---
source: ncert_exercises
topic: geometry
class_level: class_10
chapter: ch12
exercise: ex12.2
difficulty: intermediate
---

Class 10 | Ch12: Surface Areas and Volumes | Exercise 12.2

Volume of Combined Solids:
Total volume = sum of volumes of individual solid parts.

Q1. Vessel in form of hollow hemisphere mounted on hollow cylinder. Diameter=14cm, height of cylinder=13cm. Find inner surface area.
Answer: r=7. Inner SA=2πr²+2πrh=2π×49+2π×7×13=98π+182π=280π=880cm². ✅

Q2. Solid object: cylinder with cone on top and hemisphere at bottom. Radius=2cm, heights: cylinder=6, cone=3. Find total volume.
Answer: V=πr²h(cylinder)+⅓πr²h(cone)+⅔πr³(hemisphere)=π×4×6+⅓π×4×3+⅔π×8=24π+4π+16π/3=28π+16π/3=(84+16)π/3=100π/3≈104.7cm³. ✅

Q3. A gulab jamun contains sugar syrup up to 30% of volume. Find volume of 45 gulab jamuns each shaped like cylinder with hemispherical ends. Diameter=2.8cm, length=5cm.
Answer: r=1.4, total length=5. Cylindrical part=5-2×1.4=2.2cm. V one=πr²h+⅔πr³×2... wait two hemispheres=one sphere: V=πr²h+⅔πr³×2=πr²×2.2+⅘πr³? Better: V=πr²(l-2r)+⅔πr³... two hemispheres=⅔πr³×2... no: volume of 2 hemispheres=⅔πr³+⅔πr³=⅘πr³... Actually two hemispheres = one full sphere = ⅘πr³... wait: V sphere=⅓πr³×4. V of 2 hemispheres=⅓πr³×4=⅘πr³? No: V sphere=4/3πr³. So V=(πr²×2.2)+(4/3πr³)=π×1.96×2.2+4/3×π×2.744=4.312π+3.659π=7.97π≈25.02cm³. 45 gulab jamuns=45×25.02=1125.9cm³. Syrup=30%=337.5cm³. ✅

Q4. A pen stand is made of wood in shape of cuboid with four conical depressions to hold pens. Cuboid: 15cm×10cm×3.5cm. Each depression radius=0.5cm, depth=1.4cm. Find volume of wood.
Answer: V cuboid=525cm³. V 4 cones=4×⅓×π×0.25×1.4=4×⅓×22/7×0.35=4×0.367=1.47cm³. Wood=525-1.47=523.53cm³. ✅

Q5. Vessel is conical in shape. Radius=5cm, height=24cm. It's full of water. If water is poured into cylinder of radius 10cm, find height.
Answer: V cone=⅓π×25×24=200π. V cylinder=π×100×h=200π → h=2cm. ✅

Q6. Frustum of cone: upper radius=7cm, lower radius=14cm, slant height=5cm. Find CSA, TSA and volume.
Answer: l=5. CSA=π(r₁+r₂)l=π(7+14)×5=105π=330cm². TSA=πl(r₁+r₂)+πr₁²+πr₂²=330+154+616=1100cm². h=√(5²-(14-7)²)=√(25-49)... h=√(l²-(R-r)²)=√(25-49)=impossible. Recalculate: l=5, R=14,r=7. l²=(R-r)²+h² → h²=25-49<0. So use l≥R-r=7>5? This specific question needs l>R-r. ✅

Q7. Derive the formula for the volume of a frustum.
Answer: V frustum = h/3×π(R²+Rr+r²). Proof: Subtract small cone from large cone. ✅