---
title: "NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2"
url: https://www.swavid.com/maths/class/10/chapter/surface-areas-and-volumes/ncert-solutions/exercise-12-2
dateModified: 2026-10-07T15:55:53+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2

Chapter 12: Surface Areas and Volumes. Every question from Exercise 12.2, with full working and the final answer.

Free PDF (8 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-12-surface-areas-and-volumes-1cfc67625e.pdf

## EXERCISE 12.2

### Question 1

*3 marks · Short answer*

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1\text{ cm}$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.

**Solution**

1. Radius of hemisphere $r = 1\text{ cm}$, radius of cone $r = 1\text{ cm}$, and height of cone $h = 1\text{ cm}$.
2. Volume of the solid = Volume of the hemisphere + Volume of the cone = $\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h$.
3. Volume = $\frac{2}{3}\pi (1)^3 + \frac{1}{3}\pi (1)^2 (1) = \frac{2}{3}\pi + \frac{1}{3}\pi = \pi\text{ cm}^3$.

**Answer:** $\pi\text{ cm}^3$

> Common mistake: Forgetting to add the two volumes or substituting incorrect values for height and radius.

### Question 2

*3 marks · Short answer*

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is $3\text{ cm}$ and its length is $12\text{ cm}$. If each cone has a height of $2\text{ cm}$, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

**Solution**

1. Given diameter $= 3\text{ cm}$, so radius $r = 1.5\text{ cm}$, length of model $= 12\text{ cm}$, and height of each cone $= 2\text{ cm}$.
2. Height of the cylindrical part $h = 12 - 2 - 2 = 8\text{ cm}$.
3. Volume of air in the model = Volume of cylinder + $2 \times$ Volume of cone $= \pi r^2 h + 2 \times \left(\frac{1}{3} \pi r^2 h_{\text{cone}}\right) = \pi r^2 \left(h + \frac{2}{3} h_{\text{cone}}\right)$.
4. Substituting the values: $\frac{22}{7} \times (1.5)^2 \times \left(8 + \frac{2}{3} \times 2\right) = \frac{22}{7} \times 2.25 \times \left(8 + \frac{4}{3}\right) = \frac{22}{7} \times 2.25 \times \frac{28}{3} = 66\text{ cm}^3$.

**Answer:** $66\text{ cm}^3$

> Common mistake: Taking the height of the cylinder equal to the total length of the model without subtracting the heights of the two cones.

### Question 3

*3 marks · Short answer*

A gulab jamun, contains sugar syrup up to about $30\%$ of its volume. Find approximately how much syrup would be found in $45$ gulab jamuns, each shaped like a cylinder with two hemispherical ends with length $5\text{ cm}$ and diameter $2.8\text{ cm}$ (see Fig. 12.15).

**Solution**

1. Radius of each hemispherical end $r = \frac{2.8}{2} = 1.4\text{ cm}$.
2. Length of the cylindrical part $h = 5 - 2(1.4) = 5 - 2.8 = 2.2\text{ cm}$.
3. Volume of one gulab jamun = Volume of cylinder + $2 \times$ Volume of hemisphere = $\pi r^2 h + \frac{4}{3} \pi r^3 = \pi r^2 \left(h + \frac{2}{3}r\right)$.
4. Volume of one gulab jamun = $\frac{22}{7} \times 1.4 \times 1.4 \times \left(2.2 + \frac{2}{3} \times 1.4\right) = 6.16 \times \left(2.2 + \frac{2.8}{3}\right) = 6.16 \times \frac{9.4}{3} = 25.0493\text{ cm}^3$.
5. Volume of $45$ gulab jamuns = $45 \times 25.0493 = 1127.22\text{ cm}^3$.
6. Amount of sugar syrup = $30\%$ of total volume = $0.30 \times 1127.22 = 338.166\text{ cm}^3 \approx 338\text{ cm}^3$.

**Answer:** $338\text{ cm}^3$ (or $338.18\text{ cm}^3$ depending on $\pi$ approximation)

> Common mistake: Taking the height of the cylinder as the total length of $5\text{ cm}$ without subtracting the radii of the two hemispheres.

### Question 4

*3 marks · Short answer*

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are $15\text{ cm}$ by $10\text{ cm}$ by $3.5\text{ cm}$. The radius of each of the depressions is $0.5\text{ cm}$ and the depth is $1.4\text{ cm}$. Find the volume of wood in the entire stand (see Fig. 12.16).

**Solution**

1. Dimensions of cuboid are $l = 15\text{ cm}$, $b = 10\text{ cm}$, $h = 3.5\text{ cm}$. Volume of cuboid = $15 \times 10 \times 3.5 = 525\text{ cm}^3$.
2. Radius of each conical depression $r = 0.5\text{ cm}$, depth $h = 1.4\text{ cm}$. Volume of one cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4 = \frac{11}{30}\text{ cm}^3$.
3. Volume of wood = Volume of cuboid $-$ 4 $\times$ Volume of one cone = $525 - 4 \times \frac{11}{30} = 525 - 1.47 = 523.53\text{ cm}^3$.

**Answer:** $523.53\text{ cm}^3$

> Common mistake: Adding the volume of the cones instead of subtracting them from the cuboid volume.

### Question 5

*3 marks · Short answer*

A vessel is in the form of an inverted cone. Its height is $8\text{ cm}$ and the radius of its top, which is open, is $5\text{ cm}$. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius $0.5\text{ cm}$ are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

**Solution**

1. Given height of cone $h = 8\text{ cm}$ and radius $r = 5\text{ cm}$.
2. Volume of water in the conical vessel = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (5)^2 (8) = \frac{200}{3} \pi\text{ cm}^3$.
3. Volume of one spherical lead shot of radius $r_1 = 0.5\text{ cm}$ = $\frac{4}{3} \pi (0.5)^3 = \frac{4}{3} \pi \left(\frac{1}{2}\right)^3 = \frac{1}{6} \pi\text{ cm}^3$.
4. Volume of water that flows out = $\frac{1}{4} \times \text{Volume of cone} = \frac{1}{4} \times \frac{200}{3} \pi = \frac{50}{3} \pi\text{ cm}^3$.
5. Number of lead shots = $\frac{\text{Volume of water that flows out}}{\text{Volume of one lead shot}} = \frac{\frac{50}{3} \pi}{\frac{1}{6} \pi} = \frac{50}{3} \times 6 = 100$.

**Answer:** $100$

> Common mistake: Forgetting to multiply by $\frac{1}{4}$ for the water that flows out.

### Question 6

*3 marks · Short answer*

A solid iron pole consists of a cylinder of height $220\text{ cm}$ and base diameter $24\text{ cm}$, which is surmounted by another cylinder of height $60\text{ cm}$ and radius $8\text{ cm}$. Find the mass of the pole, given that $1\text{ cm}^3$ of iron has approximately $8\text{g}$ mass. (Use $\pi = 3.14$)

**Solution**

1. Given first cylinder height $h_1 = 220\text{ cm}$, radius $r_1 = \frac{24}{2} = 12\text{ cm}$.
2. Given second cylinder height $h_2 = 60\text{ cm}$, radius $r_2 = 8\text{ cm}$.
3. Volume of the first cylinder = $\pi r_1^2 h_1 = 3.14 \times (12)^2 \times 220 = 3.14 \times 144 \times 220 = 99475.2\text{ cm}^3$.
4. Volume of the second cylinder = $\pi r_2^2 h_2 = 3.14 \times (8)^2 \times 60 = 3.14 \times 64 \times 60 = 12057.6\text{ cm}^3$.
5. Total volume of the pole = $99475.2 + 12057.6 = 111532.8\text{ cm}^3$.
6. Mass of the pole = $111532.8 \times 8\text{ g} = 892262.4\text{ g} = 892.26\text{ kg}$ (or approx. $892.27\text{ kg}$ depending on $\pi$).

**Answer:** $892.26\text{ kg}$

> Common mistake: Using diameter instead of radius for the first cylinder.

### Question 7

*3 marks · Short answer*

A solid consisting of a right circular cone of height $120\text{ cm}$ and radius $60\text{ cm}$ standing on a hemisphere of radius $60\text{ cm}$ is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is $60\text{ cm}$ and its height is $180\text{ cm}$.

**Solution**

1. Given cone height $h = 120\text{ cm}$, radius $r = 60\text{ cm}$, and hemisphere radius $r = 60\text{ cm}$.
2. Cylinder radius $R = 60\text{ cm}$ and height $H = 180\text{ cm}$.
3. Volume of the solid = Volume of cone + Volume of hemisphere = $\frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3 = \frac{1}{3} \pi (60)^2 (120) + \frac{2}{3} \pi (60)^3$.
4. Volume of solid = $\frac{1}{3} \pi (3600)(120) + \frac{2}{3} \pi (216000) = 144000\pi + 144000\pi = 288000\pi\text{ cm}^3$.
5. Volume of the cylinder = $\pi R^2 H = \pi (60)^2 (180) = 648000\pi\text{ cm}^3$.
6. Volume of water left = Volume of cylinder $-$ Volume of solid = $648000\pi - 288000\pi = 360000\pi\text{ cm}^3$.
7. Volume of water left = $360000 \times \frac{22}{7} = \frac{7920000}{7}\text{ cm}^3 = 1131428.57\text{ cm}^3 = 1.131\text{ m}^3$.

**Answer:** $1.13\text{ m}^3$ (or $1131428.57\text{ cm}^3$)

> Common mistake: Subtracting the height of the cone and hemisphere incorrectly from the cylinder height.

### Question 8

*3 marks · Short answer*

A spherical glass vessel has a cylindrical neck $8\text{ cm}$ long, $2\text{ cm}$ in diameter; the diameter of the spherical part is $8.5\text{ cm}$. By measuring the amount of water it holds, a child finds its volume to be $345\text{ cm}^3$. Check whether she is correct, taking the above as the inside measurements, and $\pi = 3.14$.

**Solution**

1. Given cylindrical neck length $h = 8\text{ cm}$, diameter $d = 2\text{ cm}$ (radius $r = 1\text{ cm}$).
2. Diameter of spherical part $d_1 = 8.5\text{ cm}$ (radius $r_1 = 4.25\text{ cm}$).
3. Volume of the cylindrical neck = $\pi r^2 h = 3.14 \times (1)^2 \times 8 = 25.12\text{ cm}^3$.
4. Volume of the spherical part = $\frac{4}{3} \pi r_1^3 = \frac{4}{3} \times 3.14 \times (4.25)^3 = \frac{4}{3} \times 3.14 \times 76.765625 = 321.39\text{ cm}^3$.
5. Total volume of the vessel = $25.12 + 321.39 = 346.51\text{ cm}^3$.
6. Since $346.51\text{ cm}^3 \neq 345\text{ cm}^3$, the child is incorrect.

**Answer:** She is incorrect; the correct volume is $346.51\text{ cm}^3$.

> Common mistake: Using radius as 8.5 cm instead of 4.25 cm for the spherical part.

## Related pages

- [All Chapter 12 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/surface-areas-and-volumes/ncert-solutions)
- [Exercise 12.1](https://www.swavid.com/maths/class/10/chapter/surface-areas-and-volumes/ncert-solutions/exercise-12-1)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student