CBSE Class 10 · Chapter 12 · Mensuration · 2026-27

Surface Areas and Volumes Class 10: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceSingle solids (review for combinations)

The volume of a hemisphere of radius $3$ cm is

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Answer: (a)

18 π {cm}^{3}

Why: V = (2/3)πr³ = (2/3)π × 27.

$\frac{2}{3} π (3)^{3} = \frac{2}{3} \times 27 π = 18 π {cm}^{3}$ .

Q2

·1 mark·Multiple choiceSingle solids (review for combinations)

The curved surface area of a cylinder of radius $7$ cm and height $10$ cm is (Take $π = \frac{22}{7}$ .)

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Answer: (b)

440 {cm}^{2}

Why: CSA = 2πrh = 2 × (22/7) × 7 × 10.

$2 π r h = 2 \times \frac{22}{7} \times 7 \times 10 = 440 {cm}^{2}$ .

Q3

·1 mark·Multiple choiceSurface area of combinations of solids

A solid is made of a cylinder of radius $r$ and height $h$ with a hemisphere of radius $r$ fixed on its top. Its total surface area is

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Answer: (c)

2 π r h + 3 π r^{2}

Why: Cylinder's curved surface + its bottom disc + the hemisphere's curved surface.

Exposed parts: curved surface of the cylinder $2 π r h$ , its base $π r^{2}$ and the curved surface of the hemisphere $2 π r^{2}$ . Total $= 2 π r h + 3 π r^{2}$ . (The joined circle is not on the outside.)

Q4

·1 mark·Multiple choiceVolume of combinations of solids

A toy is a cone of radius $3$ cm and height $4$ cm mounted on a hemisphere of the same radius. The volume of the toy is

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Answer: (a)

30 π {cm}^{3}

Why: Cone (1/3)π × 9 × 4 = 12π plus hemisphere (2/3)π × 27 = 18π.

$\frac{1}{3} π (3)^{2} (4) + \frac{2}{3} π (3)^{3} = 12 π + 18 π = 30 π {cm}^{3}$ .

Q5

·1 mark·Multiple choiceSingle solids (review for combinations)

A cone has base radius $7$ cm and slant height $25$ cm. Its vertical height is

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Answer: (b)

24

Why: h = √(l² − r²) = √(625 − 49).

$h = \sqrt{25^{2} - 7^{2}} = \sqrt{576} = 24$ cm.

Q6

·1 mark·Multiple choiceSurface area of combinations of solids

A solid cuboid of size $8$ cm $\times$ $4$ cm $\times$ $4$ cm is cut into two equal cubes. The increase in the total surface area is

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Answer: (b)

32 {cm}^{2}

Why: The cut creates two new 4 cm × 4 cm faces.

Cuboid: $2 (32 + 16 + 32) = 160 {cm}^{2}$ . Two cubes: $2 \times 6 \times 16 = 192 {cm}^{2}$ . Increase $= 32 {cm}^{2}$ (two new faces of $16 {cm}^{2}$ ).

Q7

·1 mark·Multiple choiceSingle solids (review for combinations)

The surface area of a sphere of radius $7$ cm is (Take $π = \frac{22}{7}$ .)

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Answer: (d)

616 {cm}^{2}

Why: S = 4πr² = 4 × (22/7) × 49.

$4 \times \frac{22}{7} \times 49 = 616 {cm}^{2}$ .

Q8

·1 mark·Multiple choiceSurface area of combinations of solids

A solid is a cylinder of radius $7$ cm with a cone of the same radius and height $24$ cm on top. The curved surface area of the conical part is (Take $π = \frac{22}{7}$ .)

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Answer: (a)

550 {cm}^{2}

Why: Slant height √(49 + 576) = 25; CSA = πrl.

$l = \sqrt{7^{2} + 24^{2}} = 25$ cm. CSA $= π r l = \frac{22}{7} \times 7 \times 25 = 550 {cm}^{2}$ .

Q9

·1 mark·Multiple choiceVolume of combinations of solids

A hemisphere and a cone have the same radius $r$ and the same volume. The height of the cone is

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Answer: (b)

2 r

Why: (2/3)πr³ = (1/3)πr²h gives h = 2r.

$\frac{2}{3} π r^{3} = \frac{1}{3} π r^{2} h \Rightarrow h = 2 r$ .

Q10

·1 mark·Multiple choiceVolume of combinations of solids

The largest sphere is carved out of a solid cube of edge $14$ cm. The volume of the sphere is (Take $π = \frac{22}{7}$ .)

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Answer: (c)

\frac{4312}{3} {cm}^{3}

Why: The sphere's diameter equals the edge, so r = 7.

$r = 7$ cm. $V = \frac{4}{3} \times \frac{22}{7} \times 343 = \frac{4312}{3} {cm}^{3}$ .

Q11

·1 mark·Multiple choiceVolume of combinations of solids

A solid cylinder of radius $3$ cm and height $10$ cm has a hemispherical hollow of radius $3$ cm scooped out of one end. The volume of the remaining solid is

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Answer: (c)

72 π {cm}^{3}

Why: Cylinder 90π minus hemisphere 18π.

$π (3)^{2} (10) - \frac{2}{3} π (3)^{3} = 90 π - 18 π = 72 π {cm}^{3}$ .

Q12

·1 mark·Multiple choiceSurface area of combinations of solids

A tent is a cylinder of radius $7$ m and height $3$ m with a conical top of the same radius and slant height $10$ m. The area of canvas needed (no floor) is (Take $π = \frac{22}{7}$ .)

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Answer: (d)

352 m^{2}

Why: Canvas = 2πrh + πrl = (22/7) × 7 × (6 + 10).

$2 π r h + π r l = \frac{22}{7} \times 7 \times (2 \times 3 + 10) = 22 \times 16 = 352 m^{2}$ .

Q13

·1 mark·Multiple choiceVolume of combinations of solids

A solid cylinder of radius $2$ cm and height $6$ cm has a cone of the same radius fixed on top. The volume of the whole solid is $28 π {cm}^{3}$ . The height of the cone is

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Answer: (d)

3

Why: Cylinder 24π, so the cone is 4π = (1/3)π × 4 × h.

Cylinder $= π (2)^{2} (6) = 24 π$ . Cone $= 28 π - 24 π = 4 π = \frac{1}{3} π (2)^{2} h \Rightarrow h = 3$ cm.

Q14

·1 mark·Multiple choiceSingle solids (review for combinations)

If the radius of a sphere is doubled, its surface area becomes

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Answer: (b)

4

times

Why: Surface area is proportional to r².

$\frac{4 π (2 r)^{2}}{4 π r^{2}} = 4$ : the surface area becomes $4$ times.

Q15

·1 mark·Multiple choiceVolume of combinations of solids

A solid is a cone of radius $r$ and height $h$ standing on a cylinder of radius $r$ and height $h$ . Its volume is

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Answer: (c)

\frac{4}{3} π r^{2} h

Why: πr²h + (1/3)πr²h.

$π r^{2} h + \frac{1}{3} π r^{2} h = \frac{4}{3} π r^{2} h$ .

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 10 case study questions by chapter.

Case study 1: Rocket model (4 marks)

For the school science fair, Zoya builds a solid model rocket: a cylinder of radius $3$ cm and height $12$ cm with a cone of the same radius and height $4$ cm fixed on top. She will paint the whole outside, including the flat circular base. Leave answers in terms of $π$ .

(i) Find the slant height of the cone. [1 mark]

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Answer:

5

l = \sqrt{3^{2} + 4^{2}} = 5

cm. A1

(ii) Find the curved surface area of the cone. [1 mark]

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Answer:

15 π {cm}^{2}

π r l = π \times 3 \times 5 = 15 π {cm}^{2}

. A1

(iii) Find the total area Zoya has to paint. [2 marks]

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Answer:

96 π {cm}^{2}

2 π (3) (12) + 15 π + π (3)^{2}

= 72 π + 15 π + 9 π = 96 π {cm}^{2}

. A1

OR Find the volume of the model rocket. [2 marks]

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Answer:

120 π {cm}^{3}

π (3)^{2} (12) + \frac{1}{3} π (3)^{2} (4)

= 108 π + 12 π = 120 π {cm}^{3}

. A1

Case study 2: Spinning top (4 marks)

A craftsman in Channapatna turns a wooden spinning top. It is a hemisphere of radius $7$ cm with a cone of the same radius and height $24$ cm on its flat face, so that the cone points upwards. (Take $π = \frac{22}{7}$ .)

(i) Find the slant height of the cone. [1 mark]

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Answer:

25

l = \sqrt{7^{2} + 24^{2}} = \sqrt{625} = 25

cm. A1

(ii) Find the curved surface area of the hemisphere. [1 mark]

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Answer:

308 {cm}^{2}

2 π r^{2} = 2 \times \frac{22}{7} \times 49 = 308 {cm}^{2}

. A1

(iii) Find the total surface area of the top that is lacquered (the whole outside). [2 marks]

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Answer:

858 {cm}^{2}

Cone:

π r l = \frac{22}{7} \times 7 \times 25 = 550

M1; total

= 308 + 550 = 858 {cm}^{2}

. A1

OR Find the volume of wood in the top. [2 marks]

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Answer:

\frac{5852}{3} {cm}^{3}

Hemisphere

\frac{2}{3} \times \frac{22}{7} \times 343 = \frac{2156}{3}

; cone

\frac{1}{3} \times \frac{22}{7} \times 49 \times 24 = 1232

M1; total

= \frac{2156 + 3696}{3} = \frac{5852}{3} {cm}^{3}

. A1

Next steps for Surface Areas and Volumes

This free set is separate from the chapter's question bank. On the Surface Areas and Volumes chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 46 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Surface Areas and Volumes: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.