CBSE • Class 9 • Mathematics • Chapter 11 (Surface Areas and Volumes) • Exercise 12.2

Exercise 12.2: Surface Areas and Volumes — NCERT Solutions

Volumes of solids — cones, spheres, hemispheres and capacity conversions.

Aligned to the latest NCERT 2024-25 edition • 11 questions in this exercise • Free plan, no credit card

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What this exercise covers

Volume formulasLitres conversionMixed solids

Step-by-step solutions — Exercise 12.2

6 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 12.2 Q1 • 2 marks

Find the volume of a cone whose base radius is 66 cm and height is 1414 cm. (Use π=227\pi = \tfrac{22}{7}.)
Hint (Socratic — try this first)▾
How does the volume of a cone compare with that of a cylinder of the same base and height?
Step-by-step solution▾

Given: r=6r = 6 cm, h=14h = 14 cm.

Formula: Volume of a cone =13πr2h= \tfrac{1}{3}\pi r^2 h.

Substitute: =13×227×62×14= \frac{1}{3} \times \frac{22}{7} \times 6^2 \times 14 =13×227×36×14= \frac{1}{3} \times \frac{22}{7} \times 36 \times 14 =13×22×36×2=1584...= \frac{1}{3} \times 22 \times 36 \times 2 = \frac{1584}{...} Compute: 227×14=44\frac{22}{7}\times 14 = 44, so =13×44×36=15843=528= \frac{1}{3}\times 44 \times 36 = \frac{1584}{3} = 528 cm³.

Volume =528= 528 cm³.

Common mistake:
Forgetting the factor 13\tfrac{1}{3} and computing the cylinder volume instead.
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Exercise 12.2 Q2 • 3 marks

Find the volume of a sphere of radius 3.53.5 cm. (Use π=227\pi = \tfrac{22}{7}.)
Hint (Socratic — try this first)▾
Which power of the radius appears in the sphere's volume formula?
Step-by-step solution▾

Given: r=3.5r = 3.5 cm =72= \tfrac{7}{2} cm.

Formula: Volume of sphere =43πr3= \tfrac{4}{3}\pi r^3.

Substitute: =43×227×(72)3= \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3 =43×227×3438= \frac{4}{3} \times \frac{22}{7} \times \frac{343}{8} =43×22×498=43×10788=431224≈179.67 cm3= \frac{4}{3} \times \frac{22 \times 49}{8} = \frac{4}{3} \times \frac{1078}{8} = \frac{4312}{24} \approx 179.67 \text{ cm}^3

Volume ≈179.67\approx 179.67 cm³.

Common mistake:
Using r2r^2 instead of r3r^3, or forgetting to cube the radius fully.
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Exercise 12.2 Q3 • 3 marks

A hemispherical bowl has inner radius 10.510.5 cm. Find the volume of water it can hold. (Use π=227\pi = \tfrac{22}{7}.)
Hint (Socratic — try this first)▾
A hemisphere is what fraction of a full sphere?
Step-by-step solution▾

Given: r=10.5r = 10.5 cm =212= \tfrac{21}{2} cm.

Formula: Volume of hemisphere =23πr3= \tfrac{2}{3}\pi r^3.

Substitute: =23×227×(212)3= \frac{2}{3} \times \frac{22}{7} \times \left(\frac{21}{2}\right)^3 =23×227×92618= \frac{2}{3} \times \frac{22}{7} \times \frac{9261}{8} =23×22×13238=23×291068= \frac{2}{3} \times \frac{22 \times 1323}{8} = \frac{2}{3} \times \frac{29106}{8} =2×2910624=5821224=2425.5 cm3= \frac{2 \times 29106}{24} = \frac{58212}{24} = 2425.5 \text{ cm}^3

Volume of water =2425.5= 2425.5 cm³.

Common mistake:
Using 43πr3\tfrac{4}{3}\pi r^3 (full sphere) instead of 23πr3\tfrac{2}{3}\pi r^3 for a hemisphere.
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Exercise 12.2 Q4 • 3 marks

A conical tent has base radius 77 m and height 2424 m. Find its capacity in litres. (Use π=227\pi = \tfrac{22}{7}; 11 m³ =1000= 1000 L.)
Hint (Socratic — try this first)▾
Once you find the volume in cubic metres, how do you convert to litres?
Step-by-step solution▾

Given: r=7r = 7 m, h=24h = 24 m.

Volume =13πr2h= \tfrac{1}{3}\pi r^2 h: =13×227×72×24= \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24 =13×227×49×24= \frac{1}{3} \times \frac{22}{7} \times 49 \times 24 =13×22×7×24=13×3696=1232 m3= \frac{1}{3} \times 22 \times 7 \times 24 = \frac{1}{3} \times 3696 = 1232 \text{ m}^3

Convert: 1232×1000=1 232 0001232 \times 1000 = 1\,232\,000 litres.

Capacity =1,232,000= 1{,}232{,}000 litres.

Common mistake:
Converting incorrectly (e.g. multiplying by 100 instead of 1000) or forgetting the 13\tfrac{1}{3} factor.
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Exercise 12.2 Q5 • 3 marks

A solid metallic sphere of radius 66 cm is melted and recast into small cones each of radius 22 cm and height 33 cm. How many cones are formed?
Hint (Socratic — try this first)▾
When one solid is recast into another, which quantity stays the same?
Step-by-step solution▾

Key idea: Volume is conserved on melting and recasting.

Volume of sphere =43πr3=43π(6)3=43π×216=288π= \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi (6)^3 = \tfrac{4}{3}\pi \times 216 = 288\pi cm³.

Volume of one cone =13πr2h=13π(2)2(3)=13π×12=4π= \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi (2)^2 (3) = \tfrac{1}{3}\pi \times 12 = 4\pi cm³.

Number of cones =Volume of sphereVolume of one cone=288π4π=72= \dfrac{\text{Volume of sphere}}{\text{Volume of one cone}} = \dfrac{288\pi}{4\pi} = 72.

72 cones are formed.

Common mistake:
Comparing surface areas instead of volumes, or cancelling π\pi incorrectly.
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Exercise 12.2 Q6 • 3 marks

A cylindrical tank of radius 1.41.4 m and height 33 m is full of water. Find how many litres it holds. (Use π=227\pi = \tfrac{22}{7}.)
Hint (Socratic — try this first)▾
What is the volume formula for a cylinder, and how do cubic metres relate to litres?
Step-by-step solution▾

Given: r=1.4r = 1.4 m, h=3h = 3 m.

Volume =πr2h= \pi r^2 h: =227×(1.4)2×3= \frac{22}{7} \times (1.4)^2 \times 3 =227×1.96×3= \frac{22}{7} \times 1.96 \times 3 =227×5.88=22×0.84=18.48 m3= \frac{22}{7} \times 5.88 = 22 \times 0.84 = 18.48 \text{ m}^3

Convert: 18.48×1000=18 48018.48 \times 1000 = 18\,480 litres.

The tank holds 18,48018{,}480 litres.

Common mistake:
Squaring the diameter instead of the radius, or forgetting to convert m³ to litres.
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How to approach Exercise 12.2

  1. Re-read the chapter summary first. Open Surface Areas and Volumes and refresh the key concepts: Cuboid, Cube, Cylinder, Cone.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Surface Areas and Volumes

  1. Exercise 12.1 — Surface areas of cuboid, cube, cylinder and combinations.
  2. Exercise 12.2 — Volumes of solids — cones, spheres, hemispheres and capacity conversions.

← Exercise 12.1

Surface areas of cuboid, cube, cylinder and combinations.

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