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SparkEd Math offers 60+ practice questions for Surface Areas & Volumes across 3 difficulty levels (Easy, Medium, Hard) plus a timed test mode. Each question comes with visual step-by-step solutions and AI coaching hints.

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SparkEd Math uses visual step-by-step solutions for every Surface Areas & Volumes problem, making complex concepts easy to understand. The platform includes 3 difficulty levels for progressive learning, a Super Power Help AI coach for instant doubt-clearing, and interactive practice rather than passive reading.

Is SparkEd's Surface Areas & Volumes content aligned with the CBSE syllabus?

Yes, all Surface Areas & Volumes questions and lessons on SparkEd Math are fully aligned with the Central Board of Secondary Education (CBSE) syllabus for Class 10. The content follows the official curriculum structure and exam pattern.

Surface Areas & Volumes — Class 10 CBSE (2025-26)

About Surface Areas & Volumes — Class 10 CBSE

Find surface area and volume of combinations of solids and conversion between solids. This topic is part of the CBSE Class 10 mathematics syllabus (chapter: Chapter 12). On this page you can practice 61 questions across three difficulty levels — 20 easy, 21 medium, and 20 hard — each with a visual step-by-step solution, plus a timed 36-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

What you'll learn in Surface Areas & Volumes

Introduction to Surface Areas & Volumes: The Basics
Surface Areas of Combinations of Solids
Volumes of Combinations of Solids
Conversion of Solids: Reshaping and Recasting
Summary, Connections, and Preparation for Practice

Interactive lesson · about 15 minutes · checkpoint question after every unit

Surface Areas & Volumes — solved examples for Class 10 CBSE

Example 1easy

A toy is in the shape of a cone mounted on a hemisphere of the same base radius. Which expression correctly represents the total surface area of the toy?

A)CSA of cone + CSA of hemisphere
B)CSA of cone + TSA of hemisphere
C)TSA of cone + CSA of hemisphere
D)TSA of cone + TSA of hemisphere - area of common base

Step-by-step solution

The total surface area of a combined solid is the sum of the curved (or exposed) surface areas of its individual components.
When a cone is mounted on a hemisphere, the circular base of the cone and the circular top of the hemisphere are joined together, becoming internal surfaces.
Therefore, the exposed surface area consists only of the curved surface area of the cone and the curved surface area of the hemisphere.

Answer: CSA of cone + CSA of hemisphere

Example 2medium

A decorative block is made of two solids: a cube and a hemisphere. The base of the block is a cube with edge length 5 cm, and the hemispherical part is fixed on the top of the cube. If the diameter of the hemisphere is 4.2 cm, find the total surface area of the block. (Use π = 22/7)

A)163.86 cm²
B)166.32 cm²
C)169.88 cm²
D)172.44 cm²

Step-by-step solution

Edge length of cube (a) = 5 cm. Surface area of cube = 6a² = 6 × (5)² = 6 × 25 = 150 cm².
Diameter of hemisphere = 4.2 cm, so radius (r) = 4.2 / 2 = 2.1 cm. Area of base of hemisphere = πr² = (22/7) × (2.1)² = (22/7) × 4.41 = 22 × 0.63 = 13.86 cm².
Curved surface area of hemisphere = 2πr² = 2 × (22/7) × (2.1)² = 2 × 13.86 = 27.72 cm².
Total surface area of the block = Surface area of cube - Area of base of hemisphere + Curved surface area of hemisphere = 150 - 13.86 + 27.72 = 163.86 cm².

Answer: 163.86 cm²

Example 3hard

A solid is formed by placing a hemisphere on top of a cylinder. The radius of the hemisphere is equal to the radius of the cylinder. If the total surface area of the combined solid is 4πr² and 'r' is the common radius, what is the height 'h' of the cylinder?

A)r
B)2r
C)r/2
D)r/3

Step-by-step solution

Let 'r' be the common radius of the cylinder and hemisphere, and 'h' be the height of the cylinder.
Total surface area (TSA) of the combined solid = Curved Surface Area (CSA) of cylinder + CSA of hemisphere + Area of base of cylinder.
TSA = 2πrh + 2πr² + πr² = 2πrh + 3πr².
Given TSA = 4πr². Equating the two expressions: 2πrh + 3πr² = 4πr². Subtracting 3πr² from both sides gives 2πrh = πr². Dividing by 2πr (since r ≠ 0) yields h = r/2.

Answer: r/2

Practice questions on Surface Areas & Volumes

Q1.easy
A solid is formed by placing a cylinder on top of another larger cylinder. To find the total volume of this combined solid, what approach should be used?
1. A)Add the Curved Surface Areas of both cylinders.
2. B)Add the Volumes of both cylinders.
3. C)Add the Total Surface Areas of both cylinders.
4. D)Add the Volume of the larger cylinder and the Curved Surface Area of the smaller cylinder.
Show answer
Answer: Add the Volumes of both cylinders.
Hint: Volume measures the space occupied by a solid. When combining solids, their individual volumes simply add up.
Q2.easy
Ravi is calculating the total surface area of a cubical block surmounted by a hemisphere. He uses the formula: Total Surface Area = (Total Surface Area of Cube) + (Curved Surface Area of Hemisphere). What mistake, if any, did Ravi make?
1. A)No mistake, the formula is correct.
2. B)He should have subtracted the area of the base of the hemisphere from the surface area of the cube.
3. C)He should have added the Total Surface Area of the hemisphere instead of the Curved Surface Area.
4. D)He should have used only the Curved Surface Area of the cube.
Show answer
Answer: He should have subtracted the area of the base of the hemisphere from the surface area of the cube.
Hint: When the hemisphere is placed on the cubical block, the area of the top face of the cube covered by the hemisphere is no longer exposed.
Q3.easy
A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 6 cm and the diameter of the base is 4 cm. What is the volume of the toy? (Use π = 22/7)
1. A)25.14 cm³
2. B)33.52 cm³
3. C)41.90 cm³
4. D)50.28 cm³
Show answer
Answer: 41.90 cm³
Hint: Calculate the radius from the diameter. Then, find the volume of the cone and the hemisphere separately and add them.
Q4.medium
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. (Use π = 3.14)
1. A)22.12 cm³
2. B)25.12 cm³
3. C)28.26 cm³
4. D)30.14 cm³
Show answer
Answer: 25.12 cm³
Hint: The volume of the toy is the sum of the volume of the hemisphere and the volume of the cone. Ensure you use the correct radius for both parts.
Q5.medium
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. What is the height of the cylinder? (Use π = 22/7)
1. A)2.744 cm
2. B)3.120 cm
3. C)3.872 cm
4. D)4.016 cm
Show answer
Answer: 2.744 cm
Hint: When a solid is melted and recast into another shape, its volume remains constant. Equate the volume of the sphere to the volume of the cylinder.
Q6.medium
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. If the radius of the cylinder is 60 cm and its height is 180 cm, find the volume of water left in the cylinder. (Use π = 22/7)
1. A)102432 cm³
2. B)108432 cm³
3. C)110880 cm³
4. D)113142.86 cm³
Show answer
Answer: 113142.86 cm³
Hint: The volume of water left in the cylinder is the volume of the cylinder minus the volume of the solid toy (cone + hemisphere) submerged in it.
Q7.hard
A solid metallic cuboid of dimensions 20 cm × 10 cm × 8 cm is melted and recast into a solid cylinder of radius 5 cm. If only 80% of the metal is used in the recasting process, what is the height of the cylinder formed? (Assume π = 22/7)
1. A)14 cm
2. B)16.29 cm
3. C)18.5 cm
4. D)20 cm
Show answer
Answer: 16.29 cm
Hint: First, calculate the volume of the cuboid. Then, find 80% of this volume, which will be the volume of the cylinder. Finally, use the cylinder's volume formula to find its height.
Q8.hard
A solid right circular cylinder of height 'H' and radius 'R' is melted and recast into 'N' identical cones, each of height 'h' and radius 'r'. If h = H/2 and r = R/2, what is the value of N?
1. A)8
2. B)12
3. C)16
4. D)24
Show answer
Answer: 24
Hint: The total volume of the 'N' cones must be equal to the volume of the cylinder. Express the volume of a single cone in terms of H and R using the given ratios.
Q9.hard
Water flows through a cylindrical pipe of internal diameter 7 cm at the rate of 192.5 litres per minute. In what time (in minutes) will this pipe fill a conical tank whose base radius is 1.4 m and height is 3 m? (Use π = 22/7)
1. A)25 minutes
2. B)30 minutes
3. C)32 minutes
4. D)35 minutes
Show answer
Answer: 32 minutes
Hint: Ensure all dimensions are in consistent units (e.g., cm). Convert litres to cm³ and then calculate the volume of the conical tank. Time will be total volume divided by the flow rate.

These are 9 of the 61 questions available for Surface Areas & Volumes. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.