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| source: ncert_exercises |
| topic: geometry |
| class_level: class_9 |
| chapter: ch11 |
| exercise: ex11.3 |
| difficulty: intermediate |
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| Class 9 | Ch11: Surface Areas and Volumes | Exercise 11.3 (Volume of Cone) |
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| Key Formula — Cone: Volume = (1/3)πr²h |
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| 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. ✅ |
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| 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. ✅ |
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| 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. ✅ |
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| 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. ✅ |
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| 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. ✅ |
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| Q6. Rope wound on a cylinder has volume. Find length. (Application problem.) |
| Answer: Rope=cylinder. V=πr²l (length). Solve using given values. ✅ |
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| 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. ✅ |
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| 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. ✅ |
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| 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. ✅ |
