source: ncert_exercises
topic: geometry
class_level: class_10
chapter: ch12
exercise: ex12.3
difficulty: intermediate
Class 10 | Ch12: Surface Areas and Volumes | Exercise 12.3
Frustum of Cone: CSA = πl(R+r), where l = slant height = √[h²+(R-r)²] TSA = πl(R+r)+πR²+πr² Volume = πh/3 × (R²+Rr+r²)
Q1. Small sphere of radius r₁=3cm is placed inside larger sphere of radius r₂=4cm. Find surface area of glass sphere (both together). Answer: Not standard frustum — this is concentric spheres. Outer sphere SA=4π×16=64π≈201cm². ✅
Q2. Metallic sphere of radius 4.2cm is melted and recast into small spheres of radius 0.6cm. Find number. Answer: n=R³/r³=(4.2)³/(0.6)³=(4.2/0.6)³=7³=343. ✅
Q3. Container in shape of frustum of cone. Bottom radius=8cm, top radius=20cm, depth=16cm. Find capacity in litres. Answer: V=πh/3(R²+Rr+r²)=π×16/3(400+160+64)=π×16/3×624=16×624π/3=3328π/3≈3482cm³≈3.48 litres. ✅
Q4. 21 cylindrical pillars of building, each 3.5m diameter and 10m high. Find cost of painting at ₹25.50/m². Answer: CSA each=2π×1.75×10=35π=110m². 21 pillars=2310m². Cost=2310×25.50=₹58905. ✅
Q5. Cone of radius 10cm divided into two parts at midpoint of height. Find ratio of volumes. Answer: Small cone (top): radius=5cm (by similar triangles), height=h/2. V_small=⅓π×25×h/2. V_large cone=⅓π×100×h. V_frustum=V_large-V_small=⅓πh(100-25/2)=⅓πh×175/2. Ratio=V_small:V_frustum=(25h/2):(175/2h/... Actually ratio V_small/V_total=(r/R)³=(5/10)³=1/8. V_frustum=7/8×V_total. Ratio 1:7. ✅
Q6. Frustum: top radius=28m, bottom radius=7m, slant height=4m. Find CSA. Answer: CSA=πl(R+r)=22/7×4×(28+7)=22/7×4×35=22×20=440m². ✅