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NCERT Solutions · Class 10 Maths · Chapter 12

Chapter 12: Surface Areas and Volumes (Mensuration)

Step-by-step NCERT solutions for Class 10 Maths Chapter 12, Surface Areas and Volumes (2026-27 reprint): Exercise 12.1 surface areas of combined solids (vessel, toy, capsule, tent, scooped cylinder) and Exercise 12.2 volumes of combinations (gulab jamuns, pen stand, lead shots, iron pole, glass vessel). All 17 questions are answered, with the key answer highlighted.

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-10-maths/chapter-12-surface-areas-and-volumes

Formulas ( $r$ radius, $h$ height, $l$ slant height): cylinder CSA $2 π r h$ , volume $π r^{2} h$ ; cone CSA $π r l$ with $l = r^{2} + h^{2}$ , volume $\frac{1}{3} π r^{2} h$ ; sphere surface $4 π r^{2}$ , volume $\frac{4}{3} π r^{3}$ ; hemisphere CSA $2 π r^{2}$ , volume $\frac{2}{3} π r^{3}$ . For combined solids, add only the surfaces that remain visible; volumes simply add (or subtract for cavities). Take $π = \frac{22}{7}$ unless stated otherwise.

Exercise 12.1

Two cubes, each of volume 64 cm³, are joined end to end. Find the surface area of the resulting cuboid.

Solution

Each cube has side $364 = 4$ cm, so the cuboid is $8 \times 4 \times 4$ cm.

$S A = 2 (l b + bh + h l) = 2 (32 + 16 + 32) = 160$

160 cm²

A vessel is a hollow hemisphere with a hollow cylinder on it. The hemisphere's diameter is 14 cm and the total height is 13 cm. Find the inner surface area.

Solution

$r = 7$ cm; the cylinder's height is $13 - 7 = 6$ cm.

Inner SA $= 2 π r^{2} + 2 π r h = 2 \times \frac{22}{7} \times 49 + 2 \times \frac{22}{7} \times 7 \times 6 = 308 + 264 = 572$

572 cm²

A toy is a cone of radius 3.5 cm on a hemisphere of the same radius, with total height 15.5 cm. Find its total surface area.

Solution

Cone height $h = 15.5 - 3.5 = 12$ cm, so $l = 1 2^{2} + 3. 5^{2} = 156.25 = 12.5$ cm.

TSA $= π r l + 2 π r^{2} = \frac{22}{7} \times 3.5 \times 12.5 + 2 \times \frac{22}{7} \times 12.25 = 137.5 + 77 = 214.5$

214.5 cm²

A cube of side 7 cm is surmounted by a hemisphere. What is the greatest diameter of the hemisphere? Find the surface area of the solid.

Solution

The hemisphere must sit on a face of the cube, so its greatest diameter is 7 cm ( $r = 3.5$ ).

SA $= 6 a^{2} - π r^{2} + 2 π r^{2} = 294 + π r^{2} = 294 + \frac{22}{7} \times 12.25 = 294 + 38.5 = 332.5$

Greatest diameter 7 cm; surface area 332.5 cm².

A hemispherical depression is cut from one face of a cube, with the hemisphere's diameter l equal to the cube's edge. Find the surface area of the remaining solid.

Solution

Radius $= \frac{l}{2}$ . Remove the circle of the opening and add the inside of the hemisphere:

$S A = 6 l^{2} - π (\frac{l}{2})^{2} + 2 π (\frac{l}{2})^{2} = 6 l^{2} + \frac{π l ^{2}}{4} = \frac{l ^{2}}{4} (π + 24)$

(l²/4)(π + 24) square units

A capsule is a cylinder with two hemispheres at its ends. The capsule is 14 mm long and 5 mm in diameter. Find its surface area.

Solution

$r = 2.5$ mm; the cylinder part is $14 - 5 = 9$ mm long.

SA $= 4 π r^{2} + 2 π r h = 2 π r (2 r + h) = 2 \times \frac{22}{7} \times 2.5 \times 14 = 220$

220 mm²

A tent is a cylinder (height 2.1 m, diameter 4 m) with a conical top of slant height 2.8 m. Find the canvas area and its cost at ₹500 per m² (the base is not covered).

Solution

$r = 2$ m. Canvas $= 2 π r h + π r l = π r (2 h + l) = \frac{22}{7} \times 2 \times (4.2 + 2.8) = 44$ m².

Cost $= 44 \times 500 =$ ₹22,000

44 m²; ₹22,000

A conical cavity of the same height (2.4 cm) and diameter (1.4 cm) is hollowed out of a solid cylinder. Find the total surface area of the remaining solid to the nearest cm².

Solution

$r = 0.7$ , $h = 2.4$ , so $l = 0.49 + 5.76 = 2.5$ cm.

TSA $=$ curved surface of the cylinder $+$ its top $+$ the inside of the cone $= 2 π r h + π r^{2} + π r l = π r (2 h + r + l)$

$= \frac{22}{7} \times 0.7 \times (4.8 + 0.7 + 2.5) = 2.2 \times 8 = 17.6$

17.6 cm², i.e. about 18 cm²

A hemisphere is scooped out of each end of a solid cylinder of height 10 cm and radius 3.5 cm. Find the total surface area.

Solution

TSA $= 2 π r h + 2 \times 2 π r^{2} = 2 \times \frac{22}{7} \times 3.5 \times 10 + 4 \times \frac{22}{7} \times 12.25 = 220 + 154 = 374$

374 cm²

Exercise 12.2

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 π.

Solution

$V = \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = \frac{1}{3} π + \frac{2}{3} π = π$

π cm³

A model is a cylinder with a cone at each end. Its diameter is 3 cm, its length 12 cm and each cone is 2 cm high. Find the volume of air inside.

Solution

$r = 1.5$ cm; the cylinder is $12 - 4 = 8$ cm long.

$V = π r^{2} \times 8 + 2 \times \frac{1}{3} π r^{2} \times 2 = π r^{2} (8 + \frac{4}{3}) = \frac{22}{7} \times 2.25 \times \frac{28}{3} = 66$

66 cm³

A gulab jamun (a cylinder with two hemispherical ends, length 5 cm, diameter 2.8 cm) contains syrup up to about 30% of its volume. How much syrup is in 45 of them?

Solution

$r = 1.4$ cm; the cylinder part is $5 - 2.8 = 2.2$ cm long.

One gulab jamun $= π r^{2} (2.2) + \frac{4}{3} π r^{3} = \frac{22}{7} \times 1.96 \times (2.2 + \frac{4 \times 1.4}{3}) = 6.16 \times 4.0 \overline{6} \approx 25.05$ cm³

45 gulab jamuns $\approx 1127.28$ cm³; syrup $\approx 30%$ of this $\approx 338$ cm³.

About 338 cm³

A pen stand is a 15 × 10 × 3.5 cm cuboid with four conical depressions, each of radius 0.5 cm and depth 1.4 cm. Find the volume of wood.

Solution

Cuboid $= 15 \times 10 \times 3.5 = 525$ cm³
Four cones $= 4 \times \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 1.4 = 1.47$ cm³

Wood $= 525 - 1.47 = 523.53$ cm³

523.53 cm³

An inverted cone of height 8 cm and top radius 5 cm is full of water. Lead shots (spheres of radius 0.5 cm) are dropped in until one-fourth of the water flows out. Find the number of shots.

Solution

Cone $= \frac{1}{3} \times \frac{22}{7} \times 25 \times 8 = \frac{4400}{21}$ cm³; one-fourth $= \frac{1100}{21}$ cm³
One shot $= \frac{4}{3} \times \frac{22}{7} \times \frac{1}{8} = \frac{11}{21}$ cm³

Number $= \frac{1100/21}{11/21} = 100$

100 lead shots

An iron pole is a cylinder of height 220 cm and base diameter 24 cm, with a cylinder of height 60 cm and radius 8 cm on top. Find its mass if 1 cm³ of iron is about 8 g (π = 3.14).

Solution

$V = 3.14 (1 2^{2} \times 220 + 8^{2} \times 60) = 3.14 \times (31680 + 3840) = 3.14 \times 35520 = 111532.8 cm^{3}$

Mass $= 111532.8 \times 8 = 892262.4$ g

≈ 892.26 kg

A solid cone (height 120 cm, radius 60 cm) on a hemisphere of radius 60 cm is placed upright in a cylinder of radius 60 cm and height 180 cm full of water, touching the bottom. Find the volume of water left.

Solution

Cylinder $= π \times 3600 \times 180 = 648000 π$
Cone $= \frac{1}{3} π \times 3600 \times 120 = 144000 π$
Hemisphere $= \frac{2}{3} π \times 216000 = 144000 π$

Water left $= 648000 π - 288000 π = 360000 π$ cm³ $= 0.36 π$ m³ $\approx 1.131$ m³

About 1.131 m³

A spherical glass vessel (inside diameter 8.5 cm) has a cylindrical neck 8 cm long and 2 cm in diameter. A child finds its volume to be 345 cm³. Is she correct? (π = 3.14)

Solution

Sphere $= \frac{4}{3} \times 3.14 \times 4.2 5^{3} = 321.39$ cm³
Neck $= 3.14 \times 1^{2} \times 8 = 25.12$ cm³

Total $= 346.51$ cm³

Not quite: the volume is 346.51 cm³, not 345 cm³.

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