$ 103x + 48 $$ \\times $$ 1010t)$$\\widehat k$$ V/m)\n\nGiven $$\\overrightarrow B $$ = 3 $$ \\times $$ 10-8 sin(1.6 $$ \\times $$ 103x + 48 $$ \\times $$ 1010t)$$\\widehat j$$ T\n
Unit vector along $$\\overrightarrow E \\times \\overrightarrow B $$ = $${{\\left( {2\\widehat j + 2\\widehat i} \\right)} \\over {2\\sqrt 2 }}$$\n
= $${1 \\over {\\sqrt 2 }}\\left( {\\widehat i + \\widehat j} \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9034,
"subject": "Physics",
"question": "An electromagnetic wave of frequency 5 GHz, is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2. Its velocity in this medium is ____________ $$\\times$$ 107 m/s.",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\nGiven, $$\\mu$$r = $$\\varepsilon $$r = 2
where, $$\\mu$$r is relative permeability, $$\\varepsilon $$r is relative permittivity.
Speed of electromagnetic wave v is given by
$$v = {c \\over n}$$
where, n = refractive index = $$\\sqrt {{\\mu _r}{\\varepsilon _r}} = \\sqrt 4 = 2$$
$$ \\Rightarrow v = {{3 \\times {{10}^8}} \\over 2}$$ = 15 $$\\times$$ 107 m/s
$$\\because$$ x $$\\times$$ 107 = 15 $$\\times$$ 107
$$\\Rightarrow$$ x = 15",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9035,
"subject": "Physics",
"question": "An electromagnetic wave of frequency 3 GHz enters a dielectric medium of relative electric permittivity 2.25 from vacuum. The wavelength of this wave in that medium will be _________ $$\\times$$ 10$$-$$2 cm.",
"options": [],
"answer": "667",
"solution": "**Answer:** 667\n\nGiven, frequency of wave, f = 3 GHz = 3 $$\\times$$ 109 Hz
= 6.67 cm = 667 $$\\times$$ 10$$-$$2 cm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9036,
"subject": "Physics",
"question": "The peak electric field produced by the radiation coming from the 8W bulb at a distance of 10 m is $${x \\over {10}}\\sqrt {{{{\\mu _0}c} \\over \\pi }} {V \\over m}$$. The efficiency of the bulb is 10% and it is a point source. The value of x is ___________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nFirstly, we know that the intensity (I) of a wave is defined as the power (P) per unit area (A). For a spherical wave emanating from a point source, the area of the sphere is $4\\pi r^2$ where r is the distance from the source. So, \n\n
Comparing with the expression in the question, we find that $x = 2$.\n
Therefore, the answer is $x = 2$.\n",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9037,
"subject": "Physics",
"question": "A radiation is emitted by 1000W bulb and it generates an electric field and magnetic field at P, placed at a distance of 2m. The efficiency of the bulb is 1.25%. The value of peak electric field at P is x $$\\times$$ 10$$-$$1 V/m. Value of x is ___________. (Rounded off to the nearest integer) [Take $${\\varepsilon _0} = 8.85 \\times {10^{ - 12}}$$ C2N$$-$$1 m$$-$$2, c = $$3 \\times {10^8}$$ ms$$-$$1]",
"options": [],
"answer": "137",
"solution": "**Answer:** 137\n\nIntensity of electro magnetic wave is,
$${E_0} \\approx 137 \\times {10^{ - 1}}$$ v/m",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9038,
"subject": "Physics",
"question": "A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-direction. At a particular point in space and time, $$\\overrightarrow B $$ = 8.0 $$\\times$$ 10$$-$$8 $$\\widehat z$$T. The value of electric field at this point is :
(speed of light = 3 $$\\times$$ 108 ms$$-$$1)
$$\\widehat x$$, $$\\widehat y$$, $$\\widehat z$$ are unit vectors along x, y and z directions.",
"options": [
{
"text": "2.6 $$\\widehat x$$ V/m"
},
{
"text": "$$-$$24 $$\\widehat x$$ V/m"
},
{
"text": "24 $$\\widehat x$$ V/m"
},
{
"text": "$$-$$2.6 $$\\widehat y$$ V/m"
}
],
"answer": "$$-$$24 $$\\widehat x$$ V/m",
"solution": "**Answer:** $$-$$24 $$\\widehat x$$ V/m\n\n$${E_0} = B.C$$
Direction of wave travelling is in $$\\overrightarrow E \\times \\overrightarrow B $$
So, $$( - \\widehat x) \\times \\widehat z = + \\widehat y$$
$$ \\therefore $$ $$\\widehat E = -24\\widehat x$$ V/m",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9039,
"subject": "Physics",
"question": "For an electromagnetic wave travelling in free space, the relation between average energy densities due to electric (Ue) and magnetic (Um) fields is :",
"options": [
{
"text": "Ue = Um"
},
{
"text": "Ue $$\\ne$$ Um"
},
{
"text": "Ue < Um"
},
{
"text": "Ue > Um"
}
],
"answer": "Ue = Um",
"solution": "**Answer:** Ue = Um\n\nIn EMW, average energy density due to electric field (Ue) and magnetic field (Um) is same.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9040,
"subject": "Physics",
"question": "If 2.5 $$\\times$$ 10$$-$$6 N average force is exerted by a light wave on a non-reflecting surface of 30 cm2 area during 40 minutes of time span, the energy flux of light just before it falls on the surface is ___________ W/cm2. (Round off to the Nearest Integer)
(Assume complete absorption and normal incidence conditions are there)",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\nPressure = $${{Intensity} \\over C}$$ (for absorbing surface)
I = P $$\\times$$ C
I = $${{2.5 \\times {{10}^{ - 6}}} \\over {30c{m^2}}}$$ N $$\\times$$ 3 $$\\times$$ 108 m/s
I = 25 W/cm2",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9041,
"subject": "Physics",
"question": "The electric field intensity produced by the radiation coming from a 100 W bulb at a distance of 3 m is E. The electric field intensity produced by the radiation coming from 60W at the same distance is $$\\sqrt {{x \\over 5}} $$E. Where the value of x = ____________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n$$I = {1 \\over 2}C{ \\in _0}{E^2}$$
So the value of x = 3",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9042,
"subject": "Physics",
"question": "Seawater at a frequency f = 9 $$\\times$$ 102 Hz, has permittivity $$\\varepsilon $$ = 80$$\\varepsilon $$0 and resistivity $$\\rho$$ = 0.25 $$\\Omega$$m. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source V(t) = V0 sin(2$$\\pi$$ft). Then the conduction current density becomes 10x times the displacement current density after time t = $${1 \\over {800}}$$s. The value of x is _____________. (Given : $${1 \\over {4\\pi {\\varepsilon _0}}} = 9 \\times {10^9}$$ Nm2C$$-$$2)",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\n\n Given f = 9 $$\\times$$ 102 Hz
So x = 6",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9043,
"subject": "Physics",
"question": "A plane electromagnetic wave of frequency 100 MHz is travelling in vacuum along the x-direction. At a particular point in space and time, $$\\overrightarrow B = 2.0 \\times {10^{ - 8}}\\widehat kT$$. (where, $$\\widehat k$$ is unit vector along z-direction) What is $$\\overrightarrow E $$ at this point?",
"options": [
{
"text": "0.6 $$\\widehat j$$ V/m"
},
{
"text": "6.0 $$\\widehat k$$ V/m"
},
{
"text": "6.0 $$\\widehat j$$ V/m"
},
{
"text": "0.6 $$\\widehat k$$ V/m"
}
],
"answer": "6.0 $$\\widehat j$$ V/m",
"solution": "**Answer:** 6.0 $$\\widehat j$$ V/m\n\nf = 100 MHz
$$\\overrightarrow B = 2 \\times {10^{ - 8}}T$$
$$\\overrightarrow E = \\overrightarrow B \\times \\overrightarrow V $$
$$ \\Rightarrow $$ $$\\overrightarrow E = 6\\widehat j$$ V/m",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9044,
"subject": "Physics",
"question": "A plane electromagnetic wave propagating along y-direction can have the following pair of electric field $$\\left( {\\overrightarrow E } \\right)$$ and magnetic field $$\\left( {\\overrightarrow B } \\right)$$ components.",
"options": [
{
"text": "Ex, Bz or Ez, Bx"
},
{
"text": "Ex, By or Ey, Bx"
},
{
"text": "Ey, By or Ez, Bz"
},
{
"text": "Ey, Bx or Ex, By"
}
],
"answer": "Ex, Bz or Ez, Bx",
"solution": "**Answer:** Ex, Bz or Ez, Bx\n\n\n $$ \\because $$ $$\\widehat E \\times \\widehat B = \\widehat C$$
$$ \\therefore $$ $$\\widehat E \\times \\widehat B$$ should point in the direction of propagation of wave (y-direction here)
$$ \\therefore $$ possible combinations are (Ex, Bz) or (Ez, Bx)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9045,
"subject": "Physics",
"question": "In an electromagnetic wave the electric field vector and magnetic field vector are given as $$\\overrightarrow E = {E_0}\\widehat i$$ and $$\\overrightarrow B = {B_0}\\widehat k$$ respectively. The direction of propagation of electromagnetic wave is along :",
"options": [
{
"text": "$$\\left( {\\widehat k} \\right)$$"
},
{
"text": "$$\\widehat j$$"
},
{
"text": "$$\\left( { - \\widehat k} \\right)$$"
},
{
"text": "$$\\left( { - \\widehat j} \\right)$$"
}
],
"answer": "$$\\left( { - \\widehat j} \\right)$$",
"solution": "**Answer:** $$\\left( { - \\widehat j} \\right)$$\n\nDirection of propagation = $$\\overrightarrow E \\times \\overrightarrow B = \\widehat i \\times \\widehat k = - \\widehat j$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 9046,
"subject": "Physics",
"question": "Intensity of sunlight is observed as 0.092 Wm$$-$$2 at a point in free space. What will be the peak value of magnetic field at the point?
is incident normally on a perfectly reflecting wall at z = a. Choose the correct option",
"options": [
{
"text": "The wavelength is 5.4 m"
},
{
"text": "The frequency of electromagnetic wave is 54 $$\\times$$ 104 Hz."
},
{
"text": "The transmitted wave will be $$3.1\\cos \\left[ {(1.8)z - (5.4 \\times {{10}^6})t} \\right]\\widehat iN/C$$"
},
{
"text": "The reflected wave will be $$3.1\\cos \\left[ {(1.8)z + (5.4 \\times {{10}^6})t} \\right]\\widehat iN/C$$"
}
],
"answer": "The reflected wave will be $$3.1\\cos \\left[ {(1.8)z + (5.4 \\times {{10}^6})t} \\right]\\widehat iN/C$$",
"solution": "**Answer:** The reflected wave will be $$3.1\\cos \\left[ {(1.8)z + (5.4 \\times {{10}^6})t} \\right]\\widehat iN/C$$\n\nReflected wave will have direction opposite to incident wave.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9048,
"subject": "Physics",
"question": "The relative permittivity of distilled water is 81. The velocity of light in it will be :
$$ v = \\frac{3 \\times 10^8 \\, \\text{m/s}}{9} $$
\n
$$ v = 3.33 \\times 10^7 \\, \\text{m/s} $$
\n
So, the speed of light in distilled water is $3.33 \\times 10^7$ m/s.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9049,
"subject": "Physics",
"question": "The electric field in a plane electromagnetic wave is given by
$$\\overrightarrow E = 200\\cos \\left[ {\\left( {{{0.5 \\times {{10}^3}} \\over m}} \\right)x - \\left( {1.5 \\times {{10}^{11}}{{rad} \\over s} \\times t} \\right)} \\right]{V \\over m}\\widehat j$$. If this wave falls normally on a perfectly reflecting surface having an area of 100 cm2. If the radiation pressure exerted by the E.M. wave on the surface during a 10 minute exposure is $${x \\over {{{10}^9}}}{N \\over {{m^2}}}$$. Find the value of x .",
"options": [],
"answer": "354",
"solution": "**Answer:** 354\n\nE0 = 200
$$ = {{354} \\over {{{10}^9}}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9050,
"subject": "Physics",
"question": "A light beam is described by $$E = 800\\sin \\omega \\left( {t - {x \\over c}} \\right)$$. An electron is allowed to move normal to the propagation of light beam with a speed of 3 $$\\times$$ 107 ms$$-$$1. What is the maximum magnetic force exerted on the electron?",
"options": [
{
"text": "1.28 $$\\times$$ 10$$-$$18 N"
},
{
"text": "1.28 $$\\times$$ 10$$-$$21 N"
},
{
"text": "12.8 $$\\times$$ 10$$-$$17 N"
},
{
"text": "12.8 $$\\times$$ 10$$-$$18 N"
}
],
"answer": "12.8 $$\\times$$ 10$$-$$18 N",
"solution": "**Answer:** 12.8 $$\\times$$ 10$$-$$18 N\n\n$${{{E_0}} \\over C} = {B_0}$$
$$ = 12.8 \\times {10^{ - 18}}$$ N",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9051,
"subject": "Physics",
"question": "Electric field in a plane electromagnetic wave is given by E = 50 sin(500x $$-$$ 10 $$\\times$$ 1010 t) V/m The velocity of electromagnetic wave in this medium is :
$$V = {{2C} \\over 3}$$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9052,
"subject": "Physics",
"question": "A plane electromagnetic wave with frequency of 30 MHz travels in free space. At particular point in space and time, electric field is 6 V/m. The magnetic field at this point will be x $$\\times$$ 10$$-$$8 T. The value of x is ___________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n$$|B|\\, = {{|E|} \\over C} = {6 \\over {3 \\times {{10}^8}}}$$
= 2 $$\\times$$ 10$$-$$8 T
$$\\therefore$$ x = 2",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9053,
"subject": "Physics",
"question": "The electric field in an electromagnetic wave is given by E = (50 NC$$-$$1) sin$$\\omega$$ (t $$-$$ x/c)
The energy contained in a cylinder of volume V is 5.5 $$\\times$$ 10$$-$$12 J. The value of V is _____________ cm3. (given $$\\in$$0 = 8.8 $$\\times$$ 10$$-$$12C2N$$-$$1m$$-$$2)",
"options": [],
"answer": "500",
"solution": "**Answer:** 500\n\n$$E = 50\\sin \\left( {\\omega t - {\\omega \\over c}.\\,x} \\right)$$
Energy density = $${1 \\over 2}{ \\in _0}E_0^2$$
Energy of volume $$V = {1 \\over 2}{ \\in _0}E_0^2.\\,V = 5.5 \\times {10^{ - 12}}$$
= 500 (c.m)3",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9054,
"subject": "Physics",
"question": "The magnetic field vector of an electromagnetic wave is given by $$B = {B_0}{{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}\\cos (kz - \\omega t)$$; where $$\\widehat i,\\widehat j$$ represents unit vector along x and y-axis respectively. At t = 0s, two electric charges q1 of 4$$\\pi$$ coulomb and q2 of 2$$\\pi$$ coulomb located at $$\\left( {0,0,{\\pi \\over k}} \\right)$$ and $$\\left( {0,0,{{3\\pi } \\over k}} \\right)$$, respectively, have the same velocity of 0.5 c $$\\widehat i$$, (where c is the velocity of light). The ratio of the force acting on charge q1 to q2 is :-",
"options": [
{
"text": "$$2\\sqrt 2 :1$$"
},
{
"text": "$$1:\\sqrt 2 $$"
},
{
"text": "2 : 1"
},
{
"text": "$$\\sqrt 2 :1$$"
}
],
"answer": "2 : 1",
"solution": "**Answer:** 2 : 1\n\n$$\\overrightarrow F = q\\left( {\\overrightarrow V \\times \\overrightarrow B } \\right)$$
$${\\overrightarrow F _1} = 4\\pi \\left[ {0.5c\\widehat i \\times {B_0}\\left( {{{\\widehat i + \\widehat j} \\over 2}} \\right)\\cos \\left( {K.{\\pi \\over K} - 0} \\right)} \\right]$$
$${\\overrightarrow F _2} = 2\\pi \\left[ {0.5c\\widehat i \\times {B_0}\\left( {{{\\widehat i + \\widehat j} \\over 2}} \\right)\\cos \\left( {K.{{3\\pi } \\over K} - 0} \\right)} \\right]$$
$$\\cos \\pi = - 1$$, $$\\cos 3\\pi = - 1$$
$$\\therefore$$ $${{{F_1}} \\over {{F_2}}} = 2$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 9055,
"subject": "Physics",
"question": "Electric field of plane electromagnetic wave propagating through a non-magnetic medium is given by
E = 20cos(2 $$\\times$$ 1010 t $$-$$ 200x) V/m. The dielectric constant of the medium is equal to : (Take $$\\mu$$r = 1)",
"options": [
{
"text": "9"
},
{
"text": "2"
},
{
"text": "$${1 \\over 3}$$"
},
{
"text": "3"
}
],
"answer": "9",
"solution": "**Answer:** 9\n\nGiven, electric field,
As we know the relation between the refractive index and dielectric constant,
$$\\mu = \\sqrt {{\\varepsilon _r}{\\mu _r}} $$
Substituting the value in the above equations, we get
$$3 = \\sqrt {{\\varepsilon _r}(1)} $$
$${\\varepsilon _r} = 9$$
Thus, the dielectric constant of the medium is 9.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9056,
"subject": "Physics",
"question": "
The intensity of the light from a bulb incident on a surface is 0.22 W/m2. The amplitude of the magnetic field in this light-wave is ______________ $$\\times$$ 10$$-$$9 T.
\n
(Given : Permittivity of vacuum $$\\in$$0 = 8.85 $$\\times$$ 10$$-$$12 C2 N$$-$$1-m$$-$$2, speed of light in vacuum c = 3 $$\\times$$ 108 ms$$-$$1)
The displacement current of 4.425 $$\\mu$$A is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 106 Vs$$-$$1. The area of each plate of the capacitor is 40 cm2. The distance between each plate of the capacitor is x $$\\times$$ 10$$-$$3 m. The value of x is __________.
An EM wave propagating in x-direction has a wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm$$-$$1. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum :
A radar sends an electromagnetic signal of electric field (E0) = 2.25 V/m and magnetic field (B0) = 1.5 $$\\times$$ 10$$-$$8 T which strikes a target on line of sight at a distance of 3 km in a medium. After that, a part of signal (echo) reflects back towards the radar with same velocity and by same path. If the signal was transmitted at time t = 0 from radar, then after how much time echo will reach to the radar?
Statement I : A time varying electric field is a source of changing magnetic field and vice-versa. Thus a disturbance in electric or magnetic field creates EM waves.
\n
Statement II : In a material medium, the EM wave travels with speed $$v = {1 \\over {\\sqrt {{\\mu _0}{ \\in _0}} }}$$. In the light of the above statements, choose the correct answer from the options given below.
",
"options": [
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Statement I is correct but Statement II is false"
},
{
"text": "Statement I is incorrect but Statement II is true"
}
],
"answer": "Statement I is correct but Statement II is false",
"solution": "**Answer:** Statement I is correct but Statement II is false\n\n
In a material medium speed of light is given by $$v = {1 \\over {\\sqrt {{\\varepsilon _0}{\\varepsilon _r}{\\mu _0}{\\mu _r}} }}$$. So statement 2 is false.
If Electric field intensity of a uniform plane electromagnetic wave is given as $$E = - 301.6\\sin (kz - \\omega t){\\widehat a_x} + 452.4\\sin (kz - \\omega t){\\widehat a_y}{V \\over m}$$. Then magnetic intensity 'H' of this wave in Am$$-$$1 will be :
\n
[Given : Speed of light in vacuum $$c = 3 \\times {10^8}$$ ms$$-$$1, Permeability of vacuum $${\\mu _0} = 4\\pi \\times {10^{ - 7}}$$ NA$$-$$2]
Also, $$\\widehat B = \\widehat C \\times \\widehat E \\Rightarrow \\widehat k \\times {{({E_{0x}}\\widehat i + {E_{0y}}\\widehat j)} \\over {\\sqrt {E_{0x}^2 + E_{0y}^2} }}$$
In free space, an electromagnetic wave of 3 GHz frequency strikes over the edge of an object of size $${\\lambda \\over {100}}$$, where $$\\lambda$$ is the wavelength of the wave in free space. The phenomenon, which happens there will be :
For reflection size of obstacle must be much larger than wavelength, for diffraction size should be order of wavelength.\n\n
Since the object is of size $\\frac{\\lambda}{100}$, much smaller than wavelength, so scattering will occur.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 9063,
"subject": "Physics",
"question": "
The electromagnetic waves travel in a medium at a speed of 2.0 $$\\times$$ 108 m/s. The relative permeability of the medium is 1.0. The relative permittivity of the medium will be :
",
"options": [
{
"text": "2.25"
},
{
"text": "4.25"
},
{
"text": "6.25"
},
{
"text": "8.25"
}
],
"answer": "2.25",
"solution": "**Answer:** 2.25\n\nThe speed of electromagnetic waves in a medium is given by the formula:\n\n
$$v = \\frac{1}{\\sqrt{\\mu \\varepsilon}}$$\n\n
where $\\mu$ and $\\varepsilon$ are the absolute permeability and absolute permittivity of the medium, respectively.\n\n
Given that $\\mu = \\mu_0 \\mu_r$ and $\\varepsilon = \\varepsilon_0 \\varepsilon_r$, we can rewrite the above equation as :\n\n
So, the relative permittivity of the medium is 2.25.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9064,
"subject": "Physics",
"question": "
The electric field in an electromagnetic wave is given by E = 56.5 sin $$\\omega$$(t $$-$$ x/c) NC$$-$$1. Find the intensity of the wave if it is propagating along x-axis in the free space.
An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.
",
"options": [
{
"text": "1.19 $$\\times$$ 10$$-$$8T"
},
{
"text": "1.71 $$\\times$$ 10$$-$$8T"
},
{
"text": "0.84 $$\\times$$ 10$$-$$8T"
},
{
"text": "3.36 $$\\times$$ 10$$-$$8T"
}
],
"answer": "1.71 $$\\times$$ 10$$-$$8T",
"solution": "**Answer:** 1.71 $$\\times$$ 10$$-$$8T\n\nThe total power (PT) of the light bulb is given as 200 W, but only 3.5% of this power is actually emitted as radiation, which we will call P. \n
A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 $$\\times$$ 10$$-$$2 Am$$-$$1 at a point, what will be the approximate magnitude of electric field intensity at that point?
\n
(Given : Permeability of free space $$\\mu$$0 = 4$$\\pi$$ $$\\times$$ 10$$-$$7 NA$$-$$2, speed of light in vacuum c = 3 $$\\times$$ 108 ms$$-$$1)
An expression for oscillating electric field in a plane electromagnetic wave is given as Ez = 300 sin(5$$\\pi$$ $$\\times$$ 103x $$-$$ 3$$\\pi$$ $$\\times$$ 1011t) Vm$$-$$1
\n
Then, the value of magnetic field amplitude will be :
\n
(Given : speed of light in Vacuum c = 3 $$\\times$$ 108 ms$$-$$1)
The amplitude of the electric field ($E_0$) is $300 \\, \\text{V/m}$.
\n\n
The velocity ($v$) of the wave in the medium is given by the ratio of the coefficients of time and displacement in the wave equation, which can be calculated as:
The rms value of conduction current in a parallel plate capacitor is $$6.9 \\,\\mu \\mathrm{A}$$. The capacity of this capacitor, if it is connected to $$230 \\mathrm{~V}$$ ac supply with an angular frequency of $$600 \\,\\mathrm{rad} / \\mathrm{s}$$, will be :
Light wave travelling in air along x-direction is given by $${E_y} = 540\\sin \\pi \\times {10^4}(x - ct)\\,V{m^{ - 1}}$$. Then, the peak value of magnetic field of wave will be (Given c = 3 $$\\times$$ 108 ms$$-$$1)
The oscillating magnetic field in a plane electromagnetic wave is given by
$$B_{y}=5 \\times 10^{-6} \\sin 1000 \\pi\\left(5 x-4 \\times 10^{8} t\\right) T$$. The amplitude of electric field will be :
",
"options": [
{
"text": "$$15 \\times 10^{2} \\,\\mathrm{Vm}^{-1}$$"
},
{
"text": "$$5 \\times 10^{-6} \\,\\mathrm{Vm}^{-1}$$"
},
{
"text": "$$16 \\times 10^{12} \\,\\mathrm{Vm}^{-1}$$"
},
{
"text": "$$4 \\times 10^{2} \\,\\mathrm{Vm}^{-1}$$"
}
],
"answer": "$$4 \\times 10^{2} \\,\\mathrm{Vm}^{-1}$$",
"solution": "**Answer:** $$4 \\times 10^{2} \\,\\mathrm{Vm}^{-1}$$\n\nIn an electromagnetic wave, the magnitude of the electric field (E) and the magnetic field (B) are related by the speed of the wave (v), which can be represented as :\n\n
$$E = Bv$$\n\n
Here, B is the magnetic field, E is the electric field, and v is the speed of the electromagnetic wave. \n\n
In the given problem, the peak value of the magnetic field is given as :\n\n
$$B_0 = 5 \\times 10^{-6} T$$\n\n
The wave speed (v) can be calculated from the given equation for the magnetic field, using the relationship between frequency (f), wave number (k), and speed :\n\n
A velocity selector consists of electric field $$\\vec{E}=E \\,\\hat{k}$$ and magnetic field $$\\vec{B}=B \\,\\hat{j}$$ with $$B=12 \\,m T$$. The value of $$E$$ required for an electron of energy $$728 \\,\\mathrm{e} V$$ moving along the positive $$x$$-axis to pass undeflected is :
\n
(Given, mass of electron $$=9.1 \\times 10^{-31} \\mathrm{~kg}$$ )
A beam of light travelling along $$X$$-axis is described by the electric field $$E_{y}=900 \\sin \\omega(\\mathrm{t}-x / c)$$. The ratio of electric force to magnetic force on a charge $$\\mathrm{q}$$ moving along $$Y$$-axis with a speed of $$3 \\times 10^{7} \\mathrm{~ms}^{-1}$$ will be :
\n
(Given speed of light $$=3 \\times 10^{8} \\mathrm{~ms}^{-1}$$)
Identify the correct statements from the following descriptions of various properties of electromagnetic waves.
\n
(A) In a plane electromagnetic wave electric field and magnetic field must be perpendicular to each other and direction of propagation of wave should be along electric field or magnetic field.
\n
(B) The energy in electromagnetic wave is divided equally between electric and magnetic fields.
\n
(C) Both electric field and magnetic field are parallel to each other and perpendicular to the direction of propagation of wave.
\n
(D) The electric field, magnetic field and direction of propagation of wave must be perpendicular to each other.
\n
(E) The ratio of amplitude of magnetic field to the amplitude of electric field is equal to speed of light.
\n
Choose the most appropriate answer from the options given below :
",
"options": [
{
"text": "(D) only"
},
{
"text": "(B) and (D) only"
},
{
"text": "(B), (C) and (E) only"
},
{
"text": "(A), (B) and (E) only"
}
],
"answer": "(B) and (D) only",
"solution": "**Answer:** (B) and (D) only\n\n
In an EM wave :
\n
1. $$\\overrightarrow E \\,\\, \\bot \\,\\,\\overrightarrow B $$
\n
2. $$\\overrightarrow V \\equiv \\overrightarrow E \\times \\overrightarrow B $$
\n
3. Energy is equally divided
\n
4. $$\\left| {\\overrightarrow V } \\right| = \\left| {\\overrightarrow E } \\right|/\\left| {\\overrightarrow B } \\right|$$
Sun light falls normally on a surface of area $$36 \\mathrm{~cm}^{2}$$ and exerts an average force of $$7.2 \\times 10^{-9} \\mathrm{~N}$$ within a time period of 20 minutes. Considering a case of complete absorption, the energy flux of incident light is
Nearly 10% of the power of a $$110 \\mathrm{~W}$$ light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of $$1 \\mathrm{~m}$$ from the bulb to a distance of $$5 \\mathrm{~m}$$ is $$a \\times 10^{-2} \\mathrm{~W} / \\mathrm{m}^{2}$$. The value of 'a' will be _________.
In a medium the speed of light wave decreases to $$0.2$$ times to its speed in free space The ratio of relative permittivity to the refractive index of the medium is $$x: 1$$. The value of $$x$$ is _________.
\n
(Given speed of light in free space $$=3 \\times 10^{8} \\mathrm{~m} \\mathrm{~s}^{-1}$$ and for the given medium $$\\mu_{\\mathrm{r}}=1$$)
$\\Rightarrow$ Required ratio $=\\frac{\\varepsilon_{r}}{n}=\\frac{\\varepsilon_{r}}{\\sqrt{\\varepsilon_{r}}}=\\sqrt{\\varepsilon_{r}}=5$\n\n
$\\Rightarrow x=5$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9079,
"subject": "Physics",
"question": "A point source of $100 \\mathrm{~W}$ emits light with $5 \\%$ efficiency. At a distance of $5 \\mathrm{~m}$ from the source, the intensity produced by the electric field component is:",
"options": [
{
"text": "$\\frac{1}{40 \\pi} \\frac{W}{m^2}$"
},
{
"text": "$\\frac{1}{10 \\pi} \\frac{W}{m^2}$"
},
{
"text": "$\\frac{1}{20 \\pi} \\frac{W}{m^2}$"
},
{
"text": "$\\frac{1}{2 \\pi} \\frac{W}{m^2}$"
}
],
"answer": "$\\frac{1}{40 \\pi} \\frac{W}{m^2}$",
"solution": "**Answer:** $\\frac{1}{40 \\pi} \\frac{W}{m^2}$\n\n
A point source of $100 \\mathrm{~W}$ emits light with $5 \\%$ efficiency. At a distance of $5 \\mathrm{~m}$ from the source, the intensity produced by the electric field component is evaluated as follows:
\n\n
\n\n
The intensity at a distance of $5$ meters can be calculated as:
\n\n
Intensity at 5 m $$= \\frac{5}{4\\pi \\times 5^2}\\left(\\frac{W}{m^2}\\right)$$
A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of $$24 \\mathrm{~W}$$. The radius of curvature of hemisphere is $$10 \\mathrm{~cm}$$ and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is ____________ $$\\times~10^{-8} \\mathrm{~N}$$.
Statement I : Electromagnetic waves are not deflected by electric and magnetic field.
\n
Statement II : The amplitude of electric field and the magnetic field in electromagnetic waves are related to each other as $${E_0} = \\sqrt {{{{\\mu _0}} \\over {{\\varepsilon _0}}}} {B_0}$$.
\n
In the light of the above statements, choose the correct answer from the options given below :
",
"options": [
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Statement I is true and Statement II is false"
},
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Statement I is false but Statement II is true"
}
],
"answer": "Statement I is true and Statement II is false",
"solution": "**Answer:** Statement I is true and Statement II is false\n\n
Statement I is correct as photon do not carry any charge, hence cannot feel force from either fields.
A. Speed of light in vacuum is dependent on the direction of propagation.
\n
B. Speed of light in a medium is independent of the wavelength of light.
\n
C. The speed of light is independent of the motion of the source.
\n
D. The speed of light in a medium is independent of intensity.
\n
Choose the correct answer from the options given below:
",
"options": [
{
"text": "B and D only"
},
{
"text": "B and C only"
},
{
"text": "A and C only"
},
{
"text": "C and D only"
}
],
"answer": "C and D only",
"solution": "**Answer:** C and D only\n\nThe speed of light does not depend on the motion of the source as well as intensity. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9083,
"subject": "Physics",
"question": "
Match List I with List II
\n
\n
\n
\n
\n
\n
\n
\n
\n\n
\n
\n
List I
\n
\n
List II
\n
\n\n\n
\n
A.
\n
Gauss's Law in Electrostatics
\n
I.
\n
$$\\oint {\\overrightarrow E \\,.\\,d\\overrightarrow l = - {{d{\\phi _B}} \\over {dt}}} $$
\n
\n
\n
B.
\n
Faraday's Law
\n
II.
\n
$$\\oint {\\overrightarrow B \\,.\\,d\\overrightarrow A = 0} $$
\n
\n
\n
C.
\n
Gauss's Law in Magnetism
\n
III.
\n
$$\\oint {\\overrightarrow B \\,.\\,d\\overrightarrow l = {\\mu _0}{i_c} + {\\mu _0}{ \\in _0}{{d{\\phi _E}} \\over {dt}}} $$
\n
\n
\n
D.
\n
Ampere-Maxwell Law
\n
IV.
\n
$$\\oint {\\overrightarrow E \\,.\\,d\\overrightarrow s = {q \\over {{ \\in _0}}}} $$
\n
\n\n
\n
Choose the correct answer from the options given below :
An electromagnetic wave is transporting energy in the negative $$z$$ direction. At a certain point and certain time the direction of electric field of the wave is along positive $$y$$ direction. What will be the direction of the magnetic field of the wave at that point and instant?
",
"options": [
{
"text": "Negative direction of $$y$$"
},
{
"text": "Positive direction of $$z$$"
},
{
"text": "Positive direction of $$x$$"
},
{
"text": "Negative direction $$x$$"
}
],
"answer": "Positive direction of $$x$$",
"solution": "**Answer:** Positive direction of $$x$$\n\nAs, poynting vector
\nHence according to vector cross product magnetic field should be positive $\\mathrm{x}$ direction.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9085,
"subject": "Physics",
"question": "
The electric field and magnetic field components of an electromagnetic wave going through vacuum is described by
\n
$$\\mathrm{{E_x} = {E_o}\\sin (kz - \\omega t)}$$
\n
$$\\mathrm{{B_y} = {B_o}\\sin (kz - \\omega t)}$$
\n
Then the correct relation between E$$_0$$ and B$$_0$$ is given by
In $$\\overrightarrow E $$ and $$\\overrightarrow K $$ represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by :
\n
($$\\omega$$ - angular frequency) :
",
"options": [
{
"text": "$${1 \\over \\omega }\\left( {\\overline K \\times \\overline E } \\right)$$"
},
{
"text": "$$\\overline K \\times \\overline E $$"
},
{
"text": "$$\\omega \\left( {\\overline K \\times \\overline E } \\right)$$"
},
{
"text": "$$\\omega \\left( {\\overline E \\times \\overline K } \\right)$$"
}
],
"answer": "$${1 \\over \\omega }\\left( {\\overline K \\times \\overline E } \\right)$$",
"solution": "**Answer:** $${1 \\over \\omega }\\left( {\\overline K \\times \\overline E } \\right)$$\n\nMagnetic field vector will be in the direction of $\\hat{\\mathrm{K}} \\times \\hat{\\mathrm{E}}$
\nmagnitude of $B=\\frac{E}{C}=\\frac{K}{\\omega} E$
In an electromagnetic wave, at an instant and at particular position, the electric field is along the negative $$z$$-axis and magnetic field is along the positive $$x$$-axis. Then the direction of propagation of electromagnetic wave is:
",
"options": [
{
"text": "at $$45^{\\circ}$$ angle from positive y-axis"
},
{
"text": " positive $$y$$-axis"
},
{
"text": "negative $$\\mathrm{y}$$-axis"
},
{
"text": "positive z-axis"
}
],
"answer": "negative $$\\mathrm{y}$$-axis",
"solution": "**Answer:** negative $$\\mathrm{y}$$-axis\n\nAs the electric field is along the negative $z$-axis and the magnetic field is along the positive $x$-axis, the direction of propagation of the wave is given by the vector product $\\overrightarrow{\\mathrm{E}} \\times \\overrightarrow{\\mathrm{B}}$, which is in the direction of the negative $y$-axis. \n
\nUsing the right-hand rule, if you point your right thumb in the direction of the vector product $\\overrightarrow{\\mathrm{E}} \\times \\overrightarrow{\\mathrm{B}}$, your fingers will curl in the direction of the propagation of the wave. In this case, the fingers would curl in the direction of the negative $y$-axis.\n
\nTherefore, the direction of propagation of the electromagnetic wave in this scenario is along the negative $y$-axis.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9088,
"subject": "Physics",
"question": "
Which of the following Maxwell's equation is valid for time varying conditions but not valid for static conditions :
",
"options": [
{
"text": "$$\\oint \\overrightarrow{\\mathrm{E}} \\cdot \\overrightarrow{d l}=0$$"
},
{
"text": "$$\\oint \\vec{B} \\cdot \\overrightarrow{d l}=\\mu_{0} I$$"
},
{
"text": "$$\\oint \\vec{E} \\cdot \\overrightarrow{d l}=-\\frac{\\partial \\phi_{B}}{\\partial t}$$"
},
{
"text": "$$\\oint \\vec{D} \\cdot \\overrightarrow{d A}=Q$$"
}
],
"answer": "$$\\oint \\vec{E} \\cdot \\overrightarrow{d l}=-\\frac{\\partial \\phi_{B}}{\\partial t}$$",
"solution": "**Answer:** $$\\oint \\vec{E} \\cdot \\overrightarrow{d l}=-\\frac{\\partial \\phi_{B}}{\\partial t}$$\n\nMaxwell's equations describe the behavior of electric and magnetic fields. There are four equations, and each has a specific role. In the given options, Option C refers to Faraday's Law of Electromagnetic Induction, which is the only equation among the options that is not valid for static conditions.\n
\nOption C: Faraday's Law of Electromagnetic Induction:
\nThis equation states that a time-varying magnetic field (changing magnetic flux, $\\phi_B$) induces an electromotive force (EMF) in a closed conducting loop, creating an electric field. In static conditions, the magnetic field doesn't change over time, and there is no induced EMF. Therefore, Faraday's Law is valid for time-varying conditions but not for static conditions.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 9089,
"subject": "Physics",
"question": "
A plane electromagnetic wave of frequency $$20 ~\\mathrm{MHz}$$ propagates in free space along $$\\mathrm{x}$$-direction. At a particular space and time, $$\\overrightarrow{\\mathrm{E}}=6.6 \\hat{j} \\mathrm{~V} / \\mathrm{m}$$. What is $$\\overrightarrow{\\mathrm{B}}$$ at this point?
",
"options": [
{
"text": "$$-2.2 \\times 10^{-8} \\hat{i} T$$"
},
{
"text": "$$2.2 \\times 10^{-8} \\hat{i} T$$"
},
{
"text": "$$2.2 \\times 10^{-8} \\hat{k} T$$"
},
{
"text": "$$-2.2 \\times 10^{-8} \\hat{k} T$$"
}
],
"answer": "$$2.2 \\times 10^{-8} \\hat{k} T$$",
"solution": "**Answer:** $$2.2 \\times 10^{-8} \\hat{k} T$$\n\nIn free space, the relationship between the electric field (E) and the magnetic field (B) in an electromagnetic wave is given by:\n
\n$$\nB = \\frac{E}{c}\n$$\n
\nwhere c is the speed of light in a vacuum, approximately equal to $$3 \\times 10^8 \\mathrm{~m} / \\mathrm{s}$$. We are given that the electric field is $$\\overrightarrow{\\mathrm{E}}=6.6 \\hat{j} \\mathrm{~V} / \\mathrm{m}$$. To find the magnetic field, we can first calculate the magnitude of B:\n
\nNow, we need to find the direction of the magnetic field. Since the electromagnetic wave propagates in the x-direction, and the electric field is in the y-direction (j), the magnetic field should be in the z-direction (k) to satisfy the right-hand rule for electromagnetic waves. The direction of the magnetic field will be positive k (counter-clockwise rotation from x to y).\n
where $$\\omega$$ and $$c$$ are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of $$5 \\times 10^{-4} \\mathrm{~m}^{3}$$ will be
The amplitude of magnetic field in an electromagnetic wave propagating along y-axis is $$6.0 \\times 10^{-7} \\mathrm{~T}$$. The maximum value of electric field in the electromagnetic wave is
",
"options": [
{
"text": "$$6.0 \\times 10^{-7} ~\\mathrm{Vm}^{-1}$$"
},
{
"text": "$$5 \\times 10^{14} ~\\mathrm{Vm}^{-1}$$"
},
{
"text": "$$180 ~\\mathrm{Vm}^{-1}$$"
},
{
"text": "$$2 \\times 10^{15} ~\\mathrm{Vm}^{-1}$$"
}
],
"answer": "$$180 ~\\mathrm{Vm}^{-1}$$",
"solution": "**Answer:** $$180 ~\\mathrm{Vm}^{-1}$$\n\nIn an electromagnetic wave, the maximum value of the electric field E is related to the maximum value of the magnetic field B by the equation\n
\n$E = cB$\n
\nwhere c is the speed of light in a vacuum, which is approximately $3 × 10^8 m/s$.\n
\nGiven that the amplitude of the magnetic field B is $6.0 × 10^{-7} T$, we can substitute these values into the equation to find the maximum value of the electric field:\n
\nTherefore, 180 V/m is the correct answer.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9092,
"subject": "Physics",
"question": "
The energy of an electromagnetic wave contained in a small volume oscillates with
",
"options": [
{
"text": "double the frequency of the wave"
},
{
"text": "the frequency of the wave"
},
{
"text": "half the frequency of the wave"
},
{
"text": "zero frequency"
}
],
"answer": "double the frequency of the wave",
"solution": "**Answer:** double the frequency of the wave\n\n
The energy of an electromagnetic wave contained in a small volume oscillates with double the frequency of the wave.
\n
Here's why: The electromagnetic wave consists of oscillating electric and magnetic fields. The energy density of the wave is proportional to the square of the amplitude of these fields. Since the square of a sinusoidal function oscillates with twice the frequency of the original function, the energy density of the electromagnetic wave also oscillates with twice the frequency of the wave.
For the plane electromagnetic wave given by $$E=E_{0} \\sin (\\omega t-k x)$$ and $$B=B_{0} \\sin (\\omega t-k x)$$, the ratio of average electric energy density to average magnetic energy density is
The average energy density of an electromagnetic wave is equally shared between the electric field and the magnetic field. This means that the average electric energy density is equal to the average magnetic energy density.
\n
The average electric energy density ($u_E$) is given by:
\n
$ u_E = \\frac{1}{2} \\varepsilon_0 E^2 $
\n
and the average magnetic energy density ($u_B$) is given by:
\n
$ u_B = \\frac{1}{2\\mu_0} B^2 $
\n
where $\\varepsilon_0$ is the permittivity of free space, $\\mu_0$ is the permeability of free space, $E$ is the electric field strength, and $B$ is the magnetic field strength.
\n
In an electromagnetic wave, $E$ and $B$ are related by the equation:
\n
$ E = cB $
\n
where $c$ is the speed of light in vacuum.
\n
Substituting this into the energy density equations gives:
\n
$ u_E = \\frac{1}{2} \\varepsilon_0 (cB)^2 $
\n
and
\n
$ u_B = \\frac{1}{2\\mu_0} B^2 $
\n
Since $c = \\frac{1}{\\sqrt{\\varepsilon_0 \\mu_0}}$, we can rewrite $u_E$ as:
The energy density associated with electric field $$\\vec{E}$$ and magnetic field $$\\vec{B}$$ of an electromagnetic wave in free space is given by $$\\left(\\epsilon_{0}-\\right.$$ permittivity of free space, $$\\mu_{0}-$$ permeability of free space)
The energy density associated with the electric field $$\\vec{E}$$ and magnetic field $$\\vec{B}$$ of an electromagnetic wave in free space is given by:
\n
For the electric field:\n$$U_{E} = \\frac{1}{2} \\epsilon_{0} E^2$$
\n
For the magnetic field:\n$$U_{B} = \\frac{1}{2} \\frac{B^2}{\\mu_{0}}$$
\n",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9095,
"subject": "Physics",
"question": "If frequency of electromagnetic wave is $60 \\mathrm{~MHz}$ and it travels in air along $z$ direction then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other and the wavelength of the wave (in $\\mathrm{m}$ ) is :",
"options": [
{
"text": "2.5"
},
{
"text": "5"
},
{
"text": "10"
},
{
"text": "2"
}
],
"answer": "5",
"solution": "**Answer:** 5\n\n
The speed of electromagnetic waves in air (and in a vacuum) is approximately the speed of light, which we denote as $$ c $$. The speed of light $$ c $$ is $$ 3 \\times 10^8 $$ meters per second. The relationship between the speed of light $$ c $$, the frequency $$ f $$, and the wavelength $$ \\lambda $$ of an electromagnetic wave is given by the equation:
\n
$$ c = f \\lambda $$
\n
Here, we are given that the frequency $$ f $$ of the electromagnetic wave is 60 MHz (megahertz), which can be converted to hertz (Hz) by multiplying by $$ 10^6 $$ (because 1 MHz = $$ 10^6 $$ Hz).
\n
$$ f = 60 \\times 10^6 \\text{ Hz} $$
\n
To find the wavelength $$ \\lambda $$ in meters, we can rearrange the wave equation to solve for $$ \\lambda $$:
\n
$$ \\lambda = \\frac{c}{f} $$
\n
Substitute the given values of $$ c $$ and $$ f $$ into the equation to find the wavelength:
Therefore, the wavelength of the wave is 5 meters, corresponding to option B. Additionally, in an electromagnetic wave, the electric and magnetic field vectors are indeed mutually perpendicular to each other and to the direction of propagation, which in this case is the $$ z $$ direction.
An object is placed in a medium of refractive index 3 . An electromagnetic wave of intensity $$6 \\times 10^8 \\mathrm{~W} / \\mathrm{m}^2$$ falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space $$=3 \\times 10^8 \\mathrm{~m} / \\mathrm{s}$$ ) :
Statement I: Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields.
\n
Statement II: When electromagnetic waves strike a surface, a pressure is exerted on the surface.
\n
In the light of the above statements, choose the most appropriate answer from the options given below:
",
"options": [
{
"text": "Statement I is incorrect but Statement II is correct.\n"
},
{
"text": "Both Statement I and Statement II are correct.\n"
},
{
"text": "Statement I is correct but Statement II is incorrect.\n"
},
{
"text": "Both Statement I and Statement II are incorrect."
}
],
"answer": "Both Statement I and Statement II are correct.\n",
"solution": "**Answer:** Both Statement I and Statement II are correct.\n\n\n
In a plane EM wave, the electric field oscillates sinusoidally at a frequency of $$5 \\times 10^{10} \\mathrm{~Hz}$$ and an amplitude of $$50 \\mathrm{~Vm}^{-1}$$. The total average energy density of the electromagnetic field of the wave is : [Use $$\\varepsilon_0=8.85 \\times 10^{-12} \\mathrm{C}^2 / \\mathrm{Nm}^2$$ ]
For a plane electromagnetic wave, the electric field (E) and the magnetic field (B) contribute equally to the energy density. Therefore, we can focus on just one part to find the total energy density. The formula for the energy density due to the electric field is:
\n\n
$$U_E = \\frac{1}{2} \\varepsilon_0 E^2$$
\n\n
Given that the electric field amplitude ($E$) is $50 \\, \\mathrm{Vm}^{-1}$ and the permittivity of free space ($\\varepsilon_0$) is $8.85 \\times 10^{-12} \\, \\mathrm{C}^2/\\mathrm{Nm}^2$, we can calculate the energy density due to the electric field as follows:
Since the energy density contributions from the electric and magnetic fields are equal, the total average energy density of the electromagnetic wave is just this value.
A plane electromagnetic wave of frequency $$35 \\mathrm{~MHz}$$ travels in free space along the $$X$$-direction. At a particular point (in space and time) $$\\vec{E}=9.6 \\hat{j} \\mathrm{~V} / \\mathrm{m}$$. The value of magnetic field at this point is :
The electric field of an electromagnetic wave in free space is represented as $$\\overrightarrow{\\mathrm{E}}=\\mathrm{E}_0 \\cos (\\omega \\mathrm{t}-\\mathrm{kz}) \\hat{i}$$. The corresponding magnetic induction vector will be :
The magnetic field in a plane electromagnetic wave is $$\\mathrm{B}_{\\mathrm{y}}=\\left(3.5 \\times 10^{-7}\\right) \\sin \\left(1.5 \\times 10^3 x+0.5 \\times 10^{11} t\\right) \\mathrm{T}$$. The corresponding electric field will be :
For an electromagnetic wave propagating in free space, the relationship between the magnitudes of the electric field ($$E$$) and the magnetic field ($$B$$) can be described using the equation:
\n\n\n\n
$E = cB$
\n\n\n\n
where
\n\n
\n\n
$c$ is the speed of light in vacuum, approximately $3.0 \\times 10^8 \\, \\text{m/s}$.
\n\n
$B$ is the magnitude of the magnetic field.
\n\n
\n\n
Given the magnetic field $B_y = (3.5 \\times 10^{-7}) \\sin (1.5 \\times 10^3 x + 0.5 \\times 10^{11} t) \\, \\text{T}$, we can calculate the corresponding electric field magnitude using the formula above:
Thus, the magnitude of the electric field associated with the given magnetic field is $105 \\, \\text{Vm}^{-1}$. The direction of the electric field is perpendicular to both the magnetic field and the direction of propagation. Given $B_y$, this means $E$ will have components in the $x-z$ plane. Since electromagnetic waves are transverse, and given that the magnetic field is specified to be in the $y$-direction, the corresponding electric field component must lie in a plane perpendicular to the $y$-axis, which could be either the $x$ or the $z$ direction.
\n\n
However, knowing electromagnetic wave properties, if the wave is propagating along the $x$-axis and the magnetic field ($B_y$) is along the $y$-axis, then by right-hand rule, the electric field ($E$) must be along the $z$-axis to maintain the orthogonal relationship among the direction of propagation, electric field, and magnetic field vector directions.
A plane EM wave is propagating along $$x$$ direction. It has a wavelength of $$4 \\mathrm{~mm}$$. If electric field is in $$y$$ direction with the maximum magnitude of $$60 \\mathrm{~Vm}^{-1}$$, the equation for magnetic field is :
To find the correct equation for the magnetic field of the plane electromagnetic wave given its parameters, we can use a couple of known relationships from electromagnetism.
\n\n
Firstly, the wavelength ($ \\lambda $) of the wave is given as $4 \\mathrm{~mm} = 4 \\times 10^{-3} \\mathrm{~m}$. The speed of light (and all electromagnetic waves in vacuum) is $ c = 3 \\times 10^8 \\mathrm{~m/s} $. Using these values, we can find the frequency ($ f $) of the wave using the relationship:
The angular frequency ($ \\omega $) which appears in wave equations is related to the frequency by $ \\omega = 2\\pi f $. However, in this context, what we need is the wave vector ($ k $), which defines how the phase of the wave changes with space. The wave vector $ k = \\frac{2\\pi}{\\lambda} $. So, for this wave, $ k = \\frac{2\\pi}{4 \\times 10^{-3}} = \\frac{\\pi}{2} \\times 10^3 \\, \\mathrm{m}^{-1} $.
\n\n
Knowing that the electric field ($ \\mathbf{E} $) and magnetic field ($ \\mathbf{B} $) are related as $ E = cB $ in a vacuum, where $ E $ and $ B $ are the magnitudes of the electric and magnetic fields, respectively, we can calculate the magnitude of the magnetic field using the provided maximum electric field magnitude ($ E = 60 \\, \\mathrm{Vm}^{-1} $).
Therefore, the wave equation for the magnetic field, considering it propagates in the $ x $-direction and oscillates in a direction perpendicular to both the $ x $-direction and the direction of the electric field (thus, in the $ z $-direction if $ \\mathbf{E} $ is in the $ y $-direction), is:
\n\n
$ \\mathbf{B} = B \\sin(kx - \\omega t) \\hat{\\mathbf{k}} $
The electric field in an electromagnetic wave is given by $$\\overrightarrow{\\mathrm{E}}=\\hat{i} 40 \\cos \\omega(\\mathrm{t}-z / \\mathrm{c}) \\mathrm{NC}^{-1}$$. The magnetic field induction of this wave is (in SI unit) :
To determine the magnetic field induction of the given electromagnetic wave, we need to use the relationship between the electric field $$\\overrightarrow{\\mathrm{E}}$$ and the magnetic field $$\\overrightarrow{\\mathrm{B}}$$ in an electromagnetic wave. For an electromagnetic wave propagating in vacuum, the following relation holds:
The wave is propagating in the $$z$$-direction, so $$\\hat{\\mathrm{k}} = \\hat{z}$$. The unit vector $$\\hat{\\mathrm{i}}$$ represents the $$x$$-direction.
Average force exerted on a non-reflecting surface at normal incidence is $$2.4 \\times 10^{-4} \\mathrm{~N}$$. If $$360 \\mathrm{~W} / \\mathrm{cm}^2$$ is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:
To solve for the area of the surface, we need to understand the relationship between the force exerted by the light, the light energy flux, and the area of the surface. The pressure exerted by the light on a non-reflecting surface is given by the formula:
\n\n
\n\n
$$ P = \\frac{F}{A} $$
\n\n\n\n
where $$P$$ is the pressure, $$F$$ is the force, and $$A$$ is the area. The pressure due to the light can also be related to the energy flux $$I$$ by the relationship:
\n\n
\n\n
$$ P = \\frac{I}{c} $$
\n\n\n\n
where $$I$$ is the light energy flux and $$c$$ is the speed of light in a vacuum ($$3 \\times 10^8 \\mathrm{~m/s}$$).
\n\n
Given that the average force $$F$$ is $$2.4 \\times 10^{-4} \\mathrm{~N}$$ and the light energy flux $$I$$ is $$360 \\mathrm{~W/cm}^2$$, we first convert the flux to $$\\mathrm{W/m}^2$$:
In the given electromagnetic wave $$\\mathrm{E}_{\\mathrm{y}}=600 \\sin (\\omega t-\\mathrm{kx}) \\mathrm{Vm}^{-1}$$, intensity of the associated light beam is (in $$\\mathrm{W} / \\mathrm{m}^2$$ : (Given $$\\epsilon_0=9 \\times 10^{-12} \\mathrm{C}^2 \\mathrm{~N}^{-1} \\mathrm{~m}^{-2}$$ )
Electromagnetic waves travel in a medium with speed of $$1.5 \\times 10^8 \\mathrm{~m} \\mathrm{~s}^{-1}$$. The relative permeability of the medium is 2.0. The relative permittivity will be:
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9108,
"subject": "Physics",
"question": "Which of the following are not electromagnetic waves? ",
"options": [
{
"text": "cosmic rays "
},
{
"text": "gamma rays "
},
{
"text": "$$\\beta $$-rays "
},
{
"text": "$$X$$-rays "
}
],
"answer": "$$\\beta $$-rays ",
"solution": "**Answer:** $$\\beta $$-rays \n\n$$\\beta $$ -rays are fast moving beam of electrons.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9109,
"subject": "Physics",
"question": "Match the List - I (Phenomenon associated with electromagnetic radiation) with List - II (Part of electromagnetic spectrum) and select the correct code from the choices given the lists : \n
\n
\n
\n
List - I
\n
List - II
\n
\n
\n
(I)
\n
Doublet of sodium
\n
(A)
\n
Visible radiation
\n
\n
\n
(II)
\n
Wavelength corresponding to temperature associated with the isotropic radiation filling all space
\n
(B)
\n
Microwave
\n
\n
\n
(III
\n
Wavelength emitted by atomic hydrogen in interstellar space
\n
(C)
\n
Short radiowave
\n
\n
\n
(IV)
\n
Wavelength of radiation arising from two close energy levels in hydrogen
(I) These have a wavelength of 589 nm – 589.6 nm that corresponds to visible light in the yellow region.\nThe correct matching is (I)-(A), Visible radiation. The given wavelength range falls within the visible spectrum.
\n
(II) The temperature is 2.7 K, which corresponds to the emission of microwaves.\nThe correct matching is (II)-(B), Microwaves. The given temperature corresponds to the cosmic microwave background radiation.
\n
(III) The wavelength associated is 21 cm, which corresponds to radiowaves of short wavelength.\nThe correct matching is (III)-(C), Short radiowave. The given wavelength of 21 cm falls within the range of radiowaves.
\n
(IV) This radiation has a frequency of 1,057 MHz, which corresponds to radiowaves of high frequency or short wavelength.\nThe correct matching is (IV)-(B), Microwaves. The given frequency falls within the microwave range.
\n
Therefore, the correct answer is Option B : (I)-(A), (II)-(B), (III)-(C), (IV)-(B).
\n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9110,
"subject": "Physics",
"question": "If microwaves, X rays, infracted, gamma rays, ultra-violet, radio waves and visible parts of the electromagnetic spectrum by M, X, I, G, U, R and V, the following is the arrangement in ascending order of wavelength : ",
"options": [
{
"text": "R, M, I, V, U, X and G"
},
{
"text": "M, R, V, X, U, G and I"
},
{
"text": "G, X, U, V, I, M and R"
},
{
"text": "I, M, R, U, V, X and G"
}
],
"answer": "G, X, U, V, I, M and R",
"solution": "**Answer:** G, X, U, V, I, M and R\n\nThe arrangement in ascending order of wavelength will be\n
$$\n\\lambda_{\\mathrm{G}}<\\lambda_{\\mathrm{X}}<\\lambda_{\\mathrm{U}}<\\lambda_{\\mathrm{v}}<\\lambda_{\\mathrm{I}}<\\lambda_{\\mathrm{M}}<\\lambda_{\\mathrm{R}}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9111,
"subject": "Physics",
"question": "Microwave oven acts on the principle of :",
"options": [
{
"text": "transferring electrons from lower to higher energy levels in water molecule"
},
{
"text": "giving rotational energy to water molecules "
},
{
"text": "giving vibrational energy to water molecules"
},
{
"text": "giving translational energy to water molecules "
}
],
"answer": "giving vibrational energy to water molecules",
"solution": "**Answer:** giving vibrational energy to water molecules\n\nMicrowave over use the principle of giving vibrational energy to water molecule.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9112,
"subject": "Physics",
"question": "Arrange the following electromagnetic\nradiations per quantum in the order of\nincreasing energy :\n
A : Blue light B : Yellow light\n
C : X-ray D : Radiowave.\n",
"options": [
{
"text": "C, A, B, D"
},
{
"text": "B, A, D, C"
},
{
"text": "D, B, A, C"
},
{
"text": "A, B, D, C"
}
],
"answer": "D, B, A, C",
"solution": "**Answer:** D, B, A, C\n\n
Energy is in terms of increasing wavelength, we have
Radio wave > Microwave > Infrared > ROYGBIV(Visiable Resion) > Ultraviolet > X rays > $$\\gamma $$ rays\n
So the correct order -\n
$$\\lambda $$radio waves > $$\\lambda $$micro waves > $$\\lambda $$visible > $$\\lambda $$x-rays",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9114,
"subject": "Physics",
"question": "The correct match between the entries in\ncolumn I and column II are :\n
Radio wave > Microwave > Infrared > ROYGBIV(Visiable Resion) > Ultraviolet > X rays > $$\\gamma $$ rays\n
$$ \\therefore $$ Correct order is \n
(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9115,
"subject": "Physics",
"question": "Match List - I with List - II.
\n\n
\n
\n
List I
\n
\n
List II
\n
\n\n\n
\n
(a)
\n
Source of microwave frequency
\n
(i)
\n
Radioactive decay of nucleus
\n
\n
\n
(b)
\n
Source of infrared frequency
\n
(ii)
\n
Magnetron
\n
\n
\n
(c)
\n
Source of Gamma Rays
\n
(iii)
\n
Inner shell electrons
\n
\n
\n
(d)
\n
Source of X-rays
\n
(iv)
\n
Vibration of atoms and molecules
\n
\n
\n
\n
\n
(v)
\n
LASER
\n
\n
\n
\n
\n
(vi)
\n
RC circuit
\n
\n\n
Choose the correct answer from the options given below :",
"options": [
{
"text": "(a)-(vi), (b)-(v), (c)-(i), (d)-(iv)"
},
{
"text": "(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)"
},
{
"text": "(a)-(ii), (b)-(iv), (c)-(vi), (d)-(iii)"
},
{
"text": "(a)-(vi), (b)-(iv), (c)-(i), (d)-(v)"
}
],
"answer": "(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)",
"solution": "**Answer:** (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)\n\n(a) Source of microwave frequency - (ii) Magnetron
(b) Source of infra red frequency - (iv) Vibration of atom and molecules
(c) Source of gamma ray - (i) Radio active decay of nucleus
(d) Source of X-ray - (iii) inner shell electron",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9116,
"subject": "Physics",
"question": "The wavelength of an X-ray beam is 10$$\\mathop A\\limits^o $$. The mass of a fictitious particle having the same energy as that of the X-ray photons is $${x \\over 3}h$$ kg. The value of x is __________. (h = Planck's constant)",
"options": [],
"answer": "10",
"solution": "**Answer:** 10\n\nGiven wavelength of an x-ray beam = 10$$\\mathop A\\limits^o $$
Infrared wave - Improving visibility in foggy days.
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9120,
"subject": "Physics",
"question": "Match List I with List II
\n\n
\n\n
\n
LIST I
\n
LIST II
\n
\n\n\n
\n
A.
\n
Microwaves
\n
I.
\n
Physiotherapy
\n
\n
\n
B.
\n
UV rays
\n
II.
\n
Treatment of cancer
\n
\n
\n
C.
\n
Infra-red light
\n
III.
\n
Lasik eye surgery
\n
\n
\n
D.
\n
X-ray
\n
IV.
\n
Aircraft navigation
\n
\n\n
\nChoose the correct answer from the options given below:",
"options": [
{
"text": "A - IV, B - III, C - I, D - II"
},
{
"text": "A - II, B - IV, C - III, D - I"
},
{
"text": "A - III, B - II, C - I, D - IV"
},
{
"text": "A - IV, B - I, C - II, D - III"
}
],
"answer": "A - IV, B - III, C - I, D - II",
"solution": "**Answer:** A - IV, B - III, C - I, D - II\n\nA. Microwave → IV \n
3. Radio wave $\\rightarrow$ Rapid acceleration and deacceleration of electrons in aerials\n\n
4. X-ray $\\rightarrow$ Inner shell electrons",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9122,
"subject": "Physics",
"question": "Match List I with List II of Electromagnetic waves with corresponding wavelength range :
\n\n
\n\n
\n
List I
\n
List II
\n
\n\n\n
\n
(A) Microwave
\n
(I) $400 \\mathrm{~nm}$ to $1 \\mathrm{~nm}$
\n
\n
\n
(B) Ultraviolet
\n
(II) $1 \\mathrm{~nm}$ to $10^{-3} \\mathrm{~nm}$
\n
\n
\n
(C) X-Ray
\n
(III) $1 \\mathrm{~mm}$ to $700 \\mathrm{~nm}$
\n
\n
\n
(D) Infra-red
\n
(IV) $0.1 \\mathrm{~m}$ to $1 \\mathrm{~mm}$
\n
\n\n
\nChoose the correct answer from the options given below:",
"options": [
{
"text": "(A)-(I), (B)-(IV), (C)-(II), (D)-(III)"
},
{
"text": "(A)-(IV), (B)-(I), (C)-(II), (D)-(III)"
},
{
"text": "(A)-(IV), (B)-(I), (C)-(III), (D) -(II)"
},
{
"text": "(A)-(IV), (B)-(II), (C)-(I), (D)-(III)"
}
],
"answer": "(A)-(IV), (B)-(I), (C)-(II), (D)-(III)",
"solution": "**Answer:** (A)-(IV), (B)-(I), (C)-(II), (D)-(III)\n\nThe correct matching of the electromagnetic waves with their corresponding wavelength ranges is:\n
Given below are two statements: one is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$
\n
Assertion A : EM waves used for optical communication have longer wavelengths than that of microwave, employed in Radar technology.
\n
Reason R : Infrared EM waves are more energetic than microwaves, (used in Radar)
\n
In the light of given statements, choose the correct answer from the options given below.
",
"options": [
{
"text": "Both $$\\mathrm{A}$$ and $$\\mathrm{R}$$ are true but $$\\mathrm{R}$$ is NOT the correct explanation of $$\\mathrm{A}$$"
},
{
"text": "$$\\mathrm{A}$$ is true but $$\\mathrm{R}$$ is false"
},
{
"text": "Both $$\\mathrm{A}$$ and $$\\mathrm{R}$$ are true and $$\\mathrm{r}$$ is the correct explanation of $$\\mathrm{A}$$"
},
{
"text": "$$\\mathrm{A}$$ is false but $$\\mathrm{R}$$ is true"
}
],
"answer": "$$\\mathrm{A}$$ is false but $$\\mathrm{R}$$ is true",
"solution": "**Answer:** $$\\mathrm{A}$$ is false but $$\\mathrm{R}$$ is true\n\nOptical communication is performed in the frequency range of $1 \\mathrm{THz}$ to $1000 \\mathrm{THz}$.\n(Microwave to UV)
\nSo, EM waves used for optical communication have shorter wavelength than that of microwaves used in RADAR.
To match the given List I (EM-Wave) with List II (Wavelength Range), we need to understand the typical wavelength ranges associated with each type of electromagnetic (EM) wave mentioned in List I.
\n\n
Here's the matching based on known wavelength ranges:
\n\n
\n
Infra-red: Infrared radiation lies between the visible and microwave portions of the electromagnetic spectrum. The wavelength range for infrared is typically from about 700 nm (nanometers) to 1 mm (millimeter), which means it matches with option (III) - 1 mm to 700 nm.
\n
Ultraviolet: Ultraviolet (UV) light sits between visible light and X-rays along the electromagnetic spectrum, with wavelengths shorter than visible light but longer than X-rays. The typical wavelength range for UV light is from 400 nm down to 10 nm, so it matches with option (II) - 400 nm to 1 nm (even though the strict lower bound should be 10 nm, this is the closest option provided).
\n
X-rays: X-rays have a wavelength range shorter than UV rays and longer than gamma rays, typically ranging from 10 nm down to 0.01 nm (or $$10^{-2}$$ nm). Thus, X-rays match with option (IV) - 1 nm to $$10^{-3}$$ nm, as this is the closest range provided that fits the typical scope of X-ray wavelengths.
\n
Gamma rays: Gamma rays have the shortest wavelength on the electromagnetic spectrum, with wavelength less than 10 picometers (pm), which equates to less than $$10^{-3}$$ nm. This corresponds to option (I) - $$<10^{-3}$$ nm.
\n
\n
So, based on the above analysis, the correct option for the matching would be:
\n\n
\n
(A)-(III)
\n
(B)-(II)
\n
(C)-(IV)
\n
(D)-(I)
\n
\n
Therefore, the correct answer is Option B: (A)-(III), (B)-(II), (C)-(IV), (D)-(I).
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9126,
"subject": "Physics",
"question": "The truth table given in fig. represents :\n \n
(1) A = 0 and B = 1 $$\\left( {\\overline A B} \\right)$$\n
(2) A = 1 and B = 0 $$\\left( {A\\,\\overline B } \\right)$$\n
(3) A = 1 and B = 1 $$\\left( {A\\,B} \\right)$$\n
$$ \\therefore $$ Y = $${\\overline A }$$ B + A $${\\overline B }$$ + AB\n
= $${\\overline A }$$ B + A ($${\\overline B }$$ + B)\n
= $${\\overline A }$$ B + A [as $${\\overline B }$$ + B = 1]\n
= (A + $${\\overline A }$$) (A + B)\n
= A + B\n
So, this represented OR gate.\n
Note :\n
X + $${\\overline X }$$ Y = (X + $${\\overline X }$$) (X + Y)\n
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9127,
"subject": "Physics",
"question": "At absolute zero, Si acts as ",
"options": [
{
"text": "non-metal "
},
{
"text": "metal "
},
{
"text": "insulator "
},
{
"text": "none of these "
}
],
"answer": "insulator ",
"solution": "**Answer:** insulator \n\nPure silicon, at absolute zero, will contain all the electrons in bounded state. The conduction band will be empty. So there will be no free electrons (in conduction band) and holes (in valence band) due to thermal agitation. Pure silicon will act as insulator.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9128,
"subject": "Physics",
"question": "By increasing the temperature, the specific resistance of a conductor and a semiconductor ",
"options": [
{
"text": "increases for both "
},
{
"text": "decreases for both "
},
{
"text": "increases, decreases "
},
{
"text": "decreases, increases "
}
],
"answer": "increases, decreases ",
"solution": "**Answer:** increases, decreases \n\nSpecific resistance is resistivity which is given by \n
$$\\rho = {m \\over {m{e^2}\\,\\tau }}$$\n
where $$n=no.$$ of free electrons per unit volume and $$\\tau $$ $$=$$ average relaxation time\n
For a conductor with rise in temperature $$n$$ increases and $$\\tau $$ decreases. But decrease in $$\\tau $$ is more dominant than increase in $$n$$ resulting an increase in the value of $$\\rho $$\n
For a semiconductor with rise in temperature, $$n$$ increases and $$\\tau $$ decreases. But the increase in $$n$$ is more dominant than decrease in $$\\tau $$ resulting in decrease in the value of $$\\rho .$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9129,
"subject": "Physics",
"question": "The energy band gap is maximum in ",
"options": [
{
"text": "metals "
},
{
"text": "superconductors "
},
{
"text": "insulator "
},
{
"text": "semiconductor "
}
],
"answer": "insulator ",
"solution": "**Answer:** insulator \n\nThe energy band gap is maximum in insulators.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9130,
"subject": "Physics",
"question": "A strip of copper and another of germanium are cooled from room temperature to $$80K.$$ The resistance of ",
"options": [
{
"text": "each of these decreases "
},
{
"text": "copper strip increases and that of germanium decreases "
},
{
"text": "copper strip decreases and that of germanium increases "
},
{
"text": "each of these increases "
}
],
"answer": "copper strip decreases and that of germanium increases ",
"solution": "**Answer:** copper strip decreases and that of germanium increases \n\nThe resistance of metal (like $$Cu$$) decreases with decrease in temperature whereas the resistance of a semi-conductor (like $$Ge$$) increases with decrease in temperature.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9131,
"subject": "Physics",
"question": "The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the ",
"options": [
{
"text": "crystal structure "
},
{
"text": "variation of the number of charge carries with temperature "
},
{
"text": "type of bonding "
},
{
"text": "variation of scattering mechanism with temperature "
}
],
"answer": "variation of the number of charge carries with temperature ",
"solution": "**Answer:** variation of the number of charge carries with temperature \n\nWhen the temperature increases, certain bounded electrons become free which tend to promote conductivity. Simultaneously number of collisions between electrons and positive kernels increases ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9132,
"subject": "Physics",
"question": "In the middle of the depletion layer of a reverse- biased $$p$$-$$n$$ junction, the ",
"options": [
{
"text": "electric field is zero "
},
{
"text": "potential is maximum "
},
{
"text": "electric field is maximum "
},
{
"text": "potential is zero "
}
],
"answer": "electric field is zero ",
"solution": "**Answer:** electric field is zero \n\nAs in reverse bias, the current through the $$0000$$ is zero through the electric field is also zero.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9133,
"subject": "Physics",
"question": "A piece of copper and another of germanium are cooled from room temperature to $$77K,$$ the resistance of ",
"options": [
{
"text": "copper increases and germanium decreases "
},
{
"text": "each of them decreases "
},
{
"text": "each of them increases "
},
{
"text": "copper decreases and germanium increases "
}
],
"answer": "copper decreases and germanium increases ",
"solution": "**Answer:** copper decreases and germanium increases \n\nCopper is a conductor, so its resistance decreases on decreasing temperature as thermal agitation decreases; whereas germanium is semiconductor therefore on decreasing temperature resistance increases.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9134,
"subject": "Physics",
"question": "The manifestation of band structure in solids is due to ",
"options": [
{
"text": "Bohr's correspondence principle "
},
{
"text": "Pauli's exclusion principle "
},
{
"text": "Heisenberg's uncertainty principle "
},
{
"text": "Boltzmann's law"
}
],
"answer": "Pauli's exclusion principle ",
"solution": "**Answer:** Pauli's exclusion principle \n\nPauli's exclusion principle. ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9135,
"subject": "Physics",
"question": "When $$p$$-$$n$$ junction diode is forward biased then ",
"options": [
{
"text": "both the depletion region and barrier height are reduced "
},
{
"text": "the depletion region is widened and barrier height is reduced "
},
{
"text": "the depletion region is reduced and barrier height is increased "
},
{
"text": "both the depletion region and barrier height are increased "
}
],
"answer": "both the depletion region and barrier height are reduced ",
"solution": "**Answer:** both the depletion region and barrier height are reduced \n\nBoth the depletion region and barrier height is reduced.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9136,
"subject": "Physics",
"question": "The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than $$2480$$ $$nm$$ is incident on it. The band gap in $$(eV)$$ for the semiconductor is ",
"options": [
{
"text": "$$2.5$$ $$eV$$ "
},
{
"text": "$$1.1$$ $$eV$$ "
},
{
"text": "$$0.7$$ $$eV$$ "
},
{
"text": "$$0.5$$ $$eV$$ "
}
],
"answer": "$$0.5$$ $$eV$$ ",
"solution": "**Answer:** $$0.5$$ $$eV$$ \n\nBand gap $$=$$ energy of photon of wavelength $$2480$$ $$nm.$$ So,\n
$$ = 0.5\\,eV$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9137,
"subject": "Physics",
"question": "In a full wave rectifier circuit operating from $$50$$ $$Hz$$ mains frequency, the fundamental frequency in the ripple would be ",
"options": [
{
"text": "$$25$$ $$Hz$$ "
},
{
"text": "$$50$$ $$Hz$$ "
},
{
"text": "$$70.7$$ $$Hz$$ "
},
{
"text": "$$100$$ $$Hz$$"
}
],
"answer": "$$100$$ $$Hz$$",
"solution": "**Answer:** $$100$$ $$Hz$$\n\nInput frequency, $$f = 50\\,Hz \\Rightarrow T = {1 \\over {50}}$$\n
For full wave rectifier, $${T_1} = {T \\over 2} = {1 \\over {100}} \\Rightarrow {f_1} = 100\\,Hz.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9138,
"subject": "Physics",
"question": "If the ratio of the concentration of electrons to that of holes in a semiconductor is $${7 \\over 5}$$ and the ratio of currents is $${7 \\over 4},$$ then what is the ratio of their drift velocities? ",
"options": [
{
"text": "$${5 \\over 8}$$ "
},
{
"text": "$${4 \\over 5}$$"
},
{
"text": "$${5 \\over 4}$$"
},
{
"text": "$${4 \\over 7}$$"
}
],
"answer": "$${5 \\over 4}$$",
"solution": "**Answer:** $${5 \\over 4}$$\n\n$${{{I_e}} \\over {{I_h}}} = {{n{}_eeA{v_e}} \\over {n{}_heA{v_h}}}$$\n
$$ \\Rightarrow {{{v_e}} \\over {{v_h}}} = {5 \\over 4}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9139,
"subject": "Physics",
"question": "A solid which is not transparent to visible light and whose conductivity increases with temperature is formed by ",
"options": [
{
"text": "Ionic bonding "
},
{
"text": "Covalent bonding"
},
{
"text": "Vander Waals bonding "
},
{
"text": "Metallic bonding "
}
],
"answer": "Covalent bonding",
"solution": "**Answer:** Covalent bonding\n\nVan der Waal's bonding is attributed to the attractive forces between molecules of a liquid. The conductivity of semiconductors (covalent bonding) and insulators (ionic bonding) increases with increase in temperature while that of metals (metallic bonding) decreases. ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9140,
"subject": "Physics",
"question": "Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate ? ",
"options": [
{
"text": "The number of free electrons for conduction is significant only in $$Si$$ and $$Ge$$ but small in $$C.$$ "
},
{
"text": "The number of free conduction electrons is significant in $$C$$ but small in $$Si$$ and $$Ge.$$ "
},
{
"text": "The number of free conduction electrons is negligibly small in all the three."
},
{
"text": "The number of free electrons for conduction is significant in all the three "
}
],
"answer": "The number of free electrons for conduction is significant only in $$Si$$ and $$Ge$$ but small in $$C.$$ ",
"solution": "**Answer:** The number of free electrons for conduction is significant only in $$Si$$ and $$Ge$$ but small in $$C.$$ \n\n$$Si$$ and $$Ge$$ are semiconductors but $$C$$ is an insulator. Also, the conductivity of $$Si$$ and $$Ge$$ is more than $$C$$ because the valence electrons of $$Si, Ge$$ and $$C$$ lie in third, fourth and second orbit respectively. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9141,
"subject": "Physics",
"question": "The current voltage relation of diode is given by $${\\rm I} = \\left( {{e^{100V/T}} - 1} \\right)mA,$$ where the applied voltage $$V$$ is in volts and the temperature $$T$$ is in degree kelvin. If a student makes an error measuring $$ \\pm 0.01\\,V$$ while measuring the current of $$5$$ $$mA$$ at $$300$$ $$K,$$ what will be the error in the value of current on $$mA$$? ",
"options": [
{
"text": "$$0.2$$ $$mA$$ "
},
{
"text": "$$0.02$$ $$mA$$ "
},
{
"text": "$$0.5$$ $$mA$$ "
},
{
"text": "$$0.05$$ $$mA$$ "
}
],
"answer": "$$0.2$$ $$mA$$ ",
"solution": "**Answer:** $$0.2$$ $$mA$$ \n\nThe current voltage relation of diode is \n
$$=0.2$$ $$mA$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9142,
"subject": "Physics",
"question": "A red $$LED$$ emits light at $$0.1$$ watt uniformly around it. The amplitude of the electric field of the light at a distance of $$1$$ $$m$$ from the diode is : ",
"options": [
{
"text": "$$5.48$$ $$V/m$$ "
},
{
"text": "$$7.75$$ $$V/m$$ "
},
{
"text": "$$1.73$$ $$V/m$$ "
},
{
"text": "$$2.45$$ $$V/m$$ "
}
],
"answer": "$$2.45$$ $$V/m$$ ",
"solution": "**Answer:** $$2.45$$ $$V/m$$ \n\n
Intensity of light at a distance r, $$I = {P \\over {4\\pi {r^2}}}$$ [P = power]
\n
Again, if the amplitude of the electric field is E0
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9143,
"subject": "Physics",
"question": "The temperature dependence of resistance of $$Cu$$ and undoped $$Si$$ in the temperature range $$300-400$$ $$K,$$ is best described by : ",
"options": [
{
"text": "Linear increases for $$Cu,$$ exponential decrease of $$Si.$$ "
},
{
"text": "Linear decrease for $$Cu,$$ linear decrease for $$Si$$"
},
{
"text": "Linear increase for $$Cu,$$ linear increase for $$Si.$$ "
},
{
"text": "Linear increase for $$Cu,$$ exponential increase for $$Si$$"
}
],
"answer": "Linear increases for $$Cu,$$ exponential decrease of $$Si.$$ ",
"solution": "**Answer:** Linear increases for $$Cu,$$ exponential decrease of $$Si.$$ \n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9144,
"subject": "Physics",
"question": "An experiment is performed to determine the I - V characteristics of a Zener diode, which has a protective resistance of R = 100 $$\\Omega $$, and a maximum power of dissipation rating of 1 W. The minimum voltage range of the DC source in the circuit is :",
"options": [
{
"text": "0 $$-$$ 5 V "
},
{
"text": "0 $$-$$ 8 V "
},
{
"text": "0 $$-$$ 12 V "
},
{
"text": "0 $$-$$ 24 V "
}
],
"answer": "0 $$-$$ 24 V ",
"solution": "**Answer:** 0 $$-$$ 24 V \n\n\n
Potential drop accross zener diode.\n
Vz = V $$-$$ 100 I\n
$$ \\therefore $$ Power dissiption = Vz I\n
= (V $$-$$ 100 I) I\n
Given that, \n
(V $$-$$ 100 I) I = 1\n
$$ \\Rightarrow $$ VI $$-$$ 100 I2 = 1\n
$$ \\Rightarrow $$ 100 I2 $$-$$ VI + 1 = 0\n
As As I is real, \n
So, b2 $$-$$ 4ac $$ \\ge $$ far this quadratic equation. \n
$$ \\therefore $$ V2 $$-$$ 4(100)1 $$ \\ge $$ 0\n
$$ \\Rightarrow $$ V $$ \\ge $$ 20 V\n
$$ \\therefore $$ Voltage range should be 0 $$-$$ 24 V",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9145,
"subject": "Physics",
"question": "What is the conductivity of a semiconductor sale having electron concentration of $$5 \\times {10^{18}}\\,\\,{m^{ - 3}},$$ hole concentration of $$5 \\times {10^{19}}\\,\\,{m^{ - 3}},$$ electron mobility of 2.0 m2 V$$-$$1 s-1 and hole mobility of 0.01 m2 V$$-$$1 s$$-$$1 ? \n
= 1.68",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9146,
"subject": "Physics",
"question": "Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019 m$$-$$3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to : ",
"options": [
{
"text": "$$2\\,\\Omega $$m"
},
{
"text": "4$$\\,\\Omega $$m"
},
{
"text": "0.4 $$\\,\\Omega $$m"
},
{
"text": "0.2 $$\\,\\Omega $$m"
}
],
"answer": "0.4 $$\\,\\Omega $$m",
"solution": "**Answer:** 0.4 $$\\,\\Omega $$m\n\nFor semiconductor, \n
= 0.4 $$\\Omega $$ m",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9147,
"subject": "Physics",
"question": "With increasing biasing voltage of a photodiode,\nthe photocurrent magnitude :",
"options": [
{
"text": "Increases initially and after attaining\ncertain value, it decreases"
},
{
"text": "Increases linearly"
},
{
"text": "Increases initially and saturates finally"
},
{
"text": "Remains constant"
}
],
"answer": "Increases initially and saturates finally",
"solution": "**Answer:** Increases initially and saturates finally\n\nIn photodiode, photocurrent increases with\nincreasing biasing voltage and then becomes\nsaturated.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9148,
"subject": "Physics",
"question": "Which of the following will NOT be observed when a multimeter (operating in resistance measuring\nmode) probes connected across a component, are just reversed?",
"options": [
{
"text": "Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the\nchosen component is metal wire."
},
{
"text": "Multimeter shows a deflection, accompanied by a splash of light out of connected component in one direction and NO deflection on reversing the probes if the chosen component is LED."
},
{
"text": "Multimeter shows an equal deflection in both cases i.e. before and after reversing the probes if\nthe chosen component is resistor."
},
{
"text": "Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the\nchosen component is capacitor."
}
],
"answer": "Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the\nchosen component is capacitor.",
"solution": "**Answer:** Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the\nchosen component is capacitor.\n\n(1) Multimeter shows deflection when it connects with capacitor.\n
(2) If we assume that LED has negligible resistance then multimeter shows no deflection for the forward bias but when it connects in reverse direction, it break down occurs so splash of light out.\n
(3) The resistance of metal wire may be taken zero, so no deflection in multimeter.\n
(4) No matter, how we connect the resistance across multimeter. It shows same deflection.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9149,
"subject": "Physics",
"question": "If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm, then\nits band gap energy is :\n
= 3.1 eV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9150,
"subject": "Physics",
"question": "When a diode is forward biased, it has a\nvoltage drop of 0.5 V. The safe limit of current\nthrough the diode is 10 mA. If a battery of emf\n1.5 V is used in the circuit, the value of\nminimum resistance to be connected in series\nwith the diode so that the current does not\nexceed the safe limit is",
"options": [
{
"text": "50 $$\\Omega $$"
},
{
"text": "200 $$\\Omega $$"
},
{
"text": "300 $$\\Omega $$"
},
{
"text": "100 $$\\Omega $$"
}
],
"answer": "100 $$\\Omega $$",
"solution": "**Answer:** 100 $$\\Omega $$\n\n\n
1.5 – 0.5 – R $$ \\times $$ 10–2\n = 0\n
$$ \\Rightarrow $$ R = 100 $$\\Omega $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9151,
"subject": "Physics",
"question": "Zener breakdown occurs in a p $$-$$ n junction having p and n both :",
"options": [
{
"text": "heavily doped and have wide depletion layer."
},
{
"text": "lightly doped and have narrow depletion layer."
},
{
"text": "heavily doped and have narrow depletion layer."
},
{
"text": "lightly doped and have wide depletion layer."
}
],
"answer": "heavily doped and have narrow depletion layer.",
"solution": "**Answer:** heavily doped and have narrow depletion layer.\n\nThe zener breakdown occurs in the heavily doped p-n junction diode. Heavily doped p-n junction diodes have narrow depletion region.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9152,
"subject": "Physics",
"question": "For extrinsic semiconductors; when doping level is increased;",
"options": [
{
"text": "Fermi-level of both p-type and n-type semiconductors will go upward for T > TFK and downward for T < TFK, where TF is Fermi temperature."
},
{
"text": "Fermi-level of p-type semiconductor will go upward and Fermi-level of n-type semiconductors will go downward"
},
{
"text": "Fermi-level of p and n-type semiconductors will not be affected."
},
{
"text": "Fermi-level of p-type semiconductors will go downward and Fermi-level of n-type semiconductor will go upward."
}
],
"answer": "Fermi-level of p-type semiconductors will go downward and Fermi-level of n-type semiconductor will go upward.",
"solution": "**Answer:** Fermi-level of p-type semiconductors will go downward and Fermi-level of n-type semiconductor will go upward.\n\nIn n-type semiconductor pentavalent impurity is added. Each pentavalent impurity donates a free electron. So the Fermi-level of n-type semiconductor will go upward.
And In p-type semiconductor trivalent impurity is added. Each trivalent impurity creates a hole in the valence band. So the Fermi-level of p-type semiconductor will go downward.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9153,
"subject": "Physics",
"question": "LED is constructed from Ga-As-P semiconducting material. The energy gap of this LED is 1.9 eV. Calculate the wavelength of light emitted and its colour.
[h = 6.63 $$\\times$$ 10$$-$$34 Js and c = 3 $$\\times$$ 108 ms$$-$$1]",
"options": [
{
"text": "654 nm and orange colour"
},
{
"text": "654 nm and red colour"
},
{
"text": "1046 nm and red colour"
},
{
"text": "1046 nm and blue colour"
}
],
"answer": "654 nm and red colour",
"solution": "**Answer:** 654 nm and red colour\n\nWe know that $$E = {{hc} \\over \\lambda }$$
Wavelength of red light is 620 nm to 750 nm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9154,
"subject": "Physics",
"question": "Consider a situation in which reserve biased current of a particular P-N junction increases when it is exposed to a light of wavelength $$\\le$$ 621 nm. During this process, enhancement in carrier concentration takes place due to generation of hole-electron pairs. The value of band gap is nearly.",
"options": [
{
"text": "1 eV"
},
{
"text": "4 eV"
},
{
"text": "0.5 eV"
},
{
"text": "2 eV"
}
],
"answer": "2 eV",
"solution": "**Answer:** 2 eV\n\nBand gap = $${{hc} \\over \\lambda } = 2$$ eV",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9155,
"subject": "Physics",
"question": "In a semiconductor, the number density of intrinsic charge carries at 27$$^\\circ$$C is 1.5 $$\\times$$ 1016/m3. If the semiconductor is doped with impurity atom, the hole density increases to 4.5 $$\\times$$ 1022/m3. The electron density in the doped semiconductor is ___________ $$\\times$$ 109/m3.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n$${n_e}{n_h} = {n_i}^2$$
$$ = 5 \\times {10^9}$$/m3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9156,
"subject": "Physics",
"question": "Statement I : By doping silicon semiconductor with pentavalent material, the electrons density increases.
Statement II : The n-type semiconductor has net negative charge.
In the light of the above statements, choose the most appropriate answer from the options given below :",
"options": [
{
"text": "Statement - I is true but Statement - II is false."
},
{
"text": "Statement - I is false but Statement - II is true."
},
{
"text": "Both Statement I and Statement II are true."
},
{
"text": "Both Statement I and Statement II are false."
}
],
"answer": "Statement - I is true but Statement - II is false.",
"solution": "**Answer:** Statement - I is true but Statement - II is false.\n\nPentavalent activities have excess free e$$-$$, so e$$-$$ density increases but overall semiconductor is neutral.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9157,
"subject": "Physics",
"question": "Statement - I :
To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect a capacitor across the output parallel to the load RL.
Statement - II :
To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect an inductor in series with RL.
In the light of the above statements, choose the most appropriate answer from the options given below :",
"options": [
{
"text": "Statement I is true but Statement II is false"
},
{
"text": "Statement I is false but Statement II is true"
},
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Both Statement I and Statement II are true"
}
],
"answer": "Both Statement I and Statement II are true",
"solution": "**Answer:** Both Statement I and Statement II are true\n\nTo convert pulsating dc into steady dc both of mentioned method are correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9158,
"subject": "Physics",
"question": "
A potential barrier of 0.4 V exists across a p-n junction. An electron enters the junction from the n-side with a speed of 6.0 $$\\times$$ 105 ms$$-$$1. The speed with which electron enters the p side will be $${x \\over 3} \\times {10^5}$$ ms$$-$$1 the value of x is _____________.
\n
(Given mass of electron = 9 $$\\times$$ 10$$-$$31 kg, charge on electron = 1.6 $$\\times$$ 10$$-$$19 C.)
For using a multimeter to identify diode from electrical components, choose the correct statement out of the following about the diode :
",
"options": [
{
"text": "It is two terminal device which conducts current in both directions."
},
{
"text": "It is two terminal device which conducts current in one direction only."
},
{
"text": "It does not conduct current gives an initial deflection which decays to zero."
},
{
"text": "It is three terminal device which conducts current in one direction only between central terminal and either of the remaining two terminals"
}
],
"answer": "It is two terminal device which conducts current in one direction only.",
"solution": "**Answer:** It is two terminal device which conducts current in one direction only.\n\n
A diode is a two terminal device which conducts current in forward bias only.
The photodiode is used to detect the optical signals. These diodes are preferably operated in reverse biased mode because :
",
"options": [
{
"text": "fractional change in majority carriers produce higher forward bias current"
},
{
"text": "fractional change in majority carriers produce higher reverse bias current"
},
{
"text": "fractional change in minority carriers produce higher forward bias current"
},
{
"text": "fractional change in minority carriers produce higher reverse bias current"
}
],
"answer": "fractional change in minority carriers produce higher reverse bias current",
"solution": "**Answer:** fractional change in minority carriers produce higher reverse bias current\n\n
A photodiode is reverse biased. When light falling on it produces charge carriers, the fractional change, in minority carriers is high since the original current is very small.
The energy band gap of semiconducting material to produce violet (wavelength = 4000$$\\mathop A\\limits^o $$ ) LED is ______________ $$\\mathrm{eV}$$. (Round off to the nearest integer).
If the potential barrier across a p-n junction is $$0.6 \\mathrm{~V}$$. Then the electric field intensity, in the depletion region having the width of $$6 \\times 10^{-6} \\mathrm{~m}$$, will be __________ $$\\times 10^{5} \\mathrm{~N} / \\mathrm{C}$$.
",
"options": [
{
"text": "It works as a voltage regulator in reverse bias and behaves like simple pn junction diode in forward bias. "
},
{
"text": "It works as a voltage regulator in both forward and reverse bias."
},
{
"text": "It works as a voltage regulator only in forward bias."
},
{
"text": "It works as a voltage regulator in forward bias and behaves like simple pn junction diode in reverse bias."
}
],
"answer": "It works as a voltage regulator in reverse bias and behaves like simple pn junction diode in forward bias. ",
"solution": "**Answer:** It works as a voltage regulator in reverse bias and behaves like simple pn junction diode in forward bias. \n\nOption A is the correct statement about Zener diode. It works as a voltage regulator in reverse bias and behaves like a simple pn junction diode in forward bias. When a Zener diode is reverse-biased, it operates in the breakdown region, where a relatively constant voltage is maintained across the diode, regardless of the current flowing through it. This property makes it useful as a voltage regulator. In forward bias, the voltage applied across the diode is in the same direction as the normal direction of current flow. In this condition, the Zener diode behaves like a simple pn junction diode and allows current to flow in the forward direction.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9164,
"subject": "Physics",
"question": "
Match List I with List II:
\n
\n
\n
\n
\n
\n
\n
\n
\n\n
\n
\n
List I
\n
\n
List II
\n
\n\n\n
\n
A.
\n
Intrinsic semiconductor
\n
I.
\n
Fermi-level near the valence bond
\n
\n
\n
B.
\n
n-type semiconductor
\n
II.
\n
Fermi-level in the middle of valence and conduction band.
\n
\n
\n
C.
\n
p-type semiconductor
\n
III.
\n
Fermi-level near the conduction band
\n
\n
\n
D.
\n
Metals
\n
IV.
\n
Fermi-level inside the conduction band
\n
\n\n
\n
Choose the correct answer from the options given below :
Which one of the following statement is not correct in the case of light emitting diodes?
\n
A. It is a heavily doped p-n junction.
\n
B. It emits light only when it is forward biased.
\n
C. It emits light only when it is reverse biased.
\n
D. The energy of the light emitted is equal to or slightly less than the energy gap of the semiconductor used.
\n
Choose the correct answer from the options given below:
",
"options": [
{
"text": "A"
},
{
"text": "B"
},
{
"text": "C and D"
},
{
"text": "C"
}
],
"answer": "C",
"solution": "**Answer:** C\n\nThe correct answer is C. It is not correct that a light-emitting diode (LED) emits light only when it is reverse biased. In fact, an LED emits light only when it is forward biased, which is stated correctly in option B.\n\n
Option A is also correct, as an LED is indeed a heavily doped p-n junction. Option D is also correct, as the energy of the light emitted by an LED is equal to or slightly less than the energy gap of the semiconductor material used in the device.\n\n
Therefore, the statement that is not correct in the case of light-emitting diodes is C.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9167,
"subject": "Physics",
"question": "
Statement I : When a Si sample is doped with Boron, it becomes P type and when doped by Arsenic it becomes N-type semi conductor such that P-type has excess holes and N-type has excess electrons.
\n
Statement II : When such P-type and N-type semi-conductors, are fused to make a junction, a current will automatically flow which can be detected with an externally connected ameter.
\n
In the light of above statements, choose the most appropriate answer from the options given below
",
"options": [
{
"text": "Statement I is incorrect but statement II is correct"
},
{
"text": "Both Statement I and statement II are correct"
},
{
"text": "Statement I is correct but statement II is incorrect"
},
{
"text": "Both Statement I and Statement II are incorrect"
}
],
"answer": "Statement I is correct but statement II is incorrect",
"solution": "**Answer:** Statement I is correct but statement II is incorrect\n\n
Statement I is correct but in statement II we cannot\ndetect the current through ammeter thus the\nstatement II is incorrect.
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R
\n
Assertion A : Photodiodes are used in forward bias usually for measuring the light intensity.
\n
Reason R : For a p-n junction diode, at applied voltage V the current in the forward bias is more than the current in the reverse bias for $$\\mathrm{|{V_z}| > \\pm v \\ge |{v_0}|}$$ where $$\\mathrm{v_0}$$ is the threshold voltage and $$\\mathrm{V_z}$$ is the breakdown voltage.
\n
In the light of the above statements, choose the correct answer from the options given below
",
"options": [
{
"text": "Both A and R are true but R is NOT the correct explanation of A"
},
{
"text": "A is true but R is false"
},
{
"text": "Both A and R are true and R is the correct explanation of A"
},
{
"text": "A is false but R is true"
}
],
"answer": "A is false but R is true",
"solution": "**Answer:** A is false but R is true\n\nPhotodiodes are used in reverse bias therefore the assertion is incorrect.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9169,
"subject": "Physics",
"question": "
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R
\n
Assertion A : Photodiodes are preferably operated in reverse bias condition for light intensity measurement.
\n
Reason R : The current in the forward bias is more than the current in the reverse bias for a $$p-n$$ junction diode.
\n
In the light of the above statements, choose the correct answer from the options given below :
",
"options": [
{
"text": "Both A and R are true and R is the correct explanation of A"
},
{
"text": "A is false but R is true"
},
{
"text": "A is true but R is false"
},
{
"text": "Both A and R are true but R is NOT the correct explanation of A"
}
],
"answer": "Both A and R are true but R is NOT the correct explanation of A",
"solution": "**Answer:** Both A and R are true but R is NOT the correct explanation of A\n\nPhotodiodes are preferably operated in reverse\nbias condition for light intensity measurement\nbecause it increases the width of depletion layer,\ntherefore both are correct but not the correct\nexplanation.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9170,
"subject": "Physics",
"question": "
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
\n
Assertion A: Diffusion current in a p-n junction is greater than the drift current in magnitude if the junction is forward biased.
\n
Reason R: Diffusion current in a p-n junction is from the $$\\mathrm{n}$$-side to the p-side if the junction is forward biased.
\n
In the light of the above statements, choose the most appropriate answer from the options given below
",
"options": [
{
"text": "Both A and R are correct but R is NOT the correct explanation of A"
},
{
"text": "A is correct but R is not correct"
},
{
"text": "A is not correct but R is correct"
},
{
"text": "Both A and R are correct and R is the correct explanation of A"
}
],
"answer": "A is correct but R is not correct",
"solution": "**Answer:** A is correct but R is not correct\n\nA p-n junction consists of a p-type semiconductor (which has an excess of holes) in contact with an n-type semiconductor (which has an excess of electrons). In a forward-biased p-n junction, an external voltage is applied such that the positive terminal is connected to the p-side and the negative terminal is connected to the n-side. This configuration promotes the flow of majority charge carriers (holes from the p-side and electrons from the n-side) across the junction.\n
\nThere are two types of currents in a p-n junction: drift current and diffusion current. Drift current is caused by the electric field due to the built-in potential, which opposes the flow of majority charge carriers. Diffusion current is caused by the concentration gradient of the charge carriers, which promotes the flow of majority charge carriers.\n
\nWhen the p-n junction is forward biased, the applied voltage reduces the potential barrier, allowing more majority charge carriers to flow across the junction. This results in an increase in the diffusion current. In a forward-biased p-n junction, the diffusion current is indeed greater than the drift current in magnitude, so Assertion A is correct.\n
\nRegarding Reason R: The diffusion current in a p-n junction is actually from the p-side to the n-side, as holes (majority charge carriers in the p-side) move from the p-side to the n-side, and electrons (majority charge carriers in the n-side) move from the n-side to the p-side. Therefore, Reason R is incorrect.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9171,
"subject": "Physics",
"question": "
A light emitting diode (LED) is fabricated using GaAs semiconducting material whose band gap is $$1.42 \\mathrm{~eV}$$. The wavelength of light emitted from the LED is :
The wavelength of light emitted by a Light Emitting Diode (LED) fabricated using a semiconducting material can be determined by the energy band gap of the material. The energy of the photon emitted, which corresponds to the band gap energy, is given by the equation:
\n\n
$$E = \\frac{hc}{\\lambda}$$
\n\n
Where:
\n\n
\n\n
$E$ is the energy of the emitted photon (in joules when $h$ and $c$ are in SI units), corresponding to the band gap energy of the material.
\n\n
$h$ is Planck's constant ($6.626 \\times 10^{-34} \\, \\mathrm{m^2\\,kg/s}$).
\n\n
$c$ is the speed of light in vacuum ($3.0 \\times 10^{8} \\,\\mathrm{m/s}$).
\n\n
$\\lambda$ is the wavelength of the emitted light (in meters).
\n\n
\n\n
However, since the energy band gap given is in electronvolts (eV), and we are looking for the wavelength in nanometers (nm), we can use the energy formula directly in terms of eV and then do the unit conversion conveniently. The conversion between energy (in eV) and wavelength (in nm) without needing to convert eV to Joules is facilitated by the equation:
The acceptor level of a p-type semiconductor is $$6 \\mathrm{~eV}$$. The maximum wavelength of light which can create a hole would be : Given $$\\mathrm{hc}=1242 \\mathrm{~eV} \\mathrm{~nm}$$.
The energy required to create a hole in a p-type semiconductor can be directly related to the acceptor level because this energy level represents the minimum energy required to excite an electron from the valence band into the acceptor level, effectively creating a hole. The acceptor level is given as $$6 \\, \\text{eV}$$.
\n\n
To find the maximum wavelength of light that can excite an electron into this level, we use the equation that relates the energy ($$E$$) of a photon to its wavelength ($$\\lambda$$):
\n\n
$$E = \\frac{hc}{\\lambda}$$
\n\n
Where:
\n\n
\n\n
$$E$$ is the energy in electronvolts (eV),
\n\n
$$h$$ is Planck's constant,
\n\n
$$c$$ is the speed of light, and
\n\n
$$\\lambda$$ is the wavelength in nanometers (nm).
\n\n
\n\n
The product of $$hc$$ is given as $$1242 \\, \\text{eV nm}$$, allowing us to solve for $$\\lambda$$ directly:
Thus, the maximum wavelength of light that can create a hole in the semiconductor by exciting an electron into the acceptor level is $$207 \\, \\text{nm}$$. Therefore, the correct answer is Option D: 207 nm.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9173,
"subject": "Physics",
"question": "The part of a transistor which is most heavily doped to produce large number of majority carriers is ",
"options": [
{
"text": "emmiter "
},
{
"text": "base "
},
{
"text": "collector "
},
{
"text": "can be any of the above three"
}
],
"answer": "emmiter ",
"solution": "**Answer:** emmiter \n\nEmitter sends the majority charge carries towards the collector. Therefore emitter is most heavily doped.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9174,
"subject": "Physics",
"question": "When $$npn$$ transistor is used as an amplifer",
"options": [
{
"text": "electrons move from collector to base "
},
{
"text": "holes move from emitter to base "
},
{
"text": "electrons move from base to collector"
},
{
"text": "holes move from base to emitter "
}
],
"answer": "electrons move from base to collector",
"solution": "**Answer:** electrons move from base to collector\n\nElectrons move from base to emmitter. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9175,
"subject": "Physics",
"question": "For a transistor amplifier in common emitter configuration for load impedance of $$1k\\,\\Omega $$ $$\\left( {{h_{fe}} = 50} \\right.$$ and $$\\left. {{h_{oe}} = 25} \\right)$$ the current gain is ",
"options": [
{
"text": "$$-24.8$$ "
},
{
"text": "$$-15.7$$ "
},
{
"text": "$$-5.2$$ "
},
{
"text": "$$-48.78$$ "
}
],
"answer": "$$-48.78$$ ",
"solution": "**Answer:** $$-48.78$$ \n\nIn common emitter configuration current gain \n
$$ = - 48.78$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9176,
"subject": "Physics",
"question": "In a common base amplifier, the phase difference between the input signal voltage and output voltage is ",
"options": [
{
"text": "$$\\pi $$ "
},
{
"text": "$${\\pi \\over 4}$$ "
},
{
"text": "$${\\pi \\over 2}$$"
},
{
"text": "$$0$$ "
}
],
"answer": "$$0$$ ",
"solution": "**Answer:** $$0$$ \n\nZero; In common base amplifier circuit, input and output voltage are in the same phase.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9177,
"subject": "Physics",
"question": "In a common base mode of a transistor, the collector current is $$5.488$$ $$mA$$ for an emitter current of $$5.60mA.$$ The value of the base current amplification factor $$\\left( \\beta \\right)$$ will be ",
"options": [
{
"text": "$$49$$ "
},
{
"text": "$$50$$ "
},
{
"text": "$$51$$ "
},
{
"text": "$$48$$ "
}
],
"answer": "$$49$$ ",
"solution": "**Answer:** $$49$$ \n\n$${I_C} = 5.488\\,mA,\\,\\,{I_e} = 5.6\\,mA,\\,{I_B} = {I_E} - {I_C}$$\n
$$\\beta = {{{I_c}} \\over {{I_B}}} = {{5.488} \\over {5.6 - 5.485}} = 49$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 9178,
"subject": "Physics",
"question": "A working transistor with its three legs marked $$P,Q$$ and $$R$$ is tested using a multi-meter. No conduction is found between $$P$$ and $$Q$$. By connecting the common (negative) terminal of the multi-meter to $$R$$ and the other (positive) terminal to $$P$$ or $$Q,$$ some resistance is seen on the multi-meter. Which of the following is true for the transistor? ",
"options": [
{
"text": "It is an $$npn$$ transistor with $$R$$ as base "
},
{
"text": "It is an $$pnp$$ transistor with $$R$$ as collector"
},
{
"text": "It is an $$pnp$$ transistor with $$R$$ as emitter "
},
{
"text": "It is an $$npn$$ transistor with $$R$$ as collector "
}
],
"answer": "It is an $$npn$$ transistor with $$R$$ as base ",
"solution": "**Answer:** It is an $$npn$$ transistor with $$R$$ as base \n\nIt is a $$n$$-$$p$$-$$n$$ transistor with $$R$$ as base. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9179,
"subject": "Physics",
"question": "The ratio (R) of output resistance r0, and the input resistance ri in measurements of input and output characteristics of a transistor is typically in the range:",
"options": [
{
"text": "R ~ 102 $$-$$ 103"
},
{
"text": "R ~ 1 $$-$$ 10"
},
{
"text": "R ~ 0.1 $$-$$ 0.01"
},
{
"text": "R ~ 0.1 $$-$$ 1.0"
}
],
"answer": "R ~ 1 $$-$$ 10",
"solution": "**Answer:** R ~ 1 $$-$$ 10\n\nR $$=$$ $${{{r_0}} \\over {{r_i}}} \\equiv 1 - 10$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9180,
"subject": "Physics",
"question": "An unknown transistor needs to be identified as a npn or pnp type. A multimeter, with + ve and − ve terminals, is used to measure resistance between different terminals of transistor. If terminal 2 is the base of the transistor then which of the following is correct for a pnp transistor ?",
"options": [
{
"text": "+ ve termial 1, $$-$$ve terminal 2, resistance high"
},
{
"text": "+ ve termial 2, $$-$$ve terminal 1, resistance high"
},
{
"text": "+ ve termial 3, $$-$$ve terminal 2, resistance high"
},
{
"text": "+ ve termial 2, $$-$$ve terminal 3, resistance low"
}
],
"answer": "+ ve termial 1, $$-$$ve terminal 2, resistance high",
"solution": "**Answer:** + ve termial 1, $$-$$ve terminal 2, resistance high\n\n\n
+ ve terminal at 1, $$-$$ ve terminal at 2 and resistance high for pnp transistor.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9181,
"subject": "Physics",
"question": "The current gain of a common emitter amplifier is 69. If the emitter current is 7.0 mA, collector current is :",
"options": [
{
"text": "9.6 mA"
},
{
"text": "6.9 mA"
},
{
"text": "0.69 mA"
},
{
"text": "69 mA"
}
],
"answer": "6.9 mA",
"solution": "**Answer:** 6.9 mA\n\nHere, $$\\beta $$ = 69, Ie\n = 7 mA, Ic\n = ?\n
$$ \\Rightarrow $$ Ic = $${{69} \\over {70}} \\times 7$$ = 6.9 mA",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9182,
"subject": "Physics",
"question": "In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and\nthe output voltages will be: ",
"options": [
{
"text": "180°"
},
{
"text": "45°"
},
{
"text": "90°"
},
{
"text": "135°"
}
],
"answer": "180°",
"solution": "**Answer:** 180°\n\nIn common emitter configuration for n-p-n transistor\ninput and output signals are 180° out of phase i.e., phase\ndifference between output and input voltage is 180°.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9183,
"subject": "Physics",
"question": "In a common emitter configuration with suitable bias, it is given that $${R_L}$$ is the load resistance and $${R_{BE}}$$ is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by : \n
Voltage gain = $${{{V_{CE}}} \\over {{V_{BE}}}} = \\beta {{{R_L}} \\over {{R_{BE}}}}$$\n
Power gain = voltage gain x current gain = $${\\beta ^2}{{{R_L}} \\over {{R_{BE}}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9184,
"subject": "Physics",
"question": "An NPN transistor is used in common emitter\nconfiguration as an amplifier with 1 k$$\\Omega $$ load\nresistance. Signal voltage of 10 mV is applied\nacross the base-emitter. This produces a 3 mA\nchange in the collector current and 15μA\nchange in the base current of the amplifier. The\ninput resistance and voltage gain are :",
"options": [
{
"text": "0.67 kW, 200"
},
{
"text": "0.33 kW, 1.5"
},
{
"text": "0.67 kW, 300"
},
{
"text": "0.33 kW, 300"
}
],
"answer": "0.67 kW, 300",
"solution": "**Answer:** 0.67 kW, 300\n\nInput current = 15 × 10–6
Statement I : PN junction diodes can be used to function as transistor, simply by connecting two diodes, back to back, which acts as the base terminal.
Statement II : In the study of transistor, the amplification factor $$\\beta$$ indicates ratio of the collector current to the base current.
In the light of the above statements, choose the correct answer from the options given below.",
"options": [
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Statement I is false but Statement II is true"
},
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Statement I is true but Statement II is false"
}
],
"answer": "Statement I is false but Statement II is true",
"solution": "**Answer:** Statement I is false but Statement II is true\n\nS-1 :
Statement 1 is false because in case of two discrete back to back connected diodes, there are four doped regions instead of three and there is nothing that resembles a thin base region between an emitter and a collector.
S-2 :
Statement-2 is true, as we know that, amplification factor ($$\\beta$$) is the ratio of collector current to base current.
$$\\beta = {{{I_C}} \\over {{I_B}}}$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 9188,
"subject": "Physics",
"question": "An npn transistor operates as a common emitter amplifier with a power gain of 106. The input circuit resistance is 100$$\\Omega$$ and the output load resistance is 10 K$$\\Omega$$. The common emitter current gain '$$\\beta$$' will be ________. (Round off to the Nearest Integer).",
"options": [],
"answer": "100",
"solution": "**Answer:** 100\n\nPower gain = 106
Input resistance = 100$$\\Omega$$
Output load resistance = 10K$$\\Omega$$
Power gain = $${\\beta^2} \\times {{{r_{out}}} \\over {{R_{in}}}}$$
$$ \\Rightarrow \\alpha = {\\beta \\over {1 + \\beta }}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9190,
"subject": "Physics",
"question": "A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage drop across a resistor of 1000 $$\\Omega$$ in the collector circuit is 0.6 V. If the current gain factor ($$\\beta$$) is 24, then the base current is _____________ $$\\mu$$A. (Round off to the Nearest Integer)",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\n$$\\beta = {{{I_C}} \\over {{I_B}}} = 24;$$
RC = 1000
$$\\Delta$$V = 0.6
$${I_C} = {{0.6} \\over {1000}}$$
IC = 6 $$\\times$$ 10$$-$$4
$${I_B} = {{{I_C}} \\over \\beta } = {{6 \\times {{10}^{ - 4}}} \\over {24}} = 25\\mu A$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 9191,
"subject": "Physics",
"question": "For a transistor in CE mode to be used as an amplifier, it must be operated in :",
"options": [
{
"text": "Both cut-off and Saturation"
},
{
"text": "Saturation region only"
},
{
"text": "Cut-off region only"
},
{
"text": "The active region only"
}
],
"answer": "The active region only",
"solution": "**Answer:** The active region only\n\nActive region of the CE transistor is linear region and is best suited for its use as an amplifier.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9192,
"subject": "Physics",
"question": "For a transistor $$\\alpha$$ and $$\\beta$$ are given as $$\\alpha = {{{I_C}} \\over {{I_E}}}$$ and $$\\beta = {{{I_C}} \\over {{I_B}}}$$. Then the correct relation between $$\\alpha$$ and $$\\beta$$ will be :",
"options": [
{
"text": "$$\\alpha = {{1 - \\beta } \\over \\beta }$$"
},
{
"text": "$$\\beta = {\\alpha \\over {1 - \\alpha }}$$"
},
{
"text": "$$\\alpha \\beta = 1$$"
},
{
"text": "$$\\alpha = {\\beta \\over {1 - \\beta }}$$"
}
],
"answer": "$$\\beta = {\\alpha \\over {1 - \\alpha }}$$",
"solution": "**Answer:** $$\\beta = {\\alpha \\over {1 - \\alpha }}$$\n\n$$\\alpha = {{{I_C}} \\over {{I_E}}}$$, $$\\beta = {{{I_C}} \\over {{I_B}}}$$; $${I_E} = {I_C} + {I_B}$$
A transistor is used in an amplifier circuit in common emitter mode. If the base current changes by 100 $$\\mu$$A, it brings a change of 10 mA in collector current. If the load resistance is 2 k$$\\Omega$$ and input resistance is 1 k$$\\Omega$$, the value of power gain is x $$\\times$$ 104. The value of x is _____________.
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.
\n
Assertion A : n-p-n transistor permits more current than a p-n-p transistor.
\n
Reason R : Electrons have greater mobility as a charge carrier.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "Both A and R are true, and R is correct explanation of A."
},
{
"text": "Both A and R are true but R is NOT the correct explanation of A."
},
{
"text": "A is true but R is false."
},
{
"text": "A is false but R is true."
}
],
"answer": "Both A and R are true, and R is correct explanation of A.",
"solution": "**Answer:** Both A and R are true, and R is correct explanation of A.\n\n
(A) is true as n-p-n transistor permits more current than p-n-p transistor as electrons which are majority charge carriers in n-p-n have higher mobility than holes which are majority carriers in p-n-p transistor.
\n
$$\\Rightarrow$$ Statement R is correct explanation of statement A.
The positive feedback is required by an amplifier to act an oscillator. The feedback here means :
",
"options": [
{
"text": "External input is necessary to sustain ac signal in output."
},
{
"text": "A portion of the output power is returned back to the input."
},
{
"text": "Feedback can be achieved by LR network."
},
{
"text": "The base-collector junction must be forward biased."
}
],
"answer": "A portion of the output power is returned back to the input.",
"solution": "**Answer:** A portion of the output power is returned back to the input.\n\n
Feedback means a portion of the output power is fed to the inputs.
A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10 $$\\mu$$A and the collector current changes by 1.5 mA. The load resistance is 5 k$$\\Omega$$. The voltage gain of the transistor will be _________.
For a constant collector-emitter voltage of $$8 \\mathrm{~V}$$, the collector current of a transistor reached to the value of $$6 \\mathrm{~mA}$$ from $$4 \\mathrm{~mA}$$, whereas base current changed from $$20 \\,\\mu \\mathrm{A}$$ to $$25 \\,\\mu \\mathrm{A}$$ value. If transistor is in active state, small signal current gain (current amplification factor) will be :
Statement I: In a typical transistor, all three regions emitter, base and collector have same doping level.
\n
Statement II: In a transistor, collector is the thickest and base is the thinnest segment.
\n
In the light of the above statements, choose the most appropriate answer from the options given below.
",
"options": [
{
"text": "Statement I is correct but Statement II is incorrect"
},
{
"text": "Both Statement I and Statement II are incorrect"
},
{
"text": "Statement I is incorrect but Statement II is correct"
},
{
"text": "Both Statement I and Statement II are correct"
}
],
"answer": "Statement I is incorrect but Statement II is correct",
"solution": "**Answer:** Statement I is incorrect but Statement II is correct\n\n\n
In an n-p-n common emitter (CE) transistor the collector current changes from 5 $$\\mathrm{mA}$$ to $$16 \\mathrm{~mA}$$ for the change in base current from $$100~ \\mu \\mathrm{A}$$ and $$200 ~\\mu \\mathrm{A}$$, respectively. The current gain of transistor is __________.
",
"options": [
{
"text": "210"
},
{
"text": "0.9"
},
{
"text": "9"
},
{
"text": "110"
}
],
"answer": "110",
"solution": "**Answer:** 110\n\nThe current gain of a transistor in common emitter configuration is given by:
\n$$\\beta = \\frac{I_C}{I_B}$$
\nwhere $$I_C$$ is the collector current and $$I_B$$ is the base current.\n
\nIn this case, the collector current changes from $$5 \\mathrm{~mA}$$ to $$16 \\mathrm{~mA}$$ for the change in base current from $$100~ \\mu \\mathrm{A}$$ to $$200 ~\\mu \\mathrm{A}$$. Therefore, we have:
\nTherefore, the current gain of the transistor is 110.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9201,
"subject": "Physics",
"question": "If a charge $$q$$ is placed at the center of the line joining two equal charges $$Q$$ such that the system is in equilibrium then the value of $$q$$ is ",
"options": [
{
"text": "$$Q/2$$ "
},
{
"text": "$$ - Q/2$$"
},
{
"text": "$$Q/4$$"
},
{
"text": "$$ - Q/4$$"
}
],
"answer": "$$ - Q/4$$",
"solution": "**Answer:** $$ - Q/4$$\n\nFor equilibrium of charge $$Q$$\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9202,
"subject": "Physics",
"question": "A charge $$Q$$ is placed at each of the opposite corners of a square. A charge $$q$$ is placed at each of the other two corners. If the net electrical force on $$Q$$ is zero, then $$Q/q$$ equals: ",
"options": [
{
"text": "$$-1$$ "
},
{
"text": "$$1$$ "
},
{
"text": "$$ - {1 \\over {\\sqrt 2 }}$$ "
},
{
"text": "$$ - 2\\sqrt 2 $$ "
}
],
"answer": "$$ - 2\\sqrt 2 $$ ",
"solution": "**Answer:** $$ - 2\\sqrt 2 $$ \n\n\n
Let $$F$$ be the force between $$Q$$ and $$Q.$$ The force between $$q$$ and $$Q$$ should be attractive for net force on $$Q$$ to be zero. Let $$F'$$ be the force between $$Q$$ and $$q.$$ For equilibrium\n
$$ \\Rightarrow {Q \\over q} = - 2\\sqrt 2 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9203,
"subject": "Physics",
"question": "Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $${30^ \\circ }$$ with each other. When suspended in a liquid of density $$0.8g$$ $$c{m^{ - 3}},$$ the angle remains the same. If density of the material of the sphere is $$1.6$$ $$g$$ $$c{m^{ - 3}},$$ the dielectric constant of the liquid is ",
"options": [
{
"text": "$$4$$ "
},
{
"text": "$$3$$ "
},
{
"text": "$$2$$ "
},
{
"text": "$$1$$ "
}
],
"answer": "$$2$$ ",
"solution": "**Answer:** $$2$$ \n\n\n $${F_e} = T\\sin {15^ \\circ }\\,\\,;$$\n $$mg = T\\cos {15^ \\circ }$$\n $$ \\Rightarrow \\tan {15^ \\circ } = {{{F_e}} \\over {mg}}$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...(i)$$\n In liquid, $${F_e}' = T'\\sin {15^ \\circ }$$ $$\\,\\,\\,\\,\\,\\,...(ii)$$\n
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9204,
"subject": "Physics",
"question": "Two identical charged spheres suspended from a common point by two massless strings of length $$l$$ are initially a distance $$d\\left( {d < < 1} \\right)$$ apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result charges approach each other with a velocity $$v$$. Then as a function of distance $$x$$ between them, ",
"options": [
{
"text": "$$v\\, \\propto \\,{x^{ - 1}}$$ "
},
{
"text": "$$y\\, \\propto \\,{x^{{\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}}$$ "
},
{
"text": "$$v\\, \\propto \\,x$$ "
},
{
"text": "$$v\\, \\propto \\,{x^{ - {\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}}$$ "
}
],
"answer": "$$v\\, \\propto \\,{x^{ - {\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}}$$ ",
"solution": "**Answer:** $$v\\, \\propto \\,{x^{ - {\\raise0.5ex\\hbox{$\\scriptstyle 1$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle 2$}}}}$$ \n\nAt any instant \n
$$T\\cos \\theta = mg\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( i \\right)$$\n
$$ \\Rightarrow v \\propto {x^{ - 1/2}}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9205,
"subject": "Physics",
"question": "Two charges, each equals to $$q,$$ are kept at $$x=-a$$ and $$x=a$$ on the $$x$$-axis. A particle of mass $$m$$ and charge $${q_0} = {q \\over 2}$$ is placed at the origin. If charge $${q_0}$$ is given a small displacement $$\\left( {y < < a} \\right)$$ along the $$y$$-axis, the net force acting on the particle is proportional to ",
"options": [
{
"text": "$$y$$ "
},
{
"text": "$$-y$$ "
},
{
"text": "$${1 \\over y}$$ "
},
{
"text": "$$-{1 \\over y}$$"
}
],
"answer": "$$y$$ ",
"solution": "**Answer:** $$y$$ \n\n \n
S0, $$F \\propto y$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9206,
"subject": "Physics",
"question": "Two identical conducting spheres A and B, carry equal charge. They are separated by a distance much larger than their diameters, and the force between theis F. A third identical conducting sphere, C, is uncharged. Sphere C is first touhed to A, then to B, and then removed. As a result, the force between A and B would be equal to : ",
"options": [
{
"text": "F "
},
{
"text": "$${{3F} \\over 4}$$"
},
{
"text": "$${{3F} \\over 8}$$"
},
{
"text": "$${{F} \\over 2}$$"
}
],
"answer": "$${{3F} \\over 8}$$",
"solution": "**Answer:** $${{3F} \\over 8}$$\n\nLet, change of A and B = q\n
$$\\therefore\\,\\,\\,$$ Force between them, F = $${{k \\times q \\times q} \\over {{r^2}}} = {{k{q^2}} \\over {{r^2}}}$$\n
When C touched with A then charge of A. Will fl;ow to C and divide into half parts. \n
= $${3 \\over 8}\\,F$$ ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9207,
"subject": "Physics",
"question": "Three charges + Q, q, + Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by + Q, placed at x = 0, is zero, then value of q is : ",
"options": [
{
"text": "$$-$$ $${Q \\over 4}$$"
},
{
"text": "+ $${Q \\over 2}$$"
},
{
"text": "+ $${Q \\over 4}$$"
},
{
"text": "$$-$$ $${Q \\over 2}$$"
}
],
"answer": "$$-$$ $${Q \\over 4}$$",
"solution": "**Answer:** $$-$$ $${Q \\over 4}$$\n\n\n
Force on + Q charge at x = 0 due to q charge, F1 = $${{KQq} \\over {{{\\left( {{d \\over 2}} \\right)}^2}}}$$\n
Force on +Q charge at x = 0 due to + Q charge at x = d is, \n
$$\\therefore$$ $$\\omega = {\\left( {{1 \\over {2\\pi {\\varepsilon _0}}}{{{q^2}} \\over {m{d^3}}}} \\right)^{1/2}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9210,
"subject": "Physics",
"question": "Two identical conducting spheres with negligible volume have 2.1 nC and $$-$$0.1 nC charges, respectively. They are brought into contact and then separated by a distance of 0.5 m. The electrostatic force acting between the spheres is __________ $$\\times$$ 10$$-$$9 N.
[Given : $$4\\pi {\\varepsilon _0} = {1 \\over {9 \\times {{10}^9}}}$$ SI unit]",
"options": [],
"answer": "36",
"solution": "**Answer:** 36\n\n\n When they brought into contact & then separated by a distance = 0.5 m
$$ \\Rightarrow $$ $${F_e} = 36 \\times {10^{ - 9}}N$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9211,
"subject": "Physics",
"question": "Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m. The charge on each of the sphere is $${a \\over {21}} \\times {10^{ - 8}}$$C. The value of 'a' will be ___________. [Given g = 10 ms$$-$$2]",
"options": [],
"answer": "20",
"solution": "**Answer:** 20\n\n\n T sin$$\\theta$$ = $${{k{q^2}} \\over {{r^2}}}$$
after comparison from the given equation a = 20",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9212,
"subject": "Physics",
"question": "An infinite number of point charges, each carrying 1 $$\\mu$$C charge, are placed along the y-axis at y = 1 m, 2 m, 4 m, 8 m ...............
The total force on a 1C point charge, placed at the origin, is x $$\\times$$ 103 N.
The value of x, to the nearest integer, is __________. [Take $${1 \\over {4\\pi {\\varepsilon _0}}} = 9 \\times {10^9}$$ Nm2/C2]",
"options": [],
"answer": "12",
"solution": "**Answer:** 12\n\n\n $${F_{total}} = {{k{q_1}{q_2}} \\over {r_1^2}} + {{k{q_1}{q_3}} \\over {r_2^2}} + {{k{q_1}{q_4}} \\over {r_3^2}} + .....$$
$$ = 9 \\times {10^3} \\times {4 \\over 3}$$ = 12 $$\\times$$ 103 N",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9213,
"subject": "Physics",
"question": "A certain charge Q is divided into two parts q and (Q $$-$$ q). How should the charges Q and q be divided so that q and (Q $$-$$ q) placed at a certain distance apart experience maximum electrostatic repulsion?",
"options": [
{
"text": "Q = 2q"
},
{
"text": "Q = 4q"
},
{
"text": "Q = 3q"
},
{
"text": "Q = $${q \\over 2}$$"
}
],
"answer": "Q = 2q",
"solution": "**Answer:** Q = 2q\n\nLet's say the charge q and (Q $$-$$ q) are at r distance from each other. This can be shown as
According to Coulomb's law, force between both the parts can be given as
$$F = {{kq(Q - q)} \\over {{r^2}}}$$
$$F = {k \\over {{r^2}}}(qQ - {q^2})$$
As we know that $${{dF} \\over {dq}} = 0$$, for maximum force.
$$ \\Rightarrow {k \\over {{r^2}}}(Q - 2q) = 0 \\Rightarrow Q = 2q$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9214,
"subject": "Physics",
"question": "A particle of mass 1 mg and charge q is lying at the mid-point of two stationary particles kept at a distance '2 m' when each is carrying same charge 'q'. If the free charged particle is displaced from its equilibrium position through distance 'x' (x < < 1 m). The particle executes SHM. Its angular frequency of oscillation will be ____________ $$\\times$$ 105 rad/s if q2 = 10 C2.",
"options": [],
"answer": "6000",
"solution": "**Answer:** 6000\n\n
Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is :
Two identical charged particles each having a mass 10 g and charge 2.0 $$\\times$$ 10$$-$$7C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10 ms$$-$$2]
A long cylindrical volume contains a uniformly distributed charge of density $$\\rho$$. The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic energy of the particle is :
The volume charge density of a sphere of radius $$6 \\mathrm{~m}$$ is $$2 \\,\\mu \\mathrm{C} \\,\\mathrm{cm}^{-3}$$. The number of lines of force per unit surface area coming out from the surface of the sphere is _______________ $$\\times 10^{10} \\,\\mathrm{NC}^{-1}$$.
Three point charges of magnitude $$5 \\mu \\mathrm{C}, 0.16 \\mu \\mathrm{C}$$ and $$0.3 \\mu \\mathrm{C}$$ are located at the vertices $$A, B, C$$ of a right angled triangle whose sides are $$A B=3 \\mathrm{~cm}, B C=3 \\sqrt{2} \\mathrm{~cm}$$ and $$C A=3 \\mathrm{~cm}$$ and point $$A$$ is the right angle corner. Charge at point $$\\mathrm{A}$$ experiences ____________ $$\\mathrm{N}$$ of electrostatic force due to the other two charges.
Two identical positive charges $$Q$$ each are fixed at a distance of '2a' apart from each other. Another point charge $$q_{0}$$ with mass 'm' is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge $$\\mathrm{q}_{0}$$ executes $$\\mathrm{SHM}$$. The time period of oscillation of charge $$\\mathrm{q}_{0}$$ will be :
",
"options": [
{
"text": "$$\\sqrt{\\frac{4 \\pi^{3} \\varepsilon_{0} m a^{3}}{q_{0} Q}}$$"
},
{
"text": "$$\\sqrt{\\frac{q_{0} Q}{4 \\pi^{3} \\varepsilon_{0} m a^{3}}}$$"
},
{
"text": "$$\\sqrt{\\frac{2 \\pi^{2} \\varepsilon_{0} m a^{3}}{q_{0} Q}}$$"
},
{
"text": "$$\\sqrt{\\frac{8 \\pi^{3} \\varepsilon_{0} m a^{3}}{q_{0} Q}}$$"
}
],
"answer": "$$\\sqrt{\\frac{4 \\pi^{3} \\varepsilon_{0} m a^{3}}{q_{0} Q}}$$",
"solution": "**Answer:** $$\\sqrt{\\frac{4 \\pi^{3} \\varepsilon_{0} m a^{3}}{q_{0} Q}}$$\n\n
A charge of $$4 \\,\\mu \\mathrm{C}$$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be :
",
"options": [
{
"text": "$$1 \\,\\mu \\mathrm{C}$$ and $$3 \\,\\mu\\mathrm{C}$$"
},
{
"text": "$$2 \\,\\mu \\mathrm{C}$$ and $$2\\, \\mu \\mathrm{C}$$"
},
{
"text": "0 and $$4\\, \\mu\\, \\mathrm{C}$$"
},
{
"text": "$$1.5 \\,\\mu \\mathrm{C}$$ and $$2.5\\, \\mu \\mathrm{C}$$"
}
],
"answer": "$$2 \\,\\mu \\mathrm{C}$$ and $$2\\, \\mu \\mathrm{C}$$",
"solution": "**Answer:** $$2 \\,\\mu \\mathrm{C}$$ and $$2\\, \\mu \\mathrm{C}$$\n\n
Two identical metallic spheres $$\\mathrm{A}$$ and $$\\mathrm{B}$$ when placed at certain distance in air repel each other with a force of $$\\mathrm{F}$$. Another identical uncharged sphere $$\\mathrm{C}$$ is first placed in contact with $$\\mathrm{A}$$ and then in contact with $$\\mathrm{B}$$ and finally placed at midpoint between spheres A and B. The force experienced by sphere C will be:
Two isolated metallic solid spheres of radii $$\\mathrm{R}$$ and $$2 \\mathrm{R}$$ are charged such that both have same charge density $$\\sigma$$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $$\\sigma^{\\prime}$$. The ratio $$\\frac{\\sigma^{\\prime}}{\\sigma}$$ is :
A point charge $$q_1=4q_0$$ is placed at origin. Another point charge $$q_2=-q_0$$ is placed at $$x=12$$ cm. Charge of proton is $$q_0$$. The proton is placed on $$x$$ axis so that the electrostatic force on the proton is zero. In this situation, the position of the proton from the origin is ___________ cm.
",
"options": [],
"answer": "24",
"solution": "**Answer:** 24\n\n\n Let a proton having charge $q_0$ on the $x$ axis at distance $x$ from $q_2$ and $(12+x)$ distance from $q_1$.\n
Now, balance the force between them $\\vec{F}_1+\\vec{F}_2=0$\n
If two charges q$$_1$$ and q$$_2$$ are separated with distance 'd' and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic force?
A $$10 ~\\mu \\mathrm{C}$$ charge is divided into two parts and placed at $$1 \\mathrm{~cm}$$ distance so that the repulsive force between them is maximum. The charges of the two parts are:
",
"options": [
{
"text": "$$9 ~\\mu\\mathrm{C}, 1 ~\\mu \\mathrm{C}$$"
},
{
"text": "$$5 ~\\mu\\mathrm{C}, 5 ~\\mu \\mathrm{C}$$"
},
{
"text": "$$8 ~\\mu\\mathrm{C}, 2 ~\\mu \\mathrm{C}$$"
},
{
"text": "$$7 ~\\mu\\mathrm{C}, 3 ~\\mu \\mathrm{C}$$"
}
],
"answer": "$$5 ~\\mu\\mathrm{C}, 5 ~\\mu \\mathrm{C}$$",
"solution": "**Answer:** $$5 ~\\mu\\mathrm{C}, 5 ~\\mu \\mathrm{C}$$\n\nThe repulsive force between the two charges is given by Coulomb's law:
\nwhere $F$ is the force, $q_1$ and $q_2$ are the charges, $r$ is the distance between them, and $\\epsilon_0$ is the electric constant.\n
\nTo maximize the force, we need to maximize the product $q_1q_2$. Let $q_1$ be the charge on one part and $q_2$ be the charge on the other part. Then we have $q_1+q_2=10\\mu\\mathrm{C}$, since the total charge is $10\\mu\\mathrm{C}$.\n
\nThe distance between the two charges is $r=1\\mathrm{~cm}=0.01\\mathrm{~m}$. To maximize the force, we need to maximize $q_1q_2$, subject to the constraint that $q_1+q_2=10\\mu\\mathrm{C}$.\n
\nWe can use the method of Lagrange multipliers to find the values of $q_1$ and $q_2$ that maximize $q_1q_2$ subject to the constraint $q_1+q_2=10\\mu\\mathrm{C}$. The Lagrangian is given by\n$$\\mathcal{L}=q_1q_2-\\lambda(q_1+q_2-10\\mu\\mathrm{C}),$$\nwhere $\\lambda$ is the Lagrange multiplier.\n
\nTaking the partial derivatives of $\\mathcal{L}$ with respect to $q_1$, $q_2$, and $\\lambda$, and setting them equal to zero, we get:
\n$$q_2-\\lambda=0$$
\n$$q_1-\\lambda=0$$
\n$$q_1+q_2=10\\mu\\mathrm{C}$$\n
\nSolving for $q_1$ and $q_2$, we get $q_1=q_2=5\\mu\\mathrm{C}$.
Therefore, the charges of the two parts are both $5\\mu\\mathrm{C}$.\n
\nTherefore, to maximize the repulsive force between the two charges, we need to divide the $10\\mu\\mathrm{C}$ charge into two equal parts of $5\\mu\\mathrm{C}$ each, and place them $1\\mathrm{~cm}$ apart.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9228,
"subject": "Physics",
"question": "Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle $\\theta$ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is $1.5 \\mathrm{~g} / \\mathrm{cc}$, the dielectric constant of water will be __________.
(Take density of water $=1 \\mathrm{~g} / \\mathrm{cc}$ )",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n\n
In air $\\tan \\frac{\\theta}{2}=\\frac{F}{m g}=\\frac{q^2}{4 \\pi \\varepsilon_0 r^2 m g}$\n
In water $\\tan \\frac{\\theta}{2}=\\frac{\\mathrm{F}^{\\prime}}{\\mathrm{mg}^{\\prime}}=\\frac{\\mathrm{q}^2}{4 \\pi \\varepsilon_0 \\varepsilon_{\\mathrm{r}} \\mathrm{r}^2 \\mathrm{mg}_{\\text {eff }}}$\n
A thin metallic wire having cross sectional area of $$10^{-4} \\mathrm{~m}^2$$ is used to make a ring of radius $$30 \\mathrm{~cm}$$. A positive charge of $$2 \\pi \\mathrm{~C}$$ is uniformly distributed over the ring, while another positive charge of 30 $$\\mathrm{pC}$$ is kept at the centre of the ring. The tension in the ring is ______ $$\\mathrm{N}$$; provided that the ring does not get deformed (neglect the influence of gravity). (given, $$\\frac{1}{4 \\pi \\epsilon_0}=9 \\times 10^9$$ SI units)
Force between two point charges $$q_1$$ and $$q_2$$ placed in vacuum at '$$r$$' cm apart is $$F$$. Force between them when placed in a medium having dielectric constant $$K=5$$ at '$$r / 5$$' $$\\mathrm{cm}$$ apart will be:
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $$37^{\\circ}$$ with each other. When suspended in a liquid of density $$0.7 \\mathrm{~g} / \\mathrm{cm}^3$$, the angle remains same. If density of material of the sphere is $$1.4 \\mathrm{~g} / \\mathrm{cm}^3$$, the dielectric constant of the liquid is _______ $$\\left(\\tan 37^{\\circ}=\\frac{3}{4}\\right)$$
Two charged conducting spheres of radii $$a$$ and $$b$$ are connected to each other by a conducting wire. The ratio of charges of the two spheres respectively is:
When two conducting spheres of radii $$a$$ and $$b$$ are connected by a conducting wire, they come to the same potential because conductors in contact share charges until their potentials become equal. The potential of a charged sphere is given by $$V = \\frac{kQ}{R}$$, where $$V$$ is the potential, $$k$$ is Coulomb's constant, $$Q$$ is the charge on the sphere, and $$R$$ is the radius of the sphere.
\n\n
For the two spheres at the same potential, we have:
\n\n
$$\\frac{kQ_a}{a} = \\frac{kQ_b}{b}$$
\n\n
Where:
\n\n
\n\n
$$Q_a$$ and $$Q_b$$ are the charges on the spheres with radii $$a$$ and $$b$$, respectively.
\n\n
$$k$$ cancels out from both sides as it is a constant.
\n\n
\n\n
From the above equation, to find the ratio of charges $$\\frac{Q_a}{Q_b}$$, we rearrange it as follows:
\n\n
$$\\frac{Q_a}{Q_b} = \\frac{a}{b}$$
\n\n
Therefore, the ratio of the charges of the two spheres respectively is $$\\frac{a}{b}$$.
\n\n
So, the correct answer is Option C: $$\\frac{a}{b}$$.
To find the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system, we use the formulae for both forces and then divide them.
\n\n
The Coulombic (electrostatic) force, $F_C$, between two charges is given by Coulomb's law:
\n\n
$F_C = k \\frac{|q_1 q_2|}{r^2}$
\n\n
where $k$ is Coulomb's constant ($8.987 \\times 10^9 \\, \\text{Nm}^2\\text{C}^{-2}$), $q_1$ and $q_2$ are the magnitudes of the charges, and $r$ is the distance between the charges. For a proton and an electron, $q_1 = q_2 = e$, where $e$ is the elementary charge ($1.602 \\times 10^{-19} \\, \\text{C}$).
\n\n
The gravitational force, $F_G$, between two masses is given by Newton's law of universal gravitation:
\n\n
$F_G = G \\frac{m_1 m_2}{r^2}$
\n\n
where $G$ is the gravitational constant ($6.674 \\times 10^{-11} \\, \\text{Nm}^2\\text{kg}^{-2}$), and $m_1$ and $m_2$ are the masses of the two objects. For a proton and an electron, $m_p \\approx 1.673 \\times 10^{-27} \\, \\text{kg}$ and $m_e \\approx 9.109 \\times 10^{-31} \\, \\text{kg}$, respectively.
\n\n
The ratio of the Coulombic to the gravitational force is therefore:
Therefore, the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system is on the order of $10^{39}$, making Option B the correct answer.
The vehicles carrying inflammable fluids usually have metallic chains touching the ground:
",
"options": [
{
"text": "To protect tyres from catching dirt from ground\n"
},
{
"text": "It is a custom\n"
},
{
"text": "To alert other vehicles\n"
},
{
"text": "To conduct excess charge due to air friction to ground and prevent sparking"
}
],
"answer": "To conduct excess charge due to air friction to ground and prevent sparking",
"solution": "**Answer:** To conduct excess charge due to air friction to ground and prevent sparking\n\n
The correct option is Option D: To conduct excess charge due to air friction to ground and prevent sparking.
\n\n
This method is grounded in the principles of physics, particularly relating to static electricity and grounding. As vehicles move through the air, especially at high speeds, friction between the air and the vehicle can lead to the accumulation of static electricity on the vehicle. This is a common phenomenon and is more pronounced in dry conditions where humidity is low, as moisture in the air can help dissipate static charge more effectively.
\n\n
In the case of vehicles carrying inflammable fluids, the presence of static electricity poses a significant risk. This is because a static discharge (or sparking) in the presence of flammable vapors can ignite those vapors, leading to potentially catastrophic fires or explosions. The metallic chains that touch the ground serve a crucial safety function by providing a path for the static electrical charge to safely dissipate into the earth, a process known as grounding. The earth acts as a vast reservoir that can absorb large amounts of electrical charge. By grounding the vehicle in this way, the risk of static discharge into the surrounding atmosphere is significantly reduced, enhancing safety by preventing ignition of inflammable materials.
\n\n
The effectiveness of this safety measure hinges on the electrical conductivity of the chain and its contact with the ground. The chain must be made of a material with good electrical conductivity (such as metal) and maintain adequate contact with the ground to ensure that the static charge can be continuously dissipated. This safety precaution is a practical application of electrostatic principles, showcasing how understanding and harnessing the laws of physics can provide effective solutions to real-world problems.
Two identical conducting spheres P and S with charge Q on each, repel each other with a force $$16 \\mathrm{~N}$$. A third identical uncharged conducting sphere $$\\mathrm{R}$$ is successively brought in contact with the two spheres. The new force of repulsion between $$\\mathrm{P}$$ and $$\\mathrm{S}$$ is :
Final charges on spheres are $$\\frac{Q}{2}$$ and $$\\frac{3 Q}{4}$$.
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9236,
"subject": "Physics",
"question": "An electric dipole is placed at an angle of $${30^ \\circ }$$ to a non-uniform electric field. The dipole will experience ",
"options": [
{
"text": "a translation force only in the direction of the field "
},
{
"text": "a translation force only in a direction normal to the direction of the field "
},
{
"text": "a torque as well as a translational force"
},
{
"text": "a torque only "
}
],
"answer": "a torque as well as a translational force",
"solution": "**Answer:** a torque as well as a translational force\n\n \n
The electric field will be different at the location of the two charges. Therefore the two forces will be unequal. This will result in a force as well as torque.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9237,
"subject": "Physics",
"question": "An electric dipole has a fixed dipole moment $$\\overrightarrow p $$, which makes angle $$\\theta$$ with respect to x-axis. When\nsubjected to an electric field $$\\mathop {{E_1}}\\limits^ \\to = E\\widehat i$$ , it experiences a torque $$\\overrightarrow {{T_1}} = \\tau \\widehat k$$ . When subjected to another electric\nfield $$\\mathop {{E_2}}\\limits^ \\to = \\sqrt 3 {E_1}\\widehat j$$ it experiences a torque $$\\mathop {{T_2}}\\limits^ \\to = \\mathop { - {T_1}}\\limits^ \\to $$ . The angle $$\\theta$$ is: ",
"options": [
{
"text": "90o"
},
{
"text": "45o"
},
{
"text": "30o"
},
{
"text": "60o"
}
],
"answer": "60o",
"solution": "**Answer:** 60o\n\n
Torque experienced by the dipole in an\nelectric field, \n
$$T $$ = pE sin$$\\theta $$ \n
$$\\overrightarrow T = \\overrightarrow p \\times \\overrightarrow E $$\n
$$\\overrightarrow p = p\\cos \\theta \\widehat i + p\\sin \\theta \\widehat j$$\n
$$ \\Rightarrow $$ $$\\theta $$ = 60o",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9238,
"subject": "Physics",
"question": "An electric field of 1000 V/m is applied to an electric dipole at angle of 45o. The value of electric dipole moment is 10–29 C.m. What is the potential energy of the electric dipole? ",
"options": [
{
"text": "- 7 $$ \\times $$ 10–27 J"
},
{
"text": "$$-$$ 9 $$ \\times $$ 10–20 J"
},
{
"text": "$$-$$ 10 $$ \\times $$ 10–29 J"
},
{
"text": "$$-$$ 20 $$ \\times $$ 10–18 J"
}
],
"answer": "- 7 $$ \\times $$ 10–27 J",
"solution": "**Answer:** - 7 $$ \\times $$ 10–27 J\n\nU = $$-$$ $$\\overrightarrow P .\\overrightarrow E $$\n
= $$-$$ PE cos $$\\theta $$\n
= $$-$$ (10$$-$$29) (103) cos 45o\n
= $$-$$ 0.707 $$ \\times $$ 10$$-$$26 J\n
= $$-$$ 7 $$ \\times $$ 10$$-$$27 J.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9239,
"subject": "Physics",
"question": "A point dipole $$\\overrightarrow p = - {p_0}\\widehat x$$\nis kept at the origin. The potential and electric field due to this dipole on the\ny-axis at a distance d are, respectively: (Take V= 0 at infinity)",
"options": [
{
"text": "$${{\\left| {\\overrightarrow p } \\right|} \\over {4\\pi { \\in _0}{d^2}}},{{ - \\overrightarrow p } \\over {4\\pi { \\in _0}{d^3}}}$$"
},
{
"text": "$$0,{{\\overrightarrow p } \\over {4\\pi { \\in _0}{d^3}}}$$"
},
{
"text": "$${{\\left| {\\overrightarrow p } \\right|} \\over {4\\pi { \\in _0}{d^2}}},{{\\overrightarrow p } \\over {4\\pi { \\in _0}{d^3}}}$$"
},
{
"text": "$$0,{{ - \\overrightarrow p } \\over {4\\pi { \\in _0}{d^3}}}$$"
}
],
"answer": "$$0,{{ - \\overrightarrow p } \\over {4\\pi { \\in _0}{d^3}}}$$",
"solution": "**Answer:** $$0,{{ - \\overrightarrow p } \\over {4\\pi { \\in _0}{d^3}}}$$\n\nV = 0
\n$$E = - {{K\\overrightarrow P } \\over {{r^3}}}$$
\n$$ = - {{\\overrightarrow p } \\over {4\\pi {\\varepsilon _0}{d^3}}}$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9240,
"subject": "Physics",
"question": "An electric dipole of moment\n $$\\overrightarrow p = \\left( { - \\widehat i - 3\\widehat j + 2\\widehat k} \\right) \\times {10^{ - 29}} $$ C.m is at the origin\n(0, 0, 0). The electric field due to this dipole at\n $$\\overrightarrow r = + \\widehat i + 3\\widehat j + 5\\widehat k$$ (note that $$\\overrightarrow r .\\overrightarrow p = 0$$ ) is parallel to :",
"options": [
{
"text": "$$\\left( { + \\widehat i + 3\\widehat j - 2\\widehat k} \\right)$$"
},
{
"text": "$$\\left( { + \\widehat i - 3\\widehat j - 2\\widehat k} \\right)$$"
},
{
"text": "$$\\left( { - \\widehat i + 3\\widehat j - 2\\widehat k} \\right)$$"
},
{
"text": "$$\\left( { - \\widehat i - 3\\widehat j + 2\\widehat k} \\right)$$"
}
],
"answer": "$$\\left( { + \\widehat i + 3\\widehat j - 2\\widehat k} \\right)$$",
"solution": "**Answer:** $$\\left( { + \\widehat i + 3\\widehat j - 2\\widehat k} \\right)$$\n\nSince $$\\overrightarrow r $$\nand $$\\overrightarrow p $$\nare perpendicular to each other\ntherefore point lies on the equitorial plane.\nTherefore electric field at the point will be\nantiparallel to the dipole moment.\n
$$ \\therefore $$ $$\\overrightarrow E || - \\overrightarrow p $$\n
$$ \\Rightarrow $$ $$\\overrightarrow E ||\\left( {\\widehat i + 3\\widehat j - 2\\widehat k} \\right)$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9241,
"subject": "Physics",
"question": "Two identical electric point dipoles have dipole moments $${\\overrightarrow p _1} = p\\widehat i$$ and $${\\overrightarrow p _2} = - p\\widehat i$$ and are held on the x\naxis at distance '$$a$$' from each other. When released, they move along the x-axis with the direction\nof their dipole moments remaining unchanged. If the mass of each dipole is 'm', their speed when\nthey are infinitely far apart is :",
"options": [
{
"text": "$${p \\over a}\\sqrt {{3 \\over {2\\pi { \\in _0}ma}}} $$"
},
{
"text": "$${p \\over a}\\sqrt {{1 \\over {\\pi { \\in _0}ma}}} $$"
},
{
"text": "$${p \\over a}\\sqrt {{1 \\over {2\\pi { \\in _0}ma}}} $$"
},
{
"text": "$${p \\over a}\\sqrt {{2 \\over {\\pi { \\in _0}ma}}} $$"
}
],
"answer": "$${p \\over a}\\sqrt {{1 \\over {2\\pi { \\in _0}ma}}} $$",
"solution": "**Answer:** $${p \\over a}\\sqrt {{1 \\over {2\\pi { \\in _0}ma}}} $$\n\n\n
Statement I : An electric dipole is placed at the center of a hollow sphere. The flux of the electric field through the sphere is zero but the electric field is not zero anywhere in the sphere.
Statement II : If R is the radius of a solid metallic sphere and Q be the total charge on it. The electric field at any point on the spherical surface of radius r (< R) is zero but the electric flux passing through this closed spherical surface of radius r is not zero..
In the light of the above statements, choose the correct answer from the options given below :",
"options": [
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Statement I is false but Statement II is true"
},
{
"text": "Statement I is true but Statement II is false"
},
{
"text": "Both Statement I and Statement II are false"
}
],
"answer": "Statement I is true but Statement II is false",
"solution": "**Answer:** Statement I is true but Statement II is false\n\n
Electric field due to electric dipole is non-zero and varies at point to point.
\n
Hence, statement I is true.
\n
Electric field due to charged solid sphere at a distance r from centre.
\n$$E = {1 \\over {4\\pi {\\varepsilon _0}}}\\,.\\,{{Qr} \\over {{R^3}}}$$ [when r < R, R $$\\to$$ radius] which is non-zero.\n
Hence, statement II is false.
\n
Hence, option (c) is the correct.
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9243,
"subject": "Physics",
"question": "An electric dipole is placed on x-axis in proximity to a line charge of linear charge density 3.0 $$\\times$$ 10$$-$$6 C/m. Line charge is placed on z-axis and positive and negative charge of dipole is at a distance of 10 mm and 12 mm from the origin respectively. If total force of 4N is exerted on the dipole, find out the amount of positive or negative charge of the dipole.",
"options": [
{
"text": "0.485 mC"
},
{
"text": "815.1 nC"
},
{
"text": "8.8 $$\\mu$$C"
},
{
"text": "4.44 $$\\mu$$C"
}
],
"answer": "4.44 $$\\mu$$C",
"solution": "**Answer:** 4.44 $$\\mu$$C\n\n
Given below two statements : One is labelled as Assertion (A) and other is labelled as Reason (R).
\n
Assertion (A) : Non-polar materials do not have any permanent dipole moment.
\n
Reason (R) : When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it's individual atom or molecule coincides with the centre of the negative charge distribution.
\n
In the light of above statements, choose the most appropriate answer from the options given below.
",
"options": [
{
"text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)."
},
{
"text": "Both (A) and (R) are correct and (R) is not the correct explanation of (A)."
},
{
"text": "(A) is correct but (R) is not correct."
},
{
"text": "(A) is not correct but (R) is correct."
}
],
"answer": "(A) is correct but (R) is not correct.",
"solution": "**Answer:** (A) is correct but (R) is not correct.\n\n
Non-polar bonds do not have any net dipole moment and are generally formed in compound where there is presence of symmetry.
\n
When non polar material placed in electric field, due to redistribution of charges dipole is formed.
Two electric dipoles of dipole moments $$1.2 \\times 10^{-30} \\,\\mathrm{Cm}$$ and $$2.4 \\times 10^{-30} \\,\\mathrm{Cm}$$ are placed in two different uniform electric fields of strengths $$5 \\times 10^{4} \\,\\mathrm{NC}^{-1}$$ and $$15 \\times 10^{4} \\,\\mathrm{NC}^{-1}$$ respectively. The ratio of maximum torque experienced by the electric dipoles will be $$\\frac{1}{x}$$. The value of $$x$$ is __________.
Two charges each of magnitude $$0.01 ~\\mathrm{C}$$ and separated by a distance of $$0.4 \\mathrm{~mm}$$ constitute an electric dipole. If the dipole is placed in an uniform electric field '$$\\vec{E}$$' of 10 dyne/C making $$30^{\\circ}$$ angle with $$\\vec{E}$$, the magnitude of torque acting on dipole is:
",
"options": [
{
"text": "$$4 \\cdot 0 \\times 10^{-10} ~\\mathrm{Nm}$$"
},
{
"text": "$$1.5 \\times 10^{-9} ~\\mathrm{Nm}$$"
},
{
"text": "$$1.0 \\times 10^{-8} ~\\mathrm{Nm}$$"
},
{
"text": "$$2.0 \\times 10^{-10} ~\\mathrm{Nm}$$"
}
],
"answer": "$$2.0 \\times 10^{-10} ~\\mathrm{Nm}$$",
"solution": "**Answer:** $$2.0 \\times 10^{-10} ~\\mathrm{Nm}$$\n\nGiven two charges each of magnitude $$0.01 \\,\\text{C}$$ and separated by a distance of $$0.4 \\,\\text{mm} = 0.4 \\times 10^{-3} \\,\\text{m}$$, we can calculate the dipole moment $$\\vec{p}$$:\n
\nThe dipole is placed in a uniform electric field $$\\vec{E}$$ of magnitude $$10 \\,\\text{dyne/C}$$, which is equivalent to $$10 \\times 10^{-5} \\,\\text{N/C}$$.\n
\nThe angle between the dipole moment and the electric field is $$30^{\\circ}$$.\n
\nThe torque acting on the dipole can be calculated using the formula:\n
\nThe magnitude of the torque acting on the dipole is $$2.0 \\times 10^{-10} \\,\\text{Nm}$$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9247,
"subject": "Physics",
"question": "
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
\n
Assertion A : If an electric dipole of dipole moment $$30 \\times 10^{-5} ~\\mathrm{C} ~\\mathrm{m}$$ is enclosed by a closed surface, the net flux coming out of the surface will be zero.
\n
Reason R : Electric dipole consists of two equal and opposite charges.
\n
In the light of above, statements, choose the correct answer from the options given below.
",
"options": [
{
"text": "A is true but R is false"
},
{
"text": "A is false but R is true"
},
{
"text": "Both A and R are true but R is NOT the correct explanation of A"
},
{
"text": "Both A and R are true and R is the correct explanation of A"
}
],
"answer": "Both A and R are true and R is the correct explanation of A",
"solution": "**Answer:** Both A and R are true and R is the correct explanation of A\n\nAssertion A: If an electric dipole of dipole moment $$30 \\times 10^{-5} ~\\mathrm{C} ~\\mathrm{m}$$ is enclosed by a closed surface, the net flux coming out of the surface will be zero.
\nThis statement is true. According to Gauss's Law, the electric flux through a closed surface is proportional to the net charge enclosed by the surface. Since an electric dipole consists of two equal and opposite charges, the net charge enclosed by the surface is zero, and therefore, the net flux coming out of the surface will also be zero.\n
\nReason R: Electric dipole consists of two equal and opposite charges.
\nThis statement is also true. An electric dipole is formed by two equal and opposite charges separated by a fixed distance.\n
\nThe reason R correctly explains the assertion A because the net charge enclosed by the surface is zero due to the presence of equal and opposite charges in the electric dipole, resulting in zero net electric flux coming out of the surface.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9248,
"subject": "Physics",
"question": "
An electric dipole of dipole moment is $$6.0 \\times 10^{-6} ~\\mathrm{C m}$$ placed in a uniform electric field of $$1.5 \\times 10^{3} ~\\mathrm{NC}^{-1}$$ in such a way that dipole moment is along electric field. The work done in rotating dipole by $$180^{\\circ}$$ in this field will be ___________ $$\\mathrm{m J}$$.
The work done $W$ in rotating an electric dipole in a uniform electric field is given by:
\n
$W = pE(1 - \\cos\\theta)$,
\n
where $p$ is the dipole moment, $E$ is the strength of the electric field, and $\\theta$ is the angle the dipole is rotated through.
\n
In this case, the dipole moment $p$ is $6.0 \\times 10^{-6} ~\\mathrm{C m}$, the electric field $E$ is $1.5 \\times 10^{3} ~\\mathrm{NC}^{-1}$, and the angle $\\theta$ is $180^{\\circ}$.
A dipole comprises of two charged particles of identical magnitude $$q$$ and opposite in nature. The mass 'm' of the positive charged particle is half of the mass of the negative charged particle. The two charges are separated by a distance '$$l$$'. If the dipole is placed in a uniform electric field '$$\\bar{E}$$'; in such a way that dipole axis makes a very small angle with the electric field, '$$\\bar{E}$$'. The angular frequency of the oscillations of the dipole when released is given by:
",
"options": [
{
"text": "$$\\sqrt{\\frac{3 q E}{2 m l}}$$"
},
{
"text": "$$\\sqrt{\\frac{4 q E}{m l}}$$"
},
{
"text": "$$\\sqrt{\\frac{8 q E}{3 m l}}$$"
},
{
"text": "$$\\sqrt{\\frac{8 q E}{m l}}$$"
}
],
"answer": "$$\\sqrt{\\frac{3 q E}{2 m l}}$$",
"solution": "**Answer:** $$\\sqrt{\\frac{3 q E}{2 m l}}$$\n\n If released, it will oscillate about centre of mass.
Two charges of $$-4 \\mu \\mathrm{C}$$ and $$+4 \\mu \\mathrm{C}$$ are placed at the points $$\\mathrm{A}(1,0,4) \\mathrm{m}$$ and $$\\mathrm{B}(2,-1,5) \\mathrm{m}$$ located in an electric field $$\\overrightarrow{\\mathrm{E}}=0.20 \\hat{i} \\mathrm{~V} / \\mathrm{cm}$$. The magnitude of the torque acting on the dipole is $$8 \\sqrt{\\alpha} \\times 10^{-5} \\mathrm{Nm}$$, where $$\\alpha=$$ _________.
$$ \\Rightarrow q = {Q \\over 4}\\left( {2\\sqrt 2 + 1} \\right).$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9253,
"subject": "Physics",
"question": "A charged oil drop is suspended in a uniform field of $$3 \\times {10^4}$$ $$v/m$$ so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge $$ = 9.9 \\times {10^{ - 15}}\\,\\,kg$$ and $$g = 10\\,m/{s^2}$$) ",
"options": [
{
"text": "$$1.6 \\times {10^{ - 18}}\\,C$$ "
},
{
"text": "$$3.2 \\times {10^{ - 18}}\\,C$$"
},
{
"text": "$$3.3 \\times {10^{ - 18}}\\,C$$"
},
{
"text": "$$4.8 \\times {10^{ - 18}}\\,C$$"
}
],
"answer": "$$3.3 \\times {10^{ - 18}}\\,C$$",
"solution": "**Answer:** $$3.3 \\times {10^{ - 18}}\\,C$$\n\nAt equilibrium, electric force on drop balances weight of drop.\n
$$ = 3.3 \\times {10^{ - 18}}C$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9254,
"subject": "Physics",
"question": "Two point charges $$+8q$$ and $$-2q$$ are located at $$x=0$$ and $$x=L$$ respectively. The location of a point on the $$x$$ axis at which the net electric field due to these two point charges is zero is ",
"options": [
{
"text": "$${L \\over 4}$$ "
},
{
"text": "$$2$$ $$L$$ "
},
{
"text": "$$4$$ $$L$$ "
},
{
"text": "$$8$$ $$L$$"
}
],
"answer": "$$2$$ $$L$$ ",
"solution": "**Answer:** $$2$$ $$L$$ \n\n$${{ - K2q} \\over {{{\\left( {x - L} \\right)}^2}}} + {{K8q} \\over {{x^2}}} = 0 \\Rightarrow {1 \\over {{{\\left( {x - L} \\right)}^2}}} = {4 \\over {{x^2}}}$$\n
or, $$x=2L$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9255,
"subject": "Physics",
"question": "Let $$P\\left( r \\right) = {Q \\over {\\pi {R^4}}}r$$ be the change density distribution for a solid sphere of radius $$R$$ and total charge $$Q$$. For a point $$'p'$$ inside the sphere at distance $${r_1}$$ from the center of the sphere, the magnitude of electric field is : ",
"options": [
{
"text": "$${Q \\over {4\\pi \\,{ \\in _0}\\,r_1^2}}$$ "
},
{
"text": "$${{Qr_1^2} \\over {4\\pi \\,{ \\in _0}\\,{R^4}}}$$"
},
{
"text": "$${{Qr_1^2} \\over {3\\pi \\,{ \\in _0}\\,{R^4}}}$$ "
},
{
"text": "$$0$$"
}
],
"answer": "$${{Qr_1^2} \\over {4\\pi \\,{ \\in _0}\\,{R^4}}}$$",
"solution": "**Answer:** $${{Qr_1^2} \\over {4\\pi \\,{ \\in _0}\\,{R^4}}}$$\n\n\n
Let us consider a spherical shell of thickness $$dx$$ and radius $$x.$$ The volume of this spherical shell $$ = 4\\pi {r^2}dr.$$\n
$$ = {{{\\rho _0}r} \\over {4{ \\in _0}}}\\left( {{5 \\over 3} - {r \\over R}} \\right)$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9257,
"subject": "Physics",
"question": "A body of mass $$M$$ and charge $$q$$ is connected to spring of spring constant $$k.$$ It is oscillating along $$x$$-direction about its equilibrium position, taken to be at $$x=0,$$ with an amplitude $$A$$. An electric field $$E$$ is applied along the $$x$$-direction. Which of the following statements is correct ?",
"options": [
{
"text": "The new equilibrium position is at a distance $${{qE} \\over {2k}}$$ from $$x=0.$$"
},
{
"text": "The total energy of the system is $${1 \\over 2}m{\\omega ^2}{A^2} + {1 \\over 2}{{{q^2}{E^2}} \\over k}.$$"
},
{
"text": "The total energy of the system is $${1 \\over 2}m{\\omega ^2}{A^2} - {1 \\over 2}{{{q^2}{E^2}} \\over k}.$$"
},
{
"text": "The new equilibrium position is at a distance $${{2qE} \\over k}$$ from $$x=0.$$"
}
],
"answer": "The total energy of the system is $${1 \\over 2}m{\\omega ^2}{A^2} + {1 \\over 2}{{{q^2}{E^2}} \\over k}.$$",
"solution": "**Answer:** The total energy of the system is $${1 \\over 2}m{\\omega ^2}{A^2} + {1 \\over 2}{{{q^2}{E^2}} \\over k}.$$\n\nOn the body of charge q a electric fied E is applied, because of this equilibrium position of body will shift to a point where resulttant force is zero. \n
From eqns. (i) and (ii), E = $${{{\\rho _o}{R^3}} \\over {12{ \\in _o}{r^2}}}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9259,
"subject": "Physics",
"question": "For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its center. Then value of h is : ",
"options": [
{
"text": "$${R \\over {\\sqrt 5 }}$$"
},
{
"text": "$${R \\over {\\sqrt 2 }}$$"
},
{
"text": "R"
},
{
"text": "R$$\\sqrt 2 $$"
}
],
"answer": "$${R \\over {\\sqrt 2 }}$$",
"solution": "**Answer:** $${R \\over {\\sqrt 2 }}$$\n\n
\n$$ = {{9kq{d^2}} \\over {{D^4}}}$$\n",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9261,
"subject": "Physics",
"question": "The bob of a simple pendulum has mass 2g and\na charge of 5.0 μC. It is at rest in a uniform\nhorizontal electric field of intensity 2000 V/m.\nAt equilibrium, the angle that the pendulum\nmakes with the vertical is : (take g = 10 m/s2)",
"options": [
{
"text": "tan–1(5.0)"
},
{
"text": "tan–1(2.0)"
},
{
"text": "tan–1(0.5)"
},
{
"text": "tan–1(0.2)"
}
],
"answer": "tan–1(0.5)",
"solution": "**Answer:** tan–1(0.5)\n\nTcos$$\\theta $$ = mg
$$\\overrightarrow E $$ = $${\\overrightarrow E _1}$$ + $${\\overrightarrow E _2}$$\n
$$ = \\left( {63\\widehat i - 27\\widehat j} \\right) \\times {10^2}\\,\\,V/m$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9263,
"subject": "Physics",
"question": "Consider the force F on a charge 'q' due to a uniformly charged spherical shell of radius R carrying\ncharge Q distributed uniformly over it. Which one of the following statements is true for F, if 'q' is\nplaced at distance r from the centre of the shell?",
"options": [
{
"text": "$${1 \\over {4\\pi {\\varepsilon _0}}}{{qQ} \\over {{R^2}}} > F > 0$$ for r < R"
},
{
"text": "$$F = {1 \\over {4\\pi {\\varepsilon _0}}}{{qQ} \\over {{r^2}}}$$ for r > R"
},
{
"text": "$$F = {1 \\over {4\\pi {\\varepsilon _0}}}{{qQ} \\over {{r^2}}}$$ for all r"
},
{
"text": "$$F = {1 \\over {4\\pi {\\varepsilon _0}}}{{qQ} \\over {{R^2}}}$$ for r < R"
}
],
"answer": "$$F = {1 \\over {4\\pi {\\varepsilon _0}}}{{qQ} \\over {{r^2}}}$$ for r > R",
"solution": "**Answer:** $$F = {1 \\over {4\\pi {\\varepsilon _0}}}{{qQ} \\over {{r^2}}}$$ for r > R\n\n\n Inside the shell for r < R\n
E = 0\n
hence F = 0\n
Outside the shell\n
E = $${1 \\over {4\\pi {\\varepsilon _0}}}{Q \\over {{r^2}}}$$\n
hence F = $${1 \\over {4\\pi {\\varepsilon _0}}}{{Qq} \\over {{r^2}}}$$ for r > R",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9264,
"subject": "Physics",
"question": "A particle of charge q and mass m is subjected to an electric field E = E0\n (1 – $$a$$x2) in the x-direction,\nwhere $$a$$ and E0\n are constants. Initially the particle was at rest at x = 0. Other than the initial\nposition the kinetic energy of the particle becomes zero when the distance of the particle from the\norigin is :",
"options": [
{
"text": "$$a$$"
},
{
"text": "$$\\sqrt {{2 \\over a}} $$"
},
{
"text": "$$\\sqrt {{3 \\over a}} $$"
},
{
"text": "$$\\sqrt {{1 \\over a}} $$"
}
],
"answer": "$$\\sqrt {{3 \\over a}} $$",
"solution": "**Answer:** $$\\sqrt {{3 \\over a}} $$\n\n$$W = \\Delta KE$$\n
As inital and final kinetic energy both are zero so $$\\Delta KE$$ = 0\n
$$ \\Rightarrow $$ $${x = \\sqrt {{3 \\over a}} }$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9265,
"subject": "Physics",
"question": "Find out the surface charge density at the intersection of point x = 3 m plane and x-axis, in the region of uniform line charge of 8 nC/m lying along the z-axis in free space.",
"options": [
{
"text": "0.424 nC m$$-$$2"
},
{
"text": "4.0 nC m$$-$$2"
},
{
"text": "47.88 C/m"
},
{
"text": "0.07 nC m$$-$$2"
}
],
"answer": "0.424 nC m$$-$$2",
"solution": "**Answer:** 0.424 nC m$$-$$2\n\nElectric field due to wire is given by $$E = {{2k\\lambda } \\over r}$$
Electric field with surface charge density $$E = {\\sigma \\over {{ \\in _0}}}$$
$$ \\Rightarrow \\sigma = 0.424$$ n Cm$$-$$2",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9266,
"subject": "Physics",
"question": "An oil drop of radius 2 mm with a density 3g cm$$-$$3 is held stationary under a constant electric field 3.55 $$\\times$$ 105 V m$$-$$1 in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (consider g = 9.81 m/s2)",
"options": [
{
"text": "48.8 $$\\times$$ 1011"
},
{
"text": "1.73 $$\\times$$ 1010"
},
{
"text": "17.3 $$\\times$$ 1010"
},
{
"text": "1.73 $$\\times$$ 1012"
}
],
"answer": "1.73 $$\\times$$ 1010",
"solution": "**Answer:** 1.73 $$\\times$$ 1010\n\n\n Fe = qE = (ne)E
A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength 1 $$\\times$$ 105 NC$$-$$1. If the charge on the particle is 40 $$\\mu$$C and the initial velocity is 200 ms$$-$$1, how much distance it will travel before coming to the rest momentarily :
Two point charges A and B of magnitude +8 $$\\times$$ 10$$-$$6 C and $$-$$8 $$\\times$$ 10$$-$$6 C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4 $$\\times$$ 104 NC$$-$$1. The distance 'd' between the point charges A and B is :
Statement I : A point charge is brought in an electric field. The value of electric field at a point near to the charge may increase if the charge is positive.
\n
Statement II : An electric dipole is placed in a non-uniform electric field. The net electric force on the dipole will not be zero.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "Both Statement I and Statement II are true."
},
{
"text": "Both Statement I and Statement II are false."
},
{
"text": "Statement I is true but Statement II is false."
},
{
"text": "Statement I is false but Statement II is true."
}
],
"answer": "Both Statement I and Statement II are true.",
"solution": "**Answer:** Both Statement I and Statement II are true.\n\n
As one moves closer to a positive charge (isolated) the density of electric field line increases and so does the electric field intensity
\n
$$\\Rightarrow$$ Statement I is true
\n
As opposite poles of an electric dipole would experience equal and opposite forces so net force on a dipole in a uniform electric field will be zero
A vertical electric field of magnitude 4.9 $$\\times$$ 105 N/C just prevents a water droplet of a mass 0.1 g from falling. The value of charge on the droplet will be :
Two uniformly charged spherical conductors $$A$$ and $$B$$ of radii $$5 \\mathrm{~mm}$$ and $$10 \\mathrm{~mm}$$ are separated by a distance of $$2 \\mathrm{~cm}$$. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere $$A$$ and $$B$$ will be :
Two equal positive point charges are separated by a distance $$2 a$$. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge $$\\mathrm{q}_{0}$$ becomes maximum is $$\\frac{a}{\\sqrt{x}}$$. The value of $$x$$ is __________.
Electric field in a certain region is given by $$\\overrightarrow{\\mathrm{E}}=\\left(\\frac{\\mathrm{A}}{x^{2}} \\hat{i}+\\frac{\\mathrm{B}}{y^{3}} \\hat{j}\\right) \\text {. The } \\mathrm{SI} \\text { unit of } \\mathrm{A} \\text { and } \\mathrm{B}$$ are :
A point charge $$2\\times10^{-2}~\\mathrm{C}$$ is moved from P to S in a uniform electric field of $$30~\\mathrm{NC^{-1}}$$ directed along positive x-axis. If coordinates of P and S are (1, 2, 0) m and (0, 0, 0) m respectively, the work done by electric field will be
A point charge of 10 $$\\mu$$C is placed at the origin. At what location on the X-axis should a point charge of 40 $$\\mu$$C be placed so that the net electric field is zero at $$x=2$$cm on the X-axis?
A stream of a positively charged particles having $${q \\over m} = 2 \\times {10^{11}}{C \\over {kg}}$$ and velocity $${\\overrightarrow v _0} = 3 \\times {10^7}\\widehat i\\,m/s$$ is deflected by an electric field $$1.8\\widehat j$$ kV/m. The electric field exists in a region of 10 cm along $$x$$ direction. Due to the electric field, the deflection of the charge particles in the $$y$$ direction is _________ mm.
",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n \n$$\n\\begin{aligned}\n& F_y=\\frac{q E_y}{m} \\\\\\\\\n& a_y=2 \\times 10^{11} \\times 1800 \\\\\\\\\n& =36 \\times 10^{13} \\mathrm{~m} / \\mathrm{s}^2 \\\\\\\\\n& \\text { Time }=\\frac{10 \\times 10^{-2}}{v_0}=\\frac{0.1}{3 \\times 10^7}=\\left(\\frac{1}{3} \\times 10^{-8}\\right) \\mathrm{sec} \\text {. } \\\\\\\\\n& \\therefore \\quad y=\\frac{1}{2} a t^2 \\\\\\\\\n& \\Rightarrow y=\\frac{1}{2} \\times 36 \\times 10^{13} \\times\\left(\\frac{1}{3} \\times 10^{-8}\\right)^2 \\\\\\\\\n& =2 \\times 10^{-3} \\mathrm{~m} \\\\\\\\\n& =2 \\mathrm{~mm} \\\\\\\\\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9278,
"subject": "Physics",
"question": "The electric field due to a short electric dipole at a large distance $(r)$ from center of dipole on the equatorial plane varies with distance as :",
"options": [
{
"text": "$\\frac{1}{r^{2}}$"
},
{
"text": "$\\frac{1}{r}$"
},
{
"text": "$r$"
},
{
"text": "$\\frac{1}{r^{3}}$"
}
],
"answer": "$\\frac{1}{r^{3}}$",
"solution": "**Answer:** $\\frac{1}{r^{3}}$\n\nAt a large distance $r$ from the center of a short electric dipole, the electric field on the equatorial plane can be approximated as:\n
\nwhere $p$ is the dipole moment of the electric dipole, and $\\epsilon_0$ is the permittivity of free space.\n
\nThis formula is derived using the concept of electric dipole moment, which is defined as:\n
\n$$\n\\vec{p} = q\\vec{d}\n$$\n
\nwhere $q$ is the magnitude of the electric charge, and $\\vec{d}$ is the separation vector between the positive and negative charges of the dipole. The electric field at a point on the equatorial plane of the dipole is due to the electric field of the positive and negative charges at that point.
Since the charges are equal in magnitude and opposite in sign, their electric fields at a point on the equatorial plane cancel out along the axis of the dipole, leaving only the component perpendicular to the axis.
This perpendicular component of the electric field is proportional to the dipole moment $p$ and inversely proportional to the cube of the distance $r$ from the center of the dipole.\n
\nTherefore, the electric field due to a short electric dipole at a large distance $r$ from the center of the dipole on the equatorial plane varies with distance as:\n
\n$$\n\\boxed{E \\propto \\frac{1}{r^3}}\n$$\n
\nwhere the proportionality constant is $\\frac{1}{4\\pi\\epsilon_0}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9279,
"subject": "Physics",
"question": "
A thin infinite sheet charge and an infinite line charge of respective charge densities $$+\\sigma$$ and $$+\\lambda$$ are placed parallel at $$5 \\mathrm{~m}$$ distance from each other. Points 'P' and 'Q' are at $$\\frac{3}{\\pi}$$ m and $$\\frac{4}{\\pi}$$ m perpendicular distances from line charge towards sheet charge, respectively. '$$\\mathrm{E}_{\\mathrm{P}}$$' and '$$\\mathrm{E}_{\\mathrm{Q}}$$' are the magnitudes of resultant electric field intensities at point 'P' and 'Q', respectively. If $$\\frac{E_{p}}{E_{0}}=\\frac{4}{a}$$ for $$2|\\sigma|=|\\lambda|$$, then the value of $$a$$ is ___________.
",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\nWith the given equations:\n
\nAs given in the question, $$\\frac{E_P}{E_0} = \\frac{4}{a}$$, and since $$\\frac{E_P}{E_Q} = \\frac{2}{3}$$, we can say $$\\frac{E_P}{E_Q} = \\frac{E_P}{2E_Q} = \\frac{4}{2 \\times 3}$$ = $\\frac{4}{ 6}$ .\n
\nSo, the value of $$a$$ is $$\\boxed{6}$$.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9280,
"subject": "Physics",
"question": "
In a metallic conductor, under the effect of applied electric field, the free electrons of the conductor
",
"options": [
{
"text": "move in the straight line paths in the same direction"
},
{
"text": "move with the uniform velocity throughout from lower potential to higher potential"
},
{
"text": "drift from higher potential to lower potential."
},
{
"text": "move in the curved paths from lower potential to higher potential"
}
],
"answer": "move in the curved paths from lower potential to higher potential",
"solution": "**Answer:** move in the curved paths from lower potential to higher potential\n\nElectron drifts from lower potential to higher potential on curved path.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9281,
"subject": "Physics",
"question": "Suppose a uniformly charged wall provides a uniform electric field of $2 \\times 10^4 \\mathrm{~N} / \\mathrm{C}$ normally. A charged particle of mass $2 \\mathrm{~g}$ being suspended through a silk thread of length $20 \\mathrm{~cm}$ and remain stayed at a distance of $10 \\mathrm{~cm}$ from the wall.
Then the charge on the particle will be $\\frac{1}{\\sqrt{x}} \\mu \\mathrm{C}$ where $x=$ ___________ . [use $\\mathrm{g}=10 \\mathrm{~m} / \\mathrm{s}^2$ ]
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
\n
Assertion (A) : Work done by electric field on moving a positive charge on an equipotential surface is always zero.
\n
Reason (R) : Electric lines of forces are always perpendicular to equipotential surfaces.
\n
In the light of the above statements, choose the most appropriate answer from the options given below :
",
"options": [
{
"text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)\n"
},
{
"text": "(A) is correct but (R) is not correct\n"
},
{
"text": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)\n"
},
{
"text": "(A) is not correct but (R) is correct"
}
],
"answer": "Both (A) and (R) are correct and (R) is the correct explanation of (A)\n",
"solution": "**Answer:** Both (A) and (R) are correct and (R) is the correct explanation of (A)\n\n\n
The most appropriate answer from the options given would be Option A: Both (A) and (R) are correct, and (R) is the correct explanation of (A).
\n
Here is the reasoning for this answer:
\n
Assertion (A) states that the work done by an electric field on moving a positive charge on an equipotential surface is always zero. This statement is true because by definition, an equipotential surface is a surface over which the electric potential is constant. When a charge moves along an equipotential surface, there is no change in its electric potential energy since potential difference $$ \\Delta V $$ is zero. Work done ($$ W $$) is defined as the product of charge ($$ q $$), potential difference ($$ \\Delta V $$), and the cosine of the angle between the field and direction of motion ($$ \\cos \\theta $$), which can be written as:
\n
$$ W = q \\Delta V \\cos \\theta $$
\n
Because $$ \\Delta V = 0 $$ on an equipotential surface, irrespective of the value of $$ \\cos \\theta $$, the work $$ W $$ will be zero. Hence, the Assertion (A) is correct.
\n
Reason (R) says that electric lines of forces are always perpendicular to equipotential surfaces. This statement is also correct as the electric field lines, by definition, are directed such that they are tangent to the electric field vector at any point in space. Since the electric potential is constant on an equipotential surface, there can be no component of the electric field parallel to the surface, as that would imply a force and potential change along the surface. The electric field thus must be perpendicular to the equipotential surface, which means that the electric field lines must also be perpendicular to the equipotential surface. In other words, the electric field does no work when a charge moves along an equipotential surface because the motion is perpendicular to the force.
\n
Therefore, Reason (R) is not only correct, but it is also the correct explanation for Assertion (A), making Option A the right choice.
Two charges $$q$$ and $$3 q$$ are separated by a distance '$$r$$' in air. At a distance $$x$$ from charge $$q$$, the resultant electric field is zero. The value of $$x$$ is :
An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet $$\\mathrm{S}$$ having surface charge density $$+\\sigma$$. The electron at $$t=0$$ is at a distance of $$1 \\mathrm{~m}$$ from $$S$$ and has a speed of $$1 \\mathrm{~m} / \\mathrm{s}$$. The maximum value of $$\\sigma$$ if the electron strikes $$S$$ at $$t=1 \\mathrm{~s}$$ is $$\\alpha\\left[\\frac{m \\epsilon_0}{e}\\right] \\frac{C}{m^2}$$, the value of $$\\alpha$$ is ___________.
An infinite plane sheet of charge having uniform surface charge density $$+\\sigma_{\\mathrm{s}} \\mathrm{C} / \\mathrm{m}^2$$ is placed on $$x$$-$$y$$ plane. Another infinitely long line charge having uniform linear charge density $$+\\lambda_e \\mathrm{C} / \\mathrm{m}$$ is placed at $$z=4 \\mathrm{~m}$$ plane and parallel to $$y$$-axis. If the magnitude values $$\\left|\\sigma_{\\mathrm{s}}\\right|=2\\left|\\lambda_{\\mathrm{e}}\\right|$$ then at point $$(0,0,2)$$, the ratio of magnitudes of electric field values due to sheet charge to that of line charge is $$\\pi \\sqrt{n}: 1$$. The value of $$n$$ is _________.
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9286,
"subject": "Physics",
"question": "If the electric flux entering and leaving an enclosed surface respectively is $${\\phi _1}$$ and $${\\phi _2},$$ the electric charge inside the surface will be",
"options": [
{
"text": "$$\\left( {{\\phi _2} - {\\phi _1}} \\right){\\varepsilon _0}$$ "
},
{
"text": "$$\\left( {{\\phi _2} + {\\phi _1}} \\right)/{\\varepsilon _0}$$"
},
{
"text": "$$\\left( {{\\phi _1} - {\\phi _2}} \\right)/{\\varepsilon _0}$$ "
},
{
"text": "$$\\left( {{\\phi _1} + {\\phi _2}} \\right){\\varepsilon _0}$$ "
}
],
"answer": "$$\\left( {{\\phi _2} - {\\phi _1}} \\right){\\varepsilon _0}$$ ",
"solution": "**Answer:** $$\\left( {{\\phi _2} - {\\phi _1}} \\right){\\varepsilon _0}$$ \n\nThe flux entering an enclosed surface is taken as negative and the flux leaving the surface is taken as positive, by convention. Therefore the net flux leaving the enclosed surface $$ = {\\phi _2} - {\\phi _1}$$\n
$$\\therefore$$ the change enclosed in the surface by Gauss's law is $$q = { \\varepsilon _0}\\,\\left( {{\\phi _2} - {\\phi _1}} \\right)$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9287,
"subject": "Physics",
"question": "Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by $$\\rho $$(r) = kr, where\nr is the distance from the centre. Two charges A and B, of –Q each, are placed on diametrically opposite\npoints, at equal distance, $$a$$ from the centre. If A and B do not experience any force, then :",
"options": [
{
"text": "$$a = {8^{ - 1/4}}R$$"
},
{
"text": "$$a = {2^{ - 1/4}}R$$"
},
{
"text": "$$a = {{3R} \\over {{2^{1/4}}}}$$"
},
{
"text": "$$a = {R \\over {\\sqrt 3 }}$$"
}
],
"answer": "$$a = {8^{ - 1/4}}R$$",
"solution": "**Answer:** $$a = {8^{ - 1/4}}R$$\n\nTotal charge = 2Q\n Charging density $$\\rho $$ = kr\n Radius = R\n\nCharge enclosed in the sphere,\n qin = $$\\int\\limits_0^R {\\rho dV} $$\n
$$ \\Rightarrow $$ $$a = {8^{ - 1/4}}R$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9288,
"subject": "Physics",
"question": "In finding the electric field using Gauss Law\nthe formula $$\\left| {\\overrightarrow E } \\right| = {{{q_{enc}}} \\over {{\\varepsilon _0}\\left| A \\right|}}$$ is applicable. In the\nformula $${{\\varepsilon _0}}$$ is permittivity of free space, A is the\narea of Gaussian surface and qenc is charge\nenclosed by the Gaussian surface. The equation\ncan be used in which of the following situation?",
"options": [
{
"text": "Only when $$\\left| {\\overrightarrow E } \\right|$$ = constant on the surface."
},
{
"text": "For any choice of Gaussian surface."
},
{
"text": "Only when the Gaussian surface is an\nequipotential surface."
},
{
"text": "Only when the Gaussian surface is an\nequipotential surface and $$\\left| {\\overrightarrow E } \\right|$$ is constant on\nthe surface."
}
],
"answer": "Only when the Gaussian surface is an\nequipotential surface and $$\\left| {\\overrightarrow E } \\right|$$ is constant on\nthe surface.",
"solution": "**Answer:** Only when the Gaussian surface is an\nequipotential surface and $$\\left| {\\overrightarrow E } \\right|$$ is constant on\nthe surface.\n\nBy Gauss law\n
$$\\oint {\\overrightarrow E .d\\overrightarrow A } = {{{q_{in}}} \\over {{\\varepsilon _0}}}$$\n
When $${\\overrightarrow E ||\\overrightarrow A }$$ and $${\\left| {\\overrightarrow E } \\right|}$$ is constant then\n
$$ \\therefore $$ $${\\left| {\\overrightarrow E } \\right|}$$ should be constant on the surface and the\nsurface should be equipotential.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9289,
"subject": "Physics",
"question": "The electric field in a region is given by $$\\overrightarrow E = \\left( {{3 \\over 5}{E_0}\\widehat i + {4 \\over 5}{E_0}\\widehat j} \\right){N \\over C}$$. The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y $$-$$ z plane) to that of the surface of area 0.3 m2 (parallel to x $$-$$ z plane) is a : b, where a = __________ [Here $${\\widehat i}$$, $${\\widehat j}$$ and $${\\widehat k}$$ are unit vectors along x, y and z-axes respectively.]",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n$$\\phi = \\overrightarrow E \\,.\\,\\overrightarrow A $$
$${a \\over b} = {{{\\phi _a}} \\over {{\\phi _b}}} = {{{3 \\over 5}{E_0} \\times 0.2} \\over {{4 \\over 5}{E_0} \\times 0.3}} = {6 \\over {12}} = 0.5$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9290,
"subject": "Physics",
"question": "The electric field in a region is given by $$\\overrightarrow E = {2 \\over 5}{E_0}\\widehat i + {3 \\over 5}{E_0}\\widehat j$$ with $${E_0} = 4.0 \\times {10^3}{N \\over C}$$. The flux of this field through a rectangular surface area 0.4 m2 parallel to the Y-Z plane is __________ Nm2C$$-$$1.",
"options": [],
"answer": "640",
"solution": "**Answer:** 640\n\n$$\\phi = \\overrightarrow E \\,.\\,\\overrightarrow A $$
$$ = 640$$ Nm2 C$$-$$1",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9291,
"subject": "Physics",
"question": "The total charge enclosed in an incremental volume of 2 $$\\times$$ 10$$-$$9 m3 located at the origin is ___________ nC, if electric flux density of its field is found as
D = e$$-$$x sin y $$\\widehat i$$ $$-$$ e$$-$$x cos y $$\\widehat j$$ + 2z $$\\widehat k$$ C/m2",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n$$\\overline D = {\\varepsilon _0}\\overline E $$
$$Div.\\,\\overline E = {\\rho \\over {{\\varepsilon _0}}}$$
A cubical volume is bounded by the surfaces $$\\mathrm{x}=0, x=\\mathrm{a}, y=0, y=\\mathrm{a}, \\mathrm{z}=0, z=\\mathrm{a}$$. The electric field in the region is given by $$\\overrightarrow{\\mathrm{E}}=\\mathrm{E}_{0} x \\hat{i}$$. Where $$\\mathrm{E}_{0}=4 \\times 10^{4} ~\\mathrm{NC}^{-1} \\mathrm{~m}^{-1}$$. If $$\\mathrm{a}=2 \\mathrm{~cm}$$, the charge contained in the cubical volume is $$\\mathrm{Q} \\times 10^{-14} \\mathrm{C}$$. The value of $$\\mathrm{Q}$$ is ________________.
In a cuboid of dimension $$2 \\mathrm{~L} \\times 2 \\mathrm{~L} \\times \\mathrm{L}$$, a charge $$q$$ is placed at the center of the surface '$$\\mathrm{S}$$' having area of $$4 \\mathrm{~L}^{2}$$. The flux through the opposite surface to '$$\\mathrm{S}$$' is given by
",
"options": [
{
"text": "$$\\frac{q}{2 \\epsilon_{0}}$$"
},
{
"text": "$$\\frac{q}{3 \\epsilon_{0}}$$"
},
{
"text": "$$\\frac{q}{12 \\epsilon_{0}}$$"
},
{
"text": "$$\\frac{q}{6 \\in_{0}}$$"
}
],
"answer": "$$\\frac{q}{6 \\in_{0}}$$",
"solution": "**Answer:** $$\\frac{q}{6 \\in_{0}}$$\n\n \nIf we consider a similar box above this box then it becomes cube of side length $2 L$\n
\n$\\phi$ through a surface $=\\frac{q}{6 \\varepsilon_{0}}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9295,
"subject": "Physics",
"question": "
Two charges of $$5 Q$$ and $$-2 Q$$ are situated at the points $$(3 a, 0)$$ and $$(-5 a, 0)$$ respectively. The electric flux through a sphere of radius '$$4 a$$' having center at origin is :
$$Q_{\\text{enc}}$$ = total charge enclosed by the surface
\n
$$\\varepsilon_0$$ = permittivity of free space (electric constant)
\n
\n\n
In this scenario, we have a sphere of radius $$4a$$ with its center at the origin and two charges, $$5Q$$ at point (3a, 0) and $$-2Q$$ at point (-5a, 0). Since the sphere's radius is $$4a$$, the charge $$5Q$$, which is located at (3a, 0), lies inside the sphere, whereas the charge $$-2Q$$, located at (-5a, 0), lies outside the sphere. Gauss's law only considers charges that are enclosed within the surface.
\n\n
Thus, the charge $$-2Q$$ has no impact on the electric flux through the sphere because it is not enclosed by the sphere. Only the charge $$5Q$$ contributes to the electric flux inside the sphere.
\n\n
The electric flux through the sphere is therefore given simply by the enclosed charge:
An electric field is given by $$(6 \\hat{i}+5 \\hat{j}+3 \\hat{k}) \\mathrm{N} / \\mathrm{C}$$. The electric flux through a surface area $$30 \\hat{i} \\mathrm{~m}^2$$ lying in YZ-plane (in SI unit) is :
A particle of charge '$$-q$$' and mass '$$m$$' moves in a circle of radius '$$r$$' around an infinitely long line charge of linear charge density '$$+\\lambda$$'. Then time period will be given as :
\n
(Consider $$k$$ as Coulomb's constant)
",
"options": [
{
"text": "$$T^2=\\frac{4 \\pi^2 m}{2 k \\lambda q} r^3$$\n"
},
{
"text": "$$T=\\frac{1}{2 \\pi r} \\sqrt{\\frac{m}{2 k \\lambda q}}$$\n"
},
{
"text": "$$T=\\frac{1}{2 \\pi} \\sqrt{\\frac{2 k \\lambda q}{m}}$$"
},
{
"text": "$$T=2 \\pi r \\sqrt{\\frac{m}{2 k \\lambda q}}$$"
}
],
"answer": "$$T=2 \\pi r \\sqrt{\\frac{m}{2 k \\lambda q}}$$",
"solution": "**Answer:** $$T=2 \\pi r \\sqrt{\\frac{m}{2 k \\lambda q}}$$\n\n
When considering the electric flux linked with a cube due to a charge placed at one of its surfaces, it's important to apply Gauss's law. Gauss's law states that the total electric flux through a closed surface is equal to $$\\frac{q_{\\text{enc}}}{\\epsilon_0}$$, where $$q_{\\text{enc}}$$ is the charge enclosed by the surface and $$\\epsilon_0$$ is the permittivity of free space.
\n\n
In the given scenario, the charge $$q$$ is placed at the center of one of the surfaces of the cube. Conceptually, we can think of this arrangement as part of a larger situation where if we had a larger, imaginary cube that encompasses the entire setup such that the point charge is at its geometric center, the charge would then be uniformly distributing its flux through all six faces of this larger cube. However, since the actual setup involves only one cube with one face adjacent to the charge, we essentially have only one-half of the total possible geometry through which the flux from this charge can emerge - implying the flux through our actual cube is a fraction of the total flux that would emanate from the charge if it were centrally placed within a larger, encompassing cube.
\n\n
Therefore, only half of the flux emanating from the charge will pass through the actual cube because the charge is placed directly on one of its surfaces, effectively distributing its influence through 180 degrees (half of the space around it) instead of the full 360 degrees. Hence, the flux linked with the cube will be half of the total flux $$\\frac{q}{\\epsilon_0}$$, which is calculated using Gauss's law for a point charge.
\n\n
Therefore, the correct answer is $$\\frac{q}{2\\epsilon_0}$$.
\n\n
Option A $$\\frac{q}{2 \\epsilon_0}$$ is the correct choice.
An electric field, $$\\overrightarrow{\\mathrm{E}}=\\frac{2 \\hat{i}+6 \\hat{j}+8 \\hat{k}}{\\sqrt{6}}$$ passes through the surface of $$4 \\mathrm{~m}^2$$ area having unit vector $$\\hat{n}=\\left(\\frac{2 \\hat{i}+\\hat{j}+\\hat{k}}{\\sqrt{6}}\\right)$$. The electric flux for that surface is _________ $$\\mathrm{Vm}$$.
where $$\\overrightarrow{\\mathrm{E}}$$ is the electric field, $$\\overrightarrow{\\mathrm{A}}$$ is the area vector (with magnitude equal to the area of the surface and direction perpendicular to the surface, defined by the unit vector $$\\hat{n}$$), and $$\\theta$$ is the angle between $$\\overrightarrow{\\mathrm{E}}$$ and $$\\overrightarrow{\\mathrm{A}}$$. However, when using unit vectors to describe the directions of $$\\overrightarrow{\\mathrm{E}}$$ and $$\\hat{n}$$, the dot product can be used to simplify the calculation as follows:
Given that $$\\overrightarrow{\\mathrm{E}}=\\frac{2 \\hat{i}+6 \\hat{j}+8 \\hat{k}}{\\sqrt{6}}$$ and $$\\hat{n}=\\left(\\frac{2 \\hat{i}+\\hat{j}+\\hat{k}}{\\sqrt{6}}\\right)$$, and the area $$A = 4 \\mathrm{m}^2$$, we can substitute them into our formula. Note that since $$\\overrightarrow{\\mathrm{A}} = A\\hat{n}$$, the magnitude of the area vector is the area of the surface itself. First, let's find $$\\overrightarrow{\\mathrm{E}} \\cdot \\hat{n}$$:
$$\\sigma$$ is the uniform surface charge density of a thin spherical shell of radius R. The electric field at any point on the surface of the spherical shell is :
The question is about calculating the electric field at the surface of a thin spherical shell with a uniform surface charge density denoted by $$\\sigma$$. To determine the electric field at any point on the surface of the shell, we can use Gauss's law, which is particularly useful for systems with high symmetry like a spherical shell.
\n\n
Gauss's law in its integral form states that the electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space ($$ \\epsilon_0 $$), mathematically represented as:
In the case of a spherical shell of radius $$R$$, the total charge $$Q_{\\text{enc}}$$ on the shell can be written in terms of the surface charge density $$\\sigma$$ as:
\n\n
$$ Q_{\\text{enc}} = \\sigma \\times 4\\pi R^2 $$
\n\n
Now, we apply Gauss's law using a Gaussian surface that coincides with the surface of the spherical shell. The electric field $$E$$ at the surface of the shell is uniform over the Gaussian surface, and the area of the Gaussian surface (which is also the area of the spherical shell) is $$4\\pi R^2$$. Thus, the electric flux $$\\Phi_E$$ through the Gaussian surface is:
Electric potential due to charge $$q$$ on the surface of the spherical shell at any point inside the shell is \n
$${V_2} = {1 \\over {4\\pi {\\varepsilon _0}}}{q \\over R}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9303,
"subject": "Physics",
"question": "Two spherical conductors $$B$$ and $$C$$ having equal radii and carrying equal charges on them repel each other with a force $$F$$ when kept apart at some distance. A third spherical conductor having same radius as that $$B$$ but uncharged is brought in contact with $$B,$$ then brought in correct with $$C$$ and finally removed away from both. The new force of repulsion between $$B$$ and $$C$$ is ",
"options": [
{
"text": "$$F/8$$ "
},
{
"text": "$$3$$ $$F/4$$ "
},
{
"text": "$$F/4$$ "
},
{
"text": "$$3$$ $$F/8$$ "
}
],
"answer": "$$3$$ $$F/8$$ ",
"solution": "**Answer:** $$3$$ $$F/8$$ \n\n$$F \\propto {{{Q_A}{Q_C}} \\over {{x^2}}}$$\n
$$x$$ is distance between the spheres. After first operation charge on $$B$$ is halved i.e $${Q \\over 2}.$$ \n
and charge on third sphere becomes $${Q \\over 2}.$$ Now it is touched to $$C$$, charge then equally \n
distributes them selves to make potential same, hence charge on $$C$$ becomes \n
or, $${F_{new}} = {3 \\over 8}F$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9304,
"subject": "Physics",
"question": "A charge particle $$'q'$$ is shot towards another charged particle $$'Q'$$ which is fixed, with a speed $$'v'$$. It approaches $$'Q'$$ upto a closest distance $$r$$ and then returns. If $$q$$ were given a speed of $$'2v'$$ the closest distances of approaches would be ",
"options": [
{
"text": "$$r/2$$ "
},
{
"text": "$$2r$$ "
},
{
"text": "$$r$$ "
},
{
"text": "$$r/4$$ "
}
],
"answer": "$$r/4$$ ",
"solution": "**Answer:** $$r/4$$ \n\n$${1 \\over 2}m{v^2} = {{kQq} \\over r}$$\n
$$ = {1 \\over {2\\pi {\\varepsilon _0}}}\\left[ {{q \\over R} - {q \\over {\\sqrt {{R^2} + {d^2}} }}} \\right]$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9306,
"subject": "Physics",
"question": "Two spherical conductors $$A$$ and $$B$$ of radii $$1$$ $$mm$$ and $$2$$ $$mm$$ are separated by a distance of $$5$$ $$cm$$ and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres $$A$$ and $$B$$ is ",
"options": [
{
"text": "$$4:1$$ "
},
{
"text": "$$1:2$$ "
},
{
"text": "$$2:1$$ "
},
{
"text": "$$1:4$$ "
}
],
"answer": "$$2:1$$ ",
"solution": "**Answer:** $$2:1$$ \n\n\n
Since the distance between the spheres is large as compared to their diameters, the induced effects may be ignored.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9307,
"subject": "Physics",
"question": "An electric charge $${10^{ - 3}}\\,\\,\\mu \\,C$$ is placed at the origin $$(0,0)$$ of $$X-Y$$ co-ordinate system. Two points $$A$$ and $$B$$ are situated at $$\\left( {\\sqrt 2 ,\\sqrt 2 } \\right)$$ and $$\\left( {2,0} \\right)$$ respectively. The potential difference between the points $$A$$ and $$B$$ will be",
"options": [
{
"text": "$$4.5$$ volts "
},
{
"text": "$$9$$ volts"
},
{
"text": "zero "
},
{
"text": "$$2$$ volts"
}
],
"answer": "zero ",
"solution": "**Answer:** zero \n\n \n
The distance of point $$A\\left( {\\sqrt 2 ,\\sqrt 2 } \\right)$$ from the origin, \n
$$\\therefore$$ Potential difference between the points $$A$$ and $$B$$ is zero.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9308,
"subject": "Physics",
"question": "The potential at a point $$x$$ (measured in $$\\mu \\,m$$) due to some charges situated on the $$x$$-axis is given by $$V\\left( x \\right) = 20/\\left( {{x^2} - 4} \\right)$$ volt\n The electric field $$E$$ at $$x = 4\\,\\mu \\,m$$ is given by ",
"options": [
{
"text": "$$(10/9)$$ volt / $$\\mu $$ $$m$$ and in the $$ + ve$$ $$x$$ direction"
},
{
"text": "$$\\left( {5/3} \\right)$$ volt/ $$\\mu $$ $$m$$ and in the $$-ve$$ $$x$$ direction"
},
{
"text": "$$\\left( {5/3} \\right)$$ volt/$$\\mu $$ $$m$$ and in the $$+ve$$ $$x$$ direction "
},
{
"text": "$$\\left( {10/9} \\right)$$ volt/ $$\\mu \\,m$$ and in the $$-ve$$ $$x$$ direction "
}
],
"answer": "$$(10/9)$$ volt / $$\\mu $$ $$m$$ and in the $$ + ve$$ $$x$$ direction",
"solution": "**Answer:** $$(10/9)$$ volt / $$\\mu $$ $$m$$ and in the $$ + ve$$ $$x$$ direction\n\nHere, $$V\\left( x \\right) = {{20} \\over {{x^2} - 4}}volt$$\n
Positive sign indicates that $$\\overrightarrow E $$ is in $$+ve$$ $$x$$-direction.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9309,
"subject": "Physics",
"question": "(This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.)\n
Statement-1 : For a charged particle moving from point $$P$$ to point $$Q$$, the net work done by an electrostatic field on the particle is independent of the path connecting point $$P$$ to point $$Q.$$\n Statement-2 : The net work done by a conservative force on an object moving along a closed loop is zero.
",
"options": [
{
"text": "Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1."
},
{
"text": "Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1."
},
{
"text": "Statement- 1 is false, Statement- 2 is true."
},
{
"text": "Statement- 1 true, Statement- 2 is false "
}
],
"answer": "Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.",
"solution": "**Answer:** Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.\n\nStatement $$1$$ is true.\n
Statement $$2$$ is true and is the correct explanation of $$(1)$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9310,
"subject": "Physics",
"question": "Two points $$P$$ and $$Q$$ are maintained at the potentials of $$10$$ $$V$$ and $$-4$$ $$V$$, respectively. The work done in moving $$100$$ electrons from $$P$$ to $$Q$$ is : ",
"options": [
{
"text": "$$9.60 \\times {10^{ - 17}}J$$ "
},
{
"text": "$$ - 2.24 \\times {10^{ - 16}}J$$ "
},
{
"text": "$$ 2.24 \\times {10^{ - 16}}J$$ "
},
{
"text": "$$- 9.60 \\times {10^{ - 17}}J$$ "
}
],
"answer": "$$ 2.24 \\times {10^{ - 16}}J$$ ",
"solution": "**Answer:** $$ 2.24 \\times {10^{ - 16}}J$$ \n\n$$ {{{W_{PQ}}} \\over q} = \\left( {{V_Q} - {V_P}} \\right)$$\n
Charge density, $$\\rho = {{dq} \\over {4\\pi {r^2}dr}} = - 6{\\varepsilon _0}a$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9312,
"subject": "Physics",
"question": "This question has statement- $$1$$ and statement- $$2.$$ Of the four choices given after the statements, choose the one that best describe the two statements.\n An insulating solid sphere of radius $$R$$ has a uniformly positive charge density $$\\rho $$. As a result of this uniform charge distribution there is a finite value of electric potential at the center of the sphere, at the surface of the sphere and also at a point out side the sphere. The electric potential at infinite is zero. \n
Statement- $$1:$$ When a charge $$q$$ is take from the centre of the surface of the sphere its potential energy changes by $${{q\\rho } \\over {3{\\varepsilon _0}}}$$ \n Statement- $$2:$$ The electric field at a distance $$r\\left( {r < R} \\right)$$ from the center of the sphere is $${{\\rho r} \\over {3{\\varepsilon _0}}}.$$
",
"options": [
{
"text": "Statement- $$1$$ is true, Statement- $$2$$ is true; Statement- $$2$$ is not the correct explanation of Statement- $$1$$."
},
{
"text": "Statement $$1$$ is true, Statement $$2$$ is false."
},
{
"text": "Statement $$1$$ is false, Statement $$2$$ is true."
},
{
"text": "Statement- $$1$$ is true, Statement- $$2$$ is true; Statement- $$2$$ is the correct explanation of Statement- $$1$$."
}
],
"answer": "Statement $$1$$ is false, Statement $$2$$ is true.",
"solution": "**Answer:** Statement $$1$$ is false, Statement $$2$$ is true.\n\nThe electric field inside a uniformly charged sphere is \n
= $${{\\rho .r} \\over {3{ \\in _0}}}$$\n
The electric potential inside a uniformly charged sphere\n
$$\\Delta U = {{q\\rho {R^2}} \\over {6{ \\in _0}}}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9313,
"subject": "Physics",
"question": "Assume that an electric field $$\\overrightarrow E = 30{x^2}\\widehat i$$ exists in space. Then the potential difference $${V_A} - {V_O},$$ where $${V_O}$$ is the potential at the origin and $${V_A}$$ the potential at $$x=2$$ $$m$$ is :",
"options": [
{
"text": "$$120$$ $$J/C$$ "
},
{
"text": "$$-120$$ $$J/C$$ "
},
{
"text": "$$-80$$ $$J/C$$ "
},
{
"text": "$$80$$ $$J/C$$ "
}
],
"answer": "$$-80$$ $$J/C$$ ",
"solution": "**Answer:** $$-80$$ $$J/C$$ \n\nPotential difference between any two points in an electric field is given by,\n
$$dV = - \\overrightarrow E .\\overrightarrow {dx} $$\n
$${V_A} - {V_O} = \\left[ {10{x^3}} \\right]_{0}^2 = - 80J/C$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9314,
"subject": "Physics",
"question": "The potential (in volts) of a charge distribution is given by.\n
V(z) = 30 $$-$$ 5x2 for $$\\left| z \\right|$$ $$ \\le $$ 1 m.\n V(z) = 35 $$-$$ 10 $$\\left| z \\right|$$ for $$\\left| z \\right|$$ $$ \\ge $$1 m.\n
V(z) does not depend on x and y. If this potential is generated by a constant charge per unit volume $${\\rho _0}$$ (in units of $${\\varepsilon _0}$$) which is spread over a certain region, then choose the correct statement.",
"options": [
{
"text": "$${\\rho _0}$$ = 10 $${\\varepsilon _0}$$ for $$\\left| z \\right|$$ $$ \\le $$ 1 m and $${\\rho _0} = 0$$ elsewhere"
},
{
"text": "$${\\rho _0}$$ = 20 $${\\varepsilon _0}$$ in the entire region"
},
{
"text": "$${\\rho _0}$$ = 40 $${\\varepsilon _0}$$ in the entire region"
},
{
"text": "$${\\rho _0}$$ = 20 $${\\varepsilon _0}$$ for $$\\left| z \\right|$$ $$ \\le $$ 1 m and $${\\rho _0} = 0$$ elsewhere"
}
],
"answer": "$${\\rho _0}$$ = 10 $${\\varepsilon _0}$$ for $$\\left| z \\right|$$ $$ \\le $$ 1 m and $${\\rho _0} = 0$$ elsewhere",
"solution": "**Answer:** $${\\rho _0}$$ = 10 $${\\varepsilon _0}$$ for $$\\left| z \\right|$$ $$ \\le $$ 1 m and $${\\rho _0} = 0$$ elsewhere\n\nWe know, \n
E(z) = $$-$$ $${{dv} \\over {dz}}$$\n
$$ \\therefore $$ E(z) = $$-$$ 10 z for $$\\left| z \\right| \\le 1$$ m\n
and E(z) = 10 for $$\\left| z \\right| \\ge 1$$ m\n
$$ \\therefore $$ The source is an infinity large non conducting thick of thickness z = 2 m.\n
$$ \\Rightarrow $$ $$\\rho $$ = $${10\\,{\\varepsilon _0}}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9315,
"subject": "Physics",
"question": "Within a spherical charge distribution of charge density $$\\rho $$(r), N equipotential surfaces of potential V0, V0 + $$\\Delta $$V, V0 + 2$$\\Delta $$V, .......... V0 + N$$\\Delta $$V ($$\\Delta $$ V > 0), are drawn and have increasing radii r0, r1, r2,..........rN, respectively. If the difference in the radii of the surfaces is constant for all values of V0 and $$\\Delta $$V then :",
"options": [
{
"text": "$$\\rho $$ (r) $$\\alpha $$ r"
},
{
"text": "$$\\rho $$ (r) = constant "
},
{
"text": "$$\\rho $$ (r) $$\\alpha $$ $${1 \\over r}$$ "
},
{
"text": "$$\\rho $$ (r) $$\\alpha $$ $${1 \\over {{r^2}}}$$ "
}
],
"answer": "$$\\rho $$ (r) $$\\alpha $$ $${1 \\over r}$$ ",
"solution": "**Answer:** $$\\rho $$ (r) $$\\alpha $$ $${1 \\over r}$$ \n\n\n
Here, $$\\Delta $$v and $$\\Delta $$r are same for any pair of surface.\n
we know, \n
Electric field, E = $$-$$ $${{dv} \\over {dr}}$$\n
$$ \\therefore $$ E = constant [As dv and dr are constant]\n
Electric field inside the spherical charge distribution. \n
E = $${{\\rho r} \\over {3{\\varepsilon _0}}}$$\n
Now, as E = constant \n
$$ \\therefore $$ $$\\rho $$ r = constant\n
$$ \\Rightarrow $$ $$\\rho $$ (r) $$ \\propto $$ $${1 \\over r}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9316,
"subject": "Physics",
"question": "There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at $$P,$$ in the region, is found to vary between the limits 589.0 V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of 60o with the direction of the field ?",
"options": [
{
"text": "589.5 V"
},
{
"text": "589.2 V"
},
{
"text": "589.4 V"
},
{
"text": "589.6 V"
}
],
"answer": "589.4 V",
"solution": "**Answer:** 589.4 V\n\nPotential gradient, \n
$$\\Delta $$V = E. d\n
$$ \\Rightarrow $$$$\\,\\,\\,$$ 589.8 $$-$$ 589.0 = (E d)max\n
$$ \\Rightarrow $$$$\\,\\,\\,$$ (E d)max = 0.8\n
$$\\therefore\\,\\,\\,$$ $$\\Delta $$V = E d cos$$\\theta $$\n
= 0.8 $$ \\times $$ cos60o\n
= 0.4\n
$$\\therefore\\,\\,\\,$$ Maximum potential on the sphere = 589.4 V",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9317,
"subject": "Physics",
"question": "Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge\ndensities $$ + \\sigma $$, $$ - \\sigma $$ and $$ + \\sigma $$ respectively. The potential of shell B is :",
"options": [
{
"text": "$${\\sigma \\over { \\in {}_0}}\\left[ {{{{b^2} - {c^2}} \\over c} + a} \\right]$$ "
},
{
"text": "$${\\sigma \\over { \\in {}_0}}\\left[ {{{{a^2} - {b^2}} \\over a} + c} \\right]$$ "
},
{
"text": "$${\\sigma \\over { \\in {}_0}}\\left[ {{{{a^2} - {b^2}} \\over b} + c} \\right]$$ "
},
{
"text": "$${\\sigma \\over { \\in {}_0}}\\left[ {{{{b^2} - {c^2}} \\over b} + a} \\right]$$ "
}
],
"answer": "$${\\sigma \\over { \\in {}_0}}\\left[ {{{{a^2} - {b^2}} \\over b} + c} \\right]$$ ",
"solution": "**Answer:** $${\\sigma \\over { \\in {}_0}}\\left[ {{{{a^2} - {b^2}} \\over b} + c} \\right]$$ \n\n\n
Let charge of shell A, B and C are QA, QB and QC respectively.\n
Potential of B shell will be due to charge QA, QB and QC.\n
Here charge QA is inside of the shell B and QB is on the surface of the shell B in both cases you have to take the radius of the shell B, while calculating potential of shell B. \n
Charge QC is outside of the shell B so take radius of shell C for calculating potential of shell B. \n
= $${\\sigma \\over {{\\varepsilon _0}}}\\left[ {{{{a^2} - {b^2}} \\over b} + c} \\right]$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9318,
"subject": "Physics",
"question": "In free space, a particle A of charge 1$$\\mu $$C is held fixed at a point P. Another particle B of the same charge and\nmass 4$$\\mu $$g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P\nis :\n$$\\left[ {Take\\,{1 \\over {4\\pi { \\in _0}}} = 9 \\times {{10}^9}N{m^2}{C^{ - 2}}} \\right]$$",
"options": [
{
"text": "1.0 m/s"
},
{
"text": "6.32 $$ \\times $$ 104 m/s"
},
{
"text": "2.0 $$ \\times $$ 103 m/s"
},
{
"text": "1.5 $$ \\times $$ 102 m/s"
}
],
"answer": "6.32 $$ \\times $$ 104 m/s",
"solution": "**Answer:** 6.32 $$ \\times $$ 104 m/s\n\nqA = 1 $$\\mu $$c ; qB = 1 $$\\mu $$c, mB = 4 × 10–9 kg, rAB = 10–3 m
\n$$ \\therefore $$ v = $$\\sqrt {{2 \\over m}} {\\left( {{{2{q^2}} \\over {15 \\times 4\\pi {\\varepsilon _0}a}}} \\right)^{{1 \\over 2}}}$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9320,
"subject": "Physics",
"question": "The electric field in a region is given by\n$$\\mathop E\\limits^ \\to = \\left( {Ax + B} \\right)\\mathop i\\limits^ \\wedge $$\n, where E is in NC–1 and x is in\nmetres. The values of constants are\nA = 20 SI unit and B = 10 SI unit. If the potential\nat x = 1 is V1 and that at x = –5 is V2, then\nV1 – V2 is :-",
"options": [
{
"text": "–520 V"
},
{
"text": "180 V"
},
{
"text": "–48 V"
},
{
"text": "320 V"
}
],
"answer": "180 V",
"solution": "**Answer:** 180 V\n\n$$\\overrightarrow E = (20x + 10)\\widehat i$$
\n$${V_1} - {V_2} = 180\\,V$$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 9321,
"subject": "Physics",
"question": "A solid conducting sphere, having a charge Q,\nis surrounded by an uncharged conducting\nhollow spherical shell. Let the potential\ndifference between the surface of the solid\nsphere and that of the outer surface of the\nhollow shell be V. If the shell is now given a\ncharge of –4 Q, the new potential difference\nbetween the same two surfaces is :",
"options": [
{
"text": "V"
},
{
"text": "2V"
},
{
"text": "–2V"
},
{
"text": "4V"
}
],
"answer": "V",
"solution": "**Answer:** V\n\n
Initially when uncharged shell encloses charge Q, charge distribution due to induction will be as shown,
Hence, we obtain that potential difference does not depend on the charge of outer sphere, hence potential difference remains same.
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9322,
"subject": "Physics",
"question": "A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be -",
"options": [
{
"text": "$${{Q\\left( {{a^2} + {b^2} + {c^2}} \\right)} \\over {4\\pi {\\varepsilon _0}\\left( {{a^3} + {b^3} + {c^3}} \\right)}}$$"
},
{
"text": "$${Q \\over {4\\pi {\\varepsilon _0}\\left( {a + b + c} \\right)}}$$"
},
{
"text": "$${Q \\over {12\\pi {\\varepsilon _0}}}{{ab + bc + ca} \\over {abc}}$$"
},
{
"text": "$${{Q\\left( {a + b + c} \\right)} \\over {4\\pi {\\varepsilon _0}\\left( {{a^2} + {b^2} + {c^2}} \\right)}}$$"
}
],
"answer": "$${{Q\\left( {a + b + c} \\right)} \\over {4\\pi {\\varepsilon _0}\\left( {{a^2} + {b^2} + {c^2}} \\right)}}$$",
"solution": "**Answer:** $${{Q\\left( {a + b + c} \\right)} \\over {4\\pi {\\varepsilon _0}\\left( {{a^2} + {b^2} + {c^2}} \\right)}}$$\n\n\n
Potential at point P, V = $${{k{Q_a}} \\over a} + {{k{Q_b}} \\over b} + {{k{Q_c}} \\over c}$$\n
$$ \\therefore $$ Work required to put a fifth charge Q \n
at origin is equal to $${{{Q^2}} \\over {4\\pi {\\varepsilon _0}}}\\left( {1 + {1 \\over {\\sqrt 5 }}} \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9324,
"subject": "Physics",
"question": "Consider two charged metallic spheres S1 and\nS2 of radii R1 and R2, respectively. The electric\nfields E1 (on S1) and E2 (on S2) on their surfaces\nare such that E1/E2 = R1/R2. Then the ratio\nV1 (on S1) / V2 (on S2) of the electrostatic\npotentials on each sphere is :",
"options": [
{
"text": "(R1/R2)2"
},
{
"text": "(R2/R1)"
},
{
"text": "(R1/R2)3"
},
{
"text": "R1/R2"
}
],
"answer": "(R1/R2)2",
"solution": "**Answer:** (R1/R2)2\n\nWe know,\n
Now $${{{V_1}} \\over {{V_2}}} = {{{{K{Q_1}} \\over {{R_1}}}} \\over {{{K{Q_2}} \\over {{R_2}}}}}$$ = $${{R_1^2} \\over {R_2^2}}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9325,
"subject": "Physics",
"question": "Two isolated conducting spheres S1 and S2 of\nradius $${2 \\over 3}R$$ and $${1 \\over 3}R$$ have 12 $$\\mu $$C and –3 $$\\mu $$C\ncharges, respectively, and are at a large\ndistance from each other. They are now\nconnected by a conducting wire. A long time\nafter this is done the charges on S1 and S2 are\nrespectively :",
"options": [
{
"text": "4.5 $$\\mu $$C on both"
},
{
"text": "+4.5 $$\\mu $$C and –4.5 $$\\mu $$C"
},
{
"text": "6 $$\\mu $$C and 3 $$\\mu $$C"
},
{
"text": "3 $$\\mu $$C and 6 $$\\mu $$C"
}
],
"answer": "6 $$\\mu $$C and 3 $$\\mu $$C",
"solution": "**Answer:** 6 $$\\mu $$C and 3 $$\\mu $$C\n\n\n
$$ = {{3K{Q_1}} \\over {4R}} = {{3{Q_1}} \\over {16\\pi {\\varepsilon _0}R}}$$ [As K = $${1 \\over {4\\pi {\\varepsilon _0}}}$$ ]",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9327,
"subject": "Physics",
"question": "Ten charges are placed on the circumference\nof a circle of radius R with constant angular\nseparation between successive charges.\nAlternate charges 1, 3, 5, 7, 9 have charge (+q)\neach, while 2, 4, 6, 8, 10 have charge (–q) each.\nThe potential V and the electric field E at the\ncentre of the circle are respectively.\n (Take V = 0 at infinity)",
"options": [
{
"text": "V = 0; E = 0"
},
{
"text": "$$V = {{10q} \\over {4\\pi {\\varepsilon _0}R}}$$; $$E = {{10q} \\over {4\\pi {\\varepsilon _0}{R^2}}}$$"
},
{
"text": "$$V = {{10q} \\over {4\\pi {\\varepsilon _0}R}}$$; E = 0"
},
{
"text": "V = 0; $$E = {{10q} \\over {4\\pi {\\varepsilon _0}{R^2}}}$$"
}
],
"answer": "V = 0; E = 0",
"solution": "**Answer:** V = 0; E = 0\n\nNet charge = 5q - 5q = 0\n
Potential of centre = V = $${{K\\sum q } \\over r}$$\n
VC = $${{K\\left( 0 \\right)} \\over r}$$ = 0\n\n Let E be electric field produced by each charge at the centre, then resultant electric field will be\n
\n
EC = 0, Since equal electric field vectors are acting at equal angle so their resultant is equal to zero.( From symmetric property of vector)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9328,
"subject": "Physics",
"question": "512 identical drops of mercury are charged to a potential of 2V each. The drops are joined to form a single drop. The potential of this drop is ________ V.",
"options": [],
"answer": "128",
"solution": "**Answer:** 128\n\nLet charge on each drop = q
$$ = 128V$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9329,
"subject": "Physics",
"question": "27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is __________ times that of a smaller drop.",
"options": [],
"answer": "243",
"solution": "**Answer:** 243\n\n$$(27)\\left( {{4 \\over 3}\\pi {r^3}} \\right) = {4 \\over 3}\\pi {R^3}$$
Sixty four conducting drops each of radius 0.02 m and each carrying a charge of 5 $$\\mu$$C are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be :
",
"options": [
{
"text": "1 : 4"
},
{
"text": "4 : 1"
},
{
"text": "1 : 8"
},
{
"text": "8 : 1"
}
],
"answer": "4 : 1",
"solution": "**Answer:** 4 : 1\n\nSurface charge density of a spherical conductor is given by,\n
$$\n\\sigma=\\frac{q}{4 \\pi r^2}\n$$\n
When all smaller drops combine, the radius of the bigger drop is given by \n
Eight similar drops of mercury are maintained at 12 V each. All these spherical drops combine into a single big drop. The potential energy of bigger drop will be ____________ E. Where E is the potential energy of a single smaller drop.
",
"options": [],
"answer": "32",
"solution": "**Answer:** 32\n\nFrom law of conservation of charge\n
Statement I : Electric potential is constant within and at the surface of each conductor.
\n
Statement II : Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point.
\n
In the light of the above statements, choose the most appropriate answer from the options given below.
",
"options": [
{
"text": "Both Statement I and Statement II are correct"
},
{
"text": "Both Statement I and Statement II are incorrect"
},
{
"text": "Statement I is correct but Statement II is incorrect"
},
{
"text": "Statement I is incorrect but Statement II is correct"
}
],
"answer": "Both Statement I and Statement II are correct",
"solution": "**Answer:** Both Statement I and Statement II are correct\n\n
Since $${\\overrightarrow E _{net}} = \\overrightarrow 0 $$ in the bulk of a conductor
\n
$$\\Rightarrow$$ Potential would be constant.
\n
$$\\Rightarrow$$ Statement I is correct
\n
Since a conductor's surface is equipotential, $$\\overrightarrow E $$ just outside is perpendicular to the surface.
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9336,
"subject": "Physics",
"question": "Considering a group of positive charges, which of the following statements is correct ?",
"options": [
{
"text": "Net potential of the system cannot be zero at a point but net electric field can be zero at that point"
},
{
"text": "Net potential of the system at a point can be zero but net electric field can't be zero at that point."
},
{
"text": "Both the net potential and the net electric field cannot be zero at a point."
},
{
"text": "Both the net potential and the net field can be zero at a point."
}
],
"answer": "Net potential of the system cannot be zero at a point but net electric field can be zero at that point",
"solution": "**Answer:** Net potential of the system cannot be zero at a point but net electric field can be zero at that point\n\n$V=\\frac{\\sum K Q_{i}}{r_{i}}$\n\n
Here, $Q_{i}$ and $r_{i}$ are positive.\n\n
$\\therefore V > 0$\n
The correct statement is:\n\n
(A) Net potential of the system cannot be zero at a point but net electric field can be zero at that point.\n\n
Explanation:\n
In a group of positive charges, the net potential at a point is the sum of the potentials due to each individual charge. The potential due to a point charge is given by the Coulomb's law, which is non-zero except at the location of the charge itself. Therefore, the net potential due to a group of positive charges can never be zero at a point.\n\n
On the other hand, the net electric field at a point is the vector sum of the electric fields due to each individual charge. If the charges are arranged in such a way that their electric fields cancel out at a particular point, then the net electric field at that point can be zero, even though the charges are present. This can happen, for example, in a symmetrical arrangement of charges.\n\n
So, statement (A) is the correct statement.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9337,
"subject": "Physics",
"question": "
For a charged spherical ball, electrostatic potential inside the ball varies with $$r$$ as $$\\mathrm{V}=2ar^2+b$$.
\n
Here, $$a$$ and $$b$$ are constant and r is the distance from the center. The volume charge density inside the ball is $$-\\lambda a\\varepsilon$$. The value of $$\\lambda$$ is ____________.
64 identical drops each charged upto potential of $$10 ~\\mathrm{mV}$$ are combined to form a bigger drop. The potential of the bigger drop will be __________ $$\\mathrm{mV}$$.
",
"options": [],
"answer": "160",
"solution": "**Answer:** 160\n\nThe problem involves 64 identical drops, each charged up to a potential of $$10 \\mathrm{~mV}$$, that are combined to form a bigger drop. We are asked to find the potential of the bigger drop.\n
\nThe potential of each drop is given by:
\n$$V = \\frac{Kq}{r}$$
\nwhere $$K$$ is the Coulomb constant, $$q$$ is the charge on each drop, and $$r$$ is the radius of each drop.\n
\nThe radius of the bigger drop is:
\n$$R = 4r$$
\nsince the 64 identical drops combine to form a bigger drop.\n
\nTherefore, the potential of the bigger drop is $$160 \\mathrm{~mV}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9339,
"subject": "Physics",
"question": "
Three concentric spherical metallic shells X, Y and Z of radius a, b and c respectively [a < b < c] have surface charge densities $$\\sigma,-\\sigma$$ and $$\\sigma$$ respectively. The shells X and Z are at same potential. If the radii of X & Y are 2 cm and 3 cm, respectively. The radius of shell Z is _________ cm.
Given three concentric spherical shells X, Y, and Z with radii a, b, and c respectively, and with surface charge densities ( $\\sigma$ ), ( $-\\sigma$ ), and ( $\\sigma$ ) respectively, we know that the potential at the surface of a sphere due to a uniform surface charge is given by:
\n
$ V = \\frac{1}{4\\pi\\epsilon_0} \\frac{Q}{r} $
\n
where ( $\\epsilon_0$ ) is the permittivity of free space, ( Q ) is the total charge on the sphere, and ( r ) is the radius of the sphere.
\n
However, in this case, the total charge on each sphere is given by its surface charge density ( $\\sigma$ ) times its surface area ( $4\\pi r^2$ ). Substituting this into the formula for ( Q ) gives:
\n
$ Q = \\sigma 4\\pi r^2 $
\n
So the potential at the surface of each sphere is given by:
However, we also need to take into account the effect of the charge on shell Y on the potentials at X and Z. The potential at any point due to a charged shell is the same everywhere outside the shell, so we can add the potential due to shell Y at X to both sides of the equation. This gives:
Electric potential at a point '$$\\mathrm{P}$$' due to a point charge of $$5 \\times 10^{-9} \\mathrm{C}$$ is $$50 \\mathrm{~V}$$. The distance of '$$\\mathrm{P}$$' from the point charge is:
An electric charge $$10^{-6} \\mu \\mathrm{C}$$ is placed at origin $$(0,0)$$ $$\\mathrm{m}$$ of $$\\mathrm{X}-\\mathrm{Y}$$ co-ordinate system. Two points $$\\mathrm{P}$$ and $$\\mathrm{Q}$$ are situated at $$(\\sqrt{3}, \\sqrt{3}) \\mathrm{m}$$ and $$(\\sqrt{6}, 0) \\mathrm{m}$$ respectively. The potential difference between the points $\\mathrm{P}$ and $\\mathrm{Q}$ will be :
The electric potential at the surface of an atomic nucleus $$(z=50)$$ of radius $$9 \\times 10^{-13} \\mathrm{~cm}$$ is __________ $$\\times 10^6 \\mathrm{~V}$$.
At the centre of a half ring of radius $$\\mathrm{R}=10 \\mathrm{~cm}$$ and linear charge density $$4 \\mathrm{~nC} \\mathrm{~m}^{-1}$$, the potential is $$x \\pi \\mathrm{V}$$. The value of $$x$$ is _________.
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9344,
"subject": "Physics",
"question": "A plano convex lens of refractive index $$1.5$$ and radius of curvature $$30$$ $$cm$$. Is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of size of the object ",
"options": [
{
"text": "$$60$$ $$cm$$ "
},
{
"text": "$$30$$ $$cm$$ "
},
{
"text": "$$20$$ $$cm$$ "
},
{
"text": "$$80$$ $$cm$$ "
}
],
"answer": "$$20$$ $$cm$$ ",
"solution": "**Answer:** $$20$$ $$cm$$ \n\nKEY CONCEPT : The focal length $$\\left( F \\right)$$ of the final mirror\n
$$\\therefore$$ $$f = - \\left( {{1 \\over {10}} \\times 100} \\right)cm = - 10\\,cm.$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 9347,
"subject": "Physics",
"question": "In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance $$u$$ and the image distance $$v,$$ from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of $${45^ \\circ }$$ with the $$x$$-axis meets the experimental curve at $$P.$$ The coordinates of $$P$$ will be : ",
"options": [
{
"text": "$$\\left( {{f \\over 2},{f \\over 2}} \\right)$$ "
},
{
"text": "$$\\left( {f,f} \\right)$$ "
},
{
"text": "$$\\left( {4f,4f} \\right)$$ "
},
{
"text": "$$\\left( {2f,2f} \\right)$$ "
}
],
"answer": "$$\\left( {2f,2f} \\right)$$ ",
"solution": "**Answer:** $$\\left( {2f,2f} \\right)$$ \n\n \n
Here $$u = - 2f,v = 2f$$\n
As $$|u|$$ increases, $$v$$ decreases for $$|u| > f.$$ The graph between $$|v|$$ and $$|u|$$ is shown in the figure. A straight line passing through the origin and making an angle of $${45^ \\circ }$$ with the $$x$$-axis meets the experimental curve at $$P\\left( {2f,2f} \\right).$$",
"topic": "Discrete Mathematics",
"subtopic": "Other"
},
{
"id": 9348,
"subject": "Physics",
"question": "An object $$2.4$$ $$m$$ in front of a lens forms a sharp image on a film $$12$$ $$cm$$ behind the lens. A glass plate $$1$$ $$cm$$ thick, of refractive index $$1.50$$ is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus of film? ",
"options": [
{
"text": "$$7.2$$ $$m$$ "
},
{
"text": "$$24$$ $$m$$ "
},
{
"text": "$$3.2$$ $$m$$ "
},
{
"text": "$$5.6$$ $$m$$ "
}
],
"answer": "$$5.6$$ $$m$$ ",
"solution": "**Answer:** $$5.6$$ $$m$$ \n\nThe focal length of the lens\n
$$ \\Rightarrow {f_2} = \\left( - \\right)ve$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9351,
"subject": "Physics",
"question": "To find the focal length of a convex mirror, a student records the following data :\n
\n
\n
\n
Object Pin
\n
Convex Lens
\n
Convex Mirror
\n
Image Pin
\n
\n
\n
22.2 cm
\n
32.2 cm
\n
45.8 cm
\n
71.2 cm
\n
\n
\n
The focal length of the convex lens is f1 and that of mirror is f2. Then taking index correction to be negligibly small, f1 and f2 are close to :",
"options": [
{
"text": "f1 = 12.7 cm f2 = 7.8 cm"
},
{
"text": "f1 = 7.8 cm f2 = 12.7 cm"
},
{
"text": "f1 = 7.8 cm f2 = 25.4 cm"
},
{
"text": "f1 = 15.6 cm f2 = 25.4 cm"
}
],
"answer": "f1 = 7.8 cm f2 = 12.7 cm",
"solution": "**Answer:** f1 = 7.8 cm f2 = 12.7 cm\n\nFor lens : \n
$$ \\therefore $$ f2 = $${R \\over 2}$$ = $${{25.4} \\over 2}$$ = 12.7 cm",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9352,
"subject": "Physics",
"question": "In an experiment a convex lens of focal length 15 cm is placed coaxially on an optical bench in front of a convex mirror at a distance of 5 cm from it. It is found that an object and its image coincide, if the object is placed at a distance of 20 cm from the lens. The focal length of the convex mirror is : ",
"options": [
{
"text": "27.5 cm"
},
{
"text": "20.0 cm"
},
{
"text": "25.0 cm"
},
{
"text": "30.5 cm"
}
],
"answer": "27.5 cm",
"solution": "**Answer:** 27.5 cm\n\n
The given optical situation is depicted in the following ray diagram:
\n
\n
In this case, the image forms 55 cm behind the convex mirror and then the reflection takes place due to the mirror image that forms at a distance (v) behind the mirror. Now, this image acts as an object for lens and the final image forms at a distance of 20 m from the lens if the focal length of convex mirror is as follows:
\n
$${{55} \\over 2} = 27.5$$ cm
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9353,
"subject": "Physics",
"question": "A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15cm from a converging\nlens of magnitude of focal length 20cm. A beam of parallel light falls on the diverging lens. The final\nimage formed is:",
"options": [
{
"text": "real and at a distance of 6 cm from the convergent lens."
},
{
"text": "real and at a distance of 40 cm from convergent lens."
},
{
"text": "virtual and at a distance of 40 cm from convergent lens."
},
{
"text": "real and at a distance of 40 cm from the divergent lens."
}
],
"answer": "real and at a distance of 40 cm from convergent lens.",
"solution": "**Answer:** real and at a distance of 40 cm from convergent lens.\n\nAs parallel beam incident on diverging lens so the image will be formed\nat the focus of diverging lens.\n
$$ \\therefore $$ v = –25 cm\n\n The image formed by diverging lens is used as an object for\nconverging lens.\n
V is positive so image will be real and will form at right side of converging lens at 40 cm.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9354,
"subject": "Physics",
"question": "A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is $$2$$ cm. The focal lengths of the component lenses are : ",
"options": [
{
"text": "10 cm, 12 cm"
},
{
"text": "12 cm, 14 cm"
},
{
"text": "16 cm, 18 cm"
},
{
"text": "18 cm, 20cm"
}
],
"answer": "18 cm, 20cm",
"solution": "**Answer:** 18 cm, 20cm\n\n
For a convergent doublet of separated lens, we have
For f1 = 18 cm and f2 = 20 cm, the above equation satisfies.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9355,
"subject": "Physics",
"question": "A plano-convex lens (focal length f2, refractive index $$\\mu $$2, radius of curvature R) fits exactly into a plano-concave lens (focal length f1, refractive index $$\\mu $$1, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be : \n",
"options": [
{
"text": "f1 + f2"
},
{
"text": "f1 $$-$$ f2"
},
{
"text": "$${R \\over {{\\mu _2} - {\\mu _1}}}$$"
},
{
"text": "$${{2{f_1}{f_2}} \\over {{f_1} + {f_2}}}$$"
}
],
"answer": "$${R \\over {{\\mu _2} - {\\mu _1}}}$$",
"solution": "**Answer:** $${R \\over {{\\mu _2} - {\\mu _1}}}$$\n\n\n
$${1 \\over F} = {1 \\over {{f_1}}} + {1 \\over {{f_2}}} = {{1 - {\\mu _1}} \\over R} + {{{\\mu _2} - 1} \\over R}$$",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 9356,
"subject": "Physics",
"question": "A convex lens of focal length 20 cm produces\nimages of the same magnification 2 when an\nobject is kept at two distances x1 and x2\n(x1 > x2) from the lens. The ratio of x1 and x2\nis :-",
"options": [
{
"text": "3 : 1"
},
{
"text": "2 : 1"
},
{
"text": "5 : 3"
},
{
"text": "4 : 3"
}
],
"answer": "3 : 1",
"solution": "**Answer:** 3 : 1\n\nMagnification is 2
\nIf image is real, $${x_1} = {{3f} \\over 2}$$
\nIf image is virtual, $${x_2} = {f \\over 2}$$
\n$${{{x_1}} \\over {{x_2}}} = 3:1$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9357,
"subject": "Physics",
"question": "A convex lens (of focal length 20 cm) and a\nconcave mirror, having their principal axes\nalong the same lines, are kept 80 cm apart from\neach other. The concave mirror is to the right\nof the convex lens. When an object is kept at\na distance of 30 cm to the left of the convex\nlens, its image remains at the same position\neven if the concave mirror is removed. The\nmaximum distance of the object for which this\nconcave mirror, by itself would produce a\nvirtual image would be :-",
"options": [
{
"text": "25 cm"
},
{
"text": "10 cm"
},
{
"text": "20 cm"
},
{
"text": "30 cm"
}
],
"answer": "10 cm",
"solution": "**Answer:** 10 cm\n\n
\nv = +60 cm \nIf image position does not change even\nwhen mirror is removed it means image\nformed by lens is formed at centre of\ncurvature of spherical mirror. \nRadius of curvature of mirror = 80 – 60 = 20\ncm.
\n$$ \\Rightarrow $$ Focal length of mirror f = 10 cm for\nvirtual image, object is to be kept\nbetween focus and pole.
\n$$ \\Rightarrow $$ Maximum distance of object from\nspherical mirror for which virtual image\nis formed, is 10 cm. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9358,
"subject": "Physics",
"question": "An upright object is placed at a distance of\n40 cm in front of a convergent lens of focal\nlength 20 cm. A convergent mirror of focal\nlength 10 cm is placed at a distance of 60 cm\non the other side of the lens. The position and\nsize of the final image will be :",
"options": [
{
"text": "20 cm from the convergent mirror, same\nsize as the object"
},
{
"text": "40 cm from the convergent mirror, same\nsize as the object"
},
{
"text": "40 cm from the convergent lens, same\nsize as the object"
},
{
"text": "20 cm from the convergent mirror, twice the\nsize of the object"
}
],
"answer": "40 cm from the convergent lens, same\nsize as the object",
"solution": "**Answer:** 40 cm from the convergent lens, same\nsize as the object\n\n
In given system of lens and mirror, position of object O in front of lens is at a distance 2f. i.e. u = 2f = 40 cm
\n
\n
So, image (I1) formed is real, inverted and at a distance, v = 2f = 2 $$\\times$$ 20 = 40 cm, (behind lens) magnification, $${m_1} = {v \\over u} = {{40} \\over {40}} = 1$$
\n
Thus, size of image is same as that of that of object. This image (I1) acts like a real object for mirror.
\n
\n
As object distance for mirror is
\n
u = C = 2f = $$-$$ 20 cm
\n
where, C = centre of curvature.
\n
So, image (I2) formed by mirror is at 2f.
\n
$$\\therefore$$ For mirror v = 2f = 2($$-$$ 10) = $$-$$ 20 cm
The image I2 formed by the mirror will act like an object for lens.
\n
\n
As the object is at 2f distance from lens, so image (I3) will be formed at a distance 2f or 40 cm. Thus, magnification,
\n
$${m_3} = {v \\over u} = {{40} \\over {40}} = 1$$
\n
So, final magnification, $$m = {m_1} \\times {m_2} \\times {m_3} = - 1$$
\n
Hence, final image (I3) is real, inverted of same size as that of object and coinciding with object.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9359,
"subject": "Physics",
"question": "An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image of the object. If the object moves away from the lens at a speed of 5 m/s, the speed and direction of the image will be : ",
"options": [
{
"text": "1.16 $$ \\times $$ 10–3 m/s towards the lens "
},
{
"text": "2.26 $$ \\times $$ 10–3 m/s away from the lens"
},
{
"text": "3.22 × 10–3 m/s towards the lens"
},
{
"text": "0.92 $$ \\times $$ 10$$-$$3 m/s away from the lens"
}
],
"answer": "1.16 $$ \\times $$ 10–3 m/s towards the lens ",
"solution": "**Answer:** 1.16 $$ \\times $$ 10–3 m/s towards the lens \n\nFrom lens equation\n
direction of velocity of image is same as that of object\n
vO/L = 5 m/s",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9360,
"subject": "Physics",
"question": "The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea (7.8 mm). This surface separateds two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus -",
"options": [
{
"text": "2 cm"
},
{
"text": "3.1 cm"
},
{
"text": "4.0 cm"
},
{
"text": "1 cm"
}
],
"answer": "3.1 cm",
"solution": "**Answer:** 3.1 cm\n\n\n
$$2{\\mu _1} - {\\mu _2} = 1$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9362,
"subject": "Physics",
"question": "A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is : ",
"options": [
{
"text": "1.1 cm away from the lens"
},
{
"text": "0"
},
{
"text": "0.55 cm towards the lens"
},
{
"text": "0.55 cm away from the lens"
}
],
"answer": "0.55 cm away from the lens",
"solution": "**Answer:** 0.55 cm away from the lens\n\n\n
So, to get sharp image screen should shift away (10.55 $$-$$ 10) = 0.55 cm from the lens. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9363,
"subject": "Physics",
"question": "A thin lens made of glass (refractive index = 1.5) of focal length f = 16 cm is immersed in a liquid\nof refractive index 1.42. If its focal length in liquid is f1\n, then the ratio $${{{f_1}} \\over f}$$ is closest to the\ninteger :",
"options": [
{
"text": "17"
},
{
"text": "1"
},
{
"text": "9"
},
{
"text": "5"
}
],
"answer": "9",
"solution": "**Answer:** 9\n\nUsing formula\n $${1 \\over f} = \\left( {{{{\\mu _2}} \\over {{\\mu _1}}} - 1} \\right)\\left( {{1 \\over {{R_1}}} - {1 \\over {{R_2}}}} \\right)$$\n
$${{{f_1}} \\over f} = {{0.5} \\over {0.056}}$$ = 8.93 $$ \\approx $$ 9 ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9364,
"subject": "Physics",
"question": "A point object in air is in front of the curved\nsurface of a plano-convex lens. The radius of\ncurvature of the curved surface is 30 cm and\nthe refractive index of the lens material is 1.5,\nthen the focal length of the lens (in cm)\nis ______.",
"options": [],
"answer": "60",
"solution": "**Answer:** 60\n\nLens-maker formula\n
$$ \\therefore $$ f = $${R \\over {\\mu - 1}}$$ = $${{30} \\over {1.5 - 1}}$$ = 60 cm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9365,
"subject": "Physics",
"question": "The distance between an object and a screen is 100 cm. A lens can produce real image of the\nobject on the screen for two different positions between the screen and the object. The distance\nbetween these two positions is 40 cm. If the power of the lens is close to $$\\left( {{N \\over {100}}} \\right)D$$ where N is an\ninteger, the value of N is _________.",
"options": [],
"answer": "476",
"solution": "**Answer:** 476\n\nUsing displacement method
$$N = 476$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9366,
"subject": "Physics",
"question": "A point like object is placed at a distance of 1 m in front of a convex lens of focal length 0.5 m. A\nplane mirror is placed at a distance of 2 m behind the lens. The position and nature of the final\nimage formed by the system is :",
"options": [
{
"text": "2.6 m from the mirror, real"
},
{
"text": "1 m from the mirror, real"
},
{
"text": "2.6 m from the mirror, virtual"
},
{
"text": "1 m from the mirror, virtual"
}
],
"answer": "2.6 m from the mirror, real",
"solution": "**Answer:** 2.6 m from the mirror, real\n\n\n
Object is at 2f. So image will also be at 2f. (I1).\n
Image(I2) of I1 will be 1 m behind mirror.\n
Now I2 will be object for lens.\n
$$ \\therefore $$ u = -3 m\n
f = + 0.5 m\n
$${1 \\over v} = {1 \\over f} + {1 \\over u}$$\n
= $${1 \\over {0.5}} + {1 \\over { - 3}}$$\n
$$ \\Rightarrow $$ v = $${3 \\over 5}$$ = 0.6 m\n
So total distance from mirror = 2 + 0.6 = 2.6 m and real image.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9367,
"subject": "Physics",
"question": "A double convex lens has power P and same radii of curvature R of both the surfaces. The radius of\ncurvature of a surface of a plano-convex lens made of the same material with power 1.5 P is :",
"options": [
{
"text": "$${R \\over 3}$$"
},
{
"text": "$${{3R} \\over 2}$$"
},
{
"text": "$${R \\over 2}$$"
},
{
"text": "2R"
}
],
"answer": "$${R \\over 3}$$",
"solution": "**Answer:** $${R \\over 3}$$\n\nAssume refractive index = $$\\mu $$$$l$$\n
$$ \\Rightarrow $$ R1 = $${R \\over 3}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9368,
"subject": "Physics",
"question": "The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of convex lens is __________ cm.",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\n$${1 \\over v} - {1 \\over u} = {1 \\over f}$$ .... (1)\n
$$ \\Rightarrow $$ f = 15 cm ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9369,
"subject": "Physics",
"question": "The refractive index of a converging lens is 1.4. What will be the focal length of this lens if it is placed in a medium of same refractive index? Assume the radii of curvature of the faces of lens are R1 and R2 respectively.",
"options": [
{
"text": "Zero"
},
{
"text": "Infinite"
},
{
"text": "1"
},
{
"text": "$${{{R_1}{R_2}} \\over {{R_1} - {R_2}}}$$"
}
],
"answer": "Infinite",
"solution": "**Answer:** Infinite\n\nGiven, initially refractive index (n1) = 1.4\n
Then it placed in medium of same refractive index.\n
$$ \\Rightarrow $$ f = infinite",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9370,
"subject": "Physics",
"question": "The thickness at the centre of a plane convex lens is 3 mm and the diameter is 6 cm. If the speed of light in the material of the lens is 2 $$\\times$$ 108 ms$$-$$1. The focal length of the lens is ____________.",
"options": [
{
"text": "0.30 cm"
},
{
"text": "30 cm"
},
{
"text": "15 cm"
},
{
"text": "1.5 cm"
}
],
"answer": "30 cm",
"solution": "**Answer:** 30 cm\n\n
A parallel beam of light is allowed to fall on a transparent spherical globe of diameter 30 cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is _____________ mm.
Two identical thin biconvex lens of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is __________ cm.
For an object placed at a distance 2.4 m from a lens, a sharp focused image is observed on a screen placed at a distance 12 cm from the lens. A glass plate of refractive index 1.5 and thickness 1 cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?
So final image must be produced at $$\\left( {12 - {1 \\over 3}} \\right)$$ cm $$ = {{35} \\over 3}$$ cm from lens so that glass plate must shift it to produce image at screen. So
A convex lens of focal length 20 cm is placed in front of a convex mirror with principal axis coinciding each other. The distance between the lens and mirror is 10 cm. A point object is placed on principal axis at a distance of 60 cm from the convex lens. The image formed by combination coincides the object itself. The focal length of the convex mirror is ____________ cm.
The power of a lens (biconvex) is $$1.25 \\mathrm{~m}^{-1}$$ in particular medium. Refractive index of the lens is 1.5 and radii of curvature are $$20 \\mathrm{~cm}$$ and $$40 \\mathrm{~cm}$$ respectively. The refractive index of surrounding medium:
In an experiment with a convex lens, The plot of the image distance $$\\left(v^{\\prime}\\right)$$ against the object distance ($$\\left.\\mu^{\\prime}\\right)$$ measured from the focus gives a curve $$v^{\\prime} \\mu^{\\prime}=225$$. If all the distances are measured in $$\\mathrm{cm}$$. The magnitude of the focal length of the lens is ___________ cm.
A person has been using spectacles of power $$-1.0$$ dioptre for distant vision and a separate reading glass of power $$2.0$$ dioptres. What is the least distance of distinct vision for this person :
A convex lens of refractive index 1.5 and focal length 18cm in air is immersed in water. The change in focal length of the lens will be ___________ cm.
\n
(Given refractive index of water $$=\\frac{4}{3}$$)
A bi convex lens of focal length $$10 \\mathrm{~cm}$$ is cut in two identical parts along a plane perpendicular to the principal axis. The power of each lens after cut is ____________ D.
",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nWhen a biconvex lens is cut into two identical parts along a plane perpendicular to the principal axis, each part becomes a plano-convex lens. To find the new power of each plano-convex lens, we can use the lensmaker's formula:\n
\nFor the original biconvex lens, both radii of curvature have the same magnitude but opposite signs, so let's denote them as ±R. The focal length of the original lens is given as 10 cm, and the refractive index (n) is constant for both the original lens and the new plano-convex lenses. \n
\nFor the original lens, the lensmaker's formula becomes:\n
\nFor each plano-convex lens, one radius of curvature (R1) is infinite (the flat side), and the other radius (R2) is the same as the original lens (R). The lensmaker's formula for the plano-convex lens becomes:\n
\nConverting the focal length to meters and calculating the power:\n
\n$$P = \\frac{1}{0.2}$$
\n$$P = 5 \\text{ D}$$
\nSo, the power of each plano-convex lens after the cut is 5 diopters (D).",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9381,
"subject": "Physics",
"question": "
Two convex lenses of focal length $$20 \\mathrm{~cm}$$ each are placed coaxially with a separation of $$60 \\mathrm{~cm}$$ between them. The image of the distant object formed by the combination is at _____________ $$\\mathrm{cm}$$ from the first lens.
",
"options": [],
"answer": "100",
"solution": "**Answer:** 100\n\n\n\n 1. First refraction in L1 (lens 1):\n
When considering the first lens (L1), the object is at infinity, so the image I1 formed by this lens is at its focal point. Using the lensmaker's equation:\n\n
Given that the object is at infinity, $$d_o = \\infty$$, and the focal length of the first lens is $$f = 20 \\mathrm{~cm}$$. Plugging in these values, we get:\n\n
As $$\\frac{1}{\\infty}$$ is essentially 0, we have:\n\n
$$\\frac{1}{20} = \\frac{1}{d_i}$$\n\n
This implies that $$d_i = 20 \\mathrm{~cm}$$, meaning that the image I1 is formed 20 cm from the first lens L1.\n
\n\n2. Second refraction in L2 (lens 2):\n\n
Now, the image I1 formed by L1 becomes the object for L2. The distance between the lenses is 60 cm, so the object distance for L2 (u) is -40 cm, because the object is to the left of the lens (u is negative). The focal length of L2 (f) is 20 cm. We can use the lensmaker's equation to find the image distance (v) for L2:\n\n
Since the image distance (v) is positive, the image I2 is formed on the right side of L2 at a distance of 40 cm. To find the distance of the final image from L1, we add the distance between the lenses (60 cm) and the image distance (v) from L2.\n\n
Final image distance from L1 = 60 cm + 40 cm = 100 cm\n\n
Thus, the correct answer is 100 cm.",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 9382,
"subject": "Physics",
"question": "
The radius of curvature of each surface of a convex lens having refractive index 1.8 is $$20 \\mathrm{~cm}$$. The lens is now immersed in a liquid of refractive index 1.5 . The ratio of power of lens in air to its power in the liquid will be $$x: 1$$. The value of $$x$$ is _________.
where $$f$$ is the focal length, $$\\mu$$ is the refractive index of the lens material, and $$R_1$$ and $$R_2$$ are the radii of curvature of the lens surfaces.
\n
Since the lens is convex, both surfaces have the same radius of curvature (positive), so $$R_1 = R_2 = 20 \\mathrm{~cm}$$.
\n
First, let's find the focal length of the lens in air:
Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature $$30 \\mathrm{~cm}$$. The centre of curvature of surface is towards denser medium and a point object is placed on the principle axis in rarer medium at a distance of $$15 \\mathrm{~cm}$$ from the pole of the surface. The distance of image from the pole of the surface is ____________ $$\\mathrm{cm}$$.
Therefore, the image is formed at a distance of 30 cm from the pole of the surface, on the same side as the object. The negative sign indicates that the image is virtual and is formed on the same side of the surface as the light is coming from.
A 2 meter long scale with least count of $$0.2 \\mathrm{~cm}$$ is used to measure the locations of objects on an optical bench. While measuring the focal length of a convex lens, the object pin and the convex lens are placed at $$80 \\mathrm{~cm}$$ mark and $$1 \\mathrm{~m}$$ mark, respectively. The image of the object pin on the other side of lens coincides with image pin that is kept at $$180 \\mathrm{~cm}$$ mark. The $$\\%$$ error in the estimation of focal length is:
In this problem, you are asked to find the percentage error in the estimation of the focal length of a convex lens using a 2-meter long scale with a least count of 0.2 cm.
\n
First, let's determine the object distance (u), image distance (v), and focal length (f) of the lens.
\n\n
Object distance (u): It's the distance between the object pin and the convex lens. The object pin is at the 80 cm mark, and the convex lens is at the 1 m (100 cm) mark, so the object distance is $$u = 100 - 80 = 20~cm$$.
\n
\n
Image distance (v): It's the distance between the image pin and the convex lens. The image pin is at the 180 cm mark, and the convex lens is at the 1 m (100 cm) mark, so the image distance is $$v = 180 - 100 = 80~cm$$.
\n
\n
Focal length (f): Using the lens formula, we can calculate the focal length:
Now, we will calculate the error in the focal length (df) using the given least count (0.2 cm). The error in the object distance and image distance will both be 0.2 cm.
\n\n
Error in the focal length (df): We can use the formula for the error in the focal length:
So, the percentage error in the estimation of the focal length is 1.70%.
\n",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9385,
"subject": "Physics",
"question": "The distance between object and its 3 times magnified virtual image as produced by a convex lens is $20 \\mathrm{~cm}$. The focal length of the lens used is __________ $\\mathrm{cm}$.",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\n\n
A convex lens of focal length $$40 \\mathrm{~cm}$$ forms an image of an extended source of light on a photoelectric cell. A current I is produced. The lens is replaced by another convex lens having the same diameter but focal length $$20 \\mathrm{~cm}$$. The photoelectric current now is :
A biconvex lens of refractive index 1.5 has a focal length of $$20 \\mathrm{~cm}$$ in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:
The distance between object and its two times magnified real image as produced by a convex lens is $$45 \\mathrm{~cm}$$. The focal length of the lens used is _______ cm.
Statement (I) : When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side.
\n
Statement (II) : Concave lens always forms a virtual and erect image.
\n
In the light of the above statements, choose the correct answer from the options given below :
",
"options": [
{
"text": "Statement I is false but Statement II is true\n"
},
{
"text": "Both Statement I and Statement II are false\n"
},
{
"text": "Both Statement I and Statement II are true\n"
},
{
"text": "Statement I is true but Statement II is false"
}
],
"answer": "Statement I is false but Statement II is true\n",
"solution": "**Answer:** Statement I is false but Statement II is true\n\n\n
Let's analyze each statement carefully:
\n\n
Statement (I): When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side.
\n\n
To understand this statement, let's recall the nature of images formed by concave lenses. A concave lens (diverging lens) always forms images that are virtual, erect, and smaller than the object, regardless of the object's position. The term \"centre of curvature\" is typically associated with mirrors rather than lenses. Therefore, this statement is incorrect because a concave lens does not form a real image that could be said to be at the centre of curvature on the other side.
\n\n
Statement (II): Concave lens always forms a virtual and erect image.
\n\n
Now, considering the second statement, a concave lens (diverging lens) indeed always forms virtual, erect, and diminished images. This is a fundamental property of concave lenses.
\n\n
Given this analysis:
\n\n
The correct answer is Option A: Statement I is false but Statement II is true.
An effective power of a combination of 5 identical convex lenses which are kept in contact along the principal axis is $$25 \\mathrm{D}$$. Focal length of each of the convex lens is:
When we have a combination of identical lenses in contact, the effective power ($$P_{\\text{eff}}$$) of the combination can be calculated as the sum of the powers of all the individual lenses. This is because the lenses are in direct contact, and their powers effectively add up.
\n\n
Given that the effective power of a combination of 5 identical convex lenses is $$25 \\mathrm{D}$$, we can use the formula for the effective power of the combination:
\n\n
$$P_{\\text{eff}} = nP$$
\n\n
where:
\n\n
\n\n
$P_{\\text{eff}}$ is the effective power of the combination,
\n\n
$n$ is the number of lenses, and
\n\n
$P$ is the power of each individual lens.
\n\n
\n\n
Given $n = 5$ and $P_{\\text{eff}} = 25 \\mathrm{D}$, we can solve for $P$, the power of each lens:
\n\n
$$25 \\mathrm{D} = 5P$$
\n\n
Dividing both sides by 5:
\n\n
$$P = \\frac{25 \\mathrm{D}}{5} = 5 \\mathrm{D}$$
\n\n
The power of a lens ($P$) is related to its focal length ($f$) by the equation:
\n\n
$$P = \\frac{1}{f}$$
\n\n
where $P$ is in diopters (D) and $f$ is in meters. Thus:
In an experiment to measure focal length ($$f$$) of convex lens, the least counts of the measuring scales for the position of object (u) and for the position of image (v) are $$\\Delta u$$ and $$\\Delta v$$, respectively. The error in the measurement of the focal length of the convex lens will be:
First, let's understand the relationship between the object distance ($$u$$), the image distance ($$v$$), and the focal length ($$f$$) of a convex lens, which is given by the lens formula:
Now, to find the error in the focal length ($$\\Delta f$$) due to the errors in the measurements of $$u$$ and $$v$$ ($$\\Delta u$$ and $$\\Delta v$$, respectively), we have to differentiate the lens formula with respect to $$u$$ and $$v$$, keeping in mind the propagation of error.
\n\n
By differentiating both sides of the lens formula with respect to $$v$$ and $$u$$, and also considering the negative reciprocal relation (given $$f$$ is a constant for a specific lens), we have:
Therefore, the correct option showing the error in the measurement of the focal length of the convex lens, taking into account the least counts ($$\\Delta u$$ and $$\\Delta v$$) of the measuring scales for the position of the object ($$u$$) and for the position of the image ($$v$$), is:
For the thin convex lens, the radii of curvature are at $$15 \\mathrm{~cm}$$ and $$30 \\mathrm{~cm}$$ respectively. The focal length the lens is $$20 \\mathrm{~cm}$$. The refractive index of the material is :
$\\mu$ is the refractive index of the material of the lens,
\n\n
$R_1$ is the radius of curvature of the first surface (convex side, positive),
\n\n
$R_2$ is the radius of curvature of the second surface (concave side, negative for convex lens).
\n\n
\n\n
Given in the question, the radii of curvature are $15 \\, \\mathrm{cm}$ and $30 \\, \\mathrm{cm}$ respectively, and the focal length $f = 20\\, \\mathrm{cm}$.
\n\n
It's important to pay attention to the signs of the radii of curvature according to the lens maker's convention. For convex lenses, $R_1$ is positive and $R_2$ is negative; however, since the problem doesn't specify which curvature corresponds to which side in relation to the direction of light travel, we'll assume the light travels from left to right: thus, $R_1 = +15 \\, \\mathrm{cm}$ and $R_2 = -30 \\, \\mathrm{cm}$.
\n\n
Substituting the given values into the Lensmaker's Formula, we get:
Hence, the refractive index of the material of the lens is 1.5, which corresponds to Option B.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9393,
"subject": "Physics",
"question": "Wavelength of light used in an optical instrument are $${\\lambda _1} = 4000\\mathop A\\limits^ \\circ $$ and $${\\lambda _2} = 5000\\mathop A\\limits^ \\circ ,$$ then ratio of their respective resolving powers (corresponding to $${\\lambda _1}$$ and $${\\lambda _2}$$ ) is :",
"options": [
{
"text": "$$16:25$$ "
},
{
"text": "$$9:1$$ "
},
{
"text": "$$4:5$$ "
},
{
"text": "$$5:4$$"
}
],
"answer": "$$5:4$$",
"solution": "**Answer:** $$5:4$$\n\n
The resolving power (RP) of an optical instrument is inversely proportional to the wavelength (λ) of light used. So, if we denote the resolving powers corresponding to λ₁ and λ₂ as RP₁ and RP₂ respectively, we can express this relationship as :
\n
RP $$ \\propto $$ $${1 \\over \\lambda }$$
\n
Therefore, the ratio of the resolving powers corresponding to λ₁ and λ₂ is given by the inverse of the ratio of the wavelengths. In LaTeX notation, this relationship can be written as :
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9394,
"subject": "Physics",
"question": "An astronomical telescope has a large aperture to ",
"options": [
{
"text": "reduce spherical aberration "
},
{
"text": "have high resolution "
},
{
"text": "increase span of observation "
},
{
"text": "have low dispersion"
}
],
"answer": "have high resolution ",
"solution": "**Answer:** have high resolution \n\nKEY CONCEPT : The resolving power of a telescope \n
$$R.P = {D \\over {122\\lambda }}$$ where $$D=$$ diameter of the objective lens \n
$$\\lambda = $$ wavelength of light.\n
Clearly, larger the aperture, larger is the value of $$D,$$\n
more is the resolving power or resolution.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9395,
"subject": "Physics",
"question": "The image formed by an objective of a compound microscope is ",
"options": [
{
"text": "virtual and diminished "
},
{
"text": "real an diminished "
},
{
"text": "real and enlarged "
},
{
"text": "virtual and enlarged "
}
],
"answer": "real and enlarged ",
"solution": "**Answer:** real and enlarged \n\nA real, inverted and enlarged image of the object is formed by the objective lens of a compound microscope.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9396,
"subject": "Physics",
"question": "An observer looks at a distant tree of height $$10$$ $$m$$ with a telescope of magnifying power of $$20.$$ To the observer the tree appears: ",
"options": [
{
"text": "$$20$$ times taller "
},
{
"text": "$$20$$ times nearer "
},
{
"text": "$$10$$ times taller "
},
{
"text": "$$10$$ times nearer "
}
],
"answer": "$$20$$ times nearer ",
"solution": "**Answer:** $$20$$ times nearer \n\nA telescope magnifies by making the object appearing closer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9397,
"subject": "Physics",
"question": "If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece, should be close to :",
"options": [
{
"text": "22 mm"
},
{
"text": "12 mm"
},
{
"text": "33 mm"
},
{
"text": "2 mm"
}
],
"answer": "22 mm",
"solution": "**Answer:** 22 mm\n\nCase 1 : Near – point adjustment\n
$$ \\Rightarrow $$ fe = 20 mm\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9398,
"subject": "Physics",
"question": "The magnifying power of a telescope with tube\n60 cm is 5. What is the focal length of its eye\npiece ?",
"options": [
{
"text": "40 cm"
},
{
"text": "10 cm"
},
{
"text": "30 cm"
},
{
"text": "20 cm"
}
],
"answer": "10 cm",
"solution": "**Answer:** 10 cm\n\nFor telescope\n
Tube length (L) = f0 + fe\n
and magnification (m) = $${{{f_0}} \\over {{f_e}}}$$ = 5\n
where fo and fe are focal length of objective\nand eyepiece.\n
$$ \\therefore $$ f0 + fe = 60\n
$$ \\therefore $$ fo = 50 cm\n
fe = 10 cm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9399,
"subject": "Physics",
"question": "The aperture diameter of a telescope is 5m. The\nseparation between the moon and the earth is\n4 × 105 km. With light of wavelength of\n5500 $$\\mathop A\\limits^o $$, the minimum separation between\nobjects on the surface of moon, so that they are\njust resolved, is close to :",
"options": [
{
"text": "20 m"
},
{
"text": "200 m"
},
{
"text": "600 m"
},
{
"text": "60 m"
}
],
"answer": "60 m",
"solution": "**Answer:** 60 m\n\n\n
= 53.68 m\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9400,
"subject": "Physics",
"question": "In a compound microscope, the magnified virtual image is formed at a distance of 25 cm from the\neye-piece. The focal length of its objective lens is 1 cm. If the magnification is 100 and the tube\nlength of the microscope is 20 cm, then the focal length of the eye-piece lens (in cm) is __________.",
"options": [],
"answer": "6.25",
"solution": "**Answer:** 6.25\n\nL = 20, f0 = 1cm, M = 100
fe = 6.25 cm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9401,
"subject": "Physics",
"question": "A compound microscope consists of an\nobjective lens of focal length 1 cm and an\neyepiece of focal length 5 cm with a separation\nof 10 cm.\n The distance between an object and the\nobjective lens, at which the strain on the eye is\nminimum is $${n \\over {40}}$$ cm. The value of n is _____.",
"options": [],
"answer": "50",
"solution": "**Answer:** 50\n\n\n
$$ \\therefore $$ n = 50",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9402,
"subject": "Physics",
"question": "Given below are two statements : one is labeled as Assertion A and the other is labeled as Reason R.
Assertion A : For a simple microscope, the angular size of the object equals the angular size of the image.
Reason R : Magnification is achieved as the small object can be kept much closer to the eye than 25 cm and hence it subtends a large angle.
In the light of the above statements, choose the most appropriate answer from the options given below :",
"options": [
{
"text": "A is true but R is false"
},
{
"text": "A is false but R is true"
},
{
"text": "Both A and R are true and R is the correct explanation of A"
},
{
"text": "Both A and R are true but R is NOT the correct explanation of A"
}
],
"answer": "Both A and R are true and R is the correct explanation of A",
"solution": "**Answer:** Both A and R are true and R is the correct explanation of A\n\n
The formation of image with simple microscope is shown below.
Hence, if u0 < D (25 cm), hence, the value of $$\\theta$$' will obtain large.
\n
So, option (c) is the correct.
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9403,
"subject": "Physics",
"question": "Your friend is having eye sight problem. She is not able to see clearly a distant uniform window mesh and it appears to her as non-uniform and distorted. The doctor diagnosed the problem as :",
"options": [
{
"text": "Astigmatism"
},
{
"text": "Myopia with Astigmatism"
},
{
"text": "Presbyopia with Astigmatism"
},
{
"text": "Myopia and Hypermetropia"
}
],
"answer": "Myopia with Astigmatism",
"solution": "**Answer:** Myopia with Astigmatism\n\nMyopia with Astigmatism causes distant objects to\nbe blurry and distorted.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9404,
"subject": "Physics",
"question": "An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification '6', gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to __________ cm.",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\nGiven, magnification, M = 6
Since we know that magnifying power of a simple microscope is given by
$$M = 1 + {D \\over {{f_0}}}$$
where, D = least distance of distinct vision = 25 cm
This is the required focal length of eyepiece.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9405,
"subject": "Physics",
"question": "
The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a light of wavelength 2440 $$\\mathop A\\limits^o $$ is used to see th object will be :
A microscope was initially placed in air (refractive index 1). It is then immersed in oil (refractive index 2). For a light whose wavelength in air is $$\\lambda$$, calculate the change of microscope's resolving power due to oil and choose the correct option.
",
"options": [
{
"text": "Resolving power will be $$\\frac{1}{4}$$ in the oil than it was in the air."
},
{
"text": "Resolving power will be twice in the oil than it was in the air."
},
{
"text": "Resolving power will be four times in the oil than it was in the air."
},
{
"text": "Resolving power will be $$\\frac{1}{2}$$ in the oil than it was in the air."
}
],
"answer": "Resolving power will be twice in the oil than it was in the air.",
"solution": "**Answer:** Resolving power will be twice in the oil than it was in the air.\n\n
The resolving power of a microscope is determined by the Rayleigh criterion, which states that it is inversely proportional to the wavelength of light used in the medium in which the microscopy is being performed. Mathematically, the resolving power (RP) can be represented as:
\n\n
$ RP = \\frac{1}{\\lambda_n} $
\n\n
where $\\lambda_n$ is the wavelength of light in the medium, which can be found using the formula:
\n\n
$ \\lambda_n = \\frac{\\lambda}{n} $
\n\n
Here, $\\lambda$ is the wavelength of light in vacuum (or air, for practical purposes, since their refractive indices are close enough), and $n$ is the refractive index of the medium.
\n\n
In air, the refractive index $n = 1$, so the wavelength of light in air ($\\lambda_{air}$) is equal to $\\lambda$.
\n\n
In oil, the refractive index $n = 2$, so the wavelength of light in oil ($\\lambda_{oil}$) is $\\lambda / 2$.
\n\n
Therefore, the change in resolving power when moving from air to oil can be calculated as the ratio of resolving powers in oil to air:
In normal adujstment, for a refracting telescope, the distance between objective and eye piece is $$30 \\mathrm{~cm}$$. The focal length of the objective, when the angular magnification of the telescope is 2 , will be :
A scientist is observing a bacteria through a compound microscope. For better analysis and to improve its resolving power he should. (Select the best option)
",
"options": [
{
"text": "Decrease the focal length of the eye piece."
},
{
"text": "Increase the wave length of the light"
},
{
"text": "Increase the refractive index of the medium between the object and objective lens"
},
{
"text": "Decrease the diameter of the objective lens"
}
],
"answer": "Increase the refractive index of the medium between the object and objective lens",
"solution": "**Answer:** Increase the refractive index of the medium between the object and objective lens\n\n
Resolving power of microscope $$ = \\left( {{{2n\\sin \\theta } \\over \\lambda }} \\right)$$
In a reflecting telescope, a secondary mirror is used to:
",
"options": [
{
"text": "make chromatic aberration zero"
},
{
"text": "remove spherical aberration"
},
{
"text": "reduce the problem of mechanical support"
},
{
"text": "move the eyepiece outside the telescopic tube"
}
],
"answer": "move the eyepiece outside the telescopic tube",
"solution": "**Answer:** move the eyepiece outside the telescopic tube\n\n
Reflecting telescopes, also known as reflectors, use a set of mirrors instead of lenses to gather and focus light. The primary mirror (usually a concave mirror) gathers the incoming light and reflects it to a focus point. The secondary mirror is used to redirect this focused light out to where it can be conveniently observed.
\n
In other words, the secondary mirror in a reflecting telescope is used to move the eyepiece outside the telescopic tube, where the image can be comfortably viewed. This is especially important for large telescopes, where the focus point may be inside the telescope tube and inaccessible without a secondary mirror.
Identify the physical quantity that cannot be measured using spherometer :
",
"options": [
{
"text": "Radius of curvature of concave surface\n"
},
{
"text": "Specific rotation of liquids\n"
},
{
"text": "Thickness of thin plates\n"
},
{
"text": "Radius of curvature of convex surface\n"
}
],
"answer": "Specific rotation of liquids\n",
"solution": "**Answer:** Specific rotation of liquids\n\n\n
Spherometer can be used to measure curvature of surface.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9411,
"subject": "Physics",
"question": "If two mirrors are kept at $${60^ \\circ }$$ to each other, then the number of images formed by them is ",
"options": [
{
"text": "$$5$$ "
},
{
"text": "$$6$$ "
},
{
"text": "$$7$$ "
},
{
"text": "$$8$$ "
}
],
"answer": "$$5$$ ",
"solution": "**Answer:** $$5$$ \n\nKEY CONCEPT : When two plane mirrors are inclined a each other at an angle $$\\theta $$ then the number of the images of a point object placed between the plane mirrors is \n
$$ = {{{{360}^ \\circ }} \\over {{{60}^ \\circ }}} - 1 = 5$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9412,
"subject": "Physics",
"question": "To get three images of a single object, one should have two plane mirrors at an angle of ",
"options": [
{
"text": "$${60^ \\circ }$$ "
},
{
"text": "$${90^ \\circ }$$"
},
{
"text": "$${120^ \\circ }$$"
},
{
"text": "$${30^ \\circ }$$"
}
],
"answer": "$${90^ \\circ }$$",
"solution": "**Answer:** $${90^ \\circ }$$\n\nWhen $$\\theta = {90^ \\circ }$$ then $${{360} \\over \\theta } = {{360} \\over {90}} = 4$$\n
is an even number. The number of images formed is given by\n
$$n = {{360} \\over \\theta } - 1 = {{360} \\over {90}} - 1 = 4 - 1 = 3$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9413,
"subject": "Physics",
"question": "A car is fitted with a convex side-view mirror of focal length $$20$$ $$cm$$. A second car $$2.8m$$ behind the first car is overtaking the first car at a relative speed of $$15$$ $$m/s$$. The speed of the image of the second car as seen in the mirror of the first one is : ",
"options": [
{
"text": "$${1 \\over {15}}\\,m/s$$ "
},
{
"text": "$$10\\,m/s$$ "
},
{
"text": "$$15\\,m/s$$ "
},
{
"text": "$${1 \\over {10}}\\,m/s$$"
}
],
"answer": "$${1 \\over {15}}\\,m/s$$ ",
"solution": "**Answer:** $${1 \\over {15}}\\,m/s$$ \n\nFrom mirror formula\n
$$ \\Rightarrow {{dv} \\over {dt}} = {1 \\over {15}}m/s$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9414,
"subject": "Physics",
"question": "Two plane mirrors are inclined to each other such that a ray of light incident on the first mirror (M1) and parallel to the second mirror (M2) is finally reflected from the second mirror (M2) parallel to the first mirror (M1). The angle between the two mirrors will be : ",
"options": [
{
"text": "45o"
},
{
"text": "60o"
},
{
"text": "75o"
},
{
"text": "90o"
}
],
"answer": "60o",
"solution": "**Answer:** 60o\n\n\n Let the angle between two mirrors are $$\\theta $$.\n
We know sum of angles of triangle = 180o\n
$$ \\therefore $$ 3$$\\theta $$ = 180o\n
$$ \\Rightarrow $$ $$\\theta $$ = 60o",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9415,
"subject": "Physics",
"question": "A concave mirror for face viewing has focal\nlength of 0.4 m. The distance at which you hold\nthe mirror from your face in order to see your\nimage upright with a magnification of 5 is :",
"options": [
{
"text": "0.24 m"
},
{
"text": "0.32 m"
},
{
"text": "1.60 m"
},
{
"text": "0.16 m"
}
],
"answer": "0.32 m",
"solution": "**Answer:** 0.32 m\n\n$$m = {f \\over {f - u}}$$
\n$$5 = {{ - 40} \\over { - 40 - u}};\\,u = - 32cm$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9416,
"subject": "Physics",
"question": "When an object is kept at a distance of 30 cm from a concave mirror, the image is formed at a\ndistance of 10 cm from the mirror. If the object is moved with a speed of 9 cms–1, the speed\n(in cms–1) with which image moves at that instant is ____.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nVI = Velocity of image with respect to mirror
$$ = 1$$ cm/s\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9417,
"subject": "Physics",
"question": "The focal length f is related to the radius of curvature r of the spherical convex mirror by :",
"options": [
{
"text": "f = r"
},
{
"text": "f = $$-$$ r"
},
{
"text": "f = +$${{1 \\over 2}}$$ r"
},
{
"text": "f = $$-$$$${{1 \\over 2}}$$ r"
}
],
"answer": "f = +$${{1 \\over 2}}$$ r",
"solution": "**Answer:** f = +$${{1 \\over 2}}$$ r\n\nFor convex mirror, the focal length (f) and radius of curvature (r) are related as $$f = + {r \\over 2}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9418,
"subject": "Physics",
"question": "A short straight object of height 100 cm lies before the central axis of a spherical mirror whose focal length has absolute value | f | = 40 cm. The image of object produced by the mirror is of height 25 cm and has the same orientation of the object. One may conclude from the information :",
"options": [
{
"text": "Image is real, same side of convex mirror."
},
{
"text": "Image is virtual, opposite side of convex mirror."
},
{
"text": "Image is virtual, opposite side of concave mirror."
},
{
"text": "Image is real, same side of concave mirror."
}
],
"answer": "Image is virtual, opposite side of convex mirror.",
"solution": "**Answer:** Image is virtual, opposite side of convex mirror.\n\n\n
Same orientation so image is virtual. It is combination of real object and virtual image using height, it is possible only from convex mirror.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9419,
"subject": "Physics",
"question": "The incident ray, reflected ray and the outward drawn normal are denoted by the unit vectors $$\\overrightarrow a $$, $$\\overrightarrow b $$ and $$\\overrightarrow c $$ respectively. Then choose the correct relation for these vectors.",
"options": [
{
"text": "$$\\overrightarrow b $$ = $$\\overrightarrow a $$ + 2$$\\overrightarrow c $$"
},
{
"text": "$$\\overrightarrow b $$ = $$\\overrightarrow a $$ $$-$$ 2 ($$\\overrightarrow a $$ . $$\\overrightarrow c $$)$$\\overrightarrow c $$"
},
{
"text": "$$\\overrightarrow b $$ = 2$$\\overrightarrow a $$ + $$\\overrightarrow c $$"
},
{
"text": "$$\\overrightarrow b $$ = $$\\overrightarrow a $$ $$-$$ $$\\overrightarrow c $$"
}
],
"answer": "$$\\overrightarrow b $$ = $$\\overrightarrow a $$ $$-$$ 2 ($$\\overrightarrow a $$ . $$\\overrightarrow c $$)$$\\overrightarrow c $$",
"solution": "**Answer:** $$\\overrightarrow b $$ = $$\\overrightarrow a $$ $$-$$ 2 ($$\\overrightarrow a $$ . $$\\overrightarrow c $$)$$\\overrightarrow c $$\n\n\n
Here $$\\overrightarrow a = \\left| {\\overrightarrow a } \\right|\\sin \\theta \\widehat i - \\left| {\\overrightarrow a } \\right|\\cos \\theta \\widehat j$$\n
As $$\\overrightarrow a $$ is an unit vector, so $$\\left| {\\overrightarrow a } \\right|$$ = 1\n
$$ \\therefore $$ $$\\overrightarrow a = \\left| {\\overrightarrow a } \\right|\\sin \\theta \\widehat i - \\left| {\\overrightarrow a } \\right|\\cos \\theta \\widehat j$$\n
Similarly $$\\overrightarrow b = \\sin \\theta \\widehat i + \\cos \\theta \\widehat j$$\n
and $$\\overrightarrow c = \\widehat j$$\n
From option (B),\n
$$\\overrightarrow a $$ $$-$$ 2 ($$\\overrightarrow a $$ . $$\\overrightarrow c $$)$$\\overrightarrow c $$\n
= $$\\sin \\theta \\widehat i + \\cos \\theta \\widehat j$$ = $$\\overrightarrow b $$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 9420,
"subject": "Physics",
"question": "Car B overtakes another car A at a relative speed of 40 ms$$-$$1. How fast will the image of car B appear to move in the mirror of focal length 10 cm fitted in car A, when the car B is 1.9 m away from the car A?",
"options": [
{
"text": "4 ms$$-$$1"
},
{
"text": "0.2 ms$$-$$1"
},
{
"text": "40 ms$$-$$1"
},
{
"text": "0.1 ms$$-$$1"
}
],
"answer": "0.1 ms$$-$$1",
"solution": "**Answer:** 0.1 ms$$-$$1\n\nHere, $${1 \\over v} + {1 \\over u} = {1 \\over f}$$ .......(1)\n
$$ = {1 \\over {400}} \\times 40 = 0.1$$ m/s",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9421,
"subject": "Physics",
"question": "An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d1 from C and the distance of the image formed is d2 from C, the radius of curvature of this mirror is :",
"options": [
{
"text": "$${{2{d_1}{d_2}} \\over {{d_1} - {d_2}}}$$"
},
{
"text": "$${{2{d_1}{d_2}} \\over {{d_1} + {d_2}}}$$"
},
{
"text": "$${{{d_1}{d_2}} \\over {{d_1} + {d_2}}}$$"
},
{
"text": "$${{{d_1}{d_2}} \\over {{d_1} - {d_2}}}$$"
}
],
"answer": "$${{2{d_1}{d_2}} \\over {{d_1} - {d_2}}}$$",
"solution": "**Answer:** $${{2{d_1}{d_2}} \\over {{d_1} - {d_2}}}$$\n\nUsing Newton's formula
Two objects A and B are placed at 15 cm and 25 cm from the pole in front of a concave mirror having radius of curvature 40 cm. The distance between images formed by the mirror is _______________.
A thin cylindrical rod of length $$10 \\mathrm{~cm}$$ is placed horizontally on the principle axis of a concave mirror of focal length $$20 \\mathrm{~cm}$$. The rod is placed in a such a way that mid point of the rod is at $$40 \\mathrm{~cm}$$ from the pole of mirror. The length of the image formed by the mirror will be $$\\frac{x}{3} \\mathrm{~cm}$$. The value of $$x$$ is _____________.
In an experiment for estimating the value of focal length of converging mirror, image of an object placed at $$40 \\mathrm{~cm}$$ from the pole of the mirror is formed at distance $$120 \\mathrm{~cm}$$ from the pole of the mirror. These distances are measured with a modified scale in which there are 20 small divisions in \n$$1 \\mathrm{~cm}$$. The value of error in measurement of focal length of the mirror is $$\\frac{1}{\\mathrm{~K}} \\mathrm{~cm}$$. The value of $$\\mathrm{K}$$ is __________.
The light rays from an object have been reflected towards an observer from a standard flat mirror, the image observed by the observer are :-
\n
A. Real
\n
B. Erect
\n
C. Smaller in size then object
\n
D. Laterally inverted
\n
Choose the most appropriate answer from the options given below :
",
"options": [
{
"text": "B and D only"
},
{
"text": "A, C, and D only"
},
{
"text": "A and D only"
},
{
"text": "B and C only"
}
],
"answer": "B and D only",
"solution": "**Answer:** B and D only\n\nPlane mirror forms erect, same sized, laterally\ninverted and virtual image of real object.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9426,
"subject": "Physics",
"question": "
When one light ray is reflected from a plane mirror with $$30^{\\circ}$$ angle of reflection, the angle of deviation of the ray after reflection is :
",
"options": [
{
"text": "$$140^{\\circ}$$"
},
{
"text": "$$130^{\\circ}$$"
},
{
"text": "$$120^{\\circ}$$"
},
{
"text": "$$110^{\\circ}$$"
}
],
"answer": "$$120^{\\circ}$$",
"solution": "**Answer:** $$120^{\\circ}$$\n\nWhen a light ray is reflected from a plane mirror, the angle of incidence (i) is equal to the angle of reflection (r). In this case, the angle of reflection is given as $$30^{\\circ}$$, so the angle of incidence is also $$30^{\\circ}$$.\n
\nThe angle of deviation (D) is the angle between the incident ray and the reflected ray. To find this angle, consider the fact that the angle between the incident ray and the normal to the mirror and the angle between the reflected ray and the normal add up to $$180^{\\circ}$$, since they are supplementary angles.\n
\nThus, we have:
\n$$i + r + D = 180^{\\circ}$$
\nSince $$i = r$$, we can rewrite the equation as:
\nSo, the angle of deviation of the ray after reflection is $$120^{\\circ}$$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9427,
"subject": "Physics",
"question": "
An object is placed at a distance of 12 cm in front of a plane mirror. The virtual and erect image is formed by the mirror. Now the mirror is moved by 4 cm towards the stationary object. The distance by which the position of image would be shifted, will be
",
"options": [
{
"text": "4 cm towards mirror"
},
{
"text": "2 cm towards mirror"
},
{
"text": "8 cm away from mirror"
},
{
"text": "8 cm towards mirror"
}
],
"answer": "8 cm towards mirror",
"solution": "**Answer:** 8 cm towards mirror\n\n \n \n$$ \\therefore $$ Shifting of image will be 8 cm towards mirror. ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9428,
"subject": "Physics",
"question": "
A convex mirror of radius of curvature $$30 \\mathrm{~cm}$$ forms an image that is half the size of the object. The object distance is :
If the distance between object and its two times magnified virtual image produced by a curved mirror is $$15 \\mathrm{~cm}$$, the focal length of the mirror must be:
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9430,
"subject": "Physics",
"question": "Which of the following is used in optical fibres? ",
"options": [
{
"text": "total internal reflection "
},
{
"text": "scattering "
},
{
"text": "diffracttion"
},
{
"text": "refraction "
}
],
"answer": "total internal reflection ",
"solution": "**Answer:** total internal reflection \n\n
Optical fibers work on the principle of total internal reflection. When light is transmitted through the fiber, it is reflected off the inner walls of the fiber in such a way that it remains within the fiber, allowing it to carry the light signal over great distances with minimal loss.
\n
So, the correct answer is :
\n
Option A : total internal reflection.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9431,
"subject": "Physics",
"question": "A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is $${4 \\over 3}$$ and the fish is $$12$$ $$cm$$ below the surface, the radius of this circle in $$cm$$ is ",
"options": [
{
"text": "$${{36} \\over {\\sqrt 7 }}$$ "
},
{
"text": "$${36\\sqrt 7 }$$ "
},
{
"text": "$${4\\sqrt 5 }$$"
},
{
"text": "$${36\\sqrt 5 }$$"
}
],
"answer": "$${{36} \\over {\\sqrt 7 }}$$ ",
"solution": "**Answer:** $${{36} \\over {\\sqrt 7 }}$$ \n\n$$\\sin {\\theta _c} = {1 \\over \\mu } = {3 \\over 4}$$\n
$$ \\Rightarrow R = {{36} \\over {\\sqrt 7 }}\\,cm$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9432,
"subject": "Physics",
"question": "The refractive index of a glass is $$1.520$$ for red light and $$1.525$$ for blue light. Let $${D_1}$$ and $${D_2}$$ be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then, ",
"options": [
{
"text": "$${D_1} < {D_2}$$ "
},
{
"text": "$${D_1} = {D_2}$$ "
},
{
"text": "$${D_1}$$ can be less than or greater than $${D_2}$$ depending upon the angle of prism"
},
{
"text": "$${D_1} > {D_2}$$ "
}
],
"answer": "$${D_1} < {D_2}$$ ",
"solution": "**Answer:** $${D_1} < {D_2}$$ \n\nFor a thin prism, $$D = \\left( {\\mu - 1} \\right)A$$\n
Since $${\\lambda _b} < {\\lambda _r} \\Rightarrow {\\mu _r} < {\\mu _b} \\Rightarrow {D_1} < {D_2}$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9433,
"subject": "Physics",
"question": "An initially parallel cylindrical beam travels in a medium of refractive index $$\\mu \\left( I \\right) = {\\mu _0} + {\\mu _2}\\,I,$$ where $${\\mu _0}$$ and $${\\mu _2}$$ are positive constants and $$I$$ is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.\n
The speed of light in the medium is
",
"options": [
{
"text": "minimum on the axis of the beam "
},
{
"text": "the same everywhere in the beam "
},
{
"text": "directly proportional to the intensity $$I$$ "
},
{
"text": "maximum on the axis of the beam "
}
],
"answer": "minimum on the axis of the beam ",
"solution": "**Answer:** minimum on the axis of the beam \n\nThe speed of light $$(c)$$ in a medium of refractive index $$\\left( \\mu \\right)$$ is given by\n
$$\\mu = {{{c_0}} \\over c},$$ where $${c_0}$$ is the speed of light in vacuum\n
As $$I$$ is decreasing with increasing radius, it is maximum \n
on the axis of the beam. Therefore, $$c$$ is minimum on the axis of the beam.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9434,
"subject": "Physics",
"question": "An initially parallel cylindrical beam travels in a medium of refractive index $$\\mu \\left( I \\right) = {\\mu _0} + {\\mu _2}\\,I,$$ where $${\\mu _0}$$ and $${\\mu _2}$$ are positive constants and $$I$$ is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.\n
As the beam enters the medium, it will
",
"options": [
{
"text": "diverge "
},
{
"text": "converge "
},
{
"text": "diverge near the axis and converge near the periphery "
},
{
"text": "travel as a cylindrical beam "
}
],
"answer": "converge ",
"solution": "**Answer:** converge \n\nIn the medium, the refractive index will decreases from the axis forwards the periphery of the beam.\n
Therefore, the beam will move as one move from the axis to the periphery and hence the beam will converge.\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9435,
"subject": "Physics",
"question": "Let $$x$$-$$z$$ plane be the boundary between two transparent media. Medium $$1$$ in $$z \\ge 0$$ has a refractive index of $$\\sqrt 2 $$ and medium $$2$$ with $$z < 0$$ has a refractive index of $$\\sqrt 3 .$$ A ray of light in medium $$1$$ given by the vector $$\\overrightarrow A = 6\\sqrt 3 \\widehat i + 8\\sqrt 3 \\widehat j - 10\\widehat k$$ is incident on the plane of separation. The angle of refraction in medium $$2$$ is: ",
"options": [
{
"text": "$${45^ \\circ }$$ "
},
{
"text": "$${60^ \\circ }$$"
},
{
"text": "$${75^ \\circ }$$"
},
{
"text": "$${30^ \\circ }$$"
}
],
"answer": "$${45^ \\circ }$$ ",
"solution": "**Answer:** $${45^ \\circ }$$ \n\n \n
From Snell's law, $$\\sqrt 2 \\sin i = \\sqrt 3 \\sin r$$\n
$$ \\Rightarrow $$ $$\\angle r = {45^ \\circ }$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9436,
"subject": "Physics",
"question": "A green light is incident from the water to the air - water interface at the critical angle $$\\left( \\theta \\right)$$. Select the correct statement. ",
"options": [
{
"text": "The entire spectrum of visible light will come out of the water at an angle of $${90^ \\circ }$$ to the normal. "
},
{
"text": "The spectrum of visible light whose frequency is less than that of green light will come out to the air medium. "
},
{
"text": "The spectrum of visible light whose frequency is more than that of green light will come out to the air medium. "
},
{
"text": "The entire spectrum of visible light will come out of the water at various angles to the normal. "
}
],
"answer": "The spectrum of visible light whose frequency is less than that of green light will come out to the air medium. ",
"solution": "**Answer:** The spectrum of visible light whose frequency is less than that of green light will come out to the air medium. \n\nFor critical angle $${\\theta _c},$$\n
$$\\sin {\\theta _c} = {1 \\over \\mu }$$\n
For greater wavelength or lesser frequency $$\\mu $$ is less.\n
\n
So, critical angle would be more, So, they will not suffer reflection and come out at angles less then $${90^ \\circ }.$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9437,
"subject": "Physics",
"question": "In an experiment for determination of refractive index of glass of a prism by $$i - \\delta ,$$ plot it was found thata ray incident at angle $${35^ \\circ }$$, suffers a deviation of $${40^ \\circ }$$ and that it emerges at angle $${79^ \\circ }.$$ In that case which of the following is closest to the maximum possible value of the refractive index? ",
"options": [
{
"text": "$$1.7$$ "
},
{
"text": "$$1.8$$ "
},
{
"text": "$$1.5$$ "
},
{
"text": "$$1.6$$ "
}
],
"answer": "$$1.5$$ ",
"solution": "**Answer:** $$1.5$$ \n\nWe know that $$i + e - A = \\delta $$\n
Therefore, the minimum of three readings are required.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9439,
"subject": "Physics",
"question": "Let the refractive index of a denser medium with respect to a rarer medium be n12 and its critical angle be θC . At an angle of incidence A when light is travelling from\ndenser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is 90o. Angle A is given by :\n",
"options": [
{
"text": "$${1 \\over {{{\\cos }^{ - 1}}\\left( {\\sin {\\theta _C}} \\right)}}$$ "
},
{
"text": "$${1 \\over {{{\\tan }^{ - 1}}\\left( {\\sin {\\theta _C}} \\right)}}$$ "
},
{
"text": "$${\\cos ^{ - 1}}\\,\\left( {\\sin {\\theta _C}} \\right)$$ "
},
{
"text": "$${\\tan ^{ - 1}}\\,\\left( {\\sin {\\theta _C}} \\right)$$ "
}
],
"answer": "$${\\tan ^{ - 1}}\\,\\left( {\\sin {\\theta _C}} \\right)$$ ",
"solution": "**Answer:** $${\\tan ^{ - 1}}\\,\\left( {\\sin {\\theta _C}} \\right)$$ \n\n\n
$$ \\Rightarrow $$$$\\,\\,\\,$$ A = tan$$-$$1 (sin $$\\theta $$c)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9440,
"subject": "Physics",
"question": "A ray of light is incident at an angle of 60o on one face of a prism of angle 30o. The emergent ray of light makes an angle of 30o with incident ray. The angle made by the emergent ray with second face of prism will be : ",
"options": [
{
"text": "0o "
},
{
"text": "90o "
},
{
"text": "45o "
},
{
"text": "30o "
}
],
"answer": "90o ",
"solution": "**Answer:** 90o \n\n
Given : $${i_1} = 60^\\circ $$; $$A = 30^\\circ $$
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9441,
"subject": "Physics",
"question": "A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is $$\\sqrt 3 $$, then the angle of incidence is:",
"options": [
{
"text": "60o"
},
{
"text": "45o"
},
{
"text": "90o"
},
{
"text": "30o"
}
],
"answer": "60o",
"solution": "**Answer:** 60o\n\ni = e\n
r1 = r2 = $${A \\over 2}$$ = 30o\n
by Snell's law\n
1 $$ \\times $$ sin i = $$\\sqrt 3 \\times {1 \\over 2} = {{\\sqrt 3 } \\over 2}$$\n
i = 60",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9442,
"subject": "Physics",
"question": "The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability $${4 \\over 3}$$ for this wavelength, will be :",
"options": [
{
"text": "45°"
},
{
"text": "15°"
},
{
"text": "30°"
},
{
"text": "60°"
}
],
"answer": "30°",
"solution": "**Answer:** 30°\n\nn = $$\\sqrt {{\\varepsilon _r}{\\mu _r}} = \\sqrt {3 \\times {4 \\over 3}} $$ = 2\n
n sin c = 1 sin 90o\n
sin c = $${1 \\over 2}$$\n
$$ \\Rightarrow $$ c = 30o",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9443,
"subject": "Physics",
"question": "A prism of angle A = 1o has a refractive index\n$$\\mu $$ = 1.5. A good estimate for the minimum angle\nof deviation (in degrees) is close to $${N \\over {10}}$$.\n Value of N is ____.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n
Deviation for small-angled prism is given by
\n
$$\\delta$$ = ($$\\mu$$ $$-$$ 1) A
\n
Given, A = 1$$^\\circ$$, $$\\mu$$ = 1.5
\n
Substituting these values in above equation, we get
\n
$$\\delta$$ = (1.5 $$-$$ 1)1 $$\\delta$$ = 0.5
\n
According to question, $$\\delta = {N \\over {10}}$$
\n
$$ \\Rightarrow 0.5 = {N \\over {10}} = N = 5$$
\n
Hence, the value of N is 5.
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9444,
"subject": "Physics",
"question": "A light ray enters a solid glass sphere of\nrefractive index $$\\mu $$ = $$\\sqrt 3 $$ at an angle of incidence\n60o. The ray is both reflected and refracted at\nthe farther surface of the sphere. The angle (in\ndegrees) between the reflected and refracted\nrays at this surface is ________.",
"options": [],
"answer": "90",
"solution": "**Answer:** 90\n\n\n
By Snell's law at S1 :\n
1 $$ \\times $$ sin 60° = $$\\sqrt 3 $$ sin r\n
$$ \\Rightarrow $$ r = 30o\n
As S2 :\n
$$\\sqrt 3 \\sin r = 1\\sin e$$\n
$$ \\Rightarrow $$ e = 60o\n
Now, r\n + $$\\theta $$ + e = 180o",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9445,
"subject": "Physics",
"question": "There is a small source of light at some depth\nbelow the surface of water (refractive\nindex = $${4 \\over 3}$$) in a tank of large cross sectional\nsurface area. Neglecting any reflection from the\nbottom and absorption by water, percentage of\nlight that emerges out of surface is (nearly) :\n [Use the fact that surface area of a spherical cap\nof height h and radius of curvature r is 2$$\\pi $$rh]:",
"options": [
{
"text": "17%"
},
{
"text": "34%"
},
{
"text": "50%"
},
{
"text": "21%"
}
],
"answer": "17%",
"solution": "**Answer:** 17%\n\n\n $${4 \\over 3}$$ sin $$\\theta $$ = 1sin90o\n
sin $$\\theta $$ = $${3 \\over 4}$$\n
cos $$\\theta $$ = $${{\\sqrt 7 } \\over 4}$$\n
Surface area in solid angle d$$\\Omega $$ = 2$$\\pi $$R2(1 - cos $$\\theta $$)\n
= $${3 \\over 4}h\\sqrt 2 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9447,
"subject": "Physics",
"question": "Red light differs from blue light as they have :",
"options": [
{
"text": "Different frequencies and same wavelengths"
},
{
"text": "Different frequencies and different wavelengths"
},
{
"text": "Same frequencies and different wavelengths"
},
{
"text": "Same frequencies and same wavelengths"
}
],
"answer": "Different frequencies and different wavelengths",
"solution": "**Answer:** Different frequencies and different wavelengths\n\nRed light and blue light have different wavelength and different frequency.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9448,
"subject": "Physics",
"question": "A deviation of 2$$^\\circ$$ is produced in the yellow ray when prism of crown and flint glass are achromatically combined. Taking dispersive powers of crown and flint glass as 0.02 and 0.03 respectively and refractive index for yellow light for these glasses are 1.5 and 1.6 respectively. The refracting angles for crown glass prism will be ____________$$^\\circ$$ (in degree). (Round off to the Nearest Integer)",
"options": [],
"answer": "12",
"solution": "**Answer:** 12\n\n$${\\delta _{net}} = ({\\mu _1} - 1){A_1} - ({\\mu _2} - 1){A_2}$$
$${A_1} = 12^\\circ $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9449,
"subject": "Physics",
"question": "The image of an object placed in air formed by a convex refracting surface is at a distance of 10 m behind the surface. The image is real and is at $${{{2^{rd}}} \\over 3}$$ of the distance of the object from the surface. The wavelength of light inside the surface is $${2 \\over 3}$$ times the wavelength in air. The radius of the curved surface is $${x \\over {13}}$$ m. The value of 'x' is ___________.",
"options": [],
"answer": "30",
"solution": "**Answer:** 30\n\n$${{{\\mu _2}} \\over v} - {{{\\mu _1}} \\over u} = {{{\\mu _2} - {\\mu _1}} \\over R}$$
$$R = {{30} \\over {13}}$$ m",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9450,
"subject": "Physics",
"question": "A ray of light passing through a prism ($$\\mu$$ = $$\\sqrt 3 $$) suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is _____________ (in degrees).",
"options": [],
"answer": "60",
"solution": "**Answer:** 60\n\nFor minimum deviation r1 = r2 = A/2
$$\\Rightarrow$$ A = 60$$^\\circ$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9451,
"subject": "Physics",
"question": "A ray of laser of a wavelength 630 nm is incident at an angle of 30$$^\\circ$$ at the diamond-air interface. It is going from diamond to air. The refractive index of diamond is 2.42 and that of air is 1. Choose the correct option.",
"options": [
{
"text": "angle of refraction is 24.41$$^\\circ$$"
},
{
"text": "angle of refraction is 30$$^\\circ$$"
},
{
"text": "refraction is not possible"
},
{
"text": "angle of refraction is 53.4$$^\\circ$$"
}
],
"answer": "refraction is not possible",
"solution": "**Answer:** refraction is not possible\n\n$$\\sin {\\theta _C} = {1 \\over \\mu } = {1 \\over {2{\\mu _2}}} < \\sin {\\theta _C}$$
sin$$\\theta$$ > sin$$\\theta$$C
$$\\theta$$ > $$\\theta$$C
Total internal reflection will happen.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9452,
"subject": "Physics",
"question": "A prism of refractive index $$\\mu$$ and angle of prism A is placed in the position of minimum angle of deviation. If minimum angle of deviation is also A, then in terms of refractive index",
"options": [
{
"text": "$$2{\\cos ^{ - 1}}\\left( {{\\mu \\over 2}} \\right)$$"
},
{
"text": "$${\\sin ^{ - 1}}\\left( {{\\mu \\over 2}} \\right)$$"
},
{
"text": "$${\\sin ^{ - 1}}\\left( {\\sqrt {{{\\mu - 1} \\over 2}} } \\right)$$"
},
{
"text": "$${\\cos ^{ - 1}}\\left( {{\\mu \\over 2}} \\right)$$"
}
],
"answer": "$$2{\\cos ^{ - 1}}\\left( {{\\mu \\over 2}} \\right)$$",
"solution": "**Answer:** $$2{\\cos ^{ - 1}}\\left( {{\\mu \\over 2}} \\right)$$\n\n$$\\mu = {{\\sin \\left( {{{A + {\\delta _{\\min }}} \\over 2}} \\right)} \\over {\\sin \\left( {{A \\over 2}} \\right)}}$$
$$A = 2{\\cos ^{ - 1}}\\left( {{\\mu \\over 2}} \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9453,
"subject": "Physics",
"question": "A glass tumbler having inner depth of 17.5 cm is kept on a table. A student starts pouring water ($$\\mu$$ = 4/3) into it while looking at the surface of water from the above. When he feels that the tumbler is half filled, he stops pouring water. Up to what height, the tumbler is actually filled?",
"options": [
{
"text": "11.7 cm"
},
{
"text": "10 cm"
},
{
"text": "7.5 cm"
},
{
"text": "8.75 cm"
}
],
"answer": "10 cm",
"solution": "**Answer:** 10 cm\n\n Consider the actual height of the tumbler be H.
The speed of light in media 'A' and 'B' are $$2.0 \\times {10^{10}}$$ cm/s and $$1.5 \\times {10^{10}}$$ cm/s respectively. A ray of light enters from the medium B to A at an incident angle '$$\\theta$$'. If the ray suffers total internal reflection, then
Consider a light ray travelling in air is incident into a medium of refractive index $$\\sqrt{2n}$$. The incident angle is twice that of refracting angle. Then, the angle of incidence will be :
A small bulb is placed at the bottom of a tank containing water to a depth of $$\\sqrt7$$ m. The refractive index of water is $${4 \\over 3}$$. The area of the surface of water through which light from the bulb can emerge out is x$$\\pi$$ m2. The value of x is __________.
A light wave travelling linearly in a medium of dielectric constant 4, incidents on the horizontal interface separating medium with air. The angle of incidence for which the total intensity of incident wave will be reflected back into the same medium will be :
\n
(Given : relative permeability of medium $$\\mu$$r = 1)
The difference of speed of light in the two media A and B (vA $$-$$ vB) is 2.6 $$\\times$$ 107 m/s. If the refractive index of medium B is 1.47, then the ratio of refractive index of medium B to medium A is : (Given : speed of light in vacuum c = 3 $$\\times$$ 108 ms$$-$$1)
A ray of light is incident at an angle of incidence 60$$^\\circ$$ on the glass slab of refractive index $$\\sqrt3$$. After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is 4$$\\sqrt3$$ cm. The thickness of the glass slab is __________ cm.
The refractive index of an equilateral prism is $$\\sqrt 2 $$. The angle of emergence under minimum deviation position of prism, in degree, is ___________.
",
"options": [
{
"text": "In primary rainbow, observer sees red colour on the top and violet on the bottom"
},
{
"text": "In primary rainbow, observer sees violet colour on the top and red on the bottom"
},
{
"text": "In primary rainbow, light wave suffers total internal reflection twice before coming out of water drops."
},
{
"text": "Primary rainbow is less bright than secondary rainbow."
}
],
"answer": "In primary rainbow, observer sees red colour on the top and violet on the bottom",
"solution": "**Answer:** In primary rainbow, observer sees red colour on the top and violet on the bottom\n\n
In primary rainbow, observer sees red colour on the top and violet on the bottom.
Time taken by light to travel in two different materials $$A$$ and $$B$$ of refractive indices $$\\mu_{A}$$ and $$\\mu_{B}$$ of same thickness is $$t_{1}$$ and $$t_{2}$$ respectively. If $$t_{2}-t_{1}=5 \\times 10^{-10}$$ s and the ratio of $$\\mu_{A}$$ to $$\\mu_{B}$$ is $$1: 2$$. Then, the thickness of material, in meter is: (Given $$v_{\\mathrm{A}}$$ and $$v_{\\mathrm{B}}$$ are velocities of light in $$A$$ and $$B$$ materials respectively.)
Light travels in two media $$M_{1}$$ and $$M_{2}$$ with speeds $$1.5 \\times 10^{8} \\mathrm{~ms}^{-1}$$ and $$2.0 \\times 10^{8} \\mathrm{~ms}^{-1}$$ respectively. The critical angle between them is :
The X-Y plane be taken as the boundary between two transparent media $$\\mathrm{M}_{1}$$ and $$\\mathrm{M}_{2}$$. $$\\mathrm{M}_{1}$$ in $$Z \\geqslant 0$$ has a refractive index of $$\\sqrt{2}$$ and $$M_{2}$$ with $$Z<0$$ has a refractive index of $$\\sqrt{3}$$. A ray of light travelling in $$\\mathrm{M}_{1}$$ along the direction given by the vector $$\\overrightarrow{\\mathrm{P}}=4 \\sqrt{3} \\hat{i}-3 \\sqrt{3} \\hat{j}-5 \\hat{k}$$, is incident on the plane of separation. The value of difference between the angle of incident in $$\\mathrm{M}_{1}$$ and the angle of refraction in $$\\mathrm{M}_{2}$$ will be __________ degree.
Light enters from air into a given medium at an angle of $$45^{\\circ}$$ with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of \n$$15^{\\circ}$$ from its original direction. The refractive index of the medium is:
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9467,
"subject": "Physics",
"question": "A microscope is focused on an object at the bottom of a bucket. If liquid with refractive index $\\frac{5}{3}$\n\nis poured inside the bucket, then the microscope has to be raised by $30 \\mathrm{~cm}$ to focus the object again.\n\nThe height of the liquid in the bucket is :",
"options": [
{
"text": "$50 \\mathrm{~cm}$"
},
{
"text": "$18 \\mathrm{~cm}$"
},
{
"text": "$75 \\mathrm{~cm}$"
},
{
"text": "$12 \\mathrm{~cm}$"
}
],
"answer": "$75 \\mathrm{~cm}$",
"solution": "**Answer:** $75 \\mathrm{~cm}$\n\nShift $=\\left(d-\\frac{d}{\\mu}\\right)=30 \\mathrm{~cm}$\n\n
A thin prism $P_1$ with an angle $6^{\\circ}$ and made of glass of refractive index $1.54$ is combined with another prism $P_2$ made from glass of refractive index $1.72$ to produce dispersion without average deviation. The angle of prism $P_2$ is
In an experiment of measuring the refractive index of a glass slab using travelling microscope in physics lab, a student measures real thickness of the glass slab as 5.25 mm and apparent thickness of the glass slab as 5.00 mm. Travelling microscope has 20 divisions in one cm on main scale and 20 divisions on vernier scale is equal to 49 divisions on main scale. The estimated uncertainty in the measurement of refractive index of the slab is $$\\frac{x}{10}\\times10^{-3}$$, where $$x$$ is ___________
A ray of light is incident from air on a glass plate having thickness $$\\sqrt3$$ cm and refractive index $$\\sqrt2$$. The angle of incidence of a ray is equal to the critical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is ____________ $$\\times$$ 10$$^{-2}$$ cm. (given $$\\sin 15^\\circ = 0.26$$)
When a beam of white light is allowed to pass through convex lens parallel to principal axis, the different colours of light converge at different point on the principle axis after refraction. This is called :
The phenomenon described in the statement is called chromatic aberration.
\n
\nChromatic aberration is a type of optical aberration that occurs when a lens is unable to focus different colors of light at the same point after refraction. This causes the different colors of light to converge at different points on the principal axis, resulting in blurred or distorted images.
\n\n
In the case of a convex lens, the refractive index of the lens is different for different colors of light. The blue light, having the shortest wavelength, refracts the most and the red light, having the longest wavelength, refracts the least. This causes the blue light to converge at a point closer to the lens than the red light, resulting in a blurred image.\n
\n
Spherical aberration, on the other hand, is a type of optical aberration that occurs when a spherical lens is unable to focus all the incident light rays at a single point, resulting in a blurred image. Polarization refers to the orientation of the electric field of light waves, and scattering refers to the phenomenon of light being redirected in different directions due to interaction with matter.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9472,
"subject": "Physics",
"question": "The refractive index of a transparent liquid filled in an equilateral hollow prism is $\\sqrt{2}$. The angle of minimum deviation for the liquid will be ___________ $$^\\circ$$.",
"options": [],
"answer": "30",
"solution": "**Answer:** 30\n\nTo find the angle of minimum deviation for the liquid in the equilateral hollow prism, we can use the formula for the angle of minimum deviation based on the refractive index $n$ and the prism angle $A$:\n
\nIn this case, the refractive index $n = \\sqrt{2}$, and since the prism is equilateral, the prism angle $A = 60^\\circ$. Now, we can substitute these values into the formula and solve for the angle of minimum deviation $\\delta_m$:\n
\nThe angle of minimum deviation for the liquid in the equilateral hollow prism is $30^\\circ$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9473,
"subject": "Physics",
"question": "
A vessel of depth '$$d$$' is half filled with oil of refractive index $$n_{1}$$ and the other half is filled with water of refractive index $$n_{2}$$. The apparent depth of this vessel when viewed from above will be-
",
"options": [
{
"text": "$$\\frac{2 d\\left(n_{1}+n_{2}\\right)}{n_{1} n_{2}}$$"
},
{
"text": "$$\\frac{d\\left(n_{1}+n_{2}\\right)}{2 n_{1} n_{2}}$$"
},
{
"text": "$$\\frac{d n_{1} n_{2}}{2\\left(n_{1}+n_{2}\\right)}$$"
},
{
"text": "$$\\frac{d n_{1} n_{2}}{\\left(n_{1}+n_{2}\\right)}$$"
}
],
"answer": "$$\\frac{d\\left(n_{1}+n_{2}\\right)}{2 n_{1} n_{2}}$$",
"solution": "**Answer:** $$\\frac{d\\left(n_{1}+n_{2}\\right)}{2 n_{1} n_{2}}$$\n\nTo find the apparent depth of the vessel when viewed from above, we can calculate the apparent depths of the oil and water separately and then add them together. \n
\nThe formula to find the apparent depth ($$h_{apparent}$$) is:\n
\n$$h_{apparent} = \\frac{h_{real}}{n}$$\n
\nWhere $$h_{real}$$ is the actual depth, and $$n$$ is the refractive index of the medium.\n
\nFor the oil (with depth $$\\frac{d}{2}$$ and refractive index $$n_1$$):\n
\n$$h_{oil} = \\frac{\\frac{d}{2}}{n_{1}}$$\n
\nFor the water (with depth $$\\frac{d}{2}$$ and refractive index $$n_2$$):\n
\n$$h_{water} = \\frac{\\frac{d}{2}}{n_{2}}$$\n
\nNow, add the two apparent depths together to find the total apparent depth:\n
A fish rising vertically upward with a uniform velocity of $$8 \\mathrm{~ms}^{-1}$$, observes that a bird is diving vertically downward towards the fish with the velocity of $$12 \\mathrm{~ms}^{-1}$$. If the refractive index of water is $$\\frac{4}{3}$$, then the actual velocity of the diving bird to pick the fish, will be __________ $$\\mathrm{ms}^{-1}$$.
",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nThe bird's diving velocity is given relative to the fish. In order to find the actual velocity of the bird, we need to consider the refractive index of the water.\n
\nThe fish sees the bird diving with a velocity of $$12 \\ \\text{m/s}$$. We can write the equation considering the velocities relative to the fish:\n
\nSo, the actual velocity of the diving bird to pick the fish relative to the fish is $$3 \\ \\text{m/s}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9475,
"subject": "Physics",
"question": "
An ice cube has a bubble inside. When viewed from one side the apparent distance of the bubble is $$12 \\mathrm{~cm}$$. When viewed from the opposite side, the apparent distance of the bubble is observed as $$4 \\mathrm{~cm}$$. If the side of the ice cube is $$24 \\mathrm{~cm}$$, the refractive index of the ice cube is
",
"options": [
{
"text": "$$\\frac{3}{2}$$"
},
{
"text": "$$\\frac{4}{3}$$"
},
{
"text": "$$\\frac{2}{3}$$"
},
{
"text": "$$\\frac{6}{5}$$"
}
],
"answer": "$$\\frac{3}{2}$$",
"solution": "**Answer:** $$\\frac{3}{2}$$\n\nLet's denote the true distance of the bubble from one side of the ice cube as $$x$$ and the refractive index of the ice cube as $$n$$. We will use the formula for apparent depth, which states that the ratio of the true depth to the apparent depth is equal to the refractive index:\n
\nWhen viewing the bubble from one side, the true depth is $$x$$ and the apparent depth is $$12 \\mathrm{~cm}$$. Using the formula:\n
\n$$n = \\frac{x}{12}$$\n
\nWhen viewing the bubble from the opposite side, the true depth is $$24 - x$$ (since the side of the ice cube is $$24 \\mathrm{~cm}$$) and the apparent depth is $$4 \\mathrm{~cm}$$. Using the formula:\n
\n$$n = \\frac{24 - x}{4}$$\n
\nNow we have a system of two equations with two variables:\n
\n1) $$n = \\frac{x}{12}$$
\n2) $$n = \\frac{24 - x}{4}$$\n
\nWe can solve this system by setting the two expressions for $$n$$ equal to each other:\n
\n$$\\frac{x}{12} = \\frac{24 - x}{4}$$\n
\nTo solve for $$x$$, first multiply both sides by $$12$$:\n
\n$$x = 3(24 - x)$$\n
\n$$x = 72 - 3x$$\n
\nAdd $$3x$$ to both sides:\n
\n$$4x = 72$$\n
\nDivide by $$4$$:\n
\n$$x = 18$$\n
\nNow that we have the value of $$x$$, we can find the refractive index $$n$$ using either equation 1 or 2. Using equation 1:\n
\n$$n = \\frac{18}{12} = \\frac{3}{2}$$\n
\nTherefore, the refractive index of the ice cube is $$\\frac{3}{2}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9476,
"subject": "Physics",
"question": "
The critical angle for a denser-rarer interface is $$45^{\\circ}$$. The speed of light in rarer medium is $$3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}$$. The speed of light in the denser medium is:
To find the speed of light in the denser medium, we can use Snell's Law at the critical angle, where the angle of refraction is $$90^{\\circ}$$. Snell's Law states:
\n
$$n_1 \\sin{\\theta_1} = n_2 \\sin{\\theta_2}$$
\n
where $$n_1$$ and $$n_2$$ are the indices of refraction for the denser and rarer media, respectively, and $$\\theta_1$$ and $$\\theta_2$$ are the angles of incidence and refraction, respectively.
The critical angle is given as $$45^{\\circ}$$, and the speed of light in the rarer medium is given as $$3 \\times 10^8 \\mathrm{~m/s}$$. We can find the index of refraction of the rarer medium using the formula:
\n
$$n = \\frac{c}{v}$$
\n
where $$c$$ is the speed of light in a vacuum ($$3 \\times 10^8 \\mathrm{~m/s}$$), and $$v$$ is the speed of light in the medium.
A monochromatic light wave with wavelength $$\\lambda_{1}$$ and frequency $$v_{1}$$ in air enters another medium. If the angle of incidence and angle of refraction at the interface are $$45^{\\circ}$$ and $$30^{\\circ}$$ respectively, then the wavelength $$\\lambda_{2}$$ and frequency $$v_{2}$$ of the refracted wave are:
When a light wave moves from one medium to another, its speed and wavelength may change, but its frequency remains constant because it is determined by the source of the light. This is because frequency depends on the oscillations of the source, which do not change when entering a different medium.
\n
Snell's law, which describes the relationship between the angles of incidence and refraction, can be written as:
\n
$ n_1 \\sin(\\theta_1) = n_2 \\sin(\\theta_2), $
\n
where $n_1$ and $n_2$ are the refractive indices of the two media, and $\\theta_1$ and $\\theta_2$ are the angles of incidence and refraction, respectively.
\n
The refractive index of a medium is also related to the speed of light in that medium:
\n
$ n = \\frac{c}{v}, $
\n
where $c$ is the speed of light in a vacuum, and $v$ is the speed of light in the medium.
\n
From the above relations, we see that as light enters a medium with a higher refractive index (and hence a lower speed), its wavelength decreases. The frequency remains the same.
\n
Given that the angle of incidence is $45^\\circ$ and the angle of refraction is $30^\\circ$, the ratio of the refractive indices is:
This indicates that the speed of light (and hence the wavelength) decreases by a factor of $\\sqrt{2}$ as it enters the second medium. The frequency remains the same.
A pole is vertically submerged in swimming pool, such that it gives a length of shadow $$2.15 \\mathrm{~m}$$ within water when sunlight is incident at angle of $$30^{\\circ}$$ with the surface of water. If swimming pool is filled to a height of $$1.5 \\mathrm{~m}$$, then the height of the pole above the water surface in centimeters is $$\\left(n_{w}=4 / 3\\right)$$ ____________.
The pole is vertically submerged in the swimming pool, and the length of the shadow is due to the sunlight that is incident at an angle of $30^{\\circ}$ with the surface of water. Hence, the angle of incidence, $i$, is $60^{\\circ}$ (since the angle of incidence is measured from the normal to the surface, and the normal is perpendicular to the surface).
\n
Using Snell's law, we find the angle of refraction, $r$, as you correctly did:
\n
$$\\sin r = \\frac{n_{1}}{n_{2}} \\sin i = \\frac{3}{4} \\sin 60^{\\circ} = \\frac{3 \\sqrt{3}}{8}$$
\n
This gives $\\tan r = \\frac{3 \\sqrt{3}}{\\sqrt{37}}$.
\n
Now, the shadow of the pole in the water forms a right triangle, with the submerged part of the pole as one side, the shadow as the hypotenuse, and the line segment from the water surface to the end of the shadow as the other side. The angle at the water surface is $r$. Thus, we can write the following relationships:
\n
The length of the submerged part of the pole (which I'll call $x$), is given by:
\n
$$x = 2.15 \\, \\text{m} \\cdot \\cos r$$
\n
And the length of the line segment from the water surface to the end of the shadow (which I'll call $y$), is given by:
\n
$$y = 2.15 \\, \\text{m} \\cdot \\sin r$$
\n
The total length of the pole above the water surface is then the sum of $x$ and the depth of the swimming pool (1.5 m), which gives us the equation:
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
\n
Assertion A: The phase difference of two light waves change if they travel through different media having same thickness, but different indices of refraction.
\n
Reason R: The wavelengths of waves are different in different media.
\n
In the light of the above statements, choose the most appropriate answer from the options given below
",
"options": [
{
"text": "Both A and R are correct but R is NOT the correct explanation of A"
},
{
"text": "A is correct but R is not correct"
},
{
"text": "A is not correct but R is correct"
},
{
"text": "Both A and R are correct and R is the correct explanation of A"
}
],
"answer": "Both A and R are correct and R is the correct explanation of A",
"solution": "**Answer:** Both A and R are correct and R is the correct explanation of A\n\n
\nAssertion A is correct. When light waves travel through different media with different indices of refraction, the speed of light changes, which in turn changes the phase of the light waves. This is because the phase of a wave is directly related to the distance it has traveled, which in this case is affected by the speed of light in the different media.
\n
Reason R is also correct. The speed of light in a medium is determined by its refractive index. When light enters a medium with a different refractive index, its speed, and therefore its wavelength, changes. This is described by the equation v = c/n, where v is the speed of light in the medium, c is the speed of light in a vacuum, and n is the refractive index of the medium.
\n
Moreover, R is indeed the correct explanation of A. The change in wavelength (and therefore phase) of the light waves in different media (as explained in R) is the reason why the phase difference changes when light waves travel through different media with the same thickness but different indices of refraction (as stated in A).
If the refractive index of the material of a prism is $$\\cot \\left(\\frac{A}{2}\\right)$$, where $$A$$ is the angle of prism then the angle of minimum deviation will be
Light from a point source in air falls on a convex curved surface of radius $$20 \\mathrm{~cm}$$ and refractive index 1.5. If the source is located at $$100 \\mathrm{~cm}$$ from the convex surface, the image will be formed at ________ $$\\mathrm{cm}$$ from the object.
In the problem, you're dealing with refraction at a convex surface. The light is coming from a point source located in air (with a refractive index, $$ \\mu_1 = 1.0 $$) and entering a medium with a refractive index of $$ \\mu_2 = 1.5 $$. The convex surface has a radius of curvature $$ R = 20 \\, \\text{cm} $$, and the source is placed $$ 100 \\, \\text{cm} $$ from the convex surface.
\n
The formula used to find the image distance $$ v $$ due to refraction at a spherical surface is:
This equation allows us to solve for $$ v $$, the distance from the convex surface to the image. Upon solving, we found that $$ v = 100 \\, \\text{cm} $$, which means the image forms $$ 100 \\, \\text{cm} $$ on the other side of the convex surface, away from the point of refraction.
\n
The distance from the object to the image isn't just $$ v $$, the distance from the surface to where the image forms. Since the object is $$ 100 \\, \\text{cm} $$ from the convex surface and the image also forms $$ 100 \\, \\text{cm} $$ from the convex surface but on the opposite side, the total distance between the object and the image is the sum of these distances:
\n
\n
Distance from the object to the convex surface: $$ 100 \\, \\text{cm} $$
\n
Distance from the convex surface to the image: $$ 100 \\, \\text{cm} $$
\n
\n
Therefore, the total distance between the object and the image is $$ 100 \\, \\text{cm} + 100 \\, \\text{cm} = 200 \\, \\text{cm} $$.
\n
This calculation accounts for the physical layout where the object and the image are on opposite sides of the convex surface, and to determine the distance between them, you sum the distances from each to the surface. This results in the image being formed $$ 200 \\, \\text{cm} $$ from the object, which means if you were to measure directly from the object to its image, the total distance covered would be $$ 200 \\, \\text{cm} $$.
A light ray is incident on a glass slab of thickness $$4 \\sqrt{3} \\mathrm{~cm}$$ and refractive index $$\\sqrt{2}$$ The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of ray after passing through glass slab is ______ $$\\mathrm{cm}$$.
The critical angle is the angle of incidence at which light is refracted along the boundary, meaning the angle of refraction is $$90^{\\circ}$$. The relationship between the critical angle and the refractive indices of two media can be understood using Snell's Law, given as:
Here, $$n_1$$ is the refractive index of the first medium, $$n_2$$ is the refractive index of the second medium, and $$\\theta_c$$ is the critical angle. Since $$\\sin(90^{\\circ}) = 1$$, the equation simplifies to:
\n\n
$$ n_1 \\sin(\\theta_c) = n_2 $$
\n\n
Given that the critical angle $$\\theta_c$$ is $$45^{\\circ}$$, we have:
\n\n
$$ \\sin(45^{\\circ}) = \\frac{\\sqrt{2}}{2} $$
\n\n
Substituting this value in the simplified Snell's Law equation, we get:
The refractive index of prism is $$\\mu=\\sqrt{3}$$ and the ratio of the angle of minimum deviation to the angle of prism is one. The value of angle of prism is _________$$^\\circ$$.
To find the value of the angle of the prism given the refractive index of the prism $$\\mu = \\sqrt{3}$$ and the ratio of the angle of minimum deviation ($$\\delta_m$$) to the angle of the prism ($$A$$) is 1 (i.e., $$\\frac{\\delta_m}{A} = 1 \\Rightarrow \\delta_m = A$$), we can use the prism formula relating these variables.
\n\n
The formula that relates the angle of deviation $$\\delta$$, refractive index $$\\mu$$, angle of prism $$A$$, and angle of minimum deviation $$\\delta_m$$ is given by:
This corresponds to an angle $$\\frac{A}{2}$$ of $$30^\\circ$$ since the cosine of 30 degrees is $$\\frac{\\sqrt{3}}{2}$$. Thus:
\n\n
$$\\frac{A}{2} = 30^\\circ$$
\n\n
So, the angle of the prism $$A$$ is:
\n\n
$$A = 2 \\times 30^\\circ = 60^\\circ$$.
\n\n
Therefore, the value of the angle of the prism is $$60^\\circ$$.
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9486,
"subject": "Physics",
"question": "The change in the value of $$g$$ at a height $$h$$ above the surface of the earth is the same as at a depth $$d$$ below the surface of earth. When both $$d$$ and $$h$$ are much smaller than the radius of earth, then which one of the following is correct? ",
"options": [
{
"text": "$$d = {{3h} \\over 2}$$ "
},
{
"text": "$$d = {h \\over 2}$$ "
},
{
"text": "$$d = h$$ "
},
{
"text": "$$d = 2\\,h$$ "
}
],
"answer": "$$d = 2\\,h$$ ",
"solution": "**Answer:** $$d = 2\\,h$$ \n\nAt height h acceleration due to gravity, $${g_h} = g\\left[ {1 - {{2h} \\over R}} \\right];$$ \n
At depth d acceleration due to gravity, $${g_d} = g\\left[ {1 - {d \\over R}} \\right]$$\n
$$ \\Rightarrow $$ $$d=2h$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9487,
"subject": "Physics",
"question": "If $${g_E}$$ and $${g_M}$$ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio \n $${{electro\\,\\,ch\\arg e\\,\\,on\\,\\,the\\,\\,moon} \\over {electronic\\,\\,ch\\arg e\\,\\,on\\,\\,the\\,\\,earth}}\\,\\,to\\,be$$ ",
"options": [
{
"text": "$${g_M}/{g_E}$$ "
},
{
"text": "$$1$$ "
},
{
"text": "$$0$$ "
},
{
"text": "$${g_E}/{g_M}$$ "
}
],
"answer": "$$1$$ ",
"solution": "**Answer:** $$1$$ \n\nelectronic charge does does not depend on acceleration due to gravity as it is a universal constant.\n
So, electronic charge on earth\n
$$=$$ electronic charge on moon\n
$$\\therefore$$ Required ratio $$=1.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9488,
"subject": "Physics",
"question": "The height at which the acceleration due to gravity becomes $${g \\over 9}$$ (where $$g=$$ the acceleration due to gravity on the surface of the earth) in terms of $$R,$$ the radius of the earth, is: ",
"options": [
{
"text": "$${R \\over {\\sqrt 2 }}$$ "
},
{
"text": "$$R/2$$ "
},
{
"text": "$$\\sqrt 2 \\,\\,R$$ "
},
{
"text": "$$2\\,R$$ "
}
],
"answer": "$$2\\,R$$ ",
"solution": "**Answer:** $$2\\,R$$ \n\nGiven that, at height h from ground the acceleration due to gravity becomes $${g \\over 9}$$.\n
We know acceleration at earth surface due to gravity g = $${{GM} \\over {{R^2}}}$$\n
and acceleration at height h due to gravity g' = $${{GM} \\over {{{\\left( {R + h} \\right)}^2}}}$$\n
$$\\therefore$$ $$h = 2R$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9489,
"subject": "Physics",
"question": "If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh $${3 \\over 4}$$ W. Radius of the Earth is 6400 km and g=10 m/s2.",
"options": [
{
"text": "1.1 $$ \\times $$ 10−3 rad/s "
},
{
"text": "0.83 $$ \\times $$ 10−3 rad/s "
},
{
"text": "0.63 $$ \\times $$ 10−3 rad/s "
},
{
"text": "0.28 $$ \\times $$ 10−3 rad/s "
}
],
"answer": "0.63 $$ \\times $$ 10−3 rad/s ",
"solution": "**Answer:** 0.63 $$ \\times $$ 10−3 rad/s \n\nInitially when earth is not rotating then weight of the person is $$w$$.\n
When earth rotares about it's axis then weight = $${{3\\omega } \\over 4}$$ \n
= 0.63 $$ \\times $$ 10$$-$$3 rad/s",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9490,
"subject": "Physics",
"question": "Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence : ",
"options": [
{
"text": "Weight of the object, everywhere on the earth, will increase. "
},
{
"text": "Weight of the object, everywhere on the earth, will decrease. "
},
{
"text": "There will be no change in weight anywhere on the earth."
},
{
"text": "Except at poles, weight of the object on the earth will decrease."
}
],
"answer": "Except at poles, weight of the object on the earth will decrease.",
"solution": "**Answer:** Except at poles, weight of the object on the earth will decrease.\n\nWith rotation of earth the effect on acceleration due to gravity vary as \n
g' = g $$-$$ $$\\omega $$2 R cos2 $$\\theta $$\n
Here $$\\theta $$ is lattitude. \n
At poles $$\\theta $$ = 90o so those will be no change in gravity. \n
At all other point $$\\omega $$ will increase so g' will decrease. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9491,
"subject": "Physics",
"question": "The value of acceleration due to gravity at\nEarth's surface is 9.8 ms–2. The altitude above\nits surface at which the acceleration due to\ngravity decreases to 4.9 ms–2, is close to :\n(Radius of earth = 6.4 × 106 m)",
"options": [
{
"text": "1.6 × 106 m"
},
{
"text": "9.0 × 106 m"
},
{
"text": "6.4 × 106 m"
},
{
"text": "2.6 × 106 m"
}
],
"answer": "2.6 × 106 m",
"solution": "**Answer:** 2.6 × 106 m\n\n$${{GM} \\over {{{\\left( {R + h} \\right)}^2}}} = {{GM} \\over {2{R^2}}}$$
\n$$R + h = \\sqrt 2 R$$
\n$$h = \\left( {\\sqrt 2 - 1} \\right)R \\simeq 2.6 \\times {10^6}m$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9492,
"subject": "Physics",
"question": "A box weight 196 N on a spring balance at the north pole. Its weight recorded on the same\nbalance if it is shifted to the equator is close to (Take g = 10 ms–2 at the north pole and the radius\nof the earth = 6400 km) :",
"options": [
{
"text": "194.32 N"
},
{
"text": "195.66 N"
},
{
"text": "195.32 N"
},
{
"text": "194.66 N"
}
],
"answer": "195.32 N",
"solution": "**Answer:** 195.32 N\n\nAt equator, weight\n
= 195.32 N",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9493,
"subject": "Physics",
"question": "A ball is dropped from the top of a 100 m high\ntower on a planet. In the last $${1 \\over 2}s$$ before hitting\nthe ground, it covers a distance of 19 m.\nAcceleration due to gravity (in ms–2) near the\nsurface on that planet is _____.",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nLet time to travel 81 m is t sec.\n
Time to travel 100 m is t + $${1 \\over 2}$$ sec.\n
$$ \\Rightarrow $$ a = 8 m/s2",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9494,
"subject": "Physics",
"question": "The height ‘h’ at which the weight of a body will\nbe the same as that at the same depth ‘h’ from\nthe surface of the earth is (Radius of the earth\nis R and effect of the rotation of the earth is\nneglected)",
"options": [
{
"text": "$${R \\over 2}$$"
},
{
"text": "$${{\\sqrt 5 R - R} \\over 2}$$"
},
{
"text": "$${{\\sqrt 3 R - R} \\over 2}$$"
},
{
"text": "$${{\\sqrt 5 } \\over 2}R - R$$"
}
],
"answer": "$${{\\sqrt 5 R - R} \\over 2}$$",
"solution": "**Answer:** $${{\\sqrt 5 R - R} \\over 2}$$\n\n\n
$$ \\therefore $$ $$r = \\sqrt {{5 \\over 9}} R$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9496,
"subject": "Physics",
"question": "On the x-axis and at a distance x from the origin, the gravitational field due a mass distribution is\ngiven by $${{Ax} \\over {{{\\left( {{x^2} + {a^2}} \\right)}^{3/2}}}}$$ in the x-direction. The magnitude of gravitational potential on the x-axis at a\ndistance x, taking its value to be zero at infinity, is:",
"options": [
{
"text": "$${A{{\\left( {{x^2} + {a^2}} \\right)}^{3/2}}}$$"
},
{
"text": "$${A{{\\left( {{x^2} + {a^2}} \\right)}^{1/2}}}$$"
},
{
"text": "$${A \\over {{{\\left( {{x^2} + {a^2}} \\right)}^{1/2}}}}$$"
},
{
"text": "$${A \\over {{{\\left( {{x^2} + {a^2}} \\right)}^{3/2}}}}$$"
}
],
"answer": "$${A \\over {{{\\left( {{x^2} + {a^2}} \\right)}^{1/2}}}}$$",
"solution": "**Answer:** $${A \\over {{{\\left( {{x^2} + {a^2}} \\right)}^{1/2}}}}$$\n\nGiven $${E_x} = {{Ax} \\over {{{({x^2} + {a^2})}^{3/2}}}}$$
$$ \\Rightarrow $$ $$V = {A \\over {{{({x^2} + {a^2})}^{1/2}}}}$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9497,
"subject": "Physics",
"question": "The value of the acceleration due to gravity is\ng1 at a height h = $${R \\over 2}$$ (R = radius of the earth) from the surface of the earth. It is again equal\nto g1 at a depth d below the surface of the\nearth. The ratio $$\\left( {{d \\over R}} \\right)$$ equals :",
"options": [
{
"text": "$${5 \\over 9}$$"
},
{
"text": "$${1 \\over 9}$$"
},
{
"text": "$${7 \\over 9}$$"
},
{
"text": "$${4 \\over 9}$$"
}
],
"answer": "$${5 \\over 9}$$",
"solution": "**Answer:** $${5 \\over 9}$$\n\nGiven, $${g_{at\\,high}} = {g_{at\\,depth}}$$
$$ \\Rightarrow $$ $${d \\over R} = 1 - {4 \\over 9} = {5 \\over 9}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9498,
"subject": "Physics",
"question": "The acceleration due to gravity on the earth’s\nsurface at the poles is g and angular velocity of\nthe earth about the axis passing through the\npole is $$\\omega $$. An object is weighed at the equator\nand at a height h above the poles by using a\nspring balance. If the weights are found to be\nsame, then h is (h << R, where R is the radius\nof the earth)\n",
"options": [
{
"text": "$${{{R^2}{\\omega ^2}} \\over {2g}}$$"
},
{
"text": "$${{{R^2}{\\omega ^2}} \\over g}$$"
},
{
"text": "$${{{R^2}{\\omega ^2}} \\over {8g}}$$"
},
{
"text": "$${{{R^2}{\\omega ^2}} \\over {4g}}$$"
}
],
"answer": "$${{{R^2}{\\omega ^2}} \\over {2g}}$$",
"solution": "**Answer:** $${{{R^2}{\\omega ^2}} \\over {2g}}$$\n\nAt equator, g1 = g - R$${\\omega ^2}$$\n
At height h, g2 = $$g\\left( {1 - {{2h} \\over R}} \\right)$$ [as given h << R]\n
$$ \\because $$ Weight same at poles and at h (so g1\n = g2)\n
$$ \\Rightarrow $$ $${h = {{{R^2}{\\omega ^2}} \\over {2g}}}$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9499,
"subject": "Physics",
"question": "A body weights 49N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?
[Use $$g = {{GM} \\over {{R^2}}}$$ = 9.8 ms$$-$$2 and radius of earth, R = 6400 km.]",
"options": [
{
"text": "49 N"
},
{
"text": "49.83 N"
},
{
"text": "48.83 N"
},
{
"text": "49.17 N"
}
],
"answer": "48.83 N",
"solution": "**Answer:** 48.83 N\n\nGiven, weight of body at North pole,
wp = mg = 49 N
Radius of Earth, R = 6400 km
Let weight of body at equator be we.
At equator, ge = g $$-$$ R$$\\omega$$2
$$\\therefore$$ we = mge = m(g $$-$$ R$$\\omega$$2)
Since, wp > we $$\\Rightarrow$$ we < 49 N
Hence, above condition is satisfied by only option (b).",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9500,
"subject": "Physics",
"question": "If the angular velocity of earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately : [Take g = 10 ms$$-$$2, the radius of earth, R = 6400 $$\\times$$ 103 m, Take $$\\pi$$ = 3.14]",
"options": [
{
"text": "84 minutes"
},
{
"text": "1200 minutes"
},
{
"text": "60 minutes"
},
{
"text": "does not change"
}
],
"answer": "84 minutes",
"solution": "**Answer:** 84 minutes\n\nFor objects to float
mg = 2$$\\omega$$2R
$$\\omega$$ = angular velocity of earth.
R = Radius of earth
$$\\omega = \\sqrt {{g \\over R}} $$ ..... (1)
Duration of day = T
$$T = {{2\\pi } \\over \\omega }$$ ..... (2)
$$ \\Rightarrow T = 2\\pi \\sqrt {{R \\over g}} $$
$$ \\simeq $$ 84 minuites",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9501,
"subject": "Physics",
"question": "Consider a planet in some solar system which has a mass double the mass of earth and density equal to the average density of earth. If the weight of an object on earth is W, the weight of the same object on that planet will be :",
"options": [
{
"text": "2W"
},
{
"text": "W"
},
{
"text": "$${2^{{1 \\over 3}}}$$W"
},
{
"text": "$$\\sqrt 2 $$W"
}
],
"answer": "$${2^{{1 \\over 3}}}$$W",
"solution": "**Answer:** $${2^{{1 \\over 3}}}$$W\n\nDensity is same
Statement I : The law of gravitation holds good for any pair of bodies in the universe.
\n
Statement II : The weight of any person becomes zero when the person is at the centre of the earth.
\n
In the light of the above statements, choose the correct answer from the options given below.
",
"options": [
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Statement I is true but Statement II is false"
},
{
"text": "Statement I is false but Statement II is true"
}
],
"answer": "Both Statement I and Statement II are true",
"solution": "**Answer:** Both Statement I and Statement II are true\n\n
Statement I is true. Newton's law of universal gravitation states that every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between their centers. This law applies to any pair of bodies in the universe.
\n
Statement II is also true. The weight of an object is the force of gravity acting on it. If a person were at the center of the Earth, the gravitational pull from all the surrounding mass of the Earth would cancel out, resulting in zero net gravitational force and therefore zero weight.
\n
Therefore, Option A: Both Statement I and Statement II are true, is the correct answer.
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.
\n
Assertion A : If we move from poles to equator, the direction of acceleration due to gravity of earth always points towards the center of earth without any variation in its magnitude.
\n
Reason R : At equator, the direction of acceleration due to the gravity is towards the center of earth.
\n
In the light of above statements, choose the correct answer from the options given below:
",
"options": [
{
"text": "Both A and R are true and R is the correct explanation of A."
},
{
"text": "Both A and R are true but R is NOT the correct explanation of A."
},
{
"text": "A is true but R is false."
},
{
"text": "A is false but R is true."
}
],
"answer": "A is false but R is true.",
"solution": "**Answer:** A is false but R is true.\n\n
Assertion A is false. While the direction of acceleration due to gravity does always point towards the center of the Earth, its magnitude actually varies from the poles to the equator. This is due to the Earth's rotation and the fact that Earth is not a perfect sphere but an oblate spheroid, meaning it's slightly flattened at the poles and bulging at the equator.
\n
Reason R is true. At the equator, the direction of acceleration due to gravity is towards the center of the Earth.
The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be : (Given g = acceleration due to gravity at the surface of earth).
The acceleration due to gravity (g) at a distance (r) from the center of a planet is given by:
\n
$$g = \\frac{G M}{r^2}$$
\n
where:
\n
\n
G is the gravitational constant,
\n
M is the mass of the planet,
\n
r is the distance from the center of the planet.
\n
\n
If the height h of a point P above the surface of the Earth is equal to the diameter of the Earth, then the distance r from the center of the Earth to the point P is 3 times the radius of the Earth. Substituting this into the equation for g gives:
\n
$$g' = g \\left(\\frac{R}{3R}\\right)^2 = \\frac{g}{9}$$
\n
where:
\n
\n
g' is the acceleration due to gravity at the point P,
\n
R is the radius of the Earth.
\n
\n
Therefore, the value of acceleration due to gravity at point P is g/9.
The radii of two planets A and B are in the ratio 2 : 3. Their densities are 3$$\\rho$$ and 5$$\\rho$$ respectively. The ratio of their acceleration due to gravity is :
An object is taken to a height above the surface of earth at a distance $${5 \\over 4}$$ R from the centre of the earth. Where radius of earth, R = 6400 km. The percentage decrease in the weight of the object will be :
The weight of an object at a distance d from the center of the Earth is given by:
\n
$$W' = W \\left(\\frac{R}{d}\\right)^2$$
\n
where:
\n
\n
W' is the weight at distance d,
\n
W is the weight at the Earth's surface,
\n
R is the radius of the Earth,
\n
d is the distance from the center of the Earth.
\n
\n
In this problem, we are asked to find the percentage decrease in weight, which can be calculated by:
\n
$$\\Delta W = \\frac{W - W'}{W} \\times 100\\%$$
\n
where ΔW is the percentage change in weight. Substituting the weight formula into the percentage change formula gives:
\n
$$\\Delta W = \\left(1 - \\left(\\frac{R}{d}\\right)^2\\right) \\times 100\\%$$
\n
The radius of the Earth R is 6400 km and the distance d from the center of the Earth is given as $\\frac{5}{4}R$. Substituting these values into the equation gives:
If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a depth $$\\alpha \\mathrm{h}\\left(\\mathrm{h}<<\\mathrm{R}_{\\mathrm{e}}\\right)$$ from the earth surface. The value of $$\\alpha$$ will be _________.
For a body projected at an angle with the horizontal from the ground, choose the correct statement.
",
"options": [
{
"text": "Gravitational potential energy is maximum at the highest point."
},
{
"text": "The vertical component of momentum is maximum at the highest point."
},
{
"text": "The horizontal component of velocity is zero at the highest point."
},
{
"text": "The Kinetic Energy (K.E.) is zero at the highest point of projectile motion."
}
],
"answer": "Gravitational potential energy is maximum at the highest point.",
"solution": "**Answer:** Gravitational potential energy is maximum at the highest point.\n\nAt highest point height is maximum and vertical component of velocity is zero.\n
So momentum is zero.\n
At highest point horizontal component of velocity will not be zero but vertical component of velocity is equal to zero and because of this K.E. will not be equal to zero.\n
Gravitational potential energy is maximum at highest point and equal to $\\mathrm{mgH}=$$$mg\\left( {{{{u^2}{{\\sin }^2}\\theta } \\over {2g}}} \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9514,
"subject": "Physics",
"question": "A body weight $\\mathrm{W}$, is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be :",
"options": [
{
"text": "$\\frac{W}{91}$"
},
{
"text": "$\\frac{\\mathrm{W}}{3}$"
},
{
"text": "$\\frac{\\mathrm{W}}{100}$"
},
{
"text": "$\\frac{\\mathrm{W}}{9}$"
}
],
"answer": "$\\frac{\\mathrm{W}}{100}$",
"solution": "**Answer:** $\\frac{\\mathrm{W}}{100}$\n\n
The weight of an object varies with altitude due to the change in gravitational force. The force of gravity decreases with the square of the distance from the center of the Earth.
\n
The gravitational force (weight) at a height h from the surface of the Earth is given by:
\n
$$W' = W \\left(\\frac{R}{{R + h}}\\right)^2$$
\n
where:
\n
\n
W' is the weight at height h,
\n
W is the weight at the Earth's surface,
\n
R is the radius of the Earth,
\n
h is the height above the Earth's surface.
\n
\n
In this problem, the height h is given as nine times the radius of the Earth (h = 9R). Substituting h = 9R into the equation gives:
\n
$$W' = W \\left(\\frac{R}{{R + 9R}}\\right)^2 = W \\left(\\frac{1}{10}\\right)^2$$
\n
Simplifying this gives:
\n
$$W' = \\frac{W}{100}$$
\n
Therefore, the weight of the body at that height will be 1/100th of its weight on the surface of the Earth.
Statement I: Acceleration due to gravity is different at different places on the surface of earth.
\n
Statement II: Acceleration due to gravity increases as we go down below the earth's surface.
\n
In the light of the above statements, choose the correct answer from the options given below
",
"options": [
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Statement I is false but Statement II is true"
},
{
"text": "Statement I is true but Statement II is false"
}
],
"answer": "Statement I is true but Statement II is false",
"solution": "**Answer:** Statement I is true but Statement II is false\n\n
Statement I is true. The acceleration due to gravity varies slightly over the surface of the Earth, being affected by factors such as latitude (because of the Earth's oblate shape or its equatorial bulge), altitude (it decreases with height above the Earth's surface), and local geological variations in the Earth's density.
\n
Statement II is false. The acceleration due to gravity actually decreases as we go down below the Earth's surface. This is because the mass that is "above" or outside the location begins to pull the object away from the Earth's center, and the net gravitational acceleration decreases.
\n
Therefore, Option D: Statement I is true but Statement II is false, is the correct answer.
At a certain depth \"d \" below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $$\\mathrm{3 R}$$ above earth surface. Where $$\\mathrm{R}$$ is Radius of earth (Take $$\\mathrm{R}=6400 \\mathrm{~km}$$ ). The depth $$\\mathrm{d}$$ is equal to
",
"options": [
{
"text": "5260 km"
},
{
"text": "2560 km"
},
{
"text": "640 km"
},
{
"text": "4800 km"
}
],
"answer": "4800 km",
"solution": "**Answer:** 4800 km\n\nThe acceleration due to gravity $$g$$ at a distance $$d$$ below the surface of the earth is given by :\n
Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is 100 g. The time period of the motion of the particle will be (approximately)
\n
(Take g = 10 m s$$^{-2}$$ , radius of earth = 6400 km)
",
"options": [
{
"text": "12 hours"
},
{
"text": "1 hour 24 minutes"
},
{
"text": "24 hours"
},
{
"text": "1 hour 40 minutes"
}
],
"answer": "1 hour 24 minutes",
"solution": "**Answer:** 1 hour 24 minutes\n\nGravitational acceleration at a distance of $r$ from centre of earth is given by\n
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
\n
Assertion A : A pendulum clock when taken to Mount Everest becomes fast.
\n
Reason R : The value of g (acceleration due to gravity) is less at Mount Everest than its value on the surface of earth.
\n
In the light of the above statements, choose the most appropriate answer from the options given below
",
"options": [
{
"text": "Both A and R are correct but R is NOT the correct explanation of A"
},
{
"text": "A is correct but R is not correct"
},
{
"text": "Both A and R are correct and R is the correct explanation of A"
},
{
"text": "A is not correct but R is correct"
}
],
"answer": "A is not correct but R is correct",
"solution": "**Answer:** A is not correct but R is correct\n\nWhen we go on the Mount Everest the value of gravitational acceleration decreases $\\left(g=\\frac{g_{0}}{\\left(1+\\frac{h}{R_{e}}\\right)^{2}}\\right)$. Therefore, the time period of oscillation $\\left(T=2 \\pi \\sqrt{\\frac{I}{g}}\\right)$ increases and the pendulum clock becomes slow thus the assertion is wrong but reason is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9519,
"subject": "Physics",
"question": "
Given below are two statements:
\n
Statement I : Acceleration due to earth's gravity decreases as you go 'up' or 'down' from earth's surface.
\n
Statement II : Acceleration due to earth's gravity is same at a height 'h' and depth 'd' from earth's surface, if h = d.
\n
In the light of above statements, choose the most appropriate answer from the options given below
",
"options": [
{
"text": "Statement I is correct but statement II is incorect"
},
{
"text": "Both Statement I and II are correct"
},
{
"text": "Statement I is incorrect but statement II is correct"
},
{
"text": "Both Statement I and Statement II are incorrect"
}
],
"answer": "Statement I is correct but statement II is incorect",
"solution": "**Answer:** Statement I is correct but statement II is incorect\n\n
The most appropriate answer is Statement I is correct but statement II is incorrect.
\n
\nStatement I is correct as acceleration due to Earth's gravity decreases as you move away from its surface either upward or downward. This is because the gravity of the Earth follows an inverse square law, which means that the gravitational force decreases as the square of the distance between two objects increases.
\n\n
However, statement II is incorrect because acceleration due to Earth's gravity is not the same at a height and depth from Earth's surface, even if they are equal in magnitude. This is because the Earth is not a perfect sphere and has a non-uniform distribution of mass, which causes variations in the strength of gravity at different locations. Therefore, the acceleration due to Earth's gravity will be different at a certain height and depth, even if they are equal.
The weight of a body at the surface of earth is 18 N. The weight of the body at an altitude of 3200 km above the earth's surface is (given, radius of earth $$\\mathrm{R_e=6400~km}$$) :
The weight of an object at a height h from the Earth's surface is given by:
\n
$$W' = W \\left(\\frac{R}{{R + h}}\\right)^2$$
\n
where:
\n
\n
W' is the weight at height h,
\n
W is the weight at the Earth's surface,
\n
R is the radius of the Earth,
\n
h is the height above the Earth's surface.
\n
\n
In this problem, the weight at the Earth's surface W is given as 18 N, the radius of the Earth R is given as 6400 km, and the height h is given as 3200 km. Substituting these values into the equation gives:
\n
$$W' = 18 N \\left(\\frac{6400~km}{{6400~km + 3200~km}}\\right)^2 = 18 N \\left(\\frac{2}{3}\\right)^2 = 18 N \\cdot \\frac{4}{9} = 8 N$$
\n
Therefore, the weight of the body at an altitude of 3200 km above the Earth's surface is 8 N.
Two planets A and B of radii $$\\mathrm{R}$$ and 1.5 R have densities $$\\rho$$ and $$\\rho / 2$$ respectively. The ratio of acceleration due to gravity at the surface of $$\\mathrm{B}$$ to $$\\mathrm{A}$$ is:
",
"options": [
{
"text": "2 : 1"
},
{
"text": "2 : 3"
},
{
"text": "4 : 3"
},
{
"text": "3 : 4"
}
],
"answer": "3 : 4",
"solution": "**Answer:** 3 : 4\n\nThe acceleration due to gravity at the surface of a planet is given by the formula:\n
\n$$g=\\frac{GM}{R^2}$$\n
\nwhere $$G$$ is the gravitational constant, $$M$$ is the mass of the planet, and $$R$$ is the radius of the planet.\n
\nIn this problem, we are given two planets A and B, with radii $$R_A$$ and $$R_B=1.5R_A$$, and densities $$\\rho_A$$ and $$\\rho_B=\\frac{\\rho_A}{2}$$, respectively. We can use the formula above to calculate the acceleration due to gravity at the surface of each planet:\n
\nTo find the ratio of the accelerations due to gravity at the surfaces of planets A and B, we can divide the expression for $$g_B$$ by the expression for $$g_A$$:\n
\nSo, we can see that the acceleration due to gravity at the surface of planet B is $$\\frac{3}{2}$$ times larger than that at the surface of planet A.\n
\nHowever, the question asks for the ratio of the gravitational accelerations at the surfaces of the planets, which is equal to the ratio of the densities times the ratio of the radii:\n
\nTherefore, the ratio of the gravitational accelerations at the surfaces of the planets is $$3/4$$, which is our final answer.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9522,
"subject": "Physics",
"question": "
The radii of two planets 'A' and 'B' are 'R' and '4R' and their densities are $$\\rho$$ and $$\\rho / 3$$ respectively. The ratio of acceleration due to gravity at their surfaces $$\\left(g_{A}: g_{B}\\right)$$ will be:
",
"options": [
{
"text": "3 : 16"
},
{
"text": "4 : 3"
},
{
"text": "1 : 16"
},
{
"text": "3 : 4"
}
],
"answer": "3 : 4",
"solution": "**Answer:** 3 : 4\n\n The acceleration due to gravity at the surface of a planet can be expressed as:\n
\n$$g \\propto \\rho R$$\n
\nNow let's find the ratio of acceleration due to gravity at the surfaces of planets A and B:\n
\nSo, the ratio of acceleration due to gravity at their surfaces is 3 : 4",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9523,
"subject": "Physics",
"question": "
Given below are two statements:
\n
Statement I : Rotation of the earth shows effect on the value of acceleration due to gravity (g)
\n
Statement II : The effect of rotation of the earth on the value of 'g' at the equator is minimum and that at the pole is maximum.
\n
In the light of the above statements, choose the correct answer from the options given below
",
"options": [
{
"text": "Statement I is false but statement II is true"
},
{
"text": "Statement I is true but statement II is false"
},
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Both Statement I and Statement II are false"
}
],
"answer": "Statement I is true but statement II is false",
"solution": "**Answer:** Statement I is true but statement II is false\n\n
Let's analyze both statements:
\n\n
Statement I: Rotation of the earth shows an effect on the value of acceleration due to gravity (g).
\n
This is true. The value of $ g $ is affected by the rotation of the Earth. The centripetal force due to the Earth's rotation causes a reduction in the perceived gravitational acceleration for objects on the surface. This effect is zero at the poles and increases towards the equator. The formula for the effective acceleration due to gravity at a latitude $ \\theta $ is:
\n$$ g_{\\text{effective}} = g - R\\omega^2\\cos^2(\\theta) $$\n
where:
\n
\n
$ g $ is the theoretical acceleration due to gravity without Earth's rotation,
\n
$ R $ is the radius of the Earth,
\n
$ \\omega $ is the angular velocity of the Earth's rotation, and
\n
$ \\theta $ is the latitude.
\n
\n\n
Statement II: The effect of rotation of the earth on the value of 'g' at the equator is minimum and that at the pole is maximum.
\n
This is false. For clarification, the effect of the Earth's rotation on the value of $ g $ is maximum at the equator and minimum at the poles. At the equator, the centrifugal force is highest because of the maximum velocity due to Earth's rotation, which decreases the effect of gravity more than at any other latitude. At the poles, the centrifugal force is zero since there is no rotational velocity contributing to a centripetal effect; therefore, the effect of rotation on $ g $ is minimum (zero).
\n\n
With these considerations, the correct option is:
\n\n
Option B: Statement I is true but Statement II is false.
Assuming the earth to be a sphere of uniform mass density, the weight of a body at a depth $$d=\\frac{R}{2}$$ from the surface of earth, if its weight on the surface of earth is 200 N, will be:
The gravitational field strength, or equivalently, the weight of an object, decreases linearly from its surface value to zero at the center of a sphere of uniform mass density. This is because only the mass inside the radius at which the object is located contributes to the gravitational force at that location.
\n
If ( $d = \\frac{R}{2}$ ) is the depth below the surface of the Earth, then the radius of the sphere contributing to the gravitational force at that depth is ( $R - d = R - \\frac{R}{2} = \\frac{R}{2}$ ).
\n
Since the gravitational force (or weight) decreases linearly with the radius in a sphere of uniform density, the weight of the object at depth ( $d = \\frac{R}{2} $) is half its weight at the surface of the Earth.
\n
So, if the weight of the body on the surface of the Earth is 200 N, its weight at depth ( $d = \\frac{R}{2} $) is half of that, or 100 N.
The acceleration due to gravity ($g$) at the surface of Earth is approximately $9.81 \\, \\text{m/s}^2$. This acceleration decreases as we move away from the Earth's surface because gravity is a force that attracts objects towards the center of the Earth. This force decreases with the square of the distance from the center of the Earth, due to the inverse square law.
\n
However, if we are at a height ($h$) much less than the radius of the Earth ($R$), we can use a linear approximation to calculate the new acceleration due to gravity ($g'$) at this height. This is because for small heights compared to the radius of Earth, the decrease in $g$ can be approximated to be linear. This is represented by the following formula:
\n
$$g' = g\\left(1 - \\frac{2h}{R}\\right)$$
\n
Here's how this formula is derived:
\n
The force of gravity ($F$) is given by the universal law of gravitation:
\n
$$F = G \\frac{m_1 m_2}{d^2}$$
\n
where:
\n
\n
$G$ is the gravitational constant,
\n
$m_1$ and $m_2$ are the masses of the two objects (in this case, the Earth and the object we're considering),
\n
$d$ is the distance between the centers of the two objects.
\n
\n
If we consider an object of mass $m$ at the surface of Earth, the force it experiences due to gravity is:
\n
$$F = G \\frac{mM}{R^2} = mg$$
\n
where:
\n
\n
$M$ is the mass of Earth,
\n
$R$ is the radius of Earth,
\n
$g = \\frac{GM}{R^2}$ is the acceleration due to gravity at the Earth's surface.
\n
\n
Now, if the object is at a height $h$ above the Earth's surface, the force it experiences is:
\n
$$F' = G \\frac{mM}{(R + h)^2}$$
\n
We can write $(R + h)^2$ as $R^2 (1 + \\frac{h}{R})^2$, and because $h << R$, we can use the binomial approximation $(1 + x)^2 \\approx 1 + 2x$ for small $x$ to get:
The gravitational field inside a uniform spherical body varies linearly with distance from the center. If we consider the Earth to be such a body, then the gravitational field strength (and hence weight) of an object would decrease linearly as we go deeper inside the Earth.
\n
The weight of an object at a depth $d$ from the Earth's surface is given by:
\n
$W_d = W_e (1 - \\frac{d}{R})$,
\n
where:
\n
\n
$W_d$ is the weight at depth $d$,
\n
$W_e$ is the weight at the Earth's surface (i.e., the weight of the object), and
\n
$R$ is the radius of the Earth.
\n
\n
In this case, we're given that $W_e = 400 \\, \\text{N}$, and we're asked to find the weight at a depth of $d = \\frac{R}{2}$. Substituting these values into the formula, we get:
The weight of a body on the surface of the earth is $$100 \\mathrm{~N}$$. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:
To find the gravitational force on the body when taken at a height equal to one-fourth the radius of the Earth, we can use the formula for gravitational force:
\n
$$F = G \\frac{m_1 m_2}{r^2}$$
\n
where $$F$$ is the gravitational force, $$G$$ is the gravitational constant, $$m_1$$ and $$m_2$$ are the masses of the two objects, and $$r$$ is the distance between their centers.
\n
The weight of the body on the surface of the Earth is given as $$100\\,\\text{N}$$, which is also the gravitational force acting on it:
\n
$$F_\\text{surface} = G \\frac{m_\\text{body} m_\\text{earth}}{R_\\text{earth}^2}$$
\n
When the body is taken to a height equal to one-fourth the radius of the Earth, the distance between the centers of the body and the Earth becomes $$r_\\text{new} = R_\\text{earth} + \\frac{1}{4} R_\\text{earth} = \\frac{5}{4} R_\\text{earth}$$.
\n
Now the gravitational force acting on the body at this new height is:
\n
$$F_\\text{new} = G \\frac{m_\\text{body} m_\\text{earth}}{r_\\text{new}^2}$$
\n
$$F_\\text{new} = G \\frac{m_\\text{body} m_\\text{earth}}{\\left(\\frac{5}{4} R_\\text{earth}\\right)^2}$$
\n
To find the ratio between the new gravitational force and the original force on the surface, we can write:
So, the gravitational force on the body when taken at a height equal to one-fourth the radius of the Earth is $$64\\,\\text{N}$$.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9528,
"subject": "Physics",
"question": "If $\\mathrm{R}$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g=\\pi^2 \\mathrm{~m} / \\mathrm{s}^2$, then the length of the second's pendulum at a height $\\mathrm{h}=2 R$ from the surface of earth will be, :",
"options": [
{
"text": "$\\frac{1}{9} \\mathrm{~m}$"
},
{
"text": "$\\frac{8}{9} \\mathrm{~m}$"
},
{
"text": "$\\frac{2}{9} \\mathrm{~m}$"
},
{
"text": "$\\frac{4}{9} \\mathrm{~m}$"
}
],
"answer": "$\\frac{1}{9} \\mathrm{~m}$",
"solution": "**Answer:** $\\frac{1}{9} \\mathrm{~m}$\n\n
To find the length of the second's pendulum at a height $h = 2R$ from the surface of the Earth, we must first understand that the length of a second's pendulum, $L$, is related to the gravitational acceleration, $g$, and the period, $T$, by the formula: $$ T = 2\\pi\\sqrt{\\frac{L}{g}} $$ Since we are talking about a second's pendulum, the period, $T$, is 2 seconds (since it takes one second for the pendulum to swing in one direction and another second to swing back), thus $T = 2 \\text{ seconds}$.
\n\n
Now let's find the gravitational acceleration at height $h = 2R$ where $R$ is the radius of the earth. The general formula for gravitational acceleration at a height $h$ above the surface is: $$ g_h = \\frac{g}{{\\left(1 + \\frac{h}{R}\\right)}^2} $$ Plugging $h = 2R$ into the formula, we get:\n\n
$$ g_h = \\frac{g}{{(1 + \\frac{2R}{R})}^2} $$
\n\n
$$ g_h = \\frac{g}{{(1 + 2)}^2} $$
\n\n
$$ g_h = \\frac{g}{3^2} = \\frac{g}{9} $$
\n\n
So the gravitational acceleration at height $h$ is one-ninth of the gravitational acceleration at the surface of the Earth. Given that $g = \\pi^2 \\text{ m/s}^2$, we get: $$ g_h = \\frac{\\pi^2}{9} \\text{ m/s}^2 $$
\n\n
Now knowing the gravitational acceleration at height $h$ and with the period $T$ of 2 seconds, we can rearrange the formula for the second's pendulum to solve for the length $L_h$:\n\n
Multiplying both sides by $g_h$ gives us the length $L_h$:
\n\n
$$ L_h = \\frac{g_h}{\\pi^2} $$
\n\n
Substituting $g_h$ into the equation yields:\n\n
$$ L_h = \\frac{\\pi^2}{9\\pi^2} $$
\n\n
$$ L_h = \\frac{1}{9} \\text{ m} $$
\n\n
Therefore, the length of the second's pendulum at a height $h = 2R$ from the surface of the Earth is $\\frac{1}{9}$ meters. The correct answer is Option A.
The acceleration due to gravity on the surface of earth is $$\\mathrm{g}$$. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be :
A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is $$4 m$$, then the time period of small oscillations will be __________ s. [take $$g=\\pi^2 m s^{-2}$$]
The gravitational force $$F$$ that an object of mass $$m$$ placed at a distance $$r$$ (from the center of Earth) experiences can be calculated using the universal law of gravitation, given by:
\n\n
$$F = \\frac{G M m}{r^2}$$
\n\n
Where:
\n\n
\n\n
$$G$$ is the gravitational constant ($$6.674 \\times 10^{-11} \\text{Nm}^2/\\text{kg}^2$$)
\n\n
$$M$$ is the mass of the Earth (approximately $$5.972 \\times 10^{24} \\text{kg}$$)
\n\n
$$m$$ is the mass of the object
\n\n
$$r$$ is the distance from the center of the Earth to the object
\n\n
\n\n
However, when the object is at a distance $$2R$$ from the surface of the Earth, the total distance from the center of the Earth $$r'$$ becomes $$R + 2R = 3R$$. This is because the radius of the Earth $$R$$ is the distance from the Earth's center to its surface, so if the object is $$2R$$ above the surface, the total distance from the center is $$3R$$.
\n\n
At the surface of the Earth, the gravitational force ($$F_{earth}$$) that acts on an object is given by its weight, which can be calculated using the formula $$F_{earth} = m \\cdot g$$, where $$g$$ is the acceleration due to gravity on the surface of the Earth. Given that $$g = 10 \\text{m/s}^2$$ and the mass of the body $$m = 90 \\text{kg}$$, we get:
To find the gravitational pull at a distance $$2R$$ from the Earth's surface, we use the fact that gravitational force varies inversely as the square of the distance from the center of the Earth. Since the distance triples ($$3R$$ from $$R$$), the gravitational force becomes $$\\frac{1}{3^2} = \\frac{1}{9}$$ of the force at the surface.
\n\n
Therefore, the gravitational pull $$F_{2R}$$ on the body when placed at $$2R$$ from the Earth's surface is:
Assuming the earth to be a sphere of uniform mass density, a body weighed $$300 \\mathrm{~N}$$ on the surface of earth. How much it would weigh at R/4 depth under surface of earth ?
To solve this question, we first need to understand how gravitational force (and hence weight) changes with depth under the surface of the Earth. The gravitational force at a depth $d$ is given by the formula:
\n\n
$$ F = F_0 \\left(1 - \\frac{d}{R}\\right) $$
\n\n
where $F_0$ is the gravitational force (or the weight) at the surface, $R$ is the radius of the Earth, and $d$ is the depth below the Earth's surface.
\n\n
In your question, the body weighs 300 N on the surface, so $F_0 = 300$ N. It is taken to a depth of $\\frac{R}{4}$ under the surface. Therefore, $d = \\frac{R}{4}$.
\n\n
Substituting these values into our formula, we get:
\n\n
$$ F = 300 \\left(1 - \\frac{\\frac{R}{4}}{R}\\right) $$
\n\n
$$ F = 300 \\left(1 - \\frac{1}{4}\\right) $$
\n\n
$$ F = 300 \\left(\\frac{3}{4}\\right) $$
\n\n
$$ F = 225 \\, \\text{N} $$
\n\n
Therefore, at a depth of $\\frac{R}{4}$ under the surface of the Earth, the body would weigh 225 N. The correct option is Option D.
If the radius of earth is reduced to three-fourth of its present value without change in its mass then value of duration of the day of earth will be ________ hours 30 minutes.
Given that the radius of the Earth is decreased to three-fourths of its current value while its mass remains unchanged, we need to determine the new duration of the Earth's day.
\n\n
First, using the principle of conservation of angular momentum, we have:
\n\n
$ \\tau_{\\text{ext}} = 0 \\implies \\text{Angular momentum is conserved} $
\n\n
Therefore,
\n\n
$ \\frac{2}{5} M R^2 \\cdot \\omega_i = \\frac{2}{5} M \\left(\\frac{3R}{4}\\right)^2 \\cdot \\omega_f $
Where $$M,R$$ are the mass and radius of the planet respectively. In this expression the mass of the body $$(m)$$ is not present. The escape velocity is independent of the mass m or it depends on m0.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9536,
"subject": "Physics",
"question": "The kinetic energy needed to project a body of mass $$m$$ from the earth surface (radius $$R$$) to infinity is ",
"options": [
{
"text": "$$mgR/2$$ "
},
{
"text": "$$2mgR$$ "
},
{
"text": "$$mgR$$ "
},
{
"text": "$$mgR/4$$"
}
],
"answer": "$$mgR$$ ",
"solution": "**Answer:** $$mgR$$ \n\nThe required velocity is called escape velocity ($${v_e}$$) to leave the earth surface of a body.\n
$$\\therefore$$ $$K.E = {1 \\over 2}m \\times 2gR = mgR$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9537,
"subject": "Physics",
"question": "The escape velocity for a body projected vertically upwards from the surface of earth is $$11$$ $$km/s.$$ If the body is projected at an angle of $${45^ \\circ }$$ with the vertical, the escape velocity will be ",
"options": [
{
"text": "$$11\\sqrt 2 \\,\\,km/s$$ "
},
{
"text": "$$22$$ $$km/s$$ "
},
{
"text": "$$11$$ $$km/s$$ "
},
{
"text": "$${{11} \\over {\\sqrt 2 }}km/s$$ "
}
],
"answer": "$$11$$ $$km/s$$ ",
"solution": "**Answer:** $$11$$ $$km/s$$ \n\nWe know, Escape velocity, $${v_e} = \\sqrt {2gR} $$\n
So the escape velocity is independent of the angle at which the body is projected, hence it will remain same as 11 km/s.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9538,
"subject": "Physics",
"question": "Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius $$R$$ around the sun will be proportional to ",
"options": [
{
"text": "$${R^n}$$ "
},
{
"text": "$${R^{\\left( {{{n - 1} \\over 2}} \\right)}}$$ "
},
{
"text": "$${R^{\\left( {{{n + 1} \\over 2}} \\right)}}$$"
},
{
"text": "$${R^{\\left( {{{n - 2} \\over 2}} \\right)}}$$"
}
],
"answer": "$${R^{\\left( {{{n + 1} \\over 2}} \\right)}}$$",
"solution": "**Answer:** $${R^{\\left( {{{n + 1} \\over 2}} \\right)}}$$\n\nFor moving a planet around the sun in the circular orbit,\n
The necessary centripetal force = Gravitational force exerted on it\n
$$\\therefore$$ $$T \\propto {R^{{{ \\left( {n + 1} \\right)} \\over 2}}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9539,
"subject": "Physics",
"question": "A planet in a distant solar system is $$10$$ times more massive than the earth and its radius is $$10$$ times smaller. Given that the escape velocity from the earth is $$11\\,\\,km\\,{s^{ - 1}},$$ the escape velocity from the surface of the planet would be ",
"options": [
{
"text": "$$1.1\\,\\,km\\,{s^{ - 1}}$$ "
},
{
"text": "$$100\\,\\,km\\,{s^{ - 1}}$$ "
},
{
"text": "$$110\\,\\,km\\,{s^{ - 1}}$$ "
},
{
"text": "$$0.11\\,\\,km\\,{s^{ - 1}}$$ "
}
],
"answer": "$$110\\,\\,km\\,{s^{ - 1}}$$ ",
"solution": "**Answer:** $$110\\,\\,km\\,{s^{ - 1}}$$ \n\nLet Me is mass of earth then mass of planet Mp = 10Me.\n
And let Re is radius of earth then radius of planet Rp = $${{{R_e}} \\over {10}}$$\n
$$ = 10 \\times 11 = 110\\,km/s$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9540,
"subject": "Physics",
"question": "The mass of a spaceship is $$1000$$ $$kg.$$ It is to be launched from the earth's surface out into free space. The value of $$g$$ and $$R$$ (radius of earth ) are $$10\\,m/{s^2}$$ and $$6400$$ $$km$$ respectively. The required energy for this work will be:",
"options": [
{
"text": "$$6.4 \\times {10^{11}}\\,$$ Joules"
},
{
"text": "$$6.4 \\times {10^8}\\,$$ Joules "
},
{
"text": "$$6.4 \\times {10^9}\\,$$ Joules "
},
{
"text": "$$6.4 \\times {10^{10}}\\,$$ Joules"
}
],
"answer": "$$6.4 \\times {10^{10}}\\,$$ Joules",
"solution": "**Answer:** $$6.4 \\times {10^{10}}\\,$$ Joules\n\nPotential energy at earth surface = $$ - {{GMm} \\over R}$$\n
and at free space potential energy = 0\n
Work done for this = $$0 - \\left( { - {{GMm} \\over R}} \\right)$$ = $${{{GMm} \\over R}}$$\n
$$ = 6.4 \\times {10^{10}}\\,\\,$$ Joules",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9541,
"subject": "Physics",
"question": "What is the minimum energy required to launch a satellite of mass $$m$$ from the surface of a planet of mass $$M$$ and radius $$R$$ in a circular orbit at an altitude of $$2R$$? ",
"options": [
{
"text": "$${{5GmM} \\over {6R}}$$ "
},
{
"text": "$${{2GmM} \\over {3R}}$$ "
},
{
"text": "$${{GmM} \\over {2R}}$$ "
},
{
"text": "$${{GmM} \\over {3R}}$$ "
}
],
"answer": "$${{5GmM} \\over {6R}}$$ ",
"solution": "**Answer:** $${{5GmM} \\over {6R}}$$ \n\nEnergy of the satellite on the surface of the planet \n
= $${{5GMm} \\over {6R}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9542,
"subject": "Physics",
"question": "A satellite is revolving in a circular orbit at a height $$'h'$$ from the earth's surface (radius of earth $$R;h < < R$$). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to : (Neglect the effect of atmosphere.) ",
"options": [
{
"text": "$$\\sqrt{2 g R}$$"
},
{
"text": "$$\\sqrt{g R}$$"
},
{
"text": "$$\\sqrt{g R / 2}$$"
},
{
"text": "$$\\sqrt{g R}(\\sqrt{2}-1)$$"
}
],
"answer": "$$\\sqrt{g R}(\\sqrt{2}-1)$$",
"solution": "**Answer:** $$\\sqrt{g R}(\\sqrt{2}-1)$$\n\nOrbital velocity of satellite,\n
$${v_0} = \\sqrt {{{GM} \\over {R + h}}} $$ \n
= $$\\sqrt {{{GM} \\over R}} $$ [ As $$h < < R$$ then R + h = R ]\n
$$\\therefore$$ The minimum increase in its orbital velocity required to escape from the earth gravitational field\n
$$ = \\sqrt {2gR} - \\sqrt {gR} = \\sqrt {gR} \\left( {\\sqrt 2 - 1} \\right)$$ ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9543,
"subject": "Physics",
"question": "A test particle is moving in a circular orbit in\nthe gravitational field produced by a mass\ndensity $$\\rho (r) = {K \\over {{r^2}}}$$ . Identify the correct relation\nbetween the radius R of the particle's orbit and\nits period T",
"options": [
{
"text": "T2/R3 is a constant"
},
{
"text": "TR is a constant"
},
{
"text": "T/R2 is a constant"
},
{
"text": "T/R is a constant"
}
],
"answer": "T/R is a constant",
"solution": "**Answer:** T/R is a constant\n\nFor circular motion of particle: \n$${{m{V^2}} \\over r} = mE$$ \n$$ = m\\left( {{{GM} \\over {{r^2}}}} \\right)$$
\n$$ \\Rightarrow $$ $${T \\over R}$$ = Constant",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9544,
"subject": "Physics",
"question": "A rocket has to be launched from earth in such\na way that it never returns. If E is the minimum\nenergy delivered by the rocket launcher, what\nshould be the minimum energy that the\nlauncher should have if the same rocket is to\nbe launched from the surface of the moon ?\nAssume that the density of the earth and the\nmoon are equal and that the earth's volume is\n64 times the volume of the moon :-",
"options": [
{
"text": "E/32"
},
{
"text": "E/16"
},
{
"text": "E/4"
},
{
"text": "E/64"
}
],
"answer": "E/16",
"solution": "**Answer:** E/16\n\nMinimum energy required (E) = – (Potential energy of\nobject at surface of earth)
\n$$ \\Rightarrow {E_{moon}} = {E \\over {16}}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9545,
"subject": "Physics",
"question": "Two satellites, A and B, have masses m and 2m respectively. A is in a circular orbit of radius R, and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, TA/TB, is ;\n",
"options": [
{
"text": "2"
},
{
"text": "$${{1 \\over 2}}$$"
},
{
"text": "$$\\sqrt {{1 \\over 2}} $$"
},
{
"text": "1"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\nOrbital velocity V = $$\\sqrt {{{GMe} \\over r}} $$\n
$$ \\Rightarrow $$ $${{{T_A}} \\over {{T_B}}}$$ = 1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9546,
"subject": "Physics",
"question": "A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be : ",
"options": [
{
"text": "in the same circular orbit of radius R"
},
{
"text": "such that it escapes to infinity"
},
{
"text": "in a circular orbit of a different radius "
},
{
"text": "in an elliptical orbit"
}
],
"answer": "in an elliptical orbit",
"solution": "**Answer:** in an elliptical orbit\n\n\n mv$$\\widehat i$$ + mv$$\\widehat j$$\n
= 2m$${\\overrightarrow v ^1}$$\n
$$\\overrightarrow v $$ = $${1 \\over {\\sqrt 2 }} \\times \\sqrt {{{GM} \\over R}} $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9547,
"subject": "Physics",
"question": "Two stars of masses 3 $$ \\times $$ 1031 kg each, and at distance 2 $$ \\times $$ 1011 m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is - (Take Gravitational constant; G = 6.67 $$ \\times $$ 10–11 Nm2 kg–2) ",
"options": [
{
"text": "2.4 $$ \\times $$ 104 m/s"
},
{
"text": "1.4 $$ \\times $$ 105 m/s"
},
{
"text": "3.8 $$ \\times $$ 104 m/s"
},
{
"text": "2.8 $$ \\times $$ 105 m/s"
}
],
"answer": "2.8 $$ \\times $$ 105 m/s",
"solution": "**Answer:** 2.8 $$ \\times $$ 105 m/s\n\nBy energy convervation between 0 & $$\\infty $$.\n
$$ \\Rightarrow $$ v $$ = \\sqrt {{{4GM} \\over r}} = 2.8 \\times {10^5}$$m/s",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9548,
"subject": "Physics",
"question": "A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is -\n",
"options": [
{
"text": "mv2"
},
{
"text": "$${1 \\over 2}$$ mv2"
},
{
"text": "$${3 \\over 2}$$ mv2"
},
{
"text": "2 mv2"
}
],
"answer": "mv2",
"solution": "**Answer:** mv2\n\nAt height r from center of earth. orbital velocity\n
= $$\\sqrt {{{GM} \\over r}} $$\n
$$ \\therefore $$ By energy conservation\n
KE of 'm' + $$\\left( { - {{GMm} \\over r}} \\right)$$ = 0 + 0\n
(At infinity, PE = KE = 0)\n
$$ \\Rightarrow $$ KE of 'm' = $${{{GMm} \\over r}}$$ = $${\\left( {\\sqrt {{{GM} \\over r}} } \\right)^2}$$ m = mv2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9549,
"subject": "Physics",
"question": "The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 $$ \\times $$ 103 km) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal, is ",
"options": [
{
"text": "1.6 $$ \\times $$ 103 km"
},
{
"text": "3.2 $$ \\times $$ 103 km"
},
{
"text": "6.4 $$ \\times $$ 103 km"
},
{
"text": "1.28 $$ \\times $$ 104 km"
}
],
"answer": "3.2 $$ \\times $$ 103 km",
"solution": "**Answer:** 3.2 $$ \\times $$ 103 km\n\nEnergy required to move a satellite from earth surface to height h is, \n
$$ \\therefore $$ h = $${{6.4 \\times {{10}^3}} \\over 2}$$\n
= 3.2 $$ \\times $$ 103 km",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9550,
"subject": "Physics",
"question": "A satellite is revolving in a circular orbit at a height h form the earth surface, such that h < < R where R is the earth. Assuming that the effect of earth's atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is : ",
"options": [
{
"text": "$$\\sqrt {gR} \\left( {\\sqrt 2 - 1} \\right)$$"
},
{
"text": "$$\\sqrt {2gR} $$"
},
{
"text": "$$\\sqrt {gR} $$"
},
{
"text": "$${{\\sqrt {gR} } \\over 2}$$"
}
],
"answer": "$$\\sqrt {gR} \\left( {\\sqrt 2 - 1} \\right)$$",
"solution": "**Answer:** $$\\sqrt {gR} \\left( {\\sqrt 2 - 1} \\right)$$\n\nv0 = $$\\sqrt {g(R + h)} \\approx \\sqrt {gR} $$\n
ve = $$\\sqrt {2g(R + h)} \\approx \\sqrt {2gR} $$\n
$$\\Delta $$v=ve $$-$$ v0 = $$\\left( {\\sqrt 2 - 1} \\right)\\sqrt {gR} $$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9551,
"subject": "Physics",
"question": "A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to\nthat of the earth’s radius Re\n. By firing rockets attached to it, its speed is instantaneously increased\nin the direction of its motion so that it become $$\\sqrt {{3 \\over 2}} $$\n times larger. Due to this the farthest distance\nfrom the centre of the earth that the satellite reaches is R. Value of R is :",
"options": [
{
"text": "2Re"
},
{
"text": "3Re"
},
{
"text": "4Re"
},
{
"text": "2.5Re"
}
],
"answer": "3Re",
"solution": "**Answer:** 3Re\n\n\n
$$ \\Rightarrow $$ R = 3Re",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9552,
"subject": "Physics",
"question": "A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the\norbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape\nvelocity from the planet is:\n",
"options": [
{
"text": "2"
},
{
"text": "1"
},
{
"text": "$$\\sqrt 2 $$"
},
{
"text": "$${1 \\over {\\sqrt 2 }}$$"
}
],
"answer": "$${1 \\over {\\sqrt 2 }}$$",
"solution": "**Answer:** $${1 \\over {\\sqrt 2 }}$$\n\n$${V_0} = \\sqrt {{{GM} \\over r}} $$\n
$${V_e} = \\sqrt {{{2GM} \\over r}} $$\n
$$ \\therefore $$ $${{{V_0}} \\over {{V_e}}} = \\sqrt {{{GM} \\over r} \\times {{2GM} \\over r}} = {1 \\over {\\sqrt 2 }}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9553,
"subject": "Physics",
"question": "The mass density of a spherical galaxy varies\nas\n$${K \\over r}$$ over a large distance ‘r’ from its centre.\nIn that region, a small star is in a circular orbit\nof radius R. Then the period of revolution, T\ndepends on R as :",
"options": [
{
"text": "T2 $$ \\propto $$ R"
},
{
"text": "T2 $$ \\propto $$ R3"
},
{
"text": "T $$ \\propto $$ R"
},
{
"text": "T2 $$ \\propto $$ $${1 \\over {{R^3}}}$$"
}
],
"answer": "T2 $$ \\propto $$ R",
"solution": "**Answer:** T2 $$ \\propto $$ R\n\n\n
$$ \\Rightarrow $$ T2 $$ \\propto $$ R",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9554,
"subject": "Physics",
"question": "Planet A has mass M and radius R. Planet B has\nhalf the mass and half the radius of Planet A.\nIf the escape velocities from the Planets A and\nB are vA and vB, respectively, then $${{{v_A}} \\over {{v_B}}} = {n \\over 4}$$.\nThe value of n is :",
"options": [
{
"text": "1"
},
{
"text": "2"
},
{
"text": "4"
},
{
"text": "3"
}
],
"answer": "4",
"solution": "**Answer:** 4\n\nEscape velocity, Ve = $$\\sqrt {{{2GM} \\over R}} $$\n
Given $${{{V_A}} \\over {{V_B}}} = {n \\over 4}$$\n
$$ \\therefore $$ $${n \\over 4}$$ = 1\n
$$ \\Rightarrow $$ n = 4",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9555,
"subject": "Physics",
"question": "An asteroid is moving directly towards the\ncentre of the earth. When at a distance of\n10R (R is the radius of the earth) from the earths\ncentre, it has a speed of 12 km/s. Neglecting\nthe effect of earths atmosphere, what will be the\nspeed of the asteroid when it hits the surface\nof the earth (escape velocity from the earth is\n11.2 km/s) ? Give your answer to the nearest\ninteger in kilometer/s _____.",
"options": [],
"answer": "16",
"solution": "**Answer:** 16\n\nU1 + K1 = U2 + K2\n
$$ \\Rightarrow $$ V2 = 16 km/s",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9556,
"subject": "Physics",
"question": "A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R (R = radius of the earth), it ejects a rocket\nof mass $${m \\over {10}}$$\nso that subsequently the\nsatellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant; M is the mass of the earth) :\n",
"options": [
{
"text": "$${{3m} \\over 8}{\\left( {u + \\sqrt {{{5GM} \\over {6R}}} } \\right)^2}$$"
},
{
"text": "$${m \\over {20}}\\left( {{u^2} + {{113} \\over {100}}{{GM} \\over R}} \\right)$$"
},
{
"text": "$$5m\\left( {{u^2} - {{119} \\over {100}}{{GM} \\over R}} \\right)$$"
},
{
"text": "$${m \\over {20}}{\\left( {u - \\sqrt {{{2GM} \\over {3R}}} } \\right)^2}$$"
}
],
"answer": "$$5m\\left( {{u^2} - {{119} \\over {100}}{{GM} \\over R}} \\right)$$",
"solution": "**Answer:** $$5m\\left( {{u^2} - {{119} \\over {100}}{{GM} \\over R}} \\right)$$\n\n\n Using energy conservation\n
After ejecting a rocket of mass $${m \\over {10}}$$ the remaining part of mass $${{9m} \\over {10}}$$ will rotate the earth with orbital velocity v0.\n
= $$5m\\left( {{u^2} - {{119GM} \\over {200R}}} \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9557,
"subject": "Physics",
"question": "Consider two satellites S1 and S2 with periods of revolution 1 hr. and 8 hr. respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S1 to the angular velocity of satellite S2 is :",
"options": [
{
"text": "1 : 4"
},
{
"text": "8 : 1"
},
{
"text": "2 : 1"
},
{
"text": "1 : 8"
}
],
"answer": "8 : 1",
"solution": "**Answer:** 8 : 1\n\nGiven, period of revolution of first satellite,
or $${\\omega _1}:{\\omega _2} = 8:1$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9558,
"subject": "Physics",
"question": "Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The escape velocities of planet A and B are same. But A and B are of unequal mass.
Reason R : The product of their mass and radius must be same. M1R1 = M2R2
In the light of the above statements, choose the most appropriate answer from the options given below :",
"options": [
{
"text": "Both A and R are correct and R is the correct explanation of A"
},
{
"text": "Both A and R are correct but R is NOT the correct explanation of A"
},
{
"text": "A is correct but R is not correct"
},
{
"text": "A is not correct but R is correct"
}
],
"answer": "A is correct but R is not correct",
"solution": "**Answer:** A is correct but R is not correct\n\n$${v_e}$$ = escape velocity
$${v_e} = \\sqrt {{{2GM} \\over R}} $$
so for same $${v_e},{{{M_1}} \\over {{R_1}}} = {{{M_2}} \\over {{R_2}}}$$
A is true but R is false",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9559,
"subject": "Physics",
"question": "The initial velocity vi required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity ve such that $${v_i} = \\sqrt {{x \\over y}} \\times {v_e}$$. The value of x will be ____________.",
"options": [],
"answer": "10",
"solution": "**Answer:** 10\n\nHere R = radius of the earth
Then the value of x = 10",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9560,
"subject": "Physics",
"question": "The radius in kilometer to which the present radius of earth (R = 6400 km) to be compressed so that the escape velocity is increased 10 times is ___________.",
"options": [],
"answer": "64",
"solution": "**Answer:** 64\n\n$${V_{es}} = \\sqrt {{{2GM} \\over R}} $$
$${V_{es}}\\sqrt R $$ = const
$$ \\therefore $$ $${V_{es}}.\\sqrt R = 10{V_{es}}\\sqrt {R'} $$
$$ \\Rightarrow $$ $$R' = {R \\over {100}}$$ = 64 km",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9561,
"subject": "Physics",
"question": "A geostationary satellite is orbiting around an arbitrary planet 'P' at a height of 11R above the surface of 'P', R being the radius of 'P'. The time period of another satellite in hours at a height of 2R from the surface of 'P' is _________. 'P' has the time period of 24 hours.",
"options": [
{
"text": "3"
},
{
"text": "5"
},
{
"text": "$$6\\sqrt 2 $$"
},
{
"text": "$${6 \\over {\\sqrt 2 }}$$"
}
],
"answer": "3",
"solution": "**Answer:** 3\n\n\n From Kepler's law
T = 3 sec",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9562,
"subject": "Physics",
"question": "A satellite is launched into a circular orbit of radius R around earth, while a second satellite is launched into a circular orbit of radius 1.02 R. The percentage difference in the time periods of the two satellites is : ",
"options": [
{
"text": "1.5"
},
{
"text": "2.0"
},
{
"text": "0.7"
},
{
"text": "3.0"
}
],
"answer": "3.0",
"solution": "**Answer:** 3.0\n\n$${T^2} \\propto {R^3}$$
$$T = k{R^{3/2}}$$
$${{dT} \\over T} = {3 \\over 2}{{dR} \\over R}$$
$$ = {3 \\over 2} \\times 0.02 = 0.03$$
% Change = 3%",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9563,
"subject": "Physics",
"question": "The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of 9.0 $$\\times$$ 103 km. Find the mass of Mars.
The escape velocity of a body on a planet 'A' is 12 kms$$-$$1. The escape velocity of the body on another planet 'B', whose density is four times and radius is half of the planet 'A', is :
Two satellites S1 and S2 are revolving in circular orbits around a planet with radius R1 = 3200 km and R2 = 800 km respectively. The ratio of speed of satellite S1 to be speed of satellite S2 in their respective orbits would be $${1 \\over x}$$ where x = ___________.
A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be :
\n
(Take radius of earth $$=6400 \\mathrm{~km}$$ and $$\\mathrm{g}=10 \\mathrm{~ms}^{-2}$$ )
Two satellites $$\\mathrm{A}$$ and $$\\mathrm{B}$$, having masses in the ratio $$4: 3$$, are revolving in circular orbits of radii $$3 \\mathrm{r}$$ and $$4 \\mathrm{r}$$ respectively around the earth. The ratio of total mechanical energy of $$\\mathrm{A}$$ to $$\\mathrm{B}$$ is :
A body of mass $$\\mathrm{m}$$ is projected with velocity $$\\lambda \\,v_{\\mathrm{e}}$$ in vertically upward direction from the surface of the earth into space. It is given that $$v_{\\mathrm{e}}$$ is escape velocity and $$\\lambda<1$$. If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be :
The escape velocities of two planets $$\\mathrm{A}$$ and $$\\mathrm{B}$$ are in the ratio $$1: 2$$. If the ratio of their radii respectively is $$1: 3$$, then the ratio of acceleration due to gravity of planet A to the acceleration of gravity of planet B will be :
",
"options": [
{
"text": "$$\\frac{4}{3}$$"
},
{
"text": "$$\\frac{2}{3}$$"
},
{
"text": "$$\\frac{3}{4}$$"
},
{
"text": "$$\\frac{3}{2}$$"
}
],
"answer": "$$\\frac{3}{4}$$",
"solution": "**Answer:** $$\\frac{3}{4}$$\n\nThe escape velocity of a planet is given by the formula:\n\n
If earth has a mass nine times and radius twice to that of a planet P. Then $$\\frac{v_{e}}{3} \\sqrt{x} \\mathrm{~ms}^{-1}$$ will be the minimum velocity required by a rocket to pull out of gravitational force of $$\\mathrm{P}$$, where $$v_{e}$$ is escape velocity on earth. The value of $$x$$ is
$$\n\\begin{aligned}\n& \\text { For } P, v_e=\\sqrt{\\frac{\\frac{2 G M_E}{9}}{\\frac{R_E}{2}}}=\\sqrt{\\frac{2 G M_E}{R_E} \\times \\frac{2}{9}} \\\\\\\\\n& =\\frac{v_e \\sqrt{2}}{3}\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9572,
"subject": "Physics",
"question": "
Given below are two statements:
\n
Statement I : For a planet, if the ratio of mass of the planet to its radius increases, the escape velocity from the planet also increases.
\n
Statement II : Escape velocity is independent of the radius of the planet.
\n
In the light of above statements, choose the most appropriate answer form the options given below
",
"options": [
{
"text": "Both Statement I and Statement II are correct"
},
{
"text": "Statement I is correct but statement II is incorrect"
},
{
"text": "Both Statement I and Statement II are incorrect"
},
{
"text": "Statement I is incorrect but statement II is correct"
}
],
"answer": "Statement I is correct but statement II is incorrect",
"solution": "**Answer:** Statement I is correct but statement II is incorrect\n\nStatement I suggests that the escape velocity of a planet increases with an increase in the ratio of its mass to its radius. This statement is consistent with the formula for escape velocity, which shows that the escape velocity of a planet is directly proportional to the square root of its mass and inversely proportional to the square root of its radius.\n
\nOn the other hand, Statement II suggests that the escape velocity is independent of the radius of the planet. This statement is not consistent with the formula for escape velocity, which clearly shows that the escape velocity is inversely proportional to the square root of the radius of the planet.\n
\nHence, the most appropriate answer is: Statement I is true, but Statement II is false.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9573,
"subject": "Physics",
"question": "
A planet having mass $$9 \\mathrm{Me}$$ and radius $$4 \\mathrm{R}_{\\mathrm{e}}$$, where $$\\mathrm{Me}$$ and $$\\mathrm{Re}$$ are mass and radius of earth respectively, has escape velocity in $$\\mathrm{km} / \\mathrm{s}$$ given by:
",
"options": [
{
"text": "33.6"
},
{
"text": "11.2"
},
{
"text": "16.8"
},
{
"text": "67.2"
}
],
"answer": "16.8",
"solution": "**Answer:** 16.8\n\nThe escape velocity on a planet is given by the following formula:\n
\n$$v_{esc} = \\sqrt{\\frac{2GM}{R}}$$\n
\nwhere $v_{esc}$ is the escape velocity, $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet.\n
\nFor Earth, we are given that $v_e = 11.2 \\times 10^3$ m/s. We know that the mass of the planet in question is $9M_e$ and its radius is $4R_e$. Let's find the escape velocity of this planet:\n
\n$$v_{esc} = \\sqrt{\\frac{2G(9M_e)}{4R_e}}$$\n
\nDivide both sides by the Earth's escape velocity formula:\n
\nTherefore, the ratio of the escape velocity of the planet to that of earth is 2 : 1.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9575,
"subject": "Physics",
"question": "
Two satellites $$\\mathrm{A}$$ and $$\\mathrm{B}$$ move round the earth in the same orbit. The mass of $$\\mathrm{A}$$ is twice the mass of $$\\mathrm{B}$$. The quantity which is same for the two satellites will be
",
"options": [
{
"text": "Potential energy"
},
{
"text": " Kinetic energy"
},
{
"text": "Total energy"
},
{
"text": "Speed"
}
],
"answer": "Speed",
"solution": "**Answer:** Speed\n\nThe quantity which is same for the two satellites will be speed.\n
\nThe orbital speed of a satellite is independent of the mass of the satellite, but it depends on the radius of the orbit. Potential energy, kinetic energy and total energy depend on the mass of the the satellite.\n
\nTherefore, the only quantity that is the same for the two satellites is speed.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9576,
"subject": "Physics",
"question": "
A space ship of mass $$2 \\times 10^{4} \\mathrm{~kg}$$ is launched into a circular orbit close to the earth surface. The additional velocity to be imparted to the space ship in the orbit to overcome the gravitational pull will be (if $$g=10 \\mathrm{~m} / \\mathrm{s}^{2}$$ and radius of earth $$=6400 \\mathrm{~km}$$ ):
To find the additional velocity required to overcome the gravitational pull and launch the spaceship into orbit, we first need to find the orbital velocity. The formula for orbital velocity (v) is given by:
\n
$$v = \\sqrt{\\frac{GM}{r}}$$
\n
where G is the gravitational constant, M is the mass of Earth, and r is the distance from the center of the Earth to the spaceship (which is the sum of the Earth's radius and the altitude of the spaceship's orbit).
\n
In this problem, the spaceship is orbiting close to Earth's surface, so we can approximate r as the Earth's radius. Given that g = 10 m/s² and Earth's radius R = 6400 km, we can relate the gravitational constant G and the mass of Earth M through the formula:
For a satellite orbiting the Earth at a height equal to Earth's radius, its distance from the center of the Earth will be 2R, where R is the radius of the Earth.
\n
Using the formula for the gravitational force:
\n
$$F = G\\frac{Mm}{r^2}$$
\n
where F is the gravitational force, G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance from the center of the Earth.
\n
The centripetal force acting on the satellite is given by:
\n
$$F_c = \\frac{mv^2}{r}$$
\n
Equating the gravitational force and the centripetal force, we get:
\n
$$G\\frac{Mm}{(2R)^2} = \\frac{mv^2}{2R}$$
\n
Solving for the orbital speed v, we get:
\n
$$v^2 = \\frac{GM}{2R}$$
\n
The circumference of the satellite's orbit is given by:
\n
$$C = 2\\pi(2R) = 4\\pi R$$
\n
The time period T of the satellite's orbit can be calculated as the ratio of the circumference to the orbital speed:
Given that the acceleration due to gravity at the Earth's surface is $$g = \\pi^2 \\,\\mathrm{m/s^2}$$, we can express the gravitational constant G in terms of the Earth's radius R and mass M:
Two satellites of masses m and 3m revolve around the earth in circular orbits of radii r & 3r respectively. The ratio of orbital speeds of the satellites respectively is
The orbital speed of an object moving in a circular orbit around Earth (or any other celestial body) is given by the formula:
\n
$ v = \\sqrt{\\frac{GM}{r}} $
\n
where (v) is the orbital speed, (G) is the gravitational constant, (M) is the mass of the central body (Earth, in this case), and (r) is the radius of the orbit.
\n
For the two satellites of masses (m) and (3m) in orbits of radii (r) and (3r) respectively, the ratio of their orbital speeds ($v_1/v_2$) is:
So, the ratio of the orbital speeds of the satellites is ($\\sqrt{3} : 1$), which corresponds to Option B.
\n
The inclusion of two different masses for the satellites, $m$ and $3m$, in the problem might initially seem to suggest that the masses would influence their orbital speeds. However, when it comes to circular orbital motion, especially around a much larger body like the Earth, the mass of the orbiting satellite does not directly affect its orbital speed. This is because the orbital speed equation:\n
\n$ v = \\sqrt{\\frac{GM}{r}} $\n
\nonly takes into account the mass of the central body (in this case, Earth's mass $M$), and the radius of the orbit $(r)$, where $G$ is the gravitational constant.
The orbital angular momentum of a satellite is L, when it is revolving in a circular orbit at height h from earth surface. If the distance of satellite from the earth centre is increased by eight times to its initial value, then the new angular momentum will be -
If we take the velocity (v) of the satellite to be $v = \\sqrt{GM/r}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the distance from the center of the Earth to the satellite, then the orbital angular momentum (L) is:
\n
$ L = mvr = m \\sqrt{GMr} $
\n
If the distance (r) from the Earth's center is increased by a factor of 8, the new angular momentum $L'$ is:
\n
$ L' = m \\sqrt{GM(8r)} = m \\sqrt{8GMr} = 2 \\sqrt{2} L $
\n
However, the distance from the earth's surface is given, so we have to take into account the radius of the Earth ($R$) in our calculations. When the height from the earth's surface is increased eight times, the distance from the earth's center is $r = R + 8h$, which is approximately $9R$ (because the height of the satellite above the Earth is generally much less than the radius of the Earth).
Statement I: If $$\\mathrm{E}$$ be the total energy of a satellite moving around the earth, then its potential energy will be $$\\frac{E}{2}$$.
\n
Statement II: The kinetic energy of a satellite revolving in an orbit is equal to the half the magnitude of total energy $$\\mathrm{E}$$.
\n
In the light of the above statements, choose the most appropriate answer from the options given below
",
"options": [
{
"text": "Both Statement I and Statement II are incorrect"
},
{
"text": "Statement I is incorrect but Statement II is correct"
},
{
"text": "Statement I is correct but Statement II is incorrect"
},
{
"text": "Both Statement I and Statement II are correct"
}
],
"answer": "Both Statement I and Statement II are incorrect",
"solution": "**Answer:** Both Statement I and Statement II are incorrect\n\n
A satellite in orbit around a planet is subject to two main forces: gravitational force, which is trying to pull it towards the planet, and its own kinetic energy or inertia, which is trying to keep it moving in a straight line. The balance of these two forces results in the satellite moving in a circular or elliptical orbit.
\n
The gravitational potential energy ($U$) of the satellite is given by the formula:
\n
$U = -\\frac{GMm}{R}$
\n
where $G$ is the gravitational constant, $M$ is the mass of the Earth, $m$ is the mass of the satellite, and $R$ is the radius of the orbit. The negative sign indicates that work would have to be done to remove the satellite from the Earth's gravitational influence.
\n
The kinetic energy ($K$) of the satellite is given by the formula:
\n
$K = \\frac{GMm}{2R}$
\n
This is obtained from the fact that for a satellite in stable orbit, the gravitational force must be equal to the centripetal force required to keep the satellite moving in a circle. From this, we can derive an expression for the velocity of the satellite, and hence its kinetic energy.
\n
The total mechanical energy ($E$) of the satellite, which is the sum of its kinetic and potential energy, is therefore:
\n
$E = K + U = \\frac{GMm}{2R} - \\frac{GMm}{R} = -\\frac{GMm}{2R}$
\n
So the potential energy $U$ is $-2E$, and the kinetic energy $K$ is $-E$. Thus, the statement "If $E$ be the total energy of a satellite moving around the earth, then its potential energy will be $2E$" is correct, and the statement "The kinetic energy of a satellite revolving in an orbit is equal to the half the magnitude of total energy $E$" is incorrect.
Given below are two statements : one is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$.
\n
Assertion A : Earth has atmosphere whereas moon doesn't have any atmosphere.
\n
Reason R : The escape velocity on moon is very small as compared to that on earth.
\n
In the light of the above statements, choose the correct answer from the options given below:
",
"options": [
{
"text": "$$\\mathbf{A}$$ is false but $$\\mathbf{R}$$ is true"
},
{
"text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct but $$\\mathbf{R}$$ is NOT the correct explanation of $$\\mathbf{A}$$"
},
{
"text": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$"
},
{
"text": "$$\\mathbf{A}$$ is true but $$\\mathbf{R}$$ is false"
}
],
"answer": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$",
"solution": "**Answer:** Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct and $$\\mathbf{R}$$ is the correct explanation of $$\\mathbf{A}$$\n\n
The assertion (A) is true: Earth does have an atmosphere, while the Moon does not have a significant atmosphere.
\n
The reason (R) is also true: The escape velocity on the Moon is indeed smaller than that on Earth. The escape velocity is the minimum velocity an object must have to escape the gravitational pull of a planet or moon. A smaller escape velocity means it's easier for particles (such as the particles that make up an atmosphere) to escape into space.
\n
Additionally, the reason (R) is a correct explanation for the assertion (A). The fact that the Moon's escape velocity is smaller than the Earth's is a major reason why the Moon doesn't have a significant atmosphere. Over time, particles that could have made up an atmosphere have escaped the Moon's gravitational pull and dispersed into space.
Choose the incorrect statement from the following:
",
"options": [
{
"text": "The linear speed of a planet revolving around the sun remains constant."
},
{
"text": "When a body falls towards earth, the displacement of earth towards the body is negligible."
},
{
"text": "The speed of satellite in a given circular orbit remains constant."
},
{
"text": "For a planet revolving around the sun in an elliptical orbit, the total energy of the planet remains constant."
}
],
"answer": "The linear speed of a planet revolving around the sun remains constant.",
"solution": "**Answer:** The linear speed of a planet revolving around the sun remains constant.\n\nThe linear speed of a planet revolving around the sun does not remain constant, as planets follow an elliptical orbit around the sun. According to Kepler's second law, the areal velocity of a planet remains constant, which means that the planet moves faster when it is closer to the sun (at perihelion) and slower when it is farther away (at aphelion).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9583,
"subject": "Physics",
"question": "
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
\n
Assertion (A) : The angular speed of the moon in its orbit about the earth is more than the angular speed of the earth in its orbit about the sun.
\n
Reason (R) : The moon takes less time to move around the earth than the time taken by the earth to move around the sun.
\n
In the light of the above statements, choose the most appropriate answer from the options given below :
",
"options": [
{
"text": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)"
},
{
"text": "(A) is correct but (R) is not correct"
},
{
"text": "Both (A) and (R) are correct and (R) is the correct explanation of (A)"
},
{
"text": "(A) is not correct but (R) is correct"
}
],
"answer": "Both (A) and (R) are correct and (R) is the correct explanation of (A)",
"solution": "**Answer:** Both (A) and (R) are correct and (R) is the correct explanation of (A)\n\n
The angular speed $$ \\omega $$ of an object in circular motion is given by the equation
\n
$ \\omega = \\frac{2\\pi}{T} $
\n
where $$ T $$ is the period of the motion - the time it takes to make one complete revolution.
\n
Assertion (A) states that the angular speed of the Moon in its orbit around the Earth is more than the angular speed of the Earth in its orbit around the Sun. We can compare these angular speeds by comparing their periods.
\n
The Moon takes approximately 27.3 days to orbit the Earth (this is its sidereal period, not its synodic period, which accounts for the Earth's motion around the Sun as well). The Earth takes approximately 365.25 days to orbit the Sun. Since the period of the Moon’s orbit is much less than the period of the Earth’s orbit, the angular speed of the Moon is much larger than that of the Earth. This makes Assertion (A) correct.
\n
Reason (R) states that the Moon takes less time to move around the Earth than the time taken by the Earth to move around the Sun. This is also correct, as the previously mentioned periods demonstrate: 27.3 days for the Moon to orbit Earth vs. 365.25 days for Earth to orbit the Sun.
\n
Furthermore, the reason (R) directly explains why assertion (A) is true. Since angular speed is inversely proportional to the period of orbit, the fact that the Moon takes less time to complete an orbit implies it has a higher angular speed relative to the Earth’s angular speed in its orbit around the Sun.
\n
Therefore, the correct answer is:
\n
Option C\nBoth (A) and (R) are correct and (R) is the correct explanation of (A)
The mass of the moon is $$\\frac{1}{144}$$ times the mass of a planet and its diameter is $$\\frac{1}{16}$$ times the diameter of a planet. If the escape velocity on the planet is $$v$$, the escape velocity on the moon will be :
Escape velocity of a body from earth is $$11.2 \\mathrm{~km} / \\mathrm{s}$$. If the radius of a planet be onethird the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is :
A satellite of $$10^3 \\mathrm{~kg}$$ mass is revolving in circular orbit of radius $$2 R$$. If $$\\frac{10^4 R}{6} \\mathrm{~J}$$ energy is supplied to the satellite, it would revolve in a new circular orbit of radius
\n
(use $$g=10 \\mathrm{~m} / \\mathrm{s}^2, R=$$ radius of earth)
To determine the new radius of the orbit after the energy is supplied to the satellite, we need to compare the initial and final energies of the satellite in its orbit around the Earth. We will use the formula for total energy (the sum of kinetic and potential energy) of a satellite in a circular orbit:
\n\n
The total energy, $$E$$, of a satellite of mass $$m$$ orbiting at a distance $$r$$ from the center of the Earth is given by:
\n\n
$$ E = -\\frac{GMm}{2r} $$
\n\n
where:
\n\n
\n\n
$$G$$ is the gravitational constant
\n\n
$$M$$ is the mass of the Earth
\n\n
$$r$$ is the radius of the orbit
\n\n
\n\n
Given, the initial radius of orbit is $$2R$$. Therefore, the initial total energy $$E_i$$ is:
Now, the energy supplied to the satellite is given as $$\\frac{10^4 R}{6} \\mathrm{~J}$$. The new total energy $$E_f$$ will be the sum of the initial energy and the supplied energy:
\n\n
$$ E_f = E_i + \\text{Energy Supplied} $$
\n\n
Substituting the values:
\n\n
$$ E_f = -\\frac{GMm}{4R} + \\frac{10^4 R}{6} $$
\n\n
The final energy formula for a new orbit radius $$r_f$$ is similar to the initial energy formula:
\n\n
$$ E_f = -\\frac{GMm}{2r_f} $$
\n\n
By equating the two expressions for $$E_f$$, we get:
An astronaut takes a ball of mass $$m$$ from earth to space. He throws the ball into a circular orbit about earth at an altitude of $$318.5 \\mathrm{~km}$$. From earth's surface to the orbit, the change in total mechanical energy of the ball is $$x \\frac{\\mathrm{GM}_{\\mathrm{e}} \\mathrm{m}}{21 \\mathrm{R}_{\\mathrm{e}}}$$. The value of $$x$$ is (take $$\\mathrm{R}_{\\mathrm{e}}=6370 \\mathrm{~km})$$ :
A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is :
\n
(Given $$=$$ Radius of geo-stationary orbit for earth is $$4.2 \\times 10^4 \\mathrm{~km}$$)
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9590,
"subject": "Physics",
"question": "Energy required to move a body of mass $$m$$ from an orbit of radius $$2R$$ to $$3R$$ is ",
"options": [
{
"text": "$${{GMm} \\over {12{R^2}}}$$"
},
{
"text": "$${{GMm} \\over {3{R^2}}}$$"
},
{
"text": "$${{GMm} \\over {8R}}$$ "
},
{
"text": "$${{GMm} \\over {6R}}$$ "
}
],
"answer": "$${{GMm} \\over {6R}}$$ ",
"solution": "**Answer:** $${{GMm} \\over {6R}}$$ \n\nGravitational potential energy E = $$ - {{GMm} \\over r}$$\n
where M = mass of earth\n
m = mass of body\n
r = radius of earth\n
Energy required to move a body of mass $$m$$ from an orbit of radius $$2R$$ to $$3R$$\n
$$=$$ (Potential energy of the Earth-mass system when mass is at distance $$3R$$ ) $$-$$ (Potential energy of the Earth-mass system when mass is at distance $$2R$$) \n
$$ = {{ - 2GMm + 3GMm} \\over {6R}} = {{GMm} \\over {6R}}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9591,
"subject": "Physics",
"question": "If $$g$$ is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass $$m$$ raised from the surface of the earth to a height equal to the radius $$R$$ of the earth is ",
"options": [
{
"text": "$${1 \\over 4}mgR$$ "
},
{
"text": "$$2mgR$$ "
},
{
"text": "$${1 \\over 2}mgR$$ "
},
{
"text": "$$mgR$$ "
}
],
"answer": "$${1 \\over 2}mgR$$ ",
"solution": "**Answer:** $${1 \\over 2}mgR$$ \n\nGravitational potential energy on the earth surface of a body \n
U = $$-{{GmM} \\over R}$$\n
And at the height h from the earth surface the potential energy\n
$${U_h} = - {{GmM} \\over {R + h}}$$ = $$ - {{GmM} \\over {2R}}$$ [ as h = R ]\n
$$\\therefore$$ $$\\Delta U = {1 \\over 2}mgR$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9592,
"subject": "Physics",
"question": "A particle of mass $$10$$ $$g$$ is kept on the surface of a uniform sphere of mass $$100$$ $$kg$$ and radius $$10$$ $$cm.$$ Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take $$G$$ $$ = 6.67 \\times {10^{ - 11}}\\,\\,N{m^2}/k{g^2}$$) ",
"options": [
{
"text": "$$3.33 \\times {10^{ - 10}}\\,J$$ "
},
{
"text": "$$13.34 \\times {10^{ - 10}}\\,J$$ "
},
{
"text": "$$6.67 \\times {10^{ - 10}}\\,J$$ "
},
{
"text": "$$6.67 \\times {10^{ - 9}}\\,J$$ "
}
],
"answer": "$$6.67 \\times {10^{ - 10}}\\,J$$ ",
"solution": "**Answer:** $$6.67 \\times {10^{ - 10}}\\,J$$ \n\nWe know, Work done = Difference in potential energy\n
$$V = - {{Gm} \\over {{r \\over 3}}} - {{4Gm} \\over {{{2r} \\over 3}}} = -{{9Gm} \\over r}$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9594,
"subject": "Physics",
"question": "Two planets have masses M and 16 M and their radii are $$a$$ and 2$$a$$, respectively. The separation\nbetween the centres of the planets is 10$$a$$. A body of mass m is fired from the surface of the larger\nplanet towards the smaller planet along the line joining their centres. For the body to be able to\nreach at the surface of smaller planet, the minimum firing speed needed is :",
"options": [
{
"text": "$$2\\sqrt {{{GM} \\over a}} $$"
},
{
"text": "$$\\sqrt {{{G{M^2}} \\over {ma}}} $$"
},
{
"text": "$${3 \\over 2}\\sqrt {{{5GM} \\over a}} $$"
},
{
"text": "$$4\\sqrt {{{GM} \\over a}} $$"
}
],
"answer": "$${3 \\over 2}\\sqrt {{{5GM} \\over a}} $$",
"solution": "**Answer:** $${3 \\over 2}\\sqrt {{{5GM} \\over a}} $$\n\n\n
Let at point P, net gravitational force = 0\n
$$ \\therefore $$ $${{G\\left( M \\right)\\left( m \\right)} \\over {{{\\left( {10a - x} \\right)}^2}}} = {{G\\left( {16M} \\right)\\left( m \\right)} \\over {{x^2}}}$$\n
$$ \\Rightarrow $$ v = $$\\sqrt {{{90GM} \\over {8a}}} $$ = $${3 \\over 2}\\sqrt {{{5GM} \\over a}} $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9595,
"subject": "Physics",
"question": "If one wants to remove all the mass of the earth to infinity in order to break it up completely.
The amount of energy that needs to be supplied will be $${x \\over 5}{{G{M^2}} \\over R}$$ where x is __________ (Round off to the Nearest Integer) (M is the mass of earth, R is the radius of earth, G is the gravitational constant)",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n
We know that binding energy of earth,
\n
$$BE = - {3 \\over 5}{{G{M^2}} \\over R}$$
\n
$$\\therefore$$ Energy required to break the earth into pieces
(4) the gravitational potential is same everywhere
(5) all of the above
Choose the most appropriate answer from the options given below :",
"options": [
{
"text": "(1), (3) and (4) only"
},
{
"text": "(5) only"
},
{
"text": "(1), (2) and (3) only"
},
{
"text": "(2), (3) and (4) only"
}
],
"answer": "(1), (3) and (4) only",
"solution": "**Answer:** (1), (3) and (4) only\n\nInside a spherical shell, gravitational field is zero and hence potential remains same everywhere.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9598,
"subject": "Physics",
"question": "A mass of 50 kg is placed at the centre of a uniform spherical shell of mass 100 kg and radius 50 m. If the gravitational potential at a point, 25 m from the centre is V kg/m. The value of V is :",
"options": [
{
"text": "$$-$$60 G"
},
{
"text": "+2 G"
},
{
"text": "$$-$$20 G"
},
{
"text": "$$-$$4 G"
}
],
"answer": "$$-$$4 G",
"solution": "**Answer:** $$-$$4 G\n\n $${V_A} = \\left[ { - {{G{M_1}} \\over r} - {{G{M_2}} \\over R}} \\right]$$
Water falls from a 40 m high dam at the rate of 9 $$\\times$$ 104 kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydroelectric energy number of 100 W lamps, that can be lit, is :
An object of mass $$1 \\mathrm{~kg}$$ is taken to a height from the surface of earth which is equal to three times the radius of earth. The gain in potential energy of the object will be [If, $$\\mathrm{g}=10 \\mathrm{~ms}^{-2}$$ and radius of earth $$=6400 \\mathrm{~km}$$ ]
$\n\\begin{aligned}\n&=-\\frac{G M m}{4 R}+\\frac{G M m}{R} \\\\\\\\\n&=\\frac{3 G M m}{4 R}=\\frac{3}{4} m g R \\\\\\\\\n&=48 \\mathrm{MJ}\n\\end{aligned}\n$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9601,
"subject": "Physics",
"question": "An object is allowed to fall from a height $R$ above the earth, where $R$ is the radius of earth. Its velocity when it strikes the earth's surface, ignoring air resistance, will be",
"options": [
{
"text": "$\\sqrt{\\frac{g R}{2}}$"
},
{
"text": "$\\sqrt{g R}$"
},
{
"text": "$\\sqrt{2 g R}$"
},
{
"text": "$2 \\sqrt{g R}$"
}
],
"answer": "$\\sqrt{g R}$",
"solution": "**Answer:** $\\sqrt{g R}$\n\n
If the gravitational field in the space is given as $$\\left(-\\frac{K}{r^{2}}\\right)$$. Taking the reference point to be at $$\\mathrm{r}=2 \\mathrm{~cm}$$ \nwith gravitational potential $$\\mathrm{V}=10 \\mathrm{~J} / \\mathrm{kg}$$. Find the gravitational potential at $$\\mathrm{r}=3 \\mathrm{~cm}$$ in SI unit (Given, that $$\\mathrm{K}=6 \\mathrm{~Jcm} / \\mathrm{kg}$$)
\n$$\n\\begin{aligned}\n& V_{\\text {surface }}=-\\left(\\frac{G M m}{R_{e}}\\right) \\\\\\\\\n& V_{p}=-\\frac{G M m}{3 R_{e}} \\\\\\\\\n& \\Delta V=\\frac{G M m}{R_{e}}\\left(1-\\frac{1}{3}\\right) \\\\\\\\\n& =\\frac{2}{3} \\frac{G M m}{\\left(R_{e}^{2}\\right)} \\times R_{e} \\\\\\\\\n& \\Delta V=\\frac{2}{3} m g R_{e}\n\\end{aligned}\n$$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9604,
"subject": "Physics",
"question": "A body is released from a height equal to the radius $(\\mathrm{R})$ of the earth. The velocity of the body when it strikes the surface of the earth will be\n
\n(Given $g=$ acceleration due to gravity on the earth.)",
"options": [
{
"text": "$\\sqrt{\\frac{g R}{2}}$"
},
{
"text": "$\\sqrt{4 g R}$"
},
{
"text": "$\\sqrt{2 g R}$"
},
{
"text": "$\\sqrt{g R}$"
}
],
"answer": "$\\sqrt{g R}$",
"solution": "**Answer:** $\\sqrt{g R}$\n\n By conservation of mechanical energy
If $$\\mathrm{V}$$ is the gravitational potential due to sphere of uniform density on it's surface, then it's value at the center of sphere will be:-
",
"options": [
{
"text": "$$\\frac{3 \\mathrm{~V}}{2}$$"
},
{
"text": "$$\\frac{\\mathrm{V}}{2}$$"
},
{
"text": "$$\\frac{4}{3} \\mathrm{~V}$$"
},
{
"text": "$$\\mathrm{V}$$"
}
],
"answer": "$$\\frac{3 \\mathrm{~V}}{2}$$",
"solution": "**Answer:** $$\\frac{3 \\mathrm{~V}}{2}$$\n\nThe gravitational potential (V) due to a sphere of uniform density at a distance r from its center is given by:\n
The gravitational potential at a point above the surface of earth is $$-5.12 \\times 10^7 \\mathrm{~J} / \\mathrm{kg}$$ and the acceleration due to gravity at that point is $$6.4 \\mathrm{~m} / \\mathrm{s}^2$$. Assume that the mean radius of earth to be $$6400 \\mathrm{~km}$$. The height of this point above the earth's surface is :
If $$\\mathrm{G}$$ be the gravitational constant and $$\\mathrm{u}$$ be the energy density then which of the following quantity have the dimensions as that of the $$\\sqrt{\\mathrm{uG}}$$ :
",
"options": [
{
"text": "Gravitational potential\n"
},
{
"text": "pressure gradient per unit mass\n"
},
{
"text": "Energy per unit mass\n"
},
{
"text": "Force per unit mass"
}
],
"answer": "Force per unit mass",
"solution": "**Answer:** Force per unit mass\n\n
To determine the dimension of the quantity $$\\sqrt{uG}$$, we first need to understand the dimensions of both the gravitational constant (G) and the energy density (u).
\n\n
The gravitational constant $$G$$ has dimensions given by:\n\n
$$[G] = M^{-1}L^{3}T^{-2}$$
\n\n
where $$M$$ stands for mass, $$L$$ for length, and $$T$$ for time.
\n\n
Energy density $$u$$ is defined as the energy per unit volume. Since energy has dimensions of $$ML^{2}T^{-2}$$ (from the dimension of work or energy, which is force times distance, and force itself has dimension $$MLT^{-2}$$), and volume has dimensions of $$L^{3}$$, the dimensions of energy density would be:\n\n
$$[u] = \\frac{ML^{2}T^{-2}}{L^{3}} = M L^{-1} T^{-2}$$
\n\n
Now, we find the dimensions of $$\\sqrt{uG}$$ by multiplying the dimensions of $$u$$ and $$G$$, and then taking the square root:
So, the dimension of $$\\sqrt{uG}$$ is $$LT^{-2}$$, which corresponds to acceleration (length per square time).
\n\n
Now, let's match this with the provided options:
\n\n
\n\n
Option A (Gravitational potential) has dimensions of $$[L^{2}T^{-2}]$$, not matching our target of $$LT^{-2}$$.
\n\n
Option B (Pressure gradient per unit mass) would have dimensions of $$[M^{-1}L^{-2}T^{-2}][L^{-1}]$$ ($$Pressure\\ Gradient = \\frac{Pressure}{Length} = \\frac{ML^{-1}T^{-2}}{L}$$, and then divided by mass, $$M$$), which simplifies to $$L^{-3}T^{-2}M^{-1}$$, not matching.
\n\n
Option C (Energy per unit mass) has dimensions $$ML^{2}T^{-2}M^{-1}$$ which simplifies to $$L^{2}T^{-2}$$, also not a match for the target dimension.
\n\n
Option D (Force per unit mass) has dimensions $$MLT^{-2}M^{-1}$$ which simplifies directly to $$LT^{-2}$$, an exact match for our target dimension.
\n\n
\n\n
Thus, the correct answer is Option D (Force per unit mass), which has the same dimensions as that of $$\\sqrt{\\mathrm{uG}}$$.
To project a body of mass $$m$$ from earth's surface to infinity, the required kinetic energy is (assume, the radius of earth is $$R_E, g=$$ acceleration due to gravity on the surface of earth):
",
"options": [
{
"text": "$$1 / 2 m g R_E$$\n"
},
{
"text": "$$4 m g R_E$$\n"
},
{
"text": "$$m g R_E$$\n"
},
{
"text": "$$2 m g R_E$$"
}
],
"answer": "$$m g R_E$$\n",
"solution": "**Answer:** $$m g R_E$$\n\n\n
The kinetic energy required to project a body of mass $m$ from the Earth's surface to infinity, also known as the escape kinetic energy, can be calculated using the concept of gravitational potential energy. The escape velocity $v_e$ is the velocity a body must have to escape the gravitational field of the Earth without any further propulsion. The formula for escape velocity is:
\n\n
$v_e = \\sqrt{2gR_E}$
\n\n
Where $g$ is the acceleration due to gravity on the surface of Earth and $R_E$ is the radius of the Earth. The kinetic energy $K$ required for this is given by:
\n\n
$K = \\frac{1}{2}mv_e^2$
\n\n
Substituting the escape velocity formula into the kinetic energy formula gives:
\n\n
$K = \\frac{1}{2}m\\left(2gR_E\\right)$
\n\n
$K = \\frac{1}{2} \\times 2 \\times mgR_E$
\n\n
$K = mgR_E$
\n\n
Therefore, the required kinetic energy to project a body of mass $m$ from Earth's surface to infinity is $mgR_E$. So, the correct answer is:
\n\n
Option C: $mgR_E$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9610,
"subject": "Physics",
"question": "If suddenly the gravitational force of attraction between Earth and a satellite revolving around it becomes zero, then the satellite will ",
"options": [
{
"text": "continue to move in its orbit with same velocity"
},
{
"text": "move tangentially to the original orbit with the same velocity"
},
{
"text": "become stationary in its orbit"
},
{
"text": "move towards the earth"
}
],
"answer": "move tangentially to the original orbit with the same velocity",
"solution": "**Answer:** move tangentially to the original orbit with the same velocity\n\nWhen gravitational force of attraction between Earth and a satellite revolving around it becomes zero, then the centripetal force becomes zero. So the satellite will move tangentially to the original orbit with the same velocity as it has at the instant when gravitational force becomes zero.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9611,
"subject": "Physics",
"question": "The time period of satellite of earth is $$5$$ hours. If the separation between the earth and the satellite is increased to $$4$$ times the previous value, the new time period will become ",
"options": [
{
"text": "$$10$$ hours "
},
{
"text": "$$80$$ hours "
},
{
"text": "$$40$$ hours "
},
{
"text": "$$20$$ hours "
}
],
"answer": "$$40$$ hours ",
"solution": "**Answer:** $$40$$ hours \n\nAccording to kepler's law,\n
$$ = 5 \\times {2^3} = 40\\,\\,$$ hour\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9612,
"subject": "Physics",
"question": "Two spherical bodies of mass $$M$$ and $$5M$$ & radii $$R$$ & $$2R$$ respectively are released in free space with initial separation between their centers equal to $$12R$$. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is ",
"options": [
{
"text": "$$2.5$$ $$R$$ "
},
{
"text": "$$4.5$$ $$R$$ "
},
{
"text": "$$7.5$$ $$R$$ "
},
{
"text": "$$1.5$$ $$R$$ "
}
],
"answer": "$$7.5$$ $$R$$ ",
"solution": "**Answer:** $$7.5$$ $$R$$ \n\n \n
Let $$t$$ be the time taken for the two masses to collide and $${x_{5M,}}\\,{x_M}$$ be the distance travelled by the mass $$5M$$ and $$M$$ respectively.\n
The gravitational force acting between two sphere when the distance between them (12R - x) where x is a variable,\n
So two sphere collide when the sphere of mass M covered the distance of 7.5R.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9613,
"subject": "Physics",
"question": "The time period of an earth satellite in circular orbit is independent of ",
"options": [
{
"text": "both the mass and radius of the orbit"
},
{
"text": "radius of its orbit "
},
{
"text": "the mass of the satellite "
},
{
"text": "neither the mass of the satellite nor the radius of its orbit "
}
],
"answer": "the mass of the satellite ",
"solution": "**Answer:** the mass of the satellite \n\nFor satellite, gravitational force = centripetal force\n
which is independent of mass of satellite",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9614,
"subject": "Physics",
"question": "A satellite of mass $$m$$ revolves around the earth of radius $$R$$ at a height $$x$$ from its surface. If $$g$$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is ",
"options": [
{
"text": "$${{g{R^2}} \\over {R + x}}$$ "
},
{
"text": "$${{gR} \\over {R - x}}$$ "
},
{
"text": "$${gx}$$ "
},
{
"text": "$${\\left( {{{g{R^2}} \\over {R + x}}} \\right)^{1/2}}$$ "
}
],
"answer": "$${\\left( {{{g{R^2}} \\over {R + x}}} \\right)^{1/2}}$$ ",
"solution": "**Answer:** $${\\left( {{{g{R^2}} \\over {R + x}}} \\right)^{1/2}}$$ \n\nGravitational force applied on the satellite,\n
$$\\therefore$$ $$g \\propto \\rho $$ or $$\\rho \\propto g$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9616,
"subject": "Physics",
"question": "This question contains Statement - $$1$$ and Statement - $$2$$. of the four choices given after the statements, choose the one that best describes the two statements.\n
Statement - $$1$$:\n
For a mass $$M$$ kept at the center of a cube of side $$'a'$$, the flux of gravitational field passing through its sides $$4\\,\\pi \\,GM.$$ \n
Statement - 2:\n
If the direction of a field due to a point source is radial and its dependence on the distance $$'r'$$ from the source is given as $${1 \\over {{r^2}}},$$ its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface. ",
"options": [
{
"text": "Statement - $$1$$ is false, Statement - $$2$$ is true"
},
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is a correct explanation for Statement - $$1$$"
},
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is not a correct explanation for Statement - $$1$$"
},
{
"text": "Statement - $$1$$ is true, Statement - $$2$$ is false"
}
],
"answer": "Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is a correct explanation for Statement - $$1$$",
"solution": "**Answer:** Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is a correct explanation for Statement - $$1$$\n\nGravitational field $$\\overrightarrow g $$ = $$ - {{GM} \\over {{r^2}}}$$\n
where, $$M=$$ mass enclosed in the closed surface\n
Gravitational flux through a closed surface is given by \n
$${\\left| {\\overrightarrow g .d\\overrightarrow S } \\right|}$$ = $$4\\pi {r^2}.{{GM} \\over {{r^2}}}$$ = $$4\\pi GM$$ \n
So Statement - 1 is correct.\n
Statement - 2 is also correct because when the shape of the earth is spherical, area of the Gaussian surface is $$4\\pi {r^2}$$. This proves inverse square law.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9617,
"subject": "Physics",
"question": "Four particles, each of mass $$M$$ and equidistant from each other, move along a circle of radius $$R$$ under the action of their mutual gravitational attraction. The speed of each particle is : ",
"options": [
{
"text": "$$\\sqrt {{{GM} \\over R}} $$ "
},
{
"text": "$$\\sqrt {2\\sqrt 2 {{GM} \\over R}} $$ "
},
{
"text": "$$\\sqrt {{{GM} \\over R}\\left( {1 + 2\\sqrt 2 } \\right)} $$ "
},
{
"text": "$${1 \\over 2}\\sqrt {{{GM} \\over R}\\left( {1 + 2\\sqrt 2 } \\right)} $$ "
}
],
"answer": "$${1 \\over 2}\\sqrt {{{GM} \\over R}\\left( {1 + 2\\sqrt 2 } \\right)} $$ ",
"solution": "**Answer:** $${1 \\over 2}\\sqrt {{{GM} \\over R}\\left( {1 + 2\\sqrt 2 } \\right)} $$ \n\n\nAll those particles are moving due to their mutual gravitational attraction.\n
The force between each masses are repulsive force.\n
On mass M at C, due to mass at D the repulsive force is F in the vertical direction.\n
On mass M at C, due to mass at B the repulsive force is F in the horizontal direction.\n
On mass M at C, due to mass at A the repulsive force is F'\n
Net force acting on particle at C,\n
= $$2F\\,\\cos \\,{45^ \\circ } + F'$$\n
Where $$F = {{G{M^2}} \\over {{{\\left( {\\sqrt 2 R} \\right)}^2}}}$$ and $$F' = {{G{M^2}} \\over {4{R^2}}}$$ \n
$$ = {1 \\over 2}\\sqrt {{{Gm} \\over R}\\left( {1 + 2\\sqrt 2 } \\right)} $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9618,
"subject": "Physics",
"question": "A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely\nproportional to the nth power of R. If the period of rotation of the particle is T, then :",
"options": [
{
"text": "T $$ \\propto $$ Rn/2 "
},
{
"text": "T $$ \\propto $$ R3/2 for any n "
},
{
"text": "T $$ \\propto $$ Rn/2 +1"
},
{
"text": "T $$ \\propto $$ R(n+1)/2"
}
],
"answer": "T $$ \\propto $$ R(n+1)/2",
"solution": "**Answer:** T $$ \\propto $$ R(n+1)/2\n\nWe know, Central force in circular motion, F = $$m{\\omega ^2}R$$\n
$$T \\propto {R^{{{n + 1} \\over 2}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9619,
"subject": "Physics",
"question": "A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius $${R \\over 2},$$ and the other mass, in a circular orbit of radius $${3R \\over 2}$$. The difference between the final and initial total energies is : ",
"options": [
{
"text": "$$ - {{GMm} \\over {2R}}$$ "
},
{
"text": "$$ + {{GMm} \\over {6R}}$$"
},
{
"text": "$${{GMm} \\over {2R}}$$"
},
{
"text": "$$ - {{GMm} \\over {6R}}$$"
}
],
"answer": "$$ - {{GMm} \\over {6R}}$$",
"solution": "**Answer:** $$ - {{GMm} \\over {6R}}$$\n\nInitially gravitational potenrial energy \n
= $$-$$ $${{GMm} \\over {6R}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9620,
"subject": "Physics",
"question": "Take the mean distance of the moon and the sun from the earth to be $$0.4 \\times {10^6}$$ km and $$150 \\times {10^6}$$ km respectively. Their masses are $$8 \\times {10^{22}}$$ kg and $$2 \\times {10^{30}}$$ kg respectively. The radius of the earth is $$6400$$ km. Let $$\\Delta {F_1}$$ be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and $$\\Delta {F_2}$$ be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to $${{\\Delta {F_1}} \\over {\\Delta {F_2}}}$$ is :",
"options": [
{
"text": "$$2$$"
},
{
"text": "$${10^{ - 2}}$$"
},
{
"text": "$$0.6$$"
},
{
"text": "$$6$$"
}
],
"answer": "$$2$$",
"solution": "**Answer:** $$2$$\n\nAs gravitational force of attraction, \n
F = $${{GMm} \\over {{R^2}}}$$\n
$$\\therefore\\,\\,\\,\\,$$ Force of attraction berween earth and moon\n
= 2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9621,
"subject": "Physics",
"question": "The ratio of the weights of a body on the Earth’s surface to that on the surface of a planets is 9 : 4. The mass\nof the planet is\n$${1 \\over 9}$$\nth of that of the Earth. If 'R' is the radius of the Earth, what is the radius of the planet ?\n(Take the planets to have the same mass density)",
"options": [
{
"text": "$${R \\over 9}$$"
},
{
"text": "$${R \\over 2}$$"
},
{
"text": "$${R \\over 3}$$"
},
{
"text": "$${R \\over 4}$$"
}
],
"answer": "$${R \\over 2}$$",
"solution": "**Answer:** $${R \\over 2}$$\n\nW = mg\n
$${R_p} = {{{R_E}} \\over 2}$$ = $${R \\over 2}$$ [Here RE = R ]",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9622,
"subject": "Physics",
"question": "A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational\nfield of the planet acts on the spaceship, what will be the number of complete revolutions made by the\nspaceship in 24 hours around the planet?\n
= GM$$\\left[ {A\\left( {{1 \\over a} - {1 \\over {a + L}}} \\right) + BL} \\right]$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9625,
"subject": "Physics",
"question": "A body A of mass m is moving in a circular orbit\nof radius R about a planet. Another body B of\nmass\n$${m \\over 2}$$\ncollides with A with a velocity which is half $$\\left( {{{\\overrightarrow v } \\over 2}} \\right)$$ the instantaneous velocity$${\\overrightarrow v }$$\nof A.\nThe collision is completely inelastic. Then, the\ncombined body :",
"options": [
{
"text": "starts moving in an elliptical orbit around\nthe planet."
},
{
"text": "Falls vertically downwards towards the\nplanet"
},
{
"text": "Escapes from the Planet's Gravitational field."
},
{
"text": "continues to move in a circular orbit"
}
],
"answer": "starts moving in an elliptical orbit around\nthe planet.",
"solution": "**Answer:** starts moving in an elliptical orbit around\nthe planet.\n\nOrbital speed for of A is v = $$\\sqrt {{{GM} \\over R}} $$\n
After collision, let the combined mass moves\nwith speed v1\n
Since after collision, the speed is not equal to\norbital speed at that point. So motion cannot be\ncircular. Since velocity will remain tangential,\nso it cannot fall vertically towards the planet.\nTheir speed after collision is less than escape\nspeed $$\\sqrt 2 v$$, so they cannot escape\ngravitational field.\nSo their motion will be elliptical around the\nplanet.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9626,
"subject": "Physics",
"question": "A satellite is in an elliptical orbit around a planet P. It is observed that the velocity of the satellite\nwhen it is farthest from the planet is 6 times less than that when it is closest to the planet. The\nratio of distances between the satellite and the planet at closest and farthest points is:",
"options": [
{
"text": "1 : 2"
},
{
"text": "1 : 3"
},
{
"text": "1 : 6"
},
{
"text": "3 : 4"
}
],
"answer": "1 : 6",
"solution": "**Answer:** 1 : 6\n\n\nBy angular momentum conservation\n
Li\n = Lf\n
rminvmax = rmaxvmin .....(i)\n
Given, vmax = 6vmin\n
From equation (i),\n
$${{{r_{\\min }}} \\over {{r_{\\max }}}}$$ = $${{{v_{\\min }}} \\over {{v_{\\max }}}} = {1 \\over 6}$$",
"topic": "Geometry",
"subtopic": "Non-Euclidean Geometry"
},
{
"id": 9627,
"subject": "Physics",
"question": "Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be :",
"options": [
{
"text": "$$\\sqrt {{G \\over 2}(1 + 2\\sqrt 2 )} $$"
},
{
"text": "$$\\sqrt {{G \\over 2}(2\\sqrt 2 - 1)} $$"
},
{
"text": "$$\\sqrt {G(1 + 2\\sqrt 2 )} $$"
},
{
"text": "$${1\\over2}\\sqrt {G(1 + 2\\sqrt 2 )} $$"
}
],
"answer": "$${1\\over2}\\sqrt {G(1 + 2\\sqrt 2 )} $$",
"solution": "**Answer:** $${1\\over2}\\sqrt {G(1 + 2\\sqrt 2 )} $$\n\n Given, m = 1 kg, R = 1 m
B. has the least velocity when it is nearest to the sun.
C. its areal velocity is directly proportional to its velocity.
D. areal velocity is inversely proportional to its velocity.
E. to follow a trajectory such that the areal velocity is constant.
Choose the correct answer from the options given below :",
"options": [
{
"text": "D only"
},
{
"text": "E only"
},
{
"text": "C only"
},
{
"text": "A only"
}
],
"answer": "E only",
"solution": "**Answer:** E only\n\n
According to Kepler’s second law of planetary motion, areal velocity of every planet moving around the sun should remain constant in elliptical orbit.
",
"topic": "Geometry",
"subtopic": "Non-Euclidean Geometry"
},
{
"id": 9630,
"subject": "Physics",
"question": "The maximum and minimum distances of a comet from the Sun are 1.6 $$\\times$$ 1012 m and 8.0 $$\\times$$ 1010 m respectively. If the speed of the comet at the nearest point is 6 $$\\times$$ 104 ms$$-$$1, the speed at the farthest point is :",
"options": [
{
"text": "3.0 $$\\times$$ 103 m/s"
},
{
"text": "6.0 $$\\times$$ 103 m/s"
},
{
"text": "1.5 $$\\times$$ 103 m/s"
},
{
"text": "4.5 $$\\times$$ 103 m/s"
}
],
"answer": "3.0 $$\\times$$ 103 m/s",
"solution": "**Answer:** 3.0 $$\\times$$ 103 m/s\n\n\n v1 = 6 $$\\times$$ 104 m/s
Let point 1 is nearest point,
and point 2 is farthest point.
Given, r1 = 8 $$\\times$$ 1010 m & r2 = 1.6 $$\\times$$ 1012 m
$$ \\Rightarrow $$ v2 = 3.0 $$\\times$$ 103 m/s",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9631,
"subject": "Physics",
"question": "The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is :",
"options": [
{
"text": "9 T"
},
{
"text": "27 T"
},
{
"text": "12 T"
},
{
"text": "3 T"
}
],
"answer": "27 T",
"solution": "**Answer:** 27 T\n\n
Kepler's Third Law states that the square of the period of a satellite's orbit is proportional to the cube of the semi-major axis of its orbit. This relationship can be written as:
\n
$$T^2 \\propto r^3$$
\n
where:
\n
\n
$T$ is the orbital period
\n
$r$ is the radius of the circular orbit (which serves as the semi-major axis in this case)
\n
\n
Considering two satellites, one with period $T$ and radius $R$, and another with unknown period $T'$ and radius $9R$, we can form an equation:
\n
$$\\frac{{T'}^2}{T^2} = \\frac{(9R)^3}{R^3}$$
\n
This simplifies to:
\n
$$\\frac{{T'}^2}{T^2} = 729$$
\n
Taking the square root of both sides, we get:
\n
$$T' = T \\times \\sqrt{729} = T \\times 27$$
\n
So, the period of another satellite in a circular orbit of radius $9R$ is $27T$.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9632,
"subject": "Physics",
"question": "The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is $${\\overrightarrow L }$$. The magnitude of the areal velocity of the planet is :",
"options": [
{
"text": "$${{2L} \\over M}$$"
},
{
"text": "$${{L} \\over 2M}$$"
},
{
"text": "$${{L} \\over M}$$"
},
{
"text": "$${{4L} \\over M}$$"
}
],
"answer": "$${{L} \\over 2M}$$",
"solution": "**Answer:** $${{L} \\over 2M}$$\n\n\n\n Gravitational force line passes through the sun so torque about sun always zero for the planet.
$$ \\therefore $$ Angular momentum about sum is constant.
$$\\overrightarrow L $$ = Constant
we know, L = M$${v_ \\bot }$$ r
Now,
$$dA = {1 \\over 2}$$ $$\\times$$ base $$\\times$$ height
$$ \\Rightarrow {{dA} \\over {dt}} = {{r{v_ \\bot }} \\over 2} = {r \\over 2} \\times {L \\over {Mr}} = {L \\over {2M}}$$",
"topic": "Geometry",
"subtopic": "Non-Euclidean Geometry"
},
{
"id": 9633,
"subject": "Physics",
"question": "Consider a binary star system of star A and star B with masses mA and mB revolving in a circular orbit of radii rA an rB, respectively. If TA and TB are the time period of star A and star B, respectively,
In a binary star system, the two stars orbit around a common center of mass. When considering periods of revolution, Kepler's Third Law comes into play. This law states that the square of the period of revolution (T) is proportional to the cube of the semi-major axis (r) of the orbit. It's often written in the following form for a single object orbiting another:
\n
T² ∝ r³
\n
For a binary star system, this would still hold true. The periods of revolution for both stars A and B will be the same because they are both orbiting the same common center of mass, regardless of their individual masses or individual orbital radii. In other words, star A and star B complete one orbit in the same amount of time.
\n
So, Option B: TA = TB is correct.
\n
The other options (Option A, C, and D) would not be correct. Kepler's Third Law is not a ratio between the periods and radii of two different bodies, and the periods do not depend on the masses of the individual stars or their individual distances from the center of mass.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9634,
"subject": "Physics",
"question": "The minimum and maximum distances of a planet revolving around the sun are x1 and x2. If the minimum speed of the planet on its trajectory is v0 then its maximum speed will be :",
"options": [
{
"text": "$${{{v_0}x_1^2} \\over {x_2^2}}$$"
},
{
"text": "$${{{v_0}x_2^2} \\over {x_1^2}}$$"
},
{
"text": "$${{{v_0}x_1^{}} \\over {x_2^{}}}$$"
},
{
"text": "$${{{v_0}x_2^{}} \\over {x_1^{}}}$$"
}
],
"answer": "$${{{v_0}x_2^{}} \\over {x_1^{}}}$$",
"solution": "**Answer:** $${{{v_0}x_2^{}} \\over {x_1^{}}}$$\n\nAngular momentum conservation equation $${v_0}{x_2} = {v_1}{x_1}$$
$${v_1} = {{{v_0}{x_2}} \\over {{x_1}}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9635,
"subject": "Physics",
"question": "Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is :",
"options": [
{
"text": "$$\\sqrt {{G \\over {2{R^3}}}} $$"
},
{
"text": "$${1 \\over 2}\\sqrt {{G \\over {{R^3}}}} $$"
},
{
"text": "$${1 \\over {2R}}\\sqrt {{1 \\over G}} $$"
},
{
"text": "$${{2G} \\over {{R^3}}}$$"
}
],
"answer": "$${1 \\over 2}\\sqrt {{G \\over {{R^3}}}} $$",
"solution": "**Answer:** $${1 \\over 2}\\sqrt {{G \\over {{R^3}}}} $$\n\n
The problem describes two identical particles of mass m=1kg each moving in a circle of radius R under the action of their mutual gravitational attraction. This means that the particles are moving around a common center, and the distance between the particles is 2R (as the diameter of the circle).
\n
In this case, the force providing the centripetal force for each particle to move in a circular path is the gravitational force between the particles.
\n
The gravitational force between two masses m1 and m2 separated by a distance r is given by Newton's law of universal gravitation:
\n
$$F = G \\frac{{m_1 m_2}}{{r^2}}$$
\n
Since the two particles are identical, m1=m2=m=1kg. And the distance between them r is 2R. Substituting these into the above equation gives the gravitational force between the two particles:
\n
$$F = G \\frac{{m^2}}{{(2R)^2}} = G \\frac{{1}}{{4R^2}}$$
\n
This gravitational force is also equal to the centripetal force needed for each particle to move in a circular path of radius R. The centripetal force is given by:
\n
$$F = m R ω^2$$
\n
Setting these two equations equal to each other gives:
\n
$$G \\frac{{1}}{{4R^2}} = 1 \\cdot R \\cdot ω^2$$
\n
Rearranging this to solve for ω (the angular speed) gives:
\n
$$ω = \\frac{1}{2} \\sqrt{\\frac{G}{R^3}}$$
\n
So, the angular speed of each particle is: $\\frac{1}{2} \\sqrt{\\frac{G}{R^3}}$.
\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9636,
"subject": "Physics",
"question": "Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hours respectively. The radius of the orbit of nearer satellite is 2 $$\\times$$ 103 km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $${\\pi \\over x}rad\\,{h^{ - 1}}$$ where x is ____________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n T1 = 1 hour
The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three times its previous value, then approximate new time period of the satellite will be
Two objects of equal masses placed at certain distance from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other object, then the new force will be :
Two planets A and B of equal mass are having their period of revolutions TA and TB such that TA = 2TB. These planets are revolving in the circular orbits of radii rA and rB respectively. Which out of the following would be the correct relationship of their orbits?
The distance of the Sun from earth is 1.5 $$\\times$$ 1011 m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be :
Three identical particles $$\\mathrm{A}, \\mathrm{B}$$ and $$\\mathrm{C}$$ of mass $$100 \\mathrm{~kg}$$ each are placed in a straight line with $$\\mathrm{AB}=\\mathrm{BC}=13 \\mathrm{~m}$$. The gravitational force on a fourth particle $$\\mathrm{P}$$ of the same mass is $$\\mathrm{F}$$, when placed at a distance $$13 \\mathrm{~m}$$ from the particle $$\\mathrm{B}$$ on the perpendicular bisector of the line $$\\mathrm{AC}$$. The value of $$\\mathrm{F}$$ will be approximately :
The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.
Two particles of equal mass '$$m$$' move in a circle of radius '$$r$$' under the action of their mutual gravitational attraction. The speed of each particle will be :
Every planet revolves around the sun in an elliptical orbit :-
\n
A. The force acting on a planet is inversely proportional to square of distance from sun.
\n
B. Force acting on planet is inversely proportional to product of the masses of the planet and the sun.
\n
C. The Centripetal force acting on the planet is directed away from the sun.
\n
D. The square of time period of revolution of planet around sun is directly proportional to cube of semi-major axis of elliptical orbit.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "C and D only"
},
{
"text": "B and C only"
},
{
"text": "A and D only"
},
{
"text": "A and C only"
}
],
"answer": "A and D only",
"solution": "**Answer:** A and D only\n\n
A. The force acting on a planet is inversely proportional to the square of the distance from the sun. This is known as the inverse square law and is described by Newton's law of gravitation.
\n\n
B. Force acting on a planet is inversely proportional to the product of the masses of the planet and the sun. This is also described by Newton's law of gravitation, but it is not directly related to the planet's elliptical orbit.
\n
\nC. The centripetal force acting on the planet is directed towards the sun, not away from it.
\n\n
D. The square of the time period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of the elliptical orbit. This is known as Kepler's third law.
\n\nTherefore, only A and D are correct.",
"topic": "Geometry",
"subtopic": "Non-Euclidean Geometry"
},
{
"id": 9646,
"subject": "Physics",
"question": "
If the distance of the earth from Sun is 1.5 $$\\times$$ 10$$^6$$ km. Then the distance of an imaginary planet from Sun, if its period of revolution is 2.83 years is :
",
"options": [
{
"text": "$$6\\times10^6$$ km"
},
{
"text": "$$3\\times10^7$$ km"
},
{
"text": "$$6\\times10^7$$ km"
},
{
"text": "$$3\\times10^6$$ km"
}
],
"answer": "$$3\\times10^6$$ km",
"solution": "**Answer:** $$3\\times10^6$$ km\n\nWe can use Kepler's third law to solve this problem. Kepler's third law states that the square of the period of revolution of a planet around the Sun is proportional to the cube of its average distance from the Sun. Let $T$ be the period of revolution of the imaginary planet in years, and let $d$ be its average distance from the Sun in kilometers. We can use the following equation to solve for $d$:\n
\nwhere $T_1$ and $d_1$ are the period of revolution and an average distance of the Earth from the Sun, respectively, and $T_2$ is the period of revolution of the imaginary planet.\n
\nSubstituting the given values, we get:
\n$$\n\\begin{aligned}\n& \\frac{T_{1}}{T_{2}}=\\left(\\frac{d_{1}}{d_{2}}\\right)^{\\frac{3}{2}} \\\\\\\\\n& \\Rightarrow\\left(\\frac{1 \\text { year }}{2.83 \\text { year }}\\right)^{\\frac{2}{3}}=\\left(\\frac{1.5 \\times 10^{6} \\mathrm{~km}}{d_{2}}\\right) \\\\\\\\\n& \\Rightarrow \\frac{1}{2}=\\frac{1.5 \\times 10^{6} \\mathrm{~km}}{d_{2}} \\\\\\\\\n& d_{2}=3 \\times 10^{6} \\mathrm{~km}\n\\end{aligned}\n$$\n\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9647,
"subject": "Physics",
"question": "Two identical particles each of mass ' $m$ ' go round a circle of radius $a$ under the action of their mutual gravitational attraction. The angular speed of each particle will be :",
"options": [
{
"text": "$\\sqrt{\\frac{G m}{2 a^{3}}}$"
},
{
"text": "$\\sqrt{\\frac{G m}{a^{3}}}$\n"
},
{
"text": "$\\sqrt{\\frac{G m}{8 a^{3}}}$"
},
{
"text": "$\\sqrt{\\frac{G m}{4 a^{3}}}$"
}
],
"answer": "$\\sqrt{\\frac{G m}{4 a^{3}}}$",
"solution": "**Answer:** $\\sqrt{\\frac{G m}{4 a^{3}}}$\n\nThe gravitational force between two particles of mass $m$ separated by a distance $r$ is given by:\n
\n$$\nF = \\frac{Gm^2}{r^2}\n$$\n
\nwhere $G$ is the gravitational constant. In this problem, the two particles are moving in a circular orbit of radius $a$ under the influence of their mutual gravitational attraction. Therefore, the gravitational force between the two particles provides the necessary centripetal force to keep them in circular motion.\n
\nThe centripetal force required for a particle of mass $m$ moving in a circle of radius $a$ with angular speed $\\omega$ is given by:\n
\n$$\nF_{\\text{centripetal}} = m\\omega^2a\n$$\n
\nSetting the gravitational force equal to the centripetal force, we get:\n
\n$$\n\\frac{Gm^2}{r^2} = m\\omega^2a\n$$\n
\nSubstituting $r = 2a$ (since the two particles are separated by a distance equal to twice the radius of the circle), we get:\n
\n$$\n\\frac{Gm^2}{(2a)^2} = m\\omega^2a\n$$\n
\nSimplifying, we get:\n
\n$$\n\\omega^2 = \\frac{Gm}{4a^3}\n$$\n
\nTaking the square root of both sides, we get:\n
\n$$\n\\omega = \\sqrt{\\frac{Gm}{4a^3}}\n$$\n
\nTherefore, the angular speed of each particle is $\\sqrt{\\frac{Gm}{4a^3}}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9648,
"subject": "Physics",
"question": "
If the earth suddenly shrinks to $$\\frac{1}{64}$$th of its original volume with its mass remaining the same, the period of rotation of earth becomes $$\\frac{24}{x}$$h. The value of x is __________.
So, if the Earth suddenly shrinks to ( $\\frac{1}{64}$ )th of its original volume with its mass remaining the same, the period of rotation of Earth becomes ( $\\frac{24}{16}$ )h, or 1.5 hours. Therefore, the value of (x) is 16.
A planet has double the mass of the earth. Its average density is equal to that of the earth. An object weighing $$\\mathrm{W}$$ on earth will weigh on that planet:
The weight of an object on a planet is given by the equation $W = mg$, where $m$ is the mass of the object and $g$ is the acceleration due to gravity.
\n
The acceleration due to gravity on a planet is given by the equation $g = \\frac{GM}{R^2}$, where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet.
\n
In this case, the mass of the planet is double that of Earth ($M = 2M_E$), but the density is the same. Density is defined as mass divided by volume, so if the mass is doubled and the density stays the same, the volume must also double.
\n
Since the volume of a sphere (like a planet) is given by the equation $V = \\frac{4}{3}\\pi R^3$, a doubling of the volume implies that the radius of the planet is $R = 2^{1/3}R_E$.
\n
Substituting these values back into the equation for $g$, we get:
\n
$g_{\\text{planet}} = G * \\frac{2M_E}{(2^{1/3}R_E)^2} = 2^{1/3} * g_E$
\n
So the weight of the object on the planet is $W_{\\text{planet}} = m*g_{\\text{planet}} = m2^{1/3} * g_E = 2^{1/3}W_E$
\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9650,
"subject": "Physics",
"question": "A light planet is revolving around a massive star in a circular orbit of radius $\\mathrm{R}$ with a period of revolution T. If the force of attraction between planet and star is proportional to $\\mathrm{R}^{-3 / 2}$ then choose the correct option :",
"options": [
{
"text": "$\\mathrm{T}^2 \\propto \\mathrm{R}^{7 / 2}$"
},
{
"text": "$\\mathrm{T}^2 \\propto \\mathrm{R}^3$"
},
{
"text": "$\\mathrm{T}^2 \\propto \\mathrm{R}^{5 / 2}$"
},
{
"text": "$\\mathrm{T}^2 \\propto \\mathrm{R}^{3 / 2}$"
}
],
"answer": "$\\mathrm{T}^2 \\propto \\mathrm{R}^{5 / 2}$",
"solution": "**Answer:** $\\mathrm{T}^2 \\propto \\mathrm{R}^{5 / 2}$\n\n
To find the correct option for the relationship between the period of revolution T and the radius of the orbit R, we will consider the force of attraction and its proportionality to $\\mathrm{R}^{-3 / 2}$.
\n
According to Newton's law of universal gravitation, the force of attraction $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by\n$$ F = \\frac{G m_1 m_2}{r^2}, $$\nwhere $G$ is the gravitational constant.
\n
However, in this particular case, the force of attraction is given to be proportional to $\\mathrm{R}^{-3 / 2}$, so we can write\n$$ F \\propto \\frac{1}{R^{3/2}}. $$
\n
Since the planet is in a circular orbit around the star, the centripetal force required to keep the planet in orbit must be provided by this gravitational force. Hence, we can write that\n$$ \\frac{m v^2}{R} \\propto \\frac{1}{R^{3/2}}, $$\nwhere $m$ is the mass of the planet and $v$ is its orbital speed.
\n
Simplifying this, we get\n$$ v^2 \\propto \\frac{1}{R^{1/2}}. $$
\n
Now, the speed $v$ can be related to the period T through the circumference of the orbit, which is given by $2\\pi R$. The orbital speed is the circumference divided by the period:\n$$ v = \\frac{2\\pi R}{T}. $$
\n
Substituting this into our proportionality, we get\n
Four identical particles of mass $$m$$ are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is $$\\left(\\frac{2 \\sqrt{2}+1}{32}\\right) \\frac{\\mathrm{Gm}^2}{L^2}$$, the length of the sides of the square is
A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution :
A metal wire of uniform mass density having length $$L$$ and mass $$M$$ is bent to form a semicircular arc and a particle of mass $$\\mathrm{m}$$ is placed at the centre of the arc. The gravitational force on the particle by the wire is :
Two satellite A and B go round a planet in circular orbits having radii 4R and R respectively. If the speed of $$\\mathrm{A}$$ is $$3 v$$, the speed of $$\\mathrm{B}$$ will be :
To solve this question, we will use the fact that for an object in a circular orbit, the centripetal force required to keep the object in orbit is provided by the gravitational force between the object and the planet it is orbiting. This principle gives us the relationship between the speed of the satellite, its orbital radius, and the mass of the planet.
\n\n
The formula for the gravitational force is given by:
\n\n
$$ F = \\frac{G M m}{r^2} $$
\n\n
where:
\n\n
\n\n
$F$ is the gravitational force between the two masses,
\n\n
$G$ is the gravitational constant,
\n\n
$M$ is the mass of the planet,
\n\n
$m$ is the mass of the satellite, and
\n\n
$r$ is the distance between the center of the planet and the satellite (radius of the orbit).
\n\n
\n\n
The centripetal force required to keep the satellite in orbit is given by:
\n\n
$$ F_c = \\frac{m v^2}{r} $$
\n\n
where:
\n\n
\n\n
$F_c$ is the centripetal force,
\n\n
$v$ is the orbital speed of the satellite, and
\n\n
$r$ is the radius of the orbit.
\n\n
\n\n
Since the gravitational force is providing the centripetal force, we can set the two forces equal to each other to find the relationship between speed and radius:
\n\n
$$ \\frac{G M m}{r^2} = \\frac{m v^2}{r} $$
\n\n
By simplifying this, we find the equation for the orbital speed of a satellite:
\n\n
$$ v = \\sqrt{\\frac{G M}{r}} $$
\n\n
Now, we can compare the speeds of satellites $A$ and $B$ based on their radii. Satellite $A$ orbits at a radius of $4R$ with a speed of $3v$, and we need to find the speed of satellite $B$ which orbits at a radius of $R$.
\n\n
Substituting the respective radii into the speed equation:
\n\n
\n\n
For satellite $A$, $v_A = 3v = \\sqrt{\\frac{G M}{4R}}$
\n\n
For satellite $B$, we want to find $v_B$ given that its radius is $R$: $v_B = \\sqrt{\\frac{G M}{R}}$
\n\n
\n\n
To find the ratio of $v_B$ to $3v$ (or the speed of $A$), we can write:
Two planets $$A$$ and $$B$$ having masses $$m_1$$ and $$m_2$$ move around the sun in circular orbits of $$r_1$$ and $$r_2$$ radii respectively. If angular momentum of $$A$$ is $$L$$ and that of $$B$$ is $$3 \\mathrm{~L}$$, the ratio of time period $$\\left(\\frac{T_A}{T_B}\\right)$$ is:
$$ \\Rightarrow $$$$\\,\\,\\,$$ n = 2 moles.\n\n\n\n\n\n\n\n\n\n\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9660,
"subject": "Physics",
"question": "A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. considering only translational and rotational modes, the total internal energy of the system is :",
"options": [
{
"text": "12 RT"
},
{
"text": "20 RT"
},
{
"text": "4 RT"
},
{
"text": "15 RT"
}
],
"answer": "15 RT",
"solution": "**Answer:** 15 RT\n\nU $$ = {{{f_1}} \\over 2}{n_1}RT + {{{f_2}} \\over 2}{n_2}RT$$\n
U $$ = 15RT$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9661,
"subject": "Physics",
"question": "An ideal gas occupies a volume of 2m3 at a pressure of 3 $$ \\times $$ 106 Pa. The energy of the gas is : ",
"options": [
{
"text": "6 $$ \\times $$ 104 J"
},
{
"text": "9$$ \\times $$ 106 J"
},
{
"text": "3 $$ \\times $$ 102 J"
},
{
"text": "108 J"
}
],
"answer": "9$$ \\times $$ 106 J",
"solution": "**Answer:** 9$$ \\times $$ 106 J\n\nEnergy = $${1 \\over 2}$$ nRT = $${f \\over 2}$$PV\n
= $${f \\over 2}$$ (3 $$ \\times $$ 106) (2)\n
= f $$ \\times $$ 3 $$ \\times $$ 106\n
Considering gas is monoatomic i.e. f = 3\n
E. = 9 $$ \\times $$ 106 J",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9662,
"subject": "Physics",
"question": "An HCl molecule has rotational, translational\nand vibrational motions. If the rms velocity of\nHCl molecules in its gaseous phase is $$\\overline v $$ , m is\nits mass and kB is Boltzmann constant, then its\ntemperature will be :",
"options": [
{
"text": "$${{m{{\\overline v }^2}} \\over {5{k_B}}}$$"
},
{
"text": "$${{m{{\\overline v }^2}} \\over {6{k_B}}}$$"
},
{
"text": "$${{m{{\\overline v }^2}} \\over {7{k_B}}}$$"
},
{
"text": "$${{m{{\\overline v }^2}} \\over {3{k_B}}}$$"
}
],
"answer": "$${{m{{\\overline v }^2}} \\over {7{k_B}}}$$",
"solution": "**Answer:** $${{m{{\\overline v }^2}} \\over {7{k_B}}}$$\n\n
An HCl molecule, being diatomic, has:
\n
\n
3 translational degrees of freedom
\n
2 rotational degrees of freedom
\n
2 vibrational degrees of freedom
\n
\n
The total number of degrees of freedom is $3 + 2 + 2 = 7$.
\n
According to the equipartition theorem, each degree of freedom contributes $\\frac{1}{2} k_B T$ to the total energy. So the total energy is given by:\n$\\frac{7}{2} k_B T$
\n
The translational kinetic energy is related to the root-mean-square (rms) speed $\\overline{v}$ by:\n$\\frac{1}{2} m \\overline{v}^2 = \\frac{7}{2} k_B T$
\n
Rearranging to solve for the temperature, we find:\n$T = \\frac{m \\overline{v}^2}{7k_B}$
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9663,
"subject": "Physics",
"question": "The specific heats, CP and CV of a gas of\ndiatomic molecules, A, are given (in units of\nJ mol–1 K–1) by 29 and 22, respectively.\nAnother gas of diatomic molecules, B, has the\ncorresponding values 30 and 21. If they are\ntreated as ideal gases, then :-",
"options": [
{
"text": "A is rigid but B has a vibrational mode"
},
{
"text": "A has a vibrational mode but B has none"
},
{
"text": "A has one vibrational mode and B has two"
},
{
"text": "Both A and B have a vibrational mode each"
}
],
"answer": "A has a vibrational mode but B has none",
"solution": "**Answer:** A has a vibrational mode but B has none\n\nFor A: \n$${{{C_p}} \\over {{C_v}}} = \\gamma = 1 + {2 \\over f} = {{29} \\over {22}}$$ \nIt gives f = 6.3 $$ \\approx $$ 6 (3 translational, 2 rotational and 1 vibrational)
\nFor B: \n$${{{C_p}} \\over {{C_v}}} = \\gamma = 1 + {2 \\over f} = {{30} \\over {21}}$$ \n$$ \\therefore $$ f = 4.67 $$ \\approx $$ 5 (3 translational, 2 rotational, no vibrational)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9664,
"subject": "Physics",
"question": "To raise the temperature of a certain mass of gas by 50oC at a constant pressure, 160 calories of\nheat is required. When the same mass of gas is cooled by 100oC at constant volume, 240 calories\nof heat is released. How many degrees of freedom does each molecule of this gas have (assume\ngas to be ideal)?",
"options": [
{
"text": "6"
},
{
"text": "7"
},
{
"text": "5"
},
{
"text": "3"
}
],
"answer": "6",
"solution": "**Answer:** 6\n\n$$160 = n{C_p}50$$ ....(i)
$$ \\Rightarrow f = 6$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9665,
"subject": "Physics",
"question": "Molecules of an ideal gas are known to have three translational degrees of freedom and two\nrotational degrees of freedom.The gas is maintained at a temperature of T. The total internal\nenergy, U of a mole of this gas, and the value of $$\\gamma \\left( { = {{{C_p}} \\over {{C_v}}}} \\right)$$ are given, respectively by:",
"options": [
{
"text": "U = $${5 \\over 2}RT$$ and $$\\gamma = {7 \\over 5}$$"
},
{
"text": "U = 5RT and $$\\gamma = {6 \\over 5}$$"
},
{
"text": "U = 5RT and $$\\gamma = {7 \\over 5}$$"
},
{
"text": "U = $${5 \\over 2}RT$$ and $$\\gamma = {6 \\over 5}$$"
}
],
"answer": "U = $${5 \\over 2}RT$$ and $$\\gamma = {7 \\over 5}$$",
"solution": "**Answer:** U = $${5 \\over 2}RT$$ and $$\\gamma = {7 \\over 5}$$\n\nTotal degree of freedom (f) = 3 + 2 = 5\n
U = $${{nfRT} \\over 2}$$ = $${{5RT} \\over 2}$$\n
$$\\gamma $$ = $${{{C_p}} \\over {{C_v}}}$$ = $$1 + {2 \\over f}$$ = $$1 + {2 \\over 5}$$ = $${7 \\over 5}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9666,
"subject": "Physics",
"question": "An engine takes in 5 moles of air at 20oC and\n1 atm, and compresses it adiabatically to\n1/10th of the original volume. Assuming air to\nbe a diatomic ideal gas made up of rigid\nmolecules, the change in its internal energy\nduring this process comes out to be X kJ. The\nvalue of X to the nearest integer is________.",
"options": [],
"answer": "46",
"solution": "**Answer:** 46\n\nFor diatomic ideal gas :\n
f = 5\n
$$\\gamma $$ = $${7 \\over 5}$$\n
Ti\n = T = 273 + 20 = 293 K\n
Vi\n = V\n
Vf = $${V \\over {10}}$$\n
For adiabatic process TV$$\\gamma $$ - 1 = constant\n
$$ \\simeq $$ 46 kJ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9667,
"subject": "Physics",
"question": "A gas mixture consists of 3 moles of oxygen\nand 5 moles of argon at temperature T.\nAssuming the gases to be ideal and the oxygen\nbond to be rigid, the total internal energy (in\nunits of RT) of the mixture is :",
"options": [
{
"text": "11"
},
{
"text": "20"
},
{
"text": "15"
},
{
"text": "13"
}
],
"answer": "15",
"solution": "**Answer:** 15\n\nU $$ = {{{f_1}} \\over 2}{n_1}RT + {{{f_2}} \\over 2}{n_2}RT$$\n
$$ \\therefore $$ U $$ = 15RT$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9668,
"subject": "Physics",
"question": "Match the $${{{C_P}} \\over {{C_V}}}$$ ratio for ideal gases with different type of molecules :\n
$${\\gamma _D} = 1 + {2 \\over 6} = 4/3$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9669,
"subject": "Physics",
"question": "Consider two ideal diatomic gases A and B at\nsome temperature T. Molecules of the gas A are\nrigid, and have a mass m. Molecules of the gas\nB have an additional vibrational mode, and\nhave a mass $${m \\over 4}$$\n. The ratio of the specific heats ($$C_V^A$$ and $$C_V^B$$ ) of gas A and B, respectively is :",
"options": [
{
"text": "7 : 9"
},
{
"text": "5 : 7"
},
{
"text": "3 : 5"
},
{
"text": "5 : 9"
}
],
"answer": "5 : 7",
"solution": "**Answer:** 5 : 7\n\nDegree of freedom of a diatomic molecule if\nvibration is absent = 5\n
Degree of freedom of a diatomic molecule if\nvibration is present = 7\n
Statement I : In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution.
Statement II : In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule.
In the light of the above statements, choose the correct answer from the options given below :",
"options": [
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Statement I is true but Statement II is false."
},
{
"text": "Statement I is false but Statement II is true."
}
],
"answer": "Statement I is true but Statement II is false.",
"solution": "**Answer:** Statement I is true but Statement II is false.\n\nThe translational kinetic energy & rotational kinetic energy both obey Maxwell's distribution independent of each other.
T.K.E. of diatomic molecules = $${3 \\over 2}kT$$
R.K.E. of diatomic molecules = $${2 \\over 2}kT$$
So statement I is true but statement II is false.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9672,
"subject": "Physics",
"question": "The internal energy (U), pressure (P) and volume (V) of an ideal gas are related as U $$=$$ 3PV + 4. The gas is :",
"options": [
{
"text": "either monoatomic or diatomic."
},
{
"text": "monoatomic only."
},
{
"text": "polyatomic only."
},
{
"text": "diatomic only."
}
],
"answer": "polyatomic only.",
"solution": "**Answer:** polyatomic only.\n\nU = 3PV + 4
Since degree of freedom is more than 6 therefore gas is polyatomic.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9673,
"subject": "Physics",
"question": "A polyatomic ideal gas has 24 vibrational modes. What is the value of $$\\gamma$$?",
"options": [
{
"text": "1.37"
},
{
"text": "1.30"
},
{
"text": "1.03"
},
{
"text": "10.3"
}
],
"answer": "1.03",
"solution": "**Answer:** 1.03\n\nf = 3T + 3R + 24V
= 30
$$\\gamma$$ = 1 + $${2 \\over f}$$
$$\\gamma$$ = 1 + $${2 \\over 30}$$
= 1.066
Nearest Ans. = 1.03",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9674,
"subject": "Physics",
"question": "Two ideal polyatomic gases at temperatures T1 and T2 are mixed so that there is no loss of energy. If F1 and F2, m1 and m2, n1 and n2 be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is :",
"options": [
{
"text": "$${{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \\over {{F_1} + {F_2}}}$$"
},
{
"text": "$${{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \\over {{n_1}{F_1} + {n_2}{F_2}}}$$"
},
{
"text": "$${{{n_1}{T_1} + {n_2}{T_2}} \\over {{n_1} + {n_2}}}$$"
},
{
"text": "$${{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \\over {{n_1} + {n_2}}}$$"
}
],
"answer": "$${{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \\over {{n_1}{F_1} + {n_2}{F_2}}}$$",
"solution": "**Answer:** $${{{n_1}{F_1}{T_1} + {n_2}{F_2}{T_2}} \\over {{n_1}{F_1} + {n_2}{F_2}}}$$\n\nInitial internal energy = Final internal energy
$$ \\Rightarrow $$ $$T = {{{F_1}{n_1}{T_1} + {F_2}{n_2}{T_2}} \\over {{F_1}{n_1} + {F_2}{n_2}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9675,
"subject": "Physics",
"question": "If one mole of the polyatomic gas is having two vibrational modes and $$\\beta$$ is the ratio of molar specific heats for polyatomic gas $$\\left( {\\beta = {{{C_P}} \\over {{C_V}}}} \\right)$$ then the value of $$\\beta$$ is :",
"options": [
{
"text": "1.02"
},
{
"text": "1.35"
},
{
"text": "1.2"
},
{
"text": "1.25"
}
],
"answer": "1.2",
"solution": "**Answer:** 1.2\n\nFor polyatomic gas molecule has 3 rotational degrees of freedom,\n3 translational degrees of freedom, and 2 vibrational modes.\n
So, number of vibrational degrees of freedom = 2 $$ \\times $$ 2 = 4\n
Degree of freedom of polyatomic gas
f = T + R + V
f = 3 + 3 + 4 = 10
$$\\beta = 1 + {2 \\over f} = 1 + {2 \\over 10}$$
$$\\beta = {{12} \\over 10} = 1.2$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9676,
"subject": "Physics",
"question": "What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature T? (kB is Boltzmann constant)",
"options": [
{
"text": "$${1 \\over 2}{k_B}T$$"
},
{
"text": "$${2 \\over 3}{k_B}T$$"
},
{
"text": "$${3 \\over 2}{k_B}T$$"
},
{
"text": "$${k_B}T$$"
}
],
"answer": "$${1 \\over 2}{k_B}T$$",
"solution": "**Answer:** $${1 \\over 2}{k_B}T$$\n\nEnergy associated with each digress of freedom is $${1 \\over 2}{k_B}T$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9677,
"subject": "Physics",
"question": "The correct relation between the degrees of freedom f and the ratio of specific heat $$\\gamma$$ is :",
"options": [
{
"text": "$$f = {2 \\over {\\gamma - 1}}$$"
},
{
"text": "$$f = {2 \\over {\\gamma + 1}}$$"
},
{
"text": "$$f = {{\\gamma + 1} \\over 2}$$"
},
{
"text": "$$f = {1 \\over {\\gamma + 1}}$$"
}
],
"answer": "$$f = {2 \\over {\\gamma - 1}}$$",
"solution": "**Answer:** $$f = {2 \\over {\\gamma - 1}}$$\n\n$$\\gamma = 1 + {2 \\over f}$$
$$ \\Rightarrow $$ $$f = {2 \\over {\\gamma - 1}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9678,
"subject": "Physics",
"question": "What will be the average value of energy for a monoatomic gas in thermal equilibrium at temperature T?",
"options": [
{
"text": "$${3 \\over 2}{k_B}T$$"
},
{
"text": "$${k_B}T$$"
},
{
"text": "$${2 \\over 3}{k_B}T$$"
},
{
"text": "$${1 \\over 2}{k_B}T$$"
}
],
"answer": "$${3 \\over 2}{k_B}T$$",
"solution": "**Answer:** $${3 \\over 2}{k_B}T$$\n\n
For a monoatomic ideal gas, the average kinetic energy per molecule is determined by the equipartition theorem. This theorem states that the energy is equally distributed among all the available degrees of freedom.
\n
A monoatomic gas has three translational degrees of freedom, corresponding to motion in the x, y, and z directions. Each degree of freedom contributes an average energy of $\\frac{1}{2} k_BT$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
\n
So for a monoatomic gas with three translational degrees of freedom, the average energy per molecule is: $$\\frac{3}{2} k_BT.$$
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9679,
"subject": "Physics",
"question": "The temperature of 3.00 mol of an ideal diatomic gas is increased by 40.0$$^\\circ$$C without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is $${x \\over {10}}$$. Then the value of x (round off to the nearest integer) is ___________. (Given R = 8.31 J mol$$-$$1 K$$-$$1)",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\nGiven, the number of diatomic moles, n = 3 mol
The increase in temperature of the diatomic mole,
$$\\Delta$$T = 40$$^\\circ$$C
Now, the degree of freedom
f = linear + rotational + no oscillation
f = 3 + 2 + 0 $$\\Rightarrow$$ f = 5
Change in internal energy,
$$\\Delta$$U = nCv$$\\Delta$$T .... (i)
where, $${C_v} = {f \\over 2}R = {5 \\over 2}R$$
Substituting the value in Eq. (i), we get
$$\\Delta U = {{5R} \\over 2}nR\\Delta T$$
Now, work done by the gas for isobaric process,
$$W = p\\Delta V = nR\\Delta T$$
The ratio of the change in internal energy to the work done by the gas,
A. The motion of the gas molecules freezes at 0$$^\\circ$$C.
\n
B. The mean free path of gas molecules decreases if the density of molecules is increased.
\n
C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant.
\n
D. Average kinetic energy per molecule per degree of freedom is $${3 \\over 2}{k_B}T$$ (for monoatomic gases).
\n
Choose the most appropriate answer from the options given below :
",
"options": [
{
"text": "A and C only"
},
{
"text": "B and C only"
},
{
"text": "A and B only"
},
{
"text": "C and D only"
}
],
"answer": "B and C only",
"solution": "**Answer:** B and C only\n\n
According to kinetic theory of gases,
\n
A. The motion of the gas molecules freezes at 0 K.
\n
B. The mean free path decreases on increasing the number density of the molecules as $$\\mu = {1 \\over {\\sqrt 2 \\pi n{d^2}}} \\Rightarrow \\mu \\propto {1 \\over n}$$.
\n
C. The mean free path increases on increasing the volume. Now if temperature is increased by keeping the pressure constant the volume should increase that is mean free path increases.
\n
D. K.E.avg per molecule per degree of freedom is $${1 \\over 2}{k_B}T$$.
Which statements are correct about degrees of freedom ?
\n
(A) A molecule with n degrees of freedom has n$$^{2}$$ different ways of storing energy.
\n
(B) Each degree of freedom is associated with $$\\frac{1}{2}$$ RT average energy per mole.
\n
(C) A monatomic gas molecule has 1 rotational degree of freedom where as diatomic molecule has 2 rotational degrees of freedom.
\n
(D) $$\\mathrm{CH}_{4}$$ has a total of 6 degrees of freedom.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "(B) and (C) only"
},
{
"text": "(B) and (D) only"
},
{
"text": "(A) and (B) only"
},
{
"text": "(C) and (D) only"
}
],
"answer": "(B) and (D) only",
"solution": "**Answer:** (B) and (D) only\n\n
Statement A is incorrect, statement B is correct by equipartition of energy. Statement C is incorrect as monoatomic does not have any rotational degree of freedom and CH4 is a polyatomic gas so it has 6 degree of freedom. So only B and D are correct.
One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is $$\\frac{\\alpha^{2}}{4} \\mathrm{R} \\,\\mathrm{J} / \\mathrm{mol} \\,\\mathrm{K}$$; then the value of $$\\alpha$$ will be _________. (Assume that the given diatomic gas has no vibrational mode).
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9686,
"subject": "Physics",
"question": "Heat energy of $735 \\mathrm{~J}$ is given to a diatomic gas allowing the gas to expand at constant pressure.\n\nEach gas molecule rotates around an internal axis but do not oscillate. The increase in the internal\n\nenergy of the gas will be :",
"options": [
{
"text": "$572 \\mathrm{~J}$"
},
{
"text": " $441 \\mathrm{~J}$"
},
{
"text": "$525 \\mathrm{~J}$"
},
{
"text": "$735 \\mathrm{~J}$"
}
],
"answer": "$525 \\mathrm{~J}$",
"solution": "**Answer:** $525 \\mathrm{~J}$\n\n$\\Delta Q=n C_{P} \\Delta T= $ \n
According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is :-
",
"options": [
{
"text": "$$\\frac{9}{2}R$$"
},
{
"text": "$$\\frac{5}{2}R$$"
},
{
"text": "$$\\frac{3}{2}R$$"
},
{
"text": "$$\\frac{7}{2}R$$"
}
],
"answer": "$$\\frac{7}{2}R$$",
"solution": "**Answer:** $$\\frac{7}{2}R$$\n\nDiatomic gas molecules have three translational degree of freedom, two rotational degree of freedom \\& it is given that it has one vibrational mode so there are two additional degree of freedom corresponding to one vibrational mode, so total degree of freedom $=7$
Let $$\\gamma_1$$ be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and $$\\gamma_2$$ be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio, $$\\frac{\\gamma_1}{\\gamma_2}$$ is :
The mean free path of molecules of a certain gas at STP is $$1500 \\mathrm{~d}$$, where $$\\mathrm{d}$$ is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at $$373 \\mathrm{~K}$$ is approximately:
",
"options": [
{
"text": "$$750 \\mathrm{~d}$$"
},
{
"text": "$$1500 \\mathrm{~d}$$"
},
{
"text": "$$\\mathrm{2049~ d}$$"
},
{
"text": "$$1098 \\mathrm{~d}$$"
}
],
"answer": "$$\\mathrm{2049~ d}$$",
"solution": "**Answer:** $$\\mathrm{2049~ d}$$\n\nThe mean free path (λ) of molecules in a gas is given by the formula:\n
\nwhere k is the Boltzmann constant, T is the temperature in Kelvin, d is the diameter of the gas molecules, and P is the pressure.\n
\nAt STP (standard temperature and pressure), the temperature is $$273\\mathrm{~K}$$ and the pressure is $$1\\mathrm{~atm}$$. We are given that the mean free path at STP is $$1500d$$. Let's denote the mean free path at $$373\\mathrm{~K}$$ as λ':\n
\nThus, the mean free path of the molecules at $$373\\mathrm{~K}$$ while maintaining the standard pressure is approximately $$2049d$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9690,
"subject": "Physics",
"question": "
The rms speed of oxygen molecule in a vessel at particular temperature is $$\\left(1+\\frac{5}{x}\\right)^{\\frac{1}{2}} v$$, where $$v$$ is the average speed of the molecule. The value of $$x$$ will be:
\n
$$\\left(\\right.$$ Take $$\\left.\\pi=\\frac{22}{7}\\right)$$
The relationship between the root-mean-square (rms) speed ($$v_{rms}$$) and the average speed ($$v_{avg}$$) of molecules in a gas can be found using the Maxwell-Boltzmann distribution. The rms speed and average speed are related as follows:
So, the value of $$x$$ is $$\\boxed{28}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9691,
"subject": "Physics",
"question": "
Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafloride (polyatomic). Arrange these on the basis of their root mean square speed $$\\left(v_{\\mathrm{rms}}\\right)$$ and choose the correct answer from the options given below:
Since the temperature and the Boltzmann constant are the same for all gases, the root mean square speed is inversely proportional to the square root of the molar mass:
\n
$$v_{rms} \\propto \\frac{1}{\\sqrt{m}}$$
\n
The lighter the gas, the higher its root mean square speed. Comparing the molar masses of the gases, we find that neon is the lightest, followed by chlorine, and uranium hexafluoride is the heaviest. Therefore, the root mean square speeds will be:
Monoatomic gases possess only translational degrees of freedom. So, option (I) matches with (A).
\n
Polyatomic gases have translational and rotational degrees of freedom. So, option (II) matches with (D).
\n
Rigid diatomic gases have translational, rotational and vibrational degrees of freedom. So, option (III) matches with (B).
\n
Non-rigid diatomic gases possess translational, rotational and vibrational degrees of freedom. So, option (IV) matches with (C).
\n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9693,
"subject": "Physics",
"question": "Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :",
"options": [
{
"text": "$\\frac{3}{2} \\mathrm{R}$"
},
{
"text": "$\\frac{7}{4} \\mathrm{R}$"
},
{
"text": "$\\frac{5}{2} \\mathrm{R}$"
},
{
"text": "$\\frac{9}{4} \\mathrm{R}$"
}
],
"answer": "$\\frac{9}{4} \\mathrm{R}$",
"solution": "**Answer:** $\\frac{9}{4} \\mathrm{R}$\n\n
To calculate the molar specific heat at constant volume of a gas mixture, we can use a weighted average based on the molar specific heats of the individual gases and their respective amounts (moles).
\n\n
Let's call $C_{V,m}$ the molar specific heat at constant volume of the mixture, $n_1$ the number of moles of the monoatomic gas, $n_2$ the number of moles of the diatomic gas, $C_{V,m1}$ the molar specific heat at constant volume of the monoatomic gas, and $C_{V,m2}$ the molar specific heat at constant volume of the diatomic gas.
\n\n
For a monoatomic ideal gas, the molar specific heat at constant volume is:
\n\n
$$ C_{V,m1} = \\frac{3}{2}R $$
\n\n
For a diatomic ideal gas, if we assume the gas is rigid and does not exhibit vibrational modes, the molar specific heat at constant volume is:
\n\n
$$ C_{V,m2} = \\frac{5}{2}R $$
\n\n
The weighted average of the molar specific heat for the mixture is:
To find the temperature at which the average kinetic energy of a monatomic molecule is $$0.414 \\, \\text{eV}$$, we use the equation for the average kinetic energy of a molecule in terms of temperature:
\n
\n
$$K_{\\text{avg}} = \\frac{3}{2} k_B T$$
\n
\n
Where:
\n
\n
$$K_{\\text{avg}}$$ is the average kinetic energy
\n
$$k_B$$ is the Boltzmann constant, given as $$1.38 \\times 10^{-23} \\, \\text{J/K}$$
\n
$$T$$ is the temperature in Kelvin
\n
\n
First, we need to convert the average kinetic energy from eV to Joules since the Boltzmann constant is in Joules. The conversion factor is $$1 \\, \\text{eV} = 1.6 \\times 10^{-19} \\, \\text{J}$$, so:
A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:
To determine the total internal energy of the gas mixture, we adhere to the equipartition theorem, which dictates that each degree of freedom contributes $$\\frac{1}{2} RT$$ to the internal energy per mole, where $R$ is the gas constant and $T$ is the temperature.
\n
An argon atom, being a noble gas, is monoatomic, with 3 translational degrees of freedom. Since we neglect all vibrational modes, and monoatomic gases have no rotational or vibrational degrees of freedom that contribute to energy at our specified temperature, each mole of argon has $$3 \\times \\frac{1}{2} RT = \\frac{3}{2} RT$$ of energy.
\n
Oxygen, on the other hand, is a diatomic molecule. This implies that under normal conditions, it has 3 translational and 2 rotational degrees of freedom. So, each mole of oxygen has $$5 \\times \\frac{1}{2} RT = \\frac{5}{2} RT$$ of energy as we are neglecting vibrational modes which typically become relevant only at higher temperatures.
\n
Hence, the total internal energy ($U$) of the gas mixture is:
\n
$$\nU = (\\text{Energy per mole of Ar} \\times \\text{Number of moles of Ar}) + (\\text{Energy per mole of O}_{2} \\times \\text{Number of moles of O}_{2})\n$$
$$N$$ moles of a polyatomic gas $$(f=6)$$ must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of $$N$$ is :
If three moles of monoatomic gas $$\\left(\\gamma=\\frac{5}{3}\\right)$$ is mixed with two moles of a diatomic gas $$\\left(\\gamma=\\frac{7}{5}\\right)$$, the value of adiabatic exponent $$\\gamma$$ for the mixture is
The translational degrees of freedom $$\\left(f_t\\right)$$ and rotational degrees of freedom $$\\left(f_r\\right)$$ of $$\\mathrm{CH}_4$$ molecule are:
",
"options": [
{
"text": "$$f_t=2$$ and $$f_r=2$$\n"
},
{
"text": "$$f_t=3$$ and $$f_r=3$$\n"
},
{
"text": "$$f_t=3$$ and 4$f_r=2$$\n"
},
{
"text": "$$f_t=2$$ and $$f_r=3$$"
}
],
"answer": "$$f_t=3$$ and $$f_r=3$$\n",
"solution": "**Answer:** $$f_t=3$$ and $$f_r=3$$\n\n\n
For non-linear polyatomic molecules, both \ntranslational and rotational degree of freedom have \nsame value and is equal to 3.
A mixture of one mole of monoatomic gas and one mole of a diatomic gas (rigid) are kept at room temperature $$(27^{\\circ} \\mathrm{C})$$. The ratio of specific heat of gases at constant volume respectively is:
To find the ratio of specific heats at constant volume ($C_V$) of the gases, we need to understand the degrees of freedom each type of gas molecule has, as this determines their specific heat capacity at constant volume. Degrees of freedom refer to the number of independent ways in which a molecule can store energy.
\n\n
A monoatomic gas molecule has 3 translational degrees of freedom. This is because it can move in three dimensions: x, y, and z. A diatomic gas, if we assume it to be rigid for this context, has 5 degrees of freedom: 3 translational like the monoatomic gas and 2 rotational since it can rotate around two axes perpendicular to the bond axis connecting the two atoms. Vibrational modes are not considered at room temperature for a rigid diatomic gas, as these require higher energy to become accessible.
\n\n
The specific heat capacity at constant volume for a monoatomic gas is given by:
\n\n
$$C_{V, mono} = \\frac{3}{2} R$$
\n\n
And for a diatomic gas, it's:
\n\n
$$C_{V, diatomic} = \\frac{5}{2} R$$
\n\n
Where $R$ is the ideal gas constant.
\n\n
To find the ratio of their specific heats at constant volume, we divide the specific heat of the monoatomic gas by that of the diatomic gas:
The energy of a diatomic molecule depends on the degrees of freedom it has. For a non-rigid diatomic molecule, there are more degrees of freedom compared to a rigid diatomic molecule. Specifically, a non-rigid diatomic molecule has translational, rotational, and vibrational degrees of freedom. The translational and rotational degrees of freedom are the same for both rigid and non-rigid diatomic molecules, which include:
\n\n
\n\n
3 translational degrees of freedom,
\n\n
2 rotational degrees of freedom (since linear molecules do not have the third rotational freedom because it requires rotating around the bond axis, which doesn't change the energy for linear molecules).
\n\n
\n\n
Additionally, non-rigid diatomic molecules have vibrational degrees of freedom. For a simple diatomic molecule, there is 1 vibrational degree of freedom (since vibration along the bond axis is possible). However, considering the energy distribution across these vibrations requires accounting for both the potential and kinetic energy associated with vibrations, effectively doubling the vibrational degrees of freedom for energy calculations to 2 (one for kinetic and one for potential energy).
\n\n
Thus, the total degrees of freedom for a non-rigid diatomic molecule are:
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9701,
"subject": "Physics",
"question": "''Heat cannot by itself flow from a body at lower temperature to a body at higher temperature'' is a statement or consequence of :",
"options": [
{
"text": "second law of thermodynamics"
},
{
"text": "conservation of momentum"
},
{
"text": "conservation of mass "
},
{
"text": "first law of thermodynamics "
}
],
"answer": "second law of thermodynamics",
"solution": "**Answer:** second law of thermodynamics\n\nThe statement \"Heat cannot by itself flow from a body at lower temperature to a body at higher temperature\" is a statement or consequence of the second law of thermodynamics. This law essentially states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible. In simpler terms, it implies that heat energy cannot spontaneously transfer from a colder body to a hotter one. \n
Therefore, the correct option is :\n\n
Option A : second law of thermodynamics.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9702,
"subject": "Physics",
"question": "A solid body of constant heat capacity $$1$$ $$J/{}^ \\circ C$$ is being heated by keeping it in contact with reservoirs in two ways:\n $$(i)$$ Sequentially keeping in contact with $$2$$ reservoirs such that each reservoir \n $$\\,\\,\\,\\,\\,\\,\\,\\,$$supplies same amount of heat. \n $$(ii)$$ Sequentially keeping in contact with $$8$$ reservoirs such that each reservoir \n $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$supplies same amount of heat. \n In both the cases body is brought from initial temperature $${100^ \\circ }C$$ to final temperature $${200^ \\circ }C$$. Entropy change of the body in the two cases respectively is : ",
"options": [
{
"text": "$$ln2, 2ln2$$ "
},
{
"text": "$$2ln2, 8ln2$$ "
},
{
"text": "$$ln2, 4ln2$$ "
},
{
"text": "$$ln2, ln2$$ "
}
],
"answer": "$$ln2, ln2$$ ",
"solution": "**Answer:** $$ln2, ln2$$ \n\nThe entropy change of the body in the two cases is same as entropy is a state function.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9703,
"subject": "Physics",
"question": "An ideal gas in a cylinder is separated by a piston in such a way that the entropy of one part is S1 and that of the other part is S2. Given that S1 > S2. If the piston is removed then the total entropy of the system will be :",
"options": [
{
"text": "S1 $$-$$ S2"
},
{
"text": "$${{{S_1}} \\over {{S_2}}}$$"
},
{
"text": "S1 $$\\times$$ S2"
},
{
"text": "S1 + S2"
}
],
"answer": "S1 + S2",
"solution": "**Answer:** S1 + S2\n\n\n
for gas 1, S1 = $${f \\over 2}{n_1}R$$
for gas 2, S2 = $${f \\over 2}{n_2}R$$
after removal of piston,
S = $${f \\over 2}({n_1} + {n_2})R = {S_1} + {S_2}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9704,
"subject": "Physics",
"question": "The entropy of any system is given by $$S = {\\alpha ^2}\\beta \\ln \\left[ {{{\\mu kR} \\over {J{\\beta ^2}}} + 3} \\right]$$ where $$\\alpha$$ and $$\\beta$$ are the constants. $$\\mu$$, J, k and R are no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively. [Take $$S = {{dQ} \\over T}$$] Choose the incorrect option from the following :",
"options": [
{
"text": "$$\\alpha$$ and J have the same dimensions."
},
{
"text": "S and $$\\alpha$$ have different dimensions"
},
{
"text": "S, $$\\beta$$, k and $$\\mu$$R have the same dimensions"
},
{
"text": "$$\\alpha$$ and k have the same dimensions"
}
],
"answer": "$$\\alpha$$ and k have the same dimensions",
"solution": "**Answer:** $$\\alpha$$ and k have the same dimensions\n\nSince, entropy of the system is given by
So, from Eqs. (iii) and (viii), we can say that $$\\alpha$$ and k have different dimensions.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9705,
"subject": "Physics",
"question": "Which of the following is more close to a black body? ",
"options": [
{
"text": "black board paint "
},
{
"text": "green leaves "
},
{
"text": "black holes "
},
{
"text": "red roses "
}
],
"answer": "black board paint ",
"solution": "**Answer:** black board paint \n\nBlack board paint is quite approximately equal to black bodies.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9706,
"subject": "Physics",
"question": "If mass-energy equivalence is taken into account, when water is cooled to form ice, the mass of water should ",
"options": [
{
"text": "increase "
},
{
"text": "remain unchanged "
},
{
"text": "decrease "
},
{
"text": "first increase then decrease "
}
],
"answer": "decrease ",
"solution": "**Answer:** decrease \n\nWhen water is cooled to form ice, energy is released from water in the form of heat. As energy is equivalent to mass therefore when water is cooled to ice, its mass decreases. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9707,
"subject": "Physics",
"question": "Two spheres of the same material have radii $$1$$ $$m$$ and $$4$$ $$m$$ and temperatures $$4000$$ $$K$$ and $$2000$$ $$K$$ respectively. The ratio of the energy radiated per second by the first sphere to that by the second is ",
"options": [
{
"text": "$$1:1$$ "
},
{
"text": "$$16:1$$ "
},
{
"text": "$$4:1$$ "
},
{
"text": "$$1:9$$ "
}
],
"answer": "$$1:1$$ ",
"solution": "**Answer:** $$1:1$$ \n\nThe energy radiated per second is given by $$E = e\\sigma {T^4}A$$\n For same material $$e$$ is same. $$\\sigma $$ is stefan's constant\n $$\\therefore$$ $${{{E_1}} \\over {{E_2}}} = {{T_1^4{A_1}} \\over {T_2^4{A_2}}} = {{T_1^44\\pi r_1^2} \\over {T_2^44\\pi r_2^2}}$$\n $$ = {{{{\\left( {4000} \\right)}^4} \\times {1^2}} \\over {{{\\left( {2000} \\right)}^4} \\times {4^2}}} = {1 \\over 1}$$ ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9708,
"subject": "Physics",
"question": "Infrared radiation is detected by ",
"options": [
{
"text": "spectrometer "
},
{
"text": "pyrometer "
},
{
"text": "nanometer "
},
{
"text": "photometer "
}
],
"answer": "pyrometer ",
"solution": "**Answer:** pyrometer \n\nPyrometer is used to detect infra-red radiation.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9709,
"subject": "Physics",
"question": "The earth radiates in the infra-red region of the spectrum. The spectrum is correctly given by ",
"options": [
{
"text": "Rayleigh Jeans law "
},
{
"text": "Planck's law of radiation "
},
{
"text": "Stefan's law of radiation"
},
{
"text": "Wien's law "
}
],
"answer": "Wien's law ",
"solution": "**Answer:** Wien's law \n\nWein's law correctly explanations the spectrum",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9710,
"subject": "Physics",
"question": "If the temperature of the sun were to increase from $$T$$ to $$2T$$ and its radius from $$R$$ to $$2R$$, then the ratio of the radiant energy received on earth to what it was previously will be ",
"options": [
{
"text": "$$32$$ "
},
{
"text": "$$16$$ "
},
{
"text": "$$4$$ "
},
{
"text": "$$64$$ "
}
],
"answer": "$$64$$ ",
"solution": "**Answer:** $$64$$ \n\n$$E = \\sigma A{T^4};\\,\\,A \\propto {R^2}$$ \n $$\\therefore$$ $$E \\propto {R^2}{T^4}$$\n $$\\therefore$$ $${{{E_2}} \\over {{E_1}}} = {{R_2^2T_2^4} \\over {R_1^2T_1^4}}$$\n $$ \\Rightarrow {{{E_2}} \\over {{E_1}}}$$\n $$ = {{{{\\left( {2R} \\right)}^2}{{\\left( {2T} \\right)}^4}} \\over {{R^2}{T^4}}}$$\n $$ = 64$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9711,
"subject": "Physics",
"question": "Assuming the Sun to be a spherical body of radius $$R$$ at a temperature of $$TK$$, evaluate the total radiant powered incident of Earth at a distance $$r$$ from the Sun\n
Where r0 is the radius of the Earth and $$\\sigma $$ is Stefan's constant.
The intensity of power at earth's surface $$ = {{\\sigma {T^4} \\times 4\\pi {R^2}} \\over {4\\pi {r^2}}}$$\n
Total power received by Earth $$ = {{\\sigma {T^4}{R^2}} \\over {{r^2}}}\\left( {\\pi r_0^2} \\right)$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9712,
"subject": "Physics",
"question": "Three rods of Copper, Brass and Steel are welded together to form a $$Y$$ shaped structure. Area of cross - section of each rod $$ = 4c{m^2}.$$ End of copper rod is maintained at $${100^ \\circ }C$$ where as ends of brass and steel are kept at $${0^ \\circ }C$$. Lengths of the copper, brass and steel rods are $$46,$$ $$13$$ and $$12$$ $$cms$$ respectively. The rods are thermally insulated from surroundings excepts at ends. Thermal conductivities of copper, brass and steel are $$0.92, 0.26$$ and $$0.12$$ $$CGS$$ units respectively. Rate of heat flow through copper rod is:",
"options": [
{
"text": "$$1.2$$ $$cal/s$$"
},
{
"text": "$$2.4$$ $$cal/s$$"
},
{
"text": "$$4.8$$ $$cal/s$$"
},
{
"text": "$$6.0$$ $$cal/s$$ "
}
],
"answer": "$$4.8$$ $$cal/s$$",
"solution": "**Answer:** $$4.8$$ $$cal/s$$\n\nRate of heat flow is given by,\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9713,
"subject": "Physics",
"question": "A heat source at T = 103 K is connected to another heat reservoir at T = 102 K by a copper slab which is 1 mthick. Given that the thermal conductivity of copper is 0.1 WK–1m–1, the energy flux through it in the steady state is -",
"options": [
{
"text": "200 Wm$$-$$2"
},
{
"text": "65 Wm$$-$$2"
},
{
"text": "120 Wm$$-$$2"
},
{
"text": "90 Wm$$-$$2"
}
],
"answer": "90 Wm$$-$$2",
"solution": "**Answer:** 90 Wm$$-$$2\n\n\n
$$ \\Rightarrow $$ $${1 \\over A}\\left( {{{dQ} \\over {dt}}} \\right) = {{\\left( {0.1} \\right)\\left( {900} \\right)} \\over 1} = 90W/{m^2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9714,
"subject": "Physics",
"question": "A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K1 and the of the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is : \n",
"options": [
{
"text": "K1 + K2 "
},
{
"text": "$${{{K_1} + 3{K_2}} \\over 4}$$"
},
{
"text": "$${{{K_1} + {K_2}} \\over 2}$$"
},
{
"text": "$${{2{K_1} + 3{K_2}} \\over 5}$$"
}
],
"answer": "$${{{K_1} + 3{K_2}} \\over 4}$$",
"solution": "**Answer:** $${{{K_1} + 3{K_2}} \\over 4}$$\n\n\n Keq = $${{{K_1}{A_1} + {K_2}{A_2}} \\over {{A_1} + {A_2}}}$$\n
Two coils require 20 minutes and 60 minutes respectively to produce same amount of heat energy when connected separately to the same source. If they are connected in parallel arrangement to the same source; the time required to produce same amount of heat by the combination of coils, will be ___________ min.
An ice cube of dimensions $$60 \\mathrm{~cm} \\times 50 \\mathrm{~cm} \\times 20 \\mathrm{~cm}$$ is placed in an insulation box of wall thickness $$1 \\mathrm{~cm}$$. The box keeping the ice cube at $$0^{\\circ} \\mathrm{C}$$ of temperature is brought to a room of temperature $$40^{\\circ} \\mathrm{C}$$. The rate of melting of ice is approximately :
\n
(Latent heat of fusion of ice is $$3.4 \\times 10^{5} \\mathrm{~J} \\mathrm{~kg}^{-1}$$ and thermal conducting of insulation wall is $$0.05 \\,\\mathrm{Wm}^{-1 \\circ} \\mathrm{C}^{-1}$$ )
A. When small temperature difference between a liquid and its surrounding is doubled, the rate of loss of heat of the liquid becomes twice.
\n
B. Two bodies $$P$$ and $$Q$$ having equal surface areas are maintained at temperature $$10^{\\circ} \\mathrm{C}$$ and $$20^{\\circ} \\mathrm{C}$$. The thermal radiation emitted in a given time by $$\\mathrm{P}$$ and $$\\mathrm{Q}$$ are in the ratio $$1: 1.15$$.
\n
C. A Carnot Engine working between $$100 \\mathrm{~K}$$ and $$400 \\mathrm{~K}$$ has an efficiency of $$75 \\%$$.
\n
D. When small temperature difference between a liquid and its surrounding is quadrupled, the rate of loss of heat of the liquid becomes twice.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "A, B, C only"
},
{
"text": "A, B only"
},
{
"text": "A, C only"
},
{
"text": "B, C, D only"
}
],
"answer": "A, B, C only",
"solution": "**Answer:** A, B, C only\n\n
From Newton's cooling law $${{dQ} \\over {dt}} = - k(T - {T_s})$$ the statement A is correct.
Two plates $$\\mathrm{A}$$ and $$\\mathrm{B}$$ have thermal conductivities $$84 ~\\mathrm{Wm}^{-1} \\mathrm{~K}^{-1}$$ and $$126 ~\\mathrm{Wm}^{-1} \\mathrm{~K}^{-1}$$ respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of $$\\mathrm{A}$$ and $$\\mathrm{B}$$ are kept at $$100^{\\circ} \\mathrm{C}$$ and $$0{ }^{\\circ} \\mathrm{C}$$ respectively, then the temperature of the surface of contact in steady state is _____________ $${ }^{\\circ} \\mathrm{C}$$.
",
"options": [],
"answer": "40",
"solution": "**Answer:** 40\n\nLet's denote the temperature at the surface of contact as T. We can find this temperature by considering the heat transfer through each plate when the system reaches steady state. At steady state, the rate of heat transfer through both plates A and B is the same.\n
\nWe can use the formula for heat transfer rate through a plate:\n
\n$$Q = kA\\frac{T_2 - T_1}{d}$$\n
\nWhere Q is the heat transfer rate, k is the thermal conductivity, A is the surface area, T1 and T2 are the temperatures on either side of the plate, and d is the thickness of the plate.\n
\nFor plate A:\n
\n$$Q_A = k_A A\\frac{T_A - T}{d}$$\n
\nFor plate B:\n
\n$$Q_B = k_B A\\frac{T - T_B}{d}$$\n
\nSince the heat transfer rate is the same through both plates in steady state:\n
\n$$Q_A = Q_B$$\n
\nWe can now substitute the given values for thermal conductivities and temperatures:\n
\nNotice that the surface area (A) and thickness (d) are the same for both plates, so they cancel out:\n
\n$$84(100 - T) = 126T$$\n
\nNow, we can solve for T:\n
\n$$8400 - 84T = 126T$$
\n$$210T = 8400$$
\n$$T = \\frac{8400}{210}$$
\n$$T = 40$$\n
\nSo, the temperature of the surface of contact in steady state is $$40{ }^{\\circ} \\mathrm{C}$$.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9721,
"subject": "Physics",
"question": "
A body cools in 7 minutes from $$60^{\\circ} \\mathrm{C}$$ to $$40^{\\circ} \\mathrm{C}$$. The temperature of the surrounding is $$10^{\\circ} \\mathrm{C}$$. The temperature of the body after the next 7 minutes will be:
Newton's law of cooling states that the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings. The average rate of cooling can be represented as:
\n
$$\\frac{T_1-T_2}{t} = K \\left(\\frac{T_1+T_2}{2} - T_s\\right)$$
\n
where:
\n
\n
$T_1$ and $T_2$ are the initial and final temperatures of the body,
\n
$t$ is the time it takes for the body to cool from $T_1$ to $T_2$,
\n
$T_s$ is the temperature of the surroundings, and
\n
$K$ is a constant of proportionality.
\n
\n
In the first 7 minutes, the body cools from $60^\\circ C$ to $40^\\circ C$, and the surrounding temperature is $10^\\circ C$. So, the first equation is:
\n
$$\\frac{60-40}{7} = K \\left(\\frac{60+40}{2} - 10\\right) \\tag{1}$$
\n
In the next 7 minutes, the body cools from $40^\\circ C$ to $T^\\circ C$, with the same surrounding temperature. So, the second equation is:
\n
$$\\frac{40-T}{7} = K \\left(\\frac{40+T}{2} - 10\\right) \\tag{2}$$
\n
Solving equation (1) for $K$ and substituting into equation (2) will give the temperature $T$ after the next 7 minutes:
To find the increase in temperature on the Fahrenheit scale, we use the relationship between the Celsius and Fahrenheit temperature scales. The formula to convert Celsius to Fahrenheit is:
\n\n
$$F = \\frac{9}{5}C + 32$$
\n\n
However, since we are interested in the increase in temperature, we can ignore the \"+ 32\" part of the formula, because this constant does not affect the change in temperature, only the absolute temperatures. Thus, to find the increase in temperature on the Fahrenheit scale, we can use:
\n\n
$$\\Delta F = \\frac{9}{5}\\Delta C$$
\n\n
Given that the increase in temperature is $$40^{\\circ}C$$, we can substitute this value into the equation:
\n\n
$$\\Delta F = \\frac{9}{5} \\times 40^{\\circ}C$$
\n\n
$$\\Delta F = 9 \\times 8$$
\n\n
$$\\Delta F = 72^{\\circ}F$$
\n\n
Therefore, the increase in temperature on the Fahrenheit scale is $$72^{\\circ}F$$. So, the correct answer is:
\n\n
Option C $$72^{\\circ} \\mathrm{F}$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9723,
"subject": "Physics",
"question": "At what temperature is the $$r.m.s$$ velocity of a hydrogen molecule equal to that of an oxygen molecule at $${47^ \\circ }C?$$ ",
"options": [
{
"text": "$$80K$$ "
},
{
"text": "$$-73$$ $$K$$ "
},
{
"text": "$$3$$ $$K$$ "
},
{
"text": "$$20$$ $$K$$ "
}
],
"answer": "$$20$$ $$K$$ ",
"solution": "**Answer:** $$20$$ $$K$$ \n\n$${v_{rms}} = $$$$\\sqrt {{{RT} \\over M}} $$\n
For $${v_{rms}}$$ to be equal $${{{T_{{H_2}}}} \\over {{M_{{H_2}}}}} = {{{T_{{O_2}}}} \\over {{M_{{O_2}}}}}$$\n
Here $${M_{{H_2}}} = 2;\\,\\,{M_{{O_2}}} = 32;$$\n
$$\\Rightarrow {T_{{H_2}}} = 20K$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9724,
"subject": "Physics",
"question": "Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will ",
"options": [
{
"text": "increase "
},
{
"text": "decrease "
},
{
"text": "remain same "
},
{
"text": "decrease for some, while increase for others "
}
],
"answer": "remain same ",
"solution": "**Answer:** remain same \n\nSince pressure and volume are not changing, so temperature remains same.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9725,
"subject": "Physics",
"question": "1 mole of a gas with $$\\gamma = 7/5$$ is mixed with $$1$$ mole of a gas with $$\\gamma = 5/3,$$ then the value of $$\\gamma $$ for the resulting mixture is ",
"options": [
{
"text": "$$7/5$$ "
},
{
"text": "$$2/5$$ "
},
{
"text": "$$3/2$$ "
},
{
"text": "$$12/7$$"
}
],
"answer": "$$3/2$$ ",
"solution": "**Answer:** $$3/2$$ \n\nIf $${n_1}$$ moles of adiabatic exponent $${\\gamma _1}$$ is mixed with $${n_2}$$ moles of adiabatic exponent $${\\gamma _2}$$ then the adiabatic component of the resulting mixture is given by\n
= – 2.5 $$ \\times $$ 1025",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9733,
"subject": "Physics",
"question": "CP and Cv are specific heats at constant pressure and constant volume respectively. It is observed that CP – Cv = a for hydrogen gas CP – Cv = b for nitrogen gas The correct relation between a and b is ",
"options": [
{
"text": "a = 28 b"
},
{
"text": "a = 1/14 b"
},
{
"text": "a = b"
},
{
"text": "a = 14 b"
}
],
"answer": "a = 14 b",
"solution": "**Answer:** a = 14 b\n\nAs we know, for 1 g mole of a gas,\n
Cp – Cv = R where Cp and Cv are molar\nspecific heat capacities.\n
$$ \\Rightarrow $$ $$a$$ = 14 $$b$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9734,
"subject": "Physics",
"question": "N moles of a diatomic gas in a cylinder are at a temperature T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monoatomic gas. What is the change in the total kinetic energy of the gas ?",
"options": [
{
"text": "$${1 \\over 2}$$ nRT"
},
{
"text": "0"
},
{
"text": "$${3 \\over 2}$$ nRT"
},
{
"text": "$${5 \\over 2}$$ nRT"
}
],
"answer": "$${1 \\over 2}$$ nRT",
"solution": "**Answer:** $${1 \\over 2}$$ nRT\n\nInitial kinetic energy of N mole of diatomic gas, \n
Ki = N$${5 \\over 2}$$ RT\n
Kinetic energy of n mole of monoatomic gas \n = n $${3 \\over 2}$$ RT\n
When n mole of diatomic gas converted into monoatomic gas then remaining diatomic gas = (N $$-$$ n)\n
\n= $${{2 \\times {{10}^{ - 3}}} \\over {3 \\times 8.3}} \\times 125.44 \\times {10^6} = {10^4}K$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9743,
"subject": "Physics",
"question": "A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the piston is $$\\ell $$1, and that below the piston is $$\\ell $$2, such that $$\\ell $$1 > $$\\ell $$2. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass, m, will be given by : (R is universal gas constant and g is the acceleration due to gravity) ",
"options": [
{
"text": "$${{nRT} \\over g}\\left[ {{{{\\ell _1} - {\\ell _2}} \\over {{\\ell _1}{\\ell _2}}}} \\right]$$"
},
{
"text": "$${{RT} \\over g}\\left[ {{{2{\\ell _1} + {\\ell _2}} \\over {{\\ell _1}{\\ell _2}}}} \\right]$$"
},
{
"text": "$${{nRT} \\over g}\\left[ {{1 \\over {{\\ell _2}}} + {1 \\over {{\\ell _1}}}} \\right]$$"
},
{
"text": "$${{RT} \\over {ng}}\\left[ {{{{\\ell _1} - 3{\\ell _2}} \\over {{\\ell _1}{\\ell _2}}}} \\right]$$"
}
],
"answer": "$${{nRT} \\over g}\\left[ {{{{\\ell _1} - {\\ell _2}} \\over {{\\ell _1}{\\ell _2}}}} \\right]$$",
"solution": "**Answer:** $${{nRT} \\over g}\\left[ {{{{\\ell _1} - {\\ell _2}} \\over {{\\ell _1}{\\ell _2}}}} \\right]$$\n\n\n
m = $${{nRT} \\over g}\\left( {{{{\\ell _1} - {\\ell _2}} \\over {{\\ell _1}.{\\ell _2}}}} \\right)$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9744,
"subject": "Physics",
"question": "An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature 300 K. The mean time between two successive collisions is 6 $$ \\times $$ 10–8 s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to",
"options": [
{
"text": "0.5 $$ \\times $$ 10$$-$$8 s"
},
{
"text": "4 $$ \\times $$ 10$$-$$8 s"
},
{
"text": "3 $$ \\times $$ 10$$-$$6 s"
},
{
"text": "2 $$ \\times $$ 10$$-$$7 s"
}
],
"answer": "4 $$ \\times $$ 10$$-$$8 s",
"solution": "**Answer:** 4 $$ \\times $$ 10$$-$$8 s\n\nt $$ \\propto $$ $${{Volume} \\over {velocity}}$$\n
volume $$ \\propto $$ $${T \\over P}$$\n
$$ \\therefore $$ t $$ \\propto $$ $${{\\sqrt T } \\over P}$$\n
$$ \\approx $$ 4 $$ \\times $$ 10$$-$$8",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9745,
"subject": "Physics",
"question": "Two kg of a monoatomic gas is at a pressure of 4 $$ \\times $$ 104 N/m2. The density of the gas is 8 kg/m3. What is the order of energy of the gas due to its thermal motion ?",
"options": [
{
"text": "104 J"
},
{
"text": "103 J"
},
{
"text": "105 J"
},
{
"text": "106 J"
}
],
"answer": "104 J",
"solution": "**Answer:** 104 J\n\nThermal energy of N molecule\n
order will 104",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9746,
"subject": "Physics",
"question": "A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27oC. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about : [Take R = 8.3 J/K mole]",
"options": [
{
"text": "0.9 kJ"
},
{
"text": "6 kJ "
},
{
"text": "10 kJ "
},
{
"text": "14 kJ"
}
],
"answer": "10 kJ ",
"solution": "**Answer:** 10 kJ \n\nWe know,\n
Vrms $$ \\propto $$ $$\\sqrt T $$\n
So, to make Vrms double we have to make temperature 4 times.\n
$$ \\therefore $$ Final temperature = 300 $$ \\times $$ 4 = 1200 K\n
= 10 kJ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9747,
"subject": "Physics",
"question": "A mixture of 2 moles of helium gas (atomic mass = 4 u), and 1 mole of argon gas (atomic mass = 40 u) is kept at 300 K in a container. The ratio of their rms speeds $$\\left[ {{{{V_{rms}}\\,(helium)} \\over {{V_{rms}}\\,(\\arg on)}}} \\right],$$ is close to : ",
"options": [
{
"text": "3.16"
},
{
"text": "0.32"
},
{
"text": "0.45"
},
{
"text": "2.24"
}
],
"answer": "3.16",
"solution": "**Answer:** 3.16\n\nWe know, \n
Vrms = $$\\sqrt {{{3RT} \\over M}} $$\n
Where M = molar mass of the gas.\n
Here temperature is 300 K for both the gas. So temperature is constant. R is also a constant. \n
= $${{19} \\over {13}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9749,
"subject": "Physics",
"question": "In a dilute gas at pressure P and temperature T, the mean time between successive collisions of a\nmolecule varies with T as :",
"options": [
{
"text": "$$\\sqrt T $$"
},
{
"text": "T"
},
{
"text": "$${1 \\over T}$$"
},
{
"text": "$${1 \\over {\\sqrt T }}$$"
}
],
"answer": "$${1 \\over {\\sqrt T }}$$",
"solution": "**Answer:** $${1 \\over {\\sqrt T }}$$\n\nTime (t) = $${V \\over {4\\pi \\sqrt 2 {r^2}vN}}$$ ....(1)\n
Here, v = most probable speed \n = $$\\sqrt {{{2RT} \\over {\\pi M}}} $$\n
$$ \\Rightarrow $$ v $$ \\propto $$ $$\\sqrt T $$\n
$$ \\therefore $$ From (1),\n
t $$ \\propto $$ $${1 \\over {\\sqrt T }}$$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9750,
"subject": "Physics",
"question": "Initially a gas of diatomic molecules is contained in a cylinder of volume V1\n at a pressure P1\n and\ntemperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in\nnumber of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a\nvolume 2V1\n is given by P2\n. The ratio $${{{P_2}} \\over {{P_1}}}$$\n is ________.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nWe know, PV = nRT\n $$ \\therefore $$ P1V1 = nR (250)\n
and P2(2V1) = $${{5n} \\over 4}R \\times \\left( {2000} \\right)$$\n
$$ \\Rightarrow $$ $${{{P_2}} \\over {{P_1}}} = 5$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9751,
"subject": "Physics",
"question": "Nitrogen gas is at 300oC temperature. The\ntemperature (in K) at which the rms speed of a\nH2 molecule would be equal to the rms speed\nof a nitrogen molecule, is _______.\n (Molar mass of N2 gas 28 g).",
"options": [],
"answer": "40TO41",
"solution": "**Answer:** 40TO41\n\nVrms = $$\\sqrt {{{3RT} \\over M}} $$\n
$$ \\Rightarrow $$ $$N = 4 \\times {10^{18}}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9753,
"subject": "Physics",
"question": "An ideal gas in a closed container is slowly\nheated. As its temperature increases, which of\nthe following statements are true?\n (A) the mean free path of the molecules\ndecreases.\n (B) the mean collision time between the\nmolecules decreases.\n (C) the mean free path remains unchanged.\n (D) the mean collision time remains unchanged.\n",
"options": [
{
"text": "(C) and (D)\n"
},
{
"text": "(A) and (D)"
},
{
"text": "(B) and (C)"
},
{
"text": "(A) and (B)"
}
],
"answer": "(B) and (C)",
"solution": "**Answer:** (B) and (C)\n\nThe mean free path of molecules of an ideal gas is given as:\n
where : V = Volume of container\n N = No of molecules\n
Mean free path is independent of temperature hence with increasing temp since volume of container does not change (closed container), so mean free path\nis unchanged.\n
Average collision time = \n
$$\\lambda $$
\n
Vav
\n\n
and Vav $$ \\propto $$ $$\\sqrt T $$\n
$$ \\therefore $$ Average collision time $$ \\propto $$ $${1 \\over {\\sqrt T }}$$\n
Hence with increase in temperature the average collision time decreases.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9754,
"subject": "Physics",
"question": "Two gases-argon (atomic radius 0.07 nm,\natomic weight 40) and xenon (atomic radius\n0.1 nm, atomic weight 140) have the same\nnumber density and are at the same\ntemperature. The raito of their respective mean\nfree times is closest to :",
"options": [
{
"text": "2.3"
},
{
"text": "1.83"
},
{
"text": "4.67"
},
{
"text": "3.67"
}
],
"answer": "1.83",
"solution": "**Answer:** 1.83\n\n$$\\lambda = {1 \\over {\\sqrt 2 \\pi {d^2}n}}$$\n
Mean free time, t = $${\\lambda \\over v}$$\n
Also v $$ \\propto $$ $$\\sqrt {{T \\over M}} $$\n
$$ \\therefore $$ t $$ \\propto $$ $${{\\sqrt M } \\over d}$$\n
$$ \\therefore $$ $${{{\\tau _2}} \\over {{\\tau _1}}}$$ = $${\\left( {{{2V} \\over V}} \\right)^{^{{{\\gamma + 1} \\over 2}}}}$$ = $${\\left( 2 \\right)^{{{1 + \\gamma } \\over 2}}}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9756,
"subject": "Physics",
"question": "The change in the magnitude of the volume of an ideal gas when a small additional pressure $$\\Delta $$P is\napplied at a constant temperature, is the same as the change when the temperature is reduced by\na small quantity $$\\Delta $$T at constant pressure. The initial temperature and pressure of the gas were 300\nK and 2 atm. respectively. If |$$\\Delta $$T| = C|$$\\Delta $$P| then value of C in (K/atm.) is _________.\n",
"options": [],
"answer": "150",
"solution": "**Answer:** 150\n\nWe know, $$PV = nRT$$
$$ \\therefore $$ $$P\\Delta V + V\\Delta P = 0$$ (for constant temp.)
and $$P\\Delta V$$ = $$nR\\Delta T$$ (for constant pressure)
$$\\Delta T = {{P\\Delta V} \\over {nR}}$$
$$\\Delta P = - {{P\\Delta V} \\over V}$$ ($$\\Delta V$$ is same in both cases)
$$ \\therefore $$ $$\\left| {{{\\Delta T} \\over {\\Delta P}}} \\right| = \\left| {{{ - 300} \\over 2}} \\right| = 150$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9757,
"subject": "Physics",
"question": "On the basis of kinetic theory of gases, the gas exerts pressure because its molecules :",
"options": [
{
"text": "continuously lose their energy till it reaches wall."
},
{
"text": "are attracted by the walls of container."
},
{
"text": "suffer change in momentum when impinge on the walls of container."
},
{
"text": "continuously stick to the walls of container."
}
],
"answer": "suffer change in momentum when impinge on the walls of container.",
"solution": "**Answer:** suffer change in momentum when impinge on the walls of container.\n\nOn the basis of kinetic theory of gases, the gas exerts pressure\nbecause its molecules contain uniform speed, random motion and\nperform elastic collision with each other, as well as with the walls of\ncontainer. As a result of which gaseous molecules suffer change in\nmomentum when impinge on the walls of container",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9758,
"subject": "Physics",
"question": "The root mean square speed of molecules of a given mass of a gas at 27$$^\\circ$$C and 1 atmosphere pressure is 200 ms$$-$$1. The root mean square speed of molecules of the gas at 127$$^\\circ$$C and 2 atmosphere pressure is $${{x \\over {\\sqrt 3 }}}$$ ms$$-$$1. The value of x will be _________.",
"options": [],
"answer": "400",
"solution": "**Answer:** 400\n\nGiven, T1 = 27$$^\\circ$$C = 27 + 273 = 300K, p1 = 1 atm, v1 = 200 ms$$-$$1, T2 = 127$$^\\circ$$C = 400 K, p2 = 12 atm, v2 = ?
$$ \\Rightarrow {x \\over {\\sqrt 3 }} = {{400} \\over {\\sqrt 3 }} \\Rightarrow x = 400$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9759,
"subject": "Physics",
"question": "A container is divided into two chambers by a partition. The volume of first chamber is 4.5 litre and second chamber is 5.5 litre. The first chamber contain 3.0 moles of gas at pressure 2.0 atm and second chamber contain 4.0 moles of gas at pressure 3.0 atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x $$\\times$$ 10$$-$$1 atm. Value of x is ________.",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\nBy energy conservation
$$ \\approx 25 \\times {10^{ - 1}}$$ atm",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9760,
"subject": "Physics",
"question": "The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas constant. The pressure of the mixture of gases is :",
"options": [
{
"text": "$${{3RT} \\over V}$$"
},
{
"text": "$${{4RT} \\over V}$$"
},
{
"text": "$${{88RT} \\over V}$$"
},
{
"text": "$${5 \\over 2}{{RT} \\over V}$$"
}
],
"answer": "$${5 \\over 2}{{RT} \\over V}$$",
"solution": "**Answer:** $${5 \\over 2}{{RT} \\over V}$$\n\nNo. of moles of O2 : \n n1 = $${{16} \\over {32}}$$ = 0.5 mole
No. of moles of N2 : \n n2 = $${{28} \\over {28}}$$ = 1 mole
No. of moles of CO2 : \n n3 = $${{44} \\over {44}}$$ = 1 mole
Total no. of moles in container : n = n1 + n2 + n3
$$ = 102$$ nm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9762,
"subject": "Physics",
"question": "Consider a sample of oxygen behaving like an ideal gas. At 300 K, the ratio of root mean square (rms) velocity to the average velocity of gas molecule would be :
where, m = molar mass of the gas in kilograms per mole,
R = molar gas constant,
and T = temperature in kelvin.
According to question,
mA < mB < mC
From Eq. (i),
$${v_{rms}} \\propto {1 \\over {\\sqrt m }}$$
$$\\therefore$$ We can write,
vA > vB > vC or $${1 \\over {{v_A}}} < {1 \\over {{v_B}}} < {1 \\over {{v_C}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9764,
"subject": "Physics",
"question": "For a gas CP $$-$$ CV = R in a state P and CP $$-$$ CV = 1.10 R in a state Q, TP and TQ are the temperatures in two different states P and Q respectively. Then",
"options": [
{
"text": "TP = TQ"
},
{
"text": "TP < TQ"
},
{
"text": "TP = 0.9 TQ"
},
{
"text": "TP > TQ"
}
],
"answer": "TP > TQ",
"solution": "**Answer:** TP > TQ\n\nCP $$-$$ CV = R for ideal gas and gas behaves as ideal gas at high temperature, so TP > TQ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9765,
"subject": "Physics",
"question": "A system consists of two types of gas molecules A and B having same number density 2 $$\\times$$ 1025/m3. The diameter of A and B are 10 $$\\mathop A\\limits^o $$ and 5 $$\\mathop A\\limits^o $$ respectively. They suffer collision at room temperature. The ratio of average distance covered by the molecule A to that of B between two successive collision is ____________ $$\\times$$ 10$$-$$2",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\n$$\\because$$ mean free path
$$ = {\\left( {{5 \\over {10}}} \\right)^2} = 0.25 = 25 \\times {10^{ - 2}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9766,
"subject": "Physics",
"question": "The number of molecules in one litre of an ideal gas at 300 K and 2 atmospheric pressure with mean kinetic energy 2 $$\\times$$ 10$$-$$9 J per molecules is :",
"options": [
{
"text": "0.75 $$\\times$$ 1011"
},
{
"text": "3 $$\\times$$ 1011"
},
{
"text": "1.5 $$\\times$$ 1011"
},
{
"text": "6 $$\\times$$ 1011"
}
],
"answer": "1.5 $$\\times$$ 1011",
"solution": "**Answer:** 1.5 $$\\times$$ 1011\n\nKE = $${3 \\over 2}kT$$
PV = $${N \\over {{N_A}}}RT$$
N = $${{PV} \\over {kT}}$$
= N = 1.5 $$\\times$$ 1011",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9767,
"subject": "Physics",
"question": "The rms speeds of the molecules of Hydrogen, Oxygen and Carbon dioxide at the same temperature are VH, VO and VC respectively then :",
"options": [
{
"text": "VH > VO > VC"
},
{
"text": "VC > VO > VH"
},
{
"text": "VH = VO > VC"
},
{
"text": "VH = VO = VC"
}
],
"answer": "VH > VO > VC",
"solution": "**Answer:** VH > VO > VC\n\n$${V_{RMS}} = \\sqrt {{{3RT} \\over {{M_W}}}} $$
At the same temperature $${V_{RMS}} \\propto {1 \\over {\\sqrt {{M_W}} }}$$
$$\\Rightarrow$$ VH > VO > VC
Option (a)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9768,
"subject": "Physics",
"question": "A cylindrical container of volume 4.0 $$\\times$$ 10$$-$$3 m3 contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is 400 K. The pressure of the mixture of gases is :
= 24.9 $$\\times$$ 105 Pa",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9769,
"subject": "Physics",
"question": "A balloon carries a total load of 185 kg at normal pressure and temperature of 27$$^\\circ$$C. What load will the balloon carry on rising to a height at which the barometric pressure is 45 cm of Hg and the temperature is $$-$$7$$^\\circ$$C. Assuming the volume constant?",
"options": [
{
"text": "181.46 kg"
},
{
"text": "214.15 kg"
},
{
"text": "219.07 kg"
},
{
"text": "123.54 kg"
}
],
"answer": "123.54 kg",
"solution": "**Answer:** 123.54 kg\n\nPm = $$\\rho$$RT
$$ = 640$$ m/s",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9772,
"subject": "Physics",
"question": "For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation $${{dp} \\over {dv}} = - ap$$. If p = p0 at v =0 is the given boundary condition, then the maximum temperature one mole of gas can attain is : (Here R is the gas constant)",
"options": [
{
"text": "$${{{p_0}} \\over {aeR}}$$"
},
{
"text": "$${{a{p_0}} \\over {eR}}$$"
},
{
"text": "infinity"
},
{
"text": "0$$^\\circ$$C"
}
],
"answer": "$${{{p_0}} \\over {aeR}}$$",
"solution": "**Answer:** $${{{p_0}} \\over {aeR}}$$\n\n$$\\int\\limits_{{p_0}}^p {{{dp} \\over P} = - a\\int\\limits_0^v {dv} } $$
$${{{M_2}} \\over {{M_1}}} = {{16} \\over 3}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9774,
"subject": "Physics",
"question": "The average translational kinetic energy of N2 gas molecules at .............$$^\\circ$$C becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given kB = 1.38 $$\\times$$ 10$$-$$23 J/K) (Fill the nearest integer).",
"options": [],
"answer": "500",
"solution": "**Answer:** 500\n\nGiven, the average translational kinetic energy of dinitrogen (N2) = Kinetic energy of an electron .... (i)
Translational kinetic energy of dinitrogen (N2)
$$KE = {3 \\over 2}{K_B}T$$
Here, T = temperature of the gas,
and KB = Boltzmann constant.
Kinetic energy of an electron = eV
Given, the potential differential of an electron, V = 0.1 V
A cylinder of fixed capacity of 44.8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0$$^\\circ$$C will be :
In van der Waal equation $$\\left[ {P + {a \\over {{V^2}}}} \\right]$$ [V $$-$$ b] = RT; P is pressure, V is volume, R is universal gas constant and T is temperature. The ratio of constants $${a \\over b}$$ is dimensionally equal to :
What will be the effect on the root mean square velocity of oxygen molecules if the temperature is doubled and oxygen molecule dissociates into atomic oxygen?
",
"options": [
{
"text": "The velocity of atomic oxygen remains same"
},
{
"text": "The velocity of atomic oxygen doubles"
},
{
"text": "The velocity of atomic oxygen becomes half"
},
{
"text": "The velocity of atomic oxygen becomes four times"
}
],
"answer": "The velocity of atomic oxygen doubles",
"solution": "**Answer:** The velocity of atomic oxygen doubles\n\n
As $${v_{rms}} = \\sqrt {{{3RT} \\over {{M_0}}}} $$
\n
T is doubled and oxygen molecule is dissociated into atomic oxygen molar mass is halved.
A mixture of hydrogen and oxygen has volume 2000 cm3, temperature 300 K, pressure 100 kPa and mass 0.76 g. The ratio of number of moles of hydrogen to number of moles of oxygen in the mixture will be:
A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats 1.4. Vessel is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by :
A flask contains argon and oxygen in the ratio of 3 : 2 in mass and the mixture is kept at 27$$^\\circ$$C. The ratio of their average kinetic energy per molecule respectively will be :
When a gas filled in a closed vessel is heated by raising the temperature by 1$$^\\circ$$C, its pressure increases by 0.4%. The initial temperature of the gas is ___________ K.
The relation between root mean square speed (vrms) and most probable sped (vp) for the molar mass M of oxygen gas molecule at the temperature of 300 K will be :
A monoatomic gas performs a work of $${Q \\over {4}}$$ where Q is the heat supplied to it. The molar heat capacity of the gas will be ______________ R during this transformation. Where R is the gas constant.
0.056 kg of Nitrogen is enclosed in a vessel at a temperature of 127$$^\\circ$$C. Th amount of heat required to double the speed of its molecules is ____________ k cal.
The pressure of the gas in a constant volume gas thermometer is 100 cm of mercury when placed in melting ice at 1 atm. When the bulb is placed in a liquid, the pressure becomes 180 cm of mercury. Temperature of the liquid is :
Sound travels in a mixture of two moles of helium and n moles of hydrogen. If rms speed of gas molecules in the mixture is $$\\sqrt2$$ times the speed of sound, then the value of n will be :
7 mol of a certain monoatomic ideal gas undergoes a temperature increase of $$40 \\mathrm{~K}$$ at constant pressure. The increase in the internal energy of the gas in this process is :
Same gas is filled in two vessels of the same volume at the same temperature. If the ratio of the number of molecules is $$1: 4$$, then
\n
A. The r.m.s. velocity of gas molecules in two vessels will be the same.
\n
B. The ratio of pressure in these vessels will be $$1: 4$$.
\n
C. The ratio of pressure will be $$1: 1$$.
\n
D. The r.m.s. velocity of gas molecules in two vessels will be in the ratio of $$1: 4$$.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "A and C only"
},
{
"text": "B and D only"
},
{
"text": "A and B only"
},
{
"text": "C and D only"
}
],
"answer": "A and B only",
"solution": "**Answer:** A and B only\n\n
$${v_{rms}} = \\sqrt {{{3RT} \\over {{M_0}}}} $$ because T is same
Statement I : The average momentum of a molecule in a sample of an ideal gas depends on temperature.
\n
Statement II : The rms speed of oxygen molecules in a gas is $$v$$. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become $$2 v$$.
\n
In the light of the above statements, choose the correct answer from the options given below :
",
"options": [
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Both Statement I and Statement II are false"
},
{
"text": "Statement I is true but Statement II is false"
},
{
"text": "Statement I is false but Statement II is true"
}
],
"answer": "Statement I is false but Statement II is true",
"solution": "**Answer:** Statement I is false but Statement II is true\n\n
Average momentum $$ = \\left\\langle {\\overrightarrow P } \\right\\rangle = 0$$
\n
$${v_{rms}} = \\sqrt {{{3RT} \\over M}} $$
\n
If temperature is doubled and oxygen atoms are used then
A vessel contains $$14 \\mathrm{~g}$$ of nitrogen gas at a temperature of $$27^{\\circ} \\mathrm{C}$$. The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be :
\n
Take $$\\mathrm{R}=8.32 \\mathrm{~J} \\mathrm{~mol}^{-1} \\,\\mathrm{k}^{-1}$$.
The root mean square speed of smoke particles of mass $$5 \\times 10^{-17} \\mathrm{~kg}$$ in their Brownian motion in air at NTP is approximately. [Given $$\\mathrm{k}=1.38 \\times 10^{-23} \\mathrm{JK}^{-1}$$]
$$\\left(P+\\frac{a}{V^{2}}\\right)(V-b)=R T$$ represents the equation of state of some gases. Where $$P$$ is the pressure, $$V$$ is the volume, $$T$$ is the temperature and $$a, b, R$$ are the constants. The physical quantity, which has dimensional formula as that of $$\\frac{b^{2}}{a}$$, will be:
The average kinetic energy of a molecule of the gas is
",
"options": [
{
"text": "proportional to volume"
},
{
"text": "dependent on the nature of the gas"
},
{
"text": "proportional to absolute temperature"
},
{
"text": "proportional to pressure"
}
],
"answer": "proportional to absolute temperature",
"solution": "**Answer:** proportional to absolute temperature\n\nAverage kinetic energy of a molecule of gas\n\n
$$\n=\\frac{f}{2} k_{B} T\n$$\n
f is degree of freedom.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9795,
"subject": "Physics",
"question": "
The correct relation between $$\\gamma = {{{c_p}} \\over {{c_v}}}$$ and temperature T is :
At low temperature $(T), \\gamma$ is independent of $T$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9796,
"subject": "Physics",
"question": "A flask contains hydrogen and oxygen in the ratio of $2: 1$ by mass at temperature $27^{\\circ} \\mathrm{C}$. The ratio of average kinetic energy per molecule of hydrogen and oxygen respectively is:",
"options": [
{
"text": "1 : 1"
},
{
"text": "4 : 1"
},
{
"text": "1 : 4"
},
{
"text": "2 : 1"
}
],
"answer": "1 : 1",
"solution": "**Answer:** 1 : 1\n\n
K.E. per molecule $$ = \\left( {{f \\over 2}KT} \\right)$$
The pressure $$(\\mathrm{P})$$ and temperature ($$\\mathrm{T})$$ relationship of an ideal gas obeys the equation\n\n$$\\mathrm{PT}^{2}=$$ constant. The volume expansion coefficient of the gas will be :
At 300 K, the rms speed of oxygen molecules is $$\\sqrt {{{\\alpha + 5} \\over \\alpha }} $$ times to that of its average speed in the gas. Then, the value of $$\\alpha$$ will be
The root mean square velocity of molecules of gas is
",
"options": [
{
"text": "Proportional to temperature ($$T$$)"
},
{
"text": "Inversely proportional to square root of temperature $$\\left( {\\sqrt {{1 \\over T}} } \\right)$$"
},
{
"text": "Proportional to square of temperature ($$T^2$$)"
},
{
"text": "Proportional to square root of temperature ($$\\sqrt T$$)"
}
],
"answer": "Proportional to square root of temperature ($$\\sqrt T$$)",
"solution": "**Answer:** Proportional to square root of temperature ($$\\sqrt T$$)\n\nThe rms speed of a gas molecule is
Statement I : The temperature of a gas is $$-73^\\circ$$C. When the gas is heated to $$527^\\circ$$C, the root mean square speed of the molecules is doubled.
\n
Statement II : The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules.
\n
In the light of the above statements, choose the correct answer from the option given below :
",
"options": [
{
"text": "Statement I is true but Statement II is false"
},
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Statement I is false but Statement II is true"
},
{
"text": "Both Statement I and Statement II are false"
}
],
"answer": "Statement I is true but Statement II is false",
"solution": "**Answer:** Statement I is true but Statement II is false\n\n$T_{i}=200 \\mathrm{~K} \\quad\\quad v_{\\mathrm{rms}} \\propto \\sqrt{T}$\n
\nTranslational K.E. $=\\left(\\frac{3}{2} P V\\right)$\n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9801,
"subject": "Physics",
"question": "A flask contains Hydrogen and Argon in the ratio $2: 1$ by mass. The temperature of the mixture is $30^{\\circ} \\mathrm{C}$. The ratio of average kinetic energy per molecule of the two gases ( $\\mathrm{K}$ argon/K hydrogen) is :\n
\n(Given: Atomic Weight of $\\mathrm{Ar}=39.9$ )\n",
"options": [
{
"text": "$\\frac{39.9}{2}$"
},
{
"text": "2"
},
{
"text": "39.9"
},
{
"text": "1"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\nThe average kinetic energy per molecule of a gas is given by the expression $\\frac{3}{2}kT$, where $k$ is the Boltzmann constant and $T$ is the temperature of the gas in kelvins. Since the temperature of the mixture is given in Celsius, we need to convert it to kelvins by adding 273.15 to get $303.15$ K.\n
\nLet us assume that the total mass of the mixture is $3x$ (where $x$ is a constant), then the mass of hydrogen and argon in the mixture will be $2x$ and $x$ respectively, according to the given ratio.\n
\nThe number of moles of hydrogen and argon can be calculated using their respective masses and molar masses, which are 1 g/mol and 39.9 g/mol respectively. Therefore:\n
\nNumber of moles of hydrogen = $\\frac{2x}{1~\\mathrm{g/mol}} = 2x$ mol\n
\nNumber of moles of argon = $\\frac{x}{39.9~\\mathrm{g/mol}} = \\frac{x}{39.9}$ mol\n
\nThe total number of moles of gas in the mixture is the sum of the number of moles of hydrogen and argon:\n
\nTotal number of moles of gas = $2x + \\frac{x}{39.9} = \\frac{79.9x}{39.9}$ mol\n
\nThe average kinetic energy per molecule of hydrogen is:\n
\nHence, the ratio of average kinetic energy per molecule of argon to hydrogen is $1$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9802,
"subject": "Physics",
"question": "
The initial pressure and volume of an ideal gas are P$$_0$$ and V$$_0$$. The final pressure of the gas when the gas is suddenly compressed to volume $$\\frac{V_0}{4}$$ will be :
\n
(Given $$\\gamma$$ = ratio of specific heats at constant pressure and at constant volume)
",
"options": [
{
"text": "P$$_0$$(4)$$^{\\frac{1}{\\gamma}}$$"
},
{
"text": "P$$_0$$"
},
{
"text": "4P$$_0$$"
},
{
"text": "P$$_0$$(4)$$^{\\gamma}$$"
}
],
"answer": "P$$_0$$(4)$$^{\\gamma}$$",
"solution": "**Answer:** P$$_0$$(4)$$^{\\gamma}$$\n\nWhen the gas is compressed suddenly, it undergoes an adiabatic process where no heat is exchanged with the surroundings.
Therefore, we can use the adiabatic equation of state to relate the initial and final pressure and volume of the gas: $$P_0V_0^\\gamma=P_fV_f^\\gamma$$ where $$P_f$$ and $$V_f$$ are the final pressure and volume of the gas, respectively.
Since the gas is compressed to $$\\frac{V_0}{4}$$, we have: $$V_f=\\frac{V_0}{4}$$ Substituting this into the adiabatic equation of state, we get: $$P_f=P_0\\left(\\frac{V_0}{V_f}\\right)^\\gamma=P_0\\left(\\frac{4}{1}\\right)^\\gamma=4^\\gamma P_0$$
Therefore, the final pressure of the gas when it is suddenly compressed to $$\\frac{V_0}{4}$$ is $$4^\\gamma P_0$$.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9803,
"subject": "Physics",
"question": "
If the r. m.s speed of chlorine molecule is $$490 \\mathrm{~m} / \\mathrm{s}$$ at $$27^{\\circ} \\mathrm{C}$$, the r. m. s speed of argon molecules at the same temperature will be (Atomic mass of argon $$=39.9 \\mathrm{u}$$, molecular mass of chlorine $$=70.9 \\mathrm{u}$$ )
\nThe rms speed of argon molecules at the same temperature as the chlorine molecules is approximately $$651.7 \\mathrm{~m} / \\mathrm{s}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9804,
"subject": "Physics",
"question": "
The root mean square speed of molecules of nitrogen gas at $$27^{\\circ} \\mathrm{C}$$ is approximately : (Given mass of a nitrogen molecule $$=4.6 \\times 10^{-26} \\mathrm{~kg}$$ and take Boltzmann constant $$\\mathrm{k}_{\\mathrm{B}}=1.4 \\times 10^{-23} \\mathrm{JK}^{-1}$$ )
",
"options": [
{
"text": "91 m/s"
},
{
"text": "1260 m/s"
},
{
"text": "27.4 m/s"
},
{
"text": "523 m/s"
}
],
"answer": "523 m/s",
"solution": "**Answer:** 523 m/s\n\nTo find the root mean square speed of molecules of nitrogen gas, we can use the formula:\n
\n$$\nv_{rms} = \\sqrt{\\frac{3k_BT}{m}}\n$$\n
\nwhere $$v_{rms}$$ is the root mean square speed, $$k_B$$ is the Boltzmann constant, $$T$$ is the temperature in Kelvin, and $$m$$ is the mass of a nitrogen molecule.\n
\nFirst, we need to convert the temperature from Celsius to Kelvin:\n
\nSimplify and calculate the root mean square speed:\n
\n$$\nv_{rms} = 523\\,\\mathrm{m/s}\n$$\n
\nThe root mean square speed of molecules of nitrogen gas at $$27^{\\circ}\\mathrm{C}$$ is 523 m/s.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9805,
"subject": "Physics",
"question": "
A gas mixture consists of 2 moles of oxygen and 4 moles of neon at temperature T. Neglecting all vibrational modes, the total internal energy of the system will be,
The internal energy (U) of a gas depends on its degrees of freedom (f).
For a monatomic gas like neon, the degrees of freedom are f = 3 (translational). For a diatomic gas like oxygen, the degrees of freedom are f = 5 (3 translational + 2 rotational).
\n
The internal energy for each component of the gas mixture can be calculated using the formula:
\n
$$U = \\frac{f}{2}nRT$$
\n
where n is the number of moles, R is the universal gas constant, and T is the temperature.
\n
For the 4 moles of neon (monatomic):
\n
$$U_{Ne} = \\frac{3}{2} \\cdot 4RT = 6RT$$
\n
For the 2 moles of oxygen (diatomic):
\n
$$U_{O_2} = \\frac{5}{2} \\cdot 2RT = 5RT$$
\n
Now, to find the total internal energy, we sum the internal energies of the individual components:
The kinetic energy of an ideal gas is given by the equation:
\n
$KE = \\frac{3}{2} kT$
\n
where (k) is Boltzmann's constant and (T) is the absolute temperature in kelvins. Therefore, the kinetic energy of a gas is directly proportional to its temperature.
\n
If the kinetic energy doubles, the temperature must also double. The original temperature is given as ($27^\\circ C$), which is equal to (300 K) in absolute terms. Therefore, the final temperature ($T_f$) in kelvins is:
The number of air molecules per cm$$^3$$ increased from $$3\\times10^{19}$$ to $$12\\times10^{19}$$. The ratio of collision frequency of air molecules before and after the increase in number respectively is:
",
"options": [
{
"text": "1.25"
},
{
"text": "0.25"
},
{
"text": "0.50"
},
{
"text": "0.75"
}
],
"answer": "0.25",
"solution": "**Answer:** 0.25\n\n1. The collision frequency (f) is given by the formula :\n\n
$$ f = \\sqrt{2} \\pi d^2 v n_v $$\n\n
Where:\n \n
- $d$ is the diameter of the molecule,\n
- $v$ is the average velocity of the molecules, and\n
- $n_v$ is the number density (number of molecules per unit volume).\n\n
2. From this equation, we can see that the collision frequency (f) is directly proportional to the number density ($n_v$), because all the other variables ($d$ and $v$) are constant :\n\n
$$ f \\propto n_v $$\n\n
3. Therefore, the ratio of two different collision frequencies ($f_1$ and $f_2$) is equal to the ratio of their corresponding number densities ($n_{v1}$ and $n_{v2}$):\n\n
4. Given that the number density increased from $3 \\times 10^{19}$ to $12 \\times 10^{19}$, we can substitute these values into the equation to find the ratio of the collision frequencies:\n\n
So, the ratio of the collision frequency of air molecules before and after the increase in number is 0.25.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9808,
"subject": "Physics",
"question": "
The temperature of an ideal gas is increased from $$200 \\mathrm{~K}$$ to $$800 \\mathrm{~K}$$. If r.m.s. speed of gas at $$200 \\mathrm{~K}$$ is $$v_{0}$$. Then, r.m.s. speed of the gas at $$800 \\mathrm{~K}$$ will be:
The root-mean-square (r.m.s) speed of an ideal gas is given by the formula:
\n
$$v_\\mathrm{rms} = \\sqrt{\\frac{3RT}{M}}$$
\n
where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas.
\n
In this case, we are given that the initial temperature is $$200 \\, K$$ and the final temperature is $$800 \\, K$$. Let the r.m.s speed at $$200 \\, K$$ be $$v_0$$, then:
\n
$$v_0 = \\sqrt{\\frac{3R \\cdot 200}{M}}$$
\n
Now, we want to find the r.m.s speed at $$800 \\, K$$, let's call this $$v_1$$:
The speed of sound in a gas is given by the formula:
\n
$$v = \\sqrt{\\frac{\\gamma RT}{M}}$$
\n
where $$v$$ is the speed of sound, $$\\gamma$$ is the adiabatic index, $$R$$ is the universal gas constant, $$T$$ is the temperature, and $$M$$ is the molar mass of the gas.
\n
For diatomic gases, such as hydrogen (H₂) and oxygen (O₂), the adiabatic index $$\\gamma$$ is the same, approximately equal to $$\\frac{7}{5}$$, and the temperature is given as the same for both gases.
\n
Let's denote the speed of sound in hydrogen as $$v_\\text{H}$$ and in oxygen as $$v_\\text{O}$$. The ratio of the speeds can be calculated as:
This means the ratio of the speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is $$4:1$$.
\n
Therefore, the correct answer is $$4:1$$.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9810,
"subject": "Physics",
"question": "If the root mean square velocity of hydrogen molecule at a given temperature and pressure is $2 \\mathrm{~km} / \\mathrm{s}$, the root mean square velocity of oxygen at the same condition in $\\mathrm{km} / \\mathrm{s}$ is :",
"options": [
{
"text": "1.0"
},
{
"text": "1.5"
},
{
"text": "2.0"
},
{
"text": "0.5"
}
],
"answer": "0.5",
"solution": "**Answer:** 0.5\n\n
Here is your text with LaTeX notation and paragraph tags converted as requested:
\n
To calculate the root mean square (rms) velocity of gas molecules, we can use the formula:
\n
$$ v_{\\text{rms}} = \\sqrt{\\frac{3kT}{m}} $$
\n
where:
\n
\n
$$ v_{\\text{rms}} $$ is the root mean square velocity of the gas molecules,
\n
$$ k $$ is the Boltzmann constant,
\n
$$ T $$ is the absolute temperature in Kelvin, and
\n
$$ m $$ is the mass of one molecule of the gas.
\n
\n
Since the temperature and pressure are the same for hydrogen and oxygen, we can ignore the constant and temperature parts of the equation because they will cancel out in the comparison between the two gases.
\n
Now, we need to compare the mass of one molecule of hydrogen to that of one molecule of oxygen. The molecular mass of hydrogen (H₂) is approximately 2 g/mol, while the molecular mass of oxygen (O₂) is approximately 32 g/mol.
\n
We know the rms speed of hydrogen is 2 km/s, so let's find the mass ratio and then determine the speed of oxygen molecules. Using the rms velocity formula and considering the ratio of masses:
Therefore, the rms velocity of oxygen molecules at the same temperature and pressure conditions as that of hydrogen with a rms velocity of 2 km/s is 0.5 km/s.
\n
The correct answer is Option D : 0.5.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9811,
"subject": "Physics",
"question": "The pressure and volume of an ideal gas are related as $\\mathrm{PV}^{\\frac{3}{2}}=\\mathrm{K}$ (Constant). The work done when the gas is taken from state $A\\left(P_1, V_1, T_1\\right)$ to state $B\\left(P_2, V_2, T_2\\right)$ is :",
"options": [
{
"text": "$2\\left(\\mathrm{P}_2 \\sqrt{\\mathrm{V}_2}-\\mathrm{P}_1 \\sqrt{\\mathrm{V}_1}\\right)$"
},
{
"text": "$2\\left(\\sqrt{\\mathrm{P}_1} \\mathrm{~V}_1-\\sqrt{\\mathrm{P}_2} \\mathrm{~V}_2\\right)$"
},
{
"text": "$2\\left(\\mathrm{P}_2 \\mathrm{~V}_2-\\mathrm{P}_1 \\mathrm{~V}_1\\right)$"
},
{
"text": "$2\\left(\\mathrm{P}_1 \\mathrm{~V}_1-\\mathrm{P}_2 \\mathrm{~V}_2\\right)$"
}
],
"answer": "$2\\left(\\mathrm{P}_1 \\mathrm{~V}_1-\\mathrm{P}_2 \\mathrm{~V}_2\\right)$",
"solution": "**Answer:** $2\\left(\\mathrm{P}_1 \\mathrm{~V}_1-\\mathrm{P}_2 \\mathrm{~V}_2\\right)$\n\n
To find the work done by the gas when it goes from state A to state B, we can look at the definition of work done on or by a gas in a thermodynamic process. For a quasi-static process, the work done $$ W $$ is given by the integral of the pressure $$ P $$ with respect to the volume $$ V $$:
\n
$$ W = \\int_{V_1}^{V_2} P dV $$
\n
Given the relationship $$ PV^{\\frac{3}{2}} = K $$, we can solve for $$ P $$:
\n
$$ P = \\frac{K}{V^{\\frac{3}{2}}} $$
\n
Now, substitute $$ P $$ into the work integral and evaluate it:
\n
$$ W = \\int_{V_1}^{V_2} \\frac{K}{V^{\\frac{3}{2}}} dV = K \\int_{V_1}^{V_2} V^{-\\frac{3}{2}} dV $$
\n
To integrate this, we'll use the power rule for integration. The integral of $$ V^{-\\frac{3}{2}} $$ is $$ -2V^{-\\frac{1}{2}} $$, so the work done is:
\n
$$ W = K \\left[-2V^{-\\frac{1}{2}}\\right]_{V_1}^{V_2} = K \\left(-2V_2^{-\\frac{1}{2}} + 2V_1^{-\\frac{1}{2}} \\right) $$
\n
We can rewrite $$ V^{-\\frac{1}{2}} $$ as $$ \\frac{1}{\\sqrt{V}} $$:
\n
$$ W = K \\left(-2\\frac{1}{\\sqrt{V_2}} + 2\\frac{1}{\\sqrt{V_1}} \\right) $$
\n
Since $$ P_2V_2^{\\frac{3}{2}} = K $$ and $$ P_1V_1^{\\frac{3}{2}} = K $$, we can express $$ K $$ in terms of $$ P_1 $$ and $$ V_1 $$ (or $$ P_2 $$ and $$ V_2 $$, but we’ll use $$ P_1 $$ and $$ V_1 $$ for now):
\n
$$ W = P_1V_1^{\\frac{3}{2}} \\left(-2\\frac{1}{\\sqrt{V_2}} + 2\\frac{1}{\\sqrt{V_1}} \\right) $$
\n
$$ W = -2P_1V_1\\sqrt{V_1}\\frac{1}{\\sqrt{V_2}} + 2P_1V_1\\sqrt{V_1}\\frac{1}{\\sqrt{V_1}} $$
\n
$$ W = -2P_1V_1\\frac{\\sqrt{V_1}}{\\sqrt{V_2}} + 2P_1V_1 $$
\n
Since $$ \\frac{\\sqrt{V_1}}{\\sqrt{V_2}} = \\sqrt{\\frac{V_1}{V_2}} $$, and recalling $$ P_2V_2^{\\frac{3}{2}} = K $$ once more:
\n
$$ W = -2P_1V_1\\sqrt{\\frac{V_1}{V_2}} + 2P_1V_1 $$
\n
$$ W = -2 \\frac{P_1V_1^{\\frac{3}{2}}}{\\sqrt{V_2}} + 2P_1V_1 $$
\n
Putting $$ P_1V_1^{\\frac{3}{2}} $$ back as $$ K $$:
\n
$$ W = -2\\frac{K}{\\sqrt{V_2}} + 2P_1V_1 $$
\n
$$ W = -2P_2V_2 \\sqrt{V_2}\\frac{1}{\\sqrt{V_2}} + 2P_1V_1 $$
\n
$$ W = -2P_2V_2 + 2P_1V_1 $$
\n
So the work done is:
\n
$$ W = 2P_1V_1 - 2P_2V_2 $$
\n
Therefore, the correct answer matching the given options is:
The equation of state of a real gas is given by $$\\left(\\mathrm{P}+\\frac{\\mathrm{a}}{\\mathrm{V}^2}\\right)(\\mathrm{V}-\\mathrm{b})=\\mathrm{RT}$$, where $$\\mathrm{P}, \\mathrm{V}$$ and $$\\mathrm{T}$$ are pressure, volume and temperature respectively and $$\\mathrm{R}$$ is the universal gas constant. The dimensions of $$\\frac{\\mathrm{a}}{\\mathrm{b}^2}$$ is similar to that of :
The total kinetic energy of a mole of an ideal gas can be determined by using the equipartition theorem, which states that the energy is equally distributed among degrees of freedom. For a diatomic molecule such as oxygen ($$O_2$$), there are 5 degrees of freedom (3 translational and 2 rotational - assuming the vibrational modes are not excited at room temperature), so each degree of freedom has an average energy of $$\\frac{1}{2} kT$$ per molecule, where $$k$$ is the Boltzmann constant and $$T$$ is the temperature in kelvins.
\n\n
However, since we are dealing with moles, we'll use the universal gas constant $$R$$ instead of the Boltzmann constant $$k$$, because $$R = k \\cdot N_A$$ where $$N_A$$ is the Avogadro constant (the number of molecules in a mole). Therefore, the average energy per mole for each degree of freedom is $$\\frac{1}{2}RT$$.
\n\n
To find the total energy, we multiply the energy per degree of freedom by the number of degrees of freedom for the diatomic gas:
\n\n$$ E_{\\text{total}} = \\text{degrees of freedom} \\times \\frac{1}{2} R T $$\n\n
For diatomic oxygen:
\n\n$$ E_{\\text{total}} = 5 \\times \\frac{1}{2} R T $$\n\n
Given that temperature $$ T $$ is $$ 27^{\\circ} \\mathrm{C} $$, we first convert it to kelvins:
The speed of sound in a gas at standard temperature and pressure (STP) can be calculated using the following formula derived from the ideal gas law and the speed of sound relation in a gas:
\n\n
$$ v = \\sqrt{\\gamma \\frac{R T}{M}} $$
\n\n
Where:
\n
\n
$$ v $$ = speed of sound in the gas
\n
$$ \\gamma $$ = adiabatic index (ratio of specific heats, $$ C_p/C_v $$)
\n
$$ R $$ = universal gas constant
\n
$$ T $$ = temperature in Kelvin
\n
$$ M $$ = molar mass of the gas in kilograms per mole (kg/mol)
\n
\n\n
Given:
\n
\n
$$ R = 8.3 \\, J \\cdot mol^{-1} \\cdot K^{-1} $$
\n
$$ \\gamma = 1.4 $$ (For diatomic gases such as oxygen)
\n
$$ T = 273.15 \\, K $$ (Standard temperature, 0°C in Kelvin)
\n
Molar mass of Oxygen ($$ O_2 $$) $$ M = 32 \\times 10^{-3} \\, kg/mol $$ (32 g/mol converted to kg/mol)
Two vessels $$A$$ and $$B$$ are of the same size and are at same temperature. A contains $$1 \\mathrm{~g}$$ of hydrogen and $$B$$ contains $$1 \\mathrm{~g}$$ of oxygen. $$\\mathrm{P}_{\\mathrm{A}}$$ and $$\\mathrm{P}_{\\mathrm{B}}$$ are the pressures of the gases in $$\\mathrm{A}$$ and $$\\mathrm{B}$$ respectively, then $$\\frac{P_A}{P_B}$$ is:
The temperature of a gas having $$2.0 \\times 10^{25}$$ molecules per cubic meter at $$1.38 \\mathrm{~atm}$$ (Given, $$\\mathrm{k}=1.38 \\times 10^{-23} \\mathrm{JK}^{-1}$$) is :
The temperature of a gas is $$-78^{\\circ} \\mathrm{C}$$ and the average translational kinetic energy of its molecules is $$\\mathrm{K}$$. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes $$2 \\mathrm{~K}$$ is :
The average translational kinetic energy ($K_{avg}$) of a molecule is directly proportional to the absolute temperature (T) of the gas, as described by the equation:
\n\n
$K_{avg} = \\frac{3}{2}kT$
\n\n
Where:
\n\n
\n\n
$K_{avg}$ is the average kinetic energy,
\n\n
$k$ is the Boltzmann's constant, and
\n\n
$T$ is the temperature in Kelvin.
\n\n
\n\n
From the given problem, if the temperature of the gas is $-78^{\\circ}C$, which in Kelvin is $T_1 = -78 + 273 = 195K$, and the average translational kinetic energy is $K$, when the energy becomes $2K$, we need to find the new temperature $(T_2)$.
\n\n
Using the direct proportionality relation, we can set up the following equation:
\n\n
$K \\propto T$
\n\n
$2K = K_2 = \\frac{3}{2}kT_2$
\n\n
Given that at $T_1$, the kinetic energy is $K$, and at $T_2$, it's $2K$, we can use the ratio as follows:
To find the temperature in Celsius, we convert $390K$ back to Celsius:
\n\n
$T_{2(Celsius)} = 390 - 273 = 117^{\\circ}C$
\n\n
Therefore, the temperature at which the average translational kinetic energy of the molecules of the same gas becomes $2K$ is $117^{\\circ}C$. The correct answer is:
Statement (I) : The mean free path of gas molecules is inversely proportional to square of molecular diameter.
\n
Statement (II) : Average kinetic energy of gas molecules is directly proportional to absolute temperature of gas.
\n
In the light of the above statements, choose the correct answer from the options given below :
",
"options": [
{
"text": "Statement I is false but Statement II is true\n"
},
{
"text": "Both Statement I and Statement II are true\n"
},
{
"text": "Statement I is true but Statement II is false\n"
},
{
"text": "Both Statement I and Statement II are false"
}
],
"answer": "Both Statement I and Statement II are true\n",
"solution": "**Answer:** Both Statement I and Statement II are true\n\n\n
Let's analyze the given statements one by one in detail:
\n\n
Statement (I): The mean free path of gas molecules is inversely proportional to the square of molecular diameter.
\n\n
The mean free path ($ \\lambda $) of gas molecules is the average distance a molecule travels before colliding with another molecule. The formula for the mean free path in terms of molecular diameter ($ d $) is given by:
Here, $ k_B $ is the Boltzmann constant, $ T $ is the temperature, and $ P $ is the pressure. From this equation, it is evident that the mean free path ($ \\lambda $) is inversely proportional to the square of the molecular diameter ($ d^2 $). Hence, Statement I is true.
\n\n
Statement (II): Average kinetic energy of gas molecules is directly proportional to absolute temperature of gas.
\n\n
The average kinetic energy ($ E_{\\text{avg}} $) of gas molecules is given by:
\n\n
$$E_{\\text{avg}} = \\frac{3}{2} k_B T$$
\n\n
where $ k_B $ is the Boltzmann constant and $ T $ is the absolute temperature. This shows that the average kinetic energy is directly proportional to the absolute temperature of the gas. Hence, Statement II is true.
\n\n
Considering the analysis above, the correct answer is:
\n\n
Option B Both Statement I and Statement II are true
If the collision frequency of hydrogen molecules in a closed chamber at $$27^{\\circ} \\mathrm{C}$$ is $$\\mathrm{Z}$$, then the collision frequency of the same system at $$127^{\\circ} \\mathrm{C}$$ is :
The collision frequency ($Z$) of gas molecules is proportional to the square root of the absolute temperature ($T$) of the system. Mathematically, it can be represented as $Z \\propto \\sqrt{T}$. This implies that when the temperature changes, the collision frequency changes as well according to the formula:
\n\n
$$Z_1 = Z_0 \\sqrt{\\frac{T_1}{T_0}}$$
\n\n
where:
\n\n
\n\n
$Z_1$ is the collision frequency at temperature $T_1$,
\n\n
$Z_0$ is the initial collision frequency at temperature $T_0$,
\n\n
$T_1$ is the final absolute temperature, and
\n\n
$T_0$ is the initial absolute temperature.
\n\n
\n\n
To solve the given problem, we first need to convert the provided temperatures from Celsius to Kelvin (since absolute temperature in Kelvin should be used):
\n\n
\n\n
$T_0 = 27^{\\circ}C = 27 + 273 = 300$ K
\n\n
$T_1 = 127^{\\circ}C = 127 + 273 = 400$ K
\n\n
\n\n
Using the formula for collision frequency change and substituting the given values:
Thus, the collision frequency of the system at $127^{\\circ}C$ is $\\frac{2}{\\sqrt{3}}$ times the collision frequency at $27^{\\circ}C$, making Option B the correct answer:
If $$\\mathrm{n}$$ is the number density and $$\\mathrm{d}$$ is the diameter of the molecule, then the average distance covered by a molecule between two successive collisions (i.e. mean free path) is represented by :
$$n$$ is the number density of the molecules, i.e., the number of molecules per unit volume,
\n\n
$$d$$ is the diameter of the molecules, and
\n\n
$$\\sqrt{2}$$ arises from considering the relative motion of a pair of particles in a gas, taking into account that either particle can be moving towards the other, which effectively doubles the cross-sectional area through which they can collide.
\n\n
\n\n
This formula shows that the mean free path is inversely proportional to the number density $$n$$ of the molecules and the square of the diameter $$d$$ of the molecules. It also takes into account that collisions are more likely as the cross-sectional area of the molecules increases, or as the density of the molecules increases.
A total of $$48 \\mathrm{~J}$$ heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $$2^{\\circ} \\mathrm{C}$$. The work done by the gas is:\nGiven, $$\\mathrm{R}=8.3 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$$.
To solve this problem, we can use the first law of thermodynamics, which states:
\n\n
$$\\Delta Q = \\Delta U + W$$
\n\n
where:
\n\n
\n\n
$$\\Delta Q$$ is the heat transferred to the system, in this case, $$48 \\mathrm{~J}$$.
\n\n
$$\\Delta U$$ is the change in internal energy of the system.
\n\n
$$W$$ is the work done by the system.
\n\n
\n\n
For one mole of an ideal gas, the change in internal energy can be calculated using the equation:
\n\n
$$\\Delta U = nC_{v}\\Delta T$$
\n\n
where:
\n\n
\n\n
$$n$$ is the number of moles, which is $$1$$ in this case.
\n\n
$$C_{v}$$ is the molar specific heat capacity at constant volume.
\n\n
$$\\Delta T$$ is the change in temperature, in Kelvins.
\n\n
\n\n
Given that we are dealing with helium, a monatomic ideal gas, we know that:
\n\n
$$C_{v} = \\frac{3}{2}R$$
\n\n
Substituting the given values, we get:
\n\n
$$\\Delta U = (1) \\frac{3}{2}(8.3 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1})(2^{\\circ} \\mathrm{C})$$
\n\n
Since a change in temperature of $$1^{\\circ} \\mathrm{C}$$ is equivalent to a change in temperature of $$1 \\mathrm{~K}$$, we can directly substitute $$2^{\\circ} \\mathrm{C}$$ as a $$2 \\mathrm{~K}$$ change without needing to convert Celsius to Kelvin as both scales have the same magnitude of degree.
( $$\\therefore$$ $$\\Delta Q = $$ heat supplied in time $$t$$ for heating $$1L$$ water from $${10^ \\circ }C$$ to $${40^ \\circ }C$$ )\n
also $$\\Delta Q = 836 \\times t \\Rightarrow t = {{4180 \\times 30} \\over {836}} = 150\\,s$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9827,
"subject": "Physics",
"question": "Assume that a drop of liquid evaporates by decreases in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible ? The surface tension is $$T,$$ density of liquid is $$\\rho $$ and $$L$$ is its latent heat of vaporization. ",
"options": [
{
"text": "$$\\rho L/T$$ "
},
{
"text": "$$\\sqrt {T/\\rho L} $$ "
},
{
"text": "$$T/\\rho L$$ "
},
{
"text": "$$2T/\\rho L$$ "
}
],
"answer": "$$2T/\\rho L$$ ",
"solution": "**Answer:** $$2T/\\rho L$$ \n\nWhen radius is decrease by $$\\Delta R,$$\n $$4\\pi {R^2}\\Delta R\\rho L = 4\\pi T\\left[ {{R^2} - {{\\left( {R - \\Delta R} \\right)}^2}} \\right]$$\n $$ \\Rightarrow \\rho {R^2}\\Delta RL = T\\left[ {{R^2} - {R^2} + 2R\\Delta R - \\Delta {R^2}} \\right]$$\n $$ \\Rightarrow \\rho {R^2}\\Delta RL = T2R\\Delta R\\,\\,$$ [ $$\\Delta R$$ is very small ]\n $$ \\Rightarrow R = {{2T} \\over {\\rho L}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9828,
"subject": "Physics",
"question": "200 g water is heated from 40oC to 60oC. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water = 4184 J/kg/K) :",
"options": [
{
"text": "8.4 kJ"
},
{
"text": "4.2 kJ"
},
{
"text": "16.7 kJ"
},
{
"text": "167.4 kJ"
}
],
"answer": "16.7 kJ",
"solution": "**Answer:** 16.7 kJ\n\nAccording to the first law of thermodynamics, \n
Q = $$\\Delta $$u + w\n
For isochoric process Q = $$\\Delta $$U = ms$$\\Delta $$t\n
$$\\Delta $$T = (233 $$-$$ 213) = 20 k\n
$$ \\therefore $$ $$\\Delta $$u = 0.2 $$ \\times $$ 4184 $$ \\times $$ 20 = 16.7 kJ",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9829,
"subject": "Physics",
"question": "A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm,\nfilled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to\nbe 75oC. T is given by: (Given : room temperature = 30oC, specific heat of copper = 0.1 cal/gmoC) ",
"options": [
{
"text": "825oC "
},
{
"text": "800oC"
},
{
"text": "885oC"
},
{
"text": "1250oC"
}
],
"answer": "885oC",
"solution": "**Answer:** 885oC\n\nAccording to principle of calorimetry,\n
T = 885°C",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9830,
"subject": "Physics",
"question": "In an experiment, a sphere of aluminium of mass 0.20 kg is heated upto 150oC. Immediately, it is put into water of volume 150 cc at 27oC kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is 40oC. The specific heat of aluminium is : (take 4.2 Joule = 1 calorie)\n",
"options": [
{
"text": "378 J/kg $$-$$oC"
},
{
"text": "315 J/kg $$-$$oC"
},
{
"text": "476 J/kg $$-$$oC"
},
{
"text": "434 J/kg $$-$$oC"
}
],
"answer": "434 J/kg $$-$$oC",
"solution": "**Answer:** 434 J/kg $$-$$oC\n\nLet specific heat of aluminium = S, \n
$$ \\Rightarrow $$$$\\,\\,\\,$$ S = 434 J/kg - oC",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9831,
"subject": "Physics",
"question": "An unknown metal of mass 192 g heated to a temperature of 100oC was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4oC. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5oC. (Specific heat of brass is 394 J kg–1 K–1)",
"options": [
{
"text": "458 J kg–1 K–1"
},
{
"text": "1232 J kg–1 K–1"
},
{
"text": "654 J kg–1 K–1"
},
{
"text": "916 J kg–1 K–1"
}
],
"answer": "916 J kg–1 K–1",
"solution": "**Answer:** 916 J kg–1 K–1\n\n192 $$ \\times $$ S $$ \\times $$ (100 $$-$$ 21.5)\n
$$ \\Rightarrow $$ S = 916",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9832,
"subject": "Physics",
"question": "Ice at –20oC is added to 50 g of water at 40oC. When the temperature of the mixture reaches 0oC, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water = 4.2J/g/oC Specific heat of Ice = 2.1J/g/oC Heat of fusion of water at 0oC= 334J/g)\n",
"options": [
{
"text": "100 g"
},
{
"text": "60 g"
},
{
"text": "50 g"
},
{
"text": "40 g"
}
],
"answer": "40 g",
"solution": "**Answer:** 40 g\n\nLet amount of ice is m gm.\n
According to principal of calorimeter heat taken by ice = heat given by water\n
m = 40.1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9833,
"subject": "Physics",
"question": "When 100 g of a liquid A at 100oC is added to 50 g of a liquid B at temperature 75oC, the temperature of the mixture becomes 90oC. The temperature of the mixture, if 100 g of liquid A at 100oC is added to 50 g of liquid B at 50oC, will be : ",
"options": [
{
"text": "60oC"
},
{
"text": "70oC"
},
{
"text": "85oC"
},
{
"text": "80oC"
}
],
"answer": "80oC",
"solution": "**Answer:** 80oC\n\n100 $$ \\times $$ SA $$ \\times $$ [100 $$-$$ 90] = 50 $$ \\times $$ SB $$ \\times $$ (90 $$-$$ 75)\n
T = 80",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9834,
"subject": "Physics",
"question": "A metal ball of mass 0.1 kg is heated upto 500oC and dropped into a vessel of heat capacity 800 JK–1 and containing 0.5 kg water. The initial temperature of water and vessel is 30oC. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, 4200 Jkg–1 and 400 Jkg–1 K–1\n",
"options": [
{
"text": "20%"
},
{
"text": "25%"
},
{
"text": "15%"
},
{
"text": "30%"
}
],
"answer": "20%",
"solution": "**Answer:** 20%\n\n0.1 $$ \\times $$ 400 $$ \\times $$ (500 $$-$$ T) = 0.5 $$ \\times $$ 4200 $$ \\times $$ (T $$-$$ 30) + 800 (T $$-$$ 30)\n
$$ \\simeq $$ 20%",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9835,
"subject": "Physics",
"question": "A thermally insulated vessel contains 150g of\nwater at 0°C. Then the air from the vessel is\npumped out adiabatically. A fraction of water\nturns into ice and the rest evaporates at 0°C\nitself. The mass of evaporated water will be\nclosest to :\n(Latent heat of vaporization of water\n= 2.10 × 106 J kg–1 and Latent heat of Fusion\nof water = 3.36 × 105 J kg–1)",
"options": [
{
"text": "35 g"
},
{
"text": "130 g"
},
{
"text": "20 g"
},
{
"text": "150 g"
}
],
"answer": "20 g",
"solution": "**Answer:** 20 g\n\n
Let x grams of water is evaporated.
\n
According to the principle of calorimetry,
\n
Heat lost by freezing water (that turns into ice) = Heat gained by evaporated water
$$ \\Rightarrow x = {{150} \\over {7.25}} = 20.6$$
\n
$$\\therefore$$ $$x \\approx 20g$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9836,
"subject": "Physics",
"question": "A massless spring (k = 800 N/m), attached with\na mass (500 g) is completely immersed in 1 kg\nof water. The spring is stretched by 2 cm and\nreleased so that it starts vibrating. What would\nbe the order of magnitude of the change in the\ntemperature of water when the vibrations stop\ncompletely ? (Assume that the water container\nand spring receive negligible heat and specific\nheat of mass = 400 J/kg K, specific heat of\nwater = 4184 J/kg K)",
"options": [
{
"text": "10–3 K"
},
{
"text": "10–1 K"
},
{
"text": "10–5K"
},
{
"text": "10–4 K"
}
],
"answer": "10–5K",
"solution": "**Answer:** 10–5K\n\nBy law of conservation of energy
\n$$\\Delta T = {{16 \\times {{10}^{ - 2}}} \\over {4384}} = 3.65 \\times {10^{ - 5}}$$ K",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9837,
"subject": "Physics",
"question": "When M1 gram of ice at –10oC (specific heat = 0.5 cal g–1\noC–1\n) is added to M2 gram of water at 50C, finally\nno ice is left and the water is at 0°C. The value of latent heat of ice, in cal g–1\n is :",
"options": [
{
"text": "$${{50{M_2}} \\over {{M_1}}} - 5$$"
},
{
"text": "$${{50{M_2}} \\over {{M_1}}}$$"
},
{
"text": "$${{5{M_2}} \\over {{M_1}}} - 5$$"
},
{
"text": "$${{5{M_1}} \\over {{M_2}}} - 50$$"
}
],
"answer": "$${{50{M_2}} \\over {{M_1}}} - 5$$",
"solution": "**Answer:** $${{50{M_2}} \\over {{M_1}}} - 5$$\n\nHeat lost = Heat gain
\n$$ = {{50{M_2}} \\over {{M_1}}} - 5$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9838,
"subject": "Physics",
"question": "One kg of water, at 20oC, heated in an electric kettle whose heating element has a mean (temperature\naveraged) resistance of 20 $$\\Omega $$. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time\ntaken for water to evaporate fully, is close to :\n[Specific heat of water = 4200 J/(kg oC), Latent heat of water = 2260 kJ/kg]",
"options": [
{
"text": "10 minutes"
},
{
"text": "22 minutes"
},
{
"text": "3 minutes"
},
{
"text": "16 minutes"
}
],
"answer": "22 minutes",
"solution": "**Answer:** 22 minutes\n\nP$$ \\times $$t = mS$$\\Delta $$t + mLv\n
$$ \\Rightarrow $$ t = 1298 s = 21.6 min $$ \\simeq $$ 22 min\n",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9839,
"subject": "Physics",
"question": "A bullet of mass 5 g, travelling with a speed of\n210 m/s, strikes a fixed wooden target. One half\nof its kinetic energy is converted into heat in\nthe bullet while the other half is converted into\nheat in the wood. The rise of temperature of\nthe bullet if the specific heat of its material is\n0.030 cal/(g – oC) (1 cal = 4.2 × 107 ergs) close\nto :",
"options": [
{
"text": "87.5 oC"
},
{
"text": "83.3 oC"
},
{
"text": "38.4 oC"
},
{
"text": "119.2 oC"
}
],
"answer": "87.5 oC",
"solution": "**Answer:** 87.5 oC\n\n$${1 \\over 2}m{v^2} \\times {1 \\over 2} = ms\\Delta T$$
$$ = 87.5^\\circ C$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9840,
"subject": "Physics",
"question": "A calorimeter of water equivalent 20 g contains 180 g of water at 25oC. ‘m’ grams of steam at\n100oC is mixed in it till the temperature of the mixure is 31oC. The value of ‘m’ is close to : (Latent\nheat of water = 540 cal g–1, specific heat of water = 1 cal g–1oC–1)",
"options": [
{
"text": "2.6"
},
{
"text": "2"
},
{
"text": "4"
},
{
"text": "3.2"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\nGiven Temp of mixture = 31°C\n
$$ \\Rightarrow $$ m = $${{1200} \\over {609}}$$ = 1.97",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9841,
"subject": "Physics",
"question": "The specific heat of water = 4200 J kg-1K-1 and the latent heat of ice = 3.4 $$ \\times $$ 105 J kg–1. 100 grams of ice at 0oC is placed in 200 g of water at 25oC. The amount of ice that will melt as the temperature of water reaches 0oC is close to (in grams) :",
"options": [
{
"text": "63.8"
},
{
"text": "61.7"
},
{
"text": "69.3"
},
{
"text": "64.6"
}
],
"answer": "61.7",
"solution": "**Answer:** 61.7\n\nHeat loss by water
This heat will absorbed by the ice and let mass $$\\Delta $$mi got melted.\n
So $$\\Delta {m_i}L = 21000$$
$$\\Delta {m_i} = {{21000} \\over {3.4 \\times {{10}^5}}} \\times {10^3}\\,gm = 61.7\\,grams$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9842,
"subject": "Physics",
"question": "M grams of steam at 100oC is mixed with 200 g of ice at its melting point in a thermally insulated\ncontainer. If it produced liquid water at 40oC [heat of vaporization of water is 540 cal/g and heat\nof fusion of ice is 80 cal/g] the value of M is ____",
"options": [],
"answer": "40",
"solution": "**Answer:** 40\n\nM × 540 + M + 60 = 200 × 80 + 200 × 1× (40– 0)\n
$$ \\Rightarrow $$ 600 M = 24000\n
$$ \\Rightarrow $$ M = 40",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9843,
"subject": "Physics",
"question": "Two moles of an ideal gas with $${{{C_P}} \\over {{C_V}}} = {5 \\over 3}$$\nare mixed with 3 moles of another ideal gas\nwith $${{{C_P}} \\over {{C_V}}} = {4 \\over 3}$$. The value of $${{{C_P}} \\over {{C_V}}}$$ for the\nmixture is :\n",
"options": [
{
"text": "1.50"
},
{
"text": "1.45"
},
{
"text": "1.47"
},
{
"text": "1.42"
}
],
"answer": "1.42",
"solution": "**Answer:** 1.42\n\nCp = $${{{n_1}{C_{{p_1}}} + {n_2}{C_{{p_2}}}} \\over {{n_1} + {n_2}}}$$\n
= $${{5 + 12} \\over {3 + 9}}$$ = 1.42",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9844,
"subject": "Physics",
"question": "The amount of heat needed to raise the temperature of 4 moles of rigid diatomic gas from 0$$^\\circ$$ C to 50$$^\\circ$$ C when no work is done is ___________. (R is the universal gas constant).",
"options": [
{
"text": "500 R"
},
{
"text": "250 R"
},
{
"text": "750 R"
},
{
"text": "175 R"
}
],
"answer": "500 R",
"solution": "**Answer:** 500 R\n\nAccording to first law of thermodynamics,
where, $$\\Delta$$Q = quantity of heat energy supplied to the system, $$\\Delta$$U = change in the internal energy of a closed system and $$\\Delta$$W = work done by the system on its surroundings.
CV = specific heat capacity at constant volume for diatomic gas = $${{5R} \\over 2}$$
$$\\Delta$$T = change in temperature = (50 $$-$$ 0) = 50$$^\\circ$$C
n = number of moles = 4
$$\\Rightarrow$$ $$\\Delta$$Q = nCV$$\\Delta$$T
= $$4 \\times {{5R} \\over 2} \\times (50)$$ = 500 R = 500 R",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9845,
"subject": "Physics",
"question": "The temperature of equal masses of three different liquids x, y and z are 10$$^\\circ$$C, 20$$^\\circ$$C and 30$$^\\circ$$C respectively. The temperature of mixture when x is mixed with y is 16$$^\\circ$$C and that when y is mixed with z is 26$$^\\circ$$C. The temperature of mixture when x and z are mixed will be :",
"options": [
{
"text": "28.32$$^\\circ$$C"
},
{
"text": "25.62$$^\\circ$$C"
},
{
"text": "23.84$$^\\circ$$C"
},
{
"text": "20.28$$^\\circ$$C"
}
],
"answer": "23.84$$^\\circ$$C",
"solution": "**Answer:** 23.84$$^\\circ$$C\n\n
Tf = 23.84$$^\\circ$$C",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9846,
"subject": "Physics",
"question": "The height of victoria falls is 63 m. What is the difference in temperature of water at the top and at the bottom of fall?
[Given 1 cal = 4.2 J and specific heat of water = 1 cal g$$-$$1 $$^\\circ$$0C$$-$$1]",
"options": [
{
"text": "0.147$$^\\circ$$ C"
},
{
"text": "14.76$$^\\circ$$ C"
},
{
"text": "1.476$$^\\circ$$ C"
},
{
"text": "0.014$$^\\circ$$ C"
}
],
"answer": "0.147$$^\\circ$$ C",
"solution": "**Answer:** 0.147$$^\\circ$$ C\n\nChange in P.E. = Heat energy
A lead bullet penetrates into a solid object and melts. Assuming that 40% of its kinetic energy is used to heat it, the initial speed of bullet is :
\n
(Given : initial temperature of the bullet = 127$$^\\circ$$C, Melting point of the bullet = 327$$^\\circ$$C, Latent heat of fusion of lead = 2.5 $$\\times$$ 104 J kg$$-$$1, Specific heat capacity of lead = 125 J/kg K)
A geyser heats water flowing at a rate of 2.0 kg per minute from 30$$^\\circ$$C to 70$$^\\circ$$C. If geyser operates on a gas burner, the rate of combustion of fuel will be ___________ g min$$-$$1.
\n
[Heat of combustion = 8 $$\\times$$ 103 Jg$$-$$1, Specific heat of water = 4.2 Jg$$-$$1 $$^\\circ$$C$$-$$1]
A copper block of mass 5.0 kg is heated to a temperature of 500$$^\\circ$$C and is placed on a large ice block. What is the maximum amount of ice that can melt? [Specific heat of copper : 0.39 J g$$-$$1 $$^\\circ$$C$$-$$1 and latent heat of fusion of water : 335 J g$$-$$1]
A 100 g of iron nail is hit by a 1.5 kg hammer striking at a velocity of 60 ms$$-$$1. What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail?
\n
[Specific heat capacity of iron = 0.42 Jg$$-$$1 $$^\\circ$$C$$-$$1]
A block of ice of mass 120 g at temperature 0$$^\\circ$$C is put in 300 g of water at 25$$^\\circ$$C. The x g of ice melts as the temperature of the water reaches 0$$^\\circ$$C. The value of x is _____________.
\n
[Use specific heat capacity of water = 4200 Jkg$$-$$1K$$-$$1, Latent heat of ice = 3.5 $$\\times$$ 105 Jkg$$-$$1]
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9852,
"subject": "Physics",
"question": "A water heater of power $2000 \\mathrm{~W}$ is used to heat water. The specific heat capacity of water is $4200 \\mathrm{~J}$ $\\mathrm{kg}^{-1} \\mathrm{~K}^{-1}$. The efficiency of heater is $70 \\%$. Time required to heat $2 \\mathrm{~kg}$ of water from $10^{\\circ} \\mathrm{C}$ to $60^{\\circ} \\mathrm{C}$ is _________ s.
(Assume that the specific heat capacity of water remains constant over the temperature range of the water).",
"options": [],
"answer": "300",
"solution": "**Answer:** 300\n\n
The amount of heat energy required to raise the temperature of a substance can be calculated as:
\n\n
Q = m $$ \\times $$ c $$ \\times $$ ΔT
\n\n
where Q is the heat energy required, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.
\n\n
The time required to heat a substance can be calculated as :
\n\n
t = $${Q \\over P}$$
\n\n
where t is the time required, and P is the power of the heating device.
\n\n
The actual power output of the heating device can be calculated as:
\n\n
Pactual = Pinput $$ \\times $$ efficiency
\n\n
where P_input is the input power to the device and efficiency is the fraction of input power that is actually converted to useful power output.\n
Heat energy of 184 kJ is given to ice of mass 600 g at $$-12^\\circ \\mathrm{C}$$. Specific heat of ice is $$\\mathrm{2222.3~J~kg^{-1^\\circ}~C^{-1}}$$ and latent heat of ice in 336 $$\\mathrm{kJ/kg^{-1}}$$
\n
A. Final temperature of system will be 0$$^\\circ$$C.
\n
B. Final temperature of the system will be greater than 0$$^\\circ$$C.
\n
C. The final system will have a mixture of ice and water in the ratio of 5 : 1.
\n
D. The final system will have a mixture of ice and water in the ratio of 1 : 5.
\n
E. The final system will have water only.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "A and E only"
},
{
"text": "B and D only"
},
{
"text": "A and C only"
},
{
"text": "A and D only"
}
],
"answer": "A and D only",
"solution": "**Answer:** A and D only\n\n
Heat required to raise the temperature of ice to 0$$^\\circ$$C is
$$0.08 \\mathrm{~kg}$$ air is heated at constant volume through $$5^{\\circ} \\mathrm{C}$$. The specific heat of air at constant volume is $$0.17 \\mathrm{~kcal} / \\mathrm{kg}^{\\circ} \\mathrm{C}$$ and $$\\mathrm{J}=4.18$$ joule/$$\\mathrm{~cal}$$. The change in its internal energy is approximately.
To find the change in the internal energy of air when it is heated at constant volume, we use the formula for heat transfer at constant volume, which is given by:
\n
$$\\Delta U = m c_v \\Delta T$$
\n
Where:
\n
\n
$$\\Delta U$$ is the change in internal energy,
\n
$$m$$ is the mass of the substance (in this case, air),
\n
$$c_v$$ is the specific heat at constant volume,
\n
$$\\Delta T$$ is the change in temperature.
\n
\n
Given that:
\n
\n
The mass of air, $$m = 0.08 \\, \\text{kg},$$
\n
The specific heat of air at constant volume, $$c_v = 0.17 \\, \\text{kcal/kg}^{\\circ}\\text{C},$$
\n
The change in temperature, $$\\Delta T = 5^{\\circ} \\text{C},$$
\n
And the conversion factor from calories to Joules, $$1 \\, \\text{cal} = 4.18 \\, \\text{J}.$$
\n
\n
First, convert the specific heat from kcal to Joules:
Statement (I) : Dimensions of specific heat is $$[\\mathrm{L}^2 \\mathrm{~T}^{-2} \\mathrm{~K}^{-1}]$$.
\n
Statement (II) : Dimensions of gas constant is $$[\\mathrm{M} \\mathrm{L}^2 \\mathrm{~T}^{-1} \\mathrm{~K}^{-1}]$$.
\n
In the light of the above statements, choose the most appropriate answer from the options given below.
",
"options": [
{
"text": "Statement (I) is incorrect but statement (II) is correct\n"
},
{
"text": "Both statement (I) and statement (II) are incorrect\n"
},
{
"text": "Both statement (I) and statement (II) are correct\n"
},
{
"text": "Statement (I) is correct but statement (II) is incorrect"
}
],
"answer": "Statement (I) is correct but statement (II) is incorrect",
"solution": "**Answer:** Statement (I) is correct but statement (II) is incorrect\n\n
To evaluate the veracity of the given statements, we need to understand the physical quantities involved and their dimensional formulas. Specifically, we're looking at the dimensions of specific heat and gas constant.
\n\n
Specific Heat:
\n\n
Specific heat (c) is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin, since the increment is the same in both scales). Its formula is $$q = mc\\Delta T$$, where $$q$$ is the heat added, $$m$$ is the mass, $$c$$ is the specific heat, and $$\\Delta T$$ is the change in temperature.
\n\n
From the formula, we can deduce the dimensions of specific heat as follows:
\n\n
$$[q] = [m][c][\\Delta T]$$
\n\n
Knowing that the dimension of heat (q) is equivalent to energy, which is $$[\\mathrm{M} \\mathrm{L}^2 \\mathrm{T}^{-2}]$$, the mass (m) is $$[\\mathrm{M}]$$, and temperature ($\\Delta T$) is $$[\\mathrm{K}]$$, we can solve for $$[c]$$:
So, $$[c] = [\\mathrm{L}^2 \\mathrm{T}^{-2} \\mathrm{K}^{-1}]$$
\n\n
This reveals that Statement (I) is correct.
\n\n
Gas Constant:
\n\n
The gas constant (R) appears in the ideal gas law, represented as $$PV = nRT$$, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature. The dimensions of the gas constant can be derived from this relation.
\n\n
Pressure (P) has dimensions $$[\\mathrm{M} \\mathrm{L}^{-1} \\mathrm{T}^{-2}]$$, volume (V) has dimensions $$[\\mathrm{L}^3]$$, and temperature (T) has dimensions $$[\\mathrm{K}]$$.
This indicates that Statement (II) has stated the dimensions incorrectly, presenting them as $$[\\mathrm{M} \\mathrm{L}^2 \\mathrm{T}^{-1} \\mathrm{K}^{-1}]$$ when it should be $$[\\mathrm{M} \\mathrm{L}^2 \\mathrm{T}^{-2} \\mathrm{K}^{-1}]$$, making Statement (II) incorrect.
\n\n
Based on the analysis:
\n\n
Option A, \"Statement (I) is incorrect but statement (II) is correct,\" is wrong because Statement (I) is correct.
\n\n
Option B, \"Both statement (I) and statement (II) are incorrect,\" is wrong because Statement (I) is correct.
\n\n
Option C, \"Both statement (I) and statement (II) are correct,\" is wrong because Statement (II) is incorrect.
\n\n
Option D, \"Statement (I) is correct but statement (II) is incorrect,\" is the correct choice, reflecting the true nature of the statements provided.
$$\\begin{aligned}\n& \\because \\quad P V^2=R T \\\\\n& P(2 v d v)+V^2(d P)=R d T\n\\end{aligned}$$
\n
at $$P=$$ const.
\n
$$P d v=\\frac{R d T}{2 V} \\quad \\text{... (i)}$$
\n
Now, for $$n=1$$
\n
$$\\begin{aligned}\n& d \\theta=d v+d w \\\\\n& C_P d T=C_v d T+P d v \\quad \\text{... (ii)}\n\\end{aligned}$$
\n
from (i) and (ii)
\n
$$C_P=C_V+\\frac{R}{2 V}$$
",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 9857,
"subject": "Physics",
"question": "During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio $${C_p}/{C_V}$$ for the gas is ",
"options": [
{
"text": "$${4 \\over 3}$$ "
},
{
"text": "$$2$$ "
},
{
"text": "$${5 \\over 3}$$"
},
{
"text": "$${3 \\over 2}$$"
}
],
"answer": "$${3 \\over 2}$$",
"solution": "**Answer:** $${3 \\over 2}$$\n\n$$P \\propto {T^3} \\Rightarrow P{T^{ - 3}} = $$ constant ....$$(i)$$\n But for an adiabatic process, the pressure temperature relationship is given by\n $${P^{1 - \\gamma }}\\,\\,{T^\\gamma } = $$ constant $$ \\Rightarrow P{T^{{\\gamma \\over {1 - \\gamma }}}} = $$ constant. ....$$(ii)$$\n From $$(i)$$ and $$(ii)$$ $${\\gamma \\over {1 - \\gamma }} = - 3 \\Rightarrow \\gamma = - 3 + 3\\gamma \\Rightarrow \\gamma = {3 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9858,
"subject": "Physics",
"question": "The work of $$146$$ $$kJ$$ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by $${7^ \\circ }C.$$ The gas is $$\\left( {R = 8.3J\\,\\,mo{l^{ - 1}}\\,{K^{ - 1}}} \\right)$$ ",
"options": [
{
"text": "diatomic "
},
{
"text": "triatomic "
},
{
"text": "a mixture of monoatomic and diatomic "
},
{
"text": "monoatomic"
}
],
"answer": "diatomic ",
"solution": "**Answer:** diatomic \n\n$$W = {{nR\\Delta T} \\over {1 - \\gamma }} \\Rightarrow - 146000$$\n $$ = {{1000 \\times 8.3 \\times 7} \\over {1 - \\gamma }}$$\n or $$1 - \\gamma = - {{58.1} \\over {146}} \\Rightarrow \\gamma $$\n $$ = 1 + {{58.1} \\over {146}} = 1.4$$ \n Hence the gas is diatomic.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9859,
"subject": "Physics",
"question": "Consider a spherical shell of radius $$R$$ at temperature $$T$$. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume $$u = {U \\over V}\\, \\propto \\,{T^4}$$ and pressure $$p = {1 \\over 3}\\left( {{U \\over V}} \\right)$$ . If the shell now undergoes an adiabatic expansion the relation between $$T$$ and $$R$$ is: ",
"options": [
{
"text": "$$T\\, \\propto {1 \\over R}$$ "
},
{
"text": "$$T\\, \\propto {1 \\over {{R^3}}}$$ "
},
{
"text": "$$T\\, \\propto \\,{e^{ - R}}$$ "
},
{
"text": "$$T\\, \\propto \\,{e^{ - 3R}}$$ "
}
],
"answer": "$$T\\, \\propto {1 \\over R}$$ ",
"solution": "**Answer:** $$T\\, \\propto {1 \\over R}$$ \n\nAs, $$P = {1 \\over 3}\\left( {{U \\over V}} \\right)$$\n
But $$\\,\\,\\,\\,$$ $${U \\over V} = KT{}^4$$\n
So, $$\\,\\,\\,\\,\\,P = {1 \\over 3}K{T^4}$$\n
or $$\\,\\,\\,\\,{{uRT} \\over V} = {1 \\over 3}K{T^4}\\,\\,\\,\\,$$ \n
$$\\left[ \\, \\right.$$ As $$PV = uRT$$ $$\\left. \\, \\right]$$\n
[Cp = $${5 \\over 2}$$ R for monoatomic gas]",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9862,
"subject": "Physics",
"question": "Two moles of an ideal monatomic gas occupies a volume V at 27oC. The gas expands adiabatically to a\nvolume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.",
"options": [
{
"text": "(a) 195 K (b) 2.7 kJ"
},
{
"text": "(a) 189 K (b) 2.7 kJ"
},
{
"text": "(a) 195 K (b) –2.7 kJ"
},
{
"text": "(a) 189 K (b) – 2.7 kJ"
}
],
"answer": "(a) 189 K (b) – 2.7 kJ",
"solution": "**Answer:** (a) 189 K (b) – 2.7 kJ\n\nFor adiabatic process, \n
pv$$\\gamma $$ = constant.\n
and we know, pv = nRT\n
$$\\therefore\\,\\,\\,$$ p = $${{nRT} \\over v}$$\n
= $${5 \\over 4}{{{P_0}{V_0}} \\over R}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9865,
"subject": "Physics",
"question": "n moles of an ideal gas with constant volume\nheat capcity CV undergo an isobaric expansion\nby certain volume. The ratio of the work done\nin the process, to the heat supplied is :",
"options": [
{
"text": "$${{nR} \\over {{C_V} - nR}}$$"
},
{
"text": "$${{4nR} \\over {{C_V} - nR}}$$"
},
{
"text": "$${{4nR} \\over {{C_V} + nR}}$$"
},
{
"text": "$${{nR} \\over {{C_V} + nR}}$$"
}
],
"answer": "$${{nR} \\over {{C_V} + nR}}$$",
"solution": "**Answer:** $${{nR} \\over {{C_V} + nR}}$$\n\nw = nR$$\\Delta $$T \n$$\\Delta $$H = (Cv + nR)$$\\Delta $$T \n$${\\omega \\over {\\Delta H}} = {{nR} \\over {{C_v} + nR}}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9866,
"subject": "Physics",
"question": "In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by $$\\Delta $$T. The\namount of heat absorbed by gas is (R is gas constant) : ",
"options": [
{
"text": "$${1 \\over 2}$$ KR$$\\Delta $$T"
},
{
"text": "$${1 \\over 2}$$ R$$\\Delta $$T"
},
{
"text": "$${3 \\over 2}$$ R$$\\Delta $$T"
},
{
"text": "$${2K \\over 3}$$ $$\\Delta $$T"
}
],
"answer": "$${1 \\over 2}$$ R$$\\Delta $$T",
"solution": "**Answer:** $${1 \\over 2}$$ R$$\\Delta $$T\n\nVT = K\n
= $${R \\over 2} \\times \\Delta $$T",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9867,
"subject": "Physics",
"question": "A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is TVx = constant, then x is : ",
"options": [
{
"text": "$${5 \\over 3}$$"
},
{
"text": "$${2 \\over 5}$$"
},
{
"text": "$${3 \\over 5}$$"
},
{
"text": "$${2 \\over 3}$$"
}
],
"answer": "$${2 \\over 5}$$",
"solution": "**Answer:** $${2 \\over 5}$$\n\nFor adiabatic process : TV$$\\gamma $$$$-$$1 = constant\n
For diatomic process : $$\\gamma $$$$-$$1 = $${7 \\over 5} - 1$$\n
$$ \\therefore $$ x = $${2 \\over 5}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9868,
"subject": "Physics",
"question": "In an adiabatic process, the density of a\ndiatomic gas becomes 32 times its initial value.\nThe final pressure of the gas is found to be n\ntimes the initial pressure. The value of n is :",
"options": [
{
"text": "128"
},
{
"text": "32"
},
{
"text": "326"
},
{
"text": "$${1 \\over {32}}$$"
}
],
"answer": "128",
"solution": "**Answer:** 128\n\nIn adiabatic process\n
[ For diatomic gas $$\\gamma $$ = $${{7 \\over 5}}$$ ]",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9869,
"subject": "Physics",
"question": "Match the thermodynamic processes taking place in a system with the correct conditions. In the\ntable : $$\\Delta $$Q is the heat supplied, $$\\Delta $$W is the work done and $$\\Delta $$U is change in internal energy of the\nsystem.\n
(IV) Isobaric process $$ \\Rightarrow $$ Pressure remains constant\n
W = P.$$\\Delta $$V $$ \\ne $$ 0\n
$$\\Delta $$U $$ \\ne $$ 0\n
$$\\Delta $$Q = nCp$$\\Delta $$T $$ \\ne $$ 0",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9870,
"subject": "Physics",
"question": "A balloon filled with helium (32oC and 1.7 atm.)\nbursts. Immediately afterwards the expansion\nof helium can be considered as",
"options": [
{
"text": "Irreversible adiabatic"
},
{
"text": "Reversible adiabatic"
},
{
"text": "Irreversible isothermal"
},
{
"text": "Reversible isothermal"
}
],
"answer": "Irreversible adiabatic",
"solution": "**Answer:** Irreversible adiabatic\n\nBursting of helium ballon is irreversible adiabatic because Energy can not be restored.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9871,
"subject": "Physics",
"question": "Starting at temperature 300 K, one mole of an\n ideal diatomic gas ($$\\gamma $$ = 1.4) is first compressed\n adiabatically from volume V1 to V2 = $${{{V_1}} \\over {16}}$$. It is\n then allowed to expand isobarically to volume\n2V2. If all the processes are the quasi-static then\n the final temperature of the gas (in oK) is (to\nthe nearest integer) _____.",
"options": [],
"answer": "1818TO1819",
"solution": "**Answer:** 1818TO1819\n\nT1V1$$\\gamma $$–1 = T2V2\n$$\\gamma $$–1\n
A $$\\to$$ 2, B $$\\to$$ 3, C $$\\to$$ 4, D $$\\to$$ 1",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9874,
"subject": "Physics",
"question": "In a certain thermodynamical process, the pressure of a gas depends on its volume as kV3. The work done when the temperature changes from 100$$^\\circ$$C to 300$$^\\circ$$C will be ___________ nR, where n denotes number of moles of a gas.",
"options": [],
"answer": "50",
"solution": "**Answer:** 50\n\n$$P = k{v^3}$$
$$ \\Rightarrow $$ $$p{v^{ - 3}} = k$$
$$ \\Rightarrow $$ $$x = - 3$$
$$w = {{nR({T_1} - {T_2})} \\over {x - 1}}$$
$$ = {{nR(100 - 300)} \\over { - 3 - 1}}$$
$$ = {{nR( - 200)} \\over { - 4}}$$
$$ = 50$$ nR",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9875,
"subject": "Physics",
"question": "The volume V of a given mass of monoatomic gas changes with temperature T according to the relation $$V = K{T^{{2 \\over 3}}}$$. The workdone when temperature changes by 90K will be xR. The value of x is _________. [R = universal gas constant]",
"options": [],
"answer": "60",
"solution": "**Answer:** 60\n\nWe know that work done is
So, W = 60R",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9876,
"subject": "Physics",
"question": "For an adiabatic expansion of an ideal gas, the fractional change in its pressure is equal to (where $$\\gamma$$ is the ratio of specific heats) :",
"options": [
{
"text": "$$ - {1 \\over \\gamma }{{dV} \\over V}$$"
},
{
"text": "$$ - \\gamma {V \\over {dV}}$$"
},
{
"text": "$$ - \\gamma {{dV} \\over V}$$"
},
{
"text": "$${{dV} \\over V}$$"
}
],
"answer": "$$ - \\gamma {{dV} \\over V}$$",
"solution": "**Answer:** $$ - \\gamma {{dV} \\over V}$$\n\nfor adiabatic expansion :
PV$$\\gamma$$ = const.
$$ \\Rightarrow $$ ln P + $$\\gamma$$ln v = const.
$$ \\Rightarrow $$ differentiating both sides;
$${{dp} \\over p} + \\gamma {{dv} \\over v} = 0$$
$$ \\Rightarrow {{dp} \\over p} = - \\gamma {{dv} \\over V}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9877,
"subject": "Physics",
"question": "A monoatomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If l1 and l2 are the lengths of the gas column, before and after the expansion respectively, then the value of $${{{T_1}} \\over {{T_2}}}$$ will be :",
"options": [
{
"text": "$${\\left( {{{{l_1}} \\over {{l_2}}}} \\right)^{{2 \\over 3}}}$$"
},
{
"text": "$${\\left( {{{{l_2}} \\over {{l_1}}}} \\right)^{{2 \\over 3}}}$$"
},
{
"text": "$${{{l_2}} \\over {{l_1}}}$$"
},
{
"text": "$${{{l_1}} \\over {{l_2}}}$$"
}
],
"answer": "$${\\left( {{{{l_2}} \\over {{l_1}}}} \\right)^{{2 \\over 3}}}$$",
"solution": "**Answer:** $${\\left( {{{{l_2}} \\over {{l_1}}}} \\right)^{{2 \\over 3}}}$$\n\nPVr = const.
TVr $$-$$ 1 = const.
$$T{(l)^{{5 \\over 3} - 1}}$$ = const.
$${{{T_1}} \\over {{T_2}}} = {\\left( {{{{l_2}} \\over {{l_1}}}} \\right)^{{2 \\over 3}}}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9878,
"subject": "Physics",
"question": "One mole of an ideal gas is taken through an adiabatic process where the temperature rises from 27$$^\\circ$$ C to 37$$^\\circ$$ C. If the ideal gas is composed of polyatomic molecule that has 4 vibrational modes, which of the following is true? [R = 8.314 J mol$$-$$1 k$$-$$1]",
"options": [
{
"text": "work done by the gas is close to 332 J"
},
{
"text": "work done on the gas is close to 582 J"
},
{
"text": "work done by the gas is close to 582 J"
},
{
"text": "work done on the gas is close to 332 J"
}
],
"answer": "work done on the gas is close to 582 J",
"solution": "**Answer:** work done on the gas is close to 582 J\n\nSince, each vibrational mode, corresponds to two degrees of freedom, hence, f = 3 (trans.) + 3 (rot.) + 4 $$ \\times $$ 2 (vib.) = 14
As W < 0. work is done on the gas.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9879,
"subject": "Physics",
"question": "A sample of gas with $$\\gamma$$ = 1.5 is taken through an adiabatic process in which the volume is compressed from 1200 cm3 to 300 cm3. If the initial pressure is 200 kPa. The absolute value of the workdone by the gas in the process = _____________ J.",
"options": [],
"answer": "480",
"solution": "**Answer:** 480\n\nv = 1.5
Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2, if the process is purely adiabatic and W3 if the process is purely isobaric. Then, choose the correct option
Statement I : When $$\\mu$$ amount of an ideal gas undergoes adiabatic change from state (P1, V1, T1) to state (P2, V2, T2), then work done is $$W = {{\\mu R({T_2} - {T_1})} \\over {1 - \\gamma }}$$, where $$\\gamma = {{{C_p}} \\over {{C_v}}}$$ and R = universal gas constant.
\n
Statement II : In the above case, when work is done on the gas, the temperature of the gas would rise.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "Both Statement I and Statement II are true."
},
{
"text": "Both Statement I and Statement II are false."
},
{
"text": "Statement I is true but Statement II is false."
},
{
"text": "Statement I is false but Statement II is true."
}
],
"answer": "Both Statement I and Statement II are true.",
"solution": "**Answer:** Both Statement I and Statement II are true.\n\n
$$W = {{\\mu R({T_2} - {T_1})} \\over {1 - r}}$$ for a polytropic process for adiabatic process r = $$\\gamma$$
\n
$$\\Rightarrow$$ Statement I is true.
\n
In an adiabatic process
\n
$$\\Delta$$U = $$-$$ $$\\Delta$$W
\n
$$\\Rightarrow$$ If work is done on the gas
\n
$$\\Rightarrow$$ $$\\Delta$$W is negative
\n
$$\\Rightarrow$$ $$\\Delta$$U is positive or temperature increases
A sample of monoatomic gas is taken at initial pressure of 75 kPa. The volume of the gas is then compressed from 1200 cm3 to 150 cm3 adiabatically. In this process, the value of workdone on the gas will be :
A certain amount of gas of volume $$\\mathrm{V}$$ at $$27^{\\circ} \\mathrm{C}$$ temperature and pressure $$2 \\times 10^{7} \\mathrm{Nm}^{-2}$$ expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use $$\\gamma=1.5)$$ :
A monoatomic gas at pressure $$\\mathrm{P}$$ and volume $$\\mathrm{V}$$ is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be :\n
The pressure $$\\mathrm{P}_{1}$$ and density $$\\mathrm{d}_{1}$$ of diatomic gas $$\\left(\\gamma=\\frac{7}{5}\\right)$$ changes suddenly to $$\\mathrm{P}_{2}\\left(>\\mathrm{P}_{1}\\right)$$ and $$\\mathrm{d}_{2}$$ respectively during an adiabatic process. The temperature of the gas increases and becomes ________ times of its initial temperature. (given $$\\frac{\\mathrm{d}_{2}}{\\mathrm{~d}_{1}}=32$$)
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9887,
"subject": "Physics",
"question": "A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 litres. If the ratio of final pressure of the gas to initial pressure of the gas is $\\frac{16}{81}$. Then the ratio of $\\frac{\\mathrm{Cp}}{\\mathrm{Cv}}$ will be.",
"options": [
{
"text": "$\\frac{3}{1}$"
},
{
"text": "$\\frac{4}{3}$"
},
{
"text": "$\\frac{1}{2}$"
},
{
"text": "$\\frac{3}{2}$"
}
],
"answer": "$\\frac{4}{3}$",
"solution": "**Answer:** $\\frac{4}{3}$\n\nLet $\\gamma$ be the ratio of $\\frac{C_{p}}{C_{v}}$\n\n
A sample of gas at temperature $$T$$ is adiabatically expanded to double its volume. The work done by the gas in the process is $$\\left(\\mathrm{given}, \\gamma=\\frac{3}{2}\\right)$$ :
Heat is given to an ideal gas in an isothermal process.
\n
A. Internal energy of the gas will decrease.
\n
B. Internal energy of the gas will increase.
\n
C. Internal energy of the gas will not change.
\n
D. The gas will do positive work.
\n
E. The gas will do negative work.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "B and D only"
},
{
"text": "C and E only"
},
{
"text": "A and E only"
},
{
"text": "C and D only"
}
],
"answer": "C and D only",
"solution": "**Answer:** C and D only\n\n
\nHeat absorbed goes partly to increase internal energy and partly to do work.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9891,
"subject": "Physics",
"question": "
1 g of a liquid is converted to vapour at 3 $$\\times$$ 10$$^5$$ Pa pressure. If 10% of the heat supplied is used for increasing the volume by 1600 cm$$^3$$ during this phase change, then the increase in internal energy in the process will be :
The Thermodynamic process, in which internal energy of the system remains constant is
",
"options": [
{
"text": "Isobaric"
},
{
"text": "Isochoric"
},
{
"text": "Adiabatic"
},
{
"text": "Isothermal"
}
],
"answer": "Isothermal",
"solution": "**Answer:** Isothermal\n\nIf the temperature (T) remains constant, the internal energy (U) also remains constant, since the internal energy of an ideal gas depends only on its temperature.\n
\nIn this case, the thermodynamic process in which the internal energy of the system remains constant is an isothermal process. Isothermal processes occur at constant temperature, and for an ideal gas, this means that the internal energy remains constant as well.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9893,
"subject": "Physics",
"question": "
Consider two containers A and B containing monoatomic gases at the same Pressure (P), Volume (V) and Temperature (T). The gas in A is compressed isothermally to $$\\frac{1}{8}$$ of its original volume while the gas in B is compressed adiabatically to $$\\frac{1}{8}$$ of its original volume. The ratio of final pressure of gas in B to that of gas in A is
The final pressure of gas in container A after isothermal compression can be found using the equation of state for an ideal gas, $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature. For an isothermal process, the temperature $T$ is constant, so the equation becomes $P_1V_1 = P_2V_2$.
The final volume is $\\frac{1}{8}$ of the initial volume, so $P_2 = P_1 \\times \\frac{V_1}{V_2} = P_1 \\times 8 = 8P_1$.
\n
The final pressure of gas in container B after adiabatic compression can be found using the adiabatic equation for an ideal gas, $PV^\\gamma = \\text{constant}$, where $\\gamma$ is the ratio of the heat capacities, which is $\\frac{5}{3}$ for a monoatomic gas.
Since $V_2 = \\frac{V_1}{8}$, we have $P_2 = P_1 \\times \\left(\\frac{V_1}{V_2}\\right) ^ \\gamma = P_1 \\times 8^\\frac{5}{3} = P_1 \\times 2^5 = 32P_1$.
\n
The ratio of the final pressure of gas in B to that of gas in A is therefore $\\frac{32P_1}{8P_1} = 4$.
Statement I: If heat is added to a system, its temperature must increase.
\n
Statement II: If positive work is done by a system in a thermodynamic process, its volume must increase.
\n
In the light of the above statements, choose the correct answer from the options given below
",
"options": [
{
"text": "Both Statement I and Statement II are true"
},
{
"text": "Statement I is false but Statement II is true"
},
{
"text": "Statement I is true but Statement II is false"
},
{
"text": "Both Statement I and Statement II are false"
}
],
"answer": "Statement I is false but Statement II is true",
"solution": "**Answer:** Statement I is false but Statement II is true\n\n
Statement I: If heat is added to a system, its temperature must increase. This statement is not necessarily true. For example, in a phase transition (like melting or boiling), heat can be added to a system without increasing its temperature. The added heat energy is used to break intermolecular bonds and change the phase of the substance, not to increase the kinetic energy of the particles (which would raise the temperature).
\n
Statement II: If positive work is done by a system in a thermodynamic process, its volume must increase. This statement is generally true, as positive work being done by a system often involves expansion against an external pressure, thus increasing its volume.
\n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9895,
"subject": "Physics",
"question": "A diatomic gas $(\\gamma=1.4)$ does $200 \\mathrm{~J}$ of work when it is expanded isobarically. The heat given to the gas in the process is :",
"options": [
{
"text": "$800 \\mathrm{~J}$"
},
{
"text": "$600 \\mathrm{~J}$"
},
{
"text": "$700 \\mathrm{~J}$"
},
{
"text": "$850 \\mathrm{~J}$"
}
],
"answer": "$700 \\mathrm{~J}$",
"solution": "**Answer:** $700 \\mathrm{~J}$\n\n$\\begin{aligned} & \\gamma=1+\\frac{2}{\\mathrm{f}}=1.4 \\Rightarrow \\frac{2}{\\mathrm{f}}=0.4 \\\\\\\\ & \\Rightarrow \\mathrm{f}=5 \\\\\\\\ & \\mathrm{~W}=\\mathrm{nR} \\Delta \\mathrm{T}=200 \\mathrm{~J} \\\\\\\\ & \\mathrm{Q}=\\left(\\frac{\\mathrm{f}+2}{2}\\right) \\mathrm{nR} \\Delta \\mathrm{T} \\\\\\\\ & =\\frac{7}{2} \\times 200=700 \\mathrm{~J}\\end{aligned}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9896,
"subject": "Physics",
"question": "
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio of $$\\frac{\\mathrm{Cp}}{\\mathrm{Cv}}$$ for the gas is :
To find the ratio of the initial pressure to the final pressure of an ideal gas undergoing an adiabatic process, we can use the adiabatic equation for an ideal gas, which relates pressure (P) and volume (V) as follows:
\n\n
$P_{1}V_{1}^{\\gamma} = P_{2}V_{2}^{\\gamma}$
\n\n
Here, $P_{1}$ and $V_{1}$ are the initial pressure and volume, respectively, $P_{2}$ and $V_{2}$ are the final pressure and volume, respectively, and $\\gamma$ is the heat capacity ratio of the gas.
\n\n
Given in the question, $\\gamma = 1.5$, $V_{1} = 5$ litres and $V_{2} = 4$ litres. We need to find the ratio $\\frac{P_{1}}{P_{2}}$.
\n\n
Rearranging the adiabatic equation for the ratio $\\frac{P_{1}}{P_{2}}$, we get:
A sample of 1 mole gas at temperature $$T$$ is adiabatically expanded to double its volume. If adiab constant for the gas is $$\\gamma=\\frac{3}{2}$$, then the work done by the gas in the process is :
For an adiabatic process, the work done by the gas can be found using the formula:
\n\n
$$ W = \\frac{P_1 V_1 - P_2 V_2}{\\gamma - 1} $$
\n\n
Given that the volume is doubled ($$V_2 = 2V_1$$) and the adiabatic constant $$\\gamma = \\frac{3}{2}$$, we can manipulate the ideal gas law and the adiabatic process relationship to find an expression for work done in terms of the initial conditions and $$\\gamma$$.
\n\n
Recall, for an adiabatic process, $$PV^{\\gamma} = \\text{constant}$$, so we can write:
\n\n
$$ P_1V_1^{\\gamma} = P_2V_2^{\\gamma} $$
\n\n
Using the fact that $$V_2 = 2V_1$$, we can express $$P_2$$ in terms of $$P_1$$ and $$V_1$$ as follows:
A sample of gas at temperature $$T$$ is adiabatically expanded to double its volume. Adiabatic constant for the gas is $$\\gamma=3 / 2$$. The work done by the gas in the process is:
To find the heat given to a diatomic gas during an isobaric (constant pressure) expansion, we can use the formula that relates the work done by the gas, the heat added to the system, and the change in the internal energy of the system. The first law of thermodynamics states that:
\n\n
$$\\Delta Q = \\Delta U + W$$
\n\n
where:
\n\n
\n\n
$\\Delta Q$ is the heat added to the system,
\n\n
$\\Delta U$ is the change in internal energy of the system, and
\n\n
$W$ is the work done by the gas.
\n\n
\n\n
For an isobaric process, the work done $W$ is given by:
\n\n
$$W = P \\Delta V$$
\n\n
We're given that $W = 100 \\ \\mathrm{J}$ for this process, so:
\n\n
$$W = 100 \\ \\mathrm{J}$$
\n\n
The change in internal energy $\\Delta U$ for an ideal gas can also be related to the temperature change and the specific heat capacity at constant volume $C_v$. Using the equation:
\n\n
$$\\Delta U = nC_v\\Delta T$$
\n\n
However, without direct values for $n$, $\\Delta T$, or $C_v$, we need to rely on the relation between the provided work and the heat capacity ratio $\\gamma$ to find the heat added. For a diatomic gas, $\\gamma = C_p/C_v$. The heat added at constant pressure can also be described as:
\n\n
$$\\Delta Q = nC_p\\Delta T$$
\n\n
Since we know that for an ideal gas, the work done on the gas during an isobaric process is related to the heat added by the ratio of the specific heats ($\\gamma$), we can use the fact that $C_p = \\gamma C_v$, and relating that to the work done, we get:
\n\n
$$\\Delta Q = \\frac{\\gamma}{\\gamma - 1} W$$
\n\n
Substituting the known values, with $\\gamma = 1.4$, and $W = 100 \\ \\mathrm{J}$, we find:
During an adiabatic process, if the pressure of a gas is found to be proportional to the cube of its absolute temperature, then the ratio of $$\\frac{\\mathrm{C}_{\\mathrm{P}}}{\\mathrm{C}_{\\mathrm{V}}}$$ for the gas is :
To begin with, we're told that during an adiabatic process, the pressure $P$ of a gas is directly proportional to the cube of its absolute temperature $T$, that is, $P \\propto T^3$. From this, we can express the relation as $P = kT^3$, where $k$ is a constant.
\n\n
In an adiabatic process, $PV^\\gamma = \\text{constant}$, where $\\gamma = \\frac{C_P}{C_V}$, $P$ is the pressure, $V$ is the volume, $C_P$ is the heat capacity at constant pressure, and $C_V$ is the heat capacity at constant volume.
\n\n
Also, the ideal gas equation is $PV = nRT$, where $n$ is the amount of substance, $R$ is the ideal gas constant, and $T$ is the absolute temperature. This can be rearranged to express $P$ in terms of $T$ and $V$, resulting in $P = \\frac{nRT}{V}$.
\n\n
Given that $P = kT^3$, we can equate this to the expression we got from the ideal gas law: $\\frac{nRT}{V} = kT^3$. Simplifying, we get $V = \\frac{nR}{kT^2}$, which shows that $V$ is inversely proportional to $T^2$.
\n\n
Now, let's recall the adiabatic condition $PV^\\gamma = \\text{constant}$. Substituting the proportionalities $P \\propto T^3$ and $V \\propto T^{-2}$ into this equation, it becomes $T^3(T^{-2})^\\gamma = \\text{constant}$, simplifying to $T^{3-2\\gamma} = \\text{constant}$.
\n\n
For the expression $T^{3-2\\gamma} = \\text{constant}$ to hold true in an adiabatic process, where the only variable is temperature $T$, the exponent must equal to zero (as the quantity of $T$ to some power equals to a constant suggests that the change in $T$ does not alter the value of the expression). This implies $3 - 2\\gamma = 0$, solving for $\\gamma$:\n\n
Therefore, the ratio of $\\frac{C_P}{C_V}$ for the gas is $\\frac{3}{2}$, which matches with Option B $\\frac{3}{2}$.
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9902,
"subject": "Physics",
"question": "Statement - 1: The temperature dependence of resistance is usually given as $$R = {R_0}\\left( {1 + \\alpha \\,\\Delta t} \\right).$$ The resistance of wire changes from $$100\\Omega $$ to $$150\\Omega $$ when its temperature is increased from $${27^ \\circ }C$$ to $${227^ \\circ }C$$. This implies that $$\\alpha = 2.5 \\times {10^{ - 3}}/C.$$ \n
Statement - 2: $$R = {R_0}\\left( {1 + \\alpha \\,\\Delta t} \\right)$$ is valid only when the change in the temperature $$\\Delta T$$ is small and $$\\Delta T = \\left( {R - {R_0}} \\right) < < {R_0}.$$
",
"options": [
{
"text": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is the correct explanation of Statement - 1"
},
{
"text": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is not the correct explanation of Statement - 1"
},
{
"text": "Statement - 1 is false, Statement - 2 is true"
},
{
"text": "Statement - 1 is true, Statement - 2 is false"
}
],
"answer": "Statement - 1 is false, Statement - 2 is true",
"solution": "**Answer:** Statement - 1 is false, Statement - 2 is true\n\nThe relation $$R = {R_0}\\left( {1 + \\alpha \\,\\Delta t} \\right)$$ is valid for small values of $$\\Delta t$$ and $${R_0}$$ is resistance at $${0^ \\circ }C$$ and also $$\\left( {R - {R_0}} \\right)$$ should be much smaller than $${R_0}.$$ So, statement $$(1)$$ is wrong but statement $$(2)$$ is correct. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9903,
"subject": "Physics",
"question": "A pendulum clock loses $$12$$ $$s$$ a day if the temperature is $${40^ \\circ }C$$ and gains $$4$$ $$s$$ a day if the temperature is $${20^ \\circ }C.$$ The temperature at which the clock will show correct time, and the co-efficient of linear expansion $$\\left( \\alpha \\right)$$ of the metal of the pendulum shaft are respectively : ",
"options": [
{
"text": "$${30^ \\circ }C;\\,\\,\\alpha = 1.85 \\times {10^{ - 3}}/{}^ \\circ C$$ "
},
{
"text": "$${55^ \\circ }C;\\,\\,\\alpha = 1.85 \\times {10^{ - 2}}/{}^ \\circ C$$"
},
{
"text": "$${25^ \\circ }C;\\,\\,\\alpha = 1.85 \\times {10^{ - 5}}/{}^ \\circ C$$"
},
{
"text": "$${60^ \\circ }C;\\,\\,\\alpha = 1.85 \\times {10^{ - 4}}/{}^ \\circ C$$"
}
],
"answer": "$${25^ \\circ }C;\\,\\,\\alpha = 1.85 \\times {10^{ - 5}}/{}^ \\circ C$$",
"solution": "**Answer:** $${25^ \\circ }C;\\,\\,\\alpha = 1.85 \\times {10^{ - 5}}/{}^ \\circ C$$\n\nTime lost/gained per day $$ = {1 \\over 2} \\propto \\Delta \\theta \\times 86400$$ second\n
On dividing we get, $$\\,\\,\\,3 = {{40 - \\theta } \\over {\\theta - 20}}$$\n
$$3\\theta - 60 = 40 - \\theta $$\n
$$4\\theta = 100 \\Rightarrow \\theta = {25^ \\circ }C$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9904,
"subject": "Physics",
"question": "A simple pendulum made of a bob of mass m and a metallic wire of negligible mass has time period 2 s at T=0oC. If the temperature of the wire is increased and the corresponding change in its time period is\nplotted against its temperature, the resulting graph is a line of slope S. If the\ncoefficient of linear expansion of metal is $$\\alpha $$ then the value of S is :",
"options": [
{
"text": "$$\\alpha $$"
},
{
"text": "$${\\alpha \\over 2}$$"
},
{
"text": "2$$\\alpha $$"
},
{
"text": "$${1 \\over \\alpha }$$"
}
],
"answer": "$$\\alpha $$",
"solution": "**Answer:** $$\\alpha $$\n\nChange of length of wire with temperature, \n
$${{\\Delta T} \\over {\\Delta \\theta }}$$ is the shape of $$\\Delta $$T and $${\\Delta \\theta }$$\n
curve = S (given)\n
$$ \\therefore $$ S = $$\\alpha $$",
"topic": "Discrete Mathematics",
"subtopic": "Other"
},
{
"id": 9905,
"subject": "Physics",
"question": "An external pressure P is applied on a cube at 0oC so that it is equally compressed from all sides. K is the\nbulk modulus of the material of the cube and $$\\alpha$$ is its coefficient of linear expansion. Suppose we want to\nbring the cube to its original size by heating. The temperature should be raised by: ",
"options": [
{
"text": "$${P \\over {3\\alpha K}}$$ "
},
{
"text": "$${P \\over {\\alpha K}}$$"
},
{
"text": "$${3 \\alpha \\over {P K}}$$"
},
{
"text": "3PK$$\\alpha$$"
}
],
"answer": "$${P \\over {3\\alpha K}}$$ ",
"solution": "**Answer:** $${P \\over {3\\alpha K}}$$ \n\nAs we know, Bulk modulus\n
$$ \\Rightarrow $$ $${\\Delta t = {P \\over {\\gamma K}}}$$ = $${{P \\over {3\\alpha K}}}$$\n",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9906,
"subject": "Physics",
"question": "A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by $$\\Delta $$T. The net change in its length is zero. Let $$\\ell $$ be the length of the rod, A its area of cross-section,Y its Young’s modulus, and $$\\alpha $$ its coefficient of linear expansion. Then, F is equal to :",
"options": [
{
"text": "$$\\ell $$2 Y$$\\alpha $$ $$\\Delta $$T"
},
{
"text": "$$\\ell $$A Y$$\\alpha $$ $$\\Delta $$T"
},
{
"text": "A Y$$\\alpha $$ $$\\Delta $$T"
},
{
"text": "$${{AY} \\over {\\alpha \\,\\Delta T}}$$ "
}
],
"answer": "A Y$$\\alpha $$ $$\\Delta $$T",
"solution": "**Answer:** A Y$$\\alpha $$ $$\\Delta $$T\n\nBecause of thermal expansion, change in length \n
$$ \\simeq $$ 1 $$ \\times $$ 105 N",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9908,
"subject": "Physics",
"question": "A rod, of length L at room temperature and uniform area of cross section A, is made of a metal having coefficient of linear expansion $$\\alpha $$/oC. It is observed that an external compressive force F, is applied on each of its ends, prevents any change in the length of the rod, when its temperature rises by $$\\Delta $$TK. Young's modulus, Y, for this metal is : ",
"options": [
{
"text": "$${F \\over {A\\alpha \\Delta T}}$$"
},
{
"text": "$${F \\over {A\\alpha (\\Delta T - 273)}}$$"
},
{
"text": "$${F \\over {2A\\alpha \\Delta T}}$$"
},
{
"text": "$${{2F} \\over {A\\alpha \\Delta T}}$$"
}
],
"answer": "$${F \\over {A\\alpha \\Delta T}}$$",
"solution": "**Answer:** $${F \\over {A\\alpha \\Delta T}}$$\n\nWe know, \n
Young's Modulus, Y = $${{Stress} \\over {Strain}}$$\n
= $${F \\over {A\\alpha \\Delta T}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9909,
"subject": "Physics",
"question": "Two rods A and B of identical dimensions are at temperature 30°C. If A is heated upto 180oC and B upto ToC, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is ",
"options": [
{
"text": "200oC"
},
{
"text": "270oC"
},
{
"text": "230oC"
},
{
"text": "250oC"
}
],
"answer": "230oC",
"solution": "**Answer:** 230oC\n\n$$\\Delta {\\ell _1} = \\Delta {\\ell _2}$$\n
$$T = {230^o}C$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9910,
"subject": "Physics",
"question": "A thermometer graduated according to a linear scale reads a value x0 when in contact with boiling water, and x0/3 when in contact with ice. What is the temperature of an object in oC, if this thermometer in the contact with the object reads x0/2 ?",
"options": [
{
"text": "60"
},
{
"text": "35"
},
{
"text": "25"
},
{
"text": "40"
}
],
"answer": "25",
"solution": "**Answer:** 25\n\n\n
$$ \\Rightarrow $$ ToC = $${{150} \\over 6}$$ = 25oC",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9911,
"subject": "Physics",
"question": "At 40o C, a brass wire of 1 mm radius is hung from the ceiling. A small mass, M is hung from the free end of\nthe wire. When the wire is cooled down from 40oC to 20oC it regains its original length of 0.2 m. The value\nof M is close to :\n(Coefficient of linear expansion and Young’s modulus of brass are 10–5\n/oC and 1011 N/m\n2\n, respectively; g=\n10 ms–2\n)",
"options": [
{
"text": "1.5 kg"
},
{
"text": "0.5 kg"
},
{
"text": "9 kg"
},
{
"text": "0.9 kg"
}
],
"answer": "9 kg",
"solution": "**Answer:** 9 kg\n\n$$Mg = \\left( {{{Ay} \\over \\ell }} \\right)\\Delta \\ell $$
\n$$Mg = \\left( {Ay} \\right)\\alpha \\Delta T = 2\\pi $$
\nIt is closest to 9",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9912,
"subject": "Physics",
"question": "A bakelite beaker has volume capacity of 500 cc at 30oC. When it is partially filled with Vm\n volume\n(at 30oC) of mercury, it is found that the unfilled volume of the beaker remains constant as\ntemperature is varied. If $$\\gamma $$(beaker) = 6 × 10–6oC–1 and $$\\gamma $$(mercury) = 1.5 × 10–4oC–1, where $$\\gamma $$ is the\ncoefficient of volume expansion, then Vm\n (in cc) is close to ____.",
"options": [],
"answer": "20",
"solution": "**Answer:** 20\n\n$$\\Delta $$V = V$$\\gamma $$$$\\Delta $$T\n
= 0.04 + 0.02 = 0.06 %",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9915,
"subject": "Physics",
"question": "A non-isotropic solid metal cube has coefficients of linear expansion as : 5 $$ \\times $$ 10-5/oC along the x-axis and 5 $$ \\times $$ 10-6/oC along the y and the z-axis. If the coefficient of volume expansion of the solid is C $$ \\times $$ 10-6/oC then the value of C is",
"options": [],
"answer": "60",
"solution": "**Answer:** 60\n\n$$\\gamma $$ = $$\\alpha $$x + $$\\alpha $$y + $$\\alpha $$z\n
$$ \\Rightarrow $$ C = 60",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9916,
"subject": "Physics",
"question": "Each side of a box made of metal sheet in cubic shape is 'a' at room temperature 'T', the coefficient of linear expansion of the metal sheet is '$$\\alpha$$'. The metal sheet is heated uniformly, by a small temperature $$\\Delta$$T, so that its new temperature is T + $$\\Delta$$T. Calculate the increase in the volume of the metal box.",
"options": [
{
"text": "3a3$$\\alpha$$$$\\Delta$$T"
},
{
"text": "4$$\\pi$$a3$$\\alpha$$$$\\Delta$$T"
},
{
"text": "$${{4 \\over 3}}$$$$\\pi$$a3$$\\alpha$$$$\\Delta$$T"
},
{
"text": "4a3$$\\alpha$$$$\\Delta$$T"
}
],
"answer": "3a3$$\\alpha$$$$\\Delta$$T",
"solution": "**Answer:** 3a3$$\\alpha$$$$\\Delta$$T\n\nWe know that, $$\\gamma = 3\\alpha $$ .... (i)
where, $$\\alpha$$ is the coefficient of linear expansion and $$\\gamma$$ is the coefficient of volume expansion.
$$\\Delta V = 3{a^3}\\alpha \\Delta T$$ [ $$\\because$$ volume of cube = a3 ]",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9917,
"subject": "Physics",
"question": "Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : When a rod lying freely is heated, no thermal stress is developed in it.
Reason R : On heating, the length of the rod increases.
In the light of the above statements, choose the correct answer from the options given below :",
"options": [
{
"text": "A is true but R is false"
},
{
"text": "A is false but R is true"
},
{
"text": "Both A and B are true but R is NOT the correct explanation of A"
},
{
"text": "Both A and R are true and R is the correct explanation of A"
}
],
"answer": "Both A and B are true but R is NOT the correct explanation of A",
"solution": "**Answer:** Both A and B are true but R is NOT the correct explanation of A\n\nWhen a rod is free and it is heated then there is no thermal stress produced in it.\n
The rod will expand due to increase in temperature.\n
So both A & R are true.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9918,
"subject": "Physics",
"question": "The area of cross-section of a railway track is 0.01 m2. The temperature variation is 10$$^\\circ$$C. Coefficient of liner expansion of material of track is 10$$-$$5/$$^\\circ$$C. The energy stored per meter in the track is ____________ J/m.
(Young's modulus of material of track is 1011 Nm$$-$$2)",
"options": [],
"answer": "05",
"solution": "**Answer:** 05\n\nAs the tracks won't be allowed to expand linearly, the rise in temperature would lead to developing thermal stress in track.
At what temperature a gold ring of diameter 6.230 cm be heated so that it can be fitted on a wooden bangle of diameter 6.241 cm ? Both the diameters have been measured at room temperature (27$$^\\circ$$C).
\n
(Given : coefficient of linear thermal expansion of gold $$\\alpha$$L = 1.4 $$\\times$$ 10$$-$$5 K$$-$$1)
A solid metallic cube having total surface area 24 m2 is uniformly heated. If its temperature is increased by 10$$^\\circ$$C, calculate the increase in volume of the cube. (Given $$\\alpha$$ = 5.0 $$\\times$$ 10$$-$$4 $$^\\circ$$C$$-$$1).
A unit scale is to be prepared whose length does not change with temperature and remains $$20 \\mathrm{~cm}$$, using a bimetallic strip made of brass and iron each of different length. The length of both components would change in such a way that difference between their lengths remains constant. If length of brass is $$40 \\mathrm{~cm}$$ and length of iron will be __________ $$\\mathrm{cm}$$. $$\\left(\\alpha_{\\text {iron }}=1.2 \\times 10^{-5} \\mathrm{~K}^{-1}\\right.$$ and $$\\left.\\alpha_{\\text {brass }}=1.8 \\times 10^{-5} \\mathrm{~K}^{-1}\\right)$$.
$${L_1} = {{1.8} \\over {1.2}} \\times 40 = 60$$ cm
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9922,
"subject": "Physics",
"question": "A faulty thermometer reads $5^{\\circ} \\mathrm{C}$ in melting ice and $95^{\\circ} \\mathrm{C}$ in stream. The correct temperature on absolute scale will be __________ $\\mathrm{K}$ when the faulty thermometer reads $41^{\\circ} \\mathrm{C}$.",
"options": [],
"answer": "313",
"solution": "**Answer:** 313\n\n
A hole is drilled in a metal sheet. At $$\\mathrm{27^\\circ \n C}$$, the diameter of hole is 5 cm. When the sheet is heated to $$\\mathrm{177^\\circ \n C}$$, the change in the diameter of hole is $$\\mathrm{d\\times10^{-3}}$$ cm. The value of d will be __________ if coefficient of linear expansion of the metal is $$1.6\\times10^{-5}/^\\circ$$C.
On a temperature scale '$$\\mathrm{X}$$', the boiling point of water is $$65^{\\circ} \\mathrm{X}$$ and the freezing point is $$-15^{\\circ} \\mathrm{X}$$. Assume that the $$\\mathrm{X}$$ scale is linear. The equivalent temperature corresponding to $$-95^{\\circ} \\mathrm{X}$$ on the Farenheit scale would be:
",
"options": [
{
"text": "$$-148^{\\circ} \\mathrm{F}$$"
},
{
"text": "$$-48^{\\circ} \\mathrm{F}$$"
},
{
"text": "$$-63^{\\circ} \\mathrm{F}$$"
},
{
"text": "$$-112^{\\circ} \\mathrm{F}$$"
}
],
"answer": "$$-148^{\\circ} \\mathrm{F}$$",
"solution": "**Answer:** $$-148^{\\circ} \\mathrm{F}$$\n\nWe are given two temperature scales: the X-scale and the Celsius scale. The relationship between two linear temperature scales can be expressed as follows:\n
\nHere, $$X_{\\text{freeze}}$$ and $$X_{\\text{boil}}$$ are the freezing and boiling points of water on the X-scale, while $$C_{\\text{freeze}}$$ and $$C_{\\text{boil}}$$ are the freezing and boiling points of water on the Celsius scale.\n
\nStep 2: Convert $$-100^{\\circ} \\mathrm{C}$$ to Fahrenheit:\n
\nUse the conversion formula between Celsius and Fahrenheit:\n
\n$$T_F = \\frac{9}{5}T_C + 32$$\n
\nSubstitute the Celsius temperature:\n
\n$$T_F = \\frac{9}{5} \\times (-100) + 32$$
\n$$T_F = -180 + 32$$
\n$$T_F = -148^{\\circ} \\mathrm{F}$$\n
\nSo, the equivalent temperature corresponding to $$-95^{\\circ} \\mathrm{X}$$ on the Fahrenheit scale is $$-148^{\\circ} \\mathrm{F}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9925,
"subject": "Physics",
"question": "
A steel rod of length $$1 \\mathrm{~m}$$ and cross sectional area $$10^{-4} \\mathrm{~m}^{2}$$ is heated from $$0^{\\circ} \\mathrm{C}$$ to $$200^{\\circ} \\mathrm{C}$$ without being allowed to extend or bend. The compressive tension produced in the rod is ___________ $$\\times 10^{4} \\mathrm{~N}$$. (Given Young's modulus of steel $$=2 \\times 10^{11} \\mathrm{Nm}^{-2}$$, coefficient of linear expansion $$=10^{-5} \\mathrm{~K}^{-1}$$ )
The rod is not allowed to extend or bend, so a stress is created in the rod. This stress can be calculated using Young's modulus (Y), which is the ratio of the stress (force per unit area, F/A) to the strain (change in length per unit length, $\\Delta L / L_0$):
The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are $$8 \\Omega$$ and $$10 \\Omega$$ respectively. After inserting in a hot bath of temperature $$400^{\\circ} \\mathrm{C}$$, the resistance of platinum wire is :
Hence, the resistance of the platinum wire at $$400^{\\circ}C$$ is $$16 \\Omega$$. The correct option is:
\n\n
Option B: 16 $$\\Omega$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9927,
"subject": "Physics",
"question": "Which of the following parameters does not characterize the thermodynamic state of mattter? ",
"options": [
{
"text": "Temperature "
},
{
"text": "Pressure "
},
{
"text": "Work "
},
{
"text": "Volume "
}
],
"answer": "Work ",
"solution": "**Answer:** Work \n\nWork is a path function. The remaining three parameters are state function.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9928,
"subject": "Physics",
"question": "Two thermally insulated vessels $$1$$ and $$2$$ are filled with air at temperatures $$\\left( {{T_1},{T_2}} \\right),$$ volume $$\\left( {{V_1},{V_2}} \\right)$$ and pressure $$\\left( {{P_1},{P_2}} \\right)$$ respectively. If the value joining the two vessels is opened, the temperature inside the vessel at equilibrium will be",
"options": [
{
"text": "$${T_1}{T_2}\\left( {{P_1}{V_1} + {P_2}{V_2}} \\right)/\\left( {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}} \\right)$$"
},
{
"text": "$$\\left( {{T_1} + {T_2}} \\right)/2$$ "
},
{
"text": "$${{T_1} + {T_2}}$$ "
},
{
"text": "$${T_1}{T_2}\\left( {{P_1}{V_1} + {P_2}{V_2}} \\right)/\\left( {{P_1}{V_1}{T_1} + {P_2}{V_2}{T_2}} \\right)$$ "
}
],
"answer": "$${T_1}{T_2}\\left( {{P_1}{V_1} + {P_2}{V_2}} \\right)/\\left( {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}} \\right)$$",
"solution": "**Answer:** $${T_1}{T_2}\\left( {{P_1}{V_1} + {P_2}{V_2}} \\right)/\\left( {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}} \\right)$$\n\nHere $$Q=0$$ and $$W=0.$$ Therefore from first law of thermodynamics $$\\Delta U = Q + W = 0$$\n $$\\therefore$$ Internal energy of the system with partition $$=$$ Internal energy of the system without partition.\n $${n_1}{C_v}\\,{T_1} + {n_2}\\,{C_v}{T_2} = \\left( {{n_1} + {n_2}} \\right){C_v}\\,T$$\n $$\\therefore$$ $$T = {{{n_1}{T_1} + {n_2}T{}_2} \\over {{n_1} + {n_2}}}$$\n But $${n_1} = {{{P_1}{V_1}} \\over {R{T_1}}}$$ and $${n_2} = {{{P_2}{V_2}} \\over {R{T_2}}}$$\n $$\\therefore$$ $$T = {{{{{P_1}{V_1}} \\over {R{T_1}}} \\times {T_1} + {{{P_2}{V_2}} \\over {R{T_2}}} \\times {T_2}} \\over {{{{P_1}{V_1}} \\over {R{T_1}}} + {{{P_2}{V_2}} \\over {R{T_2}}}}}$$\n $$ = {{{T_1}{T_2}\\left( {{P_1}{V_1} + {P_2}{V_2}} \\right)} \\over {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}$$ ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9929,
"subject": "Physics",
"question": "Which of the following statements is correct for any thermodynamic system ? ",
"options": [
{
"text": "The change in entropy can never be zero"
},
{
"text": "Internal energy and entropy and state functions "
},
{
"text": "The internal energy changes in all processes "
},
{
"text": "The work done in an adiabatic process is always zero, "
}
],
"answer": "Internal energy and entropy and state functions ",
"solution": "**Answer:** Internal energy and entropy and state functions \n\nInternal energy and entropy are state function, they do not depend upon path taken.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9930,
"subject": "Physics",
"question": "An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $${V_1}$$ and contains ideal gas at pressure $${P_1}$$ and temperature $${T_1}$$. The other chamber has volume $${V_2}$$ and contains ideal gas at pressure $${P_2}$$ and temperature $${T_2}$$. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be ",
"options": [
{
"text": "$${{{T_1}{T_2}\\left( {{P_1}{V_1} + {P_2}{V_2}} \\right)} \\over {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}$$ "
},
{
"text": "$${{{P_1}{V_1}{T_1} + {P_2}{V_2}{T_2}} \\over {{P_1}{V_1} + {P_2}{V_2}}}$$ "
},
{
"text": "$${{{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}} \\over {{P_1}{V_1} + {P_2}{V_2}}}$$"
},
{
"text": "$${{{T_1}{T_2}\\left( {{P_1}{V_1} + {P_2}{V_2}} \\right)} \\over {{P_1}{V_1}{T_1} + {P_2}{V_2}{T_2}}}$$ "
}
],
"answer": "$${{{T_1}{T_2}\\left( {{P_1}{V_1} + {P_2}{V_2}} \\right)} \\over {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}$$ ",
"solution": "**Answer:** $${{{T_1}{T_2}\\left( {{P_1}{V_1} + {P_2}{V_2}} \\right)} \\over {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}$$ \n\nSince, no work is done and system is thermally insulated from surrounding. Therefore, total internal energy is constant, that is, $U=U_1+U_2$ Assuming gas have same degree of freedom, we have\n
\n$$ \\approx 748\\,J$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 9934,
"subject": "Physics",
"question": "A closed vessel contains 0.1 mole of a monoatomic ideal gas at 200 K. If 0.05 mole of the same gas\nat 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close\nto _______.",
"options": [],
"answer": "266",
"solution": "**Answer:** 266\n\nAs work done on gas and heat supplied to the gas are zero,\n
$$ \\therefore $$ Total internal energy of gases remain same.\n
$$ \\Rightarrow $$ Tf = 266.67",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9935,
"subject": "Physics",
"question": "A monoatomic gas of mass 4.0 u is kept in an insulated container. Container is moving with velocity 30 m/s. If container is suddenly stopped then change in temperature of the gas (R = gas constant) is $${x \\over {3R}}$$. Value of x is ___________.",
"options": [],
"answer": "3600",
"solution": "**Answer:** 3600\n\n$$\\Delta {K_E} = \\Delta U$$
$$ \\Rightarrow $$ $$x = 3600$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9936,
"subject": "Physics",
"question": "1 mole of rigid diatomic gas performs a work of $${Q \\over 5}$$ when heat Q is supplied to it. The molar heat capacity of the gas during this transformation is $${xR \\over 8}$$. The value of x is _________. [R = universal gas constant]",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\nFrom thermodynamics law:\n
$$X = 25$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9937,
"subject": "Physics",
"question": "In thermodynamics, heat and work are :",
"options": [
{
"text": "Path functions"
},
{
"text": "Point functions"
},
{
"text": "Extensive thermodynamics state variables"
},
{
"text": "Intensive thermodynamic state variables"
}
],
"answer": "Path functions",
"solution": "**Answer:** Path functions\n\nHeat and work are path function.
Heat and work depends on the path taken to reach the final state from initial state.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9938,
"subject": "Physics",
"question": "An electric appliance supplies 6000 J/min heat to the system. If the system delivers a power of 90W. How long it would take to increase the internal energy by 2.5 $$\\times$$ 103 J ?",
"options": [
{
"text": "2.5 $$\\times$$ 102 s"
},
{
"text": "4.1 $$\\times$$ 101 s"
},
{
"text": "2.4 $$\\times$$ 103 s"
},
{
"text": "2.5 $$\\times$$ 101 s"
}
],
"answer": "2.5 $$\\times$$ 102 s",
"solution": "**Answer:** 2.5 $$\\times$$ 102 s\n\n$$\\Delta$$Q = $$\\Delta$$U + $$\\Delta$$W
At a certain temperature, the degrees of freedom per molecule for gas is 8. The gas performs 150 J of work when it expands under constant pressure. The amount of heat absorbed by the gas will be _________ J.
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
\n
Assertion A: If $$d Q$$ and $$d W$$ represent the heat supplied to the system and the work done on the system respectively. Then according to the first law of thermodynamics $$d Q=d U-d W$$.
\n
Reason R: First law of thermodynamics is based on law of conservation of energy.
\n
In the light of the above statements, choose the correct answer from the options given below:
",
"options": [
{
"text": "Both A and R are correct but R is not the correct explanation of A"
},
{
"text": "Both A and R are correct and R is the correct explanation of A"
},
{
"text": "A is correct but R is not correct"
},
{
"text": "A is not correct but R is correct"
}
],
"answer": "Both A and R are correct and R is the correct explanation of A",
"solution": "**Answer:** Both A and R are correct and R is the correct explanation of A\n\nHeat supplied to the system $=d Q$ \n
Work done on the system $=-d W$ \n
By the first law of thermodynamics, $d Q=d U+d W$ \n
We done the work, therefore,\n
$$\nd Q=d U+(-d W) = d U-d W\n$$\n
So, both A and R are correct and R is the correct explanation\nof A.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9941,
"subject": "Physics",
"question": "
$$1 \\mathrm{~kg}$$ of water at $$100^{\\circ} \\mathrm{C}$$ is converted into steam at $$100^{\\circ} \\mathrm{C}$$ by boiling at atmospheric pressure. The volume of water changes from $$1.00 \\times 10^{-3} \\mathrm{~m}^{3}$$ as a liquid to $$1.671 \\mathrm{~m}^{3}$$ as steam. The change in internal energy of the system during the process will be
Next, let's calculate the work done ($W$) on the system during the process. The work done is given by:
\n
$$W = -P \\Delta V$$
\n
where $$P$$ is the atmospheric pressure and $$\\Delta V$$ is the change in volume. We are given that the atmospheric pressure is $$P = 1 \\times 10^5 \\mathrm{~Pa}$$, and the change in volume is
A gas is compressed adiabatically, which one of the following statement is NOT true.
",
"options": [
{
"text": "There is no heat supplied to the system"
},
{
"text": "The temperature of the gas increases."
},
{
"text": "There is no change in the internal energy"
},
{
"text": "The change in the internal energy is equal to the work done on the gas."
}
],
"answer": "There is no change in the internal energy",
"solution": "**Answer:** There is no change in the internal energy\n\n
An adiabatic process is one in which there is no heat exchange between a system (in this case, the gas) and its surroundings. This happens because the system is perfectly insulated or the process occurs very quickly, not allowing for heat exchange.
\n
Given the options:
\n
Option A: There is no heat supplied to the system.
\n
\n
This statement is TRUE. In an adiabatic process, there is no heat exchange between the system and its surroundings, as mentioned above.
\n
\n
Option B: The temperature of the gas increases.
\n
\n
This statement is TRUE. When a gas is compressed adiabatically, the work done on the gas increases its internal energy, which in turn increases the temperature of the gas.
\n
\n
Option C: There is no change in the internal energy.
\n
\n
This statement is NOT TRUE. In an adiabatic process, the change in internal energy of the system is equal to the work done on the system. When the gas is compressed, work is done on the gas, which increases its internal energy.
\n
\n
Option D: The change in the internal energy is equal to the work done on the gas.
\n
\n
This statement is TRUE. As mentioned above, in an adiabatic process, the change in internal energy of the system is equal to the work done on the system.
\n
\n
So, Option C is the statement that is not true for an adiabatic process.
A source supplies heat to a system at the rate of $$1000 \\mathrm{~W}$$. If the system performs work at a rate of $$200 \\mathrm{~W}$$. The rate at which internal energy of the system increases is
The rate of increase of internal energy of a system can be found using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
\n
In this case, the heat being supplied to the system is 1000 W and the work being done by the system is 200 W.
\n
Therefore, the rate at which the internal energy of the system increases is:
\n
1000 W (heat supplied) - 200 W (work done) = 800 W
\n
So, the correct answer is 800 W.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9944,
"subject": "Physics",
"question": "When forces $${F_1},\\,\\,{F_2},\\,\\,{F_3}$$ are acting on a particle of mass $$m$$ such that $${F_2}$$ and $${F_3}$$ are mutually perpendicular, then the particle remains stationary. If the force $${F_1}$$ is now removed then the acceleration of the particle is ",
"options": [
{
"text": "$${F_1}/m$$ "
},
{
"text": "$${F_2}{F_3}/m{F_1}$$ "
},
{
"text": "$$\\left( {F{}_2 - {F_3}} \\right)/m$$ "
},
{
"text": "$${F_2}/m$$ "
}
],
"answer": "$${F_1}/m$$ ",
"solution": "**Answer:** $${F_1}/m$$ \n\nWhen $${F_1},{F_2}$$ and $${F_3}$$ are acting on a particle then the particle remains stationary. This means that the resultant of $${F_1},{F_2}$$ and $${F_3}$$ is zero. When $${F_1}$$ is removed then particle will start moving due to the force $${F_2}$$ and $${F_3}$$ in the resultant of $${F_2}$$ and $${F_3}$$ and it should be equal and opposite to $${F_1}.$$ $$i.e.$$ $$\\left| {{{\\overrightarrow F }_2} + {{\\overrightarrow F }_3}} \\right| = \\left| {{{\\overrightarrow F }_1}} \\right|$$ \n $$\\therefore$$ $$\\,\\,\\,\\,\\,a = {{\\left| {{{\\overrightarrow F }_2} + {{\\overrightarrow F }_3}} \\right|} \\over m} \\Rightarrow a = {{{F_1}} \\over m}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9945,
"subject": "Physics",
"question": "A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point. the rope deviated at an angle of 45o at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms$$-$$2",
"options": [
{
"text": "200 N"
},
{
"text": "140 N"
},
{
"text": "70 N"
},
{
"text": "100 N"
}
],
"answer": "100 N",
"solution": "**Answer:** 100 N\n\n\n tan 45o = $${F \\over {mg}}$$\n
$$ \\therefore $$ F = mg\n
= 10 $$ \\times $$ 10\n
= 100 N",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9946,
"subject": "Physics",
"question": "A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied\nhorizontally at the mid point of the rope such that the top half of the rope makes an angle of 45o\nwith the vertical. Then F equal : (Take g = 10 ms–2 and the rope to be massless)",
"options": [
{
"text": "100 N"
},
{
"text": "75 N"
},
{
"text": "90 N"
},
{
"text": "70 N"
}
],
"answer": "100 N",
"solution": "**Answer:** 100 N\n\n\n For equilibrium,\n
T sin 45o = F ....(1)\n
and T cos 45o = 10g ....(2)\n
Performing (1)/(2)\n
we get F = 10g = 100 N",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9947,
"subject": "Physics",
"question": "Statement : 1
If three forces $${\\overrightarrow F _1},{\\overrightarrow F _2}$$ and $${\\overrightarrow F _3}$$ are represented by three sides of a triangle and $${\\overrightarrow F _1} + {\\overrightarrow F _2} = - {\\overrightarrow F _3}$$, then these three forces are concurrent forces and satisfy the condition for equilibrium.
Statement : 2
A triangle made up of three forces $${\\overrightarrow F _1}$$, $${\\overrightarrow F _2}$$ and $${\\overrightarrow F _3}$$ as its sides taken in the same order, satisfy the condition for translatory equilibrium.
In the light of the above statements, choose the most appropriate answer from the options given below :",
"options": [
{
"text": "Statement - I is false but Statement - II is true"
},
{
"text": "Statement - I is true but Statement - II is false"
},
{
"text": "Both Statement-I and Statement-II are false"
},
{
"text": "Both Statement-I and Statement-II are true"
}
],
"answer": "Both Statement-I and Statement-II are true",
"solution": "**Answer:** Both Statement-I and Statement-II are true\n\n
Here, $${\\overrightarrow F _1} + {\\overrightarrow F _2} + {\\overrightarrow F _3} = 0$$
$${\\overrightarrow F _1} + {\\overrightarrow F _2} = - {\\overrightarrow F _3}$$
Since $${\\overrightarrow F _{net}} = 0$$ (equilibrium)
A mass of 10 kg is suspended vertically by a rope of length 5 m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $$\\theta$$ = tan$$-$$1 (x $$\\times$$ 10$$-$$1). The value of x is ____________.
Three forces $$F_{1}=10 \\mathrm{~N}, F_{2}=8 \\mathrm{~N}, \\mathrm{~F}_{3}=6 \\mathrm{~N}$$ are acting on a particle of mass $$5 \\mathrm{~kg}$$. The forces $$\\mathrm{F}_{2}$$ and $$\\mathrm{F}_{3}$$ are applied perpendicularly so that particle remains at rest. If the force $$F_{1}$$ is removed, then the acceleration of the particle is:
Since the particle is initially at rest, the net force acting on it is zero. This means that the forces $$F_1$$, $$F_2$$, and $$F_3$$ are balanced. Given that $$F_2$$ and $$F_3$$ are acting perpendicularly, we can represent the balance of forces as follows:
\n
$$F_1 = \\sqrt{F_2^2 + F_3^2}$$
\n
We can now calculate the equivalent force resulting from $$F_2$$ and $$F_3$$:
Since the particle is initially at rest, when the force $$F_1$$ is removed, only $$F_2$$ and $$F_3$$ remain. The net force acting on the particle is the equivalent force resulting from $$F_2$$ and $$F_3$$, which we have calculated to be $$10 \\mathrm{~N}$$.
\n
Now, we can use Newton's second law to find the acceleration of the particle:
A heavy iron bar, of weight $$W$$ is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle $$\\theta$$ with the horizontal. The weight experienced by the person is :
Two forces $$\\overline{\\mathrm{F}}_1$$ and $$\\overline{\\mathrm{F}}_2$$ are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between $$\\vec{F}_1$$ and $$\\vec{F}_2$$ is $$\\cos ^{-1}\\left(\\frac{1}{n}\\right)$$. The value of $$|n|$$ is _______.
Let's denote the magnitude of the smaller force as $$F$$, hence the magnitude of the larger force is $$3F$$. The resultant force $$\\vec{R}$$ is equal in magnitude to the larger force, which means $$|\\vec{R}| = 3F$$. When two forces $$\\vec{F}_1$$ and $$\\vec{F}_2$$ act on a body, the magnitude of their resultant $$\\vec{R}$$ can be found using the law of vector addition:
where $$\\theta$$ is the angle between $$\\vec{F}_1$$ and $$\\vec{F}_2$$. Given that in our case $$|\\vec{R}| = 3F$$, $$|\\vec{F}_1| = F$$ and $$|\\vec{F}_2| = 3F$$, by substituting these values into the equation, we get:
Simplifying this equation by subtracting $$10F^2$$ from both sides gives:
\n\n
$$-F^2 = 6F^2\\cos\\theta$$
\n\n
Dividing both sides by $$-F^2$$ gives:
\n\n
$$-1 = -6\\cos\\theta$$
\n\n
Therefore, $$\\cos\\theta = \\frac{1}{6}$$.
\n\n
It is given that the angle between $$\\vec{F}_1$$ and $$\\vec{F}_2$$ is $$\\cos^{-1}\\left(\\frac{1}{n}\\right)$$, hence comparing this with the above result, we find that $$n = 6$$. Therefore, $$|n| = 6$$.
A body of weight $$200 \\mathrm{~N}$$ is suspended from a tree branch through a chain of mass $$10 \\mathrm{~kg}$$. The branch pulls the chain by a force equal to (if $$\\mathrm{g}=10 \\mathrm{~m} / \\mathrm{s}^2$$) :
To determine the force that the branch pulls on the chain, we need to consider the combined weight of the body and the chain, as this is the total force that the branch must support due to gravity.
\n\n
The weight of the body is given as $$200\\, \\mathrm{N}$$.
\n\n
To find the weight of the chain, we use the formula for weight, which is the mass of an object multiplied by the acceleration due to gravity ($$g$$). The mass of the chain is given as $$10\\, \\mathrm{kg}$$ and $$g = 10\\, \\mathrm{m/s}^2$$, so:
\n\n
$$\\text{Weight of the chain} = \\text{Mass of the chain} \\times g$$\n\n
The total force that the branch must support is the sum of the weight of the body and the weight of the chain:
\n\n
$$\\text{Total force} = \\text{Weight of the body} + \\text{Weight of the chain}$$\n\n
$$= 200\\, \\mathrm{N} + 100\\, \\mathrm{N}$$
\n\n
$$= 300\\, \\mathrm{N}$$
\n\n
Therefore, the branch pulls the chain by a force of $$300\\, \\mathrm{N}$$, which corresponds to Option A.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9953,
"subject": "Physics",
"question": "A marble block of mass $$2$$ $$kg$$ lying on ice when given a velocity of $$6$$ $$m/s$$ is stopped by friction in $$10$$ $$s.$$ Then the coefficient of friction is ",
"options": [
{
"text": "$$0.02$$ "
},
{
"text": "$$0.03$$ "
},
{
"text": "$$0.04$$ "
},
{
"text": "$$0.06$$ "
}
],
"answer": "$$0.06$$ ",
"solution": "**Answer:** $$0.06$$ \n\nThe retarding force is created by the frictional force.\n
$$\\therefore$$ $$\\mu = 0.06$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9954,
"subject": "Physics",
"question": "A block rests on a rough inclined plane `making an angle of $${30^ \\circ }$$ with the horizontal. The coefficient of static friction between the block and the plane is $$0.8.$$ If the frictionless force on the block is $$10$$ $$N,$$ the mass of the block (in $$kg$$) is $$\\left( {take\\,\\,\\,g\\, = \\,10\\,\\,m/{s^2}} \\right)$$ ",
"options": [
{
"text": "$$1.6$$ "
},
{
"text": "$$4.0$$ "
},
{
"text": "$$2.0$$"
},
{
"text": "$$2.5$$ "
}
],
"answer": "$$2.0$$",
"solution": "**Answer:** $$2.0$$\n\n\n For equilibrum of block,\n
$$ \\Rightarrow s = 1000\\,m$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9957,
"subject": "Physics",
"question": "A block of mass $$m$$ is placed on a surface with a vertical cross section given by $$y = {{{x^3}} \\over 6}.$$ If the coefficient of friction is $$0.5,$$ the maximum height above the ground at which the block can be placed without slipping is:",
"options": [
{
"text": "$${1 \\over 6}m$$ "
},
{
"text": "$${2 \\over 3}m$$ "
},
{
"text": "$${1 \\over 3}m$$ "
},
{
"text": "$${1 \\over 2}m$$ "
}
],
"answer": "$${1 \\over 6}m$$ ",
"solution": "**Answer:** $${1 \\over 6}m$$ \n\nAt limiting equilibrium, $$\\mu = \\tan \\theta $$\n
Now, $$y = {{{x^3}} \\over 6} = {1 \\over 6}m$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9958,
"subject": "Physics",
"question": "A rocket is fired vertically from the earth with an acceleration of 2g, where g is the\ngravitational acceleration. On an inclined plane inside the rocket, making an angle $$\\theta $$ with the horizontal, a point object of mass m is kept. The minimum coefficient of friction $$\\mu $$min between the mass and the inclined surface such that the mass does not move is :",
"options": [
{
"text": "tan$$\\theta $$"
},
{
"text": "2tan$$\\theta $$"
},
{
"text": "3tan$$\\theta $$"
},
{
"text": "tan2$$\\theta $$"
}
],
"answer": "tan$$\\theta $$",
"solution": "**Answer:** tan$$\\theta $$\n\nRocket is moving upward with acceleration 2g and gravitation acceleration is g downward direction.\n
So, acceleration experienced by the point object, \n
$$ \\Rightarrow $$ $$\\mu $$ = tan$$\\theta $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9959,
"subject": "Physics",
"question": "A given object takes n times more time to slide down a $${45^ \\circ }$$ rough inclined plane as it takes to slide down a perfectly smooth $${45^ \\circ }$$ incline. The coefficient of kinetic friction between the object and the incline is :",
"options": [
{
"text": "$${1 \\over {2 - {n^2}}}$$ "
},
{
"text": "$$1 - {1 \\over {{n^2}}}$$"
},
{
"text": "$$\\sqrt {1 - {1 \\over {{n^2}}}} $$"
},
{
"text": "$$\\sqrt {{1 \\over {1 - {n^2}}}} $$ "
}
],
"answer": "$$1 - {1 \\over {{n^2}}}$$",
"solution": "**Answer:** $$1 - {1 \\over {{n^2}}}$$\n\nLet, t1 and t2 are time taken to move on the smooth and rough surface for smooth surface, \n
$$ \\Rightarrow $$$$\\,\\,\\,\\,$$ $${\\mu _k}$$ = 1 $$-$$ $${1 \\over {{n^2}}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9960,
"subject": "Physics",
"question": "A body of mass 2 kg slides down with an acceleration of 3 m/s2 on a rough inclined plane having a slope of $${30^o}$$. The external force required to take the same body up the plane with the same acceleration will be : (g = 10 m/s2)",
"options": [
{
"text": "14 N"
},
{
"text": "20 N"
},
{
"text": "6 N"
},
{
"text": "4 N"
}
],
"answer": "20 N",
"solution": "**Answer:** 20 N\n\nWhen mass slide down then, \n
= 20 N ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9961,
"subject": "Physics",
"question": "An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls up the ditch but starts\nslipping after it is at height h from the bottom. If the coefficient of friction between the ground and\nthe insect is 0.75, then h is : \n (g = 10 ms–2)",
"options": [
{
"text": "0.45 m"
},
{
"text": "0.60 m"
},
{
"text": "0.20 m"
},
{
"text": "0.80 m"
}
],
"answer": "0.20 m",
"solution": "**Answer:** 0.20 m\n\n\n
$$ \\therefore $$ h = $${R \\over 5}$$ = $${1 \\over 5}$$ = 0.2 m",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9962,
"subject": "Physics",
"question": "An inclined plane is bent in such a way that the vertical cross-section is given by $$y = {{{x^2}} \\over 4}$$ where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction $$\\mu$$ = 0.5, the maximum height in cm at which a stationary block will not slip downward is _________ cm.",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\nThe graph for given equation is shown below
$$\\therefore$$ $$y = {{{x^2}} \\over 4} = {1 \\over 4}$$ = 0.25 m = 25 cm",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9963,
"subject": "Physics",
"question": "The coefficient of static friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall will be _________ N. [g = 10 ms$$-$$2]",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\nGiven, coefficient of static friction, $$\\mu$$s = 0.2
Various forces acting on block are shown below
Frictional force $$\\le$$ mg
$$\\Rightarrow$$ N $$\\times$$ 0.2 $$\\le$$ 5
$$\\Rightarrow$$ N $$\\le$$ 25
$$\\therefore$$ Magnitude of horizontal force, F = N = 25 N",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9964,
"subject": "Physics",
"question": "A body of mass 1 kg rests on a horizontal floor with which it has a coefficient of static friction $${1 \\over {\\sqrt 3 }}$$. It is desired to make the body move by applying the minimum possible force F N. The value of F will be ____________. (Round off to the Nearest Integer) [Take g = 10 ms$$-$$2 ]",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n\n Minimum possible force $$ \\Rightarrow $$
Fmin = 5N",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9965,
"subject": "Physics",
"question": "A body of mass 'm' is launched up on a rough inclined plane making an angle of 30$$^\\circ$$ with the horizontal. The coefficient of friction between the body and plane is $${{\\sqrt x } \\over 5}$$ if the time of ascent is half of the time of descent. The value of x is __________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n$${t_a} = {1 \\over 2}{t_d}$$
using the above values of aa and ad and putting in equation (i) we will gate $$\\mu = {{\\sqrt 3 } \\over 5}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9966,
"subject": "Physics",
"question": "When a body slides down from rest along a smooth inclined plane making an angle of 30$$^\\circ$$ with the horizontal, it takes time T. When the same body slides down from the rest along a rough inclined plane making the same angle and through the same distance, it takes time $$\\alpha$$T, where $$\\alpha$$ is a constant greater than 1. The co-efficient of friction between the body and the rough plane is $${1 \\over {\\sqrt x }}\\left( {{{{\\alpha ^2} - 1} \\over {{\\alpha ^2}}}} \\right)$$ where x = __________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n On smooth incline
Comparing with $$\\mu = \\left( {{{{\\alpha ^2} - 1} \\over {{\\alpha ^2}}}} \\right){1 \\over {\\sqrt x }}$$
The value of the x = 3.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9967,
"subject": "Physics",
"question": "
A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 ms$$-$$1. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is :
A bag is gently dropped on a conveyor belt moving at a speed of $$2 \\mathrm{~m} / \\mathrm{s}$$. The coefficient of friction between the conveyor belt and bag is $$0.4$$. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion, is : [Take $$\\mathrm{g}=10 \\mathrm{~m} / \\mathrm{s}^{-2}$$ ]
If $s$ be the distance travelled by the bag on the belt during slipping motion, then\n
$\n\\ v^2 \\ =u^2-2 a s $
$\n\\Rightarrow \\ 0=v^2-2 a s $
$\n\\Rightarrow s \\ =\\frac{v^2}{2 a}=\\frac{2^2}{2 \\times 4}=0.5 \\mathrm{~m}\n$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9969,
"subject": "Physics",
"question": "A body of mass $10 \\mathrm{~kg}$ is moving with an initial speed of $20 \\mathrm{~m} / \\mathrm{s}$. The body stops after $5 \\mathrm{~s}$ due to friction between body and the floor. The value of the coefficient of friction is:
A block of mass $$5 \\mathrm{~kg}$$ is placed at rest on a table of rough surface. Now, if a force of $$30 \\mathrm{~N}$$ is applied in the direction parallel to surface of the table, the block slides through a distance of $$50 \\mathrm{~m}$$ in an interval of time $$10 \\mathrm{~s}$$. Coefficient of kinetic friction is (given, $$g=10 \\mathrm{~ms}^{-2}$$):
The time taken by an object to slide down 45$$^\\circ$$ rough inclined plane is n times as it takes to slide down a perfectly smooth 45$$^\\circ$$ incline plane. The coefficient of kinetic friction between the object and the incline plane is :
A block of mass $m$ slides down the plane inclined at angle $$30^{\\circ}$$ with an acceleration $$\\frac{g}{4}$$. The value of coefficient of kinetic friction will be:
Statement (I) : The limiting force of static friction depends on the area of contact and independent of materials.
\n
Statement (II) : The limiting force of kinetic friction is independent of the area of contact and depends on materials.
\n
In the light of the above statements, choose the most appropriate answer from the options given below :
",
"options": [
{
"text": "Statement I is incorrect but Statement II is correct\n"
},
{
"text": "Both Statement I and Statement II are correct\n"
},
{
"text": "Both Statement I and Statement II are incorrect\n"
},
{
"text": "Statement I is correct but Statement II is incorrect"
}
],
"answer": "Statement I is incorrect but Statement II is correct\n",
"solution": "**Answer:** Statement I is incorrect but Statement II is correct\n\n\n
Let's analyze both statements:
\n
Statement (I): The limiting force of static friction depends on the area of contact and independent of materials.\n
This statement is incorrect. The limiting force of static friction does not depend on the area of contact but is dependent on the materials in contact. According to the law of static friction, the maximum static frictional force $$ f_s $$ that can occur before motion commences is given by the product of the coefficient of static friction $$ \\mu_s $$ and the normal reaction force $$ N $$:
\n
$$ f_s = \\mu_s \\times N $$
\n
The coefficient $$ \\mu_s $$ is a property that depends on the materials in contact, not on the area of contact. The normal force $$ N $$ is the force perpendicular to the surfaces in contact, influenced by the weight of the object and any other perpendicular forces acting upon it.
\n
Statement (II): The limiting force of kinetic friction is independent of the area of contact and depends on materials.\n
This statement is correct. Once an object is in motion, the kinetic frictional force $$ f_k $$ opposing its motion is given by the product of the coefficient of kinetic friction $$ \\mu_k $$ and the normal force $$ N $$:
\n
$$ f_k = \\mu_k \\times N $$
\n
The coefficient $$ \\mu_k $$, like $$ \\mu_s $$, is also a property dependent on the materials of the surfaces in contact. It generally has a lower value than $$ \\mu_s $$, which is why objects tend to be easier to keep moving once they've started. The kinetic frictional force is independent of the area of contact between the two surfaces.
\n
Given these explanations, the correct answer would be:\n
Option A: Statement I is incorrect but Statement II is correct.
A block of mass $$m$$ is placed on a surface having vertical crossection given by $$y=x^2 / 4$$. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:
A given object takes $$\\mathrm{n}$$ times the time to slide down $$45^{\\circ}$$ rough inclined plane as it takes the time to slide down an identical perfectly smooth $$45^{\\circ}$$ inclined plane. The coefficient of kinetic friction between the object and the surface of inclined plane is :
To determine the coefficient of kinetic friction, let's analyze the time taken by the object to slide down each plane and use the equations of motion for both scenarios.
\n\n
First, consider the perfectly smooth $$45^{\\circ}$$ inclined plane (no friction). The acceleration of the object on this plane can be calculated using the component of gravitational force parallel to the incline. Since there is no friction, the only force acting down the plane is the component of the gravitational force:
\n\n
$$ a_{\\text{smooth}} = g \\sin 45^{\\circ} = \\frac{g}{\\sqrt{2}} $$
\n\n
Let the time taken to slide down this smooth plane be $$t_{\\text{smooth}}$$. The distance $$d$$ covered by the object can be expressed using the equation of motion:
Now, consider the rough $$45^{\\circ}$$ inclined plane. The acceleration down the rough plane can be found by considering both the component of gravitational force and the kinetic friction force. Here, the friction force is $$f_k = \\mu_k N$$, where $$N = mg \\cos 45^\\circ = \\frac{mg}{\\sqrt{2}}$$. Thus the frictional force is:
\n\n
$$ f_k = \\mu_k \\cdot \\frac{mg}{\\sqrt{2}} $$
\n\n
The net force acting on the object down the plane would be:
A heavy box of mass $$50 \\mathrm{~kg}$$ is moving on a horizontal surface. If co-efficient of kinetic friction between the box and horizontal surface is 0.3 then force of kinetic friction is :
To find the force of kinetic friction acting on the box, we can use the formula for kinetic friction, which is given by:
\n\n
$F_{\\text{friction}} = \\mu_{k} \\cdot N$
\n\n
where:
\n\n
\n\n
$F_{\\text{friction}}$ is the force of friction,
\n\n
$\\mu_{k}$ is the coefficient of kinetic friction,
\n\n
$N$ is the normal force exerted by the surface onto the box.
\n\n
\n\n
Since the box is moving on a horizontal surface, the normal force $N$ would be equal to the weight of the box, which is calculated by $mg$, where $m$ is the mass of the box and $g$ is the acceleration due to gravity. For most calculations, $g$ is approximated as $$9.8 \\, \\text{m/s}^2$$.
\n\n
Substituting the given values:
\n\n
\n\n
$m = 50 \\, \\text{kg}$,
\n\n
$\\mu_k = 0.3$,
\n\n
$g = 9.8 \\, \\text{m/s}^2$.
\n\n
\n\n
First, calculate the weight of the box, which is the normal force:
Thus, the force of kinetic friction acting on the box is 147 N. The correct answer is Option B.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9977,
"subject": "Physics",
"question": "A light string passing over a smooth light pulley connects two blocks of masses $${m_1}$$ and $${m_2}$$ (vertically). If the acceleration of the system is $$g/8$$, then the ratio of the masses is ",
"options": [
{
"text": "$$8:1$$ "
},
{
"text": "$$9:7$$ "
},
{
"text": "$$4:3$$ "
},
{
"text": "$$5:3$$ "
}
],
"answer": "$$9:7$$ ",
"solution": "**Answer:** $$9:7$$ \n\n \n Assume that, mass m1 is greater than mass m2, so the heavier mass m1 is accelerating downward and the lighter mass m2 is accelerating upwards.\n
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9978,
"subject": "Physics",
"question": "A block of mass $$M$$ is pulled along a horizontal frictionless surface by a rope of mass $$m.$$ If a force $$P$$ is applied at the free end of the rope, the force exerted by the rope on the block is ",
"options": [
{
"text": "$${{Pm} \\over {M + m}}$$ "
},
{
"text": "$${{Pm} \\over {M - m}}$$ "
},
{
"text": "$$P$$ "
},
{
"text": "$${{PM} \\over {M + m}}$$ "
}
],
"answer": "$${{PM} \\over {M + m}}$$ ",
"solution": "**Answer:** $${{PM} \\over {M + m}}$$ \n\nTaking the rope and the block as a system\n \n Acceleration of block($$a$$) = $${{Force\\,applied} \\over {total\\,mass}}$$\n
$$\\therefore$$ $$a = {P \\over {m + M}}$$ \n
Force on block(T) = Mass of block $$ \\times $$ $$a$$\n
A light unstretchable string passing over a smooth light pulley connects two blocks of masses $$m_1$$ and $$m_2$$. If the acceleration of the system is $$\\frac{g}{8}$$, then the ratio of the masses $$\\frac{m_2}{m_1}$$ is :
To find the ratio of the masses $$\\frac{m_2}{m_1}$$ given that the acceleration of the system is $$\\frac{g}{8}$$, where $$g$$ is the acceleration due to gravity, we need to analyze the forces acting on each mass and apply Newton's second law of motion.
\n\n
For mass $$m_1$$, the force acting downward (towards the centre of the Earth) is the gravitational force $$m_1g$$, and the tension $$T$$ acts upward. Its equation of motion, considering downward as positive, can be given by:\n\n
$m_1g - T = m_1a$
\n\n
For mass $$m_2$$, the force acting downward is $$m_2g$$, but since it is on the opposite side of the pulley, the tension $$T$$ is upwards (considering upward movement as positive direction), so we have:\n\n
$T - m_2g = m_2a$
\n\n
Since the pulley is light and smooth, the tension $$T$$ is the same on both sides of the pulley. Also, the system accelerates together, so $$a = \\frac{g}{8}$$.
\n\n
Adding the two equations to eliminate $$T$$ gives us:\n\n
$m_1g - m_2g = (m_1 - m_2)a$
\n\n
$g(m_1 - m_2) = (m_1 - m_2)\\frac{g}{8}$
\n\n
Canceling out $$g$$ and dividing both sides by $$m_1 - m_2$$ (assuming $$m_1 \\neq m_2$$), we get:\n\n
$1 = \\frac{1}{8}$
\n\n
This simplification doesn't align with finding the ratio directly, indicating a mistake in handling the simultaneous equations relation to $$a$$ and $$T$$. Let's correct our approach to finding the ratio of the masses based on the acceleration and tension.
\n\n
Since the acceleration $$a = \\frac{g}{8}$$, and considering the system as a whole, the net force causing the acceleration is the difference in gravitational forces on the two masses. This net force provides the system's entire acceleration. Thus, correctly setting up the equations for the system should give:\n\n
$m_2g - m_1g = (m_1 + m_2)\\frac{g}{8}$
\n\n
$g(m_2-m_1) = \\frac{g}{8}(m_1 + m_2)$
\n\n
$8(m_2 - m_1) = m_1 + m_2$
\n\n
$8m_2 - 8m_1 = m_1 + m_2$
\n\n
$7m_2 = 9m_1$
\n\n
$\\frac{m_2}{m_1} = \\frac{9}{7}$
\n\n
Therefore, the correct ratio of the masses $$\\frac{m_2}{m_1}$$ is $$9:7$$, which corresponds to Option C.
A light string passing over a smooth light pulley connects two blocks of masses $$m_1$$ and $$m_2\\left(\\right.$$ where $$\\left.m_2>m_1\\right)$$. If the acceleration of the system is $$\\frac{g}{\\sqrt{2}}$$, then the ratio of the masses $$\\frac{m_1}{m_2}$$ is:
Given that the acceleration of the system is $$\\frac{g}{\\sqrt{2}}$$, let's analyze the forces acting on both masses to find the ratio $$\\frac{m_1}{m_2}$$.
\n\n
The tension force in the string is equal for both masses since the pulley and string are light and smooth. Assuming the pulley only changes the direction of the tension force without affecting its magnitude, we can analyze the forces in the vertical direction for both masses.
\n\n
The net force acting on $$m_1$$ (towards the pulley) would be the tension $$T$$ minus its weight component in the direction of motion, which we will assume is $$m_1g$$, and for $$m_2$$, it would be its weight component in the direction of motion, which we can assume is $$m_2g$$ minus the tension $$T$$. Given that $$m_2$$ is moving downwards and $$m_1$$ is moving upwards, and considering that $$m_2>m_1$$, the acceleration $$a$$ of both masses will be the same and can be described as:
\n\n
For $$m_1$$:\n\n
$$T - m_1g = m_1a$$
\n\n
For $$m_2$$:\n\n
$$m_2g - T = m_2a$$
\n\n
Adding these equations to eliminate $$T$$ gives:\n\n
$$m_2g - m_1g = m_1a + m_2a$$
\n\n
Given that the acceleration $$a$$ of the system is $$\\frac{g}{\\sqrt{2}}$$, we can substitute $$a$$ with $$\\frac{g}{\\sqrt{2}}$$ in the equation:\n\n
This corresponds to Option C $$\\frac{\\sqrt{2}-1}{\\sqrt{2}+1}$$.
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9981,
"subject": "Physics",
"question": "A car is moving on a plane inclined at 30$$^\\circ$$ to the horizontal with an acceleration of 10 ms$$-$$2 parallel to the plane upward. A bob is suspended by a string from the roof of the car. The angle in degrees which the string makes with the vertical is ______________. (Take g = 10 ms$$-$$2)",
"options": [],
"answer": "30",
"solution": "**Answer:** 30\n\n
Given,
\n
Angle of inclination, $$\\theta$$ = 30$$^\\circ$$
\n
Acceleration, a = 10 ms$$-$$2
\n
Acceleration due to gravity, g = 10 ms$$-$$2
\n
According to the question the car and bob is as shown below,
\n
\n
Here, F' is the pseudo force acting on the bob when we considered it from car's frame and T is the tension on the string.
\n
In equilibrium, $$\\Sigma$$Fx = 0 and $$\\Sigma$$Fy = 0
\n
$$\\Rightarrow$$ F' cos30$$^\\circ$$ = T sin$$\\alpha$$
A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is $$\\theta$$. The magnitude of the contact force will be :
A block 'A' takes 2 s to slide down a frictionless incline of 30$$^\\circ$$ and length 'l', kept inside a lift going up with uniform velocity 'v'. If the incline is changed to 45$$^\\circ$$, the time taken by the block, to slide down the incline, will be approximately :
A $$2 \\mathrm{~kg}$$ brick begins to slide over a surface which is inclined at an angle of $$45^{\\circ}$$ with respect to horizontal axis. The co-efficient of static friction between their surfaces is:
$$\\begin{aligned}\n& m g \\sin \\theta=\\mu \\cdot m g \\cos \\theta \\\\\n& \\Rightarrow \\tan \\theta=\\mu=1\n\\end{aligned}$$
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9985,
"subject": "Physics",
"question": "A lift is moving down with acceleration $$a.$$ A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively",
"options": [
{
"text": "$$g,g$$ "
},
{
"text": "$$g-a, g-a$$"
},
{
"text": "$$g-a, g$$ "
},
{
"text": "$$a, g$$ "
}
],
"answer": "$$g-a, g$$ ",
"solution": "**Answer:** $$g-a, g$$ \n\nLet acceleration of ball = $${\\overrightarrow a _b}$$ and acceleration of man is = $${\\overrightarrow a _m}$$\n
With respect to the man standing in the lift, the acceleration of the ball \n
$${\\overrightarrow a _{bm}} = {\\overrightarrow a _b} - {\\overrightarrow a _m}$$\n
$$ \\Rightarrow {a_{bm}} = g - a$$\n
Where $$a$$ is the acceleration of the man as the acceleration of the lift is $$a$$.\n
With respect to the man standing on the ground the acceleration of the ball\n
$${\\overrightarrow a _{bm}} = {\\overrightarrow a _b} - {\\overrightarrow a _m}$$ \n
$$ \\Rightarrow {a_{bm}} = g - 0$$\n
$$ = g$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9986,
"subject": "Physics",
"question": "Two forces are such that the sum of their magnitudes is $$18$$ $$N$$ and their resultant is $$12$$ $$N$$ which is perpendicular to the smaller force. Then the magnitudes of the forces are ",
"options": [
{
"text": "$$12N,$$ $$6N$$ "
},
{
"text": "$$13N,$$ $$5N$$"
},
{
"text": "$$10N,$$ $$8N$$"
},
{
"text": "$$16N$$, $$2N.$$ "
}
],
"answer": "$$13N,$$ $$5N$$",
"solution": "**Answer:** $$13N,$$ $$5N$$\n\nLet the two forces be $${F_1}$$ and $${F_2}$$ and let $${F_2}$$ is smaller than $$ {F_1} $$ and assume $$R$$ is the resultant force.\n
Given $${F_1} + {F_2} = 18$$ $$\\,\\,\\,\\,\\,\\,$$ ....$$(i)$$\n
From the right angle triangle, $$F_2^2 + {R^2} = F_1^2$$\n
or $$\\left( {{F_1} - {F_2}} \\right) = 8$$ $$\\,\\,\\,\\,\\,\\,$$ ....$$(ii)$$\n
By solving equation $$(i)$$ and $$(ii)$$ we get,\n
$${{F_1} = 13\\,N}$$ and $${{F_2} = 5\\,N}$$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9987,
"subject": "Physics",
"question": "A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads $$49$$ $$N,$$ when the lift is stationary. If the lift moves downward with an acceleration of $$5 m/{s^2}$$, the reading of the spring balance will be ",
"options": [
{
"text": "$$24$$ $$N$$ "
},
{
"text": "$$74$$ $$N$$ "
},
{
"text": "$$15$$ $$N$$ "
},
{
"text": "$$49$$ $$N$$ "
}
],
"answer": "$$24$$ $$N$$ ",
"solution": "**Answer:** $$24$$ $$N$$ \n\n \n When lift is stationary then, \n
T1 = $$mg$$ = 49 N\n
$$m$$ = 5\n
For the bag accelerating down \n
$$mg-T=ma$$\n
$$\\therefore$$ $$T=m(g-a)$$\n
$$ = 5\\left( {9.8 - 5} \\right) = 24\\,N$$ ",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9988,
"subject": "Physics",
"question": "A rocket with a lift-off mass $$3.5 \\times {10^4}\\,\\,kg$$ is blasted upwards with an initial acceleration of $$10m/{s^2}.$$ Then the initial thrust of the blast is ",
"options": [
{
"text": "$$3.5 \\times {10^5}N$$ "
},
{
"text": "$$7.0 \\times {10^5}N$$ "
},
{
"text": "$$14.0 \\times {10^5}N$$ "
},
{
"text": "$$1.75 \\times {10^5}N$$ "
}
],
"answer": "$$3.5 \\times {10^5}N$$ ",
"solution": "**Answer:** $$3.5 \\times {10^5}N$$ \n\nHere, thrust force is responsible to accelerate the rocket, \n
So initial thrust of the blast\n
= (Lift-off mass) × acceleration\n
= (3.5 × 104) × (10)\n
= 3.5 × 105 N",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9989,
"subject": "Physics",
"question": "A light spring balance hangs from the hook of the other light spring balance and a block of mass $$M$$ $$kg$$ hangs from the former one. Then the true statement about the scale reading is ",
"options": [
{
"text": "Both the scales read $$M$$ $$kg$$ each"
},
{
"text": "The scale of the lower one reads $$M$$ $$kg$$ and of the upper one zero "
},
{
"text": "The reading of the two scales can be anything but the sum of the reading will be $$M$$ $$kg$$ "
},
{
"text": "Both the scales read $$M/2$$ $$kg$$ each "
}
],
"answer": "Both the scales read $$M$$ $$kg$$ each",
"solution": "**Answer:** Both the scales read $$M$$ $$kg$$ each\n\n Question says, both spring balance are light that means springs are massless.\n \n The Earth pulls the block by a force $$Mg.$$ The block in turn exerts a force $$Mg$$ on the spring of spring balance $${S_1}$$ which therefore shows a reading of $$Mkgf.$$ \n
The spring $${S_1}$$ is massless. Therefore it exerts a force of $$Mg$$ on the spring of spring balance $${S_2}$$ which shows the reading of $$Mkgf.$$
\n",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 9990,
"subject": "Physics",
"question": "A particle of mass 0.3 kg subjected to a force $$F=-kx$$ with $$k=15$$ $$N/m$$. What will be its initial acceleration if it is released from a point 20 cm away from the origin?",
"options": [
{
"text": "$$15\\,\\,\\,\\,m/{s^2}$$ "
},
{
"text": "$$3\\,\\,\\,m/{s^2}$$ "
},
{
"text": "$$10\\,\\,\\,m/{s^2}$$ "
},
{
"text": "$$5\\,\\,\\,m/{s^2}$$ "
}
],
"answer": "$$10\\,\\,\\,m/{s^2}$$ ",
"solution": "**Answer:** $$10\\,\\,\\,m/{s^2}$$ \n\n
$$\\Rightarrow$$ a = $${3 \\over m}$$ = $${3 \\over {0.3}}$$ = 10 m/s2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9991,
"subject": "Physics",
"question": "A particle of mass m is acted upon by a force F given by the empirical law\nF =$${R \\over {{t^2}}}\\,v\\left( t \\right).$$ If this law is to be tested experimentally by observing the motion starting from rest, the best way is to plot :",
"options": [
{
"text": "$$\\upsilon $$(t) against t2"
},
{
"text": "log $$\\upsilon $$(t) against $${1 \\over {{t^2}}}$$ "
},
{
"text": "log $$\\upsilon $$(t) against t"
},
{
"text": "log $$\\upsilon $$(t) against $${1 \\over {{t}}}$$ "
}
],
"answer": "log $$\\upsilon $$(t) against $${1 \\over {{t}}}$$ ",
"solution": "**Answer:** log $$\\upsilon $$(t) against $${1 \\over {{t}}}$$ \n\nGiven, \n
Graph between lnv and $${{1 \\over t}}$$ will be straight line curve.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9992,
"subject": "Physics",
"question": "Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle $$\\theta $$ with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle $$\\theta $$ is -",
"options": [
{
"text": "90o"
},
{
"text": "60o"
},
{
"text": "30o"
},
{
"text": "120o"
}
],
"answer": "120o",
"solution": "**Answer:** 120o\n\n4F2 + 9F2 + 12F2 cos $$\\theta $$ = R2\n
cos $$\\theta $$ = $$-$$ $${{12{F^2}} \\over {24{F^2}}}$$ = $$-$$ $${1 \\over 2}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9993,
"subject": "Physics",
"question": "A ball is thrown upward with an initial velocity\nV0 from the surface of the earth. The motion\nof the ball is affected by a drag force equal to\nm$$\\gamma $$u2 (where m is mass of the ball, u is its\ninstantaneous velocity and $$\\gamma $$ is a constant).\nTime taken by the ball to rise to its zenith is :",
"options": [
{
"text": "$${1 \\over {\\sqrt {\\gamma g} }}{\\tan ^{ - 1}}\\left( {\\sqrt {{\\gamma \\over g}} {V_0}} \\right)$$"
},
{
"text": "$${1 \\over {\\sqrt {\\gamma g} }}{ln}\\left( 1+ {\\sqrt {{\\gamma \\over g}} {V_0}} \\right)$$"
},
{
"text": "$${1 \\over {\\sqrt {\\gamma g} }}{\\sin ^{ - 1}}\\left( {\\sqrt {{\\gamma \\over g}} {V_0}} \\right)$$"
},
{
"text": "$${1 \\over {\\sqrt {2\\gamma g} }}{\\tan ^{ - 1}}\\left( {\\sqrt {{2\\gamma \\over g}} {V_0}} \\right)$$"
}
],
"answer": "$${1 \\over {\\sqrt {\\gamma g} }}{\\tan ^{ - 1}}\\left( {\\sqrt {{\\gamma \\over g}} {V_0}} \\right)$$",
"solution": "**Answer:** $${1 \\over {\\sqrt {\\gamma g} }}{\\tan ^{ - 1}}\\left( {\\sqrt {{\\gamma \\over g}} {V_0}} \\right)$$\n\nGiven, drag force, $F=m \\gamma v^2$ ......(i)\n
As we know, general equation of force\n
$$\n=m a\n$$ .........(ii)\n
Comparing Eqs. (i) and (ii), we get\n
$$\na=\\gamma v^2\n$$\n
\nThe net retardation of the ball when thrown vertically upward is therefore \n
$a_{\\text{net}} = - (g + \\gamma v^2) = \\frac{dv}{dt}$, where $g$ is the acceleration due to gravity.\n\n
Rearranging terms gives us :\n\n
$$\\frac{dv}{g + \\gamma v^2} = - dt$$\n\n
We now need to integrate both sides of this equation.\n\n
When the ball is thrown upward with velocity $v_0$ and reaches its zenith ( \"zenith\" refers to the highest point that the ball reaches in its upward trajectory.), the velocity is $0$. The time to reach the zenith is $t$.\n\n
$$\\frac{1}{\\gamma} \\int\\limits_{v_0}^{0} \\frac{1}{\\left(\\frac{g}{\\gamma}+v^2\\right)} dv = - \\int\\limits_{0}^{t} dt$$\n\n
Recognizing the integral as the standard form $\\frac{1}{x^2 + a^2} = \\frac{1}{a} \\tan^{-1}\\left(\\frac{x}{a}\\right)$, we write the integral in terms of the arctangent function.\n\n
Therefore, the time taken by the ball to rise to its zenith, considering the drag force, is given by \n\n
$$t = \\frac{1}{\\sqrt{\\gamma g}} \\tan^{-1}\\left(\\sqrt{\\frac{\\gamma}{g}} V_0\\right)$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9994,
"subject": "Physics",
"question": "A bullet of mass 20 g has an initial speed of 1 ms–1\n, just before it starts penetrating a mud wall of thickness\n20 cm. If the wall offers a mean resistance of 2.5 × 10–2 N, the speed of the bullet after emerging from the\nother side of the wall is close to : ",
"options": [
{
"text": "0.3 ms-1"
},
{
"text": "0.1 ms-1"
},
{
"text": "0.7 ms-1"
},
{
"text": "0.4 ms-1"
}
],
"answer": "0.7 ms-1",
"solution": "**Answer:** 0.7 ms-1\n\nGiven, resistance offered by the wall\n
$$\n\\begin{array}{ll}\n\\Rightarrow v^2=\\frac{1}{2} \\\\\\\\\n\\therefore v=\\frac{1}{\\sqrt{2}} \\approx 0.7 \\mathrm{~ms}^{-1}\n\\end{array}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9995,
"subject": "Physics",
"question": "A spaceship in space sweeps stationary\ninterplanetary dust. As a result, its mass\n increases at a rate $${{dM\\left( t \\right)} \\over {dt}}$$ = bv2(t), where v(t) is\nits instantaneous velocity. The instantaneous\nacceleration of the satellite is :",
"options": [
{
"text": "-bv3(t)"
},
{
"text": "$$ - {{2b{v^3}} \\over {M\\left( t \\right)}}$$"
},
{
"text": "$$ - {{b{v^3}} \\over {M\\left( t \\right)}}$$"
},
{
"text": "$$ - {{b{v^3}} \\over {2M\\left( t \\right)}}$$"
}
],
"answer": "$$ - {{b{v^3}} \\over {M\\left( t \\right)}}$$",
"solution": "**Answer:** $$ - {{b{v^3}} \\over {M\\left( t \\right)}}$$\n\nGiven $${{dM\\left( t \\right)} \\over {dt}}$$ = bv2(t)\n
In free space\nno external force\nso there in only thrust force on rocket.\n
Fthrust = v$${{dm} \\over {dt}}$$\n
Force on satellite = $$ - \\overrightarrow v {{dm\\left( t \\right)} \\over {dt}}$$\n
M(t)a = – v (bv2)\n
$$ \\Rightarrow $$ a = $$ - {{b{v^3}} \\over {M\\left( t \\right)}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 9996,
"subject": "Physics",
"question": "A particle moving in the xy plane experiences a velocity dependent force\n $$\\overrightarrow F = k\\left( {{v_y}\\widehat i + {v_x}\\widehat j} \\right)$$\n, where vx\nand vy\n are the x and y components of its velocity $$\\overrightarrow v $$\n. If $$\\overrightarrow a $$\n is the acceleration of the particle, then\n which of the following statements is true for the particle?",
"options": [
{
"text": "kinetic energy of particle is constant in time"
},
{
"text": "quantity $$\\overrightarrow v \\times \\overrightarrow a $$\n is constant in time"
},
{
"text": "quantity $$\\overrightarrow v .\\overrightarrow a $$\n is constant in time"
},
{
"text": "$$\\overrightarrow F $$ arises due to a magnetic field\n"
}
],
"answer": "quantity $$\\overrightarrow v \\times \\overrightarrow a $$\n is constant in time",
"solution": "**Answer:** quantity $$\\overrightarrow v \\times \\overrightarrow a $$\n is constant in time\n\nGiven $$\\overrightarrow F = k\\left( {{v_y}\\widehat i + {v_x}\\widehat j} \\right)$$\n
$$ \\Rightarrow $$ m$$\\overrightarrow a $$ = $$k\\left( {{v_y}\\widehat i + {v_x}\\widehat j} \\right)$$\n
$$ \\Rightarrow $$ $$\\overrightarrow a = {k \\over m}\\left( {{v_y}\\widehat i + {v_x}\\widehat j} \\right)$$\n
Also $${{d{v_x}} \\over {dt}} = {k \\over m}{v_x}$$\n
and $${{d{v_y}} \\over {dt}} = {k \\over m}{v_y}$$\n
$$ \\Rightarrow 0 - v_0^2/2 = - {\\alpha \\over m}{{{x^3}} \\over 3} \\Rightarrow x = {\\left( {{{3m} \\over 2}{{v_0^2} \\over \\alpha }} \\right)^{1/3}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9999,
"subject": "Physics",
"question": "A person standing on a spring balance inside a stationary lift measures 60 kg. The weight of that person if the lift descends with uniform downward acceleration of 1.8 m/s2 will be ______________ N. [g = 10 m/s2]",
"options": [],
"answer": "492",
"solution": "**Answer:** 492\n\n
The apparent weight $W_{\\text{app}}$ of a person in an elevator moving with acceleration is given by:
\n
$$W_{\\text{app}} = m(g - a)$$
\n
where:
\n
\n
$W_{\\text{app}}$ is the apparent weight,
\n
$m$ is the mass of the person,
\n
$g$ is the acceleration due to gravity,
\n
$a$ is the acceleration of the elevator.
\n
\n
Given that the person's mass is 60 kg, the acceleration due to gravity is 10 m/s², and the acceleration of the lift is 1.8 m/s², we can substitute these values into the formula:
So, the apparent weight of the person when the lift descends with a uniform downward acceleration of 1.8 m/s² will be 492 N.
\n",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 10000,
"subject": "Physics",
"question": "A boy pushes a box of mass 2 kg with a force $$\\overrightarrow F = \\left( {20\\widehat i + 10\\widehat j} \\right)N$$ on a frictionless surface. If the box was initially at rest, then ___________ m is displacement along the x-axis after 10s.",
"options": [],
"answer": "500",
"solution": "**Answer:** 500\n\n$$\\overrightarrow F = 20\\widehat i + 10\\widehat j$$
$$\\overrightarrow a = {{\\overrightarrow F } \\over m} = {{20\\widehat i + 10\\widehat j} \\over 2} = 10\\widehat i + 5\\widehat j$$
$$ \\therefore $$ $$\\overrightarrow s = {1 \\over 2}\\overrightarrow a {t^2} = {1 \\over 2}\\left( {10\\widehat i + 5\\widehat j} \\right) \\times {\\left( {10} \\right)^2}$$
$$ = 50\\left( {10\\widehat i + 5\\widehat j} \\right)m$$
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwjvlw/dec041c9-e406-4f98-a94d-783370c1f55d/ddd44840-ca2d-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwjvlx.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnkfs1i7/86b3f2b6-ecef-40f3-8533-8213186f85af/f96ad2f0-677b-11ee-aeb9-a1910acfd49b/file-6y3zli1lnkfs1i8.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8s48w/7106c58a-521f-4564-b530-e2a9d2152dfd/2d124800-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8s48x.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwk4845t/9c9451b9-d46c-4c1c-ab08-287bac97c8b0/bd468810-197c-11ef-8e15-2107d8c35500/file-1lwk4845u.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264686/exam_images/ug1vcgctwnpzuwlv3nqn.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265351/exam_images/bpha9tfo2ojwynaftia1.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267479/exam_images/wjrxugubrcihnruos7c4.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263683/exam_images/w8qnmvdmubr7jibhtirj.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwjjyhu6/768a07f6-81f2-4c0b-ad88-e9817b38c6d7/b186a5f0-5075-11ec-8f7b-45f8e5a35226/file-1kwjjyhu7.png\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7518,
"subject": "General Science",
"question": "Let A be a fixed point (0, 6) and B be a moving point (2t, 0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is : ",
"options": [
{
"text": "3x2 $$-$$ 2y $$-$$ 6 = 0"
},
{
"text": "3x2 + 2y $$-$$ 6 = 0"
},
{
"text": "2x2 + 3y $$-$$ 9 = 0"
},
{
"text": "2x2 $$-$$ 3y + 9 = 0"
}
],
"answer": "2x2 + 3y $$-$$ 9 = 0",
"solution": "**Answer:** 2x2 + 3y $$-$$ 9 = 0\n\nA(0, 6) and B(2t, 0)![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265928/exam_images/t38wivroym0l1mk7udq7.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lhu7fnjs/31373c2a-3f37-47d1-bed5-e74cffaeb86b/e21d2070-f611-11ed-8eab-4b3832376002/file-1lhu7fnjt.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lvnxhk5n/61340bd7-15c2-4d49-8787-6b1d30c56e5c/4f3a81b0-07c9-11ef-afdb-fb69ed1077de/file-6y3zli1lvnxhk5o.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwagdtgz/adc2b178-39c7-4610-ae95-7cf96da59f71/72edac30-142c-11ef-983f-65b4bd1ca415/file-1lwagdth0.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264578/exam_images/cudjhzff4ynokmmza2f9.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266344/exam_images/ctbame8e1bgmyuficejz.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1u0w1z/5e71e275-9890-4f83-b0e3-dce0674de6b5/173b5570-d40a-11ee-b9d5-0585032231f0/file-1lt1u0w20.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263525/exam_images/hrqnkqyxxwu0ays0itme.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264412/exam_images/wsbio4fzzawvxpgqqiq9.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5oc3byi/c98d69fd-398a-42cd-a87f-3ee77ee795a4/cca73aa0-0544-11ed-987f-3938cfc0f7f1/file-1l5oc3byj.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw8pr3ta/992548a6-ea30-4544-b8d1-c7f041deb6d3/854cb2e0-1337-11ef-9f8d-838c388c326d/file-1lw8pr3tb.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265012/exam_images/xq65o2ooxotlgbornwg4.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266358/exam_images/znwolrv8chfin230lej7.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264559/exam_images/k1bzxqvdfbjwl4zlissa.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266026/exam_images/irq1wiy4pougehf2jvx7.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265670/exam_images/k7tcjadeqglnae8jaifx.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263835/exam_images/thzbx3newaxtuygg2q1m.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266256/exam_images/wqywhib9gce5rptudaka.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266409/exam_images/r46trc585juw1pxqfahv.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263333/exam_images/sgkffsp4mfuv6idliumw.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263689/exam_images/shi6jmtimy8lhgilqtoi.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263884/exam_images/aumkosbopsf2ogx1v0ai.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264527/exam_images/ysrpqzdcmjighfj5a4xy.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxoe9q9z/76596a6c-2785-45d0-8cbb-72347daf83fe/95cbec80-66eb-11ec-b4c9-97a7fc3f3aad/file-1kxoe9qa0.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264667/exam_images/acnl84vwiakold3way0b.webp\")
\n\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266102/exam_images/h1fc0mfvmn1izevgxbxm.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l97rw5km/2f99cd4f-8200-402b-aaa3-f8d0d73a633d/21992d60-4b5a-11ed-bfde-e1cb3fafe700/file-1l97rw5kn.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1ldtzo4m3/b85082b3-e539-44fe-bf5f-e63a3546f4f7/a7383ba0-a6c2-11ed-8d92-0101c7cb78f8/file-1ldtzo4m4.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnmwb3jr/502d615e-8cf5-4478-9d73-0b9837e70b02/2ebbe560-68d6-11ee-a3a1-07c5c60fca90/file-6y3zli1lnmwb3js.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwcdv7/6c17fcea-c801-4445-8b27-227c5944b078/0d75d830-ca2d-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwcdv8.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsnjikpx/f32556f0-5b8e-46de-b6e9-9f078c18e4de/05872950-cc2e-11ee-b20d-39b621d226e3/file-6y3zli1lsnjikpy.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnkh52/7d853a24-3c49-423c-95b3-38a63c841203/38d89660-cde4-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnkh53.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw37w9ht/b3e9aa4d-c4a4-4387-bd7f-990e7790d425/c5298810-1031-11ef-b980-477f779c8c59/file-1lw37w9hu.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwd2p5t3/fdf80bf0-c51d-4c22-865c-bd4b3e3d59be/487f8460-159d-11ef-aabb-5de744d2ad81/file-1lwd2p5t4.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwey5iox/7c0cf8ef-bef6-4412-bab2-d9c808c7d020/130d0510-16a5-11ef-aafd-dd0edda8277a/file-1lwey5ioy.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265710/exam_images/vm4amwtny6bublz2trzy.webp\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 7591,
"subject": "General Science",
"question": "Let $$\\cos \\left( {\\alpha + \\beta } \\right) = {4 \\over 5}$$ and $$\\sin \\,\\,\\,\\left( {\\alpha - \\beta } \\right) = {5 \\over {13}},$$ where $$0 \\le \\alpha ,\\,\\beta \\le {\\pi \\over 4}.$$ \n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264103/exam_images/a6qatzrn0wehdulviizi.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264283/exam_images/duu7crwpjpg3itzy9gd0.webp\")
",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 7645,
"subject": "General Science",
"question": "Let $$a, b$$ and $$c$$ be distinct non-negative numbers. If the vectors $$a\\widehat i + a\\widehat j + c\\widehat k,\\,\\,\\widehat i + \\widehat k$$ and $$c\\widehat i + c\\widehat j + b\\widehat k$$ lie in a plane, then $$c$$ is :",
"options": [
{
"text": "the Geometric Mean of $$a$$ and $$b$$"
},
{
"text": "the Arithmetic Mean of $$a$$ and $$b$$"
},
{
"text": "equal to zero "
},
{
"text": "the Harmonic Mean of $$a$$ and $$b$$"
}
],
"answer": "the Geometric Mean of $$a$$ and $$b$$",
"solution": "**Answer:** the Geometric Mean of $$a$$ and $$b$$\n\nVector $$a\\overrightarrow i + a\\overrightarrow j + c\\overrightarrow k ,\\,\\,\\overrightarrow i + \\overrightarrow k $$ \n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264248/exam_images/rtiinzvghawcvhtcjcms.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267236/exam_images/mxsolnud9xur1rrhm47n.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266760/exam_images/elmya39omzdt3xm8r3m7.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lfhv3u5p/8ecc7bc1-f988-47d9-9271-a3d8261cd426/d568c1d0-c7af-11ed-b61f-698c390a560e/file-1lfhv3u5q.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnbavvev/3506a77b-c00f-43d4-873d-aea0b917a5f0/c874fd70-6275-11ee-af05-bf9054b926c1/file-6y3zli1lnbavvew.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1v5c9e/0c2c210a-ab17-46c0-813d-6d826d2adf6f/7c1dc320-d40e-11ee-b9d5-0585032231f0/file-1lt1v5c9f.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263647/exam_images/vxui8byefbrtoqgpopbv.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265163/exam_images/odwfpjssxxiwn1znqs07.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266031/exam_images/ube74chtyryryxq4ef6j.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263298/exam_images/rovhevula1wlt2hu6mar.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264070/exam_images/ep6a24u2iigywhjyduzo.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263795/exam_images/vuqsv6zh5ecsiacx3auh.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265482/exam_images/mcpi8ysmwwzqgrff8js2.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267524/exam_images/mzqj6yentbg3ruppcugm.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l97sapyl/779dd912-c32e-40a3-8f62-d0f26f1743e1/b6acf9d0-4b5b-11ed-bfde-e1cb3fafe700/file-1l97sapym.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lh1phdby/e74db975-74e1-4987-80c4-b8ae4380dc69/bb54dde0-e665-11ed-9cfb-9bcc868d407f/file-1lh1phdbz.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1vv91f/efa66592-d248-4f32-8e31-2416eaa514c4/4caa5d30-d411-11ee-b9d5-0585032231f0/file-1lt1vv91g.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsnjezh5/8e16ffb2-7d5b-416d-bc65-2de0bc58b08c/a1b0b090-cc2d-11ee-b20d-39b621d226e3/file-6y3zli1lsnjezh6.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8xnpy/836457b4-b43d-438c-b5ce-783fa67cdc3d/c728ff60-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8xnpz.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lw3cizlj/32ca14d9-c7d9-4752-a76d-bc023959c231/e1a4ec70-1043-11ef-9f6c-75804a813f04/file-jaoe38c1lw3cizlk.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lw3cka9k/5af601d2-c9f5-428f-ad0f-ea9fdc1cb8f2/05b19780-1044-11ef-9f6c-75804a813f04/file-jaoe38c1lw3cka9l.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwgfpgex/f708b26f-9a53-4063-a888-8e61f5ee5884/8151b490-1776-11ef-bded-616e4abb66b9/file-1lwgfpgey.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/EyI9jwNfZakNSnx6N/RG2tUhYR6L2pFM5r986jI6AkPlnGj/HFxkeghaZOPffeRH5uOmx1/image.svg\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l36s2v57/7fb9ba2f-2acb-4af2-95e3-05d177745de7/c748d6c0-d404-11ec-b808-5752a3163b13/file-1l36s2v58.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l331fpnn/3c8a124b-c725-4252-85f9-e46acf79e535/3c20df30-d1f6-11ec-b83f-ebfea682138a/file-1l331fpno.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263369/exam_images/ajrqexexn67nnrokgcbn.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263802/exam_images/i0j0zbm9qpkivxppcwe2.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264162/exam_images/lrbug02qowhjyevbsblu.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264054/exam_images/f7nywumtoax65vva4edj.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264378/exam_images/oupn0b4afamqmsslzyc0.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6z6wn42/89ab94f0-99d2-49fe-acfa-1f6b868af330/5b7bf350-1f09-11ed-8ba4-19601952a51f/file-1l6z6wn45.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqmaruy/fe41a022-2b85-4305-a918-b331d99425f6/41e7dea0-cddf-11ee-a0d3-7b75c4537559/file-6y3zli1lsqmaruz.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7e9lrut/71df2bd1-d468-4054-adb5-aa2728ee9596/aa5ffe60-2753-11ed-a077-1f1e3989e798/file-1l7e9lruu.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7e9mb2y/a94941c7-3bc0-4deb-8977-65554f010b7d/b939dfa0-2753-11ed-a077-1f1e3989e798/file-1l7e9mb2z.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266433/exam_images/sj7otzqwlwlgwdcfoecv.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264800/exam_images/jvb96sku7yx0jplifu85.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264527/exam_images/e9xb0kf3dr4kfgm1qhb9.webp\")
\n$${{hc} \\over \\lambda } = 13.6\\,ev(g)\\left\\{ {1 - {1 \\over {16}}} \\right\\}$$![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt2v9xlh/fd15bcdb-72be-482e-91f5-33b849e89d1e/c47bc850-d49b-11ee-8574-81dc091f6c89/file-1lt2v9xli.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266887/exam_images/ync9uilcrpsh67oseyar.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265008/exam_images/cvxnwh6xtmsk8yd94ycf.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxpz7lve/098683e2-ad9d-4fbe-b0a9-a193566a87fa/4322eda0-67ca-11ec-9473-1365722e3eb6/file-1kxpz7lvf.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5jy57cb/ed42b5eb-8159-4e9f-ad0a-ea7c25361b2a/1e4cb3b0-02db-11ed-b2c8-3573db55a50d/file-1l5jy57cc.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1ledmh918/65776890-780e-4809-8019-30f315388ea4/be47c3c0-b18e-11ed-a682-13f364283dca/file-1ledmh919.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265977/exam_images/zlkeaudseiga8sxodwgb.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266139/exam_images/vsnellriwflmnvy8txrs.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6umubk6/bb6034f0-8796-4414-a5a4-1fb6bd4e7762/c1c29170-1c87-11ed-b633-f353ad3cb8e4/file-1l6umubk7.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/dCrxyqN2t9qic3vxL/3sLhZsfK0H88LZCPg6LGhLCmdtHkX/NsSodhvhgkZE8gIkM7A6dJ/image.svg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264396/exam_images/y2xokb3qgd0nfpixreoq.webp\")
\nInitially Q = CV(1 + n)![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264847/exam_images/qxzicl0hvoglymt0fdpf.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265309/exam_images/ko7eif0kpubtporx7vlq.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264082/exam_images/b5etp3tyjqbdfjzwwg4i.webp\")
\n\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264372/exam_images/seugbn3rslsguohlfjs9.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263682/exam_images/mq7hbh1khr7bl8km9wyb.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266215/exam_images/akxhw7oe3vgt5873yxio.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6z6ite4/3a105f96-a760-4f30-8d6f-d7a5e79bc293/db0148c0-1f07-11ed-8ba4-19601952a51f/file-1l6z6ite5.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1ledm9s54/971e9f03-de7f-4bb1-892a-71705e3b5fb2/ee915380-b18d-11ed-a682-13f364283dca/file-1ledm9s55.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5k7nwf/9d447182-d3e9-4a54-a3bf-bb97dcd4e0ce/9c3824f0-ad1f-11ed-8bc1-d3bd0941e5b5/file-1le5k7nwg.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw4xr87f/2bfc35fc-cdad-4ec5-bc19-fb1602f5ce82/af20e1b0-1123-11ef-a3e8-ff0446908720/file-1lw4xr87g.png?format=png\")
![\"JEE](\"https://imagex.cdn.examgoal.net/3hjN6VPUKrhF8wmJe/9nSSVr34CsR1q8aQVzIsZy6bxcbnx/gCgSrRrj2yv32OSaUAlwzw/image.svg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265616/exam_images/se6f7getrznbywz5wib4.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266447/exam_images/vbuasjqxftln1akirgfi.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt2xedjy/60f44ef6-b510-41ec-ab7a-933525632e2d/124e23e0-d4a4-11ee-bdd1-01c80c3e2d9a/file-1lt2xedjz.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263537/exam_images/yspdppo1hkbkogn8bc74.webp\")
\nCharges at inner plates are 1 $$\\mu $$C and –1 $$\\mu $$C.![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264354/exam_images/beubqj2qzbmludro0org.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l79fmjbf/beef1ae2-10b1-4672-9c56-1efa220d1b8b/4b1d91b0-24ab-11ed-8d2e-5f0df5271c2d/file-1l79fmjbg.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/jc161ivQcbdq0bi67/mIw0nNGl9d4hY8XmNmdEH1ag2TFCU/3cSxm5dD2HtlXDAQt3pMeb/image.svg\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l91ofg7j/6a0d62ec-2d96-49cf-9855-a36685260c79/e1c685f0-47ff-11ed-8757-0f869593f41f/file-1l91ofg7k.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/MmP0ascyNLHCptltH/PafY176kaTqUDIfLmTGROM52Rc6tA/jYuZQjt2765EWW34bkSVE4/image.svg\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263717/exam_images/el3kjnrjr1uz8y3o9hry.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6ummxzq/f571ac43-99e4-45d3-856a-a0bab6271d11/f49d5360-1c86-11ed-b633-f353ad3cb8e4/file-1l6ummxzr.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l91oguxy/85d124a8-cb21-4c9b-ad8d-99746c3176de/08f77b70-4800-11ed-8757-0f869593f41f/file-1l91oguxz.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263536/exam_images/hyykblaoxaijojoyenag.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l32mth8k/035418a3-4c67-48c4-8a2a-6921067bdf76/12040440-d1bd-11ec-9218-efb2cf12c71b/file-1l32mth8l.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l32mu62c/397c94a2-021d-4f5b-ae17-11932eb8142c/2531af40-d1bd-11ec-9218-efb2cf12c71b/file-1l32mu62d.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265484/exam_images/vr7aeezmxuya5yrhjcrk.webp\")
\n$$ \\therefore 2{{mv} \\over {\\sqrt 2 }} = 4mv$$![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lk26wd5q/84e70eea-d2a2-4655-a938-b7c9295d5dac/7daab4d0-220e-11ee-8e56-71f4c7b5eb20/file-6y3zli1lk26wd5r.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264972/exam_images/pnpxc6iwgd5vdrx0lrey.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265728/exam_images/qmtmemrdworv9ooluskm.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264437/exam_images/luzo4llr3tdlts6frj7c.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263593/exam_images/ofwdg54bhgf9i5vpwwdd.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267164/exam_images/d9xgpotrwc6q37lnbm13.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267066/exam_images/oubyomaxlt5cjmbgbozw.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266619/exam_images/sypjio5qe7rurgxwnyps.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265717/exam_images/nwijf9rwaqeaylnr89dj.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxfolmp7/259289fe-e6ab-4d89-9d92-47e703ed19eb/ee306fb0-6220-11ec-bbdc-cb29b93c10b9/file-1kxfolmp8.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267752/exam_images/hynytorlgqmtqikmstpj.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263745/exam_images/ljpavkqhq5dilkime3yy.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264069/exam_images/tjslxrzg7ryb7tnzp55m.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kypoz6o1/630aca9f-74a0-4ed4-a616-de2652c7aedb/75019620-7b6e-11ec-92f3-61d366101705/file-1kypoz6o2.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kypp0g6t/dbc4f204-1472-45e3-b6d8-b10a5a9d6e06/982b8250-7b6e-11ec-a2d2-9980f7a7c35c/file-1kypp0g6u.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264319/exam_images/mcdijut1xrwp3d57papi.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1les28d5t/a0b69fe4-7a4d-42f1-b0f1-c67f1d5a4141/52583110-b97f-11ed-981f-89b0ac23a016/file-1les28d5u.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1les295g1/34b2071c-d979-412e-918d-9e34e02505ca/68317410-b97f-11ed-981f-89b0ac23a016/file-1les295g2.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lslupoi3/cc2aecc8-c56e-451b-bdf9-359e235929a2/3e3c51b0-cb40-11ee-ad47-a16d1086e690/file-6y3zli1lslupoi4.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/3T7YjPhEZDsqaIC1J/y2vNdG9oZcJCnYzyzEJWNr8riMB2X/v3D5rnZX3TA0kk8ruDjhrU/image.svg\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lk25nvxa/233cb955-9742-46f3-9b57-ec5058f9932c/a8c077e0-2209-11ee-92b9-c1cea67c1a68/file-6y3zli1lk25nvxb.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lk25n7e3/29a2d2ae-b219-49ab-8824-df6d442d684b/95ccefb0-2209-11ee-92b9-c1cea67c1a68/file-6y3zli1lk25n7e4.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5jhmqf4/db9636cb-1bea-4bce-8bbd-a227247c026f/8a98c400-029a-11ed-a9b8-43edceee002f/file-1l5jhmqf5.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/c84SriojYvqSujHdA/kNl0cD8ffQJBhayktDgBQx195q9Rx/7JEIpgVEtSC3j4ZogLIPh7/image.svg\")
\nThe velocity of center of mass of two particle system is\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265914/exam_images/cfqf4om8fmsgxgbo9rrf.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5ifjvd9/19bdec24-cf90-42bb-9f11-b9b2b39390ea/a01583d0-0205-11ed-aa88-69f11483e075/file-1l5ifjvda.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l9zxsgak/53381e88-7018-4a98-a85a-8506f2484124/239a0bc0-5ad7-11ed-8fe8-b7743a8ec2da/file-1l9zxsgal.png?format=png\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265200/exam_images/efzjn8yxwoqfk3d0ulad.webp\")
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 8409,
"subject": "Physics",
"question": "Two cars of masses m1 and m2 are moving in circles of radii r1 and r2, respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal acceleration is",
"options": [
{
"text": "m1r1 : m2r2"
},
{
"text": "m1 : m2"
},
{
"text": "r1 : r2"
},
{
"text": "1 : 1"
}
],
"answer": "r1 : r2",
"solution": "**Answer:** r1 : r2\n\nWe know, $$a = r\\,{w^2} = r \\times {\\left( {{{2\\pi } \\over T}} \\right)^2}$$\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l32myl9a/2586242a-e474-4384-ab16-c8fa737101cc/a02ab3e0-d1bd-11ec-9218-efb2cf12c71b/file-1l32myl9b.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266911/exam_images/vpuji6pk4hocdy19ts4u.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266050/exam_images/j66lwsw8rlnrpkauyaww.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264526/exam_images/sv3rgdbszslmmxryuipw.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxofybix/398719e0-82dc-4102-981e-6996580a7696/2ac253a0-66f2-11ec-9866-0df874a38238/file-1kxofybiy.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266199/exam_images/hklugbxghe7cp73jqegz.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5kpo2ny/38be5e44-dc63-4c88-be0c-45cfdf5a0089/c18c12d0-0346-11ed-8429-27c3de0e24ff/file-1l5kpo2nz.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5l6hdcv/7e2054d6-02e0-4296-bf08-104db7f196ee/8256b6f0-0388-11ed-bfd1-873560f9d960/file-1l5l6hdcw.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lebeg1j0/b8d95c72-61ba-4228-898f-b8c4675e71f6/c25718b0-b055-11ed-b17f-e3e679a04034/file-1lebeg1j1.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lebue8j4/7194737a-103a-428c-9211-c9d357a44a3e/2232b3f0-b094-11ed-a9ff-1951bbe50b73/file-1lebue8j5.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1let35sjj/457e1035-e46d-40a0-b724-d3dc666cfc52/bc939de0-ba0f-11ed-bf62-29125e0e0d77/file-1let35sjk.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5jfhmh/21063d45-0704-4c91-a742-33934be23a35/8cb70e90-ad1c-11ed-8bc1-d3bd0941e5b5/file-1le5jfhmi.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5jhpbr/0d94830e-6fea-4c2f-a400-9ced2e09681a/ca487870-ad1c-11ed-8bc1-d3bd0941e5b5/file-1le5jhpbs.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le4a1hbu/cf5132ff-fd3d-4892-a446-59d2b498fbb0/0c58ae90-ac6b-11ed-8b11-cb59760cfd30/file-1le4a1hbv.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lscudf4i/53002a9a-3b22-4fee-9d31-e8f7de906502/db6ffb20-c64b-11ee-9d8b-f1be86a1b2f3/file-6y3zli1lscudf4j.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsluc3m3/d6f96f5a-f4c4-4f9b-a47b-e687d115c288/c494a1b0-cb3e-11ee-ad47-a16d1086e690/file-6y3zli1lsluc3m4.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr7p1jk/c47a89d4-e629-47c6-95c2-1f208c4603e0/ee6d4810-ce32-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr7p1jl.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263539/exam_images/txcxddg3y9nz52bi3dio.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264554/exam_images/me7nyal89af3qde93tyd.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266908/exam_images/r83onjxtatbp4zew98ss.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6tc4h2y/fe8dff0e-e551-4dae-8c49-554ce7caec00/0f0bcda0-1bd1-11ed-bd9f-71db3a4811b2/file-1l6tc4h2z.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265501/exam_images/xkrsjni0xrpnrah1p5cq.webp\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8493,
"subject": "Physics",
"question": "A modulated signal Cm(t) has the form Cm(t) = 30 sin 300 $$\\pi $$t + 10 (cos 200 $$\\pi $$t −cos 400 $$\\pi $$t). The carrier frequency c, the modulating frequency (message frequency) f$$\\omega $$, and the modulation index $$\\mu $$ are respectively given by :",
"options": [
{
"text": "fc = 200 Hz; f$$\\omega $$ = 50 Hz; $$\\mu $$ = $${1 \\over 2}$$"
},
{
"text": "fc = 150 Hz; f$$\\omega $$ = 50 Hz; $$\\mu $$ = $${2 \\over 3}$$ "
},
{
"text": "fc = 150 Hz; f$$\\omega $$ = 30 Hz; $$\\mu $$ = $${1 \\over 3}$$"
},
{
"text": "fc = 200 Hz; f$$\\omega $$ = 30 Hz; $$\\mu $$ = $${1 \\over 2}$$"
}
],
"answer": "fc = 150 Hz; f$$\\omega $$ = 50 Hz; $$\\mu $$ = $${2 \\over 3}$$ ",
"solution": "**Answer:** fc = 150 Hz; f$$\\omega $$ = 50 Hz; $$\\mu $$ = $${2 \\over 3}$$ \n\nGiven, \n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l39reki4/bfab760e-888d-4fc1-ae30-0742f566d94a/7d97b8c0-d5a8-11ec-a99d-9f93f26c7ff1/file-1l39reki5.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266192/exam_images/cui8xqelobmon8hkehmk.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264406/exam_images/fs8jz4onnrsunnejjp2j.webp\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/Isfifx7pjt7J58Si7/iCKpyiWHecEpyQtlMiwzDmWCHDkQ6/6qo1FAQpkipEZTGRDNIPjP/image.svg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264047/exam_images/t8ovjuqhs3at5ofuad56.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264453/exam_images/sedvwibkhijjrdzbay1u.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265899/exam_images/f6xutcotw2gnkr6gsjjp.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6tdw0df/66410f4a-df19-4dd2-9891-92bd65da2bc2/f5ec8330-1bd7-11ed-a7b0-d7e65118bbd8/file-1l6tdw0dg.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267597/exam_images/zcqqat174duzg8hy57t5.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5gh8rty/9f7be829-74c4-4a2a-91bd-244864fc7747/ac694860-00f2-11ed-ba34-71a54c393c2b/file-1l5gh8rtz.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5kpqvui/e64e8ecd-4932-4da0-8ac2-457e4b1ec585/0fb561a0-0347-11ed-8429-27c3de0e24ff/file-1l5kpqvuj.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6xtgjgt/df9bb4a8-274b-4f7e-9993-5a6f469898f5/fc6433d0-1e47-11ed-9c61-4529b721806b/file-1l6xtgjgu.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1leh3iv36/459bc693-e3a2-4716-af92-d39b71f2d6f6/c04e7200-b377-11ed-aaca-cb4b36b68d7d/file-1leh3iv4x.png?format=png\")
\n\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1leh3k4gy/997a8a4d-5f00-479f-b4e4-b73453b4124c/e3539820-b377-11ed-b883-ed230c2572e2/file-1leh3k4gz.png?format=png\")
\n\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l32mjai6/b190ee43-aa8c-418c-ba9e-86a1ff895f27/f6b920e0-d1bb-11ec-9218-efb2cf12c71b/file-1l32mjai7.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/uKKPmrvAB9zP9OIov/3XAQYM47v2Cp3Do2vEQkn7cwJerbt/pLK0N1xvtov0dMu8TqRAOZ/image.svg\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/Su5iQgFpRruKB9bSm/2PNpI3661XhvAaNHQ2H61TmWEK7xM/vgMSh5TViNN6U85fqIRfiB/image.svg\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/V9QIUmohYHBXTlUl5/f8mxRp2jX0WzAnk0PybL6xM0kBUZX/aSP2euKBoceZJXT5aRdmUA/image.svg\")
\n![\"JEE](\"https://imagex.cdn.examgoal.net/O5zR4337zfTSNyDRm/hWmAcYLkn35cX3r5lK4TjhXK06PmG/x0LvodpPTlpqZnvNwCCuq4/image.svg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264618/exam_images/ea2nafn2gmwz6ppiff4c.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264155/exam_images/wwxrp1djjw35p2zrgkwe.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263355/exam_images/du3rwcbudypfrggki8u3.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264358/exam_images/ywczrpvunyf45azc53kd.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266264/exam_images/vxnh6e6twqsuiy6jmfoq.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266082/exam_images/tjnvxgx1lebnzycejswf.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lj1zmexe/1c7aa9ec-09df-4229-b070-17a125b93818/d1a89810-0e25-11ee-b193-e7c493386590/file-6y3zli1lj1zmexf.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264080/exam_images/neabkssdo9r8ikvqvdeg.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263293/exam_images/ircp5doxcaunxz6jbmqc.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267064/exam_images/xhoeqte0df7ysa5k6q1q.webp\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 8657,
"subject": "Physics",
"question": "A galvanometer is used in laboratory for\ndetecting the null point in electrical\nexperiments. If, on passing a current of 6 mA it\nproduces a deflection of 2o, its figure of merit\nis close to :",
"options": [
{
"text": "6 $$ \\times $$ 10–3 A/div"
},
{
"text": "666o A/div"
},
{
"text": "3 $$ \\times $$ 10–3 A/div"
},
{
"text": "333o A/div"
}
],
"answer": "3 $$ \\times $$ 10–3 A/div",
"solution": "**Answer:** 3 $$ \\times $$ 10–3 A/div\n\nfigure of merit = $${I \\over \\theta }$$\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266174/exam_images/imyxmqkvtiuxojy8nxgb.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266900/exam_images/mrj18lewipk8cejnns6s.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5jqemdq/c70f58cd-41c8-412b-8e71-6f5c0feb722d/db2a8ee0-02bc-11ed-95db-c3fa9a0f41ba/file-1l5jqemdr.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5kpt3dk/2d747c27-da40-45f5-81ac-b01f5f127587/4d249d80-0347-11ed-8429-27c3de0e24ff/file-1l5kpt3dl.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5ky536y/839f6d87-1d8b-48ef-b90e-58dc536e3270/e3be5590-0367-11ed-af51-374257f3e209/file-1l5ky536z.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5le9d7o/bb316064-5929-4f7d-9df2-1a7193aea921/ecc98e40-03a6-11ed-b731-675fd498ca49/file-1l5le9d7p.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le4beg6w/f8e46281-d0b4-4633-967d-41b176518f4c/5e1dcf80-ac70-11ed-a551-772121b6ed7f/file-1le4beg6x.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt2toclz/fdc1cef0-c338-481f-b621-896a016be062/8326ae70-d495-11ee-a613-e11ae62575eb/file-1lt2tocm0.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt2tqvvt/bf1d6741-5b7b-461d-bde1-af7b8f468aec/c9a874a0-d495-11ee-a613-e11ae62575eb/file-1lt2tqvvu.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt2thne5/2f5730dc-195f-4232-8eab-2f5be126e174/c8d0f5d0-d494-11ee-a613-e11ae62575eb/file-1lt2thne6.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoiwwz2/a7d1ced9-1f16-481d-ba5f-11d472167116/73d790e0-ccb8-11ee-8ef5-472d1767d2da/file-6y3zli1lsoiwwz3.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/jr80KlmMStRLVuRaq/YUFjVqeck4OZIBsyzblcm0nH7RhKK/nheUeoJQOmHkvoJmCI3rct/image.svg\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/0RRrcFySZ6JcNyEdf/1H3KDfGLejP8HICSLTHRp6q7v3uW4/qrXU67CSqnnQhJWBuZMeWj/image.svg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265437/exam_images/f6vvbufkzrapliec7lo3.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265659/exam_images/u9kwtelgpeoesooze2a9.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264619/exam_images/kup8anqwrhk1lxlcsrrj.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266981/exam_images/vk2z5nfou4fbv2hwae5b.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265120/exam_images/t1yqpwefn9imrvoyubql.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsiiqszq/b75469fd-7157-4fe4-9ed7-2c53636080e8/1603de60-c96b-11ee-b416-eff853096672/file-6y3zli1lsiiqszr.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265790/exam_images/w6xlxbddqdjpsipdd2ho.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5jqaucn/96f18807-6859-40ab-bd14-b64bed306520/72161370-02bc-11ed-95db-c3fa9a0f41ba/file-1l5jqauco.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5l6kgp9/4955abcb-b0c1-4312-b9c2-87b467ab097c/d85973d0-0388-11ed-bfd1-873560f9d960/file-1l5l6kgpa.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6wir8x4/ff3dc388-37a3-4dc9-b20b-2dfaeb9bbc0c/59131680-1d91-11ed-b1e3-c3a54149d5d6/file-1l6wir8x5.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7m6bkq7/a6d47508-08e1-44c8-ad64-3b608b9c37ae/1895a3f0-2bad-11ed-bad3-59534b1d6f8c/file-1l7m6bkq8.png?format=png\")
\n$$\n\\frac{\\mathrm{R}_{\\mathrm{X}}}{\\mathrm{R}_{\\mathrm{Y}}}=\\frac{\\ell_{\\mathrm{X}}}{\\ell_{\\mathrm{Y}}}\n$$\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1ledln4ww/f3c91b62-ea60-4cdd-8190-c3805dd18cc4/78d2ba00-b18b-11ed-a682-13f364283dca/file-1ledln4wx.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1ledlod87/fe50f982-2220-409b-853d-9e7cdffac554/9b0e0070-b18b-11ed-a682-13f364283dca/file-1ledlod88.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267599/exam_images/xf5ezrbsu9exsycwrfqp.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263265/exam_images/wxewoixyz1w4tkswhhrb.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264132/exam_images/h9aj4cl5tbtovut5jkru.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5jhw3yo/4d242564-1414-4300-8c7c-e97049b43a97/8f574100-029b-11ed-a9b8-43edceee002f/file-1l5jhw3yp.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lslumatj/08a420dc-d051-43e9-90c2-11fdf1acba69/e03d5c80-cb3f-11ee-ad47-a16d1086e690/file-6y3zli1lslumatk.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt2vwplj/57f7acea-fc42-4254-b91c-29d33a047dfc/3dea4a70-d49e-11ee-8574-81dc091f6c89/file-1lt2vwplk.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1ledl0tks/6f0abe55-b10a-42f6-b8a8-2562be02de8e/0c43cfc0-b189-11ed-94f5-239612b3abb5/file-1ledl0tkt.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsrmpn5k/8bb0421b-5dee-44c6-a49e-5ddfb61af2af/a80e9870-ce6d-11ee-991d-5322ace9b43e/file-6y3zli1lsrmpn5l.png?format=png\")
\n\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266142/exam_images/hdccrflkxnnaahjcbqtb.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267214/exam_images/gjb7lailvc3finrye5we.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265979/exam_images/fm55y9ao1bcrdgrr7ts0.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265694/exam_images/xqklknithqotrk70vj6s.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1leoi23xo/0eb20d87-a1d6-4001-9b1f-f87c1a57bca1/13d4dac0-b78a-11ed-8877-d5d81ac5ff88/file-1leoi23xp.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lepxxcug/c3ca6ed1-a3c6-4b09-ab77-d8f46565abdd/ea627c80-b854-11ed-8195-4f3c56fa1eb5/file-1lepxxcuh.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l86i49st/88fabaf6-1f84-46c9-ae16-6c0f212e152e/e80474d0-36da-11ed-b5bb-c51481bdd3e4/file-1l86i49su.png?format=png\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9144,
"subject": "Physics",
"question": "An experiment is performed to determine the I - V characteristics of a Zener diode, which has a protective resistance of R = 100 $$\\Omega $$, and a maximum power of dissipation rating of 1 W. The minimum voltage range of the DC source in the circuit is :",
"options": [
{
"text": "0 $$-$$ 5 V "
},
{
"text": "0 $$-$$ 8 V "
},
{
"text": "0 $$-$$ 12 V "
},
{
"text": "0 $$-$$ 24 V "
}
],
"answer": "0 $$-$$ 24 V ",
"solution": "**Answer:** 0 $$-$$ 24 V \n\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266481/exam_images/ni5xya4ak6ucpf3q7qda.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264087/exam_images/lwixd0dpakgiutyuslmo.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265236/exam_images/apllg3zofsznrk6kgtfe.webp\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/uwMfIPegkqq5unwbi/9lyy3aPlgjMFqhnOnTdGTX07SuN0j/mFUPqvUV9I6wzTnHPaGlmL/image.svg\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9202,
"subject": "Physics",
"question": "A charge $$Q$$ is placed at each of the opposite corners of a square. A charge $$q$$ is placed at each of the other two corners. If the net electrical force on $$Q$$ is zero, then $$Q/q$$ equals: ",
"options": [
{
"text": "$$-1$$ "
},
{
"text": "$$1$$ "
},
{
"text": "$$ - {1 \\over {\\sqrt 2 }}$$ "
},
{
"text": "$$ - 2\\sqrt 2 $$ "
}
],
"answer": "$$ - 2\\sqrt 2 $$ ",
"solution": "**Answer:** $$ - 2\\sqrt 2 $$ \n\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/BzLlRzn0u5SWvCQIN/QF7s2wmRe0lUFoqR2BFoI3loWBMP7/5S1b9Dscj5TnQnS3RNsVJO/image.svg\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/paBaDvpbGvFTYIrc8/FIM2CuSQMrqmccIKacsvnOe2dAQUd/u1zfSTfN8FoeY8jq4BOqVP/image.svg\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/nYkxgNrCOEUTv7dJO/r9sPqa2SXPQiMMHbURqJrDYoSluEc/i8My6X7r5x1onqzNJKWnNW/image.svg\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/8wMAjSfnsrTpd0rS9/TqxHFnNrrIYW3j3e8ZtYxQA0WMF9j/WTEQieh2LpLxP7qBWo9qh2/image.svg\")
\n![\"JEE](\"https://imagex.cdn.examgoal.net/RGZ4HWH6bnTOIkap8/0h5nBN8m8itaD1JAG2e57Iy2tJISM/gubYshaILBzO3nqHnlFawN/image.svg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266898/exam_images/mx6xqj9dyicnobspdecp.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265571/exam_images/j7witirl6mpdk1vjuhz7.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxkjbam7/db054aff-6fa5-4bc9-b885-bc089d50f444/2c5d70f0-64cc-11ec-9ade-1d5c0ad4149d/file-1kxkjbam8.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265402/exam_images/yzqpv0wg8rtbnb1ll7xj.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263678/exam_images/fhfy19gajsghu3gpq8so.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266307/exam_images/dphpuza4bcv0gjeagma2.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264869/exam_images/k1gias56dnhyv1lyuoxv.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxujo649/7d917cd4-3560-4963-86ae-fdedb5d78653/68e72e90-6a4d-11ec-b083-77e72b2a3e30/file-1kxujo64a.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264611/exam_images/xydpt7zuxjr7u6rdoj1n.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264709/exam_images/yw9zpstjtnvowwwyhi8e.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6wilsbc/22019c3f-5fd5-4e87-9071-dbc4ccb0dd15/c133b180-1d90-11ed-b1e3-c3a54149d5d6/file-1l6wilsbd.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6xteu2a/47b4cfe2-54c6-471c-9b57-3e894f8e9072/ccf56920-1e47-11ed-9c61-4529b721806b/file-1l6xteu2b.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6yvpl4z/f567a908-d37f-4672-8f56-4c0e1c02e1d8/92df5930-1edd-11ed-8851-99197a0d39b5/file-1l6yvpl50.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7e9e2ip/7ad5cc7e-7a59-4e8b-9ba9-345351172840/d4237b10-2752-11ed-a077-1f1e3989e798/file-1l7e9e2iq.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7e9ep9c/5965f3b9-8c69-4efc-b7b6-a7969ef15cbd/e5b46600-2752-11ed-a077-1f1e3989e798/file-1l7e9ep9d.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lezr8qcy/4cbe3d9b-35fe-4402-b44f-4635f13e79f5/9d508520-bdba-11ed-9388-79c0aa407a4b/file-1lezr8qcz.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le1k19i6/4d596806-bd65-4927-9344-83ff4829221a/c7a577e0-aaeb-11ed-9a71-b7e1b547314d/file-1le1k19i7.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsrmbvjf/d3342755-760a-4a06-b677-ad719d5d05b9/2934cca0-ce6c-11ee-bec8-1b5b622d3f34/file-6y3zli1lsrmbvjg.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lscubuue/3ed41737-f019-4c7a-8f0d-bac514d096af/aff65f70-c64b-11ee-9d8b-f1be86a1b2f3/file-6y3zli1lscubuuf.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoisih0/d9aade95-6651-4b12-86db-5e337763a237/f9674440-ccb7-11ee-8ef5-472d1767d2da/file-6y3zli1lsoisih1.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/lSSuraUDM0v2yQv6N/vxj8MNbpFMfDbgUclOxathUPNYfpg/PFyUvUiYaSVEWiGxY485K2/image.svg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267542/exam_images/r7jew9joncb4qsxk3vq0.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264882/exam_images/alv8eapcojfiw5tesv9t.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lho7gb66/b1ac5890-9696-4252-9e45-79d3bb1d99c8/40771ae0-f2c5-11ed-8447-35082643c601/file-1lho7gb67.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1jkx8u/8b1b83ad-240e-48dd-8158-10998e53159f/3fe9e6e0-d3e1-11ee-b732-196aad4e0551/file-1lt1jkx8v.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/3tdsc44Gi1jrSfYF3/3jwAtqrMCIeFdFwF6jxjV7SVyWP1Y/z260H5cYeVP5poJc3GTCTn/image.svg\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/tiqW9aKqmddagbGAr/b4V9twI4tWYWLeyOCDAebisNjE4uF/YYD86LfwchXuGfn5rM0nDR/image.svg\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/kvzezHFULJwKbPuOf/fPGmKYccDI61W2mAR2B1YFqVOJwB1/QsSPLM6oozaTN9yDFVztgQ/image.svg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263523/exam_images/imqgnxixrtktf3zmqwry.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267838/exam_images/q5jroq8rpfxgyk8ca7nc.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267619/exam_images/zcpjnrsnrk6jjvw4n6pz.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264370/exam_images/nmdoxex4shlc8eg0kpol.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265987/exam_images/vkiwxwi0hvmzwe61bo4s.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265676/exam_images/z50dgjy3i2h18x8nrf7x.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266111/exam_images/bowbou8oq94cjbuq8clm.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5hwhjpc/e4eb5f33-6b70-496e-b551-33898f5bfaaa/12008710-01bb-11ed-85a8-43d162d2b7e8/file-1l5hwhjpd.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le7bzmtq/c9c497e4-0d22-40c1-a9ed-d9d25c205d11/052e76e0-ae19-11ed-89cc-995c981e21dd/file-1le7bzmtr.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1let3h9fn/25ba725f-606b-445d-8cf6-b491ca398815/fb86c530-ba10-11ed-bf62-29125e0e0d77/file-1let3h9fo.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1ledkim7h/2a5846c1-6725-40f5-a507-26fbada88443/1201c2c0-b187-11ed-8feb-a5a20e534682/file-1ledkim7l.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le1lduvl/ce1453ce-9f24-417e-b0cb-3f2d000af47e/0efec010-aaf1-11ed-9454-6baf9e2c8206/file-1le1lduvm.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsc0tkq8/ba672f20-8640-4879-9e31-ae8d9fb385cc/4ad67300-c5d8-11ee-baf2-61ef5327b65c/file-6y3zli1lsc0tkq9.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsluewb7/95253c3b-95ab-4fdb-8ed5-a7454583f0ab/125df630-cb3f-11ee-ad47-a16d1086e690/file-6y3zli1lsluewb8.png?format=png\")
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![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264089/exam_images/f7dyggcbmhlz6ow9zfwl.webp\")
\nCharge enclosed in the sphere,\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267700/exam_images/omgt7jjvzjbvmodmuinj.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263509/exam_images/hmkg5pqf7ygwkfgwox9p.webp\")
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\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lei36rxe/59d0fd3a-a039-4cc9-9db6-6ed6319a699d/38c48c20-b403-11ed-bf7e-c52177c53cde/file-1lei36rxf.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/Fulzm6VzYUnKLFd1t/EiKKTBSgrpk8gsOFRhaqoW2Movt5n/hhYoHwIjbmTHwJyBQybZ1K/image.svg\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/uOJw6w4SfAqNabRTV/N95JCOcDGN2bQYEKqFnqT54ek75cG/OrLdXR5Y7T7WtU6kHxf07S/image.svg\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/2qqwh79Y5h5q9NopZ/1vI8hKGGYyCiulbpOnRC08WUqHPpq/fZdPzlNUDsRJgcOxBYvLjq/image.svg\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/fqys8IgQx5sirqNo5/4fpjKxCE5mFnAcorJQgCo2iExZ2wG/hqHbxL4MkqNWBlsmF5OPBv/image.svg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265214/exam_images/bdzjl6n2cwc2grx7szo3.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263826/exam_images/qr2kas4ypl7ja1klnal1.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l38tc07h/e54ee83d-87e2-4c93-ac59-36910f46e007/3fe88f20-d523-11ec-aeec-6fd1cdec7420/file-1l38tc07n.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264540/exam_images/ioij8ebwku9u9alns7km.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267351/exam_images/oo3hui5vr6c5bsnpqzbm.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263347/exam_images/d5xyaku5zjfxr4duxpwk.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265460/exam_images/aeu7ki4snictajaatgwf.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264883/exam_images/jzyrlwuzyu2q8gsea9oj.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266407/exam_images/nyhkvzsxxrfg7ayrw4fa.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264682/exam_images/e64mwnxui8pfey2pftd6.webp\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265852/exam_images/c4pi5km2sotdchk0vdjn.webp\")
\n![\"JEE](\"https://imagex.cdn.examgoal.net/PX9LeYP8KWRWDFYVw/7s4aVYynlCAzCLApMDxTss02CcRuY/YJrGi0uEP7jzJtcydwLIqj/image.svg\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l0b0kbts/b7915c52-39cd-425f-ab1e-3c7b5a11f01d/726bae00-9af4-11ec-b7c4-9d40ef99bd3a/file-1l0b0kbtt.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266075/exam_images/tad0rleixsoein2adytk.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264151/exam_images/mojklubuscboti9ko6j0.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265550/exam_images/jypm5yq892xxv7umqtpr.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l35inazs/a048151d-56b2-4eba-8b91-8feeb29d77cf/1bc563b0-d353-11ec-ada9-55bd22062331/file-1l35inazv.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l35ios53/4ce31f1e-9507-4046-ba27-9b404657e621/44d30870-d353-11ec-ada9-55bd22062331/file-1l35ios54.png?format=png\")
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![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266065/exam_images/ua5kkh2hltbggjl5mrfb.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267378/exam_images/dqzebqidyn1wn19qpyr5.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263458/exam_images/vukfskgvsnmbrvyt3ads.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267189/exam_images/u2imgop0es6bxcb9q976.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266212/exam_images/osor7k0wtqqcp5iin9uu.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263869/exam_images/j5kj3zbqnpkfn4wd4ipc.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5grbhxq/331088c1-e677-4427-83c5-a1030fb4cbdc/137c8ae0-011a-11ed-9784-f3e7c455b437/file-1l5grbhxr.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6umex33/b64b6a21-31fd-4076-9f18-72715abb5a3c/1570e8f0-1c86-11ed-b633-f353ad3cb8e4/file-1l6umex34.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lguyz197/be80220e-4d6f-4336-abee-5a7998e060c0/5c2f0eb0-e2b1-11ed-b448-73bf9320498f/file-1lguyz198.png?format=png\")
\n\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqfs9pm/e4b3d7cc-7bac-404d-a994-0f5252a57a48/c88c18a0-cdc5-11ee-979b-67a81185a3d4/file-6y3zli1lsqfs9pn.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266491/exam_images/jqmo1keeva32zbonxaud.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266381/exam_images/ejkxqjbkymy0cf9ngwou.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l382x8d4/96a0a072-ff7e-490e-9a36-a1e6d7468b91/f7b61180-d4bb-11ec-ad28-abe411cf4979/file-1l382x8d5.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l382y00m/2da84c40-0d16-4a4b-a4d3-48d900becd14/0d128360-d4bc-11ec-ad28-abe411cf4979/file-1l382y00n.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264788/exam_images/esnw36gyccfzd7jqpo90.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263271/exam_images/km8tw62tqklutnw8arx3.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263905/exam_images/krpywiiwkfihbrbw7lnm.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1ldwx1iyu/3e34de25-f3d2-4bb2-9e86-861b45dcf7b0/ba635e60-a85e-11ed-b881-afa398664315/file-1ldwx1iyv.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le7cg4sv/5786aa17-f64f-438a-b4a2-840791af1975/d0038df0-ae1a-11ed-89cc-995c981e21dd/file-1le7cg4sw.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgyn09jf/74002ae0-d0b3-4092-b65b-efa035bbfc8b/b370a1b0-e4b5-11ed-a899-5b4605ce4b6b/file-1lgyn09jg.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgyn0pgv/8dd0a09d-0f7d-4f2c-859c-d429d0078a12/bfbea7f0-e4b5-11ed-80c5-11e224e2f0f4/file-1lgyn0pgw.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt2t02mt/73956b35-a36c-4cf5-ae05-55c1a46ecbe3/e0065850-d492-11ee-a613-e11ae62575eb/file-1lt2t02mu.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr7hjzk/4e6a2686-6c33-409b-bbbb-9246b5f9730b/1e337700-ce32-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr7hjzl.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266373/exam_images/uowdivblw0wvz9kswkcu.webp\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/KdjDfWp5V9zzsn0Lo/QBvMvL102dqdbkr35AdPrnhhdPDUl/rAV5pJWaudrIlBvXhD4f4O/image.svg\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9435,
"subject": "Physics",
"question": "Let $$x$$-$$z$$ plane be the boundary between two transparent media. Medium $$1$$ in $$z \\ge 0$$ has a refractive index of $$\\sqrt 2 $$ and medium $$2$$ with $$z < 0$$ has a refractive index of $$\\sqrt 3 .$$ A ray of light in medium $$1$$ given by the vector $$\\overrightarrow A = 6\\sqrt 3 \\widehat i + 8\\sqrt 3 \\widehat j - 10\\widehat k$$ is incident on the plane of separation. The angle of refraction in medium $$2$$ is: ",
"options": [
{
"text": "$${45^ \\circ }$$ "
},
{
"text": "$${60^ \\circ }$$"
},
{
"text": "$${75^ \\circ }$$"
},
{
"text": "$${30^ \\circ }$$"
}
],
"answer": "$${45^ \\circ }$$ ",
"solution": "**Answer:** $${45^ \\circ }$$ \n\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/q3PyNgFUzgj5GeNnm/l9vtPNEOzTGkyDNTp1hHvUs2efj7P/H3QIVJJPnzKOJEOLNG5xGl/image.svg\")
\n![\"JEE](\"https://imagex.cdn.examgoal.net/hhTdAjobHgPLKc3tR/b9n6f2hvnjvtQP2Ebix5jGOwiVq64/jnwPywrJPbbmZ0FUGzPY0Q/image.svg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265838/exam_images/bt5kenfxh7cgn6fgj50x.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l38thq3y/35e9c61f-1a3f-4eb1-8bdb-dd03005644c1/def60700-d523-11ec-aeec-6fd1cdec7420/file-1l38thq40.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265254/exam_images/nj2n4djqc8xc3pjn11vp.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265954/exam_images/hufvgakbidtjllnqcbnl.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263436/exam_images/cjks174r2n5zrmiowomg.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwoqx8kt/edda6e91-d67c-425f-9b50-c288d43f2a8a/d9b6d5d0-5350-11ec-9cbb-695a838b20fb/file-1kwoqx8ku.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5jy7jwr/75b07127-5f22-444c-99f6-b3fc9391277c/5fa055b0-02db-11ed-b2c8-3573db55a50d/file-1l5jy7jws.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5l6mlm3/d52b88c8-3275-4d8b-a4f7-93bcc25985be/13c342c0-0389-11ed-bfd1-873560f9d960/file-1l5l6mlm4.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7myl3p1/cc381127-5ea0-492f-8313-a032aac1ca92/a1229250-2c1b-11ed-a609-572a60403bd8/file-1l7myl3p2.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7e9j0tr/70a4befa-e3ac-4a19-8966-4b086c719e70/5de067f0-2753-11ed-a077-1f1e3989e798/file-1l7e9j0ts.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5l49yu/73016b8f-e271-4dd7-8ef6-7403f7b71fbc/27288160-ad23-11ed-8bc1-d3bd0941e5b5/file-1le5l49yv.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264180/exam_images/clcbi7mxzhetqrqs6zq7.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266273/exam_images/oeq8e819cefrue1iqjgh.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt2v1ilo/5ca0b32d-d791-4661-a910-3d04300a949e/da6d6ac0-d49a-11ee-8574-81dc091f6c89/file-1lt2v1ilp.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266684/exam_images/ic8lr1ryal9qbn2uyhyy.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265123/exam_images/hszefzg6upwkvwdi4fej.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265575/exam_images/tghangpmtivh6aceem4q.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267353/exam_images/mly9ped4ovwxylglilgi.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264696/exam_images/tapsi3c6rlflbm27qaj2.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266402/exam_images/iqkrykzyb4jm0zvdos04.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267561/exam_images/wjummzime8hsxnszt3lv.webp\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/F96G5bLlzCo2IyU6l/8xgm4Q1VZzWEvfkCeFRSXybAqL3PO/NwRFsKbOQpdNYndqqiEXC5/image.svg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265795/exam_images/s4q8yayuwtszongry8fp.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267842/exam_images/liy4vlvgeuij7zkczva8.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1ledl4rmn/eddbd67a-5fa1-405c-a90c-053d6dba7a7e/79ff16f0-b189-11ed-94f5-239612b3abb5/file-1ledl4rmo.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgrfzyob/3ce53df2-7a14-45dd-8f83-57e6063a144d/ce9901b0-e0c0-11ed-9768-bd0baf44cdee/file-1lgrfzyoc.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265771/exam_images/ur6rvgpqywvuyclxwos4.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267664/exam_images/irzeijsu6dn5wjuqjdz2.webp\")
\nAll those particles are moving due to their mutual gravitational attraction.\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263895/exam_images/r8nkxkrbt46tcn6aoxdy.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266297/exam_images/lhsxksjsnoud9wdnhmx4.webp\")
\nBy angular momentum conservation\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxehgyer/c578dd8a-84ff-4cd5-87c2-8f654e73200d/43a2b830-6178-11ec-90ef-97830a7fdb59/file-1kxehgyes.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxehbywy/aea2594a-3ff3-49a2-8b56-6e4bb066e705/b8facf10-6177-11ec-97a3-cfb8e5d281d3/file-1kxehbywz.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267811/exam_images/q0j574ymlqyfvbkzluoh.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265259/exam_images/o2uqxj2u2a5o2yefppde.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264301/exam_images/gq1uq8ad7giyxsvrk6xv.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwosl7k8/b883e894-890a-4a8b-a93e-ee5cd706cd6a/5d7a7790-5357-11ec-b443-85f16d0c41b6/file-1kwosl7ka.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwosm9cu/efdb8c6e-09fd-4e14-bc6b-aa090a1d8f3b/7aac5fe0-5357-11ec-b443-85f16d0c41b6/file-1kwosm9cv.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5hwg3n0/f13a1ec3-1130-4a3f-980d-b49d7ef40d09/e9c85ac0-01ba-11ed-85a8-43d162d2b7e8/file-1l5hwg3n1.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6tcunfv/d5d44637-c5a5-4a91-b83c-43a82e39ede4/e6feefb0-1bd3-11ed-bd9f-71db3a4811b2/file-1l6tcunfw.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lei2xrln/9034b0ff-b12e-40cd-b333-548731e8cda9/3e3b95a0-b402-11ed-8ab2-65ca6f80c54f/file-1lei2xrlo.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgzrc0c0/488210fe-6508-4f7a-9a44-b65f6e8b45f8/673423f0-e553-11ed-a972-b5db7753e054/file-1lgzrc0c1.png?format=png\")
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\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265438/exam_images/wvpivtsierqol0p9xqwj.webp\")
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\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266841/exam_images/crlq65aurxrs21vnmphd.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267684/exam_images/bi1mkhum4a5moycv0iae.webp\")
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\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265074/exam_images/qfvtqjmgutbpwe3mydhn.webp\")
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\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265304/exam_images/acu1grvcj9ng8qxarkx0.webp\")
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\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwosodfn/ed9bd245-b1ab-439f-891c-65071a8882e6/b5711530-5357-11ec-b443-85f16d0c41b6/file-1kwosodfo.png?format=png\")
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\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l14pcg0g/000193f2-3a8b-4ffc-9785-e6f77ffb0f24/163dd9f0-ab48-11ec-a9df-f9e258adcdc5/file-1l14pcg0h.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/1CNJwyGbISt5nPqgl/i7MgjesnEtVzMfw5MnCPYW4hxjgUZ/SMFQ3DkUGLfdwqNBAwTzrv/image.svg\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 9987,
"subject": "Physics",
"question": "A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads $$49$$ $$N,$$ when the lift is stationary. If the lift moves downward with an acceleration of $$5 m/{s^2}$$, the reading of the spring balance will be ",
"options": [
{
"text": "$$24$$ $$N$$ "
},
{
"text": "$$74$$ $$N$$ "
},
{
"text": "$$15$$ $$N$$ "
},
{
"text": "$$49$$ $$N$$ "
}
],
"answer": "$$24$$ $$N$$ ",
"solution": "**Answer:** $$24$$ $$N$$ \n\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267763/exam_images/u60jjsap0qzr5mddcca3.webp\")
\n![\"AIEEE](\"https://imagex.cdn.examgoal.net/uBJp8wPSVEFPVdWYD/gnPAttI5mMPIeJMwsOhiywab3sCr2/7kdhGAfV5xfeOTTe8gUA0D/image.svg\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265973/exam_images/ndu9oh9g3o6khs3ynqdu.webp\")
\n