Identify the correct order of standard enthalpy of formation of sodium halides :
",
"options": [
{
"text": "$$\\mathrm{NaI}<\\mathrm{NaBr}<\\mathrm{NaCl}<\\mathrm{NaF}$$"
},
{
"text": "$$\\mathrm{NaI}<\\mathrm{NaBr}<\\mathrm{NaF}<\\mathrm{NaCl}$$"
},
{
"text": "$$\\mathrm{NaF}<\\mathrm{NaCl}<\\mathrm{NaBr}<\\mathrm{NaI}$$"
},
{
"text": "$$\\mathrm{NaCl}<\\mathrm{NaF}<\\mathrm{NaBr}<\\mathrm{NaI}$$"
}
],
"answer": "$$\\mathrm{NaI}<\\mathrm{NaBr}<\\mathrm{NaCl}<\\mathrm{NaF}$$",
"solution": "**Answer:** $$\\mathrm{NaI}<\\mathrm{NaBr}<\\mathrm{NaCl}<\\mathrm{NaF}$$\n\nLattice energy is directly proportional to the charges of the ions and inversely proportional to the sum of their radii:\n
\nLattice energy ∝ (Q₁Q₂) / (r₁ + r₂)\n
\nAs we move down the group, the size of the halide ions increases, resulting in a decrease in lattice energy. This leads to a decrease in the exothermicity of the lattice energy release during the formation of sodium halides.\n
\nThus, the correct order of standard enthalpy of formation of sodium halides, considering both lattice energy and electron affinity, is:\n
Let's consider each item in List-I and match it with an appropriate item from List-II:
\n
A. K (Potassium): Potassium ions are crucial for the operation of the sodium-potassium pump in nerve cells. So, A matches with III.
\n
B. KCl (Potassium Chloride): Potassium chloride is widely used in agriculture as a fertilizer. So, B matches with II.
\n
C. KOH (Potassium Hydroxide): Potassium hydroxide is a strong base, but it is not typically used in a sodium-potassium pump, as a fertilizer, for thermonuclear reactions, or as a CO₂ absorbent. However, as per the given options and its closest possible use, KOH is often used to absorb CO₂ in certain applications like rebreathing gas masks. So, C matches with IV.
\n
D. Li (Lithium): Lithium is used in thermonuclear reactions, specifically in fusion reactions. So, D matches with I.
\n
Therefore, the correct option is Option B: A-III, B-II, C-IV, D-I.
Number of water molecules in washing soda and soda ash respectively are :
",
"options": [
{
"text": "1 and 0"
},
{
"text": "10 and 1"
},
{
"text": "1 and 10"
},
{
"text": "10 and 0"
}
],
"answer": "10 and 0",
"solution": "**Answer:** 10 and 0\n\n
\n
Washing soda is a hydrated form of sodium carbonate, with the chemical formula Na2CO3·10H2O. This means that each molecule of washing soda contains 10 water molecules.
\n
Soda ash is the anhydrous form of sodium carbonate, with the chemical formula Na2CO3. This means that it does not contain any water molecules.
\n
\n
Soda ash is typically produced by heating washing soda to drive off the water molecules. Washing soda can be produced by dissolving soda ash in water and then recrystallizing it.
The electrolysis of brine, which is a solution of sodium chloride (NaCl) in water, involves the migration of ions towards electrodes. When an electric current is passed through the solution, positive sodium ions ($Na^+$) move towards the cathode (negative electrode), and negative chloride ions ($Cl^-$) move towards the anode (positive electrode).
At the same time, water ($H_2O$), which also dissociates into $H^+$ and $OH^-$ ions, also contribute to the reactions at the electrodes.
\n
At the anode, chloride ions are oxidized to chlorine gas:
\n
$2Cl^- \\rightarrow Cl_2(g) + 2e^-$
\n
At the cathode, water is reduced to hydrogen gas and hydroxide ions, instead of sodium ions being reduced to sodium metal. This is because water is more easily reduced than sodium ions:
\n
$2H_2O + 2e^- \\rightarrow H_2(g) + 2OH^-$
\n
The sodium ions remain in the solution and combine with the hydroxide ions to form sodium hydroxide:
\n
$Na^+ + OH^- \\rightarrow NaOH$
\n
So the products of the electrolysis of brine are chlorine gas, hydrogen gas, and sodium hydroxide. Hydrochloric acid (HCl) is not produced during this process because the hydrogen and chloride ions do not combine in the electrolyte solution to form HCl. They are separately reduced and oxidized at the cathode and anode respectively.
For a good quality cement, the ratio of lime to the total of the oxides of $$\\mathrm{Si}, \\mathrm{Al}$$ and $\\mathrm{Fe}$ should be as close as to :
",
"options": [
{
"text": "3"
},
{
"text": "2"
},
{
"text": "1"
},
{
"text": "4"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\nFor good quality cement, the ratio of lime (CaO) to the sum of the oxides of silicon (SiO₂), aluminum (Al₂O₃), and iron (Fe₂O₃) should be approximately 2:1.\n
\nThis proportion ensures that the cement has good strength and setting properties, as well as resistance to chemical attack.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2507,
"subject": "Chemistry",
"question": "The metallic sodium disolves in liquid ammonia to form a deep blue coloured solution. The deep blue colour is due to formation of :",
"options": [
{
"text": "solvated electron, $$e(NH_3)^-_x$$"
},
{
"text": "solvated atomic sodium, Na(NH3)y"
},
{
"text": "(Na+ + Na-)"
},
{
"text": "NaNH2 + H2"
}
],
"answer": "solvated electron, $$e(NH_3)^-_x$$",
"solution": "**Answer:** solvated electron, $$e(NH_3)^-_x$$\n\nThe alkali metals dissolve in liquid ammonia without evolution of hydrogen. The metal loses electrons and combine with ammonia molecule. \n
It is ammoniated electron which is responsible for color.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2508,
"subject": "Chemistry",
"question": "Based on lattice energy and other considerations which one of the following alkali\nmetal chlorides is expected to have the highest melting point :",
"options": [
{
"text": "LiCl"
},
{
"text": "NaCl "
},
{
"text": "KCl"
},
{
"text": "RbCl"
}
],
"answer": "NaCl ",
"solution": "**Answer:** NaCl \n\n$$LiCl$$ has partly covalent character. Other halides are ionic in nature. Lattice energy decreases with increases of ionic radius of cation, anion being the same. Larger is the lattice energy, the higher will be $$m.$$ $$pt.$$ hence $$NaCl$$ will have highest lattice energy. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2509,
"subject": "Chemistry",
"question": "The ionic mobility of alkali metal ions in aqueous solution is maximum for :",
"options": [
{
"text": "K+"
},
{
"text": "Rb+"
},
{
"text": "Li+"
},
{
"text": "Na+"
}
],
"answer": "Rb+",
"solution": "**Answer:** Rb+\n\nSince the mobility of of ions is inversely proportional to the size of their hydrated ions, hence the increasing order of ionic mobility is\n
The first ionization energy is the energy required to remove the most loosely held electron from an atom or a cation. It is measured in units of electron volts (eV) or kilojoules per mole (kJ/mol).
\n\n
Among the given options, Scandium (Sc) has the highest first ionization energy because it has the smallest atomic radius and the highest nuclear charge. As we move from left to right across a period in the periodic table, the atomic radius decreases while the nuclear charge increases. This results in a stronger attraction between the positively charged nucleus and the negatively charged electrons in the outermost energy level, which requires more energy to remove the most loosely held electron.
\n\n
The first ionization energy of the given options in increasing order is:\n
Na < K < Rb < Sc
\n\n
Therefore, Scandium (Sc) has the highest first ionization energy among the given options.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2511,
"subject": "Chemistry",
"question": "The main oxides formed on combustion of Li, Na and K in excess of air are, respectively :",
"options": [
{
"text": "LiO2, Na2O2 and K2O"
},
{
"text": "Li2O2, Na2O2 and KO2"
},
{
"text": "Li2O, Na2O2 and KO2"
},
{
"text": "Li2O, Na2O and KO2"
}
],
"answer": "Li2O, Na2O2 and KO2",
"solution": "**Answer:** Li2O, Na2O2 and KO2\n\nOn heating with excess of air $$Li,$$ $$Na$$ and $$K$$ forms following oxides \n
$$ \\therefore $$ Smaller the size more is hydration energy.\n
Li+ having minimum radius so maximum hydration energy.\n
Correct order is :\n
Li+ > Na+ > K+ > Rb+ > Cs+",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2515,
"subject": "Chemistry",
"question": "The INCORRECT statement is :",
"options": [
{
"text": "Lithium is least reactive with water among the alkali metals"
},
{
"text": "LiCl crystallises from aqueous solution as LiCl.2H2O"
},
{
"text": "Lithium is the strongest reducing agent among the alkali metals"
},
{
"text": "LiNO3 decomposes on heating to give LiNO2 and O2\n"
}
],
"answer": "LiNO3 decomposes on heating to give LiNO2 and O2\n",
"solution": "**Answer:** LiNO3 decomposes on heating to give LiNO2 and O2\n\n\nLiNO3 $$\\buildrel \\Delta \\over\n \\longrightarrow $$ Li2O + NO2 + O2\n
Ionic potential of Li is high that is why on thermal demoposition of nitrate form of Li will give oxide of Li and NO2 and O2.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2516,
"subject": "Chemistry",
"question": "Among the statements (a)-(d) the correct ones\nare :
\n(a) Lithium has the highest hydration enthalpy\namong the alkali metals.
\n(b) Lithium chloride is insoluble in pyridine.
\n(c) Lithium cannot form ethynide upon its\nreaction with ethyne.
\n(d) Both lithium and magnesium react slowly\nwith H2O.",
"options": [
{
"text": "(b) and (c) only"
},
{
"text": "(a) and (d) only"
},
{
"text": "(a), (b) and (d) only"
},
{
"text": "(a), (c) and (d) only"
}
],
"answer": "(a), (c) and (d) only",
"solution": "**Answer:** (a), (c) and (d) only\n\nLithium has highest hydration enthalpy among\nalkali metals due to its small size.\n
LiCl is soluble in pyridine because LiCl have\nmore covalent character.\n
Li does not form ethynide with ethyne.\n
Both Li and Mg reacts slowly with H2O.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2517,
"subject": "Chemistry",
"question": "The metal mainly used in devising\nphotoelectric cells is :",
"options": [
{
"text": "Na"
},
{
"text": "Rb"
},
{
"text": "Li"
},
{
"text": "Cs"
}
],
"answer": "Cs",
"solution": "**Answer:** Cs\n\nCs has least work function value because of\nless electronegativity thats why it is used in\nphotoelectric cells.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2518,
"subject": "Chemistry",
"question": "On combustion of Li, Na and K in excess of air,\nthe major oxides formed, respectively, are :",
"options": [
{
"text": "Li2O, Na2O2 and K2O"
},
{
"text": "Li2O2, Na2O2 and K2O2"
},
{
"text": "Li2O, Na2O2 and KO2"
},
{
"text": "Li2O, Na2O and K2O2"
}
],
"answer": "Li2O, Na2O2 and KO2",
"solution": "**Answer:** Li2O, Na2O2 and KO2\n\n4Li + O2 $$ \\to $$ 2Li2O\n
2Na + O2 $$ \\to $$ Na2O2\n
K + O2 $$ \\to $$ KO2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2519,
"subject": "Chemistry",
"question": "Match List - I with List - II.
Alkali metals have very low value of ionisation energy as compared to other metals. So, alkali metals easily get excited and impart colour to flame.
\n
Hence, Rb is most excited and having high value of wavelength in all alkali metals.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2520,
"subject": "Chemistry",
"question": "The correct order of conductivity of ions in water is :",
"options": [
{
"text": "K+ > Na+ > Cs+ > Rb+"
},
{
"text": "Cs+ > Rb+ > K+ > Na+"
},
{
"text": "Na+ > K+ > Rb+ > Cs+"
},
{
"text": "Rb+ > Na+ > K+ > Li+"
}
],
"answer": "Cs+ > Rb+ > K+ > Na+",
"solution": "**Answer:** Cs+ > Rb+ > K+ > Na+\n\nAs the size of gaseous ion decreases, it get\nmore hydrated in water and hence, the size of\naqueous ion increases. When this bulky ion\nmove in solution, it experience greater\nresistance and hence lower conductivity.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2521,
"subject": "Chemistry",
"question": "Given below are two statements : One is labelled as Assertion A and the other labelled as Reason R.
Assertion A : Lithium halides are some what covalent in nature.
Reason R : Lithium posses high polarisation capability.
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 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\nLithium due to small size has very high polarization capability and thus increases covalent nature of Halides.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2522,
"subject": "Chemistry",
"question": "Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Lithium salts are hydrated.
Reason (R) : Lithium has higher polarising power than other alkali metal group members.
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 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\nLithium salts are hydrated due to high hydration energy of Li+
Li+ due to smallest size in IA group has highest polarizing power.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 2523,
"subject": "Chemistry",
"question": "
Arrange the following in the decreasing order of their covalent character :
\n
(A) LiCl
\n
(B) NaCl
\n
(C) KCl
\n
(D) CsCl
\n
Choose the most appropriate answer from the options given below :
In the industrial production of which of the following, molecular hydrogen is obtained as a byproduct?
",
"options": [
{
"text": "NaOH"
},
{
"text": "NaCl"
},
{
"text": "Na metal"
},
{
"text": "Na2CO3"
}
],
"answer": "NaOH",
"solution": "**Answer:** NaOH\n\nMolecular hydrogen is produced as a byproduct in the industrial production of $$\\mathrm{NaOH}$$ by electrolysis of aq $$\\mathrm{NaCl}$$ solution
\n$$\\mathrm{NaOH}$$ is crystallised from the remaining part of electrolyte.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2526,
"subject": "Chemistry",
"question": "
Which of the following statements are correct?
\n
(A) Both LiCl and MgCl2 are soluble in ethanol.
\n
(B) The oxides Li2O and MgO combine with excess of oxygen to give superoxide.
\n
(C) LiF is less soluble in water than other alkali metal fluorides.
\n
(D) Li2O is more soluble in water than other alkali metal oxides.
\n
Choose the most appropriate answer from the options given below :
",
"options": [
{
"text": "(A) and (C) only"
},
{
"text": "(A), (C) and (D) only"
},
{
"text": "(B) and (C) only"
},
{
"text": "(A) and (D) only"
}
],
"answer": "(A) and (C) only",
"solution": "**Answer:** (A) and (C) only\n\n(A) Both LiCl and MgCl2 are soluble in ethanol
\n(B) Li and Mg do not form superoxide
\n(C) LiF has high lattice energy
\n(D) Li2O is least soluble in water than another alkali metal oxides ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2527,
"subject": "Chemistry",
"question": "
Choose the correct order of density of the alkali metals :
",
"options": [
{
"text": "Li < K < Na < Rb < Cs"
},
{
"text": "Li < Na < K < Rb < Cs"
},
{
"text": "Cs < Rb < K < Na < Li"
},
{
"text": "Li < Na < K < Cs < Rb"
}
],
"answer": "Li < K < Na < Rb < Cs",
"solution": "**Answer:** Li < K < Na < Rb < Cs\n\nThe increasing order of density of alkali metals as\n
\n' $\\mathrm{K}$ ' metal has less density as compare to ' $\\mathrm{Na}$ ' metal. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2528,
"subject": "Chemistry",
"question": "
Which of the following statement is incorrect?
",
"options": [
{
"text": "Low solubility of LiF in water is due to its small hydration enthalpy."
},
{
"text": "$$\\mathrm{KO}_{2}$$ is paramagnetic."
},
{
"text": "Solution of sodium in liquid ammonia is conducting in nature."
},
{
"text": "Sodium metal has higher density than potassium metal."
}
],
"answer": "Low solubility of LiF in water is due to its small hydration enthalpy.",
"solution": "**Answer:** Low solubility of LiF in water is due to its small hydration enthalpy.\n\nLow solubility of LiF in water is due to the fact that though $\\mathrm{Li}^{+}$ is having high hydration enthalpy but it has higher lattice enthalpy when present in LiF. Due to higher lattice enthalpy its solubility is less.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2529,
"subject": "Chemistry",
"question": "The number of alkali metal(s), from $\\mathrm{Li}, \\mathrm{K}, \\mathrm{Cs}$, Rb having ionization enthalpy greater than $400 \\mathrm{~kJ}$ $\\mathrm{mol}^{-1}$ and forming stable super oxide is _____________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n\n
Li does not form superoxide.\n
K, Rb and Cs form stable super oxides but Cs has\nionisation enthalpy less than 400 kJ. While K and Rb\nhave ionisation enthalpy greater than 400 kJ mol–1.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2530,
"subject": "Chemistry",
"question": "
Which one among the following metals is the weakest reducing agent?
",
"options": [
{
"text": "K"
},
{
"text": "Na"
},
{
"text": "Li"
},
{
"text": "Rb"
}
],
"answer": "Na",
"solution": "**Answer:** Na\n\nSodium has the lowest oxidation potential in alkali metals. Hence it is the weakest reducing agent among alkali metals.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2531,
"subject": "Chemistry",
"question": "
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
\n
Assertion A : The alkali metals and their salts impart characteristic colour to reducing flame.
\n
Reason R : Alkali metals can be detected using flame tests.
\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": "A is correct but R is not 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": "A is not correct but R is correct",
"solution": "**Answer:** A is not correct but R is correct\n\nAssertion is not correct because alkali metals and\ntheir salts impart characteristic colour to oxidising\npart of flame and not reducing part of flame.\nReason is correct because all alkali metals can be\ndetected by their flame tests. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2532,
"subject": "Chemistry",
"question": "
Identify the correct statements about alkali metals.
\n
A. The order of standard reduction potential (M$$^+$$ \n | M) for alkali metal ions is Na > Rb > Li.
\n
B. CsI is highly soluble in water.
\n
C. Lithium carbonate is highly stable to heat.
\n
D. Potassium dissolved in concentrated liquid ammonia is blue in colour and paramagnetic.
\n
E. All the alkali metal hydrides are ionic solids.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "C and E only"
},
{
"text": "A and E only"
},
{
"text": "A, B, D only"
},
{
"text": "A, B and E only"
}
],
"answer": "A and E only",
"solution": "**Answer:** A and E only\n\nA. The order of standard reduction potential (M+ | M) for alkali metal ions is Na > Rb > Li. This statement is correct. The standard reduction potential generally increases as we move down the group from Li to Cs so order should be Cs > Rb > K > Na > Li but here Na and K are exception, Na has highest standard reduction potential in the entire group and then K. So right order is Na > K > Cs > Rb > Li. That is why right order is Na > Rb > Li.\n
Standard reduction potential (M+ | M) for alkali metal ions are -\n
(1). Li = -3.04 V\n
(2). Na = -2.714 V\n
(3). K = -2.925 V\n
(4). Rb = -2.930 V\n
(5). Cs = -2.927 V\n
\nB. CsI is highly soluble in water. This statement is not correct. CsI is the least soluble alkali metal halide in water, and its solubility decreases as the size of the anion increases.\n
\nC. Lithium carbonate is highly stable to heat. This statement is not correct. Lithium carbonates decomposes easily on heating to form the lithium oxide and CO2.\n
\nD. Potassium dissolved in concentrated liquid ammonia is blue in colour and paramagnetic. This statement is not correct. In concentrated solution, the blue colour changes to bronze colour and the solution becomes dimagnetic.\n
\nE. All the alkali metal hydrides are ionic solids. This statement is correct. All alkali metal hydrides are ionic solids, and their properties are determined by the ionic nature of the compounds.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2533,
"subject": "Chemistry",
"question": "
Alkali metals are part of Group 1 on the periodic table and include lithium (Li), sodium (Na), potassium (K), rubidium (Rb), and cesium (Cs). As you move down the group, the atomic radius increases due to the addition of energy levels (shells). While you might think that the increased atomic mass would correspond to an increase in density, the increase in atomic radius generally outpaces the increase in atomic mass, leading to a decrease in density.
\n
The densities (at room temperature) of the mentioned alkali metals are :
The alkali metals are the elements in Group 1 of the periodic table. They are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), and cesium (Cs).
\n
The melting point of these metals decreases down the group as the size of the atoms increases. This is because the metallic bonding weakens as the atomic size increases, resulting in a lower melting point.
\n
Therefore, the alkali metal with the lowest melting point is cesium (Cs).
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
\n
Assertion A : Sodium is about 30 times as abundant as potassium in the oceans.
\n
Reason R : Potassium is bigger in size than sodium.
\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 false but R is true"
},
{
"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\nDue to bigger size of potassium, it forms more\nefficient lattices as compared to sodium with\nsilicates.
\nThe abundance of sodium in ocean is more due to\nthe more soluble nature of salt of sodium as\ncompared to potassium salts. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2536,
"subject": "Chemistry",
"question": "A metal M readily forms its sulphate MSO4, which is water-soluble. It forms its oxide MO which becomes inert on heating. It forms an insoluble hydroxide M(OH)2 which is soluble in NaOH solution. The M is :",
"options": [
{
"text": "Mg"
},
{
"text": "Ba"
},
{
"text": "Ca"
},
{
"text": "Be"
}
],
"answer": "Be",
"solution": "**Answer:** Be\n\nSulphate of alkaline earth metal are sparingly soluble or almost not soluble in water whereas $$BeS{O_4}$$ is soluble in water due to high degree of solvation. $$Be{\\left( {OH} \\right)_2}$$ is insoluble in water but soluble in $$NaOH.$$ \n
$$BeO + 2NaOH\\,\\,$$ $$ \\to \\,\\,N{a_2}Be{O_2} + {H_2}O$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2537,
"subject": "Chemistry",
"question": "The solubilities of carbonates decrease down the magnesium group due to a decrease in :",
"options": [
{
"text": "hydration energy of cations"
},
{
"text": "inter-ionic attraction"
},
{
"text": "entropy of solution formation"
},
{
"text": "lattice energies of solids"
}
],
"answer": "hydration energy of cations",
"solution": "**Answer:** hydration energy of cations\n\nAs we move down the group, the lattice energies of carbonates remain approximately the same. However the hydration energies of the metal cation decreases from $$\\,\\,B{e^{ + + }}\\,\\,$$ to $$\\,\\,B{a^{ + + }},\\,$$ hence the solubilities of carbonates of the alkaline earth metal decreases down the group mainly due to decreasing hydration energies of the cations from $$\\,\\,B{e^{ + + }}\\,\\,$$ to $$\\,\\,B{a^{ + + }}.$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 2538,
"subject": "Chemistry",
"question": "One mole of magnesium nitride on the reaction with an excess of water gives :",
"options": [
{
"text": "one mole of ammonia"
},
{
"text": "two moles of nitric acid "
},
{
"text": "two moles of ammonia "
},
{
"text": "one mole of nitric acid "
}
],
"answer": "two moles of ammonia ",
"solution": "**Answer:** two moles of ammonia \n\n$$M{g_3}{N_2} + 6{H_2}O\\,\\,\\rightleftharpoons\\,\\,3Mg{\\left( {OH} \\right)_2} + 2N{H_3}$$ \n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2539,
"subject": "Chemistry",
"question": "The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is :",
"options": [
{
"text": "Cu"
},
{
"text": "Cr"
},
{
"text": "Ag"
},
{
"text": "Ca"
}
],
"answer": "Ca",
"solution": "**Answer:** Ca\n\nOn electrolysis of aqueous solution of $$s$$-block elements $${H_2}$$ gas discharge at cathode.\n
At cathode: $${H_2}O + {e^ - } \\to {1 \\over 2}{H_2} + O{H^ - }$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2540,
"subject": "Chemistry",
"question": "Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice\nenthalpy?",
"options": [
{
"text": "BeSO4"
},
{
"text": "BaSO4"
},
{
"text": "SrSO4"
},
{
"text": "CaSO4"
}
],
"answer": "BeSO4",
"solution": "**Answer:** BeSO4\n\nIn alkaline earth metals, ionic size increases down the group. The lattice energy remains constant because sulphate ion is so large, so that small change in cationic size does not make any difference. On moving down the group the degree of hydration of metal ions decreases very much leading to decrease in solubility. \n
$$\\therefore$$ $$\\,\\,\\,BeS{O_4} > MgS{O_4} > CaS{O_4} > SrS{O_4} > BaS{O_4}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2541,
"subject": "Chemistry",
"question": "The correct order of the solubility of alkaline-earth metal sulphates in water is :",
"options": [
{
"text": "Mg < Ca < Sr < Ba"
},
{
"text": "Mg < Sr < Ca < Ba"
},
{
"text": "Mg > Sr > Ca > Ba"
},
{
"text": "Mg > Ca > Sr > Ba"
}
],
"answer": "Mg > Ca > Sr > Ba",
"solution": "**Answer:** Mg > Ca > Sr > Ba\n\nThe solubility of the sulphates in water decreases down\nthe group. Mg > Ca > Sr > Ba. MgSO4\n is soluble, but\nCaSO4 is sparingly soluble, and the sulphates of Sr, Ba and Ra are virtually insoluble. The significantly high solubility of MgSO4 is due to the high enthalpy of solvation\nof the smaller Mg2+ ion.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2542,
"subject": "Chemistry",
"question": "Both lithium and magnesium display several similar properties due to the diagonal relationship; however,\nthe one which is incorrect, is :",
"options": [
{
"text": "both form soluble bicarbonates"
},
{
"text": "both from nitrides\n"
},
{
"text": "nitrates of both Li and Mg yield NO2 and O2 on heating "
},
{
"text": "both form basic carbonates "
}
],
"answer": "both form basic carbonates ",
"solution": "**Answer:** both form basic carbonates \n\nMg can form basic carbonate like\n
3MgCO3 . Mg(OH)2 . 3H2O\n
While Li can not.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2543,
"subject": "Chemistry",
"question": "The correct sequence of thermal stability of the following carbonates is :",
"options": [
{
"text": "BaCO3 < SrCO3 < CaCO3 < MgCO3"
},
{
"text": "MgCO3 < CaCO3 < SrCO3 < BaCO3"
},
{
"text": "BaCO3 < CaCO3 < SrCO3 < MgCO3"
},
{
"text": "MgCO3 < SrCO3 < CaCO3 < BaCO3"
}
],
"answer": "MgCO3 < CaCO3 < SrCO3 < BaCO3",
"solution": "**Answer:** MgCO3 < CaCO3 < SrCO3 < BaCO3\n\nStability of alkaline earth metal carbonates\nincreases down the group as size of cation increases ans polarization decreases down the group :\n
MgCO3 < CaCO3 < SrCO3 < BaCO3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2544,
"subject": "Chemistry",
"question": "The alloy used in the construction of aircrafts is :",
"options": [
{
"text": "Mg – Zn"
},
{
"text": "Mg – Al "
},
{
"text": "Mg – Mn"
},
{
"text": "Mg – Sn"
}
],
"answer": "Mg – Al ",
"solution": "**Answer:** Mg – Al \n\nAn alloy of Mg and Al called magnalium is used\nin manufacturing of aircraft due to its light\nweight and high strength.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2545,
"subject": "Chemistry",
"question": "The amphoteric hydroxide is :",
"options": [
{
"text": "Ca(OH)2"
},
{
"text": "Be(OH)2"
},
{
"text": "Mg(OH)2"
},
{
"text": "Sr(OH)2"
}
],
"answer": "Be(OH)2",
"solution": "**Answer:** Be(OH)2\n\nBe(OH)2\n is amphoteric in nature while rest all alkaline earth metal hydroxide are basic in nature.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2546,
"subject": "Chemistry",
"question": "The metal used for marking X–ray tube window is : ",
"options": [
{
"text": "Ca"
},
{
"text": "Na"
},
{
"text": "Mg"
},
{
"text": "Be"
}
],
"answer": "Be",
"solution": "**Answer:** Be\n\n''Be'' Metal is used in x-ray window is due to transparent to x-rays. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2547,
"subject": "Chemistry",
"question": "The alkaline earth metal nitrate that does not crystallise with water molecules, is : ",
"options": [
{
"text": "Mg(NO3)2"
},
{
"text": "Sr(NO3)2"
},
{
"text": "Ca(NO3)2"
},
{
"text": "Ba(NO3)2"
}
],
"answer": "Ba(NO3)2",
"solution": "**Answer:** Ba(NO3)2\n\nWhen central atom size is smaller then more water molecule will crystallize. Ba(NO3)2 does not ctystallize with water as size of Ba+2 ion is very large. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2548,
"subject": "Chemistry",
"question": "Two elements A and B have similar chemical\nproperties. They don’t form solid\nhydrogencarbonates, but react with nitrogen to\nform nitrides. A and B, respectively, are :",
"options": [
{
"text": "Li and Mg"
},
{
"text": "Cs and Ba"
},
{
"text": "Na and Rb"
},
{
"text": "Na and Ca"
}
],
"answer": "Li and Mg",
"solution": "**Answer:** Li and Mg\n\nSolid hydrogencarbonates are not formed\nby lithium and magnesium.\n
– 6Li + N2 $$ \\to $$ 2Li3N\n
– 3Mg + N2 $$ \\to $$ Mg3N2",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2549,
"subject": "Chemistry",
"question": "An alkaline earth metal ‘M’ readily forms water\nsoluble sulphate and water insoluble hydroxide.\nIts oxide MO is very stable to heat and does not\nhave rock-salt structure. M is :",
"options": [
{
"text": "Ca"
},
{
"text": "Mg"
},
{
"text": "Sr"
},
{
"text": "Be"
}
],
"answer": "Be",
"solution": "**Answer:** Be\n\nBeSO4\n is water soluble.\n Be(OH)2\n is water insoluble.\n BeO is stable to heat.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2550,
"subject": "Chemistry",
"question": "Among the sulphates of alkaline earth metals,\nthe solubilities of BeSO4 and MgSO4 in water,\nrespectively, are :",
"options": [
{
"text": "poor and poor"
},
{
"text": "poor and high"
},
{
"text": "high and poor"
},
{
"text": "high and high"
}
],
"answer": "high and high",
"solution": "**Answer:** high and high\n\nOrder of solubility of sulphate of Alkaline earth metals\n
BeSO4\n > MgSO4\n > CaSO4\n > SrSO4\n > BaSO4\n
The greater hydration enthalpies of\nBe2+ and Mg2+ ions dominate over their lattice\nenthalpies and therefore their sulphates are\nhighly soluble.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2551,
"subject": "Chemistry",
"question": "Given below are two statements :
Statement I : Both CaCl2 . 6H2O and MgCl2 . 8H2O undergo dehydration on heating.
Statement II : BeO is amphoteric whereas the oxides of other elements in the same group are acidic.
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": "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 false",
"solution": "**Answer:** Both statement I and statement II are false\n\nCaCl2.6H2O $$\\buildrel \\Delta \\over\n \\longrightarrow $$ CaCl2 + 6H2O
\nAmong alkaline earth metal BeO is amphoteric & rest are basic oxide",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 2552,
"subject": "Chemistry",
"question": "The set of elements that differ in mutual relationship from those of the other sets is :",
"options": [
{
"text": "Li - Na"
},
{
"text": "Be - Al"
},
{
"text": "B - Si"
},
{
"text": "Li - Mg"
}
],
"answer": "Li - Na",
"solution": "**Answer:** Li - Na\n\n \nLi–Mg, B–Si, Be–Al show diagonal\nrelationship but Li and Na do not show\ndiagonal relationship as both belongs to same\ngroup and not placed diagonally.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2553,
"subject": "Chemistry",
"question": "Match List - I with List - II :
\n\n
\n
\n
List - I
\n
\n
List - II
\n
\n\n\n
\n
(a)
\n
Be
\n
(i)
\n
treatment of cancer
\n
\n
\n
(b)
\n
Mg
\n
(ii)
\n
extraction of metals
\n
\n
\n
(c)
\n
Ca
\n
(iii)
\n
incendiary bombs and signals
\n
\n
\n
(d)
\n
Ra
\n
(iv)
\n
windows of X-ray tubes
\n
\n
\n
\n
\n
(v)
\n
bearings for motor engines
\n
\n\n
Choose the most appropriate answer from the option given below :",
"options": [
{
"text": "(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)"
},
{
"text": "(a)-(iii), (b)-(iv), (c)-(v), (d)-(ii)"
},
{
"text": "(a)-(iii), (b)-(iv), (c)-(ii), (d)-(v)"
},
{
"text": "(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)"
}
],
"answer": "(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)",
"solution": "**Answer:** (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)\n\nBeryllium is used in the manufacture of alloys. Metallic beryllium is used for making windows of\nX-ray tubes. Radium salts are used in radiotherapy, for example, in the treatment of cancer. Magnesium\n(powder and ribbon) is used in flash powders and bulbs, incendiary bombs and signals. Calcium is used\nin the extraction of metals from oxides which are difficult to reduce with carbon.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 2554,
"subject": "Chemistry",
"question": "The set in which compounds have different nature is :",
"options": [
{
"text": "B(OH)3 and H3PO3"
},
{
"text": "B(OH)3 and Al(OH)3"
},
{
"text": "NaOH and Ca(OH)2"
},
{
"text": "Be(OH)2 and Al(OH)3"
}
],
"answer": "B(OH)3 and Al(OH)3",
"solution": "**Answer:** B(OH)3 and Al(OH)3\n\n
(A) Both are acidic in nature.\n\n\n\n B(OH)3 is orthoboric acid. It can accept a pair of electrons from Lewis bases and hence, act as Lewis acid.\n\n
H3PO3 is (phosphorus acid) diprotic acid as it readily ionises two protons.
\n
(B) B(OH)3 is acidic and Al(OH)3 is amphoteric in nature.\n\n
(C) Both NaOH and Ca(OH)2 are basic hydroxides.\n\n
(D) Both Be(OH)2 and Al(OH)3 are amphoteric hydroxides.
Hence, only B(OH)3 and Al(OH)3 have different nature.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2555,
"subject": "Chemistry",
"question": "Given below are two statements :
Statement I : None of the alkaline earth metal hydroxides dissolve in alkali.
Statement II : Solubility of alkaline earth metal hydroxides in water increases down the group.
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": "Statement I is incorrect but Statement II is correct."
},
{
"text": "Statement I and Statement II both are incorrect."
},
{
"text": "Statement I and Statement II both are correct."
}
],
"answer": "Statement I is incorrect but Statement II is correct.",
"solution": "**Answer:** Statement I is incorrect but Statement II is correct.\n\nStatement - I is incorrect
Be(OH)2 dissolve in alkali due to it's amphoteric nature.
Statement - II is correct
Solubility of alkaline earth metal hydroxide in water increases down the group due to rapid decreases in lattice energy as compared to hydration energy.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 2556,
"subject": "Chemistry",
"question": "
s-block element which cannot be qualitatively confirmed by the flame test is :
",
"options": [
{
"text": "Li"
},
{
"text": "Na"
},
{
"text": "Rb"
},
{
"text": "Be"
}
],
"answer": "Be",
"solution": "**Answer:** Be\n\nBeryllium does not give flame test because of its\nsmall size and high ionization energy the energy of\nflame is not sufficient to excite the electrons to\nhigher energy level.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2558,
"subject": "Chemistry",
"question": "
Which one of the following alkaline earth metal ions has the highest ionic mobility in its aqueous solution?
",
"options": [
{
"text": "Be2+"
},
{
"text": "Mg2+"
},
{
"text": "Ca2+"
},
{
"text": "Sr2+"
}
],
"answer": "Sr2+",
"solution": "**Answer:** Sr2+\n\nIn aqueous solution, the ionic mobility is inversely\nproportional to the charge density on the ion. Hence\nSr2+ has highest ionic mobility in water.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2559,
"subject": "Chemistry",
"question": "
The correct order of density is :
",
"options": [
{
"text": "Be > Mg > Ca > Sr"
},
{
"text": "Sr > Ca > Mg > Be"
},
{
"text": "Sr > Be > Mg > Ca"
},
{
"text": "Be > Sr > Mg > Ca"
}
],
"answer": "Sr > Be > Mg > Ca",
"solution": "**Answer:** Sr > Be > Mg > Ca\n\nDensity of $$\\mathrm{Sr}=2.63 \\mathrm{~g} / \\mathrm{cm}^{3}$$\n
\nDensity of $$\\mathrm{Be}=1.84 \\mathrm{~g} / \\mathrm{cm}^{3}$$\n
\nDensity of $$\\mathrm{Mg}=1.74 \\mathrm{~g} / \\mathrm{cm}^{3}$$\n
The correct decreasing order for metallic character is :
",
"options": [
{
"text": "Na > Mg > Be > Si > P"
},
{
"text": "P > Si > Be > Mg > Na"
},
{
"text": "Si > P > Be > Na > Mg"
},
{
"text": "Be > Na > Mg > Si > P"
}
],
"answer": "Na > Mg > Be > Si > P",
"solution": "**Answer:** Na > Mg > Be > Si > P\n\nMetallic character increases top to bottom in group and decreases left to right in a period.\n
\n$\\mathrm{Mg}$ is from second group it will be less metallic than $\\mathrm{Na}$. Be comes above $\\mathrm{Mg}$ hence less metallic than Mg. Si is more metallic than phosphorous. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2563,
"subject": "Chemistry",
"question": "
Statement I : An alloy of lithium and magnesium is used to make aircraft plates.
\n
Statement II : The magnesium ions are important for cell-membrane integrity.
\n
In the light 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 false",
"solution": "**Answer:** Both Statement I and Statement II are false\n\nAlloy of Li and Mg is used to make armour plates\nand not aircraft plates.
\nCalcium plays important roles in neuromuscular\nfunction, interneuronal transmission and cell\nmembrane integrity .",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2564,
"subject": "Chemistry",
"question": "The element playing significant role in neuromuscular function and interneuronal transmission is :",
"options": [
{
"text": "Be"
},
{
"text": "$\\mathrm{Ca}$"
},
{
"text": "$\\mathrm{Mg}$"
},
{
"text": "$\\mathrm{Li}$"
}
],
"answer": "$\\mathrm{Ca}$",
"solution": "**Answer:** $\\mathrm{Ca}$\n\nCalcium plays important role in neuromuscular\nfunction, interneuronal transmission, cell\nmembrane etc.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2565,
"subject": "Chemistry",
"question": "
Choose the correct statement(s) :
\n
A. Beryllium oxide is purely acidic in nature.
\n
B. Beryllium carbonate is kept in the atmosphere of $$\\mathrm{CO_2}$$.
\n
C. Beryllium sulphate is readily soluble in water.
\n
D. Beryllium shows anomalous behaviour.
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "A only"
},
{
"text": "B, C and D only"
},
{
"text": "A, B and C only"
},
{
"text": "A and B only"
}
],
"answer": "B, C and D only",
"solution": "**Answer:** B, C and D only\n\n
A. Beryllium oxide is amphoteric in nature.
\n
B. Beryllium carbonate is kept in the atmosphere \nof CO2 because it is thermally less stable.
\n
C. Beryllium sulphate is readily soluble in water \ndue to high degree of hydration.
\n
D. Beryllium shows anomalous behaviour due to \nsmall size, high ionization energy and high value \nof $$\\phi $$ (polarising power).
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2566,
"subject": "Chemistry",
"question": "Chlorides of which metal are soluble in organic solvents :",
"options": [
{
"text": "K"
},
{
"text": "Be"
},
{
"text": "Ca"
},
{
"text": "Mg"
}
],
"answer": "Be",
"solution": "**Answer:** Be\n\n
$$\\mathrm{BeCl_2}$$ is a covalent molecule
\n
So, it is soluble in organic solvents, rest are ionic compounds.
The alkaline earth metal sulphate(s) which are readily soluble in water is/are :
\n
A. $$\\mathrm{BeSO}_{4}$$
\n
B. $$\\mathrm{MgSO}_{4}$$
\n
C. $$\\mathrm{CaSO}_{4}$$
\n
D. $$\\mathrm{SrSO}_{4}$$
\n
E. $$\\mathrm{BaSO}_{4}$$
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "A and B"
},
{
"text": "B and C"
},
{
"text": "B only"
},
{
"text": "A only"
}
],
"answer": "A and B",
"solution": "**Answer:** A and B\n\n
BeSO$$_4$$ and MgSO$$_4$$ are readily soluble in water.
Choose the correct answer from the options given below :
",
"options": [
{
"text": "E > C > A > B > D"
},
{
"text": "C > A > E > B > D"
},
{
"text": "C > E > A > D > B"
},
{
"text": "C > E > A > B > D"
}
],
"answer": "C > E > A > B > D",
"solution": "**Answer:** C > E > A > B > D\n\n Hydration enthalpy $\\propto$ charge density\n
Charge density means $${{Magnitude\\,of\\,Charge} \\over {Ionic\\,Radius}}$$.\n
\n$\\therefore$ The correct order of charge density is\n
Which of the following represents the correct order of metallic character of the given elements?
",
"options": [
{
"text": "K < Mg < Be < Si"
},
{
"text": "Si < Be < Mg < K"
},
{
"text": "Be < Si < Mg < K"
},
{
"text": "Be < Si < K < Mg"
}
],
"answer": "Si < Be < Mg < K",
"solution": "**Answer:** Si < Be < Mg < K\n\nMetallic character of an element is directly proportional to its electropositivity. Of the given elements silicon is least electro positive and potassium is most electropositive whereas beryllium and magnesium have intermediate values in the increasing order. Therefore, correct order of metallic character is $\\mathrm{Si} < \\mathrm{Be} < \\mathrm{Mg} < \\mathrm{K}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2570,
"subject": "Chemistry",
"question": "
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R
\n
Assertion A : Beryllium has less negative value of reduction potential compared to the other alkaline earth metals.
\n
Reason R : Beryllium has large hydration energy due to small size of Be$$^{2+}$$ but relatively large value of atomization enthalpy
\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": "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\nThe most appropriate answer is Both A and R are correct and R is the correct explanation of A.\n
\nAssertion A states that beryllium has a less negative value of reduction potential compared to the other alkaline earth metals. This assertion is correct, as beryllium has higher ionization energy and a smaller atomic radius than the other alkaline earth metals, which makes it more difficult to reduce Be2+ to Be metal.\n
\nReason R provides an explanation for Assertion A. Beryllium has a relatively large value of atomization enthalpy due to the strong Be-Be bonds in the solid state, but it also has a large hydration energy due to the small size of Be2+ ions. The strong Be-Be bonds make it difficult to reduce Be2+ to Be metal, and the small size of Be2+ ions leads to a large hydration energy. This makes it energetically unfavorable to remove the Be2+ ion from the aqueous phase and reduce it to Be metal, which results in a less negative value of reduction potential for beryllium.\n
\nTherefore, both Assertion A and Reason R are correct, and Reason R provides the correct explanation for Assertion A.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2571,
"subject": "Chemistry",
"question": "Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R :\n
\nAssertion (A) : $\\mathrm{BeCl}_{2}$ and $\\mathrm{MgCl}_{2}$ produce characteristic flame
\nReason (R) : The excitation energy is high in $\\mathrm{BeCl}_{2}$ and $\\mathrm{MgCl}_{2}$\n
\nIn the light of the above statements, choose the correct answer from the options given below :",
"options": [
{
"text": "$\\mathrm{(A)}$ is true but $(\\mathrm{R})$ is false"
},
{
"text": "Both (A) and $(\\mathrm{R})$ are true and $(\\mathrm{R})$ is the correct explanation of $(A)$"
},
{
"text": "(A) is false but (R) is true\n"
},
{
"text": "Both (A) and (R) are true but (R) is NOT the correct explanation of (A)"
}
],
"answer": "(A) is false but (R) is true\n",
"solution": "**Answer:** (A) is false but (R) is true\n\n\nBeCl2 and MgCl2 do not produce characteristic\nflame because excitation energy is high in BeCl2\nand MgCl2. Hence, the correct answer is option (C).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2572,
"subject": "Chemistry",
"question": "
The correct order of metallic character is :
",
"options": [
{
"text": "K > Be > Ca"
},
{
"text": "Ca > K > Be"
},
{
"text": "K > Ca > Be"
},
{
"text": "Be > Ca > K"
}
],
"answer": "K > Ca > Be",
"solution": "**Answer:** K > Ca > Be\n\n
As you move from top to bottom in a group (vertical column) on the periodic table, the metallic character increases. This is because elements at the bottom of a group have more electron shells than those at the top. These extra shells shield the outermost electrons from the pull of the nucleus, which makes it easier for these electrons to be lost, a characteristic of metals.
\n
On the other hand, as you move from left to right across a period (horizontal row) on the periodic table, the metallic character decreases. This is because the nuclear charge increases as you move across a period, which pulls the electrons closer to the nucleus and makes them less likely to be lost.
\n
Given this, for the elements K (Potassium), Be (Beryllium), and Ca (Calcium), the order of metallic character should be K > Ca > Be. Potassium (K) is furthest down (group 1, period 4), calcium (Ca) is above potassium (group 2, period 4), and beryllium (Be) is above calcium (group 2, period 2). So, the correct order should be Option C: K > Ca > Be.
Statement I : Lithium and Magnesium do not form superoxide
\n
Statement II : The ionic radius of $$\\mathrm{Li}^{+}$$ is larger than ionic radius of $$\\mathrm{Mg}^{2+}$$
\n
In the light of the above statements, choose the most appropriate answer\n\nfrom the options given below :
",
"options": [
{
"text": "Statement I is incorrect but Statement II is correct"
},
{
"text": "Both Statement I and Statement II are incorrect"
},
{
"text": "Both Statement I and Statement II are correct"
},
{
"text": "Statement I is correct but Statement II is incorrect"
}
],
"answer": "Both Statement I and Statement II are correct",
"solution": "**Answer:** Both Statement I and Statement II are correct\n\n
Statement I : "Lithium and Magnesium do not form superoxide"
\n
This statement is correct. Lithium and magnesium are smaller in size compared to other elements in their respective groups. Although lithium can form superoxide under certain conditions, in general, the smaller size of these two elements makes it harder for them to form stable superoxides.
\n
Statement II : "The ionic radius of Li+ is larger than the ionic radius of Mg2+"
\n
This statement is still correct. Despite both lithium and magnesium being small in size, when they lose electrons to form ions, the Mg2+ ion is smaller than the Li+ ion due to the loss of two electrons (and thus an energy level) in magnesium compared to the loss of one electron in lithium.
\n
So, based on the information, the answer is :
\n
Option C : Both Statement I and Statement II are correct.
Hydration enthalpy is the amount of energy released when one mole of gaseous ions is completely surrounded by water molecules. The process is exothermic, hence the enthalpy change is negative.
\n
Generally, smaller ions and ions with higher charge have larger hydration enthalpies due to the stronger attraction to the water molecules. This is because the hydration enthalpy depends on the charge density of the ion.
\n
Among the given options, $$\\mathrm{Be}^{2+}$$ is the smallest ion and it has the same charge as the other ions. Hence, it has the highest charge density and will have the highest hydration enthalpy.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2575,
"subject": "Chemistry",
"question": "In a compound atoms of element Y from ccp lattice and those of element X occupy 2/3rd of tetrahedral\nvoids. The formula of the compound will be :",
"options": [
{
"text": "X4Y3"
},
{
"text": "X2Y3"
},
{
"text": "X2Y"
},
{
"text": "X3Y4"
}
],
"answer": "X4Y3",
"solution": "**Answer:** X4Y3\n\nFrom the given data, we have \n
From the above we get the formula of the compound \n
as $${X_{16/3}}{Y_4}\\,\\,$$ or $${X_4}{Y_3}$$ ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2576,
"subject": "Chemistry",
"question": "Percentages of free space in cubic close packed structure and in body centred packed structure are\nrespectively :",
"options": [
{
"text": "30% and 26%"
},
{
"text": "26% and 32%"
},
{
"text": "32% and 48%"
},
{
"text": "48% and 26%"
}
],
"answer": "26% and 32%",
"solution": "**Answer:** 26% and 32%\n\nPacking fraction is defined as the ratio of the volume of the unit cell that is occupied by the spheres to the volume of the unit cell. \n
P.F. for cpp and bcc are $$0.74$$ and $$0.68$$ respectively. \n
So, the free space in ccp and bcc are $$26\\% $$ & $$32\\% $$ respectively.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2577,
"subject": "Chemistry",
"question": "A compound of formula A2B3 has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms : ",
"options": [
{
"text": "hcp lattice - A, $${1 \\over 3}$$ Tetrahedral voids-B"
},
{
"text": "hcp lattice - B, $${1 \\over 3}$$ Tetrahedral voids - A"
},
{
"text": "hcp lattice - A, $${2 \\over 3}$$ Tetrahedral voids - B"
},
{
"text": "hcp lattice - B, $${2 \\over 3}$$ Tetrahedral voids - A"
}
],
"answer": "hcp lattice - B, $${1 \\over 3}$$ Tetrahedral voids - A",
"solution": "**Answer:** hcp lattice - B, $${1 \\over 3}$$ Tetrahedral voids - A\n\nA2B3 has HCP lattice\n
If A form HCP, \n then $${{{3^{th}}} \\over 4}$$ of THV must occupied by B to form A2B3\n
If B form HCP, then $${{{1^{th}}} \\over 3}$$ of THV must occupied by A to form A2B3",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2578,
"subject": "Chemistry",
"question": "Ga (atomic mass 70 u) crystallizes in a hexagonal close packed structure. The total number of voids in 0.581 g of Ga is __________ $$\\times$$ 1021. (Round off to the Nearest Integer). [Given : NA = 6.023 $$\\times$$ 1023]",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\nFor HCP structure Z = 6
Tetrahedral Void = (Z $$\\times$$ 2) = 12
Octahedral Void = [Z $$\\times$$ 1] = 6
No. of void per unit cell = 18
No. of unit cell = $$\\left( {{{0.581 \\times {N_A}} \\over {70 \\times 6}}} \\right)$$
$$ = 14.99 \\times {10^{21}}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2579,
"subject": "Chemistry",
"question": "In a binary compound, atoms of element A form a hcp structure and those of element M occupy 2/3 of the tetrahedral voids of the hcp structure. The formula of the binary compound is :",
"options": [
{
"text": "M2A3"
},
{
"text": "M4A3"
},
{
"text": "M4A"
},
{
"text": "MA3"
}
],
"answer": "M4A3",
"solution": "**Answer:** M4A3\n\nIn HCP unit cell,
Z = 6 so A = 6
Also we know, in HCP
Tetrahedral voids = 2Z = 12
$$ \\therefore $$ No. of M = $${2 \\over 3}$$ [TV] = $${2 \\over 3}$$ $$\\times$$ 12 = 8
$$ \\therefore $$ Formula = M8A6 = M4A3",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2580,
"subject": "Chemistry",
"question": "The empirical formula for a compound with a cubic close packed arrangement of anions and with cations occupying all the octahedral sites in AxB. The value of x is _____________. (Integer answer)",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nAnions forms CCP or FCC(A$$-$$) = 4A$$-$$ per unit cell Cations occupy all octahedral voids (B+) = 4B+ per unit cell
In a solid AB, A atoms are in ccp arrangement and B atoms occupy all the octahedral sites. If two atoms from the opposite faces are removed, then the resultant stoichiometry of the compound is AxBy. The value of x is ____________. [nearest integer]
",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n$A$ atoms are in CCP contribution of $A$ is
Ionic radii of cation $$\\mathrm{A}^{+}$$ and anion $$\\mathrm{B}^{-}$$ are 102 and $$181 \\,\\mathrm{pm}$$ respectively. These ions are allowed to crystallize into an ionic solid. This crystal has cubic close packing for $$\\mathrm{B}^{-}$$. $$\\mathrm{A}^{+}$$ is present in all octahedral voids. The edge length of the unit cell of the crystal AB is __________ pm. (Nearest Integer)
",
"options": [],
"answer": "566",
"solution": "**Answer:** 566\n\nIn cubic close packing, octahedral voids form at\nedge centers and body center of the cube
A compound is formed by two elements $$\\mathrm{X}$$ and $$\\mathrm{Y}$$. The element $$\\mathrm{Y}$$ forms cubic close packed arrangement and those of element $$\\mathrm{X}$$ occupy one third of the tetrahedral voids. What is the formula of the compound?
A compound is formed by two elements $$\\mathrm{X}$$ and $$\\mathrm{Y}$$. The element $$\\mathrm{Y}$$ forms a cubic close-packed (CCP) arrangement, and element $$\\mathrm{X}$$ occupies one third of the tetrahedral voids.
\n
In a CCP structure, there are 4 atoms of $$\\mathrm{Y}$$ in a unit cell. This means there are 8 tetrahedral voids in the CCP structure.
\n
Since $$\\mathrm{X}$$ occupies one third of the tetrahedral voids, there are $$\\frac{1}{3} \\times 8 = \\frac{8}{3}$$ atoms of $$\\mathrm{X}$$ in the formula unit.
\n
The ratio of $$\\mathrm{X}$$ to $$\\mathrm{Y}$$ in the formula unit is $$\\frac{8}{3} : 4 = \\frac{2}{3} : 1$$. Multiplying both parts of the ratio by 3, we get $$2 : 3$$.
\n
So, the formula of the compound is $$\\mathrm{X_2Y_{3}}$$
\n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2585,
"subject": "Chemistry",
"question": "Na and Mg crystallize in BCC and FCC type crystals respectively, then the number of atoms of\nNa and Mg present in the unit cell of their respective crystal is :",
"options": [
{
"text": "4 and 2"
},
{
"text": "9 and 14"
},
{
"text": "14 and 9"
},
{
"text": "2 and 4"
}
],
"answer": "2 and 4",
"solution": "**Answer:** 2 and 4\n\nIn bcc - points are at corners and one in the center of the unit cell. \n
Number of atoms per unit cell $$ = 8 \\times {1 \\over 8} + 1 = 2.$$ \n
In fcc - points are at the corners and also centre of the six faces of each cell. \n
Number of atoms per unit cell $$ = 8 \\times {1 \\over 8} + 6 \\times {1 \\over 2} = 4.$$ ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2586,
"subject": "Chemistry",
"question": "Total volume of atoms present in a face-centre cubic unit cell of a metal is (r is atomic radius) :",
"options": [
{
"text": "20/3 $$\\pi r^3$$"
},
{
"text": "24/3 $$\\pi r^3$$"
},
{
"text": "16/3 $$\\pi r^3$$"
},
{
"text": "12/3 $$\\pi r^3$$"
}
],
"answer": "16/3 $$\\pi r^3$$",
"solution": "**Answer:** 16/3 $$\\pi r^3$$\n\nThe face centered cubic unit cell contains $$4$$ atom\n
$$\\therefore$$ Total volume of atoms \n
$$ = 4 \\times {4 \\over 3}\\pi {r^3} = {{16} \\over 3}\\pi {r^3}$$ ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2587,
"subject": "Chemistry",
"question": "Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom?",
"options": [
{
"text": "108 pm "
},
{
"text": "127 pm"
},
{
"text": "157 pm"
},
{
"text": "181 pm"
}
],
"answer": "127 pm",
"solution": "**Answer:** 127 pm\n\nFor $$fcc$$ unit cell, \n
$$4r = \\sqrt 2 \\,\\,a;\\,r = {{\\sqrt 2 \\times 361} \\over 4} = 127\\,\\,pm$$ \n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2588,
"subject": "Chemistry",
"question": "In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre\npositions. If one atom of B is missing from one of the face centred points, the formula of the compound is :",
"options": [
{
"text": "AB2"
},
{
"text": "A2B3"
},
{
"text": "A2B5"
},
{
"text": "A2B"
}
],
"answer": "A2B5",
"solution": "**Answer:** A2B5\n\nNo. of atoms in the corners \n
$$ \\Rightarrow $$ M = 0.0305 kg mol–1",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 2596,
"subject": "Chemistry",
"question": "The radius of the largest sphere which fits properly at the centre of the edge of a body centred cubic unit cell is : (Edge length is represented by 'a')\n",
"options": [
{
"text": "0.134 a "
},
{
"text": "0.067 a"
},
{
"text": "0.047 a"
},
{
"text": "0.027 a "
}
],
"answer": "0.067 a",
"solution": "**Answer:** 0.067 a\n\na = 2(R + r)\n
Number of moles in 200 gm = $${{{200} \\over {50}}}$$ = 4\n
$$ \\therefore $$ Number of molecules = 4$$ \\times $$NA",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 2602,
"subject": "Chemistry",
"question": "An element with molar mass 2.7 $$ \\times $$ 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its\ndensity is 2.7 $$ \\times $$ 103\n kg m-3, the radius of the element is approximately ______ $$ \\times $$ 10-12 m (to the\nnearest integer).",
"options": [],
"answer": "143",
"solution": "**Answer:** 143\n\nMolar mass of an element (M) = 27 gm mol–1\n
Edge length of a cubic unit cell (a) = 405 pm = 4.05 × 10–8 cm\n
density of the element (d) = 2.7 gm/cc\n
d = $${{Z \\times M} \\over {{N_A} \\times {{\\left( a \\right)}^3}}}$$\n
$$ \\Rightarrow $$ r = $${{1.414 \\times 405} \\over 4}$$\n
= 143 pm\n
= 143 × 10–12 m",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 2603,
"subject": "Chemistry",
"question": "The coordination number of an atom in a body-centered cubic structure is _______. \n[Assume that the lattice is made up of atoms.]",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nCoordination number is the number of nearest neighbours of a central atom in the structure.
bcc has a coordination number of 8 and contains 2 atoms per unit cell.
This is because each atom touches four atoms in the layer above it, four in the layer below it and none in its own layer.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2604,
"subject": "Chemistry",
"question": "The unit cell of copper corresponds to a face centered cube of edge length 3.596 $$\\mathop A\\limits^o $$ with one copper atom at each lattice point. The calculated density of copper in kg/m3 is ___________. [Molar mass of Cu : 63.54 g; Avogadro Number = 6.022 $$\\times$$ 1023]",
"options": [],
"answer": "9077",
"solution": "**Answer:** 9077\n\n
Density of copper, $$d = {{Z \\times M} \\over {{a^3} \\times {N_A}}}$$
\n
Given, Z = 4, for fcc lattice,
\n
M = 63.54 g mol$$-$$1
\n
= 63.54 $$\\times$$ 10$$-$$3 kg mol$$-$$1,
\n
a = 3.596 $$\\mathop A\\limits^o $$ = 3.596 $$\\times$$ 10$$-$$10 m,
",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 2605,
"subject": "Chemistry",
"question": "A certain element crystallises in a bcc lattice of unit cell edge length 27$$\\mathop A\\limits^o $$. If the same element under the same conditions crystallises in the fcc lattice, the edge length of the unit cel in $$\\mathop A\\limits^o $$ will be ____________. (Round off to the Nearest Integer).
$$ \\Rightarrow $$ $$a = 33.12 \\approx 33$$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 2606,
"subject": "Chemistry",
"question": "Diamond has a three dimensional structure of C atoms formed by covalent bonds. The structure of diamond has face centred cubic lattice where 50% of the tetrahedral voids are also occupied by carbon atoms. The number of carbon atoms present per unit cell of diamond is ____________.",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nCarbon atoms occupy FCC lattice points as well as half of the tetrahedral voids.
Therefore number of carbon atoms per unit cell = 8",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2607,
"subject": "Chemistry",
"question": "A copper complex crystallising in a CCP lattice with a cell edge of 0.4518 nm has been revealed by employing X-ray diffraction studies. The density of a copper complex is found to be 7.62 g cm$$-$$3. The molar mass of copper complex is ____________ g mol$$-$$1. (Nearest integer)
$$\\Rightarrow $$ M = 105.8 g/mol",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 2608,
"subject": "Chemistry",
"question": "The parameters of the unit cell of a substance are a = 2.5, b = 3.0, c = 4.0, $$\\alpha$$ = 90$$^\\circ$$, $$\\beta$$ = 120$$^\\circ$$, $$\\gamma$$ = 90$$^\\circ$$. The crystal system of the substance is :",
"options": [
{
"text": "Hexagonal "
},
{
"text": "Orthorhombic"
},
{
"text": "Monoclinic"
},
{
"text": "Triclinic"
}
],
"answer": "Monoclinic",
"solution": "**Answer:** Monoclinic\n\na $$\\ne$$ b $$\\ne$$ c and $$\\alpha$$ = $$\\gamma$$ = 90$$^\\circ$$ $$\\ne$$ $$\\beta$$ are parameters of monoclinic unit cell.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2609,
"subject": "Chemistry",
"question": "
Atoms of element X form hcp lattice and those of element Y occupy $${2 \\over 3}$$ of its tetrahedral voids. The percentage of element X in the lattice is ____________. (Nearest integer)
",
"options": [],
"answer": "43",
"solution": "**Answer:** 43\n\nSince $X$ occupies hop lattice,\n\nNumber of particles of type $X$ in a unit cell $=6$\n
\nNumber of particles of type $Y=\\frac{2}{3} \\times 12=8$\n
\n$\\therefore$ Percentage of element $X=\\frac{6}{14} \\times 100$\n
An element X has a body centred cubic (bcc) structure with a cell edge of 200 pm. The density of the element is 5 g cm$$-$$3. The number of atoms present in 300 g of the element X is _______________.
\n
Given : Avogadro constant, NA = 6.0 $$\\times$$ 1023 mol$$-$$1.
An element M crystallises in a body centred cubic unit cell with a cell edge of $$300 \\,\\mathrm{pm}$$. The density of the element is $$6.0 \\mathrm{~g} \\mathrm{~cm}^{-3}$$. The number of atoms present in $$180 \\mathrm{~g}$$ of the element is ____________ $$\\times 10^{23}$$. (Nearest integer)
",
"options": [],
"answer": "22",
"solution": "**Answer:** 22\n\n$M$ is body certred cubic, $\\therefore Z=2$
Metal $$\\mathrm{M}$$ crystallizes into a fcc lattice with the edge length of $$4.0 \\times 10^{-8} \\mathrm{~cm}$$. The atomic mass of the metal is ____________ $$\\mathrm{g} / \\mathrm{mol}$$. (Nearest integer)
\n
$$\\left(\\right.$$ Use : $$\\mathrm{N}_{\\mathrm{A}}=6.02 \\times 10^{23} \\mathrm{~mol}^{-1}$$, density of metal, $$\\mathrm{M}=9.03 \\mathrm{~g} \\mathrm{~cm}^{-3}$$ )
A metal M crystallizes into two lattices :- face centred cubic (fcc) and body centred cubic (bcc) with unit cell edge length of $$2.0$$ and $$2.5\\,\\mathop A\\limits^o $$ respectively. The ratio of densities of lattices fcc to bcc for the metal M is ____________. (Nearest integer)
",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n$\\begin{aligned} & d_1 \\text {, Density of fcc lattice of metal } M=\\frac{4 \\times M}{N_0\\left(a_{\\mathrm{fcc}}\\right)^3} \\\\\\\\ & d_2 \\text {, Density of bcc lattice of metal } M=\\frac{2 \\times M}{N_0\\left(a_{\\mathrm{bcc}}\\right)^3} \\\\\\\\ & \\frac{d_1}{d_2}=\\frac{4}{2}\\left(\\frac{a_{\\mathrm{bcc}}}{a_{\\mathrm{fcc}}}\\right)^3=2\\left(\\frac{2.5}{2}\\right)^3=3.90 \\simeq 4\\end{aligned}$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 2614,
"subject": "Chemistry",
"question": "Iron oxide FeO, crystallises in a cubic lattice with a unit cell edge length of 5.0 Å. If density of the $\\mathrm{FeO}$ in the crystal is $4.0 \\mathrm{~g} \\mathrm{~cm}^{-3}$, then the number of $\\mathrm{FeO}$ units present per unit cell is __________. (Nearest integer)\n
\nGiven: Molar mass of $\\mathrm{Fe}$ and $\\mathrm{O}$ is 56 and $16 \\mathrm{~g} \\mathrm{~mol}^{-1}$ respectively. $\\mathrm{N}_{\\mathrm{A}}=6.0 \\times 10^{23} \\mathrm{~mol}^{-1}$",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n
A cubic solid is made up of two elements X and Y. Atoms of X are present on every alternate corner and one at the center of cube. Y is at $$\\frac{1}{3}^{\\mathrm{rd}}$$ of the total faces. The empirical formula of the compound is :
",
"options": [
{
"text": "$$\\mathrm{{X_2}{Y_{1.5}}}$$"
},
{
"text": "$$\\mathrm{{X_{3}}{Y_2}}$$"
},
{
"text": "$$\\mathrm{X{Y_{2.5}}}$$"
},
{
"text": "$$\\mathrm{{X_{2.5}}Y}$$"
}
],
"answer": "$$\\mathrm{{X_{3}}{Y_2}}$$",
"solution": "**Answer:** $$\\mathrm{{X_{3}}{Y_2}}$$\n\n$\\begin{aligned} & \\text { Number of } X \\text { particles }=4 \\times \\frac{1}{8}+1=1.5 \\\\\\\\ & \\text { Number of } Y \\text { particles }=6 \\times \\frac{1}{3} \\times \\frac{1}{2}=1 \\\\\\\\ & \\therefore \\text { Empirical formula }=X_{1.5} Y_1=X_3 Y_2\\end{aligned}$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2616,
"subject": "Chemistry",
"question": "Which of the following expressions is correct in case of a $\\mathrm{CsCl}$ unit cell (edge length 'a')?",
"options": [
{
"text": "$\\mathrm{r}_{\\mathrm{Cs}^{+}}+\\mathrm{r}_{\\mathrm{Cl}^{-}}=\\frac{\\sqrt{3}}{2} \\mathrm{a}$"
},
{
"text": "$\\mathrm{r}_{\\mathrm{Cs}^{+}}+\\mathrm{r}_{\\mathrm{Cl}^{-}}=\\frac{\\mathrm{a}}{\\sqrt{2}}$"
},
{
"text": "$\\mathrm{r}_{\\mathrm{Cs}^{+}}+\\mathrm{r}_{\\mathrm{Cl}^{-}}=\\mathrm{a}$"
},
{
"text": "$\\mathrm{r}_{\\mathrm{Cs}^{+}}+\\mathrm{r}_{\\mathrm{Cl}^{-}}=\\frac{\\mathrm{a}}{2}$"
}
],
"answer": "$\\mathrm{r}_{\\mathrm{Cs}^{+}}+\\mathrm{r}_{\\mathrm{Cl}^{-}}=\\frac{\\sqrt{3}}{2} \\mathrm{a}$",
"solution": "**Answer:** $\\mathrm{r}_{\\mathrm{Cs}^{+}}+\\mathrm{r}_{\\mathrm{Cl}^{-}}=\\frac{\\sqrt{3}}{2} \\mathrm{a}$\n\n$\\mathrm{CsCl}$ has body centered type structure in which $\\mathrm{Cs}^{+}$ occupies at corner of a cube and $\\mathrm{Cl}^{-}$occupies the centre of the cube.
\n$2 \\mathrm{r}_{\\mathrm{Cs}^{+}}+2 \\mathrm{r}_{\\mathrm{Cl}^{-}}=\\sqrt{3} \\mathrm{a}$ (where a is the edge length of the cube)
Sodium metal crystallizes in a body centred cubic lattice with unit cell edge length of $$4~\\mathop A\\limits^o $$. The radius of sodium atom is __________ $$\\times ~10^{-1}$$ $$\\mathop A\\limits^o $$ (Nearest integer)
",
"options": [],
"answer": "17",
"solution": "**Answer:** 17\n\nIn a body-centered cubic (BCC) lattice, the relationship between the edge length (a) and the atomic radius (r) is given by :\n
\n$$\\sqrt{3}a = 4r$$\n
\nGiven the unit cell edge length (a) of sodium metal as 4 Å :\n
\n$$a = 4 ~\\mathop A\\limits^o$$\n
\nWe can now solve for the radius (r) of the sodium atom :\n
\nSo, the radius of the sodium atom is 17.32 × 10⁻¹ $\\mathop A\\limits^o$.",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 2618,
"subject": "Chemistry",
"question": "
An atomic substance A of molar mass $$12 \\mathrm{~g} \\mathrm{~mol}^{-1}$$ has a cubic crystal structure with edge length of $$300 ~\\mathrm{pm}$$. The no. of atoms present in one unit cell of $$\\mathrm{A}$$ is ____________. (Nearest integer)
\n
Given the density of $$\\mathrm{A}$$ is $$3.0 \\mathrm{~g} \\mathrm{~mL}^{-1}$$ and $$\\mathrm{N}_{\\mathrm{A}}=6.02 \\times 10^{23} \\mathrm{~mol}^{-1}$$
The correct relationships between unit cell edge length '$$a$$' and radius of sphere '$$r$$' for face-centred and body-centred cubic structures respectively are :
In a face-centered cubic (FCC) unit cell, atoms are present at the corners as well as at the centers of the faces. Hence, the diagonal of the face of the unit cell is equal to 4 times the radius of an atom.
\n
This gives us the equation:
\n$$\\sqrt{2} a = 4r$$
\nWhich simplifies to:
\n$$a = 2\\sqrt{2}r$$
\n
In a body-centered cubic (BCC) unit cell, atoms are present at the corners and at the center of the unit cell. The body diagonal of the unit cell is equal to 4 times the radius of an atom.
\n
This gives us the equation:
\n$$\\sqrt{3} a = 4r$$
\nWhich simplifies to:
\n$$a = \\frac{4}{\\sqrt{3}}r$$
\n
Therefore, Option D is correct:\n$$2\\sqrt{2}r = a$$ and $$4r = \\sqrt{3}a$$
Body-centered unit cells can be found in the following crystal systems among those listed:
\n\n
Cubic: Body-centered cubic (BCC) is one of the lattice structures that cubic systems can have.
\n
Tetragonal: A body-centered tetragonal system is also a possibility.
\n
Orthorhombic: The orthorhombic system can have a body-centered orthorhombic lattice.
\n\n
Therefore, the number of crystal systems from the list where a body-centered unit cell can be found is 3.
\n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2621,
"subject": "Chemistry",
"question": "Which type of ‘defect’ has the presence of cations in the interstitial sites?",
"options": [
{
"text": "Metal deficiency defect"
},
{
"text": "Schottky defect"
},
{
"text": "Vacancy defect"
},
{
"text": "Frenkel defect "
}
],
"answer": "Frenkel defect ",
"solution": "**Answer:** Frenkel defect \n\nIn Frenkel defect in a molecule an atom or ion (normally the cation) leave their original site and places itself in the interstitial site (area between all other cations and anions). Which is shown below \n
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2622,
"subject": "Chemistry",
"question": "Which of the following compounds is likely to\nshow both Frenkel and Schottky defects in its\ncrystalline form?",
"options": [
{
"text": "CsCl"
},
{
"text": "AgBr"
},
{
"text": "ZnS"
},
{
"text": "KBr"
}
],
"answer": "AgBr",
"solution": "**Answer:** AgBr\n\nSince AgBr has intermediate radius ratio. AgBr shows both, Frenkel as well as Schottky\ndefects.\n
ZnS only shows Frenkel defects.\n
KBr, CsCl only shows Schottky defects.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2623,
"subject": "Chemistry",
"question": "KBr is doped with 10$$-$$5 mole percent of SrBr2. The number of cationic vacancies in 1g of KBr crystal is ____________ 1014. (Round off to the Nearest Integer).
[Atomic Mass : K : 39.1 u, Br : 79.9 u NA = 6.023 $$\\times$$ 1023]",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nFor every Sr+2 ion, 1 cationic vacancy is created. Hence, no. of Sr+2 ion = Number of cationic vacancies
Since mole percentage of SrBr2 dropped is 10$$-$$5 to that of total moles of KBr.
Statement I : Frenkel defects are vacancy as well as interstitial defects.
Statement II : Frenkel defect leads to colour in ionic solids due to presence of F-centres.
Choose the most appropriate answer for the statements from the options given below :",
"options": [
{
"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"
},
{
"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
To answer this question, let's analyze each statement separately.
\n\n
Statement I: Frenkel defects are vacancy as well as interstitial defects.
\n\n
This statement is true. A Frenkel defect, also known as a dislocation defect, occurs in a crystalline solid when an atom or ion leaves its normal site and moves to an interstitial site, creating a vacancy defect at its original position and an interstitial defect at the new position. However, it's crucial to note that the vacancy and interstitial are created by the same atom or ion moving from one site to another within the crystal. So, the Frenkel defect involves both a vacancy and an interstitial defect, but they are correlated because they involve the displacement of the same atom or ion.
\n\n
Statement II: Frenkel defect leads to color in ionic solids due to presence of F-centres.
\n\n
This statement is false. The color in ionic solids is usually caused by F-centers (color centers), which are defects formed when an anion vacancy is occupied by one or more electrons. These electrons can absorb visible light, giving the crystal a characteristic color. However, the Frenkel defect, involving an atom or ion moving from its lattice site to an interstitial site, does not directly lead to the creation of F-centers. Instead, the coloration associated with F-centers is most commonly linked with Schottky defects (which involve paired vacancies of cations and anions but do not include displaced atoms to interstitial sites) or directly with anion vacancies that are not necessarily related to Frenkel defects.
\n\n
Conclusion: Given the explanations above, the correct answer is Option C: Statement I is true but Statement II is false.
The incorrect statement about the imperfections in solids is :
",
"options": [
{
"text": "Schottky defect decreases the density of the substance."
},
{
"text": "Interstitial defect increases the density of the substance."
},
{
"text": "Frenkel defect does not alter the density of the substance."
},
{
"text": "Vacancy defect increases the density of the substance."
}
],
"answer": "Vacancy defect increases the density of the substance.",
"solution": "**Answer:** Vacancy defect increases the density of the substance.\n\nThe vacancy defect increases the density of substance.
\nIt does not change the density of the crystal. It only creates cationic vacancies. Frenkel\nDefect causes vacancy defect at its original site and an interstitial defect at its new location. Therefore, it\ndoes not change the density of the solid.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2626,
"subject": "Chemistry",
"question": "
Metal deficiency defect is shown by Fe0.93O. In the crystal, some Fe2+ cations are missing and loss of positive charge is compensated by the presence of Fe3+ ions. The percentage of Fe2+ ions in the Fe0.93O crystals is __________. (Nearest integer)
\nLet the number of $\\mathrm{O}^{-2}$ ions be 100 and the number of $\\mathrm{Fe}^{+2}$ ions be $\\mathrm{X}$ The number of $\\mathrm{Fe}^{+3}$ ions be $(93-\\mathrm{X})$\n
\n$$\n\\begin{aligned}\n&\\therefore \\,X(2)+(93-X) 3=200 \\\\\\\\\n&279-X=200 \\\\\\\\\n&X=79 \\\\\\\\\n&\\therefore \\quad \\% \\text { of } F^{+2} \\text { ions } =\\frac{79}{93} \\times 100 \\\\\\\\\n&\\simeq 85 \\%\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2627,
"subject": "Chemistry",
"question": "An element crystallises in a face-centred cubic (fcc) unit cell with cell edge $$a$$. The distance\nbetween the centres of two nearest octahedral voids in the crystal lattice is :",
"options": [
{
"text": "$$a$$"
},
{
"text": "$${a \\over 2}$$"
},
{
"text": "$${a \\over {\\sqrt 2 }}$$"
},
{
"text": "$${\\sqrt 2 a}$$"
}
],
"answer": "$${a \\over {\\sqrt 2 }}$$",
"solution": "**Answer:** $${a \\over {\\sqrt 2 }}$$\n\n\n
Distance between nearest octahedral voids (O.V.)\n
= $$\\sqrt {{{\\left( {{a \\over 2}} \\right)}^2} + {{\\left( {{a \\over 2}} \\right)}^2}} $$ = $${a \\over {\\sqrt 2 }}$$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 2628,
"subject": "Chemistry",
"question": "The number of octahedral voids per lattice site in a lattice is _________. (Rounded off to the nearest integer)",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n
Let us assume, the crystal has fcc or ccp lattice which has octahedral voids.
\n
Number of lattice sites occupied = 8 corner + 6 face centres = 14
\n
Number of octahedral voids = 12 edge centres + 1 body centre = 13
\n
Number of octahedral void(s) per lattice site
\n
$$ = {{13} \\over {14}} = 0.928 \\simeq 1$$
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2629,
"subject": "Chemistry",
"question": "How many unit cells are present in a cubeshaped ideal crystal of NaCl of mass 1.00 g?\n[Atomic masses: Na = 23, Cl = 35.5]",
"options": [
{
"text": "5.14 $$\\times$$ 1021 unit cells"
},
{
"text": "1.28 $$\\times$$ 1021 unit cells"
},
{
"text": "1.71 $$\\times$$ 1021 unit cells"
},
{
"text": "2.57 $$\\times$$ 1021 unit cells"
}
],
"answer": "2.57 $$\\times$$ 1021 unit cells",
"solution": "**Answer:** 2.57 $$\\times$$ 1021 unit cells\n\nSince in $$NaCl$$ type of structure $$4$$ formula units form a cell. Number of formulas in cube shaped crystals \n
$$ = 2.57 \\times {10^{21}}$$ unit cels.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2630,
"subject": "Chemistry",
"question": "An ionic compound has a unit cell consisting of A ions at the corners of a cube and B\nions on the centres of the faces of the cube. The empirical formula for this compound\nwould be :",
"options": [
{
"text": "AB"
},
{
"text": "A2B "
},
{
"text": "AB3 "
},
{
"text": "A3B"
}
],
"answer": "AB3 ",
"solution": "**Answer:** AB3 \n\nNumber of A ions in the unit cell. $$ = {1 \\over 8} \\times 8 = 1$$ \n
Number of $$B$$ ions in the unit cell $$ = {1 \\over 2} \\times 6 = 3$$ \n
Hence empirical formula of the compound $$ = A{B_3}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2631,
"subject": "Chemistry",
"question": "The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the\ncation is 110 pm, the radius of the anion is :",
"options": [
{
"text": "288 pm"
},
{
"text": "398 pm"
},
{
"text": "618 pm"
},
{
"text": "144 pm"
}
],
"answer": "144 pm",
"solution": "**Answer:** 144 pm\n\nFor an Fcc crystal\n
\nfor $\\mathrm{NaCl}$, distance between $\\mathrm{Na}^{+}$and $\\mathrm{Cl}^{-}=\\frac{a}{2}$\n
\n$=1.04 \\times 10^{-10} \\mathrm{~m}$\n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2635,
"subject": "Chemistry",
"question": "Which of the following exists as covalent crystals in the solid state? ",
"options": [
{
"text": "Silicon "
},
{
"text": "Sulphur "
},
{
"text": "Phosphorous "
},
{
"text": "Iodine "
}
],
"answer": "Silicon ",
"solution": "**Answer:** Silicon \n\nAmong the given crystals only silicon is as a covalent solid. ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2636,
"subject": "Chemistry",
"question": "0.27 g of a long chain fatty acid was dissolved\nin 100 cm3 of hexane. 10 mL of this solution\nwas added dropwise to the surface of water in\na round watch glass. Hexane evaporates and a\nmonolayer is formed. The distance from edge\nto centre of the watch glass is 10 cm. What is\nthe height of the monolayer? \n[Density of fatty acid = 0.9 g cm–3, $$\\pi $$ = 3]",
"options": [
{
"text": "10–8 m"
},
{
"text": "10–2 m"
},
{
"text": "10–4 m"
},
{
"text": "10–6 m"
}
],
"answer": "10–6 m",
"solution": "**Answer:** 10–6 m\n\nIn 100 ml of hexane solution contains 0.27 g of fatty acid.\n
$$ \\therefore $$ In 10 ml of hexane solution contains 0.027 g of fatty acid.\n
Volume of fatty acid present on the round glass = $${{0.027} \\over {0.9}}$$\n
As here Area of fatty acid layer = Area of round plate = $$\\pi {r^2}$$\n
$$ \\Rightarrow $$ h = 10-4 cm = 10-6 m\n",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 2637,
"subject": "Chemistry",
"question": "A hard substance melts at high temperature and is an insulator in both solid and in molten state. This solid is most likely to be a/an :",
"options": [
{
"text": "Covalent solid"
},
{
"text": "Molecular solid"
},
{
"text": "Ionic solid"
},
{
"text": "Metallic solid"
}
],
"answer": "Covalent solid",
"solution": "**Answer:** Covalent solid\n\nCovalent or network solid are insulator (except graphite) and have very high melting point.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2638,
"subject": "Chemistry",
"question": "Given below are two statements. One is labelled as Assertion A nd the other is labelled as Reason R.
Assertion A : Sharp glass edge becomes smooth on heating it upto its melting point.
Reason R : The viscosity of glass decreases on melting.
Choose the most appropriate answer from the options given below. ",
"options": [
{
"text": "A is true but R is false"
},
{
"text": "Both A and R are true but R is NOT the correct explanation of A."
},
{
"text": "A is false but R is true."
},
{
"text": "Both A and R are true and R is 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\n
On heating the glass, it melts and takes up rounded shape at the\nedges, which has minimum surface area. This is due to the property\nof surface tension of liquids and not due to decrease in viscosity.
\n
Viscosity generally decreases as the temperature increases.
\n
Hence, both A and R are true but R is not the correct explanation\nof A.
(C) Amorphous solid are sometimes called pseudo solids.
(D) Amorphous solids soften over a range of temperatures.
(E) Amorphous solids have a definite heat of fusion.
Choose the most appropriate answer from the options given below.",
"options": [
{
"text": "(A), (B), (E) only"
},
{
"text": "(B), (D) only"
},
{
"text": "(C), (D) only"
},
{
"text": "(A), (C), (D) only"
}
],
"answer": "(A), (C), (D) only",
"solution": "**Answer:** (A), (C), (D) only\n\n(A) Crystalline solids have definite arrangement of constituent particles and have long range order.
(C), (D) Different constituent particles of an amorphous solid have different bond strengths and soften over a range of temperatures.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2640,
"subject": "Chemistry",
"question": "In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3. Taking kf for water as 1.85, the freezing point of the solution will be nearest to",
"options": [
{
"text": "-0.360oC"
},
{
"text": "-0.260oC"
},
{
"text": "+0.480oC"
},
{
"text": "-0.480oC"
}
],
"answer": "-0.480oC",
"solution": "**Answer:** -0.480oC\n\n$$\\Delta {T_f} = {K_f} \\times m \\times i;$$ \n
$$\\mathop {\\left( {HX} \\right.}\\limits_{1 - 0.3} \\,\\,\\rightleftharpoons\\,\\,\\mathop {{H^ + }}\\limits_{0.3} + \\mathop {{X^ - }}\\limits_{0.3} ,i = 1.\\left. 3 \\right)$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2641,
"subject": "Chemistry",
"question": "If liquids A and B form an ideal solution",
"options": [
{
"text": "the entropy of mixing is zero"
},
{
"text": "the free energy of mixing is zero"
},
{
"text": "the free energy as well as the entropy of mixing are each zero"
},
{
"text": "the enthalpy of mixing is zero"
}
],
"answer": "the enthalpy of mixing is zero",
"solution": "**Answer:** the enthalpy of mixing is zero\n\nWhen $$A$$ and $$B$$ from an ideal solution, $$\\Delta {H_{mix}} = 0$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2642,
"subject": "Chemistry",
"question": "Which one of the following statements is false?",
"options": [
{
"text": "Raoult’s law states that the vapour pressure of a components over a solution is\nproportional to its mole fraction "
},
{
"text": "Two sucrose solutions of same molality prepared in different solvents will have the same\nfreezing point depression "
},
{
"text": "The correct order of osmotic pressure for 0.01 M aqueous solution of each compound is\n BaCl2 > KCl > CH3COOH > sucrose"
},
{
"text": "The osmotic pressure ($$\\pi$$) = MRT, where M is the molarity of the solution "
}
],
"answer": "Two sucrose solutions of same molality prepared in different solvents will have the same\nfreezing point depression ",
"solution": "**Answer:** Two sucrose solutions of same molality prepared in different solvents will have the same\nfreezing point depression \n\n$$\\Delta {T_f} = {K_f} \\times m \\times i.$$ Since $${K_f}$$ has different values for different solvents, hence even if the $$m$$ is the same $$\\Delta {T_f}$$ will be different ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2643,
"subject": "Chemistry",
"question": "Which one of the following aqueous solutions will exhibit highest boiling point? ",
"options": [
{
"text": "0.01 M Na2SO4 "
},
{
"text": "0.015 M glucose"
},
{
"text": "0.015 M urea "
},
{
"text": "0.01 M KNO3"
}
],
"answer": "0.01 M Na2SO4 ",
"solution": "**Answer:** 0.01 M Na2SO4 \n\nAs$$\\,\\,\\,\\,\\Delta {T_b}^b = {T_b} - {T_b}^o$$ \n
Where $${T_b} = b.pt$$ of solution\n
$$T_b^o = b.pt$$ of solvent or \n
$${T_b} = T_b^o + \\Delta {T_b}$$\n
NOTE : Elevation in boiling point is a colligative property, which depends upon the no. of particles. Thus greater the number of particles, greater is it elevation and hence greater will be its boiling point. \n$$$N{a_2}S{O_4}\\,\\,\\rightleftharpoons\\,\\,2N{a^ + } + SO_4^{2 - }$$$\n Since $$N{a_2}S{O_4}\\,\\,$$ has maximum number of particles $$(3)$$ hence has maximum boiling point. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2644,
"subject": "Chemistry",
"question": "If $$\\alpha$$ is the degree of dissociation of Na2SO4, the vant Hoff’s factor (i) used for\ncalculating the molecular mass is :",
"options": [
{
"text": "1 + $$\\alpha$$"
},
{
"text": "1 + $$2\\alpha$$"
},
{
"text": "1 - $$\\alpha$$"
},
{
"text": "1 - $$2\\alpha$$"
}
],
"answer": "1 + $$2\\alpha$$",
"solution": "**Answer:** 1 + $$2\\alpha$$\n\n$$N{a_2}\\,S{O_4}\\,\\rightleftharpoons\\mathop {2N{a^ + }}\\limits_{2\\alpha } \\,\\, + \\,\\,\\mathop {SO_4^{ - - }}\\limits_\\alpha $$ \n
Vant. Hoff's factor $$\\,\\,\\,i = {{1 - \\alpha + 2\\alpha + \\alpha } \\over 1} = 1 + 2\\alpha $$ ",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 2645,
"subject": "Chemistry",
"question": "Among the following mixtures, dipole-dipole as the major interaction, is present in ",
"options": [
{
"text": "benzene and ethanol"
},
{
"text": "acetonitrile and acetone "
},
{
"text": "KCl and water"
},
{
"text": "benzene and carbon tetrachloride"
}
],
"answer": "acetonitrile and acetone ",
"solution": "**Answer:** acetonitrile and acetone \n\nAcetonitrile $$\\mathop {\\left( {C{H_3}} \\right.}\\limits^{\\delta + } \\,$$ $$\\,\\, - \\,\\,\\,C \\equiv \\,\\,$$ $$\\mathop {\\left. N \\right)}\\limits^{\\delta - } $$ and acetone\n
\n dipole-dipole interaction exist between them. Between $$KCl$$ and water ion-dipole interaction is found and in Benzene - ethanol,\n Benzene is non polar but ethanol is polar so it create dipole -\n induced dipole interaction Benzene$$-$$Carbon tetra chloride, both are non polar so one create instantaneous dipole and other gets induced by it. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2646,
"subject": "Chemistry",
"question": "If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous\nsolution, the change in freezing point of water (∆Tf), when 0.01 mol of sodium sulphate is dissolved\nin 1 kg of water, is (Kf = 1.86 K kg mol–1)",
"options": [
{
"text": "0.0372 K"
},
{
"text": "0.0558 K"
},
{
"text": "0.0744 K"
},
{
"text": "0.0186 K"
}
],
"answer": "0.0558 K",
"solution": "**Answer:** 0.0558 K\n\nSodium sulphate dissociates as \n
Total no. of moles $$(i)$$ $$ = 1 - \\alpha + x\\alpha + y\\alpha $$\n
$$i - 1 = x\\alpha + y\\alpha - \\alpha $$\n
$$ = \\alpha \\left( {x + y - 1} \\right)$$\n
$$\\therefore$$ $$\\,\\,\\,\\,\\,\\,\\,\\alpha = {{i - 1} \\over {\\left( {x + y - 1} \\right)}}$$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 2648,
"subject": "Chemistry",
"question": "Kf for water is 1.86K kg mol–1. If your automobile radiator holds 1.0 kg of water, how many grams of\nethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8oC ?",
"options": [
{
"text": "72 g"
},
{
"text": "93 g"
},
{
"text": "39 g"
},
{
"text": "27 g"
}
],
"answer": "93 g",
"solution": "**Answer:** 93 g\n\n$$\\Delta {T_f} = i \\times {K_f} \\times m$$\n
or $$\\,\\,\\,\\,\\,\\,x = {{2.8 \\times 62} \\over {1.86}} = 93gm$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2649,
"subject": "Chemistry",
"question": "Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and 0.125\nM Na3PO4(aq) at 25oC. Which statement is true about these solutions, assuming all salts to be strong electrolytes?",
"options": [
{
"text": "0.125 M Na3PO4(aq) has the highest osmotic pressure."
},
{
"text": "0.500 M C2H5OH(aq) has the highest osmotic pressure"
},
{
"text": "They all have the same osmotic pressure"
},
{
"text": "0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure."
}
],
"answer": "They all have the same osmotic pressure",
"solution": "**Answer:** They all have the same osmotic pressure\n\n$$\\pi = i\\,CRT$$\n
$$^\\pi {C_2}{H_3}OH$$\n
$$ = 1 \\times 0.500 \\times R \\times T = 0.5{\\mkern 1mu} RT$$\n
Since the osmotic pressure of all the given solutions is equal.\n
Hence all are isotonic solution. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2650,
"subject": "Chemistry",
"question": "An aqueous solution of a salt MX2 at certain temperature has a van’t Hoff factor of 2. The degree of dissociation for this solution of the salt is :",
"options": [
{
"text": "0.33"
},
{
"text": "0.50"
},
{
"text": "0.67"
},
{
"text": "0.80"
}
],
"answer": "0.50",
"solution": "**Answer:** 0.50\n\n
Let us assume that degree of dissociation is $$\\alpha$$.
",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 2651,
"subject": "Chemistry",
"question": "The freezing point of benzene decreases by 0.450C when 0.2 g of acetic acid is added to 20g of benzene. If\nacetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be:\n(Kf for benzene = 5.12 K kg mol–1) ",
"options": [
{
"text": "80.4 %"
},
{
"text": "74.6 %"
},
{
"text": "94.6 %"
},
{
"text": "64.6 %"
}
],
"answer": "94.6 %",
"solution": "**Answer:** 94.6 %\n\n$$\\Delta $$Tf = i $$ \\times $$ Kf $$ \\times $$ m\n
$$ \\therefore $$ % dissociation is 94.6%.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2652,
"subject": "Chemistry",
"question": "5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be \n3.82oC. If Na2SO4 is 81.5% ionised,\nthe value of x\n
(Kf for water=1.86oC kg mol−1) is approximately :\n
(molar mass of S = 32 g mol−1 and that of Na = 23 g mol−1)\n",
"options": [
{
"text": "15 g"
},
{
"text": "25 g"
},
{
"text": "45 g"
},
{
"text": "65 g"
}
],
"answer": "45 g",
"solution": "**Answer:** 45 g\n\n
\n
\n
\n
Na2SO4
\n
$$\\rightleftharpoons$$
\n
2Na+
\n
+
\n
SO4-2
\n
\n
\n
Initial
\n
1 mol
\n
\n
0
\n
\n
0
\n
\n
\n
After dissociation
\n
1 - x
\n
\n
2x
\n
\n
x
\n
\n
\n
$$\\therefore\\,\\,\\,$$ Total no of Moles after dissociation = 1 + 2x \n
$$ \\Rightarrow $$$$\\,\\,\\,$$ x = 45 gm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2653,
"subject": "Chemistry",
"question": "For 1 molal aqueous solution of the following compounds, which one will show the highest freezing\npoint?",
"options": [
{
"text": "[Co(H2O)3 Cl3].3H2O "
},
{
"text": "[Co(H2O)6] Cl3"
},
{
"text": "[Co(H2O)5 Cl] Cl2.H2O"
},
{
"text": "[Co(H2O)4 Cl2] Cl.2H2O"
}
],
"answer": "[Co(H2O)3 Cl3].3H2O ",
"solution": "**Answer:** [Co(H2O)3 Cl3].3H2O \n\nWe know, \n
$$\\Delta {T_f}\\, = \\,im{K_f}$$\n
Where\n
$$\\Delta {T_f}\\, = \\,$$ Depression in freezing point\n
i = no of ions or molecule\n
m = molality of solute\n
Kf = freezing point depression constant.\n
In this question m = 1 and Kf is constant.\n
So, $$\\Delta {T_f}\\,\\, \\propto \\,\\,i$$\n
Which solutions have more i, those solutions $$\\Delta {T_f}\\,$$ will be more.\n
Now,\n
Freezing point of a aqueous solution $${f_p} = \\, - \\,\\Delta {T_f}$$\n
So, those solutions have more $$\\Delta {T_f}$$ will have less freezing point \n
and $$\\Delta {T_f}$$ will be more when value of i is more.\n
Now, for,\n
[ Co (H2O)6] Cl3 $$\\rightleftharpoons$$ [ Co(H2O)6]3+ + 3Cl$$-$$\n
Value of i = 4 (no. of ions), 3 ions of Cl$$-$$ \n
and one ion of [ Co (H2O)6]3+ \n
[ Co (H2O)5 Cl ] Cl2 . H2O $$\\rightleftharpoons$$ [ Co (H2O)5 Cl ]2+ + 2Cl$$-$$ \n
Value of i = 3\n
[ Co (H2O)4 Cl2 ] Cl . 2H2O $$\\rightleftharpoons$$ [ Co (H2O)4 Cl2 ] + + Cl$$-$$\n
Value of i = 2\n
and [ Co(H2O)3 Cl ] . 3H2O do not show ionization.\n
So, i = 1.\n
As, [ Co (H2O)3 Cl ]. 3H2O have lowest value of i, \n
So, it's freezing point will be maximum.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2654,
"subject": "Chemistry",
"question": "Molecules of benzoic acid (C6H5COOH) dimerise in benzene. 'w' g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2K. If the percentage association of the acid to form dimmer in the solution is 80, then w is – (Its given that Kf = 5 K kg mol–1, Molar mass of benzoic acid = 122 g mol–1) ",
"options": [
{
"text": "1.5 g"
},
{
"text": "1.8 g"
},
{
"text": "1.0 g"
},
{
"text": "2.4 g"
}
],
"answer": "2.4 g",
"solution": "**Answer:** 2.4 g\n\n\n We know, \n
= 1.8",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 2657,
"subject": "Chemistry",
"question": "The freezing point of a diluted milk sample is found to be –0.2oC, while it should have been –0.5oC for pure milk. How much water has been added to pure milk to make the diluted sample?",
"options": [
{
"text": "1 cup of water to 2 cups of pure milk "
},
{
"text": "2 cups of water to 3 cups of pure milk"
},
{
"text": "3 cups of water to 2 cups of pure milk"
},
{
"text": "1 cup of water to 3 cups of pure milk"
}
],
"answer": "3 cups of water to 2 cups of pure milk",
"solution": "**Answer:** 3 cups of water to 2 cups of pure milk\n\nWe know,\n
Which means 3 cups of water has been added to 2 cups of pure milk.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2658,
"subject": "Chemistry",
"question": "How much amount of NaCl should be added\nto 600 g of water ($$\\rho $$ = 1.00 g/mL) to decrease\nthe freezing point of water to – 0.2 °C ?\n______. \n(The freezing point depression constant for water = 2K kg mol–1)",
"options": [],
"answer": "1.74to1.76",
"solution": "**Answer:** 1.74to1.76\n\n$$\\Delta $$Tf = 0.2o C\n
$$ \\Rightarrow $$ WNaCl = $${{58.5 \\times 600 \\times 0.2} \\over {4 \\times 1000}}$$ = 1.76 g",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2659,
"subject": "Chemistry",
"question": "The elevation of boiling point of 0.10 m\naqueous CrCl3.xNH3 solution is two times that\nof 0.05 m aqueous CaCl2 solution. The value of\nx is ______.\n
[Assume 100% ionisation of the complex and\nCaCl2, coordination number of Cr as 6, and that\nall NH3 molecules are present inside the\ncoordination sphere]",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nMolality of CaCl2 solution = 0.05 m\n
$$\\Delta $$Tb = i Kb m = 3 × Kb × 0.05 = 0.15 Kb\n
$$ \\Rightarrow $$ x = 5\n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2660,
"subject": "Chemistry",
"question": "When 9.45 g of CICH2COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of CICH2COOH is x $$ \\times $$ 10-3. The value of x is ________. (Rounded\noff to the nearest integer) \n[Kf(H20) = 1.86 K kg mol-1]",
"options": [],
"answer": "34.4",
"solution": "**Answer:** 34.4\n\n Here, $$i = {{Final\\,moles} \\over {Initial\\,moles}}$$
i = 1 $$-$$ $$\\alpha$$ + $$\\alpha$$ + $$\\alpha$$ $$\\Rightarrow$$ i = 1 + $$\\alpha$$
Formula used for freezing point; $$\\Delta$$Tf = i Kfm
Percentage of dissociation = 34.4%",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2661,
"subject": "Chemistry",
"question": "C6H6 freezes at 5.5$$^\\circ$$C. The temperature at which a solution of 10g of C4H10 in 200g of C6H6 freeze is __________ $$^\\circ$$C. (The molal freezing point depression constant of C6H6 is 5.12$$^\\circ$$C/m.)",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nPure solvent : C6H6(l)
Given,
$$T_f^o = 5.5^\\circ C$$
$${K_f} = 5.12^\\circ C/m \\Rightarrow m = 200g$$
$${m_{solute}} = 10g$$
Molar mass of solute $${C_4}{H_{10}} = 12 \\times 4 + 10 = 58$$
or $$T_f^1 \\approx 1^\\circ C$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2662,
"subject": "Chemistry",
"question": "1 molal aqueous solution of an electrolyte A2B3 is 60% ionised. The boiling point of the solution at 1 atm is _________ K. (Rounded off to the nearest integer)
[Given Kb for (H2O) = 0.52 K kg mol$$-$$1]",
"options": [],
"answer": "375",
"solution": "**Answer:** 375\n\n$${A_2}{B_3}\\buildrel {} \\over\n \\longrightarrow 2{A^{ + 3}} + 3{B^{ - 2}}$$
No. of ions = 2 + 3 = 5
i = 1 + (n $$-$$ 1) $$\\alpha $$
= 1 + (5 $$-$$ 1) $$\\times$$ 0.6
= 1 + 4 $$\\times$$ 0.6 = 1 + 2.4 = 3.4
$$\\Delta {T_b} = {K_b} \\times m \\times i = 0.52 \\times 1 \\times 3.4 = 1.768^\\circ $$ C
$$ = 375$$ K",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2663,
"subject": "Chemistry",
"question": "If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5 K rise in the boiling point of the solution is ____________ molal. (Rounded off to the nearest integer)
[Kb = 0.52 K kg mol$$-$$1]",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n
As AB is a binary electrolyte,
\n
$$\\therefore$$ AB $$\\rightleftharpoons$$ A+ + B$$-$$, n = 2
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2664,
"subject": "Chemistry",
"question": "224 mL of SO2(g) at 298 K and 1 atm is passed through 100 mL of 0.1 M NaOH solution. The non-volatile solute produced is dissolved in 36g of water. The lowering of vapour pressure of solution (assuming the solution in dilute) (P$$_{({H_2}O)}^o$$ $$-$$ 24 mm of Hg) is x $$\\times$$ 10$$-$$2 mm of Hg, the value of x is ___________. (Integer answer)",
"options": [],
"answer": "18TO24",
"solution": "**Answer:** 18TO24\n\nmoles of SO2 = $${{224} \\over {22400}}$$ = 0.01\n
moles of NaOH = molarity × volume (in litre)\n = 0.1 × 0.1\n = 0.01 moles\n
The balanced equation is\n
SO2 + 2NaOH $$ \\to $$ Na2SO3 + H2O\n
$$ \\therefore $$ Here NaOH is limiting Reagent.\n
Lowering in pressure ($$\\Delta $$P) = 0.18 mm of Hg\n
= 18 × 10–2 mm of Hg",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 2665,
"subject": "Chemistry",
"question": "When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be $$-$$0.93$$^\\circ$$C (Kf(H2O) = 1.86 K kg mol$$-$$1). The number (n) of benzoic acid molecules associated (assuming 100% association) is ___________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n$\\underset{\\text{Benzoic acid}}{n \\mathrm{PhCOOH}} \\stackrel{\\text { Association }}{\\longrightarrow}(\\mathrm{PhCOOH})_n$
\n$\\therefore$ Number of benzoic acid molecules associated, $n=2$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2666,
"subject": "Chemistry",
"question": "AB2 is 10% dissociated in water to A2+ and B$$-$$. The boiling point of a 10.0 molal aqueous solution of AB2 is __________$$^\\circ$$C. (Round off to the Nearest Integer).
[Given : Molal elevation constant of water Kb = 0.5 K kg mol$$-$$1 boiling point of pure water = 100$$^\\circ$$C]",
"options": [],
"answer": "106",
"solution": "**Answer:** 106\n\nAB2 $$ \\leftrightharpoons $$ A+ + 2B$$-$$
$$ \\Rightarrow $$ Tb = 106",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2667,
"subject": "Chemistry",
"question": "2 molal solution of a weak acid HA has a freezing point of 3.885$$^\\circ$$C. The degree of dissociation of this acid is ___________ $$\\times$$ 10$$-$$3. (Round off to the Nearest Integer).
[Given : Molal depression constant of water = 1.85 K kg mol$$-$$1 Freezing point of pure water = 0$$^\\circ$$ C]",
"options": [],
"answer": "50",
"solution": "**Answer:** 50\n\n$$\\Delta$$Tf = Kf (im)
$$ \\Rightarrow $$ 3.885 = i $$\\times$$ 1.85 $$\\times$$ 2
$$ \\Rightarrow $$ i = 1.05
Also, we know,
i = 1 + (n $$-$$ 1) $$\\alpha$$
here n = number of particle obtained upon the dissociation of one particle.
$$HA\\rightleftharpoons H^{+}+A^{-} $$
here from one particle HA we get two particle H+ and A$$-$$.
$$ \\therefore $$ n = 2
So, i = 1 + (2 $$-$$ 1)$$\\alpha$$
$$ \\Rightarrow $$ 1.05 = 1 + $$\\alpha$$
$$ \\Rightarrow $$ $$\\alpha$$ = 0.05 = 50 $$\\times$$ 10$$-$$3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2668,
"subject": "Chemistry",
"question": "A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52$$^\\circ$$C. The percentage association of A is __________. (Round off to the Nearest Integer).
[Use : Kb for water = 0.52 K kg mol$$-$$1 Boiling point of water = 100$$^\\circ$$C]",
"options": [],
"answer": "100",
"solution": "**Answer:** 100\n\n$$\\Delta$$Tb = Boiling point of the solution $$-$$ Boiling point of the pure solvent
= 100.52 $$-$$ 100
= 0.52
$$ \\therefore $$ $$\\Delta$$Tb = Kb (iM)
$$ \\Rightarrow $$ 0.52 = i $$\\times$$ 0.52 $$\\times$$ 2
$$ \\therefore $$ % association = 100",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2669,
"subject": "Chemistry",
"question": "Which one of the following 0.06 M aqueous solutions has lowest freezing point?",
"options": [
{
"text": "Al2(SO4)3"
},
{
"text": "C6H12O6"
},
{
"text": "KI"
},
{
"text": "K2SO4"
}
],
"answer": "Al2(SO4)3",
"solution": "**Answer:** Al2(SO4)3\n\nTf $$-$$ T'f = i Kf . m
For minimum T'f
'i' should be maximum
Al2(SO4)3 i = 5
C6H12O6 i = 1
KI i = 2
K2SO4 i = 3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2670,
"subject": "Chemistry",
"question": "In a solvent 50% of an acid HA dimerizes and the rest dissociates. The van't Hoff factor of the acid is __________ $$\\times$$ 10$$-$$2.
(Round off to the nearest integer)",
"options": [],
"answer": "125",
"solution": "**Answer:** 125\n\n Now, $$i = {{final\\,moles} \\over {initial\\,moles}} = {{0.25a + 0.5a + 0.5a} \\over {0.5a + 0.5a}}$$
= 1.25 = 125 $$\\times$$ 10$$-$$2",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 2671,
"subject": "Chemistry",
"question": "Of the following four aqueous solutions, total number of those solutions whose freezing point is lower than that of 0.10 M C2H5OH is __________ (Integer answer)
(i) 0.10 M Ba3(PO4)2
(ii) 0.10 M Na2SO4
(iii) 0.10 M KCl
(iv) 0.10 M Li3PO4",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nAs 0.1 M C2H5OH is non-dissociative and rest all salt given are electrolyte so in each case effective molarity > 0.1 so each will have lower freezing point.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2672,
"subject": "Chemistry",
"question": "1.22 g of an organic acid is separately dissolved in 100 g of benzene (Kb = 2.6 K kg mol$$-$$1) and 100 g of acetone (Kb = 1.7 K kg mol$$-$$1). The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by 0.17$$^\\circ$$C. The increase in boiling point of solution in benzene in $$^\\circ$$C is x $$\\times$$ 10$$-$$2. The value of x is ______________. (Nearest integer) [Atomic mass : C = 12.0, H = 1.0, O =16.0]",
"options": [],
"answer": "13",
"solution": "**Answer:** 13\n\nWith benzene as solvent
1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198$$^\\circ$$C. The percentage of dissociation of the acid is ___________. (Nearest integer)
\n
[Given : Density of acetic acid is 1.02 g mL$$-$$1, Molar mass of acetic acid is 60 g mol$$-$$1, Kf(H2O) = 1.85 K kg mol$$-$$1]
Elevation in boiling point for 1.5 molal solution of glucose in water is 4 K. The depression in freezing point for 4.5 molal solution of glucose in water is 4 K. The ratio of molal elevation constant to molal depression constant (Kb/Kf) is _________.
2.5 g of protein containing only glycine (C2H5NO2) is dissolved in water to make 500 mL of solution. The osmotic pressure of this solution at 300 K is found to be 5.03 $$\\times$$ 10$$-$$3 bar. The total number of glycine units present in the protein is ____________.
2 g of a non-volatile non-electrolyte solute is dissolved in \n 200 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 8. The elevation in boiling points of A and B are in the ratio $${x \\over y}$$ (x : y). The value of y is ______________. (Nearest integer)
",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\n$\\Delta T b=k b m$\n
A 0.5 percent solution of potassium chloride was found to freeze at $$-$$0.24$$^\\circ$$C. The percentage dissociation of potassium chloride is ______________. (Nearest integer)
\n
(Molal depression constant for water is 1.80 K kg mol$$-$$1 and molar mass of KCl is 74.6 g mol$$-$$1)
The osmotic pressure exerted by a solution prepared by dissolving 2.0 g of protein of molar mass 60 kg mol$$-$$1 in 200 mL of water at 27$$^\\circ$$C is ______________ Pa. [integer value]
\n
(use R = 0.083 L bar mol$$-$$1 K$$-$$1)
",
"options": [],
"answer": "415",
"solution": "**Answer:** 415\n\n$\\pi=\\mathrm{i} C R T \\quad(\\mathrm{i}=1)$\n
Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point by 0.2$$^\\circ$$C. The percentage association of solute A in water, is :
\n
[Given : Molar mass of A = 93 g mol$$-$$1. Molal depression constant of water is 1.86 K kg mol$$-$$1.]
\nHence, the percentage association of solute $A$ is $80 \\%$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2680,
"subject": "Chemistry",
"question": "
The depression in freezing point observed for a formic acid solution of concentration $$0.5 \\mathrm{~mL} \\mathrm{~L}^{-1}$$ is $$0.0405^{\\circ} \\mathrm{C}$$. Density of formic acid is $$1.05 \\mathrm{~g} \\mathrm{~mL}^{-1}$$. The Van't Hoff factor of the formic acid solution is nearly : (Given for water $$\\mathrm{k}_{\\mathrm{f}}=1.86\\, \\mathrm{k} \\,\\mathrm{kg}\\,\\mathrm{mol}^{-1}$$ )
Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y, respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is
$$20 \\%$$ of acetic acid is dissociated when its $$5 \\mathrm{~g}$$ is added to $$500 \\mathrm{~mL}$$ of water. The depression in freezing point of such water is _________ $$\\times 10^{-3}{ }^{\\circ} \\mathrm{C}$$.
\n
Atomic mass of $$\\mathrm{C}, \\mathrm{H}$$ and $$\\mathrm{O}$$ are 12,1 and 16 a.m.u. respectively.
\n
[Given : Molal depression constant and density of water are $$1.86 \\mathrm{~K} \\mathrm{~kg} \\mathrm{~mol}^{-1}$$ and $$1 \\mathrm{~g} \\mathrm{~cm}^{-3}$$ respectively.]
\nA. The elevation in boiling point temperature of water will be same for $0.1 \\mathrm{M} \\, \\mathrm{NaCl}$ and $0.1 \\mathrm{M}$ urea.\n
\nB. Azeotropic mixtures boil without change in their composition.\n
\nC. Osmosis always takes place from hypertonic to hypotonic solution.\n
\nD. The density of $32 \\% \\, \\mathrm{H}_{2} \\mathrm{SO}_{4}$ solution having molarity $4.09 ~\\mathrm{M}$ is approximately $1.26 \\mathrm{~g} \\mathrm{~mL}^{-1}$\n
\nE. A negatively charged sol is obtained when KI solution is added to silver nitrate solution.\n
\nChoose the correct answer from the options given below :",
"options": [
{
"text": "A, B and D only"
},
{
"text": "A and C only"
},
{
"text": "B and D only"
},
{
"text": "B, D and E only"
}
],
"answer": "B and D only",
"solution": "**Answer:** B and D only\n\n(A) Elevation in boiling point temperature of water will\nbe higher for 0.1 M NaCl as compared to 0.1 M\nurea.\n
(B) Azeotropic mixtures boil without change in their\ncomposition\n
(C) Osmosis always takes place from hypotonic (low\nconcentration of solute) solution to hypertonic\n(high concentration of solute) solution.\n
(E) When KI solution is added to AgNO3 solution, positively charged solution is formed due to adsorption of Ag+ ions from dispersion medium. \n
AgI/ Ag+ (positively charged) \n
(D) Let the mass of H2SO4 (32%) is 100.\n
$$\n\\therefore \\text { Weight of } \\mathrm{H}_2 \\mathrm{SO}_4=32\n$$\n
Moles of $\\mathrm{H}_2 \\mathrm{SO}_4=\\frac{32}{98}$\n\n
Hence, correct answer is (C) B and D only.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2684,
"subject": "Chemistry",
"question": "
25 mL of an aqueous solution of KCl was found to require 20 mL of 1 M $$\\mathrm{AgNO_3}$$ solution when titrated using $$\\mathrm{K_2CrO_4}$$ as an indicator. What is the depression in freezing point of KCl solution of the given concentration? _________ (Nearest integer).
\n
(Given : $$\\mathrm{K_f=2.0~K~kg~mol^{-1}}$$)
\n
Assume 1) 100% ionization and 2) density of the aqueous solution as 1 g mL$$^{-1}$$
$$\n\\begin{aligned}\nM & =\\frac{20}{25}=\\frac{4}{5}=0.8 \\\\\\\\\n\\Delta T_{f} & =(\\mathrm{i})\\left(\\mathrm{K}_{\\mathrm{f}}\\right)(\\mathrm{m}) \\\\\\\\\n& =(2)(2)\\left(\\frac{4}{5}\\right)=\\frac{16}{5}=3.2\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2685,
"subject": "Chemistry",
"question": "Lead storage battery contains $38 \\%$ by weight solution of $\\mathrm{H}_{2} \\mathrm{SO}_{4}$. The van't Hoff factor is $2.67$ at this concentration. The temperature in Kelvin at which the solution in the battery will freeze is ________. (Nearest integer).\n
Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15$$^\\circ$$C. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at $$-0.8^\\circ$$ C. The solubility product of PbCl$$_2$$ formed is __________ $$\\times$$ 10$$^{-6}$$ at 298 K. (Nearest integer)
\n
Given : $$\\mathrm{K_b=0.5}$$ K kg mol$$^{-1}$$ and $$\\mathrm{K_f=1.8}$$ K kg mol$$^{-1}$$. Assume molality to the equal to molarity in all cases.
\n$$\n\\begin{aligned}\n & \\pi_1=0.5 \\times 1 \\times R T =0.5 R T \\\\\\\\\n\\mathrm{KBr} & \\rightleftharpoons \\mathrm{K}^{+}+\\mathrm{Br}^{-} \\\\\\\\\n& \\pi_2=0.25 \\times 2 \\times R T=0.5 R T\n\\end{aligned}\n$$\n
Thus, $\\pi_1=\\pi_2$\n
\n(B) $0.100 \\mathrm{M} ~\\mathrm{K}_{4}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right](\\mathrm{aq}) \\mathrm{i}=5$ and $0.100 \\mathrm{M}$ $\\mathrm{FeSO}_{4}\\left(\\mathrm{NH}_{4}\\right)_{2}(\\mathrm{aq}) \\mathrm{i}=5$\n\n
$$\n\\begin{array}{ll}\n \\mathrm{K}_4\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right] \\rightleftharpoons 4 \\mathrm{~K}^{+}+\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]^{4-} (i=5) \\\\\\\\\n\\quad \\pi_1=0.1 \\times 5 \\times R T=0.5 R T & \\\\\\\\\n\\mathrm{FeSO}_4 \\cdot\\left(\\mathrm{NH}_4\\right)_2 \\mathrm{SO}_4 \\rightleftharpoons \\mathrm{Fe}^{2+}+2 \\mathrm{NH}_4^{+}+2 \\mathrm{SO}_4^{2-} & (i=5) \\\\\\\\\n\\pi_2=0.1 \\times 5 \\times R T=0.5 R T &\n\\end{array}\n$$\n
Thus, $\\pi_1=\\pi_2$\n
C. 0.05 $$\\mathrm{M~K_4[Fe(CN)_6]~(aq)}$$ and 0.25 $$\\mathrm{M~NaCl~(aq)}$$\n
$$\n\\begin{gathered}\n {K}_4\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right] \\rightleftharpoons 4 {~K}^{+}+\\left[\\mathrm{Fe}(\\mathrm{CN})_6\\right]^{4-} (i=5) \\\\\\\\\n \\pi_1=0.05 \\times 5 \\times R T=0.25 R T \\\\\\\\\n{NaCl} \\rightleftharpoons \\mathrm{Na}^{+}+\\mathrm{Cl}^{-} \\quad(i=2) \\\\\\\\\n\\pi_2=0.25 \\times 2 R T=0.5 R T\n\\end{gathered}\n$$\n
$$\n\\begin{aligned}\n& \\pi_1=5 \\times 0.02 R T=0.1 R T \\\\\\\\\n& 0.05 \\mathrm{M} \\mathrm{KCl}, \\mathrm{KCl} \\rightleftharpoons \\mathrm{K}^{+}+\\mathrm{Cl}^{-}(i=2) \\\\\\\\\n& \\pi_2=0.05 \\times 2 R T=0.1 R T\n\\end{aligned}\n$$\n
Solution of $$12 \\mathrm{~g}$$ of non-electrolyte (A) prepared by dissolving it in $$1000 \\mathrm{~mL}$$ of water exerts the same osmotic pressure as that of $$0.05 ~\\mathrm{M}$$ glucose solution at the same temperature. The empirical formula of $$\\mathrm{A}$$ is $$\\mathrm{CH}_{2} \\mathrm{O}$$. The molecular mass of $$\\mathrm{A}$$ is __________ g. (Nearest integer)
",
"options": [],
"answer": "240",
"solution": "**Answer:** 240\n\nTo solve this problem, we will first calculate the osmotic pressure of the 0.05 M glucose solution and then use that information to determine the molecular mass of compound A.\n
\n1. Osmotic pressure equation: \n
\n$$\\Pi = iMRT$$\n
\nwhere \n$$\\Pi$$ is the osmotic pressure,\n$$i$$ is the van't Hoff factor (which is 1 for non-electrolytes),\n$$M$$ is the molarity,\n$$R$$ is the ideal gas constant (0.0821 L atm/mol K), and\n$$T$$ is the temperature in Kelvin.\n
\nSince both solutions have the same osmotic pressure at the same temperature, we can set their osmotic pressures equal to each other:\n
\n$$\\Pi_{A} = \\Pi_{glucose}$$\n
\n2. Calculate the osmotic pressure of the 0.05 M glucose solution:\n
\nGlucose is a non-electrolyte, so its van't Hoff factor is 1. We don't know the temperature, but since both solutions are at the same temperature, it will cancel out in our calculations.\n
\nThe van't Hoff factors, ideal gas constant, and temperature cancel out:\n
\n$$\\frac{12}{x} = 0.05$$\n
\n5. Solve for the molecular mass (x) of compound A:\n
\n$$x = \\frac{12}{0.05}$$\n
\n$$x = 240 ~\\mathrm{g/mol}$$\n
\nThe molecular mass of compound A is 240 g/mol.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2690,
"subject": "Chemistry",
"question": "
0.004 M K$$_2$$SO$$_4$$ solution is isotonic with 0.01 M glucose solution. Percentage dissociation of K$$_2$$SO$$_4$$ is ___________ (Nearest integer)
",
"options": [],
"answer": "75",
"solution": "**Answer:** 75\n\nGiven that isotonic solutions have the same osmotic pressure, we equate the osmotic pressures of the K2SO4 and glucose solutions :\n\n
Here, $i$ is the van't Hoff factor, which accounts for the number of ions the compound dissociates into in the solution. Solving this equation for $i$, we get $i = 2.5$.\n\n
For the compound K2SO4, which dissociates into 3 ions (2K+ and 1 SO42-), the formula for $i$ is given by\n\n
$i = 1 + (n-1)\\alpha$\n\n
where $n$ is the number of ions the solute dissociates into and $\\alpha$ is the degree of dissociation.\n\n
Substituting $n=3$ and $i=2.5$ into this formula and solving for $\\alpha$ gives\n\n
To convert this into a percentage, we multiply by 100, yielding a percentage dissociation of 75%.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2691,
"subject": "Chemistry",
"question": "
If the degree of dissociation of aqueous solution of weak monobasic acid is determined to be 0.3, then the observed freezing point will be ___________% higher than the expected/theoretical freezing point. (Nearest integer)
The degree of dissociation, often represented as $\\alpha$, is the fraction of a mole of a substance that has dissociated into ions in solution. For a weak monobasic acid, this degree of dissociation can increase the number of particles in solution, which can in turn affect colligative properties such as the freezing point.
\n
In this case, a weak monobasic acid, when it dissociates, produces two particles: one $H^{+}$ ion and one anion. So if the degree of dissociation is 0.3 ($\\alpha = 0.3$), the average number of particles per molecule of the acid ($i$), also known as the van't Hoff factor, will be $1 + \\alpha = 1 + 0.3 = 1.3$.
\n
The decrease in freezing point ($\\Delta T_{f}$) is given by the formula $\\Delta T_{f} = i \\cdot m \\cdot K_{f}$, where $m$ is the molality of the solution and $K_{f}$ is the cryoscopic constant of the solvent. Therefore, the observed freezing point depression will be 1.3 times the theoretical freezing point depression for a non-dissociating solute (where $i = 1$).
\n
So, the observed freezing point will be 30% higher than the theoretical freezing point, given the degree of dissociation of 0.3.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2692,
"subject": "Chemistry",
"question": "We have three aqueous solutions of $\\mathrm{NaCl}$ labelled as ' $\\mathrm{A}$ ', ' $\\mathrm{B}$ ' and ' $\\mathrm{C}$ ' with concentration $0.1 \\mathrm{M}$, $0.01 \\mathrm{M}$ and $0.001 \\mathrm{M}$, respectively. The value of van 't Hoff factor(i) for these solutions will be in the order :",
"options": [
{
"text": "$\\mathrm{i}_{\\mathrm{A}}<\\mathrm{i}_{\\mathrm{C}}<\\mathrm{i}_{\\mathrm{B}}$"
},
{
"text": "$\\mathrm{i}_{\\mathrm{A}}<\\mathrm{i}_{\\mathrm{B}}<\\mathrm{i}_{\\mathrm{C}}$"
},
{
"text": "$\\mathrm{i}_{\\mathrm{A}}>\\mathrm{i}_{\\mathrm{B}}>\\mathrm{i}_{\\mathrm{C}}$"
},
{
"text": "$\\mathrm{i}_{\\mathrm{A}}=\\mathrm{i}_{\\mathrm{B}}=\\mathrm{i}_{\\mathrm{C}}$"
}
],
"answer": "$\\mathrm{i}_{\\mathrm{A}}<\\mathrm{i}_{\\mathrm{B}}<\\mathrm{i}_{\\mathrm{C}}$",
"solution": "**Answer:** $\\mathrm{i}_{\\mathrm{A}}<\\mathrm{i}_{\\mathrm{B}}<\\mathrm{i}_{\\mathrm{C}}$\n\n\n\n
\n
\n
Salt
\n
Values of i (for different conc. of a Salt)
\n
\n
\n
NaCl
\n
0.1 M
\n
0.01 M
\n
0.001 M
\n
\n
\n
\n
1.87
\n
1.94
\n
1.94
\n
\n
\n
The van 't Hoff factor (i) is used to describe the number of particles a solute formula unit produces in a solution. For an electrolyte like $\\mathrm{NaCl}$, which dissociates completely in very dilute solutions, the theoretical value of $i$ is approximately 2, since $\\mathrm{NaCl}$ dissociates into $\\mathrm{Na}^+$ and $\\mathrm{Cl}^-$ ions.
In real scenarios, as the concentration of the solution decreases (making the solution more dilute), the interaction between the ions decreases, allowing more complete dissociation. Therefore, for practical purposes, the van 't Hoff factor $i$ approaches its theoretical maximum value as concentration decreases. Thus, for $\\mathrm{NaCl}$ solutions of concentrations $0.1 \\mathrm{M}$, $0.01 \\mathrm{M}$, and $0.001 \\mathrm{M}$, the fact that $\\mathrm{NaCl}$ dissociates more completely in more dilute solutions implies that $i$ increases with decreasing concentration.
Hence, the order of $i$ based on the concentration would be $\\mathrm{i}_{\\mathrm{A}} < \\mathrm{i}_{\\mathrm{B}} < \\mathrm{i}_{\\mathrm{C}}$, as concentration $0.1 \\mathrm{M} > 0.01 \\mathrm{M} > 0.001 \\mathrm{M}$, respectively. Thus, option B correctly describes the order of the van 't Hoff factors for these solutions.
",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 2693,
"subject": "Chemistry",
"question": "Freezing point of an aqueous solution is (-0.186)oC. Elevation of boiling point of the same solution is Kb = 0.512 oC, Kf = 1.86 oC, find the increase in boiling point. ",
"options": [
{
"text": "0.186 oC"
},
{
"text": "0.0512 oC"
},
{
"text": "0.092 oC"
},
{
"text": "0.2732 oC"
}
],
"answer": "0.0512 oC",
"solution": "**Answer:** 0.0512 oC\n\n$$\\Delta {T_b} = {K_b}{{{W_B}} \\over {{M_B} \\times {W_A}}} \\times 1000;$$\n
$$ = {0.0512^ \\circ }C.$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2694,
"subject": "Chemistry",
"question": "Equimolar solutions in the same solvent have",
"options": [
{
"text": "Same boiling point but different freezing point"
},
{
"text": "Same freezing point but different boiling point "
},
{
"text": "Same boiling and same freezing points "
},
{
"text": "Different boiling and different freezing points"
}
],
"answer": "Same boiling and same freezing points ",
"solution": "**Answer:** Same boiling and same freezing points \n\nEquimolar solutions of normal solutes in the same solvent will have the same boiling point and same freezing point.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2695,
"subject": "Chemistry",
"question": "Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be\nadded to 4 kg of water to prevent it from freezing at −6oC will be :\n [Kf for water = 1.86 K kg mol−1 , and molar mass of ethylene glycol = 62 g mol−1 )",
"options": [
{
"text": "204.30 g"
},
{
"text": "400.00 g "
},
{
"text": "304.60 g"
},
{
"text": "804.32 g"
}
],
"answer": "804.32 g",
"solution": "**Answer:** 804.32 g\n\nGiven $${K_r} = 1.86\\,K\\,kg\\,mo{l^{ - 1}}$$\n
$$\\,\\,\\,\\,\\,w = 0.8\\,kg = 800\\,gm$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2696,
"subject": "Chemistry",
"question": "A solution containing 62 g ethylene glycol in 250 g water is cooled to $$-$$ 10oC. If Kf for water is 1.86 K kg mol$$-$$1 , the amount of water (in g) separated as ice is : ",
"options": [
{
"text": "48"
},
{
"text": "32"
},
{
"text": "64"
},
{
"text": "16"
}
],
"answer": "64",
"solution": "**Answer:** 64\n\nHere water is solvent and ethylene glycol is solute.\n
We know, freezing point of pure water (here solvent) is 0oC. Which is represented by $$T_f^0$$.\n
We also know, $$\\Delta {T_f} = T_f^0 - {T_f}$$\n
where $$T_f^0$$ = Freezing point of pure solvent and $${T_f}$$ = Freezing point of solution\n
$$ \\therefore $$ 7.44 = 0 - $${T_f}$$\n
$$ \\Rightarrow $$ $${T_f}$$ = -7.44oC\n
So, not a single drop of water solution will become ice until temperature reaches -7.44oC. When temperature decrease more than -7.44oC then some part of the solution starts becoming ice staring from surface of the solution.\n
Now let Wl gm of solution still stays in liquid phase when temperature reaches -10oC.\n
$${k_b} = 2{k_f}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2698,
"subject": "Chemistry",
"question": "Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is - ",
"options": [
{
"text": "4A"
},
{
"text": "2A"
},
{
"text": "3A"
},
{
"text": "A"
}
],
"answer": "3A",
"solution": "**Answer:** 3A\n\nFor same freezing point,\n
$$ \\Rightarrow $$ M = 3.27A $$ \\simeq $$ 3A ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2699,
"subject": "Chemistry",
"question": "83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is ____________ K. (Nearest integer)
[Use : Molal Freezing point depression constant of water = 1.86 K kg mol$$-$$1]
Freezing Point of water = 273 K
Atomic masses : C : 12.0 u, O : 16.0 u, H : 1.0 u]",
"options": [],
"answer": "269",
"solution": "**Answer:** 269\n\nkf = 1.86 kg/mol
$$ \\Rightarrow T_f^1 = 259$$ K",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2700,
"subject": "Chemistry",
"question": "1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to $$-$$4$$^\\circ$$C before freezing. The amount of ice (in g) that will be separated out is __________. (Nearest integer)
[Given : Kf(H2O) = 1.86 K kg mol$$-$$1]",
"options": [],
"answer": "518",
"solution": "**Answer:** 518\n\nLet mass of water initially present = x gm
$$\\Rightarrow$$ Mass of sucrose = (1000 $$-$$ x) gm
$$ \\Rightarrow 4 = {{0.5969} \\over a} \\times 1.86 \\Rightarrow $$ a = 0.2775 kg
$$\\Rightarrow$$ ice separated = (795.86 $$-$$ 277.5) = 518.3 gm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2701,
"subject": "Chemistry",
"question": "40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is __________ K. (Nearest integer) [Given : Kf = 1.86 K kg mol$$-$$1; Density of water = 1.00 g cm$$-$$3; Freezing point of water = 273.15 K]",
"options": [],
"answer": "271",
"solution": "**Answer:** 271\n\nMolality = $${{\\left( {{{40} \\over {180}}} \\right)mol} \\over {0.2\\,Kg}} = \\left( {{{10} \\over 9}} \\right)$$ molal
$$ \\simeq $$ 271 K (nearest - integer)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2702,
"subject": "Chemistry",
"question": "Which one of the following 0.10 M aqueous solutions will exhibit the largest freezing point depression?",
"options": [
{
"text": "hydrazine"
},
{
"text": "glucose"
},
{
"text": "glycine"
},
{
"text": "KHSO4"
}
],
"answer": "KHSO4",
"solution": "**Answer:** KHSO4\n\nΔTf $$ \\propto $$ i × m\n
greater the value of i, greater will be the ΔTf\n value.\n
Van't Hoff factor is highest for KHSO4
$$\\therefore$$ colligative property ($$\\Delta$$Tf) will be highest for KHSO4",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2703,
"subject": "Chemistry",
"question": "
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
\n
Assertion (A) : At 10$$^\\circ$$C, the density of a 5 M solution of KCl [atomic masses of K & Cl are 39 & 35.5 g mol$$-$$1 respectively], is 'x' g ml$$-$$1. The solution is cooled to $$-$$21$$^\\circ$$C. The molality of the solution will remain unchanged.
\n
Reason (R) : The molality of a solution does not change with temperature as mass remains unaffected with temperature.
\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\nMolality and Mass are temperature Independent so on changing temp., molality and mass remain unchanged.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2704,
"subject": "Chemistry",
"question": "
The elevation in boiling point for 1 molal solution of non-volatile solute A is $$3 \\mathrm{~K}$$. The depression in freezing point for 2 molal solution of $$\\mathrm{A}$$ in the same solvent is 6 $$K$$. The ratio of $$K_{b}$$ and $$K_{f}$$ i.e., $$K_{b} / K_{f}$$ is $$1: X$$. The value of $$X$$ is [nearest integer]
",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nMolality of a solution of non volatile solute $(A)=1$
$$150 \\mathrm{~g}$$ of acetic acid was contaminated with $$10.2 \\mathrm{~g}$$ ascorbic acid $$\\left(\\mathrm{C}_{6} \\mathrm{H}_{8} \\mathrm{O}_{6}\\right)$$ to lower down its freezing point by $$\\left(x \\times 10^{-1}\\right)^{\\circ} \\mathrm{C}$$. The value of $$x$$ is ___________. (Nearest integer)
\n
[Given $$\\mathrm{K}_{f}=3.9 \\mathrm{~K} \\mathrm{~kg} \\mathrm{~mol}^{-1}$$; molar mass of ascorbic acid $$=176 \\mathrm{~g} \\mathrm{~mol}^{-1}$$]
$$1.80 \\mathrm{~g}$$ of solute A was dissolved in $$62.5 \\mathrm{~cm}^{3}$$ of ethanol and freezing point of the solution was found to be $$155.1 \\mathrm{~K}$$. The molar mass of solute A is ________ g $$\\mathrm{mol}^{-1}$$.
\n
[Given : Freezing point of ethanol is 156.0 K.
\n
Density of ethanol is 0.80 g cm$$-$$3.
\n
Freezing point depression constant of ethanol is 2.00 K kg mol$$-$$1]
",
"options": [],
"answer": "80",
"solution": "**Answer:** 80\n\nMass of solvent $$=d x v=0.8 \\times 62.5=50$$ gram
A. Vapour pressure of the solution is less than that of pure solvent
\n
B. Vapour pressure of the solution is more than that of pure solvent
\n
C. Only solute molecules solidify at the freezing point
\n
D. Only solvent molecules solidify at the freezing point
\n
Choose the most appropriate answer from the options given below :
",
"options": [
{
"text": "A and C only"
},
{
"text": "A only"
},
{
"text": "A and D only"
},
{
"text": "B and C only"
}
],
"answer": "A and D only",
"solution": "**Answer:** A and D only\n\nIn the depression of freezing point experiment\nonly solvent molecules solidify and vapour pressure of\nsolution decreases as some of the surface area is occupied\nby solute molecules, so less number of molecules will go in\nvapour form.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2708,
"subject": "Chemistry",
"question": "Mass of ethylene glycol (antifreeze) to be added to $18.6 \\mathrm{~kg}$ of water to protect the freezing point at $-24^{\\circ} \\mathrm{C}$ is ________ $\\mathrm{kg}$ (Molar mass in $\\mathrm{g} ~\\mathrm{mol}^{-1}$ for ethylene glycol $62, \\mathrm{~K}_f$ of water $=1.86 \\mathrm{~K}$ $\\mathrm{kg} ~\\mathrm{mol}^{-1}$ )",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\n
To determine the mass of ethylene glycol (antifreeze) needed to lower the freezing point of water to $-24^{\\circ} \\mathrm{C}$, we can use the freezing point depression equation, which is a colligative property given by:
\n\n$$ \\Delta T_f = i \\cdot K_f \\cdot m $$\n\n
where:
\n
\n
$\\Delta T_f$ is the depression in the freezing point.
\n
$i$ is the van't Hoff factor, which is the number of particles the solute splits into in solution. For ethylene glycol, which does not dissociate in solution, $i = 1$.
\n
$K_f$ is the cryoscopic constant (freezing point depression constant) of water, which is given as $1.86 \\mathrm{~K} ~\\mathrm{kg} ~\\mathrm{mol}^{-1}$.
\n
$m$ is the molality of the solution.
\n
\n\n
However, molality is defined as the number of moles of solute per kilogram of solvent, so we first need to find the molality that corresponds to the desired freezing point depression:
Note that $\\mathrm{C}$ and $\\mathrm{K}$ are interchangeable when calculating changes in temperature. Now, we can compute the molality:
\n\n$$ m = \\frac{-24}{1.86} $$\n\n
Next, we calculate the molality of the ethylene glycol:
\n\n$$ m \\approx -12.903 \\mathrm{~mol} ~\\mathrm{kg}^{-1} $$\n\n
Keep in mind that molality is always positive, but the negative sign indicates the direction of the temperature change. Since we're calculating the amount needed, we can consider it as 12.903 moles per kilogram of water.
\n\n
Now we calculate the moles of ethylene glycol required for $18.6 \\mathrm{~kg}$ of water:
\n\n$$ \\text{moles of ethylene glycol} = m \\cdot \\text{mass of water (in kg)} $$\n\n$$ \\text{moles of ethylene glycol} = 12.903 \\mathrm{~mol} ~\\mathrm{kg}^{-1} \\cdot 18.6 \\mathrm{~kg} $$\n\n$$ \\text{moles of ethylene glycol} \\approx 240.0158 \\mathrm{~mol} $$\n\n
Next, we find the mass of ethylene glycol needed using its molar mass:
\n\n$$ \\text{mass of ethylene glycol} = \\text{moles of ethylene glycol} \\cdot \\text{molar mass of ethylene glycol} $$\n\n
Given the molar mass of ethylene glycol is $62 \\mathrm{~g} ~\\mathrm{mol}^{-1}$, we then convert it to $\\mathrm{kg} ~\\mathrm{mol}^{-1}$ by dividing by 1000:
\n\n$$ \\text{molar mass of ethylene glycol} = 0.062 \\mathrm{~kg} ~\\mathrm{mol}^{-1} $$\n\n
Therefore, approximately $14.881 \\mathrm{~kg}$ of ethylene glycol needs to be added to $18.6 \\mathrm{~kg}$ of water to lower the freezing point to $-24^{\\circ} \\mathrm{C}$.
As per NCERT, $$\\mathrm{k}_{\\mathrm{f}}{ }^{\\prime}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)=1.86 \\mathrm{~k} \\cdot \\mathrm{~kg} \\mathrm{~mol}^{-1}$$
$$2.7 \\mathrm{~kg}$$ of each of water and acetic acid are mixed. The freezing point of the solution will be $$-x^{\\circ} \\mathrm{C}$$. Consider the acetic acid does not dimerise in water, nor dissociates in water. $$x=$$ ________ (nearest integer)
\n
[Given: Molar mass of water $$=18 \\mathrm{~g} \\mathrm{~mol}^{-1}$$, acetic acid $$=60 \\mathrm{~g} \\mathrm{~mol}^{-1}$$
Considering acetic acid dissociates in water, its dissociation constant is $$6.25 \\times 10^{-5}$$. If $$5 \\mathrm{~mL}$$ of acetic acid is dissolved in 1 litre water, the solution will freeze at $$-x \\times 10^{-2}{ }^{\\circ} \\mathrm{C}$$, provided pure water freezes at $$0{ }^{\\circ} \\mathrm{C}$$.
\n
$$x=$$ _________. (Nearest integer)
\n
$$\\begin{aligned}\n\\text{Given :} \\quad & \\left(\\mathrm{K}_{\\mathrm{f}}\\right)_{\\text {water }}=1.86 \\mathrm{~K} \\mathrm{~kg} \\mathrm{~mol}-1 \\\\\n& \\text { density of acetic acid is } 1.2 \\mathrm{~g} \\mathrm{~mol}^{-1} \\text {. } \\\\\n& \\text { molar mass of water }=18 \\mathrm{~g} \\mathrm{~mol}^{-1} \\text {. } \\\\\n& \\text { molar mass of acetic acid= } 60 \\mathrm{~g} \\mathrm{~mol}^{-1} \\text {. } \\\\\n& \\text { density of water }=1 \\mathrm{~g} \\mathrm{~cm}^{-3}\n\\end{aligned}$$
\n
Acetic acid dissociates as $$\\mathrm{CH}_3 \\mathrm{COOH} \\rightleftharpoons \\mathrm{CH}_3 \\mathrm{COO}^{\\ominus}+\\mathrm{H}^{\\oplus}$$
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 2713,
"subject": "Chemistry",
"question": "A pressure cooker reduces cooking time for food because",
"options": [
{
"text": "boiling point of water involved in cooking is increased"
},
{
"text": "the higher pressure inside the cooker crushes the food material"
},
{
"text": "cooking involves chemical changes helped by a rise in temperature"
},
{
"text": "heat is more evenly distributed in the cooking space"
}
],
"answer": "boiling point of water involved in cooking is increased",
"solution": "**Answer:** boiling point of water involved in cooking is increased\n\nNOTE : On increasing pressure, the temperature is also increased. Thus in pressure cooker due to increase in pressure the $$b.p.$$ of water increases. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2714,
"subject": "Chemistry",
"question": "1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose\nebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points, $${{\\Delta {T_b}(A)} \\over {\\Delta {T_b}(B)}}$$, is :",
"options": [
{
"text": "5 : 1"
},
{
"text": "1 : 0.2"
},
{
"text": "10 : 1"
},
{
"text": "1 : 5"
}
],
"answer": "1 : 5",
"solution": "**Answer:** 1 : 5\n\nGiven,\n
$$\\Delta {T_b}\\left( A \\right) = {k_b}\\left( A \\right) \\times $$ Molality of A\n
$$\\Delta {T_b}\\left( B \\right) = {k_b}\\left( B \\right) \\times $$ Molality of B\n
Here, Molality of A = Molality of B = m\n
$${{\\Delta {T_b}\\left( A \\right)} \\over {\\Delta {T_b}\\left( B \\right)}} = {{{{\\left( {{k_b}} \\right)}_A}} \\over {{{\\left( {{k_b}} \\right)}_B}}} \\times {m \\over m}$$ = $${1 \\over 5}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2715,
"subject": "Chemistry",
"question": "A 1 molal K4Fe(CN)6 solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute A. The molar mass of A is __________ u. (Round off to the Nearest Integer). [Density of water = 1.0 g cm$$-$$3 ]",
"options": [],
"answer": "85",
"solution": "**Answer:** 85\n\n
\n
Effective molality = 0.6 + 1.6 + 0.4 = 2.6 m
\n
As elevation in boiling point is a colligative property which depends on the amount of solute. So, to have same boiling point, the molality of two solutions should be same.
\n
Molality of non-electrolyte solution = molality of $${K_4}[Fe{(CN)_6}]$$ = 2.6 m
\n
Now, 18.1 weight per cent solution means 18.1 g solute is present in 100 g solution and hence, (100 $$-$$ 18.1) = 81.9 g water.
where, M is the molar mass of non-electrolyte solute
\n
Molar mass of solute, M = 85
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2716,
"subject": "Chemistry",
"question": "When 3.00 g of a substance 'X' is dissolved in 100 g of CCl4, it raises the boiling point by 0.60 K. The molar mass of the substance 'X' is ______________ g mol$$-$$1. (Nearest integer).
[Given Kb for CCl4 is 5.0 K kg mol$$-$$1]",
"options": [],
"answer": "250",
"solution": "**Answer:** 250\n\n$$\\Delta$$Tb = Kb $$\\times$$ molality
A solution containing 2.5 $$\\times$$ 10$$-$$3 kg of a solute dissolved in 75 $$\\times$$ 10$$-$$3 kg of water boils at 373.535 K. The molar mass of the solute is ____________ g mol$$-$$1. [nearest integer] (Given : Kb(H2O) = 0.52 K kg mol$$-$$1 and boiling point of water = 373.15 K)
Boiling point of a $$2 \\%$$ aqueous solution of a non-volatile solute A is equal to the boiling point of $$8 \\%$$ aqueous solution of a non-volatile solute B. The relation between molecular weights of A and B is
A solution containing $$2 \\mathrm{~g}$$ of a non-volatile solute in $$20 \\mathrm{~g}$$ of water boils at $$373.52 \\mathrm{~K}$$. The molecular mass of the solute is ___________ $$\\mathrm{g} ~\\mathrm{mol}^{-1}$$. (Nearest integer)
\n
Given, water boils at $$373 \\mathrm{~K}, \\mathrm{~K}_{\\mathrm{b}}$$ for water $$=0.52 \\mathrm{~K} \\mathrm{~kg} \\mathrm{~mol}^{-1}$$
Sea water contains $$29.25 \\% ~\\mathrm{NaCl}$$ and $$19 \\% ~\\mathrm{MgCl}_{2}$$ by weight of solution. The normal boiling point of the sea water is _____________ $${ }^{\\circ} \\mathrm{C}$$ (Nearest integer)
\n
Assume $$100 \\%$$ ionization for both $$\\mathrm{NaCl}$$ and $$\\mathrm{MgCl}_{2}$$
\n
Given : $$\\mathrm{K}_{\\mathrm{b}}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)=0.52 \\mathrm{~K} \\mathrm{~kg} \\mathrm{~mol}^{-1}$$
\n
Molar mass of $$\\mathrm{NaCl}$$ and $$\\mathrm{MgCl}_{2}$$ is 58.5 and 95 $$\\mathrm{g} \\mathrm{~mol}^{-1}$$ respectively.
\nThe boiling point of the sea water is the normal boiling point of water plus the change in boiling point:\n
\nBoiling point of sea water = 100 °C + 16.075 °C = 116.075 °C\n
\nRounding to the nearest integer, the normal boiling point of the sea water is approximately 116 °C.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2721,
"subject": "Chemistry",
"question": "
If the boiling points of two solvents X and Y (having same molecular weights) are in the ratio $$2: 1$$ and their enthalpy of vaporizations are in the ratio $$1: 2$$, then the boiling point elevation constant of $$\\mathrm{X}$$ is $$\\underline{\\mathrm{m}}$$ times the boiling point elevation constant of Y. The value of m is ____________ (nearest integer)
",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nThe boiling point elevation constant, also known as the ebullioscopic constant ($K_{b}$), can be determined using the formula:\n
\n- $R$ is the universal gas constant (8.31 J mol⁻¹ K⁻¹)
\n- $T_{b}$ is the boiling point of the solvent (in K)
\n- $\\Delta H_{vap}$ is the enthalpy of vaporization (in J/mol)\n
\nAccording to the problem, the ratio of boiling points for X and Y is 2:1, and the ratio of their enthalpy of vaporization is 1:2. \n
\nLet's denote the boiling point of Y as $T_{b}(Y)$ and the boiling point of X as $T_{b}(X)$, and likewise for the enthalpy of vaporization $\\Delta H_{vap}(Y)$ and $\\Delta H_{vap}(X)$. \n
\nWe then have:\n
\n$T_{b}(X) = 2T_{b}(Y)$ and $\\Delta H_{vap}(X) = 0.5\\Delta H_{vap}(Y)$\n
\nWe can substitute these values into the equation for $K_{b}$:\n
\nSo, the boiling point elevation constant of X is 8 times the boiling point elevation constant of Y. Therefore, m = 8.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2722,
"subject": "Chemistry",
"question": "
A solution of two miscible liquids showing negative deviation from Raoult's law will have :\n
Negative deviation means the intermolecular forces of attraction between the molecules of the two liquids (A-B) are stronger than the forces between molecules of the pure liquids (A-A and B-B).
\n
This stronger attraction makes it harder for molecules to escape into the vapor phase.
\n
\n
Effect on Vapor Pressure and Boiling Point\n\n
\n
Vapor Pressure : Since the molecules are held more tightly, the vapor pressure of the solution will be lower than expected from Raoult's law. Decreased vapor pressure.
\n
Boiling Point : A lower vapor pressure means you need to increase the temperature further to reach atmospheric pressure, where boiling occurs. Therefore, the boiling point of the solution will be higher than expected. Increased boiling point.
\n
\n Answer\n\n
The correct answer is Option D: decreased vapor pressure, increased boiling point.
$$2.5 \\mathrm{~g}$$ of a non-volatile, non-electrolyte is dissolved in $$100 \\mathrm{~g}$$ of water at $$25^{\\circ} \\mathrm{C}$$. The solution showed a boiling point elevation by $$2^{\\circ} \\mathrm{C}$$. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is _________ $$\\mathrm{mm}$$ of $$\\mathrm{Hg}$$ (nearest integer)
\n
[Given : Molal boiling point elevation constant of water $\\left(\\mathrm{K}_{\\mathrm{b}}\\right)=0.52 \\mathrm{~K} . \\mathrm{kg} \\mathrm{mol}^{-1}$, $$1 \\mathrm{~atm}$$ pressure $$=760 \\mathrm{~mm}$$ of $$\\mathrm{Hg}$$, molar mass of water $$=18 \\mathrm{~g} \\mathrm{~mol}^{-1}]$$
A solution containing $$10 \\mathrm{~g}$$ of an electrolyte $$\\mathrm{AB}_2$$ in $$100 \\mathrm{~g}$$ of water boils at $$100.52^{\\circ} \\mathrm{C}$$. The degree of ionization of the electrolyte $$(\\alpha)$$ is _________ $$\\times 10^{-1}$$. (nearest integer)
\n
[Given : Molar mass of $$\\mathrm{AB}_2=200 \\mathrm{~g} \\mathrm{~mol}^{-1}, \\mathrm{~K}_{\\mathrm{b}}$$ (molal boiling point elevation const. of water) $$=0.52 \\mathrm{~K} \\mathrm{~kg} \\mathrm{~mol}^{-1}$$, boiling point of water $$=100^{\\circ} \\mathrm{C} ; \\mathrm{AB}_2$$ ionises as $$\\mathrm{AB}_2 \\rightarrow \\mathrm{A}^{2+}+2 \\mathrm{~B}^{-}]$$
To find the degree of ionization $$(\\alpha)$$, we need to use the boiling point elevation formula and the van't Hoff factor. The formulas we will use are:
\n\n
\n\n
$$\\Delta T_b = i \\cdot K_b \\cdot m$$
\n\n\n\n
where:\n\n
$$\\Delta T_b$$ is the boiling point elevation,
\n\n
$$i$$ is the van't Hoff factor,
\n\n
$$K_b$$ is the molal boiling point elevation constant, and
Next, using the formula for boiling point elevation:\n\n
\n\n
$$0.52 = i \\cdot 0.52 \\cdot 0.5$$
\n\n
$$i = \\frac{0.52}{0.52 \\cdot 0.5}$$
\n\n
$$i = \\frac{1}{0.5}$$
\n\n
$$i = 2$$
\n\n\n\n
Now, the van't Hoff factor $i$ is related to the degree of ionization $$(\\alpha)$$. For the electrolyte $$\\mathrm{AB}_2$$, which ionizes as $$\\mathrm{AB}_2 \\rightarrow \\mathrm{A}^{2+}+2\\mathrm{B}^{-}$$, the van't Hoff factor $i$ can be expressed as:\n\n
\n\n
\n\n
$$i = 1 + (n-1)\\alpha$$
\n\n\n\n
where\n\n
\n\n
$$n$$ is the number of ions produced (which is 3 for $$\\mathrm{AB}_2 \\rightarrow \\mathrm{A}^{2+}+2\\mathrm{B}^{-}$$)
\n\n
$$\\alpha$$ is the degree of ionization\n\n
\n\n\n\n
Substituting $i = 2$ into the equation:\n\n
\n\n
$$2 = 1 + 2 \\alpha$$
\n\n
$$2 - 1 = 2 \\alpha$$
\n\n
$$\\alpha = \\frac{1}{2}$$
\n\n
$$\\alpha = 0.5$$
\n\n\n\n
The degree of ionization $$(\\alpha)$$ is therefore\n\n
$$0.5 \\times 10^{-1} = 5 \\times 10^{-2}$$.
\n\n\n\n
Hence, the degree of ionization of the electrolyte $$(\\alpha)$$ is approximately 5 $$\\times 10^{-1}$$ (the nearest integer is 5).
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2725,
"subject": "Chemistry",
"question": "The solubility of N2 in water at 300 K and 500 torr partial pressure is 0.01 g L−1. The solubility (in g L−1) at 750 torr partial pressure is :",
"options": [
{
"text": "0.0075"
},
{
"text": "0.015"
},
{
"text": "0.02"
},
{
"text": "0.005"
}
],
"answer": "0.015",
"solution": "**Answer:** 0.015\n\nPartial pressure = Mole fraction × Solubility\n
$$ \\Rightarrow $$ s2 = 0.015 g L-1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2726,
"subject": "Chemistry",
"question": "Which one of the following statements regarding Henry's law is not correct ? ",
"options": [
{
"text": "Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids"
},
{
"text": "Different gases have different KH (Henry's law constant) values at the same temperature."
},
{
"text": "The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution. "
},
{
"text": "The value of KH increases with increase of temperature and KH is function of the nature of the gas "
}
],
"answer": "Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids",
"solution": "**Answer:** Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids\n\nFrom Henry's law we know, \n
Pgas = KH xg\n
Where, Pgas = Pressure of undissolved gas \n
xg = Mole fraction of gas dissolved into the liquid. \n
KH = Henry's Carnot. \n
When pressure is constant then,\n
xg $$ \\propto $$ $${1 \\over {K{}_H}}$$\n
So. when KH is high then xg or solubility of gas is lower in the liquid.\n
So, option (A) is wrong.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2727,
"subject": "Chemistry",
"question": "The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is __________ $$\\times$$ 10$$-$$5 mol dm$$-$$3. (Round off to the Nearest Integer).
[Given : Henry's law constant = KH = 8.0 $$\\times$$ 104 kPa for O2. Density of water with dissolved oxygen = 1.0 kg dm$$-$$3 ]",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\nPartial pressure (Pg) $$ \\propto $$ Solubility\n
Solubility = $${{20 \\times {{10}^3}} \\over {8.0 \\times {{10}^7}}}$$ = 25 $$ \\times $$ 10-5",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2728,
"subject": "Chemistry",
"question": "CO2 gas is bubbled through water during a soft drink manufacturing process at 298 K. If CO2 exerts a partial pressure of 0.835 bar then x m mol of CO2 would dissolve in 0.9 L of water. The value of x is ____________. (Nearest integer)
(Henry's law constant for CO2 at 298 K is 1.67 $$\\times$$ 103 bar)",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\nFrom Henry's law
\nMass of $\\mathrm{CO}_{2}$ in $1 \\mathrm{~L}$ water $=1221 \\times 10^{-3} \\mathrm{~g}$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2730,
"subject": "Chemistry",
"question": "
If $$\\mathrm{O}_{2}$$ gas is bubbled through water at $$303 \\mathrm{~K}$$, the number of millimoles of $$\\mathrm{O}_{2}$$ gas that dissolve in 1 litre of water is __________. (Nearest Integer)
\n
(Given : Henry's Law constant for $$\\mathrm{O}_{2}$$ at $$303 \\mathrm{~K}$$ is $$46.82 \\,\\mathrm{k}$$ bar and partial pressure of $$\\mathrm{O}_{2}=0.920$$ bar)
\n
(Assume solubility of $$\\mathrm{O}_{2}$$ in water is too small, nearly negligible)
\n$$\\Rightarrow \\mathrm{m}\\, \\mathrm{mol}\\, \\mathrm{of}\\, \\mathrm{O}_{2}=1$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2731,
"subject": "Chemistry",
"question": "A 5.25 % solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol−1) in\nthe same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm−3, molar mass of the substance will be ",
"options": [
{
"text": "90.0 g mol−1\n "
},
{
"text": "115.0 g mol−1"
},
{
"text": "105.0 g mol−1\n "
},
{
"text": "210.0 g mol−1"
}
],
"answer": "210.0 g mol−1",
"solution": "**Answer:** 210.0 g mol−1\n\nTIPS/FORMULAE: \n
Osmotic pressure $$\\left( \\pi \\right)$$ of isotonic solutions are equal.\n
For solution of unknown substance $$\\left( {\\pi = CRT} \\right)$$\n
$$\\therefore$$ $$\\,\\,\\,\\,\\,{C_1}RT = {C_2}RT\\,\\,$$ or $$\\,\\,{C_1} = {C_2}$$\n
or $$\\,\\,\\,\\,{{5.25/M} \\over V} = {{1.8/60} \\over V}$$\n
$$\\therefore$$ $$\\,\\,\\,\\,\\,M = 210g/mol$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2732,
"subject": "Chemistry",
"question": "The osmotic pressure of a dilute solution of an\nionic compound XY in water is four times that\nof a solution of 0.01 M BaCl2 in water.\nAssuming complete dissociation of the given\nionic compounds in water, the concentration of\nXY (in mol L–1) in solution is :",
"options": [
{
"text": "4 × 10–4"
},
{
"text": "6 × 10–2"
},
{
"text": "4 × 10–2"
},
{
"text": "16 × 10–4"
}
],
"answer": "6 × 10–2",
"solution": "**Answer:** 6 × 10–2\n\nGiven that, osmotic pressure of XY($${\\pi _{XY}}$$) in water is four times that\nof a solution of 0.01 M BaCl2($${\\pi _{BaC{l_2}}}$$).\n
$$ \\therefore $$ Osmotic pressure ($$\\pi $$) = (0.2)$$ \\times $$0.08206$$ \\times $$300 = 4.92 atm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2734,
"subject": "Chemistry",
"question": "The size of a raw mango shrinks to a much\nsmaller size when kept in a concentrated salt\nsolution. Which one of the following processes\ncan explain this?",
"options": [
{
"text": "Osmosis"
},
{
"text": "Reverse osmosis"
},
{
"text": "Diffusion"
},
{
"text": "Dialysis"
}
],
"answer": "Osmosis",
"solution": "**Answer:** Osmosis\n\nRaw mango shrink in salt solution due to net transfer of water molecules from mango to salt solution due to\nphenomenon of osmosis.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2735,
"subject": "Chemistry",
"question": "If 250 cm3\n of an aqueous solution containing 0.73 g of a protein A is isotonic with one litre of\nanother aqueous solution containing 1.65 g of a protein B, at 298 K, the ratio of the molecular\nmasses of A and B is ______ × 10–2 (to the nearest integer).",
"options": [],
"answer": "177",
"solution": "**Answer:** 177\n\nLet molar mass of protein A = x g/mol\n
Let molar mass of protein B = y g/mol\n
$$\\pi $$A = osmotic pressure of protein A = $${{0.73} \\over x} \\times {{1000} \\over {250}} \\times RT$$\n
$$\\pi $$B = osmotic pressure of protein B = $${{1.65} \\over y} \\times {1 \\over 1} \\times RT$$\n
= 177 $$ \\times $$ 10-2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2736,
"subject": "Chemistry",
"question": "The osmotic pressure of a solution of NaCl is\n0.10 atm and that of a glucose solution is\n0.20 atm. The osmotic pressure of a solution\nformed by mixing 1 L of the sodium chloride\nsolution with 2 L of the glucose solution is\nx $$ \\times $$ 10–3 atm. x is _____. (nearest integer)",
"options": [],
"answer": "167",
"solution": "**Answer:** 167\n\nFor NaCl : $$\\pi $$1 = 0.1 atm, v1 = 1 L\n
= 167 $$ \\times $$ 10-3 atm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2737,
"subject": "Chemistry",
"question": "1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 $$\\times$$ 10$$-$$3 bar.
The molar mass of the biopolymer is _____________ $$\\times$$ 104 g mol$$-$$1. (Round off to the Nearest Integer)
[Use : R = 0.083 L bar mol$$-$$1 K$$-$$1]",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\n$$\\pi$$ = CRT;
The osmotic pressure of blood is 7.47 bar at 300 K. To inject glucose to a patient intravenously, it has to be isotonic with blood. The concentration of glucose solution in gL$$-$$1 is _____________.
\n
(Molar mass of glucose = 180 g mol$$-$$1, R = 0.083 L bar K$$-$$1 mol$$-$$1) (Nearest integer)
At $$27^{\\circ} \\mathrm{C}$$, a solution containing $$2.5 \\mathrm{~g}$$ of solute in $$250.0 \\mathrm{~mL}$$ of solution exerts an osmotic pressure of $$400 \\mathrm{~Pa}$$. The molar mass of the solute is ___________ $$\\mathrm{g} \\mathrm{~mol}^{-1}$$ (Nearest integer)
An aqueous solution of volume $$300 \\mathrm{~cm}^{3}$$ contains $$0.63 \\mathrm{~g}$$ of protein. The osmotic pressure of the solution at $$300 \\mathrm{~K}$$ is 1.29 mbar. The molar mass of the protein is ___________ $$\\mathrm{g} ~\\mathrm{mol}^{-1}$$
The concept we are utilizing to solve this problem is osmotic pressure. Osmotic pressure is the pressure required to stop the flow of solvent into a solution through a semipermeable membrane. For dilute solutions, osmotic pressure behaves similarly to an ideal gas, hence the formula we use is similar to the ideal gas law:
\n
$$\\Pi = \\frac{n}{V}RT$$
\n
where:
\n
\n
$\\Pi$ is the osmotic pressure,
\n
$n$ is the number of moles of solute,
\n
$V$ is the volume of the solution,
\n
$R$ is the ideal gas constant, and
\n
$T$ is the temperature in Kelvin.
\n
\n
First, we convert all given quantities to the appropriate units.
\n
\n
$\\Pi = 1.29$ mbar is equivalent to $1.29 \\times 10^{-3}$ bar (since 1 bar = 1000 mbar),
\n
$V = 300$ cm³ is equivalent to $0.3$ L (since 1 L = 1000 cm³),
\n
$T = 300$ K,
\n
$R = 0.083$ L bar K$^{-1}$ mol$^{-1}$.
\n
\n
Instead of the number of moles, we are given the mass of the solute. However, we can replace the moles with mass using the relationship $n = \\frac{m}{M}$, where $M$ is the molar mass, and $m$ is the mass of the solute.
\n
So the equation becomes:
\n
$$\\Pi = \\frac{m}{MV}RT$$
\n
We are trying to solve for $M$, so rearrange the equation to isolate $M$:
\n
$$M = \\frac{mRT}{\\Pi V}$$
\n
Finally, substitute the given values into the equation:
To find the number of pairs of solutions that are isotonic with each other, we need to consider the van 't Hoff factor ($i$) which represents the number of particles a substance dissociates into in solution. We then use this to find the osmotic pressure of each solution in the pairs, using the formula:
\n
$ \\Pi = i \\cdot C \\cdot R \\cdot T $
\n
where:
\n
\n
$ \\Pi $ is the osmotic pressure
\n
$ i $ is the van 't Hoff factor
\n
$ C $ is the concentration of the solution
\n
$ R $ is the ideal gas constant
\n
$ T $ is the temperature in Kelvin
\n
\n
Two solutions are isotonic if they have the same osmotic pressure. Therefore, the $i \\cdot C$ products for each solution in the pair should be equal.
\n
The van 't Hoff factors for the substances given are:
\n
\n
$ \\text{NaCl} $: 2 (NaCl dissociates into Na⁺ and Cl⁻)
\n
$ \\text{Urea (NH}_2\\text{CONH}_2\\text{)} $: 1 (Urea does not dissociate)
\n
$ \\text{CaCl}_2 $: 3 (CaCl₂ dissociates into Ca²⁺ and 2 Cl⁻)
\n
$ \\text{KCl} $: 2 (KCl dissociates into K⁺ and Cl⁻)
\n
$ \\text{AlCl}_3 $: 4 (AlCl₃ dissociates into Al³⁺ and 3 Cl⁻)
\n
$ \\text{Na}_2\\text{SO}_4 $: 3 (Na₂SO₄ dissociates into 2 Na⁺ and SO₄²⁻)
\n
$ \\text{Al}_2(\\text{SO}_4)_3 $: 5 (Al₂(SO₄)₃ dissociates into 2 Al³⁺ and 3 SO₄²⁻)
\n
\n
Now, we will compare the $i \\cdot C$ products for each pair:
\n\n
Pair A: $2 \\cdot 1 = 2$ and $1 \\cdot 2 = 2$, so these are isotonic.
\n
Pair B: $3 \\cdot 1 = 3$ and $2 \\cdot 1.5 = 3$, so these are isotonic.
\n
Pair C: $4 \\cdot 1.5 = 6$ and $3 \\cdot 2 = 6$, so these are isotonic.
\n
Pair D: $2 \\cdot 2.5 = 5$ and $5 \\cdot 1 = 5$, so these are isotonic.
\n\n
Therefore, all four pairs of solutions are isotonic at the same temperature. The number of pairs of solutions is 4.
The osmotic pressure of a dilute solution is $$7 \\times 10^5 \\mathrm{~Pa}$$ at $$273 \\mathrm{~K}$$. Osmotic pressure of the same solution at $$283 \\mathrm{~K}$$ is _________ $$\\times 10^4 \\mathrm{Nm}^{-2}$$.
An artificial cell is made by encapsulating $$0.2 \\mathrm{~M}$$ glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a $$0.05 \\mathrm{~M}$$ solution of $$\\mathrm{NaCl}$$ at $$300 \\mathrm{~K}$$ is ________ $$\\times 10^{-1}$$ bar. (nearest integer).
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2744,
"subject": "Chemistry",
"question": "In a mixture of A and B, components show negative deviation when :",
"options": [
{
"text": "$$\\Delta V_{mix}$$ > 0, $$\\Delta S_{mix}$$ > 0"
},
{
"text": "$$\\Delta V_{mix}$$ = 0, $$\\Delta S_{mix}$$ > 0"
},
{
"text": "A - B interaction is weaker than A - A and B - B interaction"
},
{
"text": "A - B interaction is stronger than A - A and B - B interaction"
}
],
"answer": "A - B interaction is stronger than A - A and B - B interaction",
"solution": "**Answer:** A - B interaction is stronger than A - A and B - B interaction\n\n
In the context of solutions and mixtures, a negative deviation from Raoult's law occurs when the interaction between different molecules (A-B interaction) is stronger than the interaction between similar molecules (A-A and B-B interactions). This is because the molecules prefer to interact with each other rather than themselves, leading to a decrease in the overall vapor pressure compared to what would be expected from Raoult's law.
\n
So, the correct option is :
\n
Option D : A - B interaction is stronger than A - A and B - B interaction
\n\n
Option A suggests that the change in volume ($\\Delta V_{mix}$) and change in entropy ($\\Delta S_{mix}$) of the mixture are both greater than zero. This doesn't necessarily indicate a negative deviation from Raoult's law. It's possible for these conditions to be true in both positive and negative deviations or even in ideal solutions. Therefore, this option isn't specifically associated with negative deviations.
\n
Option B suggests that there's no change in the volume of the mixture ($\\Delta V_{mix} = 0$) and the change in entropy ($\\Delta S_{mix}$) is greater than zero. Again, this doesn't specifically indicate a negative deviation from Raoult's law. The volume change upon mixing can be zero, positive, or negative depending on the nature of the interactions among the molecules.
\n
Raoult's law deviations are primarily dictated by the relative strength of the intermolecular forces among the different components in the mixture. Thus, the change in volume and entropy, while they can be related, do not directly determine whether a solution will show a positive or negative deviation.
\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2745,
"subject": "Chemistry",
"question": "Which of the following liquid pairs shows a positive deviation from Raoult’s law? ",
"options": [
{
"text": "Water – hydrochloric acid "
},
{
"text": "Acetone – chloroform"
},
{
"text": "Water – nitric acid "
},
{
"text": "Benzene – methanol "
}
],
"answer": "Benzene – methanol ",
"solution": "**Answer:** Benzene – methanol \n\nNOTE : Positive deviations are shown by such solutions in which solvent-solvent and solute-solute interactions are stronger than the solvent interactions. In such solution, the interactions among molecules becomes weaker. Therefore their escaping tendency increases which results in the increase in their partial vapour pressures. \n
In a solutions of benzene and methanol there exists inter molecular $$H-$$ bonding.\n
\n In this solution benzene molecules come between ethanol molecules which weaken intermolecular forces. This results in increase in vapour pressure. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2746,
"subject": "Chemistry",
"question": "For which of the following parameters the structural isomers C2H5OH and CH3OCH3 would\nbe expected to have the same values? (Assume ideal behaviour) ",
"options": [
{
"text": "Heat of vaporization "
},
{
"text": "Gaseous densities at the same temperature and pressure"
},
{
"text": "Boiling points"
},
{
"text": "Vapour pressure at the same temperature "
}
],
"answer": "Gaseous densities at the same temperature and pressure",
"solution": "**Answer:** Gaseous densities at the same temperature and pressure\n\nGaseous densities of ethanol and dimethyl ether would be same at same temperature and pressure. The heat of vaporisation , $$V.P.$$ and $$b.pts$$ will differ due to $$H$$-bonding in ethanol. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2747,
"subject": "Chemistry",
"question": "Benzene and toluene form nearly ideal solutions. At 20 oC, the vapour pressure of\nbenzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of\nbenzene at 20 oC for a solution containing 78 g of benzene and 46 g of toluene in\ntorr is ",
"options": [
{
"text": "50"
},
{
"text": "25"
},
{
"text": "53.5"
},
{
"text": "37.5"
}
],
"answer": "50",
"solution": "**Answer:** 50\n\nGiven, Vapour pressure of benzene $$ = 75\\,\\,torr$$\n
$$ = 752.4\\,\\,$$ Torr",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2749,
"subject": "Chemistry",
"question": "A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour\npressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure\n(in mm) at the same temperature will be ",
"options": [
{
"text": "360"
},
{
"text": "350"
},
{
"text": "300"
},
{
"text": "700"
}
],
"answer": "350",
"solution": "**Answer:** 350\n\n$$P_A^ \\circ = ?,\\,\\,$$ \n
$$\\therefore$$ $$\\,\\,\\,\\,\\,\\,\\,\\,$$ $$P_A^ \\circ = 350mm$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2750,
"subject": "Chemistry",
"question": "At 80oC, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture solution of ‘A’ and ‘B’ boils at 80oC and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg) ",
"options": [
{
"text": "52 mol percent "
},
{
"text": "34 mol percent "
},
{
"text": "48 mol percent"
},
{
"text": "50 mol percent"
}
],
"answer": "50 mol percent",
"solution": "**Answer:** 50 mol percent\n\nAt $$1$$ atmospheric pressure the boiling point of mixture is $${80^ \\circ }C.$$\n
At boiling point the vapour pressure of mixture, $${P_T} = 1$$\n
or $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$50$$ mol. percent\n
i.e., The correct answer is $$(d)$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2751,
"subject": "Chemistry",
"question": "The vapour pressure of water at 20oC is 17.5 mm Hg. If 18 g of glucose (C6H12O6) is added to 178.2 g of water at 20oC, the vapour pressure of the resulting solution will be ",
"options": [
{
"text": "17.675 mm Hg "
},
{
"text": "15.750 mm Hg "
},
{
"text": "16.500 mm Hg "
},
{
"text": "17.325 mm Hg"
}
],
"answer": "17.325 mm Hg",
"solution": "**Answer:** 17.325 mm Hg\n\nNOTE : On addition of glucose to water, vapour pressure of water will decrease. The vapour pressure of a solution of glucose in water can be calculated using the relation\n
or $$\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $${P_S} = 17.325\\,mm\\,Hg.$$\n
Hence (d) is correct answer.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2752,
"subject": "Chemistry",
"question": "A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following\nstatements is correct regarding the behaviour of the solution ? ",
"options": [
{
"text": "The solution formed is an ideal solution "
},
{
"text": "The solution is non-ideal, showing +ve deviation from Raoult’s law. "
},
{
"text": "The solution is non-ideal, showing –ve deviation from Raoult’s law. "
},
{
"text": "n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s law. "
}
],
"answer": "The solution is non-ideal, showing +ve deviation from Raoult’s law. ",
"solution": "**Answer:** The solution is non-ideal, showing +ve deviation from Raoult’s law. \n\nFor this solution intermolecular interactions between $$n$$-heptane and ethanol aare weaker than $$n$$-heptane -$$n$$- heptane & ethanol-ethanol interactions hence the solution of $$n$$-heptane and ethanol is non-ideal and shows positive deviation from Raoult's law.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2753,
"subject": "Chemistry",
"question": "Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol\nof X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this\nsolution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of X\nand Y in their pure states will be, respectively : ",
"options": [
{
"text": "200 and 300 "
},
{
"text": "300 and 400"
},
{
"text": "400 and 600 "
},
{
"text": "500 and 600 "
}
],
"answer": "400 and 600 ",
"solution": "**Answer:** 400 and 600 \n\n$${P_{total}} = P_A^ \\circ {X_A} + P_B^ \\circ {X_B};$$\n
As $$\\,\\,\\,\\,\\,\\,\\,\\,$$ $$P_A^ \\circ = 400$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2754,
"subject": "Chemistry",
"question": "On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two\nliquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of\nthe solution obtained by mixing 25.0g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol–1 and of octane = 114 g mol–1)",
"options": [
{
"text": "72.0 kPa"
},
{
"text": "36.1 kPa"
},
{
"text": "96.2 kPa"
},
{
"text": "144.5 kPa"
}
],
"answer": "72.0 kPa",
"solution": "**Answer:** 72.0 kPa\n\n$${P_{Total}} = {P_A}^ \\circ {x_A} + {P_B}^ \\circ {x_B}$$\n
$$ = {{26.25 + 13.5} \\over {0.55}} = 72\\,kPa$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2755,
"subject": "Chemistry",
"question": "The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in\n100 g of acetone at 20oC, its vapour pressure was 183 torr. The molar mass (g mol-1) of the substance is:",
"options": [
{
"text": "64"
},
{
"text": "128"
},
{
"text": "488"
},
{
"text": "32"
}
],
"answer": "64",
"solution": "**Answer:** 64\n\nUsing relation, \n
$$\\therefore$$ $$\\,\\,\\,\\,\\,$$ Molar mass of substance $$=64$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2756,
"subject": "Chemistry",
"question": "18 g glucose (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is :",
"options": [
{
"text": "76.0 "
},
{
"text": "752.4"
},
{
"text": "759.0"
},
{
"text": "7.6"
}
],
"answer": "752.4",
"solution": "**Answer:** 752.4\n\nAccording to Raoult's Law\n
$${P_s} = 752.4\\,\\,torr$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2757,
"subject": "Chemistry",
"question": "A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 . If vapour pressure of CH2Cl2\n and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is : (Molar mass of Cl = 35.5 g mol−1)\n",
"options": [
{
"text": "0.162"
},
{
"text": "0.675"
},
{
"text": "0.325"
},
{
"text": "0.486"
}
],
"answer": "0.325",
"solution": "**Answer:** 0.325\n\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2758,
"subject": "Chemistry",
"question": "Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are Mx and My, respectively where Mx = $${3 \\over 4}$$ My. The relative lowering of vapor pressure of the solution in X is ''m'' times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of the solvent, the value of ''m'' is : ",
"options": [
{
"text": "$${4 \\over 3}$$"
},
{
"text": "$${3 \\over 4}$$"
},
{
"text": "$${1 \\over 2}$$"
},
{
"text": "$${1 \\over 4}$$"
}
],
"answer": "$${3 \\over 4}$$",
"solution": "**Answer:** $${3 \\over 4}$$\n\nRelative lowering of vapour pressure, \n
$$ \\Rightarrow $$ $$\\,\\,\\,\\,$$ $$\\omega $$ = 150 g",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2760,
"subject": "Chemistry",
"question": "At room temperature, a dilute solution of urea is prepared by dissolving 0.60 of urea in 360 g of water. If the\nvapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be.\n(molar mass of urea = 60 g mol–1)",
"options": [
{
"text": "0.031 mmHg"
},
{
"text": "0.017 mmHg"
},
{
"text": "0.028 mmHg"
},
{
"text": "0.027 mmHg"
}
],
"answer": "0.017 mmHg",
"solution": "**Answer:** 0.017 mmHg\n\nGiven that,\n
$$ \\Rightarrow $$ $${{{Y_M}} \\over {{Y_N}}}$$ < $${{{X_M}} \\over {{X_N}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2762,
"subject": "Chemistry",
"question": "Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vaapor pressures of pure A and pure B are 7 $$ \\times $$ 103 Pa and 12 $$ \\times $$ 103 Pa, respectively . The composition of the vapor in equilibriumwith a solution containing 40 mole percent of A at this temperature is : ",
"options": [
{
"text": "xA = 0.76; xB = 0.24 "
},
{
"text": "xA = 0.28; xB = 0.72"
},
{
"text": "xA = 0.4; xB = 0.6"
},
{
"text": "xA = 0.37; xB = 0.63 "
}
],
"answer": "xA = 0.28; xB = 0.72",
"solution": "**Answer:** xA = 0.28; xB = 0.72\n\n$${y_A} = {{{P_A}} \\over {{P_{Total}}}} = {{P_A^0{x_A}} \\over {P_A^0{X_A} \\times P_B^0X{}_B}}$$\n
$${y_B} = 0.72$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2763,
"subject": "Chemistry",
"question": "The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K on\nmixing the two liquids, the sum of their initial volume is equal ot the volume of the final mixture.\nThe mole fraction of liquid B is 0.5 in the mixture, The vapour pressure of the final solution, the\nmole fractions of components A and B in vapour phase, respectively are :",
"options": [
{
"text": "500 mmHg. 0.5,0.5"
},
{
"text": "500 mmHg, 0.4, 0.6"
},
{
"text": "450 mmHg, 0.4,0.6"
},
{
"text": "450 mmHg.0.5,0.5"
}
],
"answer": "500 mmHg, 0.4, 0.6",
"solution": "**Answer:** 500 mmHg, 0.4, 0.6\n\nGiven \n
$$P_A^0$$ = 400 mm of Hg\n
$$P_B^0$$ = 600 mm of Hg\n
Mole fraction of B (xB) in liquid phase = 0.5\n
Mole fraction of A (xA) in liquid phase = 1 - 0.5 = 0.5\n
Pressure of solution :\n
Ps = xA$$P_A^0$$ + xB$$P_B^0$$\n
= 0.5 $$ \\times $$ 400 + 0.5 $$ \\times $$ 600\n
= 500 mm of Hg\n
From Roult's law we know,\n
Partial pressure of A (PA) = xA$$P_A^0$$ ......(1)\n
$$ \\therefore $$ mole fraction of B in vapour phase = 1 - 0.4 = 0.6",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2764,
"subject": "Chemistry",
"question": "At 35oC, the vapour pressure of CS2\n is 512 mm. Hg and that of acetone is 344 mm Hg. A solution\nof CS2\n in acetone has a total vapour pressure of 600 mm Hg. The false statement amongst the\nfollowing is :",
"options": [
{
"text": "CS2 and acetone are less attracted to each other than to themselves."
},
{
"text": "a mixture of 100 ml. CS2 and 100 ml. acetone has a volume < 200 ml."
},
{
"text": "Heat must be absorved in order to produce the solution 35oC"
},
{
"text": "Raoult's law is not obeyed by this system."
}
],
"answer": "a mixture of 100 ml. CS2 and 100 ml. acetone has a volume < 200 ml.",
"solution": "**Answer:** a mixture of 100 ml. CS2 and 100 ml. acetone has a volume < 200 ml.\n\nAs the vapour pressure of mixture increases so the force of attraction of mixture is less than the individual component.\n
Since the vapour pressure of the solution is\ngreater than individual vapour pressure of both\npure components, the solution shows a\npositive deviation from Raoult’s law.\n
$$ \\therefore $$ $$\\Delta $$SolH $$>$$ 0 \n
As $$\\Delta $$SolH $$>$$ 0 then solution is endothermic.\n
and here $$\\Delta $$Sol Volume $$>$$ 0. As $$\\Delta $$Sol Volume $$>$$ 0 then a mixture of 100 ml. CS2 and 100 ml. acetone has a volume more than 200 ml. It happens because the distance between molecules increases due to force of attraction of mixture decreases.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2765,
"subject": "Chemistry",
"question": "Two open beakers one containing a solvent and the other containing a mixture of that solvent\nwith a non voltatile solute are together selated in a container. Over time :\n",
"options": [
{
"text": "The volume of the solution increases and the volume of the solvent decreases."
},
{
"text": "The volume of solution does not change and the volume of the solvent decreases."
},
{
"text": "The volumer of the solution and the solvent does not change."
},
{
"text": "The volume of the solution decreases and the volume of the solvent increases.\n"
}
],
"answer": "The volume of the solution increases and the volume of the solvent decreases.",
"solution": "**Answer:** The volume of the solution increases and the volume of the solvent decreases.\n\nThere will be lowering in vapour pressure for\nsolution containing non-volatile solute. So,\nthere will be transfer of solvent molecules\nfrom pure solvent to solution and hence,\nvolume of beaker containing solvent (pure) will\ndecrease and volume of beaker containing\nsolution will increase.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2766,
"subject": "Chemistry",
"question": "At 300 K, the vapour pressure of a solution\ncontaining 1 mole of n-hexane and 3 moles of\nn-heptane is 550 mm of Hg. At the same\ntemperature, if one more mole of n-heptane is\nadded to this solution, the vapour pressure of\nthe solution increases by 10 mm of Hg. What is\nthe vapour pressure in mm Hg of n-heptane in\nits pure state ______?",
"options": [],
"answer": "600",
"solution": "**Answer:** 600\n\nLet X1 and P$$P_1^o$$\n are the mole fraction and vapour\npressure of n-hexane in solution and X2 and $$P_2^o$$\n\n are the mole fraction and vapour pressure of\nn-heptane in solution then\n
$$P_1^o$$ = 400, $$P_2^o$$ = 600",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2767,
"subject": "Chemistry",
"question": "A set of solutions is prepared using 180 g of\nwater as a solvent and 10 g of different nonvolatile solutes A, B and C. The relative\nlowering of vapour pressure in the presence of\nthese solutes are in the order :
[Given, molar\nmass of A = 100 g mol–1; B = 200 g mol–1;\nC = 10,000 g mol–1]",
"options": [
{
"text": "A > C > B"
},
{
"text": "C > B > A"
},
{
"text": "A > B > C"
},
{
"text": "B > C > A"
}
],
"answer": "A > B > C",
"solution": "**Answer:** A > B > C\n\nRelative lowering in vapour pressure (RLVP)\n
So, A > B > C",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 2768,
"subject": "Chemistry",
"question": "At 363 K, the vapour pressure of A is 21 kPa and that of B is 18 kPa. One mole of A and 2 moles of B are mixed. Assuming that this solution is ideal, the vapour pressure of the mixture is ___________ kPa. (Round off to the Nearest Integer).",
"options": [],
"answer": "19",
"solution": "**Answer:** 19\n\n$${X_A} = {1 \\over {1 + 2}} = {1 \\over 3}$$
$$ = 19 kPa$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2769,
"subject": "Chemistry",
"question": "At 20$$^\\circ$$ C, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20$$^\\circ$$ above an equimolar mixture of benzene and methyl benzene is _____________ $$\\times$$ 10$$-$$2. (Nearest integer)",
"options": [],
"answer": "78",
"solution": "**Answer:** 78\n\nVapour pressure of pure benzene, $$p_A^o$$ = 70 Torr
Vapour pressure of pure methyl benzene, $$p_B^o$$ = 20 Torr
This mixture is equimolar, so number of moles of benzene,
nA = number of moles of methyl benzene, nB
Mole fraction of benzene in vapour phase, yA = $${{{p_A}} \\over {{p_T}}}$$ .... (i)
yA = 0.78 = 78 $$\\times$$ 10$$-$$2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2770,
"subject": "Chemistry",
"question": "The vapour pressures of A and B at 25$$^\\circ$$C are 90 mm Hg and 15 mm Hg respectively. If A and B are mixed such that the mole fraction of A in the mixture is 0.6, then the mole fraction of B in the vapour phase is x $$\\times$$ 10$$-$$1. The value of x is ___________. (Nearest integer)",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nxA = 0.6
The vapour pressures of two volatile liquids A and B at 25$$^\\circ$$C are 50 Torr and 100 Torr, respectively. If the liquid mixture contains 0.3 mole fraction of A, then the mole fraction of liquid B in the vapour phase is $${x \\over {17}}$$. The value of x is ______________.
When a certain amount of solid A is dissolved in $$100 \\mathrm{~g}$$ of water at $$25^{\\circ} \\mathrm{C}$$ to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is $$23.76 \\,\\mathrm{mmHg}$$. The number of moles of solute A added is _____________. (Nearest Integer)
A gaseous mixture of two substances A and B, under a total pressure of $$0.8$$ atm is in equilibrium with an ideal liquid solution. The mole fraction of substance A is $$0.5$$ in the vapour phase and $$0.2$$ in the liquid phase. The vapour pressure of pure liquid $$\\mathrm{A}$$ is __________ atm. (Nearest integer)
The total pressure observed by mixing two liquids A and B is 350 mm Hg when their mole fractions are 0.7 and 0.3 respectively.
\n
The total pressure becomes 410 mm Hg if the mole fractions are changed to 0.2 and 0.8 respectively for A and B. The vapour pressure of pure A is __________ mm Hg. (Nearest integer)
\n
Consider the liquids and solutions behave ideally.
",
"options": [],
"answer": "314",
"solution": "**Answer:** 314\n\nLet $V . P$ of pure $A$ be $P_A^0$
\n$$\n\\begin{aligned}\n& \\mathrm{P}_{\\mathrm{A}}^0=314 \\mathrm{~mm} \\mathrm{Hg} \\\\\\\\\n& \\mathrm{P}_{\\mathrm{B}}^0=434 \\mathrm{~mm} \\mathrm{Hg} \\\\\\\\\n& \\therefore \\text{Answer} = 314\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2775,
"subject": "Chemistry",
"question": "The vapour pressure of $30 \\%(\\mathrm{w} / \\mathrm{v})$ aqueous solution of glucose is __________ $\\mathrm{mm} ~\\mathrm{Hg}$ at $25^{\\circ} \\mathrm{C}$.\n
\n[Given : The density of $30 \\%$ (w/v), aqueous solution of glucose is $1.2 \\mathrm{~g} \\mathrm{~cm}^{-3}$ and vapour pressure of pure water is $24 \\mathrm{~mm}~ \\mathrm{Hg}$.]\n
\n(Molar mass of glucose is $180 \\mathrm{~g} \\mathrm{~mol}^{-1}$.)",
"options": [],
"answer": "23",
"solution": "**Answer:** 23\n\nThe given solution is a $30 \\%$ (w/v) aqueous solution of glucose. This means that $30\\ \\mathrm{g}$ of glucose is dissolved in $100\\ \\mathrm{mL}$ of the solution. Since the density of the solution is $1.2\\ \\mathrm{g/mL}$, the weight of $100\\ \\mathrm{mL}$ of the solution is:\n
\nSince the solution is $30 \\%$ (w/v), the weight of glucose in $100\\ \\mathrm{mL}$ of the solution is $30\\ \\mathrm{g}$. Therefore, the weight of water in $100\\ \\mathrm{mL}$ of the solution is:\n
\nNow, we can use Raoult's law to calculate the vapour pressure of the solution. Raoult's law states that the partial vapour pressure of a solvent in a solution is proportional to its mole fraction in the solution. For a dilute solution, the mole fraction of the solvent can be approximated as the ratio of the number of moles of solvent to the total number of moles in the solution. \n
\nLet us assume that we have 1 mole of the solution. The number of moles of glucose in the solution is:\n
\nTherefore, the vapour pressure of the $30\\%$ (w/v) aqueous solution of glucose at $25^\\circ\\mathrm{C}$ is $\\boxed{23\\ \\mathrm{mmHg}}$ (rounded off to the nearest integer).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2776,
"subject": "Chemistry",
"question": "
80 mole percent of $$\\mathrm{MgCl}_{2}$$ is dissociated in aqueous solution. The vapour pressure of $$1.0 ~\\mathrm{molal}$$ aqueous solution of $$\\mathrm{MgCl}_{2}$$ at $$38^{\\circ} \\mathrm{C}$$ is ____________ $$\\mathrm{mm} ~\\mathrm{Hg.} ~\\mathrm{(Nearest} ~\\mathrm{integer)}$$
\n
Given : Vapour pressure of water at $$38^{\\circ} \\mathrm{C}$$ is $$50 \\mathrm{~mm} ~\\mathrm{Hg}$$
What weight of glucose must be dissolved in $$100 \\mathrm{~g}$$ of water to lower the vapour pressure by $$0.20 \\mathrm{~mm} ~\\mathrm{Hg}$$ ?
\n
(Assume dilute solution is being formed)
\n
Given : Vapour pressure of pure water is $$54.2 \\mathrm{~mm} ~\\mathrm{Hg}$$ at room temperature. Molar mass of glucose is $$180 \\mathrm{~g} \\mathrm{~mol}^{-1}$$
",
"options": [
{
"text": "3.69 g"
},
{
"text": "2.59 g"
},
{
"text": "3.59 g"
},
{
"text": "4.69 g"
}
],
"answer": "3.69 g",
"solution": "**Answer:** 3.69 g\n\nThe lowering of vapor pressure of a solvent by a nonvolatile solute is given by Raoult's law:\n\n
$$\\Delta P = x_{\\text{solute}} \\cdot P_0$$\n\n
where $\\Delta P$ is the change in vapor pressure, $x_{\\text{solute}}$ is the mole fraction of the solute, and $P_0$ is the vapor pressure of the pure solvent.\n\n
Rearranging the formula for $x_{\\text{solute}}$, we have:\n\n
Since the mole fraction of the solute is also equal to the number of moles of solute divided by the total number of moles, we can express $x_{\\text{solute}}$ as:\n\n
Assuming that the solution is dilute, the number of moles of water will be much larger than the number of moles of solute, so we can approximate the total number of moles as the number of moles of water.\n\n
So, the correct answer is 3.69 g.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2778,
"subject": "Chemistry",
"question": "
Mass of Urea $$\\left(\\mathrm{NH}_{2} \\mathrm{CONH}_{2}\\right)$$ required to be dissolved in $$1000 \\mathrm{~g}$$ of water in order to reduce the vapour pressure of water by $$25 \\%$$ is _________ g. (Nearest integer)
\n
Given: Molar mass of N, C, O and H are $$14,12,16$$ and $$1 \\mathrm{~g} \\mathrm{~mol}^{-1}$$ respectively
The vapor pressure of pure benzene and methyl benzene at $$27^{\\circ} \\mathrm{C}$$ is given as 80 Torr and 24 Torr, respectively. The mole fraction of methyl benzene in vapor phase, in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is _________ $$\\times 10^{-2}$$ (nearest integer)
To find the mole fraction of methyl benzene in the vapor phase when it is in equilibrium with an equimolar (equal mole) mixture of benzene and methyl benzene at $$27^{\\circ}C$$, we can apply Raoult's law for an ideal solution. Given the vapor pressures of pure benzene and methyl benzene are 80 Torr and 24 Torr, respectively.
\n\n
Raoult's law states that the partial vapor pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution ($$P_i = P_i^\\circ x_i$$), where $$P_i$$ is the partial vapor pressure of the component $$i$$, $$P_i^\\circ$$ is the vapor pressure of the pure component $$i$$, and $$x_i$$ is the mole fraction of the component $$i$$ in the solution.
\n\n
For an equimolar mixture of benzene and methyl benzene, the mole fractions of benzene ($$x_b$$) and methyl benzene ($$x_{mb}$$) in the liquid phase are both $$0.5$$, because the mixture is equimolar.
\n\n
The total pressure of the mixture ($$P_{total}$$) can be calculated using Raoult's law for each component and summing up the individual pressures:
The mole fraction of methyl benzene in the vapor phase ($$y_{mb}$$) can be calculated using Dalton's law, which states that the partial pressure of a component in a mixture is equal to the mole fraction of that component in the vapor phase times the total pressure of the mixture.
\n\n
The partial pressure of methyl benzene can be obtained from Raoult's law as we did earlier:
So, the mole fraction of methyl benzene in the vapor phase, in equilibrium with an equimolar mixture of those two liquids at $$27^{\\circ}C$$, is approximately $$23 \\times 10^{-2}$$ (rounded to the nearest integer).
KEY Concept : Those units which are volume related will be affected by the change in temperature.\n\n
The definition of molarity is - Number of moles of solute present in one litre of solution. From definition you can see it is depends on volume which increases with increasing temperature\nand decreases with decreasing temperature.
\nAll the other three options (Molality, Mole fraction, Weight fraction of solute) are mass related units and temperature has no effect on mass.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2782,
"subject": "Chemistry",
"question": "25 ml of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a litre value of 35 ml. The molarity of barium hydroxide solution was",
"options": [
{
"text": "0.14"
},
{
"text": "0.28"
},
{
"text": "0.35"
},
{
"text": "0.07"
}
],
"answer": "0.07",
"solution": "**Answer:** 0.07\n\nBa(OH)2 + 2HCl(aq) $$ \\to $$ BaCl2(aq) + 2H2O(l)\n
\n[Note : While calculating Concentration volume should always be in litre.]
\n$$ \\Rightarrow $$ $${{0.001} \\over {0.1}}$$
\n$$ \\Rightarrow $$ 0.01 M
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2785,
"subject": "Chemistry",
"question": "Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 ml of 1.2 M second solution. What is the molarity of the final mixture?",
"options": [
{
"text": "2.70 M"
},
{
"text": "1.344 M"
},
{
"text": "1.50 M"
},
{
"text": "1.20 M"
}
],
"answer": "1.344 M",
"solution": "**Answer:** 1.344 M\n\nShort Cut Method : Molarity, Normality, Average Atomic Mass, Average Molar Mass, Average Density, Average Vapour Pressure all of those values will be between the solutions which are going to be mixed.\n
Here molarity of first solution is 1.2 and molarity of second solution is 1.5.\n So the molarity of the mixture will always be more than 1.2 and less than 1.5. Remember the molarity of mixture can't be either 1.2 or 1.5.\n
So from the option you an see only 1.344 M can be the right answer.\n
Normal method : \n The formula for molarity of mixture of two substance is = $${{{M_1}{V_1} + {M_2}{V_2}} \\over {{V_1} + {V_2}}}$$\n
= 1.344",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2786,
"subject": "Chemistry",
"question": "Density of a 2.05M solution of acetic acid in water is 1.02 g/mL The molality of the solution is",
"options": [
{
"text": "2.28 mol kg-1"
},
{
"text": "0.44 mol kg-1"
},
{
"text": "1.14 mol kg-1"
},
{
"text": "3.28 mol kg-1"
}
],
"answer": "2.28 mol kg-1",
"solution": "**Answer:** 2.28 mol kg-1\n\n2.05M solution of acetic acid in water means in 1 litre solution 2.05 moles of CH3COOH present.\n
Density of solution = 1.02 g/ml (Given)\n
Assume the volume of solution = 1 litre = 1000 ml\n
$$\\therefore$$ Mass of solution = 1000 $$ \\times $$ 1.02 = 1020 gm\n
Molar mass of CH3COOH = 60\n
So Mass of CH3COOH (msolute) = 2.05 $$ \\times $$ 60 = 123\n
$$\\therefore$$ Mass of solvent (msolvent) = 1020 - 123 = 897 gm = 0.897 kg\n
Formula of molality (m) = $${{no\\,of\\,moles\\,of\\,solute} \\over {weight\\,of\\,solvent\\,in\\,kg}}$$\n
$$\\therefore$$ m = $${{2.05} \\over {0.897}}$$ = 2.28\n
Using Formula :\n
Molality (m) = $${{1000 \\times M} \\over {1000 \\times d - M \\times {M_{solute}}}}$$\n
Here M = molarity, Msolute = molecular mass of solute, d = density of solution\n
$$\\therefore$$ m = $${{1000 \\times 2.05} \\over {1000 \\times 1.02 - 2.05 \\times 60}}$$ = 2.28",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2787,
"subject": "Chemistry",
"question": "The density (in g mL-1) of a 3.60 M sulphuric acid solution that is 29% H2SO4 (molar mass = 98 g mol-1) by mass will be",
"options": [
{
"text": "1.45"
},
{
"text": "1.64"
},
{
"text": "1.88"
},
{
"text": "1.22"
}
],
"answer": "1.22",
"solution": "**Answer:** 1.22\n\nImportant formula of Molarity(M) when % w/w is given\n
M = $${{10 \\times \\% w/w \\times d} \\over {{M_{solute}}}}$$\n
Here M = 3.6, % w/w = 29, d = density, Msolute = 98\n
X = 0.086",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2789,
"subject": "Chemistry",
"question": "The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u ) in 1000g of water is\n1.15 g/mL. The molarity of this solution is :",
"options": [
{
"text": "0.50 M"
},
{
"text": "1.78 M"
},
{
"text": "1.02 M"
},
{
"text": "2.05 M"
}
],
"answer": "2.05 M",
"solution": "**Answer:** 2.05 M\n\nWe know molarity (M) = $${{no\\,of\\,moles\\,of\\,solute} \\over {volume\\,of\\,solution\\,in\\,litre}}$$\n
Moles of solute = $${{120} \\over {60}}$$ = 2\n
Mass of solution = 1000 + 120 = 1120 gm\n
Density of solution = 1.15 g/mL\n
$$\\therefore$$ volume of solution = $${{1120} \\over {1.15}}$$ = 973.9 mL = $${{973.9} \\over {1000}}$$ litre = 0.9739 litre \n
$$\\therefore$$ M = $${2 \\over {0.9739}}$$ = 2.05",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2790,
"subject": "Chemistry",
"question": "The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCl will be: ",
"options": [
{
"text": "1.00 M"
},
{
"text": "1. 75 M "
},
{
"text": "0.975 M "
},
{
"text": "0.875 M"
}
],
"answer": "0.875 M",
"solution": "**Answer:** 0.875 M\n\nThe formula for molarity of mixture of two substance is = $${{{M_1}{V_1} + {M_2}{V_2}} \\over {{V_1} + {V_2}}}$$\n
= 0.875",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2791,
"subject": "Chemistry",
"question": "Excess of NaOH (aq) was added to 100 mL of FeCl3 (aq) resulting into 2.14 g of Fe(OH)3 . The molarity of FeCl3 (aq) is :\n
(Given molar mass of Fe = 56 g mol−1 and molar mass of Cl = 35.5 g mol−1)\n",
"options": [
{
"text": "0.2 M"
},
{
"text": "03 M"
},
{
"text": "0.6 M"
},
{
"text": "1.8 M"
}
],
"answer": "0.2 M",
"solution": "**Answer:** 0.2 M\n\n3 NaOH (aq.) + FeCl3(aq) $$ \\to $$ Fe(OH)3(s)+ 3 NaCl(aq).\n
1 mole of Fe(OH)3 is obtained from = 1 mole of FeCl3\n
$$\\therefore\\,\\,\\,$$ 2 $$ \\times $$ 10$$-$$2 moles of Fe(OH)3 will obtain from \n
= 0.02 mole of FeCl3\n
Molarity of FeCl3 = $${{No.of\\,moles} \\over {Volume\\,in\\,L}}$$ = $${{2 \\times {{10}^{ - 2}}} \\over {0.1}}$$ = 0.2 M",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2792,
"subject": "Chemistry",
"question": "The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of compound CXHYOZ is ",
"options": [
{
"text": "C2H4O3"
},
{
"text": "C3H6O3"
},
{
"text": "C2H4O"
},
{
"text": "C3H4O2"
}
],
"answer": "C2H4O3",
"solution": "**Answer:** C2H4O3\n\n\n
$$\\therefore\\,\\,\\,$$ The ratio of no of atoms of C and H in one molecule of Cx Hy Oz = 1 : 2\n
$$\\therefore\\,\\,\\,$$ y = 2x\n
In one molecule of Cx Hy Oz compound contain z atoms of oxygen. \n
According to the question, no of atoms of oxygen required to burn CxHy completely should be twice of z atoms of oxygen.
Percentage purity = $${{0.17} \\over {0.2}}$$ $$ \\times $$ 100 = 85 %",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2801,
"subject": "Chemistry",
"question": "6.023 $$ \\times $$ 1022 molecules are present in 10 g of a substance 'x'. The molarity of a solution containing\n5 g of substance 'x' in 2 L solution is _____ × 10-3",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\nMass of 6.023 × 1022 molecules of a substance\n= 10 g\n
Mass of 6.023 × 1023 molecules of the\nsubstance = 100 g\n
$$ \\therefore $$ Molar mass of the substance = 100 g mol–1\n
Molarity of the solution = \n
moles of solute
\n
volume of solution(in l)
\n\n
= $${{\\left( {5/100} \\right)} \\over 2}$$ = 0.025",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2802,
"subject": "Chemistry",
"question": "10.30 mg of O2 is dissolved into a liter of sea\nwater of density 1.03 g/mL. The concentration\nof O2 in ppm is__________.",
"options": [],
"answer": "10",
"solution": "**Answer:** 10\n\n1030 gm of sea water contains = 10.3 × 10–3 gm\n
106\n gm of sea water contains = $${{10.3 \\times {{10}^{ - 3}}} \\over {1030}} \\times {10^6}$$ = 10 ppm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2803,
"subject": "Chemistry",
"question": "The strengths of 5.6 volume hydrogen peroxide\n(of density 1 g/mL) in terms of mass percentage\nand molarity (M), respectively, are:\n(Take molar mass of hydrogen peroxide as\n34 g/mol)",
"options": [
{
"text": "0.85 and 0.5"
},
{
"text": "0.85 and 0.25"
},
{
"text": "1.7 and 0.25"
},
{
"text": "1.7 and 0.5"
}
],
"answer": "1.7 and 0.5",
"solution": "**Answer:** 1.7 and 0.5\n\nVolume strength = 11.2 × molarity\n
= $${{17} \\over {1000}} \\times 100$$ = 1.7 %",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2804,
"subject": "Chemistry",
"question": "The ratio of the mass percentages of ‘C & H’\nand ‘C & O’ of a saturated acyclic organic\ncompound ‘X’ are 4 : 1 and 3 : 4 respectively.\nThen, the moles of oxygen gas required for\ncomplete combustion of two moles of organic\ncompound ‘X’ is ________.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nLet the organic compound X is = CxHyOz\n
Here moles of C = x, moles of H = y, moles of O = z\n
For 1 mole of C2H6O2 number of moles of O2 required = $${5 \\over 2}$$\n
$$ \\therefore $$ For 2 mole of C2H6O2 number of moles of O2 required = $${5 \\over 2}$$ $$ \\times $$ 2 = 5",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2805,
"subject": "Chemistry",
"question": "The molarity of HNO3 in a sample which has\ndensity 1.4 g/mL and mass percentage of 63%\nis _____. \n (Molecular Weight of HNO3 = 63)",
"options": [],
"answer": "14",
"solution": "**Answer:** 14\n\n%w/w = 63%\n
dsolution = 1.4 g/ml\n
Molarity = \n
%w/w $$ \\times $$ dsolution
\n
Moleculer weight of Solute
\n.$$ \\times $$ 10\n
= $${{63 \\times 1.4 \\times 10} \\over {63}}$$ = 14 M",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2806,
"subject": "Chemistry",
"question": "The hardness of a water sample containing\n10–3 M MgSO4 expressed as CaCO3 equivalents\n(in ppm) is ______. \n(molar mass of MgSO4 is 120.37 g/mol)",
"options": [],
"answer": "100",
"solution": "**Answer:** 100\n\nEquivalance of MgSO4 = Equivalance of CaCO3\n
$$ \\Rightarrow $$ [CaCO3] = 10-3 M = 10-3 mol/lit\n
= 10-3 $$ \\times $$ 100 g/lit\n
= 100 mg/lit\n
= 100 ppm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2807,
"subject": "Chemistry",
"question": "Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to\nachieve 10 ppm of iron in 100 kg of wheat is _______.\n Atomic weight : Fe = 55.85; S = 32.00;\nO = 16.00",
"options": [],
"answer": "4.95to4.97",
"solution": "**Answer:** 4.95to4.97\n\nFeSO4.7H2O (M = 277.85)\n
PPM = \n
Mass of Iron
\n
Mass of wheat
\n $$ \\times $$ 106\n
$$ \\Rightarrow $$ 10 = \n
Mass of Iron
\n
100 $$ \\times $$ 103
\n$$ \\times $$ 106\n
$$ \\Rightarrow $$ Mass of Iron = 1 gm\n
Molecular mass of FeSO4.7H2O is 277.85\n
55.85 gm iron is present in 277.85 gm of salt\n
1 gm iron is present in = $${{277.85} \\over {55.85}}$$ = 4.97 gm of salt.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2808,
"subject": "Chemistry",
"question": "The volume strength of 8.9 M H2O2\n solution calculated at 273 K and 1 atm is ______. (R = 0.0821 L\natm K-1 mol-1) (rounded off ot the nearest integer)",
"options": [],
"answer": "100",
"solution": "**Answer:** 100\n\nVolume strength of H2O2 at 1 atm\n273 kelvin = M × 11.2 = 8.9 × 11.2 = 99.68 $$ \\simeq $$ 100",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2809,
"subject": "Chemistry",
"question": "4.5 g of compound A (MW = 90) was used to make 250 mL of its aqueous solution. The\nmolarity of the solution in M is x $$ \\times $$ 10-1. The value of x is _______. (Rounded off to the nearest integer)",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nGiven, weight of compound A = 4.5 g
Molecular weight of compound A = 90 g/mol
Volume of solution (in mL) = 250 mL
Now, molarity is defined as number of moles of solute or compound A divided by volume of solution (in L).
x = 2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2810,
"subject": "Chemistry",
"question": "In basic medium $$Cr{O_4}^{2 - }$$ oxidises $${S_2}{O_3}^{2 - }$$ to form $$S{O_4}^{2 - }$$ and itself changes into $$Cr{(OH)_4}^ - $$. The volume of 0.154 M $$Cr{O_4}^{2 - }$$ required to react with 40 mL of 0.25 M $${S_2}{O_3}^{2 - }$$ is __________ mL. (Rounded off to the nearest integer)",
"options": [],
"answer": "173",
"solution": "**Answer:** 173\n\n$$17{H_2}O + 8Cr{O_4} + 3{S_2}{O_3}\\buildrel {} \\over\n \\longrightarrow 6S{O_4} + 8Cr{(OH)_4}^ - + 2O{H^ - }$$
v = 173 mL",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2811,
"subject": "Chemistry",
"question": "A 6.50 molal solution of KOH (aq.) has a density of 1.89 g cm$$-$$3. The molarity of the solution is ____________ mol dm$$-$$3. (Round off to the Nearest Integer).
[Atomic masses : K : 39.0 u; O : 16.0 u; H : 1.0 u]",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\n$$m = {{1000 \\times M} \\over {1000 \\times d - M \\times {M_{solute}}}}$$
$$ \\Rightarrow $$ $$M = 9$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2812,
"subject": "Chemistry",
"question": "The exact volumes of 1 M NaOH solution required to neutralise 50 mL of 1 M H3PO3 solution and 100 mL of 2 M H3PO2 solution, respectively, are :",
"options": [
{
"text": "100 mL and 50 mL"
},
{
"text": "100 mL and 200 mL"
},
{
"text": "100 mL and 100 mL"
},
{
"text": "50 mL and 50 mL"
}
],
"answer": "100 mL and 200 mL",
"solution": "**Answer:** 100 mL and 200 mL\n\n
\n
Millimoles of H3PO3 = M $$\\times$$ V = 1 $$\\times$$ 50 = 50
\n
For 1 millimole of H3PO3, we require 2 millimoles of NaOH.
\n
For 50 millimole of H3PO3, we require (2 $$\\times$$ 50) = 100 millimoles of NaOH.
\n
Millimoles of NaOH = M $$\\times$$ V = 100
\n
1 $$\\times$$ V = 100
\n
V = 100 mL
\n
\n
Millimoles of H3PO2 = M $$\\times$$ V = 2 $$\\times$$ 100 = 200
\n
For 1 millimole of H3PO2, we require 1 millimoles of NaOH.
\n
For 200 millimoles of H3PO2, we require 200 millimoles of NaOH.
\n
So, volume of NaOH = 200 mL
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2813,
"subject": "Chemistry",
"question": "15 mL of aqueous solution of Fe2+ in acidic medium completely reacted with 20 mL of 0.03 M aqueous Cr2O$$_7^{2 - }$$. The molarity of the Fe2+ solution is __________ $$\\times$$ 10-2 M. (Round off to the Nearest Integer).",
"options": [],
"answer": "24",
"solution": "**Answer:** 24\n\n$$C{r_2}{O_7}^{2 - } + F{e^{2 + }}\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{}} C{r^{3 + }} + F{e^{3 + }}$$
Valance Factor of $$C{r_2}{O_7}^{2 - }$$ = 6 \n
Valance Factor of $$F{e^{2 + }}$$ = 1
mili eq. of $$C{r_2}{O_7}^{2 - }$$ = mili eq. of Fe2+
6(0.03 $$\\times$$ 20) = 1(M $$\\times$$ 15)
M = 6(0.03)4/3
M = 0.24 M
M = 24 $$\\times$$ 10$$-$$2 M",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2814,
"subject": "Chemistry",
"question": "The mole fraction of a solute in a 100 molal aqueous solution is ___________ $$\\times$$ 10$$-$$2. (Round off to the Nearest Integer).
[Given : Atomic masses : H : 1.0 u, O : 16.0 u ]",
"options": [],
"answer": "64",
"solution": "**Answer:** 64\n\nLet weight of H2O = 1000 g
Moles of solute = 100
(mole)H2O = $${{1000} \\over {18}}$$
Mole fraction of solute = $${{mole\\,of\\,solute} \\over {Total\\,moles}}$$
$${X_{solute}} = 64 \\times {10^{ - 2}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2815,
"subject": "Chemistry",
"question": "10.0 mL of Na2CO3 solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings :
4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL
Based on these readings, and convention of titrimetric estimation the concentration of Na2CO3 solution is ___________ mM.
(Round off to the Nearest Integer).",
"options": [],
"answer": "50",
"solution": "**Answer:** 50\n\nFrom the given value of HCl, it is clear that most appropriate volume of HCl used is 5 ml because it occurs most number of times.
= 50 mM",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2816,
"subject": "Chemistry",
"question": "If the concentration of glucose (C6H12O6) in blood is 0.72 g L$$-$$1, the molarity of glucose in blood is ____________ $$\\times$$ 10$$-$$3 M. (Nearest integer)
[Given : Atomic mass of C = 12, H = 1, O = 16 u]",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n[Glucose] = $${{C(gm/l)} \\over {M(gm/mol)}} = {{0.72} \\over {180}} = 4 \\times {10^{ - 3}}$$ M",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2817,
"subject": "Chemistry",
"question": "When 10 mL of an aqueous solution of Fe2+ ions was titrated in the presence of dil H2SO4 using diphenylamine indicator, 15 mL of 0.02 M solution of K2Cr2O7 was required to get the end point. The molarity of the solution containing Fe2+ ions is x $$\\times$$ 10$$-$$2 M. The value of x is __________. (Nearest integer)",
"options": [],
"answer": "18",
"solution": "**Answer:** 18\n\nmilli-equivalents of Fe2+ = milli-equivalents of K2Cr2O7
$$ \\Rightarrow $$ M = 0.18 = 18 $$\\times$$ 10$$-$$2 M",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2818,
"subject": "Chemistry",
"question": "The density of NaOH solution is 1.2 g cm$$-$$3. The molality of this solution is _____________ m. (Round off to the Nearest Integer)
[Use : Atomic mass : Na : 23.0 u, O : 16.0 u, H : 1.0 u, Density of H2O : 1.0 g cm$$-$$3]",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nConsider 1 litre solution
mass of solution = (1.2 $$\\times$$ 1000)g = 1200 gm
Neglecting volume of NaOH
Mass of water = 1000 gm
$$\\Rightarrow$$ Mass of NaOH = (1200 $$-$$ 1000)gm = 200 gm
$$\\Rightarrow$$ molality = $${{5mol} \\over {1kg}} = 5$$ m",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2819,
"subject": "Chemistry",
"question": "An aqueous KCl solution of density 1.20 g mL$$-$$1 has a molality of 3.30 mol kg$$-$$1. The molarity of the solution in mol L$$-$$1 is ____________ (Nearest integer) [Molar mass of KCl = 74.5]",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n1000 kg solvent has 3.3 moles of KCl
Volume of solution = $${{1245.85} \\over {1.2}}$$ ml
So molarity = $${{3.3 \\times 1.2} \\over {1245.85}} \\times 1000 = 3.17$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2820,
"subject": "Chemistry",
"question": "100 mL of Na3PO4 solution contains 3.45 g of sodium. The molarity of the solution is _____________ $$\\times$$ 10$$-$$2 mol L$$-$$1. (Nearest integer)
[Atomic Masses - Na : 23.0 u, O : 16.0 u, P : 31.0 u]",
"options": [],
"answer": "50",
"solution": "**Answer:** 50\n\nMolarity of Na3PO4 Solution = $${{{n_{N{a_3}P{O_4}}}} \\over {volume\\,of\\,solution\\,in\\,L}}$$
= 0.5 = 50 $$\\times$$ 10$$-$$2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2821,
"subject": "Chemistry",
"question": "100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is x $$\\times$$ 10$$-$$2. The value of x is ____________. (Nearest integer)
[Atomic weight : H = 1.008; C = 12.00; O = 16.00]",
"options": [],
"answer": "19",
"solution": "**Answer:** 19\n\nC3H8(g) + 5O2(g) $$\\to$$ 3CO2(g) + 4H2O(l)
\n\n
\n
t = 0
\n
2.27 mole
\n
31.25 mol
\n
\n
\n
\n\n\n
\n
t = $$\\infty $$
\n
0
\n
19.9 mol
\n
6.81 mol
\n
9.08 mol
\n
\n\n
mole fraction of CO2 in the final reaction mixture (heterogenous)
$$\\Rightarrow$$ 19",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2822,
"subject": "Chemistry",
"question": "The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (H2C2O4.2H2O) in 250 mL of water in mol L$$-$$1 is x $$\\times$$ 10$$-$$2. The value of x is _____________. (Nearest integer) [Atomic mass : H : 1.0, C : 12.0, O : 16.0]",
"options": [],
"answer": "20",
"solution": "**Answer:** 20\n\n$$[{H_2}{C_2}{O_4}.2{H_2}O] = {{weight/{M_w}} \\over {V(L)}}$$
$$x = 20$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2823,
"subject": "Chemistry",
"question": "Sodium oxide reacts with water to produce sodium hydroxide. 20.0 g of sodium oxide is dissolved in 500 mL of water. Neglecting the change in volume, the concentration of the resulting NaOH solution is ______________ $$\\times$$ 10$$-$$1 M. (Nearest integer) [Atomic mass : Na = 23.0, O = 16.0, H = 1.0]",
"options": [],
"answer": "13",
"solution": "**Answer:** 13\n\nNa2O + H2O $$\\to$$ 2NaOH
$${{20} \\over {62}}$$ moles
Moles of NaOH formed = $${{20} \\over {62}}$$ $$\\times$$ 2
[NaOH] = $${{{{40} \\over {62}}} \\over {{{500} \\over {1000}}}}$$ = 1.29 M = 13 $$\\times$$ 10$$-$$1 M (Nearest integer)",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2824,
"subject": "Chemistry",
"question": "If 80 g of copper sulphate CuSO4 . 5H2O is dissolved in deionised water to make 5L of solution. The concentration of the copper sulphate solution is x $$\\times$$ 10$$-$$3 mol L$$-$$1. The value of x is _____________.
[Atomic masses Cu : 63.54u, S : 32u, O : 16u, H : 1u]",
"options": [],
"answer": "64",
"solution": "**Answer:** 64\n\nGiven, mass of CuSO4 . 5H2O = 80 g
The concentration of copper sulphate solution is x $$\\times$$ 10$$-$$3 mol/L.
A 0.166 g sample of an organic compound was digested with conc. H2SO4 and then distilled with NaOH. The ammonia gas evolved was passed through 50.0 mL of 0.5 N H2SO4. The used acid required 30.0 mL of 0.25 N NaOH for complete neutralization. The mass percentage of nitrogen in the organic compound is ____________.
",
"options": [],
"answer": "63",
"solution": "**Answer:** 63\n\nMillimoles of used acid $=\\frac{30 \\times 0.25}{2}$\n
\nMillimoles of $\\mathrm{NH}_{3}=30 \\times 0.25=7.5$\n
16 moles of $$\\mathrm{NaOH}$$ is required for the complete neutralisation of the resultant acidic mixture. The number of moles of $$\\mathrm{SO}_{2} \\mathrm{Cl}_{2}$$ used is :
\nMoles of $$\\mathrm{NaOH}$$ required for complete neutralization of resultant acidic mixture $$=16$$ moles\n
\nAnd 1 mole of $$\\mathrm{SO}_{2} \\mathrm{Cl}_{2}$$ produced 4 moles of $$\\mathrm{H}^{+}$$.\n
\n$\\therefore$ Moles of $$\\mathrm{SO}_{2} \\mathrm{Cl}_{2}$$ used will be $$=\\frac{16}{4}=4$$ moles",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2828,
"subject": "Chemistry",
"question": "
Chlorophyll extracted from the crushed green leaves was dissolved in water to make $$2 \\mathrm{~L}$$ solution of Mg of concentration $$48\\, \\mathrm{ppm}$$. The number of atoms of $$\\mathrm{Mg}$$ in this solution is $$x \\times 10^{20}$$ atoms. The value of $$x$$ is ___________. (Nearest Integer)
\n
(Given : Atomic mass of $$\\mathrm{Mg}$$ is $$24 \\mathrm{~g} \\mathrm{~mol}^{-1} ; \\mathrm{N}_{\\mathrm{A}}=6.02 \\times 10^{23} \\mathrm{~mol}^{-1}$$ )
When 800 mL of 0.5 M nitric acid is heated in a beaker, its volume is reduced to half and 11.5 g of nitric acid is evaporated. The molarity of the remaining nitric acid solution is x $$\\times$$ 10$$-$$2 M. (Nearest integer)
$$250 \\mathrm{~g}$$ solution of $$\\mathrm{D}$$-glucose in water contains $$10.8 \\%$$ of carbon by weight. The molality of the solution is nearest to
$$20 \\mathrm{~mL}$$ of $$0.02 \\,\\mathrm{M} \\,\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$$ solution is used for the titration of $$10 \\mathrm{~mL}$$ of $$\\mathrm{Fe}^{2+}$$ solution in the acidic medium.
\n
The molarity of $$\\mathrm{Fe}^{2+}$$ solution is __________ $$\\times \\,10^{-2}\\, \\mathrm{M}$$. (Nearest Integer)
",
"options": [],
"answer": "24",
"solution": "**Answer:** 24\n\nApplying the law of equivalence, milliequivalents of $\\mathrm{Fe}^{2+}=$ milliequivalents of $\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$\n
\n$\\therefore $ Answer will be 24",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2832,
"subject": "Chemistry",
"question": "
2L of 0.2M H2SO4 is reacted with 2L of 0.1M NaOH solution, the molarity of the resulting product Na2SO4 in the solution is _________ millimolar. (Nearest integer)
",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\n Molarity of $\\mathrm{Na}_2 \\mathrm{SO}_4$ is $\\frac{0.1}{4}=0.025 \\,\\mathrm{M}$
The molality of a $$10 \\%(\\mathrm{v} / \\mathrm{v})$$ solution of di-bromine solution in $$\\mathrm{CCl}_{4}$$ (carbon tetrachloride) is '$$x$$'. $$x=$$ ____________ $$\\times 10^{-2} ~\\mathrm{M}$$. (Nearest integer)
\n
[Given : molar mass of $$\\mathrm{Br}_{2}=160 \\mathrm{~g} \\mathrm{~mol}^{-1}$$
\n
atomic mass of $$\\mathrm{C}=12 \\mathrm{~g} \\mathrm{~mol}^{-1}$$
\n
atomic mass of $$\\mathrm{Cl}=35.5 \\mathrm{~g} \\mathrm{~mol}^{-1}$$
\n
density of dibromine $$=3.2 \\mathrm{~g} \\mathrm{~cm}^{-3}$$
\n
density of $$\\mathrm{CCl}_{4}=1.6 \\mathrm{~g} \\mathrm{~cm}^{-3}$$]
",
"options": [],
"answer": "139",
"solution": "**Answer:** 139\n\nMass of $10 \\mathrm{~mL}$ of $\\mathrm{Br}_2=10 \\times 3.2=32 \\mathrm{gm}$\n
Mass of $90 \\mathrm{~mL}^{\\mathrm{L}} \\mathrm{CCl}_4=90 \\times 1.6=144 \\mathrm{gm}$\n
Molality of $\\mathrm{Br}_2$ solution in $\\mathrm{CCl}_4$\n
The density of $$3 \\mathrm{M}$$ solution of $$\\mathrm{NaCl}$$ is $$1.0 \\mathrm{~g} \\mathrm{~mL}^{-1}$$. Molality of the solution is ____________ $$\\times 10^{-2} \\mathrm{~m}$$. (Nearest integer).
\n
Given: Molar mass of $$\\mathrm{Na}$$ and $$\\mathrm{Cl}$$ is $$23$$ and $$35.5 \\mathrm{~g} \\mathrm{~mol}^{-1}$$ respectively.
Nearest integer $=364$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2835,
"subject": "Chemistry",
"question": "$1 \\mathrm{~L}, 0.02 \\mathrm{M}$ solution of $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{SO}_{4}\\right]$ Br is mixed with $1 \\mathrm{~L}, 0.02 \\mathrm{M}$ solution of $\\left[\\mathrm{Co}\\left(\\mathrm{NH}_{3}\\right)_{5} \\mathrm{Br}\\right] \\mathrm{SO}_{4}$. The resulting solution is divided into two equal parts $(\\mathrm{X})$ and treated with excess of $\\mathrm{AgNO}_{3}$ solution and $\\mathrm{BaCl}_{2}$ solution respectively as shown below:\n
\nThe number of moles of $\\mathrm{Y}$ and $\\mathrm{Z}$ respectively are",
"options": [
{
"text": "$0 .01,0.01$"
},
{
"text": "$0.01,0.02$"
},
{
"text": "$0.02,0.01$"
},
{
"text": "$0.02,0.02$"
}
],
"answer": "$0 .01,0.01$",
"solution": "**Answer:** $0 .01,0.01$\n\n
On mixing both $$\\mathrm{\\left[ {Co{{(N{H_5})}_5}S{O_4}} \\right]Br}$$ and $$\\mathrm{\\left[ {Co{{(N{H_3})}_5}Br} \\right]S{O_4}}$$ becomes 0.01 molar.
\n
$$\\therefore$$ Moles of y and z formed will also be 0.01 both.
Some amount of dichloromethane $$\\left(\\mathrm{CH}_{2} \\mathrm{Cl}_{2}\\right)$$ is added to $$671.141 \\mathrm{~mL}$$ of chloroform $$\\left(\\mathrm{CHCl}_{3}\\right)$$ to prepare $$2.6 \\times 10^{-3} \\mathrm{M}$$ solution of $$\\mathrm{CH}_{2} \\mathrm{Cl}_{2}(\\mathrm{DCM})$$. The concentration of $$\\mathrm{DCM}$$ is ___________ ppm (by mass).
\n
Given :
\n
atomic mass : C = 12
\n
H = 1
\n
Cl = 35.5
\n
density of $$\\mathrm{CHCl}_{3}=1.49 \\mathrm{~g} \\mathrm{~cm}^{-3}$$
What is the mass ratio of ethylene glycol ($$\\mathrm{C_2H_6O_2}$$, molar mass = 62 g/mol) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molar aqueous solution?
",
"options": [
{
"text": "1 L marketed solution contains 75 g of H$$_2$$O$$_2$$."
},
{
"text": "1 L marketed solution contains 250 g of H$$_2$$O$$_2$$."
},
{
"text": "1 L marketed solution contains 25 g of H$$_2$$O$$_2$$."
},
{
"text": "100 mL marketed solution contains 25 g of H$$_2$$O$$_2$$."
}
],
"answer": "1 L marketed solution contains 75 g of H$$_2$$O$$_2$$.",
"solution": "**Answer:** 1 L marketed solution contains 75 g of H$$_2$$O$$_2$$.\n\nMolarity of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ soln $=\\frac{\\text { volume strength }}{11.2}$\n
\n$=\\frac{25}{11.2}=2.23 \\mathrm{M}$\n
\n$\\therefore$ amount of $\\mathrm{H}_{2} \\mathrm{O}_{2}$ in one litre $=2.23 \\times 34=75 \\mathrm{gm}$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2840,
"subject": "Chemistry",
"question": "
The number of units, which are used to express concentration of solutions from the following is _________
\n
Mass percent, Mole, Mole fraction, Molarity, ppm, Molality
",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nMass percent, mole fraction, molarity, ppm & molality is used to express concentration. So, the number of units = 5",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2841,
"subject": "Chemistry",
"question": "
5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? _____________
\n
Given : Molar Mass of Na, O and H is 23, 16 and 1 g mol$$^{-1}$$ respectively
\n$\\mathrm{V}=180 \\mathrm{~mL}$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2842,
"subject": "Chemistry",
"question": "The volume (in $\\mathrm{mL}$) of $0.1 \\mathrm{M} ~\\mathrm{AgNO}_{3}$ required for complete precipitation of chloride ions present in $20 \\mathrm{~mL}$ of $0.01 \\mathrm{M}$ solution of $\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5} \\mathrm{Cl}\\right] \\mathrm{Cl}_{2}$ as silver chloride is __________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nThe given solution contains a complex ion $\\left(\\mathrm{Cr}\\left(\\mathrm{H}_2 \\mathrm{O}\\right)_5 \\mathrm{Cl}\\right) \\mathrm{Cl}_2$ which contains one chloride ion per complex. Therefore, when this complex is treated with silver nitrate, each mole of the complex will consume 2 moles of silver nitrate to form two moles of silver chloride. \n
\nThe first step is to write the balanced chemical equation for the reaction between the complex and silver nitrate, as follows:\n
\nFrom the equation, we can see that 1 mole of the complex consumes 2 moles of silver nitrate to form 2 moles of silver chloride. Therefore, the number of millimoles of chloride ions present in the given solution can be calculated as:\n
\nTherefore, the volume of 0.1 M silver nitrate required for complete precipitation of chloride ions present in 20 mL of 0.01 M $\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_5 \\mathrm{Cl}\\right] \\mathrm{Cl}_{2}$ solution as silver chloride is $\\boxed{4\\ \\mathrm{mL}}$.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2843,
"subject": "Chemistry",
"question": "
$$20 \\mathrm{~mL}$$ of calcium hydroxide was consumed when it was reacted with $$10 \\mathrm{~mL}$$ of unknown solution of $$\\mathrm{H}_{2} \\mathrm{SO}_{4}$$. Also $$20 \\mathrm{~mL}$$ standard solution of $$0.5 ~\\mathrm{M} ~\\mathrm{HCl}$$ containing 2 drops of phenolphthalein was titrated with calcium hydroxide, the mixture showed pink colour when burette displayed the value of $$35.5 \\mathrm{~mL}$$ whereas the burette showed $$25.5 \\mathrm{~mL}$$ initially. The concentration of $$\\mathrm{H}_{2} \\mathrm{SO}_{4}$$ is _____________ M. (Nearest integer)
",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nReaction with HCl:
\nVolume of $\\mathrm{Ca}(\\mathrm{OH})_2=10 \\mathrm{ml}$
\nVolume of $\\mathrm{HCl}=20 \\mathrm{ml}$
\nConcentration of $\\mathrm{HCl}=0.5 \\mathrm{M}$.\n
\nNo. of milli moles of $\\mathrm{HCl}=10$
\nNo. of milli moles of $\\mathrm{Ca}(\\mathrm{OH})_2=5$.\n
\ni.e. $\\mathrm{M}_{\\mathrm{Ca}(\\mathrm{OH})_2}=\\frac{\\text { no. of milli moles }}{\\mathrm{V}(\\mathrm{ml})}=\\frac{5}{10}$ $=0.5 \\mathrm{M}$.\n
\nNo. of milli moles of $\\mathrm{Ca}(\\mathrm{OH})_2=20 \\times 0.5$ $=10$
\ni.e. no. of milli moles of $\\mathrm{H}_2 \\mathrm{SO}_4=10$\n
\n$$\n\\begin{aligned}\n\\Rightarrow & \\mathrm{M}_{\\mathrm{H}_2 \\mathrm{SO}_4}=\\frac{\\text { no, of milli moles }}{\\mathrm{V}(\\mathrm{ml})} \\\\\\\\\n& =\\frac{10}{10} \\\\\\\\\n& =1 \\mathrm{M}\n\\end{aligned}\n$$\n
\nSo, the concentration of $\\mathrm{H}_2 \\mathrm{SO}_4$ is 1 M.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2844,
"subject": "Chemistry",
"question": "
A solution is prepared by adding $$2 \\mathrm{~g}$$ of \"$$\\mathrm{X}$$\" to 1 mole of water. Mass percent of \"$$\\mathrm{X}$$\" in the solution is :
",
"options": [
{
"text": "20%"
},
{
"text": "10%"
},
{
"text": "2%"
},
{
"text": "5%"
}
],
"answer": "10%",
"solution": "**Answer:** 10%\n\nFirst, we need to calculate the mass of 1 mole of water (H2O). The molecular weight of water is 18 g/mol, so 1 mole of water weighs 18 g.\n
The total mass of the solution is the mass of \"X\" plus the mass of water, which is 2 g + 18 g = 20 g.\n
We can calculate the mass percent of \"X\" in the solution with the following formula:\n\n
$$\\text{Mass percent of \"X\"} = \\left(\\frac{\\text{mass of \"X\"}}{\\text{total mass of solution}}\\right) \\times 100\\%$$\n\n
Substitute the given values into the formula :\n\n
A solution of sugar is obtained by mixing $$200 \\mathrm{~g}$$ of its $$25 \\%$$ solution and $$500 \\mathrm{~g}$$ of its $$40 \\%$$ solution (both by mass). The mass percentage of the resulting sugar solution is ___________ (Nearest integer)
Given below are two statements: one is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$
\n
Assertion A : $$3.1500 \\mathrm{~g}$$ of hydrated oxalic acid dissolved in water to make $$250.0 \\mathrm{~mL}$$ solution will result in $$0.1 \\mathrm{~M}$$ oxalic acid solution.
\n
Reason $$\\mathbf{R}$$ : Molar mass of hydrated oxalic acid is $$126 \\mathrm{~g} \\mathrm{~mol}^{-1}$$
\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": "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
Firstly, let's calculate the molarity of the oxalic acid solution given in the Assertion (A).
\n
The molarity (M) is defined as the number of moles of solute per liter of solution. The number of moles of solute is given by the mass of the solute divided by its molar mass.
\n
So, the number of moles of hydrated oxalic acid is $$\\frac{3.1500 \\, \\text{g}}{126 \\, \\text{g/mol}} = 0.025 \\, \\text{mol}$$
\n
And the volume of the solution is 250.0 mL, or 0.250 L.
\n
Therefore, the molarity of the solution is $$\\frac{0.025 \\, \\text{mol}}{0.250 \\, \\text{L}} = 0.1 \\, \\text{M}$$
\n
This matches the molarity given in the Assertion (A), so the Assertion is true.
\n
The Reason (R) gives the molar mass of hydrated oxalic acid as 126 g/mol, which was used in the calculation above. Therefore, the Reason is also true.
\n
Moreover, the Reason (R) is the correct explanation of Assertion (A), as it provides the necessary information (the molar mass of hydrated oxalic acid) to calculate the molarity of the solution.
\n
So, the correct option is C: Both A and R are true and R is the correct explanation of A.
The volume of $$0.02 ~\\mathrm{M}$$ aqueous $$\\mathrm{HBr}$$ required to neutralize $$10.0 \\mathrm{~mL}$$ of $$0.01 ~\\mathrm{M}$$ aqueous $$\\mathrm{Ba}(\\mathrm{OH})_{2}$$ is (Assume complete neutralization)
The reaction between $\\mathrm{HBr}$ and $\\mathrm{Ba(OH)_2}$ is a neutralization reaction where one $\\mathrm{Ba(OH)_2}$ reacts with two $\\mathrm{HBr}$ to form $\\mathrm{BaBr_2}$ and $2\\mathrm{H_2O}$:
Therefore, the volume of the $0.02 \\, \\mathrm{M}$ $\\mathrm{HBr}$ solution required to neutralize the $\\mathrm{Ba(OH)_2}$ solution is $10.0 \\, \\mathrm{mL}$.
The quantity that changes with temperature from the given options is Molarity.
\n
Molarity, which is denoted with a capital M, measures the number of moles of solute dissolved in one liter of solution. It is expressed as:
\n
$$ Molarity (M) = \\frac{moles \\ of \\ solute}{liters \\ of \\ solution} $$
\n
Since the volume of liquids expands or contracts with temperature, the volume of the solution will change as the temperature changes, which in turn will affect the molarity. For example, if the temperature of a solution increases, its volume may expand. If the number of moles of solute remains constant, the increased volume will result in a decreased molarity because the solute is now dispersed in a greater volume of solution.
\n
In contrast, other measures of concentration such as:
\n
Molality, which is measured as the number of moles of solute per kilogram of solvent (not affected by temperature since it's based on mass, not volume),
\n
$$ Molality (m) = \\frac{moles \\ of \\ solute}{kilograms \\ of \\ solvent} $$
\n
Mole fraction, which is the ratio of the number of moles of one component to the total number of moles of all components in the solution (also not affected by temperature since the ratio of moles doesn't change with temperature),
\n
$$ Mole \\ fraction \\ (X) = \\frac{moles \\ of \\ component}{total \\ moles \\ in \\ mixture} $$
\n
and Mass percentage, which is the percentage by mass of a component in a mixture (again, not affected by temperature because it's based on mass),
\n
$$ Mass \\ percentage \\ $$(%) $$= \\frac{mass \\ of \\ component}{total \\ mass \\ of \\ mixture} \\times 100\\% $$
\n
do not change with temperature. Therefore, the correct answer to the question is:
The molarity of $$1 \\mathrm{~L}$$ orthophosphoric acid $$\\left(\\mathrm{H}_3 \\mathrm{PO}_4\\right)$$ having $$70 \\%$$ purity by weight (specific gravity $$1.54 \\mathrm{~g} \\mathrm{~cm}^{-3}$$) is __________ $$\\mathrm{M}$$.
\n
(Molar mass of $$\\mathrm{H}_3 \\mathrm{PO}_4=98 \\mathrm{~g} \\mathrm{~mol}^{-1}$$)
A solution of $$\\mathrm{H}_2 \\mathrm{SO}_4$$ is $$31.4 \\% \\mathrm{H}_2 \\mathrm{SO}_4$$ by mass and has a density of $$1.25 \\mathrm{~g} / \\mathrm{mL}$$. The molarity of the $$\\mathrm{H}_2 \\mathrm{SO}_4$$ solution is _________ $$\\mathrm{M}$$ (nearest integer)
\n
[Given molar mass of $$\\mathrm{H}_2 \\mathrm{SO}_4=98 \\mathrm{~g} \\mathrm{~mol}^{-1}$$]
If a substance '$$A$$' dissolves in solution of a mixture of '$$B$$' and '$$C$$' with their respective number of moles as $$\\mathrm{n}_{\\mathrm{A}}, \\mathrm{n}_{\\mathrm{B}}$$ and $$\\mathrm{n}_{\\mathrm{C}_3}$$. Mole fraction of $$\\mathrm{C}$$ in the solution is
The mass of sodium acetate $$\\left(\\mathrm{CH}_3 \\mathrm{COONa}\\right)$$ required to prepare $$250 \\mathrm{~mL}$$ of $$0.35 \\mathrm{~M}$$ aqueous solution is ________ g. (Molar mass of $$\\mathrm{CH}_3 \\mathrm{COONa}$$ is $$82.02 \\mathrm{~g} \\mathrm{~mol}^{-1}$$)
Molarity $$(\\mathrm{M})$$ of an aqueous solution containing $$x \\mathrm{~g}$$ of anhyd. $$\\mathrm{CuSO}_4$$ in $$500 \\mathrm{~mL}$$ solution at $$32^{\\circ} \\mathrm{C}$$ is $$2 \\times 10^{-1} \\mathrm{M}$$. Its molality will be _________ $$\\times 10^{-3} \\mathrm{~m}$$. (nearest integer). [Given density of the solution $$=1.25 \\mathrm{~g} / \\mathrm{mL}$$]
3. Find the mass of the solvent (water) in the solution by subtracting the mass of the solute (anhydrous $$\\mathrm{CuSO}_4$$) from the total mass of the solution.
\n\n
We know the number of moles of $$\\mathrm{CuSO}_4$$, and we can find its molar mass.
\n\n
$$ \\text{Molar mass of } \\mathrm{CuSO}_4 = 63.5 + 32 + 4 \\times 16 = 159.5 \\text{ g/mol} $$
The Molarity (M) of an aqueous solution containing $$5.85 \\mathrm{~g}$$ of $$\\mathrm{NaCl}$$ in $$500 \\mathrm{~mL}$$ water is : (Given : Molar Mass $$\\mathrm{Na}: 23$$ and $$\\mathrm{Cl}: 35.5 \\mathrm{~gmol}^{-1}$$)
A solution is prepared by adding 1 mole ethyl alcohol in 9 mole water. The mass percent of solute in the solution is ________ (Integer answer) (Given : Molar mass in $$\\mathrm{g} \\mathrm{~mol}^{-1}$$ Ethyl alcohol : 46 water : 18)
To determine the mass percent of the solute (ethyl alcohol) in the solution, we start by calculating the masses of ethyl alcohol and water using their respective molar masses.
\n\n
The molar mass of ethyl alcohol (C2H5OH) is given as 46 g/mol and the molar mass of water (H2O) is given as 18 g/mol.
\n\n
First, we calculate the mass of ethyl alcohol:
\n\n
\n\n
$$\\text{Mass of ethyl alcohol} = \\text{Number of moles} \\times \\text{Molar mass}$$
Since the integer answer is requested, we round 22.12% to the nearest whole number. Therefore, the mass percent of the solute (ethyl alcohol) in the solution is 22%.
Molality of an aqueous solution of urea is $$4.44 \\mathrm{~m}$$. Mole fraction of urea in solution is $$x \\times 10^{-3}$$, Value of $$x$$ is ________. (Integer answer)
To determine the mole fraction of urea in a solution where the molality is $$4.44 \\ \\mathrm{m}$$, we first need to understand the definitions and relationships involved.
\n\n
Molality ($$\\mathrm{m}$$) is given by:
\n\n
$$ \\mathrm{m} = \\frac{\\text{moles of solute}}{\\text{mass of solvent in kg}} $$
\n\n
Here, the molality is $$4.44 \\ \\mathrm{m}$$, which means there are $$4.44$$ moles of urea (solute) per kilogram of water (solvent).
\n\n
The next step involves calculating the mole fraction. Mole fraction ($$X$$) of urea is given by:
\n\n
$$ X_{\\text{urea}} = \\frac{\\text{moles of urea}}{\\text{moles of urea} + \\text{moles of water}} $$
\n\n
Since we have $$4.44$$ moles of urea, let's determine the moles of water. The molar mass of water ($$\\mathrm{H_2O}$$) is approximately $$18 \\ \\mathrm{g/mol}$$. Therefore, the moles of water in $$1 \\ \\mathrm{kg}$$ (or $$1000 \\ \\mathrm{g}$$) of water are:
Molality $$(\\mathrm{m})$$ of $$3 \\mathrm{M}$$ aqueous solution of $$\\mathrm{NaCl}$$ is :\n(Given : Density of solution $$=1.25 \\mathrm{~g} \\mathrm{~mL}^{-1}$$, Molar mass in $$\\mathrm{g} \\mathrm{~mol}^{-1}: \\mathrm{Na}-23, \\mathrm{Cl}-35.5$$)
To calculate the molality $(\\mathrm{m})$ of a 3M (molar) aqueous solution of $\\mathrm{NaCl}$, we need to follow certain steps, using the given data and understanding the definitions properly.
\n\n
Molarity (M) is defined as the number of moles of solute per liter of solution. Molality $(\\mathrm{m})$, on the other hand, is defined as the number of moles of solute per kilogram of solvent. The given density of the solution is $1.25 \\mathrm{~g} \\mathrm{~mL}^{-1}$ which allows us to calculate the mass of the solvent in a given volume of solution.
\n\n
We know:
\n\n
\n\n
Molarity (M) of NaCl = $3 \\mathrm{~M}$
\n\n
Density of NaCl solution = $1.25 \\mathrm{~g} \\mathrm{~mL}^{-1}$
\n\n
Molar mass of $\\mathrm{NaCl} = \\mathrm{Na} + \\mathrm{Cl} = 23 + 35.5 = 58.5 \\mathrm{~g} \\mathrm{~mol}^{-1}$
\n\n
\n\n
First, to find the molality, we need to find the moles of NaCl in a given mass of the solvent. Keeping in mind that $3 \\mathrm{~M}$ means there are $3 \\mathrm{~moles}$ of $\\mathrm{NaCl}$ per liter of solution, we consider 1 liter (or 1000 mL) of solution to simplify our calculation, knowing that density can convert volume to mass directly.
\n\n
The mass of 1 liter of NaCl solution is $1.25 \\mathrm{~g} \\mathrm{~mL}^{-1} \\times 1000 \\mathrm{~mL} = 1250 \\mathrm{~g}$.
\n\n
Since we have a 3 M solution of NaCl,\n\n
$3 \\mathrm{~moles} = \\frac{\\text{mass of the solution} - \\text{mass of solvent}}{\\text{molar mass of NaCl}}$
\n\n
With 3 moles of NaCl, the mass of NaCl present in 1 L solution $= 3 \\mathrm{~moles} \\times 58.5 \\mathrm{~g/mol} = 175.5 \\mathrm{~g}$
\n\n
The mass of the solvent (water) can be found by subtracting the mass of NaCl from the total mass of the solution.
The density of '$$x$$' $$\\mathrm{M}$$ solution ('$$x$$' molar) of $$\\mathrm{NaOH}$$ is $$1.12 \\mathrm{~g} \\mathrm{~mL}^{-1}$$, while in molality, the concentration of the solution is $$3 \\mathrm{~m}$$ ( 3 molal). Then $$x$$ is
\n
(Given : Molar mass of $$\\mathrm{NaOH}$$ is $$40 \\mathrm{~g} / \\mathrm{mol}$$)
To find the molarity, $$x$$, of the $$\\mathrm{NaOH}$$ solution, we need to relate the given density and molality. Molarity is defined as the number of moles of solute per liter of solution, while molality is defined as the number of moles of solute per kilogram of solvent. We are given the following:
\n\n
\n\n
Molality (m) = $$3\\ \\mathrm{mol/kg}$$ (3 molal)
\n\n
Density of solution ($$\\rho$$) = $$1.12\\ \\mathrm{g/mL}$$ = $$1120\\ \\mathrm{g/L}$$
\n\n
Molar mass of $$\\mathrm{NaOH}$$ ($$M_\\mathrm{NaOH}$$) = $$40\\ \\mathrm{g/mol}$$
\n\n
\n\n
First, let's find the number of moles of $$\\mathrm{NaOH}$$ in 1 kg of water, which is the definition of molality. Since the molality is 3 molal, we have:
\n\n
$$3\\ \\mathrm{mol}$$ of $$\\mathrm{NaOH}$$ in 1 kg of water (or 1000 g of water).
\n\n
Now, to find the molarity, we need the volume of the solution in which these moles of $$\\mathrm{NaOH}$$ are dissolved. The mass of 1 L (1000 mL) of solution, given the density, is:
So, we have 3 moles of $$\\mathrm{NaOH}$$ in 1 L of solution. Therefore, the molarity (x) of the solution is directly 3 M (since molarity is moles of solute per liter of solution). Hence, $$x = 3$$.
\n\n
Therefore, the correct answer is Option B: 3.0.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2860,
"subject": "Chemistry",
"question": "What volume of hydrogen gas at 273 K and 1 atm pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen?",
"options": [
{
"text": "67.2 L"
},
{
"text": "44.8 L"
},
{
"text": "22.4 L"
},
{
"text": "89.6 L"
}
],
"answer": "67.2 L",
"solution": "**Answer:** 67.2 L\n\nThe reaction is\n 2BCl3 + 3H2 $$ \\to $$ 2B + 6HCL\n
$$\\therefore$$ 3 moles of hydrogen produces 2 moles of boron\n
So 2$$ \\times $$10.8 gm = 21.6 gm of boron is produces from 3$$ \\times $$22.4 = 67.2 L hydrogen",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2861,
"subject": "Chemistry",
"question": "The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is",
"options": [
{
"text": "urea"
},
{
"text": "benzamide"
},
{
"text": "acetamide"
},
{
"text": "thiourea"
}
],
"answer": "urea",
"solution": "**Answer:** urea\n\nInitially total H2SO4 present = 100 mL of 0.1 M = $${{{100} \\over {1000}} \\times 0.1}$$ mole = 0.01 mole\n
2NaOH + H2SO4 $$ \\to $$ Na2SO4 + 2H2O\n Let in this reaction H2SO4 required n mole\n
% of N in urea [(NH4)2CO] = $${{14 \\times 2} \\over {60}} \\times 100$$ = 46.6 %\n [ Mol weight of urea = 60]\n
% of N in benzamide [C6H5CONH2] = $${14 \\over {121}} \\times 100$$ = 11.5 %\n [ Mol weight of benzamide [C6H5CONH2] = 121]\n
% of N in acetamide [CH3CONH2] = $${14 \\over {59}} \\times 100$$ = 23.4 %\n [ Mol weight of acetamide [CH3CONH2] = 59]\n
% of N in thiourea [NH2CONH2] = $${{14 \\times 2} \\over {76}} \\times 100$$ = 36.8 %\n [ Mol weight of thiourea [NH2CSNH2] = 76]\n
$$\\therefore$$ compound is urea.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2862,
"subject": "Chemistry",
"question": "The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of\ntheir molecule is:",
"options": [
{
"text": "1: 8"
},
{
"text": "3 : 16"
},
{
"text": "1 : 4"
},
{
"text": "7 : 32"
}
],
"answer": "7 : 32",
"solution": "**Answer:** 7 : 32\n\nGiven ratio of mass of O2 and N2 = 1 : 4\n Let mass of O2 = w\n and mass of N2 = 4w\n
$$\\therefore$$ Number of moles of O2 = $${w \\over {32}}$$\n Number of molecules of O2 = $${w \\over {32}}$$$$ \\times $$NA\n
Number of moles of N2 = $${4w \\over {28}}$$\n Number of molecules of N2 = $${4w \\over {28}}$$$$ \\times $$NA\n
$$\\therefore$$ Ratio of molecules of O2 and N2\n = $${w \\over {32}}$$$$ \\times $$NA : $${4w \\over {28}}$$$$ \\times $$NA \n
= 7 : 32 ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2863,
"subject": "Chemistry",
"question": "What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must\nbe mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid\nsolution ?\n",
"options": [
{
"text": "320"
},
{
"text": "325"
},
{
"text": "316"
},
{
"text": "330"
}
],
"answer": "316",
"solution": "**Answer:** 316\n\n
Let the volume of a monoprotic acid solution in mL be V
$${{5V} \\over {20}} = 79 \\Rightarrow V = 316$$ mL
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2864,
"subject": "Chemistry",
"question": "A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of CO2 at T = 298.15 K and p = 1 bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet ? \n [Molar mass of NaHCO3 = 84 g mol–1]",
"options": [
{
"text": "33.6"
},
{
"text": "0.84"
},
{
"text": "8.4"
},
{
"text": "16.8"
}
],
"answer": "8.4",
"solution": "**Answer:** 8.4\n\n2NaHCO3\n + (COOH)2 $$ \\to $$ (COONa)2 + 2H2O + 2CO2\n
$$ \\therefore $$ 1 mole of CO2 is produced by 1 mole of NaHCO3.\n
Given, volume of CO2 produced = 0.25 ml\n
25 L of CO2 contains 1 mol\n
$$ \\therefore $$ 0.25 ml of CO2 contains = $${1 \\over {25 \\times {{10}^3}}} \\times 0.25$$ moles\n
= 10-5 moles\n
$$ \\therefore $$ Moles of NaHCO3 = 10-5 moles\n
Mass of NaHCO3 = 10-5 $$ \\times $$ 84 g\n
$$ \\therefore $$ % of NaHCO3 in a tablet \n
= $${{84 \\times {{10}^{ - 5}}} \\over {10 \\times {{10}^{ - 3}}}} \\times 100$$ = 8.4 %",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2865,
"subject": "Chemistry",
"question": "Amongst the following statements, that which was not proposed by Dalton was :",
"options": [
{
"text": "Matter consists of indivisible atoms all the atoms of a given element have."
},
{
"text": "Chemical reactions involve reorganization of atoms. These are neither created not destroyed in\na chemical reaction."
},
{
"text": "When gases combine or reproduced in a chemical reactionn they do so in a simple ratio by\nvolume provided all gases are the same T & P."
},
{
"text": "Identical properties including identical mass. Atoms of differemt element differ in mass."
}
],
"answer": "When gases combine or reproduced in a chemical reactionn they do so in a simple ratio by\nvolume provided all gases are the same T & P.",
"solution": "**Answer:** When gases combine or reproduced in a chemical reactionn they do so in a simple ratio by\nvolume provided all gases are the same T & P.\n\nOption(3) is according to Avogadro's law of\nvolume combination.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2866,
"subject": "Chemistry",
"question": "The volume (in mL) of 0.125 M AgNO3 required to quantitatively precipitate chloride ions in 0.3 g of\n[Co(NH3)6]Cl3 is ________.\n M[Co(NH3)6Cl3] = 267.46 g/mol\n MAgNO3 = 169.87 g/mol",
"options": [],
"answer": "26.80to27.00",
"solution": "**Answer:** 26.80to27.00\n\n[Co(NH3)6]Cl3 + 3AgNO3 $$ \\to $$ 3AgCl\n
If the above equation is balanced with integer coefficients, the value of c is ___________. (Round off to the Nearest Integer).",
"options": [],
"answer": "16",
"solution": "**Answer:** 16\n\n$$2MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \\to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 2869,
"subject": "Chemistry",
"question": "250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is ___________ $$\\times$$ 1021. (Nearest integer)
(NA = 6.022 $$\\times$$ 1023)",
"options": [],
"answer": "226",
"solution": "**Answer:** 226\n\nWe know that, number of moles = VL $$\\times$$ molarity and number of millimoles = VmL $$\\times$$ molarity
So, millimoles of NaOH = 250 $$\\times$$ 0.5 = 125
Millimoles of HCl = 500 $$\\times$$ 1 = 500
Now, reaction is
125 millimoles of NaOH reacts with 125 millimoles of HCl. So, millimoles of HCl left = 375
Moles of HCl = 375 $$\\times$$ 10$$-$$3
Number of HCl molecules
= Avogadro's constant (NA) $$\\times$$ moles of HCl
A protein 'A' contains 0.30% of glycine (molecular weight 75). The minimum molar mass of the protein 'A' is __________ $$\\times$$ 103 g mol$$-$$1 [nearest integer]
Number of grams of bromine that will completely react with 5.0 g of pent-1-ene is ___________ $$\\times$$ 10$$-$$2 g. (Atomic mass of Br = 80 g/mol) [Nearest Integer]
Blister copper is produced by reaction of copper oxide with copper sulphide.
\n
2Cu2O + Cu2S $$\\to$$ 6Cu + SO2
\n
When 2.86 $$\\times$$ 103 g of Cu2O and 4.77 $$\\times$$ 103 g of Cu2S are used for reaction, the mass of copper produced is _____________ g. (nearest integer)
\n
(Atomic mass of Cu = 63.5 a.m. u, S = 32.0 a.m. u, O = 16.0 a.m. u)
56.0 L of nitrogen gas is mixed with excess hydrogen gas and it is found that 20 L of ammonia gas is produced. The volume of unused nitrogen gas is found to be _________ L.
\n$$\n=46 \\mathrm{~L}\n$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2875,
"subject": "Chemistry",
"question": "Number of atoms in 558.5 gram Fe (at. wt. of Fe = 55.85 g mol-1) is",
"options": [
{
"text": "twice that in 60 g carbon"
},
{
"text": "6.023 $$\\times$$ 1022"
},
{
"text": "half that in 8 g He"
},
{
"text": "558.5 $$\\times$$ 6.023 $$\\times$$ 1023"
}
],
"answer": "twice that in 60 g carbon",
"solution": "**Answer:** twice that in 60 g carbon\n\nNumber of moles of Fe in 558.5 gm of Fe is
\n$$ = $$ $${{558.5} \\over {55.85}}$$
\n$$ = $$ 10
\n$$\\therefore$$ Number moles of Fe in 558.5 gm Fe is = 10
\nSince here Fe is monoatomic then no. of atoms in 10 moles Fe is
\n$$ \\Rightarrow $$ 10 $$\\times$$ 1 $$\\times$$ NA
\n$$ \\Rightarrow $$ 10 NA
\nFor Option (A) Atomic Mass of Carbon is 12 g then no. of moles in 60 g Carbon is
\n$$ \\Rightarrow $$ $${{60} \\over {12}} = 5$$
\n Here Carbon is also monoatomic then no. of atoms in 5 moles of Carbon
\n$$ \\Rightarrow $$ 5 $$\\times$$ 1 $$\\times$$ NA
\n$$ \\Rightarrow $$ 5 NA
\nNow we can clearly see 10 is twice of 5, so option (A) is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2876,
"subject": "Chemistry",
"question": "If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of the substance will ",
"options": [
{
"text": "be a function of the molecular mass of the substance"
},
{
"text": "remain unchanged"
},
{
"text": "increase two fold"
},
{
"text": "decrease twice"
}
],
"answer": "decrease twice",
"solution": "**Answer:** decrease twice\n\nLet Relative Atomic Mass is in short R.A.M then
\nas 6 is half of 12 then R.A.M will decrease twice.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2877,
"subject": "Chemistry",
"question": "How many moles of magnesium phosphate Mg3(PO4)2 will contain 0.25 mole of oxygen atoms?",
"options": [
{
"text": "1.25 $$\\times$$ 10-2"
},
{
"text": "2.5 $$\\times$$ 10-2"
},
{
"text": "0.02"
},
{
"text": "3.125 $$\\times$$ 10-2"
}
],
"answer": "3.125 $$\\times$$ 10-2",
"solution": "**Answer:** 3.125 $$\\times$$ 10-2\n\nIn one Mg3(PO4)2 molecule no of oxygen atoms = 8\n
So if there is M mole of Mg3(PO4)2 molecules then no of moles of oxygen atoms is = M $$ \\times $$ 8\n
According to question,\n
M $$ \\times $$ 8 = 0.25\n
$$ \\Rightarrow $$ M = 0.03125 = 3.125 $$\\times$$ 10-2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2878,
"subject": "Chemistry",
"question": "In the reaction
\n2Al(s) + 6HCl(aq) $$\\to$$ 2Al3+ (aq) + 6Cl-(aq) + 3H2(g)",
"options": [
{
"text": "11.2 L H2(g) at STP is produced for every mole HCl(aq) consumed"
},
{
"text": "6 L HCl(aq) is consumed for every 3L H2(g) produced"
},
{
"text": "33.6 L H2(g) is produced regardless of temperature and pressure for every mole that reacts"
},
{
"text": "67.2 H2(g) at STP is produced for every mole Al that reacts"
}
],
"answer": "11.2 L H2(g) at STP is produced for every mole HCl(aq) consumed",
"solution": "**Answer:** 11.2 L H2(g) at STP is produced for every mole HCl(aq) consumed\n\nOption A :\n 6 moles of HCl(aq) produces = 3 moles of H2(g)\n
$$\\therefore$$ 1 mole of HCl(aq) produces = $${3 \\over 6}$$ moles of H2(g) = $${1 \\over 2}$$ moles of H2(g)\n
From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L.\n
According to avogadro Hypothesis, volume is directly proportional to the no of moles of gases.\n Only for gases, no of mole and volume relationship is possible i.e, for two gases if no of moles are same then their volume is also same. But here HCL is in aqueous form and H2 is in gaseous form so we can't find volume of HCL(aq) using no of moles of HCL(aq).\n
$$\\therefore$$ Option (B) is wrong.\n
Option C :\n
2 moles of Al(s) produces = 3 moles of H2(g)\n
$$\\therefore$$ 1 mole of Al(s) produces = $${3 \\over 2}$$ moles of H2(g)\n
From avogadro hypothesis we know at STP 1 mole of any as occupies 22.4 L.\n
$$\\therefore$$ Option (D) is wrong.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2879,
"subject": "Chemistry",
"question": "The molecular formula of a commercial resin used for exchanging ions in water softening is\nC8H7SO3Na (Mol. Wt. 206). What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin?",
"options": [
{
"text": "2/309"
},
{
"text": "1/412"
},
{
"text": "1/103"
},
{
"text": "1/206"
}
],
"answer": "1/412",
"solution": "**Answer:** 1/412\n\n$$2$$ mole of water softner require $$1$$ mole of $$C{a^{2 + }}\\,\\,$$ ion\n
So, $$1$$ mole of water softner require $${1 \\over 2}$$ mole of $$C{a^{2 + }}\\,$$ ion\n
Thus, $${1 \\over {2 \\times 206}} = {1 \\over {412}}mol/g\\,\\,\\,$$ will be maximum uptake",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2880,
"subject": "Chemistry",
"question": "The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid istreated with excess H2S in the presence of conc. HCl ( assuming 100% conversion) is :",
"options": [
{
"text": "0.50 mol "
},
{
"text": "0.25 mol"
},
{
"text": "0.125 mol"
},
{
"text": "0.333 mol"
}
],
"answer": "0.125 mol",
"solution": "**Answer:** 0.125 mol\n\n2H3 As O4 + 5H2S $$\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{HCl}^{conc}} $$ As2S5 + 8H2O\n
35.5 g of H3AsO4 = $${{35.5} \\over {142}}$$ = 0.25 moles\n
For this reaction, 2KClO3 $$\\buildrel \\, \\over\n \\longrightarrow $$ 2KCl + 3O2 \n
We can write, $${{{n_{KCl{O_3}}}} \\over 2} = {{{n_{KCl}}} \\over 2} = {{{n_{{O_2}}}} \\over 3}$$\n
This means \n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2883,
"subject": "Chemistry",
"question": "The most abundant elements by mass in the body of a healthy human adult are: Oxygen (61.4%); Carbon\n(22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all\n1H atoms are replaced by 2H atoms is: ",
"options": [
{
"text": "37.5 kg"
},
{
"text": "7.5 kg"
},
{
"text": "10 kg"
},
{
"text": "15 kg"
}
],
"answer": "7.5 kg",
"solution": "**Answer:** 7.5 kg\n\nGiven that weight of human adult is = 75 kg\n
Among those 75 kg, 10% is Hydrogen(1H).\n
$$\\therefore$$ Mass of 1H = $$75 \\times {{10} \\over {100}}$$ = 7.5 kg\n
Now when every 1H atom is replaced by 2H atom then weight of every atom is become double. So total weight of 2H becomes = 2$$ \\times $$7.5 = 15 kg.\n
So the weight gain by the person = 15 - 7.5 = 7.5 kg",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2884,
"subject": "Chemistry",
"question": "A sample of $$NaCl{O_3}$$ is converted by heat to $$NaCl$$ with a loss of $$0.16$$ $$g$$ of oxygen. The residue is dissolved in water and precipitated as $$AgCl.$$ The mass of $$AgCl$$ (in $$g$$) obtained will be : (Given : Molar mass of $$AgCl=143.5$$ $$g$$ $$mo{l^{ - 1}}$$)",
"options": [
{
"text": "$$0.35$$ "
},
{
"text": "$$0.41$$ "
},
{
"text": "$$0.48$$ "
},
{
"text": "$$0.54$$ "
}
],
"answer": "$$0.48$$ ",
"solution": "**Answer:** $$0.48$$ \n\nNaClO3 $$\\buildrel \\Delta \\over\n \\longrightarrow $$ 2NaCl + 3O2\n
Here O2 produced = 0.16 g\n
$$\\therefore\\,\\,\\,\\,$$ No of moles of O2 = $${{0.16} \\over {32}}$$ = 5 $$ \\times $$10$$-$$3 \n
$$\\therefore\\,\\,\\,\\,$$ no of moles in AgCl = $${1 \\over {300}}$$ \n
$$\\therefore\\,\\,\\,\\,$$ Mass of AgCl = 143.5 $$ \\times $$ $${1 \\over {300}}$$ = 0.48 g",
"topic": "Number Theory",
"subtopic": "Residues and Primitive Roots"
},
{
"id": 2885,
"subject": "Chemistry",
"question": "An organic compound is estimated through Duma's method and was found to evolve 6 moles of CO2. 4 moles of H2O and 1 mole of nitrogen gas. The formula of the compound is : ",
"options": [
{
"text": "C6H8N"
},
{
"text": "C6H8N2"
},
{
"text": "C12H8N"
},
{
"text": "C12H8N2"
}
],
"answer": "C6H8N2",
"solution": "**Answer:** C6H8N2\n\nMolar ratio of C : H : N : : 6 : 8 : 2 i.e., 3 : 4 : 1\n
Thus, the correct formula is C6H8N2.\n.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2886,
"subject": "Chemistry",
"question": "The minimum amount of O2(g) consumed per gram of reactant is for the reaction : (Given atomic mass : Fe =\n56, O = 16, Mg = 24, P = 31, C = 12, H = 1)\n",
"options": [
{
"text": "4Fe(s) + 3O2(g) $$ \\to $$ 2Fe2O3(s)"
},
{
"text": "P4(s) + 5O2(g) $$ \\to $$ P4O10(s)"
},
{
"text": "C3H8(g) + 5O2(g) $$ \\to $$ 3CO2(g) + 4H2O(l)"
},
{
"text": "2Mg(s) + O2(g) $$ \\to $$ 2MgO(s)"
}
],
"answer": "4Fe(s) + 3O2(g) $$ \\to $$ 2Fe2O3(s)",
"solution": "**Answer:** 4Fe(s) + 3O2(g) $$ \\to $$ 2Fe2O3(s)\n\n(a) 4Fe(s) + 3O2(g) $$ \\to $$ 2Fe2O3(s)\n
$$ \\Rightarrow $$ Moles of O2 = $${1 \\over 2}$$ $$ \\times $$ $${1 \\over {24}}$$ moles = $${1 \\over {48}}$$ moles = $${1 \\over {48}} \\times 32$$ g of O2 = 0.66 g of O2\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2887,
"subject": "Chemistry",
"question": "5 moles of AB2 weigh 125 × 10–3\n kg and 10 moles of A2B2 weigh 300 × 10–3\n kg. The molar mass of A (MA)\nand molar mass of B (MB) in kg mol are:\n",
"options": [
{
"text": "MA = 10 × 10–3\n and MB = 5 × 10–3"
},
{
"text": "MA = 25 × 10–3\n and MB = 50 × 10–3"
},
{
"text": "MA = 5 × 10–3\n and MB = 10 × 10–3"
},
{
"text": "MA = 50 × 10–3\n and MB = 25 × 10–3"
}
],
"answer": "MA = 5 × 10–3\n and MB = 10 × 10–3",
"solution": "**Answer:** MA = 5 × 10–3\n and MB = 10 × 10–3\n\nMolar weigh of 'A' = MA\n
Molar weigh of 'B' = MB\n
Molar mass of AB2 = MA + 2MB\n
Molar mass of A2B2 = 2(MA + MB)\n
5 mol AB2 weighs 125 g\n
$$ \\therefore $$ AB2 = 25 g/mol\n
10 mol A2B2 weighs 300 g\n
$$ \\therefore $$ A2B2 = 30 g/mol\n
So, MA + 2MB = 25 ..............(1)\n
and 2(MA + MB) = 30 ..............(2)\n
By solving (1) and (2),\n
MA = 5 gm = 5 $$ \\times $$ 10-3 kg\n
MB = 10 gm = 10 $$ \\times $$ 10-3 kg",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2888,
"subject": "Chemistry",
"question": "25 g of an unknown hydrocarbon upon burning produces 88 g of CO2 and 9 g of H2O. This unknown\nhydrocarbon contains :",
"options": [
{
"text": "18 g of carbon and 7 g of hydrogen"
},
{
"text": "20 g of carbon and 5 g of hydrogen"
},
{
"text": "22 g of carbon and 3 g of hydrogen"
},
{
"text": "24 g of carbon and 1 g of hydrogen"
}
],
"answer": "24 g of carbon and 1 g of hydrogen",
"solution": "**Answer:** 24 g of carbon and 1 g of hydrogen\n\nCxHy + $$\\left( {x + {y \\over 4}} \\right)$$O2 $$ \\to $$ xCO2 + $${y \\over 2}$$H2O\n
No of moles of CO2 produced = $${{88} \\over {44}}$$ = 2 mol\n No of moles of H2O produced = $${9 \\over {18}}$$ = 0.5 mol\n
According to stoichiometric mole mole analysis :\n
$$ \\Rightarrow $$ $${x \\over y} = {3 \\over 1}$$",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 2891,
"subject": "Chemistry",
"question": "The NaNO3 weighed out to make 50 mL of an aqueous solution containing 70.0 mg Na+ per mL is _________ g. (Rounded off to the nearest integer)
[Given : Atomic weight in g mol$$-$$1 - Na : 23; N : 14; O : 16]",
"options": [],
"answer": "13",
"solution": "**Answer:** 13\n\nNa+ = 70 mg/mL
WNa+ in 50 mL solution
= 70 $$\\times$$ 50 mg
= 3500 mg
= 3.5 gm
Moles of Na+ in 50 mL solution = $${{3.5} \\over {23}}$$
Moles of NaNO3 = moles of Na+
= $${{3.5} \\over {23}}$$ mol
Mass of NaNO3 = $${{3.5} \\over {23}} \\times 85 = 12.934$$
$$ \\simeq $$ 13 gm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2892,
"subject": "Chemistry",
"question": "A reaction of 0.1 mole of Benzylamine with bromomethane gave 23 g of Benzyl trimethyl ammonium bromide. The number of moles of bromomethane consumed in this reaction are n $$\\times$$ 10$$-$$1, when n = __________. (Round off to the Nearest Integer).
(Given : Atomic masses : C : 12.0 u, H : 1.0 u, N : 14.0 u, Br : 80.0 u]",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nNumber of moles of benzyl trimethyl
ammonium\nbromide formed = $${{23} \\over {230}}$$ = 0.1\n
$$ \\therefore $$ No. of moles of bromomethane consumed\n
= 3 $$ \\times $$ 0.1\n
= 3 $$ \\times $$ 10–1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2893,
"subject": "Chemistry",
"question": "An average person needs about 10000 kJ energy per day. The amount of glucose (molar mass = 180.0 g mol$$-$$1) needed to meet this energy requirement is ____________ g.
Hence, amount of glucose required is 667 g.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2894,
"subject": "Chemistry",
"question": "4g equimolar mixture of NaOH and Na2CO3 contains x g of NaOH and y g of Na2CO3. The value of x is ____________ g. (Nearest integer)",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nMass of NaOH = x
Moles of NaOH = $${x \\over {40}}$$
Mass of Na2CO3 = y
Moles of Na2CO3 = $${y \\over {106}}$$
$${x \\over {40}} = {y \\over {106}}$$
x + y = 4
x = 1.1, y = 2.9
x = 1.1 $$ \\approx $$ 1 (nearest integer)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2895,
"subject": "Chemistry",
"question": "The number of atoms in 8g of sodium is x $$\\times$$ 1023. The value of x is ____________. (Nearest integer)
[Given : NA = 6.02 $$\\times$$ 1023 mol$$-$$1
Atomic mass of Na = 23.0 u]",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nGiven, mass of Na = 8g
Two elements A and B which form 0.15 moles of A2B and AB3 type compounds. If both A2B and AB3 weigh equally, then the atomic weight of A is _____________ times of atomic weight of B.
1 L aqueous solution of H2SO4 contains 0.02 m mol H2SO4. 50% of this solution is diluted with deionized water to give 1 L solution (A). In solution (A), 0.01 m mol of H2SO4 are added. Total m mols of H2SO4 in the final solution is ___________ $$\\times$$ 103 m mols.
",
"options": [],
"answer": "0",
"solution": "**Answer:** 0\n\n$\\mathrm{n}_{\\mathrm{H}_2 \\mathrm{SO}_4}$ in $\\mathrm{Sol}^{\\mathrm{n}} \\mathrm{A}=50 \\%$ of original solution
$=0.01 \\mathrm{~m} \\mathrm{~mol}$.
\n$\\mathrm{n}_{\\mathrm{H}_2 \\mathrm{SO}_4}$ in Final solution $=0.01+0.01$
A sample of 4.5 mg of an unknown monohydric alcohol, R-OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be 3.1 mL. The molecular weight of the unknown alcohol is __________ g/mol. [Nearest integer]
A $$100 \\mathrm{~mL}$$ solution of $$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{MgBr}$$ on treatment with methanol produces $$2.24 \\mathrm{~mL}$$ of a gas at STP. The weight of gas produced is _____________ mg. [nearest integer]
In Carius method of estimation of halogen, $$0.45 \\mathrm{~g}$$ of an organic compound gave $$0.36 \\mathrm{~g}$$ of $$\\mathrm{AgBr}$$. Find out the percentage of bromine in the compound.
When coal of purity 60% is allowed to burn in presence of insufficient oxygen, 60% of carbon is converted into 'CO' and the remaining is converted into '$$\\mathrm{CO}_{2}$$'. The heat generated when $$0.6 \\mathrm{~kg}$$ of coal is burnt is _________.
A 1.84 mg sample of polyhydric alcoholic compound 'X' of molar mass 92.0 g/mol gave 1.344 mL of $$\\mathrm{H}_{2}$$ gas at STP. The number of alcoholic hydrogens present in compound 'X' is ________.
STP conditions define the volume of 1 mole of any gas as $22.4$ liters or $22400$ mL. The volume of hydrogen gas evolved here is $1.344$ mL, so we can calculate the number of moles of hydrogen gas ($H_2$) using the formula:
So, compound 'X' contains $6$ alcoholic hydrogens.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2905,
"subject": "Chemistry",
"question": "A sample of a metal oxide has formula $\\mathrm{M}_{0.83} \\mathrm{O}_{1.00}$. The metal $\\mathrm{M}$ can exist in two oxidation states $+2$ and $+3$.
In the sample of $\\mathrm{M}_{0.83} \\mathrm{O}_{1.00}$, the percentage of metal ions existing in $+2$ oxidation state is __________ $\\%$. (nearest integer)",
"options": [],
"answer": "59",
"solution": "**Answer:** 59\n\nLet, metal $\\mathrm{M}$ present x and (0.83 - x) in oxidation states $+2$ and $+3$ respectively.\n
Zinc reacts with hydrochloric acid to give hydrogen and zinc chloride. The volume of hydrogen gas produced at STP from the reaction of $$11.5 \\mathrm{~g}$$ of zinc with excess $$\\mathrm{HCl}$$ is __________ L (Nearest integer)
\n
(Given : Molar mass of $$\\mathrm{Zn}$$ is $$65.4 \\mathrm{~g} \\mathrm{~mol}^{-1}$$ and Molar volume of $$\\mathrm{H}_{2}$$ at $$\\mathrm{STP}=22.7 \\mathrm{~L}$$ )
On complete combustion, $$0.492 \\mathrm{~g}$$ of an organic compound gave $$0.792 \\mathrm{~g}$$ of $$\\mathrm{CO}_{2}$$. The % of carbon in the organic compound is ___________ (Nearest integer)
",
"options": [],
"answer": "44",
"solution": "**Answer:** 44\n\nWeight of $\\mathrm{C}$ in $0.792 \\mathrm{gm} $ $\\mathrm{CO}_2$\n
The volume of HCl, containing 73 g L$$^{-1}$$, required to completely neutralise NaOH obtained by reacting 0.69 g of metallic sodium with water, is __________ mL. (Nearest Integer)
\n
(Given : molar masses of Na, Cl, O, H, are 23, 35.5, 16 and 1 g mol$$^{-1}$$ respectively.)
When 0.01 mol of an organic compound containing 60% carbon was burnt completely, 4.4 g of CO$$_2$$ was produced. The molar mass of compound is _____________ g mol$$^{-1}$$ (Nearest integer).
Number of hydrogen atoms per molecule of a hydrocarbon A having 85.8% carbon is __________
(Given : Molar mass of A = 84 g mol$$^{-1}$$)
",
"options": [],
"answer": "12",
"solution": "**Answer:** 12\n\nMolar mass of a hydrocarbon $(A)=84 \\mathrm{~g} / \\mathrm{mol}$\n
\nMass of carbon in $1 \\mathrm{~mol}$ of $(A)=\\frac{85.8}{100} \\times 84$\n
\n$$\n=72 ~\\mathrm{gm}\n$$\n
\nMass of hydrogen in $1 \\mathrm{~mol}$ of $(A)=12 ~\\mathrm{gm}$\n
\n$\\therefore$ Number of $\\mathrm{H}$-atoms in a molecule of $(\\mathrm{A})=12$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2911,
"subject": "Chemistry",
"question": "
In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is ____________ (Nearest Integer)
When $$\\mathrm{Fe_{0.93}O}$$ is heated in presence of oxygen, it converts to $$\\mathrm{Fe_2O_3}$$. The number of correct statement/s from the following is ________
\n
A. The equivalent weight of $$\\mathrm{Fe_{0.93}O}$$ is $${{\\mathrm{Molecular\\,weight}} \\over {0.79}}$$
\n
B. The number of moles of Fe$$^{2+}$$ and Fe$$^{3+}$$ in 1 mole of $$\\mathrm{Fe_{0.93}O}$$ is 0.79 and 0.14 respectively
\n
C. $$\\mathrm{Fe_{0.93}O}$$ is metal deficient with lattice comprising of cubic closed packed arrangement of O$$^{2-}$$ ions
\n
D. The % composition of Fe$$^{2+}$$ and Fe$$^{3+}$$ in $$\\mathrm{Fe_{0.93}O}$$ is 85% and 15% respectively
",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nLet the number of $\\mathrm{Fe}^{2+}$ ions be $x$ and number of $\\mathrm{Fe}^{3+}$ ions be $0.93-x$.\n
According to the charge neutrality,\n
$$\n\\begin{aligned}\n& 2 x+3(0.93-x)=2 \\Rightarrow 2 x-3 x+2.79=2 \\Rightarrow x=0.79 \\\\\\\\\n& \\mathrm{Fe}^{2+} \\text { ion }=x=0.79 \\\\\\\\\n& \\mathrm{Fe}^{3+} \\text { ion }=0.93-x=0.93-0.79=0.14 \\\\\\\\\n& \\% \\text { of } \\mathrm{Fe}^{2+} \\text { ion }=\\frac{0.79}{0.93} \\times 100=84.95 \\% \\approx 85 \\% \\\\\\\\\n& \\% \\text { of } \\mathrm{Fe}^{3+} \\text { ion }=\\frac{0.14}{0.93} \\times 100=15.05 \\%\n\\end{aligned}\n$$\n
As it is a metal deficiency defect, \n
so, 1 mol of $\\mathrm{Fe}_{0.93} \\mathrm{O}$ contains $1 \\mathrm{~mol}$ of oxide ions and $0.93 \\mathrm{~mol}$ of iron ions as $\\mathrm{Fe}^{2+}$ and $\\mathrm{Fe}^{3+}$.\n
$\\therefore $ For $0.79 \\,\\mathrm{Fe}^{2+}, n$-factor $=0.79$\n
Out of $0.93 \\mathrm{~mol}$, there are $0.79 \\mathrm{~mol} \\,\\mathrm{Fe}^{2+}$ ions are present.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2913,
"subject": "Chemistry",
"question": "
$$1 \\mathrm{~g}$$ of a carbonate $$\\left(\\mathrm{M}_{2} \\mathrm{CO}_{3}\\right)$$ on treatment with excess $$\\mathrm{HCl}$$ produces $$0.01 \\mathrm{~mol}$$ of $$\\mathrm{CO}_{2}$$. The molar mass of $$\\mathrm{M}_{2} \\mathrm{CO}_{3}$$ is __________ $$\\mathrm{g} ~\\mathrm{mol}^{-1}$$. (Nearest integer)
",
"options": [],
"answer": "100",
"solution": "**Answer:** 100\n\nWhen the carbonate reacts with HCl, it produces CO2, as shown in the following balanced equation:\n
\nFrom the problem, we know that 1 g of the carbonate produces 0.01 mol of CO2.
To determine the molar mass of the carbonate, we can use the stoichiometry of the reaction:\n
\n1 mol of $$\\mathrm{M}_2\\mathrm{CO}_3$$ produces 1 mol of CO2.\n
\nSo, 0.01 mol of CO2 corresponds to 0.01 mol of $$\\mathrm{M}_2\\mathrm{CO}_3$$.\n
\nNow, we can find the molar mass of $$\\mathrm{M}_2\\mathrm{CO}_3$$:\n
\n$$\\frac{\\text{mass of }\\mathrm{M}_2\\mathrm{CO}_3}{\\text{moles of }\\mathrm{M}_2\\mathrm{CO}_3} = \\text{molar mass of }\\mathrm{M}_2\\mathrm{CO}_3$$\n
\n$$\\frac{1 \\text{ g}}{0.01 \\text{ mol}} = 100 \\text{ g mol}^{-1}$$\n
\nSo, the molar mass of $$\\mathrm{M}_2\\mathrm{CO}_3$$ is approximately 100 g/mol.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2914,
"subject": "Chemistry",
"question": "
A metal chloride contains $$55.0 \\%$$ of chlorine by weight . $$100 \\mathrm{~mL}$$ vapours of the metal chloride at STP weigh $$0.57 \\mathrm{~g}$$. The molecular formula of the metal chloride is
\n
(Given: Atomic mass of chlorine is $$35.5 \\mathrm{u}$$)
The weight percent of chlorine in the compound is given as 55.0%. This implies that the weight percent of the metal is 45.0%.
\n
Now, we need to find the molar mass of the compound. At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 L. We know that 100 mL (or 0.1 L) of the metal chloride vapor weighs 0.57 g. Therefore, the molar mass of the compound (MClx) is:
The molar mass of $\\mathrm{CH_4}$ (Methane) is 16 g/mol. Therefore, 16 g of $\\mathrm{CH_4}$ is equivalent to 1 mole of $\\mathrm{CH_4}$. Furthermore, each molecule of $\\mathrm{CH_4}$ has 10 electrons (6 from Carbon and 4 from Hydrogen). As a result, 1 mole of $\\mathrm{CH_4}$ (or $6.022 \\times 10^{23}$ molecules of $\\mathrm{CH_4}$) would have $10 \\times 6.022 \\times 10^{23} = 60.2 \\times 10^{23}$ electrons. This matches with (II).
\n
B. 1 g of $\\mathrm{H_2~(g)}$
\n
The molar mass of $\\mathrm{H_2}$ (Hydrogen) is 2 g/mol. Therefore, 1 g of $\\mathrm{H_2}$ is equivalent to 0.5 moles of $\\mathrm{H_2}$. The volume that a given quantity of gas occupies is proportional to the number of moles of gas. At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters. Therefore, 0.5 moles of gas would occupy $0.5 \\times 22.4 = 11.2$ liters. The closest match is (IV) with 11.4 liters (the small discrepancy may be due to rounding or slightly different conditions than STP).
\n
C. 1 mole of $\\mathrm{N_2~(g)}$
\n
The molar mass of $\\mathrm{N_2}$ (Nitrogen) is 28 g/mol. Therefore, 1 mole of $\\mathrm{N_2}$ weighs 28 g. This matches with (I).
\n
D. 0.5 mol of $\\mathrm{SO_2~(g)}$
\n
The molar mass of $\\mathrm{SO_2}$ (Sulfur Dioxide) is 64 g/mol. Therefore, 0.5 moles of $\\mathrm{SO_2}$ weigh $0.5 \\times 64 = 32$ g. This matches with (III).
The number of molecules and moles in 2.8375 litres of O$$_2$$ at STP are respectively
",
"options": [
{
"text": "1.505 $$\\times$$ 10$$^{23}$$ and 0.250 mol"
},
{
"text": "7.527 $$\\times$$ 10$$^{22}$$ and 0.250 mol"
},
{
"text": "7.527 $$\\times$$ 10$$^{23}$$ and 0.125 mol"
},
{
"text": "7.527 $$\\times$$ 10$$^{22}$$ and 0.125 mol"
}
],
"answer": "7.527 $$\\times$$ 10$$^{22}$$ and 0.125 mol",
"solution": "**Answer:** 7.527 $$\\times$$ 10$$^{22}$$ and 0.125 mol\n\n
At STP, one mole of any gas occupies 22.4 liters. Therefore, 2.8375 liters of oxygen gas at STP is equal to 0.125 moles. The number of molecules of oxygen gas in 0.125 moles is calculated by multiplying the number of moles by Avogadro's number, which is $$6.022 × 10^{23} $$ molecules/mol. This gives us a total of $$7.527 × 10^{22}$$ molecules of oxygen gas in 2.8375 liters at STP.
$$0.5 \\mathrm{~g}$$ of an organic compound $$(\\mathrm{X})$$ with $$60 \\%$$ carbon will produce __________ $$\\times 10^{-1} \\mathrm{~g}$$ of $$\\mathrm{CO}_{2}$$ on complete combustion.
\nMolar Mass of $\\mathrm{CO}_2=12+(16 \\times 2)=44$
\nMass of Compound $=0.5 \\mathrm{~g}$
\n$$\n\\begin{aligned}\n&\\% \\text { of } \\mathrm{C} =\\frac{\\text { Molar mass of } \\mathrm{C} \\times \\text { Mass Of } \\mathrm{CO}_2}{\\text { Mass Of Compound } \\times \\text { Molar Mass } \\mathrm{Of} ~ \\mathrm{CO}_2} \\\\\\\\\n&\\frac{60}{100} =\\frac{12 \\times x}{0.5 \\times 44} \\\\\\\\\n&1.1 =x \\\\\\\\\n&x =11 \\times 10^{-1}\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2918,
"subject": "Chemistry",
"question": "
If 5 moles of $$\\mathrm{BaCl}_{2}$$ is mixed with 2 moles of $$\\mathrm{Na}_{3} \\mathrm{PO}_{4}$$, the maximum number of moles of $$\\mathrm{Ba}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$$ formed is ___________ (Nearest integer)
We can see that 3 moles of $\\mathrm{BaCl}_2$ react with 2 moles of $\\mathrm{Na}_3\\mathrm{PO}_4$ to produce 1 mole of $\\mathrm{Ba}_3\\mathrm{(PO}_4)_2$.
\n
Given that you have 5 moles of $\\mathrm{BaCl}_2$ and 2 moles of $\\mathrm{Na}_3\\mathrm{PO}_4$, let's calculate the maximum number of moles of $\\mathrm{Ba}_3\\mathrm{(PO}_4)_2$ that can be formed:
\n
From $\\mathrm{BaCl}_2$, using the molar ratio from the balanced equation:
The limiting reactant is $\\mathrm{Na}_3\\mathrm{PO}_4$ because it produces 1 mole of $\\mathrm{Ba}_3\\mathrm{(PO}_4)_2$, which is less than the $\\frac{5}{3}$ moles produced by $\\mathrm{BaCl}_2$.
\n
Therefore, the maximum number of moles of $\\mathrm{Ba}_3\\mathrm{(PO}_4)_2$ formed is 1 mole. Rounding to the nearest integer, the answer is 1.
Volume of $$3 \\mathrm{M} \\mathrm{~NaOH}$$ (formula weight $$40 \\mathrm{~g} \\mathrm{~mol}^{-1}$$ ) which can be prepared from $$84 \\mathrm{~g}$$ of $$\\mathrm{NaOH}$$ is __________ $$\\times 10^{-1} \\mathrm{dm}^3$$.
First, let's calculate the number of moles of NaOH that can be prepared from $$84\\ \\text{g}$$ of NaOH. The molar mass of NaOH is given as $$40\\ \\text{g/mol}$$.
\n\n
The number of moles (n) is calculated using the formula:
\n\n$$ n = \\frac{mass}{molar\\ mass} $$\n\n
So for our case:
\n\n$$ n = \\frac{84\\ \\text{g}}{40\\ \\text{g/mol}} = 2.1\\ \\text{mol} $$\n\n
Now that we know the number of moles, we can find out the volume of a $$3\\ \\text{M}$$ NaOH solution that can be prepared from it. The concentration (C) of a solution is related to the number of moles (n) and volume (V) by the following formula:
\n\n$$ C = \\frac{n}{V} $$\n\n
Where:\nC = concentration in molarity (M)\nn = number of moles\nV = volume in liters (L) - Note that $$1\\ \\text{dm}^3 = 1\\ \\text{L}$$.
\n\n
Since we want to find the volume (V), we can rearrange the formula to solve for V:
\n\n$$ V = \\frac{n}{C} $$\n\n
Using the moles of NaOH and the concentration for the preparation:
\n\n$$ V = \\frac{2.1\\ \\text{mol}}{3\\ \\text{M}} $$\n\n
Calculate the volume:
\n\n$$ V = \\frac{2.1}{3} = 0.7\\ \\text{L} $$\n\n
To convert this volume to $$\\text{dm}^3$$ (which is equivalent to liters), we use the conversion factor $$1\\ \\text{L} = 1\\ \\text{dm}^3$$. Therefore:
\n\n$$ V = 0.7\\ \\text{dm}^3 $$\n\n
To express this volume as $$\\times 10^{-1}\\ \\text{dm}^3$$, we can write:
\n\n$$ V = 7 \\times 10^{-1}\\ \\text{dm}^3 $$\n\n
Therefore, the volume of $$3\\ \\text{M}$$ NaOH solution which can be prepared from $$84\\ \\text{g}$$ of NaOH is $$7 \\times 10^{-1}\\ \\text{dm}^3$$.
A sample of $$\\mathrm{CaCO}_3$$ and $$\\mathrm{MgCO}_3$$ weighed $$2.21 \\mathrm{~g}$$ is ignited to constant weight of $$1.152 \\mathrm{~g}$$. The composition of mixture is :
\n
(Given molar mass in $$\\mathrm{g} \\mathrm{~mol}^{-1} \\mathrm{CaCO}_3: 100, \\mathrm{MgCO}_3: 84$$)
Molar mass of the salt from $$\\mathrm{NaBr}, \\mathrm{NaNO}_3, \\mathrm{KI}$$ and $$\\mathrm{CaF}_2$$ which does not evolve coloured vapours on heating with concentrated $$\\mathrm{H}_2 \\mathrm{SO}_4$$ is ________ $$\\mathrm{g} \\mathrm{~mol}{ }^{-1}$$.
If $$50 \\mathrm{~mL}$$ of $$0.5 \\mathrm{M}$$ oxalic acid is required to neutralise $$25 \\mathrm{~mL}$$ of $$\\mathrm{NaOH}$$ solution, the amount of $$\\mathrm{NaOH}$$ in $$50 \\mathrm{~mL}$$ of given $$\\mathrm{NaOH}$$ solution is ______ g.
$$0.05 \\mathrm{~cm}$$ thick coating of silver is deposited on a plate of $$0.05 \\mathrm{~m}^2$$ area. The number of silver atoms deposited on plate are ________ $$\\times 10^{23}$$. (At mass $$\\mathrm{Ag}=108, \\mathrm{~d}=7.9 \\mathrm{~g} \\mathrm{~cm}^{-3}$$)
$$\\mathrm{Xg}$$ of ethylamine is subjected to reaction with $$\\mathrm{NaNO}_2 / \\mathrm{HCl}$$ followed by water; evolved dinitrogen gas which occupied $$2.24 \\mathrm{~L}$$ volume at STP. X is _________ $$\\times 10^{-1} \\mathrm{~g}$$.
To determine the maximum amount of acetanilide that can be prepared from 6.55 g of aniline, we need to use stoichiometry. Let's go through the process step by step.
\n\n
1. Molecular weights calculation:
\n\n
The molecular weight of aniline (C6H5NH2) is calculated as follows:
An organic compound has $$42.1 \\%$$ carbon, $$6.4 \\%$$ hydrogen and remainder is oxygen. If its molecular weight is 342 , then its molecular formula is :
To determine the molecular formula of the compound, we first calculate the empirical formula based on the given percentages of carbon, hydrogen, and oxygen. Then we use the molecular weight to find the molecular formula.
\n\n
Step 1: Calculating the moles of each element per 100 g of the compound
\n\n
\n\n
For carbon (%C = 42.1%), the moles of carbon are calculated as follows:\n\n
$$\\text{Moles of C} = \\frac{\\% \\text{ by weight of C}}{\\text{Atomic weight of C}} = \\frac{42.1 \\text{ g}}{12.01 \\text{ g/mol}}$$
\n\n
For hydrogen (%H = 6.4%), the moles of hydrogen are:\n\n
$$\\text{Moles of H} = \\frac{\\% \\text{ by weight of H}}{\\text{Atomic weight of H}} = \\frac{6.4 \\text{ g}}{1.008 \\text{ g/mol}}$$
\n\n
For oxygen, since it makes up the remainder, we find its percentage by subtracting the percentages of carbon and hydrogen from 100%:\n\n
$$\\% \\text{O} = 100 - 42.1 - 6.4 = 51.5\\%$$
\n\n
Then the moles of oxygen are:
\n\n
$$\\text{Moles of O} = \\frac{\\% \\text{ by weight of O}}{\\text{Atomic weight of O}} = \\frac{51.5 \\text{ g}}{16.00 \\text{ g/mol}}$$
\n\n
\n\n\n\n
Step 2: Dividing the calculated moles by the smallest number of moles to find the simplest whole number ratio:
\n\n
This involves dividing each mole value by the smallest of the mole values obtained. However, since we didn't calculate the exact number of moles for each element in the above step, let's proceed conceptually based on given percentages:
\n\n
To find the simplest whole number ratio, we would perform the following calculations:
\n\n
\n\n
$$\\text{Moles of C} \\approx \\frac{42.1}{12}$$
\n\n\n\n
\n\n
$$\\text{Moles of H} \\approx \\frac{6.4}{1}$$
\n\n\n\n
\n\n
$$\\text{Moles of O} \\approx \\frac{51.5}{16}$$
\n\n\n\n\n\n
Step 3: Determining the empirical formula:
\n\n
Given the lack of exact numbers from the prior step, let's assume each mole value is divided by the smallest mole value among the C, H, and O calculations. This ratio generally defines the empirical formula, which constitutes the smallest whole number ratio of atoms in the compound.
\n\n
Step 4: Using the molecular weight to find the molecular formula:
\n\n
Once we have the empirical formula, we would calculate its mass and compare it to the molecular weight provided (342). The molecular formula is a multiple of the empirical formula that matches the given molecular weight. Since we haven't calculated the exact empirical formula mass, let's examine the options directly considering a molecular weight of 342:
\n\n
Option A: $$C_{14}H_{20}O_{10}$$ - The molar mass would be $$14 \\times 12 + 20 \\times 1 + 10 \\times 16 = 168 + 20 + 160 = 348 \\text{, which is close but not 342.}$$
\n\n
Option B: $$C_{12}H_{22}O_{11}$$ - The molar mass would be $$12 \\times 12 + 22 \\times 1 + 11 \\times 16 = 144 + 22 + 176 = 342 \\text{, which matches the given molecular weight.}$$
\n\n
Thus, the correct molecular formula based on the molecular weight and composition given would be Option B: $$\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}$$.
The number of moles of methane required to produce $$11 \\mathrm{~g} \\mathrm{~CO}_2(\\mathrm{g})$$ after complete combustion is : (Given molar mass of methane in $$\\mathrm{g} \\mathrm{~mol}^{-1}: 16$$ )
To find the number of moles of methane ($CH_4$) required to produce 11g of $CO_2$ upon complete combustion, we first need the balanced chemical equation for the combustion of methane:
\n\n
$$CH_4 + 2O_2 \\rightarrow CO_2 + 2H_2O$$
\n\n
From the balanced equation, we see that 1 mole of $CH_4$ produces 1 mole of $CO_2$. Therefore, the moles of $CH_4$ needed to produce a certain amount of $CO_2$ will be equal to the moles of $CO_2$ produced.
\n\n
To find the moles of $CO_2$ produced from 11g of $CO_2$, we use the formula:
The molar mass of $CO_2$ is $44 \\mathrm{~g/mol}$ (12 from carbon and 16$\\times$2 from oxygen). Then, the moles of $CO_2$ produced from 11g of $CO_2$ can be calculated as:
Since the stoichiometry of the reaction between methane and oxygen is 1:1 for $CH_4$ and $CO_2$, the moles of $CH_4$ required to produce 11g of $CO_2$ is also 0.25 moles.
\n\n
Therefore, the correct answer is Option B: 0.25.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2928,
"subject": "Chemistry",
"question": "In a compound C, H and N atoms are present in 9 : 1 : 3.5 by weight . Molecular weight of the compound is 108 g mol-1 . Molecular formula of compound is",
"options": [
{
"text": "C2H6N2"
},
{
"text": "C3H4N"
},
{
"text": "C6H8N2"
},
{
"text": "C9H12N3"
}
],
"answer": "C6H8N2",
"solution": "**Answer:** C6H8N2\n\n
\n
\n
Element
\n
Weight ratio
\n
No of atoms of element
\n
Simplest Ratio
\n
\n
\n
C
\n
9
\n
9/12 = 3/4
\n
3
\n
\n
\n
H
\n
1
\n
1/1 = 1
\n
4
\n
\n
\n
N
\n
3.5
\n
3.5/14 = 1/4
\n
1
\n
\n
\n \nEmpherical Formula of the compound (EF) will be C3H4N
\nNow to find molecular formula at first find number of moles of given mass
\nNo. of moles (n) = \n
Given mass
\n
Molar mass
\n
\n$$ \\Rightarrow $$ $${{108} \\over {54}}$$
\n$$ \\Rightarrow $$ 2
\n$$\\therefore$$ No. of Moles is 2
\nMolecular formula (MF) = n (no. of moles) $$\\times$$ Empherical Formula (EF)
\n$$ \\Rightarrow $$ 2 $$\\times$$ C3H4N
\n$$ \\Rightarrow $$ C6H8N2
\n$$\\therefore$$ Option (C) is correct",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2929,
"subject": "Chemistry",
"question": "A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2. The empirical formula of the hydrocarbon is",
"options": [
{
"text": "C2H4 "
},
{
"text": "C3H4 "
},
{
"text": "C6H5 "
},
{
"text": "C7H8"
}
],
"answer": "C7H8",
"solution": "**Answer:** C7H8\n\nRequired reaction,\n
0.72 gm of H2O = $${{{0.72} \\over {18}}}$$ mole of H2O = 0.04 mole of H2O\n
In one H2O molecule 2 hydrogen atoms present.\n
So in 0.04 mole of H2O molecules 2$$ \\times $$0.04 = 0.08 moles of H atoms present.\n
0.72 gm of CO2 = $${{{3.08} \\over {44}}}$$ mole of CO2 = 0.07 mole of CO2\n
And in one CO2 molecule 1 C atom present.\n
So in 0.07 mole of CO2 molecules 0.07$$\\times$$1 = 0.07 moles of C atoms present\n
$$\\therefore$$ C : H = 0.07 : 0.08 = 7 : 8\n
$$\\therefore$$ Empirical formula of hydrocarbon = C7H8",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2930,
"subject": "Chemistry",
"question": "Experimentally it was found that a metal oxide has formula M0.98O. Metal M, present as M2+ and M3+ in its oxide. Fraction of the metal which exists as M3+ would be",
"options": [
{
"text": "7.01%"
},
{
"text": "4.08%"
},
{
"text": "6.05%"
},
{
"text": "5.08%"
}
],
"answer": "4.08%",
"solution": "**Answer:** 4.08%\n\nGiven metal oxide = M0.98O\n We know oxidation number of O = (-2)\n Now assume oxidaton no of M = $$x$$\n
$$\\therefore$$ 0.98$$x$$ + 1$$ \\times $$ (-2) = 0\n $$ \\Rightarrow x =$$ $${{200} \\over {98}}$$\n This represent the charge in one atom of M. As you can see the charge of M is in the range $$2 < {{200} \\over {98}} < 3$$.\n So we can say in M mixture of M+2 and M+3 present.\n
Assume total no of atoms present in M is 100.\n Let M+3 present in M = y atoms\n So M+2 present in M = (100 - y) atoms\n
In 1 atom of M+3 charge present = +3\n So in y atoms of M+3 charge present = +3y\n
Similarly in 1 atom of M+2 charge present = +2\n So in (100 - y) atoms of M+2 charge present = +2(100 - y)\n
In 100 atoms of M total charge = 200 + y\n So in 1 atoms of M total charge = $${{200 + y} \\over {100}}$$\n
Earlier we found that charge in one atom of M is = $${{200} \\over {98}}$$\n
So we can write,\n $${{200 + y} \\over {100}}$$ = $${{200} \\over {98}}$$\n $$ \\Rightarrow y = 4.08$$\n
So Option (B) is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2931,
"subject": "Chemistry",
"question": "For per gram of reactant, the maximum quantity of N2 gas is produced in which of the following thermal decomposition reactions ? \n (Given : Atomic wt. - Cr = 52 u, Ba = 137 u)",
"options": [
{
"text": "(NH4)2Cr2O7(s) $$ \\to $$ N2(g) + 4H2O(g) + Cr2O3(s)"
},
{
"text": "2NH4NO3(s) $$ \\to $$ 2 N2(g) + 4H2O(g) + O2(g)"
},
{
"text": "Ba(N3)2(s) $$ \\to $$ Ba(s) + 3N2(g)"
},
{
"text": "2NH3(g) $$ \\to $$ N2(g) + 3H2(g)"
}
],
"answer": "2NH3(g) $$ \\to $$ N2(g) + 3H2(g)",
"solution": "**Answer:** 2NH3(g) $$ \\to $$ N2(g) + 3H2(g)\n\n(a) Molar mass of (NH4)2 Cr2O7 = 252 g/mol.\n
252g of (NH4)2 Cr2 O7 produce 28g mole of N2\n
$$\\therefore\\,\\,\\,$$ 1 g of (NH4)2 Cr2 O7 Produce = $${{28} \\over {252}} = 0.111$$ g N2\n
(b) Molar mass of NH4 NO3 = 80 g/mol\n
2 $$ \\times $$ 80 g of NH4 NO3 produce 28 $$ \\times $$ 2g of N2\n
$$\\therefore\\,\\,\\,$$ 1 g of NH4 NO3 produce = $${{28 \\times 2} \\over {2 \\times 80}}$$ = 0.35 g N2\n
(c) Molar mass of Ba(N3)2 = 221 g/mol\n
221 g of Ba(N3)2 produce 3 $$ \\times $$ 28 g of N2\n
$$\\therefore\\,\\,\\,$$ 1 g of Ba(N3)2 produce = $${{3 \\times 28} \\over {221}}$$ = 0.38 g of N2\n
(d) Molar mass of NH3 = 17 g/mol\n
17 $$ \\times $$ 2 NH3 produce 28 g of N2 \n
$$\\therefore\\,\\,\\,$$ 1 g of NH3 produce = $${{28} \\over {17 \\times 2}}$$ = 0.823 g of N2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2932,
"subject": "Chemistry",
"question": "An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine atoms present in 1 g of chlorohydrocarbon are : \n (Atomic wt. of Cl = 35.5 u; Avogadro constant = 6.023 $$ \\times $$ 1023 mol-1)",
"options": [
{
"text": "6.023 $$ \\times $$ 1020"
},
{
"text": "6.023 $$ \\times $$ 109"
},
{
"text": "6.023 $$ \\times $$ 1021"
},
{
"text": "6.023 $$ \\times $$ 1023"
}
],
"answer": "6.023 $$ \\times $$ 1020",
"solution": "**Answer:** 6.023 $$ \\times $$ 1020\n\n% of Cl = 3.55\n
$$\\therefore\\,\\,\\,\\,$$ In 100 g chlorohydrocarbon 3.55 gm Cl present. \n
In 1 gm chlorohydrocarbon Cl present\n
= $${{3.55} \\over {100}}$$\n
= 0.0355 gm\n
$$\\therefore\\,\\,\\,\\,$$ No of Moles of Cl = $${{0.0355} \\over {35.5}}$$\n
= 0.001 mole\n
$$\\therefore\\,\\,\\,\\,$$ no of Cl atoms = 0.001 $$ \\times $$ 6.023 $$ \\times $$ 103 \n
= 6.023 $$ \\times $$ 1020 ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2933,
"subject": "Chemistry",
"question": "For the following reaction, in the mass of water produced from 445 g of C57H110O6 is :\n
= 495 gm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2934,
"subject": "Chemistry",
"question": "The percentage composition of carbon by mole\nin methane is :",
"options": [
{
"text": "80%"
},
{
"text": "20%"
},
{
"text": "75%"
},
{
"text": "25%"
}
],
"answer": "20%",
"solution": "**Answer:** 20%\n\nIn CH4, there are 5 atoms (1C + 4H).\n
$$ \\therefore $$ % composition of C by mole in CH4\n
= $${1 \\over 5}$$ $$ \\times $$ 100 = 20 %\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2935,
"subject": "Chemistry",
"question": "For a reaction, \nN2(g) + 3H2(g) $$ \\to $$ 2NH3(g) ; \nidentify dihydrogen (H2) as a limiting reagent\nin the following reaction mixtures.",
"options": [
{
"text": "56g of N2 + 10g of H2"
},
{
"text": "14g of N2 + 4g of H2"
},
{
"text": "28g of N2 + 6g of H2"
},
{
"text": "35g of N2 + 8g of H2"
}
],
"answer": "56g of N2 + 10g of H2",
"solution": "**Answer:** 56g of N2 + 10g of H2\n\nHere you have to check every options to find answer.\n
N2(g) + 3H2(g) $$ \\to $$ 2NH3(g)\n
For dihydrogen (H2) to become a limiting reagent,\n
$${{moles\\,of\\,{H_2}} \\over {stoichiometric\\,coefficient\\,of\\,{H_2}}}$$ should be \n
less than $${{moles\\,of\\,{N_2}} \\over {stoichiometric\\,coefficient\\,of\\,{N_2}}}$$.\n
Here in 56g of N2 + 10g of H2,\n
Moles of N2 = $${{56} \\over {28}}$$ = 2\n
and Moles of H2 = $${{10} \\over 2}$$ = 5\n
Now by dividing stoichiometric coefficient we get,\n
For N2 = $${2 \\over 1}$$ = 2\n
For H2 = $${5 \\over 3}$$ = 1.67\n
$$ \\therefore $$ Here H2 is the limiting reagent.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2936,
"subject": "Chemistry",
"question": "At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O2 for complete\ncombustion, and 40 mL of CO2 is formed. The formula of the hydrocarbon is : \n",
"options": [
{
"text": "C4H7Cl"
},
{
"text": "C4H6"
},
{
"text": "C4H8"
},
{
"text": "C4H10"
}
],
"answer": "C4H6",
"solution": "**Answer:** C4H6\n\nCxHy + $$\\left( {x + {y \\over 4}} \\right)$$O2 $$ \\to $$ xCO2 + $${y \\over 2}$$H2O\n
According to the above equation,\n
1 mL of hydrocarbon = x mL of CO2 is produced\n
$$ \\therefore $$ 10 mL of hydrocarbon = 10x mL of CO2 is produced.\n
According to question,\n
From 10 mL of hydrocarbon 40 mL of CO2 is produced.\n
$$ \\therefore $$ 10x = 40\n
$$ \\Rightarrow $$ x = 4\n
Also from the above equation,\n
1 mL of CxHy react with $$\\left( {x + {y \\over 4}} \\right)$$ mL of O2.\n
$$ \\therefore $$ 10 mL of CxHy react with 10$$\\left( {x + {y \\over 4}} \\right)$$ mL of O2.\n
$$ \\therefore $$ Hydrocarbon is C4H6",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2937,
"subject": "Chemistry",
"question": "Thermal decomposition of a Mn compound (X) at 513 K results in compound Y, MnO2 and gaseous product.\nMnO2 reacts with NaCl and concentrated H2O4 to give a pungent gas Z. X, Y and Z, respectively, are :",
"options": [
{
"text": "KMnO4, K2MnO4 and Cl2\n"
},
{
"text": "K2MnO4, KMnO4 and SO2"
},
{
"text": "K3MnO4, K2MnO4 and Cl2"
},
{
"text": "K2MnO4, KMnO4 and Cl2"
}
],
"answer": "KMnO4, K2MnO4 and Cl2\n",
"solution": "**Answer:** KMnO4, K2MnO4 and Cl2\n\n\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2938,
"subject": "Chemistry",
"question": "The mass of ammonia in grams produced when\n2.8 kg of dinitrogen quantitatively reacts with 1\nkg of dihydrogen is _______.",
"options": [],
"answer": "3400",
"solution": "**Answer:** 3400\n\nN2(g) + 3H2(g) $$ \\to $$ 2NH3(g)\n
Number of moles of N2 = $${{2.8 \\times {{10}^3}} \\over {28}}$$ = 100\n
Number of moles of H2 = $${{1000} \\over 2}$$ = 500\n
Here N2 is limiting reagent.\n
$$ \\therefore $$ Number of moles of NH3 produced = 2 $$ \\times $$ 100 = 200\n
Mass of NH3 produced = 200 × 17 = 3400 gm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2939,
"subject": "Chemistry",
"question": "The mole fraction of glucose (C6H12O6\n) in an aqueous binary solution is 0.1. The mass percentage of\nwater in it, to the nearest integer, is _______.",
"options": [],
"answer": "47",
"solution": "**Answer:** 47\n\nMole fraction of glucose in aqueous solution\n= 0.1\n
Let total mole is 1 mol then mole of glucose will be 0.1 and mole of water will be 0.9.\n
So mass % of water = $${{0.9 \\times 18} \\over {0.1 \\times 180 + 0.9 \\times 18}}$$ $$ \\times $$ 100\n
= 47.37 $$ \\simeq $$ 47",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2940,
"subject": "Chemistry",
"question": "The flocculation value of HCl for arsenic sulphide sol. is 30 m mol L-1 If H2SO4 is used for the flocculatiopn of arsenic sulphide, the amount in grams, of H2SO4 in 250 ml required for the above purposed is ______. \n(molecular mass of H2SO4 = 98 g/mol)",
"options": [],
"answer": "0.36to0.38",
"solution": "**Answer:** 0.36to0.38\n\nArsenic sulphide sol is negatively charged, so for flocculation\n positive ion required. Here positive ion H+ present.\n
for 1 L, 30 mm moles of H+ is required\n
for 250 ml, $${{30} \\over 4}$$ mm moles H+ is required\n
$$ \\therefore $$ for 250 ml, $${{30} \\over {4 \\times 2}}$$ mm moles H2SO4 is required.\n
= 0.3675 g",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2941,
"subject": "Chemistry",
"question": "The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by :",
"options": [
{
"text": "200 ml of 0.02 N HCl"
},
{
"text": "100 ml of 0.2 N HCl "
},
{
"text": "100 ml of 0.1 HCl"
},
{
"text": "200 ml of 0.4 N HCl"
}
],
"answer": "100 ml of 0.2 N HCl ",
"solution": "**Answer:** 100 ml of 0.2 N HCl \n\nNH2CONH2 $$ \\to $$ NH3\n
Using Principle of Atom Conservation\n
2 $$ \\times $$ moles of urea = 1 $$ \\times $$ moles of NH3\n
Also moles of NH3 = moles of HCl, because they react in 1 : 1 ratio.\n
100 ml of 0.2 N HCl = $${{100 \\times 0.2} \\over {1000}}$$ = 0.02 mole of HCl\n
So option (B) is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2942,
"subject": "Chemistry",
"question": "The formula of a gaseous hydrocarbon which requires 6 times of its own volume of O2 for complete oxidation and produces 4 times its own volume of CO2 is CxHy. The value of y is _____________.",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nCombustion reaction :
$$4 + {y \\over 4} = 6 \\Rightarrow y = 8$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2943,
"subject": "Chemistry",
"question": "1.86 g of aniline completely reacts to form acetanilide. 10% of the product is lost during purification. Amount of acetanilide obtained after purification (in g) is __________ $$\\times$$ 10$$-$$2.",
"options": [],
"answer": "243",
"solution": "**Answer:** 243\n\n Given, weight = 18.6 g
Here, 1 mole of aniline gives 1 mole of acetanilide
$$\\therefore$$ mole of aniline = mole of acetanilide
Hence, mass of acetanilide produced $$ = 2.70 \\times {{90} \\over {100}}g = 2.43g = 243 \\times {10^2}g$$
x = 243",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 2944,
"subject": "Chemistry",
"question": "Complete combustion of 1.80 g of an oxygen containing compound (CxHyOz) gave 2.64 g of CO2 and 1.08 g of H2O. The percentage of oxygen in the organic compound is :",
"options": [
{
"text": "51.63"
},
{
"text": "50.33"
},
{
"text": "63.53"
},
{
"text": "53.33"
}
],
"answer": "53.33",
"solution": "**Answer:** 53.33\n\nCxHyOz + O2 $$ \\to $$ xCO2 + $${y \\over 2}{H_2}O$$\n
2.64 g of CO2 contains 0.72 g C.\n
1.08 g of H2O contains 0.12 g H.\n
$$ \\therefore $$ Mass of oxygen present = 1.80 – (0.72\n+0.12) = 0.96 g\n
% of O = $${{0.96} \\over {1.80}} \\times 100$$ = 53.33 %",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2945,
"subject": "Chemistry",
"question": "Complete combustion of 750 g of an organic compound provides 420 g of CO2 and 210 g of H2O. The percentage composition of carbon and hydrogen in organic compound is 15.3 and ___________ respectively. (Round off to the Nearest Integer).",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n18 gm H2O $$ \\Rightarrow $$ 2 gm H2
So, % H2 = $${{23.33} \\over {750}} \\times 100$$ = 3.11% $$ \\approx $$ 3%",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2946,
"subject": "Chemistry",
"question": "When 35 mL of 0.15 M lead nitrate solution is mixed with 20 mL of 0.12 M chromic sulphate solution, _________ $$\\times$$ 10$$-$$5 moles of lead sulphate precipitate out. (Round off to the Nearest Integer).",
"options": [],
"answer": "525",
"solution": "**Answer:** 525\n\n\n
For 3 moles of Pb(NO3)2 , we require 1 mole of Cr2(SO4)3
\n
For 5.25 moles of Pb(NO3)2, we require $$\\frac{1}{3} \\times 5.25 $$ mole of Cr2(SO4)3 = 1.75 moles
\n\n
But we have 2.4 moles. So, Cr2(SO4)3 is excess reagent and Pb(NO3)2 is limiting reagent, (LR)
\n\n\n\n
Moles of PbSO4 formed = moles of Pb(NO3)2 consumed\n\n\n
= 5.25 m mol = 525 $$ \\times $$ 10-5 moles
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2947,
"subject": "Chemistry",
"question": "__________ grams of 3-Hydroxy propanal (MW = 74) must be dehydrated to produce 7.8 g of acrolein (MW = 56) (C3H4O) if the percentage yield is 64. (Round off to the Nearest Integer).
[Given : Atomic masses : C : 12.0 u, H : 1.0 u, O : 16.0 u ]",
"options": [],
"answer": "16",
"solution": "**Answer:** 16\n\n
$$ \\therefore $$ Mass of acrolein = $$\\left[ {{x \\over {74}} \\times 0.64} \\right] \\times 56 = 78$$
$$ \\Rightarrow x = 16.1gm \\simeq 16gm$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2948,
"subject": "Chemistry",
"question": "Complete combustion of 3g of ethane gives x $$\\times$$ 1022 molecules of water. The value of x is __________. (Round off to the Nearest Integer). [Use : NA = 6.023 $$\\times$$ 1023; Atomic masses in u : C : 12.0; O : 16.0; H : 1.0]",
"options": [],
"answer": "18",
"solution": "**Answer:** 18\n\n$${C_2}{H_6} + {7 \\over 2}{O_2}\\buildrel {} \\over\n \\longrightarrow 2C{O_{\\,2}} + 3{H_2}O$$
= 18.069 $$\\times$$ 1022 molecules",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2949,
"subject": "Chemistry",
"question": "Consider the complete combustion of butane, the amount of butane utilized to produce 72.0 g of water is ___________ $$\\times$$ 10$$-$$1 g. (in nearest integer)",
"options": [],
"answer": "464",
"solution": "**Answer:** 464\n\n$${C_4}{H_{10}} + {{13} \\over 2}{O_2}\\buildrel {} \\over\n \\longrightarrow 4C{O_2} + 5{H_2}O$$
Moles of $${H_2}O = {{72} \\over {18}} = 4$$
Moles of C4H10 used $$ = {1 \\over 5} \\times 4$$
Wight of C4H10 used $$ = {4 \\over 5} \\times 58$$
The above reaction is carried out in a vessel starting with partial pressure PSO2 = 250 m bar, PO2 = 750 m bar and PSO3 = 0 bar. When the reaction is complete, the total pressure in the reaction vessel is _______ m bar. (Round off of the nearest integer).",
"options": [],
"answer": "875",
"solution": "**Answer:** 875\n\n2SO2(g) + O2(g) $$\\to$$ 2SO3(g)
$$\\therefore$$ Final total pressure = 625 + 250 = 875 m bar",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2951,
"subject": "Chemistry",
"question": "In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is ______________. (Nearest integer)
Molar mass of $$F{e_3}{O_4}$$ = 56 $$\\times$$ 3 + 16 $$\\times$$ 4 = 232 g
\n
$$\\therefore$$ Moles of $$F{e_3}{O_4} = {{4640} \\over {232}} = 20$$
\n
Also, given
\n
Mass of CO = 2.520 kg = 2520 gm
\n
Molar mass of CO = 12 + 16 = 28 gm
\n
$$\\therefore$$ Molar of $$CO = {{2520} \\over {28}} = 90$$
\n
\n
Here Fe3O4 is limiting reagent as to find limiting reagent, divide the given moles of reactants with their respective stoichiometric coefficient and reactant for which this ratio is minimum will be limiting reagent
\n
For $$F{e_3}{O_4},{{moles} \\over {stoichiometric\\,coefficient}} = {{20} \\over 1}$$
\n
For $$CO,{{moles} \\over {stoichiometric\\,coefficient}} = {{90} \\over 4} = 22.5$$
\n
$$\\therefore$$ Fe3O4 is limiting reagent.
\n
Now produced Fe = 20 $$\\times$$ 3 = 60 mol
\n
$$\\therefore$$ Weight of Fe = 60 $$\\times$$ 56 = 3360 g
The complete combustion of 0.492 g of an organic compound containing 'C', 'H' and 'O' gives 0.793 g of CO2 and 0.442 g of H2O. The percentage of oxygen composition in the organic compound is ______________. (nearest integer)
116 g of a substance upon dissociation reaction, yields 7.5 g of hydrogen, 60 g of oxygen and 48.5 g of carbon. Given that the atomic masses of H, O and C are 1, 16 and 12, respectively. The data agrees with how many formulae of the following?
On complete combustion 0.30 g of an organic compound gave 0.20 g of carbon dioxide and 0.10 g of water. The percentage of carbon in the given organic compound is _____________. (Nearest integer)
CNG is an important transportation fuel. When 100 g CNG is mixed with 208 g oxygen in vehicles, it leads to the formation of CO2 and H2O and produces large quantity of heat during this combustion, then the amount of carbon dioxide, produced in grams is ____________. [nearest integer]
120 g of an organic compound that contains only carbon and hydrogen gives 330 g of CO2 and 270 g of water on complete combustion. The percentage of carbon and hydrogen, respectively are
",
"options": [
{
"text": "25 and 75"
},
{
"text": "40 and 60"
},
{
"text": "60 and 40"
},
{
"text": "75 and 25"
}
],
"answer": "75 and 25",
"solution": "**Answer:** 75 and 25\n\n
If a rocket runs on a fuel (C15H30) and liquid oxygen, the weight of oxygen required and CO2 released for every litre of fuel respectively are :
\n
(Given : density of the fuel is 0.756 g/mL)
",
"options": [
{
"text": "1188 g and 1296 g"
},
{
"text": "2376 g and 2592 g"
},
{
"text": "2592 g and 2376 g"
},
{
"text": "3429 g and 3142 g"
}
],
"answer": "2592 g and 2376 g",
"solution": "**Answer:** 2592 g and 2376 g\n\n
if one mole of each of $$X$$ and $$Y$$ with $$0.05 \\mathrm{~mol}$$ of $$Z$$ gives compound $$X Y Z_{3}$$. (Given : Atomic masses of $$X, Y$$ and $$Z$$ are 10, 20 and 30 amu, respectively.) The yield of $$X YZ_{3}$$ is _____________ g. (Nearest integer)
",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n$\\underset{n_{\\text {moles }}=1}{X}+\\underset{1}{Y}+\\underset{0.05}{3 Z} \\rightleftharpoons X Y Z_{3}$\n
\nLimiting reagent is $Z=\\frac{0.05}{3}=.016$\n
\n3 moles of $Z \\rightarrow 1$ mole of $X Y Z_{3}$\n
\n$0.05$ mole of $Z \\rightarrow \\frac{1}{3} \\times 0.05$ mole of $X Y Z_{3}$\n
On complete combustion of $$0.492 \\mathrm{~g}$$ of an organic compound containing $$\\mathrm{C}, \\mathrm{H}$$ and $$\\mathrm{O}$$, $$0.7938 \\mathrm{~g}$$ of $$\\mathrm{CO}_{2}$$ and $$0.4428 \\mathrm{~g}$$ of $$\\mathrm{H}_{2} \\mathrm{O}$$ was produced. The % composition of oxygen in the compound is ___________.
",
"options": [],
"answer": "46",
"solution": "**Answer:** 46\n\n$\\%$ of $\\mathrm{H}=\\frac{2}{18} \\times \\frac{\\text { wt. of } \\mathrm{H}_{2} \\mathrm{O}}{\\text { wt. of organic compound }} \\times 100$\n
\nwhen $12 \\mathrm{~g}$ carbon is burnt in $48 \\mathrm{~g}$ of oxygen, the volume of carbon monoxide produced is ___________ $\\times 10^{-1} \\mathrm{~L}$ at STP [nearest integer]\n
\n[Given: Assume $\\mathrm{CO}$ as ideal gas, Mass of $\\mathrm{C}$ is $12 \\mathrm{~g} \\mathrm{~mol}^{-1}$, Mass of $\\mathrm{O}$ is $16 \\mathrm{~g} \\mathrm{~mol}^{-1}$ and molar volume of an ideal gas at STP is $22.7 \\mathrm{~L} \\mathrm{~mol}^{-1}$ ]",
"options": [],
"answer": "227",
"solution": "**Answer:** 227\n\n\n
Limiting reagent is carbon. One mole carbon\nproduces one mole CO. Hence, volume at STP is 227 $$ \\times $$ 10-1 litre.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2966,
"subject": "Chemistry",
"question": "
When a hydrocarbon A undergoes combustion in the presence of air, it requires 9.5 equivalents of oxygen and produces 3 equivalents of water. What is the molecular formula of A?
An organic compound gives $$0.220 \\mathrm{~g}$$ of $$\\mathrm{CO}_{2}$$ and $$0.126 \\mathrm{~g}$$ of $$\\mathrm{H}_{2} \\mathrm{O}$$ on complete combustion. If the $$\\%$$ of carbon is 24 then the $$\\%$$ of hydrogen is __________ $$\\times 10^{-1}$$. ( Nearest integer)
",
"options": [],
"answer": "56",
"solution": "**Answer:** 56\n\nTo find the percentage of hydrogen in the compound, we first need to determine the ratio of moles of carbon and hydrogen in the products. \n
\nThe moles of CO₂ produced can be calculated by dividing the mass of CO₂ produced by its molar mass (44.01 g/mol):\n
\nNow, we are given that the percentage of carbon is 24%. Let's find the total mass of the compound:\n
\n$$\\text{Total mass of compound} = \\frac{\\text{Mass of carbon}}{\\% \\text{ of carbon}} = \\frac{0.060 \\,\\mathrm{g}}{0.24} = 0.250 \\,\\mathrm{g}$$\n
\nFinally, let's find the percentage of hydrogen:\n
\n$$\\% \\text{ of hydrogen} = \\frac{\\text{Mass of hydrogen}}{\\text{Total mass of compound}} \\times 100 = \\frac{0.014 \\,\\mathrm{g}}{0.250 \\,\\mathrm{g}} \\times 100 = 5.6$$\n
\nSo, the percentage of hydrogen is 5.6% or 56 × 10-1.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2968,
"subject": "Chemistry",
"question": "
The volume of hydrogen liberated at STP by treating $$2.4 \\mathrm{~g}$$ of magnesium with excess of hydrochloric acid is _________ $$\\times ~10^{-2} \\mathrm{~L}$$
\n
Given : Molar volume of gas is $$22.4 \\mathrm{~L}$$ at STP.
\n
Molar mass of magnesium is $$24 \\mathrm{~g} \\mathrm{~mol}^{-1}$$
",
"options": [],
"answer": "224",
"solution": "**Answer:** 224\n\nThe reaction between magnesium and hydrochloric acid is as follows :\n\n
`Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)`\n\n
From the equation, we see that 1 mole of magnesium (Mg) produces 1 mole of hydrogen gas (H2). \n\n
The molar mass of magnesium (Mg) is given as 24 g/mol. Therefore, 2.4 g of magnesium would correspond to (2.4 g)/(24 g/mol) = 0.1 mol of Mg.\n\n
Since 1 mole of Mg produces 1 mole of H2, 0.1 mol of Mg would produce 0.1 mol of H2.\n\n
At STP (Standard Temperature and Pressure), the molar volume of a gas is 22.4 L/mol.
Therefore, the volume of 0.1 mol of H2 would be (0.1 mol)*(22.4 L/mol) = 2.24 L.\n\n
So, the volume of hydrogen liberated at STP by treating 2.4 g of magnesium with excess of hydrochloric acid is 2.24 L.\n\n
In the terms of the question where the volume is expressed as ______ × 10-2 L, we convert 2.24 L into the desired form, i.e., 2.24 L = 2.24 $$ \\times $$ 102 $$ \\times $$ 10-2 L, so the answer is 224 × 10-2 L.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2969,
"subject": "Chemistry",
"question": "$10 \\mathrm{~mL}$ of gaseous hydrocarbon on combustion gives $40 \\mathrm{~mL}$ of $\\mathrm{CO}_2(\\mathrm{~g})$ and $50 \\mathrm{~mL}$ of water vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is _________ .",
"options": [],
"answer": "14",
"solution": "**Answer:** 14\n\n
To solve this problem, we will use the fact that the volume ratio of gases in a reaction at the same conditions of temperature and pressure represents their mole ratio according to Avogadro's law. Thus, since the gas volumes given are at the same conditions, we can directly relate them to their stoichiometric coefficients in the balanced equation for combustion of a hydrocarbon.
\n\n
The general equation for complete combustion of a hydrocarbon with a formula $\\mathrm{C}_x\\mathrm{H}_y$ can be represented as:
\nIf $72 ~\\mathrm{mmol}$ of $\\mathrm{PbCl}_2$ is mixed with $50 ~\\mathrm{mmol}$ of $\\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4$, then the amount of $\\mathrm{Pb}_3\\left(\\mathrm{PO}_4\\right)_2$ formed is ________ mmol (nearest integer).",
"options": [],
"answer": "24",
"solution": "**Answer:** 24\n\n
To solve this problem, we will use the stoichiometry of the balanced chemical reaction. The balanced equation shows that 3 moles of $\\mathrm{PbCl}_2$ react with 2 moles of $\\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4$ to produce 1 mole of $\\mathrm{Pb}_3\\left(\\mathrm{PO}_4\\right)_2$.
The molar ratio of $\\mathrm{PbCl}_2$ to $\\mathrm{Pb}_3\\left(\\mathrm{PO}_4\\right)_2$ is 3:1, and the molar ratio of $\\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4$ to $\\mathrm{Pb}_3\\left(\\mathrm{PO}_4\\right)_2$ is 2:1. We need to determine which reactant is the limiting reagent because it will dictate the amount of $\\mathrm{Pb}_3\\left(\\mathrm{PO}_4\\right)_2$ produced.
\n\n
The stoichiometric calculations are as follows:
\n\n
For $\\mathrm{PbCl}_2$:
\n\n$$\n\\text{Moles of }\\mathrm{Pb}_3\\left(\\mathrm{PO}_4\\right)_2\\text{ formed from } \\mathrm{PbCl}_2 = \\frac{72 \\text{ mmol of } \\mathrm{PbCl}_2}{3 \\text{ mmol of } \\mathrm{PbCl}_2\\text{/mmol of } \\mathrm{Pb}_3\\left(\\mathrm{PO}_4\\right)_2} = 24 \\text{ mmol}\n$$\n\n
For $\\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4$:
\n\n$$\n\\text{Moles of }\\mathrm{Pb}_3\\left(\\mathrm{PO}_4\\right)_2\\text{ formed from } \\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4 = \\frac{50 \\text{ mmol of } \\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4}{2 \\text{ mmol of } \\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4\\text{/mmol of } \\mathrm{Pb}_3\\left(\\mathrm{PO}_4\\right)_2} = 25 \\text{ mmol}\n$$\n\n
Now we can identify the limiting reagent by comparing the two amounts of $\\mathrm{Pb}_3\\left(\\mathrm{PO}_4\\right)_2$ that could be produced. The smaller quantity will be the actual amount produced since the limiting reagent restricts the reaction.
\n\n
Since the $\\mathrm{PbCl}_2$ can produce only 24 mmol of $\\mathrm{Pb}_3\\left(\\mathrm{PO}_4\\right)_2$ versus the 25 mmol that could be produced by $\\left(\\mathrm{NH}_4\\right)_3 \\mathrm{PO}_4$, $\\mathrm{PbCl}_2$ is the limiting reagent.
\n\n
Therefore, the amount of $\\mathrm{Pb}_3\\left(\\mathrm{PO}_4\\right)_2$ formed is 24 mmol (as a nearest integer).
To solve this problem, we can use stoichiometry. First, we need to write down the balanced chemical equation for the complete combustion of methane ($$\\mathrm{CH}_4$$).
\n\n
The balanced equation for combustion of methane is:
This equation tells us that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water.
\n\n
Next, we should find the molar mass of methane ($$\\mathrm{CH}_4$$) using the given molar masses:
\n\n
Molar mass of $$\\mathrm{CH}_4$$ = molar mass of C + 4 * molar mass of H
\n\n
Molar mass of $$\\mathrm{CH}_4$$ = $$12.0\\ \\mathrm{g/mol}$$ (for C) + $$4 * 1.0\\ \\mathrm{g/mol}$$ (for H)
\n\n
Molar mass of $$\\mathrm{CH}_4$$ = $$12.0\\ \\mathrm{g/mol} + 4.0\\ \\mathrm{g/mol}$$
\n\n
Molar mass of $$\\mathrm{CH}_4$$ = $$16.0\\ \\mathrm{g/mol}$$
\n\n
Now, to determine the mass of methane required to produce $$22\\ \\mathrm{g}$$ of $$\\mathrm{CO}_2$$, we should find out how many moles of $$\\mathrm{CO}_2$$ there are in $$22\\ \\mathrm{g}$$ and use the mole ratio from the balanced equation to find the moles of methane required.
\n\n
Moles of $$\\mathrm{CO}_2$$ = mass of $$\\mathrm{CO}_2$$ / molar mass of $$\\mathrm{CO}_2$$
\n\n
Molar mass of $$\\mathrm{CO}_2$$ = molar mass of C + 2 * molar mass of O
\n\n
Molar mass of $$\\mathrm{CO}_2$$ = $$12.0\\ \\mathrm{g/mol}$$ + $$2 * 16.0\\ \\mathrm{g/mol}$$
\n\n
Molar mass of $$\\mathrm{CO}_2$$ = $$12.0\\ \\mathrm{g/mol} + 32.0\\ \\mathrm{g/mol}$$
\n\n
Molar mass of $$\\mathrm{CO}_2$$ = $$44.0\\ \\mathrm{g/mol}$$
\n\n
Moles of $$\\mathrm{CO}_2$$ = $$\\frac{22\\ \\mathrm{g}}{44.0\\ \\mathrm{g/mol}}$$
\n\n
Moles of $$\\mathrm{CO}_2$$ = $$0.5\\ \\mathrm{moles}$$
\n\n
We will use the stoichiometric ratio from the balanced chemical equation to find the moles of $$\\mathrm{CH}_4$$ required to produce $$0.5\\ \\mathrm{moles}$$ of $$\\mathrm{CO}_2$$:
$$9.3 \\mathrm{~g}$$ of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is $$100 \\%$$ completed is _________ $$\\times 10^{-1} \\mathrm{~g}$$.
\n
(Given molar mass in $$\\mathrm{g} \\mathrm{~mol}^{-1}$$
Number of moles of methane required to produce $$22 \\mathrm{~g} \\mathrm{~CO}_{2(\\mathrm{~g})}$$ after combustion is $$\\mathrm{x} \\times 10^{-2}$$ moles. The value of $$\\mathrm{x}$$ is _________.
Combustion of glucose $$(\\mathrm{C}_6 \\mathrm{H}_{12} \\mathrm{O}_6)$$ produces $$\\mathrm{CO}_2$$ and water. The amount of oxygen (in $$\\mathrm{g}$$) required for the complete combustion of $$900 \\mathrm{~g}$$ of glucose is :
\n
[Molar mass of glucose in $$\\mathrm{g} \\mathrm{~mol}^{-1}=180$$]
From the balanced equation, we see that 1 mole of glucose reacts with 6 moles of oxygen $$(\\mathrm{O}_2)$$.
\n\n
Next, let's calculate the number of moles of glucose in 900 g. The molar mass of glucose is given as $$180 \\ \\mathrm{g} \\ \\mathrm{mol}^{-1}$$.
\n\n
Number of moles of glucose = $$\\frac{900 \\ \\mathrm{g}}{180 \\ \\mathrm{g} \\ \\mathrm{mol}^{-1}} = 5 \\ \\mathrm{mol}$$
\n\n
According to the balanced equation, 1 mole of glucose requires 6 moles of oxygen. Therefore, 5 moles of glucose will require:
\n\n
Number of moles of oxygen = $$5 \\ \\mathrm{mol} \\times 6 = 30 \\ \\mathrm{mol}$$
\n\n
Now, we need to find the mass of 30 moles of oxygen. The molar mass of oxygen $$\\left(\\mathrm{O}_2\\right)$$ is $$32 \\ \\mathrm{g} \\ \\mathrm{mol}^{-1}$$.
\n\n
Mass of oxygen = $$30 \\ \\mathrm{mol} \\times 32 \\ \\mathrm{g} \\ \\mathrm{mol}^{-1} = 960 \\ \\mathrm{g}$$
\n\n
Therefore, the amount of oxygen required for the complete combustion of 900 g of glucose is 960 g.
\n\n
So, the correct option is Option D: 960.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2975,
"subject": "Chemistry",
"question": "The number of significant figures in 50000.020 $$\\times$$ 10$$-$$3 is _____________.",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\n10$$-$$3 has no role in significant digits. Here in 50000.020, Number of significant figure = 8 as all zeroes between non zero digits are counted as significant digits and also the last zero is also significant digit as zeroes at the end or right of the number is significant only if they are present at the right side of the decimal.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2976,
"subject": "Chemistry",
"question": "The number of significant figures in 0.00340 is ____________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nNumber of significant figures = 3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2977,
"subject": "Chemistry",
"question": "
Using the rules for significant figures, the correct answer for the expression $${{0.02858 \\times 0.112} \\over {0.5702}}$$ will be
Reported answer should not be more precise than\nleast precise term (0.112 is the least precise term with three significant figures) in calculations, so there should\nbe three significant figures in reported answer.
Which of the following have same number of significant figures?
\n
A. 0.00253
\n
B. 1.0003
\n
C. 15.0
\n
D. 163
\n
Choose the correct answer from the options given below
",
"options": [
{
"text": "A, C and D only"
},
{
"text": "B and C only"
},
{
"text": "A, B and C only"
},
{
"text": "C and D only"
}
],
"answer": "A, C and D only",
"solution": "**Answer:** A, C and D only\n\n
Significant figures are the digits in a number that carry meaningful information about its precision. In other words, they are the numbers that are known with certainty.
\n
Here is how we calculate significant figures for each number:
\n
\n
A. 0.00253: There are 3 significant figures here (253). Leading zeroes do not count as significant figures.
\n
\n
B. 1.0003: There are 5 significant figures here. All non-zero digits are significant, and zeroes between non-zero digits or at the end of the number and after the decimal point are significant.
\n
\n
C. 15.0: There are 3 significant figures here. The zero after the decimal point counts as it indicates the precision of the measurement.
\n
\n
D. 163: There are 3 significant figures here. All non-zero digits are significant.
\n
\n
\n
So, comparing these, we see that options A (0.00253), C (15.0), and D (163) each have 3 significant figures.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2979,
"subject": "Chemistry",
"question": "In a hydrogen atom, if energy of an electron in ground state is -13.6 eV, then that in the 2nd excited state is",
"options": [
{
"text": "-1.51 eV"
},
{
"text": "-3.4 eV"
},
{
"text": "-6.04 eV"
},
{
"text": "-13.6 eV"
}
],
"answer": "-1.51 eV",
"solution": "**Answer:** -1.51 eV\n\nn = 1 means ground state
\n= -13.6 $$\\times$$ $$1 \\over 9$$ [ $$\\because$$ Z = atomic number = 1 for hydrogen atom]
\n= -1.51 ev ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2980,
"subject": "Chemistry",
"question": "In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen",
"options": [
{
"text": "5 $$\\to$$ 2"
},
{
"text": "4 $$\\to$$ 1"
},
{
"text": "2 $$\\to$$ 5"
},
{
"text": "3 $$\\to$$ 2"
}
],
"answer": "5 $$\\to$$ 2",
"solution": "**Answer:** 5 $$\\to$$ 2\n\nIn Bohr series of lines of hydrogen spectrum :-\n\nRed end means end of visible region which is balmer series of lines so n will be 2 \n \nNow for third lines with respect to Balmer series means n = (2 + 3) = 5th line
\n$$\\therefore$$ n1 = 2 and n2 = 5
\nFor 2 $$\\to$$ 5 electron will need energy to jump as 2 > 5 so no spectrum will release instead energy will be absorbed.\n
\n $$\\therefore$$ 5 $$\\to$$ 2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2981,
"subject": "Chemistry",
"question": "According to Bohr's theory, the angular momentum of an electron in 5th orbit is",
"options": [
{
"text": "10 $$h/\\pi$$"
},
{
"text": "2.5 $$h/\\pi$$"
},
{
"text": "25 $$h/\\pi$$"
},
{
"text": "1.0 $$h/\\pi$$"
}
],
"answer": "2.5 $$h/\\pi$$",
"solution": "**Answer:** 2.5 $$h/\\pi$$\n\nFormula of angular momentum of an electron,\n
mvr = $${{nh} \\over {2\\pi }}$$\n
here n = 5\n
$$\\therefore$$ mvr = $${{5h} \\over {2\\pi }}$$ = 2.5$${{h} \\over {\\pi }}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2982,
"subject": "Chemistry",
"question": "Ionisation energy of He+ is 19.6 x 10–18 J atom–1. The energy of the first stationary state (n = 1) of Li2+ is",
"options": [
{
"text": "4.41 x 10–16 J atom–1"
},
{
"text": "-4.41 x 10–17 J atom–1"
},
{
"text": "-2.2 x 10–15 J atom–1"
},
{
"text": "8.82 x 10–17 J atom–1"
}
],
"answer": "-4.41 x 10–17 J atom–1",
"solution": "**Answer:** -4.41 x 10–17 J atom–1\n\nIonisation energy(IE) - It is the energy required to move an electron from ground state to infinity.\n
(A) Kinetic energy of electron is $$ \\propto {{{Z^2}} \\over {{n^2}}}$$.
(B) The product of velocity (v) of electron and principal quantum number (n), $$'vn' \\propto {Z^2}$$.
(C) Frequency of revolution of electron in an orbit is $$ \\propto {{{Z^3}} \\over {{n^3}}}$$.
(D) Coulombic force of attraction on the electron is $$ \\propto {{{Z^3}} \\over {{n^4}}}$$.
Choose the most appropriate answer from the options given below :",
"options": [
{
"text": "(A), (C) and (D) only"
},
{
"text": "(A) only"
},
{
"text": "(C) only"
},
{
"text": "(A) and (D) only"
}
],
"answer": "(A) and (D) only",
"solution": "**Answer:** (A) and (D) only\n\nAccording to Bohr's theory,
I. $$KE \\propto {{{Z^2}} \\over {{n^2}}}$$ or $$13.6 \\propto {{{Z^2}} \\over {{n^2}}}{{(eV)} \\over {(atom)'}}$$
($$\\therefore$$ Correct)
II. Speed of electron $$ \\propto {Z \\over n}$$
(Here, Z = atomic number, n = number of shells)
$$\\therefore$$ $$v \\times n \\propto Z$$ ($$\\therefore$$ Incorrect)
III. Frequency of revolution of electron $$ = {v \\over {2\\pi r}}$$
Frequency $$ \\propto {{{Z^2}} \\over {{n^3}}}$$ ($$\\because$$ $$v \\propto {z \\over n},r \\propto {{{n^2}} \\over z}$$) ($$\\therefore$$ Incorrect)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2990,
"subject": "Chemistry",
"question": "Which of the following forms of hydrogen emits low energy $$\\beta$$- particles?",
"options": [
{
"text": "Tritium $$_1^3$$H"
},
{
"text": "Proton H+"
},
{
"text": "Protium $$_1^1$$H"
},
{
"text": "Deuterium $$_1^2$$H"
}
],
"answer": "Tritium $$_1^3$$H",
"solution": "**Answer:** Tritium $$_1^3$$H\n\nTritium isotope of hydrogen is radioactive and emits low energy $$\\beta ^-$$ particles. It is because of high n/p ratio of tritium which makes nucleus unstable.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2991,
"subject": "Chemistry",
"question": "Given below are two statements :
Statement I : Bohr's theory accounts for the stability and line spectrum of Li+ ion.
Statement II : Bohr's theory was unable to explain the splitting of spectral lines in the presence of a magnetic field.
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": "Both statement I and statement II are 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\nBohr’s theory is applicable for unielectronic species only\nLi+ has two electrons.
\nBohr’s theory could not explain the splitting of\nspectral lines in the presence of external magnetic\nfield (Zeeman effect)
\nStatement I – false
\nStatement II – true",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 2992,
"subject": "Chemistry",
"question": "Given below are two statements.
Statement I : According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus.
Statement II : According to Bohr's model of an atom, qualitatively the magnitude of velocity of electron increases with decrease in principal quantum number.
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 false"
},
{
"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"
}
],
"answer": "Statement I is false but Statement II is true",
"solution": "**Answer:** Statement I is false but Statement II is true\n\nVelocity of electron in Bohr's atom is given by
$$V \\propto {Z \\over n}$$
Z = atomic number of atom, corresponds to +ve charge so as Z increase velocity increases so statement I is wrong and as 'n' decreases velocity increases so statement II is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2993,
"subject": "Chemistry",
"question": "The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $${{{h^2}} \\over {xma_0^2}}$$. The value of 10x is ___________. (a0 is radius of Bohr's orbit) (Nearest integer) [Given : $$\\pi$$ = 3.14]",
"options": [],
"answer": "3155",
"solution": "**Answer:** 3155\n\n$$mvr = {{nh} \\over {2\\pi }}$$
The longest wavelength of light that can be used for the ionisation of lithium atom (Li) in its ground state is x $$\\times$$ 10$$-$$8 m. The value of x is ___________. (Nearest Integer).
\n
(Given : Energy of the electron in the first shell of the hydrogen atom is $$-$$2.2 $$\\times$$ 10$$-$$18 J ; h = 6.63 $$\\times$$ 10$$-$$34 Js and c = 3 $$\\times$$ 108 ms$$-$$1)
Bohr model is not valid for lithium atom (Li) as Bohr model is valid for only single electronic species, so it would be valid for Li+2 but not Li atom.\n
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.
\n
Assertion A: Energy of $$2 \\mathrm{s}$$ orbital of hydrogen atom is greater than that of $$2 \\mathrm{s}$$ orbital of lithium.
\n
Reason R: Energies of the orbitals in the same subshell decrease with increase in the atomic number.
\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\nAs the atomic number increases then the potential energy of electrons present in same shell becomes more and more negative. And therefore total energy also becomes more negative.\n
\n$\\therefore$ Energies of the orbitals in the same subshell decreases with increase in atomic number. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2997,
"subject": "Chemistry",
"question": "
Identify the incorrect statement from the following.
",
"options": [
{
"text": "A circular path around the nucleus in which an electron moves is proposed as Bohr's orbit."
},
{
"text": "An orbital is the one electron wave function $$(\\psi)$$ in an atom."
},
{
"text": "The existence of Bohr's orbits is supported by hydrogen spectrum."
},
{
"text": "Atomic orbital is characterised by the quantum numbers $$\\mathrm{n}$$ and $$l$$ only."
}
],
"answer": "Atomic orbital is characterised by the quantum numbers $$\\mathrm{n}$$ and $$l$$ only.",
"solution": "**Answer:** Atomic orbital is characterised by the quantum numbers $$\\mathrm{n}$$ and $$l$$ only.\n\nAtomic orbital is characterised by the quantum numbers $\\mathrm{n}, l$ and $\\mathrm{m}$.\n
\nHence option D is incorrect.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2998,
"subject": "Chemistry",
"question": "
Assume that the radius of the first Bohr orbit of hydrogen atom is 0.6 $$\\mathrm{\\mathop A\\limits^o }$$. The radius of the third Bohr orbit of He$$^+$$ is __________ picometer. (Nearest Integer)
The radius of the $$\\mathrm{2^{nd}}$$ orbit of $$\\mathrm{Li^{2+}}$$ is $$x$$. The expected radius of the $$\\mathrm{3^{rd}}$$ orbit of $$\\mathrm{Be^{3+}}$$ is
\nStatement I : According to Bohr's model of hydrogen atom, the angular momentum of an electron in a given stationary state is quantised.\n
\nStatement II : The concept of electron in Bohr's orbit, violates the Heisenberg uncertainty principle.\n
\nIn the light of the above statements, choose the most appropriate answer from the options given below:\n",
"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\nBoth Statement I and Statement II are correct\n
\nStatement I is correct. According to Bohr's model of the hydrogen atom, the angular momentum of an electron in a given stationary state is quantized. The angular momentum of an electron in the nth orbit is given by:\n
\n$L = n\\hbar = n \\frac{h}{2\\pi}$\n
\nwhere n is an integer (quantum number), and h is the Planck's constant.\n
\nStatement II is also correct. Bohr's model does not consider the Heisenberg uncertainty principle, which states that it is impossible to know the exact position and momentum of an electron simultaneously. In Bohr's model, electrons are assumed to move in well-defined orbits with quantized angular momentum, which implies knowledge of both position and momentum, thus violating the Heisenberg uncertainty principle.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3001,
"subject": "Chemistry",
"question": "
The energy of an electron in the first Bohr orbit of hydrogen atom is $$-2.18 \\times 10^{-18} \\mathrm{~J}$$. Its energy in the third Bohr orbit is ____________.
",
"options": [
{
"text": "One third of this value"
},
{
"text": "Three times of this value"
},
{
"text": "$$\\frac{1}{9}$$ th of this value"
},
{
"text": "$$\\frac{1}{27}$$ of this value"
}
],
"answer": "$$\\frac{1}{9}$$ th of this value",
"solution": "**Answer:** $$\\frac{1}{9}$$ th of this value\n\nThe energy of an electron in a hydrogen atom is given by the following formula:\n
\n$$E_n = \\frac{-13.6\\, \\mathrm{eV}}{n^2}$$\n
\nwhere n is the principal quantum number (Bohr orbit number). To find the energy in the third Bohr orbit (n = 3), we can use the formula:\n
\nThus, the energy of the electron in the third Bohr orbit is $$\\frac{1}{9}$$ th of its energy in the first Bohr orbit.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3002,
"subject": "Chemistry",
"question": "
The value of Rydberg constant $$(R_H)$$ is $$2.18 \\times 10^{-18} \\mathrm{~J}$$. The velocity of electron having mass $$9.1 \\times 10^{-31} \\mathrm{~kg}$$ in Bohr's first orbit of hydrogen atom = ________ $$\\times 10^5 \\mathrm{~ms}^{-1}$$ (nearest integer).
For hydrogen atom, energy of an electron in first excited state is $$-3.4 \\mathrm{~eV}, \\mathrm{K} . \\mathrm{E}$$. of the same electron of hydrogen atom is $$x \\mathrm{~eV}$$. Value of $$x$$ is _________ $$\\times 10^{-1} \\mathrm{~eV}$$. (Nearest integer)
To determine the kinetic energy (K.E.) of an electron in the first excited state of a hydrogen atom, we need to understand the relationship between the total energy, potential energy, and kinetic energy in an atom.\n\n
In a hydrogen atom, the total energy (E) of an electron in the nth state is given by:
\n\n
\n\n
$$E_n = - \\frac{13.6}{n^2} \\mathrm{~eV}$$
\n\n\n\n
For the first excited state, $ n = 2 $. So, plugging in the value:
This value represents the total energy (E) of the electron in the first excited state. According to the virial theorem for an electron in a Coulomb potential (as in a hydrogen atom), the kinetic energy (K.E.) is equal to the negative of the total energy:
\n\n
\n\n
$$\\mathrm{K.E.} = - E$$
\n\n\n\n
Substituting the total energy we calculated:
\n\n
\n\n
$$\\mathrm{K.E.} = - (-3.4) = 3.4 \\mathrm{~eV}$$
\n\n\n\n
Now, we need to find the value of $$x$$ in the form of $$x \\times 10^{-1} \\mathrm{~eV}$$.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3004,
"subject": "Chemistry",
"question": "The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 meters per second is approximately",
"options": [
{
"text": "10-31 meters"
},
{
"text": "10-16 meters"
},
{
"text": "10-25 meters"
},
{
"text": "10-33 meters"
}
],
"answer": "10-33 meters",
"solution": "**Answer:** 10-33 meters\n\nAs per de-Broglie wavelength of a particle is
= 494 nm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3007,
"subject": "Chemistry",
"question": "A stream of electrons from a heated filament was passed between two charged plates kept at a potential\ndifference V esu. If e and m are charge and mass of an electron, respectively, then the value of h / $$\\lambda $$ (where\n$$\\lambda $$ is wavelength associated with electron wave) is given by :",
"options": [
{
"text": "2meV "
},
{
"text": "$$\\sqrt {meV} $$ "
},
{
"text": "$$\\sqrt {2meV} $$ "
},
{
"text": "meV"
}
],
"answer": "$$\\sqrt {2meV} $$ ",
"solution": "**Answer:** $$\\sqrt {2meV} $$ \n\nWe know, Kinetic Energy (KE) = $${1 \\over 2}m{v^2}$$\n
$$\\therefore$$ mv2 = 2KE\n
$$ \\Rightarrow $$ m2v2 = 2mKE\n
$$ \\Rightarrow $$ mv = $$\\sqrt {2m{K_E}} $$\n
For an electron of charge 'e' which is passes through 'V' volt, kinetic energy of electron will be\n
KE = eV\n
$$\\therefore$$ m = $$\\sqrt {2meV} $$\n
We know de-Broglie wavelength for an electron$$\\left( \\lambda \\right)$$ = $${h \\over p} = {h \\over {mv}}$$\n
Electrons in a cathode ray tube have been emitted with a velocity of 1000 m s$$^{-1}$$. The number of following statements which is/are $$\\underline {\\mathrm{true}} $$ about the emitted radiation is ____________.
\n
Given : $$\\mathrm{h=6\\times10^{-34}~J~s,m_e=9\\times10^{-31}~kg}$$.
\n
(A) The de-Broglie wavelength of the electron emitted is 666.67 nm.
\n
(B) The characteristic of electrons emitted depend upon the material of the electrodes of the cathode ray tube.
\n
(C) The cathode rays start from cathode and move towards anode.
\n
(D) The nature of the emitted electrons depends on the nature of the gas present in cathode ray tube.
",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nThe number of true statements about the emitted radiation is 2, i.e. statements (A) and (C).\n\n
(A) is True: The de-Broglie wavelength of an electron can be calculated as lambda = h / p, where h is Planck's constant and p is the momentum of the electron. The momentum of an electron with a velocity of 1000 m/s is mv, where m is the mass of an electron. Plugging in the values, we get \n
(B) is False : The characteristic of electrons emitted is\nindependent of the material of the electrodes of \nthe cathode ray tube.\n\n
(C) is True: In a cathode ray tube, electrons are emitted from the cathode and move towards the anode under the influence of an electric field.\n\n
(D) is False: The nature of the emitted electrons does not depend on the nature of the gas present in the cathode ray tube, but rather on the material of the electrodes and the potential difference applied across them.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3009,
"subject": "Chemistry",
"question": "
The de-Broglie's wavelength of an electron in the $$4^{\\text {th }}$$ orbit is ________ $$\\pi \\mathrm{a}_0$$. ($$\\mathrm{a}_0=$$ Bohr's radius)
The de-Broglie wavelength of an electron can be expressed through its relationship with the principal quantum number $$n$$ in a Bohr orbit. According to de Broglie, the wavelength $ \\lambda $ of an electron moving in an orbit can be related to its momentum. However, when considering an electron in an atom, specifically within the Bohr model, we can relate the de-Broglie wavelength to the circumference of the orbit it moves in. The formula representing the de Broglie wavelength of an electron in the $n^{\\text{th}}$ orbit of a hydrogen-like atom is:
\n\n
$$ \\lambda_n = \\frac{2 \\pi r_n}{n} $$
\n\n
where
\n\n
\n\n
$ \\lambda_n $ is the de-Broglie wavelength of the electron in the $n^{\\text{th}}$ orbit,
\n\n
$ r_n $ is the radius of the $n^{\\text{th}}$ orbit,
\n\n
$ n $ is the principal quantum number (which is 4 in this case).
\n\n
\n\n
According to Bohr's theory, the radius of the $n^{\\text{th}}$ orbit is given by:
\n\n
$$ r_n = n^2 a_0 $$
\n\n
where $ a_0 $ is the Bohr radius.
\n\n
For the $4^{\\text{th}}$ orbit, $n = 4$, so the radius is:
\n\n
$$ r_4 = 4^2 a_0 = 16 a_0 $$
\n\n
Substituting $ r_4 $ into the de-Broglie wavelength formula, we get:
In $$CO_3^{2 - }$$ no of electrons = 6 + 24 + 2 = 32\n
In $$NO_3^{-}$$ no of electrons = 7 + 24 + 1 = 32\n
So they are not isoelectronic.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3014,
"subject": "Chemistry",
"question": "Pick out the isoelectronic structure from the following : \n
So, option $$(B)$$ is correct. ",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 3017,
"subject": "Chemistry",
"question": "The group having isoelectronic species is:",
"options": [
{
"text": "O2- , F- , Na+, Mg2+"
},
{
"text": "O2- , F- , Na, Mg2+"
},
{
"text": "O- , F- , Na+, Mg2+"
},
{
"text": "O- , F- , Na, Mg+"
}
],
"answer": "O2- , F- , Na+, Mg2+",
"solution": "**Answer:** O2- , F- , Na+, Mg2+\n\nO2- , F- , Na+ and Mg2+, all have 10 electrons each.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 3018,
"subject": "Chemistry",
"question": "The isoelectronic set of ions is :",
"options": [
{
"text": "F–, Li+, Na+ and Mg2+ "
},
{
"text": "Li+, Na+, O2- and F- "
},
{
"text": "N3–, O2-, F- and Na+ "
},
{
"text": "N3–, Li+, Mg2+ and O2-"
}
],
"answer": "N3–, O2-, F- and Na+ ",
"solution": "**Answer:** N3–, O2-, F- and Na+ \n\nAtomic numbers of N, O, F and Na are 7, 8, 9\nand 11 respectively. \n
Therefore, total number of\nelectrons in each of N3-, O2–, F– and Na+ is 10\nand hence they are isoelectronic.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3019,
"subject": "Chemistry",
"question": "The group number, number of valence electrons, and valency of an element with atomic number 15,\nrespectively, are:",
"options": [
{
"text": "16, 6 and 3\n"
},
{
"text": "15, 6 and 12"
},
{
"text": "16, 5 and 2"
},
{
"text": "15, 5 and 3"
}
],
"answer": "15, 5 and 3",
"solution": "**Answer:** 15, 5 and 3\n\nP[15] = [Ne] 3s23p3\n
Group number = 10 + valance shell electron = 10 + 5 = 15\n
number of valence electrons = 5\n
Valency = 3",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 3020,
"subject": "Chemistry",
"question": "Hydrogen has three isotopes (A), (B) and (C).\nIf the number of neutron(s) in (A), (B) and (C)\nrespectively, are (x), (y) and (z), the sum of (x),\n(y) an (z) is :",
"options": [
{
"text": "3"
},
{
"text": "1"
},
{
"text": "4"
},
{
"text": "2"
}
],
"answer": "3",
"solution": "**Answer:** 3\n\nHydrogen has three isotopes\n
(A) Protium ($${}_1^1H$$) has 0 neutron.\n
(B) Deutrium ($${}_1^2H$$) has 1 neutrons.\n
(C) Tritium ($${}_1^3H$$) has 2 neutrons.\n
Total number of neutrons in three isotopes of\nhydrogen = 0 + 1 + 2 = 3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3021,
"subject": "Chemistry",
"question": "The orbital having two radial as well as two angular nodes is :",
"options": [
{
"text": "3p"
},
{
"text": "5d"
},
{
"text": "4d"
},
{
"text": "4f"
}
],
"answer": "5d",
"solution": "**Answer:** 5d\n\nNumber of radial nodes = (n – l – 1)
\nNumber of angular nodes = l
\nfor 5d; n = 5, l = 2
\n5d orbital has two radial nodes and two angular nodes",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3022,
"subject": "Chemistry",
"question": "A certain orbital has no angular nodes and two radial nodes. The orbital is :",
"options": [
{
"text": "2s"
},
{
"text": "3s"
},
{
"text": "3p"
},
{
"text": "2p"
}
],
"answer": "3s",
"solution": "**Answer:** 3s\n\nAngular nodes = $$\\ell $$ = 0
\n So orbital is 3s.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3023,
"subject": "Chemistry",
"question": "Outermost electronic configuration of a group 13 element, E, is 4s2, 4p1. The electronic configuration of an element of p-block period-five placed diagonally to element, E is :",
"options": [
{
"text": "[Kr] 3d10 4s2 4p2"
},
{
"text": "[Ar] 3d10 4s2 4p2"
},
{
"text": "[Xe] 5d10 6s2 6p2"
},
{
"text": "[Kr] 4d10 5s2 5p2"
}
],
"answer": "[Kr] 4d10 5s2 5p2",
"solution": "**Answer:** [Kr] 4d10 5s2 5p2\n\nThe element E is Ga and the diagonal element of 5th period is $${}_{50}Sn$$ having outer electronic configuration will be [Kr] 4d10 5s2 5p2.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 3024,
"subject": "Chemistry",
"question": "
The number of radial and angular nodes in 4d orbital are, respectively
",
"options": [
{
"text": " 1 and 2"
},
{
"text": "3 and 2"
},
{
"text": "1 and 0"
},
{
"text": "2 and 1"
}
],
"answer": " 1 and 2",
"solution": "**Answer:** 1 and 2\n\n
The pair, in which ions are isoelectronic with AI3+ is :
",
"options": [
{
"text": "Br$$-$$ and Be2+"
},
{
"text": "Cl$$-$$ and Li+"
},
{
"text": "S2$$-$$ and K+"
},
{
"text": "O2$$-$$ and Mg2+"
}
],
"answer": "O2$$-$$ and Mg2+",
"solution": "**Answer:** O2$$-$$ and Mg2+\n\nO2–, Mg2+ and Al3+ are isoelectronic. All have 10 electrons.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3026,
"subject": "Chemistry",
"question": "
Consider an imaginary ion $${ }_{22}^{48} \\mathrm{X}^{3-}$$. The nucleus contains '$$a$$'% more neutrons than the number of electrons in the ion. The value of 'a' is _______________. [nearest integer]
",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nNumber of electrons in $${ }_{22}^{48} X^{3-}$$ is 25 .\n
\nNumber of neutrons $$=48-22=26$$.\n
\n$$\\%$$ increase in the number of neutrons over electrons\n
\nElements with greater electronegativity require more energy for ionisation\n
\nHence the correct order is $C>D>A>B$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3028,
"subject": "Chemistry",
"question": "
Which one of the following sets of ions represents a collection of isoelectronic species?
\n
(Given : Atomic Number : $$\\mathrm{F:9,Cl:17,Na=11,Mg=12,Al=13,K=19,Ca=20,Sc=21}$$)
",
"options": [
{
"text": "$$\\mathrm{N}^{3-}, \\mathrm{O}^{2-}, \\mathrm{F}^{-}, \\mathrm{S}^{2-}$$"
},
{
"text": "$$\\mathrm{Ba}^{2+}, \\mathrm{Sr}^{2+}, \\mathrm{K}^{+}, \\mathrm{Ca}^{2+}$$"
},
{
"text": "$$\\mathrm{Li}^{+}, \\mathrm{Na}^{+}, \\mathrm{Mg}^{2+}, \\mathrm{Ca}^{2+}$$"
},
{
"text": "$$\\mathrm{K}^{+}, \\mathrm{Cl}^{-}, \\mathrm{Ca}^{2+}, \\mathrm{Sc}^{3+}$$"
}
],
"answer": "$$\\mathrm{K}^{+}, \\mathrm{Cl}^{-}, \\mathrm{Ca}^{2+}, \\mathrm{Sc}^{3+}$$",
"solution": "**Answer:** $$\\mathrm{K}^{+}, \\mathrm{Cl}^{-}, \\mathrm{Ca}^{2+}, \\mathrm{Sc}^{3+}$$\n\nIsoelectronic species have same number of \nelectrons.\n
$$\\mathrm{K}^{+}, \\mathrm{Cl}^{-}, \\mathrm{Ca}^{2+}, \\mathrm{Sc}^{3+}$$ all have 18 electrons, \nhence these are isoelectronic.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3029,
"subject": "Chemistry",
"question": "
The number of s-electrons present in an ion with 55 protons in its unipositive state is
",
"options": [
{
"text": "10"
},
{
"text": "9"
},
{
"text": "12"
},
{
"text": "8"
}
],
"answer": "10",
"solution": "**Answer:** 10\n\nThe number of protons in an element determines its atomic number, which is the number of electrons in a neutral atom of the element. The unipositive state of an ion indicates that it has lost one electron. Therefore, the number of electrons in a unipositive ion is one less than the atomic number.
\nThe element with 55 protons is cesium (Cs), which is in group 1 of the periodic table. Group 1 elements have one valence electron in the s sublevel of their outermost energy level.\n
\nIn a neutral atom of cesium, there are 55 electrons, with the electron configuration [Xe] 6s1. When cesium loses one electron to form a unipositive ion (Cs+), the resulting ion has 54 electrons.
\nOut of this 54 electrons s subshell has 10 electrons in 1s2, 2s2, 3s2, 4s2, 5s2.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3030,
"subject": "Chemistry",
"question": "The total number of isoelectronic species from the given set is ___________.
\n$\\mathrm{O}^{2-}, \\mathrm{F}^{-}, \\mathrm{Al}, \\mathrm{Mg}^{2+}, \\mathrm{Na}^{+}, \\mathrm{O}^{+}, \\mathrm{Mg}, \\mathrm{Al}^{3+}, \\mathrm{F}$",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nIsoelectronic species are atoms, ions or molecules that have the same number of electrons. To find the isoelectronic species from the given set, we need to determine the number of electrons in each species.\n
\nThe electron configuration of each species is:\n
Radial nodes in an atomic orbital are areas where the probability of finding an electron is zero. The number of radial nodes in an orbital is given by the formula:
\n
$ \\text{number of radial nodes} = n - l - 1 $
\n
where $n$ is the principal quantum number and $l$ is the azimuthal quantum number. The azimuthal quantum number ($l$) can have values from 0 to $n-1$, and it determines the shape of the orbital (s, p, d, f, etc.). For an s orbital, $l=0$; for a p orbital, $l=1$; for a d orbital, $l=2$; and so on.
\n
Let's calculate the number of radial nodes for each given orbital:
\n\n
7s: $n=7$, $l=0$, so the number of radial nodes is $7 - 0 - 1 = 6$, not 5.
\n
7p: $n=7$, $l=1$, so the number of radial nodes is $7 - 1 - 1 = 5$.
\n
6s: $n=6$, $l=0$, so the number of radial nodes is $6 - 0 - 1 = 5$.
\n
8p: $n=8$, $l=1$, so the number of radial nodes is $8 - 1 - 1 = 6$, not 5.
\n
8d: $n=8$, $l=2$, so the number of radial nodes is $8 - 2 - 1 = 5$.
\n\n
Therefore, the orbitals with 5 radial nodes are 7p, 6s, and 8d, so there are 3 such orbitals.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3032,
"subject": "Chemistry",
"question": "The number of radial node/s for $3 p$ orbital is :",
"options": [
{
"text": "3"
},
{
"text": "2"
},
{
"text": "1"
},
{
"text": "4"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\n
The number of radial nodes in an orbital is given by the formula:
\n
$$\n\\text{Number of radial nodes} = n - l - 1\n$$
\n
where \n$$n$$ is the principal quantum number, \n$$l$$ is the azimuthal quantum number also known as the angular momentum quantum number.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3033,
"subject": "Chemistry",
"question": "Uncertainty in position of a minute particle of mass 25 g in space is 10-5 m. What is the uncertainty in its velocity (in ms-1) (h = 6.6 $$\\times$$ 10-34 Js)",
"options": [
{
"text": "2.4 $$\\times$$ 10-34"
},
{
"text": "0.5 $$\\times$$ 10-34"
},
{
"text": "2.1 $$\\times$$ 10-28"
},
{
"text": "0.5 $$\\times$$ 10-23"
}
],
"answer": "2.1 $$\\times$$ 10-28",
"solution": "**Answer:** 2.1 $$\\times$$ 10-28\n\nAccording to Hysenberg's Uncertainty Principal\n$$$\\Delta x . \\Delta p \\ge { h \\over {4\\pi}}$$$\nWhere $$\\Delta x $$ is uncertainty in position and $$\\Delta p $$ is uncertainty in momentum and as in momentum m (mass of object) is always same so we can say\n$$$\\Delta x . m. \\Delta v \\ge { h \\over {4\\pi}}$$$\nMass of the particle (m) = 25 g = 0.025 kg
\nPosition of the particle ($$\\Delta x$$) = 10-5 m
\n$$\\Delta x . \\Delta p $$ = $${ h \\over {4\\pi}}$$
\n$$\\Delta x . m . \\Delta v $$ = $${ h \\over {4\\pi}}$$
\n$$\\Delta v$$ = $${h \\over {4 \\times \\Delta x \\times m \\times \\pi }}$$
= 1.92 $$ \\times $$ 10-2 m",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3035,
"subject": "Chemistry",
"question": "In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainity\nwith which the position of the electron can be located is (h = 6.6 $$\\times$$ 10-34 kg m2s-1, mass of electron, em = 9.1 $$\\times$$ 10-31 kg) ",
"options": [
{
"text": "5.10 $$\\times$$ 10-3 m"
},
{
"text": "1.92 $$\\times$$ 10-3 m"
},
{
"text": "3.84 $$\\times$$ 10-3 m"
},
{
"text": "1.52 $$\\times$$ 10-4 m"
}
],
"answer": "1.92 $$\\times$$ 10-3 m",
"solution": "**Answer:** 1.92 $$\\times$$ 10-3 m\n\n% error in velocity = $${{\\Delta V} \\over V} \\times 100$$\n
= 1.92 $$ \\times $$ 10-3 m",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3036,
"subject": "Chemistry",
"question": "A ball weighing 10 g is moving with a velocity of 90 ms$$-$$1. If the uncertainty in its velocity is 5%, then the uncertainty in its position is ___________ $$\\times$$ 10$$-$$33 m. (Rounded off to the nearest integer)
[Given : h = 6.63 $$\\times$$ 10$$-$$34 Js]",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nm = 10 g = 10$$-$$2 Kg
x = 1.17 $$ \\simeq $$ 1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3037,
"subject": "Chemistry",
"question": "An accelerated electron has a speed of 5 $$\\times$$ 106 ms$$-$$1 with an uncertainty of 0.02%. The uncertainty in finding its location while in motion is x $$\\times$$ 10$$-$$9 m. The value of x is ____________. (Nearest integer)
[Use mass of electron = 9.1 $$\\times$$ 10$$-$$31 kg, h =6.63 $$\\times$$ 10$$-$$34 Js, $$\\pi$$ = 3.14]",
"options": [],
"answer": "58",
"solution": "**Answer:** 58\n\n$$\\Delta v = {{0.02} \\over {100}} \\times 5 \\times {10^6} = {10^3}$$ m/s
If the uncertainty in velocity and position of a minute particle in space are, 2.4 $$\\times$$ 10$$-$$26 (m s$$-$$1) and 10$$-$$7 (m) respectively. The mass of the particle in g is ____________. (Nearest integer)
The minimum uncertainty in the speed of an electron in an one dimensional region of length $$2 \\mathrm{a}_{\\mathrm{o}}$$ (Where $$\\mathrm{a}_{\\mathrm{o}}=$$ Bohr radius $$52.9 \\,\\mathrm{pm}$$) is _________ $$\\mathrm{km} \\,\\mathrm{s}^{-1}$$.
\n
(Given : Mass of electron = 9.1 $$\\times$$ 10$$-$$31 kg, Planck's constant h = 6.63 $$\\times$$ 10$$-$$34 Js)
Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter $$10^{-15} \\mathrm{~m}$$ is ________ $$\\times 10^9 \\mathrm{~ms}^{-1}$$ (nearest integer)
\n
[Given : mass of electron $$=9.1 \\times 10^{-31} \\mathrm{~kg}$$, Plank's constant $$(h)=6.626 \\times 10^{-34} \\mathrm{Js}$$] (Value of $$\\pi=3.14$$)
To find the uncertainty in the velocity of the electron within an atomic nucleus of diameter $10^{-15} \\ \\text{m}$, we can use Heisenberg's uncertainty principle. The principle is expressed as:
\n\n
$ \\Delta x \\cdot m \\Delta v \\geq \\frac{h}{4 \\pi} $
\n\n
Here, $\\Delta x$ is the uncertainty in position, $m$ is the mass of the electron, $\\Delta v$ is the uncertainty in velocity, and $h$ is Planck's constant.
\n\n
Given:
\n\n
\n
$ \\Delta x = 10^{-15} \\ \\text{m} $
\n
$ m = 9.1 \\times 10^{-31} \\ \\text{kg} $
\n
$ h = 6.626 \\times 10^{-34} \\ \\text{Js} $
\n
$ \\pi = 3.14 $
\n
\n
We need to find $\\Delta v$. Rearrange the uncertainty principle to solve for $\\Delta v$:
\n\n
$ \\Delta v \\approx \\frac{h}{4 \\pi m \\Delta x} $
$$ \\Rightarrow $$$$\\,\\,\\,$$ $$\\lambda $$ = $${{36A} \\over 7}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3045,
"subject": "Chemistry",
"question": "The electron in the hydrogen atom undergoes transition from higher orbitals\nto orbital of radius 211.6 pm. This transition is associated with :",
"options": [
{
"text": "Lyman series"
},
{
"text": "Balmer series"
},
{
"text": "Paschen series"
},
{
"text": "Brackett series"
}
],
"answer": "Balmer series",
"solution": "**Answer:** Balmer series\n\nHere electron is coming to the orbital of radius 211.6 pm. Now we have to find which series have radius of orbital 211.6 pm.\n
As n = 2 so the series is Balmer Series.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3046,
"subject": "Chemistry",
"question": "For emission line of atomic hydrogen from ni = 8 to nf = n, the plot of wave number $$\\left( {\\overline v } \\right)$$ against $$\\left( {{1 \\over {{n^2}}}} \\right)$$ will be (The Rydberg constant, RH is in wave number unit)",
"options": [
{
"text": "Linear with intercept $$-$$RH"
},
{
"text": "Non linear "
},
{
"text": "Linear with slope RH"
},
{
"text": "Linear with slope $$-$$RH"
}
],
"answer": "Linear with slope RH",
"solution": "**Answer:** Linear with slope RH\n\nWe know, \n
So, graph will be linear with slope RH. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3047,
"subject": "Chemistry",
"question": "The ground state energy of hydrogen atom is – 13.6 eV. The energy of second excited state of He+ ion in eV is : ",
"options": [
{
"text": "$$-$$ 6.04"
},
{
"text": "$$-$$ 54.4"
},
{
"text": "$$-$$ 27.2"
},
{
"text": "$$-$$ 3.4"
}
],
"answer": "$$-$$ 6.04",
"solution": "**Answer:** $$-$$ 6.04\n\n(E)nth = (EGND)H . $${{{Z^2}} \\over {{n^2}}}$$\n
E3rd (He+) = ($$-$$13.6eV) . $${{{2^2}} \\over {{3^2}}}$$ = $$-$$ 6.04 eV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3048,
"subject": "Chemistry",
"question": "Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H atom is suitable for this purpose?\n
$$ \\therefore $$ $${{\\Delta {{\\overline v }_{Lyman}}} \\over {\\Delta {{\\overline v }_{Balmer}}}}$$ = $${{{1 \\over 4}} \\over {{1 \\over 9}}}$$ = $${{9 \\over 4}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3050,
"subject": "Chemistry",
"question": "The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The\nspectral series are :",
"options": [
{
"text": "Paschen and Pfund"
},
{
"text": "Balmer and Brackett"
},
{
"text": "Lyman and Paschen"
},
{
"text": "Brackett and Pfund"
}
],
"answer": "Lyman and Paschen",
"solution": "**Answer:** Lyman and Paschen\n\nShortest wavelength can found when n2 = $$\\infty $$\n
$$ \\therefore $$ Series number of B = 3 $$ \\times $$ Series number of A\n
So,\n
If series number of A is Lyman( n = 1 )\n
then series number of B is Paschen( n = 3 ).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3051,
"subject": "Chemistry",
"question": "For the Balmer series in the spectrum of H atom,\n $$\\overline \\nu = {R_H}\\left\\{ {{1 \\over {n_1^2}} - {1 \\over {n_2^2}}} \\right\\}$$, the correct statements among (I) to (IV)\nare :\n (I) As wavelength decreases, the lines in the series converge\n (II) The integer n1 is equal to 2\n (III) The lines of longest wavelength corresponds to n2 = 3\n (IV) The ionization energy of hydrogen can be calculated from wave number of these lines",
"options": [
{
"text": "(II), (III), (IV)"
},
{
"text": "(I), (II), (III)"
},
{
"text": "(I), (III), (IV)"
},
{
"text": "(I), (II), (IV)"
}
],
"answer": "(I), (II), (III)",
"solution": "**Answer:** (I), (II), (III)\n\nFor balmer series : n1 = 2, n2 = 3, 4, 5, .....$$\\infty $$\n
As wavelength decreases the lines in the\nBalmer series converge. The correct\nstatements are (I), (II) and (III).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3052,
"subject": "Chemistry",
"question": "The shortest wavelength of H atom in the\nLyman series is $$\\lambda $$1. The longest wavelength in\nthe Balmar series of He+ is :",
"options": [
{
"text": "$${{5{\\lambda _1}} \\over 9}$$"
},
{
"text": "$${{36{\\lambda _1}} \\over 5}$$"
},
{
"text": "$${{27{\\lambda _1}} \\over 5}$$"
},
{
"text": "$${{9{\\lambda _1}} \\over 5}$$"
}
],
"answer": "$${{9{\\lambda _1}} \\over 5}$$",
"solution": "**Answer:** $${{9{\\lambda _1}} \\over 5}$$\n\nFor Shortest wavelength energy should be maximum.\n
For maximum energy transition must be form n = $$\\infty $$ to n = 1.\n
\nHence maximum number of emission lines observed are $10 .$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3055,
"subject": "Chemistry",
"question": "
Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from $$\\mathrm{n=4}$$ to $$\\mathrm{n}=2$$ of $$\\mathrm{He}^{+}$$ spectrum
",
"options": [
{
"text": "$$\\mathrm{n}=3$$ to $$\\mathrm{n}=4$$"
},
{
"text": "$$\\mathrm{n}=2$$ to $$\\mathrm{n}=1$$"
},
{
"text": "$$\\mathrm{n}=1$$ to $$\\mathrm{n}=2$$"
},
{
"text": "$$\\mathrm{n}=1$$ to $$\\mathrm{n}=3$$"
}
],
"answer": "$$\\mathrm{n}=2$$ to $$\\mathrm{n}=1$$",
"solution": "**Answer:** $$\\mathrm{n}=2$$ to $$\\mathrm{n}=1$$\n\n$\\mathrm{He}^{+}$ion :\n
If wavelength of the first line of the Paschen series of hydrogen atom is 720 nm, then the wavelength of the second line of this series is _________ nm. (Nearest integer)
Number of spectral lines obtained in $$\\mathrm{He}^{+}$$ spectra, when an electron makes transition from fifth excited state to first excited state will be
Wavenumber for a radiation having 5800 $$\\mathop A\\limits^o $$ wavelength is $$x \\times 10 \\mathrm{~cm}^{-1}$$ The value of $$x$$ is ________. (Integer answer)
The wavenumber of a radiation is defined as the number of wavelengths per unit distance and is the reciprocal of the wavelength. Wavenumber is commonly represented in units of $$\\mathrm{cm}^{-1}$$.
\n\n
First, convert the given wavelength from angstroms ($$\\mathop A\\limits^o $$) to centimeters (cm). We know that:
Here, it is given that the wavenumber is in the form of $$x \\times 10 \\, \\mathrm{cm}^{-1}$$, so $$x$$ would be the value we calculated divided by 10:
\n\n
$$x = \\frac{1.724 \\times 10^4}{10} = 1724$$
\n\n
Thus, the integer answer for the value of $$x$$ is:
To match List I (Alkali Metals) with List II (Emission Wavelength in nm), we need to know the characteristic emission lines of these elements. Here is the information for the typical emission wavelengths:
\n\n
\n\n
Lithium (Li): 670.8 nm
\n\n
Sodium (Na): 589.2 nm
\n\n
Rubidium (Rb): 780.0 nm
\n\n
Cesium (Cs): 455.5 nm
\n\n
\n\n
Now let's match these with the given options:
\n\n
\n\n
A. Li (III): 670.8 nm
\n\n
B. Na (I): 589.2 nm
\n\n
C. Rb (IV): 780.0 nm
\n\n
D. Cs (II): 455.5 nm
\n\n
\n\n
Therefore, the correct answer is:
\n\n
Option B \n\n
(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3062,
"subject": "Chemistry",
"question": "A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at\n 680 nm, the other is at :",
"options": [
{
"text": "325 nm "
},
{
"text": "743 nm "
},
{
"text": "518 nm"
},
{
"text": "1035 nm"
}
],
"answer": "743 nm ",
"solution": "**Answer:** 743 nm \n\nHere a photon is divided into two photons of different wavelength.\n
According to the law of conservation of energy,\n
Energy of photon before dividing into two parts = Energy of first photon + Energy of second photon\n
$${\\lambda _2} = 742.77$$ nm = 743 nm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3063,
"subject": "Chemistry",
"question": "Ejection of the photoelectron from metal in the photoelectric experiment can be stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is : ",
"options": [
{
"text": "4 eV"
},
{
"text": "4.5 eV"
},
{
"text": "5 eV"
},
{
"text": "5.5 eV"
}
],
"answer": "4.5 eV",
"solution": "**Answer:** 4.5 eV\n\n$$\\lambda $$ = 250 nm = 2500 $$\\mathop A\\limits^ \\circ $$ \n
= 2.0934 eV $$ \\simeq $$ 2.1 eV",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3066,
"subject": "Chemistry",
"question": "The upper stratosphere consisting of the ozone layer protects us from sun's radiation that falls in the wavelength region of - ",
"options": [
{
"text": "200-315 nm "
},
{
"text": "400-550 nm "
},
{
"text": "0.8-1.5 nm"
},
{
"text": "600-750 nm"
}
],
"answer": "200-315 nm ",
"solution": "**Answer:** 200-315 nm \n\nSun emits UV-radiations, which have the wavelength range from 1 nm to 400 nm.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3067,
"subject": "Chemistry",
"question": "If p is the momentum of the fastest electron\nejected from a metal surface after the irradiation\nof light having wavelength $$\\lambda $$, then for 1.5 p\nmomentum of the photoelectron, the\nwavelength of the light should be:\n(Assume kinetic energy of ejected\nphotoelectron to be very high in comparison\nto work function)",
"options": [
{
"text": "1/2 $$\\lambda $$"
},
{
"text": "3/4 $$\\lambda $$"
},
{
"text": "4/9 $$\\lambda $$"
},
{
"text": "2/3 $$\\lambda $$"
}
],
"answer": "4/9 $$\\lambda $$",
"solution": "**Answer:** 4/9 $$\\lambda $$\n\nFrom photoelectric effect,\n
= 222 × 10–21 J",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3069,
"subject": "Chemistry",
"question": "Electromagnetic radiation of wavelength 663 nm is just sufficient to ionise the atom of metal A. The ionization enegy of metal A in kJ mol$$-$$1 is __________. (Rounded off to the nearest integer)
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3070,
"subject": "Chemistry",
"question": "Among the following, number of metal/s which can be used as electrodes in the photoelectric cell is _________. (Integer answer)
(A) Li
(B) Na
(C) Rb
(D) Cs",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nAmong the given alkali metals, only cesium (Cs)\nis used as electrode in the photoelectric cell\ndue to its lowest ionisation energy.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3071,
"subject": "Chemistry",
"question": "When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is _______$$\\mathop A\\limits^o $$. (Round off to the Nearest Integer).
[ Use : $$\\sqrt 3 $$ = 1.73, h = 6.63 $$\\times$$ 10$$-$$34 Js
me = 9.1 $$\\times$$ 10$$-$$31 kg; c = 3.0 $$\\times$$ 108 ms$$-$$1; 1eV = 1.6 $$\\times$$ 10$$-$$19 J]",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\n$$\\lambda$$ = 248 $$\\times$$ 10$$-$$9 m
$$ \\Rightarrow $$ $$\\lambda$$ = 8.7 $$\\times$$ 10$$-$$10 = 8.7 $$\\mathop A\\limits^o $$ $$ \\approx $$ 9 $$\\mathop A\\limits^o $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3072,
"subject": "Chemistry",
"question": "A source of monochromatic radiation of wavelength 400 nm provides 1000 J of energy in 10 seconds. When this radiation falls on the surface of sodium, x $$\\times$$ 1020 electrons are ejected per second. Assume that wavelength 400 nm is sufficient for ejection of electron from the surface of sodium metal. The value of x is ______________. (Nearest integer)
(h = 6.626 $$\\times$$ 10$$-$$34 Js)",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nTotal energy provided by
Source per second = $${{1000} \\over {10}} = 100$$ J
Energy required to eject electron = $${{hc} \\over \\lambda }$$
= $$2.01 \\times {10^{20}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3073,
"subject": "Chemistry",
"question": "A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is 4.3 $$\\times$$ 1014 Hz. The velocity of ejected electron is ____________ $$\\times$$ 105 ms$$-$$1 (Nearest integer)
[Use : h = 6.63 $$\\times$$ 10$$-$$34 Js, me = 9.0 $$\\times$$ 10$$-$$31 kg]",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n$$\\upsilon $$ : speed of electron having max. K.E.
$$\\Rightarrow$$ from Einstein equation : E = $$\\phi$$ + K.E.max
$$\\Rightarrow$$ $$\\upsilon $$ = 5 $$\\times$$ 105 m/sec.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3074,
"subject": "Chemistry",
"question": "The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x $$\\times$$ 1013. The value of x is _____________. (Nearest integer)
(h = 6.63 $$\\times$$ 10$$-$$34 Js, c = 3.00 $$\\times$$ 108 ms$$-$$1)",
"options": [],
"answer": "50",
"solution": "**Answer:** 50\n\nEnergy emitted in 0.1 sec.
= 0.1 sec. $$\\times$$ $${10^{ - 3}}{J \\over s}$$
= 10$$-$$4 J
If 'n' photons of $$\\lambda$$ = 1000 nm are emitted, then 10$$-$$4 = n $$\\times$$ $${{hc} \\over \\lambda }$$
$$\\Rightarrow$$ n = 5.02 $$\\times$$ 1014 = 50.2 $$\\times$$ 1013
$$\\Rightarrow$$ 50 (nearest integer)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3075,
"subject": "Chemistry",
"question": "A 50 watt bulb emits monochromatic red light of wavelength of 795 nm. The number of photons emitted per second by the bulb is x $$\\times$$ 1020. The value of x is __________. [Given : h = 6.63 $$\\times$$ 10$$-$$34 Js and c = 3.0 $$\\times$$ 108 ms$$-$$1]",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nEnergy of photon is given as $$E = {{nhc} \\over \\lambda }$$ .... (i)
where, E = energy of photon (50 W),
n = number of photon
h = Planck's constant (6.63 $$\\times$$ 10$$-$$34 Js)
c = speed of light (3 $$\\times$$ 108 m/s)
$$\\lambda$$ = wavelength of light (795 $$\\times$$ 10$$-$$9 m)
If the work function of a metal is 6.63 $$\\times$$ 10$$-$$19J, the maximum wavelength of the photon required to remove a photoelectron from the metal is ____________ nm. (Nearest integer)
\n
[Given : h = 6.63 $$\\times$$ 10$$-$$34 J s, and c = 3 $$\\times$$ 108 m s$$-$$1]
If the wavelength for an electron emitted from $$\\mathrm{H}$$-atom is $$3.3 \\times 10^{-10} \\mathrm{~m}$$, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is _________ times. (Nearest integer)
\n
$$\\left[\\right.$$ Given $$: \\mathrm{h}=6.626 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s}$$ ]
\n
Mass of electron $$=9.1 \\times 10^{-31} \\mathrm{~kg}$$
The photoelectric effect occurs when a photon with energy greater than the work function of a metal strikes the metal surface, causing electrons to be emitted. The minimum energy required for this process is given by the equation:
where $E$ is the energy of the photon, $h$ is Planck's constant, $\\nu$ is the frequency of the photon, $c$ is the speed of light, $\\lambda$ is the wavelength of the photon.
\n
If the energy of the photon is greater than the work function ($W_0$) of the metal, then electrons will be emitted. The energy of the photon is related to its wavelength by the above equation.
\n
For the given wavelength of 400 nm, the energy of the photon is:
For the photoelectric effect to occur, this energy must be greater than the work function of the metal. The metals that will show the photoelectric effect can be determined by comparing the energy of the photon to the work function of each metal.
\n
\n
\n\n
\n
Metal
\n
$W_0$ (eV)
\n
$W_0$ (J)
\n
$E > W_0$?
\n
\n\n\n
\n
Li
\n
2.42
\n
3.88 $\\times$ 10$^{-19}$ J
\n
Yes
\n
\n
\n
Na
\n
2.3
\n
3.68 $\\times$ 10$^{-19}$ J
\n
Yes
\n
\n
\n
K
\n
2.25
\n
3.60 $\\times$ 10$^{-19}$ J
\n
Yes
\n
\n
\n
Mg
\n
3.7
\n
5.92 $\\times$ 10$^{-19}$ J
\n
No
\n
\n
\n
Cu
\n
4.8
\n
7.68 $\\times$ 10$^{-19}$ J
\n
No
\n
\n
\n
Ag
\n
4.3
\n
6.88 $\\times$ 10$^{-19}$ J
\n
No
\n
\n\n
\n
From the table, we see that three metals (Li, Na, and K) have work functions less than the energy of the photon with a wavelength of 400 nm. Therefore, these three metals will show the photoelectric effect when light of this wavelength falls on them.
\n
Thus, the number of metals which will show photoelectric effect when light of wavelength 400 nm falls on them is 3.
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R :
\n
Assertion A : In the photoelectric effect, the electrons are ejected from the metal surface as soon as the beam of light of frequency greater than threshold frequency strikes the surface.
\n
Reason R : When the photon of any energy strikes an electron in the atom, transfer of energy from the photon to the electron takes place.
\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": "A is not correct but R is correct"
},
{
"text": "A is correct but R is not correct"
},
{
"text": "Both A and R are correct but R is NOT 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\n
Given are two statements :
\n
\n
Assertion A : "In the photoelectric effect, the electrons are ejected from the metal surface as soon as the beam of light of frequency greater than threshold frequency strikes the surface."
\n
Reason R : "When the photon of any energy strikes an electron in the atom, transfer of energy from the photon to the electron takes place."
\n
\n
From the understanding of the photoelectric effect, we know that the light beam must have a frequency $v$ that is greater than the threshold frequency $v_0$ for electrons to be ejected from the metal surface. Therefore, Assertion A is correct.
\n
However, Reason R is somewhat misleading in the context of the photoelectric effect. It's true that a photon transfers its energy to an electron when it strikes the electron. However, for the electron to be ejected (i.e., the photoelectric effect to occur), the energy of the photon, given by $E_{photon} = h \\cdot v$, where $h$ is Planck's constant and $v$ is the frequency of the light, must be greater than the work function $\\phi$ (the minimum energy needed to eject an electron) of the metal. This condition is not captured in Reason R.
\n
Therefore, the correct answer is :
\n
Option C : Assertion A is correct, but Reason R is not correct.
All the statements (A, B, C, D) provided about black bodies are correct.
\n
(A) A black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. It also emits energy in the form of electromagnetic radiation.
\n
(B) The frequency distribution of the emitted radiation by a black body indeed depends on its temperature. This phenomenon is described by Planck's law.
\n
(C) The intensity vs frequency curve for a black body, also known as the Planck's curve, does indeed pass through a maximum value for a given temperature.
\n
(D) The frequency at which the maximum intensity occurs increases with the temperature of the black body. This is stated in Wien's displacement law.
\n
So, the number of incorrect statements about the black body from the above is 0.
The ionization energy of sodium in $$\\mathrm{~kJ} \\mathrm{~mol}^{-1}$$, if electromagnetic radiation of wavelength $$242 \\mathrm{~nm}$$ is just sufficient to ionize sodium atom is _______.
The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency '$$A$$' $$\\times 10^{12}$$ hertz and that has a radiant intensity in that direction of $$\\frac{1}{{ 'B'}}$$, watt per steradian.\n'$$A$$' and '$$B$$' are respectively
",
"options": [
{
"text": "450 and 683\n"
},
{
"text": "540 and $$\\frac{1}{683}$$\n"
},
{
"text": "540 and 683\n"
},
{
"text": "450 and $$\\frac{1}{683}$$"
}
],
"answer": "540 and 683\n",
"solution": "**Answer:** 540 and 683\n\n\n
The definition of the candela involves a source emitting monochromatic radiation of a specific frequency and having a precise radiant intensity in a given direction. The candela is thereby defined based on both the frequency of the monochromatic light and its radiant intensity in watts per steradian. According to the given statement, we are dealing with a frequency identified as '$$A$$' $$\\times 10^{12}$$ hertz and a radiant intensity specified as $$\\frac{1}{B}$$ watt per steradian. However, there seems to be a small confusion in how the values are represented in the question. The correct interpretation and formulation based on the International System of Units (SI) are as follows:
\n
The frequency of the monochromatic radiation that is used in defining the candela is 540 THz (tera hertz), so the value of $$A$$ should be 540. The specific radiant intensity mentioned relates to how the candela is defined in terms of a source's power output in a particular direction within a given frequency. The value that corresponds to the efficiency of human visual perception at this frequency is approximately 683 lumens per watt. Therefore, the number 683 is involved in specifying how many lumens correspond to a power of one watt of monochromatic light at 540 THz.
\n
Given this context, the correct answer derives from understanding that:
\n
\n
'$$A$$' represents the frequency in THz, which is 540 for the definition of the candela.
\n
'$$B$$' refers to the conversion factor or the luminous efficacy in lumens per watt at this frequency, which is 683 lumens per watt, not its reciprocal.
\n
\n
Therefore, the correct values for '$$A$$' and '$$B$$' are 540 and 683, respectively, making Option C the correct answer.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3084,
"subject": "Chemistry",
"question": "Which of the following statements is false ?",
"options": [
{
"text": "Photon has momentum as well as wavelength. "
},
{
"text": "Splitting of spectral lines in electrical field is called Stark effect. "
},
{
"text": "Rydberg constant has unit of energy. "
},
{
"text": "Frequency of emitted radiation from a black body goes from a lower wavelength to higher wavelength as the temperature increases. "
}
],
"answer": "Frequency of emitted radiation from a black body goes from a lower wavelength to higher wavelength as the temperature increases. ",
"solution": "**Answer:** Frequency of emitted radiation from a black body goes from a lower wavelength to higher wavelength as the temperature increases. \n\nWhen a black body is heated, black body emit high energy radiation, from higher wavelength to lower wavelength.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3085,
"subject": "Chemistry",
"question": "If the de Broglie wavelength of the electron in nth Bohr orbit in a hydrogenic atom is equal to 1.5 $$\\pi $$a0 (a0 is Bohr radius), then the value of n/z is - ",
"options": [
{
"text": "0.75"
},
{
"text": "0.40"
},
{
"text": "1.50"
},
{
"text": "1.0"
}
],
"answer": "0.75",
"solution": "**Answer:** 0.75\n\nAccording to debroglie hypothesis,\n
$$ \\Rightarrow $$ $${n \\over Z}$$ = 0.75",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3086,
"subject": "Chemistry",
"question": "Which one of the following about an electron\noccupying the 1s orbital in a hydrogen atom is\nincorrect ?
(The Bohr radius is represented\nby a0)",
"options": [
{
"text": "The probability density of finding the\nelectron is maximum at the nucleus."
},
{
"text": "The electron can be found at a distance 2a0\nfrom the nucleus"
},
{
"text": "The total energy of the electron is maximum\nwhen it is at a distance a0 from the nucleus."
},
{
"text": "The magnitude of potential energy is\ndouble that of its kinetic energy on an\naverage."
}
],
"answer": "The total energy of the electron is maximum\nwhen it is at a distance a0 from the nucleus.",
"solution": "**Answer:** The total energy of the electron is maximum\nwhen it is at a distance a0 from the nucleus.\n\nProbablity density = $${\\psi ^2}$$\n\n From the graph you can see probablity density is maximum at the nucleus where r = 0.\n
$$ \\therefore $$ Option A is correct.\n
(B) For hydrogen atom, electron can be found at a any distance distance\nfrom the nucleus.\n
$$ \\therefore $$ Option B is correct.\n
(C) Distance a0 from the nucleus = First orbital\n
$$ \\therefore $$ n = 1 at first orbital\n
As Total energy, En = -13.6 $$ \\times $$ $${{{Z^2}} \\over {{n^2}}}$$ eV\n
$$ \\therefore $$ Total energy for hydrogen at n = 1 is,\n
$$ \\therefore $$ Magnitude of potential energy is\ndouble that of its kinetic energy.\n
$$ \\therefore $$ Option D is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3087,
"subject": "Chemistry",
"question": "The correct statement about probability density (except at infinite distance from nucleus) is :",
"options": [
{
"text": "It can be zero for 1s orbital"
},
{
"text": "It can be zero for 3p orbital"
},
{
"text": "It can never be zero for 2s orbital"
},
{
"text": "It can negative for 2p orbital"
}
],
"answer": "It can be zero for 3p orbital",
"solution": "**Answer:** It can be zero for 3p orbital\n\n$$\\phi $$2 (probability density) can be zero for 3p\norbital other than infinite distance. It has one\nradial node.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3088,
"subject": "Chemistry",
"question": "A proton and a Li3+ nucleus are accelerated by the same potential. If $$\\lambda _{Li}$$ and $$\\lambda _p$$ denote the\nde Broglie wavelengths of Li3+ and proton respectively, then the value of\n$${{{\\lambda _{Li}}} \\over {{\\lambda _p}}}$$ is x $$ \\times $$ 10-1. \nThe value of x is ______. (Rounded off to the nearest integer) \n[Mass of Li3+ = 8.3 mass of proton]",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nGiven, mass of Li3+ = 8.3 times of mass of proton formula,
Hence, x $$\\times$$ 10$$-$$1, x = 2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3089,
"subject": "Chemistry",
"question": "The wavelength of electrons accelerated from rest through a potential difference of 40 kV is x $$\\times$$ 10$$-$$12 m. The value of x is ___________. (Nearest integer)
Give : Mass of electron = 9.1 $$\\times$$ 10$$-$$31 kg
Charge on an electron = 1.6 $$\\times$$ 10$$-$$19 C
Planck's constant = 6.63 $$\\times$$ 10$$-$$34 Js",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\nWavelength of electron is given by
$$\\lambda = {h \\over {\\sqrt {2mqV} }}$$
Here q = charge on electron, V = potential difference
The wavelength of an electron and a neutron will become equal when the velocity of the electron is $$x$$ times the velocity of neutron. The value of $$x$$ is ____________. (Nearest Integer)
\n
(Mass of electron is $$9.1 \\times 10^{-31} \\mathrm{~kg}$$ and mass of neutron is $$1.6 \\times 10^{-27} \\mathrm{~kg}$$ )
\n$$\\simeq 1758$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3092,
"subject": "Chemistry",
"question": "The wave function $(\\Psi)$ of $2 \\mathrm{~s}$ is given by
The energy of one mole of photons of radiation of frequency $$2 \\times 10^{12} \\mathrm{~Hz}$$ in $$\\mathrm{J} ~\\mathrm{mol}^{-1}$$ is ___________. (Nearest integer)
In Bohr's model, the angular momentum of an electron in an orbit is quantized, which can be represented as:
\n
$$mvr = n \\cdot \\frac{h}{2\\pi}$$
\n
Where:
\n
\n
(m) is the mass of the electron,
\n
(v) is the velocity of the electron,
\n
(r) is the radius of the nth orbit, and
\n
(n) is the principal quantum number ((n = 1) for the first orbit, (n = 2) for the second orbit, and so on).
\n
\n
The de Broglie wavelength ($\\lambda$) is given by:
\n
$$\\lambda = \\frac{h}{mv}$$
\n
For the nth orbit, the radius (r) is $n^2$ times the radius of the first orbit ($r_0$, or $\\alpha_0$ in your notation):
\n
$$r = n^2 \\cdot \\alpha_0$$
\n
The circumference of the nth orbit is $2\\pi r$, which can be written as:
\n
$$2\\pi r = n \\cdot \\lambda$$
\n
Substituting the expression for (r):
\n
$$2\\pi \\cdot n^2 \\cdot \\alpha_0 = n \\cdot \\lambda$$
\n
Solving for $\\lambda$ gives:
\n
$$\\lambda = 2\\pi \\cdot n \\cdot \\alpha_0$$
\n
For the 3rd orbit ((n = 3)), this simplifies to:
\n
$$\\lambda = 6\\pi \\cdot \\alpha_0$$
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3096,
"subject": "Chemistry",
"question": "In case of isoelectronic species the size of $\\mathrm{F}^{-}, \\mathrm{Ne}$ and $\\mathrm{Na}^{+}$is affected by :",
"options": [
{
"text": "Nuclear charge $(\\mathrm{z})$"
},
{
"text": "None of the factors because their size is the same"
},
{
"text": "Electron-electron interaction in the outer orbitals"
},
{
"text": "Principal quantum number (n)"
}
],
"answer": "Nuclear charge $(\\mathrm{z})$",
"solution": "**Answer:** Nuclear charge $(\\mathrm{z})$\n\n
Option A: Nuclear charge $(\\mathrm{z})$ is the correct answer.
\n\n
Isoelectronic species are atoms and ions that have the same number of electrons. When comparing the sizes of isoelectronic species, the major factor that affects their size is the nuclear charge $(\\mathrm{z})$, which is the charge of the nucleus or the number of protons in the nucleus. Here’s why:
\n\n
The more protons there are in the nucleus (higher nuclear charge), the more positively charged the nucleus is, and thus the stronger it can attract the electrons towards it. This results in a smaller radius for the species. Conversely, if there are fewer protons in the nucleus, the attraction to the electrons is weaker, leading to a larger radius. Let's compare the species given:
\n\n
\n
$\\mathrm{F}^{-}$ (fluoride ion) has 9 protons in the nucleus and 10 electrons.
\n
$\\mathrm{Ne}$ (neon atom) has 10 protons in the nucleus and 10 electrons.
\n
$\\mathrm{Na}^{+}$ (sodium ion) has 11 protons in the nucleus and 10 electrons.
\n
\n\n
Despite having the same number of electrons, the $\\mathrm{Na}^{+}$ ion will be the smallest among the three due to the higher nuclear charge of 11 protons, which attracts the same number of electrons more strongly than $\\mathrm{Ne}$ with 10 protons and $\\mathrm{F}^{-}$ with 9 protons. Hence, $\\mathrm{Na}^{+}$ has a lower atomic radius compared to $\\mathrm{Ne}$, which in turn is smaller than $\\mathrm{F}^{-}$.
\n\n
Considering Option B (size being the same), Option C (electron-electron interaction), and Option D (principal quantum number), none of these factors are as significant as the difference in nuclear charge for isoelectronic species.
\n\n
The principal quantum number (n) is the same for all these isoelectronic species since they all have their outer electrons in the second energy level (n=2) in the ground state. Electron-electron interactions do contribute to energy levels and could affect the size slightly, but the primary and determining factor for isoelectronic species remains the nuclear charge, as it directly influences the effective nuclear charge felt by the electrons. Therefore, Option A is the best answer.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3097,
"subject": "Chemistry",
"question": "The orbital angular momentum for an electron revolving in an orbit is given by $$\\sqrt {l(l + 1)} {h \\over {2\\pi }}$$. This momentum for an s-electron will be given by",
"options": [
{
"text": "zero"
},
{
"text": "$${h \\over {2\\pi }}$$"
},
{
"text": "$$\\sqrt 2 {h \\over {2\\pi }}$$"
},
{
"text": "$$ + {1 \\over 2}{h \\over {2\\pi }}$$"
}
],
"answer": "zero",
"solution": "**Answer:** zero\n\nFor s-electron l = 0
\n = $$\\sqrt 0 {h \\over {2\\pi }}$$ = 0 (zero) ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3098,
"subject": "Chemistry",
"question": "Which of the following sets of quantum numbers is correct for an electron in 4f orbital?",
"options": [
{
"text": "n = 4, l = 3, m = +1, s = + 1/2"
},
{
"text": "n = 4, l = 4, m = -4, s = - 1/2"
},
{
"text": "n = 4, l = 3, m = +4, s = + 1/2"
},
{
"text": "n = 3, l = 2, m = -2, s = + 1/2"
}
],
"answer": "n = 4, l = 3, m = +1, s = + 1/2",
"solution": "**Answer:** n = 4, l = 3, m = +1, s = + 1/2\n\nFor 4f orbital,\n
n = 4 and l = 3\n
Note : ( for s orbital l = 0, for p orbital l = 1, for d orbital l = 2, for f orbital l = 3)\n
Value of m = + l to - l\n
Here m = +3 to -3\n
Value of s = $$ + {1 \\over 2}$$ or $$ - {1 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3099,
"subject": "Chemistry",
"question": "Consider the ground state of Cr atom (Z = 24). The number of electrons with the azimuthal quantum numbers, l = 1 and 2 are respectively",
"options": [
{
"text": "16 and 4"
},
{
"text": "12 and 5"
},
{
"text": "12 and 4"
},
{
"text": "16 and 5"
}
],
"answer": "12 and 5",
"solution": "**Answer:** 12 and 5\n\nElectronic configuration of Cr (Z = 24) =\n
1s2 2s2 2p6 3s2 3p6 4s1 3d5\n
For 2p6 and 3p6 , l = 1. Here in 2p and 3p orbital total 6 + 6 = 12 electrons present.\n
For 3d5 , l = 2. Here in 3d orbital total 5 electrons present.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3100,
"subject": "Chemistry",
"question": "In a multi-electron atom, which of the following orbitals described by the three quantum members will have the same energy in the absence of magnetic and electric fields? \n(A) n = 1, l = 0, m = 0 \n(B) n = 2, l = 0, m = 0 \n(C) n = 2, l = 1, m = 1 \n(D) n = 3, l = 2, m = 1 \n(E) n = 3, l = 2, m = 0 ",
"options": [
{
"text": "(D) and (E)"
},
{
"text": "(C) and (D)"
},
{
"text": "(B) and (C)"
},
{
"text": "(A) and (B)"
}
],
"answer": "(D) and (E)",
"solution": "**Answer:** (D) and (E)\n\nAs here is no electric and magnetic field so ignore m to calculate the energy of orbital.\n
As here atom is multi-electron so ( n + l ) rule is applicable here. This rule says those orbitals which have more value of ( n + l ) will have more energy.\n
In (A), n + l = 1 + 0 = 1\n
In (B), n + l = 2 + 0 = 2\n
In (C), n + l = 2 + 1 = 3\n
In (D), n + l = 3 + 2 = 5\n
In (E), n + l = 3 + 2 = 5\n
So (D) and (E) will be of equal energy.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 3101,
"subject": "Chemistry",
"question": "Which of the following sets of quantum numbers represents the highest energy of an atom?",
"options": [
{
"text": "n = 3, l = 0, m = 0, s = +1/ 2"
},
{
"text": "n = 3, l = 1, m = 1, s = +1/ 2"
},
{
"text": "n = 3, l = 2, m = 1, s = +1/ 2"
},
{
"text": "n = 4, l = 0, m = 0, s = +1/ 2"
}
],
"answer": "n = 3, l = 2, m = 1, s = +1/ 2",
"solution": "**Answer:** n = 3, l = 2, m = 1, s = +1/ 2\n\nIn which quantum number have highest ( n + l ) value, will represent highest energy of an atom.\n
In option (C), n + l = 3 + 2 = 5 = maximum.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3102,
"subject": "Chemistry",
"question": "The electrons identified by quantum numbers n and l : \n(a) n = 4, $$l$$ = 1 \n(b) n = 4, $$ l$$ = 0 \n(c) n = 3, $$l$$ = 2 \n(d) n = 3, $$l$$ = 1 \nCan be placed in order of increasing energy as :",
"options": [
{
"text": "(c) < (d) < (b) < (a)"
},
{
"text": "(d) < (b) < (c) < (a)"
},
{
"text": "(b) < (d) < (a) < (c)"
},
{
"text": "(a) < (c) < (b) < (d)"
}
],
"answer": "(d) < (b) < (c) < (a)",
"solution": "**Answer:** (d) < (b) < (c) < (a)\n\n(a) n = 4, $$l$$ = 1 (p-subshell), so 4p \n(b) n = 4, $$l$$ = 0 (s-subshell), so 4s \n(c) n = 3, $$l$$ = 2 (d-subshell), so 3d \n(d) n = 3, $$l$$ = 1 (p-subshell), so 3p \n Accroding to the Bohr ( n + $$l$$ ) rule,\n Enery order of the subshell : 3p < 4s < 3d < 4p\n
Note: When two orbital have the same value of ( n + $$l$$ ) then the orbital which have lower value of n will be filled first.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3103,
"subject": "Chemistry",
"question": "The correct set of four quantum numbers for the valence elections of rubidium atom (Z= 37) is:",
"options": [
{
"text": "5, 1, 1, + 1/2"
},
{
"text": "5, 1, 0, + 1/2"
},
{
"text": "5, 0, 0, + 1/2"
},
{
"text": "5, 0, 1, + 1/2"
}
],
"answer": "5, 0, 0, + 1/2",
"solution": "**Answer:** 5, 0, 0, + 1/2\n\nThe electronic configuration of Rubidium (Rb = 37):\n 1s22s22p63s23p63d104s24p65s1\n
As you can see last electron or valence electron enterin 5s subshell\n
So, the quantum numbers are n = 5, $$l$$ = 0, m = 0, s = $$ \\pm {1 \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3104,
"subject": "Chemistry",
"question": "The total number of orbitals associated with the principal quantum number 5 is :",
"options": [
{
"text": "5"
},
{
"text": "10"
},
{
"text": "20"
},
{
"text": "25"
}
],
"answer": "25",
"solution": "**Answer:** 25\n\n
Total number of orbitals associated with principal quantum number n is = n2.
\n
Here, n = 5. \n
$$ \\therefore $$ The total number of orbitals associated with the principal quantum number 5 is = 52 = 25
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3105,
"subject": "Chemistry",
"question": "The highest value of the calculated spin only magnetic moment (in BM) among all the transition metal complexes is : ",
"options": [
{
"text": "5.92"
},
{
"text": "6.93"
},
{
"text": "3.87"
},
{
"text": "4.90"
}
],
"answer": "5.92",
"solution": "**Answer:** 5.92\n\nIn transition metal contains d orbital, and in d orbital maximum no of unpaired electron possible = 5.\n
$$=$$ 5.92 B. M",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3106,
"subject": "Chemistry",
"question": "Which of the following combination of statements is true regarding the interpretation of the atomic orbitals ?
\n(a) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum.
\n(b) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number.
\n(c) According to wave mechanics, the ground state angular momentum is equal to $${h \\over {2\\pi }}$$.
\n(d) The plot of $$\\psi $$ vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value. ",
"options": [
{
"text": "(a), (b)"
},
{
"text": "(a), (d)"
},
{
"text": "(b), (c)"
},
{
"text": "(a), (c)"
}
],
"answer": "(a), (d)",
"solution": "**Answer:** (a), (d)\n\n
(d) The plot of $$\\psi $$ vs $$ r $$ for various azimuthal quantum numbers, shows peak shifting towards higher $$ r $$ value. \nThis statement is typically true. The radial distribution functions of atomic orbitals (which can be related to $$ |\\psi|^2 $$ vs $$ r $$) for higher azimuthal quantum numbers generally have their peaks at larger values of the radial distance $$ r $$ from the nucleus. This is because with higher $$ l $$ values, the electrons are more likely to be found at greater distances due to the higher angular momentum.
The given plot is\n
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 3107,
"subject": "Chemistry",
"question": "The quantum number of four electrons are given below : \nn = 4, l = 2, ml =–2, ms = –1/2 \nn = 3, l = 2, ml = 1, ms = +1/2 \nn = 4, l = 1, ml = 0, ms = +1/2 \nn = 3, l = 1, ml = 1, ms = –1/2 \nThe correct order of their increasing enegies will be :",
"options": [
{
"text": "IV < II < III < I"
},
{
"text": "I < III < II < IV"
},
{
"text": "IV < III < II < I"
},
{
"text": "I < II < III < IV"
}
],
"answer": "IV < II < III < I",
"solution": "**Answer:** IV < II < III < I\n\nn = 4, l = 2, 4d orbital, n + l = 6
\nn = 3, l = 2, 3d orbital, n + l = 5
\nn = 4, l = 1, 4p orbital, n + l = 5
\nn = 3, l = 1, 3p orbital, n + l = 4 \n more is n + l value, more is energy\n
$$ \\therefore $$ 3p < 3d < 4p < 4d",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3108,
"subject": "Chemistry",
"question": "The number of orbitals associated with quantum number n = 5, ms = +$${1 \\over 2}$$ is :",
"options": [
{
"text": "11"
},
{
"text": "25"
},
{
"text": "15"
},
{
"text": "50"
}
],
"answer": "25",
"solution": "**Answer:** 25\n\nTotal number of orbitals \n= n2 = (5)2 = 25\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3109,
"subject": "Chemistry",
"question": "Consider the hypothetical situation where the\nazimuthal quantum number,\n$$l$$, takes values 0,\n1, 2, ....., n + 1, where n is the principal\nquantum number. Then, the element with\natomic number :",
"options": [
{
"text": "13 has a half-filled valence subshell"
},
{
"text": "9 is the first alkali metal"
},
{
"text": "8 is the first noble gas"
},
{
"text": "6 has a 2p-valence subshell"
}
],
"answer": "13 has a half-filled valence subshell",
"solution": "**Answer:** 13 has a half-filled valence subshell\n\nUnder hypothetical situation, the value of l is\ngreater than n which varies from 0 to n + 1.\n
For n = 1, l = 0, 1, 2\n
n = 2, l = 0, 1, 2, 3\n
Elements follow the following electronic\nconfiguration\n
1s 1p 1d 2s 2p 2d 2f\n
Atomic number (Z) = 9\n 1s2 1p6 1d1\n
Atomic number = 6\n 1s2 1p4\n
Atomic number 8\n 1s2 1p6\n
Atomic number 13\n 1s2 1p6 1d5 (half filled)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3110,
"subject": "Chemistry",
"question": "The number of subshells associated with n = 4\nand m = –2 quantum numbers is",
"options": [
{
"text": "8"
},
{
"text": "2"
},
{
"text": "16"
},
{
"text": "4"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\nFor n = 4 possible value of\nl = 0, 1, 2, 3.\n
Only l = 2 and l = 3 can have m = -2\n
$$ \\therefore $$ 4d & 4f subshell associated with n = 4, m = –2.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3111,
"subject": "Chemistry",
"question": "The number of orbitals with n = 5, m1 = +2 is ___________. (Round off to the Nearest Integer).",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n
Given, n = 5, ml = + 2
\n
For n = 5, possible value of l = 0, 1, 2, 3, 4
\n
For l = 0, ml = 0
\n
l = 1, ml = $$-$$1, 0, 1
\n
l = 2, ml = $$-$$2, $$-$$1, 0, 1, 2
\n
l = 3, ml = $$-$$3, $$-$$2, $$-$$1, 0, 1, 2, 3
\n
l = 4, ml = $$-$$4, $$-$$3, $$-$$2, $$-$$1, 0, 1, 2, 3, 4
So, number of orbitals having n = 5 and ml = $$\\pm$$ 2 are 3.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3112,
"subject": "Chemistry",
"question": "A certain orbital has n = 4 and mL = $$-$$3. The number of radial nodes in this orbital is ____________. (Round off to the Nearest Integer).",
"options": [],
"answer": "0",
"solution": "**Answer:** 0\n\nNumber of radial nodes = n – $$\\ell $$ – 1
\nn = 4, mL =–3 so $$\\ell $$ = 3
\nradial nodes = 4 – 3 – 1 = 0\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3113,
"subject": "Chemistry",
"question": "The Azimuthal quantum number for the valence electrons of Ga+ ion is ___________.
(Atomic number of Ga = 31)",
"options": [],
"answer": "0",
"solution": "**Answer:** 0\n\nGa = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1\n
Ga+\n = 1s2 2s2 2p6 3s2 3p6 3d10 4s2\n
Azimuthal Quantum number (l) for valence shell electron is 0.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3114,
"subject": "Chemistry",
"question": "Ge(Z = 32) in its ground state electronic configuration has x completely filled orbitals with ml = 0. The value of x is ___________.",
"options": [],
"answer": "7",
"solution": "**Answer:** 7\n\n
Completely filled orbital with ml = 0 are
= 1 + 1 + 1 + 1 + 1 + 1 + 1
= 7
So, answer is 7.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3115,
"subject": "Chemistry",
"question": "
Which of the following statements are correct?
\n
(A) The electronic configuration of Cr is [Ar] 3d5 4s1.
\n
(B) The magnetic quantum number may have a negative value.
\n
(C) In the ground state of an atom, the orbitals are filled in order of their increasing energies.
\n
(D) The total number of nodes are given by n $$-$$ 2.
\n
Choose the most appropriate answer from the options given below :
",
"options": [
{
"text": "(A), (C) and (D) only"
},
{
"text": "(A) and (B) only"
},
{
"text": "(A) and (C) only"
},
{
"text": "(A), (B) and (C) only"
}
],
"answer": "(A), (B) and (C) only",
"solution": "**Answer:** (A), (B) and (C) only\n\n
(A) Cr (24) = 1s2 2s2 2p6 3s2 3p6 3d5 4s1
\n
= [Ar] 3d5 4s1
\n
(B) Magnetic quantum number (m) values ranging from $$-$$ l to + l including zero.
\n
$$\\therefore$$ It can have negative value.
\n
(C) According to Aufbau rule, electrons are filled first in these orbitals which have low energy.
\n
$$\\therefore$$ Statement C is correct.
\n
(D) We know,
\n
Number of Radial nodes = n $$-$$ l $$-$$ 1
\n
and number of Angular nodes = l
\n
$$\\therefore$$ Total nodes = n $$-$$ l $$-$$ 1 + 1 = n $$-$$ 1
\nBut in option (C), the value of $$\\mathrm{l}$$ is given ' 3 ', this is not possible.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3121,
"subject": "Chemistry",
"question": "
The correct decreasing order of energy for the orbitals having, following set of quantum numbers :
Given below are two statements: One is labelled as Assertion $$\\mathbf{A}$$ and the other is labelled as Reason $$\\mathbf{R}$$
\n
Assertion $$\\mathbf{A}$$ : Zero orbital overlap is an out of phase overlap.
\n
Reason $$\\mathbf{R}$$ : It results due to different orientation / direction of approach of orbitals.
\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\nZero overlapping is something in which there is no overlapping between two orbitals. The first condition is that the two orbitals should not be symmetrical and the second condition is that both orbitals should be in different planes.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3123,
"subject": "Chemistry",
"question": "
Given below are the quantum numbers for 4 electrons.
\n
A. $$\\mathrm{n}=3,l=2, \\mathrm{~m}_{1}=1, \\mathrm{~m}_{\\mathrm{s}}=+1 / 2$$
\n
B. $$\\mathrm{n}=4,l=1, \\mathrm{~m}_{1}=0, \\mathrm{~m}_{\\mathrm{s}}=+1 / 2$$
\n
C. $$\\mathrm{n}=4,l=2, \\mathrm{~m}_{1}=-2, \\mathrm{~m}_{\\mathrm{s}}=-1 / 2$$
\n
D. $$\\mathrm{n}=3,l=1, \\mathrm{~m}_{1}=-1, \\mathrm{~m}_{\\mathrm{s}}=+1 / 2$$
\n
The correct order of increasing energy is
",
"options": [
{
"text": "D < B < A < C"
},
{
"text": "D < A < B < C"
},
{
"text": "B < D < A < C"
},
{
"text": "B < D < C < A"
}
],
"answer": "D < A < B < C",
"solution": "**Answer:** D < A < B < C\n\nEnergy of the sub-shell is given by, $(n+l)$ rule.\n
$\\mathrm{D}<\\mathrm{A}<\\mathrm{B}<\\mathrm{C}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3124,
"subject": "Chemistry",
"question": "Arrange the following orbitals in decreasing order of energy.
\nThe correct option for the order is :",
"options": [
{
"text": "$\\mathrm{B}>\\mathrm{D}>\\mathrm{C}>\\mathrm{A}$"
},
{
"text": "$\\mathrm{A}>\\mathrm{C}>\\mathrm{B}>\\mathrm{D}$"
},
{
"text": "$\\mathrm{D}>\\mathrm{B}>\\mathrm{A}>\\mathrm{C}$"
},
{
"text": "$\\mathrm{D}>\\mathrm{B}>\\mathrm{C}>\\mathrm{A}$"
}
],
"answer": "$\\mathrm{D}>\\mathrm{B}>\\mathrm{C}>\\mathrm{A}$",
"solution": "**Answer:** $\\mathrm{D}>\\mathrm{B}>\\mathrm{C}>\\mathrm{A}$\n\n(A) n = 3; l = 0; m = 0 ; 3s orbital\n
(B) n = 4; l = 0; m = 0 ; 4s orbital\n
(C) n = 3; l = 1; m = 0 ; 3p orbital\n
(D) n = 3; l = 2; m = 0 ; 3d orbital\n
As per Hund’s ruleo of energy is given by (n + l) value.\nIf value of (n + l) remains same then energy is given\nby n only.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3125,
"subject": "Chemistry",
"question": "Maximum number of electrons that can be accommodated in shell with $n=4$ are:",
"options": [
{
"text": "50"
},
{
"text": "32"
},
{
"text": "72"
},
{
"text": "16"
}
],
"answer": "32",
"solution": "**Answer:** 32\n\n
Maximum electrons accommodated in $$\\mathrm{(n = 4)}$$ is $$\\mathrm{2n^2=32}$$ electrons
",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nThe orbitals which have electron density along the axis are all the $p$-orbitals $\\left(p_x, p_y\\right.$ and $\\left.p_z\\right)$ while, out of five $d$-orbitals only two $\\left(d_{x^2-y^2}\\right.$ and $\\left.d_{z^2}\\right)$ have electron density along the axis.\n
Therefore, the total number of given orbitals which have electron density along the axis is 5.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3127,
"subject": "Chemistry",
"question": "
The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is
The orbital angular momentum of an electron in $$3 \\mathrm{~s}$$ orbital is $$\\frac{x h}{2 \\pi}$$. The value of $$x$$ is ____________ (nearest integer)
",
"options": [],
"answer": "0",
"solution": "**Answer:** 0\n\nThe orbital angular momentum (L) of an electron can be determined using the formula:\n
\n$$L = \\sqrt{l(l+1)} \\frac{h}{2\\pi}$$\n
\nWhere $$l$$ is the azimuthal quantum number (orbital angular momentum quantum number) and $$h$$ is the Planck's constant.\n
\nFor a 3s orbital, the principal quantum number (n) is 3 and the azimuthal quantum number (l) is 0, as s orbitals have l=0.\n\nNow, let's calculate the orbital angular momentum:\n
\n$$L = \\sqrt{0(0+1)} \\frac{h}{2\\pi} = 0$$\n
\nThus, the orbital angular momentum of an electron in a 3s orbital is 0. So, the value of $$x$$ is 0.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3129,
"subject": "Chemistry",
"question": "
The number of correct statements from the following is ___________.
\n
A. For $$1 \\mathrm{s}$$ orbital, the probability density is maximum at the nucleus
\n
B. For $$2 \\mathrm{s}$$ orbital, the probability density first increases to maximum and then decreases sharply to zero.
\n
C. Boundary surface diagrams of the orbitals encloses a region of $$100 \\%$$ probability of finding the electron.
\n
D. p and d-orbitals have 1 and 2 angular nodes respectively.
\n
E. probability density of p-orbital is zero at the nucleus
A. For 1s orbital, the probability density is maximum at the nucleus.
\n
This statement is correct. For a 1s orbital, the electron is most likely to be found closest to the nucleus.
\n \n
B. For 2s orbital, the probability density first increases to maximum and then decreases sharply to zero.
\n
This statement is incorrect. For a 2s orbital, the probability density is maximum at the nucleus, then decreases to zero (this is the radial node), and then rises again to a secondary peak before gradually tapering off.
\n \n
C. Boundary surface diagrams of the orbitals encloses a region of 100% probability of finding the electron.
\n
This statement is incorrect. The boundary surface usually encloses a region where there is approximately a 90% chance of finding the electron, not 100%.
\n \n
D. p and d-orbitals have 1 and 2 angular nodes respectively.
\n
This statement is correct. The number of angular nodes corresponds to the orbital angular momentum quantum number (l). For a p-orbital, l=1, thus there is one angular node. For a d-orbital, l=2, thus there are two angular nodes.
\n
E. Probability density of p-orbital is zero at the nucleus.
\n
This statement is correct. The probability density of a p-orbital is indeed zero at the nucleus because p-orbitals have one angular node at the nucleus.
\n
So, the total number of correct statements is 3 (A, D, E).
To determine the number of electrons with a spin quantum number of $$s=+\\frac{1}{2}$$ in all completely filled subshells with principal quantum number $$n=4$$, we must first identify the subshells in the n=4 shell and then calculate the electrons with the specified spin.
\n
The n=4 shell has the following subshells and their capacity for electrons:
\n
\n
$$4s$$ subshell can hold 2 electrons
\n
$$4p$$ subshell can hold 6 electrons
\n
$$4d$$ subshell can hold 10 electrons
\n
$$4f$$ subshell can hold 14 electrons
\n
\n
In quantum mechanics, each orbital within these subshells can hold 2 electrons, each with opposite spins ($$+\\frac{1}{2}$$ and $$-\\frac{1}{2}$$). Thus, in a completely filled subshell, half of the electrons will have a spin quantum number of $$+\\frac{1}{2}$$:
\n
\n
In $$4s$$, 1 out of 2 electrons will have $$s=+\\frac{1}{2}$$
\n
In $$4p$$, 3 out of 6 electrons will have $$s=+\\frac{1}{2}$$
\n
In $$4d$$, 5 out of 10 electrons will have $$s=+\\frac{1}{2}$$
\n
In $$4f$$, 7 out of 14 electrons will have $$s=+\\frac{1}{2}$$
\n
\n
Adding these together: $$1 + 3 + 5 + 7 = 16$$ electrons with $$s=+\\frac{1}{2}$$ in all completely filled subshells with $$n=4$$.
Statement (I) : The orbitals having same energy are called as degenerate orbitals.
\n
Statement (II) : In hydrogen atom, 3p and 3d orbitals are not degenerate orbitals.
\n
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\n"
},
{
"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"
}
],
"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
For single electron species the energy depends upon principal quantum number '$$n$$' only. So, statement II is false.
\n
Statement I is correct definition of degenerate orbitals.
To compare the energies of electrons in a multielectron system, we must consider both the principal quantum number ($$\\mathrm{n}$$) and the azimuthal quantum number ($$\\mathrm{l}$$). In multielectron atoms, the energy levels are influenced by electron-electron repulsions, which modify the energy ordering compared to the hydrogen atom.
\n\n
The general rule, called the \"n+l\" rule or Madelung rule, states that the energy of an electron in a multielectron atom is primarily determined by the sum of the principal quantum number ($$\\mathrm{n}$$) and the azimuthal quantum number ($$\\mathrm{l}$$). An orbital with a lower $$(\\mathrm{n} + \\mathrm{l})$$ value has lower energy. If two orbitals have the same $$(\\mathrm{n} + \\mathrm{l})$$ value, the orbital with the lower $$\\mathrm{n}$$ value has lower energy.
\n\n
Now, let's analyze each set of quantum numbers:
\n\n
(A) $$\\mathrm{n}=4, \\mathrm{l}=1$$ \n\n
$$\\mathrm{n} + \\mathrm{l} = 4 + 1 = 5$$
\n\n
(B) $$\\mathrm{n}=4, \\mathrm{l}=2$$ \n\n
$$\\mathrm{n} + \\mathrm{l} = 4 + 2 = 6$$
\n\n
(C) $$\\mathrm{n}=3, \\mathrm{l}=1$$ \n\n
$$\\mathrm{n} + \\mathrm{l} = 3 + 1 = 4$$
\n\n
(D) $$\\mathrm{n}=3, \\mathrm{l}=2$$ \n\n
$$\\mathrm{n} + \\mathrm{l} = 3 + 2 = 5$$
\n\n
(E) $$\\mathrm{n}=4, \\mathrm{l}=0$$ \n\n
$$\\mathrm{n} + \\mathrm{l} = 4 + 0 = 4$$
\n\n
Based on the $$\\mathrm{n} + \\mathrm{l}$$ values, the order of increasing energy is:
\n\n
(C) = (E) < (A) = (D) < (B)
\n\n
Among (C) and (E), (C) has lower $$\\mathrm{n}$$ value, so it is lower in energy:
\n\n
(C) < (E) < (A) = (D) < (B)
\n\n
Among (A) and (D), (D) has lower $$\\mathrm{n}$$ value, so it is lower in energy:
\n\n
Thus, the correct order is: (C) < (E) < (D) < (A) < (B)
In an atom, total number of electrons having quantum numbers $$\\mathrm{n}=4,\\left|\\mathrm{~m}_l\\right|=1$$ and $$\\mathrm{m}_{\\mathrm{s}}=-\\frac{1}{2}$$ is _________.
To find the total number of electrons with the given quantum numbers, let's first understand what each quantum number represents:
\n\n
\n\n
n (Principal Quantum Number): This specifies the energy level or shell of the electron in the atom. Here, $$n = 4$$ means we are dealing with the fourth energy level.
\n\n
l (Azimuthal Quantum Number or Orbital Angular Momentum Quantum Number): This defines the shape of the orbital and is dependent on the value of n. It can take values from 0 to $$n-1$$. For $$n = 4$$, $$l$$ can be 0 (s orbital), 1 (p orbital), 2 (d orbital), or 3 (f orbital).
\n\n
$m_l$ (Magnetic Quantum Number): This describes the orientation of the orbital in space and can take values from $$-l$$ to $$+l$$, including zero. The condition $$\\left|\\mathrm{~m}_l\\right|=1$$ implies $$m_l = -1$$ or $$+1$$.
\n\n
$m_s$ (Spin Quantum Number): This indicates the spin direction of the electron, with possible values of $$+\\frac{1}{2}$$ or $$-\\frac{1}{2}$$. Here, $$\\mathrm{m}_{\\mathrm{s}}=-\\frac{1}{2}$$ specifies electrons with a downward spin.
\n\n
\n\n
Given $$\\left|\\mathrm{~m}_l\\right|=1$$, this directly implies that we are not considering s orbitals (which have $$l = 0$$ and hence $$m_l = 0$$). We are considering p, d, and f orbitals which can have $$m_l = 1$$ or $$m_l = -1$$.
\n\n
For $l = 1$ (p orbitals):
\n\n
\n\n
This corresponds to p orbitals, where $$\\left|\\mathrm{~m}_l\\right|=1$$ can be achieved with $$m_l = -1$$ or $$m_l = +1$$.
\n\n
With $$\\mathrm{m}_{\\mathrm{s}}=-\\frac{1}{2}$$, we only take half of the available ml values since each orbital can host two electrons with different spin quantum numbers.
\n\n
So, we get 2 electrons for p orbitals with the specified conditions.
\n\n
\n\n
For $l = 2$ (d orbitals):
\n\n
\n\n
This corresponds to d orbitals, where again $$\\left|\\mathrm{~m}_l\\right|=1$$ indicates $$m_l = -1$$ or $$m_l = +1$$.
\n\n
Similarly, we consider only electrons with a spin quantum number of $$-\\frac{1}{2}$$, leading us to again have 2 electrons.
\n\n
\n\n
For $l = 3$ (f orbitals):
\n\n
\n\n
f orbitals also can have $$m_l = -1$$ or $$m_l = +1$$ satisfying the $$\\left|\\mathrm{~m}_l\\right|=1$$.
\n\n
With $$\\mathrm{m}_{\\mathrm{s}}=-\\frac{1}{2}$$, like before, this limits us to 2 electrons.
\n\n
\n\n
Adding up all the electrons from the p, d, and f orbitals that meet the given criteria, we get:
Therefore, the total number of electrons having quantum numbers with $$\\mathrm{n}=4, \\left|\\mathrm{~m}_l\\right|=1$$ and $$\\mathrm{m}_{\\mathrm{s}}=-\\frac{1}{2}$$ is 6.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3137,
"subject": "Chemistry",
"question": "Given below are two statements :
Statement I : Rutherford's gold foil experiment cannot explain the line spectrum of hydrogen atom.
Statement II : Bohr's model of hydrogen atom contradicts Heisenberg' uncertainty principle.
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\nRutherford's gold foil experiment only proved that electrons are held towards nucleus by electrostatic forces of attraction and move in circular orbits with very high speeds.
Bohr's model gave exact formula for simultaneous calculation of speed & distance of electron from the nucleus, something which was deemed impossible according to Heisenberg.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3138,
"subject": "Chemistry",
"question": "If the Thompson model of the atom was correct, then the result of Rutherford's gold foil experiment would have been :",
"options": [
{
"text": "All of the $$\\alpha$$-particles pass through the gold foil without decrease in speed."
},
{
"text": "$$\\alpha$$-particles are deflected over a wide range of angles."
},
{
"text": "All $$\\alpha$$-particles get bounced back by 180$$^\\circ$$"
},
{
"text": "$$\\alpha$$-particles pass through the gold foil deflected by small angles and with reduced speed."
}
],
"answer": "$$\\alpha$$-particles pass through the gold foil deflected by small angles and with reduced speed.",
"solution": "**Answer:** $$\\alpha$$-particles pass through the gold foil deflected by small angles and with reduced speed.\n\nAs in Thompson model, protons are diffused (charge is not centred) $$\\alpha$$-particles deviate by small angles and due to repulsion from protons, their speed decreases.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3139,
"subject": "Chemistry",
"question": "
The number of following statement/s which is/are incorrect is ___________
\n
(A) Line emission spectra are used to study the electronic structure
\n
(B) The emission spectra of atoms in the gas phase show a continuous spread of wavelength from red to violet
\n
(C) An absorption spectrum is like the photographic negative of an emission spectrum
\n
(D) The element helium was discovered in the sun by spectroscopic method
(A) This statement is correct. Line emission spectra (or atomic emission spectra) are indeed used to study the electronic structure of atoms. These spectra are produced when photons are emitted from atoms as excited electrons return to a lower energy level. Each line corresponds to a particular quantum leap between energy levels.
\n
(B) This statement is incorrect. The emission spectra of atoms in the gas phase do not show a continuous spread of wavelength from red to violet. Instead, they show distinct lines corresponding to specific electron transitions, hence they are known as line emission spectra.
\n
(C) This statement is correct. An absorption spectrum is like the photographic negative of an emission spectrum. In an absorption spectrum, light passes through a cold, dilute gas and atoms in the gas absorb at characteristic frequencies; since the re-emitted light is unlikely to be emitted in the same direction as the absorbed photon, this gives rise to dark lines (absence of light) in the spectrum.
\n
(D) This statement is correct. The element helium was indeed discovered in the sun through spectroscopy before it was found on Earth. The name helium comes from Helios, the Greek name for the sun.
Choose the Incorrect Statement about Dalton's Atomic Theory
",
"options": [
{
"text": "Compounds are formed when atoms of different elements combine in any ratio.\n"
},
{
"text": "Matter consists of indivisible atoms.\n"
},
{
"text": "chemical reactions involve reorganization of atoms\n"
},
{
"text": "All the atoms of a given element have identical properties including identical mass."
}
],
"answer": "Compounds are formed when atoms of different elements combine in any ratio.\n",
"solution": "**Answer:** Compounds are formed when atoms of different elements combine in any ratio.\n\n\n
According to Dalton’s theory, compounds are \nformed when atoms of different elements combine \nin fixed ratio.
The adsorbent used in adsorption chromatography is/are -
\n
A. silica gel
\n
B. alumina
\n
C. quick lime
\n
D. magnesia
\n
Choose the most appropriate answer from the options given below :
",
"options": [
{
"text": "C and D only\n"
},
{
"text": "A only\n"
},
{
"text": "B only\n"
},
{
"text": "A and B only"
}
],
"answer": "A and B only",
"solution": "**Answer:** A and B only\n\n
Commonly used adsorbents in adsorption \nchromatography are silica gel and alumina.
",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 3144,
"subject": "Chemistry",
"question": "Which of the following is not an example of heterogenous catalytic reaction?",
"options": [
{
"text": "Ostwald's process "
},
{
"text": "Combustion of coal "
},
{
"text": "Hydrogenation of vegetable oils "
},
{
"text": "Haber's process "
}
],
"answer": "Combustion of coal ",
"solution": "**Answer:** Combustion of coal \n\nThen is no catalyst is required for combustion of coal.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3145,
"subject": "Chemistry",
"question": "Match the catalysts (Column I) with products\n(Column II).
(B) TiCl4 + Al(Me)3 this mixture is called Ziegler-Natta catalyst which is used in the synthesis of high density polyethene from ethene.\n
(C) PdCl2 is used in ethanol formation\n
(D) Iron oxide is used in Haber process for NH3.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3146,
"subject": "Chemistry",
"question": "For the following Assertion and Reason, the\ncorrect option is \nAssertion : For hydrogenation reactions, the\ncatalytic activity increases from Group 5 to\nGroup 11 metals with maximum activity shown\nby Group 7-9 elements. \nReason : The reactants are most strongly\nadsorbed on group 7-9 elements.",
"options": [
{
"text": "Both assertion and reason are true and\nthe reason is the correct explanation for\nthe assertion."
},
{
"text": "The assertion is true, but the reason is false."
},
{
"text": "Both assertion and reason are true but the\nreason is not the correct explanation for the\nassertion."
},
{
"text": "Both assertion and reason are false."
}
],
"answer": "The assertion is true, but the reason is false.",
"solution": "**Answer:** The assertion is true, but the reason is false.\n\nReactants must not get adsorbed so strongly\nthat they are immobilised and other reactants\nare left with no space on the catalyst’s surface\nfor adsorption.",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 3147,
"subject": "Chemistry",
"question": "
\n(D) $2 \\mathrm{SO}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\stackrel{\\mathrm{NO}(\\mathrm{g})}{\\longrightarrow} 2 \\mathrm{SO}_{3}$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3148,
"subject": "Chemistry",
"question": "Alum helps in purifying water by :",
"options": [
{
"text": "forming Si complex with clay particles"
},
{
"text": "sulphate part which combines with the dirt and removes it"
},
{
"text": "coagulating the mud particles"
},
{
"text": "making mud water soluble"
}
],
"answer": "coagulating the mud particles",
"solution": "**Answer:** coagulating the mud particles\n\nAlum furnishes $$A{l^{3 + }}$$ ions which bring about coagulation of negatively charged clay particles, bacteria etc. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3149,
"subject": "Chemistry",
"question": "The disperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and\nnegatively charged, respectively, which of the following statements is NOT correct?",
"options": [
{
"text": "magnesium chloride solution coagulates, the gold sol more readily than the iron\n(III) hydroxide sol. "
},
{
"text": "sodium sulphate solution causes coagulation in both sols"
},
{
"text": "mixing the sols has no effect"
},
{
"text": "coagulation in both sols can be brought about by electrophoresis "
}
],
"answer": "mixing the sols has no effect",
"solution": "**Answer:** mixing the sols has no effect\n\nWhen oppositely charged sols are mixed their charges are neutralised. Both sols may be partially or completely precipitated. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3150,
"subject": "Chemistry",
"question": "The volume of a colloidal particle, VC as compared to the volume of a solute particle\nin a true solution VS, could be ",
"options": [
{
"text": "$${V_C \\over V_S } \\simeq 1$$"
},
{
"text": "$${V_C \\over V_S } \\simeq 10^{-3}$$"
},
{
"text": "$${V_C \\over V_S } \\simeq 10^3$$"
},
{
"text": "$${V_C \\over V_S } \\simeq 10^{23}$$"
}
],
"answer": "$${V_C \\over V_S } \\simeq 10^3$$",
"solution": "**Answer:** $${V_C \\over V_S } \\simeq 10^3$$\n\nParticle size of colloidal particle \n
hence now $${{Vc} \\over {Vs}} = {10^3}$$ \n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3151,
"subject": "Chemistry",
"question": "Gold numbers of protective colloids A, B, C and D are 0.50, 0.01, 0.10 and 0.005, respectively. The\ncorrect order of their protective powers is",
"options": [
{
"text": "D < A < C < B"
},
{
"text": "C < B < D < A "
},
{
"text": "A < C < B < D"
},
{
"text": "B < D < A < C "
}
],
"answer": "A < C < B < D",
"solution": "**Answer:** A < C < B < D\n\nFor a protective colloid $$\\mu $$ lesser the value of gold number better is the protective power. \n
Thus the correct order of protective power of $$A,B,C$$ and $$D$$ is \n
$$ \\Rightarrow $$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$\\left( A \\right)\\,\\, < \\,\\,\\left( C \\right)\\,\\, < \\,\\,\\left( B \\right)\\,\\, < \\,\\,\\left( D \\right)$$ \n
Gold number $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0.50\\,\\,\\,\\,\\,\\,0.10\\,\\,\\,\\,\\,\\,0.01\\,\\,\\,\\,\\,\\,0.005$$ \n
Hence $$(c)$$ is the correct answer ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3152,
"subject": "Chemistry",
"question": "The coagulating power of electrolytes having ions Na+, Al3+ and Ba2+ for arsenic sulphide sol increases in the order: ",
"options": [
{
"text": "Na+ < Ba2+ < Al3+"
},
{
"text": "Al3+ < Ba2+ < Na+ "
},
{
"text": "Ba2+ < Na+ < Al3+ "
},
{
"text": "Al3+ < Na+ < Ba2+"
}
],
"answer": "Na+ < Ba2+ < Al3+",
"solution": "**Answer:** Na+ < Ba2+ < Al3+\n\nAccording to Hardly Schulze rule, greater the charge or cation, greater is its coagulating power for negatively charged solution. Hence the correct order of coagulating power : $$N{a^ + } < B{a^{2 + }} < A{l^{3 + }}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3153,
"subject": "Chemistry",
"question": "The most appropriate method of making egg-albumin sol is :\n",
"options": [
{
"text": "Break an egg carefully and transfer the transparent part of the content to 100 mL of 5% w/V saline solution and stir well"
},
{
"text": "Break an egg carefully and transfer only the yellow part of the content to 100 mL of 5% w/V saline solution and stir well."
},
{
"text": "Keep the egg in boiling water for 10 minutes. After removing the shell, transfer the white part of the content to 100 mL of 5% w/V saline solution and homogenize with a mechanical shaker."
},
{
"text": "Keep the egg in boiling water for 10 minutes. After removing the shell, transfer the yellow part of the content to 100 mL of 5% w/V saline solutionand homogenize with a mechanical shaker."
}
],
"answer": "Break an egg carefully and transfer the transparent part of the content to 100 mL of 5% w/V saline solution and stir well",
"solution": "**Answer:** Break an egg carefully and transfer the transparent part of the content to 100 mL of 5% w/V saline solution and stir well\n\n
Gold number of a protective colloid is a minimum weight of it in milligrams which must be added to 10 ml of a standard red gold sol. so that no coagulation of the gold sol. (i.e. change of colour from red to blue) takes place when 1 ml of 10% sodium chloride solution is rapidly added to it. Obviously, smaller the gold number of a protective colloid, the greater is the protective actioin.
\n
It is given that the gelatin has lowest gold number thus, it would serve as a better protective colloid than colloids in the rest of the options.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3155,
"subject": "Chemistry",
"question": "The Tyndall effect is observed only when following conditions are satisfied: \n(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used. \n(b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used. \n(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude. \n(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude. ",
"options": [
{
"text": "(b) and (d)"
},
{
"text": "(a) and (c)"
},
{
"text": "(b) and (c)"
},
{
"text": "(a) and (d)"
}
],
"answer": "(b) and (d)",
"solution": "**Answer:** (b) and (d)\n\nTyndall effect is observed only when\n
(i) The diameter of the dispersed particle is not much smaller than the wavelength of the light used.\n
(ii) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.\n
So, (b) and (d) are correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3156,
"subject": "Chemistry",
"question": "Among the following, correct statement is :",
"options": [
{
"text": "Brownian movement is more pronounced for smaller particles than for bigger–particles"
},
{
"text": "Sols of metal sulphides are lyophilic."
},
{
"text": "Hardy Schulze law states that bigger the size of the ions, the greater is its\ncoagulating power."
},
{
"text": "One would expect charcoal to adsorb chlorine more than hydrogen sulphide."
}
],
"answer": "One would expect charcoal to adsorb chlorine more than hydrogen sulphide.",
"solution": "**Answer:** One would expect charcoal to adsorb chlorine more than hydrogen sulphide.\n\n(a) Brownian motion caused by the bombardment of the particles by molecules of the median. In smaller particle the bombardment is not effective enough so that it can change it's direction and perform zig zag motion. But for bigger particle zig zag motion happens as bombardment is effective.\n
(b) Inorganic sols are lyophobic. Metal sulphates are inorganic sols so it is lyophotic. \n
(c) Hardly Schulze does not depends on the size of the ion it depends on the size of the ion it depends on the charge on the ion. \n
(d) Molar mass of Cl2 is 71 and molar mass of H2S is 34 and in both the molecule Nanderwaal force of attraction present. As Cl2 has more mass than H2S So it's vanderwall force of attraction will be more and tendency to melt will also be more. So, Cl2 will get easily absorbed.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3157,
"subject": "Chemistry",
"question": "Which of the following statements about colloids is False ?",
"options": [
{
"text": "Freezing point of colloidal solution is lower than true solution at same concentration of a solute. "
},
{
"text": "Colloidal particles can pass through ordinary filter paper."
},
{
"text": "When silver nitrate solution is added to potassium iodide solution, a negatively charged colloidal solution is formed. "
},
{
"text": "When excess of electolyte is added to colloidal solution, colloidal particle will be precipitated."
}
],
"answer": "Freezing point of colloidal solution is lower than true solution at same concentration of a solute. ",
"solution": "**Answer:** Freezing point of colloidal solution is lower than true solution at same concentration of a solute. \n\nFreezing point of colloidal solution is same as true solution at same concentration of a solute.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3158,
"subject": "Chemistry",
"question": "Among the colloids cheese (C), milk (M), and smoke (S), the correct combination of the dispersed phase and dispersion medium, respectively is : ",
"options": [
{
"text": "C : liquid in solid ; M : liquid in liquid ; S : solid in gas "
},
{
"text": "C : liquid in solid ; M : liquid in solid ; S : solid in gas"
},
{
"text": "C : solid in liquid ; M : liquid in liquid ; S : gas in solid "
},
{
"text": "C : solid in liqid ; M : solid in liquid ; S : solid in gas \n"
}
],
"answer": "C : liquid in solid ; M : liquid in liquid ; S : solid in gas ",
"solution": "**Answer:** C : liquid in solid ; M : liquid in liquid ; S : solid in gas \n\nCheese (C) is liquid dispersed in solid.\n
Milk (M) is liquid dispersed in liquid.\n
Smoke (S) is solid dispersed in gas.\n",
"topic": "Discrete Mathematics",
"subtopic": "Combinatorics"
},
{
"id": 3159,
"subject": "Chemistry",
"question": "Among the following, the INCORRECT statement about colloids is :\n",
"options": [
{
"text": "The range of diameters of colloidal particles is between 1 and to 1000 nm"
},
{
"text": "they are larger than small molecules and have high molar mass"
},
{
"text": "They can scatter light"
},
{
"text": "The osmotic pressure of a colloidal solution is of higher order than the true solution at the same\nconcentration."
}
],
"answer": "The osmotic pressure of a colloidal solution is of higher order than the true solution at the same\nconcentration.",
"solution": "**Answer:** The osmotic pressure of a colloidal solution is of higher order than the true solution at the same\nconcentration.\n\nOsmotic pressure of colloidal solution is lower than true solution of same concentration.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3160,
"subject": "Chemistry",
"question": "Peptization is a :",
"options": [
{
"text": "process of converting soluble particles to form colloidal solution"
},
{
"text": "process of converting precipitate into colloidal solution"
},
{
"text": "process of bringing colloidal molecule into solution"
},
{
"text": "process of converting a colloidal solution into precipitate"
}
],
"answer": "process of converting precipitate into colloidal solution",
"solution": "**Answer:** process of converting precipitate into colloidal solution\n\nPeptisation is the process of converting a\nprecipitate into a colloidal solution by shaking it with\ndispersion medium in the presence of small\namount of electrolyte.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3161,
"subject": "Chemistry",
"question": "The correct option among the following is :",
"options": [
{
"text": "Addition of alum to water makes it unfit for drinking."
},
{
"text": "Colloidal medicines are more effective because they have small surface area."
},
{
"text": "Colloidal paraticles in lyophobic sols can be precipitated by electrophoresis."
},
{
"text": "Brownian motion in colloidal solution is faster if the viscosity of the solution is very high."
}
],
"answer": "Colloidal paraticles in lyophobic sols can be precipitated by electrophoresis.",
"solution": "**Answer:** Colloidal paraticles in lyophobic sols can be precipitated by electrophoresis.\n\nElectrophoresis is used to coagulate lyophobic\ncolloids.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3162,
"subject": "Chemistry",
"question": "The aerosol is a kind of colloid in which :",
"options": [
{
"text": "solid is dispersed in gas"
},
{
"text": "gas is dispersed in solid"
},
{
"text": "gas is dispersed in liquid"
},
{
"text": "liquid is dispersed in water"
}
],
"answer": "solid is dispersed in gas",
"solution": "**Answer:** solid is dispersed in gas\n\nIn aerosol, the dispersion medium is gas while\nthe dispersed phase can be both solid or liquid.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3163,
"subject": "Chemistry",
"question": "Among the following the false statement is -",
"options": [
{
"text": "Lyophilic sol can be coagulated by adding an electrote "
},
{
"text": "Tyndall effect can be used to distinguish between a colloidal solution and a true solution "
},
{
"text": "It is possible to cause artificial rain by throwing electrified sand carrying charge opposite to the one on clouds from an aeroplane"
},
{
"text": "Latex is a colloidal solution of rubber particles which are positively charged "
}
],
"answer": "Latex is a colloidal solution of rubber particles which are positively charged ",
"solution": "**Answer:** Latex is a colloidal solution of rubber particles which are positively charged \n\nLatex is a colloidal solution of rubber particles which are negatively charged.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3164,
"subject": "Chemistry",
"question": "An example of solid sol is : ",
"options": [
{
"text": "Butter"
},
{
"text": "Hair cream"
},
{
"text": "Gem stones"
},
{
"text": "Paint"
}
],
"answer": "Gem stones",
"solution": "**Answer:** Gem stones\n\nAn example of solid sol is gem stones.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3165,
"subject": "Chemistry",
"question": "Haemoglobin and gold sol are examples of : ",
"options": [
{
"text": "positively charged sols"
},
{
"text": "negatively charged sols"
},
{
"text": "Positively and negatively charged sols, respectively \n"
},
{
"text": "negatively and positively charged sols, respectively "
}
],
"answer": "Positively and negatively charged sols, respectively \n",
"solution": "**Answer:** Positively and negatively charged sols, respectively \n\n\nHaemoglobin $$\\buildrel \\, \\over\n \\longrightarrow $$ positive sol\n Ag $$-$$ sol $$\\buildrel \\, \\over\n \\longrightarrow $$ negative sol",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3166,
"subject": "Chemistry",
"question": "For coagulation of arsenious sulphate sol, which one of the following salt solution will be most effective ? ",
"options": [
{
"text": "BaCl2"
},
{
"text": "AlCl3"
},
{
"text": "NaCl"
},
{
"text": "Na3PO4"
}
],
"answer": "AlCl3",
"solution": "**Answer:** AlCl3\n\nSulphide is $$-$$ve charged colloid.\n
As coagulation is directly proportional to the effective charge, so cation with maximum charge will be most effective for coagulation. \n
As per Hardy Schulze rules coagulation power\nof anion follows the order :\n
Fe(CN)63– $$>$$ CrO42– $$>$$ Cl– = Br– = NO3–\n
Higher the coagulation power lower will be its\nflocculation value\ntherefore order will be :\n
Fe(CN)63– $$<$$ CrO42– $$<$$ Cl– = Br– = NO3–",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3168,
"subject": "Chemistry",
"question": "Kraft temperature is the temperature :",
"options": [
{
"text": "below which the aqueous solution of\ndetergents starts freezing."
},
{
"text": "above which the formation of micelles takes\nplace."
},
{
"text": "below which the formation of micelles takes\nplace."
},
{
"text": "above which the aqueous solution of\ndetergents starts boiling."
}
],
"answer": "above which the formation of micelles takes\nplace.",
"solution": "**Answer:** above which the formation of micelles takes\nplace.\n\nAbove Kraft temperature(Tk) the formation of\nmicelles takes place.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3169,
"subject": "Chemistry",
"question": "A sample of red ink (a colloidal suspension) is\nprepared by mixing eosin dye, egg white, HCHO\nand water. The component which ensures\nstability of the ink sample is :",
"options": [
{
"text": "HCHO"
},
{
"text": "Water"
},
{
"text": "Eosin dye"
},
{
"text": "Egg white"
}
],
"answer": "Egg white",
"solution": "**Answer:** Egg white\n\nEgg white is used to stabilise the prepared\ncolloidal sol.of red ink.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3170,
"subject": "Chemistry",
"question": "Match the following :
Emulsion $$ \\to $$ milk",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3171,
"subject": "Chemistry",
"question": "Tyndall effect is observed when:",
"options": [
{
"text": "The diameter of dispersed particles is similar to the wavelenght of light used."
},
{
"text": "The diameter of dispersed particles is much smaller than the wavelength of light used."
},
{
"text": "The diameter of dispersed particles is much larger than the wavelength of light used."
},
{
"text": "The refractive index of dispersed phase is greater than that of the dispersion medium."
}
],
"answer": "The diameter of dispersed particles is similar to the wavelenght of light used.",
"solution": "**Answer:** The diameter of dispersed particles is similar to the wavelenght of light used.\n\nDiameter of dispersed particles should not be much smaller than wavelength of light used.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3172,
"subject": "Chemistry",
"question": "Which of the following is used for the\npreparation of colloids?",
"options": [
{
"text": "Van Arkel Method"
},
{
"text": "Bredig’s Arc Method"
},
{
"text": "Mond Process"
},
{
"text": "Ostwald Process"
}
],
"answer": "Bredig’s Arc Method",
"solution": "**Answer:** Bredig’s Arc Method\n\nBredig's arc method is used for the preparation\nof colloids.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3173,
"subject": "Chemistry",
"question": "Most suitable salt which can be used for efficient clotting of blood will be :",
"options": [
{
"text": "Mg(HCO3)2"
},
{
"text": "NaHCO3"
},
{
"text": "FeCl3"
},
{
"text": "FeSO4"
}
],
"answer": "FeCl3",
"solution": "**Answer:** FeCl3\n\n
According to Hardy-Schulze rule, for negatively charged sol, most (+ve) charged ion is needed for efficient coagulation.
Blood is a negatively charged sol. Hence FeCl3 can be used for blood clotting and it from Fe3+ ion.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3174,
"subject": "Chemistry",
"question": "Which one of the following statements is FALSE for hydrophilic sols?",
"options": [
{
"text": "They do not require electrolytes for stability."
},
{
"text": "The sols cannot be easily coagulated."
},
{
"text": "These sols are reversible in nature."
},
{
"text": "Their viscosity is of the order of that of H2O."
}
],
"answer": "Their viscosity is of the order of that of H2O.",
"solution": "**Answer:** Their viscosity is of the order of that of H2O.\n\n1. Viscosity of hydrophilic sol > viscosity of H2O
\n2. Hydrophilic sol is more stable so can't be\neasily coagulated.
\n3. Hydrophilic sols are reversible sols.
\n4. No electrolytes are required to stabilise\nhydrophilic sol.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3175,
"subject": "Chemistry",
"question": "The nature of charge on resulting colloidal particles when FeCl3 is added to excess of hot water is :",
"options": [
{
"text": "neutral"
},
{
"text": "sometimes positive and sometimes negative"
},
{
"text": "positive"
},
{
"text": "negative"
}
],
"answer": "positive",
"solution": "**Answer:** positive\n\nIf FeCl3 is added to excess of hot water, a positively charged sol of hydrated ferric oxide is\nformed due to adsorption of Fe3+ ions.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3176,
"subject": "Chemistry",
"question": "The INCORRECT statements below regarding colloidal solutions is :",
"options": [
{
"text": "The flocculating power of Al3+ is more than that of Na+."
},
{
"text": "A colloidal solutions shows Brownian motion of colloidal particles."
},
{
"text": "An ordinary filter paper can stop the flow of colloidal particles."
},
{
"text": "A colloidal solutions shows colligative properties."
}
],
"answer": "An ordinary filter paper can stop the flow of colloidal particles.",
"solution": "**Answer:** An ordinary filter paper can stop the flow of colloidal particles.\n\nColloidal solutions can pass through ordinary filter paper but cannot pass through a special filter colloidal solution coated paper.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3177,
"subject": "Chemistry",
"question": "A colloidal system consisting of a gas dispersed in a solid is called a/an :",
"options": [
{
"text": "foam"
},
{
"text": "aerosol"
},
{
"text": "solid sol"
},
{
"text": "gel"
}
],
"answer": "solid sol",
"solution": "**Answer:** solid sol\n\nA colloidal system consisting of a gas dispersed in\na solid is called a ‘solid sol’.\n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3178,
"subject": "Chemistry",
"question": "For the coagulation of a negative sol, the species below, that has the highest flocculating power is :",
"options": [
{
"text": "Na+"
},
{
"text": "Ba2+"
},
{
"text": "PO$$_4^{3 - }$$"
},
{
"text": "SO$$_4^{2 - }$$"
}
],
"answer": "Ba2+",
"solution": "**Answer:** Ba2+\n\nFor a negative sol, a positive ion is required for flocculation.\nThe greater the valence of the flocculating ion added, the greater is its power to cause\nprecipitation. This is called Hardy-Schulz law.
\nSo, Ba+2 has highest flocculating power.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3179,
"subject": "Chemistry",
"question": "The charges on the colloidal CdS sol and TiO2 sol are, respectively :",
"options": [
{
"text": "positive and positive"
},
{
"text": "positive and negative"
},
{
"text": "negative and negative"
},
{
"text": "negative and positive"
}
],
"answer": "negative and positive",
"solution": "**Answer:** negative and positive\n\nOxides & Hydroxide sol are positive\nso TiO2 is positive sol.
\nSulphide sol are negatively charge\n so CdS is negatively charge sol.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3180,
"subject": "Chemistry",
"question": "The conditions given below are in the context of observing Tyndall effect in colloidal solutions :
(A) The diameter of the colloidal particles is comparable to the wavelength of light used.
(B) The diameter of the colloidal particles is much smaller than the wavelength of light used.
(C) The diameter of the colloidal particles is much larger than the wavelength of light used.
(D) The refractive indices of the dispersed phase and the dispersion medium are comparable.
(E) The dispersed phase has a very different refractive index from the dispersion medium.
Choose the most appropriate conditions from the options given below :",
"options": [
{
"text": "(A) and (E) only"
},
{
"text": "(C) and (D) only"
},
{
"text": "(A) and (D) only"
},
{
"text": "(B) and (E) only"
}
],
"answer": "(A) and (E) only",
"solution": "**Answer:** (A) and (E) only\n\nThe phenomenon of scattering of light by colloidal particles as a result of which the path of the\nbeam becomes visible is called a tyndall effect. Smaller the diameter and similar the magnitude\nof refractive indices, lesser is the scattering and hence the tyndall effect and viced-versa. The\ndiameter of the dispersed phase particle should not be smaller than the wavelength of light\nused because they won't be able to scatter the light so, therefore, the diameter of the dispersed\nparticles should be equal or not much smaller than the wavelength of the light used.\n
2. The refractive indies (i.e. the ratio of the velocity of light in vacuum to the velocity of light in\nany medium) of the dispersed phase and the dispersion medium should differ greatly\nin magnitude than only the particles will be able to scatter the light and tyndall effect will be\nobersved. On the other hand, if the refractive indices of the dispersed phase and dispersion\nmedium are almost similar in magnitude, then there will be no-scattering of light and hence,\ntherefore, no Tyndall effect effect is observed.\n
Hence answer (A) and (E) are correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3181,
"subject": "Chemistry",
"question": "100 ml of 0.0018% (w/v) solution of Cl$$-$$ ion was the minimum concentration of Cl$$-$$ required to precipitate a negative sol in one h. The coagulating value of Cl$$-$$ ion is _________ (Nearest integer)",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nCoagulation value The minimum concentration of electrolyte in millimoles required to cause coagulation of 1 L of colloidal solution.
Given, 0.0018 g Cl$$-$$ present in 100 mL solution
Coagulation value of Cl$$-$$ is $$ = {{{{0.0018} \\over {35.5}} \\times {{10}^3}} \\over {0.1}} = 0.5070 \\simeq 1$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3182,
"subject": "Chemistry",
"question": "Match List I with List II :
\n\n
\n
List-I Example of colloids
\n
List - II Classification
\n
\n\n\n
\n
(a) Cheese
\n
(i) dispersion of liquid in liquid
\n
\n
\n
(b) Pumice stone
\n
(ii) dispersion of liquid in gas
\n
\n
\n
(c) Hair cream
\n
(iii) dispersion of gas in solid
\n
\n
\n
(d) Cloud
\n
(iv) dispersion of liquid in solid
\n
\n\n
Choose the most appropriate answer from the options given below ",
"options": [
{
"text": "(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)"
},
{
"text": "(a)-(iv), (b)-(i), (c)-(iii), (d)-(ii)"
},
{
"text": "(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)"
},
{
"text": "(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)"
}
],
"answer": "(a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)",
"solution": "**Answer:** (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)\n\nCheese $$\\to$$ liquid in solid
Pumice stone $$\\to$$ gas in solid
Hair cream $$\\to$$ liquid in liquid
Cloud $$\\to$$ liquid in gas",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3183,
"subject": "Chemistry",
"question": "The sol given below with negatively charged colloidal particles is :",
"options": [
{
"text": "FeCl3 added to hot water"
},
{
"text": "KI added to AgNO3 solution"
},
{
"text": "AgNO3 added to KI solution"
},
{
"text": "Al2O3.xH2O in water"
}
],
"answer": "AgNO3 added to KI solution",
"solution": "**Answer:** AgNO3 added to KI solution\n\n
When AgNO3\nis added to KI,
\n\n
Iodide ion gets adsorbed on the surface of AgI. This will result in\nformation of negatively charged colloids.
\n
When KI is added to AgNO3\nfollowing reaction takes place.
\n\n
Silver cation gets adsorbed on surface of AgI, resulting in\nformation of positively charged colloids.
\n\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3184,
"subject": "Chemistry",
"question": "Tyndall effect is more effectively shown by : ",
"options": [
{
"text": "true solution"
},
{
"text": "lyophilic colloid"
},
{
"text": "lyophobic colloid"
},
{
"text": "suspension"
}
],
"answer": "lyophobic colloid",
"solution": "**Answer:** lyophobic colloid\n\nTyndall effect is observed in lyophobic colloids",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3185,
"subject": "Chemistry",
"question": "Lyophilic sols are more stable than lyophobic sols because :",
"options": [
{
"text": "there is a strong electrostatic repulsion between the negatively charged colloidal particles."
},
{
"text": "the colloidal particles have positive charge."
},
{
"text": "the colloidal particles have no charge."
},
{
"text": "the colloidal particles are solvated."
}
],
"answer": "the colloidal particles are solvated.",
"solution": "**Answer:** the colloidal particles are solvated.\n\nIn the lyophilic colloids, the colloidal particles are extensively solvated.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3186,
"subject": "Chemistry",
"question": "
Which of the following is a correct statement?
",
"options": [
{
"text": "Brownian motion destabilises sols."
},
{
"text": "Any amount of dispersed phase can added to emulsion without destablising it."
},
{
"text": "Mixing two oppositely charged sols in equal amount neutralises charges and stabilises colloids. "
},
{
"text": "Presence of equal and similar charges on colloidal particles provides stability to the colloidal solution."
}
],
"answer": "Presence of equal and similar charges on colloidal particles provides stability to the colloidal solution.",
"solution": "**Answer:** Presence of equal and similar charges on colloidal particles provides stability to the colloidal solution.\n\nAs equal & similar charge particle will repel each\nother, hence will never precipitate.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3187,
"subject": "Chemistry",
"question": "
A 42.12% (w/v) solution of NaCl causes precipitation of a certain sol in 10 hours. The coagulating value of NaCl for the sol in 2 hours is :
\n
[Given : Molar mass : Na = 23.0 g mol$$-$$1 ; Cl = 35.5 g mol$$-$$1]
The Zeta potential is related to which property of colloids?
",
"options": [
{
"text": "Colour"
},
{
"text": "Tyndall effect"
},
{
"text": "Charge on the surface of colloidal particles"
},
{
"text": "Brownian movement"
}
],
"answer": "Charge on the surface of colloidal particles",
"solution": "**Answer:** Charge on the surface of colloidal particles\n\nThe potential difference between the fixed layer and\nthe diffused layer of opposite charges is called zeta\npotential.
\nIt is related to the charge on the surface of colloidal particles.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3190,
"subject": "Chemistry",
"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
Lyophilic colloid
\n
(I)
\n
Liquid-liquid colloid
\n
\n
\n
(B)
\n
Emulsion
\n
(II)
\n
Protective colloid
\n
\n
\n
(C)
\n
Positively charged colloid
\n
(III)
\n
FeCl$$_3$$ + NaOH
\n
\n
\n
(D)
\n
Negatively charged colloid
\n
(IV)
\n
FeCl$$_3$$ + hot water
\n
\n\n
\n
Choose the correct answer from the options given below :
\n (B) Emulsions are liquid in liquid colloidal\nsolutions
\n (C) FeCl3 + hot water forms positively charged\ncolloidal solution of hydrated ferric oxide.
\n (D) FeCl3 + NaOH forms negatively charged\ncolloidal solution due to preferential adsorption of\nOH- ions ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3191,
"subject": "Chemistry",
"question": "
Incorrect statement for Tyndall effect is :
",
"options": [
{
"text": "The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude."
},
{
"text": "The diameter of the dispersed particles is much smaller than the wavelength of the light used."
},
{
"text": "During projection of movies in the cinemas hall, Tyndall effect is noticed."
},
{
"text": "It is used to distinguish a true solution from a colloidal solution."
}
],
"answer": "The diameter of the dispersed particles is much smaller than the wavelength of the light used.",
"solution": "**Answer:** The diameter of the dispersed particles is much smaller than the wavelength of the light used.\n\nFor Tyndall effect, the diameter of the dispersed\nparticles is not much smaller than the wavelength\nof the light used.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3192,
"subject": "Chemistry",
"question": "
Using very little soap while washing clothes, does not serve the purpose of cleaning of clothes, because :
",
"options": [
{
"text": "soap particles remain floating in water as ions."
},
{
"text": "the hydrophobic part of soap is not able to take away grease."
},
{
"text": "the micelles are not formed due to concentration of soap, below its CMC value."
},
{
"text": "colloidal structure of soap in water is completely disturbed."
}
],
"answer": "the micelles are not formed due to concentration of soap, below its CMC value.",
"solution": "**Answer:** the micelles are not formed due to concentration of soap, below its CMC value.\n\nThe little amount of using soap while washing\nclothes, does not serve the purpose of cleaning of\nclothes because the micelles are not formed due to\nthe low concentration of soap as it is below CMC.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3193,
"subject": "Chemistry",
"question": "
Given below are two statements :
\n
Statement I : Emulsions of oil in water are unstable and sometimes they separate into two layers on standing.
\n
Statement II : For stabilisation of an emulsion, excess of electrolyte is added.
\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": "Statement I is correct but Statement II is incorrect.",
"solution": "**Answer:** Statement I is correct but Statement II is incorrect.\n\nOil in water emulsions can sometimes separate into\ntwo layers on standing.
\nThe most relevant example for the above case is\nmilk, which can separate into two layers on\nstanding for a longer time. Therefore, statement (I)\nis correct.
\nOn adding excess of electrolyte, coagulation occurs\nand emulsion is further destabilised.
\nTherefore, statement (II) is incorrect.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3194,
"subject": "Chemistry",
"question": "
TLC analysis of a mixture having 3 components (A, B, C) using silica gel as the stationary phase gave following Rf values; for A = 0.72, B = 0.48, C = 0.20.
\n
Regarding the above observations, which one of the following statements is not correct for column chromatography of the mixture?
",
"options": [
{
"text": "A is the most non-polar component"
},
{
"text": "C is the most soluble component in mobile phase"
},
{
"text": "A will be eluted first"
},
{
"text": "C is the most strongly absorbed component"
}
],
"answer": "C is the most soluble component in mobile phase",
"solution": "**Answer:** C is the most soluble component in mobile phase\n\nThe $\\mathrm{R}_{f}$ (retardation factor) value is the ratio of the Solute's distance travelled to the Solvent's distance travelled.\n
\nPolar molecules will attract more toward adsorbent and cover smaller distance from the baseline. A has maximum $\\mathrm{R}_{f}$ so it will less adsorbed on the adsorbent surface and travel faster and will be eluded first.\n
\n$\\mathrm{C}$ has minimum $\\mathrm{R}_{f}$ value. So it is more strongly adsobred and less soluble in mobile phase.",
"topic": "Discrete Mathematics",
"subtopic": "Other"
},
{
"id": 3195,
"subject": "Chemistry",
"question": "
For micelle formation, which of the following statements are correct?
\n
A. Micelle formation is an exothermic process.
\n
B. Micelle formation is an endothermic process.
\n
C. The entropy change is positive.
\n
D. The entropy change is negative.
",
"options": [
{
"text": "A and D only"
},
{
"text": "A and C only"
},
{
"text": "B and C only"
},
{
"text": "B and D only"
}
],
"answer": "B and C only",
"solution": "**Answer:** B and C only\n\nMicelle formation is an endothermic process with\npositive entropy change.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3196,
"subject": "Chemistry",
"question": "
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
\n
Assertion A : Finest gold is red in colour, as the size of the particles increases, it appears purple then blue and finally gold.
\n
Reason R : The colour of the colloidal solution depends on the wavelength of light scattered by the dispersed particles.
\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 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\nFinest gold sol is red in colour; as the size of particles increases, it appears purple, then blue and finally golden.\n
\nThe colour of colloidal solution depends on the wavelength of light scattered by the dispersed particles. The wavelength of light further depends on size and nature of the particles.\n
\nHence, Both A and R are true and R is the correct explanation of A.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3197,
"subject": "Chemistry",
"question": "
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
\n
Assertion (A) : Dissolved substances can be removed from a colloidal solution by diffusion through a parchment paper.
\n
Reason (R) : Particles in a true solution cannot pass through parchment paper but the colloidal particles can pass through the parchment paper.
\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 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\nParchment paper is a semi-permeable membrane which allows particles of true solution to pass through as their size are too small.\n
\nAssertion is correct but reason is incorrect.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3198,
"subject": "Chemistry",
"question": "
$$100 \\mathrm{~mL}$$ of $$5 \\%\\,(\\mathrm{w} / \\mathrm{v})$$ solution of $$\\mathrm{NaCl}$$ in water was prepared in $$250 \\mathrm{~mL}$$ beaker. Albumin from the egg was poured into $$\\mathrm{NaCl}$$ solution and stirred well. This resulted in a/an :
",
"options": [
{
"text": "Lyophilic sol"
},
{
"text": "Lyophobic sol"
},
{
"text": "Emulsion"
},
{
"text": "Precipitate"
}
],
"answer": "Lyophilic sol",
"solution": "**Answer:** Lyophilic sol\n\nAlbumin from the egg was poured into $100 \\mathrm{~mL}$ of $5 \\%$ (w/v) $\\mathrm{NaCl}$ solution in water. This would result in the formation of lyophilic sol. Albumin molecules get dispersed in water the colloidal particles of albumin are stabilised by hydrogen bond with water molecules.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3199,
"subject": "Chemistry",
"question": "
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
\n
Assertion (A) : In expensive scientific instruments, silica gel is kept in watch-glasses or in semipermeable membrane bags.
\n
Reason (R) : Silica gel adsorbs moisture from air via adsorption, thus protects the instrument from water corrosion (rusting) and / or prevents malfunctioning.
\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 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)"
}
],
"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 is correct and Reason is correct explanation of Assertion.
\n
Silica gel adsorbs moisture and thus products the instrument from water corrosion (rusting) and prevents malfunctioning.
Which of the following salt solutions would coagulate the colloid solution formed when $$\\mathrm{FeCl_3}$$ is added to $$\\mathrm{NaOH}$$ solution, at the fastest rate?
",
"options": [
{
"text": "10 mL of 0.15 mol dm$$^{-3}$$ CaCl$$_2$$"
},
{
"text": "10 mL of 0.2 mol dm$$^{-3}$$ AlCl$$_3$$"
},
{
"text": "10 mL of 0.1 mol dm$$^{-3}$$ Na$$_2$$SO$$_4$$"
},
{
"text": "10 mL of 0.1 mol dm$$^{-3}$$ Ca$$_3$$(PO$$_4$$)$$_2$$"
}
],
"answer": "10 mL of 0.2 mol dm$$^{-3}$$ AlCl$$_3$$",
"solution": "**Answer:** 10 mL of 0.2 mol dm$$^{-3}$$ AlCl$$_3$$\n\nIn the coagulation of a negative sol, the flocculating power is in the order.\n
\n$\\therefore \\mathrm{Al}^{3+}>\\mathrm{Ba}^{2+}>\\mathrm{Na}^{+}$and $\\mathrm{FeCl}_{3}$ with $\\mathrm{NaOH}$ forms a negative sol.\n
Statement I : For colloidal particles, the values of colligative properties are of small order as compared to values shown by true solutions at same concentration.
\n
Statement II : For colloidal particles, the potential difference between the fixed layer and the diffused layer of same charges is called the electrokinetic potential or zeta potential.
\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 false"
},
{
"text": "Statement I is true and Statement II is false"
},
{
"text": "Both Statement I and Statement II are true"
},
{
"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: For colloidal particles, the values of colligative properties are of small order as compared to values shown by true solutions at the same concentration.
\n\n
This statement is true. Colloidal particles are intermediate in size between true solutions and suspensions, and their properties differ from those of true solutions. The colligative properties of colloids, such as osmotic pressure and vapor pressure, are of a smaller order of magnitude than those of true solutions with the same concentration.
\n\n
Statement II: For colloidal particles, the potential difference between the fixed layer and the diffused layer of same charges is called the electrokinetic potential or zeta potential.
\n\n
This statement is false. For colloidal particles, the potential difference between the fixed layer and the diffused layer of opposite charges is called the electrokinetic potential or zeta potential.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3203,
"subject": "Chemistry",
"question": "$20 \\mathrm{~mL}$ of $0.5 \\mathrm{M} ~\\mathrm{NaCl}$ is required to coagulate $200 \\mathrm{~mL}$ of $\\mathrm{As}_{2} \\mathrm{S}_{3}$ solution in 2 hours. The coagulating value of $\\mathrm{NaCl}$ is ____________ .",
"options": [],
"answer": "50",
"solution": "**Answer:** 50\n\nThe coagulating value of an electrolyte is defined as the minimum amount of the electrolyte required to coagulate a unit volume of the colloidal solution in a given time. In this problem, we are given that $20\\ \\mathrm{mL}$ of $0.5\\ \\mathrm{M}$ $\\mathrm{NaCl}$ solution is required to coagulate $200\\ \\mathrm{mL}$ of $\\mathrm{As}_2\\mathrm{S}_3$ solution in 2 hours. \n
\nTo find the coagulating value of $\\mathrm{NaCl}$, we need to determine the amount of $\\mathrm{NaCl}$ required to coagulate 1 liter (1000 mL) of the $\\mathrm{As}_2\\mathrm{S}_3$ solution in 2 hours. This can be calculated using the following formula:\n
\n$$\\text{Coagulating value} = \\frac{\\text{amount of electrolyte required to coagulate 1 L of sol in 2 hours}}{1\\ \\mathrm{L}}$$\n
\nTo find the amount of $\\mathrm{NaCl}$ required to coagulate 200 mL of the $\\mathrm{As}_2\\mathrm{S}_3$ solution in 2 hours, we can use the formula for the number of moles of $\\mathrm{NaCl}$ used:\n
\nwhere $C_{\\mathrm{NaCl}}$ is the concentration of $\\mathrm{NaCl}$ in moles per liter, and $V_{\\mathrm{NaCl}}$ is the volume of $\\mathrm{NaCl}$ solution used in liters. Substituting the given values, we get:\n
\nTo find the amount of $\\mathrm{NaCl}$ required to coagulate 1 L of the $\\mathrm{As}_2\\mathrm{S}_3$ solution, we need to scale up the amount of $\\mathrm{NaCl}$ used by a factor of 5 (since 5 times the given volume of $\\mathrm{NaCl}$ solution is required to make 1 L of the $\\mathrm{As}_2\\mathrm{S}_3$ solution). This gives:\n
\nTherefore, the coagulating value of $\\mathrm{NaCl}$ is $\\boxed{50}$ (in millimoles per liter).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3204,
"subject": "Chemistry",
"question": "
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.
\n
Assertion A : The diameter of colloidal particles in solution should not be much smaller than wavelength of light to show Tyndall effect.
\n
Reason R : The light scatters in all directions when the size of particles is large enough.
\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 and R is the correct explanation of A"
},
{
"text": "A is true but R is false"
},
{
"text": "Both A and R are correct but R is NOT the correct explanation of A"
},
{
"text": "A is false but R is true"
}
],
"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\nTyndall effect is observed only when the following two conditions are satisfied
\n(a) The diameter of the dispersed particle is not much smaller than the wave length of light used.
\n(b) Refractive indices of dispersed phase and dispersion medium differ greatly in magnitude.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3205,
"subject": "Chemistry",
"question": "
What happens when a lyophilic sol is added to a lyophobic sol?
",
"options": [
{
"text": "Film of lyophobic sol is formed over lyophilic sol."
},
{
"text": "Lyophilic sol is dispersed in lyophobic sol."
},
{
"text": "Film of lyophilic sol is formed over lyophobic sol."
},
{
"text": "Lyophobic sol is coagulated"
}
],
"answer": "Film of lyophilic sol is formed over lyophobic sol.",
"solution": "**Answer:** Film of lyophilic sol is formed over lyophobic sol.\n\nWhen a lyophilic sol is added to a lyophobic sol, the lyophilic sol forms a protective film around the lyophobic sol particles, preventing their aggregation or coagulation. This process is known as \"protection\" and is used to stabilize lyophobic sols.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3206,
"subject": "Chemistry",
"question": "
Ferric chloride is applied to stop bleeding because -
",
"options": [
{
"text": "$$\\mathrm{FeCl}_{3}$$ reacts with the constituents of blood which is a positively charged sol."
},
{
"text": "Blood absorbs $$\\mathrm{FeCl}_{3}$$ and forms a complex."
},
{
"text": "$$\\mathrm{Cl}^{-}$$ ions cause coagulation of blood."
},
{
"text": "$$\\mathrm{Fe}^{3+}$$ ions coagulate blood which is a negatively charged sol."
}
],
"answer": "$$\\mathrm{Fe}^{3+}$$ ions coagulate blood which is a negatively charged sol.",
"solution": "**Answer:** $$\\mathrm{Fe}^{3+}$$ ions coagulate blood which is a negatively charged sol.\n\n
The correct answer is Option D: Fe3+ ions coagulate blood which is a negatively charged sol.
\n
Blood is a colloidal solution, which means that it is made up of small particles suspended in a liquid. The particles in blood are negatively charged, and the negative charge prevents the particles from clumping together. When ferric chloride is applied to a bleeding wound, the Fe3+ ions in the ferric chloride react with the negatively charged particles in the blood. This reaction causes the particles to clump together, forming a clot that stops the bleeding.
\n
The other options are incorrect:
\n
\n
Option A is incorrect. Blood is not a positively charged sol.
\n
Option B is incorrect. Blood does not absorb ferric chloride.
\n
Option C is incorrect. Cl- ions do not cause coagulation of blood.
Coagulating value of the electrolytes $$\\mathrm{AlCl}_{3}$$ and $$\\mathrm{NaCl}$$ for $$\\mathrm{As}_{2} \\mathrm{S}_{3}$$ are 0.09 and 50.04 respectively. The coagulating power of $$\\mathrm{AlCl}_{3}$$ is $$x$$ times the coagulating power of $$\\mathrm{NaCl}$$. The value of $$\\underline{x}$$ is ___________.
Therefore, the coagulating power of AlCl₃ is $$\n\\frac{1}{0.09}\n$$
\n
and the coagulating power of NaCl is $$\n\\frac{1}{50.04}\n$$
\n
We are asked to find the factor by which the coagulating power of AlCl₃ is greater than the coagulating power of NaCl. This factor is given by the ratio
\n
$$\nx = \\frac{\\text{Coagulating power of AlCl₃}}{\\text{Coagulating power of NaCl}} = \\frac{\\frac{1}{0.09}}{\\frac{1}{50.04}} = \\frac{50.04}{0.09}\n$$
\nSo, the coagulating power of AlCl₃ is approximately 556 times the coagulating power of NaCl.
However, since the problem asks for the nearest integer, the answer would be 556.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3208,
"subject": "Chemistry",
"question": "
The number of colloidal systems from the following, which will have 'liquid' as the dispersion medium, is ___________.
",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nPaints, milk, froth, soap lather and hair cream have liquid as dispersion medium.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3209,
"subject": "Chemistry",
"question": "If an endothermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point, then :",
"options": [
{
"text": "$$\\Delta H$$ is -ve, $$\\Delta S$$ is +ve"
},
{
"text": "$$\\Delta H$$ and $$\\Delta S$$ are both +ve"
},
{
"text": "$$\\Delta H$$ and $$\\Delta S$$ are both -ve"
},
{
"text": "$$\\Delta H$$ is +ve, $$\\Delta S$$ is -ve"
}
],
"answer": "$$\\Delta H$$ and $$\\Delta S$$ are both +ve",
"solution": "**Answer:** $$\\Delta H$$ and $$\\Delta S$$ are both +ve\n\nTIPS/Formulae : \n
$$\\Delta G = \\Delta H - T\\Delta S$$\n
Since $$\\Delta G = \\Delta H - T\\Delta S$$ for an endothermic reaction,\n
$$\\Delta H = + ve$$ and at low temperature $$\\Delta S = + ve$$\n
Hence $$\\Delta G = \\left( + \\right)\\Delta H - T\\left( + \\right)\\Delta S$$ \n
and if $$T\\Delta S < \\Delta H$$ (at low temp)\n
$$\\Delta G = + ve$$ (non spontaneous)\n
But at high temperature, reaction becomes \n
spontaneous i.e. $$\\Delta G = - ve.$$\n
because at higher temperature $$T\\Delta S > \\Delta H.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3210,
"subject": "Chemistry",
"question": "In an irreversible process taking place at constant T and P and in which only pressure-volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria :",
"options": [
{
"text": "(dS)V, E > 0, (dG)T, P < 0"
},
{
"text": "(dS)V, E = 0, (dG)T, P = 0"
},
{
"text": "(dS)V, E = 0, (dG)T, P > 0"
},
{
"text": "(dS)V, E < 0, (dG)T, P < 0"
}
],
"answer": "(dS)V, E > 0, (dG)T, P < 0",
"solution": "**Answer:** (dS)V, E > 0, (dG)T, P < 0\n\nFor spontaneous reaction, $$dS > 0$$ and $$dG$$ should be negative i.e. $$ < 0.$$ ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3211,
"subject": "Chemistry",
"question": "The correct relationship between free energy change in a reaction and the corresponding equilibrium constant Kc is :",
"options": [
{
"text": "- $$\\Delta G$$ = RT ln Kc"
},
{
"text": "$$\\Delta G^o$$ = RT ln Kc"
},
{
"text": "- $$\\Delta G^o$$ = RT ln Kc"
},
{
"text": "$$\\Delta G$$ = RT ln Kc"
}
],
"answer": "- $$\\Delta G^o$$ = RT ln Kc",
"solution": "**Answer:** - $$\\Delta G^o$$ = RT ln Kc\n\n$$\\Delta {G^ \\circ } = - RT\\,1n{K_c}\\,\\,\\,$$ or $$\\,\\,\\, - \\Delta {G^ \\circ } = RT\\,1n{K_c}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3212,
"subject": "Chemistry",
"question": "In conversion of lime-stone to lime, \nCaCO3(s) $$\\to$$ CaO(s) + CO2 (g) the vales of ∆H° and ∆S° are +179.1 kJ mol−1\n and 160.2 J/K respectively at 298 K and 1 bar.\nAssuming that ∆H° do not change with temperature, temperature above which conversion of\nlimestone to lime will be spontaneous is :",
"options": [
{
"text": "1008 K "
},
{
"text": "1200 "
},
{
"text": "845 K"
},
{
"text": "1118 K "
}
],
"answer": "1118 K ",
"solution": "**Answer:** 1118 K \n\n$$\\Delta {G^ \\circ } = \\Delta {H^ \\circ } - T\\Delta {S^ \\circ }$$ \n
For a spontaneous reaction $$\\Delta {G^ \\circ } < 0$$\n
or $$\\,\\,\\,\\,\\,$$ = $$750$$ $$K$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3214,
"subject": "Chemistry",
"question": "For a particular reversible reaction at temperature T, ∆H and ∆S were found to be both +ve. If Te is\nthe temperature at equilibrium, the reaction would be spontaneous when :",
"options": [
{
"text": "Te > T"
},
{
"text": "T > Te"
},
{
"text": "Te is 5 times T"
},
{
"text": "T = Te"
}
],
"answer": "T > Te",
"solution": "**Answer:** T > Te\n\nAt equilibrium $$\\Delta G = 0$$ \n
Hence, $$\\Delta G = \\Delta H - {T_e}\\Delta S = 0$$ \n
$$\\therefore$$ $$\\,\\,\\,\\,\\,\\,\\Delta H = {T_e}\\Delta S\\,\\,\\,$$ \n
or $$\\,\\,\\,\\,{T_e} = {{\\Delta H} \\over {\\Delta S}}\\,\\,$$\n
For a spontaneous reaction\n
$$\\Delta G$$ must be negative \n
which is possible only if $$\\Delta H < T\\Delta S$$ \n
or $$\\,\\,\\,T > {{\\Delta H} \\over {\\Delta S}};{T_e} < T$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3215,
"subject": "Chemistry",
"question": "The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a\nvolume of 10 dm3 to a volume of 100 dm3 at 27oC is :",
"options": [
{
"text": "35.8 J mol-1 K−1 "
},
{
"text": "32.3 J mol-1 K−1 "
},
{
"text": "42.3 J mol-1 K−1 "
},
{
"text": "38.3 J mol-1 K−1 "
}
],
"answer": "38.3 J mol-1 K−1 ",
"solution": "**Answer:** 38.3 J mol-1 K−1 \n\nEntropy change for an isothermal reversible process is given by \n
$$\\Delta S = nR{\\mkern 1mu} ln{{{V_2}} \\over {{V_1}}}$$\n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$ = 0.5\\left[ {2 \\times 86600 - R\\left( {298} \\right)\\ln \\left( {1.6 \\times {{10}^{12}}} \\right)} \\right]$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3218,
"subject": "Chemistry",
"question": "A reaction at 1 bar is non-spontaneous at low temperature but becomes spontaneous\nat high temperature. Identify the correct statement about the reaction among the\nfollowing :",
"options": [
{
"text": "Both $$\\Delta $$H and $$\\Delta $$S are negative."
},
{
"text": "Both $$\\Delta $$H and $$\\Delta $$S are positive."
},
{
"text": "$$\\Delta $$H is positive while $$\\Delta $$S is negative."
},
{
"text": "$$\\Delta $$H is negative while $$\\Delta $$S is positive."
}
],
"answer": "Both $$\\Delta $$H and $$\\Delta $$S are positive.",
"solution": "**Answer:** Both $$\\Delta $$H and $$\\Delta $$S are positive.\n\n
We know that $$\\Delta$$G = $$\\Delta$$H $$-$$ T$$\\Delta$$S
\n
At low temperature : $$\\Delta$$G is positive (non-spontaneous process), so $$\\Delta$$H is positive and $$\\Delta$$S is positive (T$$\\Delta$$S < $$\\Delta$$H as T is low).
\n
At high temperature : $$\\Delta$$G is negative (spontaneous process), so $$\\Delta$$H is positive and $$\\Delta$$S is positive (T$$\\Delta$$S > $$\\Delta$$H as T is high).
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3219,
"subject": "Chemistry",
"question": "For the reaction,\n A(g) + B(g) $$ \\to $$ C(g) + D(g), $$\\Delta $$Ho and $$\\Delta $$So are, respectively, − 29.8 kJ mol−1 and −0.100 kJ K−1 mol−1 at 298 K. The equilibrium constant for the reaction at 298 K is :",
"options": [
{
"text": "1.0 $$ \\times $$ 10$$-$$10"
},
{
"text": "1.0 $$ \\times $$ 1010"
},
{
"text": "10"
},
{
"text": "1"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\n
Here pressure increases. When pressure increases then the molecules of will come closer and intermoleculer distance decreases, so entropy will also decreases and $$\\Delta S\\, < \\,0$$.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3222,
"subject": "Chemistry",
"question": "Given\n
(i) 2Fe2O3(s) $$ \\to $$ 4Fe(s) + 3O2(g); \n
$$\\Delta $$rGo = + 1487.0 kJ mol-1\n
(ii) 2CO(g) + O2(g) $$ \\to $$ 2CO2(g); \n
$$\\Delta $$rGo = $$-$$ 514.4 kJ mol-1\n
Free energy change, $$\\Delta $$rGo for the reaction
$$ \\Rightarrow $$$$\\,\\,\\,$$ K = 0.1 atm",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3224,
"subject": "Chemistry",
"question": "A process has $$\\Delta $$H = 200 J mol–1 and $$\\Delta $$S = 40 JK–1 mol–1. Out of the values given below, choose the minimum temperature above which the process will be spontaneous :",
"options": [
{
"text": "4 K"
},
{
"text": "20 K"
},
{
"text": "5 K"
},
{
"text": "12 K"
}
],
"answer": "5 K",
"solution": "**Answer:** 5 K\n\n$$\\Delta $$G = $$\\Delta $$H $$-$$ T$$\\Delta $$S\n
T = $${{\\Delta H} \\over {\\Delta S}} = {{200} \\over {40}} = 5K$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3225,
"subject": "Chemistry",
"question": "For the chemical reaction X $$\\rightleftharpoons$$ Y, the standard reaction Gibbs energy depends on temperature T (in K) as\n$$\\Delta $$rGo (in kJ mol–1) = 120 $$ - {3 \\over 8}$$ T. \n
The major component of the reaction mixture at T is : \n",
"options": [
{
"text": "Y if T = 300 K"
},
{
"text": "Y if T = 280 K "
},
{
"text": "X if T = 350 K"
},
{
"text": "X if T = 315 K "
}
],
"answer": "X if T = 315 K ",
"solution": "**Answer:** X if T = 315 K \n\nX $$\\rightleftharpoons$$ Y\n
Keq = $${{\\left[ Y \\right]} \\over {\\left[ X \\right]}}$$\n
If Keq > 1 $$ \\Rightarrow $$ [Y] > [X]\n
and Keq < 1 $$ \\Rightarrow $$ [Y] < [X]\n
We know, $$\\Delta $$Go = -RT ln(Keq)\n
So when Keq > 1 then $$\\Delta $$Go < 0 and Y is major.\n
And when Keq < 1 then $$\\Delta $$Go > 0 and X is major.\n
Temperature at which Go = 0 is\n
120 $$ - {3 \\over 8}$$ T = 0\n
$$ \\Rightarrow $$ T = 320 K\n
For T > 320 K\n
$$\\Delta $$Go < 0 and Y is major.\n
For T < 320 K\n
$$\\Delta $$Go > 0 and X is major.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3226,
"subject": "Chemistry",
"question": "For the equilibrium, \n 2H2O $$\\rightleftharpoons$$ H3O+ + OH$$-$$, the value of $$\\Delta $$Go at 298 K is approximately : ",
"options": [
{
"text": "$$-$$ 80 kJ mol–1"
},
{
"text": "100 kJ mol$$-$$1"
},
{
"text": "$$-$$ 100 kJ mol$$-$$1"
},
{
"text": "80 kJ mol–1"
}
],
"answer": "80 kJ mol–1",
"solution": "**Answer:** 80 kJ mol–1\n\n2H2O $$\\rightleftharpoons$$ H3O+ + OH$$-$$\n
$$ \\Rightarrow $$ T > $${{491.1 \\times 1000} \\over {198}}$$\n
$$ \\Rightarrow $$ T > 2480.3 K",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3228,
"subject": "Chemistry",
"question": "The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by\n$$\\Delta $$rGo = A – BT\n
Where A and B are non-zero constants. Which of the following is TRUE about this reaction?",
"options": [
{
"text": "Exothermic if B < 0 "
},
{
"text": "Endothermic if A > 0"
},
{
"text": "Exothermic if A > 0 and B < 0"
},
{
"text": "Endothermic if A < 0 and B > 0"
}
],
"answer": "Endothermic if A > 0",
"solution": "**Answer:** Endothermic if A > 0\n\n$$\\Delta $$Go = $$\\Delta $$Ho - T$$\\Delta $$So\n
Given that A and B are non-zero constants, i.e., A = $$\\Delta $$Ho, B = $$\\Delta $$So\n
If $$\\Delta $$Ho is positive means reaction is endothermic.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3229,
"subject": "Chemistry",
"question": "A process will be spontaneous at all temperatures if :",
"options": [
{
"text": "$$\\Delta $$H < 0 and $$\\Delta $$S > 0"
},
{
"text": "$$\\Delta $$H < 0 and $$\\Delta $$S < 0"
},
{
"text": "$$\\Delta $$H > 0 and $$\\Delta $$S < 0"
},
{
"text": "$$\\Delta $$H > 0 and $$\\Delta $$S > 0"
}
],
"answer": "$$\\Delta $$H < 0 and $$\\Delta $$S > 0",
"solution": "**Answer:** $$\\Delta $$H < 0 and $$\\Delta $$S > 0\n\nA reaction is spontaneous if $$\\Delta $$G is negative.\n
and we know that\n
$$\\Delta $$G = $$\\Delta $$H – T$$\\Delta $$S\n
If $$\\Delta $$H = –ve and $$\\Delta $$S = + ve then at all the temperature the process will be spontaneous.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3230,
"subject": "Chemistry",
"question": "For the reaction :
\nA($$l$$) $$ \\to $$ 2B(g)
\n$$\\Delta U = 2.1\\,kcal,\\,\\Delta S = 20\\,cal\\,{K^{ - 1}}$$ at 300 K
\nHence $$\\Delta $$G in kcal is :",
"options": [],
"answer": "-2.7",
"solution": "**Answer:** -2.7\n\nA($$l$$) $$ \\to $$ 2B(g)\n
We know,\n$$\\Delta $$H = $$\\Delta $$U + $$\\Delta $$ngRT\n
and $$\\Delta $$G = $$\\Delta $$H - T$$\\Delta $$S\n
= 2.1 + 1.2 - 6 = – 2.70 Kcal/mol\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3231,
"subject": "Chemistry",
"question": "For a reaction,\n 4M(s) + nO2(g) $$ \\to $$ 2M2On(s)\n the free energy change is plotted as a function\nof temperature. The temperature below which\nthe oxide is stable could be inferred from the\nplot as the point at which :",
"options": [
{
"text": "the free energy change shows a change\nfrom negative to positive value"
},
{
"text": "the slope changes from positive to negative"
},
{
"text": "the slope changes from negative to positive"
},
{
"text": "the slope changes from positive to zero"
}
],
"answer": "the free energy change shows a change\nfrom negative to positive value",
"solution": "**Answer:** the free energy change shows a change\nfrom negative to positive value\n\n$$\\Delta $$G = $$\\Delta $$H – T$$\\Delta $$S\n
= – 13537.57 Joule",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3233,
"subject": "Chemistry",
"question": "Assuming ideal behaviour, the magnitude of log K for the following reaction at 25$$^\\circ$$C is x $$\\times$$ 10$$-$$1. The value of x is ______________. (Integer answer)\n
x = 855",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3234,
"subject": "Chemistry",
"question": "For a chemical reaction A + B ⇌ C + D
($${\\Delta _r}{H^\\Theta }$$ = 80 kJ mol$$-$$1) the entropy change $${\\Delta _r}{S^\\Theta }$$ depends on the temperature T (in K) as $${\\Delta _r}{S^\\Theta }$$ = 2T (J K$$-$$1mol$$-$$1).
Minimum temperature at which it will become spontaneous is ___________ K. (Integer)",
"options": [],
"answer": "200",
"solution": "**Answer:** 200\n\nWe know, $$\\Delta G^\\circ = \\Delta H^\\circ - T\\Delta S^\\circ $$\n
For a reaction to be spontaneous\n $$\\Delta G^\\circ < 0$$\n
The minimum temperature to make it spontaneous is 200 K.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3235,
"subject": "Chemistry",
"question": "During which of the following processes, does entropy decrease?
(A) Freezing of water to ice at 0$$^\\circ$$C
(B) Freezing of water to ice at $$-$$10$$^\\circ$$C
(C) N2(g) + 3H2(g) $$ \\to $$ 2NH3(g)
(D) Adsorption of CO(g) on lead surface.
(E) Dissolution of NaCl in water
Choose the correct answer from the options given below :",
"options": [
{
"text": "(A), (C) and (E) only"
},
{
"text": "(B) and (C) only"
},
{
"text": "(A), (B), (C) and (D) only"
},
{
"text": "(A) and (E) only"
}
],
"answer": "(A), (B), (C) and (D) only",
"solution": "**Answer:** (A), (B), (C) and (D) only\n\nA, B $$ \\to $$ Freezing of water will decrease entropy as particles will move closer and forces of\nattraction will increase. This leads to decrease in randomness. So entropy decrease.
\nC $$ \\to $$ No. of molecules decreasing
\nD $$ \\to $$ Adsorption will lead to decrease in randomness of gaseous particles.
\nSo, (A, B, C, D) decreases entropy. ",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 3236,
"subject": "Chemistry",
"question": "For a given chemical reaction A $$\\to$$ B at 300 K the free energy change is $$-$$49.4 kJ mol$$-$$1 and the enthalpy of reaction is 51.4 kJ mol$$-$$1. The entropy change of the reaction is _____________ JK$$-$$1 mol$$-$$1.",
"options": [],
"answer": "336",
"solution": "**Answer:** 336\n\n$$\\Delta$$G = $$-$$49.4 kJ/mol
$$\\Delta$$H = 51.4 kJ/mol
$$\\Delta$$G = $$\\Delta$$H $$-$$ T$$\\Delta$$S
$$-$$49400 = 51400 $$-$$ 300$$\\Delta$$S
$$\\Delta S = {{ + 100800} \\over {300}} = 336$$ JK$$-$$1 mol$$-$$1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3237,
"subject": "Chemistry",
"question": "Data given for the following reaction is as follows :
The minimum temperature in K at which the reaction becomes spontaneous is ___________. (Integer answer)",
"options": [],
"answer": "964",
"solution": "**Answer:** 964\n\n$${T_{\\min }} = \\left( {{{{\\Delta ^0}H} \\over {{\\Delta ^0}S}}} \\right)$$
4Fe(s) + 3O2(g) $$\\to$$ 2Fe2O3(s) is $$-$$550 J K$$-$$1 at 298 K.
\n
[Given : The standard enthalpy change for the reaction is $$-$$165 kJ mol$$-$$1]. The temperature in K at which the reaction attains equilibrium is _____________. (Nearest Integer)
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 reduction of a metal oxide is easier if the metal formed is in liquid state than solid state.
\n
Reason $$\\mathbf{R}$$ : The value of $$\\Delta G ^\\Theta$$ becomes more on negative side as entropy is higher in liquid state than solid state.
\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 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\nReduction of a metal oxide is easier if the metal is formed in a liquid state at the temperature of reduction because the entropy is higher if the metal is in a liquid state.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3242,
"subject": "Chemistry",
"question": "
For independent process at 300 K
\n
\n
\n
\n
\n
\n
\n
\n\n
\n
Process
\n
$$\\mathrm{\\Delta H/kJ~mol^{-1}}$$
\n
$$\\mathrm{\\Delta S/J~K^{-1}}$$
\n
\n\n\n
\n
A
\n
$$-25$$
\n
$$-80$$
\n
\n
\n
B
\n
$$-22$$
\n
40
\n
\n
\n
C
\n
25
\n
$$-50$$
\n
\n
\n
D
\n
22
\n
20
\n
\n\n
\n
The number of non-spontaneous process from the following is __________
\nProcesses $\\mathrm{C}$ and $\\mathrm{D}$ are non-spontaneous.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3243,
"subject": "Chemistry",
"question": "$30.4 \\mathrm{~kJ}$ of heat is required to melt one mole of sodium chloride and the entropy change at the melting point is $28.4 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$ at 1 atm. The melting point of sodium chloride is _______________ K (Nearest Integer)",
"options": [],
"answer": "1070",
"solution": "**Answer:** 1070\n\nThe heat required to melt one mole of a substance is known as its molar enthalpy of fusion, denoted as $\\Delta H_\\mathrm{fus}$. For sodium chloride, $\\Delta H_\\mathrm{fus}$ is $30.4\\ \\mathrm{kJ/mol}$. \n
\nThe entropy change at the melting point is given by $\\Delta S = \\frac{\\Delta H_\\mathrm{fus}}{T_\\mathrm{m}}$, where $T_\\mathrm{m}$ is the melting point in Kelvin. Substituting the given values, we get:\n
\nRounding off to the nearest integer, the melting point of sodium chloride is $\\boxed{1070\\ \\mathrm{K}}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3244,
"subject": "Chemistry",
"question": "
the amount of heat produced as measured in bomb calorimeter is $$1406 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$ at $$300 \\mathrm{~K}$$. The minimum value of $$\\mathrm{T} \\Delta \\mathrm{S}$$ needed to reach equilibrium is ($$-$$) _________ $$\\mathrm{kJ}$$. (Nearest integer)
\n
Given : $$\\mathrm{R}=8.3 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$$
In the given combustion reaction, the number of moles of gaseous products ($\\Delta n_g$) is -2 (since we have 2 moles of CO₂ gas on the product side and 4 moles of gaseous reactants).
\n
The change in internal energy ($\\Delta U$) is given as -1406 kJ/mol.
We can calculate the change in enthalpy ($\\Delta H$) using the equation $\\Delta H = \\Delta U + \\Delta n_gRT$, where $R$ is the ideal gas constant and $T$ is the temperature.
The value of $$\\log \\mathrm{K}$$ for the reaction $$\\mathrm{A} \\rightleftharpoons \\mathrm{B}$$ at $$298 \\mathrm{~K}$$ is ___________. (Nearest integer)
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3246,
"subject": "Chemistry",
"question": "For a certain reaction at $300 \\mathrm{~K}, \\mathrm{~K}=10$, then $\\Delta \\mathrm{G}^{\\circ}$ for the same reaction is - ____________ $\\times 10^{-1} \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.
To determine the change in standard Gibbs free energy ($\\Delta G^{\\circ}$) for the reaction at a given temperature when the equilibrium constant ($K$) is known, we can use the following relationship:
\n\n
$$ \\Delta G^{\\circ} = -RT \\ln K $$\n
\n
Here, $R$ is the universal gas constant ($8.314 \\ J K^{-1} mol^{-1}$), $T$ is the temperature in Kelvin ($300 \\ K$), and $K$ is the equilibrium constant.
Therefore, $\\Delta G^{\\circ}$ for the reaction is approximately $-57.43914 \\times 10^{-1} \\ kJ mol^{-1}$ or $-57.4 \\times 10^{-1} \\ kJ mol^{-1}$ when rounded to three significant figures.
Consider the following reaction at $$298 \\mathrm{~K} \\cdot \\frac{3}{2} \\mathrm{O}_{2(g)} \\rightleftharpoons \\mathrm{O}_{3(g)} \\cdot \\mathrm{K}_{\\mathrm{P}}=2.47 \\times 10^{-29}$$. $$\\Delta_r G^{\\ominus}$$ for the reaction is _________ $$\\mathrm{kJ}$$. (Given $$\\mathrm{R}=8.314 \\mathrm{~JK}^{-1} \\mathrm{~mol}^{-1}$$)
",
"options": [
{
"text": "$$\\Delta \\mathrm{G}$$ is positive for a spontaneous reaction\n"
},
{
"text": "$$\\Delta \\mathrm{G}$$ is positive for a non-spontaneous reaction\n"
},
{
"text": "$$\\Delta \\mathrm{G}$$ is zero for a reversible reaction\n"
},
{
"text": "$$\\Delta \\mathrm{G}$$ is negative for a spontaneous reaction"
}
],
"answer": "$$\\Delta \\mathrm{G}$$ is positive for a spontaneous reaction\n",
"solution": "**Answer:** $$\\Delta \\mathrm{G}$$ is positive for a spontaneous reaction\n\n\n
The statement that is not correct among the given options is Option A.
\n\n
Option A states that $$\\Delta \\mathrm{G}$$ is positive for a spontaneous reaction, which is incorrect. The criterion for spontaneity in a chemical reaction is based on the Gibbs free energy change ($$\\Delta \\mathrm{G}$$) for the process. A spontaneous reaction is one that occurs without needing continuous input of energy from an external source. The correct relation is that for a spontaneous reaction, $$\\Delta \\mathrm{G}$$ is negative ($$\\Delta \\mathrm{G} < 0$$). When $$\\Delta \\mathrm{G}$$ is negative, it indicates that the process can occur spontaneously under the given conditions.
\n\n
Option B accurately states that $$\\Delta \\mathrm{G}$$ is positive for a non-spontaneous reaction. A positive value of $$\\Delta \\mathrm{G}$$ ($$\\Delta \\mathrm{G} > 0$$) indicates that the reaction is not spontaneous under the given conditions and requires external energy to proceed.
\n\n
Option C is correct in stating that $$\\Delta \\mathrm{G}$$ is zero for a reversible reaction at equilibrium. When a reaction is at equilibrium, it has reached a state where the forward and reverse reactions occur at equal rates, and there is no net change in the composition of the system. At this point, the Gibbs free energy is at its minimum for the given conditions, and $$\\Delta \\mathrm{G} = 0$$, indicating that the system is in a state of maximum stability and no further net change can occur without the input or removal of energy.
\n\n
Option D correctly states that $$\\Delta \\mathrm{G}$$ is negative for a spontaneous reaction. As explained previously, a negative Gibbs free energy change signifies that a process or reaction can proceed spontaneously in the direction written.
When $$\\Delta \\mathrm{H}_{\\mathrm{vap}}=30 \\mathrm{~kJ} / \\mathrm{mol}$$ and $$\\Delta \\mathrm{S}_{\\mathrm{vap}}=75 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$$, then the temperature of vapour, at one atmosphere is _________ K.
To find the temperature of vaporization at one atmosphere, we can use the Clausius-Clapeyron equation, which relates the enthalpy of vaporization ($$\\Delta H_{vap}$$) to the change in entropy ($$\\Delta S_{vap}$$) during the phase transition at a particular temperature (T). The relationship can be simplified under the assumption that both the enthalpy of vaporization and the entropy change of vaporization are constant with temperature to the form:
\n\n
$$\\Delta H_{vap} = T \\cdot \\Delta S_{vap}$$
\n\n
This equation states that the enthalpy change of vaporization is equal to the product of the temperature at which the phase change occurs and the entropy change of vaporization. We rearrange this equation to solve for the temperature (T):
\n\n
$$T = \\frac{\\Delta H_{vap}}{\\Delta S_{vap}}$$
\n\n
However, it's crucial to ensure the units are consistent. Given that $$\\Delta H_{vap}$$ is in kJ/mol and $$\\Delta S_{vap}$$ is in J/(mol·K), we need to convert $$\\Delta H_{vap}$$ from kJ/mol to J/mol to match units:
For the reaction at $$298 \\mathrm{~K}, 2 \\mathrm{~A}+\\mathrm{B} \\rightarrow \\mathrm{C}, \\Delta \\mathrm{H}=400 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$ and $$\\Delta S=0.2 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$$. The reaction will become spontaneous above __________ $$\\mathrm{K}$$.
To determine the temperature above which the reaction $$2A+B \\rightarrow C$$ becomes spontaneous, we can use the Gibbs free energy equation:
\n\n
$$\\Delta G = \\Delta H - T\\Delta S$$
\n\n
The reaction becomes spontaneous when $$\\Delta G$$ is negative. Therefore, we need to find the temperature at which $$\\Delta G$$ changes from positive to negative. We set $$\\Delta G$$ to zero to find the threshold temperature:
\n\n
$$0 = \\Delta H - T\\Delta S$$
\n\n
Substituting the given values of $$\\Delta H = 400 \\, \\text{kJ mol}^{-1} = 400,000 \\, \\text{J mol}^{-1}$$ (since 1 kJ = 1000 J) and $$\\Delta S = 0.2 \\, \\text{kJ mol}^{-1} K^{-1} = 200 \\, \\text{J mol}^{-1} K^{-1}$$, we get:
Therefore, the reaction will become spontaneous above $$2000 \\, \\text{K}$$. This means that at temperatures higher than 2000 K, the reaction tends towards product formation without the need for external energy to drive the process.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3252,
"subject": "Chemistry",
"question": "A heat engine absorbs heat Q1 at temperature T1 and heat Q2 at temperature T2. Work done by the engine is J (Q1 + Q2). This data :",
"options": [
{
"text": "violates 1st law of thermodynamics"
},
{
"text": "violates 1st law of thermodynamics if Q1 is +ve"
},
{
"text": "violates 1st law of thermodynamics if Q1 is -ve"
},
{
"text": "does not violates 1st law of thermodynamics"
}
],
"answer": "violates 1st law of thermodynamics",
"solution": "**Answer:** violates 1st law of thermodynamics\n\nAccording to first law of thermodynamics energy can neither be created nor destroyed although it can be converted from one form to another. \n
NOTE : Carnot cycle is based upon this principle but during the conversion of heat into work some mechanical energy is always converted to other form of energy hence this data violates $$1$$ st. law of thermodynamics.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3253,
"subject": "Chemistry",
"question": "The internal energy change when a system goes from state A to B is 40 kJ/mole. If the system goes from A to B by a reversible path and returns to state A by an irreversible path what would be the net change in internal energy?",
"options": [
{
"text": "> 40 kJ"
},
{
"text": "< 40 kJ"
},
{
"text": "Zero"
},
{
"text": "40 kJ"
}
],
"answer": "Zero",
"solution": "**Answer:** Zero\n\nFor a cyclic process the net change in the internal energy is zero because the change in internal energy does not depend on the path.\n
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3254,
"subject": "Chemistry",
"question": "An ideal gas expands in volume from 1$$\\times$$10-3 m3 to 1 $$\\times$$ 10-2 m3 at 300 K against a constant pressure of 1$$\\times$$105 Nm-2. The work done is :",
"options": [
{
"text": "-900 J "
},
{
"text": "900 kJ "
},
{
"text": "270 kJ "
},
{
"text": "-900 kJ "
}
],
"answer": "-900 J ",
"solution": "**Answer:** -900 J \n\n$$w = - P\\Delta V = - {10^{ - 5}}\\left( {1 \\times {{10}^{ - 2}} - 1 \\times {{10}^{ - 3}}} \\right)$$ \n
$$ = - 900J$$ ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3255,
"subject": "Chemistry",
"question": "An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti is the\ninitial temperature and Tf is the final temperature, which of the following statements is correct?",
"options": [
{
"text": "(Tf)irrev > (Tf)rev"
},
{
"text": "(Tf)rev = (Tf)irrev"
},
{
"text": "Tf > Ti\n for reversible process but Tf = Ti\n for irreversible process "
},
{
"text": "Tf = Ti for both reversible and irreversible processes "
}
],
"answer": "(Tf)irrev > (Tf)rev",
"solution": "**Answer:** (Tf)irrev > (Tf)rev\n\nNOTE : In a reversible process the work done is greater than in irreversible process. Hence the heat absorbed in reversible process would be greater than in the latter case. So,\n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $${T_f}\\left( {rev.} \\right) < {T_f}\\left( {irr.} \\right)$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3256,
"subject": "Chemistry",
"question": "A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant\ntemperature of 37.0oC. As it does so, it absorbs 208J of heat. The values of q and w for the process will be : \n(R = 8.314 J/mol K) ( l n 7.5 = 2.01)",
"options": [
{
"text": "q = – 208 J, w = – 208 J "
},
{
"text": "q = – 208 J, w = + 208 J"
},
{
"text": "q = + 208 J, w = + 208 J"
},
{
"text": "q = + 208 J, w = – 208 J "
}
],
"answer": "q = + 208 J, w = – 208 J ",
"solution": "**Answer:** q = + 208 J, w = – 208 J \n\nBy Ist\n law of thermodynamics, q = $$\\Delta $$E – W\n
At const T, ∆E = 0\n
q = – W\n
Heat absorbed = 208 J\n
$$ \\therefore $$ q = +208 J\n
W = – 208 J ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3257,
"subject": "Chemistry",
"question": "If 100 mole of H2O2 decompose at 1 bar and 300 K, the work done (kJ) by one mole of O2(g) as it expands against 1 bar pressure is :\n
For reaction : 2H2O2(1) $$\\rightleftharpoons$$ 2H2O(1) + O2(g)
\n
Given, p = 1 bar, T = 300 K, n = 100 mol, R = 8.3 J K$$-$$ mol$$-$$1
\n
We know that
\n
w = p$$\\Delta$$V = nRT
\n
= 100 $$\\times$$ 8.3 $$\\times$$ 300
\n
= 249000 J or 249 kJ
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3258,
"subject": "Chemistry",
"question": "$$\\Delta $$U is equal to :",
"options": [
{
"text": "Isobaric work"
},
{
"text": "Adiabatic work "
},
{
"text": "Isothermal work "
},
{
"text": "Isochoric work "
}
],
"answer": "Adiabatic work ",
"solution": "**Answer:** Adiabatic work \n\nFrom 1st law of thermodynamics\n
ΔU = q + w\n
For adiabatic process :\n
q = 0\n
$$ \\therefore $$ ΔU = w\n
So change in internal energy (ΔU) is equal to adiabatic work.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3259,
"subject": "Chemistry",
"question": "A gas undergoes change from state A to state B. In this process, the heat absorbed\nand work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to\nA by another process during which 3 J of heat is evolved. In this reverse process of\nB to A :",
"options": [
{
"text": "10 J of the work will be done by the gas. "
},
{
"text": "6 J of the work will be done by the gas."
},
{
"text": "10 J of the work will be done by the surrounding on gas.\n"
},
{
"text": "6 J of the work will be done by the surrounding on gas."
}
],
"answer": "6 J of the work will be done by the surrounding on gas.",
"solution": "**Answer:** 6 J of the work will be done by the surrounding on gas.\n\n
From first law of thermodynamics, we have
\n
$$\\Delta$$U = q + w
\n
$$\\bullet$$ For state A to B : q = +5 J, w = $$-$$8 J and
\n
$$\\Delta$$UAB = 5 + ($$-$$8) = $$-$$3 J
\n
$$\\bullet$$ For state B to A : q = $$-$$3 J
\n
Since the internal energy is a state function, we have $$\\Delta$$UBA = $$-$$ $$\\Delta$$UAB. Therefore,
\n
$$\\Delta$$UBA = q + w
\n
3 = $$-$$3 + w $$\\Rightarrow$$ w = 6 J
\n
Positive value of work indicates that 6 J of work is done by the surrounding on gas.
$$ \\therefore $$ $$\\Delta $$(PV) = nR$$\\Delta $$T = 5 $$ \\times $$ 8 $$ \\times $$ 100 = 4 kJ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3261,
"subject": "Chemistry",
"question": "An ideal gas is allowed to expand form 1 L to 10 L against a constant external pressure of I bar. The work\ndone in kJ is :",
"options": [
{
"text": "+10.0"
},
{
"text": "–0.9"
},
{
"text": "– 2.0"
},
{
"text": "– 9.0"
}
],
"answer": "–0.9",
"solution": "**Answer:** –0.9\n\nThis is an irreverseable process as gas is expanding against a constant external process.\n
= - 0.9 kJ\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3262,
"subject": "Chemistry",
"question": "During compression of a spring the work done\nis 10kJ and 2kJ escaped to the surroundings as\nheat. The change in internal energy, $$\\Delta $$U(inkJ)\nis :",
"options": [
{
"text": "- 12"
},
{
"text": "8"
},
{
"text": "- 8"
},
{
"text": "12"
}
],
"answer": "8",
"solution": "**Answer:** 8\n\nHere heat is released so q is negative. \n
$$ \\therefore $$ q = - 2 kJ\n
Work done on the system, w = 10 kJ\n
From first law of thermodynamics,\n
$$\\Delta $$U = q + w = -2 + 10 = 8 kJ\n",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 3263,
"subject": "Chemistry",
"question": "Which one of the following equations does not correctly represent the first law of thermodynamics\nfor the given processes involving an ideal gas? (Assume non-expansion work is zero)",
"options": [
{
"text": "Adiabatic process : $$\\Delta $$U= – w"
},
{
"text": "Cyclic process : q = –w"
},
{
"text": "Isochoric process : $$\\Delta $$U= q"
},
{
"text": "Isothermal process : q = – w"
}
],
"answer": "Adiabatic process : $$\\Delta $$U= – w",
"solution": "**Answer:** Adiabatic process : $$\\Delta $$U= – w\n\nFrom 1st law of thermodynamics we know,\n
$$\\Delta $$U = q + W\n
Option A :\n
In adiabatic process exchage of heat = 0\n
$$ \\therefore $$ q = 0\n
$$ \\therefore $$ From 1st law of thermodynaics, $$\\Delta $$U = W\n
So option A is wrong.\n
Option B :\n
U is a state function. In cyclic process, initial state and final state both are same. So change in all the state function in cyclic process will be zero.\n
$$ \\therefore $$ $$\\Delta $$U = 0\n
$$ \\therefore $$ From 1st law of thermodynaics, \n
q + W = 0 $$ \\Rightarrow $$ q = -W\n
So option B is correct..\n
Option C :\n
In isochoric process volume (V) is constant. So dV = 0.\n
We know, W = $$ - \\int {{P_{ex}}} dV$$\n
$$ \\therefore $$ W = 0\n
$$ \\therefore $$ From 1st law of thermodynaics, \n
$$\\Delta $$U = q\n
So option C is correct.\n
Option D :\n
In isothermal process temerature (T) is constant. So dT = 0.\n
We know, $$\\Delta $$U = nCvdT\n
$$ \\therefore $$ $$\\Delta $$U = 0\n
$$ \\therefore $$ From 1st law of thermodynaics, \n
q + W = 0 $$ \\Rightarrow $$ q = -W\n
So option D is correct.\n",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 3264,
"subject": "Chemistry",
"question": "An ideal gas undergoes isothermal compression from 5m3 to 1 m3 against a constant external pressure of 4 Nm–2. Heat released in this process is used to increase the temperature of 1 mole of Al. If molar heat capacity of Al is 24 J mol–1 K–1, the temperature of Al increases by :",
"options": [
{
"text": "$${2 \\over 3}K$$"
},
{
"text": "$${3 \\over 2}K$$"
},
{
"text": "1 K"
},
{
"text": "2 K"
}
],
"answer": "$${2 \\over 3}K$$",
"solution": "**Answer:** $${2 \\over 3}K$$\n\nWork done on isothermal irreversible for ideal gas \n
$$\\Delta $$T = $${2 \\over 3}$$K",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3265,
"subject": "Chemistry",
"question": "At constant volume, 4 mol of an ideal gas when\nheated from 300 K to 500K changes its internal\nenergy by 5000 J. The molar heat capacity at\nconstant volume is _______.",
"options": [],
"answer": "6.25",
"solution": "**Answer:** 6.25\n\n$$\\Delta $$U = nCv$$\\Delta $$T\n
$$ \\Rightarrow $$ 5000 = 4 × Cv(500 – 300)\n
$$ \\Rightarrow $$ Cv = 6.25 JK–1mol–1",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3266,
"subject": "Chemistry",
"question": "Five moles of an ideal gas at 1 bar and 298 K\nis expanded into vacuum to double the volume.\nThe work done is :",
"options": [
{
"text": "Zero"
},
{
"text": "-RT $$\\ln {{{V_2}} \\over {{V_1}}}$$"
},
{
"text": "CV (T2 – T1)"
},
{
"text": "– RT (V2 – V1)"
}
],
"answer": "Zero",
"solution": "**Answer:** Zero\n\nAs the expansion is done in vacuum that is in absence of pext so\npext = 0\n
$$ \\therefore $$ W = - pext$$\\Delta $$V\n
= 0",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3267,
"subject": "Chemistry",
"question": "The reaction of cyanamide, NH2CN(s) with oxygen was run in a bomb calorimeter and $$\\Delta$$U was found to be $$-$$742.24 kJ mol$$-$$1. The magnitude of $$\\Delta$$H298 for the reaction
$$ = -741$$ kJ/mol",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3268,
"subject": "Chemistry",
"question": "Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against at constant external pressure 4.3 MPa. The heat transferred in this process is _________ kJ mol$$-$$1. (Rounded off to the nearest integer) [Use R = 8.314 J mol$$-$$1K$$-$$1]",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\n
The gas performs isothermal irreversible work (W).
\n
where, $$\\Delta$$U = 0 (change in internal energy)
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3269,
"subject": "Chemistry",
"question": "At 25$$^\\circ$$C, 50 g of iron reacts with HCl to form FeCl2. The evolved hydrogen gas expands against a constant pressure of 1 bar. The work done by the gas during this expansion is _________ J. (Round off to the Nearest Integer).
[Given : R = 8.314 J mol$$-$$1 K$$-$$1. Assume, hydrogen is an ideal gas] [Atomic mass of Fe is 55.85 u]",
"options": [],
"answer": "2218",
"solution": "**Answer:** 2218\n\n$$Fe + 2HCl \\to FeC{l_2} + {H_2}$$
$${{50} \\over {55.85}}moles$$\n
Moles of Fe = $${{50} \\over {55.85}}moles$$ = Moles of H2\n
No. of H2 produced $$ = {{50} \\over {55.85}}moles$$
Nearest integer = 2218",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 3270,
"subject": "Chemistry",
"question": "A system does 200 J of work and at the same time absorbs 150 J of heat. The magnitude of the change in internal energy is ____________ J. (Nearest integer)",
"options": [],
"answer": "50",
"solution": "**Answer:** 50\n\nw = $$-$$200 J, q = +150 : $$\\Delta$$U = q + w
2.2 g of nitrous oxide (N2O) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1 mL to 167.75 mL. The change in internal energy of the process, $$\\Delta$$U is '$$-$$x' J. The value of 'x' is ________. [nearest integer]
\n
(Given : atomic mass of N = 14 g mol$$-$$1 and of O = 16 g mol$$-$$1. Molar heat capacity of N2O is 100 J K$$-$$1 mol$$-$$1)
4.0 L of an ideal gas is allowed to expand isothermally into vacuum until the total volume is 2.0 L. The amount of heat absorbed in this expansion is ____________ L atm.
A fish swimming in water body when taken out from the water body is covered with a film of water of weight 36 g. When it is subjected to cooking at 100$$^\\circ$$C, then the internal energy for vaporization in kJ mol$$-$$1 is ___________. [nearest integer]
\n
[Assume steam to be an ideal gas. Given $$\\Delta$$vapH$$^\\Theta $$ for water at 373 K and 1 bar is 41.1 kJ mol$$-$$1 ; R = 8.31 J K$$-$$1 mol$$-$$1]
The molar heat capacity for an ideal gas at constant pressure is $$20.785 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$$. The change in internal energy is $$5000 \\mathrm{~J}$$ upon heating it from $$300 \\mathrm{~K}$$ to $$500 \\mathrm{~K}$$. The number of moles of the gas at constant volume is ____________. [Nearest integer] (Given: $$\\mathrm{R}=8.314 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$$)
A gas (Molar mass = 280 $$\\mathrm{~g} \\mathrm{~mol}^{-1}$$) was burnt in excess $$\\mathrm{O}_{2}$$ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from $$298.0 \\mathrm{~K}$$ to $$298.45$$ $$\\mathrm{K}$$. If the heat capacity of calorimeter is $$2.5 \\mathrm{~kJ} \\mathrm{~K}^{-1}$$ and enthalpy of combustion of gas is $$9 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$ then amount of gas burnt is _____________ g. (Nearest Integer)
\n$\\therefore$ Mass of gas burnt $=\\frac{1.125}{9} \\times 280=35 \\mathrm{~g}$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3276,
"subject": "Chemistry",
"question": "1 mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27^{\\circ} \\mathrm{C}$. The work done is $3 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. The final temperature of the gas is ________ $\\mathrm{K}$ (Nearest integer).
When 2 litre of ideal gas expands isothermally into vacuum to a total volume of 6 litre, the change in internal energy is ____________ J. (Nearest integer)
An athlete is given 100 g of glucose (C$$_6$$H$$_{12}$$O$$_6$$) for energy. This is equivalent to 1800kJ of energy. The 50% of this energy gained is utilized by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is ____________ g (Nearest integer)
\n
Assume that there is no other way of consuming stored energy.
\n
Given : The enthalpy of evaporation of water is 45 kJ mol$$^{-1}$$
\n
Molar mass of C, H & O are 12, 1 and 16 g mol$$^{-1}$$.
\nExtra energy used to convert $\\mathrm{H}_2 \\mathrm{O(l)}$ into $\\mathrm{H}_2 \\mathrm{O}(\\mathrm{l})$ into $\\mathrm{H}_2 \\mathrm{O}(\\mathrm{g})$
When a $$60 \\mathrm{~W}$$ electric heater is immersed in a gas for 100 s in a constant volume container with adiabatic walls, the temperature of the gas rises by $$5^{\\circ} \\mathrm{C}$$. The heat capacity of the given gas is ___________ $$\\mathrm{J} \\mathrm{K}^{-1}$$ (Nearest integer)
So, the heat capacity of the given gas is approximately 1200 J/K.
\n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3280,
"subject": "Chemistry",
"question": "Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following :",
"options": [
{
"text": "$\\mathrm{q}=0, \\Delta \\mathrm{T}=0, \\mathrm{w}=0$"
},
{
"text": "$\\mathrm{q}=0, \\Delta \\mathrm{T} \\neq 0, \\mathrm{w}=0$"
},
{
"text": "$\\mathrm{q} \\neq 0, \\Delta \\mathrm{T}=0, \\mathrm{w}=0$"
},
{
"text": "$\\mathrm{q}=0, \\Delta \\mathrm{T}<0, \\mathrm{w} \\neq 0$"
}
],
"answer": "$\\mathrm{q}=0, \\Delta \\mathrm{T}=0, \\mathrm{w}=0$",
"solution": "**Answer:** $\\mathrm{q}=0, \\Delta \\mathrm{T}=0, \\mathrm{w}=0$\n\n
Let's analyze what happens during a free (or unrestrained) expansion of an ideal gas under adiabatic conditions:
\n\n
\n\n
Adiabatic process : By definition, in an adiabatic process, no heat is transferred between the system and its surroundings. Thus, $ q = 0 $.
\n\n
Free expansion : In a free expansion, the gas expands into a vacuum or against zero external pressure. Since work ($ w $) is defined as the product of pressure ($ P $) and volume change ($ \\Delta V $), and since the external pressure is zero, we conclude that $ w = 0 $.
\n\n
Temperature change ($ \\Delta T $) : Since work is zero and no heat is transferred, the first law of thermodynamics tells us that the change in internal energy ($ \\Delta U $) is zero as well. For an ideal gas, internal energy is a function of temperature only; thus, if the internal energy does not change, the temperature does not change. Therefore, $ \\Delta T = 0 $.
\n\n
\n\n
Based on the above points, the free expansion of an ideal gas under adiabatic conditions is characterized by no heat transfer ($ q = 0 $), no work done ($ w = 0 $), and no change in temperature ($ \\Delta T = 0 $). Therefore, the correct option is:
If three moles of an ideal gas at $$300 \\mathrm{~K}$$ expand isothermally from $$30 \\mathrm{~dm}^3$$ to $$45 \\mathrm{~dm}^3$$ against a constant opposing pressure of $$80 \\mathrm{~kPa}$$, then the amount of heat transferred is _______ J.
The process involves an ideal gas expanding isothermally, meaning temperature ($$T$$) remains constant. In an isothermal process for an ideal gas, the change in internal energy ($$\\Delta U$$) is zero:
\n
$$\\Delta U = 0$$ because temperature is constant.
\n
According to the first law of thermodynamics, $$\\Delta U = Q + W$$, where $$Q$$ is the heat transferred to the system and $$W$$ is the work done by the system. Since $$\\Delta U = 0$$ in an isothermal process, $$Q = -W$$.
\n
The work done by the gas during the isothermal expansion against a constant external pressure ($$P_{ext}$$) is given by:
Therefore, the amount of heat transferred ($$Q$$) is $$1200$$ J, taking into consideration the sign convention that work done by the system is negative.
If 5 moles of an ideal gas expands from $$10 \\mathrm{~L}$$ to a volume of $$100 \\mathrm{~L}$$ at $$300 \\mathrm{~K}$$ under isothermal and reversible condition then work, $$\\mathrm{w}$$, is $$-x \\mathrm{~J}$$. The value of $$x$$ is __________.
Three moles of an ideal gas are compressed isothermally from $$60 \\mathrm{~L}$$ to $$20 \\mathrm{~L}$$ using constant pressure of $$5 \\mathrm{~atm}$$. Heat exchange $$\\mathrm{Q}$$ for the compression is - _________ Lit. atm.
$$\\Delta_{\\text {vap }} \\mathrm{H}^{\\ominus}$$ for water is $$+40.79 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$ at 1 bar and $$100^{\\circ} \\mathrm{C}$$. Change in internal energy for this vapourisation under same condition is ________ $$\\mathrm{kJ} \\mathrm{~mol}^{-1}$$. (Integer answer) (Given $$\\mathrm{R}=8.3 \\mathrm{~JK}^{-1} \\mathrm{~mol}^{-1}$$)
To find the change in internal energy for the vaporization of water under the given conditions, we'll use the following relationship between enthalpy change ($$\\Delta_{\\text{vap}} H$$) and internal energy change ($$\\Delta_{\\text{vap}} U$$):
\n\n
$$\\Delta_{\\text{vap}} H = \\Delta_{\\text{vap}} U + P \\Delta V$$
\n\n
For vaporization, the change in volume ($$\\Delta V$$) can be approximated by considering the volume of the vapor because the volume of liquid water is relatively small compared to the volume of the vapor.
\n\n
The ideal gas law gives us:
\n\n
$$P V = n R T$$
\n\n
Since we are dealing with 1 mole of water:
\n\n
$$V = \\frac{R T}{P}$$
\n\n
Plugging this into the enthalpy change equation, we get:
\n\n
$$\\Delta_{\\text{vap}} H = \\Delta_{\\text{vap}} U + P \\left( \\frac{R T}{P} \\right)$$
\n\n
This simplifies to:
\n\n
$$\\Delta_{\\text{vap}} H = \\Delta_{\\text{vap}} U + R T$$
\n\n
Rearranging for $$\\Delta_{\\text{vap}} U$$:
\n\n
$$\\Delta_{\\text{vap}} U = \\Delta_{\\text{vap}} H - R T$$
\n\n
Given:
\n\n
$$\\Delta_{\\text{vap}} H = 40.79 \\, \\text{kJ mol}^{-1}$$ (or 40790 J/mol)
An ideal gas, $$\\overline{\\mathrm{C}}_{\\mathrm{v}}=\\frac{5}{2} \\mathrm{R}$$, is expanded adiabatically against a constant pressure of 1 atm untill it doubles in volume. If the initial temperature and pressure is $$298 \\mathrm{~K}$$ and $$5 \\mathrm{~atm}$$, respectively then the final temperature is _________ $$\\mathrm{K}$$ (nearest integer).
\n
[$$\\overline{\\mathrm{c}}_{\\mathrm{v}}$$ is the molar heat capacity at constant volume]
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3286,
"subject": "Chemistry",
"question": "The heat required to raise the temperature of body by 1 K is called :",
"options": [
{
"text": "specific heat"
},
{
"text": "thermal capacity"
},
{
"text": "water equivalent"
},
{
"text": "none of these"
}
],
"answer": "thermal capacity",
"solution": "**Answer:** thermal capacity\n\nThe heat required to raise the temperature of body by $$1K$$ is called thermal capacity or heat capacity. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3287,
"subject": "Chemistry",
"question": "An ideal gas undergoes isothermal expansion at constant pressure. During\nthe process :",
"options": [
{
"text": "enthalpy increases but entropy decreases."
},
{
"text": "enthalpy remains constant but entropy increases."
},
{
"text": "enthalpy decreases but entropy increases."
},
{
"text": "Both enthalpy and entropy remain constant.\n"
}
],
"answer": "enthalpy remains constant but entropy increases.",
"solution": "**Answer:** enthalpy remains constant but entropy increases.\n\n
In an isothermal expansion process, the temperature of the system remains constant throughout the process. Since, for an ideal gas, U depends only on temperature, we have $$\\Delta$$U = 0. The enthalpy change of the system in isothermal expansion is also zero as
",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 3288,
"subject": "Chemistry",
"question": "Among the following, the set of parameters that\nrepresents path function, is : \n(A) q + w \n(B) q \n(C) w \n(D) H–TS ",
"options": [
{
"text": "(B) and (C)"
},
{
"text": "(A) and (D)"
},
{
"text": "(B), (C) and (D)"
},
{
"text": "(A), (B) and (C)"
}
],
"answer": "(B) and (C)",
"solution": "**Answer:** (B) and (C)\n\n(A) q + w = $$\\Delta $$E, state function\n
(B) q, Path function\n
(C) w, Path function\n
(D) H – TS = G, State function",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3289,
"subject": "Chemistry",
"question": "For one mole of an ideal gas, which of these\nstatements must be true?\n (a) U and H each depends only on temperature\n (b) Compressibility factor z is not equal to 1\n (c) CP, m – CV, m = R\n (d) dU = CVdT for any process ",
"options": [
{
"text": "(a), (c) and (d)"
},
{
"text": "(a) and (c)"
},
{
"text": "(c) and (d)"
},
{
"text": "(b), (c) and (d)"
}
],
"answer": "(a), (c) and (d)",
"solution": "**Answer:** (a), (c) and (d)\n\nFor 1 mole of ideal gas :\n
1. Both internal energy (U) and Enthalpy (H)\ndepends on temperature\n
2. Compressibility factor Z = 1\n
3. CP, m – CV, m = R\n
4. dU = CVdT for all process",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 3290,
"subject": "Chemistry",
"question": "
When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the magnitude of the maximum work obtained is __________ J. [nearest integer] (Given : R = 8.3 J K$$-$$1 mol$$-$$1 and log 2 = 0.3010)
Choose the most appropriate answer from the options given below :
",
"options": [
{
"text": "A and B only"
},
{
"text": "B and C only"
},
{
"text": "C and D only"
},
{
"text": "B and D only"
}
],
"answer": "B and C only",
"solution": "**Answer:** B and C only\n\n
",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nIntensive properties are independent of the amount of the substance. They don't change with the size or mass of a sample. Examples include temperature, pressure, density, and concentration.\n\n\n
Intensive properties :\n
1. Molarity : Molarity is a measure of concentration, defined as the number of moles of solute per liter of solution. It doesn't depend on the total volume of the solution but only on the ratio of the amount of solute to the amount of solution.\n\n
2. $$\\mathrm{E}^{\\theta}$$ cell (standard cell potential) : This is a measure of the potential difference between the anode and cathode in an electrochemical cell under standard conditions. It's independent of the amount of material in the cell.\n\n
3. Molar heat capacity : This is the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius (or Kelvin). It's an intrinsic property because it's defined per mole of substance.\n\n
4. Molar mass : Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's independent of the quantity of the substance because it's defined per mole.\n\n
On the other hand, extensive properties depend on the amount of the substance. They change when the size or mass of a sample changes. For instance, mass and volume are extensive properties.\n\n
Extensive properties :\n
1. Volume: The volume of a substance depends on the amount of that substance. For instance, one liter of water has half the volume of two liters of water.\n\n
2. Gibbs free energy change (∆G) : This is the maximum reversible work that a thermodynamic system can perform at constant temperature and pressure. It depends on the number of moles of reactants and products, so it's an extensive property.\n\n
3. Mole : The mole is a unit of measurement for amount of substance in the International System of Units (SI). It's a count of a very large number of particles, typically atoms or molecules. Like mass or volume, it's an extensive property because it depends on the quantity of the substance.\n\n",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3296,
"subject": "Chemistry",
"question": "
The heat of combustion of solid benzoic acid at constant volume is $$-321.30 \\mathrm{~kJ}$$ at $$27^{\\circ} \\mathrm{C}$$. The heat of combustion at constant pressure is $$(-321.30-x \\mathrm{R}) \\mathrm{~kJ}$$, the value of $$x$$ is __________.
$$\\begin{aligned}\n& \\Delta \\mathrm{H}=\\Delta \\mathrm{U}+\\Delta \\mathrm{n}_9 R T \\\\\n& \\Delta \\mathrm{H}=-321.30-\\frac{1}{2} \\mathrm{R} \\times 300\n\\end{aligned}$$
\n
So, $$x=150$$
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3297,
"subject": "Chemistry",
"question": "For the reactions \n2C + O2 $$\\to$$ 2CO2; $$\\Delta H$$ = -393 J \n2Zn + O2 $$\\to$$ 2ZnO; $$\\Delta H$$ = -412 J",
"options": [
{
"text": "carbon can oxidise Zn"
},
{
"text": "oxidation of carbon is not feasible"
},
{
"text": "oxidation of Zn is not feasible"
},
{
"text": "Zn can oxidise carbon"
}
],
"answer": "Zn can oxidise carbon",
"solution": "**Answer:** Zn can oxidise carbon\n\n$$\\Delta H$$ negative shows that the reaction is spontaneous. Higher negative value for $$Zn$$ shows that the reaction is more feasible. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3298,
"subject": "Chemistry",
"question": "If at 298 K the bond energies of C - H, C - C, C = C and H - H bonds are respectively 414, 347, 615 and 435 kJ/mol, the value of enthalpy change for the reaction \nH2C = CH2(g) + H2(g) $$\\to$$ H3C - CH3(g) at 298 K will be :",
"options": [
{
"text": "- 250 kJ"
},
{
"text": "+ 125 kJ"
},
{
"text": "- 125 kJ"
},
{
"text": "+ 250 kJ"
}
],
"answer": "- 125 kJ",
"solution": "**Answer:** - 125 kJ\n\n$$C{H_2} = C{H_2}\\left( g \\right) + {H_2}\\left( g \\right) \\to C{H_3} - C{H_3}$$\n
Enthalpy change $$=$$ Bond energy of reactants $$-$$ Bond energy of products.\n
$$ = - 125\\,kJ.$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3299,
"subject": "Chemistry",
"question": "The enthalpy change for a reaction does not depend upon :",
"options": [
{
"text": "use of different reactants for the same product"
},
{
"text": "the nature of intermediate reaction steps"
},
{
"text": "the differences in initial or final temperature of involved substances"
},
{
"text": "the physical states of reactants and products"
}
],
"answer": "the nature of intermediate reaction steps",
"solution": "**Answer:** the nature of intermediate reaction steps\n\nEnthalpy change for a reaction does not depend upon the nature of intermediate reaction steps.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3300,
"subject": "Chemistry",
"question": "The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol-1\nrespectively. The enthalpy of formation of carbon monoxide per mole is :",
"options": [
{
"text": "110.5 kJ"
},
{
"text": "-110.5 kJ"
},
{
"text": "-676.5 kJ"
},
{
"text": "676.5 kJ"
}
],
"answer": "-110.5 kJ",
"solution": "**Answer:** -110.5 kJ\n\n$$\\left( i \\right)\\,\\,\\,C + {O_2}\\rightleftharpoons C{O_2},$$\n
$$\\Delta H - \\Delta U = {1 \\over 2} \\times 8.314 \\times 298$$ \n
$$ = 1238.78\\,Jmo{l^{ - 1}}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3305,
"subject": "Chemistry",
"question": "The standard enthalpy of formation $$\\Delta _fH^o$$ at 298 K for methane, CH4(g), is –74.8 kJ mol–1. The additional information required to determine the average energy for C – H bond formation would be :",
"options": [
{
"text": "the dissociation energy of H2 and enthalpy of sublimation of carbon"
},
{
"text": "latent heat of vapourization of methane "
},
{
"text": "the first four ionization energies of carbon and electron gain enthalpy of hydrogen"
},
{
"text": "the dissociation energy of hydrogen molecule, H2"
}
],
"answer": "the dissociation energy of H2 and enthalpy of sublimation of carbon",
"solution": "**Answer:** the dissociation energy of H2 and enthalpy of sublimation of carbon\n\nThe standard enthalpy of formation of $$C{H_4}$$ is given by the equation : \n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$C\\left( s \\right) + 2{H_2}\\left( g \\right) \\to C{H_4}\\left( g \\right)$$ \n
In order to calculate average energy for C – H bond formation\nwe should know the following data.\n
C(graphite) $$ \\to $$ C(g) ; $$\\Delta H_f^o$$ = enthalpy of sublimation of carbon\n
H2(g) $$ \\to $$ 2H(g)\n ; $$\\Delta $$H = bond dissociation energy of H2\n ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3306,
"subject": "Chemistry",
"question": "The enthalpy changes for the following processes are listed below :
\nGiven that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of\nformation for ICl(g) is :",
"options": [
{
"text": "–14.6 kJ mol–1 "
},
{
"text": "–16.8 kJ mol–1"
},
{
"text": "+16.8 kJ mol–1"
},
{
"text": "+244.8 kJ mol–1"
}
],
"answer": "+16.8 kJ mol–1",
"solution": "**Answer:** +16.8 kJ mol–1\n\n$${{\\rm{I}}_2}\\left( s \\right) + C{l_2}\\left( g \\right) \\to 2{\\rm{I}}Cl\\left( g \\right)$$\n
$$\\Delta A = \\left[ {\\Delta {{\\rm{I}}_2}\\left( s \\right) \\to {l_2}\\left( g \\right) + \\Delta {H_{{\\rm I} - l}} + \\Delta {H_{C{\\rm I} - Cl}}} \\right] - $$\n
i.e., the correct answer is $$(b)$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3309,
"subject": "Chemistry",
"question": "On the basis of the following thermochemical data : ($$\\Delta _fG^oH^+_{(aq)}$$ = 0)
$$ = - 228.88\\,kJ$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3310,
"subject": "Chemistry",
"question": "The standard enthalpy of formation of NH3 is –46.0 kJ mol–1. If the enthalpy of formation of H2 from its atoms is –436 kJ mol–1 and that of N2 is –712 kJ mol–1, the average bond enthalpy of N–H bond in NH3 is :",
"options": [
{
"text": "–964 kJ mol–1"
},
{
"text": "+352 kJ mol–1"
},
{
"text": "+ 1056 kJ mol–1"
},
{
"text": "–1102 kJ mol–1"
}
],
"answer": "+352 kJ mol–1",
"solution": "**Answer:** +352 kJ mol–1\n\n$${N_2} + 3{H_2} \\to 2N{H_3}\\,$$\n
$$\\,\\Delta H = 2 \\times - 46.0\\,\\,kJ\\,mo{l^{ - 1}}$$ \n
Let $$x$$ be the bond enthalpy of $$N-H$$ bond then\n
[Note : Enthalpy of formation or bond formation enthalpy is given which is negative but the given reaction involves bond breaking hence values should be taken as positive. ]\n
$$\\Delta H = \\sum \\, $$ Bond energies of products\n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, - \\sum \\, $$ Bond energies of reactants\n
$$\\therefore\\,\\,\\,$$ $$\\left| {\\Delta H} \\right|$$ > $$\\left| {\\Delta U} \\right|$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3314,
"subject": "Chemistry",
"question": "The enthalpy change on freezing of 1 mol of water at 5oC to ice at −5oC is :\n
Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct?",
"options": [
{
"text": "z = x + y "
},
{
"text": "x = y + z "
},
{
"text": "x = y – z "
},
{
"text": "y = 2z – x"
}
],
"answer": "x = y + z ",
"solution": "**Answer:** x = y + z \n\nIn reaction (i), the product is CO2\n
If we add reaction (ii) and (iii) we get,\n C + O2 $$ \\to $$ CO2\n here also product is CO2 from same reactant C and O2.\n
So according to hess law, we can say\n x = y + z",
"topic": "Discrete Mathematics",
"subtopic": "Other"
},
{
"id": 3319,
"subject": "Chemistry",
"question": "For silver, Cp(J K–1 mol–1) = 23 +0.01 T. If the temperature (T) of 3 moles of silver is raised from 300\nK to 1000 K at 1 atm pressure, the value of $$\\Delta H$$ will be close to :",
"options": [
{
"text": "62 KJ"
},
{
"text": "16 KJ"
},
{
"text": "13 KJ"
},
{
"text": "21 KJ"
}
],
"answer": "62 KJ",
"solution": "**Answer:** 62 KJ\n\nGive that,\n
$$ \\therefore $$ $$\\Delta $$H - $$\\Delta $$U = - 4RT",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3321,
"subject": "Chemistry",
"question": "Enthalpy of sublimation of iodine is 24 cal g–1\n at 200 oC. If specific heat of I2(s) and l2 (vap) are 0.055 and\n0.031 cal g–1K\n–1\nrespectively, then enthalpy of sublimation of iodine at 250 oC in cal g–1\n is :",
"options": [
{
"text": "2.85"
},
{
"text": "22.8"
},
{
"text": "11.4"
},
{
"text": "5.7"
}
],
"answer": "22.8",
"solution": "**Answer:** 22.8\n\nKirchoff's Equation :\n
rest of all endothermic process.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3325,
"subject": "Chemistry",
"question": "The heat of combustion of ethanol into carbon\ndioxide and water is – 327 kcal at constant\npressure. The heat evolved (in cal) at constant\nvolume and 27oC (if all gases behave ideally) is\n(R = 2 cal mol–1 K–1) ________.",
"options": [],
"answer": "326400",
"solution": "**Answer:** 326400\n\nC2H5OH (l) + 3O2(g) $$ \\to $$ 2CO2(g) + 3H2O(l)\n
$$ \\therefore $$ Heat evolved = 326400 cal.",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3326,
"subject": "Chemistry",
"question": "If enthalpy of atomisation for Br2(1) is x kJ/mol\nand bond enthalpy for Br2 is y kJ/mol, the\nrelation between them :",
"options": [
{
"text": "does not exist"
},
{
"text": "is x < y"
},
{
"text": "is x > y"
},
{
"text": "is x = y"
}
],
"answer": "is x > y",
"solution": "**Answer:** is x > y\n\n\n
$$ \\Rightarrow $$ x $$>$$ y",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3327,
"subject": "Chemistry",
"question": "The standard heat of formation $$\\left( {{\\Delta _f}H_{298}^0} \\right)$$ of ethane (in kj/mol), if the heat of combustion of\nethane, hydrogen and graphite are - 1560, -393.5 and -286 Kj/mol, respectively is :",
"options": [],
"answer": "-192.5",
"solution": "**Answer:** -192.5\n\n2C(graphite)+ 3H(g) $$ \\to $$ C2H6(g)\n
= -572 –1180.5+1560 = -192.5 kJ/mole",
"topic": "Discrete Mathematics",
"subtopic": "Other"
},
{
"id": 3328,
"subject": "Chemistry",
"question": "The ionization enthalpy of Na+ formation from Na(g) is 495.8 kJ mol$$-$$1, while the electron gain enthalpy of Br is $$-$$325.0 kJ mol$$-$$1. Given the lattice enthalpy of NaBr is $$-$$728.4 kJ mol$$-$$1. The energy for the formation of NaBr ionic solid is ($$-$$) ____________ $$\\times$$ 10$$-$$1 kJ mol$$-$$1.",
"options": [],
"answer": "5576",
"solution": "**Answer:** 5576\n\n$$Na(s)\\buildrel {} \\over\n \\longrightarrow N{a^ + }(g)$$
$$ = - 5576 \\times {10^{ - 1}}$$ kJ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3329,
"subject": "Chemistry",
"question": "An exothermic reaction X $$ \\to $$ Y has an activation energy 30 kJ mol$$-$$1. If energy change $$\\Delta$$E during the reaction is $$-$$20 kJ, then the activation energy for the reverse reaction in kJ is ___________. (Integer answer)",
"options": [],
"answer": "50",
"solution": "**Answer:** 50\n\nX $$ \\to $$ Y\n
$$\\Delta $$E = (Ea)f\n – (Ea)b\n
$$ \\Rightarrow $$ – 20 = 30 – (Ea)b\n
$$ \\Rightarrow $$ (Ea)b = 50 kJ",
"topic": "Algebra",
"subtopic": "Abstract Algebra"
},
{
"id": 3330,
"subject": "Chemistry",
"question": "The average S-F bond energy in kJ mol$$-$$1 of SF6 is _____________. (Rounded off to the nearest integer)
[Given : The values of standard enthalpy of formation of SF6(g), S(g) and F(g) are - 1100, 275 and 80 kJ mol$$-$$1 respectively.]",
"options": [],
"answer": "309",
"solution": "**Answer:** 309\n\n So, $\\Delta_f H^{\\circ}[S, g]+6 \\times \\Delta_f H^{\\circ}[F, g]=\\Delta_f H^{\\circ}\\left[SF_6, g\\right]$\n$+ 6 \\times E_{S-F}$
\n$\\left[\\therefore E_{S-F}=\\right.$ Average $S-\\mathrm{F}$ bond energy in $\\left.\\mathrm{SF}_6\\right]$
\n$275+6 \\times 80=-1100+6 \\times E_{S-F}$
\n$\\Rightarrow E_{S-F}=\\frac{275+6 \\times 80+1100}{6}$\n$=309.16 \\mathrm{~kJ} \\mathrm{~mol}^{-1}=309 \\mathrm{~kJ} \\mathrm{~mol} \\mathrm{~m}^{-1}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3331,
"subject": "Chemistry",
"question": "The standard enthalpies of formation of Al2O3 and CaO are $$-$$1675 kJ mol-1 and $$-$$635 kJ mol$$-$$1 respectively.
For the reaction
3CaO + 2Al $$ \\to $$ 3Ca + Al2O3 the standard reaction enthalpy $$\\Delta$$rH0 = _________ kJ.
(Round off to the Nearest Integer)",
"options": [],
"answer": "230",
"solution": "**Answer:** 230\n\n$$3CaO + 2Al\\mathrel{\\mathop{\\kern0pt\\longrightarrow}\n\\limits_{}} 3Ca + A{l_2}{O_3}$$
$$ = 128$$ KJ/ mol",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3333,
"subject": "Chemistry",
"question": "If the standard molar enthalpy change for combustion of graphite powder is $$-$$2.48 $$\\times$$ 102 kJ mol$$-$$1, the amount of heat generated on combustion of 1 g of graphite powder is ___________ kJ. (Nearest integer)",
"options": [],
"answer": "21",
"solution": "**Answer:** 21\n\n1 mol graphite = 12 gm C
For 1 g of graphite = $${{248} \\over {12}}$$ = 20.67 kJ/gm heat evolved.",
"topic": "Discrete Mathematics",
"subtopic": "Other"
},
{
"id": 3334,
"subject": "Chemistry",
"question": "At 298 K, the enthalpy of fusion of a solid (X) is 2.8 kJ mol$$-$$1 and the enthalpy of vaporisation of the liquid (X) is 98.2 kJ mol$$-$$1. The enthalpy of sublimation of the substance (X) in kJ mol$$-$$1 is _____________. (in nearest integer)",
"options": [],
"answer": "101",
"solution": "**Answer:** 101\n\n$$\\Delta {H_{sub}} = \\Delta {H_{fus.}} + \\Delta {H_{vap.}}$$
= 2.8 + 98.2
= 101 kJ/mol",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3335,
"subject": "Chemistry",
"question": "For water at 100$$^\\circ$$ C and 1 bar,
$$\\Delta$$vap H $$-$$ $$\\Delta$$vap U = _____________ $$\\times$$ 102 J mol$$-$$1. (Round off to the Nearest Integer)
[Use : R = 8.31 J mol$$-$$1 K$$-$$1]
[Assume volume of H2O(l) is much smaller than volume of H2O(g). Assume H2O(g) treated as an ideal gas]",
"options": [],
"answer": "31",
"solution": "**Answer:** 31\n\nH2O(l) $$\\rightleftharpoons$$ H2O(v)
= 3099.63 J $$ \\cong $$ 31 $$\\times$$ 102 J",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3336,
"subject": "Chemistry",
"question": "When 400 mL of 0.2 M H2SO4 solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is __________ $$\\times$$ 10$$-$$2 K. (Round off to the nearest integer).
= 81.96 $$\\times$$ 10$$-$$2 K $$ \\approx $$ 82 $$\\times$$ 10$$-$$2 K",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3337,
"subject": "Chemistry",
"question": "The Born-Haber cycle for KCl is evaluated with the following data :
$${\\Delta _f}{H^\\Theta }$$ for KCl = $$-$$436.7 kJ mol$$-$$1 ;
$${\\Delta _{sub}}{H^\\Theta }$$ for K = 89.2 kJ mol$$-$$1 ;
$${\\Delta _{ionization}}{H^\\Theta }$$ for K = 419.0 kJ mol$$-$$1 ;
$${\\Delta _{electron\\,gain}}{H^\\Theta }$$ for Cl(g) = $$-$$348.6 kJ mol$$-$$1 ;
$${\\Delta _{bond}}{H^\\Theta }$$ for Cl2 = 243.0 kJ mol$$-$$1
The magnitude of lattice enthalpy of KCl in kJ mol$$-$$1 is _____________ (Nearest integer)",
"options": [],
"answer": "718",
"solution": "**Answer:** 718\n\n$${\\Delta _f}H_{KCl}^\\Theta = {\\Delta _{sub}}H_{(K)}^\\Theta + {\\Delta _{ioniztion}}H_{(K)}^\\Theta + {1 \\over 2}{\\Delta _{bond}}H_{(C{l_2})}^\\Theta + {\\Delta _{electron\\,gain}}H_{(Cl)}^\\Theta + {\\Delta _{lattice}}H_{(KCl)}^\\Theta $$
The magnitude of lattice enthalpy of KCl in kJ mol$$-$$1 is 718 (Nearest integer).",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3338,
"subject": "Chemistry",
"question": "For water $$\\Delta$$vap H = 41 kJ mol$$-$$1 at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ___________ kJ mol$$-$$1
$$\\Delta$$ DU = 41 $$-$$ 3.0959 = 38 kJ/mol",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3339,
"subject": "Chemistry",
"question": "200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is $$-$$57.1 kJ. The increase in temperature in $$^\\circ$$C of the system on mixing is x $$\\times$$ 10$$-$$2. The value of x is ___________. (Nearest integer)
[Given : Specific heat of water = 4.18 J g$$-$$1 K$$-$$1, Density of water = 1.00 g cm$$-$$3]
[Assume no volume change on mixing)",
"options": [],
"answer": "82",
"solution": "**Answer:** 82\n\n$$\\Rightarrow$$ Millimoles of HCl = 200 $$\\times$$ 0.2 = 40
$$\\Rightarrow$$ Millimoles of NaOH = 300 $$\\times$$ 0.1 = 30
17.0 g of NH3 completely vapourises at $$-$$33.42$$^\\circ$$C and 1 bar pressure and the enthalpy change in the process is 23.4 kJ mol$$-$$1. The enthalpy change for the vapourisation of 85 g of NH3 under the same conditions is _________ kJ.
",
"options": [],
"answer": "117",
"solution": "**Answer:** 117\n\nNumber of moles of $\\mathrm{NH}_{3}=5$\n
For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, $$\\Delta$$CH$$\\Theta $$ = $$-$$601.70 kJ mol$$-$$1, the magnitude of change in internal energy for the reaction is __________ kJ. (Nearest integer)
the amount of heat produced as measured by bomb calorimeter is 726 kJ mol$$-$$1 at 27$$^\\circ$$C. The enthalpy of combustion for the reaction is $$-$$x kJ mol$$-$$1, where x is ___________. (Nearest integer)
At 25$$^\\circ$$C and 1 atm pressure, the enthalpy of combustion of benzene (I) and acetylene (g) are $$-$$ 3268 kJ mol$$-$$1 and $$-$$1300 kJ mol$$-$$1, respectively. The change in enthalpy for the reaction 3 C2H2(g) $$\\to$$ C6H6 (I), is :
The enthalpy of combustion of propane, graphite and dihydrogen at $$298 \\mathrm{~K}$$ are $$-2220.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1},-393.5 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$ and $$-285.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$ respectively. The magnitude of enthalpy of formation of propane $$\\left(\\mathrm{C}_{3} \\mathrm{H}_{8}\\right)$$ is _______________ $$\\mathrm{kJ} \\,\\mathrm{mol}^{-1}$$. (Nearest integer)
",
"options": [],
"answer": "104",
"solution": "**Answer:** 104\n\nEnthalpy of combustion of propane, graphite and $$\\mathrm{H}_{2}$$ at $$298 \\mathrm{~K}$$ are\n
$$2.4 \\mathrm{~g}$$ coal is burnt in a bomb calorimeter in excess of oxygen at $$298 \\mathrm{~K}$$ and $$1 \\mathrm{~atm}$$ pressure. The temperature of the calorimeter rises from $$298 \\mathrm{~K}$$ to $$300 \\mathrm{~K}$$. The enthalpy change during the combustion of coal is $$-x \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$. The value of $$x$$ is ___________. (Nearest Integer)
\n
(Given : Heat capacity of bomb calorimeter $$20.0 \\mathrm{~kJ} \\mathrm{~K}^{-1}$$. Assume coal to be pure carbon)
When 600 mL of 0.2 M HNO3 is mixed with $$400 \\mathrm{~mL}$$ of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is ___________ $$\\times 10^{-2}{ }\\,^{\\circ} \\mathrm{C}$$.
\n
(Enthalpy of neutralisation $$=57 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$ and Specific heat of water $$=4.2 \\,\\mathrm{JK}^{-1} \\mathrm{~g}^{-1}$$) (Neglect heat capacity of flask)
",
"options": [],
"answer": "54",
"solution": "**Answer:** 54\n\nHNO3 \n600 mL × 0.2 M = 120 m mol
\n(Rounded off to the nearest integer)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3350,
"subject": "Chemistry",
"question": "
$$0.3 \\mathrm{~g}$$ of ethane undergoes combustion at $$27^{\\circ} \\mathrm{C}$$ in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by $$0.5^{\\circ} \\mathrm{C}$$. The heat evolved during combustion of ethane at constant pressure is ____________ $$\\mathrm{kJ} ~\\mathrm{mol}{ }^{-1}$$. (Nearest integer)
\n
[Given : The heat capacity of the calorimeter system is $$20 \\mathrm{~kJ} \\mathrm{~K}^{-1}, \\mathrm{R}=8.3 ~\\mathrm{JK}^{-1} \\mathrm{~mol}^{-1}$$.
\n
Assume ideal gas behaviour.
\n
Atomic mass of $$\\mathrm{C}$$ and $$\\mathrm{H}$$ are 12 and $$1 \\mathrm{~g} \\mathrm{~mol}^{-1}$$ respectively]
Hence the magnitude of this will be $173 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3352,
"subject": "Chemistry",
"question": "
At $$25^{\\circ} \\mathrm{C}$$, the enthalpy of the following processes are given :
\n
\n
\n
\n
\n
\n
\n
\n
\n\n
\n
$$\\mathrm{H_2(g)+O_2(g)}$$
\n
$$\\to$$
\n
$$2\\mathrm{OH(g)}$$
\n
$$\\mathrm{\\Delta H^\\circ=78~kJ~mol^{-1}}$$
\n
\n\n\n
\n
$$\\mathrm{H_2(g)+\\frac{1}{2}O_2(g)}$$
\n
$$\\to$$
\n
$$\\mathrm{H_2O(g)}$$
\n
$$\\mathrm{\\Delta H^\\circ=-242~kJ~mol^{-1}}$$
\n
\n
\n
$$\\mathrm{H_2(g)}$$
\n
$$\\to$$
\n
$$\\mathrm{2H(g)}$$
\n
$$\\mathrm{\\Delta H^\\circ=436~kJ~mol^{-1}}$$
\n
\n
\n
$$\\frac{1}{2}\\mathrm{O_2(g)}$$
\n
$$\\to$$
\n
$$\\mathrm{O(g)}$$
\n
$$\\mathrm{\\Delta H^\\circ=249~kJ~mol^{-1}}$$
\n
\n\n
\n
What would be the value of X for the following reaction ? _____________ (Nearest integer)
The enthalpy change for the conversion of $$\\frac{1}{2} \\mathrm{Cl}_{2}(\\mathrm{~g})$$ to $$\\mathrm{Cl}^{-}$$(aq) is ($$-$$) ___________ \n$$\\mathrm{kJ} \\mathrm{mol}^{-1}$$ (Nearest integer)
\n
Given : $$\\Delta_{\\mathrm{dis}} \\mathrm{H}_{\\mathrm{Cl}_{2(\\mathrm{~g})}^{\\theta}}^{\\ominus}=240 \\mathrm{~kJ} \\mathrm{~mol}^{-1}, \\Delta_{\\mathrm{eg}} \\mathrm{H}_{\\mathrm{Cl_{(g)}}}^{\\ominus}=-350 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$,
28.0 L of CO$$_2$$ is produced on complete combustion of 16.8 L gaseous mixture of ethene and methane at 25$$^\\circ$$C and 1 atm. Heat evolved during the combustion process is ___________ kJ.
\n
Given : $$\\mathrm{\\Delta H_c~(CH_4)=-900~kJ~mol^{-1}}$$
$$\\mathrm{AB}, \\mathrm{A}_{2}$$ and $$\\mathrm{B}_{2}$$ are diatomic molecules. If the bond enthalpies of $$\\mathrm{A}_{2}, \\mathrm{~B}_{2}$$ and $$\\mathrm{AB}$$ are in the ratio $$1: 0.5: 1$$, then the bond enthalpy of $$\\mathrm{A}_{2}$$ is ____________ $$\\mathrm{kJ} ~\\mathrm{mol}^{-1}$$ (Nearest integer)
",
"options": [],
"answer": "400",
"solution": "**Answer:** 400\n\nTo find the bond enthalpy of $$\\mathrm{A}_{2}$$, we can use the given information about the reaction and the bond enthalpies' ratio. The reaction is:\n
\nThe bond enthalpies of $$\\mathrm{A}_{2}$$, $$\\mathrm{B}_{2}$$, and $$\\mathrm{AB}$$ are in the ratio of $$1: 0.5: 1$$.\n
\nLet's denote the bond enthalpies of $$\\mathrm{A}_{2}$$, $$\\mathrm{B}_{2}$$, and $$\\mathrm{AB}$$ as $$x$$, $$0.5x$$, and $$x$$, respectively.\n
\nThe enthalpy change for the reaction can be calculated using the bond enthalpies:\n
\n$$\\Delta H_{f}^{0} = \\text{(Sum of bond enthalpies of reactants)} - \\text{(Sum of bond enthalpies of products)}$$\n
\nFor the given reaction:\n
\n$$-200 = (x + 0.5x) - 2x$$\n
\nNow we can solve for $$x$$, which represents the bond enthalpy of $$\\mathrm{A}_{2}$$:\n
\n$$-200 = 1.5x - 2x$$\n
\n$$-200 = -0.5x$$\n
\n$$x = \\frac{-200}{-0.5}$$\n
\n$$x = 400$$\n
\nThe bond enthalpy of $$\\mathrm{A}_{2}$$ is $$400 ~\\mathrm{kJ} ~\\mathrm{mol}^{-1}$$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3356,
"subject": "Chemistry",
"question": "
Solid fuel used in rocket is a mixture of $$\\mathrm{Fe}_{2} \\mathrm{O}_{3}$$ and $$\\mathrm{Al}$$ (in ratio 1 : 2). The heat evolved $$(\\mathrm{kJ})$$ per gram of the mixture is ____________. (Nearest integer)
This is the heat of reaction for the above reaction. However, we are asked to find the heat evolved per gram of the mixture.
\n
To find this, we need to determine the molar mass of the reactants in the reaction. The molar mass of $$\\mathrm{Fe}_2\\mathrm{O}_3$$ is $$2 \\times 56 + 3 \\times 16 = 160\\ \\mathrm{g/mol}$$ and the molar mass of 2 moles of $$\\mathrm{Al}$$ is $$2 \\times 27 = 54\\ \\mathrm{g/mol}$$. The total molar mass of the mixture is $$160 + 54 = 214\\ \\mathrm{g/mol}$$.
\n
So, the heat evolved per gram of the mixture is $$\\frac{-860\\ \\mathrm{kJ/mol}}{214\\ \\mathrm{g/mol}} = -4.0187\\ \\mathrm{kJ/g}$$
\n
Rounded to the nearest integer, this value is approximately -4 kJ/g. The negative sign indicates that the heat is evolved (exothermic reaction).
An endothermic process is one that absorbs heat from the surroundings. Here are the types of the given reactions:
\n
A. $\\mathrm{I}_{2}(\\mathrm{~g}) \\rightarrow 2 \\mathrm{I}(\\mathrm{g})$
\n
This process involves the dissociation of iodine molecules into individual iodine atoms. This requires energy to break the bonds between the iodine atoms, so it is an endothermic process.
\n
B. $\\mathrm{HCl}(\\mathrm{g}) \\rightarrow \\mathrm{H}(\\mathrm{g})+\\mathrm{Cl}(\\mathrm{g})$
\n
This is the dissociation of hydrogen chloride into hydrogen and chlorine. Similar to the first reaction, it requires energy to break the bond between the hydrogen and chlorine atoms. Therefore, this is also an endothermic process.
\n
C. $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightarrow \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$
\n
This is the process of evaporation or vaporization, where liquid water is converted to water vapor. Evaporation is an endothermic process because it requires heat to break the hydrogen bonds between water molecules and convert it into a gas.
\n
D. $\\mathrm{C}(\\mathrm{s})+\\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{CO}_{2}(\\mathrm{~g})$
\n
This is a combustion reaction. Combustion reactions are generally exothermic because they release energy in the form of heat and light. So, this is not an endothermic process.
\n
E. Dissolution of ammonium chloride in water
\n
The dissolution of ammonium chloride ($\\mathrm{NH}_4\\mathrm{Cl}$) in water is known to be endothermic. The ionic bonds in the ammonium chloride and the hydrogen bonds in the water must be broken for the salt to dissolve, which requires heat.
\n
So, in summary, four of these processes (A, B, C, and E) are endothermic.
can be calculated using Hess's Law, which states that the total enthalpy change for a reaction is the same whether it occurs in one step or in many steps.
\n
If we modify the given reactions to reflect the formation of CO, we can add the modified reactions together to get the reaction of interest.
Heat of combustion of $$\\mathrm{H}_{2}(\\mathrm{g})\\quad\\quad=-241.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$
\n
Heat of combustion of $$\\mathrm{C}(\\mathrm{s})\\quad\\quad=-393.5 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$
\n
Heat of combustion of $$\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(\\mathrm{l})\\quad=-1234.7 \\mathrm{~kJ}~{\\mathrm{mol}}^{-1}$$
\n
The heat of formation of $$\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(\\mathrm{l})$$ is $$(-)$$ ___________ $$\\mathrm{kJ} ~\\mathrm{mol}^{-1}$$ (Nearest integer).
",
"options": [],
"answer": "278",
"solution": "**Answer:** 278\n\nThe reaction for the formation of $$\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(\\mathrm{l})$$ is given by:\n
Standard enthalpy of vapourisation for $$\\mathrm{CCl}_4$$ is $$30.5 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$. Heat required for vapourisation of $$284 \\mathrm{~g}$$ of $$\\mathrm{CCl}_4$$ at constant temperature is ________ $$\\mathrm{kJ}$$.
\n
(Given molar mass in $$\\mathrm{g} \\mathrm{mol}^{-1} ; \\mathrm{C}=12, \\mathrm{Cl}=35.5$$)
Then enthalpy change for following reaction $$3 \\mathrm{C}_{(\\mathrm{s})}+\\mathrm{Fe}_2 \\mathrm{O}_{3(\\mathrm{~s})} \\rightarrow 2 \\mathrm{Fe}_{(\\mathrm{s})}+3 \\mathrm{CO}_{(\\mathrm{g})}$$ is _______ $$\\mathrm{kJ} / \\mathrm{mol}$$.
When equal volume of $$1 \\mathrm{~M} \\mathrm{~HCl}$$ and $$1 \\mathrm{~M} \\mathrm{~H}_2 \\mathrm{SO}_4$$ are separately neutralised by excess volume of $$1 \\mathrm{M}$$ $$\\mathrm{NaOH}$$ solution. $$x$$ and $$y \\mathrm{~kJ}$$ of heat is liberated respectively. The value of $$y / x$$ is __________.
To solve this problem, we need to understand the concept of neutralization reactions and the heat evolved during these reactions.\n\n
When an acid and a base react, they undergo a neutralization reaction to produce water and a salt. The heat released in this process is known as the enthalpy of neutralization.
\n\n
Consider the neutralization of hydrochloric acid (HCl) and sulfuric acid (H2SO4) by sodium hydroxide (NaOH). The balanced chemical equations for these neutralizations are:
For the first reaction, each mole of HCl reacts with one mole of NaOH, releasing a certain amount of heat (let's denote this amount by $ x $ kJ). For the second reaction, each mole of H2SO4 reacts with two moles of NaOH. Since we are given equal volumes and molarities of HCl and H2SO4, we can infer that one mole of H2SO4 will produce twice the heat of one mole of HCl because it produces double the amount of water.\n\n
\n\n
Thus, the heat evolved in the neutralization of H2SO4 by NaOH (denoted as $ y $ kJ) will be approximately twice that of HCl. Therefore, $ y = 2x $.\n\n
The heat of solution of anhydrous $$\\mathrm{CuSO}_4$$ and $$\\mathrm{CuSO}_4 \\cdot 5 \\mathrm{H}_2 \\mathrm{O}$$ are $$-70 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$ and $$+12 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$ respectively.
\n
The heat of hydration of $$\\mathrm{CuSO}_4$$ to $$\\mathrm{CuSO}_4 \\cdot 5 \\mathrm{H}_2 \\mathrm{O}$$ is $$-x \\mathrm{~kJ}$$. The value of $$x$$ is ________. (nearest integer).
The enthalpy of formation of ethane $$(\\mathrm{C}_2 \\mathrm{H}_6)$$ from ethylene by addition of hydrogen where the bond-energies of $$\\mathrm{C}-\\mathrm{H}, \\mathrm{C}-\\mathrm{C}, \\mathrm{C}=\\mathrm{C}, \\mathrm{H}-\\mathrm{H}$$ are $$414 \\mathrm{~kJ}, 347 \\mathrm{~kJ}, 615 \\mathrm{~kJ}$$ and $$435 \\mathrm{~kJ}$$ respectively is $$-$$ __________ $$\\mathrm{kJ}$$
Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R)
\n
Assertion (A) : Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always $$-57 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$
\n
Reason (R) : Enthalpy of neutralisation is the amount of heat liberated when one mole of $$\\mathrm{H}^{+}$$ ions furnished by acid combine with one mole of $${ }^{-} \\mathrm{OH}$$ ions furnished by base to form one mole of water.
\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": "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 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
Enthalpy of neutralisation of strong acids and bases is $$-57 \\mathrm{~kJ} / \\mathrm{mol}$$. which is fixed for reaction of 1 mole of $$\\mathrm{H}^{+}$$ with 1 mole of $$\\mathrm{OH}^{-}$$ to form 1 mole of water.
The standard enthalpy of combustion of $$2 \\mathrm{~mol}$$ of benzene is $$-^{\\prime} x^{\\prime} \\mathrm{kJ}$$. $$x=$$ __________.
\n
Given :
\n
1. standard Enthalpy of formation of $$1 \\mathrm{~mol}$$ of $$\\mathrm{C}_6 \\mathrm{H}_6(\\mathrm{l})$$, for the reaction $$6 \\mathrm{C}$$ (graphite) $$+3 \\mathrm{H}_2(\\mathrm{g}) \\rightarrow \\mathrm{C}_6 \\mathrm{H}_6(\\mathrm{l})$$ is $$48.5 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$.
\n
2. Standard Enthalpy of formation of $$1 \\mathrm{~mol}$$ of $$\\mathrm{CO}_2(\\mathrm{g})$$, for the reaction $$\\mathrm{C}$$ (graphite) $$+\\mathrm{O}_2(\\mathrm{g}) \\rightarrow \\mathrm{CO}_2(\\mathrm{g})$$ is $$-393.5 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$.
\n
3. Standard and Enthalpy of formation of $$1 \\mathrm{~mol}$$ of $$\\mathrm{H}_2 \\mathrm{O}(\\mathrm{l})$$, for the reaction $$\\mathrm{H}_2(\\mathrm{g})+\\frac{1}{2} \\mathrm{O}_2(\\mathrm{g}) \\rightarrow \\mathrm{H}_2 \\mathrm{O}(\\mathrm{l})$$ is $$-286 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$$.
To determine the standard enthalpy of combustion of 2 moles of benzene, we need to use the standard enthalpy of formation values provided and apply Hess's Law. Here is a step-by-step explanation:
\n\n
Given Data:
\n\n\n
Standard enthalpy of formation of benzene ($C_6H_6(l)$):
The standard enthalpy of combustion of 2 moles of benzene is
\n\n
$ x = 6535 \\, \\text{kJ} $
\n\n
Thus, $ x = 6535 $.
",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 3367,
"subject": "Chemistry",
"question": "Identify the correct statement regarding a spontaneous process :",
"options": [
{
"text": "For a spontaneous process in an isolated system, the change in entropy is positive"
},
{
"text": "Endothermic processes are never spontaneous"
},
{
"text": "Exothermic processes are always spontaneous "
},
{
"text": "Lowering of energy in the reaction process is the only criterion for spontaneity "
}
],
"answer": "For a spontaneous process in an isolated system, the change in entropy is positive",
"solution": "**Answer:** For a spontaneous process in an isolated system, the change in entropy is positive\n\nSpontaneity of reaction depends on tendency to acquire minimum energy state and maximum randomness. For a spontaneous process in an isolated system the change in entropy is positive. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3368,
"subject": "Chemistry",
"question": "The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is : \n
(Specific heat of water liquid and water vapour are 4.2 kJ K$$-$$1 kg$$-$$1 and 2.0 kJ K$$-$$1 kg$$-$$1; heat of liquid fusion and vapourisation of water are 334 kJ$$-$$1 and 2491 kJ kg$$-$$1, respectively). (log 273 = 2.436, log 373 = 2.572, log 383 = 2.583)",
"options": [
{
"text": "7.90 kJ kg$$-$$1 K$$-$$1"
},
{
"text": "2.64 kJ kg$$-$$1 K$$-$$1"
},
{
"text": "8.49 kJ kg$$-$$1 K$$-$$1"
},
{
"text": "9.26 kJ kg$$-$$1 K$$-$$1"
}
],
"answer": "9.26 kJ kg$$-$$1 K$$-$$1",
"solution": "**Answer:** 9.26 kJ kg$$-$$1 K$$-$$1\n\n\n $$\\Delta {S_1} = {{\\Delta {H_{fusion}}} \\over {273}} = {{334} \\over {273}} = 1.22$$\n
$$\\Delta {S_{total}} = 9.26\\,$$ kJ Kg$$-$$1 K$$-$$1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3369,
"subject": "Chemistry",
"question": "The process with negative entropy change is :",
"options": [
{
"text": "Dissociation of CaSO4 (s) to CaO(s) and SO3(g) "
},
{
"text": "Dissolution of iodine in water"
},
{
"text": "Synthesis of ammonia from N2 and H2"
},
{
"text": "Sublimation of dry ice "
}
],
"answer": "Synthesis of ammonia from N2 and H2",
"solution": "**Answer:** Synthesis of ammonia from N2 and H2\n\nN2(g) + 3H2(g) $$\\rightleftharpoons$$ 2NH3(g); $$\\Delta $$ng < 0 ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3370,
"subject": "Chemistry",
"question": "Two blocks of the same metal having same mass and at temperature T1 and T2, respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, $$\\Delta $$S, for this process is : ",
"options": [
{
"text": "2Cp In $$\\left[ {{{{{\\left( {{T_1} + {T_2}} \\right)}^{{1 \\over 2}}}} \\over {{T_1}{T_2}}}} \\right]$$"
},
{
"text": "2Cp In $$\\left[ {{{\\left( {{T_1} + {T_2}} \\right)} \\over {2{T_1}{T_2}}}} \\right]$$"
},
{
"text": "Cp In $$\\left[ {{{{{\\left( {{T_1} + {T_2}} \\right)}^2}} \\over {4{T_1}{T_2}}}} \\right]$$"
},
{
"text": "2Cp In $$\\left[ {{{\\left( {{T_1} + {T_2}} \\right)} \\over {4{T_1}{T_2}}}} \\right]$$"
}
],
"answer": "Cp In $$\\left[ {{{{{\\left( {{T_1} + {T_2}} \\right)}^2}} \\over {4{T_1}{T_2}}}} \\right]$$",
"solution": "**Answer:** Cp In $$\\left[ {{{{{\\left( {{T_1} + {T_2}} \\right)}^2}} \\over {4{T_1}{T_2}}}} \\right]$$\n\nWhen two blocks comes in contact with each other and attain thermal equilibrium then\n
= Cp $$\\ln \\left( {{{{{\\left( {{T_1} + {T_2}} \\right)}^2}} \\over {4{T_1}{T_2}}}} \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3371,
"subject": "Chemistry",
"question": "The true statement amongst the following is :",
"options": [
{
"text": "S is a function of temperature but $$\\Delta $$S is not\na function of temperature."
},
{
"text": "Both S and $$\\Delta $$S are not functions of\ntemperature."
},
{
"text": "Both $$\\Delta $$S and S are functions of temperature."
},
{
"text": "S is not a function of temperature but $$\\Delta $$S is\na function of temperature."
}
],
"answer": "Both $$\\Delta $$S and S are functions of temperature.",
"solution": "**Answer:** Both $$\\Delta $$S and S are functions of temperature.\n\n$$\\Delta S = \\int {{{d{q_{rev}}} \\over T}} $$\n
S = Kln(w)\n
Both entropy and change in entropy are\nfunction of temperature.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3372,
"subject": "Chemistry",
"question": "
One mole of an ideal gas at $$350 \\mathrm{~K}$$ is in a $$2.0 \\mathrm{~L}$$ vessel of thermally conducting walls, which are in contact with the surroundings. It undergoes isothermal reversible expansion from 2.0 L to $$3.0 \\mathrm{~L}$$ against a constant pressure of $$4 \\mathrm{~atm}$$. The change in entropy of the surroundings ( $$\\Delta \\mathrm{S})$$ is ___________ $$\\mathrm{J} \\mathrm{K}^{-1}$$ (Nearest integer)
Since the process is reversible and the total entropy change in the universe should be zero for reversible processes, the change in entropy of the surroundings is equal to the negative change of the system's entropy:
But considering the heat transfer from the system to the surroundings, the sign should be positive, which means the entropy of the surroundings also increases:
For perpenedicularity of lines $$aa' + 1 + cc' = 0$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3374,
"subject": "General Science",
"question": "A line makes the same angle $$\\theta $$, with each of the $$x$$ and $$z$$ axis.\n
If the angle $$\\beta \\,$$, which it makes with y-axis, is such that $$\\,{\\sin ^2}\\beta = 3{\\sin ^2}\\theta ,$$ then $${\\cos ^2}\\theta $$ equals :",
"options": [
{
"text": "$${2 \\over 5}$$ "
},
{
"text": "$${1 \\over 5}$$"
},
{
"text": "$${3 \\over 5}$$"
},
{
"text": "$${2 \\over 3}$$"
}
],
"answer": "$${3 \\over 5}$$",
"solution": "**Answer:** $${3 \\over 5}$$\n\nConcept : If a line makes the angle $$\\alpha ,\\beta ,\\gamma $$ with x, y, z axis respectively then\n$$${\\cos ^2}\\alpha + {\\cos ^2}\\beta + {\\cos ^2}\\gamma = 1$$$\n
In this question given that the line makes angle θ with x and z-axis and β with y−axis.\n
$$ \\Rightarrow {\\cos ^2}\\theta = {3 \\over 5}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3375,
"subject": "General Science",
"question": "If a line makes an angle of $$\\pi /4$$ with the positive directions of each of $$x$$-axis and $$y$$-axis, then the angle that the line makes with the positive direction of the $$z$$-axis is :",
"options": [
{
"text": "$${\\pi \\over 4}$$"
},
{
"text": "$${\\pi \\over 2}$$ "
},
{
"text": "$${\\pi \\over 6}$$"
},
{
"text": "$${\\pi \\over 3}$$"
}
],
"answer": "$${\\pi \\over 2}$$ ",
"solution": "**Answer:** $${\\pi \\over 2}$$ \n\nLet the angle of line makes with the positive direction of $$z$$-axis is $$\\alpha $$ direction cosines of line with the $$+ve$$ directions of $$x$$-axis, $$y$$-axis, and $$z$$-axis is $$l,$$ $$m,$$ $$n$$ respectively. \n
Hence, angle with positive direction of the $$z$$-axis is $${\\pi \\over 2}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3376,
"subject": "General Science",
"question": "Let $$L$$ be the line of intersection of the planes $$2x+3y+z=1$$ and $$x+3y+2z=2.$$ If $$L$$ makes an angle $$\\alpha $$ with the positive $$x$$-axis, then cos $$\\alpha $$ equals ",
"options": [
{
"text": "$$1$$ "
},
{
"text": "$${1 \\over {\\sqrt 2 }}$$ "
},
{
"text": "$${1 \\over {\\sqrt 3 }}$$"
},
{
"text": "$${1 \\over 2}$$ "
}
],
"answer": "$${1 \\over {\\sqrt 3 }}$$",
"solution": "**Answer:** $${1 \\over {\\sqrt 3 }}$$\n\nLet the direction cosines of line $$L$$ be $$l,m,n,$$ \n
then $$2l+3m+n=0$$ $$\\,\\,\\,\\,\\,\\,\\,....\\left( i \\right)$$\n
and $$l + 3m + 2n = 0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,....\\left( {ii} \\right)$$\n
on solving equation $$(i)$$ and $$(ii),$$ we get \n
If the median through A is equally inclined to the coordinate axes, then the value of ($$\\lambda $$3 + $$\\mu $$3 + 5) is : ",
"options": [
{
"text": "1130"
},
{
"text": "1348"
},
{
"text": "676"
},
{
"text": "1077"
}
],
"answer": "1348",
"solution": "**Answer:** 1348\n\n\n
If two straight lines whose direction cosines are given by the relations $$l + m - n = 0$$, $$3{l^2} + {m^2} + cnl = 0$$ are parallel, then the positive value of c is :
Given that the direction cosines satisfy $l + m - n = 0$, we find that $n = l + m$.
\n
The other equation is $3l^2 + m^2 + cnl = 0$, and substituting $n = l + m$ gives $3l^2 + m^2 + cl(l + m) = 0$.
\n
This simplifies to $(3 + c)l^2 + clm + m^2 = 0$.
\n
As the lines are parallel, they share the same direction ratios, so we can express $l$ in terms of $m$, say $l = km$. Substituting this into our equation gives $(3 + c)(km)^2 + ckm^2 + m^2 = 0$.
\n
This simplifies to $m^2[k^2(3 + c) + kc + 1] = 0$.
\n
Since $m \\neq 0$, we must have $k^2(3 + c) + kc + 1 = 0$. Here, we consider the ratio $k = \\frac{l}{m}$ to be constant, since the lines are parallel. The equation then becomes a quadratic equation in $k$.
\n
As the lines are parallel, the discriminant of the quadratic equation must be equal to zero for the equation to have equal roots. Hence, the discriminant $D = (c^2 - 4(3 + c)) = 0$.
Let $$\\mathrm{P}(-2,-1,1)$$ and $$\\mathrm{Q}\\left(\\frac{56}{17}, \\frac{43}{17}, \\frac{111}{17}\\right)$$ be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are $$\\alpha,-1, \\beta$$, where both $$\\alpha$$ and $$\\beta$$ are integers of minimum absolute values, then $$\\alpha^{2}+\\beta^{2}$$ is equal to ____________.
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3388,
"subject": "General Science",
"question": "Consider a $\\triangle A B C$ where $A(1,3,2), B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\\angle B A C$ meets\nthe line $B C$ at $D$, then the length of the projection of the vector $\\overrightarrow{A D}$ on the vector $\\overrightarrow{A C}$ is :",
"options": [
{
"text": "$\\frac{37}{2 \\sqrt{38}}$"
},
{
"text": "$\\sqrt{19}$"
},
{
"text": "$\\frac{39}{2 \\sqrt{38}}$"
},
{
"text": "$\\frac{\\sqrt{38}}{2}$"
}
],
"answer": "$\\frac{37}{2 \\sqrt{38}}$",
"solution": "**Answer:** $\\frac{37}{2 \\sqrt{38}}$\n\n\n
Let $$O$$ be the origin and the position vectors of $$A$$ and $$B$$ be $$2 \\hat{i}+2 \\hat{j}+\\hat{k}$$ and $$2 \\hat{i}+4 \\hat{j}+4 \\hat{k}$$ respectively. If the internal bisector of $$\\angle \\mathrm{AOB}$$ meets the line $$\\mathrm{AB}$$ at $$\\mathrm{C}$$, then the length of $$O C$$ is
Let $$\\mathrm{P}(3,2,3), \\mathrm{Q}(4,6,2)$$ and $$\\mathrm{R}(7,3,2)$$ be the vertices of $$\\triangle \\mathrm{PQR}$$. Then, the angle $$\\angle \\mathrm{QPR}$$ is
Let $$P(x, y, z)$$ be a point in the first octant, whose projection in the $$x y$$-plane is the point $$Q$$. Let $$O P=\\gamma$$; the angle between $$O Q$$ and the positive $$x$$-axis be $$\\theta$$; and the angle between $$O P$$ and the positive $$z$$-axis be $$\\phi$$, where $$O$$ is the origin. Then the distance of $$P$$ from the $$x$$-axis is
If the line $$\\frac{2-x}{3}=\\frac{3 y-2}{4 \\lambda+1}=4-z$$ makes a right angle with the line $$\\frac{x+3}{3 \\mu}=\\frac{1-2 y}{6}=\\frac{5-z}{7}$$, then $$4 \\lambda+9 \\mu$$ is equal to :
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3393,
"subject": "General Science",
"question": "A plane which passes through the point $$(3,2,0)$$ and the line \n
$${{x - 4} \\over 1} = {{y - 7} \\over 5} = {{z - 4} \\over 4}$$ is :",
"options": [
{
"text": "$$x-y+z=1$$"
},
{
"text": "$$x+y+z=5$$ "
},
{
"text": "$$x+2y-z=1$$ "
},
{
"text": "$$2x-y+z=5$$"
}
],
"answer": "$$x-y+z=1$$",
"solution": "**Answer:** $$x-y+z=1$$\n\nAs the point $$\\left( {3,2,0} \\right)$$ lies on the given line \n
$$\\therefore$$ There can be infinite many planes passing through this line. But here out of the four options only first option is satisfied by the coordinates of both the points $$\\left( {3,\\,2,\\,0} \\right)$$ and $$\\left( {4,\\,7,\\,4} \\right)$$\n
$$\\therefore$$ $$x - y + z = 1$$ is the required plane.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3394,
"subject": "General Science",
"question": "If the angel $$\\theta $$ between the line $${{x + 1} \\over 1} = {{y - 1} \\over 2} = {{z - 2} \\over 2}$$ and \n
the plane $$2x - y + \\sqrt \\lambda \\,\\,z + 4 = 0$$ is such that $$\\sin \\,\\,\\theta = {1 \\over 3}$$ then value of $$\\lambda $$ is :",
"options": [
{
"text": "$${5 \\over 3}$$"
},
{
"text": "$${-3 \\over 5}$$"
},
{
"text": "$${3 \\over 4}$$"
},
{
"text": "$${-4 \\over 3}$$"
}
],
"answer": "$${5 \\over 3}$$",
"solution": "**Answer:** $${5 \\over 3}$$\n\nIf $$\\theta $$ is the angle between line and plane then $$\\left( {{\\pi \\over 2} - 0} \\right)$$ \n
is the angle between line and normal to plane given by \n
$$ \\Rightarrow 4\\lambda = 5 + \\lambda \\Rightarrow \\lambda = {5 \\over 3}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3395,
"subject": "General Science",
"question": "If the plane $$2ax-3ay+4az+6=0$$ passes through the midpoint of the line joining the centres of the spheres \n
$${x^2} + {y^2} + {z^2} - 10x + 4y - 2z = 8$$ then a equals :",
"options": [
{
"text": "$$-1$$"
},
{
"text": "$$1$$"
},
{
"text": "$$-2$$ "
},
{
"text": "$$2$$"
}
],
"answer": "$$-2$$ ",
"solution": "**Answer:** $$-2$$ \n\nCenters of given spheres are $$\\left( { - 3,4,1} \\right)$$ and $$\\left( {5, - 2,1} \\right).$$\n
Mid point of centers is $$\\left( {1,1,1} \\right).$$\n
Satisfying this in the equation of plane, we get\n
$$2a - 3a + 4a + 6 = 0$$\n
$$ \\Rightarrow a = - 2$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3396,
"subject": "General Science",
"question": "The distance between the line \n
$$\\overrightarrow r = 2\\widehat i - 2\\widehat j + 3\\widehat k + \\lambda \\left( {i - j + 4k} \\right),$$ and the plane \n
$$ \\Rightarrow $$ $$\\,\\,\\,$$ Distance $$OP = 10\\sqrt 3 .$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3405,
"subject": "General Science",
"question": "The number of distinct real values of $$\\lambda $$ for which the lines \n
$$ \\therefore $$ 3 distinct real values are possible.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3406,
"subject": "General Science",
"question": "If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line,\n
PQ = 2 PF = 2$$\\sqrt {{1^2} + {4^2} + {5^2}} $$ = 2$$\\sqrt {42} $$\n\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3407,
"subject": "General Science",
"question": "The distance of the point (1, 3, – 7) from the plane passing through the point (1, –1, – 1), having normal\nperpendicular to both the lines\n
$$ = \\sqrt {{2 \\over 3}} $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3412,
"subject": "General Science",
"question": "An angle between the plane, x + y + z = 5 and the line of intersection of the planes, 3x + 4y + z $$-$$ 1 = 0 and 5x + 8y + 2z + 14 =0, is : ",
"options": [
{
"text": "$${\\sin ^{ - 1}}\\left( {\\sqrt {{\\raise0.5ex\\hbox{$\\scriptstyle 3$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle {17}$}}} } \\right)$$ "
},
{
"text": "$${\\cos ^{ - 1}}\\left( {\\sqrt {{\\raise0.5ex\\hbox{$\\scriptstyle 3$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle {17}$}}} } \\right)$$"
},
{
"text": "$${\\cos ^{ - 1}}\\left( {{\\raise0.5ex\\hbox{$\\scriptstyle 3$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle {17}$}}} \\right)$$"
},
{
"text": "$${\\sin ^{ - 1}}\\left( {{\\raise0.5ex\\hbox{$\\scriptstyle 3$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle {17}$}}} \\right)$$"
}
],
"answer": "$${\\sin ^{ - 1}}\\left( {\\sqrt {{\\raise0.5ex\\hbox{$\\scriptstyle 3$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle {17}$}}} } \\right)$$ ",
"solution": "**Answer:** $${\\sin ^{ - 1}}\\left( {\\sqrt {{\\raise0.5ex\\hbox{$\\scriptstyle 3$}\n\\kern-0.1em/\\kern-0.15em\n\\lower0.25ex\\hbox{$\\scriptstyle {17}$}}} } \\right)$$ \n\nNormal to $$3x + 4y + z = 1$$ is $$3\\widehat i + 4\\widehat j + \\widehat k$$\n
Normal to $$5x + 8y + 2z =$$ $$ - 14$$ is $$5\\widehat i + 8\\widehat j + 2\\widehat k$$\n
The line of intersection of the planes is perpendicular to both normals, so, direction ratios of the intersection line are directly proportional to the cross product of the normal vectors.\n
Therefore the direction ratios of the line is $$ - \\widehat j + 4\\widehat k$$\n
Hence the angle between the plane x + y + z + 5 = 0 \n
and the intersection line is $${\\sin ^{ - 1}}\\left( {{{ - 1 + 4} \\over {\\sqrt {17} \\sqrt 3 }}} \\right) = {\\sin ^{ - 1}}\\left( {\\sqrt {{3 \\over {17}}} } \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3413,
"subject": "General Science",
"question": "The equation of the line passing through (–4, 3, 1), parallel \n
to the plane x + 2y – z – 5 = 0 and intersecting\n
= $${9 \\over 2}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3418,
"subject": "General Science",
"question": "The equation of a plane containing the line of\nintersection of the planes 2x – y – 4 = 0 and\ny + 2z – 4 = 0 and passing through the point\n(1, 1, 0) is :",
"options": [
{
"text": "x – 3y – 2z = –2"
},
{
"text": "2x – z = 2"
},
{
"text": "x – y – z = 0"
},
{
"text": "x + 3y + z = 4"
}
],
"answer": "x – y – z = 0",
"solution": "**Answer:** x – y – z = 0\n\nThe equation of any plane passing through the intersection of the planes 2x – y – 4 = 0 and\ny + 2z – 4 = 0 is :\n
By checking each options you can see only point (3, 2, 1) lies on the plane. ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3425,
"subject": "General Science",
"question": "If the distance between the plane,\n23x – 10y – 2z + 48 = 0 and the plane
is equal to\n$${k \\over {\\sqrt {633} }}$$, then k is equal to ______.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nRequired distance = perpendicular distance of plane 23x – 10y – 2z + 48 = 0 either from (–1, 3, –1) or (–3, –2, 1)\n
Check options (2, –10, –2) lies on L2.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3429,
"subject": "General Science",
"question": "The distance of the point (1, –2, 3) from
the plane x – y + z = 5 measured parallel to
the line $${x \\over 2} = {y \\over 3} = {z \\over { - 6}}$$ is :",
"options": [
{
"text": "7"
},
{
"text": "1"
},
{
"text": "$${1 \\over 7}$$"
},
{
"text": "$${7 \\over 5}$$"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\nEquation of line parallel to $${x \\over 2} = {y \\over 3} = {z \\over { - 6}}$$ passes through $$(1, - 2,3)$$ is
By checking each options we can see M lies on 2x + y – z = 1.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3431,
"subject": "General Science",
"question": "A plane passing through the point (3, 1, 1)\ncontains two lines whose direction ratios are 1,\n–2, 2 and 2, 3, –1 respectively. If this plane also\npasses through the point ($$\\alpha $$, –3, 5), then\n$$\\alpha $$ is\nequal to:",
"options": [
{
"text": "-10"
},
{
"text": "10"
},
{
"text": "5"
},
{
"text": "-5"
}
],
"answer": "5",
"solution": "**Answer:** 5\n\nAs normal is perpendicular to both the lines so normal vector to the plane is
\n$$\\overrightarrow n = \\left( {\\widehat i - 2\\widehat j + 2\\widehat k} \\right) \\times \\left( {2\\widehat i + 3\\widehat j - \\widehat k} \\right)$$
$$ \\Rightarrow 3(1 - y + 2 - z) - 2( - x + 2 + z - 2) = 0$$
$$ \\Rightarrow 9 - 3y - 3z + 2x - 2z = 0$$
$$ \\Rightarrow 2x - 3y - 5z + 9 = 0$$
By checking all options you can see ($$-$$2, 0, 1) satisfy the equation.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3433,
"subject": "General Science",
"question": "Let a plane P contain two lines\n $$\\overrightarrow r = \\widehat i + \\lambda \\left( {\\widehat i + \\widehat j} \\right)$$, $$\\lambda \\in R$$ and\n $$\\overrightarrow r = - \\widehat j + \\mu \\left( {\\widehat j - \\widehat k} \\right)$$, $$\\mu \\in R$$\n If Q($$\\alpha $$, $$\\beta $$, $$\\gamma $$) is the foot of the perpendicular\ndrawn from the point M(1, 0, 1) to P, then\n3($$\\alpha $$ + $$\\beta $$ + $$\\gamma $$) equals _______.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nGiven lines,
$$\\overrightarrow r = \\widehat i + \\lambda (\\widehat i + \\widehat j)$$ parallel to $$(\\widehat i + \\widehat j)$$
Let, $$\\overrightarrow {{n_1}} = (\\widehat i + \\widehat j)$$
and $$\\overrightarrow r = - \\widehat j + \\mu (\\widehat j - \\widehat k)$$ parallel to $$(\\widehat j - \\widehat k)$$
Let, $$\\overrightarrow {{n_2}} = (\\widehat j - \\widehat k)$$
$$ \\therefore $$ Normal of plane, $$\\overrightarrow n = \\overrightarrow {{n_1}} \\times \\overrightarrow {{n_2}} $$
Line $$\\overrightarrow r = \\widehat i + \\lambda (\\widehat i + \\widehat j)$$ is on the plane so, point on the line (1, 0, 0) will be also on the plane.
$$ \\therefore $$ Equation of the plane,
$$ - 1(x - 1) + 1(y - 0) + 1(z - 0) = 0$$
$$ \\Rightarrow x - y - z - 1 = 0$$
Foot of perpendicular from (x1, y1, z1) on the plane,
$$ \\Rightarrow $$ $$\\overrightarrow r\\,.\\,(\\widehat i + 7\\widehat j + 3\\widehat k) = 7$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3436,
"subject": "General Science",
"question": "Let ($$\\lambda$$, 2, 1) be a point on the plane which passes through the point (4, $$-$$2, 2). If the plane is perpendicular to the line joining the points ($$-$$2, $$-$$21, 29) and ($$-$$1, $$-$$16, 23), then $${\\left( {{\\lambda \\over {11}}} \\right)^2} - {{4\\lambda } \\over {11}} - 4$$ is equal to __________.",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\n
$$21(\\alpha + \\beta + \\gamma ) = 21\\left( {{{34} \\over 7}} \\right) = 102$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3438,
"subject": "General Science",
"question": "Let P be a plane lx + my + nz = 0 containing
the line, $${{1 - x} \\over 1} = {{y + 4} \\over 2} = {{z + 2} \\over 3}$$. If plane P divides the line segment AB joining
points A($$-$$3, $$-$$6, 1) and B(2, 4, $$-$$3) in ratio k : 1 then the value of k is equal to :",
"options": [
{
"text": "2"
},
{
"text": "3"
},
{
"text": "1.5"
},
{
"text": "4"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\n\n Line lies on plane
$$k = 2$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3439,
"subject": "General Science",
"question": "If the distance of the point (1, $$-$$2, 3) from the plane x + 2y $$-$$ 3z + 10 = 0 measured parallel to the line, $${{x - 1} \\over 3} = {{2 - y} \\over m} = {{z + 3} \\over 1}$$ is $$\\sqrt {{7 \\over 2}} $$, then the value of |m| is equal to _________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n\n Given line L,\n
$$ \\Rightarrow $$ $${m^2} = 4 \\Rightarrow |m| = 2$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3440,
"subject": "General Science",
"question": "If the equation of the plane passing through the line of intersection of the planes 2x $$-$$ 7y + 4z $$-$$ 3 = 0, 3x $$-$$ 5y + 4z + 11 = 0 and the point ($$-$$2, 1, 3) is ax + by + cz $$-$$ 7 = 0, then the value of 2a + b + c $$-$$ 7 is ____________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nEquation of plane can be written using family of planes : P1 + $$\\lambda$$P2 = 0
$$\\overrightarrow n = 7\\widehat i + 11\\widehat j + \\widehat k$$
$$ \\therefore $$ equation of plane is $$\\overrightarrow r \\,.\\,\\overrightarrow n = \\overrightarrow a \\,.\\,\\overrightarrow n $$
$$\\overrightarrow r \\,.\\,(7\\widehat i + 11\\widehat j + \\widehat k) = ( - 2\\widehat i + 4\\widehat j - 6\\widehat k).(7\\widehat i + 11\\widehat j + \\widehat k)$$
7x + 11y + z = $$-$$14 + 44 $$-$$6
7x + 11y + z = 24
$$ \\therefore $$ $$\\alpha$$ = 7
$$\\beta$$ = 11
$$\\gamma$$ = 1
$$ \\therefore $$ $$\\alpha$$ + $$\\beta$$ + $$\\gamma$$ = 19",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3442,
"subject": "General Science",
"question": "Let P be a plane containing the line $${{x - 1} \\over 3} = {{y + 6} \\over 4} = {{z + 5} \\over 2}$$ and parallel to the line $${{x - 1} \\over 4} = {{y - 2} \\over { - 3}} = {{z + 5} \\over 7}$$. If the point (1, $$-$$1, $$\\alpha$$) lies on the plane P, then the value of |5$$\\alpha$$| is equal to ____________.",
"options": [],
"answer": "38",
"solution": "**Answer:** 38\n\n
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3443,
"subject": "General Science",
"question": "Let P be a plane passing through the points (1, 0, 1), (1, $$-$$2, 1) and (0, 1, $$-$$2). Let a vector $$\\overrightarrow a = \\alpha \\widehat i + \\beta \\widehat j + \\gamma \\widehat k$$ be such that $$\\overrightarrow a $$ is parallel to the plane P, perpendicular to $$(\\widehat i + 2\\widehat j + 3\\widehat k)$$ and $$\\overrightarrow a \\,.\\,(\\widehat i + \\widehat j + 2\\widehat k) = 2$$, then $${(\\alpha - \\beta + \\gamma )^2}$$ equals ____________.",
"options": [],
"answer": "81",
"solution": "**Answer:** 81\n\nEquation of plane :
So, ($$\\alpha$$ $$-$$ $$\\beta$$ + $\\gamma$)2 = 81",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3444,
"subject": "General Science",
"question": "Consider the line L given by the equation
Let Q be the mirror image of the point (2, 3, $$-$$1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P?",
"options": [
{
"text": "($$-$$1, 1, 2)"
},
{
"text": "(1, 1, 1)"
},
{
"text": "(1, 1, 2)"
},
{
"text": "(1, 2, 2)"
}
],
"answer": "(1, 2, 2)",
"solution": "**Answer:** (1, 2, 2)\n\nPlane p is $${ \\bot ^r}$$ to line $${{x - 3} \\over 2} = {{y - 1} \\over 1} = {{z - 2} \\over 1}$$ & passes through pt. (2, 3) equation of plane p
2(x $$-$$ 2) + 1(y $$-$$ 3) + 1 (z + 1) = 0
2x + y + z $$-$$ 6 = 0
Point (1, 2, 2) satisfies above equation",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3445,
"subject": "General Science",
"question": "Let L be the line of intersection of planes $$\\overrightarrow r .(\\widehat i - \\widehat j + 2\\widehat k) = 2$$ and $$\\overrightarrow r .(2\\widehat i + \\widehat j - \\widehat k) = 2$$. If $$P(\\alpha ,\\beta ,\\gamma )$$ is the foot of perpendicular on L from the point (1, 2, 0), then the value of $$35(\\alpha + \\beta + \\gamma )$$ is equal to :",
"options": [
{
"text": "101"
},
{
"text": "119"
},
{
"text": "143"
},
{
"text": "134"
}
],
"answer": "119",
"solution": "**Answer:** 119\n\n$${P_1}:x - y + 2z = 2$$
$${P_2}:2x + y - 3 = 2$$
Let line of Intersection of planes P1 and P2 cuts xy plane in point Q.
$$ \\Rightarrow 35(\\alpha + \\beta + \\gamma ) = {{51} \\over {15}} \\times 35 = 119$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3446,
"subject": "General Science",
"question": "Let the foot of perpendicular from a point P(1, 2, $$-$$1) to the straight line $$L:{x \\over 1} = {y \\over 0} = {z \\over { - 1}}$$ be N. Let a line be drawn from P parallel to the plane x + y + 2z = 0 which meets L at point Q. If $$\\alpha$$ is the acute angle between the lines PN and PQ, then cos$$\\alpha$$ is equal to ________________.",
"options": [
{
"text": "$${1 \\over {\\sqrt 5 }}$$"
},
{
"text": "$${{\\sqrt 3 } \\over 2}$$"
},
{
"text": "$${1 \\over {\\sqrt 3 }}$$"
},
{
"text": "$${1 \\over {2\\sqrt 3 }}$$"
}
],
"answer": "$${1 \\over {\\sqrt 3 }}$$",
"solution": "**Answer:** $${1 \\over {\\sqrt 3 }}$$\n\n $$\\overrightarrow {PN} .(\\widehat i - \\widehat k) = 0$$
$$\\overrightarrow {PN} $$ = 2$$\\widehat j$$ and $$\\overrightarrow {PQ} $$ = $$2\\widehat i + 2\\widehat j - 2\\widehat k$$
$$ \\Rightarrow \\cos \\alpha = {1 \\over {\\sqrt 3 }}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3447,
"subject": "General Science",
"question": "For real numbers $$\\alpha$$ and $$\\beta$$ $$\\ne$$ 0, if the point of intersection of the straight lines
$${{x - \\alpha } \\over 1} = {{y - 1} \\over 2} = {{z - 1} \\over 3}$$ and $${{x - 4} \\over \\beta } = {{y - 6} \\over 3} = {{z - 7} \\over 3}$$, lies on the plane x + 2y $$-$$ z = 8, then $$\\alpha$$ $$-$$ $$\\beta$$ is equal to :",
"options": [
{
"text": "5"
},
{
"text": "9"
},
{
"text": "3"
},
{
"text": "7"
}
],
"answer": "7",
"solution": "**Answer:** 7\n\nFirst line is ($$\\phi$$ + $$\\alpha$$, 2$$\\phi$$ + 1, 3$$\\phi$$ + 1)
and second line is (q$$\\beta$$ + 4, 3q + 6, 3q + 7)
Now, point of intersection is ($$\\alpha$$ + 1, 3, 4)
It lies on the plane.
Hence, $$\\alpha$$ = 5 & $$\\beta$$ = $$-$$2",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3448,
"subject": "General Science",
"question": "The distance of the point P(3, 4, 4) from the point of intersection of the line joining the points. Q(3, $$-$$4, $$-$$5) and R(2, $$-$$3, 1) and the plane 2x + y + z = 7, is equal to ______________.",
"options": [],
"answer": "7",
"solution": "**Answer:** 7\n\n$$\\overrightarrow {QR} : - {{x - 3} \\over 1} = {{y + 4} \\over { - 1}} = {{z + 5} \\over { - 6}} = r$$
$$ \\Rightarrow $$ $$k = 1$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3450,
"subject": "General Science",
"question": "Let a plane P pass through the point (3, 7, $$-$$7) and contain the line, $${{x - 2} \\over { - 3}} = {{y - 3} \\over 2} = {{z + 2} \\over 1}$$. If distance of the plane P from the origin is d, then d2 is equal to ______________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n$$\\overrightarrow {BA} = (\\widehat i + 4\\widehat j - 5\\widehat k)$$
$$a\\widehat i + b\\widehat j + c\\widehat k = - 14\\widehat i - \\widehat j(14) + \\widehat k( - 14)$$
a = 1, b = 1, c = 1
Plane is (x $$-$$ 2) + (y $$-$$ 3) + (z + 2) = 0
$$ \\Rightarrow $$ x + y + z $$-$$ 3 = 0
$$ \\therefore $$ d = $$\\sqrt 3 $$ $$\\Rightarrow$$ d2 = 3",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3451,
"subject": "General Science",
"question": "A plane P contains the line $$x + 2y + 3z + 1 = 0 = x - y - z - 6$$, and is perpendicular to the plane $$ - 2x + y + z + 8 = 0$$. Then which of the following points lies on P?",
"options": [
{
"text": "($$-$$1, 1, 2)"
},
{
"text": "(0, 1, 1)"
},
{
"text": "(1, 0, 1)"
},
{
"text": "(2, $$-$$1, 1)"
}
],
"answer": "(0, 1, 1)",
"solution": "**Answer:** (0, 1, 1)\n\nEquation of plane P can be assumed as
P : x + 2y + 3z + 1 + $$\\lambda$$ (x $$-$$ y $$-$$ z $$-$$ 6) = 0
(0, 1, 1) lies on P.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3452,
"subject": "General Science",
"question": "Let the line L be the projection of the line $${{x - 1} \\over 2} = {{y - 3} \\over 1} = {{z - 4} \\over 2}$$ in the plane x $$-$$ 2y $$-$$ z = 3. If d is the distance of the point (0, 0, 6) from L, then d2 is equal to _______________.",
"options": [],
"answer": "26",
"solution": "**Answer:** 26\n\nTo find the projection let's find the foot of perpendicular from $(1,3$,\n4) to plane $x-2 y-z=3$\n
$$ \\Rightarrow \\overrightarrow r .\\left( {\\widehat j - 3\\widehat k} \\right) + 6 = 0$$ Ans.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3457,
"subject": "General Science",
"question": "Let S be the mirror image of the point Q(1, 3, 4) with respect to the plane 2x $$-$$ y + z + 3 = 0 and let R(3, 5, $$\\gamma$$) be a point of this plane. Then the square of the length of the line segment SR is ___________.",
"options": [],
"answer": "72",
"solution": "**Answer:** 72\n\n
Since R(3, 5, $$\\gamma$$) lies on the plane 2x $$-$$ y + z + 3 = 0.
$$\\Rightarrow$$ $$\\alpha$$ + $$\\beta$$ + $$\\gamma$$ = $$-$$23",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3459,
"subject": "General Science",
"question": "The square of the distance of the point of intersection
of the line $${{x - 1} \\over 2} = {{y - 2} \\over 3} = {{z + 1} \\over 6}$$ and the plane $$2x - y + z = 6$$ from the point ($$-$$1, $$-$$1, 2) is __________.",
"options": [],
"answer": "61",
"solution": "**Answer:** 61\n\n$${{x - 1} \\over 2} = {{y - 2} \\over 3} = {{z + 1} \\over 6} = \\lambda $$
Let $${P_1}:\\overrightarrow r \\,.\\,\\left( {2\\widehat i + \\widehat j - 3\\widehat k} \\right) = 4$$ be a plane. Let P2 be another plane which passes through the points (2, $$-$$3, 2), (2, $$-$$2, $$-$$3) and (1, $$-$$4, 2). If the direction ratios of the line of intersection of P1 and P2 be 16, $$\\alpha$$, $$\\beta$$, then the value of $$\\alpha$$ + $$\\beta$$ is equal to ________________.
Let $${{x - 2} \\over 3} = {{y + 1} \\over { - 2}} = {{z + 3} \\over { - 1}}$$ lie on the plane $$px - qy + z = 5$$, for some p, q $$\\in$$ R. The shortest distance of the plane from the origin is :
Let Q be the mirror image of the point P(1, 2, 1) with respect to the plane x + 2y + 2z = 16. Let T be a plane passing through the point Q and contains the line $$\\overrightarrow r = - \\widehat k + \\lambda \\left( {\\widehat i + \\widehat j + 2\\widehat k} \\right),\\,\\lambda \\in R$$. Then, which of the following points lies on T?
\nEquation of plane $\\equiv 2(x-0)+0(y-0)-1(z+1)$ $=0$\n
\n$2 x-z=1$\n
\nPoint lying on plane from the option is $(1,2,1)$ i.e., option (B)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3465,
"subject": "General Science",
"question": "
If two distinct point Q, R lie on the line of intersection of the planes $$ - x + 2y - z = 0$$ and $$3x - 5y + 2z = 0$$ and $$PQ = PR = \\sqrt {18} $$ where the point P is (1, $$-$$2, 3), then the area of the triangle PQR is equal to :
Let the plane $$P:\\overrightarrow r \\,.\\,\\overrightarrow a = d$$ contain the line of intersection of two planes $$\\overrightarrow r \\,.\\,\\left( {\\widehat i + 3\\widehat j - \\widehat k} \\right) = 6$$ and $$\\overrightarrow r \\,.\\,\\left( { - 6\\widehat i + 5\\widehat j - \\widehat k} \\right) = 7$$. If the plane P passes through the point $$\\left( {2,3,{1 \\over 2}} \\right)$$, then the value of $${{|13\\overrightarrow a {|^2}} \\over {{d^2}}}$$ is equal to :
Family of planes passing through line of intersection of P1 and P2 is given by $$x(1 - 6\\lambda ) + y(3 + 5\\lambda ) + z( - 1 - \\lambda ) - (6 + 7\\lambda ) = 0$$
\n
It passes through $$\\left( {2,3,{1 \\over 2}} \\right)$$
Let the foot of the perpendicular from the point (1, 2, 4) on the line $${{x + 2} \\over 4} = {{y - 1} \\over 2} = {{z + 1} \\over 3}$$ be P. Then the distance of P from the plane $$3x + 4y + 12z + 23 = 0$$ is :
Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x $$-$$ 3y + 5z = 8. If the mirror image of the point $$\\left( {2, - {1 \\over 2},2} \\right)$$ in the rotated plane is B(a, b, c), then :
If the lines $$\\overrightarrow r = \\left( {\\widehat i - \\widehat j + \\widehat k} \\right) + \\lambda \\left( {3\\widehat j - \\widehat k} \\right)$$ and $$\\overrightarrow r = \\left( {\\alpha \\widehat i - \\widehat j} \\right) + \\mu \\left( {2\\widehat i - 3\\widehat k} \\right)$$ are co-planar, then the distance of the plane containing these two lines from the point ($$\\alpha$$, 0, 0) is :
Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, $$-$$1, $$-$$1), parallel to the line PQ meets the plane S at R, then QR2 is equal to :
$${L_1}:\\overrightarrow r = \\lambda \\left( {\\widehat i + 2\\widehat j + 3\\widehat k} \\right),\\,\\lambda \\in R$$
\n
$${L_2}:\\overrightarrow r = \\left( {\\widehat i + 3\\widehat j + \\widehat k} \\right) + \\mu \\left( {\\widehat i + \\widehat j + 5\\widehat k} \\right);\\,\\mu \\in R$$,
\n
intersect at the point S. If a plane ax + by $$-$$ z + d = 0 passes through S and is parallel to both the lines L1 and L2, then the value of a + b + d is equal to ____________.
The distance of the point (3, 2, $$-$$1) from the plane $$3x - y + 4z + 1 = 0$$ along the line $${{2 - x} \\over 2} = {{y - 3} \\over 2} = {{z + 1} \\over 1}$$ is equal to :
Let $$\\mathrm{P}$$ be the plane containing the straight line $$\\frac{x-3}{9}=\\frac{y+4}{-1}=\\frac{z-7}{-5}$$ and perpendicular to the plane containing the straight lines $$\\frac{x}{2}=\\frac{y}{3}=\\frac{z}{5}$$ and $$\\frac{x}{3}=\\frac{y}{7}=\\frac{z}{8}$$. If $$\\mathrm{d}$$ is the distance of $$\\mathrm{P}$$ from the point $$(2,-5,11)$$, then $$\\mathrm{d}^{2}$$ is equal to :
",
"options": [
{
"text": "$$\\frac{147}{2}$$"
},
{
"text": "96"
},
{
"text": "$$\\frac{32}{3}$$"
},
{
"text": "54"
}
],
"answer": "$$\\frac{32}{3}$$",
"solution": "**Answer:** $$\\frac{32}{3}$$\n\nLet $\\langle a, b, c\\rangle$ be direction ratios of plane containing\n
\n$$\n\\begin{aligned}\n&\\text { lines } \\frac{x}{2}=\\frac{y}{3}=\\frac{z}{5} \\text { and } \\frac{x}{3}=\\frac{y}{7}=\\frac{z}{8} . \\\\\\\\\n&\\therefore \\quad 2 a+3 b+5 c=0 \\quad \\ldots \\text { (i) } \\\\\\\\\n&\\text { and } 3 a+7 b+8 c=0 \\quad \\ldots \\text { (ii) } \\\\\\\\\n&\\text { from eq. (i) and (ii) }: \\frac{a}{24-35}=\\frac{b}{15-16}=\\frac{c}{14-9} \\\\\\\\\n&\\therefore \\text { D. R's} \\text {. of plane are }<11,1,-5>\n\\end{aligned}\n$$\n
\nLet $D . R$'s of plane $P$ be $< a_{1}, b_{1}, c_{1} >$ then.\n
\n$11 a_{1}+b_{1}-5 c_{1}=0$\n
\nand $9 a_{1}-b_{1}-5 c_{1}=0$\n
\nFrom eq. (iii) and (iv) :\n
\n$$\n\\begin{aligned}\n&\\frac{a_{1}}{-5-5}=\\frac{b_{1}}{-45+55}=\\frac{c_{1}}{-11-9} \\\\\\\\\n&\\therefore \\text { D.R's } \\text { of plane } P \\text { are }< 1,-1,2 >\n\\end{aligned}\n$$\n
\nEquation plane $P$ is : $1(x-3)-1(y+4)+2(z-7)=0$\n
\n$$\n\\Rightarrow x-y+2 z-21=0\n$$\n
\nDistance from point $(2,-5,11)$ is $d=\\frac{|2+5+22-2|}{\\sqrt{6}}$
The line of shortest distance between the lines $$\\frac{x-2}{0}=\\frac{y-1}{1}=\\frac{z}{1}$$ and $$\\frac{x-3}{2}=\\frac{y-5}{2}=\\frac{z-1}{1}$$ makes an angle of $$\\cos ^{-1}\\left(\\sqrt{\\frac{2}{27}}\\right)$$ with the plane $$\\mathrm{P}: \\mathrm{a} x-y-z=0$$, $$(a>0)$$. If the image of the point $$(1,1,-5)$$ in the plane $$P$$ is $$(\\alpha, \\beta, \\gamma)$$, then $$\\alpha+\\beta-\\gamma$$ is equal to _________________.
",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nDR's of line of shortest distance
\nNo value of (a)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3475,
"subject": "General Science",
"question": "
The largest value of $$a$$, for which the perpendicular distance of the plane containing the lines $$\n\\vec{r}=(\\hat{i}+\\hat{j})+\\lambda(\\hat{i}+a \\hat{j}-\\hat{k})$$ and $$\\vec{r}=(\\hat{i}+\\hat{j})+\\mu(-\\hat{i}+\\hat{j}-a \\hat{k})$$ from the point $$(2,1,4)$$ is $$\\sqrt{3}$$, is _________.
The plane passing through the line $$L: l x-y+3(1-l) z=1, x+2 y-z=2$$ and perpendicular to the plane $$3 x+2 y+z=6$$ is $$3 x-8 y+7 z=4$$. If $$\\theta$$ is the acute angle between the line $$L$$ and the $$y$$-axis, then $$415 \\cos ^{2} \\theta$$ is equal to _____________.
Let the line $$\\frac{x-3}{7}=\\frac{y-2}{-1}=\\frac{z-3}{-4}$$ intersect the plane containing the lines $$\\frac{x-4}{1}=\\frac{y+1}{-2}=\\frac{z}{1}$$ and $$4 a x-y+5 z-7 a=0=2 x-5 y-z-3, a \\in \\mathbb{R}$$ at the point $$P(\\alpha, \\beta, \\gamma)$$. Then the value of $$\\alpha+\\beta+\\gamma$$ equals _____________.
If the line of intersection of the planes $$a x+b y=3$$ and $$a x+b y+c z=0$$, a $$>0$$ makes an angle $$30^{\\circ}$$ with the plane $$y-z+2=0$$, then the direction cosines of the line are :
If the foot of the perpendicular from the point $$\\mathrm{A}(-1,4,3)$$ on the plane $$\\mathrm{P}: 2 x+\\mathrm{m} y+\\mathrm{n} z=4$$, is $$\\left(-2, \\frac{7}{2}, \\frac{3}{2}\\right)$$, then the distance of the point A from the plane P, measured parallel to a line with direction ratios $$3,-1,-4$$, is equal to :
Let a line with direction ratios $$a,-4 a,-7$$ be perpendicular to the lines with direction ratios $$3,-1,2 b$$ and $$b, a,-2$$. If the point of intersection of the line $$\\frac{x+1}{a^{2}+b^{2}}=\\frac{y-2}{a^{2}-b^{2}}=\\frac{z}{1}$$ and the plane $$x-y+z=0$$ is $$(\\alpha, \\beta, \\gamma)$$, then $$\\alpha+\\beta+\\gamma$$ is equal to _________.
$$\\Rightarrow$$ General point on line is $$(5\\lambda - 1,\\,3\\lambda + 2,\\,\\lambda )$$ for finding point of intersection with $$x - y + z = 0$$ we get $$(5\\lambda - 1) - (3\\lambda + 2) + (\\lambda ) = 0$$
Let the plane P pass through the intersection of the planes $$2x+3y-z=2$$ and $$x+2y+3z=6$$, and be perpendicular to the plane $$2x+y-z+1=0$$. If d is the distance of P from the point ($$-$$7, 1, 1), then $$\\mathrm{d^{2}}$$ is equal to :
",
"options": [
{
"text": "$$\\frac{250}{83}$$"
},
{
"text": "$$\\frac{250}{82}$$"
},
{
"text": "$$\\frac{15}{53}$$"
},
{
"text": "$$\\frac{25}{83}$$"
}
],
"answer": "$$\\frac{250}{83}$$",
"solution": "**Answer:** $$\\frac{250}{83}$$\n\nPlane $P$, is passing through intersection of the two planes, so,\n
The point of intersection $$\\mathrm{C}$$ of the plane $$8 x+y+2 z=0$$ and the line joining the points $$\\mathrm{A}(-3,-6,1)$$ and $$\\mathrm{B}(2,4,-3)$$ divides the line segment $$\\mathrm{AB}$$ internally in the ratio $$\\mathrm{k}: 1$$. If $$\\mathrm{a}, \\mathrm{b}, \\mathrm{c}(|\\mathrm{a}|,|\\mathrm{b}|,|\\mathrm{c}|$$ are coprime) are the direction ratios of the perpendicular from the point $$\\mathrm{C}$$ on the line $$\\frac{1-x}{1}=\\frac{y+4}{2}=\\frac{z+2}{3}$$, then $$|\\mathrm{a}+\\mathrm{b}+\\mathrm{c}|$$ is equal to ___________.
Let $$\\alpha x+\\beta y+\\gamma z=1$$ be the equation of a plane passing through the point $$(3,-2,5)$$ and perpendicular to the line joining the points $$(1,2,3)$$ and $$(-2,3,5)$$. Then the value of $$\\alpha \\beta y$$ is equal to _____________.
$\\Rightarrow$ Equation of plane $\\mathrm{P}$ is $2 \\mathrm{x}-3 \\mathrm{y}+\\mathrm{z}+3=0$\n\n
$\\Rightarrow$ Distance of line L from Plane $\\mathrm{P}$ is\n\n
$=\\left|\\frac{0-3(6)+1+3}{\\sqrt{4+9+1}}\\right|$\n\n$=\\sqrt{14}$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3486,
"subject": "General Science",
"question": "The foot of perpendicular from the origin $\\mathrm{O}$ to a plane $\\mathrm{P}$ which meets the co-ordinate axes at the points $\\mathrm{A}, \\mathrm{B}, \\mathrm{C}$ is $(2, \\mathrm{a}, 4), \\mathrm{a} \\in \\mathrm{N}$. If the volume of the tetrahedron $\\mathrm{OABC}$ is 144 unit$^{3}$, then which of the following points is NOT on P ?",
"options": [
{
"text": "$(3,0,4)$"
},
{
"text": "$(0,6,3)$"
},
{
"text": "$(0,4,4)$"
},
{
"text": "$(2,2,4)$"
}
],
"answer": "$(3,0,4)$",
"solution": "**Answer:** $(3,0,4)$\n\n Equation of Plane:\n\n
$$\n\\begin{aligned}\n& (2 \\hat{i}+a \\hat{j}+4 \\hat{\\mathrm{k}}) \\cdot[(\\mathrm{x}-2) \\hat{\\mathrm{i}}+(\\mathrm{y}-\\mathrm{a}) \\hat{\\mathrm{j}}+(\\mathrm{z}-4) \\hat{\\mathrm{k}}]=0 \\\\\\\\\n& \\Rightarrow 2 \\mathrm{x}+\\mathrm{ay}+4 \\mathrm{z}=20+\\mathrm{a}^{2} \\\\\\\\\n& \\Rightarrow \\mathrm{A} \\equiv\\left(\\frac{20+\\mathrm{a}^{2}}{2}, 0,0\\right) \\\\\\\\\n& \\mathrm{B} \\equiv\\left(0, \\frac{20+\\mathrm{a}^{2}}{\\mathrm{a}}, 0\\right) \\\\\\\\\n& \\mathrm{C} \\equiv\\left(0,0, \\frac{20+\\mathrm{a}^{2}}{4}\\right) \\\\\\\\\n& \\Rightarrow \\text { Volume of tetrahedron } \\\\\\\\\n& =\\frac{1}{6}\\left[ {\\matrix{\n {\\overrightarrow a } & {\\overrightarrow b } & {\\overrightarrow c } \\cr \n\n } } \\right] \\\\\\\\\n& =\\frac{1}{6} \\overrightarrow{\\mathrm{a}} \\cdot(\\overrightarrow{\\mathrm{b}} \\times \\overrightarrow{\\mathrm{c}}) \\\\\\\\\n& \\Rightarrow \\frac{1}{6}\\left(\\frac{20+\\mathrm{a}^{2}}{2}\\right) \\cdot\\left(\\frac{20+\\mathrm{a}^{2}}{\\mathrm{a}}\\right) \\cdot\\left(\\frac{20+\\mathrm{a}^{2}}{4}\\right)=144 \\\\\\\\\n& \\Rightarrow\\left(20+\\mathrm{a}^{2}\\right)^{3}=144 \\times 48 \\times \\mathrm{a} \\\\\\\\\n& \\Rightarrow \\mathrm{a}=2 \\\\\\\\\n& \\Rightarrow \\text { Equation of plane is } 2 \\mathrm{x}+2 \\mathrm{y}+4 \\mathrm{z}=24 \\\\\\\\\n& \\text { Or } \\mathrm{x}+\\mathrm{y}+2 \\mathrm{z}=12 \\\\\\\\\n& \\Rightarrow(3,0,4) \\quad \\text { Not } \\text { lies on } \\text { the } \\\\\\\\\n& \\mathrm{x}+\\mathrm{y}+2 \\mathrm{z}=12\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3487,
"subject": "General Science",
"question": "Let $P$ be the plane, passing through the point $(1,-1,-5)$ and perpendicular to the line joining the points $(4,1,-3)$ and $(2,4,3)$. Then the distance of $P$ from the point $(3,-2,2)$ is :",
"options": [
{
"text": "5"
},
{
"text": "4"
},
{
"text": "6"
},
{
"text": "7"
}
],
"answer": "5",
"solution": "**Answer:** 5\n\nEquation of Plane :\n\n
Let the line $$L: \\frac{x-1}{2}=\\frac{y+1}{-1}=\\frac{z-3}{1}$$ intersect the plane $$2 x+y+3 z=16$$ at the point \n$$P$$. Let the point $$Q$$ be the foot of perpendicular from the point $$R(1,-1,-3)$$ on the line $$L$$. If $$\\alpha$$ is the area of triangle $$P Q R$$, then $$\\alpha^{2}$$ is equal to __________.
Let $$\\theta$$ be the angle between the planes $$P_{1}: \\vec{r} \\cdot(\\hat{i}+\\hat{j}+2 \\hat{k})=9$$ and $$P_{2}: \\vec{r} \\cdot(2 \\hat{i}-\\hat{j}+\\hat{k})=15$$. Let $$\\mathrm{L}$$ be the line that meets $$P_{2}$$ at the point $$(4,-2,5)$$ and makes an angle $$\\theta$$ with the normal of $$P_{2}$$. If $$\\alpha$$ is the angle between $$\\mathrm{L}$$ and $$P_{2}$$, then $$\\left(\\tan ^{2} \\theta\\right)\\left(\\cot ^{2} \\alpha\\right)$$ is equal to ____________.
$$\n\\begin{aligned}\n& P_{2}: \\vec{r} \\cdot(2 \\hat{i}-\\hat{j}+\\hat{k})=15 \\\\\\\\\n& \\text { then } \\cos \\theta=\\frac{3}{\\sqrt{6} \\cdot \\sqrt{6}}=\\frac{1}{2} \\\\\\\\\n& \\begin{aligned}\n& \\therefore \\theta=\\frac{\\pi}{3}, \\text { Now } \\alpha=\\frac{\\pi}{2}-\\theta \\\\\\\\\n& \\therefore \\tan ^{2} \\theta \\cdot \\cot ^{2} \\alpha=\\tan ^{4} \\theta \\\\\\\\\n&=(\\sqrt{3})^{4}=9\n\\end{aligned}\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3490,
"subject": "General Science",
"question": "A vector $\\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\\circ}$, to the $y$-axis at 45 and to the $z$-axis at an acute angle. If a plane passing through the points $(\\sqrt{2},-1,1)$ and $(a, b, c)$, is normal to $\\vec{v}$, then :",
"options": [
{
"text": "$a+b+\\sqrt{2} c=1$"
},
{
"text": "$\\sqrt{2} a+b+c=1$"
},
{
"text": "$\\sqrt{2} a-b+c=1$"
},
{
"text": "$a+\\sqrt{2} b+c=1$"
}
],
"answer": "$a+\\sqrt{2} b+c=1$",
"solution": "**Answer:** $a+\\sqrt{2} b+c=1$\n\n
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3491,
"subject": "General Science",
"question": "If a plane passes through the points $(-1, k, 0),(2, k,-1),(1,1,2)$ and is parallel to the line $\\frac{x-1}{1}=\\frac{2 y+1}{2}=\\frac{z+1}{-1}$, then the value of $\\frac{k^2+1}{(k-1)(k-2)}$ is :",
"options": [
{
"text": "$\\frac{17}{5}$"
},
{
"text": "$\\frac{6}{13}$"
},
{
"text": "$\\frac{13}{6}$"
},
{
"text": "$\\frac{5}{17}$"
}
],
"answer": "$\\frac{13}{6}$",
"solution": "**Answer:** $\\frac{13}{6}$\n\n
The line $$l_1$$ passes through the point (2, 6, 2) and is perpendicular to the plane $$2x+y-2z=10$$. Then the shortest distance between the line $$l_1$$ and the line $$\\frac{x+1}{2}=\\frac{y+4}{-3}=\\frac{z}{2}$$ is :
If the equation of the plane passing through the point $$(1,1,2)$$ and perpendicular to the line $$x-3 y+ 2 z-1=0=4 x-y+z$$ is $$\\mathrm{A} x+\\mathrm{B} y+\\mathrm{C} z=1$$, then $$140(\\mathrm{C}-\\mathrm{B}+\\mathrm{A})$$ is equal to ___________.
The plane $$2x-y+z=4$$ intersects the line segment joining the points A ($$a,-2,4)$$ and B ($$2,b,-3)$$ at the point C in the ratio 2 : 1 and the distance of the point C from the origin is $$\\sqrt5$$. If $$ab < 0$$ and P is the point $$(a-b,b,2b-a)$$ then CP$$^2$$ is equal to :
If the lines $${{x - 1} \\over 1} = {{y - 2} \\over 2} = {{z + 3} \\over 1}$$ and $${{x - a} \\over 2} = {{y + 2} \\over 3} = {{z - 3} \\over 1}$$ intersect at the point P, then the distance of the point P from the plane $$z = a$$ is :
Let the equation of the plane P containing the line $$x+10=\\frac{8-y}{2}=z$$ be $$ax+by+3z=2(a+b)$$ and the distance of the plane $$P$$ from the point (1, 27, 7) be $$c$$. Then $$a^2+b^2+c^2$$ is equal to __________.
",
"options": [],
"answer": "355",
"solution": "**Answer:** 355\n\nThe line $\\frac{x+10}{1}=\\frac{y-8}{-2}=\\frac{z}{1}$ have a point $(-10,8,0)$ with d. r. $(1,-2,1)$\n
Let the equation of the plane passing through the line $$x - 2y - z - 5 = 0 = x + y + 3z - 5$$ and parallel to the line $$x + y + 2z - 7 = 0 = 2x + 3y + z - 2$$ be $$ax + by + cz = 65$$. Then the distance of the point (a, b, c) from the plane $$2x + 2y - z + 16 = 0$$ is ____________.
",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\nLet the equation of the plane is\n
\n$(x-2 y-z-5)+\\lambda(x+y+3 z-5)=0$\n
\n$\\because \\quad$ it's parallel to the line\n
\n$x+y+2 z-7=0=2 x+3 y+z-2$
So, vector along the line $\\left|\\begin{array}{lll}\\hat{i} & \\hat{j} & \\hat{k} \\\\ 1 & 1 & 2 \\\\ 2 & 3 & 1\\end{array}\\right|$\n
If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes $$x+2y+z=0$$ and $$3y-z=3$$ is ($$\\alpha,\\beta,\\gamma$$), then $$\\alpha+\\beta+\\gamma$$ is equal to :
\n Equation of line passing through the point P$(-1,9,-16)$ and parallel to line $\\frac{x+4}{3}=\\frac{2-y}{4}=\\frac{z-3}{12}$ is $\\frac{x+1}{3}=\\frac{y-9}{-4}=\\frac{z+16}{12}=\\lambda$\n
Any point on this line (A) $=(3 \\lambda-1,-4 \\lambda+9,12 \\lambda-16)$\n
Point of intersection line and plane\n
$$\n\\begin{aligned}\n& 2(3 \\lambda-1)+3(-4 \\lambda+9)-1(12 \\lambda-16)=5 \\\\\\\\\n\\Rightarrow & 6 \\lambda-2-12 \\lambda+27-12 \\lambda+16=5 \\\\\\\\\n\\Rightarrow & \\lambda=2 \\quad \\\\\\\\\n\\therefore & \\text { Point A }=(5,1,8) \\\\\\\\\n\\therefore & \\text { Distance (AP) } \\\\\\\\\n= & \\sqrt{(5+1)^2+(9-1)^2+(-16-8)^2}\\\\\\\\\n= & \\sqrt{36+64+576}\\\\\\\\\n= & 26 \\text { units }\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3501,
"subject": "General Science",
"question": "Let the plane $P$ contain the line $2 x+y-z-3=0=5 x-3 y+4 z+9$ and be
parallel to the line $\\frac{x+2}{2}=\\frac{3-y}{-4}=\\frac{z-7}{5}$. Then the distance of the point
$\\mathrm{A}(8,-1,-19)$ from the plane $\\mathrm{P}$ measured parallel to the line $\\frac{x}{-3}=\\frac{y-5}{4}=\\frac{2-z}{-12}$
is equal to ______________.",
"options": [],
"answer": "26",
"solution": "**Answer:** 26\n\nPlane $\\mathrm{P} \\equiv \\mathrm{P}_1+\\lambda \\mathrm{P}_2=0$
The plane, passing through the points $$(0,-1,2)$$ and $$(-1,2,1)$$ and parallel to the line passing through $$(5,1,-7)$$ and $$(1,-1,-1)$$, also passes through the point :
The first step is to find the normal vector to the desired plane. Since the plane is parallel to the line passing through the points (5, 1, -7) and (1, -1, -1), the direction vector of that line is also parallel to the plane. The direction vector is the difference between the coordinates of the two points, which is (5-1, 1-(-1), -7-(-1)) = (4, 2, -6).
\n
Next, let's find another vector that is parallel to the plane. This vector can be obtained by taking the difference between the coordinates of the points (0, -1, 2) and (-1, 2, 1), which the plane passes through. This gives us a vector of (0-(-1), -1-2, 2-1) = (1, -3, 1).
\n
The normal to the plane is perpendicular to both these vectors. It can be found by taking the cross product of the two vectors.
\n
The cross product of vectors (4, 2, -6) and (1, -3, 1) is :
The distance of the point $$(-1,2,3)$$ from the plane $$\\vec{r} \\cdot(\\hat{i}-2 \\hat{j}+3 \\hat{k})=10$$ parallel to the line of the shortest distance between the lines $$\\vec{r}=(\\hat{i}-\\hat{j})+\\lambda(2 \\hat{i}+\\hat{k})$$ and $$\\vec{r}=(2 \\hat{i}-\\hat{j})+\\mu(\\hat{i}-\\hat{j}+\\hat{k})$$ is :
",
"options": [
{
"text": "$$3 \\sqrt{6}$$"
},
{
"text": "$$3 \\sqrt{5}$$"
},
{
"text": "$$2 \\sqrt{6}$$"
},
{
"text": "$$2 \\sqrt{5}$$"
}
],
"answer": "$$2 \\sqrt{6}$$",
"solution": "**Answer:** $$2 \\sqrt{6}$$\n\n1. Determine the line of shortest distance between the given two lines:\n\n\n
Direction vector of line 1: $$\\vec{d_1} = 2\\hat{i} + \\hat{k}$$\n
Direction vector of line 2: $$\\vec{d_2} = \\hat{i} - \\hat{j} + \\hat{k}$$\n\n
Since $$P$$ lies on the plane $$x - 2y + 3z = 10$$, we can substitute the coordinates of $$P$$ in terms of $$\\lambda$$ into the equation of the plane:\n\n
Let the plane P: $$4 x-y+z=10$$ be rotated by an angle $$\\frac{\\pi}{2}$$ about its line of intersection with the plane $$x+y-z=4$$. If $$\\alpha$$ is the distance of the point $$(2,3,-4)$$ from the new position of the plane $$\\mathrm{P}$$, then $$35 \\alpha$$ is equal to :
Let the line $$l: x=\\frac{1-y}{-2}=\\frac{z-3}{\\lambda}, \\lambda \\in \\mathbb{R}$$ meet the plane $$P: x+2 y+3 z=4$$ at the point $$(\\alpha, \\beta, \\gamma)$$. If the angle between the line $$l$$ and the plane $$P$$ is $$\\cos ^{-1}\\left(\\sqrt{\\frac{5}{14}}\\right)$$, then $$\\alpha+2 \\beta+6 \\gamma$$ is equal to ___________.
Let the foot of perpendicular from the point $$\\mathrm{A}(4,3,1)$$ on the plane $$\\mathrm{P}: x-y+2 z+3=0$$ be N. If B$$(5, \\alpha, \\beta), \\alpha, \\beta \\in \\mathbb{Z}$$ is a point on plane P such that the area of the triangle ABN is $$3 \\sqrt{2}$$, then $$\\alpha^{2}+\\beta^{2}+\\alpha \\beta$$ is equal to ___________.
Let two vertices of a triangle ABC be (2, 4, 6) and (0, $$-$$2, $$-$$5), and its centroid be (2, 1, $$-$$1). If the image of the third vertex in the plane $$x+2y+4z=11$$ is $$(\\alpha,\\beta,\\gamma)$$, then $$\\alpha\\beta+\\beta\\gamma+\\gamma\\alpha$$ is equal to :
",
"options": [
{
"text": "72"
},
{
"text": "74"
},
{
"text": "76"
},
{
"text": "70"
}
],
"answer": "74",
"solution": "**Answer:** 74\n\nGiven that two vertex of a $\\triangle A B C$ be $A(2,4,6)$ and
$B(0,-2,-5)$ and $G=$ centroid $=(2,1,-1)$\n[Given]\n
Let, the other vertex is $C(x, y, z)$\n
According to the question,\n
$$\n\\begin{aligned}\n& \\frac{2+0+x}{3}=2 \\\\\\\\\n&\\Rightarrow x =4\n\\end{aligned}\n$$\n
Let P be the point of intersection of the line $${{x + 3} \\over 3} = {{y + 2} \\over 1} = {{1 - z} \\over 2}$$ and the plane $$x+y+z=2$$. If the distance of the point P from the plane $$3x - 4y + 12z = 32$$ is q, then q and 2q are the roots of the equation :
Thus, $q$ and $2 q$ are roots of the equation $x^2-18 x+72=0$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3509,
"subject": "General Science",
"question": "
For $$\\mathrm{a}, \\mathrm{b} \\in \\mathbb{Z}$$ and $$|\\mathrm{a}-\\mathrm{b}| \\leq 10$$, let the angle between the plane $$\\mathrm{P}: \\mathrm{ax}+y-\\mathrm{z}=\\mathrm{b}$$ and the line $$l: x-1=\\mathrm{a}-y=z+1$$ be $$\\cos ^{-1}\\left(\\frac{1}{3}\\right)$$. If the distance of the point $$(6,-6,4)$$ from the plane P is $$3 \\sqrt{6}$$, then $$a^{4}+b^{2}$$ is equal to :
Let $$\\lambda_{1}, \\lambda_{2}$$ be the values of $$\\lambda$$ for which the points $$\\left(\\frac{5}{2}, 1, \\lambda\\right)$$ and $$(-2,0,1)$$ are at equal distance from the plane $$2 x+3 y-6 z+7=0$$. If $$\\lambda_{1} > \\lambda_{2}$$, then the distance of the point $$\\left(\\lambda_{1}-\\lambda_{2}, \\lambda_{2}, \\lambda_{1}\\right)$$ from the line $$\\frac{x-5}{1}=\\frac{y-1}{2}=\\frac{z+7}{2}$$ is ____________.
",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\nSince $\\left(\\frac{5}{2}, 1, \\lambda\\right)$ and $(-2,0,1)$ are equidistant \n
If the equation of the plane passing through the line of intersection of the planes $$2 x-y+z=3,4 x-3 y+5 z+9=0$$ and parallel to the line $$\\frac{x+1}{-2}=\\frac{y+3}{4}=\\frac{z-2}{5}$$ is $$a x+b y+c z+6=0$$, then $$a+b+c$$ is equal to :
Let the line $$\\mathrm{L}$$ pass through the point $$(0,1,2)$$, intersect the line $$\\frac{x-1}{2}=\\frac{y-2}{3}=\\frac{z-3}{4}$$ and be parallel to the plane $$2 x+y-3 z=4$$. Then the distance of the point $$\\mathrm{P}(1,-9,2)$$ from the line $$\\mathrm{L}$$ is :
Let $$A(2,3,5)$$ and $$C(-3,4,-2)$$ be opposite vertices of a parallelogram $$A B C D$$. If the diagonal $$\\overrightarrow{\\mathrm{BD}}=\\hat{i}+2 \\hat{j}+3 \\hat{k}$$, then the area of the parallelogram is equal to :
If $$A(3,1,-1), B\\left(\\frac{5}{3}, \\frac{7}{3}, \\frac{1}{3}\\right), C(2,2,1)$$ and $$D\\left(\\frac{10}{3}, \\frac{2}{3}, \\frac{-1}{3}\\right)$$ are the vertices of a quadrilateral $$A B C D$$, then its area is
$$A(3,1,-1), B\\left(\\frac{5}{3}, \\frac{7}{3}, \\frac{1}{3}\\right), C(2,2,1), D\\left(\\frac{10}{3}, \\frac{2}{3}, \\frac{-1}{3}\\right)$$ are vertices of a quadrilateral
or $$k = 0$$ or $$-3$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3517,
"subject": "General Science",
"question": "A line with direction cosines proportional to $$2,1,2$$ meets each of the lines $$x=y+a=z$$ and $$x+a=2y=2z$$ . The co-ordinates of each of the points of intersection are given by :",
"options": [
{
"text": "$$\\left( {2a,3a,3a} \\right),\\left( {2a,a,a} \\right)$$ "
},
{
"text": "$$\\left( {3a,2a,3a} \\right),\\left( {a,a,a} \\right)$$ "
},
{
"text": "$$\\left( {3a,2a,3a} \\right),\\left( {a,a,2a} \\right)$$ "
},
{
"text": "$$\\left( {3a,3a,3a} \\right),\\left( {a,a,a} \\right)$$ "
}
],
"answer": "$$\\left( {3a,2a,3a} \\right),\\left( {a,a,a} \\right)$$ ",
"solution": "**Answer:** $$\\left( {3a,2a,3a} \\right),\\left( {a,a,a} \\right)$$ \n\nLet a point on the line \n
$$x = y + a = z$$ is $$\\left( {\\lambda ,\\lambda - a,\\lambda } \\right)$$ \n
and a point on the line \n
$$x + a = 2y = 2z$$ is $$\\left( {\\mu - a,{\\mu \\over 2},{\\mu \\over 2}} \\right),$$ \n
then direction ratio of the line joining these points are \n
and $$\\left( {1 - a} \\right){5 \\over 2} + a = - {{13} \\over 2}$$\n
$$ \\Rightarrow b = 4$$ and $$a = 6$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3522,
"subject": "General Science",
"question": "If the straight lines $$\\,\\,\\,\\,\\,$$ $$\\,\\,\\,\\,\\,$$ $${{x - 1} \\over k} = {{y - 2} \\over 2} = {{z - 3} \\over 3}$$ $$\\,\\,\\,\\,\\,$$ and$$\\,\\,\\,\\,\\,$$ $${{x - 2} \\over 3} = {{y - 3} \\over k} = {{z - 1} \\over 2}$$ intersects at a point, then the integer $$k$$ is equal to ",
"options": [
{
"text": "$$-5$$"
},
{
"text": "$$5$$"
},
{
"text": "$$2$$ "
},
{
"text": "$$-2$$"
}
],
"answer": "$$-5$$",
"solution": "**Answer:** $$-5$$\n\nThe two lines intersect if shortest distance between them is zero $$i.e.$$ \n
$${{\\left( {{{\\overrightarrow a }_2} - {{\\overrightarrow a }_1}} \\right).{{\\overrightarrow b }_1} \\times {{\\overrightarrow b }_2}} \\over {\\left| {{{\\overrightarrow b }_1} \\times {{\\overrightarrow b }_2}} \\right|}} = 0$$\n
$$ \\Rightarrow \\left( {{{\\overrightarrow a }_2} - {{\\overrightarrow a }_1}} \\right).{\\overrightarrow b _1} \\times {\\overrightarrow b _2} = 0$$\n
where $${\\overrightarrow a _1} = \\widehat i + 2\\widehat j + 3\\widehat k,{\\overrightarrow b _1} = k\\widehat i + 2\\widehat j + 3\\widehat k$$\n
$$\\,\\,\\,\\,\\,\\,\\,{\\overrightarrow a _2} = 2\\widehat i + 3\\widehat j + \\widehat k,\\,\\,{\\widehat b_2} = 3\\widehat i + k\\widehat j + 2\\widehat k$$\n
As $$k$$ is an integer, therefore $$k=-5$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3523,
"subject": "General Science",
"question": "Statement - 1 : The point $$A(1,0,7)$$ is the mirror image of the point \n
$$B(1,6,3)$$ in the line : $${x \\over 1} = {{y - 1} \\over 2} = {{z - 2} \\over 3}$$ \n
Statement - 2 : The line $${x \\over 1} = {{y - 1} \\over 2} = {{z - 2} \\over 3}$$ bisects the line \n
segment joining $$A(1,0,7)$$ and $$B(1, 6, 3)$$ ",
"options": [
{
"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."
},
{
"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. "
}
],
"answer": "Statement -1 is true, Statement -2 is true; Statement -2 is not a correct explanation for Statement -1.",
"solution": "**Answer:** Statement -1 is true, Statement -2 is true; Statement -2 is not a correct explanation for Statement -1.\n\nThe directions ratios of the line segment joining points \n
$$A\\left( {1,0,7} \\right)\\,\\,$$ and $$\\,\\,\\,B\\left( {1,6,3} \\right)$$ are $$0,6, - 4.$$\n
The direction ratios of the given line are $$1,2,3.$$\n
So, the given line is perpendicular to line $$AB.$$\n
Also, the mid point of $$A$$ and $$B$$ is $$\\left( {1,3,5} \\right)$$ which lies on the given line. \n
So, the image of $$B$$ in the given line is $$A$$, because the given line is the perpendicular bisector of line segment joining points $$A$$ and $$B$$, But statement -$$2$$ is not a correct explanation for statement$$-1.$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3524,
"subject": "General Science",
"question": "If the line $${{x - 1} \\over 2} = {{y + 1} \\over 3} = {{z - 1} \\over 4}$$ and $${{x - 3} \\over 1} = {{y - k} \\over 2} = {z \\over 1}$$ intersect, then $$k$$ is equal to :",
"options": [
{
"text": "$$-1$$ "
},
{
"text": "$${2 \\over 9}$$"
},
{
"text": "$${9 \\over 2}$$"
},
{
"text": "$$0$$ "
}
],
"answer": "$${9 \\over 2}$$",
"solution": "**Answer:** $${9 \\over 2}$$\n\nGiven lines in vector form are \n
$$\\overrightarrow r = \\left( {\\widehat i - \\overrightarrow j + \\overrightarrow k } \\right) + \\lambda \\left( {2\\overrightarrow i + 3\\overrightarrow j + 4\\overrightarrow j } \\right)$$ \n
and $$\\overrightarrow r = \\left( {3\\widehat i + k\\widehat j} \\right) + \\mu \\left( {\\widehat i + 2\\widehat j + \\widehat k} \\right)$$\n
These will intersect if shortest distance between them $$=0$$ \n
i.e.$$\\left( {{{\\overrightarrow a }_2} - {{\\overrightarrow a }_1}} \\right).\\overrightarrow {{b_1}} \\times {\\overrightarrow b _2} = 0$$ \n
\n$$ \\therefore \\beta = - 1$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3529,
"subject": "General Science",
"question": "The vertices B and C of a $$\\Delta $$ABC lie on the line, \n
$${{x + 2} \\over 3} = {{y - 1} \\over 0} = {z \\over 4}$$ such that BC = 5 units.
Then the\narea (in sq. units) of this triangle, given that the\npoint A(1, –1, 2), is :",
"options": [
{
"text": "6"
},
{
"text": "$$5\\sqrt {17} $$"
},
{
"text": "$$\\sqrt {34} $$"
},
{
"text": "$$2\\sqrt {34} $$"
}
],
"answer": "$$\\sqrt {34} $$",
"solution": "**Answer:** $$\\sqrt {34} $$\n\n\nArea of $$\\Delta $$ABC = $${1 \\over 2} \\times BC \\times AD$$
\nGiven BC = 5 so we need perpendicular\ndistance of A from line BC.
\nLet a point D on BC = (3$$\\lambda $$ - 2, 1, 4$$\\lambda $$)\n
\n= $${1 \\over 2} \\times 5 \\times {{2\\sqrt {34} } \\over 5} = \\sqrt {34} $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3530,
"subject": "General Science",
"question": "The length of the perpendicular from the point\n(2, –1, 4) on the straight line,\n
$${{x + 3} \\over {10}}$$= $${{y - 2} \\over {-7}}$$ = $${{z} \\over {1}}$$\n is :\n",
"options": [
{
"text": "less than 2"
},
{
"text": "greater than 4\n"
},
{
"text": "greater than 2 but less than 3"
},
{
"text": "greater than 3 but less than 4"
}
],
"answer": "greater than 3 but less than 4",
"solution": "**Answer:** greater than 3 but less than 4\n\n\n Let $${{x + 3} \\over {10}}$$= $${{y - 2} \\over {-7}}$$ = $${{z} \\over {1}}$$ = $$\\lambda $$\n
$$ \\therefore $$ General point Q(10$$\\lambda $$ - 3, -7$$\\lambda $$ + 2, $$\\lambda $$).\n
Lines are perpendicular, so the dot product of the Dr's of two lines are zero.\n
$$ \\therefore $$ $$aa$$' + c + c' = 0",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3532,
"subject": "General Science",
"question": "If the foot of the perpendicular drawn from the point (1, 0, 3) on a line passing through ($$\\alpha $$, 7, 1)\nis\n$$\\left( {{5 \\over 3},{7 \\over 3},{{17} \\over 3}} \\right)$$, then $$\\alpha $$ is equal to______.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n\n
= $$\\sqrt {270} $$ = $$3\\sqrt {30} $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3534,
"subject": "General Science",
"question": "If (a, b, c) is the image of the point (1, 2, -3) in
the line $${{x + 1} \\over 2} = {{y - 3} \\over { - 2}} = {z \\over { - 1}}$$, then a + b + c is :",
"options": [
{
"text": "1"
},
{
"text": "2"
},
{
"text": "3"
},
{
"text": "-1"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\n\n
$$ \\therefore $$ a + b + c = 2",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3535,
"subject": "General Science",
"question": "Let a, b$$ \\in $$R. If the mirror image of the point P(a, 6, 9) with respect to the line
$${{x - 3} \\over 7} = {{y - 2} \\over 5} = {{z - 1} \\over { - 9}}$$ is (20, b, $$-$$a$$-$$9), then | a + b |, is equal to :",
"options": [
{
"text": "88"
},
{
"text": "90"
},
{
"text": "86"
},
{
"text": "84"
}
],
"answer": "88",
"solution": "**Answer:** 88\n\nGiven, P(a, 6, 9)
Distance between skew lines $$ = {{\\left[ {{{\\overrightarrow a }_2} - {{\\overrightarrow a }_1}{{\\overrightarrow b }_1}{{\\overrightarrow b }_2}} \\right]} \\over {\\left| {{{\\overrightarrow b }_1} \\times {{\\overrightarrow b }_2}} \\right|}}$$
$$ \\Rightarrow |\\lambda | = 1$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3537,
"subject": "General Science",
"question": "The equation of the line through the point (0, 1, 2) and perpendicular to the line
Any point on this line $$(2\\lambda + 1,3\\lambda - 1, - 2\\lambda + 1)$$
Direction ratio of given line $$(2,3, - 2)$$
Direction ratio of line to be found $$(2\\lambda + 1,3\\lambda - 2, - 2\\lambda - 1)$$
$$ \\therefore $$ $${\\overrightarrow d _1}\\,.\\,{\\overrightarrow d _2} = 0$$
$$ \\Rightarrow $$ $$\\lambda = 2/17$$
Direction ratio of line $$(21, - 28, - 21) \\equiv (3, - 4, - 3) \\equiv ( - 3,4,3)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3538,
"subject": "General Science",
"question": "A line 'l' passing through origin is perpendicular to the lines
$${l_1}:\\overrightarrow r = (3 + t)\\widehat i + ( - 1 + 2t)\\widehat j + (4 + 2t)\\widehat k$$
$${l_2}:\\overrightarrow r = (3 + 2s)\\widehat i + (3 + 2s)\\widehat j + (2 + s)\\widehat k$$
If the co-ordinates of the point in the first octant on 'l2‘ at a distance of $$\\sqrt {17} $$ from the point of intersection of 'l' and 'l1' are (a, b, c) then 18(a + b + c) is equal to ___________.",
"options": [],
"answer": "44",
"solution": "**Answer:** 44\n\n$${l_1}:\\overrightarrow r = (3 + t)\\widehat i + ( - 1 + 2t)\\widehat j + (4 + 2t)\\widehat k$$
$$ = 44$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3539,
"subject": "General Science",
"question": "Let the position vectors of two points P and Q be 3$$\\widehat i$$ $$-$$ $$\\widehat j$$ + 2$$\\widehat k$$ and $$\\widehat i$$ + 2$$\\widehat j$$ $$-$$ 4$$\\widehat k$$, respectively. Let R and S be two points such that the direction ratios of lines PR and QS are (4, $$-$$1, 2) and ($$-$$2, 1, $$-$$2), respectively. Let lines PR and QS intersect at T. If the vector $$\\overrightarrow {TA} $$ is perpendicular to both $$\\overrightarrow {PR} $$ and $$\\overrightarrow {QS} $$ and the length of vector $$\\overrightarrow {TA} $$ is $$\\sqrt 5 $$ units, then the modulus of a position vector of A is :",
"options": [
{
"text": "$$\\sqrt {171} $$"
},
{
"text": "$$\\sqrt {227} $$"
},
{
"text": "$$\\sqrt {482} $$"
},
{
"text": "$$\\sqrt {5} $$"
}
],
"answer": "$$\\sqrt {171} $$",
"solution": "**Answer:** $$\\sqrt {171} $$\n\n$$\\overrightarrow p = 3\\widehat i - \\widehat j + 2\\widehat k$$ & $$\\overrightarrow Q = \\widehat i + 2\\widehat j - 4\\widehat k$$
$${\\overrightarrow v _{PR}} = (4, - 1,2)$$ & $${\\overrightarrow v _{QS}} = ( - 2,1, - 2)$$
$${L_{PR}}:\\overrightarrow r = (3\\widehat i - \\widehat j + 2\\widehat k) + \\lambda (4, - 1,2)$$
$${L_{QS}}:\\overrightarrow r = (\\widehat i + 2\\widehat j - 4\\widehat k) + \\mu ( - 2,1, - 2)$$\n
Now T on PR = $$\\left\\langle {3 + 4\\lambda , - 1 - \\lambda ,2 + 2\\lambda } \\right\\rangle $$\n
Similarly T on QS = (1 $$-$$ 2$$\\mu$$, 2 + $$\\mu$$, $$-$$4 $$-$$ 2$$\\mu$$)
Let the image of the point P(1, 2, 3) in the line $$L:{{x - 6} \\over 3} = {{y - 1} \\over 2} = {{z - 2} \\over 3}$$ be Q. Let R ($$\\alpha$$, $$\\beta$$, $$\\gamma$$) be a point that divides internally the line segment PQ in the ratio 1 : 3. Then the value of 22 ($$\\alpha$$ + $$\\beta$$ + $$\\gamma$$) is equal to __________.
Let $$\\overrightarrow a = \\widehat i + \\widehat j + 2\\widehat k$$, $$\\overrightarrow b = 2\\widehat i - 3\\widehat j + \\widehat k$$ and $$\\overrightarrow c = \\widehat i - \\widehat j + \\widehat k$$ be three given vectors. Let $$\\overrightarrow v $$ be a vector in the plane of $$\\overrightarrow a $$ and $$\\overrightarrow b $$ whose projection on $$\\overrightarrow c $$ is $${2 \\over {\\sqrt 3 }}$$. If $$\\overrightarrow v \\,.\\,\\widehat j = 7$$, then $$\\overrightarrow v \\,.\\,\\left( {\\widehat i + \\widehat k} \\right)$$ is equal to :
Let l1 be the line in xy-plane with x and y intercepts $${1 \\over 8}$$ and $${1 \\over {4\\sqrt 2 }}$$ respectively, and l2 be the line in zx-plane with x and z intercepts $$ - {1 \\over 8}$$ and $$ - {1 \\over {6\\sqrt 3 }}$$ respectively. If d is the shortest distance between the line l1 and l2, then d$$-$$2 is equal to _______________.
$$d = \\left| {{{(\\overrightarrow c - \\overrightarrow a )\\,.\\,(\\overrightarrow b \\times \\overrightarrow d )} \\over {\\left| {\\overrightarrow b \\times \\overrightarrow d } \\right|}}} \\right|$$
If the shortest distance between the lines $${{x - 1} \\over 2} = {{y - 2} \\over 3} = {{z - 3} \\over \\lambda }$$ and $${{x - 2} \\over 1} = {{y - 4} \\over 4} = {{z - 5} \\over 5}$$ is $${1 \\over {\\sqrt 3 }}$$, then the sum of all possible value of $$\\lambda$$ is :
Let a line having direction ratios, 1, $$-$$4, 2 intersect the lines $${{x - 7} \\over 3} = {{y - 1} \\over { - 1}} = {{z + 2} \\over 1}$$ and $${x \\over 2} = {{y - 7} \\over 3} = {z \\over 1}$$ at the points A and B. Then (AB)2 is equal to ___________.
$$\\overrightarrow r = \\left( { - \\widehat i + 3\\widehat k} \\right) + \\lambda \\left( {\\widehat i - a\\widehat j} \\right)$$
and $$\\overrightarrow r = \\left( { - \\widehat j + 2\\widehat k} \\right) + \\mu \\left( {\\widehat i - \\widehat j + \\widehat k} \\right)$$ is $$\\sqrt {{2 \\over 3}} $$, then the integral value of a is equal to ___________.
Consider a triangle ABC whose vertices are A(0, $$\\alpha$$, $$\\alpha$$), B($$\\alpha$$, 0, $$\\alpha$$) and C($$\\alpha$$, $$\\alpha$$, 0), $$\\alpha$$ > 0. Let D be a point moving on the line x + z $$-$$ 3 = 0 = y and G be the centroid of $$\\Delta$$ABC. If the minimum length of GD is $$\\sqrt {{{57} \\over 2}} $$, then $$\\alpha$$ is equal to ____________.
The length of the perpendicular from the point $$(1,-2,5)$$ on the line passing through $$(1,2,4)$$ and parallel to the line $$x+y-z=0=x-2 y+3 z-5$$ is :
Let $$\\mathrm{Q}$$ and $$\\mathrm{R}$$ be two points on the line $$\\frac{x+1}{2}=\\frac{y+2}{3}=\\frac{z-1}{2}$$ at a distance $$\\sqrt{26}$$ from the point $$P(4,2,7)$$. Then the square of the area of the triangle $$P Q R$$ is ___________.
If the length of the perpendicular drawn from the point $$P(a, 4,2)$$, a $$>0$$ on the line $$\\frac{x+1}{2}=\\frac{y-3}{3}=\\frac{z-1}{-1}$$ is $$2 \\sqrt{6}$$ units and $$Q\\left(\\alpha_{1}, \\alpha_{2}, \\alpha_{3}\\right)$$ is the image of the point P in this line, then $$\\mathrm{a}+\\sum\\limits_{i=1}^{3} \\alpha_{i}$$ is equal to :
But all other are in terms of $k$ hence possible.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3559,
"subject": "General Science",
"question": "Let a line $L$ pass through the point $P(2,3,1)$ and be parallel to the line $x+3 y-2 z-2=0=x-y+2 z$. If the distance of $L$ from the point $(5,3,8)$ is $\\alpha$, then $3 \\alpha^2$ is equal to :",
"options": [],
"answer": "158",
"solution": "**Answer:** 158\n\n
Let the co-ordinates of one vertex of $$\\Delta ABC$$ be $$A(0,2,\\alpha)$$ and the other two vertices lie on the line $${{x + \\alpha } \\over 5} = {{y - 1} \\over 2} = {{z + 4} \\over 3}$$. For $$\\alpha \\in \\mathbb{Z}$$, if the area of $$\\Delta ABC$$ is 21 sq. units and the line segment $$BC$$ has length $$2\\sqrt{21}$$ units, then $$\\alpha^2$$ is equal to ___________.
The foot of perpendicular of the point (2, 0, 5) on the line $${{x + 1} \\over 2} = {{y - 1} \\over 5} = {{z + 1} \\over { - 1}}$$ is ($$\\alpha,\\beta,\\gamma$$). Then, which of the following is NOT correct?
If the shortest distance between the line joining the points (1, 2, 3) and (2, 3, 4), and the line $${{x - 1} \\over 2} = {{y + 1} \\over { - 1}} = {{z - 2} \\over 0}$$ is $$\\alpha$$, then 28$$\\alpha^2$$ is equal to ____________.
The distance of the point P(4, 6, $$-$$2) from the line passing through the point ($$-$$3, 2, 3) and parallel to a line with direction ratios 3, 3, $$-$$1 is equal to :
A line $$L_3$$ having direction ratios 1, $$-$$1, $$-$$2, intersects $$L_1$$ and $$L_2$$ at the points $$P$$ and $$Q$$ respectively. Then the length of line segment $$PQ$$ is
If the shortest between the lines $${{x + \\sqrt 6 } \\over 2} = {{y - \\sqrt 6 } \\over 3} = {{z - \\sqrt 6 } \\over 4}$$ and $${{x - \\lambda } \\over 3} = {{y - 2\\sqrt 6 } \\over 4} = {{z + 2\\sqrt 6 } \\over 5}$$ is 6, then the square of sum of all possible values of $$\\lambda$$ is :
",
"options": [],
"answer": "384",
"solution": "**Answer:** 384\n\nShortest distance between the lines
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3569,
"subject": "General Science",
"question": "Let $\\mathrm{S}$ be the set of all values of $\\lambda$, for which the shortest distance between
the lines $\\frac{x-\\lambda}{0}=\\frac{y-3}{4}=\\frac{z+6}{1}$ and $\\frac{x+\\lambda}{3}=\\frac{y}{-4}=\\frac{z-6}{0}$ is 13. Then $8\\left|\\sum\\limits_{\\lambda \\in S} \\lambda\\right|$ is equal to :",
"options": [
{
"text": "306"
},
{
"text": "304"
},
{
"text": "308"
},
{
"text": "302"
}
],
"answer": "306",
"solution": "**Answer:** 306\n\nGiven the two lines :\n
Let the lines $$l_{1}: \\frac{x+5}{3}=\\frac{y+4}{1}=\\frac{z-\\alpha}{-2}$$ and $$l_{2}: 3 x+2 y+z-2=0=x-3 y+2 z-13$$ be coplanar. If the point $$\\mathrm{P}(a, b, c)$$ on $$l_{1}$$ is nearest to the point $$\\mathrm{Q}(-4,-3,2)$$, then $$|a|+|b|+|c|$$ is equal to
If $$\\mathrm{P}$$ is the point of intersection of $$l$$ and $$l_{1}$$, and $$\\mathrm{Q}(\\propto, \\beta, \\gamma)$$ is the foot of perpendicular from P on $$l_{2}$$, then $$9(\\alpha+\\beta+\\gamma)$$ is equal to _____________.
One vertex of a rectangular parallelopiped is at the origin $$\\mathrm{O}$$ and the lengths of its edges along $$x, y$$ and $$z$$ axes are $$3,4$$ and $$5$$ units respectively. Let $$\\mathrm{P}$$ be the vertex $$(3,4,5)$$. Then the shortest distance between the diagonal OP and an edge parallel to $$\\mathrm{z}$$ axis, not passing through $$\\mathrm{O}$$ or $$\\mathrm{P}$$ is :
If the lines $$\\frac{x-1}{2}=\\frac{2-y}{-3}=\\frac{z-3}{\\alpha}$$ and $$\\frac{x-4}{5}=\\frac{y-1}{2}=\\frac{z}{\\beta}$$ intersect, then the magnitude of the minimum value of $$8 \\alpha \\beta$$ is _____________.
$\\therefore$ Magnitude of the minimum value of $8 \\alpha \\beta$ is 18 .",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3577,
"subject": "General Science",
"question": "Let $\\mathrm{P}$ and $\\mathrm{Q}$ be the points on the line $\\frac{x+3}{8}=\\frac{y-4}{2}=\\frac{z+1}{2}$ which are at a distance of 6 units from the point $\\mathrm{R}(1,2,3)$. If the centroid of the triangle PQR is $(\\alpha, \\beta, \\gamma)$, then $\\alpha^2+\\beta^2+\\gamma^2$ is :",
"options": [
{
"text": "18"
},
{
"text": "24"
},
{
"text": "26"
},
{
"text": "36"
}
],
"answer": "18",
"solution": "**Answer:** 18\n\nAny point on line $\\frac{x+3}{8}=\\frac{y-4}{2}=\\frac{z+1}{2}$
can be taken as $(8 \\lambda-3,2 \\lambda+4,2 \\lambda-1)$\n
Now, $14(\\alpha+\\beta+\\gamma)=14\\left(\\frac{62+4-12}{7}\\right)=108$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3579,
"subject": "General Science",
"question": "If the shortest distance between the lines
$\\frac{x-\\lambda}{-2}=\\frac{y-2}{1}=\\frac{z-1}{1}$ and $\\frac{x-\\sqrt{3}}{1}=\\frac{y-1}{-2}=\\frac{z-2}{1}$ is 1 , then the sum of all possible values of $\\lambda$ is :",
"options": [
{
"text": "0"
},
{
"text": "$2 \\sqrt{3}$"
},
{
"text": "$3 \\sqrt{3}$"
},
{
"text": "$-2 \\sqrt{3}$"
}
],
"answer": "$2 \\sqrt{3}$",
"solution": "**Answer:** $2 \\sqrt{3}$\n\n
We observe that these lines are not parallel as their directional vectors are not proportional. The directional vector for $L_1$ is $(-2,1,1)$ and for $L_2$ is $(1,-2,1)$. The shortest distance between two skew (non-intersecting and non-parallel) lines in the three-dimensional space is along the line that is perpendicular to both lines. This implies we can find a vector that is perpendicular to both directional vectors by taking their cross product.
\n\n
The directional vector for $L_1$ is $d_1 = \\langle -2, 1, 1 \\rangle$, and for $L_2$ is $d_2 = \\langle 1, -2, 1 \\rangle$. The cross product of $d_1$ and $d_2$, which will be perpendicular to both lines, is given by:
$$ d = 3\\mathbf{i} + 3\\mathbf{j} + 3\\mathbf{k} $$
\n\n
$$ d = \\langle 3, 3, 3 \\rangle $$
\n\n
The shortest distance $D$ between the two lines can then be given by the formula:
\n\n
$$ D = \\frac{\\left| (\\mathbf{a_2} - \\mathbf{a_1}) \\cdot d \\right|}{\\|d\\|} $$
\n\n
Where $\\mathbf{a_1}$ and $\\mathbf{a_2}$ are position vectors to any points on line $L_1$ and line $L_2$, respectively, and '$\\cdot$' denotes the dot product.
\n\n
From the lines' equations, we can choose a point on each line (when the parameter is zero). Thus, for $L_1$, let's choose the point $A(\\lambda, 2, 1)$, and for $L_2$, let's choose the point $B(\\sqrt{3}, 1, 2)$. These points correspond to the vectors $\\mathbf{a_1} = \\langle \\lambda, 2, 1 \\rangle$ and $\\mathbf{a_2} = \\langle \\sqrt{3}, 1, 2 \\rangle$, respectively.
Hence, option B ($2 \\sqrt{3}$) is the correct answer.
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3580,
"subject": "General Science",
"question": "Let the line of the shortest distance between the lines\n
intersect $\\mathrm{L}_1$ and $\\mathrm{L}_2$ at $\\mathrm{P}$ and $\\mathrm{Q}$ respectively. If $(\\alpha, \\beta, \\gamma)$ is the mid point of the line segment $\\mathrm{PQ}$, then $2(\\alpha+\\beta+\\gamma)$ is equal to ____________.",
"options": [],
"answer": "21",
"solution": "**Answer:** 21\n\n$\\begin{array}{ll}L_1 \\equiv \\vec{r}=(1,2,3)+\\lambda(1,-1,1) & \\left(\\vec{r}=\\vec{a}_1+\\lambda \\vec{b}_1\\right) \\\\\\\\ L_2 \\equiv \\vec{r}=(4,5,6)+\\mu(1,1,-1) & \\left(\\vec{r}=\\vec{a}_2+\\lambda \\vec{b}_2\\right)\\end{array}$\n
$$\n\\begin{aligned}\n& P \\equiv\\left(\\frac{5}{2}, \\frac{1}{2}, \\frac{9}{2}\\right) \\& Q \\equiv\\left(\\frac{5}{2}, \\frac{7}{2}, \\frac{15}{2}\\right) \\\\\\\\\n& \\alpha=\\frac{5}{2}, \\beta=\\frac{4}{2}, \\gamma=\\frac{12}{2} \\\\\\\\\n& 2(\\alpha+\\beta+\\gamma)=21\n\\end{aligned}\n$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3581,
"subject": "General Science",
"question": "The distance, of the point $(7,-2,11)$ from the line
$\\frac{x-6}{1}=\\frac{y-4}{0}=\\frac{z-8}{3}$ along the line $\\frac{x-5}{2}=\\frac{y-1}{-3}=\\frac{z-5}{6}$, is :",
"options": [
{
"text": "12"
},
{
"text": "18"
},
{
"text": "21"
},
{
"text": "14"
}
],
"answer": "14",
"solution": "**Answer:** 14\n\n
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3582,
"subject": "General Science",
"question": "If the shortest distance between the lines
$\\frac{x-4}{1}=\\frac{y+1}{2}=\\frac{z}{-3}$ and $\\frac{x-\\lambda}{2}=\\frac{y+1}{4}=\\frac{z-2}{-5}$ is $\\frac{6}{\\sqrt{5}}$, then the sum of all possible values of $\\lambda$ is :",
"options": [
{
"text": "10"
},
{
"text": "5"
},
{
"text": "7"
},
{
"text": "8"
}
],
"answer": "8",
"solution": "**Answer:** 8\n\n
Let the image of the point $$(1,0,7)$$ in the line $$\\frac{x}{1}=\\frac{y-1}{2}=\\frac{z-2}{3}$$ be the point $$(\\alpha, \\beta, \\gamma)$$. Then which one of the following points lies on the line passing through $$(\\alpha, \\beta, \\gamma)$$ and making angles $$\\frac{2 \\pi}{3}$$ and $$\\frac{3 \\pi}{4}$$ with $$y$$-axis and $$z$$-axis respectively and an acute angle with $$x$$-axis ?
The lines $$\\frac{x-2}{2}=\\frac{y}{-2}=\\frac{z-7}{16}$$ and $$\\frac{x+3}{4}=\\frac{y+2}{3}=\\frac{z+2}{1}$$ intersect at the point $$P$$. If the distance of $$\\mathrm{P}$$ from the line $$\\frac{x+1}{2}=\\frac{y-1}{3}=\\frac{z-1}{1}$$ is $$l$$, then $$14 l^2$$ is equal to __________.
Let $$(\\alpha, \\beta, \\gamma)$$ be the mirror image of the point $$(2,3,5)$$ in the line $$\\frac{x-1}{2}=\\frac{y-2}{3}=\\frac{z-3}{4}$$. Then, $$2 \\alpha+3 \\beta+4 \\gamma$$ is equal to
The shortest distance, between lines $$L_1$$ and $$L_2$$, where $$L_1: \\frac{x-1}{2}=\\frac{y+1}{-3}=\\frac{z+4}{2}$$ and $$L_2$$ is the line, passing through the points $$\\mathrm{A}(-4,4,3), \\mathrm{B}(-1,6,3)$$ and perpendicular to the line $$\\frac{x-3}{-2}=\\frac{y}{3}=\\frac{z-1}{1}$$, is
A line passes through $$A(4,-6,-2)$$ and $$B(16,-2,4)$$. The point $$P(a, b, c)$$, where $$a, b, c$$ are non-negative integers, on the line $$A B$$ lies at a distance of 21 units, from the point $$A$$. The distance between the points $$P(a, b, c)$$ and $$Q(4,-12,3)$$ is equal to __________.
Let $$\\mathrm{Q}$$ and $$\\mathrm{R}$$ be the feet of perpendiculars from the point $$\\mathrm{P}(a, a, a)$$ on the lines $$x=y, z=1$$ and $$x=-y, z=-1$$ respectively. If $$\\angle \\mathrm{QPR}$$ is a right angle, then $$12 a^2$$ is equal to _________.
Let $$P Q R$$ be a triangle with $$R(-1,4,2)$$. Suppose $$M(2,1,2)$$ is the mid point of $$\\mathrm{PQ}$$. The distance of the centroid of $$\\triangle \\mathrm{PQR}$$ from the point of intersection of the lines $$\\frac{x-2}{0}=\\frac{y}{2}=\\frac{z+3}{-1}$$ and $$\\frac{x-1}{1}=\\frac{y+3}{-3}=\\frac{z+1}{1}$$ is
A line with direction ratios $$2,1,2$$ meets the lines $$x=y+2=z$$ and $$x+2=2 y=2 z$$ respectively at the points $$\\mathrm{P}$$ and $$\\mathrm{Q}$$. If the length of the perpendicular from the point $$(1,2,12)$$ to the line $$\\mathrm{PQ}$$ is $$l$$, then $$l^2$$ is __________.
Let O be the origin, and M and $$\\mathrm{N}$$ be the points on the lines $$\\frac{x-5}{4}=\\frac{y-4}{1}=\\frac{z-5}{3}$$ and $$\\frac{x+8}{12}=\\frac{y+2}{5}=\\frac{z+11}{9}$$ respectively such that $$\\mathrm{MN}$$ is the shortest distance between the given lines. Then $$\\overrightarrow{O M} \\cdot \\overrightarrow{O N}$$ is equal to _________.
be three lines such that $$L_1$$ is perpendicular to $$L_2$$ and $$L_3$$ is perpendicular to both $$L_1$$ and $$L_2$$. Then, the point which lies on $$L_3$$ is
Let a line passing through the point $$(-1,2,3)$$ intersect the lines $$L_1: \\frac{x-1}{3}=\\frac{y-2}{2}=\\frac{z+1}{-2}$$ at $$M(\\alpha, \\beta, \\gamma)$$ and $$L_2: \\frac{x+2}{-3}=\\frac{y-2}{-2}=\\frac{z-1}{4}$$ at $$N(a, b, c)$$. Then, the value of $$\\frac{(\\alpha+\\beta+\\gamma)^2}{(a+b+c)^2}$$ equals __________.
Let $$(\\alpha, \\beta, \\gamma)$$ be the foot of perpendicular from the point $$(1,2,3)$$ on the line $$\\frac{x+3}{5}=\\frac{y-1}{2}=\\frac{z+4}{3}$$. Then $$19(\\alpha+\\beta+\\gamma)$$ is equal to :
If $$\\mathrm{d}_1$$ is the shortest distance between the lines $$x+1=2 y=-12 z, x=y+2=6 z-6$$ and $$\\mathrm{d}_2$$ is the shortest distance between the lines $$\\frac{x-1}{2}=\\frac{y+8}{-7}=\\frac{z-4}{5}, \\frac{x-1}{2}=\\frac{y-2}{1}=\\frac{z-6}{-3}$$, then the value of $$\\frac{32 \\sqrt{3} \\mathrm{~d}_1}{\\mathrm{~d}_2}$$ is :
Consider the line $$\\mathrm{L}$$ passing through the points $$(1,2,3)$$ and $$(2,3,5)$$. The distance of the point $$\\left(\\frac{11}{3}, \\frac{11}{3}, \\frac{19}{3}\\right)$$ from the line $$\\mathrm{L}$$ along the line $$\\frac{3 x-11}{2}=\\frac{3 y-11}{1}=\\frac{3 z-19}{2}$$ is equal to
The square of the distance of the image of the point $$(6,1,5)$$ in the line $$\\frac{x-1}{3}=\\frac{y}{2}=\\frac{z-2}{4}$$, from the origin is __________.
Let the line $$\\mathrm{L}$$ intersect the lines $$x-2=-y=z-1,2(x+1)=2(y-1)=z+1$$ and be parallel to the line $$\\frac{x-2}{3}=\\frac{y-1}{1}=\\frac{z-2}{2}$$. Then which of the following points lies on $$\\mathrm{L}$$ ?
$$\\therefore \\quad$$ Points will be $$\\left(\\frac{2}{3}, \\frac{4}{3}, \\frac{-1}{3}\\right)$$ and $$\\left(\\frac{-4}{3}, \\frac{2}{3}, \\frac{-5}{3}\\right)$$
\n
$$\\therefore \\quad L$$ will be $$\\frac{x-\\frac{2}{3}}{3}=\\frac{y-\\frac{4}{3}}{1}=\\frac{z+\\frac{1}{3}}{2}$$
\n
$$\\therefore \\quad\\left(\\frac{-1}{3}, 1,-1\\right)$$ will satisfy $$L$$
Let the point, on the line passing through the points $$P(1,-2,3)$$ and $$Q(5,-4,7)$$, farther from the origin and at a distance of 9 units from the point $$P$$, be $$(\\alpha, \\beta, \\gamma)$$. Then $$\\alpha^2+\\beta^2+\\gamma^2$$ is equal to :
Let $$\\mathrm{P}$$ be the point of intersection of the lines $$\\frac{x-2}{1}=\\frac{y-4}{5}=\\frac{z-2}{1}$$ and $$\\frac{x-3}{2}=\\frac{y-2}{3}=\\frac{z-3}{2}$$. Then, the shortest distance of $$\\mathrm{P}$$ from the line $$4 x=2 y=z$$ is
Consider a line $$\\mathrm{L}$$ passing through the points $$\\mathrm{P}(1,2,1)$$ and $$\\mathrm{Q}(2,1,-1)$$. If the mirror image of the point $$\\mathrm{A}(2,2,2)$$ in the line $$\\mathrm{L}$$ is $$(\\alpha, \\beta, \\gamma)$$, then $$\\alpha+\\beta+6 \\gamma$$ is equal to __________.
If the shortest distance between the lines $$\\frac{x-\\lambda}{2}=\\frac{y-4}{3}=\\frac{z-3}{4}$$ and $$\\frac{x-2}{4}=\\frac{y-4}{6}=\\frac{z-7}{8}$$ is $$\\frac{13}{\\sqrt{29}}$$, then a value of $$\\lambda$$ is :
Let $$\\mathrm{P}(\\alpha, \\beta, \\gamma)$$ be the image of the point $$\\mathrm{Q}(1,6,4)$$ in the line $$\\frac{x}{1}=\\frac{y-1}{2}=\\frac{z-2}{3}$$. Then $$2 \\alpha+\\beta+\\gamma$$ is equal to ________
Let $$\\mathrm{d}$$ be the distance of the point of intersection of the lines $$\\frac{x+6}{3}=\\frac{y}{2}=\\frac{z+1}{1}$$ and $$\\frac{x-7}{4}=\\frac{y-9}{3}=\\frac{z-4}{2}$$ from the point $$(7,8,9)$$. Then $$\\mathrm{d}^2+6$$ is equal to :
Let $$(\\alpha, \\beta, \\gamma)$$ be the image of the point $$(8,5,7)$$ in the line $$\\frac{x-1}{2}=\\frac{y+1}{3}=\\frac{z-2}{5}$$. Then $$\\alpha+\\beta+\\gamma$$ is equal to :
Let the point $$(-1, \\alpha, \\beta)$$ lie on the line of the shortest distance between the lines $$\\frac{x+2}{-3}=\\frac{y-2}{4}=\\frac{z-5}{2}$$ and $$\\frac{x+2}{-1}=\\frac{y+6}{2}=\\frac{z-1}{0}$$. Then $$(\\alpha-\\beta)^2$$ is equal to _________.
Let $$\\mathrm{P}(\\alpha, \\beta, \\gamma)$$ be the image of the point $$\\mathrm{Q}(3,-3,1)$$ in the line $$\\frac{x-0}{1}=\\frac{y-3}{1}=\\frac{z-1}{-1}$$ and $$\\mathrm{R}$$ be the point $$(2,5,-1)$$. If the area of the triangle $$\\mathrm{PQR}$$ is $$\\lambda$$ and $$\\lambda^2=14 \\mathrm{~K}$$, then $$\\mathrm{K}$$ is equal to :
If the shortest distance between the lines $$\\frac{x-\\lambda}{3}=\\frac{y-2}{-1}=\\frac{z-1}{1}$$ and $$\\frac{x+2}{-3}=\\frac{y+5}{2}=\\frac{z-4}{4}$$ is $$\\frac{44}{\\sqrt{30}}$$, then the largest possible value of $$|\\lambda|$$ is equal to _________.
Let $$P$$ be the point $$(10,-2,-1)$$ and $$Q$$ be the foot of the perpendicular drawn from the point $$R(1,7,6)$$ on the line passing through the points $$(2,-5,11)$$ and $$(-6,7,-5)$$. Then the length of the line segment $$P Q$$ is equal to _________.
So $$d.r$$ of normal area $$a,$$ $$a\\sqrt {2a} $$ i.e. $$1,$$ $$1,\\sqrt 2 .$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3614,
"subject": "General Science",
"question": "The shortest distance from the plane $$12x+4y+3z=327$$ to the sphere
$${x^2} + {y^2} + {z^2} + 4x - 2y - 6z = 155$$ is ",
"options": [
{
"text": "$$39$$ "
},
{
"text": "$$26$$ "
},
{
"text": "$$11{4 \\over {13}}$$ "
},
{
"text": "$$13$$"
}
],
"answer": "$$13$$",
"solution": "**Answer:** $$13$$\n\nShortest distance $$=$$ perpendicular distance between the plane and sphere $$=$$ distance of plane from center of sphere $$-$$ radius\n
$$ = \\sqrt {{5 \\over 2} - {3 \\over 2}} = 1.$$ ",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3620,
"subject": "General Science",
"question": "Statement-1 : The point $$A(3, 1, 6)$$ is the mirror image of the point $$B(1, 3, 4)$$ in the plane $$x-y+z=5.$$ \n
Statement-2 : The plane $$x-y+z=5$$ bisects the line segment joining $$A(3, 1, 6)$$ and $$B(1, 3, 4).$$ ",
"options": [
{
"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."
},
{
"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."
}
],
"answer": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.",
"solution": "**Answer:** Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.\n\n$$A\\left( {3,1,6} \\right);\\,\\,B = \\left( {1,3,4} \\right)$$\n
Mid-point of $$AB = \\left( {2,2,5} \\right)$$ lies on the plane. \n
and d.r's of $$AB=(2,-2,2)$$\n
d.r's of normal to plane $$ = \\left( {1, - 1,1} \\right).$$\n
Direction ratio of $$AB$$ and normal to the plane are proportional therefore, \n
$$AB$$ is perpendicular to the plane \n
$$\\therefore$$ $$A$$ is image of $$B$$ \n
Statement-$$2$$ is correct but it is not correct explanation. ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3621,
"subject": "General Science",
"question": "A equation of a plane parallel to the plane $$x-2y+2z-5=0$$ and at a unit distance from the origin is :",
"options": [
{
"text": "$$x-2y+2z-3=0$$"
},
{
"text": "$$x-2y+2z+1=0$$"
},
{
"text": "$$x-2y+2z-1=0$$ "
},
{
"text": "$$x-2y+2z+5=0$$"
}
],
"answer": "$$x-2y+2z-3=0$$",
"solution": "**Answer:** $$x-2y+2z-3=0$$\n\nGiven equation of a plane is \n
$$x - 2y + 2z - 5 = 0$$\n
So, Equation of parallel plane is given by \n
$$x - 2y + 2z + d = 0$$\n
Now, it is given that distance from origin to the parallel planes is $$1.$$ \n
$$ \\therefore $$ Distance of point (1, $$-$$2, 4) from plane x + y $$-$$ 3 = 0 is,\n
= $$\\left| {{{1 - 2 - 3} \\over {\\sqrt {1 + 1} }}} \\right|$$ = $${4 \\over {\\sqrt 2 }}$$ = 2$$\\sqrt 2 $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3624,
"subject": "General Science",
"question": "If x = a, y = b, z = c is a solution of the system of linear equations\n
x + 8y + 7z = 0\n
9x + 2y + 3z = 0\n
x + y + z = 0\n
such that the point (a, b, c) lies on the plane x + 2y + z = 6, then 2a + b + c equals : ",
"options": [
{
"text": "$$-$$ 1"
},
{
"text": "0"
},
{
"text": "1"
},
{
"text": "2"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\nGiven,\n
x + 8y + 7z = 0\n
9x + 2y + 3z = 0\n
x + y + z = 0\n
Solving those equations, we get\n
x = $$\\lambda $$, y = 6$$\\lambda $$, z = -7$$\\lambda $$\n
$$ \\Rightarrow $$ $${1 \\over {{x^2}}} + {1 \\over {{y^2}}} + {1 \\over {{z^2}}} = 1$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3626,
"subject": "General Science",
"question": "The sum of the intercepts on the coordinate axes of the plane passing through the point ($$-$$2, $$-2,$$ 2) and containing the line joining the points (1, $$-$$1, 2) and (1, 1, 1) is :",
"options": [
{
"text": "4"
},
{
"text": "$$-$$ 4"
},
{
"text": "$$-$$ 8"
},
{
"text": "12"
}
],
"answer": "$$-$$ 4",
"solution": "**Answer:** $$-$$ 4\n\nEquation of plane passing through three given points is :\n
$$ \\therefore $$ Sum of intercepts $$ = - 8 + {8 \\over 3} + {8 \\over 6} = - 4$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3627,
"subject": "General Science",
"question": "A plane bisects the line segment joining the points (1, 2, 3) and ($$-$$ 3, 4, 5) at rigt angles. Then this plane also passes through the point : ",
"options": [
{
"text": "($$-$$ 3, 2, 1)"
},
{
"text": "(3, 2, 1)"
},
{
"text": "($$-$$ 1, 2, 3)"
},
{
"text": "(1, 2, $$-$$ 3)"
}
],
"answer": "($$-$$ 3, 2, 1)",
"solution": "**Answer:** ($$-$$ 3, 2, 1)\n\nSince the plane bisects the line joining the points (1, 2, 3) and ($$-$$3, 4, 5) then the plane passes through the midpoint of the line which is : \n
As plane cuts the line segment at right angle, so the direction cosines of the normal of the plane are ($$-$$ 3 $$-$$ 1, 4 $$-$$ 2, 5 $$-$$ 3) = ($$-$$ 4, 2, 2)\n
So the equation of the plane is : $$-$$ 4x + 2y + 2z = $$\\lambda $$\n
Therefore, equation of plane is : $$-$$ 4x + 2y + 2z = 18\n
Now, only ($$-$$ 3, 2, 1) satiesfies the given plane as \n
$$-$$ 4($$-$$ 3) + 2(2) + 2(1) = 18",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3628,
"subject": "General Science",
"question": "A variable plane passes through a fixed point (3,2,1) and meets x, y and z axes at A, B and C respectively. A plane is drawn parallel to yz -plane through A, a second plane is drawn parallel zx-plane through B and a third plane is drawn parallel to xy-plane through C. Then the locus of the point of intersection of these three planes, is :",
"options": [
{
"text": "$${x \\over 3} + {y \\over 2} + {z \\over 1} = 1$$ "
},
{
"text": "x + y + z = 6"
},
{
"text": "$${1 \\over x} + {1 \\over y} + {1 \\over z} = {{11} \\over 6}$$"
},
{
"text": "$${3 \\over x} + {2 \\over y} + {1 \\over z} = 1$$"
}
],
"answer": "$${3 \\over x} + {2 \\over y} + {1 \\over z} = 1$$",
"solution": "**Answer:** $${3 \\over x} + {2 \\over y} + {1 \\over z} = 1$$\n\nIf a, b, c are the intercepts of the variable plane on the x,y,z axes respectively, then the equation of the plane is
$$ = {1 \\over {3\\sqrt 2 }}$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3630,
"subject": "General Science",
"question": "The perpendicular distance from the origin to the plane containing the two lines,
\n$$ \\Rightarrow $$ x - 3y = 2 or 3x + y + 4z = 6
\nOut of the four given points in question's options only (2, -4, 1) lies on the plane 3x + y + 4z = 6",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3632,
"subject": "General Science",
"question": "If the plane 2x – y + 2z + 3 = 0 has the distances\n$${1 \\over 3}$$\n and\n$${2 \\over 3}$$\n units from the planes 4x – 2y + 4z + $$\\lambda $$ = 0 and\n2x – y + 2z + $$\\mu $$ = 0, respectively, then the maximum value of $$\\lambda $$ + $$\\mu $$ is equal to :",
"options": [
{
"text": "13"
},
{
"text": "9"
},
{
"text": "5"
},
{
"text": "15"
}
],
"answer": "13",
"solution": "**Answer:** 13\n\nDistance formula
\n$$ \\Rightarrow {{\\sqrt {91} } \\over 2}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3634,
"subject": "General Science",
"question": "Let P be the plane, which contains the line of\nintersection of the planes, x + y + z – 6 = 0 and\n2x + 3y + z + 5 = 0 and it is perpendicular to the\nxy-plane. Then the distance of the point (0, 0, 256)\nfrom P is equal to :",
"options": [
{
"text": "205$$\\sqrt5$$"
},
{
"text": "63$$\\sqrt5$$"
},
{
"text": "11/$$\\sqrt5$$"
},
{
"text": "17/$$\\sqrt5$$"
}
],
"answer": "11/$$\\sqrt5$$",
"solution": "**Answer:** 11/$$\\sqrt5$$\n\nP1 : x + y + z – 6 = 0\n
P2 : 2x + 3y + z + 5 = 0\n
Equation of plane which passes through the line of intersection of P1 and P2 is\n
P1 + $$\\lambda $$P2 = 0\n
$$ \\Rightarrow $$ (x + y + z – 6) + $$\\lambda $$(2x + 3y + z + 5) = 0\n
Distance of the point (0, 0, 256) from this plane\n
= $$\\left| {{{0 + 0 + 11} \\over {\\sqrt 5 }}} \\right|$$ = $${{11} \\over {\\sqrt 5 }}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3635,
"subject": "General Science",
"question": "A plane passing through the points (0, –1, 0)\nand (0, 0, 1) and making an angle $${\\pi \\over 4}$$ with the\nplane y – z + 5 = 0, also passes through the\npoint",
"options": [
{
"text": "$$\\left( {\\sqrt 2 ,1,4} \\right)$$"
},
{
"text": "$$\\left(- {\\sqrt 2 ,1,4} \\right)$$"
},
{
"text": "$$\\left( -{\\sqrt 2 ,-1,-4} \\right)$$"
},
{
"text": "$$\\left( {\\sqrt 2 ,-1,4} \\right)$$"
}
],
"answer": "$$\\left( {\\sqrt 2 ,1,4} \\right)$$",
"solution": "**Answer:** $$\\left( {\\sqrt 2 ,1,4} \\right)$$\n\nLet ax + by + cz = 1 be the equation of the plane\n
it passed through point (0, –1, 0).\n
$$ \\therefore $$ -b = 1\n
$$ \\Rightarrow $$ b = -1\n
Also it passes through point (0, 0, 1)\n
$$ \\therefore $$ c = 1\n
So the plane is ax - y + z = 1.\n
This plane an angle $${\\pi \\over 4}$$ with the\nplane y – z + 5 = 0.\n
Normal to the plane ax - y + z = 1 is\n
$${\\overrightarrow a }$$ = $$a\\widehat i - \\widehat j + \\widehat k$$\n
Normal to the plane y – z + 5 = 0 is\n
$${\\overrightarrow b }$$ = $$\\widehat j - \\widehat k$$\n
cos $$\\theta $$ = $$\\left| {{{\\overrightarrow a .\\overrightarrow b } \\over {\\left| {\\overrightarrow a } \\right|\\left| {\\overrightarrow b } \\right|}}} \\right|$$\n
$$ \\Rightarrow $$ $${1 \\over {\\sqrt 2 }}$$ = $$\\left| {{{\\overrightarrow a .\\overrightarrow b } \\over {\\left| {\\overrightarrow a } \\right|\\left| {\\overrightarrow b } \\right|}}} \\right|$$\n
Now a vector $$\\overrightarrow a $$ = $$\\mathop {2i}\\limits^ \\wedge + \\mathop {3j}\\limits^ \\wedge + \\mathop k\\limits^ \\wedge $$ is given and we have to findout projection of vector $$\\overrightarrow a $$ on $$\\overrightarrow p $$.\n
$$ \\therefore $$ Projection of vector $$\\overrightarrow a $$ on $$\\overrightarrow p $$\n
= $$\\left| {\\overrightarrow a } \\right|\\cos \\theta $$\n
= $$\\left| {\\overrightarrow a } \\right| \\times {{\\overrightarrow a .\\overrightarrow p } \\over {\\left| {\\overrightarrow a } \\right|\\left| {\\overrightarrow p } \\right|}}$$\n
= $${{\\overrightarrow a .\\overrightarrow p } \\over {\\left| {\\overrightarrow p } \\right|}}$$\n
$$ \\Rightarrow $$ direction ratio (a, b, c) = ($$\\sqrt 2 $$, $$-$$1, 1) or ($$\\sqrt 2 $$, 1, $$-$$ 1)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3640,
"subject": "General Science",
"question": "The plane which bisects the line segment joining the points (–3, –3, 4) and (3, 7, 6) at right angles, passes through which one of the following points ?",
"options": [
{
"text": "(2, 1, 3)"
},
{
"text": "(4, $$-$$ 1, 2)"
},
{
"text": "(4, 1, $$-$$ 2)"
},
{
"text": "($$-$$ 2, 3, 5)"
}
],
"answer": "(4, 1, $$-$$ 2)",
"solution": "**Answer:** (4, 1, $$-$$ 2)\n\n\n p : 3(x $$-$$ 0) + 5(y $$-$$ 2) + 1 (z $$-$$ 5) = 0\n
3x + 5y + z = 15\n
On checking all the options, the option (4, 1, − 2)\nsatisfy the equation of plane.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3641,
"subject": "General Science",
"question": "Let A be a point on the line $$\\overrightarrow r = \\left( {1 - 3\\mu } \\right)\\widehat i + \\left( {\\mu - 1} \\right)\\widehat j + \\left( {2 + 5\\mu } \\right)\\widehat k$$ and B(3, 2, 6) be a point in the space. Then the value of $$\\mu $$ for which the vector $$\\overrightarrow {AB} $$ is parallel to the plane x $$-$$ 4y + 3z = 1 is -",
"options": [
{
"text": "$${1 \\over 8}$$"
},
{
"text": "$${1 \\over 2}$$"
},
{
"text": "$${1 \\over 4}$$"
},
{
"text": "$$-$$ $${1 \\over 4}$$"
}
],
"answer": "$${1 \\over 4}$$",
"solution": "**Answer:** $${1 \\over 4}$$\n\nLet point A is \n
cos$$\\theta $$ = $${{5 + 5 + 9} \\over {{{\\left( {\\sqrt {25 + 9 + 1} } \\right)}^2}}} = {{19} \\over {35}}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3645,
"subject": "General Science",
"question": "The plane which bisects the line joining, the\npoints (4, –2, 3) and (2, 4, –1) at right angles\nalso passes through the point :",
"options": [
{
"text": "(4, 0, 1)"
},
{
"text": "(0, –1, 1)"
},
{
"text": "(0, 1, –1)"
},
{
"text": "(4, 0, –1)"
}
],
"answer": "(4, 0, –1)",
"solution": "**Answer:** (4, 0, –1)\n\nDirection ratios of normal to plane are < 2, –6, 4 >\n
Also plane passes through (3, 1, 1)\n
$$ \\therefore $$ Equation of plane \n
2(x–3)–6(y–1)+4(z–1) = 0\n
$$ \\Rightarrow $$ x – 3y + 2z = 2\n
By checking all options we can see this equation passes through (4, 0, –1)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3646,
"subject": "General Science",
"question": "If the equation of a plane P, passing through the intersection of the planes, x + 4y - z + 7 = 0\nand 3x + y + 5z = 8 is ax + by + 6z = 15 for some a, b $$ \\in $$ R, then the distance of the point\n(3, 2, -1) from the plane P is...........",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nEquation of plane P is
$$(x + 4y - z + 7) + \\lambda (3x + y + 5z - 8) = 0$$
$$ \\therefore $$ (x, y, z) = (6, 5, –2)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3649,
"subject": "General Science",
"question": "The equation of the plane passing through the point (1, 2, -3) and perpendicular to the\nplanes
Then, which one of the following is true?",
"options": [
{
"text": "P1 and P2 are parallel."
},
{
"text": "P1, P2 and P3 all are parallel."
},
{
"text": "P1 and P3 are parallel."
},
{
"text": "P2 and P3 are parallel."
}
],
"answer": "P1 and P3 are parallel.",
"solution": "**Answer:** P1 and P3 are parallel.\n\nP1 : 3x + 15y + 21z = 9,
P2 : x $$-$$ 3y $$-$$ z = 5
P3 : x + 5y + 7z = 5/2\n
$$ \\therefore $$ P1 and P3 are parallel.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3652,
"subject": "General Science",
"question": "If (1, 5, 35), (7, 5, 5), (1, $$\\lambda$$, 7) and (2$$\\lambda$$, 1, 2) are coplanar, then the sum of all possible values of $$\\lambda$$ is :",
"options": [
{
"text": "$$ - {{44} \\over 5}$$"
},
{
"text": "$$ - {{39} \\over 5}$$"
},
{
"text": "$${{44} \\over 5}$$"
},
{
"text": "$${{39} \\over 5}$$"
}
],
"answer": "$${{44} \\over 5}$$",
"solution": "**Answer:** $${{44} \\over 5}$$\n\nA(1, 5, 35), B(7, 5, 5), C(1, $$\\lambda$$, 7), D(2$$\\lambda$$, 1, 2)
$$ \\therefore $$ Equation of plane is a(x $$-$$ 1) + c(z $$-$$ 3) = 0
x = 0, z = 0 also satisfy it $$-$$a $$-$$3c = 0 $$ \\Rightarrow $$ a = $$-$$3c
$$-$$3c (x $$-$$ 1) + c (z $$-$$ 3) = 0
$$-$$3 + 3 + z $$-$$ 3 = 0
3x $$-$$ z = 0",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3657,
"subject": "General Science",
"question": "Let P be an arbitrary point having sum of the squares of the distances from the planes x + y + z = 0, lx $$-$$ nz = 0 and x $$-$$ 2y + z = 0, equal to 9. If the locus of the point P is x2 + y2 + z2 = 9, then the value of l $$-$$ n is equal to _________.",
"options": [],
"answer": "0",
"solution": "**Answer:** 0\n\nLet point P is ($$\\alpha$$, $$\\beta$$, $$\\gamma$$)
Since its given that $${x^2} + {y^2} + {z^2} = 9$$
After solving l = n,
then, l $$-$$ n = 0",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3658,
"subject": "General Science",
"question": "Let the plane ax + by + cz + d = 0 bisect the line joining the points (4, $$-$$3, 1) and (2, 3, $$-$$5) at the right angles. If a, b, c, d are integers, then the minimum value of (a2 + b2 + c2 + d2) is _________.",
"options": [],
"answer": "28",
"solution": "**Answer:** 28\n\nNormal of plane = $$\\overrightarrow {PQ} = - 2\\widehat i + 6\\widehat j - 6\\widehat k$$
a = $$-$$2, b = 6, c = $$-$$6
& equation of plane is
$$-$$2x + 6y $$-$$ 6z + d = 0
$$ M(3,0, - 2)$$ is the midpoint of the line which present on the plane\n which satisfy the plane
$$ \\therefore $$ d = $$-$$6
Now equation of plane is
$$-$$2x + 6y $$-$$ 6z $$-$$ 6 = 0
x $$-$$ 3y + 3z + 3 = 0
$$ \\Rightarrow $$ (a2 + b2 + c2 + d2)min = 12 + 9 + 9 + 9 = 28",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3659,
"subject": "General Science",
"question": "The equation of the planes parallel to the plane x $$-$$ 2y + 2z $$-$$ 3 = 0 which are at unit distance from the point (1, 2, 3) is ax + by + cz + d = 0. If (b $$-$$ d) = k(c $$-$$ a), then the positive value of k is :",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nThe equation of the planes parallel to the plane x $$-$$ 2y + 2z $$-$$ 3 = 0\n
$$ \\Rightarrow k = 4$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3660,
"subject": "General Science",
"question": "Let the mirror image of the point (1, 3, a) with respect to the plane $$\\overrightarrow r .\\left( {2\\widehat i - \\widehat j + \\widehat k} \\right) - b = 0$$ be ($$-$$3, 5, 2). Then, the value of | a + b | is equal to ____________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n
Given equation of plane in vector form is $$\\overrightarrow r \\,.\\,(2\\widehat i - \\widehat j + \\widehat k) - b = 0$$
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3661,
"subject": "General Science",
"question": "Let the plane passing through the point ($$-$$1, 0, $$-$$2) and perpendicular to each of the planes 2x + y $$-$$ z = 2 and x $$-$$ y $$-$$ z = 3 be ax + by + cz + 8 = 0. Then the value of a + b + c is equal to :",
"options": [
{
"text": "3"
},
{
"text": "8"
},
{
"text": "5"
},
{
"text": "4"
}
],
"answer": "4",
"solution": "**Answer:** 4\n\nNormal of required plane $$\\left( {2\\widehat i + \\widehat j - \\widehat k} \\right) \\times \\left( {\\widehat i - \\widehat j - \\widehat k} \\right)$$
$$a + b + c = 4$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3662,
"subject": "General Science",
"question": "Let P be the plane passing through the point (1, 2, 3) and the line of intersection of the planes $$\\overrightarrow r \\,.\\,\\left( {\\widehat i + \\widehat j + 4\\widehat k} \\right) = 16$$ and $$\\overrightarrow r \\,.\\,\\left( { - \\widehat i + \\widehat j + \\widehat k} \\right) = 6$$. Then which of the following points does NOT lie on P?",
"options": [
{
"text": "(3, 3, 2)"
},
{
"text": "(6, $$-$$6, 2)"
},
{
"text": "(4, 2, 2)"
},
{
"text": "($$-$$8, 8, 6)"
}
],
"answer": "(4, 2, 2)",
"solution": "**Answer:** (4, 2, 2)\n\n$$(x + y + 4z - 16) + \\lambda ( - x + y + z - 6) = 0$$
Let d be the distance between the foot of perpendiculars of the points P(1, 2, $$-$$1) and Q(2, $$-$$1, 3) on the plane $$-$$x + y + z = 1. Then d2 is equal to ___________.
Let the plane ax + by + cz = d pass through (2, 3, $$-$$5) and is perpendicular to the planes 2x + y $$-$$ 5z = 10 and 3x + 5y $$-$$ 7z = 12. If a, b, c, d are integers d > 0 and gcd (|a|, |b|, |c|, d) = 1, then the value of a + 7b + c + 20d is equal to :
The acute angle between the planes P1 and P2, when P1 and P2 are the planes passing through the intersection of the planes $$5x + 8y + 13z - 29 = 0$$ and $$8x - 7y + z - 20 = 0$$ and the points (2, 1, 3) and (0, 1, 2), respectively, is :
Let the mirror image of the point (a, b, c) with respect to the plane 3x $$-$$ 4y + 12z + 19 = 0 be (a $$-$$ 6, $$\\beta$$, $$\\gamma$$). If a + b + c = 5, then 7$$\\beta$$ $$-$$ 9$$\\gamma$$ is equal to ______________.
If the plane $$2x + y - 5z = 0$$ is rotated about its line of intersection with the plane $$3x - y + 4z - 7 = 0$$ by an angle of $${\\pi \\over 2}$$, then the plane after the rotation passes through the point :
Let p be the plane passing through the intersection of the planes $$\\overrightarrow r \\,.\\,\\left( {\\widehat i + 3\\widehat j - \\widehat k} \\right) = 5$$ and $$\\overrightarrow r \\,.\\,\\left( {2\\widehat i - \\widehat j + \\widehat k} \\right) = 3$$, and the point (2, 1, $$-$$2). Let the position vectors of the points X and Y be $$\\widehat i - 2\\widehat j + 4\\widehat k$$ and $$5\\widehat i - \\widehat j + 2\\widehat k$$ respectively. Then the points :
",
"options": [
{
"text": "X and X + Y are on the same side of P"
},
{
"text": "Y and Y $$-$$ X are on the opposite sides of P"
},
{
"text": "X and Y are on the opposite sides of P"
},
{
"text": "X + Y and X $$-$$ Y are on the same side of P"
}
],
"answer": "X and Y are on the opposite sides of P",
"solution": "**Answer:** X and Y are on the opposite sides of P\n\n
Let the equation of required plane
\n
$$\\pi :(x + 3y - z - 5) + \\lambda (2x - y + z - 3) = 0$$
\n
$$\\because$$ (2, 1, $$-$$2) lies on it so, $$2 + \\lambda ( - 2) = 0$$
Let the points on the plane P be equidistant from the points ($$-$$4, 2, 1) and (2, $$-$$2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is :
A plane $$E$$ is perpendicular to the two planes $$2 x-2 y+z=0$$ and $$x-y+2 z=4$$, and passes through the point $$P(1,-1,1)$$. If the distance of the plane $$E$$ from the point $$Q(a, a, 2)$$ is $$3 \\sqrt{2}$$, then $$(P Q)^{2}$$ is equal to :
A vector $$\\vec{a}$$ is parallel to the line of intersection of the plane determined by the vectors $$\\hat{i}, \\hat{i}+\\hat{j}$$ and the plane determined by the vectors $$\\hat{i}-\\hat{j}, \\hat{i}+\\hat{k}$$. The obtuse angle between $$\\vec{a}$$ and the vector $$\\vec{b}=\\hat{i}-2 \\hat{j}+2 \\hat{k}$$ is :
Vector $$\\overrightarrow a $$ is parallel to $${\\overrightarrow n _1} \\times {\\overrightarrow n _2}$$
\n
i.e. $$\\overrightarrow a $$ is parallel to $$\\left| {\\matrix{\n {\\widehat i} & {\\widehat j} & {\\widehat k} \\cr \n 0 & 0 & 1 \\cr \n { - 1} & { - 1} & 1 \\cr \n\n } } \\right| = \\widehat i - \\widehat j$$
\n
Given $$\\overrightarrow b = \\widehat i - 2\\widehat j + 2\\widehat k$$
\n
Cosine of acute angle between
\n
$$\\overrightarrow a $$ and $$\\overrightarrow b = \\left| {{{\\overrightarrow a \\,.\\,\\overrightarrow b } \\over {|\\overrightarrow a |\\,.\\,|\\overrightarrow b |}}} \\right| = {1 \\over {\\sqrt 2 }}$$
\n
Obtuse angle between $$\\overrightarrow a $$ and $$\\overrightarrow b = {{3\\pi } \\over 4}$$
If the plane $$P$$ passes through the intersection of two mutually perpendicular planes $$2 x+k y-5 z=1$$ and $$3 k x-k y+z=5, k<3$$ and intercepts a unit length on positive $$x$$-axis, then the intercept made by the plane $$P$$ on the $$y$$-axis is :
The foot of the perpendicular from a point on the circle $$x^{2}+y^{2}=1, z=0$$ to the plane $$2 x+3 y+z=6$$ lies on which one of the following curves?
A plane P is parallel to two lines whose direction ratios are $$-2,1,-3$$ and $$-1,2,-2$$ and it contains the point $$(2,2,-2)$$. Let P intersect the co-ordinate axes at the points $$\\mathrm{A}, \\mathrm{B}, \\mathrm{C}$$ making the intercepts $$\\alpha, \\beta, \\gamma$$. If $$\\mathrm{V}$$ is the volume of the tetrahedron $$\\mathrm{OABC}$$, where $$\\mathrm{O}$$ is the origin, and $$\\mathrm{p}=\\alpha+\\beta+\\gamma$$, then the ordered pair $$(\\mathrm{V}, \\mathrm{p})$$ is equal to :
Let $$Q$$ be the foot of perpendicular drawn from the point $$P(1,2,3)$$ to the plane $$x+2 y+z=14$$. If $$R$$ is a point on the plane such that $$\\angle P R Q=60^{\\circ}$$, then the area of $$\\triangle P Q R$$ is equal to :
If $$\\lambda_{1} < \\lambda_{2}$$ are two values of $$\\lambda$$ such that the angle between the planes $$P_{1}: \\vec{r}(3 \\hat{i}-5 \\hat{j}+\\hat{k})=7$$ and \n$$P_{2}: \\vec{r} \\cdot(\\lambda \\hat{i}+\\hat{j}-3 \\hat{k})=9$$ is $$\\sin ^{-1}\\left(\\frac{2 \\sqrt{6}}{5}\\right)$$, then the square of the length of perpendicular from the point $$\\left(38 \\lambda_{1}, 10 \\lambda_{2}, 2\\right)$$ to the plane $$P_{1}$$ is ______________.
\nLet equation of plane is $x-z+\\lambda=0$ passes through point $A(2,-3,1) \\Rightarrow \\lambda=-1$\n
\nEquation of plane is $x-z-1=0$\n
\nDistance of point $(7,-3,-4)$ from the plane $x-z-$ $1=0$ is $5 \\sqrt{2}$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3683,
"subject": "General Science",
"question": "Let the foot of perpendicular of the point $P(3,-2,-9)$ on the plane passing through the points $(-1,-2,-3),(9,3,4),(9,-2,1)$ be $Q(\\alpha, \\beta, \\gamma)$. Then the distance of $Q$ from the origin is :",
"options": [
{
"text": "$\\sqrt{38}$"
},
{
"text": "$\\sqrt{29}$"
},
{
"text": "$\\sqrt{42}$"
},
{
"text": "$\\sqrt{35}$"
}
],
"answer": "$\\sqrt{42}$",
"solution": "**Answer:** $\\sqrt{42}$\n\n
The equation of the plane passing through points $A(-1, -2, -3)$, $B(9, 3, 4)$, and $C(9, -2, 1)$ can be written using the determinant :
Next, we find the foot of the perpendicular from point $P(3, -2, -9)$ to the plane. Using the coordinates of $P$ and the equation of the plane, we can find the ratio of the perpendicular distance to the sum of the squares of the coefficients of $x$, $y$, and $z$ :
So, the distance of the point $(-3, 2, 0)$ from the plane $2x - 2y + z = 11$ is 7.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3685,
"subject": "General Science",
"question": "
Let $$\\mathrm{N}$$ be the foot of perpendicular from the point $$\\mathrm{P}(1,-2,3)$$ on the line passing through the points $$(4,5,8)$$ and $$(1,-7,5)$$. Then the distance of $$N$$ from the plane $$2 x-2 y+z+5=0$$ is :
Let the equation of plane passing through the line of intersection of the planes $$x+2 y+a z=2$$ and $$x-y+z=3$$ be $$5 x-11 y+b z=6 a-1$$. For $$c \\in \\mathbb{Z}$$, if the distance of this plane from the point $$(a,-c, c)$$ is $$\\frac{2}{\\sqrt{a}}$$, then $$\\frac{a+b}{c}$$ is equal to :
",
"options": [
{
"text": "$$-$$2"
},
{
"text": "4"
},
{
"text": "2"
},
{
"text": "$$-$$4"
}
],
"answer": "$$-$$4",
"solution": "**Answer:** $$-$$4\n\nGiven the equation of the plane passing through the intersection of the two given planes:\n\n
$$P: (x + 2y + az - 2) + \\lambda(x - y + z - 3) = 0$$\n
Let the image of the point $$\\left(\\frac{5}{3}, \\frac{5}{3}, \\frac{8}{3}\\right)$$ in the plane $$x-2 y+z-2=0$$ be P. If the distance of the point $$Q(6,-2, \\alpha), \\alpha > 0$$, from $$\\mathrm{P}$$ is 13 , then $$\\alpha$$ is equal to ___________.
",
"options": [],
"answer": "15",
"solution": "**Answer:** 15\n\nImage of point $\\left(\\frac{5}{3}, \\frac{5}{3}, \\frac{8}{3}\\right)$\n
Let the plane $$x+3 y-2 z+6=0$$ meet the co-ordinate axes at the points A, B, C. If the orthocenter of the triangle $$\\mathrm{ABC}$$ is $$\\left(\\alpha, \\beta, \\frac{6}{7}\\right)$$, then $$98(\\alpha+\\beta)^{2}$$ is equal to ___________.
Let the line passing through the points $$\\mathrm{P}(2,-1,2)$$ and $$\\mathrm{Q}(5,3,4)$$ meet the plane $$x-y+z=4$$ at the point $$\\mathrm{R}$$. Then the distance of the point $$\\mathrm{R}$$ from the plane $$x+2 y+3 z+2=0$$ measured parallel to the line $$\\frac{x-7}{2}=\\frac{y+3}{2}=\\frac{z-2}{1}$$ is equal to :
Let P be the plane passing through the points $$(5,3,0),(13,3,-2)$$ and $$(1,6,2)$$.\n\nFor $$\\alpha \\in \\mathbb{N}$$, if the distances of the points $$\\mathrm{A}(3,4, \\alpha)$$ and $$\\mathrm{B}(2, \\alpha, a)$$ from the plane P are 2 and 3 respectively, then the positive value of a is :
Let $$(\\alpha, \\beta, \\gamma)$$ be the image of the point $$\\mathrm{P}(2,3,5)$$ in the plane $$2 x+y-3 z=6$$. Then $$\\alpha+\\beta+\\gamma$$ is equal to :
If equation of the plane that contains the point $$(-2,3,5)$$ and is perpendicular to each of the planes $$2 x+4 y+5 z=8$$ and $$3 x-2 y+3 z=5$$ is $$\\alpha x+\\beta y+\\gamma z+97=0$$ then $$\\alpha+\\beta+\\gamma=$$
",
"options": [
{
"text": "15"
},
{
"text": "16"
},
{
"text": "17"
},
{
"text": "18"
}
],
"answer": "15",
"solution": "**Answer:** 15\n\nThe equation of plane that passes through the point $(-2,3,5)$ is\n
$$\na(x+2)+b(y-3)+c(z-5)=0\n$$ ..........(i)\n
The plane is perpendicular to\n
$$\n\\begin{array}{ll}\n2 x+4 y+5 z =8 \\text { and } 3 x-2 y+3 z=5 \\\\\\\\\n\n\\end{array}\n$$\n
Let the image of the point $$\\mathrm{P}(1,2,6)$$ in the plane passing through the points $$\\mathrm{A}(1,2,0), \\mathrm{B}(1,4,1)$$ and $$\\mathrm{C}(0,5,1)$$ be $$\\mathrm{Q}(\\alpha, \\beta, \\gamma)$$. Then $$\\left(\\alpha^{2}+\\beta^{2}+\\gamma^{2}\\right)$$ is equal to :
",
"options": [
{
"text": "76"
},
{
"text": "62"
},
{
"text": "70"
},
{
"text": "65"
}
],
"answer": "65",
"solution": "**Answer:** 65\n\nEquation of plane passing through the points $A(1,2$, $0), B(1,4,1)$ and $C(0,5,1)$ is\n
Let the line $$\\frac{x}{1}=\\frac{6-y}{2}=\\frac{z+8}{5}$$ intersect the lines $$\\frac{x-5}{4}=\\frac{y-7}{3}=\\frac{z+2}{1}$$ and $$\\frac{x+3}{6}=\\frac{3-y}{3}=\\frac{z-6}{1}$$ at the points $$\\mathrm{A}$$ and $$\\mathrm{B}$$ respectively. Then the distance of the mid-point of the line segment $$\\mathrm{AB}$$ from the plane $$2 x-2 y+z=14$$ is :
Let $$\\mathrm{P}_{1}$$ be the plane $$3 x-y-7 z=11$$ and $$\\mathrm{P}_{2}$$ be the plane passing through the points $$(2,-1,0),(2,0,-1)$$, and $$(5,1,1)$$. If the foot of the perpendicular drawn from the point $$(7,4,-1)$$ on the line of intersection of the planes $$P_{1}$$ and $$P_{2}$$ is $$(\\alpha, \\beta, \\gamma)$$, then $$\\alpha+\\beta+\\gamma$$ is equal to ___________.
",
"options": [],
"answer": "11",
"solution": "**Answer:** 11\n\nEquation of plane $\\mathrm{P}_2$ passing through $(2,-1,0),(2,0$, $-1)$ and $(5,1,1)$ is\n
Let the image of the point $$\\mathrm{P}(1,2,3)$$ in the plane $$2 x-y+z=9$$ be $$\\mathrm{Q}$$. If the coordinates of the point $$\\mathrm{R}$$ are $$(6,10,7)$$, then the square of the area of the triangle $$\\mathrm{PQR}$$ is _____________.
",
"options": [],
"answer": "594",
"solution": "**Answer:** 594\n\nLet $Q(x, y, z)$ be the image of $P(1,2,3)$ in the plane\n
$\\therefore$ The square of the area of the $\\triangle P Q R=594$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3698,
"subject": "General Science",
"question": "
A plane P contains the line of intersection of the plane $$\\vec{r} \\cdot(\\hat{i}+\\hat{j}+\\hat{k})=6$$ and $$\\vec{r} \\cdot(2 \\hat{i}+3 \\hat{j}+4 \\hat{k})=-5$$. If $$\\mathrm{P}$$ passes through the point $$(0,2,-2)$$, then the square of distance of the point $$(12,12,18)$$ from the plane $$\\mathrm{P}$$ is :
$$ \\Rightarrow $$ Area of greatest rectangle is equal to $$2ab$$\n
When $$\\sin \\,2\\theta = 1.$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3703,
"subject": "General Science",
"question": "The function $$f\\left( x \\right) = {x \\over 2} + {2 \\over x}$$ has a local minimum at ",
"options": [
{
"text": "$$x=2$$ "
},
{
"text": "$$x=-2$$ "
},
{
"text": "$$x=0$$ "
},
{
"text": "$$x=1$$ "
}
],
"answer": "$$x=2$$ ",
"solution": "**Answer:** $$x=2$$ \n\n$$f\\left( x \\right) = {x \\over 2} + {2 \\over x} \\Rightarrow f'\\left( x \\right) = {1 \\over 2} - {2 \\over {{x^2}}} = 0$$ \n
$$ \\Rightarrow {x^2} = 4$$ or $$x=2,-2;$$ $$\\,\\,\\,\\,\\,f''\\left( x \\right) = {4 \\over {{x^3}}}$$\n
$$f''{\\left. {\\left( x \\right)} \\right]_{x = 2}} = + ve \\Rightarrow f\\left( x \\right)$$ \n
has local min at $$x=2.$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3704,
"subject": "General Science",
"question": "A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length $$x$$. The maximum area enclosed by the park is ",
"options": [
{
"text": "$${3 \\over 2}{x^2}$$ "
},
{
"text": "$$\\sqrt {{{{x^3}} \\over 8}} $$ "
},
{
"text": "$${1 \\over 2}{x^2}$$ "
},
{
"text": "$$\\pi {x^2}$$ "
}
],
"answer": "$${1 \\over 2}{x^2}$$ ",
"solution": "**Answer:** $${1 \\over 2}{x^2}$$ \n\nArea $$ = {1 \\over 2}{x^2}\\,\\sin \\,\\theta $$\n
\n
Maximum value of $$\\sin \\theta $$ is $$1$$ at $$\\theta = {\\pi \\over 2}$$\n
$${A_{\\max }} = {1 \\over 2}{x^2}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3705,
"subject": "General Science",
"question": "If $$p$$ and $$q$$ are positive real numbers such that $${p^2} + {q^2} = 1$$, then the maximum value of $$(p+q)$$ is",
"options": [
{
"text": "$${1 \\over 2}$$ "
},
{
"text": "$${1 \\over {\\sqrt 2 }}$$ "
},
{
"text": "$${\\sqrt 2 }$$ "
},
{
"text": "$$2$$"
}
],
"answer": "$${\\sqrt 2 }$$ ",
"solution": "**Answer:** $${\\sqrt 2 }$$ \n\nGiven that $${p^2} + {q^2} = 1$$ \n
Hence max. value of $$p + q$$ is $$\\sqrt 2 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3706,
"subject": "General Science",
"question": "Suppose the cubic $${x^3} - px + q$$ has three distinct real roots \n where $$p>0$$ and $$q>0$$. Then which one of the following holds? ",
"options": [
{
"text": "The cubic has minima at $$\\sqrt {{p \\over 3}} $$ and maxima at $$-\\sqrt {{p \\over 3}} $$"
},
{
"text": "The cubic has minima at $$-\\sqrt {{p \\over 3}} $$ and maxima at $$\\sqrt {{p \\over 3}} $$"
},
{
"text": "The cubic has minima at both $$\\sqrt {{p \\over 3}} $$ and $$-\\sqrt {{p \\over 3}} $$"
},
{
"text": "The cubic has maxima at both $$\\sqrt {{p \\over 3}} $$ and $$-\\sqrt {{p \\over 3}} $$"
}
],
"answer": "The cubic has minima at $$\\sqrt {{p \\over 3}} $$ and maxima at $$-\\sqrt {{p \\over 3}} $$",
"solution": "**Answer:** The cubic has minima at $$\\sqrt {{p \\over 3}} $$ and maxima at $$-\\sqrt {{p \\over 3}} $$\n\nLet $$y = {x^3} - px + q$$\n
$$\\therefore$$ $$y$$ has ninima at $$x = \\sqrt {{p \\over 3}} $$ \n
and maxima at $$x = - \\sqrt {{p \\over 3}} $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3707,
"subject": "General Science",
"question": "Given $$P\\left( x \\right) = {x^4} + a{x^3} + b{x^2} + cx + d$$ such that $$x=0$$ is the only \n real root of $$P'\\,\\left( x \\right) = 0.$$ If $$P\\left( { - 1} \\right) < P\\left( 1 \\right),$$ then in the interval $$\\left[ { - 1,1} \\right]:$$",
"options": [
{
"text": "$$P(-1)$$ is not minimum but $$P(1)$$ is the maximum of $$P$$"
},
{
"text": "$$P(-1)$$ is the minimum but $$P(1)$$ is not the maximum of $$P$$"
},
{
"text": "Neither $$P(-1)$$ is the minimum nor $$P(1)$$ is the maximum of $$P$$"
},
{
"text": "$$P(-1)$$ is the minimum and $$P(1)$$ is the maximum of $$P$$"
}
],
"answer": "$$P(-1)$$ is not minimum but $$P(1)$$ is the maximum of $$P$$",
"solution": "**Answer:** $$P(-1)$$ is not minimum but $$P(1)$$ is the maximum of $$P$$\n\nWe have $$P\\left( x \\right) = {x^4} + a{x^3} + b{x^2} + cx + d$$\n
This is true where $$k=-1$$ ",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 3709,
"subject": "General Science",
"question": "Let $$f:R \\to R$$ be a continuous function defined by \n$$$f\\left( x \\right) = {1 \\over {{e^x} + 2{e^{ - x}}}}$$$\n
Statement - 1 : $$f\\left( c \\right) = {1 \\over 3},$$ for some $$c \\in R$$.
\n
Statement - 2 : $$0 < f\\left( x \\right) \\le {1 \\over {2\\sqrt 2 }},$$ for all $$x \\in R$$
",
"options": [
{
"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."
},
{
"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."
}
],
"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\n$$f\\left( x \\right) = {1 \\over {{e^x} + 2{e^{ - x}}}} = {{{e^x}} \\over {{e^{2x}} + 2}}$$ \n
maximum $$f\\left( x \\right) = {{\\sqrt 2 } \\over 4} = {1 \\over {2\\sqrt 2 }}$$\n
$$0 < f\\left( x \\right) \\le {1 \\over {2\\sqrt 2 }}\\,\\,\\forall x \\in R$$\n
Since $$0 < {1 \\over 3} < {1 \\over {2\\sqrt 2 }} \\Rightarrow $$ for some $$c \\in R$$\n
$$f\\left( c \\right) = {1 \\over 3}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3710,
"subject": "General Science",
"question": "For $$x \\in \\left( {0,{{5\\pi } \\over 2}} \\right),$$ define $$f\\left( x \\right) = \\int\\limits_0^x {\\sqrt t \\sin t\\,dt.} $$ Then $$f$$ has ",
"options": [
{
"text": "local minimum at $$\\pi $$ and $$2\\pi $$"
},
{
"text": "local minimum at $$\\pi $$ and local maximum at $$2\\pi $$"
},
{
"text": "local maximum at $$\\pi $$ and local minimum at $$2\\pi $$"
},
{
"text": "local maximum at $$\\pi $$ and $$2\\pi $$"
}
],
"answer": "local maximum at $$\\pi $$ and local minimum at $$2\\pi $$",
"solution": "**Answer:** local maximum at $$\\pi $$ and local minimum at $$2\\pi $$\n\n$$f'\\left( x \\right) = \\sqrt x \\sin x$$\n
At local maxima or minima, $$f'\\left( x \\right) = 0$$\n
At $$x = 2\\pi ,\\,\\,\\,f''\\left( x \\right) > 0$$\n
Hence local minima at $$x = 2\\pi $$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3711,
"subject": "General Science",
"question": "Let $$a,b \\in R$$ be such that the function $$f$$ given by $$f\\left( x \\right) = In\\left| x \\right| + b{x^2} + ax,\\,x \\ne 0$$ has extreme values at $$x=-1$$ and $$x=2$$ \n
Statement-1 : $$f$$ has local maximum at $$x=-1$$ and at $$x=2$$.
Hence both the statements are true and statement $$2$$ is a correct explanation for $$1.$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3712,
"subject": "General Science",
"question": "A line is drawn through the point $$(1, 2)$$ to meet the coordinate axes at $$P$$ and $$Q$$ such that it forms a triangle $$OPQ,$$ where $$O$$ is the origin. If the area of the triangle $$OPQ$$ is least, then the slope of the line $$PQ$$ is :",
"options": [
{
"text": "$$-{1 \\over 4}$$"
},
{
"text": "$$-4$$ "
},
{
"text": "$$-2$$ "
},
{
"text": "$$-{1 \\over 2}$$"
}
],
"answer": "$$-2$$ ",
"solution": "**Answer:** $$-2$$ \n\nEquation of a line passing through $$\\left( {{x_1},{y_1}} \\right)$$ having \n
slope $$m$$ is given by $$y - {y_1} = m\\left( {x - {x_1}} \\right)$$\n
Since the line $$PQ$$ is passing through $$(1,2)$$ therefore its\n
$$f\\left( 2 \\right) = 2 \\times 4 - 2 \\times 8 + {1 \\over 2} \\times 16 = 0$$",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 3715,
"subject": "General Science",
"question": "A wire of length $$2$$ units is cut into two parts which are bent respectively to form a square of side $$=x$$ units and a circle of radius $$=r$$ units. If the sum of the areas of the square and the circle so formed is minimum, then: ",
"options": [
{
"text": "$$x=2r$$ "
},
{
"text": "$$2x=r$$ "
},
{
"text": "$$2x = \\left( {\\pi + 4} \\right)r$$ "
},
{
"text": "$$\\left( {4 - \\pi } \\right)x = \\pi \\,\\, r$$ "
}
],
"answer": "$$x=2r$$ ",
"solution": "**Answer:** $$x=2r$$ \n\n$$4x + 2\\pi r = 2$$ $$\\,\\,\\,$$ $$ \\Rightarrow 2x + \\pi r = 1$$\n
= $${{\\sqrt {15} } \\over 2}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3717,
"subject": "General Science",
"question": "Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the\nmaximum area (in sq. m) of the flower-bed, is :",
"options": [
{
"text": "10"
},
{
"text": "25"
},
{
"text": "30"
},
{
"text": "12.5"
}
],
"answer": "25",
"solution": "**Answer:** 25\n\n\nWe have\n
So $$f\\left( t \\right) = {{{t^2} + 2} \\over t}$$ = $$t + {2 \\over t}$$\n
then $$f'\\left( t \\right) = 1 - {2 \\over {{t^2}}}$$\n
At maximum or minimum $$f'\\left( t \\right) = 0$$.\n
$$\\therefore$$ $$1 - {2 \\over {{t^2}}} = 0$$\n
$$ \\Rightarrow t = \\pm \\sqrt 2 $$\n
Now we need to find $$f''\\left( t \\right)$$ which will tell at which point among $$ + \\sqrt 2 $$ and $$ - \\sqrt 2 $$ will be maximum and minimum.\n
$$f''\\left( t \\right) = + {4 \\over {{t^3}}}$$\n
So when $$t = + \\sqrt 2 $$, then $$f''\\left( t \\right) = + {4 \\over {{{\\left( {\\sqrt 2 } \\right)}^3}}}$$ = $$ + \\sqrt 2 $$\n
As $$f''\\left( t \\right) > 0$$ when $$t = + \\sqrt 2 $$ then at $$t = + \\sqrt 2 $$ the function $$f\\left( t \\right)$$ will be minimum.\n
Min of $$f\\left( t \\right) = {{{{\\left( {\\sqrt 2 } \\right)}^2} + 2} \\over {\\sqrt 2 }}$$ = $${4 \\over {\\sqrt 2 }}$$ = $$2\\sqrt 2 $$\n
So local minimum of $$f\\left( t \\right)$$ = $$2\\sqrt 2 $$\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3719,
"subject": "General Science",
"question": "If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm2) of this cone is : ",
"options": [
{
"text": "$$6\\sqrt 2 \\pi $$"
},
{
"text": "$$6\\sqrt 3 \\pi $$"
},
{
"text": "$$8\\sqrt 2 \\pi $$"
},
{
"text": "$$8\\sqrt 3 \\pi $$"
}
],
"answer": "$$8\\sqrt 3 \\pi $$",
"solution": "**Answer:** $$8\\sqrt 3 \\pi $$\n\nSphere of radius r = 3 cm\n
Let b, h be base radius and height of cone respectively.\n
So, volume of cone = $${1 \\over 2}$$ $$\\pi $$b2h\n
$$ \\Rightarrow $$ b = $${{2\\sqrt 2 } \\over 3}$$ r = 2 $$\\sqrt 2 \\,\\,m$$\n
Therefore curved surface area = $$\\pi bl$$\n
= $$\\pi $$b$$\\sqrt {{h^2} + {r^2}} $$ = $$\\pi $$2$$\\sqrt 2 $$ $$\\sqrt {{4^2} + 8} $$ = 8$$\\sqrt 3 $$$$\\pi $$ cm2",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3720,
"subject": "General Science",
"question": "Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f(x) = 2x3 $$-$$ 9x2 + 12x + 5 in the interval [0, 3]. Then M $$-$$m is equal to : ",
"options": [
{
"text": "5"
},
{
"text": "9"
},
{
"text": "4"
},
{
"text": "1"
}
],
"answer": "9",
"solution": "**Answer:** 9\n\n
To determine the absolute maximum (M) and absolute minimum (m) of the function $ f(x) = 2x^3 - 9x^2 + 12x + 5 $ over the interval $[0, 3]$, we need to examine its critical points and endpoints.
\n\n
First, we find the derivative of the function, $ f'(x) $, to locate the critical points:
$$ \\therefore $$ h = 6 sin$$\\theta $$ = $${6 \\over {\\sqrt 3 }}$$ = $$2\\sqrt 3 $$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3724,
"subject": "General Science",
"question": "If S1 and S2 are respectively the sets of local\nminimum and local maximum points of the function, \n
While moving\nleft to right on x-axis whenever derivative changes sign from\nnegative to positive, we get local minima, and\nwhenever derivative changes sign from positive\nto negative, we get local maxima.\n
$$ \\therefore $$ S1 = {–2, 1}\n
S2 = {0}",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3725,
"subject": "General Science",
"question": "The maximum value of the function f(x) = 3x3 – 18x2 + 27x – 40 on the set S = $$\\left\\{ {x\\, \\in R:{x^2} + 30 \\le 11x} \\right\\}$$ is :",
"options": [
{
"text": "$$-$$ 222"
},
{
"text": "$$-$$ 122"
},
{
"text": "$$122$$ "
},
{
"text": "222"
}
],
"answer": "$$122$$ ",
"solution": "**Answer:** $$122$$ \n\nS = {x $$ \\in $$ R, x2 + 30 $$-$$ 11x $$ \\le $$ 0}\n
= $$2\\sqrt 3 \\,\\pi $$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3728,
"subject": "General Science",
"question": "Let ƒ(x) be a polynomial of degree 3 such that\nƒ(–1) = 10, ƒ(1) = –6, ƒ(x) has a critical point\nat x = –1 and ƒ'(x) has a critical point at x = 1.\nThen ƒ(x) has a local minima at x = _______.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nLet f(x) = ax3\n + bx2\n + cx + d\n
$$ \\Rightarrow $$ $${{{ - b} \\over 3}}$$ + b + 3b + 2 - b = -6\n
$$ \\Rightarrow $$ b = -3\n
$$ \\therefore $$ f'(x) = 3(x2\n - 2x - 3)\n
$$ \\Rightarrow $$ f''(x) = 3(2x\n - 2)\n
At x = 3, f''(x) = 3(2.3\n - 2) = 12 > 0\n
$$ \\therefore $$ Minima\nat x = 3.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 3729,
"subject": "General Science",
"question": "The set of all real values of $$\\lambda $$ for which the\nfunction
Which is monotonic. so no maxima/minima.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3730,
"subject": "General Science",
"question": "The position of a moving car at time t is given by f(t) = at2 + bt + c, t > 0, where a, b and c are real\nnumbers greater than 1. Then the average speed of the car over the time interval [t1\n, t2\n] is\nattained at the point :\n",
"options": [
{
"text": "$${{\\left( {{t_1} + {t_2}} \\right)} \\over 2}$$"
},
{
"text": "$${{\\left( {{t_2} - {t_1}} \\right)} \\over 2}$$"
},
{
"text": "2a(t1\n + t2) + b"
},
{
"text": "a(t2\n– t1) + b"
}
],
"answer": "$${{\\left( {{t_1} + {t_2}} \\right)} \\over 2}$$",
"solution": "**Answer:** $${{\\left( {{t_1} + {t_2}} \\right)} \\over 2}$$\n\nVav = $${{f\\left( {{t_2}} \\right) - f\\left( {{t_1}} \\right)} \\over {{t_2} - {t_1}}}$$ = f'(t)\n
$$ \\Rightarrow $$ t = $${{{t_1} + {t_2}} \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3731,
"subject": "General Science",
"question": "If x = 1 is a critical point of the function\n f(x) = (3x2\n + ax – 2 – a)ex\n, then :",
"options": [
{
"text": "x = 1 is a local maxima and x = $$ - {2 \\over 3}$$ is a\nlocal minima of f."
},
{
"text": "x = 1 and x = $$ - {2 \\over 3}$$ are local maxima of f."
},
{
"text": "x = 1 and x = $$ - {2 \\over 3}$$ are local minima of f."
},
{
"text": "x = 1 is a local minima and x = $$ - {2 \\over 3}$$ is a local\nmaxima of f."
}
],
"answer": "x = 1 is a local minima and x = $$ - {2 \\over 3}$$ is a local\nmaxima of f.",
"solution": "**Answer:** x = 1 is a local minima and x = $$ - {2 \\over 3}$$ is a local\nmaxima of f.\n\nf(x) = (3x2\n + ax – 2 – a)ex\n
= ex(x – 1) (3x + 2)\n\n x = 1 is a local minima and x = $$ - {2 \\over 3}$$ is a local\nmaxima of f.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3732,
"subject": "General Science",
"question": "If the point P on the curve, 4x2 + 5y2 = 20 is farthest from the point Q(0, -4), then PQ2 is equal to:\n",
"options": [
{
"text": "36"
},
{
"text": "48"
},
{
"text": "21"
},
{
"text": "29"
}
],
"answer": "36",
"solution": "**Answer:** 36\n\nGiven ellipse is $${{{x^2}} \\over 5} + {{{y^2}} \\over 4} = 1$$
Let point P is $$(\\sqrt 5 \\cos \\theta ,\\,2\\sin \\theta )$$
Now p(0) = C = -12",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 3735,
"subject": "General Science",
"question": "Let ƒ(x) be a polynomial of degree 5 such that x = ±1 are its critical points.\n
If $$\\mathop {\\lim }\\limits_{x \\to 0} \\left( {2 + {{f\\left( x \\right)} \\over {{x^3}}}} \\right) = 4$$, then which one of the following is not true?",
"options": [
{
"text": "ƒ(1) - 4ƒ(-1) = 4."
},
{
"text": "x = 1 is a point of minima and x = -1 is a point of maxima of ƒ."
},
{
"text": "x = 1 is a point of maxima and x = -1 is a point of minimum of ƒ."
},
{
"text": "ƒ is an odd function."
}
],
"answer": "x = 1 is a point of minima and x = -1 is a point of maxima of ƒ.",
"solution": "**Answer:** x = 1 is a point of minima and x = -1 is a point of maxima of ƒ.\n\nlet f(x) = ax5 + bx4 + cx3 + dx2 + ex + f\n
Given $$\\mathop {\\lim }\\limits_{x \\to 0} \\left( {2 + {{f\\left( x \\right)} \\over {{x^3}}}} \\right) = 4$$\n
It is clear that maxima at x = 1 and minima at x\n= –1.",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 3736,
"subject": "General Science",
"question": "The minimum value of $$\\alpha $$ for which the equation $${4 \\over {\\sin x}} + {1 \\over {1 - \\sin x}} = \\alpha $$\n has at least one\nsolution in $$\\left( {0,{\\pi \\over 2}} \\right)$$ is .......",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\n$$f(x) = {4 \\over {\\sin x}} + {1 \\over {1 - \\sin x}}$$
Let sinx = t $$ \\because $$ $$x \\in \\left( {0,{\\pi \\over 2}} \\right) \\Rightarrow 0 < t < 1$$
$$ = 9$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 3737,
"subject": "General Science",
"question": "Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x6 is unity and it has extrema at x = $$-$$1 and x = 1. If $$\\mathop {\\lim }\\limits_{x \\to 0} {{f(x)} \\over {{x^3}}} = 1$$, then $$5.f(2)$$ is equal to _________.",
"options": [],
"answer": "144",
"solution": "**Answer:** 144\n\n$$f(x) = {x^6} + a{x^5} + b{x^4} + {x^3}$$
$$ \\therefore $$ Required point = (2, 2)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3739,
"subject": "General Science",
"question": "The maximum value of
We know, maximum value of acosx $$ \\pm $$ bsinx\n
= $$\\sqrt {{a^2} + {b^2}} $$\n
$$ \\therefore $$ Here maximum value = $$\\sqrt {{1^2} + {{\\left( { - 2} \\right)}^2}} $$$$ = \\sqrt 5 $$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 3740,
"subject": "General Science",
"question": "Let f : [$$-$$1, 1] $$ \\to $$ R be defined as f(x) = ax2 + bx + c for all x$$\\in$$[$$-$$1, 1], where a, b, c$$\\in$$R such that f($$-$$1) = 2, f'($$-$$1) = 1 for x$$\\in$$($$-$$1, 1) the maximum value of f ''(x) is $${{1 \\over 2}}$$. If f(x) $$ \\le $$ $$\\alpha$$, x$$\\in$$[$$-$$1, 1], then the least value of $$\\alpha$$ is equal to _________.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n$$f(x) = a{x^2} + bx + c$$
Sum $$ = - 4 + {{20} \\over 7} = - {{88} \\over {27}}$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 3742,
"subject": "General Science",
"question": "Let 'a' be a real number such that the function f(x) = ax2 + 6x $$-$$ 15, x $$\\in$$ R is increasing in $$\\left( { - \\infty ,{3 \\over 4}} \\right)$$ and decreasing in $$\\left( {{3 \\over 4},\\infty } \\right)$$. Then the function g(x) = ax2 $$-$$ 6x + 15, x$$\\in$$R has a :",
"options": [
{
"text": "local maximum at x = $$-$$ $${{3 \\over 4}}$$"
},
{
"text": "local minimum at x = $$-$$$${{3 \\over 4}}$$"
},
{
"text": "local maximum at x = $${{3 \\over 4}}$$"
},
{
"text": "local minimum at x = $${{3 \\over 4}}$$"
}
],
"answer": "local maximum at x = $$-$$ $${{3 \\over 4}}$$",
"solution": "**Answer:** local maximum at x = $$-$$ $${{3 \\over 4}}$$\n\n$${{ - B} \\over {2A}} = {3 \\over 4}$$
$$f(3) = 27 - 27 - 9 \\times 3 = - 27$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3744,
"subject": "General Science",
"question": "A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is k (meter), then $$\\left( {{4 \\over \\pi } + 1} \\right)k$$ is equal to _____________.",
"options": [],
"answer": "36",
"solution": "**Answer:** 36\n\nLet x + y = 36
x is perimeter of square and y is perimeter of circle side of square = x/4
radius of circle = $${y \\over {2\\pi }}$$
Sum Areas = $${\\left( {{x \\over 4}} \\right)^2} + \\pi {\\left( {{y \\over {2\\pi }}} \\right)^2}$$
$$ = \\ln {4 \\over {{x^2}}} - 1 = 0 \\Rightarrow x = {2 \\over {\\sqrt e }}$$
$$LM = {2 \\over {\\sqrt e }}$$
Local maximum value = $${\\left( {{2 \\over {2/\\sqrt e }}} \\right)^{{4 \\over e}}} = {e^{{2 \\over e}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3746,
"subject": "General Science",
"question": "A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is :",
"options": [
{
"text": "$${5 \\over {2 + \\sqrt 3 }}$$"
},
{
"text": "$${{10} \\over {2 + 3\\sqrt 3 }}$$"
},
{
"text": "$${5 \\over {3 + \\sqrt 3 }}$$"
},
{
"text": "$${{10} \\over {3 + 2\\sqrt 3 }}$$"
}
],
"answer": "$${{10} \\over {3 + 2\\sqrt 3 }}$$",
"solution": "**Answer:** $${{10} \\over {3 + 2\\sqrt 3 }}$$\n\nLet the wire is cut into two pieces of length x and 20 $$-$$ x.
Area of square = $${\\left( {{x \\over 4}} \\right)^2}$$
Area of regular hexagon = $$6 \\times {{\\sqrt 3 } \\over 4}{\\left( {{{20 - x} \\over 6}} \\right)^2}$$
Total area = $$A(x) = {{{x^2}} \\over {16}} + {{3\\sqrt 3 } \\over 2}{{{{(20 - x)}^2}} \\over {36}}$$
So, 4 Real Roots\n ",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 3748,
"subject": "General Science",
"question": "A box open from top is made from a rectangular sheet of dimension a $$\\times$$ b by cutting squares each of side x from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to :",
"options": [
{
"text": "$${{a + b - \\sqrt {{a^2} + {b^2} - ab} } \\over {12}}$$"
},
{
"text": "$${{a + b - \\sqrt {{a^2} + {b^2} + ab} } \\over 6}$$"
},
{
"text": "$${{a + b - \\sqrt {{a^2} + {b^2} - ab} } \\over 6}$$"
},
{
"text": "$${{a + b + \\sqrt {{a^2} + {b^2} + ab} } \\over 6}$$"
}
],
"answer": "$${{a + b - \\sqrt {{a^2} + {b^2} - ab} } \\over 6}$$",
"solution": "**Answer:** $${{a + b - \\sqrt {{a^2} + {b^2} - ab} } \\over 6}$$\n\n
V = l . b . h = (a $$-$$ 2x)(b $$-$$ 2x) x
$$\\Rightarrow$$ V(x) = (2x $$-$$ a)(2x $$-$$ b) x
$$\\therefore$$ x = $$\\beta$$ $$ = {{a + b - \\sqrt {{a^2} + {b^2} - ab} } \\over b}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3749,
"subject": "General Science",
"question": "The number of real roots of the equation
$$\\Rightarrow$$ Number of real roots = 1",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 3750,
"subject": "General Science",
"question": "Let f(x) be a cubic polynomial with f(1) = $$-$$10, f($$-$$1) = 6, and has a local minima at x = 1, and f'(x) has a local minima at x = $$-$$1. Then f(3) is equal to ____________.",
"options": [],
"answer": "22",
"solution": "**Answer:** 22\n\n
A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is :
Consider a cuboid of sides 2x, 4x and 5x and a closed hemisphere of radius r. If the sum of their surface areas is a constant k, then the ratio x : r, for which the sum of their volumes is maximum, is :
Let $$f(x) = |(x - 1)({x^2} - 2x - 3)| + x - 3,\\,x \\in R$$. If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to ____________.
If xy4 attains maximum value at the point (x, y) on the line passing through the points (50 + $$\\alpha$$, 0) and (0, 50 + $$\\alpha$$), $$\\alpha$$ > 0, then (x, y) also lies on the line :
Let $$f(x) = 4{x^3} - 11{x^2} + 8x - 5,\\,x \\in R$$. Then f :
",
"options": [
{
"text": "has a local minina at $$x = {1 \\over 2}$$"
},
{
"text": "has a local minima at $$x = {3 \\over 4}$$"
},
{
"text": "is increasing in $$\\left( {{1 \\over 2},{3 \\over 4}} \\right)$$"
},
{
"text": "is decreasing in $$\\left( {{1 \\over 2},{4 \\over 3}} \\right)$$"
}
],
"answer": "is decreasing in $$\\left( {{1 \\over 2},{4 \\over 3}} \\right)$$",
"solution": "**Answer:** is decreasing in $$\\left( {{1 \\over 2},{4 \\over 3}} \\right)$$\n\n
Given,
\n
$$f(x) = 4{x^3} - 11{x^2} + 8x - 5$$
\n
$$\\therefore$$ $$f'(x) = 12{x^2} - 22x + 8$$
\n
$$ = 2(6{x^2} - 11x + 4)$$
\n
$$ = 2(6{x^2} - 8x - 3x + 4)$$
\n
$$ = 2\\left[ {2x(3x - 4) - 1(3x - 4)} \\right]$$
\n
$$ = 2\\left[ {(3x - 4)(2x - 1)} \\right]$$
\n
\n
$$f'(x)$$ is positive before $${1 \\over 2}$$ means slope of f(x) is positive before $${1 \\over 2}$$ and f'(x) is negative after $${1 \\over 2}$$ means slope of f(x) is negative after $${1 \\over 2}$$.
\n
\n
$$\\therefore$$ At $${1 \\over 2}$$, f(x) is local maximum
\n
Also, $$f'(x)$$ is negative before $${4 \\over 3}$$ means slope of f(x) is negative before $${4 \\over 3}$$ and f'(x) is positive after $${4 \\over 3}$$ means slope of f(x) is positive after $${4 \\over 3}$$.
\n
\n
$$\\therefore$$ At $${4 \\over 3},\\,f(x)$$ is local minimum
If the absolute maximum value of the function $$f(x)=\\left(x^{2}-2 x+7\\right) \\mathrm{e}^{\\left(4 x^{3}-12 x^{2}-180 x+31\\right)}$$ in the interval $$[-3,0]$$ is $$f(\\alpha)$$, then :\n
The curve $$y(x)=a x^{3}+b x^{2}+c x+5$$ touches the $$x$$-axis at the point $$\\mathrm{P}(-2,0)$$ and cuts the $$y$$-axis at the point $$Q$$, where $$y^{\\prime}$$ is equal to 3 . Then the local maximum value of $$y(x)$$ is:\n
The sum of the maximum and minimum values of the function $$f(x)=|5 x-7|+\\left[x^{2}+2 x\\right]$$ in the interval $$\\left[\\frac{5}{4}, 2\\right]$$, where $$[t]$$ is the greatest integer $$\\leq t$$, is ______________.
If the maximum value of $$a$$, for which the function $$f_{a}(x)=\\tan ^{-1} 2 x-3 a x+7$$ is non-decreasing in $$\\left(-\\frac{\\pi}{6}, \\frac{\\pi}{6}\\right)$$, is $$\\bar{a}$$, then $$f_{\\bar{a}}\\left(\\frac{\\pi}{8}\\right)$$ is equal to :
As the function $f_a^{\\prime}(x)$ is non-decreasing\n
$$\n\\begin{aligned}\n& \\text { in }\\left(-\\frac{\\pi}{6}, \\frac{\\pi}{6}\\right), \\\\\\\\\n& f_a^{\\prime}(x) \\geq 0\n\\end{aligned}\n$$\n
$$\n\\begin{aligned}\n& \\Rightarrow \\frac{2}{1+4 x^2}-3 a \\geq 0 \\Rightarrow \\frac{2}{1+4 x^2} \\geq 3 a \\\\\\\\\n& \\Rightarrow a \\leq \\frac{2}{3\\left(1+4 x^2\\right)}, \\text { when } x \\in\\left(-\\frac{\\pi}{6}, \\frac{\\pi}{6}\\right) \\\\\\\\\n& \\because a \\text { is maximum when } x^2=\\frac{\\pi^2}{36},\n\\end{aligned}\n$$\n
Let $$f(x)=3^{\\left(x^{2}-2\\right)^{3}+4}, x \\in \\mathrm{R}$$. Then which of the following statements are true?
\n
$$\\mathrm{P}: x=0$$ is a point of local minima of $$f$$
\n
$$\\mathrm{Q}: x=\\sqrt{2}$$ is a point of inflection of $$f$$
\n
$$R: f^{\\prime}$$ is increasing for $$x>\\sqrt{2}$$
",
"options": [
{
"text": "Only P and Q"
},
{
"text": "Only P and R"
},
{
"text": "Only Q and R"
},
{
"text": "All P, Q and R"
}
],
"answer": "All P, Q and R",
"solution": "**Answer:** All P, Q and R\n\n
The sum of the absolute maximum and minimum values of the function $$f(x)=\\left|x^{2}-5 x+6\\right|-3 x+2$$ in the interval $$[-1,3]$$ is equal to :
",
"options": [
{
"text": "13"
},
{
"text": "24"
},
{
"text": "10"
},
{
"text": "12"
}
],
"answer": "10",
"solution": "**Answer:** 10\n\nThe sum of the absolute maximum and minimum values of the function $$f(x) = \\left|x^2 - 5x + 6\\right| - 3x + 2$$ in the interval $$[-1, 3]$$ can be found by finding the maximum and minimum values of $$f(x)$$ in this interval and then adding them.\n
\nFirst, let's find the critical points of $$f(x)$$. To do this, we will find the zeros of the expression inside the absolute value:\n
\n$$x^2 - 5x + 6 = 0$$\n
\nSolving this quadratic equation, we find that the zeros are $$x = 2$$ and $$x = 3$$. These are the critical points of $$f(x)$$.\n
\nNext, we evaluate $$f(x)$$ at these critical points and at the endpoints of the interval $$[-1, 3]$$:\n
\nSo, the minimum value of $$f(x)$$ is $$-7$$ and the maximum value is $$17$$.\n
\nSo, the sum of the absolute maximum and minimum values of the function is $$-7 + 17 = 10$$.\n
Note : The absolute maximum and minimum values of a function are the largest and smallest values that the function takes on a given interval, respectively. These values are also called the \"extrema\" of the function. The absolute maximum value is the highest point on the graph of the function, and the absolute minimum value is the lowest point.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3765,
"subject": "General Science",
"question": "
A wire of length $$20 \\mathrm{~m}$$ is to be cut into two pieces. A piece of length $$l_{1}$$ is bent to make a square of area $$A_{1}$$ and the other piece of length $$l_{2}$$ is made into a circle of area $$A_{2}$$. If $$2 A_{1}+3 A_{2}$$ is minimum then $$\\left(\\pi l_{1}\\right): l_{2}$$ is equal to :
Let the function $$f(x) = 2{x^3} + (2p - 7){x^2} + 3(2p - 9)x - 6$$ have a maxima for some value of $$x < 0$$ and a minima for some value of $$x > 0$$. Then, the set of all values of p is
Let $$x=2$$ be a local minima of the function $$f(x)=2x^4-18x^2+8x+12,x\\in(-4,4)$$. If M is local maximum value of the function $$f$$ in ($$-4,4)$$, then M =
\nLocal maxima occurs at $x=\\frac{-2+\\sqrt{6}}{2}=x_0$
\n$$\nf\\left(x_0\\right)=12 \\sqrt{6}-\\frac{33}{2}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3769,
"subject": "General Science",
"question": "Consider the triangles with vertices $A(2,1), B(0,0)$ and $C(t, 4), t \\in[0,4]$.\n\n
If the maximum and the minimum perimeters of such triangles are obtained at
$t=\\alpha$ and $t=\\beta$ respectively, then $6 \\alpha+21 \\beta$ is equal to ___________.",
"options": [],
"answer": "48",
"solution": "**Answer:** 48\n\nWe have a triangle with vertices $A(2,1)$, $B(0,0)$, and $C(t, 4)$, where $t$ belongs to the interval $[0,4]$.\n\n
We want to find the maximum and minimum perimeters of such triangles, which occur at $t=\\alpha$ and $t=\\beta$, respectively.\n\n
To find the minimum perimeter, we use a geometric approach. Reflect point $B$ over the line $y=4$ to get $B'(0,8)$. The line segment $AB'$ intersects the line $y=4$ (which is the $y$-coordinate of point $C$) at the point which gives the minimum perimeter.\n
The equation of the line $AB'$ is then $y - 1 = m_{AB'}(x - 2)$, or $7x + 2y = 16$.\n\n
Solving this equation for $y = 4$ yields $x = \\frac{8}{7}$. So, the minimum perimeter is achieved at point $C\\left(\\frac{8}{7}, 4\\right)$, so $\\beta = \\frac{8}{7}$.\n\n
For the maximum perimeter, we notice that it will be achieved when point $C$ is either at $(0,4)$ or at $(4,4)$, since these are the furthest points from $A(2,1)$ within the range of $t$. By calculating the perimeters at these points, we find that the maximum perimeter is achieved at $\\alpha = 4$.\n\n
Which gives us:\n$$\nx = \\frac{5\\pi}{6}, \\frac{\\pi}{6}, 0\n$$
\n
Step 5: Find the second derivative\n
Now, let's find the second derivative of the expression:\n
$$\nf''(x) = 4\\sin{2x} - 3\\sin{3x}\n$$
\n
Step 6: Analyze the critical points\nEvaluate the second derivative at the critical points:
\n\n
$$f''\\left(\\frac{5\\pi}{6}\\right) = -2\\sqrt{3} - \\sqrt{3} < 0$$, indicating a local maximum.
\n
$$f''\\left(\\frac{\\pi}{6}\\right) = 2\\sqrt{3} - \\sqrt{3} > 0$$, indicating a local minimum.
\n
$$f''(0) = 0$$, inconclusive.
\n\n
Step 7: Evaluate the function at the local maximum point and the boundary points\nSince we are looking for the maximum value of $$f(x)$$, we can now evaluate the function at the local maximum point and the boundary points:
If the local maximum value of the function $$f(x)=\\left(\\frac{\\sqrt{3 e}}{2 \\sin x}\\right)^{\\sin ^{2} x}, x \\in\\left(0, \\frac{\\pi}{2}\\right)$$ , is $$\\frac{k}{e}$$, then $$\\left(\\frac{k}{e}\\right)^{8}+\\frac{k^{8}}{e^{5}}+k^{8}$$ is equal to
$$\n\\Rightarrow \\sin \\mathrm{x}=\\frac{\\sqrt{3}}{2} \\quad\\left(\\text { as } \\mathrm{x} \\in\\left(0, \\frac{\\pi}{2}\\right)\\right)\n$$\n
$$\n\\Rightarrow \\text { Local max value }=\\left(\\frac{\\sqrt{3 \\mathrm{e}}}{\\sqrt{3}}\\right)^{3 / 4}=\\mathrm{e}^{3 / 8}=\\frac{\\mathrm{k}}{\\mathrm{e}}\n$$\n
A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in cm$$^2$$) is equal to :
",
"options": [
{
"text": "1025"
},
{
"text": "900"
},
{
"text": "800"
},
{
"text": "675"
}
],
"answer": "800",
"solution": "**Answer:** 800\n\nGiven, length of square sheet of side is $30 \\mathrm{~cm}$.\n
\n
Let, side of small square be $x \\mathrm{~cm}$.\n
Since, a square piece of tin is to be made into a box without top by cutting square from each corner and folding up the flaps to from a box. Then, this shape formed a cuboidal shape.\n
Thus, volume of the box $(V)$\n
$$\n\\begin{aligned}\n\n& =(30-2 x)(30-2 x) x \\\\\\\\\n&\\Rightarrow V =x(30-2 x)^2\n\\end{aligned}\n$$\n
On differentiating both side with respect to $x$, we get\n
If $$a_{\\alpha}$$ is the greatest term in the sequence $$\\alpha_{n}=\\frac{n^{3}}{n^{4}+147}, n=1,2,3, \\ldots$$, then $$\\alpha$$ is equal to _____________.
Let $$g(x)=3 f\\left(\\frac{x}{3}\\right)+f(3-x)$$ and $$f^{\\prime \\prime}(x)>0$$ for all $$x \\in(0,3)$$. If $$g$$ is decreasing in $$(0, \\alpha)$$ and increasing in $$(\\alpha, 3)$$, then $$8 \\alpha$$ is :
Let $$f(x)=2^x-x^2, x \\in \\mathbb{R}$$. If $$m$$ and $$n$$ are respectively the number of points at which the curves $$y=f(x)$$ and $$y=f^{\\prime}(x)$$ intersect the $$x$$-axis, then the value of $$\\mathrm{m}+\\mathrm{n}$$ is ___________.
The function $$f(x)=2 x+3(x)^{\\frac{2}{3}}, x \\in \\mathbb{R}$$, has
",
"options": [
{
"text": "exactly one point of local minima and no point of local maxima\n"
},
{
"text": "exactly one point of local maxima and exactly one point of local minima\n"
},
{
"text": "exactly two points of local maxima and exactly one point of local minima\n"
},
{
"text": "exactly one point of local maxima and no point of local minima"
}
],
"answer": "exactly one point of local maxima and exactly one point of local minima\n",
"solution": "**Answer:** exactly one point of local maxima and exactly one point of local minima\n\n\n
Let $$f(x)=(x+3)^2(x-2)^3, x \\in[-4,4]$$. If $$M$$ and $$m$$ are the maximum and minimum values of $$f$$, respectively in $$[-4,4]$$, then the value of $$M-m$$ is
The maximum area of a triangle whose one vertex is at $$(0,0)$$ and the other two vertices lie on the curve $$y=-2 x^2+54$$ at points $$(x, y)$$ and $$(-x, y)$$, where $$y>0$$, is :
Let the set of all positive values of $$\\lambda$$, for which the point of local minimum of the function $$(1+x(\\lambda^2-x^2))$$ satisfies $$\\frac{x^2+x+2}{x^2+5 x+6}<0$$, be $$(\\alpha, \\beta)$$. Then $$\\alpha^2+\\beta^2$$ is equal to _________.
Let the sum of the maximum and the minimum values of the function $$f(x)=\\frac{2 x^2-3 x+8}{2 x^2+3 x+8}$$ be $$\\frac{m}{n}$$, where $$\\operatorname{gcd}(\\mathrm{m}, \\mathrm{n})=1$$. Then $$\\mathrm{m}+\\mathrm{n}$$ is equal to :
Let $$f(x)=3 \\sqrt{x-2}+\\sqrt{4-x}$$ be a real valued function. If $$\\alpha$$ and $$\\beta$$ are respectively the minimum and the maximum values of $$f$$, then $$\\alpha^2+2 \\beta^2$$ is equal to
If the function $$f(x)=2 x^3-9 \\mathrm{ax}^2+12 \\mathrm{a}^2 x+1, \\mathrm{a}> 0$$ has a local maximum at $$x=\\alpha$$ and a local minimum at $$x=\\alpha^2$$, then $$\\alpha$$ and $$\\alpha^2$$ are the roots of the equation :
Let $$\\mathrm{A}$$ be the region enclosed by the parabola $$y^2=2 x$$ and the line $$x=24$$. Then the maximum area of the rectangle inscribed in the region $$\\mathrm{A}$$ is ________.
To find the number of critical points of the function $$f(x)=(x-2)^{2 / 3}(2 x+1)$$, we need to determine where its derivative $$f'(x)$$ is equal to zero or undefined. Critical points occur where the derivative is zero or does not exist.
\n\n
First, let's find the derivative of the function:
\n\n
$$f(x)=(x-2)^{2 / 3}(2 x+1)$$
\n\n
We apply the product rule for differentiation, which states that $$(uv)' = u'v + uv'$$, where $$u = (x-2)^{2/3}$$ and $$v = 2x + 1$$.
Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let a and b be the sides of the rectangle PQRS when its area is maximum. Then (a+b)$$^2$$ is equal to :
Let the maximum and minimum values of $$\\left(\\sqrt{8 x-x^2-12}-4\\right)^2+(x-7)^2, x \\in \\mathbf{R}$$ be $$\\mathrm{M}$$ and $$\\mathrm{m}$$, respectively. Then $$\\mathrm{M}^2-\\mathrm{m}^2$$ is equal to _________.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3788,
"subject": "General Science",
"question": "If $$2a+3b+6c=0,$$ $$\\left( {a,b,c \\in R} \\right)$$ then the quadratic equation $$a{x^2} + bx + c = 0$$ has ",
"options": [
{
"text": "at least one root in $$\\left[ {0,1} \\right]$$"
},
{
"text": "at least one root in $$\\left[ {2,3} \\right]$$"
},
{
"text": "at least one root in $$\\left[ {4,5} \\right]$$"
},
{
"text": "none of these "
}
],
"answer": "at least one root in $$\\left[ {0,1} \\right]$$",
"solution": "**Answer:** at least one root in $$\\left[ {0,1} \\right]$$\n\nLet $$f\\left( x \\right) = {{a{x^3}} \\over 3} + {{b{x^2}} \\over 2} + cx \\Rightarrow f\\left( 0 \\right) = 0$$ and $$f(1)$$\n
$$ \\Rightarrow f\\left( 6 \\right) \\ge 8.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3792,
"subject": "General Science",
"question": "A value of $$c$$ for which conclusion of Mean Value Theorem holds for the function $$f\\left( x \\right) = {\\log _e}x$$ on the interval $$\\left[ {1,3} \\right]$$ is ",
"options": [
{
"text": "$${\\log _3}e$$ "
},
{
"text": "$${\\log _e}3$$"
},
{
"text": "$$2\\,\\,{\\log _3}e$$ "
},
{
"text": "$${1 \\over 2}{\\log _3}e$$ "
}
],
"answer": "$$2\\,\\,{\\log _3}e$$ ",
"solution": "**Answer:** $$2\\,\\,{\\log _3}e$$ \n\nUsing Lagrange's Mean Value Theorem\n
Let $$f(x)$$ be a function defined on $$\\left[ {a,b} \\right]$$\n
then, $$f'\\left( c \\right) = {{f\\left( b \\right) - f\\left( a \\right)} \\over {b - a}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,....\\left( i \\right)$$\n
$$c\\,\\, \\in \\left[ {a,b} \\right]$$\n
$$\\therefore$$ Given $$f\\left( x \\right) = {\\log _e}x$$\n
$$\\therefore$$ $$f'\\left( x \\right) = {1 \\over x}$$\n
$$ \\therefore $$ $$a = 5,b = 8$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3798,
"subject": "General Science",
"question": "Let f be any function defined on R and let it satisfy the condition : $$|f(x) - f(y)|\\, \\le \\,|{(x - y)^2}|,\\forall (x,y) \\in R$$
If f(0) = 1, then :",
"options": [
{
"text": "f(x) can take any value in R"
},
{
"text": "$$f(x) < 0,\\forall x \\in R$$"
},
{
"text": "$$f(x) > 0,\\forall x \\in R$$"
},
{
"text": "$$f(x) = 0,\\forall x \\in R$$"
}
],
"answer": "$$f(x) > 0,\\forall x \\in R$$",
"solution": "**Answer:** $$f(x) > 0,\\forall x \\in R$$\n\n$$|f(x) - f(y)|\\, \\le \\,|{(x - y)^2}|$$
Let $$f:[2,4] \\rightarrow \\mathbb{R}$$ be a differentiable function such that $$\\left(x \\log _{e} x\\right) f^{\\prime}(x)+\\left(\\log _{e} x\\right) f(x)+f(x) \\geq 1, x \\in[2,4]$$ with $$f(2)=\\frac{1}{2}$$ and $$f(4)=\\frac{1}{4}$$.
\n
Consider the following two statements :
\n
(A) : $$f(x) \\leq 1$$, for all $$x \\in[2,4]$$
\n
(B) : $$f(x) \\geq \\frac{1}{8}$$, for all $$x \\in[2,4]$$
\n
Then,
",
"options": [
{
"text": "Neither statement (A) nor statement (B) is true"
},
{
"text": "Only statement (A) is true"
},
{
"text": "Only statement (B) is true"
},
{
"text": "Both the statements $$(\\mathrm{A})$$ and (B) are true"
}
],
"answer": "Both the statements $$(\\mathrm{A})$$ and (B) are true",
"solution": "**Answer:** Both the statements $$(\\mathrm{A})$$ and (B) are true\n\nGiven, $$\\left(x \\log _{e} x\\right) f^{\\prime}(x)+\\left(\\log _{e} x\\right) f(x)+f(x) \\geq 1, x \\in[2,4]$$\n
",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 3801,
"subject": "General Science",
"question": "A function is matched below against an interval where it is supposed to be\nincreasing. Which of the following pairs is incorrectly matched?",
"options": [
{
"text": "
\n
\n
Interval
\n
Function
\n
\n
\n
(- $$\\infty $$, $$\\infty $$)
\n
x3 - 3x2 + 3x + 3
\n
\n
"
},
{
"text": "
\n
\n
Interval
\n
Function
\n
\n
\n
[2, $$\\infty $$)
\n
2x3 - 3x2 - 12x + 6
\n
\n
"
},
{
"text": "
\n
\n
Interval
\n
Function
\n
\n
\n
$$\\left( { - \\infty ,{1 \\over 3}} \\right]$$
\n
3x2 - 2x + 1
\n
\n
"
},
{
"text": "
\n
\n
Interval
\n
Function
\n
\n
\n
($$ - \\infty $$, - 4 )
\n
x3 + 6x2 + 6
\n
\n
"
}
],
"answer": "
\n
\n
Interval
\n
Function
\n
\n
\n
$$\\left( { - \\infty ,{1 \\over 3}} \\right]$$
\n
3x2 - 2x + 1
\n
\n
",
"solution": "**Answer:**
\n
\n
Interval
\n
Function
\n
\n
\n
$$\\left( { - \\infty ,{1 \\over 3}} \\right]$$
\n
3x2 - 2x + 1
\n
\n
\n\nClearly function $$f\\left( x \\right) = 3{x^2} - 2x + 1$$ is increasing\n
when $$f'\\left( x \\right) = 6x - 2 \\ge 0 \\Rightarrow \\,\\,\\,\\,\\,x \\in \\left[ {1/3,\\left. \\infty \\right)} \\right.$$\n
$$ \\therefore $$ x $$ \\in $$ $$\\left( {{\\pi \\over 4},{\\pi \\over 2}} \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3805,
"subject": "General Science",
"question": "The function f defined by\n
f(x) = x3 $$-$$ 3x2 + 5x + 7 , is : ",
"options": [
{
"text": "increasing in R."
},
{
"text": "decreasing in R."
},
{
"text": "decreasing in (0, $$\\infty $$) and increasing in ($$-$$ $$\\infty $$, 0)"
},
{
"text": "increasing in (0, $$\\infty $$) and decreasing in ($$-$$ $$\\infty $$, 0)"
}
],
"answer": "increasing in R.",
"solution": "**Answer:** increasing in R.\n\n
The given function is
\n
$$f(x) = {x^2} - 3{x^2} + 5x + 7$$
\n
$$f'(x) = 3{x^2} - 6x + 5$$
\n
The discriminant of the above quadratic equation is
\n
$$\\Delta = 36 - 4(3)(5) = 36 - 60 < 0$$
\n
Therefore, $$f'(x) > 0\\,\\forall x \\in {R^ + }$$
\n
Also, $$f'(x) > 0\\,\\forall x \\in {R^ - }$$
\n
Therefore, the given function f is increasing in R.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3806,
"subject": "General Science",
"question": "Let f(x) = $${x \\over {\\sqrt {{a^2} + {x^2}} }} - {{d - x} \\over {\\sqrt {{b^2} + {{\\left( {d - x} \\right)}^2}} }},\\,\\,$$ x $$\\, \\in $$ R, where a, b and d are non-zero real constants. Then :",
"options": [
{
"text": "f is an increasing function of x"
},
{
"text": "f is neither increasing nor decreasing function of x"
},
{
"text": "f ' is not a continuous function of x"
},
{
"text": "f is a decreasing function of x"
}
],
"answer": "f is an increasing function of x",
"solution": "**Answer:** f is an increasing function of x\n\n$$f\\left( x \\right) = {x \\over {\\sqrt {{a^2} + {x^2}} }} - {{d - x} \\over {\\sqrt {{b^2} + {{\\left( {d - x} \\right)}^2}} }}$$\n
$$ \\Rightarrow $$ ƒ'(x) is an increasing function for all x $$ \\in $$ (0, 2).\n
Case 1 : When $$\\phi $$(x) is increasing function\n
So $$\\phi $$'(x) > 0\n
$$ \\Rightarrow $$ ƒ'(x) - ƒ'(2 – x) > 0\n
$$ \\Rightarrow $$ ƒ'(x) > ƒ'(2 – x)\n
$$ \\Rightarrow $$ x > 2 – x\n
$$ \\Rightarrow $$ x > 1\n
$$ \\therefore $$ $$\\phi $$(x) is increasing on (1, 2).\n
Case 2 : When $$\\phi $$(x) is decreasing function\n
So $$\\phi $$'(x) < 0\n
$$ \\Rightarrow $$ ƒ'(x) - ƒ'(2 – x) < 0\n
$$ \\Rightarrow $$ ƒ'(x) < ƒ'(2 – x)\n
$$ \\Rightarrow $$ x < 2 – x\n
$$ \\Rightarrow $$ x < 1\n
$$ \\therefore $$ $$\\phi $$(x) is decreasing on (0, 1).\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3809,
"subject": "General Science",
"question": "If m is the minimum value of k for which the function f(x) = x$$\\sqrt {kx - {x^2}} $$ is increasing in the interval [0,3]\nand M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to : ",
"options": [
{
"text": "$$\\left( {5,3\\sqrt 6 } \\right)$$"
},
{
"text": "$$\\left( {4,3\\sqrt 3 } \\right)$$"
},
{
"text": "$$\\left( {4,3\\sqrt 2 } \\right)$$"
},
{
"text": "$$\\left( {3,3\\sqrt 3 } \\right)$$"
}
],
"answer": "$$\\left( {4,3\\sqrt 3 } \\right)$$",
"solution": "**Answer:** $$\\left( {4,3\\sqrt 3 } \\right)$$\n\n$$f\\left( x \\right) = x\\sqrt {kx - {x^2}} $$
$$ \\therefore $$ f(x) is strictly increasing function. Since, it is an odd degree polynomial it will have exactly one real root.
Now, by observation.
$$f( - 1) = 3 > 0$$
$$f( - 2) = - 64 + 80 - 80 + 40 - 20 + 10$$
$$ = - 34 < 0$$
$$ \\Rightarrow f(x)$$ has at least one root in $$( - 2, - 1) \\equiv (a,a + 1)$$
$$ \\Rightarrow a = - 2$$\n
$$ \\Rightarrow $$ |a| = - 2",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 3817,
"subject": "General Science",
"question": "Let f be a real valued function, defined on R $$-$$ {$$-$$1, 1} and given by
$$ \\therefore $$ f'(x) is an oscillating function which is non-monotonic on ($$-$$$$\\infty $$, 0) and (0, $$\\infty $$).",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 3819,
"subject": "General Science",
"question": "Let f : R $$\\to$$ R be defined as
Decreasing in $$\\left( { - {\\pi \\over 6},0} \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3821,
"subject": "General Science",
"question": "If 'R' is the least value of 'a' such that the function f(x) = x2 + ax + 1 is increasing on [1, 2] and 'S' is the greatest value of 'a' such that the function f(x) = x2 + ax + 1 is decreasing on [1, 2], then the value of |R $$-$$ S| is ___________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nf(x) = x2 + ax + 1
f'(x) = 2x + a
when f(x) is increasing on [1, 2]
2x + a $$\\ge$$ 0 $$\\forall$$ x$$\\in$$[1, 2]
a $$\\ge$$ $$-$$2x $$\\forall$$ x$$\\in$$[1, 2]
R = $$-$$4
when f(x) is decreasing on [1, 2]
2x + a $$\\le$$ 0 $$\\forall$$ x$$\\in$$[1, 2]
a $$\\le$$ $$-$$2 $$\\forall$$ x$$\\in$$[1, 2]
S = $$-$$2
|R $$-$$ S| = | $$-$$4 + 2 | = 2",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3822,
"subject": "General Science",
"question": "The function $$f(x) = {x^3} - 6{x^2} + ax + b$$ is such that $$f(2) = f(4) = 0$$. Consider two statements :
Statement 1 : there exists x1, x2 $$\\in$$(2, 4), x1 < x2, such that f'(x1) = $$-$$1 and f'(x2) = 0.
Statement 2 : there exists x3, x4 $$\\in$$ (2, 4), x3 < x4, such that f is decreasing in (2, x4), increasing in (x4, 4) and $$2f'({x_3}) = \\sqrt 3 f({x_4})$$.
Then",
"options": [
{
"text": "both Statement 1 and Statement 2 are true"
},
{
"text": "Statement 1 is false and Statement 2 is true"
},
{
"text": "both Statement 1 and Statement 2 are false"
},
{
"text": "Statement 1 is true and Statement 2 is false"
}
],
"answer": "both Statement 1 and Statement 2 are true",
"solution": "**Answer:** both Statement 1 and Statement 2 are true\n\n$$f(x) = {x^3} - 6{x^2} + ax + b$$
$$f(2) = 8 - 24 + 2a + b = 0$$
$$2a + b = 16$$ .... (1)
$$f(4) = 64 - 96 + 4a + b = 0$$
$$4a + b = 32$$ .... (2)
Solving (1) and (2)
a = 8, b = 0
$$f(x) = {x^3} - 6{x^2} + 8x$$
$$f'(x) = 3{x^2} - 12x + 8$$
$$f''(x) = 6x - 12$$
$$\\Rightarrow$$ f'(x) is $$ \\uparrow $$ for x > 2, and f'(x) is $$ \\downarrow $$ for x < 2
\nx-axis at only one point",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3824,
"subject": "General Science",
"question": "
Let $$f(x) = 2{\\cos ^{ - 1}}x + 4{\\cot ^{ - 1}}x - 3{x^2} - 2x + 10$$, $$x \\in [ - 1,1]$$. If [a, b] is the range of the function f, then 4a $$-$$ b is equal to :
Let $$\\lambda$$$$^ * $$ be the largest value of $$\\lambda$$ for which the function $${f_\\lambda }(x) = 4\\lambda {x^3} - 36\\lambda {x^2} + 36x + 48$$ is increasing for all x $$\\in$$ R. Then $${f_{{\\lambda ^ * }}}(1) + {f_{{\\lambda ^ * }}}( - 1)$$ is equal to :
$$f(x) = 4{\\log _e}(x - 1) - 2{x^2} + 4x + 5,\\,x > 1$$, which one of the following is NOT correct?
",
"options": [
{
"text": "f is increasing in (1, 2) and decreasing in (2, $$\\infty$$)"
},
{
"text": "f(x) = $$-$$1 has exactly two solutions"
},
{
"text": "$$f'(e) - f''(2) < 0$$"
},
{
"text": "f(x) = 0 has a root in the interval (e, e + 1)"
}
],
"answer": "$$f'(e) - f''(2) < 0$$",
"solution": "**Answer:** $$f'(e) - f''(2) < 0$$\n\nLets draw the curve $y=f(x)=4 \\log _e(x-1)-2 x^2$ $+4 x+5, x>1$\n
\n $$\n\\begin{aligned}\n&f(x)=4 \\log _e(x-1)-2 x^2+4 x+5, x > 1 \\\\\\\\\n&f^{\\prime}(x)=\\frac{4}{x-1}-4(x-1) \\\\\\\\\n&f^{\\prime \\prime}(x)=\\frac{-4}{(x-1)^2}-4\\\\\\\\\n&\\text {For} 1 < x< 2 \\Rightarrow f^{\\prime}(x) > 0 \\\\\\\\\n&\\text {For}~ x > 2 \\Rightarrow f^{\\prime}(x)<0 \\text { (option } A \\text { is correct) } \\\\\\\\\n&f(x)=-1 \\text { has two solution (option } B \\text { is correct) } \\\\\\\\\n&f(e) > 0 \\\\\\\\\n&f(e+1) < 0 \\\\\\\\\n&f(e) \\cdot f(e+1)<0(\\text { option } D \\text { is correct) } \\\\\\\\\n&f^{\\prime}(e)-f^{\\prime \\prime}(2)=\\frac{4}{e-1}-4(e-1)+8 > 0\n\\end{aligned}\n$$\n
(option C is incorrect)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3827,
"subject": "General Science",
"question": "
Let the function $$f(x)=2 x^{2}-\\log _{\\mathrm{e}} x, x>0$$, be decreasing in $$(0, \\mathrm{a})$$ and increasing in $$(\\mathrm{a}, 4)$$. A tangent to the parabola $$y^{2}=4 a x$$ at a point $$\\mathrm{P}$$ on it passes through the point $$(8 \\mathrm{a}, 8 \\mathrm{a}-1)$$ but does not pass through the point $$\\left(-\\frac{1}{a}, 0\\right)$$. If the equation of the normal at $$P$$ is : $$\\frac{x}{\\alpha}+\\frac{y}{\\beta}=1$$, then $$\\alpha+\\beta$$ is equal to ________________.
Let $$f:(0,1)\\to\\mathbb{R}$$ be a function defined $$f(x) = {1 \\over {1 - {e^{ - x}}}}$$, and $$g(x) = \\left( {f( - x) - f(x)} \\right)$$. Consider two statements
\n
(I) g is an increasing function in (0, 1)
\n
(II) g is one-one in (0, 1)
\n
Then,
",
"options": [
{
"text": "Both (I) and (II) are true"
},
{
"text": "Neither (I) nor (II) is true"
},
{
"text": "Only (II) is true"
},
{
"text": "Only (I) is true"
}
],
"answer": "Both (I) and (II) are true",
"solution": "**Answer:** Both (I) and (II) are true\n\n$g(x)=f(-x)-f(x)$\n
Let $$\\mathrm{g}(x)=f(x)+f(1-x)$$ and $$f^{\\prime \\prime}(x) > 0, x \\in(0,1)$$. If $$\\mathrm{g}$$ is decreasing in the interval $$(0, a)$$ and increasing in the interval $$(\\alpha, 1)$$, then $$\\tan ^{-1}(2 \\alpha)+\\tan ^{-1}\\left(\\frac{1}{\\alpha}\\right)+\\tan ^{-1}\\left(\\frac{\\alpha+1}{\\alpha}\\right)$$ is equal to :
Let $$f: \\rightarrow \\mathbb{R} \\rightarrow(0, \\infty)$$ be strictly increasing function such that $$\\lim _\\limits{x \\rightarrow \\infty} \\frac{f(7 x)}{f(x)}=1$$. Then, the value of $$\\lim _\\limits{x \\rightarrow \\infty}\\left[\\frac{f(5 x)}{f(x)}-1\\right]$$ is equal to
If the function $$f:(-\\infty,-1] \\rightarrow(a, b]$$ defined by $$f(x)=e^{x^3-3 x+1}$$ is one - one and onto, then the distance of the point $$P(2 b+4, a+2)$$ from the line $$x+e^{-3} y=4$$ is :
Consider the function $$f:\\left[\\frac{1}{2}, 1\\right] \\rightarrow \\mathbb{R}$$ defined by $$f(x)=4 \\sqrt{2} x^3-3 \\sqrt{2} x-1$$. Consider the statements
\n
(I) The curve $$y=f(x)$$ intersects the $$x$$-axis exactly at one point.
\n
(II) The curve $$y=f(x)$$ intersects the $$x$$-axis at $$x=\\cos \\frac{\\pi}{12}$$.
\n
Then
",
"options": [
{
"text": "Both (I) and (II) are correct.\n"
},
{
"text": "Only (I) is correct.\n"
},
{
"text": "Both (I) and (II) are incorrect.\n"
},
{
"text": "Only (II) is correct."
}
],
"answer": "Both (I) and (II) are correct.\n",
"solution": "**Answer:** Both (I) and (II) are correct.\n\n\n
Let the set of all values of $$p$$, for which $$f(x)=\\left(p^2-6 p+8\\right)\\left(\\sin ^2 2 x-\\cos ^2 2 x\\right)+2(2-p) x+7$$ does not have any critical point, be the interval $$(a, b)$$. Then $$16 a b$$ is equal to _________.
For the function $$f(x)=(\\cos x)-x+1, x \\in \\mathbb{R}$$, between the following two statements
\n
(S1) $$f(x)=0$$ for only one value of $$x$$ in $$[0, \\pi]$$.
\n
(S2) $$f(x)$$ is decreasing in $$\\left[0, \\frac{\\pi}{2}\\right]$$ and increasing in $$\\left[\\frac{\\pi}{2}, \\pi\\right]$$.
",
"options": [
{
"text": "Both (S1) and (S2) are incorrect.\n"
},
{
"text": "Only (S1) is correct.\n"
},
{
"text": "Only (S2) is correct.\n"
},
{
"text": "Both (S1) and (S2) are correct."
}
],
"answer": "Only (S1) is correct.\n",
"solution": "**Answer:** Only (S1) is correct.\n\n\n
Let's analyze the function $$f(x) = (\\cos x) - x + 1$$ over the interval $$[0, \\pi]$$ and the statements provided.
\n\n
First, let's consider statement (S1):
\n\n
(S1) $$f(x)=0$$ for only one value of $$x$$ in $$[0, \\pi]$$.
\n\n
To examine this statement, we need to explore the zeros of the function $$f(x)$$ within the given interval. Let's define and analyze the function:
\n\n
$$f(x) = \\cos x - x + 1$$
\n\n
We seek to determine if $$f(x) = 0$$ has only one solution in the interval $$[0, \\pi]$$. To do this, we can use the Intermediate Value Theorem and the behavior of the function's derivative. First, compute the derivative of $$f(x)$$:
\n\n
$$f'(x) = \\frac{d}{dx}(\\cos x - x + 1) = -\\sin x - 1$$
\n\n
The critical points occur when $$f'(x) = 0$$:\n\n
$$- \\sin x - 1 = 0 \\Rightarrow \\sin x = -1$$.
\n\n
The equation $$\\sin x = -1$$ does not hold for any $$x$$ in $$[0, \\pi]$$. Note that:\n\n
\n\n
For $$x \\in [0, \\frac{\\pi}{2}]$$, $$\\sin x$$ ranges from 0 to 1.
\n\n
For $$x \\in [\\frac{\\pi}{2}, \\pi]$$, $$\\sin x$$ ranges from 1 to 0.
\n\n
\n\n
Since $$f'(x)$$ is always negative (i.e., $$f'(x) < 0$$), $$f(x)$$ is a strictly decreasing function in $$[0, \\pi]$$. Moreover, we evaluate:
Given the continuous and strictly decreasing nature of $$f(x)$$ in $$[0, \\pi]$$, by the Intermediate Value Theorem, there is exactly one value of $$x$$ in the interval $$[0, \\pi]$$ where $$f(x) = 0$$, confirming (S1).
\n\n
Now, let's consider statement (S2):
\n\n
(S2) $$f(x)$$ is decreasing in $$\\left[0, \\frac{\\pi}{2}\\right]$$ and increasing in $$\\left[\\frac{\\pi}{2}, \\pi\\right]$$.
\n\n
We already analyzed that $$f'(x)$$ shows that $$f'(x) = -\\sin x - 1$$ is always less than zero in $$[0, \\pi]$$. No interval exists where the derivative is positive. This means that $$f(x)$$ is strictly decreasing throughout the entire interval of $$[0, \\pi]$$, invalidating (S2).
$$f(x)=\\sin x+3 x-\\frac{2}{\\pi}\\left(x^2+x\\right), \\text { where } x \\in\\left[0, \\frac{\\pi}{2}\\right],$$
\n
consider the following two statements :
\n
(I) $$f$$ is increasing in $$\\left(0, \\frac{\\pi}{2}\\right)$$.
\n
(II) $$f^{\\prime}$$ is decreasing in $$\\left(0, \\frac{\\pi}{2}\\right)$$.
\n
Between the above two statements,
",
"options": [
{
"text": "only (I) is true.\n"
},
{
"text": "both (I) and (II) are true.\n"
},
{
"text": "only (II) is true.\n"
},
{
"text": "neither (I) nor (II) is true."
}
],
"answer": "both (I) and (II) are true.\n",
"solution": "**Answer:** both (I) and (II) are true.\n\n\n
$$\\begin{aligned}\n& f(x)=\\sin x+3 x-\\frac{2}{\\pi}\\left(x^2+x\\right), \\text { where } x \\in\\left[0, \\frac{\\pi}{2}\\right] \\\\\n& f^{\\prime}(x)=\\cos x+3-\\frac{2}{\\pi}(2 x+1) \\\\\n& =\\cos x-\\frac{4 x}{\\pi}-\\frac{2}{\\pi}+3 \\\\\n& \\text { as } x \\in\\left[0, \\frac{\\pi}{2}\\right] \\\\\n& \\frac{4 x}{\\pi} \\in[0,2]\n\\end{aligned}$$
$$\\begin{aligned}\n& f(x)=x^x \\\\\n& f(x)=x^x(\\log x+1) \\\\\n& f(x) \\geq 0 \\\\\n& \\Rightarrow 1+\\log x \\geq 0 \\\\\n& \\Rightarrow \\log x \\geq-1 \\\\\n& \\Rightarrow x \\geq e^{-1} \\\\\n& \\therefore x \\in\\left[\\frac{1}{e^{\\prime}}, \\infty\\right)\n\\end{aligned}$$
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3840,
"subject": "General Science",
"question": "A point on the parabola $${y^2} = 18x$$ at which the ordinate increases at twice the rate of the abscissa is ",
"options": [
{
"text": "$$\\left( {{9 \\over 8},{9 \\over 2}} \\right)$$ "
},
{
"text": "$$(2, -4)$$ "
},
{
"text": "$$\\left( {{-9 \\over 8},{9 \\over 2}} \\right)$$"
},
{
"text": "$$(2, 4)$$ "
}
],
"answer": "$$\\left( {{9 \\over 8},{9 \\over 2}} \\right)$$ ",
"solution": "**Answer:** $$\\left( {{9 \\over 8},{9 \\over 2}} \\right)$$ \n\n$${y^2} = 18x \\Rightarrow 2y{{dy} \\over {dx}} = 18 \\Rightarrow {{dy} \\over {dx}} = {9 \\over y}$$\n
Given $${{dy} \\over {dx}} = 2 \\Rightarrow {9 \\over 2} = 2 \\Rightarrow y = {9 \\over 2}$$ \n
Puting in $${y^2} = 18x \\Rightarrow x = {9 \\over 8}$$\n
$$\\therefore$$ Required point is $$\\left( {{9 \\over 8},{9 \\over 2}} \\right)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3841,
"subject": "General Science",
"question": "A spherical iron ball $$10$$ cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of $$50$$ cm$$^3$$ /min. When the thickness of ice is $$5$$ cm, then the rate at which the thickness of ice decreases is \n ",
"options": [
{
"text": "$${1 \\over {36\\pi }}$$ cm/min"
},
{
"text": "$${1 \\over {18\\pi }}$$ cm/min"
},
{
"text": "$${1 \\over {54\\pi }}$$ cm/min"
},
{
"text": "$${5 \\over {6\\pi }}$$ cm/min"
}
],
"answer": "$${1 \\over {18\\pi }}$$ cm/min",
"solution": "**Answer:** $${1 \\over {18\\pi }}$$ cm/min\n\nGiven that\n
(here $$r=10+5)$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3842,
"subject": "General Science",
"question": "A lizard, at an initial distance of 21 cm behind an insect moves from rest with an acceleration of $2 \\mathrm{~cm} / \\mathrm{s}^2$ and pursues the insect which is crawling uniformly along a straight line at a speed of $20 \\mathrm{~cm} / \\mathrm{s}$. Then the lizard will catch the insect after :",
"options": [
{
"text": "20 s"
},
{
"text": "1 s"
},
{
"text": "21 s"
},
{
"text": "24 s"
}
],
"answer": "21 s",
"solution": "**Answer:** 21 s\n\n
The motion of the lizard, which starts from rest and accelerates at a rate of $a = 2 \\, \\text{cm/s}^2$, can be described by the equation of motion :
\n
$D_l = \\frac{1}{2} a t^2$
\n
where $D_l$ is the distance the lizard travels, $a$ is its acceleration, and $t$ is the time.
\n
The insect, moving at a constant speed of $v = 20 \\, \\text{cm/s}$, has a motion that can be simply described as:
\n
$D_i = v t$
\n
where $D_i$ is the distance the insect travels, $v$ is its velocity, and $t$ is the time.
\n
Since the lizard starts 21 cm behind the insect and needs to catch up, it must travel the distance the insect travels plus an additional 21 cm. Equating these two distances gives us :
\n
$\\frac{1}{2} a t^2 = v t + 21$
\n
Substituting the given values into this equation gives :
\n
$t^2 = \\frac{(v t + 21) \\cdot 2}{a}$
\n
Substituting $a = 2 \\, \\text{cm/s}^2$ and $v = 20 \\, \\text{cm/s}$, we simplify to :
\n
$t^2 = 20t + 21$
\n
This simplifies to a quadratic equation :
\n
$t^2 - 20t - 21 = 0$
\n
Solving this quadratic equation for $t$ gives the solutions $t = 21 \\, \\text{s}$ and $t = -1 \\, \\text{s}$. Since time cannot be negative, we discard the -1 s solution. Therefore, the lizard will catch the insect after 21 seconds.
\n
So, the correct answer is Option C : 21 s.
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3843,
"subject": "General Science",
"question": "A spherical balloon is filled with $$4500\\pi $$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $$72\\pi $$ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases $$49$$ minutes after the leakage began is :",
"options": [
{
"text": "$${{9 \\over 7}}$$"
},
{
"text": "$${{7 \\over 9}}$$"
},
{
"text": "$${{2 \\over 9}}$$"
},
{
"text": "$${{9 \\over 2}}$$"
}
],
"answer": "$${{2 \\over 9}}$$",
"solution": "**Answer:** $${{2 \\over 9}}$$\n\nVolume of spherical balloon $$ = V = {4 \\over 3}\\pi {r^3}$$\n
$$\\therefore$$ $$72\\pi = 4\\pi \\times 9 \\times 9\\left( {{{dr} \\over {dt}}} \\right) \\Rightarrow {{dr} \\over {dt}} = \\left( {{2 \\over 9}} \\right)$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3844,
"subject": "General Science",
"question": "A water tank has the shape of an inverted right\ncircular cone, whose semi-vertical angle is\n$${\\tan ^{ - 1}}\\left( {{1 \\over 2}} \\right)$$. Water is poured into it at a constant\nrate of 5 cubic meter per minute. The the rate\n(in m/min.), at which the level of water is rising\nat the instant when the depth of water in the tank\nis 10m; is :-",
"options": [
{
"text": "$${1 \\over {15\\pi }}$$"
},
{
"text": "$${1 \\over {5\\pi }}$$"
},
{
"text": "$${1 \\over {10\\pi }}$$"
},
{
"text": "$${2 \\over \\pi }$$"
}
],
"answer": "$${1 \\over {5\\pi }}$$",
"solution": "**Answer:** $${1 \\over {5\\pi }}$$\n\n\n
$$ \\Rightarrow $$ $${{dh} \\over {dt}}$$ = $${1 \\over {5\\pi }}$$",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 3845,
"subject": "General Science",
"question": "A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of\n50 cm3\n/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice\ndecreases, is : ",
"options": [
{
"text": "$${5 \\over {6\\pi }}$$"
},
{
"text": "$${1 \\over {9\\pi }}$$"
},
{
"text": "$${1 \\over {36\\pi }}$$"
},
{
"text": "$${1 \\over {18\\pi }}$$"
}
],
"answer": "$${1 \\over {18\\pi }}$$",
"solution": "**Answer:** $${1 \\over {18\\pi }}$$\n\n \n$$V = {4 \\over 3}\\pi \\left( {{{\\left( {10 + h} \\right)}^3} - {{10}^3}} \\right)$$
\n$$ \\Rightarrow {{dh} \\over {dt}} = - {1 \\over {18\\pi }}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3846,
"subject": "General Science",
"question": "A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate\n25 cm/sec, then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the\nhorizontal ground when the top of the ladder is 1 m above the ground is : ",
"options": [
{
"text": "$${{25} \\over 3}$$"
},
{
"text": "25"
},
{
"text": "25$$\\sqrt 3 $$"
},
{
"text": "$${{25} \\over {\\sqrt 3 }}$$"
}
],
"answer": "$${{25} \\over {\\sqrt 3 }}$$",
"solution": "**Answer:** $${{25} \\over {\\sqrt 3 }}$$\n\n \nx2 + y2 = 4 $$ \\Rightarrow $$ $$2x{{dx} \\over {dt}} + 2y{{dy} \\over {dt}} = 0$$
\nWhen upper end is 1m above the ground, $${{dx} \\over {dt}} = - {1 \\over {\\sqrt 3 }}.25 = - {{25} \\over {\\sqrt 3 }}$$ cm/sec.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3847,
"subject": "General Science",
"question": "A spherical iron ball of 10 cm radius is\ncoated with a layer of ice of uniform\nthickness the melts at a rate of 50 cm3/min.\nWhen the thickness of ice is 5 cm, then the rate\n(in cm/min.) at which of the thickness of ice\ndecreases, is :",
"options": [
{
"text": "$${1 \\over {18\\pi }}$$"
},
{
"text": "$${1 \\over {36\\pi }}$$"
},
{
"text": "$${1 \\over {54\\pi }}$$"
},
{
"text": "$${5 \\over {6\\pi }}$$"
}
],
"answer": "$${1 \\over {18\\pi }}$$",
"solution": "**Answer:** $${1 \\over {18\\pi }}$$\n\nLet the thickness = h cm\n
Volume of ice = v = $${{4\\pi } \\over 3}\\left( {{{\\left( {10 + h} \\right)}^3} - {{10}^3}} \\right)$$\n
$$ \\Rightarrow $$ $${{dh} \\over {dt}} = {{50} \\over {4\\pi \\times {{15}^2}}}$$ = $${1 \\over {18\\pi }}$$ cm/min",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3848,
"subject": "General Science",
"question": "If the surface area of a cube is increasing at a\nrate of 3.6 cm2/sec, retaining its shape; then\nthe rate of change of its volume (in cm3/sec),\nwhen the length of a side of the cube is\n10 cm, is :",
"options": [
{
"text": "9"
},
{
"text": "10"
},
{
"text": "18"
},
{
"text": "20"
}
],
"answer": "9",
"solution": "**Answer:** 9\n\nFor cube of side 'a'
Water is being filled at the rate of 1 cm3 / sec in a right circular conical vessel (vertex downwards) of height 35 cm and diameter 14 cm. When the height of the water level is 10 cm, the rate (in cm2 / sec) at which the wet conical surface area of the vessel increases is
The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :
A hostel has 100 students. On a certain day (consider it day zero) it was found that two students are infected with some virus. Assume that the rate at which the virus spreads is directly proportional to the product of the number of infected students and the number of non-infected students. If the number of infected students on 4th day is 30, then number of infected students on 8th day will be __________.
A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is $$\\tan ^{-1} \\frac{3}{4}$$. Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is ______________.
$$\\text { If } f(x)=\\left|\\begin{array}{ccc}\nx^3 & 2 x^2+1 & 1+3 x \\\\\n3 x^2+2 & 2 x & x^3+6 \\\\\nx^3-x & 4 & x^2-2\n\\end{array}\\right| \\text { for all } x \\in \\mathbb{R} \\text {, then } 2 f(0)+f^{\\prime}(0) \\text { is equal to }$$
Let $$f(x)=x^5+2 x^3+3 x+1, x \\in \\mathbf{R}$$, and $$g(x)$$ be a function such that $$g(f(x))=x$$ for all $$x \\in \\mathbf{R}$$. Then $$\\frac{g(7)}{g^{\\prime}(7)}$$ is equal to :
",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 3856,
"subject": "General Science",
"question": "A function $$y=f(x)$$ has a second order derivative $$f''\\left( x \\right) = 6\\left( {x - 1} \\right).$$ If its graph passes through the point $$(2, 1)$$ and at that point the tangent to the graph is $$y = 3x - 5$$, then the function is :",
"options": [
{
"text": "$${\\left( {x + 1} \\right)^2}$$ "
},
{
"text": "$${\\left( {x - 1} \\right)^3}$$"
},
{
"text": "$${\\left( {x + 1} \\right)^3}$$"
},
{
"text": "$${\\left( {x - 1} \\right)^2}$$"
}
],
"answer": "$${\\left( {x - 1} \\right)^3}$$",
"solution": "**Answer:** $${\\left( {x - 1} \\right)^3}$$\n\n$$f''\\left( x \\right) = 6\\left( {x - 1} \\right).$$ Inegrating, \n
we get $$f'\\left( x \\right) = 3{x^2} - 6x + c$$\n
Slope at $$\\left( {2,1} \\right) = f'\\left( 2 \\right) = c = 3$$\n
$$\\left[ {\\,\\,} \\right.$$ As slope of tangent at $$(2, 1)$$ is $$3$$ $$\\left. {\\,\\,} \\right]$$\n
which always passes through $$(a, 0)$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3858,
"subject": "General Science",
"question": "The normal to the curve \n $$x = a\\left( {\\cos \\theta + \\theta \\sin \\theta } \\right),y = a\\left( {\\sin \\theta - \\theta \\cos \\theta } \\right)$$ at any point\n $$\\theta\\, '$$ is such that\n",
"options": [
{
"text": "it passes through the origin "
},
{
"text": "it makes an angle $${\\pi \\over 2} + \\theta $$ with the $$x$$-axis "
},
{
"text": "it passes through $$\\left( {a{\\pi \\over 2}, - a} \\right)$$ "
},
{
"text": "it is at a constant distance from the origin "
}
],
"answer": "it is at a constant distance from the origin ",
"solution": "**Answer:** it is at a constant distance from the origin \n\n$$x = a\\left( {\\cos \\theta + \\theta \\sin \\theta } \\right)$$\n
i.e. the tangents are perpendicular to each other. ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3860,
"subject": "General Science",
"question": "The equation of the tangent to the curve $$y = x + {4 \\over {{x^2}}}$$, that \n is parallel to the $$x$$-axis, is ",
"options": [
{
"text": "$$y=1$$ "
},
{
"text": "$$y=2$$ "
},
{
"text": "$$y=3$$ "
},
{
"text": "$$y=0$$ "
}
],
"answer": "$$y=3$$ ",
"solution": "**Answer:** $$y=3$$ \n\nSince tangent is parallel to $$x$$-axis, \n
$$ \\Rightarrow $$ $$x$$-intercept $$ = \\pm 1.$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3863,
"subject": "General Science",
"question": "The normal to the curve, $${x^2} + 2xy - 3{y^2} = 0$$, at $$(1,1)$$ ",
"options": [
{
"text": "meets the curve again in the third quadrant. "
},
{
"text": "meets the curve again in the fourth quadrant. "
},
{
"text": "does not meet the curve again."
},
{
"text": "meets the curve again in the second quadrant."
}
],
"answer": "meets the curve again in the fourth quadrant. ",
"solution": "**Answer:** meets the curve again in the fourth quadrant. \n\nGiven curve is \n
Solving eq. $$(1)$$ and $$(2),$$ we get $$x=1,3$$ \n
Point of intersection $$\\left( {1,1} \\right),\\left( {3, - 1} \\right)$$ \n
Normal cuts the curve again in $$4$$th quadrant. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3864,
"subject": "General Science",
"question": "If the tangent at a point P, with parameter t, on the curve x = 4t2 + 3, y = 8t3−1, t $$ \\in $$ R, meets the curve again at a point Q, then the coordinates of Q are :",
"options": [
{
"text": "(t2 + 3, − t3 −1)\n"
},
{
"text": "(4t2 + 3, − 8t3 −1)\n"
},
{
"text": "(t2 + 3, t3 −1)\n"
},
{
"text": "(16t2 + 3, − 64t3 −1)"
}
],
"answer": "(t2 + 3, − t3 −1)\n",
"solution": "**Answer:** (t2 + 3, − t3 −1)\n\n\nGiven, x = 4t2 + 3 and y = 8t3 $$-$$ 1\n
$$=$$ (t2 + 3, $$-$$ t3 $$-$$ 1)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3865,
"subject": "General Science",
"question": "Let C be a curve given by y(x) = 1 + $$\\sqrt {4x - 3} ,x > {3 \\over 4}.$$ If P is a point\non C, such that the tangent at P has slope $${2 \\over 3}$$, then a point through which the normal at P passes, is :",
"options": [
{
"text": "(2, 3)"
},
{
"text": "(4, $$-$$3)"
},
{
"text": "(1, 7) "
},
{
"text": "(3, $$-$$ 4), "
}
],
"answer": "(1, 7) ",
"solution": "**Answer:** (1, 7) \n\nGiven,\n
Now put each option in equation (1) and see which one does not satisfy the equation.
By verifying each points you can see ($$-$$ 2, $$-$$ 7) does not satisfy the equation. ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3869,
"subject": "General Science",
"question": "A tangent to the curve, y = f(x) at P(x, y) meets x-axis at A and y-axis at B. If AP : BP = 1 : 3 and f(1) = 1, then the curve also passes through the point :",
"options": [
{
"text": "$$\\left( {{1 \\over 3},24} \\right)$$ "
},
{
"text": "$$\\left( {{1 \\over 2},4} \\right)$$ "
},
{
"text": "$$\\left( {2,{1 \\over 8}} \\right)$$ "
},
{
"text": "$$\\left( {3,{1 \\over 28}} \\right)$$"
}
],
"answer": "$$\\left( {2,{1 \\over 8}} \\right)$$ ",
"solution": "**Answer:** $$\\left( {2,{1 \\over 8}} \\right)$$ \n\n
$$\\ln y = - 3\\ln x + C \\Rightarrow y = k{x^{ - 3}}$$
\n
$$y(1) = 1 \\Rightarrow k = 1$$
\n
$$y = {1 \\over {{x^3}}}$$
\n
When we substitute the values from the given options, only option (c) satisfies the above equation.
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3870,
"subject": "General Science",
"question": "If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other at right angles, then the value of b is :",
"options": [
{
"text": "$${9 \\over 2}$$"
},
{
"text": "6"
},
{
"text": "$${7 \\over 2}$$"
},
{
"text": "4"
}
],
"answer": "$${9 \\over 2}$$",
"solution": "**Answer:** $${9 \\over 2}$$\n\nWhen two curves intersect each other at right angle, then at the point of intersection the product of tangent of slopes = $$-1$$.\n
Let m1, and m2 are the tangent of the slope of the two curves respectively \n
$$\\therefore\\,\\,\\,$$ m1 m2 = $$-$$ 1.\n
Now let they intersect at point (x1, y1)\n
$$\\therefore\\,\\,\\,$$ $$y_1^2 = 6x,$$ and $$9x_1^2 + b\\,y_1^2 = 16$$\n
\n$$ \\Rightarrow $$ $$\\alpha $$ = 0, $$ \\pm $$ 3 ($$\\alpha \\ne $$ 0)",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3874,
"subject": "General Science",
"question": "If the tangent to the curve, y = x3 + ax – b at\nthe point (1, –5) is perpendicular to the line,\n–x + y + 4 = 0, then which one of the following\npoints lies on the curve ?",
"options": [
{
"text": "(2, –2)"
},
{
"text": "(2, –1)"
},
{
"text": "(–2, 2)"
},
{
"text": "(–2, 1)"
}
],
"answer": "(2, –2)",
"solution": "**Answer:** (2, –2)\n\nSlope of the tangent to the curve y = x3 + ax – b at point (1, –5)\n
$$ \\Rightarrow $$ c = $${e^{\\left( {{1 \\over {e - 1}}} \\right)}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3879,
"subject": "General Science",
"question": "If the lines x + y = a and x – y = b touch the\n curve y = x2\n – 3x + 2 at the points where the\ncurve intersects the x-axis, then $${a \\over b}$$ is equal\nto _______.",
"options": [],
"answer": "0.50",
"solution": "**Answer:** 0.50\n\ny = x2\n – 3x + 2\n
$$ \\Rightarrow $$ y = (x – 1)(x – 2)\n
At x-axis y = 0 \n $$ \\Rightarrow $$ x = 1, 2\n
So this curve intersects the x-axis\nat A(1, 0) and B(2, 0).\n
Only (–2, 6) lies on it.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3881,
"subject": "General Science",
"question": "If the tangent to the curve y = x + sin y at a point\n (a, b) is parallel to the line joining $$\\left( {0,{3 \\over 2}} \\right)$$ and $$\\left( {{1 \\over 2},2} \\right)$$, then :",
"options": [
{
"text": "b = a"
},
{
"text": "|b - a| = 1"
},
{
"text": "$$b = {\\pi \\over 2}$$ + a"
},
{
"text": "|a + b| = 1"
}
],
"answer": "|b - a| = 1",
"solution": "**Answer:** |b - a| = 1\n\nSlope of tangent to the curve y = x + siny\n
at (a, b) is = $${{2 - {3 \\over 2}} \\over {{1 \\over 2} - 0}}$$ = 1\n
$$ \\therefore $$ Perpendicular distance from (0,0)\n
= $$\\left| {{{0 + 0 - 4} \\over {\\sqrt 2 }}} \\right|$$ = $${2\\sqrt 2 }$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3884,
"subject": "General Science",
"question": "Let the normal at a point P on the curve\n y2 – 3x2 + y + 10 = 0 intersect the y-axis at $$\\left( {0,{3 \\over 2}} \\right)$$\n. If m is the slope of the tangent at P to\nthe curve, then |m| is equal to",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nGiven curve : y2\n– 3x2\n + y + 10 = 0\n
$$ \\Rightarrow $$ y' = 4 = slope of tangent to the curve\n
so slope of normal to the curve = - $${1 \\over 4}$$\n
Hence equation of normal at (0, 2) is\n
y - 2 = - $${1 \\over 4}$$(x - 0)\n
$$ \\Rightarrow $$ 4y – 8 = –x\n
$$ \\Rightarrow $$ x + 4y = 8",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 3886,
"subject": "General Science",
"question": "If the tangent to the curve y = x3 at the point P(t, t3) meets the curve again at Q, then the\nordinate of the point which divides PQ internally in the ratio 1 : 2 is :",
"options": [
{
"text": "0"
},
{
"text": "2t3"
},
{
"text": "-2t3"
},
{
"text": "-t3"
}
],
"answer": "-2t3",
"solution": "**Answer:** -2t3\n\nGiven $$P(t,{t^3})$$
$$3\\sqrt 3 x + 9y = 18 \\Rightarrow x + \\sqrt 3 y = 2\\sqrt 3 $$
$$ \\Rightarrow $$ option (1) is true.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3888,
"subject": "General Science",
"question": "If the curve y = ax2 + bx + c, x$$ \\in $$R, passes through the point (1, 2) and the tangent line to this curve at origin is y = x, then the possible values of a, b, c are :",
"options": [
{
"text": "a = $$-$$ 1, b = 1, c = 1"
},
{
"text": "a = 1, b = 1, c = 0"
},
{
"text": "a = $${1 \\over 2}$$, b = $${1 \\over 2}$$, c = 1"
},
{
"text": "a = 1, b = 0, c = 1"
}
],
"answer": "a = 1, b = 1, c = 0",
"solution": "**Answer:** a = 1, b = 1, c = 0\n\nGiven curve y = ax2 + bx + c, x$$ \\in $$R\n
$$ \\therefore $$ c = 0, a = 1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3889,
"subject": "General Science",
"question": "If the curves, $${{{x^2}} \\over a} + {{{y^2}} \\over b} = 1$$ and $${{{x^2}} \\over c} + {{{y^2}} \\over d} = 1$$ intersect each other at an angle of 90$$^\\circ$$, then which of the following relations is TRUE?",
"options": [
{
"text": "a $$-$$ c = b + d"
},
{
"text": "a + b = c + d "
},
{
"text": "$$ab = {{c + d} \\over {a + b}}$$"
},
{
"text": "a $$-$$ b = c $$-$$ d"
}
],
"answer": "a $$-$$ b = c $$-$$ d",
"solution": "**Answer:** a $$-$$ b = c $$-$$ d\n\n$${{{x^2}} \\over a} + {{{y^2}} \\over b} = 1$$ ..........(1)
$$ \\Rightarrow c - d = a - b$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3890,
"subject": "General Science",
"question": "If the curves x = y4 and xy = k cut at right angles, then (4k)6 is equal to __________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n$$x = {y^4}$$ and $$xy = k$$
$$ \\Rightarrow 4 = {(4k)^6}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3891,
"subject": "General Science",
"question": "Let slope of the tangent line to a curve at any point P(x, y) be given by $${{x{y^2} + y} \\over x}$$. If the curve intersects the line x + 2y = 4 at x = $$-$$2, then the value of y, for which the point (3, y) lies on the curve, is :",
"options": [
{
"text": "$$ - {{18} \\over {19}}$$"
},
{
"text": "$$ - {{4} \\over {3}}$$"
},
{
"text": "$${{18} \\over {35}}$$"
},
{
"text": "$$ - {{18} \\over {11}}$$"
}
],
"answer": "$$ - {{18} \\over {19}}$$",
"solution": "**Answer:** $$ - {{18} \\over {19}}$$\n\n$${{dy} \\over {dx}} = {{x{y^2} + y} \\over x}$$
$$ \\Rightarrow y = - {{18} \\over {19}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3892,
"subject": "General Science",
"question": "Let the normals at all the points on a given curve pass through a fixed point (a, b). If the curve passes through (3, $$-$$3) and (4, $$-$$2$$\\sqrt 2 $$), and given that a $$-$$ 2$$\\sqrt 2 $$ b = 3, then (a2 + b2 + ab) is equal to __________.",
"options": [],
"answer": "9",
"solution": "**Answer:** 9\n\nLet the equation of normal is Y $$-$$ y = $$-$$$${1 \\over m}(X - x)$$, where, m = $${{dy} \\over {dx}}$$
Let l be a line which is normal to the curve y = 2x2 + x + 2 at a point P on the curve. If the point Q(6, 4) lies on the line l and O is origin, then the area of the triangle OPQ is equal to ___________.
Let S be the set of all the natural numbers, for which the line $${x \\over a} + {y \\over b} = 2$$ is a tangent to the curve $${\\left( {{x \\over a}} \\right)^n} + {\\left( {{y \\over b}} \\right)^n} = 2$$ at the point (a, b), ab $$\\ne$$ 0. Then :
If the angle made by the tangent at the point (x0, y0) on the curve $$x = 12(t + \\sin t\\cos t)$$, $$y = 12{(1 + \\sin t)^2}$$, $$0 < t < {\\pi \\over 2}$$, with the positive x-axis is $${\\pi \\over 3}$$, then y0 is equal to:
The slope of normal at any point (x, y), x > 0, y > 0 on the curve y = y(x) is given by $${{{x^2}} \\over {xy - {x^2}{y^2} - 1}}$$. If the curve passes through the point (1, 1), then e . y(e) is equal to
Let $$\\lambda x - 2y = \\mu $$ be a tangent to the hyperbola $${a^2}{x^2} - {y^2} = {b^2}$$. Then $${\\left( {{\\lambda \\over a}} \\right)^2} - {\\left( {{\\mu \\over b}} \\right)^2}$$ is equal to :
Let $$M$$ and $$N$$ be the number of points on the curve $$y^{5}-9 x y+2 x=0$$, where the tangents to the curve are parallel to $$x$$-axis and $$y$$-axis, respectively. Then the value of $$M+N$$ equals ___________.
If the tangent to the curve $$y=x^{3}-x^{2}+x$$ at the point $$(a, b)$$ is also tangent to the curve $$y = 5{x^2} + 2x - 25$$ at the point (2, $$-$$1), then $$|2a + 9b|$$ is equal to __________.
",
"options": [],
"answer": "195",
"solution": "**Answer:** 195\n\nSlope of tangent to curve $y=5 x^{2}+2 x-25$\n\n
If the equation of the normal to the curve $$y = {{x - a} \\over {(x + b)(x - 2)}}$$ at the point (1, $$-$$3) is $$x - 4y = 13$$, then the value of $$a + b$$ is equal to ___________.
Let the quadratic curve passing through the point $$(-1,0)$$ and touching the line $$y=x$$ at $$(1,1)$$ be $$y=f(x)$$. Then the $$x$$-intercept of the normal to the curve at the point $$(\\alpha, \\alpha+1)$$ in the first quadrant is __________.
",
"options": [],
"answer": "11",
"solution": "**Answer:** 11\n\nLet the quadratic curve be $f(x)=a x^2+b x+c$\n
The slope of tangent at any point (x, y) on a curve $$y=y(x)$$ is $${{{x^2} + {y^2}} \\over {2xy}},x > 0$$. If $$y(2) = 0$$, then a value of $$y(8)$$ is :
",
"options": [
{
"text": "$$ - 4\\sqrt 2 $$"
},
{
"text": "$$2\\sqrt 3 $$"
},
{
"text": "$$4\\sqrt 3 $$"
},
{
"text": "$$ - 2\\sqrt 3 $$"
}
],
"answer": "$$4\\sqrt 3 $$",
"solution": "**Answer:** $$4\\sqrt 3 $$\n\nLet the slope of tangent at any point \n
$(x, y)$ on a curve $y=y(x)$ is $\\frac{d y}{d x}$\n
According to the question, $\\frac{d y}{d x}=\\frac{x^2+y^2}{2 x y}(x>0)$\n[Given]\n
$$\n\\begin{aligned}\n&\\text { Let } y=v x \\Rightarrow \\frac{d y}{d x}=v+x \\frac{d v}{d x} \\\\\\\\\n&\\Rightarrow v+x \\frac{d v}{d x}=\\frac{x^2+v^2 x^2}{2 v x^2}=\\frac{1+v^2}{2 v} \\\\\\\\\n&\\Rightarrow x \\frac{d v}{d x}=\\frac{1+v^2}{2 v}-v=\\frac{1-v^2}{2 v}\n\\end{aligned}\n$$\n
$$\n\\begin{aligned}\n& \\Rightarrow \\int \\frac{2 v d v}{1-v^2}=\\int \\frac{d x}{x} \\text { [Taking integration on both sides] } \\\\\\\\\n& \\Rightarrow-\\ln \\left|1-v^2\\right|=\\ln x-\\ln c \\\\\\\\\n& \\Rightarrow \\ln \\left(1-v^2\\right) x=\\ln c \\quad(c=\\text { constant }) \\\\\\\\\n& \\Rightarrow \\left(1-y^2 / x^2\\right) \\cdot x=c \\\\\\\\\n& \\Rightarrow \\left(x^2-y^2\\right)=c x\n\\end{aligned}\n$$\n
Put $y(2)=0 \\Rightarrow c=2$\n
$$\n\\begin{array}{rlrl} \n& \\therefore x^2-y^2 =2 x \\\\\\\\\n& \\therefore y^2 =x^2-2 x \\\\\\\\\n& \\Rightarrow y(x) = \\pm \\sqrt{x^2-2 x}\n\\end{array}\n$$\n
Let a curve $$y=f(x), x \\in(0, \\infty)$$ pass through the points $$P\\left(1, \\frac{3}{2}\\right)$$ and $$Q\\left(a, \\frac{1}{2}\\right)$$. If the tangent at any point $$R(b, f(b))$$ to the given curve cuts the $$\\mathrm{y}$$-axis at the point $$S(0, c)$$ such that $$b c=3$$, then $$(P Q)^{2}$$ is equal to __________.
",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\nEquation of tangent at $R(b, f(b))$\n
$$\ny-f(b)=f^{\\prime}(b)(x-b)\n$$\n
which passes through $S(0, c)$\n
$$\n\\begin{aligned}\n& \\therefore c-f(b)=f^{\\prime}(b)(0-b) \\\\\\\\\n& b f^{\\prime}(b)-f(b)=-c \\\\\\\\\n& \\Rightarrow b f^{\\prime}(b)-f(b)=\\frac{-3}{b} (\\because b c=3) \\\\\\\\\n& \\Rightarrow \\frac{b f^{\\prime}(b)-f(b)}{b^2}=\\frac{-3}{b^3} \\\\\\\\\n& \\Rightarrow d\\left(\\frac{f(b)}{b}\\right)=\\frac{-3}{b^3} \\\\\\\\\n& \\Rightarrow \\frac{f(b)}{b}=\\frac{3}{2 b^2}+c\n\\end{aligned}\n$$\n
which passes through $P\\left(1, \\frac{3}{2}\\right)$\n
$\\because$ It passes through $Q\\left(a, \\frac{1}{2}\\right)$\n
$$\n\\begin{aligned}\n& \\therefore \\frac{1}{2}=\\frac{3}{2 a} \\\\\\\\\n& \\Rightarrow a=3 \\\\\\\\\n& \\therefore P \\equiv\\left(1, \\frac{3}{2}\\right) \\text { and } Q \\equiv\\left(3, \\frac{1}{2}\\right)\\\\\\\\\n& \\therefore (P Q)^2=(3-1)^2+\\left(\\frac{1}{2}-\\frac{3}{2}\\right)^2=4+1=5\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3907,
"subject": "General Science",
"question": "The area bounded by the curves $$y = \\ln x,y = \\ln \\left| x \\right|,y = \\left| {\\ln {\\mkern 1mu} x} \\right|$$ and $$y = \\left| {\\ln \\left| x \\right|} \\right|$$ is :",
"options": [
{
"text": "$$4$$sq. units "
},
{
"text": "$$6$$sq. units "
},
{
"text": "$$10$$sq. units"
},
{
"text": "none of these "
}
],
"answer": "$$4$$sq. units ",
"solution": "**Answer:** $$4$$sq. units \n\nFirst we draw each curve as separate graph\n
\n
NOTE : Graph of $$y = \\left| {f\\left( x \\right)} \\right|$$ can be obtained from the graph of the curve $$y = f\\left( x \\right)$$ by drawing the mirror image of the portion of the graph below $$x$$-axis, with respect to $$x$$-axis. \n
Clearly the bounded area is as shown in the following figure. \n
$$ = 2\\left[ {{{{x^2}} \\over 2} - 2x} \\right]_2^3 = 1$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3910,
"subject": "General Science",
"question": "The parabolas $${y^2} = 4x$$ and $${x^2} = 4y$$ divide the square region bounded by the lines $$x=4,$$ $$y=4$$ and the coordinate axes. If $${S_1},{S_2},{S_3}$$ are respectively the areas of these parts numbered from top to bottom ; then $${S_1},{S_2},{S_3}$$ is :",
"options": [
{
"text": "$$1:2:1$$"
},
{
"text": "$$1:2:3$$"
},
{
"text": "$$2:1:2$$"
},
{
"text": "$$1:1:1$$"
}
],
"answer": "$$1:1:1$$",
"solution": "**Answer:** $$1:1:1$$\n\nIntersection points of $${x^2} = 4y$$ and $${y^2} = 4x$$ are $$\\left( {0,0} \\right)$$ and $$\\left( {4,4} \\right).$$ The graph is as shown in the figure. \n
$$ = 2 \\times {2 \\over 3} = {4 \\over 3}\\,\\,$$ sq. units.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 3913,
"subject": "General Science",
"question": "The area of the region bounded by the parabola $${\\left( {y - 2} \\right)^2} = x - 1,$$ the tangent of the parabola at the point $$(2, 3)$$ and the $$x$$-axis is :",
"options": [
{
"text": "$$6$$"
},
{
"text": "$$9$$"
},
{
"text": "$$12$$"
},
{
"text": "$$3$$"
}
],
"answer": "$$9$$",
"solution": "**Answer:** $$9$$\n\nThe given parabola is $${\\left( {y - 2} \\right)^2} = x - 1$$\n
Vertex $$\\left( {1,2} \\right)$$ and it meets $$x$$-axis at $$\\left( {5,0} \\right)$$\n
Also it gives $${y^2} - 4y - x + 5 = 0$$\n
So, that equation of tangent to the parabola at $$\\left( {2,3} \\right)$$ is \n
Now, $$\\tan {\\pi \\over 2} = \\infty $$ and $$\\tan {\\pi \\over 6} = {1 \\over {\\sqrt 3 }}$$. Therefore, the area of the smaller portion enclosed between the two curves is obtained as follows:
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3923,
"subject": "General Science",
"question": "The area (in sq. units) of the region\n
= $${5 \\over 2}$$ sq. unit",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3924,
"subject": "General Science",
"question": "The area (in sq. units) of the region \n
{x $$ \\in $$ R : x $$ \\ge $$ 0, y $$ \\ge $$ 0, y $$ \\ge $$ x $$-$$ 2 and y $$ \\le $$ $$\\sqrt x $$}, is : ",
"options": [
{
"text": "$${{13} \\over 3}$$"
},
{
"text": "$${{8} \\over 3}$$"
},
{
"text": "$${{10} \\over 3}$$"
},
{
"text": "$${{5} \\over 3}$$"
}
],
"answer": "$${{10} \\over 3}$$",
"solution": "**Answer:** $${{10} \\over 3}$$\n\ny = $$\\sqrt x $$\n
y = x $$-$$ 2\n
$$\\therefore\\,\\,\\,$$ $$\\sqrt x $$ = x $$-$$ 2\n
$$ \\Rightarrow $$$$\\,\\,\\,$$ x = x2 $$-$$ 4x + 4\n
$$ \\Rightarrow $$ t = e$${^{{2 \\over 3}}}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3926,
"subject": "General Science",
"question": "The area (in sq. units) of the region\n
\n$$ \\Rightarrow $$ $$\\lambda $$ = 24",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3928,
"subject": "General Science",
"question": "If the area (in sq. units) of the region {(x, y) : y2\n$$ \\le $$ 4x, x + y $$ \\le $$ 1, x $$ \\ge $$ 0, y $$ \\ge $$ 0} is a $$\\sqrt 2 $$ + b, then a – b is equal\nto :",
"options": [
{
"text": "$${8 \\over 3}$$"
},
{
"text": "$$ - {2 \\over 3}$$"
},
{
"text": "6"
},
{
"text": "$${{10} \\over 3}$$"
}
],
"answer": "6",
"solution": "**Answer:** 6\n\n\nLet P be the point common to x + y = 1 and y2 = 4x
$$ = {{15} \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3932,
"subject": "General Science",
"question": "The area (in sq. units) in the first quadrant bounded by the parabola, y = x2 + 1, the tangent to it at the point (2, 5) and the coordinate axes is :",
"options": [
{
"text": "$${8 \\over 3}$$"
},
{
"text": "$${{14} \\over 3}$$"
},
{
"text": "$${{187} \\over {24}}$$"
},
{
"text": "$${{37} \\over {24}}$$"
}
],
"answer": "$${{37} \\over {24}}$$",
"solution": "**Answer:** $${{37} \\over {24}}$$\n\n\n
Area $$ = \\int\\limits_0^2 {\\left( {{x^2} + 1} \\right)dx - {1 \\over 2}} \\left( {{5 \\over 4}} \\right)\\left( 5 \\right) = {{37} \\over {24}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3933,
"subject": "General Science",
"question": "The area (in sq. units) of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is :\n",
"options": [
{
"text": "$${3 \\over 4}$$"
},
{
"text": "$${5 \\over 4}$$"
},
{
"text": "$${7 \\over 8}$$"
},
{
"text": "$${9 \\over 8}$$"
}
],
"answer": "$${9 \\over 8}$$",
"solution": "**Answer:** $${9 \\over 8}$$\n\n\n x = 4y $$-$$ 2 & x2 = 4y\n
$$ \\Rightarrow $$ x2 = x + 2 $$ \\Rightarrow $$ x2 $$-$$ x $$-$$ 2 = 0\n
x = 2, $$-$$ 1\n
So, $$\\int\\limits_{ - 1}^2 {\\left( {{{x + 2} \\over 4} - {{{x^2}} \\over 4}} \\right)\\,dx = {9 \\over 8}} $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3934,
"subject": "General Science",
"question": "If the area enclosed between the curves y = kx2 and x = ky2, (k > 0), is 1 square unit. Then k is -",
"options": [
{
"text": "$$\\sqrt 3 $$"
},
{
"text": "$${{\\sqrt 3 } \\over 2}$$"
},
{
"text": "$${2 \\over {\\sqrt 3 }}$$"
},
{
"text": "$${1 \\over {\\sqrt 3 }}$$"
}
],
"answer": "$${1 \\over {\\sqrt 3 }}$$",
"solution": "**Answer:** $${1 \\over {\\sqrt 3 }}$$\n\nArea bounded by \n
$$ \\Rightarrow $$ k $$ = {1 \\over {\\sqrt 3 }}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3935,
"subject": "General Science",
"question": "The area (in sq. units) bounded by the parabolae y = x2 – 1, the tangent at the point (2, 3) to it and the y-axis is :",
"options": [
{
"text": "$$56\\over3$$"
},
{
"text": "$$32\\over3$$"
},
{
"text": "$$8\\over3$$"
},
{
"text": "$$14\\over3$$"
}
],
"answer": "$$8\\over3$$",
"solution": "**Answer:** $$8\\over3$$\n\n\n
Equation of tangent at (2, 3) on the parabola y = x2 $$-$$ 1 is \n
$${{y + 3} \\over 2} = 2x - 1$$\n
$$ \\Rightarrow $$ y + 3 = 4x $$-$$ 2\n
$$ \\Rightarrow $$ y = 4x $$-$$ 5\n
When x = 0 then for the tangent y = $$-$$ 5\n
$$ \\therefore $$ Tangent cuts x y axis at (0, $$-$$ 5) point. \n
$$ \\therefore $$ Area of the bounded region is \n
= $${{\\sqrt 3 } \\over 4} - {1 \\over 3}$$",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 3943,
"subject": "General Science",
"question": "For a > 0, let the curves C1 : y2 = ax and\nC2 : x2 = ay intersect at origin O and a point P.\nLet the line x = b (0 < b < a) intersect the chord\nOP and the x-axis at points Q and R,\nrespectively. If the line x = b bisects the area\nbounded by the curves, C1 and C2, and the area of\n $$\\Delta $$OQR = $${1 \\over 2}$$, then 'a' satisfies the equation :",
"options": [
{
"text": "x6 – 12x3 + 4 = 0"
},
{
"text": "x6 – 12x3 – 4 = 0"
},
{
"text": "x6 + 6x3 – 4 = 0"
},
{
"text": "x6 – 6x3 + 4 = 0"
}
],
"answer": "x6 – 12x3 + 4 = 0",
"solution": "**Answer:** x6 – 12x3 + 4 = 0\n\n\nC1 : y2\n = ax, C2 : x2\n = ay (a > 0)\n
P : intersection point of y2\n = ax and x2\n = ay\n
$$ = 2|\\,9x - {x^3}\\,|_0^{\\sqrt 3 } = 12\\sqrt 3 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3949,
"subject": "General Science",
"question": "The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. Then A4 is equal to __________.",
"options": [],
"answer": "64",
"solution": "**Answer:** 64\n\n
$$ \\Rightarrow {A^4} = {\\left( {2\\sqrt 2 } \\right)^4} = 64$$",
"topic": "Discrete Mathematics",
"subtopic": "Other"
},
{
"id": 3950,
"subject": "General Science",
"question": "The area bounded by the lines y = || x $$-$$ 1 | $$-$$ 2 | is ___________.",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nQuestion is incomplete it should be area bounded\nby y = || x $$-$$ 1 | $$-$$ 2 | and y = 2.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3951,
"subject": "General Science",
"question": "Let f : [$$-$$3, 1] $$ \\to $$ R be given as
If the area bounded by y = f(x) and x-axis is A, then the value of 6A is equal to ___________.",
"options": [],
"answer": "41",
"solution": "**Answer:** 41\n\n\n Area is $$\\int\\limits_{ - 3}^{ - 2} {(x + 6)dx + \\int\\limits_{ - 2}^0 {{x^2}dx + \\int\\limits_0^1 {\\sqrt {x}dx = A} } } $$
So, 6A = 41",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 3952,
"subject": "General Science",
"question": "Let T be the tangent to the ellipse E : x2 + 4y2 = 5 at the point P(1, 1). If the area of the region bounded by the tangent T, ellipse E, lines x = 1 and x = $$\\sqrt 5 $$ is $$\\alpha$$$$\\sqrt 5 $$ + $$\\beta$$ + $$\\gamma$$ cos$$-$$1$$\\left( {{1 \\over {\\sqrt 5 }}} \\right)$$, then |$$\\alpha$$ + $$\\beta$$ + $$\\gamma$$| is equal to ______________.",
"options": [],
"answer": "1.25",
"solution": "**Answer:** 1.25\n\n E : x2 + 4y2 = 5
$$ = \\left[ {{{3{x^2}} \\over 4} + 6x - {{{x^3}} \\over 4}} \\right]_{ - 2}^4 = 27$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3958,
"subject": "General Science",
"question": "Let a and b respectively be the points of local maximum and local minimum of the function f(x) = 2x3 $$-$$ 3x2 $$-$$ 12x. If A is the total area of the region bounded by y = f(x), the x-axis and the lines x = a and x = b, then 4A is equal to ______________.",
"options": [],
"answer": "114",
"solution": "**Answer:** 114\n\nf'(x) = 6x2 $$-$$ 6x $$-$$ 12 = 6(x $$-$$ 2) (x + 1)
4A = 114",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3959,
"subject": "General Science",
"question": "The area of the region bounded by the parabola (y $$-$$ 2)2 = (x $$-$$ 1), the tangent to it at the point whose ordinate is 3 and the x-axis is :",
"options": [
{
"text": "9"
},
{
"text": "10"
},
{
"text": "4"
},
{
"text": "6"
}
],
"answer": "9",
"solution": "**Answer:** 9\n\ny = 3 $$\\Rightarrow$$ x = 2
If the area of the region $$\\left\\{ {(x,y):{x^{{2 \\over 3}}} + {y^{{2 \\over 3}}} \\le 1,\\,x + y \\ge 0,\\,y \\ge 0} \\right\\}$$ is A, then $${{256A} \\over \\pi }$$ is equal to __________.
Let S be the region bounded by the curves y = x3 and y2 = x. The curve y = 2|x| divides S into two regions of areas R1, R2. If max {R1, R2} = R2, then $${{{R_2}} \\over {{R_1}}}$$ is equal to ______________.
If for some $$\\alpha$$ > 0, the area of the region $$\\{ (x,y):|x + \\alpha | \\le y \\le 2 - |x|\\} $$ is equal to $${3 \\over 2}$$, then the area of the region $$\\{ (x,y):0 \\le y \\le x + 2\\alpha ,\\,|x| \\le 1\\} $$ is equal to ____________.
\n$$\n\\begin{aligned}\n&A=\\int_{-1}^{\\frac{1}{2}}\\left(x+2-x^{2}\\right) d x+\\int_{\\frac{1}{2}}^{1}\\left(4-3 x-x^{2}\\right) d x \\\\\\\\\n&=\\left[\\frac{x^{2}}{2}+2 x-\\frac{x^{3}}{3}\\right]_{-1}^{\\frac{1}{2}}+\\left[4 x-\\frac{3 x^{2}}{2}-\\frac{x^{3}}{3}\\right]_{\\frac{1}{2}}^{1} \\\\\\\\\n&=\\left(\\frac{1}{8}+1-\\frac{1}{24}\\right)-\\left(\\frac{1}{2}-2+\\frac{1}{3}\\right)+\\left(4-\\frac{3}{2}-\\frac{1}{3}\\right)-\\left(2-\\frac{3}{8}-\\frac{1}{24}\\right) \\\\\\\\\n&=\\frac{17}{6}\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3973,
"subject": "General Science",
"question": "
Let the locus of the centre $$(\\alpha, \\beta), \\beta>0$$, of the circle which touches the circle $$x^{2}+(y-1)^{2}=1$$ externally and also touches the $$x$$-axis be $$\\mathrm{L}$$. Then the area bounded by $$\\mathrm{L}$$ and the line $$y=4$$ is:
",
"options": [
{
"text": "$$\n\\frac{32 \\sqrt{2}}{3}\n$$"
},
{
"text": "$$\n\\frac{40 \\sqrt{2}}{3}\n$$"
},
{
"text": "$$\\frac{64}{3}$$"
},
{
"text": "$$\n\\frac{32}{3}\n$$"
}
],
"answer": "$$\\frac{64}{3}$$",
"solution": "**Answer:** $$\\frac{64}{3}$$\n\n \nRadius of circle $S$ touching $x$-axis and centre $(\\alpha, \\beta)$ is $|\\beta|$. According to given conditions\n
\n$$\n\\begin{aligned}\n&\\alpha^{2}+(\\beta-1)^{2}=(|\\beta|+1)^{2} \\\\\\\\\n&\\alpha^{2}+\\beta^{2}-2 \\beta+1=\\beta^{2}+1+2|\\beta| \\\\\\\\\n&\\alpha^{2}=4 \\beta \\text { as } \\beta>0\n\\end{aligned}\n$$\n
\n$\\therefore \\quad$ Required louse is $L: x^{2}=4 y$
\n \nThe area of shaded region $=2 \\int_{0}^{4} 2 \\sqrt{y} d y$\n
Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve $$4{x^3} - 3x{y^2} + 6{x^2} - 5xy - 8{y^2} + 9x + 14 = 0$$ at the point ($$-$$2, 3) be A. Then 8A is equal to ______________.
",
"options": [],
"answer": "170",
"solution": "**Answer:** 170\n\n$$\n\\begin{aligned}\n& 4 x^3-3 x y^2+6 x^2-5 x y-8 y^2+9 x+14=0 \\text { at } P(-2,3) \\\\\\\\\n& 12 x^2-3\\left(y^2+2 y x y^{\\prime}\\right)+12 x-5\\left(x y^{\\prime}+y\\right)-16 y y^{\\prime} + 9=0 \\\\\\\\\n& 48-3\\left(9-12 y^{\\prime}\\right)-24-5\\left(-2 y^{\\prime}+3\\right)-48 y^{\\prime}+9 =0 \\\\\\\\\n& y^{\\prime}=-9 / 2 \\\\\\\\\n& \\text { Tangent } y-3=-\\frac{9}{2}(x+2) \\Rightarrow 9 x+2 y=-12 \\\\\\\\\n& \\text { Normal : } y-3=\\frac{2}{9}(x+2) \\Rightarrow 9 y-2 x=31\n\\end{aligned}\n$$
The area enclosed by the curves $$y=\\log _{e}\\left(x+\\mathrm{e}^{2}\\right), x=\\log _{e}\\left(\\frac{2}{y}\\right)$$ and $$x=\\log _{\\mathrm{e}} 2$$, above the line $$y=1$$ is:
",
"options": [
{
"text": "$$2+\\mathrm{e}-\\log _{\\mathrm{e}} 2$$"
},
{
"text": "$$1+e-\\log _{e} 2$$"
},
{
"text": "$$e-\\log _{e} 2$$"
},
{
"text": "$$1+\\log _{e} 2$$"
}
],
"answer": "$$1+e-\\log _{e} 2$$",
"solution": "**Answer:** $$1+e-\\log _{e} 2$$\n\n \nAccording to NTA, the required region $A_2$ which is shaded in crossed lines and comes out to be
\n$$\nA_2=\\int_1^2\\left(\\ln \\frac{2}{y}-e^y+e^2\\right) d y=1+e-\\ln 2\n$$
\nBut according to us the required region $A_1$ comes out to be shaded in parallel lines, which can be obtained as
\nThere is only one region which satisfies above requirement and which also lies above line $y=1$
Line $y=1$ may or may not be the boundary of the region.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3980,
"subject": "General Science",
"question": "
The area of the region
\n
$$\\left\\{(x, y):|x-1| \\leq y \\leq \\sqrt{5-x^{2}}\\right\\}$$ is equal to :
Let $$A$$ be the area bounded by the curve $$y=x|x-3|$$, the $$x$$-axis and the ordinates $$x=-1$$ and $$x=2$$. Then $$12 A$$ is equal to ____________.
$=\\frac{288}{4}=72$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3985,
"subject": "General Science",
"question": "Let $q$ be the maximum integral value of $p$ in $[0,10]$ for which the roots of the equation $x^2-p x+\\frac{5}{4} p=0$ are rational. Then the area of the region $\\left\\{(x, y): 0 \\leq y \\leq(x-q)^2, 0 \\leq x \\leq q\\right\\}$ is :",
"options": [
{
"text": "$\\frac{125}{3}$"
},
{
"text": "243"
},
{
"text": "164"
},
{
"text": "25"
}
],
"answer": "243",
"solution": "**Answer:** 243\n\n
",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 3986,
"subject": "General Science",
"question": "Let $A$ be the area of the region\n
$\\left\\{(x, y): y \\geq x^2, y \\geq(1-x)^2, y \\leq 2 x(1-x)\\right\\}$.\n
Then $540 \\mathrm{~A}$ is equal to :",
"options": [],
"answer": "25",
"solution": "**Answer:** 25\n\n
Let $$\\alpha$$ be the area of the larger region bounded by the curve $$y^{2}=8 x$$ and the lines $$y=x$$ and $$x=2$$, which lies in the first quadrant. Then the value of $$3 \\alpha$$ is equal to ___________.
Let $$\\Delta$$ be the area of the region $$\\left\\{ {(x,y) \\in {R^2}:{x^2} + {y^2} \\le 21,{y^2} \\le 4x,x \\ge 1} \\right\\}$$. Then $${1 \\over 2}\\left( {\\Delta - 21{{\\sin }^{ - 1}}{2 \\over {\\sqrt 7 }}} \\right)$$ is equal to
Let $$[x]$$ denote the greatest integer $$\\le x$$. Consider the function $$f(x) = \\max \\left\\{ {{x^2},1 + [x]} \\right\\}$$. Then the value of the integral $$\\int\\limits_0^2 {f(x)dx} $$ is
If the area enclosed by the parabolas $$\\mathrm{P_1:2y=5x^2}$$ and $$\\mathrm{P_2:x^2-y+6=0}$$ is equal to the area enclosed by $$\\mathrm{P_1}$$ and $$\\mathrm{y=\\alpha x,\\alpha > 0}$$, then $$\\alpha^3$$ is equal to ____________.
\n$$\n\\begin{aligned}\n& \\text { Area between } P_{1} \\text { and } P_{2} \\quad \\text { [Say } \\left.A_{1}\\right] \\\\\\\\\n& =\\int\\limits_{-2}^{2}\\left(x^{2}+6\\right)-\\frac{5}{2} x^{2} d x \\\\\\\\\n& =2 \\int\\limits_{0}^{2}\\left(6-\\frac{3}{2} x^{2}\\right) d x=2\\left[6 x-\\frac{x^{3}}{2}\\right]_{0}^{2}=16\n\\end{aligned}\n$$
\n$$\n\\begin{aligned}\n& a x=\\frac{5}{2} x^2 \\Rightarrow x=0, \\frac{2 a}{5} \\\\\\\\\n& \\text { Area between } P_1 \\text { and } y=a x \\quad \\text { [Say } A_2 \\text { ] } \\\\\\\\\n& =\\int\\limits_0^{\\frac{2 \\alpha}{5}} a x-\\frac{5}{2} x^2 d x \\\\\\\\\n& \\left.=\\frac{a x^2}{2}-\\frac{5}{6} x^3\\right]_0^{\\frac{2 a}{5}}: \\frac{2 a^3}{75} \\\\\\\\\n& A_1=A_2 \\Rightarrow \\frac{2 a^3}{75}=16 \\\\\\\\\n& a^3=600\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3993,
"subject": "General Science",
"question": "
If the area of the region bounded by the curves $$y^2-2y=-x,x+y=0$$ is A, then 8 A is equal to __________
= $$\n\\begin{aligned}\n&\\left|\\int_{-\\pi}^{-3 \\pi / 4} \\sin x d x\\right|+\\left|\\int_{\\pi / 4}^\\pi \\sin x d x\\right|+\\left|\\int_{-3 \\pi / 4}^{-\\pi / 2} \\cos x d x\\right| \n+\\left|\\int_{-\\pi / 2}^{\\pi / 4} \\cos x d x\\right|\n\\end{aligned}\n$$\n
We can rewrite this as $x^3 - 3x - 2 = 0$. This equation will give us the x-coordinates of the points where the line $y = 3x + 2$ intersects the curve $y = x^3$. We already know that one point of intersection is $(-1, -1)$ (as it's a point of tangency).\n\n
To find the other points of intersection, we can solve the cubic equation $x^3 - 3x - 2 = 0$. \n\n
If A is the area in the first quadrant enclosed by the curve $$\\mathrm{C: 2 x^{2}-y+1=0}$$, the tangent to $$\\mathrm{C}$$ at the point $$(1,3)$$ and the line $$\\mathrm{x}+\\mathrm{y}=1$$, then the value of $$60 \\mathrm{~A}$$ is _________.
$\\mathrm{A}=\\int\\limits_0^1\\left(2 \\mathrm{x}^2+1\\right) \\mathrm{dx}-$ area of $(\\Delta \\mathrm{QOT})-$ area of $(\\Delta \\mathrm{PQR})+$ area of $(\\Delta \\mathrm{QRS})$\n
If the area of the region $$\\left\\{(x, \\mathrm{y}):\\left|x^{2}-2\\right| \\leq y \\leq x\\right\\}$$ is $$\\mathrm{A}$$, then $$6 \\mathrm{A}+16 \\sqrt{2}$$ is equal to __________.
",
"options": [],
"answer": "27",
"solution": "**Answer:** 27\n\n$$\n\\text { We have, }\\left\\{(x, y):\\left|x^2-2\\right| \\leq y \\leq x\\right\\}\n$$\n
\n
On solving, $\\left|x^2-2\\right|=y$ and $y=x$, we get $(1,1)$ and $(2,2)$\n
$$\n\\begin{aligned}\n& A=\\int\\limits_1^{\\sqrt{2}}\\left(x-\\left(2-x^2\\right)\\right) d x+\\int\\limits_{\\sqrt{2}}^2\\left(x-\\left(x^2-2\\right)\\right) d x \\\\\\\\\n& A=\\left[\\frac{x^2}{2}-2 x+\\frac{x^3}{3}\\right]_1^{\\sqrt{2}}+\\left[\\frac{x^2}{2}-\\frac{x^3}{3}+2 x\\right]_{\\sqrt{2}}^2\n\\end{aligned}\n$$\n
Let $$y = p(x)$$ be the parabola passing through the points $$( - 1,0),(0,1)$$ and $$(1,0)$$. If the area of the region $$\\{ (x,y):{(x + 1)^2} + {(y - 1)^2} \\le 1,y \\le p(x)\\} $$ is A, then $$12(\\pi - 4A)$$ is equal to ___________.
",
"options": [],
"answer": "16",
"solution": "**Answer:** 16\n\nLet, $y=p(x)$ be the parabola passing through the points $(-1,0)(0,1)(1,0)$.\n
Now, to find the area of the region\n
$$\n\\left\\{(x, y) ;(x+1)^2+(y-1)^2 \\leq 1, y \\leq p(x)\\right\\}\n$$\n
\n
Thus, area of shaded region is\n
$$\n\\begin{aligned}\n& A =\\int\\limits_{-1}^0\\left\\{\\left(1-x^2\\right)-\\left(1-\\sqrt{1-(x+1)^2}\\right\\} d x\\right. \\\\\\\\\n& =\\int\\limits_{-1}^0\\left\\{-x^2+\\sqrt{1-(x+1)^2}\\right\\} d x \\\\\\\\\n& =\\int\\limits_{-1}^0-x^2 d x+\\int_{-1}^0 \\sqrt{1-(x+1)^2} d x \\\\\\\\\n& =-\\left(\\frac{x^3}{3}\\right)_{-1}^0+\\left[\\frac{\\sqrt{1-(x+1)^2}}{2}+\\frac{1}{2} \\sin ^{-1}(x+1)\\right]_{-1}^0 \\\\\\\\\n& =-\\frac{1}{3}+\\left[\\frac{1}{2}+\\frac{\\pi}{4}-\\frac{1}{2}\\right] \\\\\\\\\n& A =\\frac{\\pi}{4}-\\frac{1}{3}\n\\end{aligned}\n$$\n
Let the area enclosed by the lines $$x+y=2, \\mathrm{y}=0, x=0$$ and the curve $$f(x)=\\min \\left\\{x^{2}+\\frac{3}{4}, 1+[x]\\right\\}$$ where $$[x]$$ \ndenotes the greatest integer $$\\leq x$$, be $$\\mathrm{A}$$. Then the value of $$12 \\mathrm{~A}$$ is _____________.
$$\n\\text { Required area }=\\left[\\int\\limits_0^{\\frac{1}{2}}\\left(x^2+\\frac{3}{4}\\right) d x\\right]+\\left[\\frac{1}{2}\\left(\\frac{3}{2}+\\frac{1}{2}\\right) \\times 1\\right]\n$$\n
If the area of the region $$S=\\left\\{(x, y): 2 y-y^{2} \\leq x^{2} \\leq 2 y, x \\geq y\\right\\}$$ is equal to $$\\frac{n+2}{n+1}-\\frac{\\pi}{n-1}$$, then the natural number $$n$$ is equal to ___________.
Equation of curve passing through point $(1,1)$ and $(2,1)$ and $(0,3),(3,3)$\n
$\\therefore$ Area bounded by given curves\n
$$\n=\\frac{1}{2}(1+3) \\times 2(\\text { Area of trapezium })=4\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4007,
"subject": "General Science",
"question": "Three points $\\mathrm{O}(0,0), \\mathrm{P}\\left(\\mathrm{a}, \\mathrm{a}^2\\right), \\mathrm{Q}\\left(-\\mathrm{b}, \\mathrm{b}^2\\right), \\mathrm{a}>0, \\mathrm{~b}>0$, are on the parabola $y=x^2$. Let $\\mathrm{S}_1$ be the area of the region bounded by the line $\\mathrm{PQ}$ and the parabola, and $\\mathrm{S}_2$ be the area of the triangle $\\mathrm{OPQ}$. If the minimum value of $\\frac{\\mathrm{S}_1}{\\mathrm{~S}_2}$ is $\\frac{\\mathrm{m}}{\\mathrm{n}}, \\operatorname{gcd}(\\mathrm{m}, \\mathrm{n})=1$, then $\\mathrm{m}+\\mathrm{n}$ is equal to __________.",
"options": [],
"answer": "7",
"solution": "**Answer:** 7\n\n\n
$\\begin{aligned} & \\because \\frac{b}{a}+\\frac{a}{b} \\geq 2 \\\\\\\\ & \\Rightarrow\\left(\\frac{S_1}{S_2}\\right)_{\\min }=\\frac{4}{3}=\\frac{m}{n} \\\\\\\\ & \\Rightarrow m+n=7\\end{aligned}$",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 4008,
"subject": "General Science",
"question": "The sum of squares of all possible values of $k$, for which area of the region bounded by the parabolas $2 y^2=\\mathrm{k} x$ and $\\mathrm{ky}^2=2(y-x)$ is maximum, is equal to :",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nGiven $k y^2=2(y-x)$ .........(i)\n
$$\n2 y^2=k x\n$$ .........(ii)\n\n
Point of intersection of (i) and (ii)\n
$$\n\\begin{aligned}\n& k y^2=2\\left(y-\\frac{2 y^2}{k}\\right) \\\\\\\\\n& \\Rightarrow y=0, k y=2\\left(1-\\frac{2 y}{k}\\right)\n\\end{aligned}\n$$\n
$\\begin{aligned} & k y+\\frac{4 y}{k}=2 \\\\\\\\ & y=\\frac{2}{k+\\frac{4}{k}}=\\frac{2 k}{k^2+4}\\end{aligned}$\n
$\\begin{aligned} & A=\\int\\limits_6^{\\frac{2 k}{k^2+4}}\\left(\\left(y-\\frac{k y^2}{2}\\right)-\\frac{2 y^2}{k}\\right) d y \\\\\\\\ & A=\\left[\\frac{y^2}{2}-\\left(\\frac{k}{2}+\\frac{2}{k}\\right) \\frac{y^3}{3}\\right]_0^{\\frac{2 k}{k^2+4}}\\end{aligned}$\n
To find the enclosed area between the two curves $ x y+4 y=16 $ and $ x+y=6 $, we need to determine the region of intersection and integrate the difference of the functions over the interval where they intersect.
\n\n
First, let's solve the equations simultaneously to find the points of intersection.
\n\n
The second curve $ x+y=6 $ can be rewritten as $ y=6-x $.
\n\n
Substitute $ y=6-x $ into the first equation:
\n\n
$ x(6-x) + 4(6-x) = 16$
\n\n
$ 6x - x^2 + 24 - 4x = 16 $
\n\n
$ -x^2 + 2x + 8 = 0 $
\n\n
This can be simplified to:
\n\n
$ x^2 - 2x - 8 = 0 $
\n\n
Using the quadratic formula, we can find the roots of this equation:
\n\n
$ x = \\frac{-(-2) \\pm \\sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} $
\n\n
$ x = \\frac{2 \\pm \\sqrt{4 + 32}}{2} $
\n\n
$ x = \\frac{2 \\pm \\sqrt{36}}{2} $
\n\n
$ x = \\frac{2 \\pm 6}{2} $
\n\n
So we have two solutions:
\n\n
$ x_1 = \\frac{2 + 6}{2} = 4 $
\n\n
$ x_2 = \\frac{2 - 6}{2} = -2 $
\n\n
For $ x_1 = 4 $, substitute this back into the line equation $ x+y=6 $:
\n\n
$ 4+y=6 $
\n\n
$ y=2 $
\n\n
For $ x_2 = -2 $, do the same:
\n\n
$ -2+y=6 $
\n\n
$ y=8 $
\n\n
So we have two points of intersection: $ (4,2) $ and $ (-2,8) $.
\n\n\n
Now, to find the area between $ x y + 4y = 16 $ and $ x + y = 6 $, we'll integrate the top function minus the bottom function from $ x = -2 $ to $ x = 4 $. First, we need to express $ y $ from both equations in terms of $ x $:
\n\n
For $ x y + 4 y = 16 $, express $ y $ in terms of $ x $:
\n\n
$ y(x+4) = 16 $
\n\n
$ y = \\frac{16}{x+4} $
\n\n
For $ x + y = 6 $, we have:
\n\n
$ y = 6 - x $
\n\n
The area $ A $ is therefore given by:
\n\n
$ A = \\int\\limits_{-2}^{4}\\left( 6 - x - \\frac{16}{x+4} \\right) dx $
\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4010,
"subject": "General Science",
"question": "Let the area of the region $\\left\\{(x, y): x-2 y+4 \\geqslant 0, x+2 y^2 \\geqslant 0, x+4 y^2 \\leq 8, y \\geqslant 0\\right\\}$ be $\\frac{\\mathrm{m}}{\\mathrm{n}}$, where $\\mathrm{m}$ and $\\mathrm{n}$ are coprime numbers. Then $\\mathrm{m}+\\mathrm{n}$ is equal to _____________.",
"options": [],
"answer": "119",
"solution": "**Answer:** 119\n\n
\n
$$\\begin{aligned}\n& A=\\int_\\limits0^1\\left[\\left(8-4 y^2\\right)-\\left(-2 y^2\\right)\\right] d y+ \\\\\n& \\int_\\limits1^{3 / 2}\\left[\\left(8-4 y^2\\right)-(2 y-4)\\right] d y \\\\\n& =\\left[8 y-\\frac{2 y^3}{3}\\right]_0^1+\\left[12 y-y^2-\\frac{4 y^3}{3}\\right]_1^{3 / 2}=\\frac{107}{12}=\\frac{m}{n} \\\\\n& \\therefore m+n=119\n\\end{aligned}$$
If the area of the region $$\\left\\{(x, y): 0 \\leq y \\leq \\min \\left\\{2 x, 6 x-x^2\\right\\}\\right\\}$$ is $$\\mathrm{A}$$, then $$12 \\mathrm{~A}$$ is equal to ________.
$$\\begin{aligned}\n& y^2 \\leq 4 x, x<4 \\\\\n& \\frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0 \\\\\n& \\text { Case - I}: y>0 \\\\\n& \\frac{x(x-1)(x-2)}{(x-3)(x-4)}>0 \\\\\n& x \\in(0,1) \\cup(2,3) \\\\\n& \\text { Case }- \\text { II : y<0 } \\\\\n& \\frac{x(x-1)(x-2)}{(x-3)(x-4)}<0, x \\in(1,2) \\cup(3,4)\n\\end{aligned}$$
\n
\n
$$\\begin{aligned}\n& \\text { Area }=2 \\int_\\limits0^4 \\sqrt{x} d x \\\\\n& =2 \\cdot \\frac{2}{3}\\left[x^{3 / 2}\\right]_0^4=\\frac{32}{3}\n\\end{aligned}$$
The area (in sq. units) of the part of the circle $$x^2+y^2=169$$ which is below the line $$5 x-y=13$$ is $$\\frac{\\pi \\alpha}{2 \\beta}-\\frac{65}{2}+\\frac{\\alpha}{\\beta} \\sin ^{-1}\\left(\\frac{12}{13}\\right)$$, where $$\\alpha, \\beta$$ are coprime numbers. Then $$\\alpha+\\beta$$ is equal to __________.
If the points of intersection of two distinct conics $$x^2+y^2=4 b$$ and $$\\frac{x^2}{16}+\\frac{y^2}{b^2}=1$$ lie on the curve $$y^2=3 x^2$$, then $$3 \\sqrt{3}$$ times the area of the rectangle formed by the intersection points is _________.
Let the area of the region $$\\left\\{(x, y): 0 \\leq x \\leq 3,0 \\leq y \\leq \\min \\left\\{x^2+2,2 x+2\\right\\}\\right\\}$$ be A. Then $$12 \\mathrm{~A}$$ is equal to __________.
$$\n\\begin{aligned}\n& \\int\\limits_{-2}^4\\left[\\left(\\frac{y+8}{2}\\right)-\\left(\\frac{y^2+8}{4}\\right)\\right] d y \\\\\\\\\n& =\\frac{y^2}{4}+4 y-\\frac{y^3}{12}-\\left.2 y\\right|_{-2} ^4\n\\end{aligned}\n$$\n
One of the points of intersection of the curves $$y=1+3 x-2 x^2$$ and $$y=\\frac{1}{x}$$ is $$\\left(\\frac{1}{2}, 2\\right)$$. Let the area of the region enclosed by these curves be $$\\frac{1}{24}(l \\sqrt{5}+\\mathrm{m})-\\mathrm{n} \\log _{\\mathrm{e}}(1+\\sqrt{5})$$, where $$l, \\mathrm{~m}, \\mathrm{n} \\in \\mathbf{N}$$. Then $$l+\\mathrm{m}+\\mathrm{n}$$ is equal to
Let the area of the region enclosed by the curve $$y=\\min \\{\\sin x, \\cos x\\}$$ and the $$x$$ axis between $$x=-\\pi$$ to $$x=\\pi$$ be $$A$$. Then $$A^2$$ is equal to __________.
$$\\begin{aligned}\n& A=-\\int_\\limits{-\\pi}^{\\frac{-3 \\pi}{4}} \\cos x d x-\\int_\\limits{\\frac{-3 \\pi}{4}}^0 \\sin x d x+\\int_\\limits0^{\\frac{\\pi}{4}} \\sin x d x+\\int_\\limits{\\frac{\\pi}{4}}^{\\frac{\\pi}{2}} \\cos x d x \\\\\n& -\\int_\\limits{\\frac{\\pi}{2}}^\\pi \\cos x d x \\\\\n& A=4 \\\\\n& A^2=16\n\\end{aligned}$$
Point of intersections are $$(0,0)$$ & $$(-2,4)$$.
\n
$$\\begin{aligned}\n\\therefore \\quad & \\text { Area }=\\int_\\limits{-2}^0\\left(-x^2-2 x\\right) d x \\\\\n& =\\left[\\frac{-x^3}{3}-\\frac{2 x^2}{2}\\right]_{-2}^0 \\\\\n& =-\\left[\\frac{8}{3}-4\\right]=\\frac{4}{3} \\text { sq. unit }\n\\end{aligned}$$
If the area of the region $$\\left\\{(x, y): \\frac{\\mathrm{a}}{x^2} \\leq y \\leq \\frac{1}{x}, 1 \\leq x \\leq 2,0<\\mathrm{a}<1\\right\\}$$ is $$\\left(\\log _{\\mathrm{e}} 2\\right)-\\frac{1}{7}$$ then the value of $$7 \\mathrm{a}-3$$ is equal to :
$$ = {{\\sqrt 3 - 1} \\over 2}$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4032,
"subject": "General Science",
"question": "The area of the region \n
= $$4{\\left( {{4 \\over {25}}} \\right)^{{1 \\over 3}}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4035,
"subject": "General Science",
"question": "Let A1 be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y = sinx, y = cosx, x-axis and x = $${\\pi \\over 2}$$ in the first quadrant. Then,",
"options": [
{
"text": "$${A_1}:{A_2} = 1:\\sqrt 2 $$ and $${A_1} + {A_2} = 1$$"
},
{
"text": "$${A_1} = {A_2}$$ and $${A_1} + {A_2} = \\sqrt 2 $$"
},
{
"text": "$$2{A_1} = {A_2}$$ and $${A_1} + {A_2} = 1 + \\sqrt 2 $$"
},
{
"text": "$${A_1}:{A_2} = 1:2$$ and $${A_1} + {A_2} = 1$$"
}
],
"answer": "$${A_1}:{A_2} = 1:\\sqrt 2 $$ and $${A_1} + {A_2} = 1$$",
"solution": "**Answer:** $${A_1}:{A_2} = 1:\\sqrt 2 $$ and $${A_1} + {A_2} = 1$$\n\n$${A_1} + {A_2} = \\int\\limits_0^{\\pi /2} {\\cos x.\\,dx = \\left. {\\sin x} \\right|_0^{\\pi /2}} = 1$$
$$ = {{3\\pi {} } \\over 2}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4037,
"subject": "General Science",
"question": "If the line y = mx bisects the area enclosed by the lines x = 0, y = 0, x = $${3 \\over 2}$$ and the curve y = 1 + 4x $$-$$ x2, then 12 m is equal to _____________.",
"options": [],
"answer": "26",
"solution": "**Answer:** 26\n\n
$$\\Rightarrow$$ P = 2 + (positive value less than e $$-$$ 2)
$$\\Rightarrow$$ P $$\\in$$ (2, 3)
$$\\Rightarrow$$ least integer value of P is 3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4041,
"subject": "General Science",
"question": "Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively
$$\\text { Number of integral terms in the expansion of }\\left\\{7^{\\left(\\frac{1}{2}\\right)}+11^{\\left(\\frac{1}{6}\\right)}\\right\\}^{824} \\text { is equal to _________. }$$
General term in expansion of $$\\left((7)^{1 / 2}+(11)^{1 / 6}\\right)^{824}$$ is $$\\mathrm{t}_{\\mathrm{r}+1}={ }^{824} \\mathrm{C}_{\\mathrm{r}}(7)^{\\frac{824-\\mathrm{r}}{2}}(11)^{\\mathrm{r} / 6}$$
63$$ \\times $$[Some integral value] + 2 by dividing with 9 we will get 2 as remainder as 63 is multiple of 9.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4045,
"subject": "General Science",
"question": "If (27)999 is divided by 7, then the remainder is : ",
"options": [
{
"text": "1"
},
{
"text": "2"
},
{
"text": "3"
},
{
"text": "6"
}
],
"answer": "6",
"solution": "**Answer:** 6\n\nWe have, \n
So $${{\\left( {{{49}^{63}} - 1} \\right)} \\over {48}}$$ = integer\n
$$ \\therefore $$ 4963 + 1 is a factor.\n
So k = 63.",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 4047,
"subject": "General Science",
"question": "If the remainder when x is divided by 4 is 3, then the remainder when (2020 + x)2022 is divided by 8 is __________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nLet x = 4k + 3
$$ \\therefore $$ Reminder is 1.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4048,
"subject": "General Science",
"question": "The total number of two digit numbers 'n', such that 3n + 7n is a multiple of 10, is __________.",
"options": [],
"answer": "45",
"solution": "**Answer:** 45\n\n$$ \\because $$ $${7^n} = {(10 - 3)^n} = 10k + {( - 3)^n}$$
$${7^n} + {3^n} = 10k + {( - 3)^n} + {3^n}$$
$$ \\therefore $$ 3n = 32t = (10 $$-$$ 1)t
= 10p + ($$-$$1)t
= 10p $$\\pm$$ 1
$$ \\therefore $$ if n = even then 7n + 3n will not be multiply of 10
So if n is odd then only 7n + 3n will be multiply of 10
$$ \\therefore $$ n = 11, 13, 15, ..........., 99
$$ \\therefore $$ Ans : 45",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4049,
"subject": "General Science",
"question": "If (2021)3762 is divided by 17, then the remainder is __________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n2021 = 17m - 2\n
If $${1 \\over {2\\,.\\,{3^{10}}}} + {1 \\over {{2^2}\\,.\\,{3^9}}} + \\,\\,.....\\,\\, + \\,\\,{1 \\over {{2^{10}}\\,.\\,3}} = {K \\over {{2^{10}}\\,.\\,{3^{10}}}}$$, then the remainder when K is divided by 6 is :
Remainder $=9$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4060,
"subject": "General Science",
"question": "$50^{\\text {th }}$ root of a number $x$ is 12 and $50^{\\text {th }}$ root of another number $y$ is 18 . Then the remainder obtained on dividing $(x+y)$ by 25 is ____________.",
"options": [],
"answer": "23",
"solution": "**Answer:** 23\n\n
Given $${x^{{1 \\over {50}}}} = 12 \\Rightarrow x = {12^{50}}$$
\n
$${y^{{1 \\over {50}}}} = 18 \\Rightarrow y = {18^{50}}$$
Let the number $$(22)^{2022}+(2022)^{22}$$ leave the remainder $$\\alpha$$ when divided by 3 and $$\\beta$$ when divided by 7. Then $$\\left(\\alpha^{2}+\\beta^{2}\\right)$$ is equal to :
$$25^{190}-19^{190}-8^{190}+2^{190}$$ is divisible by :
",
"options": [
{
"text": "14 but not by 34"
},
{
"text": "neither 14 nor 34"
},
{
"text": "both 14 and 34"
},
{
"text": "34 but not by 14"
}
],
"answer": "34 but not by 14",
"solution": "**Answer:** 34 but not by 14\n\nThe given expression is divisible by 6 and 17 .\n
Also, $25^{190}-8^{190}$ is not divisible by 7\n
but $19^{190}-2^{190}$ is divisible by 7 ,\n
So, $25^{190}-19^{190}-8^{190}+2^{190}$ is divisible by 34 but not by 14 .",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 4066,
"subject": "General Science",
"question": "
The largest natural number $$n$$ such that $$3^{n}$$ divides $$66 !$$ is ___________.
(S1) : $$2023^{2022}-1999^{2022}$$ is divisible by 8
\n
(S2) : $$13(13)^{n}-12 n-13$$ is divisible by 144 for infinitely many $$n \\in \\mathbb{N}$$
",
"options": [
{
"text": "both (S1) and (S2) are incorrect"
},
{
"text": "only (S1) is correct"
},
{
"text": "only (S2) is correct"
},
{
"text": "both (S1) and (S2) are correct"
}
],
"answer": "both (S1) and (S2) are correct",
"solution": "**Answer:** both (S1) and (S2) are correct\n\nWe have, $S_1$ : $(2023)^{2022}-(1999)^{2022}$\n
So coefficients of $${x^p}$$ and $${x^q}$$ are equal.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4071,
"subject": "General Science",
"question": "The number of integral terms in the expansion of $${\\left( {\\sqrt 3 + \\root 8 \\of 5 } \\right)^{256}}$$ is ",
"options": [
{
"text": "35 "
},
{
"text": "32 "
},
{
"text": "33 "
},
{
"text": "34 "
}
],
"answer": "33 ",
"solution": "**Answer:** 33 \n\nGeneral term = $${}^{256}{C_r}.{\\left( {\\sqrt 3 } \\right)^{256 - r}}.{\\left( {\\root 8 \\of 5 } \\right)^r}$$\n = $${}^{256}{C_r}.{\\left( 3 \\right)^{{{256 - r} \\over 2}}}.{\\left( 5 \\right)^{{r \\over 8}}}$$\n
When $${{{256 - r} \\over 2}}$$ is integer then $${\\left( 3 \\right)^{{{256 - r} \\over 2}}}$$ is integer.\n And when $${{r \\over 8}}$$ is integer then $${\\left( 5 \\right)^{{r \\over 8}}}$$ is integer.\n
Entire general term will be integer when $${{{256 - r} \\over 2}}$$ and $${{r \\over 8}}$$ both are integer.\n
$${{{256 - r} \\over 2}}$$ is integer when r = 0, 2, 4, 6, ......, 256\n
$${{r \\over 8}}$$ is integer when r = 0, 8, 16 ,......., 256\n
Now both $${{{256 - r} \\over 2}}$$ and $${{r \\over 8}}$$ will be integer when r = 0, 8, 16, ...., 256 (This is an AP)\n
$$\\therefore$$ 256 = 0 + (n - 1)8 using formula of AP, tn = a + (n - 1)d\n
= 182",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4079,
"subject": "General Science",
"question": "The coefficient of x−5 in the binomial expansion of \n
$$ \\Rightarrow $$ x = 82 or x = $${1 \\over 8}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4085,
"subject": "General Science",
"question": "If the fourth term in the binomial expansion of \n$${\\left( {\\sqrt {{x^{\\left( {{1 \\over {1 + {{\\log }_{10}}x}}} \\right)}}} + {x^{{1 \\over {12}}}}} \\right)^6}$$ is equal to 200, and x > 1,\nthen the value of x is :",
"options": [
{
"text": "100"
},
{
"text": "103"
},
{
"text": "10"
},
{
"text": "104"
}
],
"answer": "10",
"solution": "**Answer:** 10\n\nFourth term (T4)\n
$$\n\\begin{aligned}\n 2+\\frac{10-r}{2}-2 r =2 \\\\\\\\\n\\Rightarrow \\frac{10-r}{2}-2 r =0 \\\\\\\\\n\\Rightarrow 10-r =4 r \\Rightarrow r=2\n\\end{aligned}\n$$\n
So, the coefficient of $x^2$ is ${ }^{10} C_2 \\lambda^2=720$\n
$$ \\Rightarrow $$ |k| = 3",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4091,
"subject": "General Science",
"question": "The natural number m, for which the coefficient of x in the binomial expansion of
\n$${\\left( {{x^m} + {1 \\over {{x^2}}}} \\right)^{22}}$$ is 1540, is .............",
"options": [],
"answer": "13",
"solution": "**Answer:** 13\n\nGeneral term,\n
$$ \\therefore $$ $$18k = {{21} \\over {54}} \\times 18 = 7$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4093,
"subject": "General Science",
"question": "If the number of integral terms in the expansion\n of (31/2 + 51/8)n is exactly 33, then the least value\nof n is :",
"options": [
{
"text": "264"
},
{
"text": "256"
},
{
"text": "128"
},
{
"text": "248"
}
],
"answer": "256",
"solution": "**Answer:** 256\n\nGeneral term of the expression,
We will get integral term when $${{n - r} \\over 2}$$ and $${r \\over 8}$$ are integer
$$ \\therefore $$ (1) n $$-$$ r is multiple of 2
$$ \\Rightarrow $$ n $$-$$ r = 0, 2, 4, ......
(2) r is multiple of 8
$$ \\Rightarrow $$ r = 0, 8, 16, .......
From this two conditions common values are = 0, 8, 16, ....... which will becomes integral terms.
Given that there are 33 integral terms.
Here first integral term at 0th position.
Second integral term at 8th position.
$$ \\therefore $$ 33th integral term will be at = 0 + (33 $$-$$ 1)8 = 256
So, there should be at least 256 terms.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4094,
"subject": "General Science",
"question": "Let\n$$\\alpha $$ > 0,\n$$\\beta $$ > 0 be such that\n $$\\alpha $$3 + $$\\beta $$2 = 4. If the\nmaximum value of the term independent of x in\n the binomial expansion of\n$${\\left( {\\alpha {x^{{1 \\over 9}}} + \\beta {x^{ - {1 \\over 6}}}} \\right)^{10}}$$\nis 10k,\n then k is equal to :",
"options": [
{
"text": "176"
},
{
"text": "336"
},
{
"text": "352"
},
{
"text": "84"
}
],
"answer": "336",
"solution": "**Answer:** 336\n\nGeneral term\n
$$ \\Rightarrow $$ $$x = 2,x = {2^{ - 7/3}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4101,
"subject": "General Science",
"question": "Let the coefficients of third, fourth and fifth terms in the expansion of $${\\left( {x + {a \\over {{x^2}}}} \\right)^n},x \\ne 0$$, be in the ratio 12 : 8 : 3. Then the term independent of x in the expansion, is equal to ___________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n$${T_{r + 1}} = {n_{C_r}}{x^{n - r}}.{\\left( {{a \\over {{x^2}}}} \\right)^r}$$
So total no of terms are 21.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4104,
"subject": "General Science",
"question": "If the constant term, in binomial expansion of $${\\left( {2{x^r} + {1 \\over {{x^2}}}} \\right)^{10}}$$ is 180, then r is equal to __________________.",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\n$${\\left( {2{x^r} + {1 \\over {{x^2}}}} \\right)^{10}}$$
General term $$ = {}^{10}{C_R}{(2{x^2})^{10 - R}}{x^{ - 2R}}$$
$$\\Rightarrow$$ x $$-$$ 1 = 0 (one of the possible value).
$$\\Rightarrow$$ x = 1",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4107,
"subject": "General Science",
"question": "The sum of all those terms which are rational numbers in the
= 24 + 16 = 43",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4108,
"subject": "General Science",
"question": "If the greatest value of the term independent of 'x' in the
expansion of $${\\left( {x\\sin \\alpha + a{{\\cos \\alpha } \\over x}} \\right)^{10}}$$ is $${{10!} \\over {{{(5!)}^2}}}$$, then the value of 'a' is equal to :",
"options": [
{
"text": "$$-$$1"
},
{
"text": "1"
},
{
"text": "$$-$$2"
},
{
"text": "2"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\n$${T_{r + 1}} = {}^{10}{C_r}{(x\\sin \\alpha )^{10 - r}}{\\left( {{{a\\cos \\alpha } \\over x}} \\right)^r}$$
r = 0, 1, 2, ......., 10
Tr + 1 will be independent of x when 10 $$-$$ 2r = 0 $$\\Rightarrow$$ r = 5
$$ \\Rightarrow {}^{10}{C_5}{{{a^5}} \\over {{2^5}}} = {}^{10}{C_5} \\Rightarrow a = 2$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4109,
"subject": "General Science",
"question": "If the co-efficient of x7 and x8 in the expansion of $${\\left( {2 + {x \\over 3}} \\right)^n}$$ are equal, then the value of n is equal to _____________.",
"options": [],
"answer": "55",
"solution": "**Answer:** 55\n\n$${}^n{C_7}{2^{n - 7}}{1 \\over {{3^7}}} = {}^n{C_8}{2^{n - 8}}{1 \\over {{3^8}}}$$
$$\\Rightarrow$$ n $$-$$ 7 = 48 $$\\Rightarrow$$ n = 55",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4110,
"subject": "General Science",
"question": "If the coefficients of x7 in $${\\left( {{x^2} + {1 \\over {bx}}} \\right)^{11}}$$ and x$$-$$7 in $${\\left( {{x} - {1 \\over {bx^2}}} \\right)^{11}}$$, b $$\\ne$$ 0, are equal, then the value of b is equal to :",
"options": [
{
"text": "2"
},
{
"text": "$$-$$1"
},
{
"text": "1"
},
{
"text": "$$-$$2"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\nCoefficient of x7 in $${\\left( {{x^2} + {1 \\over {bx}}} \\right)^{11}}$$ :
General Term = $${}^{11}{C_r}{({x^2})^{11 - r}}.{\\left( {{1 \\over {bx}}} \\right)^r}$$
Let the coefficients of x$$-$$1 and x$$-$$3 in the expansion of $${\\left( {2{x^{{1 \\over 5}}} - {1 \\over {{x^{{1 \\over 5}}}}}} \\right)^{15}},x > 0$$, be m and n respectively. If r is a positive integer such that $$m{n^2} = {}^{15}{C_r}\\,.\\,{2^r}$$, then the value of r is equal to __________.
The number of positive integers k such that the constant term in the binomial expansion of $${\\left( {2{x^3} + {3 \\over {{x^k}}}} \\right)^{12}}$$, x $$\\ne$$ 0 is 28 . l, where l is an odd integer, is ______________.
If the coefficient of x10 in the binomial expansion of $${\\left( {{{\\sqrt x } \\over {{5^{{1 \\over 4}}}}} + {{\\sqrt 5 } \\over {{x^{{1 \\over 3}}}}}} \\right)^{60}}$$ is $${5^k}\\,.\\,l$$, where l, k $$\\in$$ N and l is co-prime to 5, then k is equal to _____________.
The coefficient of x101 in the expression $${(5 + x)^{500}} + x{(5 + x)^{499}} + {x^2}{(5 + x)^{498}} + \\,\\,.....\\,\\, + \\,\\,{x^{500}}$$, x > 0, is
If the sum of the co-efficient of all the positive even powers of x in the binomial expansion of $${\\left( {2{x^3} + {3 \\over x}} \\right)^{10}}$$ is $${5^{10}} - \\beta \\,.\\,{3^9}$$, then $$\\beta$$ is equal to ____________.
For two positive real numbers a and b such that $${1 \\over {{a^2}}} + {1 \\over {{b^3}}} = 4$$, then minimum value of the constant term in the expansion of $${\\left( {a{x^{{1 \\over 8}}} + b{x^{ - {1 \\over {12}}}}} \\right)^{10}}$$ is :
If the maximum value of the term independent of $$t$$ in the expansion of $$\\left(\\mathrm{t}^{2} x^{\\frac{1}{5}}+\\frac{(1-x)^{\\frac{1}{10}}}{\\mathrm{t}}\\right)^{15}, x \\geqslant 0$$, is $$\\mathrm{K}$$, then $$8 \\mathrm{~K}$$ is equal to ____________.
Let for the $$9^{\\text {th }}$$ term in the binomial expansion of $$(3+6 x)^{\\mathrm{n}}$$, in the increasing powers of $$6 x$$, to be the greatest for $$x=\\frac{3}{2}$$, the least value of $$\\mathrm{n}$$ is $$\\mathrm{n}_{0}$$. If $$\\mathrm{k}$$ is the ratio of the coefficient of $$x^{6}$$ to the coefficient of $$x^{3}$$, then $$\\mathrm{k}+\\mathrm{n}_{0}$$ is equal to :
Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $$\\left(\\sqrt[4]{2}+\\frac{1}{\\sqrt[4]{3}}\\right)^{\\mathrm{n}}$$, in the increasing powers of $$\\frac{1}{\\sqrt[4]{3}}$$ be $$\\sqrt[4]{6}: 1$$. If the sixth term from the beginning is $$\\frac{\\alpha}{\\sqrt[4]{3}}$$, then $$\\alpha$$ is equal to _________.
Let the sixth term in the binomial expansion of $${\\left( {\\sqrt {{2^{{{\\log }_2}\\left( {10 - {3^x}} \\right)}}} + \\root 5 \\of {{2^{(x - 2){{\\log }_2}3}}} } \\right)^m}$$ in the increasing powers of $$2^{(x-2) \\log _{2} 3}$$, be 21 . If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of $$x$$ is __________.
",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n${ }^m C_1,{ }^m C_2,{ }^m C_3$ are first, third and fifth term of $A P$\n
Sum of squares of values of x = 02 + 22 = 4",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4123,
"subject": "General Science",
"question": "
If the term without $$x$$ in the expansion of $$\\left(x^{\\frac{2}{3}}+\\frac{\\alpha}{x^{3}}\\right)^{22}$$ is 7315 , then $$|\\alpha|$$ is equal to ___________.
Coeff of $x^{-6}={ }^{9} C_{5}\\left(\\frac{4}{5}\\right)^{4}\\left(\\frac{5}{2}\\right)^{5}$\n\n$$\n=5040\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4125,
"subject": "General Science",
"question": "If the constant term in the binomial expansion of $\\left(\\frac{x^{\\frac{5}{2}}}{2}-\\frac{4}{x^{l}}\\right)^{9}$ is $-84$ and the coefficient of $x^{-3 l}$ is\n\n$2^{\\alpha} \\beta$, where $\\beta<0$ is an odd number, then $|\\alpha l-\\beta|$ is equal to ________.",
"options": [],
"answer": "98",
"solution": "**Answer:** 98\n\nGiven binomial expansion of $\\left(\\frac{x^{\\frac{5}{2}}}{2}-\\frac{4}{x^{l}}\\right)^{9}$\n\n
Let $$\\alpha>0$$, be the smallest number such that the expansion of $$\\left(x^{\\frac{2}{3}}+\\frac{2}{x^{3}}\\right)^{30}$$ has a term $$\\beta x^{-\\alpha}, \\beta \\in \\mathbb{N}$$. Then $$\\alpha$$ is equal to ___________.
If the coefficient of $$x^{15}$$ in the expansion of $$\\left(\\mathrm{a} x^{3}+\\frac{1}{\\mathrm{~b} x^{1 / 3}}\\right)^{15}$$ is equal to the coefficient of $$x^{-15}$$ in the expansion of $$\\left(a x^{1 / 3}-\\frac{1}{b x^{3}}\\right)^{15}$$, where $$a$$ and $$b$$ are positive real numbers, then for each such ordered pair $$(\\mathrm{a}, \\mathrm{b})$$ :
Let the coefficients of three consecutive terms in the binomial expansion of $$(1+2x)^n$$ be in the ratio 2 : 5 : 8. Then the coefficient of the term, which is in the middle of those three terms, is __________.
If the co-efficient of $$x^9$$ in $${\\left( {\\alpha {x^3} + {1 \\over {\\beta x}}} \\right)^{11}}$$ and the co-efficient of $$x^{-9}$$ in $${\\left( {\\alpha x - {1 \\over {\\beta {x^3}}}} \\right)^{11}}$$ are equal, then $$(\\alpha\\beta)^2$$ is equal to ___________.
",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nCoefficient of $\\mathrm{x}^{9}$ in $\\left(\\alpha x^{3}+\\frac{1}{\\beta x}\\right)={ }^{11} C_{6} \\cdot \\frac{\\alpha^{5}}{\\beta^{6}}$\n
Let the sum of the coefficients of the first three terms in the expansion of $${\\left( {x - {3 \\over {{x^2}}}} \\right)^n},x \\ne 0.~n \\in \\mathbb{N}$$, be 376. Then the coefficient of $$x^4$$ is __________.
So, coefficient is $\\frac{80}{9}$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4133,
"subject": "General Science",
"question": "
Let $$\\alpha$$ be the constant term in the binomial expansion of $$\\left(\\sqrt{x}-\\frac{6}{x^{\\frac{3}{2}}}\\right)^{n}, n \\leq 15$$. If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of $$x^{-n}$$ is $$\\lambda \\alpha$$, then $$\\lambda$$ is equal to _____________.
The sum of the coefficients of three consecutive terms in the binomial expansion of $$(1+\\mathrm{x})^{\\mathrm{n}+2}$$, which are in the ratio $$1: 3: 5$$, is equal to :
",
"options": [
{
"text": "63"
},
{
"text": "92"
},
{
"text": "25"
},
{
"text": "41"
}
],
"answer": "63",
"solution": "**Answer:** 63\n\nThe problem asks for the sum of the coefficients of three consecutive terms in the binomial expansion of $(1+x)^{n+2}$, which are in the ratio 1 : 3 : 5. \n\n
Given that the ratios of the coefficients are 1:3:5, we let the terms be $T_r$, $T_{r+1}$, and $T_{r+2}$. The coefficients of these terms are ${ }^{n+2} C_{r-1}$, ${ }^{n+2} C_{r}$, and ${ }^{n+2} C_{r+1}$, respectively. \n\n
If the $$1011^{\\text {th }}$$ term from the end in the binominal expansion of $$\\left(\\frac{4 x}{5}-\\frac{5}{2 x}\\right)^{2022}$$ is 1024 times $$1011^{\\text {th }}$$R term from the beginning, then $$|x|$$ is equal to
If the coefficient of $${x^7}$$ in $${\\left( {ax - {1 \\over {b{x^2}}}} \\right)^{13}}$$ and the coefficient of $${x^{ - 5}}$$ in $${\\left( {ax + {1 \\over {b{x^2}}}} \\right)^{13}}$$ are equal, then $${a^4}{b^4}$$ is equal to :
The absolute difference of the coefficients of $$x^{10}$$ and $$x^{7}$$ in the expansion of $$\\left(2 x^{2}+\\frac{1}{2 x}\\right)^{11}$$ is equal to :
Let $$[t]$$ denote the greatest integer $$\\leq t$$. If the constant term in the expansion of $$\\left(3 x^{2}-\\frac{1}{2 x^{5}}\\right)^{7}$$ is $$\\alpha$$, then $$[\\alpha]$$ is equal to ___________.
",
"options": [],
"answer": "1275",
"solution": "**Answer:** 1275\n\nLet $\\mathrm{T}_{r+1}$ be the constant term.\n
If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $$\\left(\\sqrt[4]{2}+\\frac{1}{\\sqrt[4]{3}}\\right)^{\\mathrm{n}}$$ is $$\\sqrt{6}: 1$$, then the third term from the beginning is :
If the coefficient of $${x^7}$$ in $${\\left( {a{x^2} + {1 \\over {2bx}}} \\right)^{11}}$$ and $${x^{ - 7}}$$ in $${\\left( {ax - {1 \\over {3b{x^2}}}} \\right)^{11}}$$ are equal, then :
If the term independent of $$x$$ in the expansion of $$\\left(\\sqrt{\\mathrm{a}} x^2+\\frac{1}{2 x^3}\\right)^{10}$$ is 105 , then $$\\mathrm{a}^2$$ is equal to :
If the constant term in the expansion of $$\\left(1+2 x-3 x^3\\right)\\left(\\frac{3}{2} x^2-\\frac{1}{3 x}\\right)^9$$ is $$\\mathrm{p}$$, then $$108 \\mathrm{p}$$ is equal to ________.
If the constant term in the expansion of $$\\left(\\frac{\\sqrt[5]{3}}{x}+\\frac{2 x}{\\sqrt[3]{5}}\\right)^{12}, x \\neq 0$$, is $$\\alpha \\times 2^8 \\times \\sqrt[5]{3}$$, then $$25 \\alpha$$ is equal to :
If the second, third and fourth terms in the expansion of $$(x+y)^n$$ are 135, 30 and $$\\frac{10}{3}$$, respectively, then $$6\\left(n^3+x^2+y\\right)$$ is equal to __________.
Here in $${\\left( {\\sqrt 3 } \\right)^{2n - 1}}$$, 2n - 1 is odd number. So there will be always $${\\sqrt 3 }$$ in $${\\left( {\\sqrt 3 } \\right)^{2n - 1}}$$.\n
So $${\\left( {\\sqrt 3 + 1} \\right)^{2n}} - {\\left( {\\sqrt 3 - 1} \\right)^{2n}}$$ will be always irrational number.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4150,
"subject": "General Science",
"question": "If the fractional part of the number $$\\left\\{ {{{{2^{403}}} \\over {15}}} \\right\\} is \\, {k \\over {15}}$$, then k is equal to :\n",
"options": [
{
"text": "8"
},
{
"text": "14"
},
{
"text": "6"
},
{
"text": "1"
}
],
"answer": "8",
"solution": "**Answer:** 8\n\n$${{{2^{403}}} \\over {15}}$$\n
= $$\\left\\{ {{1 \\over 8} + K} \\right\\}$$ where K $$ \\in $$ Integer\n
$$ \\therefore $$ Fractional part = $${{1 \\over 8}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4152,
"subject": "General Science",
"question": "Let $x=(8 \\sqrt{3}+13)^{13}$ and $y=(7 \\sqrt{2}+9)^9$. If $[t]$ denotes the greatest integer $\\leq t$, then :",
"options": [
{
"text": "$[x]$ is odd but $[y]$ is even"
},
{
"text": "$[x]$ and $[y]$ are both odd"
},
{
"text": "$[x]+[y]$ is even"
},
{
"text": "$[x]$ is even but $[y]$ is odd"
}
],
"answer": "$[x]+[y]$ is even",
"solution": "**Answer:** $[x]+[y]$ is even\n\n
If $${I_1} + f = {(8\\sqrt 3 + 13)^{13}},f' = {(8\\sqrt 3 - 13)^{13}}$$
\n
$${I_1} + f - f'=$$ Even
\n
$${I_1} = $$ Even
\n
$${I_2} + f - f' = {(7\\sqrt 2 + 9)^9} + {(7\\sqrt 2 - 9)^9}$$
\n
= Even
\n
$${I_2} = $$ Even
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4153,
"subject": "General Science",
"question": "The coefficient of the middle term in the binomial expansion in powers of $$x$$ of $${\\left( {1 + \\alpha x} \\right)^4}$$ and $${\\left( {1 - \\alpha x} \\right)^6}$$ is the same if $$\\alpha $$ equals ",
"options": [
{
"text": "$${3 \\over 5}$$ "
},
{
"text": "$${10 \\over 3}$$"
},
{
"text": "$${{ - 3} \\over {10}}$$ "
},
{
"text": "$${{ - 5} \\over {3}}$$"
}
],
"answer": "$${{ - 3} \\over {10}}$$ ",
"solution": "**Answer:** $${{ - 3} \\over {10}}$$ \n\nFor $${\\left( {1 + \\alpha x} \\right)^4}$$ the middle term $${T_{{4 \\over 2} + 1}}$$ = $${}^4{C_2}.{\\alpha ^2}{x^2}$$\n
$$\\therefore$$ Coefficient of middle term = $${}^4{C_2}.{\\alpha ^2}$$\n
For $${\\left( {1 - \\alpha x} \\right)^6}$$ the middle term $${T_{{6 \\over 2} + 1}}$$ = $${}^6{C_3}.-{\\alpha ^3}{x^3}$$\n
$$\\therefore$$ Coefficient of middle term = $${}^6{C_3}.{-\\alpha ^3}$$\n
$$ \\Rightarrow \\alpha = - {3 \\over {10}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4154,
"subject": "General Science",
"question": "The sum of the real values of x for which the middle term in the binomial expansion of $${\\left( {{{{x^3}} \\over 3} + {3 \\over x}} \\right)^8}$$ equals 5670 is :",
"options": [
{
"text": "0"
},
{
"text": "8"
},
{
"text": "6"
},
{
"text": "4"
}
],
"answer": "0",
"solution": "**Answer:** 0\n\n$${T_5} = {}^8{C_4}{{{x^{12}}} \\over {81}} \\times {{81} \\over {{x^4}}} = 5670$$\n
$$ \\Rightarrow 70{x^8} = 5670$$\n
$$ \\Rightarrow x = \\pm \\sqrt 3 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4155,
"subject": "General Science",
"question": "The ratio of the coefficient of the middle term in the expansion of (1 + x)20 and the sum of the coefficients of two middle terms in expansion of (1 + x)19 is _____________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nCoeff. of middle term in (1 + x)20 = $${}^{20}{C_{10}}$$ & Sum of coeff. of two middle terms in (1 + x)19 = $${}^{19}{C_{9}}$$ + $${}^{19}{C_{10}}$$
Let the coefficients of the middle terms in the expansion of $$\\left(\\frac{1}{\\sqrt{6}}+\\beta x\\right)^{4},(1-3 \\beta x)^{2}$$ and $$\\left(1-\\frac{\\beta}{2} x\\right)^{6}, \\beta>0$$, respectively form the first three terms of an A.P. If d is the common difference of this A.P. , then $$50-\\frac{2 d}{\\beta^{2}}$$ is equal to __________.
Coefficients of middle terms of given expansions are $${}^4{C_2}{1 \\over 6}{\\beta ^2},\\,{}^2{C_1}( - 3\\beta ),\\,{}^6{C_3}{\\left( {{{ - \\beta } \\over 2}} \\right)^3}$$ form an A.P.
Let K be the sum of the coefficients of the odd powers of $$x$$ in the expansion of $$(1+x)^{99}$$. Let $$a$$ be the middle term in the expansion of $${\\left( {2 + {1 \\over {\\sqrt 2 }}} \\right)^{200}}$$. If $${{{}^{200}{C_{99}}K} \\over a} = {{{2^l}m} \\over n}$$, where m and n are odd numbers, then the ordered pair $$(l,\\mathrm{n})$$ is equal to
Now we are given $p_1 = 20$ and $p_2 = 210$. So,\n$$\n10a^9 b = 20 \\implies a^9 b = 2\n$$
\n
and $$\n45a^8 b^2 + 10a^9 c = 210\n$$
\n
Now, divide the second equation by $a^8$:\n$$\n45b^2 + 10ac = 210\n$$
\n
We know that $a^9 b = 2$. Taking the $9^{th}$ root of both sides:\n$$\nab = \\sqrt[9]{2}\n$$
\n
Now, let $k = ab = \\sqrt[9]{2}$. We can rewrite the equation for $x^2$ term as:\n$$\n45k^2 + 10k^9 = 210\n$$
\n
From the equation $ab = k = \\sqrt[9]{2}$, we know that $a$ and $b$ are positive integers. Thus, $k = 2$ (as both $a$ and $b$ must be factors of 2). Now we have:
\n
$$\na+b = 2\n$$
\n
and from the equation $a^9 b = 2$, we get $a = 1, b = 2$ or vice versa.
\n
Now we need to find the value of $c$. We can use the equation for the $x^2$ term again:
\n
$$\n45a^8 b^2 + 10a^9 c = 210\n$$
\n
Using $a=1$ and $b=2$, we get:
\n
$$\n45(1)^8 (2)^2 + 10(1)^9 c = 210 \\implies 180 + 10c = 210 \\implies c = 3\n$$
\n
So, $a=1$, $b=2$, and $c=3$. Now, we need to find the value of $2(a+b+c)$:
Let $$a$$ be the sum of all coefficients in the expansion of $$\\left(1-2 x+2 x^2\\right)^{2023}\\left(3-4 x^2+2 x^3\\right)^{2024}$$ and $$b=\\lim _\\limits{x \\rightarrow 0}\\left(\\frac{\\int_0^x \\frac{\\log (1+t)}{t^{2024}+1} d t}{x^2}\\right)$$. If the equation $$c x^2+d x+e=0$$ and $$2 b x^2+a x+4=0$$ have a common root, where $$c, d, e \\in \\mathbb{R}$$, then $$\\mathrm{d}: \\mathrm{c}:$$ e equals
\na = +28, b = 315 $$ \\equiv $$ (a, b) $$ \\equiv $$ (28, 315)",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 4188,
"subject": "General Science",
"question": "If some three consecutive in the binomial\nexpansion of (x + 1)n is powers of x are in the ratio\n2 : 15 : 70, then the average of these three\ncoefficient is :-",
"options": [
{
"text": "625"
},
{
"text": "227"
},
{
"text": "964"
},
{
"text": "232"
}
],
"answer": "232",
"solution": "**Answer:** 232\n\nGiven $${}^n{C_{r - 1}}:{}^n{C_r}:{}^n{C_{r + 1}} = 2:15:70$$
= 6C4 = 15",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4195,
"subject": "General Science",
"question": "The sum of the series \n
= 225",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4196,
"subject": "General Science",
"question": "If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of (1 + x)n + 5 are in the ratio 5 : 10 : 14, then the largest coefficient in this expansion is :",
"options": [
{
"text": "330"
},
{
"text": "792 "
},
{
"text": "252"
},
{
"text": "462"
}
],
"answer": "462",
"solution": "**Answer:** 462\n\nConsider the three consecutive coefficients as
Largest coefficient in the expansion is $${^{11}{C_6}}$$ = 462",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4197,
"subject": "General Science",
"question": "The coefficient of x4\n in the expansion of\n (1 + x + x2\n + x3)6\n in powers of x, is ______.",
"options": [],
"answer": "120",
"solution": "**Answer:** 120\n\n(1 + x + x2\n + x3)6\n
[As $${{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}}$$]",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4199,
"subject": "General Science",
"question": "For a positive integer n,\n $${\\left( {1 + {1 \\over x}} \\right)^n}$$ is expanded\n in increasing powers of x. If three consecutive\n coefficients in this expansion are in the ratio,\n 2 : 5 : 12, then n is equal to________.",
"options": [],
"answer": "118",
"solution": "**Answer:** 118\n\nLet, three consecutive coefficients are
$$ \\Rightarrow $$ k = 51",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4201,
"subject": "General Science",
"question": "If the sum of the coefficients of all even powers of x in the product (1 + x + x2\n + ....+ x2n)(1 - x + x2 - x3 + ...... + x2n) is 61, then n is equal to _______.",
"options": [],
"answer": "30",
"solution": "**Answer:** 30\n\n(1 + x + x2\n + ....+ x2n)(1 - x + x2 - x3 + ...... + x2n) \n
$$ \\Rightarrow $$ n + m = 45",
"topic": "Number Theory",
"subtopic": "Divisibility / Factorization"
},
{
"id": 4206,
"subject": "General Science",
"question": "Let [ x ] denote greatest integer less than or equal to x. If for n$$\\in$$N,
Clearly the above inequality is true for n $$ \\ge $$ 5\n\n
For n = 4, we have $${1 \\over 5}\\left[ {4 + {4 \\over {{{10}^2}}}} \\right] = {4 \\over 5}\\left( {{{101} \\over {100}}} \\right) < 1$$\n
$$\\Rightarrow$$ Inequality does not hold good for n = 1, 2, 3, 4\n
So, required number of elements ={5, 6, 7, ......., 100} = 96",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4214,
"subject": "General Science",
"question": "Let n$$\\in$$N and [x] denote the greatest integer less than or equal to x. If the sum of (n + 1) terms $${}^n{C_0},3.{}^n{C_1},5.{}^n{C_2},7.{}^n{C_3},.....$$ is equal to 2100 . 101, then $$2\\left[ {{{n - 1} \\over 2}} \\right]$$ is equal to _______________.",
"options": [],
"answer": "98",
"solution": "**Answer:** 98\n\n1. $${}^n{C_0} + 3.{}^n{C_1} + 5.{}^n{C_2} + ... + (2n + 1).{}^n{C_n}$$
Let n $$\\ge$$ 5 be an integer. If 9n $$-$$ 8n $$-$$ 1 = 64$$\\alpha$$ and 6n $$-$$ 5n $$-$$ 1 = 25$$\\beta$$, then $$\\alpha$$ $$-$$ $$\\beta$$ is equal to
If the sum of the coefficients of all the positive powers of x, in the Binomial expansion of $${\\left( {{x^n} + {2 \\over {{x^5}}}} \\right)^7}$$ is 939, then the sum of all the possible integral values of n is _________.
Let Cr denote the binomial coefficient of xr in the expansion of $${(1 + x)^{10}}$$. If for $$\\alpha$$, $$\\beta$$ $$\\in$$ R, $${C_1} + 3.2{C_2} + 5.3{C_3} + $$ ....... upto 10 terms $$ = {{\\alpha \\times {2^{11}}} \\over {{2^\\beta } - 1}}\\left( {{C_0} + {{{C_1}} \\over 2} + {{{C_2}} \\over 3} + \\,\\,.....\\,\\,upto\\,10\\,terms} \\right)$$ then the value of $$\\alpha$$ + $$\\beta$$ is equal to ___________.
If the coefficients of $$x$$ and $$x^{2}$$ in the expansion of $$(1+x)^{\\mathrm{p}}(1-x)^{\\mathrm{q}}, \\mathrm{p}, \\mathrm{q} \\leq 15$$, are $$-3$$ and $$-5$$ respectively, then the coefficient of $$x^{3}$$ is equal to _____________.
If $$1 + (2 + {}^{49}{C_1} + {}^{49}{C_2} + \\,\\,...\\,\\, + \\,\\,{}^{49}{C_{49}})({}^{50}{C_2} + {}^{50}{C_4} + \\,\\,...\\,\\, + \\,\\,{}^{50}{C_{50}})$$ is equal to $$2^{\\mathrm{n}} \\cdot \\mathrm{m}$$, where $$\\mathrm{m}$$ is odd, then $$\\mathrm{n}+\\mathrm{m}$$ is equal to __________.
If $$a_r$$ is the coefficient of $$x^{10-r}$$ in the Binomial expansion of $$(1 + x)^{10}$$, then $$\\sum\\limits_{r = 1}^{10} {{r^3}{{\\left( {{{{a_r}} \\over {{a_{r - 1}}}}} \\right)}^2}} $$ is equal to
Suppose $$\\sum\\limits_{r = 0}^{2023} {{r^2}{}~^{2023}{C_r} = 2023 \\times \\alpha \\times {2^{2022}}} $$. Then the value of $$\\alpha$$ is ___________
If $$\\frac{1}{n+1}{ }^{n} \\mathrm{C}_{n}+\\frac{1}{n}{ }^{n} \\mathrm{C}_{n-1}+\\ldots+\\frac{1}{2}{ }^{n} \\mathrm{C}_{1}+{ }^{n} \\mathrm{C}_{0}=\\frac{1023}{10}$$ then $$n$$ is equal to :
If the coefficients of three consecutive terms in the expansion of $$(1+x)^{n}$$ are in the ratio $$1: 5: 20$$, then the coefficient of the fourth term is
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4240,
"subject": "General Science",
"question": "If A denotes the sum of all the coefficients in the expansion of $\\left(1-3 x+10 x^2\\right)^{\\mathrm{n}}$ and B denotes the sum of all the coefficients in the expansion of $\\left(1+x^2\\right)^n$, then :",
"options": [
{
"text": "$\\mathrm{B}=\\mathrm{A}^3$"
},
{
"text": "$3 \\mathrm{A}=\\mathrm{B}$"
},
{
"text": "$A=3 B$"
},
{
"text": "$\\mathrm{A}=\\mathrm{B}^3$"
}
],
"answer": "$\\mathrm{A}=\\mathrm{B}^3$",
"solution": "**Answer:** $\\mathrm{A}=\\mathrm{B}^3$\n\n
Sum of coefficients in the expansion of $$\\left(1-3 \\mathrm{x}+10 \\mathrm{x}^2\\right)^{\\mathrm{n}}=\\mathrm{A}$$
\n
then $$A=(1-3+10)^n=8^n$$ (put $$x=1$$)\n
and sum of coefficients in the expansion of
\n
$$\\begin{aligned}\n& \\left(1+x^2\\right)^n=B \\\\\n& \\text { then } B=(1+1)^n=2^n \\\\\n& A=B^3\n\\end{aligned}$$
Let the coefficient of $$x^r$$ in the expansion of $$(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\\ldots \\ldots \\ldots .+(x+2)^{n-1}$$ be $$\\alpha_r$$. If $$\\sum_\\limits{r=0}^n \\alpha_r=\\beta^n-\\gamma^n, \\beta, \\gamma \\in \\mathbb{N}$$, then the value of $$\\beta^2+\\gamma^2$$ equals _________.
In the expansion of $$(1+x)\\left(1-x^2\\right)\\left(1+\\frac{3}{x}+\\frac{3}{x^2}+\\frac{1}{x^3}\\right)^5, x \\neq 0$$, the sum of the coefficients of $x^3$ and $$x^{-13}$$ is equal to __________.
Let $$\\alpha=\\sum_\\limits{k=0}^n\\left(\\frac{\\left({ }^n C_k\\right)^2}{k+1}\\right)$$ and $$\\beta=\\sum_\\limits{k=0}^{n-1}\\left(\\frac{{ }^n C_k{ }^n C_{k+1}}{k+2}\\right)$$ If $$5 \\alpha=6 \\beta$$, then $$n$$ equals _______.
The coefficient of $$x^{70}$$ in $$x^2(1+x)^{98}+x^3(1+x)^{97}+x^4(1+x)^{96}+\\ldots+x^{54}(1+x)^{46}$$ is $${ }^{99} \\mathrm{C}_{\\mathrm{p}}-{ }^{46} \\mathrm{C}_{\\mathrm{q}}$$. Then a possible value of $$\\mathrm{p}+\\mathrm{q}$$ is :
$$ \\Rightarrow \\alpha = - 3,\\beta = - 4,$$ So, $$Q$$ is $$\\left( { - 3, - 4} \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4251,
"subject": "General Science",
"question": "Three distinct points A, B and C are given in the 2 -dimensional coordinates plane such that the ratio of the distance of any one of them from the point $$(1, 0)$$ to the distance from the point $$(-1, 0)$$ is equal to $${1 \\over 3}$$. Then the circumcentre of the triangle ABC is at the point :",
"options": [
{
"text": "$$\\left( {{5 \\over 4},0} \\right)$$ "
},
{
"text": "$$\\left( {{5 \\over 2},0} \\right)$$"
},
{
"text": "$$\\left( {{5 \\over 3},0} \\right)$$"
},
{
"text": "$$\\left( {0,0} \\right)$$ "
}
],
"answer": "$$\\left( {{5 \\over 4},0} \\right)$$ ",
"solution": "**Answer:** $$\\left( {{5 \\over 4},0} \\right)$$ \n\nGiven that \n
$$\\therefore$$ A lies on the circle given by eq. $$(1).$$ As $$B$$ and $$C$$ \n
also follow the same condition, - they must lie on the same circle.\n
$$\\therefore$$ Center of circumcircle of $$\\Delta ABC$$\n
$$=$$ Center of circle given by $$\\left( 1 \\right) = \\left( {{5 \\over 4},0} \\right)$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4252,
"subject": "General Science",
"question": "Locus of the image of the point $$(2, 3)$$ in the line $$\\left( {2x - 3y + 4} \\right) + k\\left( {x - 2y + 3} \\right) = 0,\\,k \\in R,$$ is a :",
"options": [
{
"text": "circle of radius $$\\sqrt 2 $$."
},
{
"text": "circle of radius $$\\sqrt 3 $$."
},
{
"text": "straight line parallel to $$x$$-axis "
},
{
"text": "straight line parallel to $$y$$-axis "
}
],
"answer": "circle of radius $$\\sqrt 2 $$.",
"solution": "**Answer:** circle of radius $$\\sqrt 2 $$.\n\nIntersection point of $$2x - 3y + 4 = 0$$ \n
and $$x-2y+3=0$$ is $$(1, 2)$$ \n
Since, $$P$$ is the fixed point for given family of lines \n
Therefore, given locus is a circle with center $$(1, 2)$$ and radius $$\\sqrt 2 .$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4253,
"subject": "General Science",
"question": "The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60o. If the area of the quadrilateral is $$4\\sqrt 3 $$, then the perimeter\nof the quadrilateral is :\n",
"options": [
{
"text": "12.5 "
},
{
"text": "13.2"
},
{
"text": "12"
},
{
"text": "13"
}
],
"answer": "12",
"solution": "**Answer:** 12\n\n\n
= 2 + 5 + 2 + 3 = 12",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4254,
"subject": "General Science",
"question": "A rectangle is inscribed in a circle with a diameter\nlying along the line 3y = x + 7. If the two adjacent\nvertices of the rectangle are (–8, 5) and (6, 5), then\nthe area of the rectangle (in sq. units) is :",
"options": [
{
"text": "72"
},
{
"text": "84"
},
{
"text": "56"
},
{
"text": "98"
}
],
"answer": "84",
"solution": "**Answer:** 84\n\n\n Distance beetween point A and B is = 14 unit\n
As y coordinate of point A(-8, 5) and B(6, 5) are same. So line AB is parallel to the x axis. \n
Point O is the center of the circle and OP is the perpendicular to the line AB. And P is the mis point of line AB.\n
As OP is perpendicular to the line AB and AB is parallel to the x axis then OP is parallel to the y axis. So x coordinate of line OP is constant.\n
$$ \\therefore $$ x coordinate of O = -1\n
And point O is lies on the line 3y = x + 7\n
$$ \\therefore $$ 3y = -1 + 7\n
$$ \\Rightarrow $$ y = 2\n
$$ \\therefore $$ Center O = (-1, 2)\n
$$ \\therefore $$ OP = 3\n
Then BC = 2OP = 6\n
$$ \\therefore $$ Area of rectangle = (AB)(BC)\n
= 14 $$ \\times $$ 6\n
= 84\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4255,
"subject": "General Science",
"question": "The diameter of the circle, whose centre lies on\nthe line x + y = 2 in the first quadrant and which\ntouches both the lines x = 3 and y = 2, is\n_______ .",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\n\n $$ \\because $$ center lies on x + y = 2 and in 1st quadrant center = ($$\\alpha $$, 2 $$-$$ $$\\alpha $$)
$$ \\Rightarrow $$ 2r = 3 = diameter.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4256,
"subject": "General Science",
"question": "Let PQ be a diameter of the circle x2 + y2 = 9. If $$\\alpha $$ and $$\\beta $$ are the lengths of the perpendiculars from P and Q on the straight line, x + y = 2 respectively, then the maximum value of $$\\alpha\\beta $$ is _____.",
"options": [],
"answer": "7",
"solution": "**Answer:** 7\n\nLet $$P(3\\cos \\theta ,\\,3\\sin \\theta )$$
$$\\alpha {\\beta _{\\max }}$$$$ = {{5 + 9} \\over 2} = 7$$ (when sin2$$\\theta $$ = 1)",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4257,
"subject": "General Science",
"question": "Let a point P be such that its distance from the point (5, 0) is thrice the distance of P from the point ($$-$$5, 0). If the locus of the point P is a circle of radius r, then 4r2 is equal to ________",
"options": [],
"answer": "56",
"solution": "**Answer:** 56\n\nLet P(h, k)
After Round of 4r2 = 56",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4258,
"subject": "General Science",
"question": "Let P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set {P, Q} is equal to :",
"options": [
{
"text": "{(4, 0), (0, 6)}"
},
{
"text": "$$\\{ (2 + 2\\sqrt 2 ,3 - \\sqrt 5 ),(2 - 2\\sqrt 2 ,3 + \\sqrt 5 )\\} $$"
},
{
"text": "$$\\{ (2 + 2\\sqrt 2 ,3 + \\sqrt 5 ),(2 - 2\\sqrt 2 ,3 - \\sqrt 5 )\\} $$"
},
{
"text": "{($$-$$1, 5), (5, 1)}"
}
],
"answer": "{($$-$$1, 5), (5, 1)}",
"solution": "**Answer:** {($$-$$1, 5), (5, 1)}\n\n
$$\\left. \\matrix{\n r - {3 \\over 2} \\ge \\sqrt 5 \\hfill \\cr \n r - {3 \\over 2} \\le - \\sqrt 5 \\hfill \\cr} \\right\\}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4260,
"subject": "General Science",
"question": "The locus of a point, which moves such that the sum of squares of its distances from the points (0, 0), (1, 0), (0, 1), (1, 1) is 18 units, is a circle of diameter d. Then d2 is equal to _____________.",
"options": [],
"answer": "16",
"solution": "**Answer:** 16\n\nLet point P(x, y)\n
$$ \\Rightarrow {d^2} = 16$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4261,
"subject": "General Science",
"question": "Let Z be the set of all integers,
If the total number of relation from A $$\\cap$$ B to A $$\\cap$$ C is 2p, then the value of p is :",
"options": [
{
"text": "16"
},
{
"text": "25"
},
{
"text": "49"
},
{
"text": "9"
}
],
"answer": "25",
"solution": "**Answer:** 25\n\n
Let a triangle ABC be inscribed in the circle $${x^2} - \\sqrt 2 (x + y) + {y^2} = 0$$ such that $$\\angle BAC = {\\pi \\over 2}$$. If the length of side AB is $$\\sqrt 2 $$, then the area of the $$\\Delta$$ABC is equal to :
A rectangle R with end points of one of its sides as (1, 2) and (3, 6) is inscribed in a circle. If the equation of a diameter of the circle is 2x $$-$$ y + 4 = 0, then the area of R is ____________.
\nTotal number of points $=27$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4265,
"subject": "General Science",
"question": "Let $P\\left(a_1, b_1\\right)$ and $Q\\left(a_2, b_2\\right)$ be two distinct points on a circle with center $C(\\sqrt{2}, \\sqrt{3})$. Let $\\mathrm{O}$ be the origin and $\\mathrm{OC}$ be perpendicular to both $\\mathrm{CP}$ and $\\mathrm{CQ}$. If the area of the triangle $\\mathrm{OCP}$ is $\\frac{\\sqrt{35}}{2}$, then $a_1^2+a_2^2+b_1^2+b_2^2$ is equal to :",
"options": [],
"answer": "24",
"solution": "**Answer:** 24\n\n
The points of intersection of the line $$ax + by = 0,(a \\ne b)$$ and the circle $${x^2} + {y^2} - 2x = 0$$ are $$A(\\alpha ,0)$$ and $$B(1,\\beta )$$. The image of the circle with AB as a diameter in the line $$x + y + 2 = 0$$ is :
\nCentre of circle as $A B$ diameter is $\\left(\\frac{1}{2}, \\frac{1}{2}\\right)$ and radius $=\\frac{1}{\\sqrt{2}}$\n
\n$\\therefore$ For image of $\\left(\\frac{1}{2} ; \\frac{1}{2}\\right)$ in $x+y+z$ we get $\\frac{x-\\frac{1}{2}}{1}=\\frac{y-\\frac{1}{2}}{1}=\\frac{-2(3)}{2}$\n
Consider a circle $$C_{1}: x^{2}+y^{2}-4 x-2 y=\\alpha-5$$. Let its mirror image in the line $$y=2 x+1$$ be another circle $$C_{2}: 5 x^{2}+5 y^{2}-10 f x-10 g y+36=0$$. Let $$r$$ be the radius of $$C_{2}$$. Then $$\\alpha+r$$ is equal to _________.
Let the point $$(p, p+1)$$ lie inside the region $$E=\\left\\{(x, y): 3-x \\leq y \\leq \\sqrt{9-x^{2}}, 0 \\leq x \\leq 3\\right\\}$$. If the set of all values of $$\\mathrm{p}$$ is the interval $$(a, b)$$, then $$b^{2}+b-a^{2}$$ is equal to ___________.
Let a circle passing through $$(2,0)$$ have its centre at the point $$(\\mathrm{h}, \\mathrm{k})$$. Let $$(x_{\\mathrm{c}}, y_{\\mathrm{c}})$$ be the point of intersection of the lines $$3 x+5 y=1$$ and $$(2+\\mathrm{c}) x+5 \\mathrm{c}^2 y=1$$. If $$\\mathrm{h}=\\lim _\\limits{\\mathrm{c} \\rightarrow 1} x_{\\mathrm{c}}$$ and $$\\mathrm{k}=\\lim _\\limits{\\mathrm{c} \\rightarrow 1} y_{\\mathrm{c}}$$, then the equation of the circle is :
A square is inscribed in the circle $$x^2+y^2-10 x-6 y+30=0$$. One side of this square is parallel to $$y=x+3$$. If $$\\left(x_i, y_i\\right)$$ are the vertices of the square, then $$\\Sigma\\left(x_i^2+y_i^2\\right)$$ is equal to:
Let ABCD and AEFG be squares of side 4 and 2 units, respectively. The point E is on the line segment AB and the point F is on the diagonal AC. Then the radius r of the circle passing through the point F and touching the line segments BC and CD satisfies :
A circle is inscribed in an equilateral triangle of side of length 12. If the area and perimeter of any square inscribed in this circle are $$m$$ and $$n$$, respectively, then $$m+n^2$$ is equal to
Let a circle C : (x $$-$$ h)2 + (y $$-$$ k)2 = r2, k > 0, touch the x-axis at (1, 0). If the line x + y = 0 intersects the circle C at P and Q such that the length of the chord PQ is 2, then the value of h + k + r is equal to ___________.
Let $$A B$$ be a chord of length 12 of the circle $$(x-2)^{2}+(y+1)^{2}=\\frac{169}{4}$$. If tangents drawn to the circle at points $$A$$ and $$B$$ intersect at the point $$P$$, then five times the distance of point $$P$$ from chord $$A B$$ is equal to __________.
Let the tangents at the points $$A(4,-11)$$ and $$B(8,-5)$$ on the circle $$x^{2}+y^{2}-3 x+10 y-15=0$$, intersect at the point $$C$$. Then the radius of the circle, whose centre is $$C$$ and the line joining $$A$$ and $$B$$ is its tangent, is equal to :
If one of the diameters of the circle $$x^2+y^2-10 x+4 y+13=0$$ is a chord of another circle $$\\mathrm{C}$$, whose center is the point of intersection of the lines $$2 x+3 y=12$$ and $$3 x-2 y=5$$, then the radius of the circle $$\\mathrm{C}$$ is :
$$ \\Rightarrow {x^2} + {y^2} - 2x + 2y - 47 = 0$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4285,
"subject": "General Science",
"question": "Let $$C$$ be the circle with centre $$(0, 0)$$ and radius $$3$$ units. The equation of the locus of the mid points of the chords of the circle $$C$$ that subtend an angle of $${{2\\pi } \\over 3}$$ at its center is :",
"options": [
{
"text": "$${x^2} + {y^2} = {3 \\over 2}$$ "
},
{
"text": "$${x^2} + {y^2} = 1$$ "
},
{
"text": "$${x^2} + {y^2} = {{27} \\over 4}$$ "
},
{
"text": "$${x^2} + {y^2} = {{9} \\over 4}$$"
}
],
"answer": "$${x^2} + {y^2} = {{9} \\over 4}$$",
"solution": "**Answer:** $${x^2} + {y^2} = {{9} \\over 4}$$\n\nLet $$M\\left( {h,k} \\right)$$ be the mid point of chord $$AB$$ where \n
$$\\therefore$$ Locus of $$\\left( {h,k} \\right)$$ is \n
$${x^2} + {y^2} = {9 \\over 4}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4286,
"subject": "General Science",
"question": "If one of the diameters of the circle, given by the equation, $${x^2} + {y^2} - 4x + 6y - 12 = 0,$$ is a chord of a circle $$S$$, whose centre is at $$(-3, 2)$$, then the radius of $$S$$ is :",
"options": [
{
"text": "$$5$$ "
},
{
"text": "$$10$$"
},
{
"text": "$$5\\sqrt 2 $$ "
},
{
"text": "$$5\\sqrt 3 $$"
}
],
"answer": "$$5\\sqrt 3 $$",
"solution": "**Answer:** $$5\\sqrt 3 $$\n\n \n
Center of $$S$$ : $$O(-3, 2)$$ center of given circle $$A(2, -3)$$\n
$$ \\Rightarrow OA = 5\\sqrt 2 $$\n
Also $$AB=5$$ (as $$AB=r$$ of the given circle) \n
$$ \\Rightarrow $$ Using pythagoras theorem in $$\\Delta OAB$$ \n
$$r = 5\\sqrt 3 $$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4287,
"subject": "General Science",
"question": "A circle passes through (−2, 4) and touches the y-axis at (0, 2). Which one of the following equations can represent a diameter of this circle?",
"options": [
{
"text": "4x + 5y − 6 = 0"
},
{
"text": "2x − 3y + 10 = 0"
},
{
"text": "3x + 4y − 3 = 0"
},
{
"text": "5x + 2y + 4 = 0 "
}
],
"answer": "2x − 3y + 10 = 0",
"solution": "**Answer:** 2x − 3y + 10 = 0\n\n\n EF = perpendicular bisector of chord AB\n
BG = perpendicular to y-axis\n
Here C = center of the circle\n
mid-point of chord AB, D = ($$-$$ 1, 3)\n
slope of AB = $${{4 - 2} \\over { - 2 - 0}}$$ = $$-$$ 1\n
$$ \\because $$ EF $$ \\bot $$ AB\n
$$ \\therefore $$ Slope of EF = 1\n
Equation of EF, y $$-$$ 3 = 1 (x + 1)\n
$$ \\Rightarrow $$ y = x + 4 . . . .(i)\n
Equation of BG\n
y = 2 . . . . (ii)\n
From equations (i) and (ii)\n
x = $$-$$ 2, y = 2\n
since C be the point of intersection of EF and BG, therefore center, C = ($$-$$ 2, 2)\n
Now coordinates of center C satiesfy the equation \n
2x $$-$$ 3y + 10 = 0\n
Hence 2x $$-$$ 3y + 10 = 0 is the equation of the diameter",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4288,
"subject": "General Science",
"question": "If two parallel chords of a circle, having diameter 4units, lie on the opposite sides of the center and subtend angles $${\\cos ^{ - 1}}\\left( {{1 \\over 7}} \\right)$$ and sec$$-$$1 (7) at the center respectivey, then the distance between these chords, is : ",
"options": [
{
"text": "$${4 \\over {\\sqrt 7 }}$$ "
},
{
"text": "$${8 \\over {\\sqrt 7 }}$$ "
},
{
"text": "$${8 \\over 7}$$ "
},
{
"text": "$${16 \\over 7}$$"
}
],
"answer": "$${8 \\over {\\sqrt 7 }}$$ ",
"solution": "**Answer:** $${8 \\over {\\sqrt 7 }}$$ \n\n\n
$$ \\Rightarrow $$ r = $${{2\\sqrt 3 } \\over 5}$$\n
$$ \\Rightarrow $$ r2 = $${{12} \\over 5}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4293,
"subject": "General Science",
"question": "If one of the diameters of the circle x2 + y2 - 2x - 6y + 6 = 0 is a chord of another circle 'C',\nwhose center is at (2, 1), then its radius is ________.",
"options": [],
"answer": "3",
"solution": "**Answer:** 3\n\nCircle x2 + y2 - 2x - 6y + 6 = 0 has centre\nO1(1, 3) and radius r\n = 2.\n
Let centre O2\n (2, 1) of required circle and its\nradius being r.\n \n
Distance between (1, 3) and (2, 1) is $$\\sqrt 5 $$
$$ \\Rightarrow r = 3$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4294,
"subject": "General Science",
"question": "If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x2 + y2 = 1 is a circle of radius r, then r is equal to :",
"options": [
{
"text": "$${1 \\over 4}$$"
},
{
"text": "$${1 \\over 2}$$"
},
{
"text": "1"
},
{
"text": "$${1 \\over 3}$$"
}
],
"answer": "$${1 \\over 2}$$",
"solution": "**Answer:** $${1 \\over 2}$$\n\nLet P(h, k) and point on the circle is (cos$$\\theta$$, sin$$\\theta$$)
Radius = $$\\sqrt {{9 \\over 4} + 1 - 3} = {1 \\over 2}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4295,
"subject": "General Science",
"question": "If a line along a chord of the circle 4x2 + 4y2 + 120x + 675 = 0, passes through the point ($$-$$30, 0) and is tangent to the parabola y2 = 30x, then the length of this chord is :",
"options": [
{
"text": "5"
},
{
"text": "7"
},
{
"text": "5$${\\sqrt 3 }$$"
},
{
"text": "3$${\\sqrt 5 }$$"
}
],
"answer": "3$${\\sqrt 5 }$$",
"solution": "**Answer:** 3$${\\sqrt 5 }$$\n\nEquation of tangent to y2 = 30 x
$$\\Rightarrow$$ lAB = 30 . $$\\sqrt {{1 \\over {20}}} $$ = $${{{15} \\over {\\sqrt 5 }}}$$ = 3$${\\sqrt 5 }$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4296,
"subject": "General Science",
"question": "A circle C touches the line x = 2y at the point (2, 1) and intersects the circle
C1 : x2 + y2 + 2y $$-$$ 5 = 0 at two points P and Q such that PQ is a diameter of C1. Then the diameter of C is :",
"options": [
{
"text": "$$7\\sqrt 5 $$"
},
{
"text": "15"
},
{
"text": "$$\\sqrt {285} $$"
},
{
"text": "$$4\\sqrt {15} $$"
}
],
"answer": "$$7\\sqrt 5 $$",
"solution": "**Answer:** $$7\\sqrt 5 $$\n\n(x $$-$$ 2)2 + (y $$-$$ 1)2 + $$\\lambda$$(x $$-$$ 2y) = 0
If one of the diameters of the circle $${x^2} + {y^2} - 2\\sqrt 2 x - 6\\sqrt 2 y + 14 = 0$$ is a chord of the circle $${(x - 2\\sqrt 2 )^2} + {(y - 2\\sqrt 2 )^2} = {r^2}$$, then the value of r2 is equal to ____________.
Let the abscissae of the two points P and Q be the roots of $$2{x^2} - rx + p = 0$$ and the ordinates of P and Q be the roots of $${x^2} - sx - q = 0$$. If the equation of the circle described on PQ as diameter is $$2({x^2} + {y^2}) - 11x - 14y - 22 = 0$$, then $$2r + s - 2q + p$$ is equal to __________.
Let the abscissae of the two points $$P$$ and $$Q$$ on a circle be the roots of $$x^{2}-4 x-6=0$$ and the ordinates of $$\\mathrm{P}$$ and $$\\mathrm{Q}$$ be the roots of $$y^{2}+2 y-7=0$$. If $$\\mathrm{PQ}$$ is a diameter of the circle $$x^{2}+y^{2}+2 a x+2 b y+c=0$$, then the value of $$(a+b-c)$$ is _____________.
Abscissae of PQ are roots of $${x^2} - 4x - 6 = 0$$
\n
Ordinates of PQ are roots of $${y^2} + 2y - 7 = 0$$
\n
and PQ is diameter
\n
$$\\Rightarrow$$ Equation of circle is
\n
$${x^2} + {y^2} - 4x + 2y - 13 = 0$$
\n
But, given $${x^2} + {y^2} + 2ax + 2by + c = 0$$
\n
By comparison $$a = - 2,b = 1,c = - 13$$
\n
$$ \\Rightarrow a + b - c = - 2 + 1 + 13 = 12$$
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4300,
"subject": "General Science",
"question": "The set of all values of $a^{2}$ for which the line $x+y=0$ bisects two distinct chords drawn from a point $\\mathrm{P}\\left(\\frac{1+a}{2}, \\frac{1-a}{2}\\right)$ on the circle $2 x^{2}+2 y^{2}-(1+a) x-(1-a) y=0$, is equal to :",
"options": [
{
"text": "$(0,4]$"
},
{
"text": "$(4, \\infty)$"
},
{
"text": "$(2,12]$"
},
{
"text": "$(8, \\infty)$"
}
],
"answer": "$(8, \\infty)$",
"solution": "**Answer:** $(8, \\infty)$\n\n$x^{2}+y^{2}-\\frac{(1+a) x}{2}-\\frac{(1-a) y}{2}=0$\n\n
Centre $\\left(\\frac{1+\\mathrm{a}}{4}, \\frac{1-\\mathrm{a}}{4}\\right) \\Rightarrow(\\mathrm{h}, \\mathrm{k})$\n\n
The locus of the mid points of the chords of the circle $${C_1}:{(x - 4)^2} + {(y - 5)^2} = 4$$ which subtend an angle $${\\theta _i}$$ at the centre of the circle $$C_1$$, is a circle of radius $$r_i$$. If $${\\theta _1} = {\\pi \\over 3},{\\theta _3} = {{2\\pi } \\over 3}$$ and $$r_1^2 = r_2^2 + r_3^2$$, then $${\\theta _2}$$ is equal to :
A line segment AB of length $$\\lambda$$ moves such that the points A and B remain on the periphery of a circle of radius $$\\lambda$$. Then the locus of the point, that divides the line segment AB in the ratio 2 : 3, is a circle of radius :
So, radius is $\\frac{\\sqrt{19} \\lambda}{5}$.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4303,
"subject": "General Science",
"question": "Let the locus of the midpoints of the chords of the circle $x^2+(y-1)^2=1$ drawn from the origin intersect the line $x+y=1$ at $\\mathrm{P}$ and $\\mathrm{Q}$. Then, the length of $\\mathrm{PQ}$ is :",
"options": [
{
"text": "$\\frac{1}{2}$"
},
{
"text": "1"
},
{
"text": "$\\frac{1}{\\sqrt{2}}$"
},
{
"text": "$\\sqrt{2}$"
}
],
"answer": "$\\frac{1}{\\sqrt{2}}$",
"solution": "**Answer:** $\\frac{1}{\\sqrt{2}}$\n\nLet mid-point is $(x, y)$\n
$$\n\\begin{aligned}\n& x^2+y^2-2 y=0 \\\\\\\\\n& x x_1+y y_1-\\left(y+y_1\\right)=x_1^2+y_1^2-2 y_1\n\\end{aligned}\n$$\n
Let $$\\mathrm{C}$$ be a circle with radius $$\\sqrt{10}$$ units and centre at the origin. Let the line $$x+y=2$$ intersects the circle $$\\mathrm{C}$$ at the points $$\\mathrm{P}$$ and $$\\mathrm{Q}$$. Let $$\\mathrm{MN}$$ be a chord of $$\\mathrm{C}$$ of length 2 unit and slope $$-1$$. Then, a distance (in units) between the chord PQ and the chord $$\\mathrm{MN}$$ is
Let $$C$$ be the centre of the circle $$x^{2}+y^{2}-x+2 y=\\frac{11}{4}$$ and $$P$$ be a point on the circle. A line passes through the point $$\\mathrm{C}$$, makes an angle of $$\\frac{\\pi}{4}$$ with the line $$\\mathrm{CP}$$ and intersects the circle at the points $$Q$$ and $$R$$. Then the area of the triangle $$P Q R$$ (in unit $$^{2}$$ ) is :
Area of $$\\Delta PQR = {1 \\over 2} \\times 4 \\times \\sqrt 2 $$
\n
$$ = 2\\sqrt 2 $$ sq. units
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4306,
"subject": "General Science",
"question": "The centres of a set of circles, each of radius 3, lie on the circle $${x^2}\\, + \\,{y^2} = 25$$. The locus of any point in the set is :",
"options": [
{
"text": "$$4\\, \\le \\,\\,{x^2}\\, + \\,{y^2}\\, \\le \\,\\,64$$ "
},
{
"text": "$${x^2}\\, + \\,{y^2}\\, \\le \\,\\,25$$ "
},
{
"text": "$${x^2}\\, + \\,{y^2}\\, \\ge \\,\\,25$$ "
},
{
"text": "$$3\\, \\le \\,\\,{x^2}\\, + \\,{y^2}\\, \\le \\,\\,9$$ "
}
],
"answer": "$$4\\, \\le \\,\\,{x^2}\\, + \\,{y^2}\\, \\le \\,\\,64$$ ",
"solution": "**Answer:** $$4\\, \\le \\,\\,{x^2}\\, + \\,{y^2}\\, \\le \\,\\,64$$ \n\nFor any point $$P(x,y)$$ in the given circle, \n
$$ \\Rightarrow 4 \\le {x^2} + {y^2} \\le 64$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4307,
"subject": "General Science",
"question": "A variable circle passes through the fixed point A (p, q) and touches x-axis. The locus of the other end of the diameter through A is :",
"options": [
{
"text": "$${(y\\, - \\,q)^2} = \\,4\\,px$$ "
},
{
"text": "$${(x\\, - \\,q)^2} = \\,4\\,py$$"
},
{
"text": "$${(y\\, - \\,p)^2} = \\,4\\,qx$$"
},
{
"text": "$${(x\\, - \\,p)^2} = \\,4\\,qy$$"
}
],
"answer": "$${(x\\, - \\,p)^2} = \\,4\\,qy$$",
"solution": "**Answer:** $${(x\\, - \\,p)^2} = \\,4\\,qy$$\n\nLet the variable circle be \n
$$\\therefore$$ locus of $$\\left( {h,k} \\right)$$ is $${x^2} + {p^2} - 2xp - 4yq = 0$$\n
$$ \\Rightarrow {\\left( {x - p} \\right)^2} = 4qy$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4308,
"subject": "General Science",
"question": "A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the centre of the circle is :",
"options": [
{
"text": "an ellipse"
},
{
"text": "a circle "
},
{
"text": "a hyperbola "
},
{
"text": "a parabola "
}
],
"answer": "a parabola ",
"solution": "**Answer:** a parabola \n\nEquation of circle with center $$(0,3)$$ and radius $$2$$ is \n
$${x^2} + {\\left( {y - 3} \\right)^2} = 4$$\n
Let locus of the variable circle is $$\\left( {\\alpha ,\\beta } \\right)$$\n
$$\\therefore$$ Locus is $${x^2} = 10\\left( {y - {1 \\over 2}} \\right)$$ which is parabola. ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4309,
"subject": "General Science",
"question": "Consider a family of circles which are passing through the point $$(-1, 1)$$ and are tangent to $$x$$-axis. If $$(h, k)$$ are the coordinate of the centre of the circles, then the set of values of $$k$$ is given by the interval :",
"options": [
{
"text": "$$ - {1 \\over 2} \\le k \\le {1 \\over 2}$$ "
},
{
"text": "$$k \\le {1 \\over 2}$$ "
},
{
"text": "$$0 \\le k \\le {1 \\over 2}$$ "
},
{
"text": "$$k \\ge {1 \\over 2}$$ "
}
],
"answer": "$$k \\ge {1 \\over 2}$$ ",
"solution": "**Answer:** $$k \\ge {1 \\over 2}$$ \n\nEquation of circle whose center is $$\\left( {h,k} \\right)$$\n
$$ \\Rightarrow {\\left( {y - 2} \\right)^2}{\\left( {y'} \\right)^2} = 25 - {\\left( {y - 2} \\right)^2}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4311,
"subject": "General Science",
"question": "If $$P$$ and $$Q$$ are the points of intersection of the circles \n $${x^2} + {y^2} + 3x + 7y + 2p - 5 = 0$$ and $${x^2} + {y^2} + 2x + 2y - {p^2} = 0$$ then there is a circle passing through $$P,Q $$ and $$(1, 1)$$ for :",
"options": [
{
"text": "all except one value of $$p$$ "
},
{
"text": "all except two values of $$p$$ "
},
{
"text": "exactly one value of $$p$$ "
},
{
"text": "all values of $$p$$"
}
],
"answer": "all except one value of $$p$$ ",
"solution": "**Answer:** all except one value of $$p$$ \n\nThe given circles are \n
which does not exist for $$p=-1$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4312,
"subject": "General Science",
"question": "The centres of those circles which touch the circle, $${x^2} + {y^2} - 8x - 8y - 4 = 0$$, externally and also touch the $$x$$-axis, lie on :",
"options": [
{
"text": "a circle "
},
{
"text": "an ellipse which is not a circle "
},
{
"text": "a hyperbola"
},
{
"text": "a parabola "
}
],
"answer": "a parabola ",
"solution": "**Answer:** a parabola \n\n\n
which is parabola. ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4313,
"subject": "General Science",
"question": "A circle passes through the points (2, 3) and (4, 5). If its centre lies on the line, $$y - 4x + 3 = 0,$$ then its radius is equal to : ",
"options": [
{
"text": "2"
},
{
"text": "$$\\sqrt 5 $$"
},
{
"text": "$$\\sqrt 2 $$"
},
{
"text": "1"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\nCentre $(\\alpha, \\beta)$ lies on line $y-4 x+3=0$\n
Therefore, $\\quad$ Radius $=\\sqrt{(2-2)^2+(5-3)^2}=2$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4314,
"subject": "General Science",
"question": "Three circles of radii a, b, c (a < b < c) touch each other externally. If they have x-axis as a common tangent, then :",
"options": [
{
"text": "a, b, c are in A.P."
},
{
"text": "$$\\sqrt a ,\\sqrt b ,\\sqrt c $$ are in A.P"
},
{
"text": "$${1 \\over {\\sqrt b }} + {1 \\over {\\sqrt c }}$$ = $${1 \\over {\\sqrt a }}$$"
},
{
"text": "$${1 \\over {\\sqrt b }} = {1 \\over {\\sqrt a }} + {1 \\over {\\sqrt c }}$$"
}
],
"answer": "$${1 \\over {\\sqrt b }} + {1 \\over {\\sqrt c }}$$ = $${1 \\over {\\sqrt a }}$$",
"solution": "**Answer:** $${1 \\over {\\sqrt b }} + {1 \\over {\\sqrt c }}$$ = $${1 \\over {\\sqrt a }}$$\n\n\n
Radius = $$\\sqrt {16 + 25 - 16} = 5$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4317,
"subject": "General Science",
"question": "The circle passing through the intersection of the circles, x2 + y2 – 6x = 0 and x2 + y2 – 4y = 0, having its centre on the line, 2x – 3y + 12 = 0, also passes through the point :\n",
"options": [
{
"text": "(–3, 1)"
},
{
"text": "(1, –3) "
},
{
"text": "(–1, 3) "
},
{
"text": "(–3, 6) "
}
],
"answer": "(–3, 6) ",
"solution": "**Answer:** (–3, 6) \n\nLet S be the circle passing through point of intersection of S1 & S2
Centre of ellipse is midpoint of C1 & C2 is (1, 0)
Equation of ellipse is $${{{{(x - 1)}^2}} \\over {{2^2}}} + {{{y^2}} \\over {{{\\left( {\\sqrt 3 } \\right)}^2}}} = 1$$
Now by cross checking the option $$\\left( {2, \\pm {3 \\over 2}} \\right)$$ satisfied it.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4319,
"subject": "General Science",
"question": "Let r1 and r2 be the radii of the largest and smallest circles, respectively, which pass through the point ($$-$$4, 1) and having their centres on the circumference of the circle x2 + y2 + 2x + 4y $$-$$ 4 = 0. If $${{{r_1}} \\over {{r_2}}} = a + b\\sqrt 2 $$, then a + b is equal to :",
"options": [
{
"text": "3"
},
{
"text": "11"
},
{
"text": "5"
},
{
"text": "7"
}
],
"answer": "5",
"solution": "**Answer:** 5\n\n
so, number of integral values of p2 is 61.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4321,
"subject": "General Science",
"question": "
Let C be a circle passing through the points A(2, $$-$$1) and B(3, 4). The line segment AB s not a diameter of C. If r is the radius of C and its centre lies on the circle $${(x - 5)^2} + {(y - 1)^2} = {{13} \\over 2}$$, then r2 is equal to :
Let a circle $$C_{1}$$ be obtained on rolling the circle $$x^{2}+y^{2}-4 x-6 y+11=0$$ upwards 4 units on the tangent $$\\mathrm{T}$$ to it at the point $$(3,2)$$. Let $$C_{2}$$ be the image of $$C_{1}$$ in $$\\mathrm{T}$$. Let $$A$$ and $$B$$ be the centers of circles $$C_{1}$$ and $$C_{2}$$ respectively, and $$M$$ and $$N$$ be respectively the feet of perpendiculars drawn from $$A$$ and $$B$$ on the $$x$$-axis. Then the area of the trapezium AMNB is :
$$ \\Rightarrow - 25 < m + 10 < 25 \\Rightarrow - 35 < m < 15$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4324,
"subject": "General Science",
"question": "If a circle C, whose radius is 3, touches externally the circle, \n $${x^2} + {y^2} + 2x - 4y - 4 = 0$$ at the point (2, 2), then the length of the intercept cut by this circle C, on the x-axis is equal to : ",
"options": [
{
"text": "$$2\\sqrt 5 $$ "
},
{
"text": "$$3\\sqrt 2 $$"
},
{
"text": "$$\\sqrt 5 $$"
},
{
"text": "$$2\\sqrt 3 $$"
}
],
"answer": "$$2\\sqrt 5 $$ ",
"solution": "**Answer:** $$2\\sqrt 5 $$ \n\nGiven circle is : \n
x2 + y2 + 2x $$-$$ 4y $$-$$4 = 0\n
$$\\therefore\\,\\,\\,$$ its center is ($$-$$ 1, 2) and radius is 3 units.\n
Let A = (x, y) be the center of the circle C\n
$$ \\therefore $$$$\\,\\,\\,$$ $${{x - 1} \\over 2}$$ = 2 $$ \\Rightarrow $$ x = 5 and $${{y + 2} \\over 2}$$ = 2 $$ \\Rightarrow $$ y = 2\n
So the center of C is (5, 2) and its radius is 3\n
$$\\therefore\\,\\,\\,$$ Equation of center C is : \n
x2 + y2 $$-$$ 10x $$-$$ 4y + 20 = 0\n
$$\\therefore\\,\\,\\,$$ The length of the intercept it cuts on the x-axis \n
= 2$$\\sqrt {{g^2} - c} = 2\\sqrt {25 - 20} = 2\\sqrt 5 $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4325,
"subject": "General Science",
"question": "The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is :",
"options": [
{
"text": "$$4\\sqrt 5 $$ "
},
{
"text": "$${{\\sqrt 5 } \\over 2}$$"
},
{
"text": "$$2\\sqrt 5 $$"
},
{
"text": "$${{\\sqrt 5 } \\over 4}$$"
}
],
"answer": "$${{\\sqrt 5 } \\over 2}$$",
"solution": "**Answer:** $${{\\sqrt 5 } \\over 2}$$\n\n\n
= $${{4 + 1} \\over {2\\sqrt 5 }} = {{\\sqrt 5 } \\over 2}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4326,
"subject": "General Science",
"question": "A square is inscribed in the circle x2 + y2\n– 6x + 8y – 103 = 0 with its sides parallel to the coordinate axes.\nThen the distance of the vertex of this square which is nearest to the origin is :",
"options": [
{
"text": "$$\\sqrt {137} $$"
},
{
"text": "6"
},
{
"text": "$$\\sqrt {41} $$"
},
{
"text": "13"
}
],
"answer": "$$\\sqrt {41} $$",
"solution": "**Answer:** $$\\sqrt {41} $$\n\n\n R $$ = \\sqrt {9 + 16 + 103} = 8\\sqrt 2 $$\n
OA $$ = 13$$\n
OB $$ = \\sqrt {265} $$\n
OC $$ = \\sqrt {137} $$\n
OD $$ = \\sqrt {41} $$",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 4327,
"subject": "General Science",
"question": "If a circle of radius R passes through the origin O and intersects the coordinates axes at A and B, then the\nlocus of the foot of perpendicular from O on AB is :",
"options": [
{
"text": "(x2 + y2)2 = 4R2x2y2"
},
{
"text": "(x2 + y2) (x + y) = R2xy"
},
{
"text": "(x2 + y2)2 = 4Rx2y2"
},
{
"text": "(x2 + y2)3 = 4R2x2y2 "
}
],
"answer": "(x2 + y2)3 = 4R2x2y2 ",
"solution": "**Answer:** (x2 + y2)3 = 4R2x2y2 \n\n\n Slope of AB = $${{ - h} \\over k}$$\n
$$ \\Rightarrow $$ (x2 + y2)3 = 4R2x2y2",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4328,
"subject": "General Science",
"question": "The sum of the squares of the lengths of the chords\nintercepted on the circle, x2 + y2 = 16, by the lines,\nx + y = n, n $$ \\in $$ N, where N is the set of all natural\nnumbers, is :",
"options": [
{
"text": "210"
},
{
"text": "160"
},
{
"text": "320"
},
{
"text": "105"
}
],
"answer": "210",
"solution": "**Answer:** 210\n\n\n Let the chord x + y = n cuts the circle x2 +\ny2 = 16 at A and B.\n
$$ \\therefore $$ Possible value of n = 1, 2, 3, 4, 5\n
Length of chord AB = $$2\\sqrt {{{\\left( 4 \\right)}^2} - {{\\left( {{n \\over {\\sqrt 2 }}} \\right)}^2}} $$\n
= $$2\\sqrt {16 - {{{n^2}} \\over 2}} $$\n
= $$\\sqrt {64 - 2{n^2}} $$ = $${l}$$\n
For n = 1, $${l^2}$$ = 62\n
For n = 2, $${l^2}$$ = 56\n
For n = 3, $${l^2}$$ = 46\n
For n = 4, $${l^2}$$ = 32\n
For n = 5, $${l^2}$$ = 14\n
$$ \\therefore $$ Sum of square of length of chords\n= 62 + 56 + 46 + 32 + 14 = 210\n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4329,
"subject": "General Science",
"question": "A circle touching the x-axis at (3, 0) and making an intercept of length 8 on the y-axis passes through the\npoint :",
"options": [
{
"text": "(1, 5)"
},
{
"text": "( 2, 3)"
},
{
"text": "(3, 5)"
},
{
"text": "(3, 10)"
}
],
"answer": "(3, 10)",
"solution": "**Answer:** (3, 10)\n\n
\nFrom the above figure equation of circle is (x - 3)2 + (y - 5)5 = 52
\nSo (3, 10) will satisfy the equation.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4330,
"subject": "General Science",
"question": "Let the lengths of intercepts on x-axis and y-axis made by the circle x2 + y2 + ax + 2ay + c = 0, (a < 0) be 2$${\\sqrt 2 }$$ and 2$${\\sqrt 5 }$$, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to :",
"options": [
{
"text": "$${\\sqrt {10} }$$"
},
{
"text": "$${\\sqrt {6} }$$"
},
{
"text": "$${\\sqrt {11} }$$"
},
{
"text": "$${\\sqrt {7} }$$"
}
],
"answer": "$${\\sqrt {6} }$$",
"solution": "**Answer:** $${\\sqrt {6} }$$\n\n$$2\\sqrt {{{{a^2}} \\over 4} - c} = 2\\sqrt 2 $$
$$\\sqrt {{a^2} - 4c} = 2\\sqrt 2 $$
$${a^2} - 4c = 8$$ .... (1)
$$2\\sqrt {{a^2} - c} = 2\\sqrt 5 $$
$${a^2} - c = 5$$ .... (2)
$$(2) - (1)$$
$$3c = - 3a \\Rightarrow c = - 1$$
$${a^2} = 4 \\Rightarrow a = - 2$$ (Given a < 0)
Equation of circle\n
$${x^2} + {y^2} - 2x - 4y - 1 = 0$$
Equation of tangent which is perpendicular to the line x + 2y = 0 is\n
Distance from origin = $${{\\sqrt {30} } \\over {\\sqrt 5 }} = \\sqrt 6 $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4331,
"subject": "General Science",
"question": "Consider a circle C which touches the y-axis at (0, 6) and cuts off an intercept $$6\\sqrt 5 $$ on the x-axis. Then the radius of the circle C is equal to :",
"options": [
{
"text": "$$\\sqrt {53} $$"
},
{
"text": "9"
},
{
"text": "8"
},
{
"text": "$$\\sqrt {82} $$"
}
],
"answer": "9",
"solution": "**Answer:** 9\n\n
If the circle $$x^{2}+y^{2}-2 g x+6 y-19 c=0, g, c \\in \\mathbb{R}$$ passes through the point $$(6,1)$$ and its centre lies on the line $$x-2 c y=8$$, then the length of intercept made by the circle on $$x$$-axis is :
Two circles in the first quadrant of radii $$r_{1}$$ and $$r_{2}$$ touch the coordinate axes. Each of them cuts off an intercept of 2 units with the line $$x+y=2$$. Then $$r_{1}^{2}+r_{2}^{2}-r_{1} r_{2}$$ is equal to ___________.
",
"options": [],
"answer": "7",
"solution": "**Answer:** 7\n\n$$\n\\begin{aligned}\n& \\text { Circle }(x-a)^2+(y-a)^2=a^2 \\\\\\\\\n& x^2+y^2-2 a x-2 a y+a^2=0 \\\\\\\\\n& \\text { intercept }=2 \\\\\\\\\n& \\Rightarrow 2 \\sqrt{a^2-d^2}=2\n\\end{aligned}\n$$\n
\n
Where $\\mathrm{d}=$ perpendicular distance of centre from line $x+y=2$\n
Let A be the point $$(1,2)$$ and B be any point on the curve $$x^{2}+y^{2}=16$$. If the centre of the locus of the point P, which divides the line segment $$\\mathrm{AB}$$ in the ratio $$3: 2$$ is the point C$$(\\alpha, \\beta)$$, then the length of the line segment $$\\mathrm{AC}$$ is :
A circle passing through the point $$P(\\alpha, \\beta)$$ in the first quadrant touches the two coordinate axes at the points $$A$$ and $$B$$. The point $$P$$ is above the line $$A B$$. The point $$Q$$ on the line segment $$A B$$ is the foot of perpendicular from $$P$$ on $$A B$$. If $$P Q$$ is equal to 11 units, then the value of $$\\alpha \\beta$$ is\n ___________.
",
"options": [],
"answer": "121",
"solution": "**Answer:** 121\n\nLet equation of circle is $(x-a)^2+(y-a)^2=a^2$\n\n
Let a variable line passing through the centre of the circle $$x^2+y^2-16 x-4 y=0$$, meet the positive co-ordinate axes at the points $$A$$ and $$B$$. Then the minimum value of $$O A+O B$$, where $$O$$ is the origin, is equal to
From $$\\left( 1 \\right)$$ and $$\\left( 2 \\right),$$ $$2 < r < 8.$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4338,
"subject": "General Science",
"question": "The two circles x2 + y2 = ax, and x2 + y2 = c2 (c > 0) touch each other if :",
"options": [
{
"text": "| a | = c"
},
{
"text": "a = 2c"
},
{
"text": "| a | = 2c"
},
{
"text": "2 | a | = c"
}
],
"answer": "| a | = c",
"solution": "**Answer:** | a | = c\n\nAs center of one circle is $$\\left( {0,0} \\right)$$ and other circle passes through $$(0,0),$$ therefore\n
Also $${C_1}\\left( {{a \\over 2},0} \\right){C_2}\\left( {0,0} \\right)$$\n
If the two circles touch each other, then they must touch each other internally. ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4339,
"subject": "General Science",
"question": "Let $$C$$ be the circle with centre at $$(1, 1)$$ and radius $$=$$ $$1$$. If $$T$$ is the circle centred at $$(0, y)$$, passing through origin and touching the circle $$C$$ externally, then the radius of $$T$$ is equal to :",
"options": [
{
"text": "$${1 \\over 2}$$ "
},
{
"text": "$${1 \\over 4}$$"
},
{
"text": "$${{\\sqrt 3 } \\over {\\sqrt 2 }}$$ "
},
{
"text": "$${{\\sqrt 3 } \\over 2}$$ "
}
],
"answer": "$${1 \\over 4}$$",
"solution": "**Answer:** $${1 \\over 4}$$\n\n\n
Equation of circle $$C \\equiv {\\left( {x - 1} \\right)^2} + {\\left( {y - 1} \\right)^2} = 1$$\n
Radius of $$T = \\left| y \\right|$$ \n
$$T$$ touches $$C$$ externally\n
therefore,\n
Distance between the centers $$=$$ sum of their radii\n
\n Therefore there are three common tangents.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4341,
"subject": "General Science",
"question": "If the circles x2\n + y2\n + 5Kx + 2y + K = 0 and 2(x2\n + y2) + 2Kx + 3y –1 = 0, (K$$ \\in $$R), intersect at the points\nP and Q, then the line 4x + 5y – K = 0 passes through P and Q, for :",
"options": [
{
"text": "exactly two values of K"
},
{
"text": "no value of K"
},
{
"text": "exactly one value of K"
},
{
"text": "infinitely many values of K"
}
],
"answer": "no value of K",
"solution": "**Answer:** no value of K\n\nS1 $$ \\equiv $$ x2 + y2 + 5Kx + 2y + K = 0
so $$\\lambda \\in \\left[ {12,21} \\right]$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4345,
"subject": "General Science",
"question": "Two circles with equal radii are intersecting at the points (0, 1) and (0, –1). The tangent at the point (0, 1) to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is :",
"options": [
{
"text": "$$2\\sqrt 2 $$"
},
{
"text": "$$\\sqrt 2 $$"
},
{
"text": "2"
},
{
"text": "1"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\n\n
$$ \\Rightarrow $$ k $$ \\in $$ {7, 8, 9, ……15} since k $$ \\in $$ I\n
$$ \\therefore $$ Total 9 integral value of k.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4348,
"subject": "General Science",
"question": "Choose the incorrect statement about the two circles whose equations are given below :
x2 + y2 $$-$$ 10x $$-$$ 10y + 41 = 0 and
x2 + y2 $$-$$ 16x $$-$$ 10y + 80 = 0",
"options": [
{
"text": "Distance between two centres is the average of radii of both the circles."
},
{
"text": "Both circles pass through the centre of each other."
},
{
"text": "Circles have two intersection points."
},
{
"text": "Both circle's centers lie inside region of one another."
}
],
"answer": "Both circle's centers lie inside region of one another.",
"solution": "**Answer:** Both circle's centers lie inside region of one another.\n\nS1 $$ \\equiv $$ x2 + y2 $$-$$ 10x $$-$$ 10y + 41 = 0
Hence both circles pass through the centre of each other, have two intersection point and distance between two centres in average of radii of both the circles.
\nHence, option (d) is the incorrect statement.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4349,
"subject": "General Science",
"question": "The minimum distance between any two points P1 and P2 while considering point P1 on one circle and point P2 on the other circle for the given circles' equations
So (P1P2)min = 1",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4350,
"subject": "General Science",
"question": "Choose the correct statement about two circles whose equations are given below :
x2 + y2 $$-$$ 10x $$-$$ 10y + 41 = 0
x2 + y2 $$-$$ 22x $$-$$ 10y + 137 = 0",
"options": [
{
"text": "circles have same centre"
},
{
"text": "circles have no meeting point"
},
{
"text": "circles have only one meeting point"
},
{
"text": "circles have two meeting points"
}
],
"answer": "circles have only one meeting point",
"solution": "**Answer:** circles have only one meeting point\n\nLet $${S_1}:{x^2} + {y^2} - 10x - 10y + 41 = 0$$
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4352,
"subject": "General Science",
"question": "The number of common tangents, to the circles
Let the centre of a circle, passing through the points $$(0,0),(1,0)$$ and touching the circle $$x^2+y^2=9$$, be $$(h, k)$$. Then for all possible values of the coordinates of the centre $$(h, k), 4\\left(h^2+k^2\\right)$$ is equal to __________.
Let the circles $$C_1:(x-\\alpha)^2+(y-\\beta)^2=r_1^2$$ and $$C_2:(x-8)^2+\\left(y-\\frac{15}{2}\\right)^2=r_2^2$$ touch each other externally at the point $$(6,6)$$. If the point $$(6,6)$$ divides the line segment joining the centres of the circles $$C_1$$ and $$C_2$$ internally in the ratio $$2: 1$$, then $$(\\alpha+\\beta)+4\\left(r_1^2+r_2^2\\right)$$ equals
$$ \\therefore $$ 1 + $$\\sqrt k $$ = 5 or $$\\sqrt k $$ - 1 = 5\n
$$ \\Rightarrow $$ k = 16 or k = 36\n
So largest value of k = 36.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4359,
"subject": "General Science",
"question": "
Let the mirror image of a circle $$c_{1}: x^{2}+y^{2}-2 x-6 y+\\alpha=0$$ in line $$y=x+1$$ be $$c_{2}: 5 x^{2}+5 y^{2}+10 g x+10 f y+38=0$$. If $$\\mathrm{r}$$ is the radius of circle $$\\mathrm{c}_{2}$$, then $$\\alpha+6 \\mathrm{r}^{2}$$ is equal to ________.
$$\\left( {{1 \\over 2}.\\sqrt 2 } \\right)$$ or $$\\left( {{1 \\over 2}, - \\sqrt 2 } \\right)$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4361,
"subject": "General Science",
"question": "Two circles each of radius 5 units touch each other at the point (1, 2). If the equation of their common tangent is 4x + 3y = 10, and C1($$\\alpha$$, $$\\beta$$) and C2($$\\gamma$$, $$\\delta$$), C1 $$\\ne$$ C2 are their centres, then |($$\\alpha$$ + $$\\beta$$) ($$\\gamma$$ + $$\\delta$$)| is equal to ___________.",
"options": [],
"answer": "40",
"solution": "**Answer:** 40\n\nSlope of line joining centres of circles = $${4 \\over 3} = \\tan \\theta $$
If a line L : y = mx + c be a common tangent to C1, C2 and C3 such that C1 and C3 lie on one side of line L while C2 lies on other side, then the value of $$20({r^2} + c)$$ is equal to :
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4363,
"subject": "General Science",
"question": "The length of the diameter of the circle which touches the $$x$$-axis at the point $$(1, 0)$$ and passes through the point $$(2, 3)$$ is :",
"options": [
{
"text": "$${{10} \\over 3}$$ "
},
{
"text": "$${{3} \\over 5}$$"
},
{
"text": "$${{6} \\over 5}$$"
},
{
"text": "$${{5} \\over 3}$$"
}
],
"answer": "$${{10} \\over 3}$$ ",
"solution": "**Answer:** $${{10} \\over 3}$$ \n\nLet center of the circle be $$\\left( {1,h} \\right)$$\n
$$\\left[ {\\,\\,} \\right.$$ as circle touches $$x$$-axis at $$\\left. {\\left( {1,0} \\right)\\,\\,} \\right]$$\n
\n
Let the circle passes through the point $$B(2,3)$$\n
= 14 + 5$$\\sqrt {3/2} $$ $$ \\times $$ $$\\sqrt 2 $$ = 14 + 5$$\\sqrt 3 $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4365,
"subject": "General Science",
"question": "Let A(1, 4) and B(1, $$-$$5) be two points. Let P be a point on the circle (x $$-$$ 1)2 + (y $$-$$ 1)2 = 1 such that (PA)2 + (PB)2 have maximum value, then the points, P, A and B lie on :",
"options": [
{
"text": "a straight line"
},
{
"text": "an ellipse"
},
{
"text": "a parabola"
},
{
"text": "a hyperbola"
}
],
"answer": "a straight line",
"solution": "**Answer:** a straight line\n\nP be a point on $${(x - 1)^2} + {(y - 1)^2} = 1$$
P, A, B are collinear points.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4366,
"subject": "General Science",
"question": "For the four circles M, N, O and P, following four equations are given :
Circle M : x2 + y2 = 1
Circle N : x2 + y2 $$-$$ 2x = 0
Circle O : x2 + y2 $$-$$ 2x $$-$$ 2y + 1 = 0
Circle P : x2 + y2 $$-$$ 2y = 0
If the centre of circle M is joined with centre of the circle N, further center of circle N is joined with centre of the circle O, centre of circle O is joined with the centre of circle P and lastly, centre of circle P is joined with centre of circle M, then these lines form the sides of a :",
"options": [
{
"text": "Rhombus"
},
{
"text": "Square"
},
{
"text": "Rectangle"
},
{
"text": "Parallelogram"
}
],
"answer": "Square",
"solution": "**Answer:** Square\n\n$${C_M} = (0,0)$$
$${C_N} = (1,0)$$
$${C_O} = (1,1)$$
$${C_P} = (0,1)$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4367,
"subject": "General Science",
"question": "Let the circle S : 36x2 + 36y2 $$-$$ 108x + 120y + C = 0 be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, x $$-$$ 2y = 4 and 2x $$-$$ y = 5 lies inside the circle S, then :",
"options": [
{
"text": "$${{25} \\over 9} < C < {{13} \\over 3}$$"
},
{
"text": "100 < C < 165"
},
{
"text": "81 < C < 156"
},
{
"text": "100 < C < 156"
}
],
"answer": "100 < C < 156",
"solution": "**Answer:** 100 < C < 156\n\nS : 36x2 + 36y2 $$-$$ 108x + 120y + C = 0
100 < C < 156 Ans.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4368,
"subject": "General Science",
"question": "If the variable line 3x + 4y = $$\\alpha$$ lies between the two circles (x $$-$$ 1)2 + (y $$-$$ 1)2 = 1 and (x $$-$$ 9)2 + (y $$-$$ 1)2 = 4, without intercepting a chord on either circle, then the sum of all the integral values of $$\\alpha$$ is ___________.",
"options": [],
"answer": "165",
"solution": "**Answer:** 165\n\n
Both centers should lie on either side of the line as well as line can be tangent to circle.
Sum of integers = 165",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4369,
"subject": "General Science",
"question": "
The set of values of k, for which the circle $$C:4{x^2} + 4{y^2} - 12x + 8y + k = 0$$ lies inside the fourth quadrant and the point $$\\left( {1, - {1 \\over 3}} \\right)$$ lies on or inside the circle C, is :
For $$\\mathrm{t} \\in(0,2 \\pi)$$, if $$\\mathrm{ABC}$$ is an equilateral triangle with vertices $$\\mathrm{A}(\\sin t,-\\cos \\mathrm{t}), \\mathrm{B}(\\operatorname{cost}, \\sin t)$$ and $$C(a, b)$$ such that its orthocentre lies on a circle with centre $$\\left(1, \\frac{1}{3}\\right)$$, then $$\\left(a^{2}-b^{2}\\right)$$ is equal to :
Equations of two diameters of a circle are $$2 x-3 y=5$$ and $$3 x-4 y=7$$. The line joining the points $$\\left(-\\frac{22}{7},-4\\right)$$ and $$\\left(-\\frac{1}{7}, 3\\right)$$ intersects the circle at only one point $$P(\\alpha, \\beta)$$. Then, $$17 \\beta-\\alpha$$ is equal to _________.
If $$\\mathrm{P}(6,1)$$ be the orthocentre of the triangle whose vertices are $$\\mathrm{A}(5,-2), \\mathrm{B}(8,3)$$ and $$\\mathrm{C}(\\mathrm{h}, \\mathrm{k})$$, then the point $$\\mathrm{C}$$ lies on the circle :
Slope of $$B F=\\left(m_{B F}\\right)=\\frac{3-1}{8-6}=1$$
\n
$$\\begin{aligned}\n& m_{A C}=\\frac{k+2}{h-5}=-1(A C \\perp B F) \\\\\n& \\Rightarrow k+2=5-h \\\\\n& \\Rightarrow k=3-h \\quad \\ldots \\text { (i) } \\\\\n& m_{A D}=\\frac{1+2}{6-5}=3 \\\\\n& m_{B C}=\\frac{k-3}{h-8}=-\\frac{1}{3}(A D \\perp B C) \\\\\n& \\Rightarrow 3 k-9=8-h \\quad \\ldots \\text { (ii) }\n\\end{aligned}$$
\n
From (i) & (ii)
\n
$$h=-4, k=7$$
\n
Now, $$h^2+k^2=16+49=65$$
\n
$$h^2+k^2-65=0$$
\n
Locus is $$x^2+y^2-65=0$$
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4374,
"subject": "General Science",
"question": "If the circles $${x^2}\\, + \\,{y^2} + \\,2ax\\, + \\,cy\\, + a\\,\\, = 0$$ and $${x^2}\\, + \\,{y^2} - \\,3ax\\, + \\,dy\\, - 1\\,\\, = 0$$ intersect in two ditinct points P and Q then the line 5x + by - a = 0 passes through P and Q for :",
"options": [
{
"text": "exactly one value of a "
},
{
"text": "no value of a"
},
{
"text": "infinitely many values of a "
},
{
"text": "exactly two values of a "
}
],
"answer": "no value of a",
"solution": "**Answer:** no value of a\n\n$${s_1} = {x^2} + {y^2} + 2ax + cy + a = 0$$ \n
$${s_2} = {x^2} + {y^2} - 3ax + dy - 1 = 0$$\n
Equation of common chord of circles $${s_1}$$ and $${s_2}$$ is \n
No real value of $$a.$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4375,
"subject": "General Science",
"question": "
Consider two circles $$C_1: x^2+y^2=25$$ and $$C_2:(x-\\alpha)^2+y^2=16$$, where $$\\alpha \\in(5,9)$$. Let the angle between the two radii (one to each circle) drawn from one of the intersection points of $$C_1$$ and $$C_2$$ be $$\\sin ^{-1}\\left(\\frac{\\sqrt{63}}{8}\\right)$$. If the length of common chord of $$C_1$$ and $$C_2$$ is $$\\beta$$, then the value of $$(\\alpha \\beta)^2$$ equals _______.
Let the circle $$C_1: x^2+y^2-2(x+y)+1=0$$ and $$\\mathrm{C_2}$$ be a circle having centre at $$(-1,0)$$ and radius 2 . If the line of the common chord of $$\\mathrm{C}_1$$ and $$\\mathrm{C}_2$$ intersects the $$\\mathrm{y}$$-axis at the point $$\\mathrm{P}$$, then the square of the distance of P from the centre of $$\\mathrm{C_1}$$ is:
$$\\therefore$$ equation of circle is $${\\left( {x - 3} \\right)^2} + {y^2} - 8 = 0$$ \n
Now, from the options $$\\left( {5, - 2} \\right)$$ satisfies equation of circle. ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4378,
"subject": "General Science",
"question": "Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x − y = 1 and 2x + y = 3 is : ",
"options": [
{
"text": "4x + y − 3 = 0"
},
{
"text": "x + 4y + 3 = 0"
},
{
"text": "3x − y − 4 = 0"
},
{
"text": "x − 3y − 4 = 0"
}
],
"answer": "x + 4y + 3 = 0",
"solution": "**Answer:** x + 4y + 3 = 0\n\nPoint of intersection of lines \n
x $$-$$ y = 1 and 2x + y = 3 is $$\\left( {{4 \\over 3},{1 \\over 3}} \\right)$$\n
$$ \\Rightarrow c = 95$$ ",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4381,
"subject": "General Science",
"question": "The tangent to the circle C1 : x2 + y2 $$-$$ 2x $$-$$ 1 = 0 at the point (2, 1) cuts off a chord of length 4 from a circle C2 whose center is (3, $$-$$2). The radius of C2 is :",
"options": [
{
"text": "2"
},
{
"text": "$$\\sqrt 2 $$"
},
{
"text": "3"
},
{
"text": "$$\\sqrt 6 $$"
}
],
"answer": "$$\\sqrt 6 $$",
"solution": "**Answer:** $$\\sqrt 6 $$\n\nHere, equation of tangent on C1 at (2, 1) is : \n
2x + y $$-$$ (x + 2) $$-$$1 = 0\n
Or x + y = 3\n
If it cuts off the chord of the circle C2 then the equation of the chord is : \n x + y = 3\n
$$\\therefore\\,\\,\\,$$ distance of the chord from (3, $$-$$ 2) is :\n
x2 + y2 – 4x2y2 = 0",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4384,
"subject": "General Science",
"question": "The line x = y touches a circle at the point (1,1). If the circle also passes through the point (1, – 3), then its\nradius is :",
"options": [
{
"text": "3"
},
{
"text": "2"
},
{
"text": "2$$\\sqrt 2 $$"
},
{
"text": "3$$\\sqrt 2 $$"
}
],
"answer": "2$$\\sqrt 2 $$",
"solution": "**Answer:** 2$$\\sqrt 2 $$\n\nEquation of circle = (x – 1)2 + (y –1)2 + $$\\lambda $$(y – x) = 0
\nWhich passes through (1, –3)
\n
\nSo, 0 + 16 + $$\\lambda $$(–3 – 1) = 0
\n16 + $$\\lambda $$(–4) = 0
\n$$ \\therefore $$ $$\\lambda $$ = 4
\nNow equation of circle
\n(x – 1)2 + (y – 1)2 + 4y – 4x = 0
\n$$ \\Rightarrow $$ x2 + y2 – 6x + 2y + 2 = 0
\nradius = $$\\sqrt {9 + 1 - 2} = 2\\sqrt 2 $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4385,
"subject": "General Science",
"question": "Let the tangents drawn from the origin to the circle, x2\n + y2\n - 8x - 4y + 16 = 0 touch it at the\npoints A and B. The (AB)2\n is equal to :",
"options": [
{
"text": "$${{56} \\over 5}$$"
},
{
"text": "$${{32} \\over 5}$$"
},
{
"text": "$${{52} \\over 5}$$"
},
{
"text": "$${{64} \\over 5}$$"
}
],
"answer": "$${{64} \\over 5}$$",
"solution": "**Answer:** $${{64} \\over 5}$$\n\nEquation of chord of contact is\n \n x.0 + y.0 – 4(x + 0) –2(y + 0) + 16 = 0\n
$$ \\Rightarrow $$ c2\n + 6c + 7 = 0 \n",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4387,
"subject": "General Science",
"question": "A circle touches the y-axis at the point (0, 4)\nand passes through the point (2, 0). Which of\nthe following lines is not a tangent to this circle?",
"options": [
{
"text": "3x – 4y – 24 = 0"
},
{
"text": "4x + 3y – 8 = 0"
},
{
"text": "3x + 4y – 6 = 0"
},
{
"text": "4x – 3y + 17 = 0"
}
],
"answer": "4x + 3y – 8 = 0",
"solution": "**Answer:** 4x + 3y – 8 = 0\n\n\nEquation of family of circle touching y-axis at\n
(0, 4) is given by (x – 0)2 + (y – 4)2 + $$\\lambda $$x = 0.\n
$$ \\because $$ It passes through (2, 0)\n
$$ \\Rightarrow $$ $$\\lambda $$ = –10.\n
$$ \\Rightarrow $$ Required circle is (x – 0)2 + (y – 4)2 – 10x = 0\n
$$ \\Rightarrow $$ x2 + y2 – 10x – 8y + 16 = 0\n
$$ \\therefore $$ center of circle (5, 4) and radius = 5\n
By checking all options you can see 4x + 3y – 8 = 0 is not a tangent to the circle.\n
As distance of 4x + 3y – 8 = 0 from (5, 4) \n
= $$\\left| {{{24} \\over 5}} \\right|$$ $$ \\ne $$ radius",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4388,
"subject": "General Science",
"question": "If the area of the triangle formed by the positive x-axis, the normal and the tangent to the circle (x $$-$$ 2)2 + (y $$-$$ 3)2 = 25 at the point (5, 7) is A, then 24A is equal to _________.",
"options": [],
"answer": "1225",
"solution": "**Answer:** 1225\n\nThis question is bonus if we consider poistive x axis.If we consider only x axis for this question then it is right question.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4389,
"subject": "General Science",
"question": "The line 2x $$-$$ y + 1 = 0 is a tangent to the circle at the point (2, 5) and the centre of the circle lies on x $$-$$ 2y = 4. Then, the radius of the circle is :",
"options": [
{
"text": "5$$\\sqrt 3 $$"
},
{
"text": "4$$\\sqrt 5 $$"
},
{
"text": "3$$\\sqrt 5 $$"
},
{
"text": "5$$\\sqrt 4 $$"
}
],
"answer": "3$$\\sqrt 5 $$",
"solution": "**Answer:** 3$$\\sqrt 5 $$\n\n\n $${m_1} \\times {m_2} = - 1$$
$$ = 3\\sqrt 5 $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4390,
"subject": "General Science",
"question": "Let the tangent to the circle x2 + y2 = 25 at the point R(3, 4) meet x-axis and y-axis at points P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then r2 is equal to :",
"options": [
{
"text": "$${{585} \\over {66}}$$"
},
{
"text": "$${{625} \\over {72}}$$"
},
{
"text": "$${{529} \\over {64}}$$"
},
{
"text": "$${{125} \\over {72}}$$"
}
],
"answer": "$${{625} \\over {72}}$$",
"solution": "**Answer:** $${{625} \\over {72}}$$\n\nGiven equation of circle
Therefore, the correct answer is (B).",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4391,
"subject": "General Science",
"question": "Two tangents are drawn from a point P to the circle x2 + y2 $$-$$ 2x $$-$$ 4y + 4 = 0, such that the angle between these tangents is $${\\tan ^{ - 1}}\\left( {{{12} \\over 5}} \\right)$$, where $${\\tan ^{ - 1}}\\left( {{{12} \\over 5}} \\right)$$ $$\\in$$(0, $$\\pi$$). If the centre of the circle is denoted by C and these tangents touch the circle at points A and B, then the ratio of the areas of $$\\Delta$$PAB and $$\\Delta$$CAB is :",
"options": [
{
"text": "3 : 1"
},
{
"text": "9 : 4"
},
{
"text": "2 : 1"
},
{
"text": "11 : 4"
}
],
"answer": "9 : 4",
"solution": "**Answer:** 9 : 4\n\n\n Let $$\\theta$$ = tan$$-$$1$$\\left( {{{12} \\over 5}} \\right)$$
Therefore, the correct answer is (2).",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4392,
"subject": "General Science",
"question": "Two tangents are drawn from the point P($$-$$1, 1) to the circle x2 + y2 $$-$$ 2x $$-$$ 6y + 6 = 0. If these tangents touch the circle at points A and B, and if D is a point on the circle such that length of the segments AB and AD are equal, then the area of the triangle ABD is equal to :",
"options": [
{
"text": "2"
},
{
"text": "$$(3\\sqrt 2 + 2)$$"
},
{
"text": "4"
},
{
"text": "$$3(\\sqrt 2 - 1)$$"
}
],
"answer": "4",
"solution": "**Answer:** 4\n\n
$$\\Delta ABD = {1 \\over 2} \\times 2 \\times 4 = 4$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4393,
"subject": "General Science",
"question": "Let B be the centre of the circle x2 + y2 $$-$$ 2x + 4y + 1 = 0. Let the tangents at two points P and Q on the circle intersect at the point A(3, 1). Then 8.$$\\left( {{{area\\,\\Delta APQ} \\over {area\\,\\Delta BPQ}}} \\right)$$ is equal to _____________.",
"options": [],
"answer": "18",
"solution": "**Answer:** 18\n\n
\n
Radius = $$\\sqrt {1 + 4 - 1} = 2$$
\n
$$AB = \\sqrt {{3^2} + {2^2}} = \\sqrt {13} $$
\n
In $$\\Delta$$ABP
\n
$$A{P^2} = A{B^2} - B{P^2} = 13 - 4 = 9$$
\n
AP = 3
\n
AQ = AP = 3
\n
Let $$\\angle$$ABP = $$\\theta$$, $$\\angle$$BAP = 90$$-$$ $$\\theta$$
Let the tangent to the circle C1 : x2 + y2 = 2 at the point M($$-$$1, 1) intersect the circle C2 : (x $$-$$ 3)2 + (y $$-$$ 2)2 = 5, at two distinct points A and B. If the tangents to C2 at the points A and B intersect at N, then the area of the triangle ANB is equal to :
If the tangents drawn at the points $$O(0,0)$$ and $$P\\left( {1 + \\sqrt 5 ,2} \\right)$$ on the circle $${x^2} + {y^2} - 2x - 4y = 0$$ intersect at the point Q, then the area of the triangle OPQ is equal to :
Let the lines $$y + 2x = \\sqrt {11} + 7\\sqrt 7 $$ and $$2y + x = 2\\sqrt {11} + 6\\sqrt 7 $$ be normal to a circle $$C:{(x - h)^2} + {(y - k)^2} = {r^2}$$. If the line $$\\sqrt {11} y - 3x = {{5\\sqrt {77} } \\over 3} + 11$$ is tangent to the circle C, then the value of $${(5h - 8k)^2} + 5{r^2}$$ is equal to __________.
Let a circle C of radius 5 lie below the x-axis. The line L1 : 4x + 3y + 2 = 0 passes through the centre P of the circle C and intersects the line L2 = 3x $$-$$ 4y $$-$$ 11 = 0 at Q. The line L2 touches C at the point Q. Then the distance of P from the line 5x $$-$$ 12y + 51 = 0 is ______________.
A circle $$C_{1}$$ passes through the origin $$\\mathrm{O}$$ and has diameter 4 on the positive $$x$$-axis. The line $$y=2 x$$ gives a chord $$\\mathrm{OA}$$ of circle $$\\mathrm{C}_{1}$$. Let $$\\mathrm{C}_{2}$$ be the circle with $$\\mathrm{OA}$$ as a diameter. If the tangent to $$\\mathrm{C}_{2}$$ at the point $$\\mathrm{A}$$ meets the $$x$$-axis at $$\\mathrm{P}$$ and $$y$$-axis at $$\\mathrm{Q}$$, then $$\\mathrm{QA}: \\mathrm{AP}$$ is equal to :
Let the tangents at two points $$\\mathrm{A}$$ and $$\\mathrm{B}$$ on the circle $$x^{2}+\\mathrm{y}^{2}-4 x+3=0$$ meet at origin $$\\mathrm{O}(0,0)$$. Then the area of the triangle $$\\mathrm{OAB}$$ is :
A circle with centre (2, 3) and radius 4 intersects the line $$x+y=3$$ at the points P and Q. If the tangents at P and Q intersect at the point $$S(\\alpha,\\beta)$$, then $$4\\alpha-7\\beta$$ is equal to ___________.
Points P($$-$$3, 2), Q(9, 10) and R($$\\alpha,4$$) lie on a circle C and PR as its diameter. The tangents to C at the points Q and R intersect at the point S. If S lies on the line $$2x-ky=1$$, then k is equal to ____________.
Let the centre of a circle C be $$(\\alpha, \\beta)$$ and its radius $$r < 8$$. Let $$3 x+4 y=24$$ and $$3 x-4 y=32$$ be two tangents and $$4 x+3 y=1$$ be a normal to C. Then $$(\\alpha-\\beta+r)$$ is equal to :
Let O be the origin and OP and OQ be the tangents to the circle $$x^2+y^2-6x+4y+8=0$$ at the points P and Q on it. If the circumcircle of the triangle OPQ passes through the point $$\\left( {\\alpha ,{1 \\over 2}} \\right)$$, then a value of $$\\alpha$$ is :
If the tangents at the points $$\\mathrm{P}$$ and $$\\mathrm{Q}$$ on the circle $$x^{2}+y^{2}-2 x+y=5$$ meet at the point $$R\\left(\\frac{9}{4}, 2\\right)$$, then the area of the triangle $$\\mathrm{PQR}$$ is :
Consider a circle $$(x-\\alpha)^2+(y-\\beta)^2=50$$, where $$\\alpha, \\beta>0$$. If the circle touches the line $$y+x=0$$ at the point $$P$$, whose distance from the origin is $$4 \\sqrt{2}$$, then $$(\\alpha+\\beta)^2$$ is equal to __________.
Let a circle C of radius 1 and closer to the origin be such that the lines passing through the point $$(3,2)$$ and parallel to the coordinate axes touch it. Then the shortest distance of the circle C from the point $$(5,5)$$ is :
$$ \\Rightarrow {\\left| z \\right|_{\\max }} = \\sqrt 5 + 1$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4413,
"subject": "General Science",
"question": "The point represented by 2 + i in the Argand plane moves 1 unit eastwards, then 2 units northwards and finally from there $$2\\sqrt 2 $$ units in the south-westwardsdirection. Then its new position in the Argand plane is at the point represented by :",
"options": [
{
"text": "2 + 2i"
},
{
"text": "1 + i"
},
{
"text": "$$-$$1 $$-$$ i"
},
{
"text": "$$-$$2 $$-$$2i"
}
],
"answer": "1 + i",
"solution": "**Answer:** 1 + i\n\n\n
Here, \n
z $$-$$ (3 + 3i) = $$2\\sqrt 2 $$ (cos($$-$$135o) + i sin ($$-$$ 135o))\n
$$ \\therefore $$ z lies on line x = y",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4423,
"subject": "General Science",
"question": "Let $$S = \\left\\{ {n \\in N\\left| {{{\\left( {\\matrix{\n 0 & i \\cr \n 1 & 0 \\cr \n\n } } \\right)}^n}\\left( {\\matrix{\n a & b \\cr \n c & d \\cr \n\n } } \\right) = \\left( {\\matrix{\n a & b \\cr \n c & d \\cr \n\n } } \\right)\\forall a,b,c,d \\in R} \\right.} \\right\\}$$, where i = $$\\sqrt { - 1} $$. Then the number of 2-digit numbers in the set S is _____________.",
"options": [],
"answer": "11",
"solution": "**Answer:** 11\n\nLet $$X = \\left( {\\matrix{\n a & b \\cr \n c & d \\cr \n\n } } \\right)$$ & $$A = {\\left( {\\matrix{\n 0 & i \\cr \n 1 & 0 \\cr \n\n } } \\right)^n}$$
Hence, sin23$$\\theta$$ + cos2$$\\theta$$ = 1.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4425,
"subject": "General Science",
"question": "If $$S = \\left\\{ {z \\in C:{{z - i} \\over {z + 2i}} \\in R} \\right\\}$$, then :",
"options": [
{
"text": "S contains exactly two elements"
},
{
"text": "S contains only one element"
},
{
"text": "S is a circle in the complex plane"
},
{
"text": "S is a straight line in the complex plane"
}
],
"answer": "S is a straight line in the complex plane",
"solution": "**Answer:** S is a straight line in the complex plane\n\nGiven $${{z - i} \\over {z + 2i}} \\in R$$
Then $$\\arg \\left( {{{z - i} \\over {z + 2i}}} \\right)$$ is 0 or $$\\pi $$
$$\\Rightarrow$$ S is straight line in complex",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4426,
"subject": "General Science",
"question": "
Let $$\\alpha$$ and $$\\beta$$ be the roots of the equation x2 + (2i $$-$$ 1) = 0. Then, the value of |$$\\alpha$$8 + $$\\beta$$8| is equal to :
Let $$S = \\{ z \\in C:|z - 2| \\le 1,\\,z(1 + i) + \\overline z (1 - i) \\le 2\\} $$. Let $$|z - 4i|$$ attains minimum and maximum values, respectively, at z1 $$\\in$$ S and z2 $$\\in$$ S. If $$5(|{z_1}{|^2} + |{z_2}{|^2}) = \\alpha + \\beta \\sqrt 5 $$, where $$\\alpha$$ and $$\\beta$$ are integers, then the value of $$\\alpha$$ + $$\\beta$$ is equal to ___________.
",
"options": [],
"answer": "26",
"solution": "**Answer:** 26\n\n \n$S$ represents the shaded region shown in the diagram.\n
\nClearly $z_{1}$ will be the point of intersection of $P A$ and given circle.\n
\n$P A: 2 x+y=4$ and given circle has equation $(x-2)^{2}+y^{2}=1$\n
\nOn solving we get\n
\n$z_{1}=\\left(2-\\frac{1}{\\sqrt{5}}\\right)+\\frac{2}{\\sqrt{5}} i \\Rightarrow\\left|z_{1}\\right|^{2}=5-\\frac{4}{\\sqrt{5}}$\n
\n$z_{2}$ will be either $B$ or $C$.\n
\n$\\because P B=\\sqrt{17}$ and $P C=\\sqrt{13}$ hence $z_{2}=1$\n
Let S be the set of all $$(\\alpha, \\beta), \\pi<\\alpha, \\beta<2 \\pi$$, for which the complex number $$\\frac{1-i \\sin \\alpha}{1+2 i \\sin \\alpha}$$ is purely imaginary and $$\\frac{1+i \\cos \\beta}{1-2 i \\cos \\beta}$$ is purely real. Let $$Z_{\\alpha \\beta}=\\sin 2 \\alpha+i \\cos 2 \\beta,(\\alpha, \\beta) \\in S$$. Then $$\\sum\\limits_{(\\alpha, \\beta) \\in S}\\left(i Z_{\\alpha \\beta}+\\frac{1}{i \\bar{Z}_{\\alpha \\beta}}\\right)$$ is equal to :
Let $$\\mathrm{z}=a+i b, b \\neq 0$$ be complex numbers satisfying $$z^{2}=\\bar{z} \\cdot 2^{1-z}$$. Then the least value of $$n \\in N$$, such that $$z^{n}=(z+1)^{n}$$, is equal to __________.
For two non-zero complex numbers $$z_{1}$$ and $$z_{2}$$, if $$\\operatorname{Re}\\left(z_{1} z_{2}\\right)=0$$ and $$\\operatorname{Re}\\left(z_{1}+z_{2}\\right)=0$$, then which of the following are possible?
\n
A. $$\\operatorname{Im}\\left(z_{1}\\right)>0$$ and $$\\operatorname{Im}\\left(z_{2}\\right) > 0$$
\n
B. $$\\operatorname{Im}\\left(z_{1}\\right) < 0$$ and $$\\operatorname{Im}\\left(z_{2}\\right) > 0$$
\n
C. $$\\operatorname{Im}\\left(z_{1}\\right) > 0$$ and $$\\operatorname{Im}\\left(z_{2}\\right) < 0$$
\n
D. $$\\operatorname{Im}\\left(z_{1}\\right) < 0$$ and $$\\operatorname{Im}\\left(z_{2}\\right) < 0$$
\n
Choose the correct answer from the options given below :
",
"options": [
{
"text": "A and C"
},
{
"text": "A and B"
},
{
"text": "B and D"
},
{
"text": "B and C"
}
],
"answer": "B and C",
"solution": "**Answer:** B and C\n\n
Let $$\\mathrm{S}=\\left\\{z \\in \\mathbb{C}-\\{i, 2 i\\}: \\frac{z^{2}+8 i z-15}{z^{2}-3 i z-2} \\in \\mathbb{R}\\right\\}$$. If $$\\alpha-\\frac{13}{11} i \\in \\mathrm{S}, \\alpha \\in \\mathbb{R}-\\{0\\}$$, then\n\n$$242 \\alpha^{2}$$ is equal to _________.
$$\n\\begin{aligned}\n& \\operatorname{lm}\\left(\\frac{z^2+8 i z-15}{z^2-3 i z-2}\\right)=0 \\\\\\\\\n& \\Rightarrow-\\left(x^2-y^2-8 y-15\\right)(2 x y-3 x)+(2 x y+8 x)\\left(x^2-\\right. \\left.y^2+3 y-2\\right)=0 \\\\\\\\\n& \\Rightarrow\\left(x^2-y^2\\right)(2 x y+8 x-2 x y+3 x)+(8 y+15)(2 x y- 3 x)+(2 x y+8 x)(3 y-2)=0 \\\\\\\\\n& \\Rightarrow 11 x^3-11 x y^2+16 x y^2-24 x y+30 x y-45 x+6 x y^2 -4 x y+24 x y-16 x=0 \\\\\\\\\n& \\Rightarrow 11 x^3+11 x y^2+26 x y-61 x=0 \\\\\\\\\n& \\Rightarrow\\left(11 x^2+11 y^2+26 y-61)=0\\right. \\\\\\\\\n& \\because \\alpha \\neq 0, \\Rightarrow x=0 \\text { (neglected) } \\\\\\\\\n& \\text { Put } y=-\\frac{13}{11}, \\quad x=\\alpha \\\\\\\\\n& 11 \\alpha^2+11 \\cdot \\frac{13^2}{11^2}-26 \\cdot \\frac{13}{11}-61=0 \\\\\\\\\n& \\Rightarrow 121 \\alpha^2=840 \\\\\\\\\n& \\Rightarrow 242 \\alpha^2=1680\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4434,
"subject": "General Science",
"question": "
Let $$S = \\left\\{ {z = x + iy:{{2z - 3i} \\over {4z + 2i}}\\,\\mathrm{is\\,a\\,real\\,number}} \\right\\}$$. Then which of the following is NOT correct?
If $x=0$, then $y \\neq-\\frac{1}{2}$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4435,
"subject": "General Science",
"question": "
Let the complex number $$z = x + iy$$ be such that $${{2z - 3i} \\over {2z + i}}$$ is purely imaginary. If $${x} + {y^2} = 0$$, then $${y^4} + {y^2} - y$$ is equal to :
Let $$A=\\left\\{\\theta \\in(0,2 \\pi): \\frac{1+2 i \\sin \\theta}{1-i \\sin \\theta}\\right.$$ is purely imaginary $$\\}$$. Then the sum of the elements in $$\\mathrm{A}$$ is :
Let $$a \\neq b$$ be two non-zero real numbers. Then the number of elements in the set $$X=\\left\\{z \\in \\mathbb{C}: \\operatorname{Re}\\left(a z^{2}+b z\\right)=a\\right.$$ and $$\\left.\\operatorname{Re}\\left(b z^{2}+a z\\right)=b\\right\\}$$ is equal to :
$$\n\\begin{array}{ll}\n&\\Rightarrow z^2=x^2-y^2+2 i x y \\\\\\\\\n&\\therefore a z^2+b z \\\\\\\\\n& =a\\left(x^2-y^2+2 i x y\\right)+b(x+i y) \\\\\\\\\n& =a\\left(x^2-y^2\\right)+b x+2 a i x y+b i y\n\\end{array}\n$$\n
$\\begin{aligned} & \\text { and } b z^2+a z \\\\\\\\ & =b\\left(x^2-y^2+2 i x y\\right)+a(x+i y) \\\\\\\\ & =b\\left(x^2-y^2\\right)+a x+2 b i x y+a i y \\\\\\\\ & \\operatorname{Re}\\left(b z^2+a z\\right)=b \\\\\\\\ & \\Rightarrow b\\left(x^2-y^2\\right)+a x=b \\\\\\\\ & \\Rightarrow x^2-y^2+\\frac{a}{b} x=1 ..........(ii)\\end{aligned}$\n
On subtracting Equation (ii) from Equation (i), we get\n
$$\n\\begin{aligned}\n& \\frac{b}{a} x-\\frac{a}{b} x=0 \\\\\\\\\n& \\Rightarrow \\left(\\frac{b}{a}-\\frac{a}{b}\\right) x=0 \\\\\\\\\n& \\Rightarrow x=0 \\text { or } \\frac{b}{a}-\\frac{a}{b}=0 \\\\\\\\\n& \\Rightarrow b^2-a^2=0 \\\\\\\\\n& \\Rightarrow a= \\pm b \\\\\\\\\n& \\Rightarrow a=-b(\\text { since } a \\neq b)\n\\end{aligned}\n$$\n
From (i), when $x=0$, then\n
$$\n\\begin{aligned}\n& 0-y^2=1 \\Rightarrow y^2=-1 \\\\\\\\\n& \\Rightarrow y \\in \\phi \\Rightarrow z \\in \\phi \\text { has no solution. }\n\\end{aligned}\n$$\n\n
When, $a=-b$, then $x^2-y^2-x=1$ has infinitely many solutions.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4438,
"subject": "General Science",
"question": "Let $\\mathrm{S}=|\\mathrm{z} \\in \\mathrm{C}:| z-1 \\mid=1$ and $(\\sqrt{2}-1)(z+\\bar{z})-i(z-\\bar{z})=2 \\sqrt{2} \\mid$. Let $z_1, z_2 \\in \\mathrm{S}$ be such that $\\left|z_1\\right|=\\max\\limits_{z \\in s}|z|$ and $\\left|z_2\\right|=\\min\\limits _{z \\in S}|z|$. Then $\\left|\\sqrt{2} z_1-z_2\\right|^2$ equals :",
"options": [
{
"text": "1"
},
{
"text": "4"
},
{
"text": "3"
},
{
"text": "2"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\nLet $z=x+i y$\n
If $$z=\\frac{1}{2}-2 i$$ is such that $$|z+1|=\\alpha z+\\beta(1+i), i=\\sqrt{-1}$$ and $$\\alpha, \\beta \\in \\mathbb{R}$$, then $$\\alpha+\\beta$$ is equal to
Let $$\\alpha, \\beta$$ be the roots of the equation $$x^2-x+2=0$$ with $$\\operatorname{Im}(\\alpha)>\\operatorname{Im}(\\beta)$$. Then $$\\alpha^6+\\alpha^4+\\beta^4-5 \\alpha^2$$ is equal to ___________.
Let $$\\alpha, \\beta$$ be the roots of the equation $$x^2-\\sqrt{6} x+3=0$$ such that $$\\operatorname{Im}(\\alpha)>\\operatorname{Im}(\\beta)$$. Let $$a, b$$ be integers not divisible by 3 and $$n$$ be a natural number such that $$\\frac{\\alpha^{99}}{\\beta}+\\alpha^{98}=3^n(a+i b), i=\\sqrt{-1}$$. Then $$n+a+b$$ is equal to __________.
Let $$\\alpha$$ and $$\\beta$$ be the sum and the product of all the non-zero solutions of the equation $$(\\bar{z})^2+|z|=0, z \\in C$$. Then $$4(\\alpha^2+\\beta^2)$$ is equal to :
The sum of all possible values of $$\\theta \\in[-\\pi, 2 \\pi]$$, for which $$\\frac{1+i \\cos \\theta}{1-2 i \\cos \\theta}$$ is purely imaginary, is equal to :
Let $$z$$ be a complex number such that $$|z+2|=1$$ and $$\\operatorname{lm}\\left(\\frac{z+1}{z+2}\\right)=\\frac{1}{5}$$. Then the value of $$|\\operatorname{Re}(\\overline{z+2})|$$ is
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4448,
"subject": "General Science",
"question": "The locus of the centre of a circle which touches the circle $$\\left| {z - {z_1}} \\right| = a$$ and$$\\left| {z - {z_2}} \\right| = b\\,$$ externally \n
($$z,\\,{z_1}\\,\\& \\,{z_2}\\,$$ are complex numbers) will be :",
"options": [
{
"text": "an ellipse "
},
{
"text": "a hyperbola"
},
{
"text": "a circle "
},
{
"text": "none of these"
}
],
"answer": "a hyperbola",
"solution": "**Answer:** a hyperbola\n\n\nLet the circle be $$\\left| {z - {z_3}} \\right| = r.$$ \n
Then according to given conditions \n
$$\\left| {{z_3} - {z_1}} \\right| = r + a$$ (Shown in the image)\n
and $$\\left| {{z_3} - {z_2}} \\right| = r + b.$$ (Shown in the image)\n
Definition of hyperbola says, when difference of distance between two points is constant from a particular point then that particular point will lie on a hyperbola.\n
Here distance of z1 from z3 is = $$r + a$$ and distance of z2 from z3 is = $$r + b$$\n
Now their difference = ($$r + a$$) - ($$r + b$$) = $$a - b$$ = a constant\n
$$\\therefore$$ Locus of z3 is a hyperbola.",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4449,
"subject": "General Science",
"question": "Let $${Z_1}$$ and $${Z_2}$$ be two roots of the equation $${Z^2} + aZ + b = 0$$, Z being complex. Further , assume that the origin, $${Z_1}$$ and $${Z_2}$$ form an equilateral triangle. Then :",
"options": [
{
"text": "$${a^2} = 4b$$ "
},
{
"text": "$${a^2} = b$$ "
},
{
"text": "$${a^2} = 2b$$ "
},
{
"text": "$${a^2} = 3b$$ "
}
],
"answer": "$${a^2} = 3b$$ ",
"solution": "**Answer:** $${a^2} = 3b$$ \n\nGiven quadratic equation,\n $${Z^2} + aZ + b = 0$$\n and two roots are $${Z_1}$$ and $${Z_2}$$.\n
$$\\therefore$$ $${Z_1}$$ + $${Z_2}$$ = $$-a$$ and $${Z_1}$$$${Z_2}$$ = $$b$$\n
Question says,\n There are three complex numbers:\n 1. Origin (0)\n 2. $${Z_1}$$\n 3. $${Z_2}$$\n and they form an equilateral triangle. So They are the vertices of the triangle.\n
[ Important Point : If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -\n $$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$ ]\n
In this question,\n $${Z_1}$$ = 0, $${Z_2}$$ = $${Z_1}$$ and $${Z_3}$$ = $${Z_2}$$\n
[ Note : This question is asked to check if you know the following formula -\n
\"If $${Z_1}$$, $${Z_2}$$ and $${Z_3}$$ are the vertices of an equilateral triangle then -\n $$Z_1^2$$ + $$Z_2^2$$ + $$Z_3^2$$ = $${Z_1}{Z_2}$$ + $${Z_2}{Z_3}$$ + $${Z_3}{Z_1}$$\" ]",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4450,
"subject": "General Science",
"question": "If $$\\,\\left| {z + 4} \\right|\\,\\, \\le \\,\\,3\\,$$, then the maximum value of $$\\left| {z + 1} \\right|$$ is :",
"options": [
{
"text": "6 "
},
{
"text": "0 "
},
{
"text": "4 "
},
{
"text": "10"
}
],
"answer": "6 ",
"solution": "**Answer:** 6 \n\n$$z$$ lies on or inside the circle with center $$(-4,0)$$ and radius $$3$$ units. \n
\n
From the Argand diagram maximum value of $$\\left| {z + 1} \\right|$$ is $$6$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4451,
"subject": "General Science",
"question": "Let $$\\alpha \\,,\\beta $$ be real and z be a complex number. If $${z^2} + \\alpha z + \\beta = 0$$ has two distinct roots on the line Re z = 1, then it is necessary that :",
"options": [
{
"text": "$$\\beta \\, \\in ( - 1,0)$$ "
},
{
"text": "$$\\left| {\\beta \\,} \\right| = 1$$ "
},
{
"text": "$$\\beta \\, \\in (1,\\infty )$$ "
},
{
"text": "$$\\beta \\, \\in (0,1)$$ "
}
],
"answer": "$$\\beta \\, \\in (1,\\infty )$$ ",
"solution": "**Answer:** $$\\beta \\, \\in (1,\\infty )$$ \n\nAs real part of roots is $$1$$\n
Let roots are $$1 + pi,1 + q$$\n
$$\\therefore$$ sum of roots $$ = 1 + pi + 1 + qi = - \\alpha $$\n
$$p \\ne 0$$ as roots are distinct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4452,
"subject": "General Science",
"question": "If $$z \\ne 1$$ and $$\\,{{{z^2}} \\over {z - 1}}\\,$$ is real, then the point represented by the complex number z lies :",
"options": [
{
"text": "either on the real axis or a circle passing through the origin."
},
{
"text": "on a circle with centre at the origin"
},
{
"text": "either on real axis or on a circle not passing through the origin."
},
{
"text": "on the imaginary axis."
}
],
"answer": "either on the real axis or a circle passing through the origin.",
"solution": "**Answer:** either on the real axis or a circle passing through the origin.\n\nLet $$z = x + iy$$ \n
$$\\therefore$$ $$\\,\\,\\,\\,$$ $$z$$ lies either on real axis or on a circle through origin.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4453,
"subject": "General Science",
"question": "Let z$$ \\in $$C, the set of complex numbers. Then the equation, 2|z + 3i| $$-$$ |z $$-$$ i| = 0 represents : ",
"options": [
{
"text": "a circle with radius $${8 \\over 3}.$$"
},
{
"text": "a circle with diameter $${{10} \\over 3}.$$"
},
{
"text": "an ellipse with length of major axis $${{16} \\over 3}.$$"
},
{
"text": "an ellipse with length of minor axis $${{16} \\over 9}.$$"
}
],
"answer": "a circle with radius $${8 \\over 3}.$$",
"solution": "**Answer:** a circle with radius $${8 \\over 3}.$$\n\nGiven, \n
$$ \\therefore $$ h = $${{{\\alpha ^2} - 1} \\over {{\\alpha ^2} + 1}}$$ and k = $${{2\\alpha } \\over {{\\alpha ^2} + 1}}$$\n
By squaring and adding we get\n
h2 + k2 = 1\n
$$ \\therefore $$ This is circle whose radius is 1.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4456,
"subject": "General Science",
"question": "The equation |z – i| = |z – 1|, i = $$\\sqrt { - 1} $$, represents :",
"options": [
{
"text": "a circle of radius 1"
},
{
"text": "the line through the origin with slope – 1"
},
{
"text": "a circle of radius $${1 \\over 2}$$"
},
{
"text": "the line through the origin with slope 1\n"
}
],
"answer": "the line through the origin with slope 1\n",
"solution": "**Answer:** the line through the origin with slope 1\n\n\nLet the complex number z = x + iy
$$ \\Rightarrow $$ $$r = {3 \\over {2\\sqrt 2 }}$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4461,
"subject": "General Science",
"question": "Let z be those complex numbers which satisfy
| z + 5 | $$ \\le $$ 4 and z(1 + i) + $$\\overline z $$(1 $$-$$ i) $$ \\ge $$ $$-$$10, i = $$\\sqrt { - 1} $$.
If the maximum value of | z + 1 |2 is $$\\alpha$$ + $$\\beta$$$$\\sqrt 2 $$, then the value of ($$\\alpha$$ + $$\\beta$$) is ____________.",
"options": [],
"answer": "48",
"solution": "**Answer:** 48\n\nLet, z = x + iy
Then the set S1 $$\\cap$$ S2 $$\\cap$$ S3 :",
"options": [
{
"text": "has exactly three elements"
},
{
"text": "is a singleton"
},
{
"text": "has infinitely many elements"
},
{
"text": "has exactly two elements"
}
],
"answer": "has infinitely many elements",
"solution": "**Answer:** has infinitely many elements\n\nLet, z = x + iy
$$ \\Rightarrow $$ S1 $$\\cap$$ S2 $$\\cap$$ S3 has infinitely many elements.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4464,
"subject": "General Science",
"question": "If the equation $$a|z{|^2} + \\overline {\\overline \\alpha z + \\alpha \\overline z } + d = 0$$ represents a circle where a, d are real constants then which of the following condition is correct?",
"options": [
{
"text": "|$$\\alpha$$|2 $$-$$ ad $$\\ne$$ 0"
},
{
"text": "|$$\\alpha$$|2 $$-$$ ad > 0 and a$$\\in$$R $$-$$ {0}"
},
{
"text": "|$$\\alpha$$|2 $$-$$ ad $$ \\ge $$ 0 and a$$\\in$$R"
},
{
"text": "$$\\alpha$$ = 0, a, d$$\\in$$R+"
}
],
"answer": "|$$\\alpha$$|2 $$-$$ ad > 0 and a$$\\in$$R $$-$$ {0}",
"solution": "**Answer:** |$$\\alpha$$|2 $$-$$ ad > 0 and a$$\\in$$R $$-$$ {0}\n\n$$a|z{|^2} + \\alpha \\overline z + \\overline \\alpha z + d = 0$$
$$ \\Rightarrow {\\left| \\alpha \\right|^2} \\ge ad$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4465,
"subject": "General Science",
"question": "Let z1, z2 be the roots of the equation z2 + az + 12 = 0 and z1, z2 form an equilateral triangle with origin. Then, the value of |a| is :",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\nFor equilateral triangle with vertices z1, z2 and z3,
$$ \\Rightarrow |a|\\, = 6$$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4466,
"subject": "General Science",
"question": "Let a complex number be w = 1 $$-$$ $${\\sqrt 3 }$$i. Let another complex number z be such that |zw| = 1 and arg(z) $$-$$ arg(w) = $${\\pi \\over 2}$$. Then the area of the triangle with vertices origin, z and w is equal to :",
"options": [
{
"text": "4"
},
{
"text": "$${1 \\over 4}$$"
},
{
"text": "2"
},
{
"text": "$${1 \\over 2}$$"
}
],
"answer": "$${1 \\over 2}$$",
"solution": "**Answer:** $${1 \\over 2}$$\n\n\n Given, w = 1 $$-$$ $$\\sqrt 3 $$ i
For (AP)2 maximum $$\\theta = - {{3\\pi } \\over 4}$$
$${(AP)^2} = {5 \\over 4} + {1 \\over {\\sqrt 2 }} = {{5\\sqrt 2 + 4} \\over {4\\sqrt 2 }}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4468,
"subject": "General Science",
"question": "The equation of a circle is Re(z2) + 2(Im(z))2 + 2Re(z) = 0, where z = x + iy. A line which passes through the center of the given circle and the vertex of the parabola, x2 $$-$$ 6x $$-$$ y + 13 = 0, has y-intercept equal to ______________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nEquation of circle is (x2 $$-$$ y2) + 2y2 + 2x = 0
x2 + y2 + 2x = 0
Centre : ($$-$$1, 0)
Parabola : x2 $$-$$ 6x $$-$$ y + 13 = 0
(x $$-$$ 3)2 = y $$-$$ 4
Vertex : (3, 4)
Equation of line $$ \\equiv y - 0 = {{4 - 0} \\over {3 + 1}}(x + 1)$$
$$y = x + 1$$
y-intercept = 1",
"topic": "Discrete Mathematics",
"subtopic": "Graph Theory"
},
{
"id": 4469,
"subject": "General Science",
"question": "Let C be the set of all complex numbers. Let
$${S_1} = \\{ z \\in C||z - 3 - 2i{|^2} = 8\\} $$
$${S_2} = \\{ z \\in C|{\\mathop{\\rm Re}\\nolimits} (z) \\ge 5\\} $$ and
$${S_3} = \\{ z \\in C||z - \\overline z | \\ge 8\\} $$.
Then the number of elements in $${S_1} \\cap {S_2} \\cap {S_3}$$ is equal to :",
"options": [
{
"text": "1"
},
{
"text": "0"
},
{
"text": "2"
},
{
"text": "Infinite"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\n$${S_1}:|z - 3 - 2i{|^2} = 8$$
$$|z - 3 - 2i| = 2\\sqrt 2 $$
$${(x - 3)^2} + {(y - 2)^2} = {(2\\sqrt 2 )^2}$$
$${S_2}:x \\ge 5$$
$${S_3}:|z - \\overline z | \\ge 8$$
$$|2iy| \\ge 8$$
$$2|y| \\ge 8$$
$$\\therefore$$ $$y \\ge 4$$, $$y \\le - 4$$
$$n\\left( {{S_1} \\cap {S_2} \\cap {S_3}} \\right) = 1$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4470,
"subject": "General Science",
"question": "The equation $$\\arg \\left( {{{z - 1} \\over {z + 1}}} \\right) = {\\pi \\over 4}$$ represents a circle with :",
"options": [
{
"text": "centre at (0, $$-$$1) and radius $$\\sqrt 2 $$"
},
{
"text": "centre at (0, 1) and radius $$\\sqrt 2 $$"
},
{
"text": "centre (0, 0) and radius $$\\sqrt 2 $$"
},
{
"text": "centre at (0, 1) and radius 2"
}
],
"answer": "centre at (0, 1) and radius $$\\sqrt 2 $$",
"solution": "**Answer:** centre at (0, 1) and radius $$\\sqrt 2 $$\n\n
$$\\therefore$$ centre (0, 1); Radius = $$\\sqrt 2 $$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4471,
"subject": "General Science",
"question": "A point z moves in the complex plane such that $$\\arg \\left( {{{z - 2} \\over {z + 2}}} \\right) = {\\pi \\over 4}$$, then the minimum value of $${\\left| {z - 9\\sqrt 2 - 2i} \\right|^2}$$ is equal to _______________.",
"options": [],
"answer": "98",
"solution": "**Answer:** 98\n\nLet $$z = x + iy$$
Let arg(z) represent the principal argument of the complex number z. Then, |z| = 3 and arg(z $$-$$ 1) $$-$$ arg(z + 1) = $${\\pi \\over 4}$$ intersect :
",
"options": [
{
"text": "exactly at one point."
},
{
"text": "exactly at two points."
},
{
"text": "nowhere."
},
{
"text": "at infinitely many points."
}
],
"answer": "nowhere.",
"solution": "**Answer:** nowhere.\n\n
Let $$z = x + iy$$
\n
$$\\therefore$$ $$|z| = \\sqrt {{x^2} + {y^2}} $$
\n
Given, $$|z| = 3$$
\n
$$\\therefore$$ $$\\sqrt {{x^2} + {y^2}} = 3$$
\n
$$ \\Rightarrow {x^2} + {y^2} = 9 = {3^2}$$
\n
This represent a circle with center at (0, 0) and radius = 3
\n
Now, given
\n
$$\\arg (z - 1) - \\arg (z + 1) = {\\pi \\over 4}$$
C1 represents a circle with centre (4, 3) and radius 2 and C2 represents a ellipse with focii at (0, 0) and (4, 0) and length of major axis = 6, and length of semi-major axis 2$$\\sqrt5$$ and (4, 2) lies inside the both C1 and C2 and (4, 3) lies outside the C2
Let O be the origin and A be the point $${z_1} = 1 + 2i$$. If B is the point $${z_2}$$, $${\\mathop{\\rm Re}\\nolimits} ({z_2}) < 0$$, such that OAB is a right angled isosceles triangle with OB as hypotenuse, then which of the following is NOT true?
Let $$\\mathrm{S}=\\{z=x+i y:|z-1+i| \\geq|z|,|z|<2,|z+i|=|z-1|\\}$$. Then the set of all values of $$x$$, for which $$w=2 x+i y \\in \\mathrm{S}$$ for some $$y \\in \\mathbb{R}$$, is :
Let $$a,b$$ be two real numbers such that $$ab < 0$$. IF the complex number $$\\frac{1+ai}{b+i}$$ is of unit modulus and $$a+ib$$ lies on the circle $$|z-1|=|2z|$$, then a possible value of $$\\frac{1+[a]}{4b}$$, where $$[t]$$ is greatest integer function, is :
If the center and radius of the circle $$\\left| {{{z - 2} \\over {z - 3}}} \\right| = 2$$ are respectively $$(\\alpha,\\beta)$$ and $$\\gamma$$, then $$3(\\alpha+\\beta+\\gamma)$$ is equal to :
So locus of $w$ is ellipse $\\frac{x^{2}}{\\left(\\frac{17}{4}\\right)^{2}}+\\frac{y^{2}}{\\left(\\frac{15}{4}\\right)^{2}}=1$\n\n
Locus of $\\mathrm{z}$ is circle $\\mathrm{x}^{2}+\\mathrm{y}^{2}=16$\n\n
So intersect at 4 points.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4480,
"subject": "General Science",
"question": "
Let $$\\mathrm{z_1=2+3i}$$ and $$\\mathrm{z_2=3+4i}$$. The set $$\\mathrm{S = \\left\\{ {z \\in \\mathbb{C}:{{\\left| {z - {z_1}} \\right|}^2} - {{\\left| {z - {z_2}} \\right|}^2} = {{\\left| {{z_1} - {z_2}} \\right|}^2}} \\right\\}}$$ represents a
",
"options": [
{
"text": "hyperbola with the length of the transverse axis 7"
},
{
"text": "hyperbola with eccentricity 2"
},
{
"text": "straight line with the sum of its intercepts on the coordinate axes equals $$-18$$"
},
{
"text": "straight line with the sum of its intercepts on the coordinate axes equals $$14$$"
}
],
"answer": "straight line with the sum of its intercepts on the coordinate axes equals $$14$$",
"solution": "**Answer:** straight line with the sum of its intercepts on the coordinate axes equals $$14$$\n\n$\\left|z-z_{1}\\right|^{2}-\\left|z-z_{2}\\right|^{2}=\\left|z_{1}-z_{2}\\right|^{2}$\n
Let $$w=z \\bar{z}+k_{1} z+k_{2} i z+\\lambda(1+i), k_{1}, k_{2} \\in \\mathbb{R}$$. Let $$\\operatorname{Re}(w)=0$$ be the circle $$\\mathrm{C}$$ of radius 1 in the first quadrant touching the line $$y=1$$ and the $$y$$-axis. If the curve $$\\operatorname{Im}(w)=0$$ intersects $$\\mathrm{C}$$ at $$\\mathrm{A}$$ and $$\\mathrm{B}$$, then $$30(A B)^{2}$$ is equal to __________
",
"options": [],
"answer": "24",
"solution": "**Answer:** 24\n\nGiven the expression for $w$ as :\n\n
1. If $w = x+iy$, we can separate this into the real and imaginary parts :\n\n
The real part is: $$\\text{Re}(w) = x^2 + y^2 + k_1x - k_2y + \\lambda = 0.$$\n \n
The imaginary part is: $$\\text{Im}(w) = k_1y + k_2x + \\lambda = 0.$$\n\n
2. We know the circle C of radius 1 touches the line $y=1$ and the y-axis. The standard form of the equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. If the circle touches y-axis, then the x-coordinate of the center $h$ must be equal to the radius. Also, the circle touches the line $y=1$, so the y-coordinate of the center $k$ must be 1 unit above this line, hence $k=2$. So the circle's equation is $(x-1)^2 + (y-2)^2 = 1$.\n\n
3. By comparing this with $\\text{Re}(w) = 0$, we can see that $k_1=-2$, $k_2=4$, and $\\lambda=4$.\n\n
4. The equation for the curve $\\text{Im}(w)=0$ becomes $-2y + 4x + 4 = 0$, which simplifies to $y = 2x + 2$.\n\n
5. To find the intersection points A and B, we solve the system of equations consisting of the circle and the line. Substituting $y = 2x + 2$ into the equation of the circle gives $5x^2 - 2x = 0$, with solutions $x = 0$ and $x = 2/5$.\n\n
6. Substituting $x = 0$ and $x = 2/5$ into $y = 2x + 2$ gives $y = 2$ and $y = 6/5$ respectively. So, the intersection points are A = $(0, 2)$ and B = $(2/5, 14/5)$.\n\n
7. The distance between A and B is $AB = \\sqrt{[(2/5)^2 + (4/5)^2]} = \\sqrt{4/5}$.\n\n
Let $$\\mathrm{C}$$ be the circle in the complex plane with centre $$\\mathrm{z}_{0}=\\frac{1}{2}(1+3 i)$$ and radius $$r=1$$. Let $$\\mathrm{z}_{1}=1+\\mathrm{i}$$ and the complex number $$z_{2}$$ be outside the circle $$C$$ such that $$\\left|z_{1}-z_{0}\\right|\\left|z_{2}-z_{0}\\right|=1$$. If $$z_{0}, z_{1}$$ and $$z_{2}$$ are collinear, then the smaller value of $$\\left|z_{2}\\right|^{2}$$ is equal to :
For $$\\alpha, \\beta, z \\in \\mathbb{C}$$ and $$\\lambda > 1$$, if $$\\sqrt{\\lambda-1}$$ is the radius of the circle $$|z-\\alpha|^{2}+|z-\\beta|^{2}=2 \\lambda$$, then $$|\\alpha-\\beta|$$ is equal to __________.
$\\begin{array}{ll}\\Rightarrow |\\alpha-\\beta|^2=4 \\\\\\\\ \\Rightarrow |\\alpha-\\beta|=2\\end{array}$",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4484,
"subject": "General Science",
"question": "Let $\\mathrm{P}=\\{\\mathrm{z} \\in \\mathbb{C}:|z+2-3 i| \\leq 1\\}$ and $\\mathrm{Q}=\\{\\mathrm{z} \\in \\mathbb{C}: z(1+i)+\\bar{z}(1-i) \\leq-8\\}$. Let in $\\mathrm{P} \\cap \\mathrm{Q}$, $|z-3+2 i|$ be maximum and minimum at $z_1$ and $z_2$ respectively. If $\\left|z_1\\right|^2+2\\left|z_2\\right|^2=\\alpha+\\beta \\sqrt{2}$, where $\\alpha, \\beta$ are integers, then $\\alpha+\\beta$ equals _____________.",
"options": [],
"answer": "36",
"solution": "**Answer:** 36\n\n\n
Clearly for the shaded region $z_1$ is the intersection of the circle and the line passing through $\\mathrm{P}\\left(\\mathrm{L}_1\\right)$ and $\\mathrm{z}_2$ is intersection of line $\\mathrm{L}_1 \\& \\mathrm{~L}_2$\n
On solving $\\mathrm{L}_1$ and $\\mathrm{z}_2$ is intersection of line $\\mathrm{L}_1 \\& \\mathrm{~L}_2$ we get $z_2:\\left(\\frac{-3}{2}, \\frac{5}{2}\\right)$\n
Let the complex numbers $$\\alpha$$ and $$\\frac{1}{\\bar{\\alpha}}$$ lie on the circles $$\\left|z-z_0\\right|^2=4$$ and $$\\left|z-z_0\\right|^2=16$$ respectively, where $$z_0=1+i$$. Then, the value of $$100|\\alpha|^2$$ is __________.
The area (in sq. units) of the region $$S=\\{z \\in \\mathbb{C}:|z-1| \\leq 2 ;(z+\\bar{z})+i(z-\\bar{z}) \\leq 2, \\operatorname{lm}(z) \\geq 0\\}$$ is
Let $$S_1=\\{z \\in \\mathbf{C}:|z| \\leq 5\\}, S_2=\\left\\{z \\in \\mathbf{C}: \\operatorname{Im}\\left(\\frac{z+1-\\sqrt{3} i}{1-\\sqrt{3} i}\\right) \\geq 0\\right\\}$$ and $$S_3=\\{z \\in \\mathbf{C}: \\operatorname{Re}(z) \\geq 0\\}$$. Then the area of the region $$S_1 \\cap S_2 \\cap S_3$$ is :
When argument of a complex number is $${\\pi \\over 2}$$, it means it is making an angle of $${\\pi \\over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.\n
So, $${{z \\over \\omega }}$$ is a purely imaginary number that means there is no real part in this complex number.\n
So we can assume,\n $${{z \\over \\omega }}$$ = $$ki$$\n
When argument of a complex number is $${\\pi \\over 2}$$, it means it is making an angle of $${\\pi \\over 2}$$ with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.\n
So, $${{z \\over \\omega }}$$ is a purely imaginary number that means there is no real part in this complex number.\n
So we can assume,\n $${{z \\over \\omega }}$$ = $$ki$$\n
$$\\therefore$$ $${\\overline z \\omega }$$ is at (0, -1) on the negative side of imaginary axis and making an angle of $${\\pi \\over 2}$$ clockwise.\n\n
$$\\therefore$$ $${\\overline z \\omega }$$ = 0 + (-1)$$ \\times $$$$i$$ = $$-i$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4492,
"subject": "General Science",
"question": "Let z and w be complex numbers such that $$\\overline z + i\\overline w = 0$$ and arg zw = $$\\pi $$. Then arg z equals :",
"options": [
{
"text": "$${{5\\pi } \\over 4}$$ "
},
{
"text": "$${{\\pi } \\over 2}$$ "
},
{
"text": "$${{3\\pi } \\over 4}$$"
},
{
"text": "$${{\\pi } \\over 4}$$ "
}
],
"answer": "$${{3\\pi } \\over 4}$$",
"solution": "**Answer:** $${{3\\pi } \\over 4}$$\n\nGiven $$\\overline z + i\\overline w = 0$$\n
$$ \\Rightarrow \\overline z = - i\\overline w $$ \n
[ $$i$$ complex number represent (0, 1) point on imaginary axis and Arg($$i$$) means the angle made by the point (0, 1) with real axis which is $${\\pi \\over 2}$$]\n
$$\\,\\left| {{z_1} + {z_2}} \\right|$$ is the vector sum of $${z_1}$$ and $${z_2}$$. So $$\\,\\left| {{z_1} + {z_2}} \\right|$$ should be $$<$$ $$\\left| {{z_1}} \\right| + \\left| {{z_2}} \\right|$$ but here they are equal so $${z_1}$$ and $${z_2}$$ are collinear.\n
S if $${z_1}$$ makes an angle $$\\theta $$ with x axis then $${z_2}$$ will also make $$\\theta $$ angle.\n
So, option (2) is correct.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4499,
"subject": "General Science",
"question": "Let z1 and z2 be two complex numbers such that $$\\arg ({z_1} - {z_2}) = {\\pi \\over 4}$$ and z1, z2 satisfy the equation | z $$-$$ 3 | = Re(z). Then the imaginary part of z1 + z2 is equal to ___________.",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\nLet z1 = x1 + iy ; z2 = x2 + iy2
Let $$A = \\left\\{ {z \\in C:\\left| {{{z + 1} \\over {z - 1}}} \\right| < 1} \\right\\}$$ and $$B = \\left\\{ {z \\in C:\\arg \\left( {{{z - 1} \\over {z + 1}}} \\right) = {{2\\pi } \\over 3}} \\right\\}$$. Then A $$\\cap$$ B is :
",
"options": [
{
"text": "a portion of a circle centred at $$\\left( {0, - {1 \\over {\\sqrt 3 }}} \\right)$$ that lies in the second and third quadrants only"
},
{
"text": "a portion of a circle centred at $$\\left( {0, - {1 \\over {\\sqrt 3 }}} \\right)$$ that lies in the second quadrant only"
},
{
"text": "an empty"
},
{
"text": "a portion of a circle of radius $${2 \\over {\\sqrt 3 }}$$ that lies in the third quadrant only"
}
],
"answer": "a portion of a circle centred at $$\\left( {0, - {1 \\over {\\sqrt 3 }}} \\right)$$ that lies in the second quadrant only",
"solution": "**Answer:** a portion of a circle centred at $$\\left( {0, - {1 \\over {\\sqrt 3 }}} \\right)$$ that lies in the second quadrant only\n\n$$\n\\left|\\frac{z+1}{z-1}\\right|<1 \\Rightarrow|z+1|<|z-1| \\Rightarrow \\operatorname{Re}(z)<0\n$$
\nand $\\arg \\left(\\frac{z-1}{z+1}\\right)=\\frac{2 \\pi}{3}$ is a part of circle as shown.
Let z1 and z2 be two complex numbers such that $${\\overline z _1} = i{\\overline z _2}$$ and $$\\arg \\left( {{{{z_1}} \\over {{{\\overline z }_2}}}} \\right) = \\pi $$. Then :
Let a circle C in complex plane pass through the points $${z_1} = 3 + 4i$$, $${z_2} = 4 + 3i$$ and $${z_3} = 5i$$. If $$z( \\ne {z_1})$$ is a point on C such that the line through z and z1 is perpendicular to the line through z2 and z3, then $$arg(z)$$ is equal to :
Let $$z=1+i$$ and $$z_{1}=\\frac{1+i \\bar{z}}{\\bar{z}(1-z)+\\frac{1}{z}}$$. Then $$\\frac{12}{\\pi} \\arg \\left(z_{1}\\right)$$ is equal to __________.
Let $$w_{1}$$ be the point obtained by the rotation of $$z_{1}=5+4 i$$ about the origin through a right angle in the anticlockwise direction, and $$w_{2}$$ be the point obtained by the rotation of $$z_{2}=3+5 i$$ about the origin through a right angle in the clockwise direction. Then the principal argument of $$w_{1}-w_{2}$$ is equal to :
To solve the problem, let's break it down step by step.
\n
Step 1 : Find $w_{1}$
\n
Given $z_{1} = 5 + 4i$.
\n
When you rotate $z_{1}$ by $90^{\\circ}$ anticlockwise about the origin, the real part becomes negative of the imaginary part of $z_{1}$, and the imaginary part becomes the real part of $z_{1}$.
\n
Therefore, $w_{1}$ becomes :\n
$w_{1} = -4 + 5i$
\n
Step 2 : Find $w_{2}$
\n
Given $z_{2} = 3 + 5i$.
\n
When you rotate $z_{2}$ by $90^{\\circ}$ clockwise about the origin, the real part becomes the imaginary part of $z_{2}$, and the imaginary part becomes negative of the real part of $z_{2}$.
\n
Therefore, $w_{2}$ becomes :\n
$w_{2} = 5 - 3i$
\n
Step 3 : Calculate $w_{1} - w_{2}$
\n
$w_{1} - w_{2} = (-4 + 5i) - (5 - 3i)$\n
$w_{1} - w_{2} = -9 + 8i$
\n
Step 4 : Find the principal argument
\n
To find the principal argument, we need to compute the tangent inverse of the ratio of the imaginary part to the real part. \n
If $$\\alpha$$ denotes the number of solutions of $$|1-i|^x=2^x$$ and $$\\beta=\\left(\\frac{|z|}{\\arg (z)}\\right)$$, where $$z=\\frac{\\pi}{4}(1+i)^4\\left[\\frac{1-\\sqrt{\\pi} i}{\\sqrt{\\pi}+i}+\\frac{\\sqrt{\\pi}-i}{1+\\sqrt{\\pi} i}\\right], i=\\sqrt{-1}$$, then the distance of the point $$(\\alpha, \\beta)$$ from the line $$4 x-3 y=7$$ is __________.
Let $$\\mathrm{r}$$ and $$\\theta$$ respectively be the modulus and amplitude of the complex number $$z=2-i\\left(2 \\tan \\frac{5 \\pi}{8}\\right)$$, then $$(\\mathrm{r}, \\theta)$$ is equal to
By checking options we found b2 – b = 30.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4510,
"subject": "General Science",
"question": "If $$\\alpha$$, $$\\beta$$ $$\\in$$ R are such that 1 $$-$$ 2i (here i2 = $$-$$1) is a root of z2 + $$\\alpha$$z + $$\\beta$$ = 0, then ($$\\alpha$$ $$-$$ $$\\beta$$) is equal to :",
"options": [
{
"text": "$$-$$7"
},
{
"text": "7"
},
{
"text": "3"
},
{
"text": "$$-$$3"
}
],
"answer": "$$-$$7",
"solution": "**Answer:** $$-$$7\n\n1 $$-$$ 2i is the root of the equation. So other root is 1 $$+$$ 2i\n
$$ \\therefore $$ Sum of roots = 1 $$-$$ 2i + 1 $$+$$ 2i = 2 = -$$\\alpha $$\n
$$ \\therefore $$ $$\\alpha $$ - $$\\beta $$ = -2 - 5 = -7",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4511,
"subject": "General Science",
"question": "Let z and $$\\omega$$ be two complex numbers such that $$\\omega = z\\overline z - 2z + 2,\\left| {{{z + i} \\over {z - 3i}}} \\right| = 1$$ and Re($$\\omega$$) has minimum value. Then, the minimum value of n $$\\in$$ N for which $$\\omega$$n is real, is equal to ______________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\nLet z = x + iy
| z + i | = | z $$-$$ 3i |
$$ \\Rightarrow $$ y = 1
Now
$$\\omega$$ = x2 + y2 $$-$$ 2x $$-$$ 2iy + 2
$$\\omega$$ = x2 + 1 $$-$$ 2x $$-$$ 2i + 2
Re($$\\omega$$) = x2 $$-$$ 2x + 3
Re($$\\omega$$) = (x $$-$$ 1)2 + 2
Re($$\\omega$$)min at x = 1 $$ \\Rightarrow $$ z = 1 + i
If $$\\omega$$n is real $$ \\Rightarrow $$ n = 4",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4512,
"subject": "General Science",
"question": "Let n denote the number of solutions of the equation z2 + 3$$\\overline z $$ = 0, where z is a complex number. Then the value of $$\\sum\\limits_{k = 0}^\\infty {{1 \\over {{n^k}}}} $$ is equal to :",
"options": [
{
"text": "1"
},
{
"text": "$${4 \\over 3}$$"
},
{
"text": "$${3 \\over 2}$$"
},
{
"text": "2"
}
],
"answer": "$${4 \\over 3}$$",
"solution": "**Answer:** $${4 \\over 3}$$\n\nz2 + 3$$\\overline z $$ = 0
\nSo there will be total 3 possible values of $z$, which are $0,\\left(\\frac{-1+\\sqrt{2}}{2}\\right)-\\frac{1}{2} i$ and $\\left(\\frac{-1-\\sqrt{2}}{2}\\right)-\\frac{1}{2} i$\n
Let $$S=\\left\\{z \\in \\mathbb{C}: z^{2}+\\bar{z}=0\\right\\}$$. Then $$\\sum\\limits_{z \\in S}(\\operatorname{Re}(z)+\\operatorname{Im}(z))$$ is equal to ______________.
Let $$S=\\left\\{z \\in \\mathbb{C}: \\bar{z}=i\\left(z^{2}+\\operatorname{Re}(\\bar{z})\\right)\\right\\}$$. Then $$\\sum_\\limits{z \\in \\mathrm{S}}|z|^{2}$$ is equal to :
$$\n\\begin{aligned}\n&\\bar{z}=i\\left(z^2+\\operatorname{Re}(\\bar{z})\\right) \\\\\\\\\n&\\Rightarrow x-i y=i\\left(x^2-y^2+2 i x y+x\\right) \\\\\\\\\n& x-i y=i\\left(x^2-y^2+x\\right)-2 x y \\\\\\\\\n &x=-2 x y \\Rightarrow x(2 y+1)=0 \\\\\\\\\n&\\Rightarrow x=0, y=\\frac{-1}{2} ........(i)\\\\\\\\\n&-y=x^2-y^2+x ...........(ii)\n\\end{aligned}\n$$\n
For $$a \\in \\mathbb{C}$$, let $$\\mathrm{A}=\\{z \\in \\mathbb{C}: \\operatorname{Re}(a+\\bar{z}) > \\operatorname{Im}(\\bar{a}+z)\\}$$ and $$\\mathrm{B}=\\{z \\in \\mathbb{C}: \\operatorname{Re}(a+\\bar{z})<\\operatorname{Im}(\\bar{a}+z)\\}$$. Then among the two statements :
\n
(S1): If $$\\operatorname{Re}(a), \\operatorname{Im}(a) > 0$$, then the set A contains all the real numbers
(S2) : If $$\\operatorname{Re}(a), \\operatorname{Im}(a) < 0$$, then the set B contains all the real numbers,
",
"options": [
{
"text": "both are false"
},
{
"text": "only (S1) is true"
},
{
"text": "only (S2) is true"
},
{
"text": "both are true"
}
],
"answer": "both are false",
"solution": "**Answer:** both are false\n\nWe are given that $a \\in \\mathbb{C}$ and $z \\in \\mathbb{C}$.\n\n
Let $a = x_1 + iy_1$ and $z = x_2 + iy_2$ where $x_1, y_1, x_2, y_2 \\in \\mathbb{R}$\n\n
We are also given two sets A and B defined as follows :\n
- A is the set of all complex numbers $z$ for which the real part of $(a + \\overline{z})$ is greater than the imaginary part of $(\\overline{a} + z)$.\n
- B is the set of all complex numbers $z$ for which the real part of $(a + \\overline{z})$ is less than the imaginary part of $(\\overline{a} + z)$.\n\n
Statement (S1) says : If the real part and imaginary part of $a$ are both positive, then the set A contains all the real numbers. \n
Statement (S2) says : If the real part and imaginary part of $a$ are both negative, then the set B contains all the real numbers.\n\n
We need to determine which of these statements are true.\n\n
This can be re-written as $x_1 + x_2 < y_2 - y_1$\n \n
If we consider only real z (i.e. $y_2 = 0$) and given that $x_1, y_1 < 0$, then the condition simplifies to $x_2 < -(x_1 + y_1)$.\n \n
This indicates that B covers a part of the positive real axis, but not the entire real axis. Therefore, Statement (S2) is false.\n\n
Therefore, both (S1) and (S2) are false, so the answer is Option A : both are false.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4521,
"subject": "General Science",
"question": "If $$z = x - iy$$ and $${z^{{1 \\over 3}}} = p + iq$$, then \n
$$ \\Rightarrow a = 0,\\,b = 9$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4526,
"subject": "General Science",
"question": "If $${\\left( {{{1 + i} \\over {1 - i}}} \\right)^x} = 1$$ then :",
"options": [
{
"text": "x = 2n + 1, where n is any positive integer"
},
{
"text": "x = 4n , where n is any positive integer"
},
{
"text": "x = 2n, where n is any positive integer\n"
},
{
"text": "x = 4n + 1, where n is any positive integer.\n"
}
],
"answer": "x = 4n , where n is any positive integer",
"solution": "**Answer:** x = 4n , where n is any positive integer\n\n$${\\left( {{{1 + i} \\over {1 - i}}} \\right)^x} = 1$$\n
Let $$\\mathrm{p,q\\in\\mathbb{R}}$$ and $${\\left( {1 - \\sqrt 3 i} \\right)^{200}} = {2^{199}}(p + iq),i = \\sqrt { - 1} $$ then $$\\mathrm{p+q+q^2}$$ and $$\\mathrm{p-q+q^2}$$ are roots of the equation.
$$\\left| z \\right|$$ represent distance of $$z$$ from point (0, 0) and\n
$$\\left| {z - {1 \\over {\\sqrt 3 }}i} \\right|$$ represent distance of $$z$$ from point $$\\left( {0,{1 \\over {\\sqrt 3 }}} \\right)$$.\n
According to the equation (1) the distance of $$z$$ from point (0, 0) and $$\\left( {0,{1 \\over {\\sqrt 3 }}} \\right)$$ is equal. Only if z is on a straight line then it will be equal distance from the both the points.\n",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4542,
"subject": "General Science",
"question": "The number of complex numbers z such that $$\\left| {z - 1} \\right| = \\left| {z + 1} \\right| = \\left| {z - i} \\right|$$ equals :",
"options": [
{
"text": "1 "
},
{
"text": "2 "
},
{
"text": "$$\\infty $$ "
},
{
"text": "0 "
}
],
"answer": "1 ",
"solution": "**Answer:** 1 \n\nLet $$z=x+iy$$ \n
$$ \\Rightarrow $$ Number of complex number $$z=1$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4543,
"subject": "General Science",
"question": "If z is a complex number such that $$\\,\\left| z \\right| \\ge 2\\,$$, then the minimum value of $$\\,\\,\\left| {z + {1 \\over 2}} \\right|$$ :",
"options": [
{
"text": "is strictly greater that $${{5 \\over 2}}$$"
},
{
"text": "is strictly greater that $${{3 \\over 2}}$$ but less than $${{5 \\over 2}}$$"
},
{
"text": "is equal to $${{5 \\over 2}}$$"
},
{
"text": "lie in the interval (1, 2)"
}
],
"answer": "is strictly greater that $${{3 \\over 2}}$$ but less than $${{5 \\over 2}}$$",
"solution": "**Answer:** is strictly greater that $${{3 \\over 2}}$$ but less than $${{5 \\over 2}}$$\n\nWe know minimum value of \n
$$ \\Rightarrow {3 \\over 2} < \\left| {Z + {1 \\over 2}} \\right| < {5 \\over 2}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4544,
"subject": "General Science",
"question": "A complex number z is said to be unimodular if $$\\,\\left| z \\right| = 1$$. Suppose $${z_1}$$ and $${z_2}$$ are complex numbers such that $${{{z_1} - 2{z_2}} \\over {2 - {z_1}\\overline {{z_2}} }}$$ is unimodular and $${z_2}$$ is not unimodular. Then the point $${z_1}$$ lies on a :",
"options": [
{
"text": "circle of radius 2."
},
{
"text": "circle of radius $${\\sqrt 2 }$$."
},
{
"text": "straight line parallel to x-axis"
},
{
"text": "straight line parallel to y-axis."
}
],
"answer": "circle of radius 2.",
"solution": "**Answer:** circle of radius 2.\n\n$$\\left| {{{{z_1} - 2{z_2}} \\over {2 - {z_1}{{\\overline z }_2}}}} \\right| = 1 \\Rightarrow {\\left| {{z_1} - 2{z_2}} \\right|^2} = {\\left| {2 - {z_1}{{\\overline z }_2}} \\right|^2}$$\n
$$ \\Rightarrow \\left( {{z_1}{{\\overline z }_1}} \\right) - 2{z_1}{\\overline z _2} - 2{\\overline z _1}{z_2} + 4{z_2}{\\overline z _2}$$\n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$ = 4 - 2{\\overline z _1}{z_2} - 2{z_1}{\\overline z _2} + {z_1}{\\overline z _1}{z_2}{\\overline z _2}$$\n
$$ \\therefore $$ |z| cannot be $$\\sqrt {7} $$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4551,
"subject": "General Science",
"question": "If the least and the largest real values of a, for which the equation z + $$\\alpha $$|z – 1| + 2i = 0\n(z $$ \\in $$ C and i = $$\\sqrt { - 1} $$) has a solution, are p and q respectively; then 4(p2 + q2) is equal to __________.",
"options": [],
"answer": "10",
"solution": "**Answer:** 10\n\n$$x + iy + \\alpha \\sqrt {{{(x - 1)}^2} + {y^2}} + 2i = 0$$
$$ \\therefore $$ y + 2 = 0 and $$x + \\alpha \\sqrt {{{(x - 1)}^2} + {y^2}} = 0$$
$$ = {5 \\over {\\sqrt 2 }} - {1 \\over {\\sqrt 2 }} = 2\\sqrt 2 $$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4555,
"subject": "General Science",
"question": "If for the complex numbers z satisfying | z $$-$$ 2 $$-$$ 2i | $$\\le$$ 1, the maximum value of | 3iz + 6 | is attained at a + ib, then a + b is equal to ______________.",
"options": [],
"answer": "5",
"solution": "**Answer:** 5\n\n| z $$-$$ 2 $$-$$ 2i | $$\\le$$ 1
Let S = {z $$\\in$$ C : |z $$-$$ 3| $$\\le$$ 1 and z(4 + 3i) + $$\\overline z $$(4 $$-$$ 3i) $$\\le$$ 24}. If $$\\alpha$$ + i$$\\beta$$ is the point in S which is closest to 4i, then 25($$\\alpha$$ + $$\\beta$$) is equal to ___________.
For $$\\mathrm{n} \\in \\mathbf{N}$$, let $$\\mathrm{S}_{\\mathrm{n}}=\\left\\{z \\in \\mathbf{C}:|z-3+2 i|=\\frac{\\mathrm{n}}{4}\\right\\}$$ and $$\\mathrm{T}_{\\mathrm{n}}=\\left\\{z \\in \\mathbf{C}:|z-2+3 i|=\\frac{1}{\\mathrm{n}}\\right\\}$$. Then the number of elements in the set $$\\left\\{n \\in \\mathbf{N}: S_{n} \\cap T_{n}=\\phi\\right\\}$$ is :\n
",
"options": [
{
"text": "0"
},
{
"text": "2"
},
{
"text": "3"
},
{
"text": "4"
}
],
"answer": "4",
"solution": "**Answer:** 4\n\n$S_{n}=\\left\\{z \\in C:|z-3+2 i|=\\frac{n}{4}\\right\\}$ represents a circle with centre $C_{1}(3,-2)$ and radius $r_{1}=\\frac{n}{4}$\n
\nSimilarly $T_{n}$ represents circle with centre $C_{2}(2,-3)$ and radius $r_{2}=\\frac{1}{n}$\n
\n$\\text { As } S_{n} \\cap T_{n}=\\phi$
\n$$\n\\begin{array}{llr}\nC_{1} C_{2}>r_{1}+r_{2}& \\text { OR } & C_{1} C_{2}<\\left|r_{1}-r_{2}\\right| \\\\\\\\\n\\sqrt{2}>\\frac{n}{4}+\\frac{1}{n} & \\text { OR } & \\sqrt{2}<\\left|\\frac{n}{4}-\\frac{1}{n}\\right| \\\\\\\\\n n=1,2,3,4&& n \\text { may take infinite values }\n\\end{array}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4559,
"subject": "General Science",
"question": "
For $$z \\in \\mathbb{C}$$ if the minimum value of $$(|z-3 \\sqrt{2}|+|z-p \\sqrt{2} i|)$$ is $$5 \\sqrt{2}$$, then a value Question: of $$p$$ is _____________.
Let the minimum value $$v_{0}$$ of $$v=|z|^{2}+|z-3|^{2}+|z-6 i|^{2}, z \\in \\mathbb{C}$$ is attained at $${ }{z}=z_{0}$$. Then $$\\left|2 z_{0}^{2}-\\bar{z}_{0}^{3}+3\\right|^{2}+v_{0}^{2}$$ is equal to :
Let $$S_{1}=\\left\\{z_{1} \\in \\mathbf{C}:\\left|z_{1}-3\\right|=\\frac{1}{2}\\right\\}$$ and $$S_{2}=\\left\\{z_{2} \\in \\mathbf{C}:\\left|z_{2}-\\right| z_{2}+1||=\\left|z_{2}+\\right| z_{2}-1||\\right\\}$$. Then, for $$z_{1} \\in S_{1}$$ and $$z_{2} \\in S_{2}$$, the least value of $$\\left|z_{2}-z_{1}\\right|$$ is :
Let $$z$$ be a complex number such that $$\\left| {{{z - 2i} \\over {z + i}}} \\right| = 2,z \\ne - i$$. Then $$z$$ lies on the circle of radius 2 and centre :
$$\n\\begin{aligned}\n& \\text { Now, }(\\alpha+2)^2+\\beta^2=(\\alpha+4)^2 \\\\\\\\\n& \\Rightarrow \\alpha^2+4+4 \\alpha+\\beta^2=\\alpha^2+16+8 \\alpha \\\\\\\\\n& \\Rightarrow 4+4 \\alpha+16=16+8 \\alpha \\\\\\\\\n& \\Rightarrow 4 \\alpha=4 \\Rightarrow \\alpha=1 \\\\\\\\\n& \\text { So, } \\alpha+\\beta=-3 \\text { and } \\alpha \\beta=-4 \\\\\\\\\n& \\therefore \\text { Required equation is } \\\\\\\\\n& x^2-(-3-4) x+(-3)(-4)=0 \\\\\\\\\n& \\Rightarrow x^2+7 x+12=0\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4566,
"subject": "General Science",
"question": "If $z$ is a complex number such that $|z| \\leqslant 1$, then the minimum value of $\\left|z+\\frac{1}{2}(3+4 i)\\right|$ is :",
"options": [
{
"text": "2"
},
{
"text": "$\\frac{5}{2}$"
},
{
"text": "$\\frac{3}{2}$"
},
{
"text": "3"
}
],
"answer": "$\\frac{3}{2}$",
"solution": "**Answer:** $\\frac{3}{2}$\n\n
To find the minimum value of $\\left|z+\\frac{1}{2}(3+4i)\\right|$, where $z$ is a complex number with $|z| \\leqslant 1$, we can think of this geometrically as the distance from any point inside or on the boundary of the unit circle in the complex plane to the fixed point $\\frac{1}{2}(3+4i)$.
\n\n
To make this more clear, first write the expression as follows:\n$$\\left|z+\\frac{1}{2}(3+4i)\\right| = \\left|z+\\frac{3}{2}+\\frac{4}{2}i\\right| = \\left|z+1.5+2i\\right|$$
\n\n
This means we are looking for the distance between any point on the complex plane represented by $z$ (where $z$ has a magnitude of up to 1) and the point $1.5+2i$.
\n\n
The triangle inequality gives us the following relationship for any two complex numbers $z_1$ and $z_2$:\n$$\\left|z_1 + z_2\\right| \\geqslant \\left| \\left|z_1\\right| - \\left|z_2\\right| \\right|$$
\n\n
Applying this to our specific case ($z_1 = z$ and $z_2 = -1.5 - 2i$):\n$$\\left|z + 1.5 + 2i\\right| \\geqslant \\left| \\left|z\\right| - \\left|-1.5 - 2i\\right| \\right|$$
\n\n
Since $z$ is within the unit circle, $|z| \\leqslant 1$ and $|-1.5 - 2i| = \\sqrt{1.5^2 + 2^2} = \\sqrt{2.25 + 4} = \\sqrt{6.25} = 2.5$.
\n\n
So now we have:\n$$\\left|z + 1.5 + 2i\\right| \\geqslant \\left| 1 - 2.5 \\right|$$
\n\n
Which simplifies to:\n$$\\left|z + 1.5 + 2i\\right| \\geqslant 1.5$$
\n\n
Therefore, the minimum value of $\\left|z+\\frac{1}{2}(3+4i)\\right|$ given that $|z| \\leqslant 1$ is $\\frac{3}{2}$.
The sum of the square of the modulus of the elements in the set $$\\{z=\\mathrm{a}+\\mathrm{ib}: \\mathrm{a}, \\mathrm{b} \\in \\mathbf{Z}, z \\in \\mathbf{C},|z-1| \\leq 1,|z-5| \\leq|z-5 \\mathrm{i}|\\}$$ is __________.
Statement I: For any two non-zero complex numbers $$z_1, z_2,(|z_1|+|z_2|)\\left|\\frac{z_1}{\\left|z_1\\right|}+\\frac{z_2}{\\left|z_2\\right|}\\right| \\leq 2\\left(\\left|z_1\\right|+\\left|z_2\\right|\\right) \\text {, and }$$
\n
Statement II : If $$x, y, z$$ are three distinct complex numbers and $$\\mathrm{a}, \\mathrm{b}, \\mathrm{c}$$ are three positive real numbers such that $$\\frac{\\mathrm{a}}{|y-z|}=\\frac{\\mathrm{b}}{|z-x|}=\\frac{\\mathrm{c}}{|x-y|}$$, then $$\\frac{\\mathrm{a}^2}{y-z}+\\frac{\\mathrm{b}^2}{z-x}+\\frac{\\mathrm{c}^2}{x-y}=1$$.
\n
Between the above two statements,
",
"options": [
{
"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.\n"
},
{
"text": "both Statement I and Statement II are correct."
}
],
"answer": "Statement I is correct but Statement II is incorrect.\n",
"solution": "**Answer:** Statement I is correct but Statement II is incorrect.\n\n\n
If $$z_1, z_2$$ are two distinct complex number such that $$\\left|\\frac{z_1-2 z_2}{\\frac{1}{2}-z_1 \\bar{z}_2}\\right|=2$$, then
",
"options": [
{
"text": "either $$z_1$$ lies on a circle of radius $$\\frac{1}{2}$$ or $$z_2$$ lies on a circle of radius 1.\n"
},
{
"text": "$$z_1$$ lies on a circle of radius $$\\frac{1}{2}$$ and $$z_2$$ lies on a circle of radius 1.\n"
},
{
"text": "either $$z_1$$ lies on a circle of radius 1 or $$z_2$$ lies on a circle of radius $$\\frac{1}{2}$$.\n"
},
{
"text": "both $$z_1$$ and $$z_2$$ lie on the same circle."
}
],
"answer": "either $$z_1$$ lies on a circle of radius 1 or $$z_2$$ lies on a circle of radius $$\\frac{1}{2}$$.\n",
"solution": "**Answer:** either $$z_1$$ lies on a circle of radius 1 or $$z_2$$ lies on a circle of radius $$\\frac{1}{2}$$.\n\n\n
If the set $$R=\\{(a, b): a+5 b=42, a, b \\in \\mathbb{N}\\}$$ has $$m$$ elements and $$\\sum_\\limits{n=1}^m\\left(1-i^{n !}\\right)=x+i y$$, where $$i=\\sqrt{-1}$$, then the value of $$m+x+y$$ is
$$\n\\begin{aligned}\n&\\text { If } \\lim _{n \\rightarrow \\infty} \\frac{(n+1)^{k-1}}{n^{k+1}}[(n k+1)+(n k+2)+\\ldots+(n k+n)] \\\\\n&=33 \\cdot \\lim _{n \\rightarrow \\infty} \\frac{1}{n^{k+1}} \\cdot\\left[1^{k}+2^{k}+3^{k}+\\ldots+n^{k}\\right]\n\\end{aligned}$$,\n\nthen the integral value of $$\\mathrm{k}$$ is equal to _____________
Let $$\\lim _\\limits{n \\rightarrow \\infty}\\left(\\frac{n}{\\sqrt{n^4+1}}-\\frac{2 n}{\\left(n^2+1\\right) \\sqrt{n^4+1}}+\\frac{n}{\\sqrt{n^4+16}}-\\frac{8 n}{\\left(n^2+4\\right) \\sqrt{n^4+16}}\\right.$$ $$\\left.+\\ldots+\\frac{n}{\\sqrt{n^4+n^4}}-\\frac{2 n \\cdot n^2}{\\left(n^2+n^2\\right) \\sqrt{n^4+n^4}}\\right)$$ be $$\\frac{\\pi}{k}$$, using only the principal values of the inverse trigonometric functions. Then $$\\mathrm{k}^2$$ is equal to _________.
This limit resembles a derivative because the fraction has the form $0/0$ as $x \\to 2$ since both the numerator (integral from $6$ to $f(x)$) and the denominator ($x - 2$) are zero when $x = 2$.\n
Applying $$L'$$ Hospital rule\n
$$\\mathop {\\lim }\\limits_{x \\to 2} {{\\left[ {4f{{\\left( x \\right)}^3}f'\\left( x \\right)} \\right]} \\over 1}$$\n
$$\\int\\limits_1^x {y\\left( t \\right)dt + x\\left[ {y\\left( x \\right) - y\\left( 1 \\right)} \\right]} $$\n
$$ = \\int\\limits_1^x {ty\\left( t \\right)dt + x\\left[ {xy\\left( x \\right) - y\\left( 1 \\right)} \\right] + xy\\left( x \\right) - y\\left( 1 \\right)} $$\n
$$\\int\\limits_1^x {y\\left( t \\right)dt = \\int\\limits_1^x {ty\\left( t \\right)dt + {x^2}y\\left( x \\right) - y\\left( 1 \\right)} } $$\n
Diff. again w.r.t to x\n
$$y\\left( x \\right) - y\\left( a \\right) = xy\\left( x \\right) - y\\left( a \\right) + 2x\\,y\\left( x \\right) + {x^2}{y^3}\\left( x \\right)$$\n $$\\left( {1 - 3x} \\right)y\\left( x \\right) = {x^2}{y^3}\\,\\left( x \\right)$$\n
$${{{y^3}\\left( x \\right)} \\over {y\\left( x \\right)}} = {{1 - 3x} \\over {{x^2}}}$$\n
Then for the function F, the point x = 1 is :",
"options": [
{
"text": "a point of inflection."
},
{
"text": "a point of local maxima."
},
{
"text": "a point of local minima."
},
{
"text": "not a critical point."
}
],
"answer": "a point of local minima.",
"solution": "**Answer:** a point of local minima.\n\n$$F(x) = \\int\\limits_1^x {{t^2}g(t)dt} $$\n
$$ \\Rightarrow f(x) = 2{e^{{e^x} - 1}} - 1$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4614,
"subject": "General Science",
"question": "If the normal to the curve y(x) = $$\\int\\limits_0^x {(2{t^2} - 15t + 10)dt} $$ at a point (a, b) is parallel to the line x + 3y = $$-$$5, a > 1, then the value of | a + 6b | is equal to ___________.",
"options": [],
"answer": "406",
"solution": "**Answer:** 406\n\nNormal to the curve at point P(a, b) is parallel to the line x + 3y = $$-$$5.\n
$${{330} \\over {360}} < 1 < {{331} \\over {360}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4616,
"subject": "General Science",
"question": "Let f : (a, b) $$\\to$$ R be twice differentiable function such that $$f(x) = \\int_a^x {g(t)dt} $$ for a differentiable function g(x). If f(x) = 0 has exactly five distinct roots in (a, b), then g(x)g'(x) = 0 has at least :",
"options": [
{
"text": "twelve roots in (a, b)"
},
{
"text": "five roots in (a, b)"
},
{
"text": "seven roots in (a, b)"
},
{
"text": "three roots in (a, b)"
}
],
"answer": "seven roots in (a, b)",
"solution": "**Answer:** seven roots in (a, b)\n\n$$f(x) = \\int_a^x {g(t)dt} $$\n
$$ \\Rightarrow $$ f′(x) = g(x)\n
$$ \\Rightarrow $$ f′'(x) = g'(x)\n
Given, g(x).g'(x) = 0\n
$$ \\Rightarrow $$ f′(x).f′'(x) = 0\n
Also given f(x) has exactly 5 roots.\n
So from Rolle's theorem we can say,\n
f′(x) has 4 roots and f′'(x) has 3 roots.\n
$$ \\therefore $$ f′(x).f′'(x) = 0 has 4 + 3 = 7 roots.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4617,
"subject": "General Science",
"question": "Let $$F:[3,5] \\to R$$ be a twice differentiable function on (3, 5) such that
If m and n respectively are the number of local maximum and local minimum points of the function $$f(x) = \\int\\limits_0^{{x^2}} {{{{t^2} - 5t + 4} \\over {2 + {e^t}}}dt} $$, then the ordered pair (m, n) is equal to
Let f be a differentiable function in $$\\left( {0,{\\pi \\over 2}} \\right)$$. If $$\\int\\limits_{\\cos x}^1 {{t^2}\\,f(t)dt = {{\\sin }^3}x + \\cos x} $$, then $${1 \\over {\\sqrt 3 }}f'\\left( {{1 \\over {\\sqrt 3 }}} \\right)$$ is equal to
Let $$f$$ be a twice differentiable function on $$\\mathbb{R}$$. If $$f^{\\prime}(0)=4$$ and $$f(x) + \\int\\limits_0^x {(x - t)f'(t)dt = \\left( {{e^{2x}} + {e^{ - 2x}}} \\right)\\cos 2x + {2 \\over a}x} $$, then $$(2 a+1)^{5}\\, a^{2}$$ is equal to _______________.
Let f be a differentiable function satisfying $$f(x)=\\frac{2}{\\sqrt{3}} \\int\\limits_{0}^{\\sqrt{3}} f\\left(\\frac{\\lambda^{2} x}{3}\\right) \\mathrm{d} \\lambda, x>0$$ and $$f(1)=\\sqrt{3}$$. If $$y=f(x)$$ passes through the point $$(\\alpha, 6)$$, then $$\\alpha$$ is equal to _____________.
The minimum value of the twice differentiable function $$f(x)=\\int\\limits_{0}^{x} \\mathrm{e}^{x-\\mathrm{t}} f^{\\prime}(\\mathrm{t}) \\mathrm{dt}-\\left(x^{2}-x+1\\right) \\mathrm{e}^{x}$$, $$x \\in \\mathbf{R}$$, is :
Let $$f$$ be $$a$$ differentiable function defined on $$\\left[ {0,{\\pi \\over 2}} \\right]$$ such that $$f(x) > 0$$ and $$f(x) + \\int_0^x {f(t)\\sqrt {1 - {{({{\\log }_e}f(t))}^2}} dt = e,\\forall x \\in \\left[ {0,{\\pi \\over 2}} \\right]}$$. Then $$\\left( {6{{\\log }_e}f\\left( {{\\pi \\over 6}} \\right)} \\right)^2$$ is equal to __________.
Let $$f$$ be a continuous function satisfying $$\\int_\\limits{0}^{t^{2}}\\left(f(x)+x^{2}\\right) d x=\\frac{4}{3} t^{3}, \\forall t > 0$$. Then\n\n$$f\\left(\\frac{\\pi^{2}}{4}\\right)$$ is equal to :
Let $$S=(-1, \\infty)$$ and $$f: S \\rightarrow \\mathbb{R}$$ be defined as
\n
$$f(x)=\\int_\\limits{-1}^x\\left(e^t-1\\right)^{11}(2 t-1)^5(t-2)^7(t-3)^{12}(2 t-10)^{61} d t \\text {, }$$
\n
Let $$\\mathrm{p}=$$ Sum of squares of the values of $$x$$, where $$f(x)$$ attains local maxima on $$S$$, and $$\\mathrm{q}=$$ Sum of the values of $$\\mathrm{x}$$, where $$f(x)$$ attains local minima on $$S$$. Then, the value of $$p^2+2 q$$ is _________.
Let the slope of the line $$45 x+5 y+3=0$$ be $$27 r_1+\\frac{9 r_2}{2}$$ for some $$r_1, r_2 \\in \\mathbb{R}$$. Then $$\\lim _\\limits{x \\rightarrow 3}\\left(\\int_3^x \\frac{8 t^2}{\\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} d t\\right)$$ is equal to _________.
Let $$\\mathrm{f}: \\mathbb{R} \\rightarrow \\mathbb{R}$$ be defined as $$f(x)=a e^{2 x}+b e^x+c x$$. If $$f(0)=-1, f^{\\prime}\\left(\\log _e 2\\right)=21$$ and $$\\int_0^{\\log _e 4}(f(x)-c x) d x=\\frac{39}{2}$$, then the value of $$|a+b+c|$$ equals
Let $$f(x)=\\int_0^x\\left(t+\\sin \\left(1-e^t\\right)\\right) d t, x \\in \\mathbb{R}$$. Then, $$\\lim _\\limits{x \\rightarrow 0} \\frac{f(x)}{x^3}$$ is equal to
$$\\therefore$$ $$I = \\pi \\log 2$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4661,
"subject": "General Science",
"question": "Statement-1 : The value of the integral \n $$\\int\\limits_{\\pi /6}^{\\pi /3} {{{dx} \\over {1 + \\sqrt {\\tan \\,x} }}} $$ is equal to $$\\pi /6$$\n
Statement-2 : $$\\int\\limits_a^b {f\\left( x \\right)} dx = \\int\\limits_a^b {f\\left( {a + b - x} \\right)} dx.$$
",
"options": [
{
"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."
},
{
"text": "Statement-1 is false; Statement-2 is true."
}
],
"answer": "Statement-1 is false; Statement-2 is true.",
"solution": "**Answer:** Statement-1 is false; Statement-2 is true.\n\nLet $$I = \\int\\limits_{\\pi /6}^{\\pi /3} {{{dx} \\over {1 + \\sqrt {\\tan \\,x} }}} $$\n
$$ = {1 \\over 2}\\left[ {9.(0) - {1 \\over 3}.{1 \\over 3}(1)} \\right] = - {1 \\over {18}}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4694,
"subject": "General Science",
"question": "Let {x} and [x] denote the fractional part of x and the greatest integer $$ \\le $$ x respectively of a real number x. If $$\\int_0^n {\\left\\{ x \\right\\}dx} ,\\int_0^n {\\left[ x \\right]dx} $$ and 10(n2 – n), $$\\left( {n \\in N,n > 1} \\right)$$ are three consecutive terms of a G.P., then n is equal to_____.",
"options": [],
"answer": "21",
"solution": "**Answer:** 21\n\n$$\\int\\limits_0^n {\\left\\{ x \\right\\}} dx = n\\int\\limits_0^1 x dx = n\\left( {{{{x^2}} \\over 2}} \\right)_0^1 = {n \\over 2}$$\n
= $$2\\left[ {\\pi x - {{{x^2}} \\over 2}} \\right]_0^\\pi $$\n
= $${\\pi ^2}$$ ",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4701,
"subject": "General Science",
"question": "If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed positive real numbers, then
$$ \\Rightarrow {I_{m,n}} = \\int\\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \\over {{{(y + 1)}^{m + n}}}}dy} \\Rightarrow \\alpha = 1$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4719,
"subject": "General Science",
"question": "Let f : R $$ \\to $$ R be a continuous function such that f(x) + f(x + 1) = 2, for all x$$\\in$$R.
If $${I_1} = \\int\\limits_0^8 {f(x)dx} $$ and $${I_2} = \\int\\limits_{ - 1}^3 {f(x)dx} $$, then the value of I1 + 2I2 is equal to ____________.",
"options": [],
"answer": "16",
"solution": "**Answer:** 16\n\n$$f(x) + f(x + 1) = 2$$ .... (i)
$$ \\Rightarrow $$ $${I_1} + 2{I_2} = 16$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4720,
"subject": "General Science",
"question": "Consider the integral $$I = \\int_0^{10} {{{[x]{e^{[x]}}} \\over {{e^{x - 1}}}}dx} $$, where [x] denotes the greatest integer less than or equal to x. Then the value of I is equal to :",
"options": [
{
"text": "45 (e $$-$$ 1)"
},
{
"text": "45 (e + 1)"
},
{
"text": "9 (e + 1)"
},
{
"text": "9 (e $$-$$ 1)"
}
],
"answer": "45 (e $$-$$ 1)",
"solution": "**Answer:** 45 (e $$-$$ 1)\n\n$$I = \\int_0^{10} {[x]\\,.\\,{e^{[x] + 1 - x}}} dx$$
9(b + c) = 7",
"topic": "Algebra",
"subtopic": "Algebraic Expressions"
},
{
"id": 4722,
"subject": "General Science",
"question": "Which of the following statements is correct for the function g($$\\alpha$$) for $$\\alpha$$ $$\\in$$ R such that
$$g(\\alpha ) = \\int\\limits_{{\\pi \\over 6}}^{{\\pi \\over 3}} {{{{{\\sin }^\\alpha }x} \\over {{{\\cos }^\\alpha }x + {{\\sin }^\\alpha }x}}dx} $$",
"options": [
{
"text": "$$g(\\alpha )$$ is a strictly increasing function"
},
{
"text": "$$g(\\alpha )$$ is an even function"
},
{
"text": "$$g(\\alpha )$$ has an inflection point at $$\\alpha$$ = $$-$$$${1 \\over 2}$$"
},
{
"text": "$$g(\\alpha )$$ is a strictly decreasing function"
}
],
"answer": "$$g(\\alpha )$$ is an even function",
"solution": "**Answer:** $$g(\\alpha )$$ is an even function\n\n$$g(\\alpha ) = \\int\\limits_{\\pi /6}^{\\pi /3} {{{{{\\sin }^\\alpha }\\left( {{\\pi \\over 2} - x} \\right)} \\over {{{\\cos }^\\alpha }\\left( {{\\pi \\over 2} - x} \\right)x + {{\\sin }^\\alpha }\\left( {{\\pi \\over 2} - x} \\right)}}dx} $$
$$ \\Rightarrow $$ $$g(\\alpha ) = {\\pi \\over {12}}$$ i.e. a constant function hence an even function.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4723,
"subject": "General Science",
"question": "If [ . ] represents the greatest integer function, then the value of
\n$\\Rightarrow \\frac{1}{3} \\leq g(3) \\leq 2$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4728,
"subject": "General Science",
"question": "Let P(x) be a real polynomial of degree 3 which vanishes at x = $$-$$3. Let P(x) have local minima at x = 1, local maxima at x = $$-$$1 and $$\\int\\limits_{ - 1}^1 {P(x)dx} $$ = 18, then the sum of all the coefficients of the polynomial P(x) is equal to _________.",
"options": [],
"answer": "8",
"solution": "**Answer:** 8\n\nP'(x) = a(x + 1)(x $$-$$ 1)
where [x] is the greatest integer less than or equal to x. Which of the following is true?",
"options": [
{
"text": "f is continuous at every point in $$[0,\\infty )$$ and differentiable except at the integer points."
},
{
"text": "f is both continuous and differentiable except at the integer points in $$[0,\\infty )$$."
},
{
"text": "f is continuous everywhere except at the integer points in $$[0,\\infty )$$."
},
{
"text": "f is differentiable at every point in $$[0,\\infty )$$."
}
],
"answer": "f is continuous at every point in $$[0,\\infty )$$ and differentiable except at the integer points.",
"solution": "**Answer:** f is continuous at every point in $$[0,\\infty )$$ and differentiable except at the integer points.\n\n$$f:[0,\\infty ) \\to [0,\\infty ),f(x) = \\int_0^x {[y]dy} $$
$$\\Rightarrow$$ Not differentiable at x = 2",
"topic": "Algebra",
"subtopic": "Linear Algebra"
},
{
"id": 4739,
"subject": "General Science",
"question": "The value of the
$$\\because$$ $$\\log \\left( {x + \\sqrt {{x^2} + 1} } \\right)$$ is an odd function
$$\\therefore$$ I = 0",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4740,
"subject": "General Science",
"question": "The value of the definite integral
$$I = 0 - {1 \\over {\\sqrt 2 }}\\left( { - {\\pi \\over 2}} \\right) = {\\pi \\over {2\\sqrt 2 }}$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4741,
"subject": "General Science",
"question": "Let the domain of the function
$$f(x) = {\\log _4}\\left( {{{\\log }_5}\\left( {{{\\log }_3}(18x - {x^2} - 77)} \\right)} \\right)$$ be (a, b). Then the value of the integral $$\\int\\limits_a^b {{{{{\\sin }^3}x} \\over {({{\\sin }^3}x + {{\\sin }^3}(a + b - x)}}} dx$$ is equal to _____________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nFor domain
where [t] is the greatest integer less than or equal to t. Let m be the number of points where f is not differentiable and $$I = \\int\\limits_{ - 2}^2 {f(x)\\,dx} $$. Then the ordered pair (m, I) is equal to :
Let f : R $$\\to$$ R be a differentiable function such that $$f\\left( {{\\pi \\over 4}} \\right) = \\sqrt 2 ,\\,f\\left( {{\\pi \\over 2}} \\right) = 0$$ and $$f'\\left( {{\\pi \\over 2}} \\right) = 1$$ and
let $$g(x) = \\int_x^{\\pi /4} {(f'(t)\\sec t + \\tan t\\sec t\\,f(t))\\,dt} $$ for $$x \\in \\left[ {{\\pi \\over 4},{\\pi \\over 2}} \\right)$$. Then $$\\mathop {\\lim }\\limits_{x \\to {{\\left( {{\\pi \\over 2}} \\right)}^ - }} g(x)$$ is equal to :
\n$$\n\\begin{aligned}\n&g(x)=\\int_{x}^{\\frac{\\pi}{4}}\\left(f^{\\prime}(t) \\sec t+\\tan t \\sec t f(t)\\right) d t \\\\\\\\\n&=[\\sec t+f(t)]_{x}^{\\frac{\\pi}{4}}=2-\\sec x f(x)\n\\end{aligned}\n$$\n
Let f : R $$\\to$$ R be a continuous function satisfying f(x) + f(x + k) = n, for all x $$\\in$$ R where k > 0 and n is a positive integer. If $${I_1} = \\int\\limits_0^{4nk} {f(x)dx} $$ and $${I_2} = \\int\\limits_{ - k}^{3k} {f(x)dx} $$, then :
\n$$\n\\begin{aligned}\n&\\text { Now, } I_{1}=\\int_{0}^{4 n k} f(x) d x = 2 n \\int_{0}^{2 k} f(x) d x \\\\\\\\\n&=2 n\\left[\\int_{0}^{k} f(x) d x+\\int_{k}^{2 k} f(x) d x\\right] \\\\\\\\\n&x=t+k \\Rightarrow d x=d t \\text { (in second integral) } \\\\\\\\\n&=2 n\\left[\\int_{0}^{k} f(x) d x+\\int_{0}^{k} f(t+k) d t\\right] \\\\\\\\\n&=2 n^{2} k\n\\end{aligned}\n$$\n
\nNow, $I_2=\\int_{-k}^{3 k} f(x) d x=2 \\int_{0}^{2 k} f(x) d x$\n
Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral $$\\int\\limits_0^1 {[ - 8{x^2} + 6x - 1]dx} $$ is equal to :
\n \n$=\\int_{0}^{\\frac{1}{4}}(-1) d x+\\int_{\\frac{1}{4}}^{\\frac{3}{4}} 0 d x+\\int_{\\frac{1}{2}}^{\\frac{3}{4}}-1 d x+\\int_{\\frac{3}{4}}^{8}-2 d x+\\int_{\\frac{3+\\sqrt{17}}{8}}^{1}-3 d x$
The integral $$\\int\\limits_0^1 {{1 \\over {{7^{\\left[ {{1 \\over x}} \\right]}}}}dx} $$, where [ . ] denotes the greatest integer function, is equal to
If $${b_n} = \\int_0^{{\\pi \\over 2}} {{{{{\\cos }^2}nx} \\over {\\sin x}}dx,\\,n \\in N} $$, then
",
"options": [
{
"text": "$${b_3} - {b_2},\\,{b_4} - {b_3},\\,{b_5} - {b_4}$$ are in A.P. with common difference $$-$$2"
},
{
"text": "$${1 \\over {{b_3} - {b_2}}},{1 \\over {{b_4} - {b_3}}},{1 \\over {{b_5} - {b_4}}}$$ are in an A.P. with common difference 2 "
},
{
"text": "$${b_3} - {b_2},\\,{b_4} - {b_3},\\,{b_5} - {b_4}$$ are in a G.P."
},
{
"text": "$${1 \\over {{b_3} - {b_2}}},{1 \\over {{b_4} - {b_3}}},{1 \\over {{b_5} - {b_4}}}$$ are in an A.P. with common difference $$-$$2"
}
],
"answer": "$${1 \\over {{b_3} - {b_2}}},{1 \\over {{b_4} - {b_3}}},{1 \\over {{b_5} - {b_4}}}$$ are in an A.P. with common difference $$-$$2",
"solution": "**Answer:** $${1 \\over {{b_3} - {b_2}}},{1 \\over {{b_4} - {b_3}}},{1 \\over {{b_5} - {b_4}}}$$ are in an A.P. with common difference $$-$$2\n\n
The value of b > 3 for which $$12\\int\\limits_3^b {{1 \\over {({x^2} - 1)({x^2} - 4)}}dx = {{\\log }_e}\\left( {{{49} \\over {40}}} \\right)} $$, is equal to ___________.
\nClearly $\\alpha_{1}=18$ and $\\alpha_{2}=16$, so $\\alpha_{1}+\\alpha_{2}=34$.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4770,
"subject": "General Science",
"question": "
Let $$f(t) = \\int\\limits_0^t {{e^{{x^3}}}\\left( {{{{x^8}} \\over {{{({x^6} + 2{x^3} + 2)}^2}}}} \\right)dx} $$. If $$f(1) + f'(1) = \\alpha e - {1 \\over 6}$$, then the value of 150$$\\alpha$$ is equal to ___________.
For any real number $$x$$, let $$[x]$$ denote the largest integer less than equal to $$x$$. Let $$f$$ be a real valued function defined on the interval $$[-10,10]$$ by $$f(x)=\\left\\{\\begin{array}{l}x-[x], \\text { if }[x] \\text { is odd } \\\\ 1+[x]-x, \\text { if }[x] \\text { is even } .\\end{array}\\right.$$\n\nThen the value of $$\\frac{\\pi^{2}}{10} \\int_{-10}^{10} f(x) \\cos \\pi x \\,d x$$ is :
Let $$[t]$$ denote the greatest integer less than or equal to $$t$$. Then the value of the integral $$\\int_{-3}^{101}\\left([\\sin (\\pi x)]+e^{[\\cos (2 \\pi x)]}\\right) d x$$ is equal to
Let $${a_n} = \\int\\limits_{ - 1}^n {\\left( {1 + {x \\over 2} + {{{x^2}} \\over 3} + \\,\\,.....\\,\\, + \\,\\,{{{x^{n - 1}}} \\over n}} \\right)dx} $$ for every n $$\\in$$ N. Then the sum of all the elements of the set {n $$\\in$$ N : an $$\\in$$ (2, 30)} is ____________.
If $$\\mathrm{n}(2 \\mathrm{n}+1) \\int_{0}^{1}\\left(1-x^{\\mathrm{n}}\\right)^{2 \\mathrm{n}} \\mathrm{d} x=1177 \\int_{0}^{1}\\left(1-x^{\\mathrm{n}}\\right)^{2 \\mathrm{n}+1} \\mathrm{~d} x$$, then $$\\mathrm{n} \\in \\mathbf{N}$$ is equal to ______________.
Let $$f: \\mathbb{R} \\rightarrow \\mathbb{R}$$ be a function defined as
\n
$$f(x)=a \\sin \\left(\\frac{\\pi[x]}{2}\\right)+[2-x], a \\in \\mathbb{R}$$ where $$[t]$$ is the greatest integer less than or equal to $$t$$. If $$\\mathop {\\lim }\\limits_{x \\to -1 } f(x)$$ exists, then the value of $$\\int\\limits_{0}^{4} f(x) d x$$ is equal to
Let a function $$f: \\mathbb{R} \\rightarrow \\mathbb{R}$$ be defined as :
\n
$$f(x)= \\begin{cases}\\int\\limits_{0}^{x}(5-|t-3|) d t, & x>4 \\\\ x^{2}+b x & , x \\leq 4\\end{cases}$$
\n
where $$\\mathrm{b} \\in \\mathbb{R}$$. If $$f$$ is continuous at $$x=4$$, then which of the following statements is NOT true?
",
"options": [
{
"text": "$$f$$ is not differentiable at $$x=4$$"
},
{
"text": "$$f^{\\prime}(3)+f^{\\prime}(5)=\\frac{35}{4}$$"
},
{
"text": "$$f$$ is increasing in $$\\left(-\\infty, \\frac{1}{8}\\right) \\cup(8, \\infty)$$"
},
{
"text": "$$f$$ has a local minima at $$x=\\frac{1}{8}$$"
}
],
"answer": "$$f$$ is increasing in $$\\left(-\\infty, \\frac{1}{8}\\right) \\cup(8, \\infty)$$",
"solution": "**Answer:** $$f$$ is increasing in $$\\left(-\\infty, \\frac{1}{8}\\right) \\cup(8, \\infty)$$\n\n
$$\\int\\limits_{0}^{2}\\left(\\left|2 x^{2}-3 x\\right|+\\left[x-\\frac{1}{2}\\right]\\right) \\mathrm{d} x$$, where [t] is the greatest integer function, is equal to :
Let $$f(x)=\\min \\{[x-1],[x-2], \\ldots,[x-10]\\}$$ where [t] denotes the greatest integer $$\\leq \\mathrm{t}$$. Then $$\\int\\limits_{0}^{10} f(x) \\mathrm{d} x+\\int\\limits_{0}^{10}(f(x))^{2} \\mathrm{~d} x+\\int\\limits_{0}^{10}|f(x)| \\mathrm{d} x$$ is equal to ________________.
If $$\\int\\limits_{0}^{\\sqrt{3}} \\frac{15 x^{3}}{\\sqrt{1+x^{2}+\\sqrt{\\left(1+x^{2}\\right)^{3}}}} \\mathrm{~d} x=\\alpha \\sqrt{2}+\\beta \\sqrt{3}$$, where $$\\alpha, \\beta$$ are integers, then $$\\alpha+\\beta$$ is equal to __________.
If $$f(\\alpha)=\\int\\limits_{1}^{\\alpha} \\frac{\\log _{10} \\mathrm{t}}{1+\\mathrm{t}} \\mathrm{dt}, \\alpha>0$$, then $$f\\left(\\mathrm{e}^{3}\\right)+f\\left(\\mathrm{e}^{-3}\\right)$$ is equal to :
If $$[t]$$ denotes the greatest integer $$\\leq t$$, then the value of $$\\int_{0}^{1}\\left[2 x-\\left|3 x^{2}-5 x+2\\right|+1\\right] \\mathrm{d} x$$ is :
If $$\\int_\\limits{0}^{1}\\left(x^{21}+x^{14}+x^{7}\\right)\\left(2 x^{14}+3 x^{7}+6\\right)^{1 / 7} d x=\\frac{1}{l}(11)^{m / n}$$ where $$l, m, n \\in \\mathbb{N}, m$$ and $$n$$ are coprime then $$l+m+n$$ is equal to ____________.
Let $$f: \\mathbb{R} \\rightarrow \\mathbb{R}$$ be a differentiable function such that $$f^{\\prime}(x)+f(x)=\\int_\\limits{0}^{2} f(t) d t$$. If $$f(0)=e^{-2}$$, then $$2 f(0)-f(2)$$ is equal to ____________.
Let $$\\alpha \\in (0,1)$$ and $$\\beta = {\\log _e}(1 - \\alpha )$$. Let $${P_n}(x) = x + {{{x^2}} \\over 2} + {{{x^3}} \\over 3}\\, + \\,...\\, + \\,{{{x^n}} \\over n},x \\in (0,1)$$. Then the integral $$\\int\\limits_0^\\alpha {{{{t^{50}}} \\over {1 - t}}dt} $$ is equal to
$$\n\\begin{array}{r}\n=\\int\\limits_{\\pi / 3}^{\\pi / 2} \\frac{2}{\\sin ^3 x} d x-\\int\\limits_{\\pi / 3}^{\\pi / 2} 2 \\cot x \\operatorname{cosec}^2 x d x \n+\\int\\limits_{\\pi / 3}^{\\pi / 2} \\frac{3}{2} \\sec ^2 \\frac{x}{2} d x\n\\end{array}\n$$\n
$$\n=2 I_1-2 I_2+\\frac{3}{2} I_3\n$$\n
Now, \n
$$\n \\begin{aligned}\nI_1 & =\\int\\limits_{\\pi / 3}^{\\pi / 2} \\operatorname{cosec}^3 x d x \\\\\\\\\n& =\\int\\limits_{\\pi / 3}^{\\pi / 2} \\sqrt{1+\\cot ^2 x} \\operatorname{cosec}^2 x d x\n\\end{aligned}\n$$\n
Put $\\cot x=t \\Rightarrow-\\operatorname{cosec}^2 x d x=d t$\n
$$\n\\begin{aligned}\n\\therefore & I_1=-\\int\\limits_{\\frac{1}{\\sqrt{3}}}^0 \\sqrt{1+t^2} d t \\\\\\\\\n= & \\int\\limits_0^{\\frac{1}{\\sqrt{3}}} \\sqrt{1+t^2} d t \\\\\\\\\n= & {\\left[\\frac{t}{2} \\sqrt{1+t^2}+\\frac{1}{2} \\log \\left[t+\\sqrt{1+t^2}\\right]\\right]_0^{\\frac{1}{\\sqrt{3}}} } \\\\\\\\\n= & \\frac{1}{3}+\\frac{1}{2} \\log \\sqrt{3}\n\\end{aligned}\n$$\n
$$\nI_2=\\int\\limits_{\\pi / 3}^{\\pi / 2} \\cot x \\operatorname{cosec}^2 x d x\n$$\n
Put $\\cot x=t \\Rightarrow \\operatorname{cosec}^2 x d x=-d t$\n
When, $x=\\frac{\\pi}{3}$\n$\\Rightarrow t=\\frac{1}{\\sqrt{3}}$ and $x=\\frac{\\pi}{2} \\Rightarrow t=0$\n
$$\n\\therefore \\begin{aligned}\nI_2 & =-\\int\\limits_{\\frac{1}{\\sqrt{3}}}^0 t d t \\\\\\\\\n& =\\int\\limits_0^{\\frac{1}{\\sqrt{3}} t} t d t=\\left[\\frac{t^2}{2}\\right]_0^{\\frac{1}{\\sqrt{3}}}=\\frac{1}{6}\n\\end{aligned}\n$$\n
$$\n\\begin{aligned}\n& I_3=\\int\\limits_{\\frac{\\pi}{3}}^{\\frac{\\pi}{2}} \\sec ^2 \\frac{x}{2} d x \\\\\\\\\n= & 2\\left(\\tan \\frac{\\pi}{4}-\\tan \\frac{\\pi}{6}\\right)=2\\left(1-\\frac{1}{\\sqrt{3}}\\right)\n\\end{aligned}\n$$\n
If [t] denotes the greatest integer $$\\le \\mathrm{t}$$, then the value of $${{3(e - 1)} \\over e}\\int\\limits_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} $$ is :
If $$\\int\\limits_{{1 \\over 3}}^3 {|{{\\log }_e}x|dx = {m \\over n}{{\\log }_e}\\left( {{{{n^2}} \\over e}} \\right)} $$, where m and n are coprime natural numbers, then $${m^2} + {n^2} - 5$$ is equal to _____________.
",
"options": [],
"answer": "20",
"solution": "**Answer:** 20\n\n$I=\\int\\limits_{\\frac{1}{3}}^{3}|\\ln x| d x=-\\int\\limits_{\\frac{1}{3}}^{1} \\ln x d x+\\int\\limits_{1}^{3} \\ln x d x$\n
The value of $${8 \\over \\pi }\\int\\limits_0^{{\\pi \\over 2}} {{{{{(\\cos x)}^{2023}}} \\over {{{(\\sin x)}^{2023}} + {{(\\cos x)}^{2023}}}}dx} $$ is ___________
$$\nI=\\frac{1}{\\sqrt{11}}\\int du = \\frac{1}{\\sqrt{11}}(u+C)\n$$
\n\n
Now we need to find the limits of the integral after the substitution. If $x=0$, then $u=\\tan^{-1}\\left(-\\frac{1}{\\sqrt{11}}\\right)$. If $x=1$, then $u=\\tan^{-1}\\left(\\frac{1}{\\sqrt{11}}\\right)$. So, the integral becomes:
Notice that the integrals in the numerator and denominator have the same form. They both involve an integral of $e^{-x} \\tan^n x$ from 0 to $\\pi / 4$, where $n$ is an integer. Let's denote this integral as $I(n)$:
\n
$$I(n) = \\int_0^{\\pi / 4} e^{-x} \\tan^n x dx$$
\n
We can then rewrite the original expression in terms of $I(n)$:
\n
$$\\frac{e^{-\\pi/4} + I(50)}{I(49) + I(51)}$$
\n
Now, we'll apply the method of integration by parts, which states that for two functions $u(x)$ and $v(x)$:
\n
$$\\int u dv = uv - \\int v du$$
\n
We'll choose:
\n
$$u = \\tan^n x, \\quad dv = e^{-x} dx$$
\n
Then we get:
\n
$$du = n \\tan^{n-1} x \\sec^2 x dx, \\quad v = -e^{-x}$$
\n
Applying integration by parts, we have:
\n
$$I(n) = -e^{-x} \\tan^n x \\Bigg|_0^{\\pi / 4} + n \\int_0^{\\pi / 4} e^{-x} \\tan^{n-1} x \\sec^2 x dx$$
\n
Since $\\tan(\\pi / 4) = 1$, the first term evaluates to:
\n
$$-e^{-\\pi / 4}$$
\n
The second term becomes:
\n
$$n \\int_0^{\\pi / 4} e^{-x} \\tan^{n-1} x (1 + \\tan^2 x) dx = n \\int_0^{\\pi / 4} e^{-x} \\tan^{n-1} x dx + n \\int_0^{\\pi / 4} e^{-x} \\tan^{n+1} x dx$$
\n
This is equal to:
\n
$$n(I(n-1) + I(n+1))$$
\n
So we have:
\n
$$I(n) = -e^{-\\pi / 4} + n(I(n-1) + I(n+1))$$
\n
Now we can substitute $n = 50$ into this equation:
Let $$f_{n}=\\int_\\limits{0}^{\\frac{\\pi}{2}}\\left(\\sum_\\limits{k=1}^{n} \\sin ^{k-1} x\\right)\\left(\\sum_\\limits{k=1}^{n}(2 k-1) \\sin ^{k-1} x\\right) \\cos x d x, n \\in \\mathbb{N}$$. Then $$f_{21}-f_{20}$$ is equal to _________
",
"options": [],
"answer": "41",
"solution": "**Answer:** 41\n\nGiven, $$f_{n}=\\int_\\limits{0}^{\\frac{\\pi}{2}}\\left(\\sum_\\limits{k=1}^{n} \\sin ^{k-1} x\\right)\\left(\\sum_\\limits{k=1}^{n}(2 k-1) \\sin ^{k-1} x\\right) \\cos x d x$$\n
$$\n\\begin{aligned}\n& \\text { Put } \\sin x=t \\\\\\\\\n& \\cos x d x=d t \\\\\\\\\n& f_n=\\int_0^1\\left(\\sum_{k=1}^n(t)^{k-1}\\right)\\left(\\sum_{k=1}^n(2 k-1)(t)^{k-1}\\right) d t\n\\end{aligned}\n$$\n
$$ \\therefore $$ $$\n\\begin{aligned}\n f_{21}=\\int_0^1\\left(\\sum_{k=1}^{21}(t)^{20}\\right)\\left(\\sum_{k=1}^{21}(2 k-1)(t)^{20}\\right) d t\n\\end{aligned}\n$$\n
$$\n\\begin{aligned}\n& S_1(x)=C_1(x)+\\int_0^x S_0(t) d t \\\\\\\\\n& =C_1 x+\\frac{x^2}{2} \\\\\\\\\n& S_2(3)=3 C_2+2 \\int_0^3\\left(C_1 t+\\frac{t^2}{2}\\right) d t \\\\\\\\\n& =3 C_2+9 C_1+9\n\\end{aligned}\n$$\n
If $$f: \\mathbb{R} \\rightarrow \\mathbb{R}$$ be a continuous function satisfying $$\\int_\\limits{0}^{\\frac{\\pi}{2}} f(\\sin 2 x) \\sin x d x+\\alpha \\int_\\limits{0}^{\\frac{\\pi}{4}} f(\\cos 2 x) \\cos x d x=0$$, then the value of $$\\alpha$$ is :
",
"options": [
{
"text": "$$-\\sqrt{3}$$"
},
{
"text": "$$\\sqrt{2}$$"
},
{
"text": "$$-\\sqrt{2}$$"
},
{
"text": "$$\\sqrt{3}$$"
}
],
"answer": "$$-\\sqrt{2}$$",
"solution": "**Answer:** $$-\\sqrt{2}$$\n\nThe integral equation is given by :\n\n
$$\\int\\limits_0^{\\frac{\\pi}{2}} f(\\sin 2x) \\sin x \\, dx + \\alpha \\int\\limits_0^{\\frac{\\pi}{4}} f(\\cos 2x) \\cos x \\, dx = 0$$\n\n
Step 1 : Break the first integral into two parts :\n\n
$$I = \\int\\limits_0^{\\frac{\\pi}{4}} f(\\sin 2x) \\sin x \\, dx + \\int\\limits_{\\frac{\\pi}{4}}^{\\frac{\\pi}{2}} f(\\sin 2x) \\sin x \\, dx + \\alpha \\int\\limits_0^{\\frac{\\pi}{4}} f(\\cos 2x) \\cos x \\, dx$$\n\n
3. Apply the King's property, $\\int\\limits_a^b f(x) dx = \\int\\limits_a^b f(a+b-x) dx$, to the first integral and substitute $x - \\frac{\\pi}{4} = t$ in the second integral. \n\n
In order for this equation to hold true, either the integral of the function is zero, or the term outside the integral is zero. Since we have no reason to assume that the integral of the function is zero, we set the term outside the integral to zero, yielding the solution:\n\n
So, the correct answer to the original problem is $\\alpha = -\\sqrt{2}$, which corresponds to Option C.",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4812,
"subject": "General Science",
"question": "
Let the function $$f:[0,2] \\rightarrow \\mathbb{R}$$ be defined as
\n
$$f(x)= \\begin{cases}e^{\\min \\left\\{x^{2}, x-[x]\\right\\},} & x \\in[0,1) \\\\ e^{\\left[x-\\log _{e} x\\right]}, & x \\in[1,2]\\end{cases}$$
\n
where $$[t]$$ denotes the greatest integer less than or equal to $$t$$. Then the value of the integral $$\\int_\\limits{0}^{2} x f(x) d x$$ is :
$$\n\\begin{aligned}\n& \\int\\limits_0^2 x f(x)=\\int\\limits_0^1 x \\cdot e^{x^2} d x+\\int\\limits_1^2 x \\cdot e d x \\\\\\\\\n& x^2=t \\Rightarrow 2 x d x=d t\n\\end{aligned}\n$$\n
$$\n=\\frac{1}{2} \\int\\limits_0^1 e^t d t+\\left.e \\frac{x^2}{2}\\right|_1 ^2\n$$\n
The value of the integral $$\\int_\\limits{-\\log _{e} 2}^{\\log _{e} 2} e^{x}\\left(\\log _{e}\\left(e^{x}+\\sqrt{1+e^{2 x}}\\right)\\right) d x$$ is equal to :
For $$m, n > 0$$, let $$\\alpha(m, n)=\\int_\\limits{0}^{2} t^{m}(1+3 t)^{n} d t$$. If $$11 \\alpha(10,6)+18 \\alpha(11,5)=p(14)^{6}$$, then $$p$$ is equal to ___________.
Let $$[t]$$ denote the greatest integer function. If $$\\int_\\limits{0}^{2.4}\\left[x^{2}\\right] d x=\\alpha+\\beta \\sqrt{2}+\\gamma \\sqrt{3}+\\delta \\sqrt{5}$$, then $$\\alpha+\\beta+\\gamma+\\delta$$ is equal to __________.
",
"options": [],
"answer": "6",
"solution": "**Answer:** 6\n\n$$\n\\begin{aligned}\n\\int\\limits_0^{2.4}\\left[x^2\\right] d x & =\\int\\limits_0^1\\left[x^2\\right] d x+\\int\\limits_1^{\\sqrt{2}}\\left[x^2\\right] d x \\\\\\\\\n& +\\int\\limits_{\\sqrt{2}}^{\\sqrt{3}}\\left[x^2\\right] d x+\\int\\limits_{\\sqrt{3}}^2\\left[x^2\\right] d x+\\int\\limits_2^{\\sqrt{5}}\\left[x^2\\right] d x+\\int\\limits_{\\sqrt{5}}^{2.4}\\left[x^2\\right] d x\n\\end{aligned}\n$$\n
$$\n\\begin{aligned}\n& =\\int_0 0 . d x+\\int_1^{\\sqrt{2}} 1 \\cdot d x+\\int_{\\sqrt{2}}^{\\sqrt{3}} 2 d x+\\int_{\\sqrt{3}}^2 3 d x+\\int_2^{\\sqrt{5}} 4 d x+\\int_{\\sqrt{5}}^{2.4} 5 d x \\\\\\\\\n& = 0+[x]_1^{\\sqrt{2}}+2[x]_{\\sqrt{2}}^{\\sqrt{3}}+3[x]_{\\sqrt{3}}^2+4[x]_2^{\\sqrt{5}}+5[x]_{\\sqrt{5}}^{2.4} \\\\\\\\\n& =\\sqrt{2}-1+2 \\sqrt{3}-2 \\sqrt{2}+6-3 \\sqrt{3}+4 \\sqrt{5}-8+12-5 \\sqrt{5} \\\\\\\\\n& =-\\sqrt{2}-\\sqrt{3}-\\sqrt{5}+9 \\\\\\\\\n& \\therefore \\alpha=9, \\beta=-1, \\gamma=-1, \\delta=-1 \\\\\\\\\n& \\text { So, } \\alpha+\\beta+\\gamma+\\delta=9-1-1-1=6\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4816,
"subject": "General Science",
"question": "
Let $$[t]$$ denote the greatest integer $$\\leq t$$. Then $$\\frac{2}{\\pi} \\int_\\limits{\\pi / 6}^{5 \\pi / 6}(8[\\operatorname{cosec} x]-5[\\cot x]) d x$$ is equal to __________.
",
"options": [],
"answer": "14",
"solution": "**Answer:** 14\n\n$$\n\\begin{aligned}\n& \\text { Let } \\mathrm{I}=\\frac{2}{\\pi} \\int\\limits_{\\frac{\\pi}{6}}^{ \\frac{5\\pi}{6}}\\{8[\\operatorname{cosec} x]-5[\\cot x]\\} d x \\\\\\\\\n& =\\frac{2}{\\pi}\\left[8 \\int\\limits_{\\frac{\\pi}{6}}^{\\frac{5 \\pi}{6}}[\\operatorname{cosec} x] d x-5 \\int\\limits_{\\frac{\\pi}{6}}^{\\frac{5 \\pi}{6}}[\\cot x] d x\\right]\n\\end{aligned}\n$$\n
$$\n\\left.\\begin{array}{r}\n=\\frac{2}{\\pi}\\left[8 \\int\\limits_{\\pi / 6}^{5 \\pi / 6} d x-5\\left\\{\\int\\limits_{\\pi / 6}^{\\pi / 4} d x+\\int\\limits_{\\pi / 4}^{\\pi / 2} 0 . d x+\\int\\limits_{\\pi / 2}^{3 \\pi / 4}(-1) d x+\\right.\\right.\n\\left.\\left.+\\int\\limits_{3 \\pi / 4}^{5 \\pi / 6}(-2) d x\\right\\}\\right]\n\\end{array}\\right]\n$$\n
Let $$f(x)$$ be a function satisfying $$f(x)+f(\\pi-x)=\\pi^{2}, \\forall x \\in \\mathbb{R}$$. Then $$\\int_\\limits{0}^{\\pi} f(x) \\sin x d x$$ is equal to :
$$\n\\begin{aligned}\n& =\\int\\limits_0^\\pi f(\\pi-x) \\sin (\\pi-x) d x \\\\\\\\\n& =\\int\\limits_0^\\pi f(\\pi-x) \\sin x d x ........(ii)\n\\end{aligned}\n$$\n
On adding Equations (i) and (ii), we get\n
$$\n\\begin{aligned}\n& 2 I=\\int\\limits_0^\\pi[f(x)+f(\\pi-x)] \\sin x d x \\\\\\\\\n& \\Rightarrow 2 I=\\int\\limits_0^\\pi \\pi^2 \\sin x d x=\\pi^2 \\int\\limits_0^\\pi \\sin x d x \\\\\\\\\n& =\\pi^2(-\\cos x)_0^\\pi=-\\pi^2(-1-1)=2 \\pi^2 \\\\\\\\\n& \\Rightarrow I=\\pi^2\n\\end{aligned}\n$$",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4819,
"subject": "General Science",
"question": "If $\\int\\limits_0^{\\frac{\\pi}{3}} \\cos ^4 x \\mathrm{~d} x=\\mathrm{a} \\pi+\\mathrm{b} \\sqrt{3}$, where $\\mathrm{a}$ and $\\mathrm{b}$ are rational numbers, then $9 \\mathrm{a}+8 \\mathrm{b}$ is equal to :",
"options": [
{
"text": "2"
},
{
"text": "1"
},
{
"text": "3"
},
{
"text": "$\\frac{3}{2}$"
}
],
"answer": "2",
"solution": "**Answer:** 2\n\n
To solve the given integral $\\int\\limits_0^{\\frac{\\pi}{3}} \\cos ^4 x \\mathrm{~d} x$, we'll apply a known power-reduction formula that allows us to express even powers of sine and cosine functions in terms of cosine of multiple angles. Specifically for $\\cos^4 x$, we can write it in terms of double angles as:
\n
$$\n\\cos^4 x = \\left(\\frac{1 + \\cos(2x)}{2}\\right)^2\n$$
\n
We can then expand and simplify the integral using this formula. Let's proceed with this:
\n
$$\n\\int\\limits_0^{\\frac{\\pi}{3}} \\cos^4 x \\mathrm{~d}x = \\int\\limits_0^{\\frac{\\pi}{3}} \\left(\\frac{1 + \\cos(2x)}{2}\\right)^2 \\mathrm{~d}x\n$$
\n
Now, let's expand the integrand and then integrate term by term:
$\\begin{aligned} & \\text { Put tant }=y \\\\\\\\ & \\sec ^2 t d t=d y \\\\\\\\ & 2 I=\\frac{\\pi}{8} \\int\\limits_0^{\\infty} \\frac{\\left(1+y^2\\right) d y}{1+y^4} \\\\\\\\ & =\\frac{\\pi}{8} \\int\\limits_0^{\\infty} \\frac{1+\\frac{1}{y^2}}{y^2+\\frac{1}{y^2}-2+2} d y \\\\\\\\ & = \\frac{\\pi}{8} \\int\\limits_0^{\\infty} \\frac{\\left(1+\\frac{1}{y^2}\\right) d y}{2+\\left(y-\\frac{1}{y}\\right)^2}\\end{aligned}$\n
$$\\begin{aligned}\n& \\int_\\limits0^1 \\frac{1}{\\sqrt{3+x}+\\sqrt{1+x}} d x=\\int_\\limits0^1 \\frac{\\sqrt{3+x}-\\sqrt{1+x}}{(3+x)-(1+x)} d x \\\\\n& \\frac{1}{2}\\left[\\int_\\limits0^1 \\sqrt{3+x} d x-\\int_\\limits0^1(\\sqrt{1+x}) d x\\right]\n\\end{aligned}$$
Let $$f(x)=\\int_\\limits0^x g(t) \\log _{\\mathrm{e}}\\left(\\frac{1-\\mathrm{t}}{1+\\mathrm{t}}\\right) \\mathrm{dt}$$, where $$g$$ is a continuous odd function.\nIf $$\\int_{-\\pi / 2}^{\\pi / 2}\\left(f(x)+\\frac{x^2 \\cos x}{1+\\mathrm{e}^x}\\right) \\mathrm{d} x=\\left(\\frac{\\pi}{\\alpha}\\right)^2-\\alpha$$, then $$\\alpha$$ is equal to _________.
$$f(x)=\\int_\\limits0^x g(t) \\ln \\left(\\frac{1-t}{1+t}\\right) d t$$
\n
$$f(-x)=\\int_\\limits0^{-x} g(t) \\ln \\left(\\frac{1-t}{1+t}\\right) d t$$
\n
$$f(-x)=-\\int_\\limits0^x g(-y) \\ln \\left(\\frac{1+y}{1-y}\\right) d y$$
\n
$$=-\\int_\\limits0^x g(y) \\ln \\left(\\frac{1-y}{1+y}\\right) d y$$ (g is odd)
\n
$$f(-x)=-f(x) \\Rightarrow f$$ is also odd
\n
Now,
\n
$$\\begin{aligned}\n& I=\\int_\\limits{-\\pi / 2}^{\\pi / 2}\\left(f(x)+\\frac{x^2 \\cos x}{1+e^x}\\right) d x \\quad \\text{... (1)}\\\\\n& I=\\int_\\limits{-\\pi / 2}^{\\pi / 2}\\left(f(-x)+\\frac{x^2 e^x \\cos x}{1+e^x}\\right) d x \\quad \\text{... (2)}\\\\\n& 2 I=\\int_\\limits{-\\pi / 2}^{\\pi / 2} x^2 \\cos x d x=2 \\int_0^{\\pi / 2} x^2 \\cos x d x\n\\end{aligned}$$
\n
$$\\begin{aligned}\n& I=\\left(x^2 \\sin x\\right)_0^{\\pi / 2}-\\int_\\limits0^{\\pi / 2} 2 x \\sin x d x \\\\\n& =\\frac{\\pi^2}{4}-2\\left(-x \\cos x+\\int \\cos x d x\\right)_0^{\\pi / 2} \\\\\n& =\\frac{\\pi^2}{4}-2(0+1)=\\frac{\\pi^2}{4}-2 \\Rightarrow\\left(\\frac{\\pi}{2}\\right)^2-2 \\\\\n& \\therefore \\alpha=2\n\\end{aligned}$$
Let $$f, g:(0, \\infty) \\rightarrow \\mathbb{R}$$ be two functions defined by $$f(x)=\\int\\limits_{-x}^x\\left(|t|-t^2\\right) e^{-t^2} d t$$ and $$g(x)=\\int\\limits_0^{x^2} t^{1 / 2} e^{-t} d t$$. Then, the value of $$9\\left(f\\left(\\sqrt{\\log _e 9}\\right)+g\\left(\\sqrt{\\log _e 9}\\right)\\right)$$ is equal to :
If the integral $$525 \\int_\\limits0^{\\frac{\\pi}{2}} \\sin 2 x \\cos ^{\\frac{11}{2}} x\\left(1+\\operatorname{Cos}^{\\frac{5}{2}} x\\right)^{\\frac{1}{2}} d x$$ is equal to $$(n \\sqrt{2}-64)$$, then $$n$$ is equal to _________.
Let $$f: \\mathbb{R} \\rightarrow \\mathbb{R}$$ be a function defined by $$f(x)=\\frac{4^x}{4^x+2}$$ and $$M=\\int_\\limits{f(a)}^{f(1-a)} x \\sin ^4(x(1-x)) d x, N=\\int_\\limits{f(a)}^{f(1-a)} \\sin ^4(x(1-x)) d x ; a \\neq \\frac{1}{2}$$. If $$\\alpha M=\\beta N, \\alpha, \\beta \\in \\mathbb{N}$$, then the least value of $$\\alpha^2+\\beta^2$$ is equal to __________.
If the value of the integral $$\\int_\\limits{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}}\\left(\\frac{x^2 \\cos x}{1+\\pi^x}+\\frac{1+\\sin ^2 x}{1+e^{\\sin x^{2123}}}\\right) d x=\\frac{\\pi}{4}(\\pi+a)-2$$, then the value of $$a$$ is
If $$\\int_\\limits{\\frac{\\pi}{6}}^{\\frac{\\pi}{3}} \\sqrt{1-\\sin 2 x} d x=\\alpha+\\beta \\sqrt{2}+\\gamma \\sqrt{3}$$, where $$\\alpha, \\beta$$ and $$\\gamma$$ are rational numbers, then\n$$3 \\alpha+4 \\beta-\\gamma$$ is equal to _________.
Let $$f: \\mathbb{R} \\rightarrow \\mathbb{R}$$ be a function defined by $$f(x)=\\frac{x}{\\left(1+x^4\\right)^{1 / 4}}$$, and $$g(x)=f(f(f(f(x))))$$. Then, $$18 \\int_0^{\\sqrt{2 \\sqrt{5}}} x^2 g(x) d x$$ is equal to
Let $$y=f(x)$$ be a thrice differentiable function in $$(-5,5)$$. Let the tangents to the curve $$y=f(x)$$ at $$(1, f(1))$$ and $$(3, f(3))$$ make angles $$\\pi / 6$$ and $$\\pi / 4$$, respectively with positive $$x$$-axis. If $$27 \\int_\\limits1^3\\left(\\left(f^{\\prime}(t)\\right)^2+1\\right) f^{\\prime \\prime}(t) d t=\\alpha+\\beta \\sqrt{3}$$ where $$\\alpha, \\beta$$ are integers, then the value of $$\\alpha+\\beta$$ equals
Let $$a$$ and $$b$$ be real constants such that the function $$f$$ defined by $$f(x)=\\left\\{\\begin{array}{ll}x^2+3 x+a & , x \\leq 1 \\\\ b x+2 & , x>1\\end{array}\\right.$$ be differentiable on $$\\mathbb{R}$$. Then, the value of $$\\int_\\limits{-2}^2 f(x) d x$$ equals
To determine the integral of the piecewise function $$f$$ over the interval $$[-2, 2]$$, we first ensure that $$f$$ is differentiable on $$\\mathbb{R}$$, as given in the problem statement. Differentiability implies continuity, so $$f$$ must also be continuous at $$x=1$$.
\n
The condition for continuity at $$x=1$$ is:
\n
$$x^2 + 3x + a = bx + 2$$ at $$x=1$$.
\n
This simplifies to:
\n
$$1 + 3 + a = b(1) + 2$$
\n
$$\\Rightarrow a = b - 2$$
\n
The first derivative of $$f$$ gives us two different expressions depending on the value of $$x$$:
\n
For $$x \\leq 1$$, $$f'(x) = 2x + 3$$;\nand for $$x > 1$$, $$f'(x) = b$$.
\n
For $$f$$ to be differentiable at $$x=1$$, these derivatives must be equal at that point. Setting $$f'(1)$$ from both expressions equal to each other gives $$2(1) + 3 = b$$, thus $$b = 5$$. And from $$a = b - 2$$, we have $$a = 3$$.
\n
Now, we can calculate the integral of $$f$$ over the specified interval:
Let $$f:\\left[-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right] \\rightarrow \\mathbf{R}$$ be a differentiable function such that $$f(0)=\\frac{1}{2}$$. If the $$\\lim _\\limits{x \\rightarrow 0} \\frac{x \\int_0^x f(\\mathrm{t}) \\mathrm{dt}}{\\mathrm{e}^{x^2}-1}=\\alpha$$, then $$8 \\alpha^2$$ is equal to :
The value of $$9 \\int_\\limits0^9\\left[\\sqrt{\\frac{10 x}{x+1}}\\right] \\mathrm{d} x$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$, is
$$f(x)=\\left\\{\\begin{array}{cc}\n-2 & -2 \\leq x \\leq 0 \\\\\nx-2 & 0< x \\leq 2\n\\end{array} h(x)=f|x|+|f(x)|\\right.$$
\n
\n
$$\\begin{aligned}\n& h(x)=\\left\\{\\begin{array}{cc}\n-x-2+2=-x & -2 \\leq x \\leq 0 \\\\\n0 & 0< x \\leq 2\n\\end{array}\\right. \\\\\n& \\therefore \\int_\\limits{-2}^2 h(x) d x=\\int_\\limits{-2}^0-x d x+\\int_\\limits0^2 0 d x \\\\\n& \\left.\\frac{x^2}{2}\\right|_{-2} ^0=\\frac{4}{2}=2\n\\end{aligned}$$
If the shortest distance between the lines $$\\frac{x+2}{2}=\\frac{y+3}{3}=\\frac{z-5}{4}$$ and $$\\frac{x-3}{1}=\\frac{y-2}{-3}=\\frac{z+4}{2}$$ is $$\\frac{38}{3 \\sqrt{5}} \\mathrm{k}$$, and $$\\int_\\limits 0^{\\mathrm{k}}\\left[x^2\\right] \\mathrm{d} x=\\alpha-\\sqrt{\\alpha}$$, where $$[x]$$ denotes the greatest integer function, then $$6 \\alpha^3$$ is equal to _________.
$$\\begin{aligned}\n& {k=\\frac{57}{38}}=\\frac{3}{2} \\\\\n& \\int_\\limits0^{\\frac{3}{2}}\\left[x^2\\right] d x \\\\\n& \\int_\\limits0^1 0 d x+\\int_\\limits0^{\\sqrt{ }} 1 d x+\\int_\\limits{\\sqrt{2}}^{\\frac{3}{2}} 2 d x \\\\\n& 0+(\\sqrt{2}-1)+2\\left(\\frac{3}{2}-\\sqrt{2}\\right) \\\\\n& \\sqrt{2}-1+3-2 \\sqrt{2} \\\\\n& 2-\\sqrt{2}\n\\end{aligned}$$
If $$\\int_0^{\\frac{\\pi}{4}} \\frac{\\sin ^2 x}{1+\\sin x \\cos x} \\mathrm{~d} x=\\frac{1}{\\mathrm{a}} \\log _{\\mathrm{e}}\\left(\\frac{\\mathrm{a}}{3}\\right)+\\frac{\\pi}{\\mathrm{b} \\sqrt{3}}$$, where $$\\mathrm{a}, \\mathrm{b} \\in \\mathrm{N}$$, then $$\\mathrm{a}+\\mathrm{b}$$ is equal to _________.
$$I=\\int_\\limits0^{\\frac{\\pi}{4}} \\frac{\\sin ^2 x}{1+\\sin x \\cos x} d x=\\int_\\limits0^{\\frac{\\pi}{4}} \\frac{\\sin ^2 x}{\\sin ^2 x+\\cos ^2 x+\\sin x \\cos x} d x$$\n
$$I=\\int_\\limits0^{\\frac{\\pi}{4}} \\frac{\\tan ^2 x}{1+\\tan x+\\tan ^2 x} d x$$
\n
$$=\\int_\\limits0^{\\frac{\\pi}{4}} \\frac{\\tan x \\cdot \\sec ^2 x d x}{\\left(1+\\tan ^2 x\\right)\\left(1+\\tan x+\\tan ^2 x\\right)}$$
\n
Let $$\\tan x=t$$
\n
$$\\begin{aligned}\nI & =\\int_\\limits0^1 \\frac{t^2}{\\left(1+t^2\\right)\\left(1+t+t^2\\right)} d t \\\\\n& =\\int_\\limits0^1\\left(\\frac{x}{1+x^2}-\\frac{x}{1+x+x^2}\\right) d x\n\\end{aligned}$$
Let $$\\int_\\limits\\alpha^{\\log _e 4} \\frac{\\mathrm{d} x}{\\sqrt{\\mathrm{e}^x-1}}=\\frac{\\pi}{6}$$. Then $$\\mathrm{e}^\\alpha$$ and $$\\mathrm{e}^{-\\alpha}$$ are the roots of the equation :
The value of $$k \\in \\mathbb{N}$$ for which the integral $$I_n=\\int_0^1\\left(1-x^k\\right)^n d x, n \\in \\mathbb{N}$$, satisfies $$147 I_{20}=148 I_{21}$$ is
$$\\begin{aligned}\n& I(21)=\\int_\\limits0^1\\left(1-x^k\\right)^{21} d x \\\\\n& =\\int_\\limits0^1\\left(1-x^k\\right)\\left(1-x^k\\right)^{20} d x \\\\\n& =\\int_\\limits0^1\\left(1-x^k\\right)^{20} d x-\\int_0 x^k\\left(1-x^k\\right)^{20} d x \\\\\n& I(21)=I(20)-\\int_\\limits0^1 x^k\\left(1-x^k\\right)^{20} d x\n\\end{aligned}$$
\n
\n
$$I(21)=I(20)-\\left\\lfloor\\frac{\\left(1-x^k\\right)^{21}}{-21 k} x-\\int_\\limits0^1 \\frac{(1-x^k)^{21}}{-21 k} d x\\right\\rfloor$$
$$\\begin{aligned}\n& I=\\int_\\limits{-\\pi}^\\pi \\frac{2 y(1+\\sin y)}{1+\\cos ^2 y} d y \\\\\n& =\\int_\\limits0^\\pi\\left(\\frac{2 y(1+\\sin y)}{1+\\cos ^2 y}+\\frac{-2 y(1-\\sin y)}{1+\\cos ^2 y}\\right) d y \\\\\n& =\\int_\\limits0^\\pi\\left(\\frac{2 y+2 y \\sin y-2 y+2 y \\sin y}{1+\\cos ^2 y}\\right) d y\n\\end{aligned}$$
\n
$$I=4 \\int_0^\\pi\\left(\\frac{y \\sin }{1+\\cos ^2 y}\\right) d y \\quad \\text{... (1)}$$
\n
$$\\begin{aligned}\n& I=4 \\int_\\limits0^\\pi\\left(\\frac{(\\pi-y) \\sin (\\pi-)}{1+\\cos ^2(\\pi-y)}\\right) d y \\\\\n& I=4\\left\\lfloor\\int_\\limits0\\left(\\frac{\\pi \\sin y}{1+\\cos ^2 y}\\right) d y-\\int_\\limits0^\\pi \\frac{y \\sin y}{1+\\cos ^2 y} d y\\right\\rfloor \\quad \\text{... (2)}\n\\end{aligned}$$
\n
Adding equation (1) and (2)
\n
$$\\begin{aligned}\n& 2 I=4 \\int_\\limits0^\\pi\\left(\\frac{\\pi \\sin y}{1+\\cos ^2 y}\\right) d y \\\\\n& I=2 \\pi \\int_\\limits0^\\pi \\frac{\\sin y}{1+\\cos ^2 y} d y \\\\\n& =2 \\pi \\times \\frac{\\pi}{2} \\\\\n& =\\pi^2\n\\end{aligned}$$
To use the Beta function, let us make a substitution. Let $ x^{10} = t $. Then, $ dx = \\frac{1}{10}t^{-\\frac{9}{10}} dt $ or $ dx = \\frac{1}{10} t^{-\\frac{9}{10}} dt $. The limits of integration change as follows: when $ x = 0 $, $ t = 0 $, and when $ x = 1 $, $ t = 1 $.
If $$f(t)=\\int_\\limits0^\\pi \\frac{2 x \\mathrm{~d} x}{1-\\cos ^2 \\mathrm{t} \\sin ^2 x}, 0<\\mathrm{t}<\\pi$$, then the value of $$\\int_\\limits0^{\\frac{\\pi}{2}} \\frac{\\pi^2 \\mathrm{dt}}{f(\\mathrm{t})}$$ equals __________.
Let $$[t]$$ denote the largest integer less than or equal to $$t$$. If $$\\int_\\limits0^3\\left(\\left[x^2\\right]+\\left[\\frac{x^2}{2}\\right]\\right) \\mathrm{d} x=\\mathrm{a}+\\mathrm{b} \\sqrt{2}-\\sqrt{3}-\\sqrt{5}+\\mathrm{c} \\sqrt{6}-\\sqrt{7}$$, where $$\\mathrm{a}, \\mathrm{b}, \\mathrm{c} \\in \\mathbf{Z}$$, then $$\\mathrm{a}+\\mathrm{b}+\\mathrm{c}$$ is equal to __________.
$$\\int_\\limits0^3\\left(\\left[x^2\\right]+\\left[\\frac{x^2}{2}\\right]\\right) d x \\quad=\\int_\\limits0^1 0 d x+\\int_1^{\\sqrt{2}} 1 d x+\\int_\\limits{\\sqrt{2}}^{\\sqrt{3}} 3 d x+$$
\n
$$\n\\begin{aligned}\n& \\int_\\limits{\\sqrt{3}}^2 4 d x+\\int_\\limits2^{\\sqrt{5}} 6 d x+\\int_\\limits{\\sqrt{5}}^{\\sqrt{6}} 7 d x+\\int_\\limits{\\sqrt{6}}^{\\sqrt{7}} 9 d x+\\int_\\limits{\\sqrt{7}}^{\\sqrt{8}} 10 d x+\\int_\\limits{\\sqrt{8}}^3 12 d x \\\\\n& =31-6 \\sqrt{2}-\\sqrt{3}-\\sqrt{5}-\\sqrt{7}-2 \\sqrt{6} \\\\\n& \\Rightarrow a=31, b=-6, c=-2 \\\\\n& \\Rightarrow a+b+c=23\n\\end{aligned}$$
Let $$r_k=\\frac{\\int_0^1\\left(1-x^7\\right)^k d x}{\\int_0^1\\left(1-x^7\\right)^{k+1} d x}, k \\in \\mathbb{N}$$. Then the value of $$\\sum_\\limits{k=1}^{10} \\frac{1}{7\\left(r_k-1\\right)}$$ is equal to _________.
$$ \\Rightarrow y{\\left( {{{dy} \\over {dx}}} \\right)^2} + 2x\\left( {{{dy} \\over {dx}}} \\right) - y = 0$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4861,
"subject": "General Science",
"question": "Let a curve y = f(x) pass through the point (2, (loge2)2) and have slope $${{2y} \\over {x{{\\log }_e}x}}$$ for all positive real value of x. Then the value of f(e) is equal to ______________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\n$$y' = {{2y} \\over {x\\ln x}}$$
Let $${{dy} \\over {dx}} = {{ax - by + a} \\over {bx + cy + a}}$$, where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is :
Let $${{dy} \\over {dx}} = {{ax - by + a} \\over {bx + cy + a}},\\,a,b,c \\in R$$, represents a circle with center ($$\\alpha$$, $$\\beta$$). Then, $$\\alpha$$ + 2$$\\beta$$ is equal to :
Let a smooth curve $$y=f(x)$$ be such that the slope of the tangent at any point $$(x, y)$$ on it is directly proportional to $$\\left(\\frac{-y}{x}\\right)$$. If the curve passes through the points $$(1,2)$$ and $$(8,1)$$, then $$\\left|y\\left(\\frac{1}{8}\\right)\\right|$$ is equal to
Let a curve $$y=y(x)$$ pass through the point $$(3,3)$$ and the area of the region under this curve, above the $$x$$-axis and between the abscissae 3 and $$x(>3)$$ be $$\\left(\\frac{y}{x}\\right)^{3}$$. If this curve also passes through the point $$(\\alpha, 6 \\sqrt{10})$$ in the first quadrant, then $$\\alpha$$ is equal to ___________.
The differential equation of the family of circles passing through the points $$(0,2)$$ and $$(0,-2)$$ is :
",
"options": [
{
"text": "$$2 x y \\frac{d y}{d x}+\\left(x^{2}-y^{2}+4\\right)=0$$"
},
{
"text": "$$2 x y \\frac{d y}{d x}+\\left(x^{2}+y^{2}-4\\right)=0$$"
},
{
"text": "$$2 x y \\frac{d y}{d x}+\\left(y^{2}-x^{2}+4\\right)=0$$"
},
{
"text": "$$2 x y \\frac{d y}{d x}-\\left(x^{2}-y^{2}+4\\right)=0$$"
}
],
"answer": "$$2 x y \\frac{d y}{d x}+\\left(x^{2}-y^{2}+4\\right)=0$$",
"solution": "**Answer:** $$2 x y \\frac{d y}{d x}+\\left(x^{2}-y^{2}+4\\right)=0$$\n\n
Family of circles passing through the points (0, 2) and (0, $$-$$2)
Let $$f(x)$$ be a positive function such that the area bounded by $$y=f(x), y=0$$ from $$x=0$$ to $$x=a>0$$ is $$e^{-a}+4 a^2+a-1$$. Then the differential equation, whose general solution is $$y=c_1 f(x)+c_2$$, where $$c_1$$ and $$c_2$$ are arbitrary constants, is
$$y\\log x = 2\\left[ {x\\log x - x} \\right] + c$$\n
Put $$x=1,y.0=-2+c$$ $$ \\Rightarrow c = 2$$\n
Put $$x=e$$ \n
$$y\\log e = 2e\\left( {\\log e - 1} \\right) + c \\Rightarrow y\\left( e \\right) = c = 2$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4876,
"subject": "General Science",
"question": "The solution of the differential equation\n
",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4880,
"subject": "General Science",
"question": "Let y = y(x) be the solution of the differential equation\n
Now check option(s), Which is satisly by option (ii)",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4886,
"subject": "General Science",
"question": "The solution of the differential equation\n
So, (a, b) = (1, 1) ",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4892,
"subject": "General Science",
"question": "Let y = y(x) be the solution of the differential\nequation\n
$$ \\Rightarrow $$ y = $$\\sqrt 2 - 2$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4893,
"subject": "General Science",
"question": "If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is $${{{x^2} - 4x + y + 8} \\over {x - 2}}$$, then this curve also passes through the point :",
"options": [
{
"text": "(4, 4)"
},
{
"text": "(5, 5)"
},
{
"text": "(5, 4)"
},
{
"text": "(4, 5)"
}
],
"answer": "(5, 5)",
"solution": "**Answer:** (5, 5)\n\nGiven
Now, at x = 0, y = 0 $$ \\Rightarrow $$ C = $$-$$2
$$ \\therefore $$ y = x (x $$-$$ 2) $$-$$ 4 $$-$$ 2 (x $$-$$ 2)
$$ \\Rightarrow $$ y = x2 $$-$$ 4x
This curve passes through (5, 5)",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4894,
"subject": "General Science",
"question": "If the curve, y = y(x) represented by the solution of the differential equation (2xy2 $$-$$ y)dx + xdy = 0, passes through the intersection of the lines, 2x $$-$$ 3y = 1 and 3x + 2y = 8, then |y(1)| is equal to _________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nGiven,
[Point of intersection of $$2x - 3y = 1$$ and $$3x + 2y = 8$$]
$$ \\therefore $$ $${2 \\over 1} = 4 + c$$
$$ \\Rightarrow c = - 2$$
$$ \\Rightarrow {x \\over y} = {x^2} - 2$$
Put x = 1
$${1 \\over y} = 1 - 2 = - 1$$
$$ \\Rightarrow y(1) = - 1$$
$$ \\Rightarrow |y(1)|\\, = 1$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4895,
"subject": "General Science",
"question": "If y = y(x) is the solution of the equation
$${e^{\\sin y}}\\cos y{{dy} \\over {dx}} + {e^{\\sin y}}\\cos x = \\cos x$$, y(0) = 0; then
$$1 + y\\left( {{\\pi \\over 6}} \\right) + {{\\sqrt 3 } \\over 2}y\\left( {{\\pi \\over 3}} \\right) + {1 \\over {\\sqrt 2 }}y\\left( {{\\pi \\over 4}} \\right)$$ is equal to ____________.",
"options": [],
"answer": "1",
"solution": "**Answer:** 1\n\nesin y cos y$${{dy} \\over {dx}}$$ + esin y cos x = cos x
Put esin y = t
esin y $$\\times$$ cos y$${{dy} \\over {dx}}$$ = $${{dt} \\over {dx}}$$
$$ \\Rightarrow $$ $${{dt} \\over {dx}}$$ + t cos x = cos x
I. F. = $${e^{\\int {\\cos x\\,dx} }} = {e^{\\sin x}}$$
Let y = y(x) be the solution of the differential equation $${{dy} \\over {dx}} + {{\\sqrt 2 y} \\over {2{{\\cos }^4}x - {{\\cos }^2}x}} = x{e^{{{\\tan }^{ - 1}}(\\sqrt 2 \\cot 2x)}},\\,0 < x < {\\pi \\over 2}$$ with $$y\\left( {{\\pi \\over 4}} \\right) = {{{\\pi ^2}} \\over {32}}$$. If $$y\\left( {{\\pi \\over 3}} \\right) = {{{\\pi ^2}} \\over {18}}{e^{ - {{\\tan }^{ - 1}}(\\alpha )}}$$, then the value of 3$$\\alpha$$2 is equal to ___________.
Let y = y(x), x > 1, be the solution of the differential equation $$(x - 1){{dy} \\over {dx}} + 2xy = {1 \\over {x - 1}}$$, with $$y(2) = {{1 + {e^4}} \\over {2{e^4}}}$$. If $$y(3) = {{{e^\\alpha } + 1} \\over {\\beta {e^\\alpha }}}$$, then the value of $$\\alpha + \\beta $$ is equal to _________.
Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 $$\\tan x(\\cos x - y)$$. If the curve passes through the point $$\\left( {{\\pi \\over 4},0} \\right)$$, then the value of $$\\int\\limits_0^{\\pi /2} {y\\,dx} $$ is equal to :
Let y = y(x) be the solution of the differential equation $$x(1 - {x^2}){{dy} \\over {dx}} + (3{x^2}y - y - 4{x^3}) = 0$$, $$x > 1$$, with $$y(2) = - 2$$. Then y(3) is equal to :
If the solution curve of the differential equation
$$(({\\tan ^{ - 1}}y) - x)dy = (1 + {y^2})dx$$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is
Let $$S = (0,2\\pi ) - \\left\\{ {{\\pi \\over 2},{{3\\pi } \\over 4},{{3\\pi } \\over 2},{{7\\pi } \\over 4}} \\right\\}$$. Let $$y = y(x)$$, x $$\\in$$ S, be the solution curve of the differential equation $${{dy} \\over {dx}} = {1 \\over {1 + \\sin 2x}},\\,y\\left( {{\\pi \\over 4}} \\right) = {1 \\over 2}$$. If the sum of abscissas of all the points of intersection of the curve y = y(x) with the curve $$y = \\sqrt 2 \\sin x$$ is $${{k\\pi } \\over {12}}$$, then k is equal to _____________.
Let $$y = y(x)$$ be the solution of the differential equation $$(x + 1)y' - y = {e^{3x}}{(x + 1)^2}$$, with $$y(0) = {1 \\over 3}$$. Then, the point $$x = - {4 \\over 3}$$ for the curve $$y = y(x)$$ is :
",
"options": [
{
"text": "not a critical point"
},
{
"text": "a point of local minima"
},
{
"text": "a point of local maxima"
},
{
"text": "a point of inflection"
}
],
"answer": "a point of local minima",
"solution": "**Answer:** a point of local minima\n\n
\n$$\n\\text { If }=e^{\\int-\\frac{2}{y} d y}=e^{-2 \\ln y}=\\frac{1}{y^{2}}\n$$\n
\nSolution is given by\n
\n$$\n\\begin{aligned}\n&x \\cdot \\frac{1}{y^{2}}=\\int y^{2}(y+1) e^{y} \\cdot \\frac{1}{y^{2}} d y \\\\\\\\\n\\Rightarrow & \\frac{x}{y^{2}}=\\int(y+1) e^{y} d y \\\\\\\\\n\\Rightarrow & \\frac{x}{y^{2}}=y e^{y}+c\n\\end{aligned}\n$$
The slope of the tangent to a curve $$C: y=y(x)$$ at any point $$(x, y)$$ on it is $$\\frac{2 \\mathrm{e}^{2 x}-6 \\mathrm{e}^{-x}+9}{2+9 \\mathrm{e}^{-2 x}}$$.\n\nIf $$C$$ passes through the points $$\\left(0, \\frac{1}{2}+\\frac{\\pi}{2 \\sqrt{2}}\\right)$$ and $$\\left(\\alpha, \\frac{1}{2} \\mathrm{e}^{2 \\alpha}\\right)$$, then $$\\mathrm{e}^{\\alpha}$$ is equal to :
If $${{dy} \\over {dx}} + 2y\\tan x = \\sin x,\\,0 < x < {\\pi \\over 2}$$ and $$y\\left( {{\\pi \\over 3}} \\right) = 0$$, then the maximum value of $$y(x)$$ is :
Let the solution curve $$y=f(x)$$ of the differential equation $$\n\\frac{d y}{d x}+\\frac{x y}{x^{2}-1}=\\frac{x^{4}+2 x}{\\sqrt{1-x^{2}}}$$, $$x\\in(-1,1)$$ pass through the origin. Then $$\\int\\limits_{-\\frac{\\sqrt{3}}{2}}^{\\frac{\\sqrt{3}}{2}} f(x) d x \n$$ is equal to
Suppose $$y=y(x)$$ be the solution curve to the differential equation $$\\frac{d y}{d x}-y=2-e^{-x}$$ such that $$\\lim\\limits_{x \\rightarrow \\infty} y(x)$$ is finite. If $$a$$ and $$b$$ are respectively the $$x$$ - and $$y$$-intercepts of the tangent to the curve at $$x=0$$, then the value of $$a-4 b$$ is equal to _____________.
Let $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ be two distinct solutions of the differential equation $$\\frac{d y}{d x}=x+y$$, with $$y_{1}(0)=0$$ and $$y_{2}(0)=1$$ respectively. Then, the number of points of intersection of $$y=y_{1}(x)$$ and $$y=y_{2}(x)$$ is
Let $$y=y(x)$$ be the solution curve of the differential equation $$\n\\frac{d y}{d x}+\\frac{1}{x^{2}-1} y=\\left(\\frac{x-1}{x+1}\\right)^{1 / 2}$$, $$x >1$$ passing through the point $$\\left(2, \\sqrt{\\frac{1}{3}}\\right)$$. Then $$\\sqrt{7}\\, y(8)$$ is equal to :
Let the solution curve $$y=y(x)$$ of the differential equation $$\\left(1+\\mathrm{e}^{2 x}\\right)\\left(\\frac{\\mathrm{d} y}{\\mathrm{~d} x}+y\\right)=1$$ pass through the point $$\\left(0, \\frac{\\pi}{2}\\right)$$. Then, $$\\lim\\limits_{x \\rightarrow \\infty} \\mathrm{e}^{x} y(x)$$ is equal to :
Let $$\\alpha x=\\exp \\left(x^{\\beta} y^{\\gamma}\\right)$$ be the solution of the differential equation $$2 x^{2} y \\mathrm{~d} y-\\left(1-x y^{2}\\right) \\mathrm{d} x=0, x > 0,y(2)=\\sqrt{\\log _{e} 2}$$. Then $$\\alpha+\\beta-\\gamma$$ equals :
",
"options": [
{
"text": "1"
},
{
"text": "0"
},
{
"text": "3"
},
{
"text": "$$-1$$"
}
],
"answer": "1",
"solution": "**Answer:** 1\n\n$\\begin{aligned} & 2 x^2 y d y-\\left(1-x y^2\\right) d x=0 \\\\\\\\ & \\Rightarrow 2 x^2 y \\frac{d y}{d x}-1+x y^2=0 \\\\\\\\ & \\Rightarrow 2 y \\frac{d y}{d x}-\\frac{1}{x^2}+\\frac{y^2}{x}=0 \\\\\\\\ & \\Rightarrow 2 y \\frac{d y}{d x}+\\frac{y^2}{x}=\\frac{1}{x^2} \\text { (L.D.E) } \\\\\\\\ & Let, y^2=t \\Rightarrow 2 y \\frac{d y}{d x}=\\frac{d t}{d x} \\\\\\\\ & \\frac{d t}{d x}+\\frac{t}{x}=\\frac{1}{x^2} \\\\\\\\ & \\text { I.F }=e^{\\int \\frac{1}{x} d x}=x\\end{aligned}$\n
So, the general solution is\n
$$\n\\begin{aligned}\n& \\Rightarrow t \\times x=\\int x \\cdot \\frac{1}{x^2} d x \\\\\\\\\n& \\Rightarrow y^2 \\cdot x=\\ln x+c \\\\\\\\\n& \\text { Also,y(2) }=\\sqrt{\\log _e 2} \\\\\\\\\n& \\log _e^2 2=\\log _e 2+c \\Rightarrow c=\\log _e 2 \\\\\\\\\n& \\Rightarrow y^2 x=\\ln x+\\ln 2 \\\\\\\\\n& \\Rightarrow y^2 x=\\ln 2 x \\\\\\\\\n& 2 x=\\exp \\left(x^1 y^2\\right)\n\\end{aligned}\n$$\n
Let the solution curve $$y=y(x)$$ of the differential equation
$$\n\\frac{\\mathrm{d} y}{\\mathrm{~d} x}-\\frac{3 x^{5} \\tan ^{-1}\\left(x^{3}\\right)}{\\left(1+x^{6}\\right)^{3 / 2}} y=2 x \\exp \\left\\{\\frac{x^{3}-\\tan ^{-1} x^{3}}{\\sqrt{\\left(1+x^{6}\\right)}}\\right\\} \\text { pass through the origin. Then } y(1) \\text { is equal to : }\n$$
Let $$y=y(x)$$ be the solution of the differential equation $$x{\\log _e}x{{dy} \\over {dx}} + y = {x^2}{\\log _e}x,(x > 1)$$. If $$y(2) = 2$$, then $$y(e)$$ is equal to
Let $$y=y(t)$$ be a solution of the differential equation $${{dy} \\over {dt}} + \\alpha y = \\gamma {e^{ - \\beta t}}$$ where, $$\\alpha > 0,\\beta > 0$$ and $$\\gamma > 0$$. Then $$\\mathop {\\lim }\\limits_{t \\to \\infty } y(t)$$
Let $$y = y(x)$$ be the solution curve of the differential equation $${{dy} \\over {dx}} = {y \\over x}\\left( {1 + x{y^2}(1 + {{\\log }_e}x)} \\right),x > 0,y(1) = 3$$. Then $${{{y^2}(x)} \\over 9}$$ is equal to :
Let $$y = y(x)$$ be the solution of the differential equation $${x^3}dy + (xy - 1)dx = 0,x > 0,y\\left( {{1 \\over 2}} \\right) = 3 - \\mathrm{e}$$. Then y (1) is equal to
\nFor $y(1)$ put $x=1, c=-e^{-1}$ in equation (i) we get $y e^{-1}=e^{-1}+e^{-1}-e^{-1}$\n
\n$\\Rightarrow y=1$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4935,
"subject": "General Science",
"question": "Let $x=x(y)$ be the solution of the differential equation\n\n
$2(y+2) \\log _{e}(y+2) d x+\\left(x+4-2 \\log _{e}(y+2)\\right) d y=0, y>-1$\n\n
Let $$y=y(x), y > 0$$, be a solution curve of the differential equation $$\\left(1+x^{2}\\right) \\mathrm{d} y=y(x-y) \\mathrm{d} x$$. If $$y(0)=1$$ and $$y(2 \\sqrt{2})=\\beta$$, then
This can be simplified using the rule $e^{a+b} = e^a e^b$ to :\n\n
$$e^{3 \\beta^{-1}} = e \\cdot e^{\\ln |2 \\sqrt{2}+3|}.$$\n\n
Since $e^{\\ln(x)} = x$ for any $x$, this simplifies to :\n\n
$$e^{3 \\beta^{-1}} = e |2 \\sqrt{2}+3|.$$\n\n
Using the given value of $\\beta$, which is $\\frac{3}{1+\\ln |2 \\sqrt{2}+3|}$, we find :\n\n
$$e^{3 \\beta^{-1}} = e |2 \\sqrt{2}+3| = e (2\\sqrt{2} + 3),$$\n\n
Since $2\\sqrt{2}+3$ is positive and so the absolute value does not affect the result.",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4938,
"subject": "General Science",
"question": "
Let $$f$$ be a differentiable function such that $${x^2}f(x) - x = 4\\int\\limits_0^x {tf(t)dt} $$, $$f(1) = {2 \\over 3}$$. Then $$18f(3)$$ is equal to :
On differentiating both sides with respect to $x$, we get\n
$$\n\\begin{array}{rlrl} \n& 2 x f(x)+x^2 f^{\\prime}(x)-1 =4 x f(x) \\\\\\\\\n&\\Rightarrow x^2 f^{\\prime}(x)-2 x f(x)-1 =0 \\\\\\\\\n&\\Rightarrow x^2 \\frac{d y}{d x}-2 x y =1 ~~~~~~~(Let, y=f(x) ]\\\\\\\\\n&\\frac{d y}{d x}-\\frac{2}{x} y =\\frac{1}{x^2}\n\\end{array}\n$$\n
On comparing above equation with \n
$\\frac{d y}{d x}+P y=Q$, where $P=-\\frac{2}{x}, Q=\\frac{1}{x^2}$ \n
Now, IF $=e^{\\int(-2 / x) d x}=e^{-2 \\log x}=1 / x^2$\n
Solution is $y\\left(\\frac{1}{x^2}\\right)=\\int\\left(\\frac{1}{x^2}\\right) \\frac{1}{x^2} d x$\n
$$\n\\begin{array}{ll}\n&\\Rightarrow \\frac{y}{x^2}=\\int \\frac{1}{x^4} d x \\Rightarrow \\frac{y}{x^2}=-\\frac{1}{3 x^3}+C \\\\\\\\\n&\\Rightarrow y =-\\frac{1}{3 x}+C x^2\n\\end{array}\n$$\n
Given, $f(1)=\\frac{2}{3}$.\n
So, $\\frac{2}{3}=-\\frac{1}{3}+C \\Rightarrow C=1$\n
Let the solution curve $$x=x(y), 0 < y < \\frac{\\pi}{2}$$, of the differential equation $$\\left(\\log _{e}(\\cos y)\\right)^{2} \\cos y \\mathrm{~d} x-\\left(1+3 x \\log _{e}(\\cos y)\\right) \\sin \\mathrm{y} d y=0$$ satisfy $$x\\left(\\frac{\\pi}{3}\\right)=\\frac{1}{2 \\log _{e} 2}$$. If $$x\\left(\\frac{\\pi}{6}\\right)=\\frac{1}{\\log _{e} m-\\log _{e} n}$$, where $$m$$ and $$n$$ are coprime, then $$m n$$ is equal to __________.
",
"options": [],
"answer": "12",
"solution": "**Answer:** 12\n\n$$\n\\begin{aligned}\n& (\\cos y)(\\ln (\\cos y))^2 d x=(1+3 x \\ln \\cos y) \\sin y d y \\\\\\\\\n& \\Rightarrow \\frac{d x}{d y}=\\frac{(1+3 x \\ln \\cos y) \\sin y}{(\\ln \\cos y)^2 \\cos y} \\\\\\\\\n& =\\tan y\\left[\\frac{1}{(\\ln \\cos y)^2}+\\frac{3 x}{\\ln \\cos y}\\right] \\\\\\\\\n& \\Rightarrow \\frac{d x}{d y}-\\left(\\frac{3 \\tan y}{\\ln \\cos y}\\right) x=\\frac{\\tan y}{(\\ln \\cos y)^2}\n\\end{aligned}\n$$\n
If the solution curve of the differential equation $$\\left(y-2 \\log _{e} x\\right) d x+\\left(x \\log _{e} x^{2}\\right) d y=0, x > 1$$ passes through the points $$\\left(e, \\frac{4}{3}\\right)$$ and $$\\left(e^{4}, \\alpha\\right)$$, then $$\\alpha$$ is equal to ____________.
$$\n\\begin{aligned}\n& y \\sqrt{\\log x}=\\int \\frac{\\sqrt{\\log x}}{x} d x \\\\\\\\\n& \\log x=v \\Rightarrow \\frac{1}{x} d x=d v \\\\\\\\\n& \\Rightarrow y \\sqrt{\\log x}=\\int \\sqrt{v} d v+C \\\\\\\\\n& \\Rightarrow y \\sqrt{\\log x}=\\frac{2 v^{3 / 2}}{3}+C \\\\\\\\\n& \\Rightarrow y \\sqrt{\\log x}=\\frac{2}{3}(\\log x)^{3 / 2}+C\n\\end{aligned}\n$$\n
Now, this curve passes through $\\left(e, \\frac{4}{3}\\right)$ and $\\left(e^4, \\alpha\\right)$\n
Let $$y=y(x)$$ be a solution of the differential equation $$(x \\cos x) d y+(x y \\sin x+y \\cos x-1) d x=0,0 < x < \\frac{\\pi}{2}$$. If $$\\frac{\\pi}{3} y\\left(\\frac{\\pi}{3}\\right)=\\sqrt{3}$$, then $$\\left|\\frac{\\pi}{6} y^{\\prime \\prime}\\left(\\frac{\\pi}{6}\\right)+2 y^{\\prime}\\left(\\frac{\\pi}{6}\\right)\\right|$$ is equal to ____________.
$\\therefore$ The general solution of the given differential equation\n
$$\n\\begin{array}{rlrl}\ny \\cdot \\mathrm{IF} =\\int(Q \\times \\mathrm{IF}) d x+c \\\\\\\\\n\\Rightarrow y(x \\sec x) =\\int\\left(\\frac{1}{x \\cos x} x \\sec x\\right) d x+c \\\\\\\\\n\\Rightarrow x y \\sec x =\\int \\sec ^2 x d x+c \\\\\\\\\n\\Rightarrow x y \\sec x =\\tan x+c ........(i)\n\\end{array}\n$$\n
The temperature $$T(t)$$ of a body at time $$t=0$$ is $$160^{\\circ} \\mathrm{F}$$ and it decreases continuously as per the differential equation $$\\frac{d T}{d t}=-K(T-80)$$, where $$K$$ is a positive constant. If $$T(15)=120^{\\circ} \\mathrm{F}$$, then $$T(45)$$ is equal to
A function $$y=f(x)$$ satisfies $$f(x) \\sin 2 x+\\sin x-\\left(1+\\cos ^2 x\\right) f^{\\prime}(x)=0$$ with condition $$f(0)=0$$. Then, $$f\\left(\\frac{\\pi}{2}\\right)$$ is equal to
Let $$y=y(x)$$ be the solution of the differential equation $$\\left(1-x^2\\right) \\mathrm{d} y=\\left[x y+\\left(x^3+2\\right) \\sqrt{3\\left(1-x^2\\right)}\\right] \\mathrm{d} x, -1< x<1, y(0)=0$$. If $$y\\left(\\frac{1}{2}\\right)=\\frac{\\mathrm{m}}{\\mathrm{n}}, \\mathrm{m}$$ and $$\\mathrm{n}$$ are co-prime numbers, then $$\\mathrm{m}+\\mathrm{n}$$ is equal to __________.
Let $$\\int_\\limits0^x \\sqrt{1-\\left(y^{\\prime}(t)\\right)^2} d t=\\int_0^x y(t) d t, 0 \\leq x \\leq 3, y \\geq 0, y(0)=0$$. Then at $$x=2, y^{\\prime \\prime}+y+1$$ is equal to
For a differentiable function $$f: \\mathbb{R} \\rightarrow \\mathbb{R}$$, suppose $$f^{\\prime}(x)=3 f(x)+\\alpha$$, where $$\\alpha \\in \\mathbb{R}, f(0)=1$$ and $$\\lim _\\limits{x \\rightarrow-\\infty} f(x)=7$$. Then $$9 f\\left(-\\log _e 3\\right)$$ is equal to _________.
The solution curve, of the differential equation $$2 y \\frac{\\mathrm{d} y}{\\mathrm{~d} x}+3=5 \\frac{\\mathrm{d} y}{\\mathrm{~d} x}$$, passing through the point $$(0,1)$$ is a conic, whose vertex lies on the line :
Let the solution $$y=y(x)$$ of the differential equation $$\\frac{\\mathrm{d} y}{\\mathrm{~d} x}-y=1+4 \\sin x$$ satisfy $$y(\\pi)=1$$. Then $$y\\left(\\frac{\\pi}{2}\\right)+10$$ is equal to __________.
If $$y=y(x)$$ is the solution of the differential equation $$\\frac{\\mathrm{d} y}{\\mathrm{~d} x}+2 y=\\sin (2 x), y(0)=\\frac{3}{4}$$, then $$y\\left(\\frac{\\pi}{8}\\right)$$ is equal to :
Let $$y=y(x)$$ be the solution of the differential equation $$\\left(2 x \\log _e x\\right) \\frac{d y}{d x}+2 y=\\frac{3}{x} \\log _e x, x>0$$ and $$y\\left(e^{-1}\\right)=0$$. Then, $$y(e)$$ is equal to
Let $$y=y(x)$$ be the solution of the differential equation $$\\left(1+x^2\\right) \\frac{d y}{d x}+y=e^{\\tan ^{-1} x}$$, $$y(1)=0$$. Then $$y(0)$$ is
$$ \\therefore $$ Order = 3 and Degree = 3",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4958,
"subject": "General Science",
"question": "The degree and order of the differential equation of the family of all parabolas whose axis is $$x$$-axis, are respectively.",
"options": [
{
"text": "$$2, 3$$ "
},
{
"text": "$$2,1$$ "
},
{
"text": "$$1,2$$ "
},
{
"text": "$$3,2.$$"
}
],
"answer": "$$1,2$$ ",
"solution": "**Answer:** $$1,2$$ \n\nGeneral equation of parabola whose axis is along X-axis\n
$${y^2} = 4a\\left( {x - h} \\right)$$\n
Differentiating with respect to x, we get\n
$$2y{y_1} = 4a$$\n
$$ \\Rightarrow y{y_1} = 2a$$ \n
Again differentiating with respect to x, we get\n
$$ \\Rightarrow {y_1}^2 + y{y_2} = 0$$\n
Degree $$=1,$$ order $$=2.$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4959,
"subject": "General Science",
"question": "The differential equation representing the family of curves $${y^2} = 2c\\left( {x + \\sqrt c } \\right),$$ where $$c>0,$$ is a parameter, is of order and degree as follows: ",
"options": [
{
"text": "order $$1,$$ degree $$2$$"
},
{
"text": "order $$1,$$ degree $$1$$"
},
{
"text": "order $$1,$$ degree $$3$$"
},
{
"text": "order $$2,$$ degree $$2$$"
}
],
"answer": "order $$1,$$ degree $$3$$",
"solution": "**Answer:** order $$1,$$ degree $$3$$\n\n$${y^2} = 2c\\left( {x + \\sqrt c } \\right)\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( i \\right)$$\n
Hence equation $$(iii)$$ is of order $$1$$ and degree $$3.$$ ",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4960,
"subject": "General Science",
"question": "The differential equation whose solution is $$A{x^2} + B{y^2} = 1$$ \n where $$A$$ and $$B$$ are arbitrary constants is of ",
"options": [
{
"text": "second order and second degree"
},
{
"text": "first order and second degree "
},
{
"text": "first order and first degree"
},
{
"text": "second order and first degree"
}
],
"answer": "second order and first degree",
"solution": "**Answer:** second order and first degree\n\n$$A{x^2} + B{y^2} = 1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,...\\left( 1 \\right)$$\n
Which is $$DE$$ of order $$2$$ and degree $$1.$$ ",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4961,
"subject": "General Science",
"question": "The difference between degree and order of a differential equation that represents the family of curves given by $${y^2} = a\\left( {x + {{\\sqrt a } \\over 2}} \\right)$$, a > 0 is _________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n$${y^2} = a\\left( {x + {{\\sqrt a } \\over 2}} \\right)$$\n
$$ \\Rightarrow {1 \\over y}\\sec x = \\tan x + c$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4966,
"subject": "General Science",
"question": "Let $$I$$ be the purchase value of an equipment and $$V(t)$$ be the value after it has been used for $$t$$ years. The value $$V(t)$$ depreciates at a rate given by differential equation $${{dv\\left( t \\right)} \\over {dt}} = - k\\left( {T - t} \\right),$$ where $$k>0$$ is a constant and $$T$$ is the total life in years of the equipment. Then the scrap value $$V(T)$$ of the equipment is ",
"options": [
{
"text": "$$I - {{k{T^2}} \\over 2}$$ "
},
{
"text": "$$I - {{k{{\\left( {T - t} \\right)}^2}} \\over 2}$$ "
},
{
"text": "$${e^{ - kT}}$$ "
},
{
"text": "$${T^2} - {1 \\over k}$$ "
}
],
"answer": "$$I - {{k{T^2}} \\over 2}$$ ",
"solution": "**Answer:** $$I - {{k{T^2}} \\over 2}$$ \n\n$${{dV\\left( t \\right)} \\over {dt}} = - k\\left( {T - t} \\right)$$\n
$$V\\left( 0 \\right) = I \\Rightarrow I = {{K{T^2}} \\over 2} + C$$\n
$$ \\Rightarrow C = I - {{K{T^2}} \\over 2}$$\n
$$\\therefore$$ $$\\,\\,\\,\\,\\,V\\left( T \\right) = 0 + C = I - {{K{T^2}} \\over 2}$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4967,
"subject": "General Science",
"question": "The population $$p$$ $$(t)$$ at time $$t$$ of a certain mouse species satisfies the differential equation $${{dp\\left( t \\right)} \\over {dt}} = 0.5\\,p\\left( t \\right) - 450.\\,\\,$$ If $$p(0)=850,$$ then the time at which the population becomes zero is :",
"options": [
{
"text": "$$2ln$$ $$18$$"
},
{
"text": "$$ln$$ $$9$$"
},
{
"text": "$${1 \\over 2}$$$$ln$$ $$18$$"
},
{
"text": "$$ln$$ $$18$$"
}
],
"answer": "$$2ln$$ $$18$$",
"solution": "**Answer:** $$2ln$$ $$18$$\n\nGiven differential equation is \n
$${{dp\\left( t \\right)} \\over {dt}} = 0.5p\\left( t \\right) - 450$$\n
$$ \\Rightarrow {{dp\\left( t \\right)} \\over {dt}} = {1 \\over 2}p\\left( t \\right) - 450$$\n
$$ \\Rightarrow {{dp\\left( t \\right)} \\over {dt}} = {{p\\left( t \\right) - 900} \\over 2}$$\n
$$ \\Rightarrow 2{{dp\\left( t \\right)} \\over {dt}} = \\left[ {900 - p\\left( t \\right)} \\right]$$\n
$$ \\Rightarrow 2{{dp\\left( t \\right)} \\over {900 - p\\left( t \\right)}} = - dt$$\n
Integrate both sides, we get \n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, - 2\\int {{{dp\\left( t \\right)} \\over {900 - p\\left( t \\right)}}} = \\int {dt} $$\n
$$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,Let\\,\\,900 - p\\left( t \\right) = u$$\n
$$ \\Rightarrow - dp\\left( t \\right) = du$$\n
$$\\therefore$$ $$\\,\\,\\,\\,\\,\\,$$ We have, \n
$$\\therefore$$ $$\\,\\,\\,\\,\\,\\,{t_1} = 2\\ln 18$$ ",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4968,
"subject": "General Science",
"question": "At present, a firm is manufacturing $$2000$$ items. It is estimated that the rate of change of production P w.r.t. additional number of workers $$x$$ is given by $${{dp} \\over {dx}} = 100 - 12\\sqrt x .$$ If the firm employs $$25$$ more workers, then the new level of production of items is ",
"options": [
{
"text": "$$2500$$"
},
{
"text": "$$3000$$ "
},
{
"text": "$$3500$$"
},
{
"text": "$$4500$$"
}
],
"answer": "$$3500$$",
"solution": "**Answer:** $$3500$$\n\nGiven, Rate of change is $${{dp} \\over {dx}} = 100 - 12\\sqrt x $$\n
$$ \\therefore $$ C = e3",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4976,
"subject": "General Science",
"question": "The curve amongst the family of curves represented by the differential equation, (x2 – y2)dx + 2xy dy = 0 which passes through (1, 1) is :",
"options": [
{
"text": "a circle with centre on the y-axis"
},
{
"text": "an ellipse with major axis along the y-axis"
},
{
"text": "a circle with centre on the x-axis"
},
{
"text": "a hyperbola with transverse axis along the x-axis"
}
],
"answer": "a circle with centre on the x-axis",
"solution": "**Answer:** a circle with centre on the x-axis\n\n(x2 $$-$$ y2) dx + 2xy dy = 0\n
",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4978,
"subject": "General Science",
"question": "Let y = y(x) be the solution of the differential equation,
$$ \\Rightarrow y = {3 \\over 2}\\sqrt e $$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4985,
"subject": "General Science",
"question": "The solution curve of the differential equation,\n
$$ \\Rightarrow $$ y = 3\n",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4988,
"subject": "General Science",
"question": "Let y = y(x) be a solution of the differential\nequation,\n
$$ \\Rightarrow t = 2\\ln 18$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4993,
"subject": "General Science",
"question": "The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t = 0. The number of bacteria is increased by 20% in 2 hours. If the population of bacteria is 2000 after $${k \\over {{{\\log }_e}\\left( {{6 \\over 5}} \\right)}}$$ hours, then $${\\left( {{k \\over {{{\\log }_e}2}}} \\right)^2}$$ is equal to :",
"options": [
{
"text": "16"
},
{
"text": "8"
},
{
"text": "2"
},
{
"text": "4"
}
],
"answer": "4",
"solution": "**Answer:** 4\n\n$${{dx} \\over {dt}} \\propto x$$
$$ \\Rightarrow {\\left( {{k \\over {\\ln 2}}} \\right)^2} = 4$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4994,
"subject": "General Science",
"question": "Let the curve y = y(x) be the solution of the differential equation, $${{dy} \\over {dx}}$$ = 2(x + 1). If the numerical value of area bounded by the curve y = y(x) and x-axis is $${{4\\sqrt 8 } \\over 3}$$, then the value of y(1) is equal to _________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\nGiven, $${{dy} \\over {dx}}$$ = 2(x + 1)\n
Integrating both sides, we get\n
$$y = {x^2} + 2x + c$$\n
Let the two roots of the quadratic equation $$\\alpha $$ and $$\\beta $$\n
As parabola intercept the x axis so D > 0\n
From figure, AB = |$$\\alpha $$ - $$\\beta $$| = $${{\\sqrt D } \\over {\\left| a \\right|}}$$ = $$\\sqrt D $$\n
and BC = $$ - {D \\over {4a}}$$ = $$ - {D \\over 4}$$\n
$$ \\therefore $$ Area of rectangle (ABCD) = AB $$ \\times $$ BC = $$\\sqrt D \\times {D \\over 4}$$\n
From property we know,\n
Area of parabola with the x axis = $${2 \\over 3}$$(Area of rectangle)\n
$$= ({\\pi \\over 2} - 1)$$ sq. units",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4996,
"subject": "General Science",
"question": "Which of the following is true for y(x) that satisfies the differential equation
$$ \\therefore $$ $$y(1) = {e^{ - {1 \\over 2}}} - 1$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 4997,
"subject": "General Science",
"question": "Let y = y(x) be the solution of the differential equation
xdy $$-$$ ydx = $$\\sqrt {({x^2} - {y^2})} dx$$, x $$ \\ge $$ 1, with y(1) = 0. If the area bounded by the line x = 1, x = e$$\\pi$$, y = 0 and y = y(x) is $$\\alpha$$e2$$\\pi$$ + $$\\beta$$, then the value of 10($$\\alpha$$ + $$\\beta$$) is equal to __________.",
"options": [],
"answer": "4",
"solution": "**Answer:** 4\n\n$$xdy - ydx = \\sqrt {{x^2} - {y^2}} dx$$
$$1 - {y^2} = {({e^3}2)^2} \\Rightarrow {y^2} = 1 - 4{e^6}$$",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 5000,
"subject": "General Science",
"question": "Let a curve y = y(x) be given by the solution of the differential equation $$\\cos \\left( {{1 \\over 2}{{\\cos }^{ - 1}}({e^{ - x}})} \\right)dx = \\sqrt {{e^{2x}} - 1} dy$$. If it intersects y-axis at y = $$-$$1, and the intersection point of the curve with x-axis is ($$\\alpha$$, 0), then e$$\\alpha$$ is equal to __________________.",
"options": [],
"answer": "2",
"solution": "**Answer:** 2\n\n$$\\cos \\left( {{1 \\over 2}{{\\cos }^{ - 1}}({e^{ - x}})} \\right)dx = \\sqrt {{e^{2x}} - 1} dy$$
Put cos$$-$$1(e$$-$$x) $$\\theta$$, $$\\theta$$ $$\\in$$ [0, $$\\pi$$]
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264095/exam_images/b544kekbu57hggvawadq.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5itvxgl/85123a1a-1c60-4f89-ae17-a908bdc5c880/af37e550-023d-11ed-82d2-afb4f8196e66/file-1l5itvxgm.png?format=png\")
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2525,
"subject": "Chemistry",
"question": "![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lecssgjd/16d9b562-4c48-4482-adcb-e3b645d7f48b/a41a5a90-b11a-11ed-8d19-c1418e0c6b67/file-1lecssgje.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264651/exam_images/f8e0ml2oibv52mp4tpzq.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7oihc9m/bdf7ed88-de80-4efb-bbfd-4b02aa3446d5/37c09ba0-2cf6-11ed-a627-29822045a51b/file-1l7oihc9n.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267288/exam_images/maa8zyxwayts409seryn.webp\")
",
"topic": "Geometry",
"subtopic": "Solid Geometry"
},
{
"id": 2622,
"subject": "Chemistry",
"question": "Which of the following compounds is likely to\nshow both Frenkel and Schottky defects in its\ncrystalline form?",
"options": [
{
"text": "CsCl"
},
{
"text": "AgBr"
},
{
"text": "ZnS"
},
{
"text": "KBr"
}
],
"answer": "AgBr",
"solution": "**Answer:** AgBr\n\nSince AgBr has intermediate radius ratio. AgBr shows both, Frenkel as well as Schottky\ndefects.\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263655/exam_images/q1lnoip3a57h44bxdiqa.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266570/exam_images/iuqkwiyvbaofgz2gykqn.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263290/exam_images/rroiix1g0nd5wlzmhukh.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266616/exam_images/jdjuoyplzv6a2jvihjeu.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxfqatp2/3d8efe2b-8583-464f-a1f5-0d7506d18635/93f38760-6227-11ec-95de-59bb36457aaa/file-1kxfqatp3.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l3hf6b3o/7a43a948-6d10-4eb8-9882-d0d5582ecf46/ef80b740-d9de-11ec-8ea5-5be6b5e3368b/file-1l3hf6b3p.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266971/exam_images/w3s8emlh3l3ksxg0sr50.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264447/exam_images/omb7cmyem4cuacaeyca0.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l3l0kzzj/a9c6d660-3491-4798-86a3-64ab73f568d1/01b6dbf0-dbd9-11ec-adf6-01b3c1852b0a/file-1l3l0kzzk.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l3l0lqhi/4f834d48-6bdb-47e0-a992-ac62b22641c4/162f2f60-dbd9-11ec-adf6-01b3c1852b0a/file-1l3l0lqhj.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7vya06e/57a2bec7-f6bd-48ad-b27f-de2cadb545d6/7a5ca060-310d-11ed-9c98-ad39d53b642b/file-1l7vya06f.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l2o2fweo/571393a7-7dcc-4669-a6ba-4de1db06cd36/68bb2d00-c9ba-11ec-83f5-cb752fc7e83a/file-1l2o2fwep.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5fiey4l/9f7176f9-a5d6-487c-9f31-6b0fb1a2791b/789bc040-006a-11ed-983d-b1f8143a2272/file-1l5fiey4m.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l91qrzwr/f024bfe5-819c-4108-b4df-01b1fcee4bf6/10f635b0-4809-11ed-935b-0fd6c7d272a9/file-1l91qrzws.png?format=png\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 2883,
"subject": "Chemistry",
"question": "The most abundant elements by mass in the body of a healthy human adult are: Oxygen (61.4%); Carbon\n(22.9%), Hydrogen (10.0%); and Nitrogen (2.6%). The weight which a 75 kg person would gain if all\n1H atoms are replaced by 2H atoms is: ",
"options": [
{
"text": "37.5 kg"
},
{
"text": "7.5 kg"
},
{
"text": "10 kg"
},
{
"text": "15 kg"
}
],
"answer": "7.5 kg",
"solution": "**Answer:** 7.5 kg\n\nGiven that weight of human adult is = 75 kg\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxo9dijv/33620a10-f4f0-4c1a-a705-e165d99dff9a/716b4fb0-66d8-11ec-8163-35395675f50a/file-1kxo9dijw.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l42q8cb8/78188ad6-e755-4dd5-afd7-450b5a6ba53d/a21d0a30-e596-11ec-8082-a116589b2e55/file-1l42q8cb9.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l42qhbq6/d44e434f-b07a-469a-a3f6-e984abf1e566/9bf1cbe0-e597-11ec-8082-a116589b2e55/file-1l42qhbq7.png?format=png\")
\n\n ![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267750/exam_images/byqxl3jhugiak03bl3a2.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l2ry7os0/3e0aeb10-b629-42f4-a9e5-70bbef4f8100/02720300-cbdd-11ec-912f-ef0b3f5040a9/file-1l2ry7os1.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5fi99d1/f19a255f-06f1-4aaf-a9ea-d6803683a320/da6e9640-0069-11ed-af33-9d935aadc57e/file-1l5fi99d2.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5mip2k4/8e8a135c-20f3-4574-aaa6-9ece3a3c64b4/0eab5f40-0445-11ed-a91d-f9384673f194/file-1l5mip2k5.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lectiu96/0bf71257-e585-402c-b457-4e5bd7d5a743/81bc90a0-b11d-11ed-b393-e7fb130a46e6/file-1lectiu97.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1oojws/578f30af-ef4e-4879-806c-11e0ae4906f5/327ec0c0-d3f5-11ee-8d6b-33dde4413c62/file-1lt1oojwt.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265272/exam_images/z9jbmvhqmo6lrjecf1wo.webp\")
\nRed end means end of visible region which is balmer series of lines so n will be 2![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264926/exam_images/q9eb3hknh6523aptcv4s.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263846/exam_images/xuzlpvd0itv9joyksdst.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l3ofg2sv/8310697b-4c0d-473c-8a0e-02f2c197bb61/7bba7bf0-ddb9-11ec-b7c7-a7ac6d9334e0/file-1l3ofg2sw.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264714/exam_images/dsv1lbbkb9ywfbctiqeb.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1liblc1mx/0d535c30-8040-4f86-9d7d-513ed22dcd29/909ba390-ffa1-11ed-b152-c3d4818c5b44/file-1liblc1my.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1liblcsfc/226358ee-8c6b-48b0-b8bf-aedc04bf869f/a54d5680-ffa1-11ed-b152-c3d4818c5b44/file-1liblcsfd.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1libldehj/9a06ffb6-b6d3-4003-82fe-530f7010a03e/b6582180-ffa1-11ed-b152-c3d4818c5b44/file-1libldehk.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l09tqvxx/7f6d0d46-03ce-4be5-9a34-f82e5b2c2e07/0042a650-9a4d-11ec-bb3a-6b144b885005/file-1l09tqvxy.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l09trwsn/1ab8d378-af4d-46d3-89d3-11cdec76de71/1cba6660-9a4d-11ec-9517-4b7bd39f02f4/file-1l09trwso.png?format=png\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265867/exam_images/kdnkykcecyti11syp5sf.webp\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 3254,
"subject": "Chemistry",
"question": "An ideal gas expands in volume from 1$$\\times$$10-3 m3 to 1 $$\\times$$ 10-2 m3 at 300 K against a constant pressure of 1$$\\times$$105 Nm-2. The work done is :",
"options": [
{
"text": "-900 J "
},
{
"text": "900 kJ "
},
{
"text": "270 kJ "
},
{
"text": "-900 kJ "
}
],
"answer": "-900 J ",
"solution": "**Answer:** -900 J \n\n$$w = - P\\Delta V = - {10^{ - 5}}\\left( {1 \\times {{10}^{ - 2}} - 1 \\times {{10}^{ - 3}}} \\right)$$ \n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264997/exam_images/xnctexndos5letrbgfdr.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7rqujrj/0f6b0ef6-761a-46cc-a9ca-a65eb48d3da2/4a0b61f0-2ebd-11ed-bc10-01eb53e9c79f/file-1l7rqujrk.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266412/exam_images/xqnr8ycf8rxtzbvwmps8.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7vzh473/2c505161-f77d-4a29-9463-89bd7ef6b91f/294280f0-3112-11ed-9c98-ad39d53b642b/file-1l7vzh474.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lef4bwye/c33528a7-bcd9-4e71-a79f-57c26aae2662/56e0d360-b261-11ed-bb8c-373642a734af/file-1lef4bwyf.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266002/exam_images/q1qp2jvtfgwib9loqjwo.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267089/exam_images/rmtgi7eihcbphpv9sdfi.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7rb45xm/29b2975e-d397-4ba9-941a-4cc0bd7b15f1/c362a6a0-2e7f-11ed-8702-156c00ced081/file-1l7rb45xn.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqeg6dq/195e875b-1d41-47ba-b7a7-6f56da027c9a/8f22c6e0-cdc0-11ee-9f50-677e7e372eae/file-6y3zli1lsqeg6dr.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt2yowkb/31073c70-5b0c-4f0f-80c0-62ea5ac61070/2035eab0-d4a9-11ee-bdd1-01c80c3e2d9a/file-1lt2yowkc.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8wgd3/6b24122f-fe39-4086-be4a-889687b965f8/a5aaee70-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8wgd4.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265452/exam_images/ibx2kf4tgkt5ewlrdikt.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266411/exam_images/lc7tycmcmkjqgjziprzg.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264547/exam_images/fmk5f20pf8f2dnokx8xv.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265539/exam_images/j1l2cyuk0ctnijsktmb6.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266605/exam_images/sruywneq0jjxnahol5cq.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264981/exam_images/ehep4l3pw7lbyi0lsu7d.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267567/exam_images/staijxnqv8p4honzsnr6.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265776/exam_images/vghtupk7erbaxkhau8et.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266970/exam_images/wlg6fjohqqr0lnvjjp15.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264908/exam_images/td5ojeuzdtmnzkretbhj.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267379/exam_images/vmb11b1oodts8ory1wgg.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267715/exam_images/itpc8wihkk6ay7ozb0q7.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267697/exam_images/pcxamvzsbmuou2a9eitt.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265409/exam_images/gzyjdvs460f1b9cskjhd.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264384/exam_images/ockxunbxex6ruzroagy0.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265917/exam_images/on0qybbkpgettztu34nb.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lfg2ca35/20453aa3-5bf8-445c-837d-c8b6d848626f/8ecab100-c6b2-11ed-b4b3-b306e87ca523/file-1lfg2ca36.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263282/exam_images/vaartuqi2uokdkrcurcw.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266720/exam_images/h71pebht6whdybpiqe1o.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266901/exam_images/kkhzoknkbvrgsubbusfx.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc5qpgrl/6590d524-93d7-4f0f-8cfb-f2167895c70e/934a1200-85a0-11ed-95b6-d9b2225f1c8e/file-1lc5qpgrm.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5obyhm8/67807093-6d4d-453b-a541-6801368b15df/45fa9b00-0544-11ed-987f-3938cfc0f7f1/file-1l5obyhm9.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l65uoj8c/397fdd39-a2f5-49e7-9863-c285adbe3668/d1a0d9c0-0ee6-11ed-a7de-eff776fdb55c/file-1l65uoj8d.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5mhm7e/af25a820-04bf-4a34-b4b6-12307077c8f7/834270a0-ad28-11ed-831c-57e1cd79ae86/file-1le5mhm7f.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1let6xptn/17e2a896-c42e-490f-a47a-4034b21a373c/808505a0-ba1e-11ed-b1c7-2f0d4a78b053/file-1let6xpto.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1leyorqpr/6043abe7-b1a8-4575-ada8-a5ec43b191a3/29ee5170-bd24-11ed-8df0-25e8b2ace386/file-1leyorqpz.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgrmtdff/7b21977a-d096-41a4-aa81-a71dc07e4eb1/773bf0b0-e0db-11ed-900a-094cf67a472f/file-1lgrmtdfg.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnkifvys/3c8e033b-fa69-4b24-96b7-4df950fb761d/62d0e540-6786-11ee-8adf-57893cbbad41/file-6y3zli1lnkifvyt.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnbbqc7r/6c542301-d903-4fd1-a7ae-79fb8e9918bd/17b8dca0-6279-11ee-9a72-3138802cff89/file-6y3zli1lnbbqc8b.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnbbyfva/975f2e66-a12d-4dc9-903b-55c6281e4359/f9019760-6279-11ee-9a72-3138802cff89/file-6y3zli1lnbbyfvb.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lmyvtkfw/de189598-fad7-463c-a96b-e9cf1720e557/77a21fc0-5ba1-11ee-b31c-37f6bf9b942e/file-6y3zli1lmyvtkfx.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1locplbnw/62b80666-6a84-4264-9637-5115e1fdf309/4c1321b0-7708-11ee-a563-572884a00dbd/file-6y3zli1locplbnx.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266711/exam_images/szym4mrimiv19nlybmq9.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266241/exam_images/fhdx0dohqlxs327d7ok9.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267568/exam_images/jagkrfpsaoaaeae9ashz.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267424/exam_images/kdcf95puxmjkf83bykfg.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264765/exam_images/f91a8wrnbdvn5tmzebxr.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265215/exam_images/ugp7hf8znshbglhegrah.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264512/exam_images/catlrcudnwzjdrd5vl2e.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l6dmv11z/765801b8-1a82-4486-8b88-3108e93f1dbd/8269d270-132e-11ed-941a-4dd6502f33e3/file-1l6dmv120.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbsbor/276c38c8-004d-4184-95dc-ab0e753739ef/40a706c0-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbsbos.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7nc1t7n/ab3c067b-907d-4d61-90d4-40c5dd219f33/4878de40-2c50-11ed-9dc0-a1792fcc650d/file-1l7nc1t7o.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7qa9n3e/85e3a21e-e14a-4db6-947b-e473de8b2f29/a9e898a0-2def-11ed-a744-1fb8f3709cfa/file-1l7qa9n3f.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lekt8kqs/c8535748-1d32-4d4a-8b11-6aa2b4c12893/a97e5c40-b582-11ed-b66a-d501f51666d6/file-1lekt8kqt.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lef5vgns/9610cbeb-a0ae-4272-89c8-8cbf8a31b61b/5f9aceb0-b267-11ed-9d4d-b96eca78f2e5/file-1lef5vgnx.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lebz24vh/fd89cb9d-00d8-44b0-a190-8609bc533a04/5f4f5dd0-b0a6-11ed-8017-e30f07067392/file-1lebz24vi.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le2m97s7/2871d106-b3b3-42e5-b023-ab35f716d2cf/3fb28b70-ab81-11ed-88fd-4faade8d5a9f/file-1le2m97s8.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnq8po6g/d220c54e-806e-4b56-ac43-a5312272c160/0b7f4280-6aad-11ee-822f-a9d01571f2aa/file-6y3zli1lnq8po6h.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsohd1pb/c1eafb9f-de3f-4c37-82e0-908506938702/622ee6f0-ccb2-11ee-aa98-13f456b8f7af/file-6y3zli1lsohd1pc.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1c0zyx/3ef0ec08-6ff8-4ca7-97e0-1c5c0fde9c3c/b5ee7090-d3c3-11ee-a50b-bb659a2e1d74/file-1lt1c0zyy.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1toimr/45988a05-1bb2-47f1-9316-43cb63fa6ccd/bf271320-d408-11ee-b9d5-0585032231f0/file-1lt1toimt.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1xgjzw/7313cfc0-b5bf-4cc6-9a03-217f83a625f3/86406ed0-d417-11ee-b9d5-0585032231f0/file-1lt1xgjzx.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwr9eg/7ae9597b-3dbe-4745-92d2-f1eee5f94c62/ab25ed80-ca2e-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwr9eh.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt30pn5h/00c97644-b936-4207-92b6-5fa8af4c0084/0701ea50-d4b1-11ee-8384-811001421c41/file-1lt30pn5i.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxte3w/356f1335-5223-47e5-b5cc-ef41c9b2c27b/bac74ec0-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxte3x.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnidnf/3617c118-3469-40cf-b021-2c95fa43dbe8/fe8826b0-cde3-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnidng.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw1mpu79/db1f9115-bd83-4afa-bbc1-7bdbc6f490b3/2a21d550-0f52-11ef-91cd-f19f7dc20f18/file-1lw1mpu7a.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw1my772/a7f8567c-57ab-46bd-a498-b733b7e930a6/12a4b0e0-0f53-11ef-91cd-f19f7dc20f18/file-1lw1my773.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwhgb1g0/feeaa3f8-43fb-402d-a95e-980c0569abe3/a234b100-1805-11ef-b156-f754785ad3ce/file-1lwhgb1g1.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwhk80oi/a0341b0b-ecfa-449a-9d8f-e828635e1020/f2b47520-1814-11ef-a08c-4f00e8f9a4da/file-1lwhk80oj.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw4lkbz1/158cd7a8-1619-430b-bf76-8b2ddd399871/01d314d0-10f4-11ef-8553-fdfc6347789d/file-1lw4lkbz2.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw4mrnop/6ea8698b-9387-47c8-8aca-29425e7a83aa/b6aa6990-10f8-11ef-b50f-395513bc7af2/file-1lw4mrnoq.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwenv6x5/39fb95d9-cd80-4c68-91b8-1f5c0166f573/d8931b90-167c-11ef-9070-f523f4c6bd4b/file-1lwenv6x6.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lweuubn4/2752e8cb-54f7-4c91-8212-691518b0463e/20511700-1698-11ef-ad74-71e0a1829759/file-1lweuubn5.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwad7urv/960c0402-e139-4dd9-89a4-53df9b260717/11b135b0-1420-11ef-90c1-6758c45d5635/file-1lwad7urw.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwd5pdpk/204a0778-5148-47b3-a659-091fdb696421/09fd8280-15a9-11ef-99cc-63deb36dcc4a/file-1lwd5pdpl.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265346/exam_images/c0dqhctziwbekfugrolb.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265266/exam_images/iljyyh25rrxltbqfibh1.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265244/exam_images/mtsizoslixc1qq9ocqig.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264937/exam_images/uqar8tycbulapjs4smas.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265437/exam_images/kiidub3irc2kt9mrj7ft.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267761/exam_images/fvcqglwivzefc79a48mr.webp\")
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![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7xrxbtg/d0e01ee5-c717-48d4-b80a-6a1b9ae27182/345bdd40-320e-11ed-bab2-5faeb0367b9d/file-1l7xrxbth.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lea9v38f/9887ff8f-08e7-4f18-9711-257b854f4854/0e8003e0-afb7-11ed-a5f0-99851f9df37c/file-1lea9v38g.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lh2hc8pe/d4ba8260-fd3b-48a4-aed6-cb4a8b6ea85d/ac53ff20-e6d2-11ed-b683-39a79c7e644a/file-1lh2hc8pf.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1li9wiey7/d8727013-7167-49ce-a48e-7f381925a62f/b4de6bf0-feb3-11ed-9a3f-cb8a261721bd/file-1li9wiey8.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lifho571/8169bced-736c-4022-8d65-aab144d24688/5f25cfd0-01c6-11ee-a619-03f3ffceb323/file-1lifho572.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lig7g1t0/e607e161-c3d1-45fc-96f7-d8420574551c/2b69fe40-022b-11ee-8eea-09e354b91938/file-1lig7g1t1.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln74ubxe/c878026c-5267-42b0-b10f-7150edf4c2a2/04392c20-602b-11ee-84d4-f1c1aec911f3/file-6y3zli1ln74ubxf.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lo34mk9p/c7b53e64-218a-4775-ad23-59117bd05abd/40f9acd0-71c3-11ee-84b8-cf7c0a25cec4/file-6y3zli1lo34mk9q.png?format=png\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263702/exam_images/e7w2mcfgqs0qwtwa1jkf.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265305/exam_images/jl4uz1aheqoemq1myxxa.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264221/exam_images/up9xtippomu7kefhskum.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265826/exam_images/evqbnlvd5jv0hmjutzfj.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265674/exam_images/oasqyywx4wcmnissw76e.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265411/exam_images/o27yks7ncu1vj6sgzmlx.webp\")
\nWe have\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265946/exam_images/x9rxyugujvfepiew1pgy.webp\")
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\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265888/exam_images/flpfpj2fdznelp4hlns0.webp\")
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\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266001/exam_images/fwsvcudxolhotbhb0swn.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264037/exam_images/uk75haohyuzity3go0py.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265993/exam_images/ignyryih3n3jwrth8owo.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264724/exam_images/xhueaftoffkovgut0qew.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263573/exam_images/qgzb6vevzrjgj6qtkfoq.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266991/exam_images/mswypgvszyrm0wd5qq5t.webp\")
",
"topic": "Algebra",
"subtopic": "Equations and Inequalities"
},
{
"id": 3748,
"subject": "General Science",
"question": "A box open from top is made from a rectangular sheet of dimension a $$\\times$$ b by cutting squares each of side x from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to :",
"options": [
{
"text": "$${{a + b - \\sqrt {{a^2} + {b^2} - ab} } \\over {12}}$$"
},
{
"text": "$${{a + b - \\sqrt {{a^2} + {b^2} + ab} } \\over 6}$$"
},
{
"text": "$${{a + b - \\sqrt {{a^2} + {b^2} - ab} } \\over 6}$$"
},
{
"text": "$${{a + b + \\sqrt {{a^2} + {b^2} + ab} } \\over 6}$$"
}
],
"answer": "$${{a + b - \\sqrt {{a^2} + {b^2} - ab} } \\over 6}$$",
"solution": "**Answer:** $${{a + b - \\sqrt {{a^2} + {b^2} - ab} } \\over 6}$$\n\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266134/exam_images/u0nyg29w60wvazbqlpbt.webp\")
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![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc6nga5p/f6bd0c59-facc-4024-b756-658e8619a898/a12a41d0-8620-11ed-ba1b-3b1de775931b/file-1lc6nga5q.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5ykgpbg/402463f1-cb39-4303-b051-0d9df56ac983/6004f6c0-0ae5-11ed-a51c-73986e88f75f/file-1l5ykgpbh.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5yki2b2/272d4c44-a92f-4bfd-a59e-312affd66d01/85dcc4e0-0ae5-11ed-a51c-73986e88f75f/file-1l5yki2b3.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5ykjlo9/ac69243c-25e4-42c2-af51-acef287e6cb5/b0a1b690-0ae5-11ed-a51c-73986e88f75f/file-1l5ykjloa.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7rb2qy2/d1f9b6e9-ed06-4643-80c4-d9d8e53ea216/9bffa4a0-2e7f-11ed-8702-156c00ced081/file-1l7rb2qy3.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7sso9mz/26e78098-e521-4cb7-8650-9f6cbc3d58bd/364659b0-2f51-11ed-85dd-19dc023e9ad1/file-1l7sso9n0.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1li61mesu/46698a3c-f1f2-4c38-a7f4-cdbe7d5c6371/8f09b1d0-fc94-11ed-8452-d7b44c8805e4/file-1li61mesv.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnbdiknq/edf7bbaa-df8b-4c8e-bede-daa23c7634e0/12049450-6280-11ee-95f7-df7702d42956/file-6y3zli1lnbdiknr.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lmls8gbc/7f8dbf18-7516-4da9-acc1-29bce97edc3b/409f8a70-546d-11ee-b10a-c342ea039081/file-6y3zli1lmls8gbd.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt307bt9/dd10aeba-887c-4185-a5fe-218e2ee9b2bf/09af1ae0-d4af-11ee-8384-811001421c41/file-1lt307bta.png?format=png\")
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![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxw4nw/e2162195-84ee-4126-95ca-5b36a9ae1ecf/06e90dc0-ccf3-11ee-a330-494dca5e9a63/file-6y3zli1lsoxw4nx.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqngokt/b3ef22bc-e563-4b45-a320-95a6d263c505/cf5ad1d0-cde3-11ee-a0d3-7b75c4537559/file-6y3zli1lsqngoku.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw3i47n9/78e3808f-b6b1-4774-ad57-06e016fc0e17/bd7fae50-1059-11ef-abcd-c333ada72a30/file-1lw3i47na.png?format=png\")
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![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw4uhftn/eaae4a9d-2d8a-4b86-93c7-b8b3dfc45e4c/e388a2b0-1116-11ef-b9cb-b5e0fe4ba33b/file-1lw4uhfto.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw4uiqqc/051e5d6a-5dcd-41d3-92ba-260c0a74c2b1/07c4e940-1117-11ef-b9cb-b5e0fe4ba33b/file-1lw4uiqqd.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw4uomxr/b876dfc7-50c6-4c89-9214-81e07c07cec0/abb17af0-1117-11ef-b9cb-b5e0fe4ba33b/file-1lw4uomxs.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw8slgdc/e7ec72d2-9c20-4762-a251-c217bf88f1ea/a3884200-1342-11ef-9cb4-095599b9956f/file-1lw8slgdd.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwgejsnu/15565bcf-b8bf-4793-9f53-176aa69e81ff/facbac90-1771-11ef-910c-8da948f7ddd9/file-1lwgejsnv.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lweqs4gr/cb92f82c-cbcf-4ca2-8ad5-d05029ed7bb2/3ea356b0-1688-11ef-9ee8-13752e98d8d6/file-1lweqs4gs.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264871/exam_images/dc3lrjubrnemgdlvwzox.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264455/exam_images/u5uxawmfdatdnnbbfy6y.webp\")
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![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc8ekncq/3bb86ec8-2e21-4aa5-b7aa-771daae2cbcc/79bb29a0-8717-11ed-b3ec-0bde88094e1e/file-1lc8ekncr.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lkavtohl/5b2f07ad-6815-4447-8b5a-bda6bf0cb5b3/3c6a7290-26d6-11ee-b52b-3728f15f4ced/file-6y3zli1lkavtohm.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lt4ueirs/bdaf4cf1-dcad-4fa2-b9ed-fb23b230da96/ec3f4080-d5b1-11ee-954c-abbae7354768/file-6y3zli1lt4ueirt.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjw7x98/08591748-2ee9-41c3-9af0-f822bc19a883/916485c0-ca2c-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjw7x99.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265845/exam_images/dk7w8bditkgxgia8lokx.webp\")
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![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7qaiwsv/317e788b-9fe5-4c30-acd5-40a1d96c242d/abb0b900-2df0-11ed-a744-1fb8f3709cfa/file-1l7qaiwsw.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1locy9o68/7ed2dae1-7f5d-4feb-a54d-f2d90c2aabe3/3a30c2f0-772a-11ee-8319-49c6602bc172/file-6y3zli1locy9o69.png?format=png\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266469/exam_images/f698mokq1rh01m1s0toe.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266899/exam_images/knvyxnlyn6etgfelas4i.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267203/exam_images/eyxgnumsxyaujoadjbfq.webp\")
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![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5miy8jq/0355ce2a-ccfd-4014-9a4e-f11ecb74e789/0d940660-0446-11ed-9d9a-21bdd0f00aa4/file-1l5miy8jr.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5mj1eqq/48f3c678-5009-4ace-b7a5-84db65f774f0/65ca6c20-0446-11ed-9d9a-21bdd0f00aa4/file-1l5mj1eqr.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5mj51wi/7cbdb428-88d9-49cc-9dab-6a1d9bb5b71b/cb1bec20-0446-11ed-9d9a-21bdd0f00aa4/file-1l5mj51wj.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5mj6a8q/2a8081a0-9fe2-49f4-95f0-647c1e4d436f/ed5c3ba0-0446-11ed-9d9a-21bdd0f00aa4/file-1l5mj6a8r.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5mj7g67/ccde0452-eeab-497b-a7fe-49ce58933922/0dc02a00-0447-11ed-9d9a-21bdd0f00aa4/file-1l5mj7g68.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5oc5ao6/7ae3148f-ca07-4559-92f9-0d6735c2710d/0346fe60-0545-11ed-987f-3938cfc0f7f1/file-1l5oc5ao7.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5vb3wli/4dde9a3d-2417-4084-b660-1fce1c16846a/70032f50-091a-11ed-80eb-03ed789c0f4d/file-1l5vb3wlj.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266365/exam_images/eejvpxyhkrguzqnnbwmv.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264182/exam_images/lbtyjmjfgihwg2ehcrky.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265258/exam_images/l9bcohz8ayjblt6nujfm.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266321/exam_images/ixjlsq9plxoilqztxxsw.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267219/exam_images/qhldynplznxc2j2lyapa.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266518/exam_images/peptywbkfivgg2xjbmvn.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266232/exam_images/czdd2doug0cscdzgdhxu.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264492/exam_images/dhvwqvaj51muxjrzua8j.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267132/exam_images/qwadcct8cbggvfesgjha.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267579/exam_images/y3qu2ps3iqhwrvq5xytr.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263338/exam_images/lx0pkl74ucmclccrmyhl.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266589/exam_images/p8cxbhqzhcyvdc09huxd.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l91nf3yh/b4e886d1-7f41-4b72-ab7a-245bf94417c9/ef2e8480-47fb-11ed-8757-0f869593f41f/file-1l91nf3yi.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265211/exam_images/xedy87gs6fsevx9ehium.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264764/exam_images/zyqeh2enbhsl6lnxeebb.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l3bio1x7/04a063ba-ee8e-4754-b8f8-b564026e9f6e/e47771a0-d69f-11ec-a354-cb09522b689a/file-1l3bio1x8.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266336/exam_images/dfewqownwnzl1h0lfyw7.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263924/exam_images/cx2cz6kvnuizfpmqmwgs.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265062/exam_images/b58nqrkplxkhfj2zdyvj.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266090/exam_images/qqdptvmz6bwhqedphfk6.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266672/exam_images/z4km1mkoieljdfzlmsq9.webp\")
\ny2 = 4$$\\lambda $$x and y = $$\\lambda x$$![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267762/exam_images/pgiki0cdc9cjw0ntf4lc.webp\")
\nLet P be the point common to x + y = 1 and y2 = 4x![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263379/exam_images/pwnxyxw58zdpddygd57j.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264662/exam_images/oejbpdgknl2hbpnnncph.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264477/exam_images/utoesgjenyhl0zmw0ys8.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264058/exam_images/uhew9xrggiezdnledix0.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263553/exam_images/ajducliqewdrjwwiwzo7.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265696/exam_images/atq9a0prxhuwrfqaru4w.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267732/exam_images/mtmqm6fur1hjtylyhwhz.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264818/exam_images/s3vrvzluxopimmp3ygw5.webp\")
\nFor point of intersection \n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264666/exam_images/koeynuqfjc3pdhjdcnwf.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265543/exam_images/g6ehrq5oh7fip48nkm6y.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267143/exam_images/d1iaq40mqsj3mmajo5mu.webp\")
\n$$A = \\int\\limits_{{1 \\over 2}}^1 {({x^2} + 1)dx + } $$$$\\int\\limits_1^2 {} $$$${(x + 1)dx}$$![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263709/exam_images/njtahhijwsodoo0btdmn.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264106/exam_images/d94wa1tsnhgimjfnfsfb.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264200/exam_images/yig7dqvbpytiruqt2hnf.webp\")
\nC1 : y2\n = ax, C2 : x2\n = ay (a > 0)\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263719/exam_images/casyif6hyghpfwzxghwt.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263874/exam_images/ilutwsifhzc063d2ocbw.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266337/exam_images/yov949hdcyfww9kn13ai.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266907/exam_images/sbavak08yt3gbpyt9h8l.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264729/exam_images/jm9chvj4tml7xrg2bqui.webp\")
\n\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264150/exam_images/scswst5jdqmnovfjgtnf.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265792/exam_images/adjt0f0zaxeqso6stpct.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263552/exam_images/hlnbvgwmqwwtyylb5tjw.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266774/exam_images/u1ozidkzbjd8lcnvbnev.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264240/exam_images/tijy7o0iderdha3gedzq.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264963/exam_images/u8k9qqbaojrvwvtkepl5.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264045/exam_images/gmfzldxp4t8nl2p4kdgf.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266757/exam_images/yc56rqbdajzn1a2loepd.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263659/exam_images/k3ibititsvzikcqrsdja.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263996/exam_images/er6uqxcqynx3kojs74k9.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5niit1g/6da14760-cb2e-4c6b-b5f6-e80807f5e0ce/291c7440-04d1-11ed-93b8-936002ac8631/file-1l5niit1h.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc5rg14e/3ec2057b-efed-4095-9156-11e5f06f21d7/760d25d0-85a3-11ed-b688-91788a069596/file-1lc5rg14f.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5obth1c/3674f2a1-bb5f-42c5-bcda-0e644fc0ece5/ba7c4f10-0543-11ed-987f-3938cfc0f7f1/file-1l5obth1d.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5p7lae4/469ec989-bde2-4216-be88-8dbb75d383c8/fb02f1c0-05bf-11ed-8617-d71e6444d1a0/file-1l5p7lae5.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5q9zfw8/6cdd28b7-2353-4e56-94fa-08f93a22752d/1f7f8b80-0656-11ed-903e-c9687588b3f3/file-1l5q9zfw9.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5qa0x8z/b3f515b6-b4dd-4a2e-b38c-8edeeca6de75/48b68a30-0656-11ed-903e-c9687588b3f3/file-1l5qa0x90.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5rh7hff/4ef2bdc8-3288-4017-9611-a67305104639/27b58cb0-06ff-11ed-b821-f5ba0940c0a2/file-1l5rh7hfg.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5qp2am3/c5e50fc7-f237-4ff6-b8a1-b5d6a9b181f3/17c870b0-0691-11ed-93bf-f57702a71509/file-1l5qp2am4.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5th819e/347b9698-98c3-43fc-baeb-9ed9884fb187/c8540420-0818-11ed-98aa-f9038709a939/file-1l5th819f.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5v6294y/2f8c41d0-8e8c-4db4-b12e-8da89b735e16/b46f9a20-0906-11ed-a790-b11fa70c8a36/file-1l5v6294z.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l8lc1nby/e7dc3cf8-3c01-44b7-9f93-5093bb7a6fcf/ea0230d0-3f02-11ed-8d74-051dc2e154aa/file-1l8lc1nbz.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l65usgec/bd118792-9d01-40fb-b8a3-8435f69bbadb/3ead1c40-0ee7-11ed-a7de-eff776fdb55c/file-1l65usged.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l65uu2u5/3cd72241-ee42-4674-8663-b746043011c3/6bd1a8d0-0ee7-11ed-a7de-eff776fdb55c/file-1l65uu2u6.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l97rmv56/3d3d2137-f038-407b-8cca-4933b5be4cbf/1f426190-4b59-11ed-bfde-e1cb3fafe700/file-1l97rmv57.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l97rzcly/4fa8a86d-de28-44c0-be52-7dde80194483/7a763360-4b5a-11ed-bfde-e1cb3fafe700/file-1l97rzclz.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l97s0prs/1a087718-c94f-41f2-86cd-76314c7ef4ef/a0702f80-4b5a-11ed-bfde-e1cb3fafe700/file-1l97s0prt.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc3zqoek/4bbf8e50-e22c-45d7-936b-72d7ce06a9e2/55dcbdb0-84aa-11ed-bd29-d15f3a725cdf/file-1lc3zqoel.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7nnzsdz/1978812c-886a-41da-a99e-ba9650919e28/fdc43870-2c7e-11ed-a18d-5933e4fde865/file-1l7nnzse0.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7ps543w/90436956-c0bb-4a4d-b417-ec2e224aeb75/c7ae54c0-2da8-11ed-8542-f96181a425b5/file-1l7ps543x.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7q9zzgg/929f9d9b-01c4-4a4d-8457-02912d2b4071/9d5df400-2dee-11ed-a744-1fb8f3709cfa/file-1l7q9zzgh.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l97u3d0g/088f2998-8bd4-4720-8b55-fa1efe186501/bc448a00-4b62-11ed-80b9-4154b7faa509/file-1l97u3d0h.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7ssldo3/57b8d8ee-cd1f-4f78-8605-15da9040ba0e/e5f60c30-2f50-11ed-85dd-19dc023e9ad1/file-1l7ssldo4.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5fqsjf/90c628e8-42f9-49e3-ac52-f0f944e1d31e/2287dcb0-ad0e-11ed-a86d-8dfe0389db88/file-1le5fqsjg.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1leeiutst/47959be2-54bf-4c50-8327-bce6176bbe32/5bf957e0-b20d-11ed-9c9a-c3bd979bc154/file-1leeiutsu.png?format=png\")
\n\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le9xvce9/cce9188c-fc73-4ab5-a87d-33da42a11b53/28073f10-af88-11ed-bd02-1d2a2a7b6687/file-1le9xvcea.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lelnbcxl/c11e6b70-de76-422b-8148-ab95b681cbde/48c30290-b5f8-11ed-bc05-1be7a3b49a67/file-1lelnbcxm.png?format=png\")
\n\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1leolev6m/9d2f1ed0-3ff3-4daa-b4d0-ae63d5dd71a4/31fa37e0-b797-11ed-b103-ed967fad3dff/file-1leolev6n.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1leol5gd1/d39e5f85-8fa2-457b-8f2b-aa98ad9bb499/2c3ee450-b796-11ed-b103-ed967fad3dff/file-1leol5gd2.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1leq0fubs/f2abef4a-3124-4350-839e-3e7bfc1aa857/beb62c80-b85e-11ed-9fed-b1659a6c339b/file-1leq0fubt.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1let7mhvc/30704cc1-9134-44fb-84c5-b2ca232ae1f5/319af780-ba21-11ed-b1c7-2f0d4a78b053/file-1let7mhvd.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1leksols2/a3491772-b199-41ab-9761-0d39a13abce7/7e1c5220-b580-11ed-b843-fd540edb80bf/file-1leksols3.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1ldt0ucjn/f3408cd8-1181-40fe-aecc-1f2a69b3243c/74ae7e20-a63a-11ed-a341-61f046f8f7d7/file-1ldt0ucjo.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1ldt0xhxy/b01c4c6d-ed94-4045-ae7d-51c181e8b9a2/cc476b60-a63a-11ed-8e98-03be028140e9/file-1ldt0xhxz.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1ldt1346q/f2c4d820-3596-4684-af86-b7ffd222c8ce/6881ed20-a63b-11ed-8e98-03be028140e9/file-1ldt1346r.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lekro7yb/473e481f-1c3a-4e67-a338-a22cacc1490d/8a4d0930-b57c-11ed-9ad9-3b1cedbe69d8/file-1lekro7yc.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lekrqfr8/c116d1fd-fb30-46f1-8aaa-33c69069d967/c7f29750-b57c-11ed-860d-ed3e7068b1ec/file-1lekrqfr9.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5i5l5v/1c86c1a0-48dd-4e1e-9be0-6401a21b507d/903a5b30-ad17-11ed-8a8c-4d67f5492755/file-1le5i5l5w.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le1yf9t4/b18e1d13-83c3-4532-94a6-9b8efbcd5fcf/0d044770-ab24-11ed-b3ea-6525e5fb1f1e/file-1le1yf9t5.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgrn6cu8/fb34c18c-fa58-4f6a-a195-c84349a81234/e04c5800-e0dc-11ed-9ecd-e999028462e7/file-1lgrn6cu9.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lh2jl21d/1eaa8b9f-d025-4054-8bc8-91786b0e9ad1/73b7e010-e6db-11ed-948f-4b963ec65c15/file-1lh2jl21e.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lh3urauz/df812b2b-41d3-412b-9523-6e99a63e5a07/ee7558b0-e793-11ed-9b1f-65ec4fb7911b/file-1lh3urav0.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1li62ldz6/e3e1ec7b-7f4a-4711-90a5-23ea1380cf4d/5bb8a710-fc98-11ed-bb91-ad55392bec91/file-1li62ldz7.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lie8i8gh/9671b86e-ee06-496d-abb6-609a695b7ace/c0004c10-0115-11ee-9f57-5de63d0f488b/file-1lie8i8gi.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln5indhw/a6c6d162-df13-4d2e-9e8e-b7936ef392f6/714f6750-5f47-11ee-8783-9144a2081c14/file-6y3zli1ln5indj9.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln5jlzv2/c746028e-399f-4439-a1c3-68cff047545d/34119ee0-5f4b-11ee-8f55-db3179c35b35/file-6y3zli1ln5jlzv3.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnk7a0xn/c3e34f25-2aa9-4803-b342-54bfe7409450/bb60e0b0-675a-11ee-a06a-699a057b80c4/file-6y3zli1lnk7a0xo.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnda9au3/d6943506-1c0a-4dc0-87a7-f279a28202cc/e60f2eb0-638c-11ee-9778-43ae0741eedc/file-6y3zli1lnda9au4.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lmyue4ph/cd17397d-b57f-4fb7-a489-8f746007324f/e1368850-5b9b-11ee-83da-a3f80d422da4/file-6y3zli1lmyue4pi.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ljg19fl1/2d19d263-28f7-4e45-975b-04e64a691566/33ff3e40-15df-11ee-bf54-2dbadff89573/file-6y3zli1ljg19fl2.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnuiszjt/77b5ee73-13d9-423b-bb94-fc3ea2d14833/a598d780-6d07-11ee-a732-754c539ba542/file-6y3zli1lnuiszju.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lo8c5sm6/ed381416-4cb6-4a32-b174-452035f058cf/a2f26cd0-74a0-11ee-8723-4d48ce392782/file-6y3zli1lo8c5sm7.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsogvt13/49cb1ddc-aaf9-4143-8cc4-b6d4940befaa/82ba6c70-ccb0-11ee-aa98-13f456b8f7af/file-6y3zli1lsogvt14.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ltaoieus/c5d8f2a6-f8e6-40bf-901f-208d94d79f63/95946740-d8e7-11ee-a868-79761dafd483/file-6y3zli1ltaoieut.png?format=png\")
\n\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1g6xtj/f303b300-000e-4c93-be80-f4117fb2f938/ffa44560-d3d3-11ee-a874-bd0f61840084/file-1lt1g6xtk.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1wj8og/d813b308-3b51-4315-93b6-2504383585b1/e7ce7b00-d413-11ee-b9d5-0585032231f0/file-1lt1wj8oh.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwhl5f/5e904568-fdcb-45b9-b0dc-a84be5bf8837/9e21fa30-ca2d-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwhl5g.png?format=png\")
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![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt30icn5/b76421b2-6e9f-4b5a-94e8-92add4114de7/3c39af10-d4b0-11ee-8384-811001421c41/file-1lt30icn6.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8vkvl/77732020-1aec-484b-950e-aed3c54f53c2/8d586e10-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8vkvm.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxplac/f4bdc645-97cb-4588-be1e-e674c4cd7e92/51166c40-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxplad.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnpz3w/f37f9151-5383-482a-a0a5-619fb7183b76/d1c573c0-cde4-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnpz3x.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw1i51n7/b27b05e5-86d3-44de-bbce-888961f304b0/435f7e30-0f40-11ef-a754-b58ce0eab988/file-1lw1i51n8.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw36wee0/c97c824c-882e-4206-aafc-00ef66493663/dfd02e70-102d-11ef-a47a-5f0284b5aece/file-1lw36wee1.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lwk3rf0n/84c1aefe-3dc9-416a-bb2f-8c36b5773026/ece80d70-197a-11ef-8f7b-89e146e4f804/file-jaoe38c1lwk3rf0o.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwhbnwbo/8e96ed5b-9608-42df-8bc6-dc691e8b1a15/7a1fa240-17f3-11ef-b996-7dd2f40f5f82/file-1lwhbnwbp.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw4imme9/55025775-792e-443b-a53e-cd7d760a7686/861b5510-10e8-11ef-8bc0-edb751127fbd/file-1lw4immea.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw8v4grs/3efd55a7-6040-4108-bf47-c4327f08c8f1/86796680-134c-11ef-8bbe-1b4949638519/file-1lw8v4grt.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lweeutim/25a3e531-aad5-453f-b0c9-bc32519834c1/9c0ef0e0-1659-11ef-b26b-2993e9de41b8/file-1lweeutin.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwcyt7aa/3fbcb7c5-4c8c-4aaa-876a-5ac80c7dce41/145c0910-158e-11ef-bb3a-95d759b5a950/file-1lwcyt7ab.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwcyxj07/0b18439b-a031-4a45-9320-e6c48137f34f/8ca66870-158e-11ef-bb3a-95d759b5a950/file-1lwcyxj08.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267502/exam_images/ssxskaas7h8ixpoi326x.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267685/exam_images/if6lbcuycldilm9eq8ed.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265519/exam_images/gatckweagely4rxump82.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266159/exam_images/ttruyt20dueawjkbojg4.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266675/exam_images/bitcqno4ggj6nfusp1fi.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264467/exam_images/ar9w3rtouhitqkvqu61e.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263264/exam_images/ejh1nelzcfnzif29c6zv.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxvqwyud/3b08f599-748b-4dc3-95b3-f0ac0c10ea2e/861eb060-6af6-11ec-b350-33e20cd86462/file-1kxvqwyue.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l99thyns/e3d7926f-e2f7-4b54-82b0-67049b426481/fa7cf380-4c79-11ed-b94d-45a8040c2a81/file-1l99thynt.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266340/exam_images/jcx1zygqvh9prqa0yh9r.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265679/exam_images/bx4nbx1k931n5ip0ploo.webp\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4182,
"subject": "General Science",
"question": "The sum of the co-efficients of all odd degree terms in the expansion of\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266144/exam_images/sxwjinh9ykk33t08ossz.webp\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4183,
"subject": "General Science",
"question": "If n is the degree of the polynomial, \n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw38sytv/376c0654-e73d-4b44-b3e8-2fec8771109f/52a01a30-1035-11ef-b980-477f779c8c59/file-1lw38sytw.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267493/exam_images/qh2bpsmsygtrhhpn1feb.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266849/exam_images/earxtruibjeyrajhopbc.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l91npsqa/a856b112-8560-4eba-878d-d29d808c6c0a/1868e920-47fd-11ed-8757-0f869593f41f/file-1l91npsqb.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265487/exam_images/fnyfpg1ibpegkks9nf7r.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266384/exam_images/sluvcuupdddpwc5u5sek.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263616/exam_images/ssi5vdlnvmrlmfb1iuud.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264845/exam_images/vn5mptmseenvrcsn84v1.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263911/exam_images/cfj3zwxthsujcfnh8i2z.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267186/exam_images/huf8jytaaw2g3husjhna.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5sczyp3/e6de313f-31b7-4de8-8ad3-c03114d47130/7ab55770-077b-11ed-94f6-83604aa63acb/file-1l5sczyp4.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5qa5m7e/7e601531-e009-4eeb-99e6-ea8138a03412/cb3a40a0-0656-11ed-903e-c9687588b3f3/file-1l5qa5m7f.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc8cegbu/4182b4ab-8c51-4d29-951d-0c7f9f73f6e7/fb3108a0-870e-11ed-9cef-67b5ed764fcf/file-1lc8cegbv.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1leol1wpz/d1fe7720-2e72-4e7b-baef-e1168e337218/c9a57f70-b795-11ed-b103-ed967fad3dff/file-1leol1wq0.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lebxhw5x/9e3df15a-0f74-4149-87bf-67c2379ca244/4344ed40-b0a0-11ed-a0da-1fff956b892c/file-1lebxhw5y.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwk74s6l/678de8fc-2a12-44e3-8b6b-dd8440b79fd7/1bfb85d0-1988-11ef-a7bd-376696e028ce/file-1lwk74s6m.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lweo69as/998331bc-73f9-46f0-9260-389c06031bce/0c501540-167e-11ef-9070-f523f4c6bd4b/file-1lweo69at.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwd0dvt3/ce397cb2-87aa-4135-a838-45d211acf529/3c9d7570-1594-11ef-88c8-4b364e13ab15/file-1lwd0dvt4.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5v64hz0/fe62d372-86ba-43d3-b376-3bcc7cdf6f30/f2e103c0-0906-11ed-a790-b11fa70c8a36/file-1l5v64hz1.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7z3hbu0/c0ab874c-f40a-4393-abd9-7c5a221a294e/2d9f8370-32c8-11ed-a433-8d18e842a53d/file-1l7z3hbu1.png?format=png\")
\n\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1ldt0172l/c56cbb61-8f00-4972-aa26-7f1c65ed7300/4a06dfd0-a637-11ed-8501-8d588d737388/file-1ldt0174e.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsnjrugw/97f6ed68-0662-482a-8e94-7842336e50a8/07576000-cc2f-11ee-b20d-39b621d226e3/file-6y3zli1lsnjrugx.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265897/exam_images/tcowhr59pgnhld1bijkk.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263288/exam_images/uamjsjszwgsvyetzibag.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266458/exam_images/taibjlarssmymugj1mse.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264340/exam_images/ylrnxacrnykkxgmlb3iu.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266917/exam_images/e9ndaij1z9oqlqstuhea.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266928/exam_images/dalvkc5ol09xl6tzqcfw.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264886/exam_images/tuxfhzteere5dc5zztu5.webp\")
\nC1C2 = $$\\sqrt {{{12}^2} + {5^2}} $$ = 13![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267098/exam_images/ce8t39fixejz7c7rqvi3.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266179/exam_images/zp0tmjnahzn7ywtwha3e.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263855/exam_images/wmekhvnakbugxniy2cw2.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lefxmkpv/6d2d69d3-c054-4a9a-b9ad-79779d129ce3/e8127240-b2d3-11ed-8169-e1635469e777/file-1lefxmkpw.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1le5ga481/2ad2340b-953f-4942-9834-237efb8eb334/3bee0510-ad10-11ed-a86d-8dfe0389db88/file-1le5ga482.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lncsenht/b19528f0-d9c0-44df-a496-57a56c3fa6bf/168f1e10-6347-11ee-8e13-d93c7a44ebe2/file-6y3zli1lncsenhu.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwhfpfz5/cd7503a4-2670-4a67-8200-c99b5e7fc89f/49a14910-1803-11ef-b156-f754785ad3ce/file-1lwhfpfz6.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7rb097k/86b1d3e8-f7fd-49d7-85c5-28047f9ac8ec/56adfa00-2e7f-11ed-8702-156c00ced081/file-1l7rb097l.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267291/exam_images/wa96grbw4mjhrj5s0fuz.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263352/exam_images/a5hzdfxls1frvo9492ui.webp\")
\n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264751/exam_images/dhlchwnp8kfgccfwlmlr.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264280/exam_images/cr0skvuxufhkzumdnzhn.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264861/exam_images/znw3pusc82f7kl1xv763.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266464/exam_images/pevci3b1ygpkppdlpbgl.webp\")
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\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266267/exam_images/raqnosna6rdmei2xv6vs.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266292/exam_images/mqtptxaki1efkgjktaz3.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265002/exam_images/bk22gu7jwtdummbjsmw8.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265069/exam_images/iuwti6tm5cysz6ebxdbi.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266278/exam_images/kb34kqlzzf1i40yfp522.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1li9wqjod/6e47a7f9-a970-48fa-94e5-b8b0e7d86b93/96fe4cd0-feb4-11ed-9a3f-cb8a261721bd/file-1li9wqjoe.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnk5g4ko/0e3a3eb2-16a6-4d51-952c-ed316089a0d7/92c95a80-6753-11ee-a917-d528e09e481b/file-6y3zli1lnk5g4kp.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lo4iz7t1/8abb90cb-5a21-43f7-858e-e4e5d897d907/29530750-7288-11ee-955e-53a531d428f9/file-6y3zli1lo4iz7t2.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263502/exam_images/az5z8h8zms4j96efvsxd.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l91nnzlt/1c575b95-e553-4210-a91e-c044bf073271/e61a5710-47fc-11ed-8757-0f869593f41f/file-1l91nnzlu.png?format=png\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264280/exam_images/rjykkmfni3s0h2hccm3q.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263648/exam_images/kjqkelq5u313u7up3q3r.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263285/exam_images/yhmndiwp2ujk1aofixgb.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264982/exam_images/rpihjtwuabjqwdlwqq9s.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw8ocyvq/170f2048-377a-4fac-b220-54e6cf0f1120/13082c50-1332-11ef-9f8d-838c388c326d/file-1lw8ocyvr.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kx5g1hrm/3c7cc7eb-253c-4f70-8baa-06a68070a4ec/9eab2120-5c7f-11ec-bf3c-e32253ee0710/file-1kx5g1hrn.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5ykyyx8/21715b9a-e296-480d-b13d-bb0f765bfb24/5c010fd0-0ae7-11ed-a51c-73986e88f75f/file-1l5ykyyx9.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265134/exam_images/q6ltvxbm00bb6jj7jhhr.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266015/exam_images/tynzdnx9ipskg5hfjo3r.webp\")
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4367,
"subject": "General Science",
"question": "Let the circle S : 36x2 + 36y2 $$-$$ 108x + 120y + C = 0 be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, x $$-$$ 2y = 4 and 2x $$-$$ y = 5 lies inside the circle S, then :",
"options": [
{
"text": "$${{25} \\over 9} < C < {{13} \\over 3}$$"
},
{
"text": "100 < C < 165"
},
{
"text": "81 < C < 156"
},
{
"text": "100 < C < 156"
}
],
"answer": "100 < C < 156",
"solution": "**Answer:** 100 < C < 156\n\nS : 36x2 + 36y2 $$-$$ 108x + 120y + C = 0![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266423/exam_images/ulkyx4q33vkkr3loyvyz.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263671/exam_images/tmz0n9h2htagrkfozwru.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7raysn2/64b7b0e9-e0f5-48bc-ba4c-fe9b8ae351d3/2e1229e0-2e7f-11ed-8702-156c00ced081/file-1l7raysn3.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1dia1n/7e33c93b-3bc8-4d95-93f9-8ca9b48aaae2/7f9962b0-d3c9-11ee-a50b-bb659a2e1d74/file-1lt1dia1o.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt2z9gaq/def8db7c-9d66-41d8-85bb-325814b0bd97/5ba32a20-d4ab-11ee-bdd1-01c80c3e2d9a/file-1lt2z9gar.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwabk4bk/5b6f7bc9-3122-44bf-ae55-efc3ca540e79/9484e510-1419-11ef-b3d9-7392e0033caf/file-1lwabk4bl.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxip5f/ce7292b4-b5dd-4ed7-ad89-ffefc757a7b4/91687230-ccf1-11ee-a330-494dca5e9a63/file-6y3zli1lsoxip5g.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265119/exam_images/izz5jb334xfqht2xhhp9.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266529/exam_images/iuvzkrct7gyuczkecdmp.webp\")
",
"topic": "Geometry",
"subtopic": "Plane Geometry"
},
{
"id": 4379,
"subject": "General Science",
"question": "The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y = |x| is :",
"options": [
{
"text": "$$2\\left( {\\sqrt 2 - 1} \\right)$$"
},
{
"text": "$$4\\left( {\\sqrt 2 - 1} \\right)$$"
},
{
"text": "$$4\\left( {\\sqrt 2 + 1} \\right)$$"
},
{
"text": "$$2\\left( {\\sqrt 2 + 1} \\right)$$"
}
],
"answer": "$$4\\left( {\\sqrt 2 - 1} \\right)$$",
"solution": "**Answer:** $$4\\left( {\\sqrt 2 - 1} \\right)$$\n\nLet the radius of circle with least area be r.\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267101/exam_images/vuu1dbzfbd8ek3dtzrtw.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267262/exam_images/flmqizn0e2xqx806nx40.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264099/exam_images/hkzkp7lu4beshmbhun0w.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265199/exam_images/h8caxck7nxqpkaubgyjn.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264374/exam_images/gls1xhkqkxbk85dqfbxj.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264590/exam_images/u5rwyx3pmckseb9azywf.webp\")
\nEquation of family of circle touching y-axis at\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263455/exam_images/vmasvdkamze2tghom94b.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267221/exam_images/arrymdsbetsftxxqter6.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265875/exam_images/hsjo5h1locnmw5jsj3fw.webp\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kxvqvz06/35771b20-6f71-497a-bc46-b4444dec857f/6a6f4960-6af6-11ec-b350-33e20cd86462/file-1kxvqvz07.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5dvedrh/aec1f8d8-ed7e-48d5-8b3d-78754e048fbe/ae421dd0-ff83-11ec-8a90-6db465c3d8db/file-1l5dvedri.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5obw2pa/6ae93711-16f3-4dfc-a416-9792742da63d/02d6cce0-0544-11ed-987f-3938cfc0f7f1/file-1l5obw2pb.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5p7n1g0/160bc86a-f674-4729-9933-246027606159/2bb7d100-05c0-11ed-8617-d71e6444d1a0/file-1l5p7n1g1.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5urhu40/938cdb8a-9891-48ef-8a0d-4f13f0b7d905/bcd66b00-08cd-11ed-a060-0972f3a5152e/file-1l5urhu41.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7qace1v/32850382-0912-489c-a1fa-6994a53f61a4/f659d730-2def-11ed-a744-1fb8f3709cfa/file-1l7qace1w.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7rrgzao/8d049878-91a0-486a-bf74-ffc8968e3205/b9d7eb00-2ebf-11ed-b92e-01f1dabc9173/file-1l7rrgzap.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1leq10yl8/23860a94-17c3-4a42-9ba1-82deb3b85bcd/09ff05c0-b861-11ed-8195-4f3c56fa1eb5/file-1leq10yl9.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lef6afmm/c04cf9db-da6e-4771-935e-0ffcde0666c2/fff172f0-b268-11ed-9d4d-b96eca78f2e5/file-1lef6afmn.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lh1kl8kh/309857da-6cb3-42cd-b0c0-e753c4a2dc87/993e3110-e652-11ed-b540-cb85a096fb04/file-1lh1kl8ki.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1loconojy/39d791f1-201c-4315-8547-452eaca7958f/a485a2e0-7704-11ee-b262-010b491aebf0/file-6y3zli1loconojz.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1x38le/83e15979-b7dd-4565-bdb1-2ae72cdc0d26/13ec8220-d416-11ee-b9d5-0585032231f0/file-1lt1x38lf.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwgjwq15/c6544ab2-5387-4f8a-8398-9aeaf56b974c/efedcb90-1786-11ef-9978-292aa9baaa14/file-1lwgjwq16.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267287/exam_images/vpkjlli9hudoskpjp7ap.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266782/exam_images/fp4f81kxo2vsscgmu4ep.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266523/exam_images/toje9q4ejzuaxxxnmphc.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc8b9cup/a3a99286-f2d0-4b94-be08-ef95f78a420f/8454b500-870a-11ed-95c4-a7dd61250e50/file-1lc8b9cuq.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8r3kr/364277a2-5c20-422c-867f-2e93bd294016/10be8ab0-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8r3ks.png?format=png\")
![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264708/exam_images/uyfgmdfsahcbye8cxeuf.webp\")
\nLet the circle be $$\\left| {z - {z_3}} \\right| = r.$$ \n![\"AIEEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264288/exam_images/yur8wemdh4kwqpqnxlji.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265025/exam_images/lalecqmsx1oe6loxq0jd.webp\")
\nLet $$z = x + iy$$![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266964/exam_images/iaixfx6nsvqnz71obifn.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267682/exam_images/cpesvpbbsxejbpma67r4.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266724/exam_images/cnzpzty76azz6qp0mhza.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263484/exam_images/rzzsbpsfml9g59x3uvfs.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266723/exam_images/d8quz7o1xhunjaulq2j3.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267632/exam_images/iarlhyp1tybzbzupcjfm.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266519/exam_images/aowdrqzdcv1y1fdhfgqa.webp\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266505/exam_images/ymchjtfklcmd2dq8kelg.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5miv4n5/2aac6eff-97fb-4c96-bb71-32df8bb84dca/b7223400-0445-11ed-9d9a-21bdd0f00aa4/file-1l5miv4n6.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc8e8f52/d1800b00-4e42-4501-bb81-7122062be06f/25ad5460-8716-11ed-92bb-e57e64e1a06d/file-1lc8e8f53.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5p8atn5/a1f6e8e9-ac22-432a-959c-96b00d255605/c11c6010-05c2-11ed-b1f1-b5d4135a474f/file-1l5p8atn6.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7nbq29i/e583a584-7e22-4939-89d9-b1f38618bbd4/01c10960-2c4f-11ed-9dc0-a1792fcc650d/file-1l7nbq29j.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc8c3waq/4ba6cc0d-ceab-4ef1-baed-c4899ebff5b2/d5a07220-870d-11ed-ae18-7336d1cc7e9d/file-1lc8c3war.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc8c4zsj/031ee17e-0a26-48dd-a170-9b19fe86b156/f422a830-870d-11ed-9cef-67b5ed764fcf/file-1lc8c4zsk.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc8c6vi8/11bce565-65c8-4e70-b43c-71f126e96d89/28717710-870e-11ed-9cef-67b5ed764fcf/file-1lc8c6vi9.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsz1bfzj/da67c7ec-984a-45c9-af11-5958669a974a/42ad13f0-d280-11ee-9b77-fbceb54c8042/file-6y3zli1lsz1bfzk.png?format=png\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lt1dbkuq/8fa39b20-7b31-407e-a4e3-48710c870daf/c548b320-d3c8-11ee-a50b-bb659a2e1d74/file-1lt1dbkur.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwhf72y5/53205d83-bd1d-474f-a6fc-77a86f42ba34/4b0075d0-1801-11ef-b156-f754785ad3ce/file-1lwhf72y6.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lwekhn5c/6d7750d9-cc34-4e40-9c38-cefbaf7666fe/a4650700-166f-11ef-8416-25c08f86a011/file-1lwekhn5d.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lweku1ft/5285b6dd-899b-419d-8a11-3e28b70cd80a/fd24cb90-1670-11ef-8416-25c08f86a011/file-1lweku1fu.png?format=png\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/7qhSHs1ANpaCp0dNu/geWSekyiopJP1qh86SBj3m7vTGItc/LQ9OeUltbbmWQsskFLgKQr/image.svg\")
\n![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lc6ndrpb/ccdf08bc-7111-4568-85c5-b0eec7b3bc91/5b4a39e0-8620-11ed-91e0-bdec0331da70/file-1lc6ndrpc.png?format=png\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4501,
"subject": "General Science",
"question": "![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266283/exam_images/hlzu8arjd1x3xqvwwhpr.webp\")
![\"AIEEE](\"https://imagex.cdn.examgoal.net/9U9Fh0URP4SDcvKcK/MWBLvve80xYO19IP7fKY2TcwiUxoW/A4s0qXhuHL5WzzyZNFbBPy/image.svg\")
",
"topic": "Algebra",
"subtopic": "Intermediate Algebra"
},
{
"id": 4542,
"subject": "General Science",
"question": "The number of complex numbers z such that $$\\left| {z - 1} \\right| = \\left| {z + 1} \\right| = \\left| {z - i} \\right|$$ equals :",
"options": [
{
"text": "1 "
},
{
"text": "2 "
},
{
"text": "$$\\infty $$ "
},
{
"text": "0 "
}
],
"answer": "1 ",
"solution": "**Answer:** 1 \n\nLet $$z=x+iy$$ \n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266491/exam_images/mjmley3qs43gpsjau1nk.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264539/exam_images/tkloovrsmfpi6fjklioa.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264898/exam_images/ry4oynpllxb7fvr8cjyv.webp\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwoq4x6f/6cf72af3-199a-4296-97de-3bb71dc1f158/c6409070-534d-11ec-9cbb-695a838b20fb/file-1kwoq4x6g.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5v605na/b9bf1f33-cb41-4f86-92e8-8094f98d7ae7/7a1f0360-0906-11ed-a790-b11fa70c8a36/file-1l5v605nb.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5sd6b1c/1293a42c-142d-44c5-b0b7-b47b530256fa/2b18ad10-077c-11ed-94f6-83604aa63acb/file-1l5sd6b1d.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5scxr6i/1a1381e8-31b3-4230-9271-a7ce070fdff2/3d4976f0-077b-11ed-94f6-83604aa63acb/file-1l5scxr6n.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7bug80c/85c5d49d-7e9c-43b0-b928-429394532f1d/d5c2fcc0-25fe-11ed-b97b-f3de74335fe6/file-1l7bug80d.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7rb1odr/f03d7ef3-d4f9-4323-b266-e49dead29829/7e3573f0-2e7f-11ed-8702-156c00ced081/file-1l7rb1ods.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1lw3in2th/8eab885e-7362-426e-8f81-a5a9157aded7/ca24c350-105b-11ef-9330-f7b57aaea8e9/file-1lw3in2ti.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5p7dl91/1ede20b3-057f-48f6-b08a-8900a3a65cc1/24ed3050-05bf-11ed-8617-d71e6444d1a0/file-1l5p7dl92.png?format=png\")
![\"JEE](\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsnjdht7/4e50bf09-665e-42ec-a15b-ac37fbbf6eff/783c33b0-cc2d-11ee-b20d-39b621d226e3/file-6y3zli1lsnjdht8.png?format=png\")
![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266617/exam_images/eleslqkk6r7otnrfth9a.webp\")
\n![\"JEE](\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264874/exam_images/hg1nq2ikleda4xwios7x.webp\")
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