Surface Areas and Volumes 2-Mark Questions Class 10: CBSE PYQs with Answers

Q: Why do we subtract a common face when solids are joined?

The shared face lies inside the combined solid. Count only the exposed surfaces when finding its surface area.

Q: Does the total surface area of a hemisphere include its circular base?

Yes. Total surface area is 3πr², while curved surface area is 2πr². Use the wording of the question to choose the formula.

Q: Why are two alternatives marked older-syllabus practice?

Those melting or reshaping alternatives appear in original 2022 OR pairs. They are retained with their paired surface-area questions so the original exam choice remains intact. The current syllabus specifies surface areas and volumes of combinations of listed solids.

Q: How do I find the slant height of a cone?

For a right circular cone, use l² = r² + h², where l is the slant height, r is the base radius and h is the perpendicular height. Find l before using the curved-surface-area formula πrl.

Q: What is the surface area when two equal cubes are joined face to face?

If each cube has side a and they share one full square face, the resulting cuboid measures 2a × a × a. Its surface area is 2(2a² + a² + 2a²) = 10a².

Q: Which units should I use for surface area and volume?

Express all lengths in the same unit before substituting. Surface area uses square units such as cm² or m². Volume uses cubic units such as cm³ or m³. For capacity, 1 cm³ = 1 mL and 1000 cm³ = 1 L.

Question 1

3 cubes each of 8 cm edge are joined end to end. Find the total surface area of the cuboid so formed.

Answer

Show solution

The cuboid has dimensions 24 cm × 8 cm × 8 cm.

Total surface area = 2(lb + bh + hl)

= 2(24 × 8 + 8 × 8 + 8 × 24)

= 896 cm².

896 cm²

Question 2

(a) A cubical block of side 7 cm is surmounted by a hemisphere of largest possible diameter as shown in the figure. Find the total surface area of the solid.

Answer

Show solution

The hemisphere has radius r = $\frac{7}{2}$ cm.

Exposed area = cube surface area − covered circular base + hemisphere curved area

= 6 × 7² − πr² + 2πr² = 294 + πr²

= 294 + ( $\frac{22}{7}$ )( $\frac{7}{2}$ )² = 332.5 cm².

332.5 cm²

Question 3

(b) How many solid cones of height 3 cm and radius 2 cm can be formed by melting a solid sphere of radius 3 cm?

Answer

Show solution

Let n be the number of cones. Volume is conserved.

n × ( $\frac{1}{3}$ )π × 2² × 3 = ( $\frac{4}{3}$ )π × 3³

4πn = 36π, so n = 9.

9 cones

Question 4

(a) A cone of height 20 cm and base radius 5 cm is made of clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere.

Answer

Show solution

The cone and sphere have equal volumes.

( $\frac{1}{3}$ )π × 5² × 20 = ( $\frac{4}{3}$ )πR³

R³ = 125, so R = 5 cm.

Diameter = 2R = 10 cm.

10 cm

Question 5

(b) Three cubes each of volume 64 cm³ are joined end to end to form a cuboid. Find the surface area of the resulting cuboid.

Answer

Show solution

Each cube has side ∛64 = 4 cm.

The cuboid has dimensions 12 cm × 4 cm × 4 cm.

Surface area = 2(12 × 4 + 4 × 4 + 4 × 12) = 224 cm².

224 cm²

Question 6

(a) Two cubes each of edge 12 cm are joined edge to edge. Find the surface area of the resulting cuboid.

Answer

Show solution

The resulting cuboid has length 24 cm, breadth 12 cm and height 12 cm.

Surface area = 2(lb + bh + hl)

= 2(24 × 12 + 12 × 12 + 12 × 24)

= 1440 cm².

1440 cm²

Question 7

(b) A solid is hemispherical at the bottom and conical of same radius, above it. If the surface areas of the two parts are equal, find the ratio of its radius and the height of its conical part.

Answer

Show solution

Let the common radius be r, cone height h and slant height l. The common circular face is internal.

Equating the exposed curved areas gives 2πr² = πrl, so l = 2r.

Using l² = h² + r² gives 4r² = h² + r², hence h = √3r.

Therefore r : h = 1 : √3.

1 : √3

Question 8

A metallic hollow cylindrical pipe has outer and inner radii as 6 cm and 4 cm respectively. Find the volume of the metal used in the pipe of length 14 cm.

Answer

Show solution

Volume of metal = outer cylinder volume − inner cylinder volume

V = π(R² − r²)h

= ( $\frac{22}{7}$ )(6² − 4²) × 14

= 880 cm³.

880 cm³

Question 9

(a) A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Answer

Show solution

Common radius r = 1 cm and cone height h = 1 cm.

Volume of solid = volume of cone + volume of hemisphere

= ( $\frac{1}{3}$ )πr²h + ( $\frac{2}{3}$ )πr³

= π/3 + 2π/3 = π cm³.

π cm³

Question 10

(b) Find the volume of a solid hemisphere whose total surface area is 462 sq. m.

Answer

Show solution

Total surface area = 3πr² = 462.

r² = 462 × 7/(3 × 22) = 49, so r = 7 m.

Volume = ( $\frac{2}{3}$ )πr³

= ( $\frac{2}{3}$ )( $\frac{22}{7}$ ) × 7³ = $\frac{2156}{3}$ m³.

$\frac{2156}{3}$ m³ ≈ 718.67 m³

Question 11

(a) 150 spherical marbles, each of diameter 1.4 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, and are completely immersed in water. Find the rise in the level of water in the cylindrical vessel.

Answer

Show solution

Marble radius = 0.7 cm and vessel radius = 3.5 cm.

Volume displaced = total volume of the marbles.

π(3.5)²h = 150 × ( $\frac{4}{3}$ )π(0.7)³

h = 150 × ( $\frac{4}{3}$ ) × (0.7)³/(3.5)² = 5.6 cm.

5.6 cm

Question 12

(b) Three cubes of side 6 cm each are joined as shown in the figure. Find the total surface area of the resulting cuboid.

Answer

Show solution

The cuboid has length 18 cm, breadth 6 cm and height 6 cm.

Surface area = 2(lb + bh + hl)

= 2(18 × 6 + 6 × 6 + 6 × 18) = 504 cm².

504 cm²

Question 13

(a) Find the curved surface area of a right circular cone whose height is 15 cm and base radius is 8 cm. [Use π =

\frac{22}{7}

]

Answer

Show solution

Slant height l = √(15² + 8²) = √289 = 17 cm.

Curved surface area = πrl

= ( $\frac{22}{7}$ ) × 8 × 17 = $\frac{2992}{7}$ cm².

$\frac{2992}{7}$ cm² ≈ 427.43 cm²

Question 14

(b) The surface area of a sphere is 616 sq cm. Find its radius. [Use π =

\frac{22}{7}

]

Answer

Show solution

4πr² = 616

r² = 616 × 7/(4 × 22) = 49

r = 7 cm.

7 cm

Question 15

A conical tent has the diameter of the base 24 m and slant height 13 m. Find its volume.

Answer

Show solution

Radius r = $\frac{24}{2}$ = 12 m.

Height h = √(l² − r²) = √(13² − 12²) = 5 m.

Volume = ( $\frac{1}{3}$ )πr²h = ( $\frac{1}{3}$ )π × 12² × 5

= 240π m³ = $\frac{5280}{7}$ m³, using π = 22/7.

240π m³ = $\frac{5280}{7}$ m³

Question 16

(a) The curved surface area of a right circular cylinder is 176 sq cm and its volume is 1232 cu cm. What is the height of the cylinder?

Answer

Show solution

2πrh = 176 and πr²h = 1232.

Dividing volume by curved surface area gives r/2 = $\frac{1232}{176}$ = 7.

Therefore r = 14 cm.

h = 176/(2 × ( $\frac{22}{7}$ ) × 14) = 2 cm.

2 cm

Question 17

(b) The largest sphere is carved out of a solid cube of side 21 cm. Find the volume of the sphere.

Answer

Show solution

The sphere’s diameter equals the cube’s side, so r = $\frac{21}{2}$ = 10.5 cm.

Volume = ( $\frac{4}{3}$ )πr³

= ( $\frac{4}{3}$ )( $\frac{22}{7}$ )(10.5)³ = 4851 cm³.

4851 cm³

Question 18

The total surface area of a solid hemisphere is 462 sq. cm. Find its volume. [Take π =

\frac{22}{7}

]

Answer

Show solution

Total surface area = 3πr² = 462.

r² = 462 × 7/(3 × 22) = 49, so r = 7 cm.

Volume = ( $\frac{2}{3}$ )πr³

= ( $\frac{2}{3}$ )( $\frac{22}{7}$ ) × 7³ = $\frac{2156}{3}$ cm³.

$\frac{2156}{3}$ cm³ ≈ 718.67 cm³

What you need	Starting relationship
Cuboid surface area	2(lb + bh + hl)
Cylinder	Curved surface area = 2πrh; volume = πr²h
Cone	Curved surface area = πrl; volume = ( $\frac{1}{3}$ )πr²h
Slant height of a cone	l² = r² + h²
Sphere	Surface area = 4πr²; volume = ( $\frac{4}{3}$ )πr³
Hemisphere	Curved area = 2πr²; total area = 3πr²; volume = ( $\frac{2}{3}$ )πr³

Surface Areas and Volumes 2-Mark Questions Class 10: CBSE PYQs with Answers

Key results

Surface areas of joined solids

Question 1 2022 · Basic

Question 8 2022 · Basic

Question 10 2022 · Standard

Question 11 2026 · Standard

Volumes and displacement

Question 2 2022 · Basic

Question 4 2022 · Basic

Question 9 2022 · Standard

Dimensions, surface area and volume

Question 3 2022 · Basic

Question 5 2022 · Basic

Question 6 2022 · Standard

Question 7 2022 · Basic

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Frequently asked questions

Make combined solids clearer

Question	Answer
Q1	896 cm²
Q2	880 cm³
Q3(a)	$\frac{2992}{7}$ cm² ≈ 427.43 cm²
Q3(b)	7 cm
Q4(a)	π cm³
Q4(b)	$\frac{2156}{3}$ m³ ≈ 718.67 m³
Q5	240π m³ = $\frac{5280}{7}$ m³
Q6(a)	2 cm
Q6(b)	4851 cm³
Q7	$\frac{2156}{3}$ cm³ ≈ 718.67 cm³
Q8(a)	332.5 cm²
Q8(b)	9 cones
Q9(a)	5.6 cm
Q9(b)	504 cm²
Q10(a)	10 cm
Q10(b)	224 cm²
Q11(a)	1440 cm²
Q11(b)	1 : √3