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Surface Areas and Volumes: 3 marks Questions (CBSE Class 10)

17 different 3 marks questions on Surface Areas and Volumes from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (54)2 marks (22)3 marks (17)4 marks (27)5 marks (41)

Q303 marksShort AnswerSurface Areas and Volumes CBSE 2026 · Standard 30/1/1

A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid. $(Use π = \frac{22}{7})$

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Answer:

\frac{4081}{6}

^{3}

≈ 680.17 cm

^{3}

Radius $r = 3.5$ cm. Height of the cylindrical part $h = 20 - 2 \times 3.5 = 13$ cm.
Volume $= π r^{2} h + 2 \times \frac{2}{3} π r^{3} = π r^{2} (h + \frac{4}{3} r)$ .
$π r^{2} = \frac{22}{7} \times 12.25 = 38.5$ .
$h + \frac{4}{3} r = 13 + \frac{14}{3} = \frac{53}{3}$ .
Volume $= 38.5 \times \frac{53}{3} = \frac{2040.5}{3} \approx 680.17$ cm $^{3}$ .

Also asked in: 2026 Standard 30/1/2

Q283 marksShort AnswerSurface Areas and Volumes CBSE 2026 · Standard 30/1/3

A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surface areas are in the ratio 8 : 5, then find the ratio between the radius of their bases to their height.

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Answer: 3 : 4

Let the common radius be $r$ and height $h$ ; slant height of the cone $l = r^{2} + h^{2}$ .
$\frac{2 π r h}{π r l} = \frac{8}{5}$ , so $\frac{2 h}{l} = \frac{8}{5}$ and $l = \frac{5 h}{4}$ .
$r^{2} + h^{2} = \frac{25 h ^{2}}{16}$ , so $r^{2} = \frac{9 h ^{2}}{16}$ and $r = \frac{3 h}{4}$ .
$r : h = 3 : 4$ .

Q293 marksShort AnswerSurface Areas and Volumes CBSE 2026 · Standard 30/4/1

To protect plants from heat, a shed of iron rods covered with green cloth is made. The lower part of the shed is a cuboid mounted by semi-cylinder as shown in the figure. Find the area of the cloth required to make this shed, if dimensions of the cuboid are 14 m $\times$ 25 m $\times$ 16 m

Diagram for CBSE 2026 Class 10 Maths question 29

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Answer: 1952 m

^{2}

Cloth covers the four walls of the cuboid, the curved surface of the semi-cylinder and its two semicircular ends (not the floor).
Radius of semi-cylinder $= 7$ m, length $= 25$ m.
Four walls $= 2 (14 + 25) \times 16 = 1248$ m $^{2}$
Curved surface of semi-cylinder $= π r l = \frac{22}{7} \times 7 \times 25 = 550$ m $^{2}$
Two semicircular ends $= π r^{2} = \frac{22}{7} \times 49 = 154$ m $^{2}$
Total cloth $= 1248 + 550 + 154 = 1952$ m $^{2}$

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q29 (OR) (OR)3 marksShort AnswerSurface Areas and Volumes CBSE 2026 · Standard 30/4/1

The internal and external radii of a hollow hemisphere are $52$ cm and 10 cm respectively. A cone of height $57$ cm and radius $52$ cm is surmounted on the hemisphere as shown in the figure. Find the total surface area of the object in terms of $π$ . (Use $2 = 1.4$ )

Diagram for CBSE 2026 Class 10 Maths question 29 (OR)

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Answer:

355 π

^{2}

Outer curved surface of hemisphere $= 2 π R^{2} = 2 π \times 100 = 200 π$
Ring at the rim $= π (R^{2} - r^{2}) = π (100 - 50) = 50 π$
Slant height of cone $l = (52)^{2} + (57)^{2} = 50 + 175 = 15$ cm
Curved surface of cone $= π r l = π \times 52 \times 15 = 752 π = 75 \times 1.4 π = 105 π$
The cone closes the hollow, so the inner surface is not exposed.
Total surface area $= 200 π + 50 π + 105 π = 355 π$ cm $^{2}$

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q303 marksShort AnswerSurface Areas and Volumes CBSE 2025 · Basic 430/5/1

A spherical glass vessel has a cylindrical neck 7 cm long and 8 cm in diameter. The radius of spherical part is 10 cm. Find the volume of the vessel.

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Answer:

\frac{95392}{21}

^{3}

\approx 4542.48

^{3}

Volume of sphere $= \frac{4}{3} π (10)^{3} = \frac{4000}{3} π$ .
Volume of neck (cylinder, r = 4 cm, h = 7 cm) $= π \times 16 \times 7 = 112 π$ .
Total $= (\frac{4000}{3} + 112) π = \frac{4336}{3} \times \frac{22}{7} = \frac{95392}{21} \approx 4542.48$ cm $^{3}$ .

Also asked in: 2025 Basic 430/5/2, 2025 Basic 430/5/3

Q30 (OR) (OR)3 marksShort AnswerSurface Areas and Volumes CBSE 2025 · Basic 430/5/1

From each end of a solid cylinder of height 20 cm and base radius 7 cm, a cone of base radius 2.1 cm and height 5 cm is scooped out. Find the volume of the remaining solid.

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Answer: 3033.8 cm

^{3}

Volume of cylinder $= \frac{22}{7} \times 7^{2} \times 20 = 3080$ cm $^{3}$ .
Volume of one cone $= \frac{1}{3} \times \frac{22}{7} \times (2.1)^{2} \times 5 = 23.1$ cm $^{3}$ .
Two cones $= 46.2$ cm $^{3}$ .
Remaining volume $= 3080 - 46.2 = 3033.8$ cm $^{3}$ .

Also asked in: 2025 Basic 430/5/2, 2025 Basic 430/5/3

Q303 marksShort AnswerSurface Areas and Volumes CBSE 2025 · Standard 30/1/1

A room is in the form of a cylinder surmounted by a hemispherical dome. The base radius of the hemisphere is half of the height of the cylindrical part. If the room contains $\frac{1408}{21}$ m $^{3}$ of air, find the height of the cylindrical part. (Use $π = \frac{22}{7}$ ).

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Answer:

4

Let the radius be $r$ ; then the height of the cylinder is $h = 2 r$ .
Volume $= π r^{2} h + \frac{2}{3} π r^{3} = 2 π r^{3} + \frac{2}{3} π r^{3} = \frac{8}{3} π r^{3}$ .
$\frac{8}{3} \times \frac{22}{7} \times r^{3} = \frac{1408}{21}$
$\frac{176}{21} r^{3} = \frac{1408}{21}$ , so $r^{3} = 8$ and $r = 2$ m.
Height of cylindrical part $= 2 r = 4$ m.

Also asked in: 2025 Standard 30/1/3

Q263 marksShort AnswerSurface Areas and Volumes CBSE 2025 · Standard 30/1/2

If the radii of the bases of a cylinder and a cone are in the ratio $3 : 4$ and their heights are in the ratio $2 : 3$ , find the ratio of their volumes.

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Answer:

9 : 8

Let the radii be $3 r$ , $4 r$ and the heights $2 h$ , $3 h$ .
Volume of cylinder $= π (3 r)^{2} (2 h) = 18 π r^{2} h$ .
Volume of cone $= \frac{1}{3} π (4 r)^{2} (3 h) = 16 π r^{2} h$ .
Ratio $= 18 : 16 = 9 : 8$ .

Q283 marksShort AnswerSurface Areas and Volumes CBSE 2024 · Basic 430/1/2

A solid is in the form of a cylinder with hemi–spherical ends of same radii. The total height of the solid is 20 cm and the diameter of the cylinder is 14 cm. Find the surface area of the solid.

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Answer: 880 cm

^{2}

Radius $r = 7$ cm; height of the cylindrical part $h = 20 - 2 \times 7 = 6$ cm.
Surface area $= 2 π r h + 2 \times 2 π r^{2}$ .
$= 2 \times \frac{22}{7} \times 7 \times 6 + 4 \times \frac{22}{7} \times 49 = 264 + 616$ .
$= 880$ cm $^{2}$ .

Also asked in: 2024 Basic 430/1/3

Q28 (OR) (OR)3 marksShort AnswerSurface Areas and Volumes CBSE 2024 · Basic 430/1/2

A juice glass is cylindrical in shape with hemi–spherical raised up portion at the bottom. The inner diameter of glass is 10 cm and its height is 14 cm. Find the capacity of the glass. (use $π = 3.14$ )

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Answer: 837.33 cm

^{3}

(approx.)

Radius $r = 5$ cm, height $h = 14$ cm.
Capacity = volume of cylinder $-$ volume of hemisphere $= π r^{2} h - \frac{2}{3} π r^{3}$ .
$= 3.14 \times (25 \times 14 - \frac{2}{3} \times 125) = 3.14 \times \frac{800}{3}$ .
$= 837.33$ cm $^{3}$ (approx.).

Also asked in: 2024 Basic 430/1/3

Q283 marksShort AnswerSurface Areas and Volumes CBSE 2024 · Basic 430/3/1

A vessel is in the form of a hollow hemisphere surmounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

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Answer: 572 cm

^{2}

Radius $r = 7$ cm; height of cylinder $h = 13 - 7 = 6$ cm.
Inner surface area = CSA of hemisphere + CSA of cylinder $= 2 π r^{2} + 2 π r h$ .
$= 2 π r (r + h) = 2 \times \frac{22}{7} \times 7 \times 13$
$= 572$ cm $^{2}$ .

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3

Q28 (OR) (OR)3 marksShort AnswerSurface Areas and Volumes CBSE 2024 · Basic 430/3/1

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy.

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Answer:

8 π = \frac{176}{7}

^{3}

(about 25.14 cm

^{3}

)

Radius $r = 2$ cm, height of cone $h = 2$ cm.
Volume = volume of cone + volume of hemisphere $= \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3}$ .
$= \frac{1}{3} π \times 4 \times 2 + \frac{2}{3} π \times 8 = \frac{8 π}{3} + \frac{16 π}{3} = 8 π$ .
$= 8 \times \frac{22}{7} = \frac{176}{7} \approx 25.14$ cm $^{3}$ .

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3

Q313 marksShort AnswerSurface Areas and Volumes CBSE 2024 · Standard 30/1/1

The difference between the outer and inner radii of a hollow right circular cylinder of length 14 cm is 1 cm. If the volume of the metal used in making the cylinder is 176 $cm^{3}$ , find the outer and inner radii of the cylinder.

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Answer: Outer radius = 2.5 cm, inner radius = 1.5 cm

Let outer radius be $R$ and inner radius $r$ ; $R - r = 1$ .
Volume of metal $= π (R^{2} - r^{2}) h = \frac{22}{7} (R^{2} - r^{2}) (14) = 44 (R^{2} - r^{2}) = 176$ .
So $R^{2} - r^{2} = 4$ , i.e. $(R + r) (R - r) = 4$ , giving $R + r = 4$ .
Solving $R + r = 4$ and $R - r = 1$ : $R = 2.5$ cm, $r = 1.5$ cm.

Q293 marksShort AnswerSurface Areas and Volumes CBSE 2024 · Standard 30/1/2

A wooden toy is made by scooping out a hemisphere of same radius as of cylinder, from each end of a wooden solid cylinder. If the height of the cylinder is 20 cm and its base is of radius 7 cm, find the total surface area of the toy.

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Answer: 1496

cm^{2}

Total surface area = curved surface of cylinder + inner surfaces of the two hemispherical hollows.
$= 2 π r h + 2 \times 2 π r^{2} = 2 π r (h + 2 r)$
$= 2 \times \frac{22}{7} \times 7 \times (20 + 14) = 44 \times 34 = 1496$ $cm^{2}$ .

Q273 marksShort AnswerSurface Areas and Volumes CBSE 2024 · Standard 30/1/3

The inner and outer radii of a hollow cylinder surmounted on a hollow hemisphere of same radii are 3 cm and 4 cm respectively. If height of the cylinder is 14 cm, then find its total surface area (inner and outer).

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Answer:

253 π = \frac{5566}{7} \approx 795.14

cm^{2}

Inner radius $r = 3$ cm, outer radius $R = 4$ cm, height of cylinder $h = 14$ cm.
Outer surface: cylinder $2 π R h = 112 π$ , hemisphere $2 π R^{2} = 32 π$ ; total $144 π$ .
Inner surface: cylinder $2 π r h = 84 π$ , hemisphere $2 π r^{2} = 18 π$ ; total $102 π$ .
Ring at the open top: $π (R^{2} - r^{2}) = 7 π$ .
Total surface area $= 144 π + 102 π + 7 π = 253 π = 253 \times \frac{22}{7} = \frac{5566}{7} \approx 795.14$ $cm^{2}$ .

Q313 marksShort AnswerSurface Areas and Volumes CBSE 2023 · Standard 30/1/1

A room is in the form of cylinder surmounted by a hemi-spherical dome. The base radius of hemisphere is one-half the height of cylindrical part. Find total height of the room if it contains $(\frac{1408}{21})$ m $^{3}$ of air. $(Take π = \frac{22}{7})$

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Answer: 6 m

Let the radius be r; height of cylinder h = 2r.
Volume $= π r^{2} (2 r) + \frac{2}{3} π r^{3} = \frac{8}{3} π r^{3}$ .
$\frac{8}{3} \times \frac{22}{7} \times r^{3} = \frac{1408}{21}$ , so $\frac{176}{21} r^{3} = \frac{1408}{21}$ , $r^{3} = 8$ , r = 2 m.
Total height = h + r = 4 + 2 = 6 m.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q31 (OR) (OR)3 marksShort AnswerSurface Areas and Volumes CBSE 2023 · Standard 30/1/1

An empty cone is of radius 3 cm and height 12 cm. Ice-cream is filled in it so that lower part of the cone which is $(\frac{1}{6})^{th}$ of the volume of the cone is unfilled but hemisphere is formed on the top. Find volume of the ice-cream. (Take $π = 3.14$ )

Diagram for CBSE 2023 Class 10 Maths question 31 (OR)

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Answer: 150.72 cm

^{3}

Volume of cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times 3.14 \times 9 \times 12 = 113.04$ cm $^{3}$ .
Filled part of cone $= \frac{5}{6} \times 113.04 = 94.2$ cm $^{3}$ .
Volume of hemisphere $= \frac{2}{3} π r^{3} = \frac{2}{3} \times 3.14 \times 27 = 56.52$ cm $^{3}$ .
Volume of ice-cream = 94.2 + 56.52 = 150.72 cm $^{3}$ .

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

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