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Class 10 Surface Areas and Volumes: 8 practice questions with answers, step by step
These are 8 practice questions on NCERT Class 10 Maths Chapter 12, Surface Areas and Volumes, each with a worked answer and the units. They cover the basic solids (cylinder, cone, sphere) and the idea the chapter is about: solids made by joining two basic solids. Use π = 22/7 in every question. Try each one on paper before you open the answer. The Class 9 version of this chapter, with a cuboid, a cylinder, a cone and a sphere, is Class 9 Surface Area and Volume.
These questions are our own, drafted with AI help. Every answer was re-computed by a test script, but no practising teacher has reviewed them yet. If you find a mistake, please tell us.
What the chapter says you need
Facts from the NCERT textbook (Mathematics, Textbook for Class X, Reprint 2026-27, Chapter 12, printed pages 161 to 170):
- The chapter takes the surface areas and volumes of the basic solids (cuboid, cone, cylinder, sphere) as already known from Class IX (section 12.1, page 161). The formulas used below are the ones you know: cylinder volume πr²h and curved surface area 2πrh; cone volume (1/3)πr²h; sphere volume (4/3)πr³; hemisphere volume (2/3)πr³ and curved surface area 2πr².
- Surface area of a combination (section 12.2, page 162): the total surface area of the new solid is the sum of the curved surface areas of its parts, because some of the surface disappears where the solids are joined. The textbook's container is a cylinder with two hemispheres at its ends: TSA = CSA of one hemisphere + CSA of the cylinder + CSA of the other hemisphere.
- Volume of a combination (section 12.3, page 167): the volume of the solid formed by joining two basic solids is the sum of the volumes of its parts.
- The textbook's toy of a cone on a hemisphere uses the same radius for both, so the surface is smooth (page 163).
Basic solids
Question 1. Find the volume of a cylinder of radius 7 cm and height 10 cm, taking π = 22/7, in cubic cm.
Answer: 1,540 cm³. V = πr²h = 22/7 × 7 × 7 × 10 = 22 × 7 × 10 = 1,540 cm³.
Question 2. Find the curved surface area of the same cylinder, in square cm.
Answer: 440 cm². CSA = 2πrh = 2 × 22/7 × 7 × 10 = 2 × 22 × 10 = 440 cm².
Question 3. Find the volume of a cone of radius 7 cm and height 12 cm, taking π = 22/7, in cubic cm.
Answer: 616 cm³. V = (1/3)πr²h = 1/3 × 22/7 × 49 × 12 = 22 × 7 × 4 = 616 cm³.
Question 4. Find the volume of a sphere of radius 21 cm, taking π = 22/7, in cubic cm.
Answer: 38,808 cm³. V = (4/3)πr³ = 4/3 × 22/7 × 21 × 21 × 21 = 4 × 22 × 21 × 21 = 4 × 22 × 441 = 38,808 cm³.
Combined solids
Question 5. A toy is a cone of height 13 cm standing on a hemisphere of radius 7 cm (the cone has the same radius). Taking π = 22/7, what is the volume of the toy in cubic cm?
Answer: 1,386 cm³. The volume is the sum of the parts (page 167). Cone: 1/3 × 22/7 × 7 × 7 × 13 = 2002/3. Hemisphere: 2/3 × 22/7 × 7 × 7 × 7 = 2156/3. Total = 4158/3 = 1,386 cm³.
Question 6. A solid is made of a cylinder of radius 7 cm and height 10 cm with a hemisphere of the same radius fixed on top. Taking π = 22/7, what is the volume of the solid in cubic cm, to the nearest whole number?
Answer: 2,259 cm³. Cylinder: πr²h = 22/7 × 49 × 10 = 1,540. Hemisphere: (2/3)πr³ = 2/3 × 22/7 × 343 ≈ 718.67. Total ≈ 2,258.67, which is 2,259 cm³ to the nearest whole number.
Question 7. A container is a cylinder of radius 7 cm and length 10 cm with a hemisphere of the same radius at each end. Taking π = 22/7, find its total surface area in square cm.
Answer: 1,056 cm². Only the curved surfaces show (page 162), so TSA = CSA of cylinder + CSA of the two hemispheres. Cylinder: 2πrh = 440 cm². Each hemisphere: 2πr² = 2 × 22/7 × 49 = 308 cm², so two give 616 cm². Total = 440 + 616 = 1,056 cm².
Question 8. Find the volume of the same container in cubic cm, to two decimal places.
Answer: 2,977.33 cm³. Volume = sum of the parts (page 167). Cylinder: 1,540 cm³. The two hemispheres together make one sphere of radius 7 cm: (4/3) × 22/7 × 343 = 4,312/3 ≈ 1,437.33 cm³. Total ≈ 1,540 + 1,437.33 = 2,977.33 cm³.
Practise more, and check your marks
- Free practice, marked instantly in your browser: Class 10 Surface Areas and Volumes questions. Questions 1 to 6 above are all in this set.
- Printable worksheets with answer keys, in English, Hindi and Marathi: Class 10 worksheets.
- See which unit this chapter belongs to and how CBSE splits the 80 theory marks in our Class 10 Maths syllabus 2026-27 guide.
- The same step-by-step style for neighbouring chapters: Class 10 Triangles and Class 10 Introduction to Trigonometry.
Where CMDLearn fits
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Sources
All opened and read on 21 October 2026. Page numbers are the printed page numbers of the textbook.
- NCERT, Mathematics, Textbook for Class X, Chapter 12 "Surface Areas and Volumes" (Reprint 2026-27), printed pages 161-170 (the basic solids known from Class IX, section 12.1, page 161; the surface area of a combination of solids and the container with two hemispheres, section 12.2, page 162; the toy of a cone on a hemisphere, page 163; the volume of a combination of solids, section 12.3, page 167; summary, page 170): ncert.nic.in/textbook/pdf/jemh112.pdf
The rules for combined solids are from the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the official CBSE notices.