---
source: ncert_exercises
topic: geometry
class_level: class_9
chapter: ch11
exercise: ex11.3
difficulty: intermediate
---

Class 9 | Ch11: Surface Areas and Volumes | Exercise 11.3 (Volume of Cone)

Key Formula — Cone: Volume = (1/3)πr²h

Q1. Find volume of right circular cone with: (i) r=6 cm, h=7 cm (ii) r=3.5 cm, h=12 cm.
Answer: V=(1/3)πr²h.
(i) (1/3)×22/7×36×7=264 cm³. ✅
(ii) (1/3)×22/7×12.25×12=154 cm³. ✅

Q2. Find capacity of conical vessel: (i) r=7 cm, slant height=25 cm (ii) h=12 cm, slant height=13 cm.
Answer:
(i) h=√(l²-r²)=√(625-49)=24 cm. V=(1/3)×22/7×49×24=1232 cm³=1.232 L. ✅
(ii) r=√(l²-h²)=√(169-144)=5 cm. V=(1/3)×22/7×25×12=314.28 cm³. ✅

Q3. Height of cone=15 cm, volume=1570 cm³. Find radius.
Answer: V=(1/3)πr²h → 1570=(1/3)×3.14×r²×15 → r²=1570/(3.14×5)=100 → r=10 cm. ✅

Q4. Conical pit: top diameter=3.5 m, depth=12 m. Find capacity in kilolitres.
Answer: r=1.75 m. V=(1/3)×22/7×1.75²×12=(1/3)×22/7×3.0625×12=38.5 m³=38.5 kL. ✅

Q5. Volume of right circular cone=9856 cm³, diameter=28 cm. Find slant height.
Answer: r=14 cm. h=9856×3/(π×196)=48 cm. l=√(196+2304)=50 cm. ✅

Q6. Rope wound on a cylinder has volume. Find length. (Application problem.)
Answer: Rope=cylinder. V=πr²l (length). Solve using given values. ✅

Q7. Conical vessel can hold 21 litres. If radius=7 cm, find height.
Answer: V=21000 cm³. (1/3)×22/7×49×h=21000 → h=21000×3×7/(22×49)=90.9 cm≈91 cm. ✅

Q8. Right triangle with sides 5,12,13 cm revolved about 12 cm side. Find volume.
Answer: r=5, h=12. V=(1/3)×π×25×12=100π≈314.28 cm³. ✅
If revolved about 5 cm: r=12, h=5. V=(1/3)×π×144×5=240π≈753.6 cm³.
Ratio=100π:240π=5:12. ✅

Q9. Heap of wheat: cone, diameter=10.5 m, height=3 m. Find volume and area of canvas.
Answer: r=5.25, h=3. V=(1/3)×22/7×27.5625×3=86.625 m³.
l=√(5.25²+3²)=√(27.5625+9)=√36.5625=6.05 m.
Canvas area=πrl=22/7×5.25×6.05=99.825 m². ✅