Skip to main content
NCERT Solutions · Class 10 · Chapter 12: Surface Areas and Volumes

NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2

Exercise 12.2: Volume of combined solids. Volumes add (or subtract, for a hollow): cylinder \(\pi r^2 h\), cone \(\tfrac13\pi r^2 h\), sphere \(\tfrac43\pi r^3\), hemisphere \(\tfrac23\pi r^3\). Keep units consistent. Unless the question gives another value, take \(\pi = \tfrac{22}{7}\), as the NCERT exercises do (no calculator: CBSE is a no-calculator exam).

  • 8 questions
  • Every answer checked by computer algebra
  • Free, no sign-in

Try each question first, then open its solution. Our own step-by-step solutions, set out for step marks.

Exercise 12.2 questions and solutions

Exercise 12.2, Question 1

A solid is a cone on a hemisphere, both of radius 1 cm, with the cone's height equal to its radius. Find its volume in terms of \(\pi\).
Show solution
  1. \[\tfrac13\pi(1)^2(1) + \tfrac23\pi(1)^3 = \tfrac13\pi + \tfrac23\pi\]
Answer: \(\pi\ \text{cm}^3\)

Practise this: Step 1, Secure the basics →

Exercise 12.2, Question 2

A model is a cylinder with a cone at each end; diameter 3 cm, total length 12 cm, each cone 2 cm high. Find the volume of air inside.
Show solution
  1. \(r = 1.5\); cylinder length \(12 - 4 = 8\).
  2. \[\begin{aligned}\pi r^2(8) + 2 \times \tfrac13\pi r^2(2) &= 18\pi + 3\pi \\ &= 21\pi\end{aligned}\]
Answer: \(66\ \text{cm}^3\)

Practise this: Step 2, Board standard →

Exercise 12.2, Question 3

Each of 45 gulab jamuns is a cylinder with hemispherical ends, 5 cm long and 2.8 cm across, and holds syrup up to 30% of its volume. Find the total syrup, approximately.
Show solution
  1. \(r = 1.4\), cylinder length \(5 - 2.8 = 2.2\).
  2. One: \[\pi(1.96)(2.2) + \tfrac43\pi(1.4)^3 \approx 13.55 + 11.50 = 25.05\ \text{cm}^3\]
  3. Syrup \(= 0.3 \times 45 \times 25.05 \approx 338\).
Answer: About \(338\ \text{cm}^3\)

Practise this: Step 3, Full marks on long answers →

Exercise 12.2, Question 4

A pen stand is a cuboid \(15 \times 10 \times 3.5\) cm with four conical holes, each of radius 0.5 cm and depth 1.4 cm. Find the volume of wood.
Show solution
  1. Cuboid \(= 525\). Four cones \[= 4 \times \tfrac13 \times \dfrac{22}{7} \times 0.25 \times 1.4 = \dfrac{22}{15} \approx 1.47\]
  2. Wood \(= 525 - 1.47\).
Answer: About \(523.53\ \text{cm}^3\)

Practise this: Step 2, Board standard →

Exercise 12.2, Question 5

An inverted cone (height 8 cm, top radius 5 cm) is full of water. Lead shots of radius 0.5 cm are dropped in until one quarter of the water flows out. How many shots?
Show solution
  1. Cone \(= \tfrac13\pi(25)(8) = \dfrac{200\pi}{3}\); a quarter is \(\dfrac{50\pi}{3}\).
  2. One shot \(= \tfrac43\pi(0.5)^3 = \dfrac{\pi}{6}\). Number \(= \dfrac{50\pi/3}{\pi/6} = 100\).
Answer: 100

Practise this: Step 3, Full marks on long answers →

Exercise 12.2, Question 6

An iron pole is a cylinder (height 220 cm, diameter 24 cm) topped by a cylinder (height 60 cm, radius 8 cm). Find its mass if 1 cm³ of iron is about 8 g (\(\pi = 3.14\)).
Show solution
  1. Volume \[\begin{aligned}= 3.14(144 \times 220 + 64 \times 60) &= 3.14 \times 35520 \\ &= 111532.8\ \text{cm}^3\end{aligned}\]
  2. Mass \(= 8 \times 111532.8 = 892262.4\) g.
Answer: About \(892.26\) kg

Practise this: Step 3, Full marks on long answers →

Exercise 12.2, Question 7

A solid (a cone of height 120 cm and radius 60 cm on a hemisphere of radius 60 cm) stands in a cylinder of radius 60 cm and height 180 cm full of water, touching the bottom. Find the water left.
Show solution
  1. Cylinder \(= \pi(3600)(180) = 648000\pi\). Cone \(= \tfrac13\pi(3600)(120) = 144000\pi\). Hemisphere \(= \tfrac23\pi(216000) = 144000\pi\).
  2. Water left \[= 360000\pi = 360000 \times \dfrac{22}{7}\]
Answer: About \(1131428.57\ \text{cm}^3\) (1.131 m³)

Practise this: Step 4, 95+ stretch (HOTS) →

Exercise 12.2, Question 8

A spherical glass vessel (diameter 8.5 cm) has a cylindrical neck 8 cm long and 2 cm across. A child says it holds 345 cm³. Is she right? (Inside measurements, \(\pi = 3.14\).)
Show solution
  1. Sphere \[= \tfrac43 \times 3.14 \times 4.25^3 \approx 321.39\]; neck \(= 3.14 \times 1 \times 8 = 25.12\).
  2. Total \(\approx 346.51\ \text{cm}^3\).
Answer: No: it holds about \(346.51\ \text{cm}^3\)

Practise this: Step 3, Full marks on long answers →

Done the NCERT exercises? The board paper asks more

Surface Areas and Volumes has 37 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.

Surface Areas and Volumes: notes and questionsCreate a free accountTeachers: set it as homework

Surface Areas and Volumes in our sample papers: Sample paper 1 (questions 14, 31) · Sample paper 2 (questions 15, 16, 34) · Sample paper 3 (questions 15, 20, 35) · Sample paper 4 (questions 15, 16, 25, 36) · Sample paper 5 (questions 15, 16, 34).

Also useful: free MCQs and case studies for Surface Areas and Volumes · Class 10 formula sheet · official CBSE board and sample papers · our sample papers with marking scheme · the Route to 95 plan

Textbook: NCERT Mathematics Class 10 (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.