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CBSE CLASS 10 · CHAPTER 12

Surface Areas and Volumes: MCQs with Answers

Split a combined solid into familiar shapes, then identify the exposed surfaces or required volume. Use the dimensions of each part and count each exposed surface once. Try each question before opening its solution.

51 distinct questions · 55 verified source appearances
Reviewed by Team TutorVertex · Last reviewed: 24 October 2026

Board papers: 2023–2026 · Official sample-paper sessions: 2022–23 to 2026–27. Aligned with NCERT Chapter 12 (2026–27 reprint) and the applicable CBSE syllabus. This page contains ordinary MCQs.

Basic solids: prerequisites Dimensions and fitting Exposed and hollow surfaces Combined and remaining volumes Ratios and capacity

Basic solids: prerequisites

Key results

Cylinder: curved area = 2πrh; total area = 2πr(h + r); volume = πr²h.
Cone: curved area = πrl; total area = πr(l + r); volume = $\frac{1}{3}$ πr²h; l² = r² + h².
Sphere: surface area = 4πr²; volume = $\frac{4}{3}$ πr³.
Hemisphere: curved area = 2πr²; total area = 3πr²; volume = $\frac{2}{3}$ πr³.
Cube: surface area = 6a²; volume = a³. Cuboid: surface area = 2(LB + BH + HL); volume = LBH.
Here r is radius, h is perpendicular height, l is cone slant height, a is cube edge, and L, B, H are cuboid length, breadth and height. Specific question symbols take priority.

Question 1 · Basic

The total surface area of a solid hemisphere of radius 7 cm is:

Appeared in: CBSE 2026 second board examination · Basic · 430/8/1 · Q6 · CBSE 2022–23 official sample · Basic · Q15

Choices and order are identical across these appearances.

A. 447π cm²B. 239π cm²C. 174π cm²D. 147π cm²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

A solid hemisphere has a curved surface and a circular base.

Total surface area = 2πr² + πr² = 3πr².

= 3π × 7².

= 147π cm².

Correct answer: D — 147π cm²

Question 2 · Basic

The total surface area of a cube of side 20 cm is:

CBSE 2024 main examination · Basic · 430/2/3 · Q1

A. 240 cm²B. 160 cm²C. 2400 cm²D. 1600 cm²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

A cube has six square faces.

Total surface area = 6a².

= 6 × 20² = 2400 cm².

Correct answer: C — 2400 cm²

Question 3 · Basic

A solid cone has radius 7 cm and slant height 25 cm. Its total surface area is: (Use π = $\frac{22}{7}$ .)

CBSE 2026 main examination · Basic · 430/5/1 · Q18

A. 724 cm²B. 704 cm²C. 550 cm²D. 616 cm²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Total surface area of a cone = πrl + πr².

= πr(l + r).

= $\frac{22}{7}$ × 7 × (25 + 7).

= 704 cm².

Correct answer: B — 704 cm²

Question 4 · Basic

For which solid are the curved surface area and total surface area the same?

CBSE 2025 main examination · Basic · 430/1/1 · Q15

A. CubeB. CuboidC. HemisphereD. Sphere

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

A sphere has only a curved surface and no flat base.

Its curved surface is therefore its complete surface.

Hence the required solid is a sphere.

Correct answer: D — Sphere

Question 5 · Standard

A cone has base area 156 cm² and perpendicular height 8 cm. Its volume is:

CBSE 2023 main examination · Standard · 30/6/1 · Q12

A. 2496 cm³B. 1248 cm³C. 1664 cm³D. 416 cm³

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Volume = $\frac{1}{3}$ × base area × perpendicular height.

= $\frac{1}{3}$ × 156 × 8.

= 416 cm³.

Correct answer: D — 416 cm³

Question 6 · Basic

A cone has radius r and perpendicular height 3r. Its volume is:

CBSE 2023 main examination · Basic · 430/1/1 · Q13

\frac{πr³}{3}

B. 3πr³C. 9πr³D. πr³

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Volume of a cone = $\frac{1}{3}$ πr²h.

Substituting h = 3r gives $\frac{1}{3}$ πr² × 3r.

Therefore, volume = πr³.

Correct answer: D — πr³

Question 7 · Basic

The total surface area of a solid hemisphere of radius r is:

CBSE 2024–25 official sample · Basic · Q14

A. πr²B. 2πr²C. 3πr²D. 4πr²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The solid hemisphere has a curved surface and one flat circular face.

Curved area = 2πr².

Circular-face area = πr².

Total surface area = 2πr² + πr² = 3πr².

Correct answer: C — 3πr²

Question 8 · Standard

The volume of a solid hemisphere is $\frac{396}{7}$ cm³. Its total surface area is: (Use π = $\frac{22}{7}$ .)

CBSE 2024–25 official sample · Standard · Q13

\frac{396}{7}

cm²B.

\frac{594}{7}

cm²C.

\frac{549}{7}

cm²D.

\frac{604}{7}

cm²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Volume of a hemisphere = $\frac{2}{3}$ πr³.

$\frac{2}{3}$ × $\frac{22}{7}$ × r³ = $\frac{396}{7}$ .

Therefore, r³ = 27 and r = 3 cm.

Total surface area = 3πr².

= 3 × $\frac{22}{7}$ × 3².

= $\frac{594}{7}$ cm².

Correct answer: B — $\frac{594}{7}$ cm²

Dimensions and fitting

Key results

Use diameter = 2 × radius. A hemisphere contributes one radius to the total height.
For a right circular cone, l² = r² + h². Distinguish slant height from perpendicular height.
A sphere inscribed in a cube has diameter equal to the cube edge. For the cones fitted to cube faces below, base diameter and perpendicular height equal the cube edge.
When a rectangle is rolled into a cylinder, one side is the height and the other is the base circumference 2πr.

Question 9 · Basic

A capsule of total length 13 mm consists of a cylinder with a hemisphere at each end, all of the same radius. The cylindrical part is 8 mm long. The capsule’s diameter is:

CBSE 2026 second board examination · Basic · 430/7/1 · Q8

Diagram redrawn from the official question · not to scale.

A. 2 mmB. 5 mmC. 2.5 mmD. 4 mm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The two hemispherical ends together contribute two radii to the length.

13 = 8 + 2r.

Therefore, the diameter 2r = 13 − 8 = 5 mm.

Correct answer: B — 5 mm

Common trap: Subtract the cylindrical length to obtain the diameter, not the radius.

Question 10 · Basic

A conical birthday cap has diameter 16 cm and perpendicular height 24 cm. Its slant height is:

CBSE 2026 second board examination · Basic · 430/7/1 · Q12

A. 8√10 cmB. 25 cmC. 10√8 cmD. √80 cm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Radius = $\frac{16}{2}$ = 8 cm.

The radius, perpendicular height and slant height form a right triangle.

l² = 8² + 24² = 640.

Therefore, l = √640 = 8√10 cm.

Correct answer: A — 8√10 cm

Question 11 · Basic

A sphere of maximum diameter is carved from a cube of side 14 cm. The sphere’s radius is:

CBSE 2024 main examination · Basic · 430/3/1 · Q16

A. 7 cmB. 14 cmC.

\frac{7}{2}

cmD. √14 cm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The largest sphere has diameter equal to the cube’s side.

Diameter = 14 cm.

Radius = $\frac{14}{2}$ = 7 cm.

Correct answer: A — 7 cm

Question 12 · Basic

A cone of slant height 10 cm and radius 6 cm is mounted on a hemisphere of the same radius. The total height of the toy is:

CBSE 2026 main examination · Basic · 430/4/1 · Q11

Diagram redrawn from the official question · not to scale.

A. 12 cmB. 16 cmC. 14 cmD. 10 cm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Let the cone’s perpendicular height be h cm.

h² = 10² − 6² = 64.

Therefore, h = 8 cm.

The hemisphere contributes its radius, 6 cm, to the height.

Total height = 8 + 6 = 14 cm.

Correct answer: C — 14 cm

Question 13 · Basic

A cone of base radius 7 cm has curved surface area 550 cm². Its slant height is: (Use π = $\frac{22}{7}$ .)

CBSE 2025 main examination · Basic · 430/4/1 · Q15

A. 25 cmB. 14 cmC. 20 cmD. 24 cm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Curved surface area = πrl.

550 = $\frac{22}{7}$ × 7 × l.

Thus 22l = 550.

l = 25 cm.

Correct answer: A — 25 cm

Question 14 · Basic

The largest possible cone is fitted inside a cube of edge 25 cm, with its circular base inscribed in a face. The radius of the base is:

CBSE 2025 main examination · Basic · 430/5/1 · Q10

A. 5 cmB. 12.5 cmC. 25 cmD. 10 cm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The cone’s base diameter equals the side of the cube.

Diameter = 25 cm.

Radius = $\frac{25}{2}$ = 12.5 cm.

Correct answer: B — 12.5 cm

Question 15 · Basic

Two identical cones touch at their vertices, as shown. Each has base radius 4 cm and slant height 6 cm. The total height is:

CBSE 2025 main examination · Basic · 430/6/1 · Q4

Diagram redrawn from the official question · not to scale.

A. 8 cmB. 4√5 cmC. 2√5 cmD. 12 cm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Let the perpendicular height of one cone be h cm.

h² = 6² − 4² = 20.

Therefore, h = 2√5 cm.

The two cones are arranged along the same axis, tip to tip.

Total height = 2h = 4√5 cm.

Correct answer: B — 4√5 cm

Question 16 · Standard

A cone of perpendicular height 12 cm and slant height 13 cm is mounted on a hemisphere of the same radius. The total height of the solid is:

CBSE 2025 main examination · Standard · 30/5/1 · Q15

A. 17 cmB. 18 cmC. 22 cmD. 23 cm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

For the cone, r² = 13² − 12² = 25.

Thus r = 5 cm.

The hemisphere’s height equals its radius, 5 cm.

Total height = 12 + 5 = 17 cm.

Correct answer: A — 17 cm

Question 17 · Standard

A cylinder of height 5 cm has curved surface area 94.2 cm². Its radius is: (Use π = 3.14.)

CBSE 2023 main examination · Standard · 30/1/1 · Q10

A. 2 cmB. 3 cmC. 2.9 cmD. 6 cm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Curved surface area = 2πrh.

94.2 = 2 × 3.14 × r × 5.

r = $\frac{94.2}{31.4}$ .

Therefore, r = 3 cm.

Correct answer: B — 3 cm

Question 18 · Standard

A cone has base area 51 cm² and volume 85 cm³. Its perpendicular height is:

Appeared in: CBSE 2023 main examination · Standard · 30/2/3 · Q6 · CBSE 2025–26 official sample · Standard · Q11

Choices and order are identical across these appearances.

\frac{5}{6}

cmB.

\frac{5}{3}

cmC.

\frac{5}{2}

cmD. 5 cm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Volume of a cone = $\frac{1}{3}$ × base area × height.

85 = $\frac{1}{3}$ × 51 × h.

h = $\frac{85 × 3}{51}$ .

Therefore, h = 5 cm.

Correct answer: D — 5 cm

Question 19 · Basic

A cylinder of height 14 cm has curved surface area 88 cm². Its base diameter is: (Use π = $\frac{22}{7}$ .)

CBSE 2023 main examination · Basic · 430/2/1 · Q8

A. 2 cmB. 1 cmC. 4 cmD. 7 cm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Curved surface area = 2πrh.

88 = 2 × $\frac{22}{7}$ × r × 14.

Thus 88r = 88, so r = 1 cm.

Diameter = 2r = 2 cm.

Correct answer: A — 2 cm

Question 20 · Basic

A rectangular sheet of paper 40 cm × 22 cm is rolled to form a hollow cylinder of height 40 cm. Its radius is: (Use π = $\frac{22}{7}$ .)

CBSE 2023–24 official sample · Basic · Q17

A. 3.5 cmB. 7 cmC.

\frac{80}{7}

cmD. 5 cm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The 40 cm side forms the height, so the 22 cm side forms the circumference.

2πr = 22.

2 × $\frac{22}{7}$ × r = 22.

Therefore, r = 3.5 cm.

Correct answer: A — 3.5 cm

Common trap: Identify which side becomes the circumference; the height is already given.

Exposed and hollow surfaces

Key results

Add only surfaces exposed to the outside or specified in the question. Exclude shared faces at solid joins.
Two cubes joined face to face hide two square faces: combined exposed area = 12a² − 2a² = 10a².
A hollow hemispherical bowl has inner curved area, outer curved area and an annular rim. The rim area is π(R² − r²), for outer radius R and inner radius r.
An open vessel has no lid. A point of contact hides no finite area.

Question 21 · Standard

The largest right circular cone is carved from a cube of edge 4 cm, with its base fitted to a face of the cube. Its curved surface area is:

CBSE 2026 second board examination · Standard · 30/7/1 · Q17

A. 4√5 π cm²B. (4 + √5)π cm²C. (4 − √5)π cm²D. 20π cm²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The cone has base diameter 4 cm and height 4 cm.

Therefore, r = 2 cm and h = 4 cm.

l = √(2² + 4²) = 2√5 cm.

Curved surface area = πrl.

= π × 2 × 2√5 = 4√5 π cm².

Correct answer: A — 4√5 π cm²

Question 22 · Standard

A solid cone of radius r and height h is placed on a solid cylinder of the same radius and height. The total exposed surface area is:

CBSE 2026 second board examination · Standard · 30/7/3 · Q18

A. 4πrh + 4πr²B. 4πrh + πr²C. πr(√(r² + h²) + 2h + r)D. πr²(√(r² + h²) + 2h + r)

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The shared circular face is hidden.

The exposed parts are the cone’s curved surface, the cylinder’s curved surface and its lower circular base.

Cone slant height l = √(r² + h²).

Total exposed area = πrl + 2πrh + πr².

= πr(√(r² + h²) + 2h + r).

Correct answer: C — πr(√(r² + h²) + 2h + r)

Common trap: Do not count the circular face where the cone and cylinder meet.

TutorVertex illustration · added to explain the solution; not to scale.

Question 23 · Basic

An open hemispherical bowl has inner radius r₁ and outer radius r₂, where r₂ > r₁. Its total internal and external surface area, including the rim, is:

CBSE 2023 supplementary examination · Basic · 430/C/1 · Q7

Diagram redrawn from the official question · not to scale.

A. 2π(r₁² + r₂²)B. π(r₁² + r₂²)C. 3πr₂² + πr₁²D. 3πr₁² + πr₂²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Inner curved area = 2πr₁².

Outer curved area = 2πr₂².

The flat rim is an annulus of area π(r₂² − r₁²).

Total area = 2πr₁² + 2πr₂² + π(r₂² − r₁²).

= 3πr₂² + πr₁².

Correct answer: C — 3πr₂² + πr₁²

Common trap: Include the annular rim, but do not treat the bowl’s opening as a filled circular base.

Question 24 · Basic

Two cubes of edge 5 cm are joined face to face to form a cuboid. The surface area of the cuboid is:

CBSE 2023 supplementary examination · Basic · 430/C/1 · Q12

A. 200 cm²B. 300 cm²C. 125 cm²D. 250 cm²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The resulting cuboid has dimensions 10 cm × 5 cm × 5 cm.

Surface area = 2(lb + bh + hl).

= 2(10 × 5 + 5 × 5 + 5 × 10).

= 2 × 125 = 250 cm².

Correct answer: D — 250 cm²

TutorVertex illustration · added to explain the solution; not to scale.

Question 25 · Basic + Standard

Two cubes, each of volume 64 cm³, are joined face to face to form a cuboid. Its total surface area is:

Appeared in: CBSE 2024 main examination · Basic · 430(B) · Q18 · CBSE 2026–27 official sample · Basic · Q18 · CBSE 2026–27 official sample · Standard · Q11

Choices shown as in CBSE 2024 main examination · Basic · 430(B) · Q18. Other appearances use a different choice order or distractors.

A. 192 cm²B. 160 cm²C. 128 cm²D. 176 cm²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

For each cube, a³ = 64, so a = 4 cm.

The cuboid has dimensions 8 cm × 4 cm × 4 cm.

Surface area = 2(8 × 4 + 4 × 4 + 4 × 8).

= 2(32 + 16 + 32).

= 160 cm².

Correct answer: B — 160 cm²

Question 26 · Standard

A solid sphere is cut into two hemispheres. The ratio of the sphere’s surface area to the combined total surface area of the two hemispheres is:

CBSE 2024 main examination · Standard · 30/1/2 · Q15

A. 1 : 1B. 1 : 4C. 2 : 3D. 3 : 2

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Sphere surface area = 4πr².

Each solid hemisphere has total area 3πr².

Combined area of the two hemispheres = 6πr².

The required ratio is 4πr² : 6πr².

= 2 : 3.

Correct answer: C — 2 : 3

Common trap: Cutting exposes two new circular faces, so the total surface area increases.

Question 27 · Standard

Two identical solid cubes of side a are joined face to face to form a cuboid. Its total surface area is:

CBSE 2024 main examination · Standard · 30/4/3 · Q8

A. 6a²B. 10a²C. 5a²D. 4a²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Before joining, the two cubes have total surface area 12a².

Two square faces, each of area a², become hidden at the join.

Exposed surface area = 12a² − 2a².

= 10a².

Correct answer: B — 10a²

Question 28 · Basic

Two solid hemispheres of the same radius r touch at a point, as shown. The total surface area of the object is:

CBSE 2026 main examination · Basic · 430/4/1 · Q3

Diagram redrawn from the official question · not to scale.

A. 4πr²B. 6πr²C. 3πr²D. 5πr²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Each solid hemisphere has total surface area 3πr².

A point of contact has zero area, so no finite surface area is hidden.

Total area = 2 × 3πr².

= 6πr².

Correct answer: B — 6πr²

Common trap: The hemispheres meet at a point, not along their flat circular faces.

Question 29 · Basic

An open vessel consists of a cylinder of radius r above a hemisphere of the same radius. Its total height is 13 cm, as shown. Its inner surface area is:

CBSE 2026 main examination · Basic · 430/5/1 · Q4

Diagram redrawn from the official question · not to scale.

A. 2πr(r + 13)B. 13πrC. 2π(13 + r)²D. 26πr

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The hemisphere’s height is r, so the cylinder’s height is 13 − r.

Inner area = curved area of cylinder + curved area of hemisphere.

= 2πr(13 − r) + 2πr².

= 26πr − 2πr² + 2πr².

= 26πr.

Correct answer: D — 26πr

Common trap: The top is open. Do not add a circular lid or a disc at the cylinder–hemisphere join.

Question 30 · Standard

A conical cavity of maximum volume is carved from a solid hemisphere of radius 10 cm, as shown. The curved surface area of the cavity is: (Use π = 3.14.)

CBSE 2026 main examination · Standard · 30/5/1 · Q7

Diagram redrawn from the official question · not to scale.

A. 314√2 cm²B. 314 cm²C.

\frac{3140}{3}

cm²D. 3140√2 cm²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The cavity’s base radius and perpendicular height are both 10 cm.

Slant height l = √(10² + 10²) = 10√2 cm.

Curved surface area = πrl.

= 3.14 × 10 × 10√2.

= 314√2 cm².

Correct answer: A — 314√2 cm²

Question 31 · Standard

A hemispherical camping tent of radius 1.4 m has a door opening of area 0.50 m². The area of its outer covering is: (Use π = $\frac{22}{7}$ .)

CBSE 2026 main examination · Standard · 30/5/1 · Q12

A. 11.78 m²B. 12.32 m²C. 11.82 m²D. 12.86 m²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The tent’s covering is its hemispherical curved surface, excluding the door.

Curved area = 2 × $\frac{22}{7}$ × 1.4² = 12.32 m².

Required area = 12.32 − 0.50.

= 11.82 m².

Correct answer: C — 11.82 m²

Question 32 · Standard

A toy top consists of a cone mounted on a hemisphere of the same radius. The hemisphere’s radius is 3.5 cm and the total height is 15.5 cm. Its total surface area is: (Use π = $\frac{22}{7}$ .)

CBSE 2023 main examination · Standard · 30(B) · Q17

A. 2145 cm²B. 21.45 cm²C. 215.4 cm²D. 214.5 cm²

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Cone height = 15.5 − 3.5 = 12 cm.

Cone slant height = √(12² + 3.5²) = 12.5 cm.

The common circular face is internal.

Exposed area = πrl + 2πr².

= $\frac{22}{7}$ × 3.5 × 12.5 + 2 × $\frac{22}{7}$ × 3.5².

= 137.5 + 77.

= 214.5 cm².

Correct answer: D — 214.5 cm²

Combined and remaining volumes

Key results

For solids joined without overlap, add the volumes of their parts.
For a cavity or empty space, subtract the removed or occupied volume from the enclosing volume.
For a thick shell, material volume = outer-solid volume − inner-solid volume. Outer radius = inner radius + thickness.
For fully submerged objects, displaced water volume equals submerged volume. In a cylindrical vessel, rise = displaced volume ÷ base area, provided no water overflows.

Question 33 · Basic

Two solid spheres, each of radius 3 cm, are completely submerged in water in a cylindrical vessel of radius 4 cm. The rise in water level is:

CBSE 2026 second board examination · Basic · 430(B) · Q16

A. 4 cmB. 4.5 cmC. 4.25 cmD. 4.15 cm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The displaced water volume equals the combined volume of the two spheres.

Displaced volume = 2 × $\frac{4}{3}$ π × 3³ = 72π cm³.

Let the rise in water level be x cm.

π × 4² × x = 72π.

x = $\frac{72}{16}$ = 4.5 cm.

Correct answer: B — 4.5 cm

Question 34 · Standard

A hemispherical steel bowl has inner radius 5 cm and uniform thickness 1 cm. The volume of steel used is:

CBSE 2023 supplementary examination · Standard · 30(B)/C · Q15

A. 182π cm³B.

\frac{182π}{3}

cm³C.

\frac{682π}{3}

cm³D.

\frac{364π}{3}

cm³

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Outer radius = 5 + 1 = 6 cm.

Steel volume = outer-hemisphere volume − inner-hemisphere volume.

= $\frac{2}{3}$ π(6³ − 5³).

= $\frac{2}{3}$ π(216 − 125).

= $\frac{182π}{3}$ cm³.

Correct answer: B — $\frac{182π}{3}$ cm³

Question 35 · Basic

A solid hemisphere of radius r is surmounted by a cone of the same base radius. The cone’s height equals its base diameter. The combined volume is:

CBSE 2023 supplementary examination · Basic · 430(B)/C · Q17

A. πr³B.

\frac{2πr³}{3}

\frac{πr³}{3}

\frac{4πr³}{3}

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The cone’s height is 2r.

Combined volume = hemisphere volume + cone volume.

= $\frac{2}{3}$ πr³ + $\frac{1}{3}$ πr² × 2r.

= $\frac{2}{3}$ πr³ + $\frac{2}{3}$ πr³.

= $\frac{4πr³}{3}$ .

Correct answer: D — $\frac{4πr³}{3}$

Question 36 · Standard

A hemispherical steel bowl has inner radius 3 cm and thickness 0.30 cm. The volume of steel used is:

CBSE 2024 supplementary examination · Standard · 30(B) · Q10

A. 595.8π cm³B. 5.958π cm³C. 6π cm³D. 59.58π cm³

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Outer radius = 3 + 0.30 = 3.3 cm.

Steel volume = $\frac{2}{3}$ π(3.3³ − 3³).

= $\frac{2}{3}$ π(35.937 − 27).

= $\frac{2}{3}$ π × 8.937.

= 5.958π cm³.

Correct answer: B — 5.958π cm³

Question 37 · Standard

Three tennis balls of radius r are just packed in a cylindrical jar, as shown. The volume of air inside the jar is:

CBSE 2026 main examination · Standard · 30/4/1 · Q13

Diagram redrawn from the official question · not to scale.

A. 2πr³B. 3πr³C. 5πr³D. 4πr³

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The jar has radius r and height 3 × 2r = 6r.

Jar volume = πr² × 6r = 6πr³.

The three balls have volume 3 × $\frac{4}{3}$ πr³ = 4πr³.

Air volume = 6πr³ − 4πr³.

= 2πr³.

Correct answer: A — 2πr³

Question 38 · Standard

A cone of maximum size is carved from a cube of edge l as shown. The cone’s volume is:

CBSE 2026 main examination · Standard · 30/3/1 · Q7

Diagram redrawn from the official question · not to scale.

\frac{πl³}{12}

\frac{πl³}{3}

C. l³(1 −

\frac{π}{3}

)D.

\frac{πl³}{8}

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The cone’s base is fitted to a face of the cube.

Its radius is $\frac{l}{2}$ and its perpendicular height is l.

Volume = $\frac{1}{3}$ π( $\frac{l}{2}$ )² × l.

= $\frac{πl³}{12}$ .

Correct answer: A — $\frac{πl³}{12}$

Question 39 · Basic

The volume of air in a hollow cylinder is 450 cm³. A cone of the same height and radius is placed inside it, as shown. The volume of empty space is:

CBSE 2025 main examination · Basic · 430/6/1 · Q18

Diagram redrawn from the official question · not to scale.

A. 225 cm³B. 150 cm³C. 250 cm³D. 300 cm³

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

For equal base radius and height, cone volume is one-third of cylinder volume.

Cone volume = $\frac{450}{3}$ = 150 cm³.

Empty volume = 450 − 150.

= 300 cm³.

Correct answer: D — 300 cm³

Ratios and capacity

Key results

For spheres, surface areas are proportional to r² and volumes to r³.
For equal-radius cones and cylinders, equal volumes give cone height = 3 × cylinder height.
Number of containers filled = available liquid volume ÷ capacity of one container. Use consistent units.
Follow the specified value of π. Give surface areas in square units and volumes in cubic units.

Question 40 · Basic

A solid has a hemispherical bottom and a conical top of the same radius. Their curved surface areas are equal. The ratio of the common radius to the height of the cone is:

CBSE 2024 supplementary examination · Basic · 430(B)/S · Q18

A. 1 : 3B. 3 : 1C. √3 : 1D. 1 : √3

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Let the common radius be r, the cone’s height be h and its slant height be l.

Equal curved areas give 2πr² = πrl.

Hence l = 2r.

By Pythagoras, h² = l² − r² = 4r² − r² = 3r².

Therefore, h = √3 r.

The required ratio r : h = 1 : √3.

Correct answer: D — 1 : √3

Question 41 · Basic

The volume of a sphere of radius R is 16 times the volume of a hemisphere of radius r. R : r is:

CBSE 2024 supplementary examination · Basic · 430/S/1 · Q18

A. 1 : 2B. 2 : 1C. 8 : 1D. 1 : 8

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

$\frac{4}{3}$ πR³ = 16 × $\frac{2}{3}$ πr³.

Cancelling common factors gives R³ = 8r³.

Therefore, R = 2r.

Hence R : r = 2 : 1.

Correct answer: B — 2 : 1

Question 42 · Standard

A cap consists of a cylinder with a hemispherical top of the same radius. The two parts have equal volumes. The ratio of the cylinder’s height to the hemisphere’s height is:

CBSE 2026 second board examination · Standard · 30/7/1 · Q16

A. 2 : 3B. 1 : 3C. 2 : 1D. 3 : 1

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Let the common radius be r and the cylinder’s height be h.

The hemisphere’s height is r.

Equal volumes give πr²h = $\frac{2}{3}$ πr³.

Thus h = $\frac{2r}{3}$ .

Therefore, h : r = 2 : 3.

Correct answer: A — 2 : 3

Question 43 · Standard

A cone and a cylinder have equal base radii and equal curved surface areas. The ratio of the cone’s slant height to the cylinder’s height is:

CBSE 2026 second board examination · Standard · 30/8/2 · Q4

A. 2 : 1B. 1 : 3C. 1 : 2D. 3 : 1

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Let the common radius be r, the cone’s slant height be l and the cylinder’s height be h.

πrl = 2πrh.

Cancelling πr gives l = 2h.

Therefore, l : h = 2 : 1.

Correct answer: A — 2 : 1

Question 44 · Basic

The volumes of two spheres are in the ratio 125 : 64. Their surface areas are in the ratio:

CBSE 2023 supplementary examination · Basic · 430(B)/C · Q18

A. 5 : 4B. 4 : 5C. 16 : 25D. 25 : 16

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Sphere volume is proportional to the cube of its radius.

Radius ratio = ∛125 : ∛64 = 5 : 4.

Surface-area ratio = 5² : 4².

= 25 : 16.

Correct answer: D — 25 : 16

Question 45 · Standard

A cap consists of a cylinder surmounted by a cone of the same base radius. The two parts have equal volumes. The ratio of the cylinder’s height to the cone’s height is:

CBSE 2024 supplementary examination · Standard · 30/S/2 · Q17

A. 1 : 2B. 1 : 3C. 2 : 1D. 3 : 1

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Let the common radius be r, cylinder height be h₁ and cone height be h₂.

πr²h₁ = $\frac{1}{3}$ πr²h₂.

Thus h₁ = $\frac{h₂}{3}$ .

Therefore, h₁ : h₂ = 1 : 3.

Correct answer: B — 1 : 3

Question 46 · Standard

A cylindrical bottle of radius 6 cm and height 8 cm is full of liquid. How many conical bottles of radius 2 cm and height 1.2 cm can it fill?

CBSE 2026 main examination · Standard · 30(B) · Q14

A. 60B. 130C. 180D. 18

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Liquid volume = π × 6² × 8 = 288π cm³.

Capacity of each conical bottle = $\frac{1}{3}$ π × 2² × 1.2.

= 1.6π cm³.

Number of bottles = $\frac{288π}{1.6π}$ .

= 180.

Correct answer: C — 180

Question 47 · Basic

Two spheres have volumes in the ratio 64 : 27. The ratio of their surface areas is:

CBSE 2026 main examination · Basic · 430(B) · Q16

A. 4 : 3B. 3 : 4C. 16 : 9D. 9 : 16

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Radius ratio = ∛64 : ∛27 = 4 : 3.

Surface area is proportional to the square of radius.

Area ratio = 4² : 3² = 16 : 9.

Correct answer: C — 16 : 9

Question 48 · Standard

Two spherical ice-cream scoops, each of radius $\frac{r}{2}$ , completely fill a cone of radius r and height h. The ratio h : 2r is:

CBSE 2026 main examination · Standard · 30/4/1 · Q16

A. 1 : 8B. 1 : 2C. 1 : 1D. 2 : 1

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

The combined volume of the two scoops is 2 × $\frac{4}{3}$ π( $\frac{r}{2}$ )³.

= $\frac{πr³}{3}$ .

Equating this to the cone volume gives $\frac{1}{3}$ πr²h = $\frac{1}{3}$ πr³.

Thus h = r.

Therefore, h : 2r = 1 : 2.

Correct answer: B — 1 : 2

Question 49 · Basic

A cone and a cylinder have the same base radius and equal volumes. The ratio of the cone’s height to the cylinder’s height is:

CBSE 2025 main examination · Basic · 430/2/1 · Q15

A. 1 : 1B. 3 : 1C. 1 : 3D. 3 : 2

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Let the cone’s height be h₁ and cylinder’s height be h₂.

Equal volumes give $\frac{1}{3}$ πr²h₁ = πr²h₂.

Thus h₁ = 3h₂.

Hence h₁ : h₂ = 3 : 1.

Correct answer: B — 3 : 1

Question 50 · Basic

Two cylinders of equal volume have heights in the ratio 1 : 2. The ratio of their radii is:

CBSE 2025 main examination · Basic · 430/4/1 · Q6

A. √2 : 1B. 1 : 2C. 1 : 4D. 1 : √2

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Equal volumes give πr₁²h₁ = πr₂²h₂.

Therefore, $\frac{r₁²}{r₂²}$ = $\frac{h₂}{h₁}$ = 2.

Taking positive square roots gives r₁ : r₂ = √2 : 1.

Correct answer: A — √2 : 1

Question 51 · Standard

A right circular cone and a sphere each have radius 5 cm. Their volumes are equal. The height of the cone is:

CBSE 2024–25 official sample · Standard · Q5

A. 5 cmB. 20 cmC. 10 cmD. 4 cm

Show Answer & Explanation

Official CBSE answer · Solution by Team TutorVertex

Let the cone’s perpendicular height be h cm.

Equal volumes give $\frac{1}{3}$ π × 5² × h = $\frac{4}{3}$ π × 5³.

Cancelling common factors gives h = 4 × 5.

Therefore, h = 20 cm.

Correct answer: B — 20 cm

Common questions

How do I find the surface area of a combined solid?

Identify the exposed surfaces of each part, calculate their areas, and add them. Shared circular or square faces inside the solid are excluded.

What is the difference between curved and total surface area?

Curved surface area counts only the curved part. Total surface area also includes any flat faces of the solid.

Does a hemisphere add its radius or diameter to the height?

Its height is its radius. A hemisphere mounted below a cone adds one radius to the cone’s perpendicular height.

How do I find the volume of material in a hollow bowl?

Subtract the inner hemisphere’s volume from the outer hemisphere’s volume. Add thickness to the inner radius to obtain the outer radius.

How do I find a cone’s slant height?

Use l² = r² + h², where r is radius and h is perpendicular height.

What changes when two cubes are joined face to face?

Their volumes add, while two square faces become internal. For cubes of edge a, the resulting surface area is 10a².

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Official sources and marking schemes

Each question’s source label links to the official file or archive. Archives may contain several sets; the paper code and question number identify the appearance used.

Board papers and marking schemes

2023 board papers

2024 board papers

2025 board papers

2026 board papers

Official sample papers and marking schemes

2022–23 sample papers

2023–24 sample papers

2024–25 sample papers

2025–26 sample papers

2026–27 sample papers

NCERT textbook and syllabus

How this question bank is prepared

This selected bank contains 51 distinct questions across 55 verified source appearances: 47 board-paper appearances and 8 sample-paper appearances. Selected repeated stems are merged and each retained appearance is listed. This is not a complete transcription of every set or language version.

Answers were checked against corresponding official marking schemes and checked mathematically. Ten original question diagrams are redrawn after inspecting the source figures; two additional teaching illustrations are labelled separately. Figures are not to scale.

Official CBSE answers are accompanied by worked explanations written by Team TutorVertex in NCERT style, with one calculation or reasoning step per paragraph. No incorrect-choice explanations are included. Minor grammatical and unit typography corrections retain the mathematical data and intended choices.

NCERT Chapter 12 (2026–27 reprint) and the CBSE 2026–27 syllabus were used to check scope, terminology and methods. Combined solids are the chapter focus. Basic-solid formulas, fitting, capacity and displacement are included as prerequisites or related applications from the selected official papers. No frustum question is included.

This page contains ordinary MCQs. Assertion–Reason and written case-study questions are excluded. The checked 2022 written board papers have not been converted into invented MCQs. Official sample-paper sessions run from 2022–23 to 2026–27.

Source screening uses text extraction and OCR where needed. Unclear or mathematically inconsistent entries are excluded. Only verified selected appearances are listed; the list does not claim to enumerate every repetition in every available set.

Surface Areas and Volumes: MCQs with Answers

Basic solids: prerequisites

Key results

The total surface area of a solid hemisphere of radius 7 cm is:

The total surface area of a cube of side 20 cm is:

A solid cone has radius 7 cm and slant height 25 cm. Its total surface area is: (Use π = 227.)

For which solid are the curved surface area and total surface area the same?

A cone has base area 156 cm² and perpendicular height 8 cm. Its volume is:

A cone has radius r and perpendicular height 3r. Its volume is:

The total surface area of a solid hemisphere of radius r is:

The volume of a solid hemisphere is 3967 cm³. Its total surface area is: (Use π = 227.)

Dimensions and fitting

Key results

A capsule of total length 13 mm consists of a cylinder with a hemisphere at each end, all of the same radius. The cylindrical part is 8 mm long. The capsule’s diameter is:

A conical birthday cap has diameter 16 cm and perpendicular height 24 cm. Its slant height is:

A sphere of maximum diameter is carved from a cube of side 14 cm. The sphere’s radius is:

A cone of slant height 10 cm and radius 6 cm is mounted on a hemisphere of the same radius. The total height of the toy is:

A cone of base radius 7 cm has curved surface area 550 cm². Its slant height is: (Use π = 227.)

The largest possible cone is fitted inside a cube of edge 25 cm, with its circular base inscribed in a face. The radius of the base is:

Two identical cones touch at their vertices, as shown. Each has base radius 4 cm and slant height 6 cm. The total height is:

A cone of perpendicular height 12 cm and slant height 13 cm is mounted on a hemisphere of the same radius. The total height of the solid is:

A cylinder of height 5 cm has curved surface area 94.2 cm². Its radius is: (Use π = 3.14.)

A cone has base area 51 cm² and volume 85 cm³. Its perpendicular height is:

A cylinder of height 14 cm has curved surface area 88 cm². Its base diameter is: (Use π = 227.)

A rectangular sheet of paper 40 cm × 22 cm is rolled to form a hollow cylinder of height 40 cm. Its radius is: (Use π = 227.)

Exposed and hollow surfaces

Key results

The largest right circular cone is carved from a cube of edge 4 cm, with its base fitted to a face of the cube. Its curved surface area is:

A solid cone of radius r and height h is placed on a solid cylinder of the same radius and height. The total exposed surface area is:

An open hemispherical bowl has inner radius r₁ and outer radius r₂, where r₂ > r₁. Its total internal and external surface area, including the rim, is:

Two cubes of edge 5 cm are joined face to face to form a cuboid. The surface area of the cuboid is:

Two cubes, each of volume 64 cm³, are joined face to face to form a cuboid. Its total surface area is:

A solid sphere is cut into two hemispheres. The ratio of the sphere’s surface area to the combined total surface area of the two hemispheres is:

Two identical solid cubes of side a are joined face to face to form a cuboid. Its total surface area is:

Two solid hemispheres of the same radius r touch at a point, as shown. The total surface area of the object is:

An open vessel consists of a cylinder of radius r above a hemisphere of the same radius. Its total height is 13 cm, as shown. Its inner surface area is:

A conical cavity of maximum volume is carved from a solid hemisphere of radius 10 cm, as shown. The curved surface area of the cavity is: (Use π = 3.14.)

A hemispherical camping tent of radius 1.4 m has a door opening of area 0.50 m². The area of its outer covering is: (Use π = 227.)

A toy top consists of a cone mounted on a hemisphere of the same radius. The hemisphere’s radius is 3.5 cm and the total height is 15.5 cm. Its total surface area is: (Use π = 227.)

Combined and remaining volumes

Key results

Two solid spheres, each of radius 3 cm, are completely submerged in water in a cylindrical vessel of radius 4 cm. The rise in water level is:

A hemispherical steel bowl has inner radius 5 cm and uniform thickness 1 cm. The volume of steel used is:

A solid hemisphere of radius r is surmounted by a cone of the same base radius. The cone’s height equals its base diameter. The combined volume is:

A hemispherical steel bowl has inner radius 3 cm and thickness 0.30 cm. The volume of steel used is:

Three tennis balls of radius r are just packed in a cylindrical jar, as shown. The volume of air inside the jar is:

A cone of maximum size is carved from a cube of edge l as shown. The cone’s volume is:

The volume of air in a hollow cylinder is 450 cm³. A cone of the same height and radius is placed inside it, as shown. The volume of empty space is:

Ratios and capacity

Key results

A solid has a hemispherical bottom and a conical top of the same radius. Their curved surface areas are equal. The ratio of the common radius to the height of the cone is:

The volume of a sphere of radius R is 16 times the volume of a hemisphere of radius r. R : r is:

A cap consists of a cylinder with a hemispherical top of the same radius. The two parts have equal volumes. The ratio of the cylinder’s height to the hemisphere’s height is:

A cone and a cylinder have equal base radii and equal curved surface areas. The ratio of the cone’s slant height to the cylinder’s height is:

The volumes of two spheres are in the ratio 125 : 64. Their surface areas are in the ratio:

A cap consists of a cylinder surmounted by a cone of the same base radius. The two parts have equal volumes. The ratio of the cylinder’s height to the cone’s height is:

A cylindrical bottle of radius 6 cm and height 8 cm is full of liquid. How many conical bottles of radius 2 cm and height 1.2 cm can it fill?

Two spheres have volumes in the ratio 64 : 27. The ratio of their surface areas is:

Two spherical ice-cream scoops, each of radius r2, completely fill a cone of radius r and height h. The ratio h : 2r is:

A cone and a cylinder have the same base radius and equal volumes. The ratio of the cone’s height to the cylinder’s height is:

Two cylinders of equal volume have heights in the ratio 1 : 2. The ratio of their radii is:

A right circular cone and a sphere each have radius 5 cm. Their volumes are equal. The height of the cone is:

Common questions

Board papers and marking schemes

Official sample papers and marking schemes

NCERT textbook and syllabus

A solid cone has radius 7 cm and slant height 25 cm. Its total surface area is: (Use π = $\frac{22}{7}$ .)

The volume of a solid hemisphere is $\frac{396}{7}$ cm³. Its total surface area is: (Use π = $\frac{22}{7}$ .)

A cone of base radius 7 cm has curved surface area 550 cm². Its slant height is: (Use π = $\frac{22}{7}$ .)

A cylinder of height 14 cm has curved surface area 88 cm². Its base diameter is: (Use π = $\frac{22}{7}$ .)

A rectangular sheet of paper 40 cm × 22 cm is rolled to form a hollow cylinder of height 40 cm. Its radius is: (Use π = $\frac{22}{7}$ .)

A hemispherical camping tent of radius 1.4 m has a door opening of area 0.50 m². The area of its outer covering is: (Use π = $\frac{22}{7}$ .)

A toy top consists of a cone mounted on a hemisphere of the same radius. The hemisphere’s radius is 3.5 cm and the total height is 15.5 cm. Its total surface area is: (Use π = $\frac{22}{7}$ .)

Two spherical ice-cream scoops, each of radius $\frac{r}{2}$ , completely fill a cone of radius r and height h. The ratio h : 2r is: