Surface Area and Volumes for Class 10: Important Questions and Answers

Important questions for Class 10 Maths Surface Area and Volume are available in this article, carefully prepared according to the latest CBSE exam pattern and syllabus guidelines. Students preparing for the CBSE Class 10 board exams can practice these questions to strengthen their conceptual understanding and improve their performance in the examination. This chapter is important from the board exam perspective, as it includes formula-based and application-oriented problems that are frequently asked in exams. Students can also refer to the detailed solutions provided by our subject experts to understand the step-by-step methods clearly. By practicing these important questions for Class 10 Maths Surface Area and Volume, students can revise the chapter thoroughly and become familiar with the types of questions expected in the final Maths paper.

Table of Contents

• Key Concepts of Surface Area and Volume for Class 10
• Topic 1: Surface Area & Volume of a Cylinder
• Topic 2: Surface Area & Volume of a Cone
• Topic 3: Surface Area & Volume of a Sphere / Hemisphere
• Topic 4: Frustum of a Cone
• Topic 5: Combination of Solids
• Topic 6: Conversion / Melting & Recasting
• Case-Based Questions

Key Concepts of Surface Area and Volume for Class 10

Topic 1: Surface Area & Volume of a Cylinder

The most repeated question types in cylinders involve finding the surface area of hollow cylinders (like pipes) and calculating volumes in water-flow problems.

Q1: 2 Marks

The diameter of a cylindrical water tank is 1.4 m and its height is 2.1 m. Find the capacity of the tank in litres. (Use π = 22/7)

Solution: Radius r = 1.4/2 = 0.7 m, height h = 2.1 m

Volume = πr²h = (22/7) × 0.7 × 0.7 × 2.1

= (22/7) × 0.49 × 2.1 = 22 × 0.49 × 0.3 = 3.234 m³

1 m³ = 1000 litres

⇒ Capacity = 3234 litres

The capacity of water tank is 3234 liters.

Q2: 3 Marks

A canal is 300 cm wide and 120 cm deep. Water flows through it at a speed of 20 km/h. How much area (in m²) will it irrigate in 20 minutes if 8 cm of standing water is required?

Solution: Given, Width = 3 m, Depth = 1.2 m, Speed = 20,000 m/h

Volume of water flowing per hour = 3 × 1.2 × 20,000 = 72,000 m³

⇒ Volume in 20 minutes = 72,000 × (20/60) = 24,000 m³

Standing water depth = 8 cm = 0.08 m

Area irrigated = Volume / depth = 24,000 / 0.08 = 3,00,000 m²

Area irrigated in 20 minutes with 8cm standing water  = 3,00,000 m²

Q3: 2 Marks

If the diameter of a wire is decreased by 5%, by what percentage must its length be increased so that the volume remains constant?

Solution: Let original radius = R. 

New radius = 0.95R (5% decrease in diameter means 5% decrease in radius).

For volume to be constant: πR²H = π(0.95R)²h

⇒ h = H / (0.95)² = H / 0.9025

⇒ Increase = h − H = H(1/0.9025 − 1) = H × 0.1080

Therefore, % increase ≈ 10.8%

Topic 2: Surface Area & Volume of a Cone

Frequently asked questions about cones usually involve topics like slant height, CSA, or volume of Cone

Q1: 2 Marks

A solid right circular cone is cut into two parts at the middle of its height by a plane parallel to the base. Find the ratio of the volumes of the smaller cone (top) to the larger piece (frustum).

Solution: Let the original cone have height H and radius R. 

At mid-height (H/2), by similar triangles, the radius of the top cone = R/2.

Volume of small cone = (1/3)π(R/2)²(H/2) = πR²H/24

Volume of original cone = (1/3)πR²H = πR²H/3

⇒ Volume of frustum = πR²H/3 − πR²H/24 = (8 − 1)πR²H/24 = 7πR²H/24

⇒ Ratio = (πR²H/24) : (7πR²H/24) = 1 : 7

Q2: 3 Marks

A conical vessel with internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height to which water rises in the cylindrical vessel.

Solution: Volume of water in cone = (1/3)πr²h 

= (1/3) × π × 25 × 24 = 200π cm³

Let H = height water rises in cylinder.

Volume of water in cylinder = π × 10² × H = 100πH cm³

Equating: 100πH = 200π

⇒ H = 2 cm

Q3: 3 Marks

A right circular cone of height 3.6 cm and base radius 1.6 cm is melted and recast into a right circular cone with base radius 1.2 cm. Find the height of the new cone.

Solution: Volume of original cone = (1/3)π × (1.6)² × 3.6 

= (1/3)π × 2.56 × 3.6

Let new height = H.

⇒ Volume of new cone = (1/3)π × (1.2)² × H 

= (1/3)π × 1.44 × H

Since volumes are equal:

2.56 × 3.6 = 1.44 × H

⇒ H = (2.56 × 3.6) / 1.44 

= 9.216 / 1.44 = 6.4 cm

Topic 3: Surface Area & Volume of a Sphere / Hemisphere

Q1: 2 Marks

A sphere of diameter 18 cm is dropped into a cylindrical vessel of diameter 36 cm, partly filled with water. If the sphere is completely submerged, find the rise in water level.

Solution: Given, Sphere: radius r = 9 cm

Cylinder: radius R = 18 cm, rise = h

Volume of sphere = Volume of water risen in cylinder

⇒ (4/3)πr³ = πR²h

⇒ (4/3) × 9³ = 18² × h

⇒ (4/3) × 729 = 324h

⇒ 972 = 324h

⇒ h = 3 cm

Q2: 2 Marks

A solid sphere of radius 3 cm is melted and recast into small balls each of radius 0.3 cm. Find the number of balls formed.

Solution: Given radius of the solid sphere = 3 cm and radius of small balls = 0.3 cm

⇒ Volume of large sphere = (4/3)π × 3³ = 36π cm³

Volume of each small ball = (4/3)π × (0.3)³ = (4/3)π × 0.027 = 0.036π cm³

⇒ Number of balls = 36π / 0.036π = 1000 balls

Therefore the number of balls formed  = 1000 balls

Q3: 2 Marks

The diameter of a sphere is 28 cm. Find the cost of painting its entire surface at ₹0.10 per cm². (Use π = 22/7)

Solution: Given, Radius = 28/2 = 14 cm

⇒ Surface area of sphere = 4πr² = 4 × (22/7) × 14 × 14 

= 4 × 22 × 2 × 14 = 2464 cm²

⇒ Cost = 2464 × 0.10 = ₹246.40

Therefore, the cost of painting the entire surface is ₹246.40.

Q4: 3 Marks

A sphere of diameter 6 cm is dropped into a right circular cylindrical vessel with diameter 12 cm, partly filled with water. The sphere is completely submerged. By how much will the water level rise?

Solution: Given: 

Sphere: r = 3 cm

Cylinder: R = 6 cm, rise = h

Volume of sphere = volume of water risen

⇒ (4/3)π(3)³ = π(6)²h

⇒(4/3) × 27 = 36h

⇒36 = 36h

⇒h = 1 cm

Topic 4: Frustum of a Cone

Q1: 3 Marks

An open metal container is in the form of a frustum of a cone with height 8 cm, lower radius 4 cm and upper radius 10 cm. Find the cost of oil that can completely fill it at ₹50 per litre. Also find the cost of metal used if it costs ₹5 per 100 cm². (Use π = 3.14)

Solution: Given h = 8 cm, R = 10 cm, r = 4 cm

Slant height l = √[8² + (10−4)²] = √[64 + 36] = √100 = 10 cm

Volume (capacity):

V = (πh/3)(R² + r² + Rr) = (3.14 × 8/3)(100 + 16 + 40)

= (25.12/3) × 156 = 8.373 × 156 = 1306.2 cm³

= 1.3062 litres

Cost of oil = 1.3062 × 50 = ₹65.31

 

CSA of frustum:

CSA = π(R + r)l = 3.14 × 14 × 10 = 439.6 cm²

Area of base = πr² = 3.14 × 16 = 50.24 cm²

Total metal area = 439.6 + 50.24 = 489.84 cm²

Cost of metal = (489.84/100) × 5 = ₹24.49

Q2: 3 Marks

The radii of the top and bottom of a bucket (in the form of a frustum of a cone) are 10 cm and 6 cm respectively, and its height is 8 cm. Find the curved surface area and total surface area of the bucket.

Solution: Given: R = 10 cm, r = 6 cm, h = 8 cm

l = √[h² + (R−r)²] = √[64 + 16] 

= √80 = 4√5 cm ≈ 8.94 cm

CSA = π(R + r)l = π × 16 × 4√5 

= 64√5 π ≈ 449.9 cm²

TSA = CSA + πR² (bottom only, open top)

= 64√5 π + π × 100

= π(64√5 + 100) ≈ π × (143.1 + 100)

≈ 763.4 cm²

Topic 5: Combination of Solids

The rule to remember: when calculating surface area of a combined solid, you only count the exposed surfaces.

Q1: 5 Marks

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The cone's height is 4 cm and the diameter of the base is 8 cm. Find: (a) the total volume of the toy, and (b) the total surface area of the toy. (Use π = 3.14)

Solution: Given: r = 4 cm (common radius), height of cone h = 4 cm

Slant height of cone l = √(4² + 4²) = √32 = 4√2 cm

 

(a) Volume of toy:

Volume = Volume of cone + Volume of hemisphere

= (1/3)πr²h + (2/3)πr³

= (1/3) × 3.14 × 16 × 4 + (2/3) × 3.14 × 64

= 66.99 + 134.04 = ≈ 201.06 cm³

 

(b) Total Surface Area:

TSA = CSA of cone + CSA of hemisphere (flat circle is not exposed)

= πrl + 2πr²

= 3.14 × 4 × 4√2 + 2 × 3.14 × 16

= 3.14 × 16√2 + 100.48

= 3.14 × 22.63 + 100.48

= 71.05 + 100.48 = ≈ 171.53 cm²

Q2: 5 Marks

An iron pillar has a cylindrical part of height 240 cm and radius 8 cm, and a conical part of height 36 cm with the same base radius. Find the weight of the pillar if 1 cm³ of iron weighs 7.8 g. (Use π = 22/7)

Solution: Volume of cylinder = πr²h = (22/7) × 64 × 240 

= (22 × 64 × 240)/7 = 48,274.3 cm³

Volume of cone = (1/3)πr²h = (1/3) × (22/7) × 64 × 36 

= (22 × 64 × 36)/(3 × 7) = 2,413.7 cm³

⇒ Total Volume = 48,274.3 + 2,413.7 = 50,688 cm³

⇒ Weight = 50,688 × 7.8 = 395,366.4 g = ≈ 395.37 kg

Therefore, weight of the pillar is ≈ 395.37 kg

Topic 6: Conversion / Melting & Recasting

Q1: 2 Marks

Three metallic solid cubes with edges 3 cm, 4 cm and 5 cm are melted and formed into a single cube. Find the edge of the new cube formed.

Solution:

Volume of new cube = 3³ + 4³ + 5³ = 27 + 64 + 125 = 216 cm³

Edge of new cube = ∛216 = 6 cm

Q2: 2 Marks

Find the number of spherical bullets each of diameter 4 cm that can be made from a solid lead cube with edge 44 cm.

Solution: Given diameter = 4cm ⇒ radius = 2cm.

Volume of cube = 44³ = 85,184 cm³

Volume of each bullet = (4/3)π × 2³ = (4/3) × (22/7) × 8 = 33.52 cm³

⇒ Number of bullets = 85,184 / 33.52 ≈ 2541 bullets

Q3: 3 Marks

Find the number of coins, each 1.5 cm in diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

Solution: Each coin is a small cylinder.

Volume of one coin = π × (0.75)² × 0.2 = 0.1125π cm³

Volume of large cylinder = π × (2.25)² × 10 = 50.625π cm³

⇒ Number of coins = 50.625π / 0.1125π 

= 50.625/0.1125 = 450 coins

Q4: 3 Marks

A well 7 m in diameter is dug 22.5 m deep. The earth dug out is spread evenly around the well to form an embankment of height 1.5 m. Find the width of the embankment. (Use π = 22/7)

Solution: Given, diameter = 7m and h = 22.5m

Volume of earth dug = π × (3.5)² × 22.5 = (22/7) × 12.25 × 22.5 = 866.25 m³

Let width of embankment = x m. 

Outer radius = (3.5 + x) m, inner radius = 3.5 m.

⇒ Volume of embankment = π[(3.5 + x)² − 3.5²] × 1.5

⇒ (22/7) × [(3.5+x)² − 12.25] × 1.5 = 866.25

⇒ [(3.5+x)² − 12.25] = 866.25 × 7 / (22 × 1.5) = 6063.75/33 = 183.75

⇒ (3.5+x)² = 183.75 + 12.25 = 196

⇒ 3.5 + x = 14

⇒ x = 10.5 m

Therefore, the width of the embankment is 10.5 m

Case-Based Questions

Case Study 1: 4 Marks

A school is building an adventure rock-climbing structure. The structure is in the form of a cylinder surmounted by a cone. The cylinder has a height of 12 m and base radius of 3 m. The cone on top has a height of 4 m with the same base radius.

(i) Find the slant height of the conical top.

Solution: l = √(r² + h²) = √(9 + 16) = √25 = 5 m

(ii) Find the total volume of the structure.

Solution: Volume of cylinder = π × 3² × 12 = 108π m³

Volume of cone = (1/3) × π × 3² × 4 = 12π m³

Total Volume = 120π = 120 × 3.14 ≈ 376.8 m³

(iii) Find the total curved surface area to be painted.

Solution: CSA of cylinder = 2πrh = 2π × 3 × 12 = 72π m²

CSA of cone = πrl = π × 3 × 5 = 15π m²

Total CSA = 87π ≈ 273.18 m²

Case Study 2: 4 Marks

A candle manufacturer makes decorative candles in the shape of a cylinder with a hemispherical top. Each candle has a total height of 10 cm and the radius of both the cylinder and hemisphere is 3 cm.

(i) What is the height of the cylindrical part?

Solution: Total height = 10 cm, radius of hemisphere = 3 cm

Height of cylinder = 10 − 3 = 7 cm

(ii) Find the total volume of wax in one candle. 

Solution: Volume of cylinder = π × 9 × 7 = 63π cm³

Volume of hemisphere = (2/3)π × 27 = 18π cm³

Total = 81π = 81 × 3.14 ≈ 254.34 cm³

(iii) If 10 such candles are made from a large cylindrical block of wax of diameter 6 cm and height 300 cm, is there enough wax? 

Solution: Volume of block = π × 9 × 300 = 2700π cm³

Volume for 10 candles = 10 × 81π = 810π cm³

2700π > 810π 

⇒ there is more than enough wax.

Case Study 3: 4 Marks

A village has a water distribution system where water is stored in a large hemispherical tank of radius 7 m. It is pumped into a cylindrical overhead tank of radius 3.5 m. (Use π = 22/7)

(i) Find the volume of the hemispherical storage tank. 

Solution: V = (2/3)πr³ = (2/3) × (22/7) × 343 = 718.67 m³

(ii) If the entire water from the hemispherical tank fills the cylindrical tank, find the height of water in the cylindrical tank. 

Solution: πR²h = (2/3)πr³

⇒ (22/7) × 12.25 × h = 718.67

⇒ 38.5h = 718.67

⇒ h ≈ 18.67 m

(iii) What is the curved surface area of the cylindrical overhead tank if its height is 18.67 m?

Solution: CSA = 2πrh = 2 × (22/7) × 3.5 × 18.67 = 22 × 18.67 ≈ 410.74 m²