Practice Worksheets for Class 10 Mathematics: Chapter 12 Surface Area and Volume
Review targeted practice sets for Class 10 Mathematics Chapter 12 Surface Area and Volume. Built according to official educational guidelines for the 2026-27 academic year, these downloadable worksheets support daily revision and core concept reinforcement.
Practice Chapter 12 Surface Area and Volume Worksheets for Class 10 Mathematics
View or download the dedicated Chapter 12 Surface Area and Volume practice resource below. Engaging with these objective and subjective questions daily ensures continuous academic progress and mastery of the 2026-27 curriculum.
Questions
Part A: Short Answer Questions
Q1. The surface area of a sphere is 616 cm². Find its radius. (π = 22/7)
View Answer & Explanation
Answer: 4πr² = 616, so r² = 49 and r = 7 cm.
Teacher's Note:
1. Substitute π = 22/7 in the sphere's surface area formula.
2. Take the square root.
Q2. A cylinder and a cone have the same base radius r and the same height r. Find the ratio of the curved surface area of the cylinder to that of the cone.
View Answer & Explanation
Answer: Cylinder CSA = 2πr × r = 2πr². Slant height of the cone = √(r² + r²) = √2 r, so cone CSA = πr × √2 r = √2 πr². Ratio = 2 : √2 = √2 : 1.
Teacher's Note:
1. Find the slant height of the cone first.
2. Divide the two curved surface areas.
Q3. Two cones have heights in the ratio 1 : 3 and base radii in the ratio 3 : 1. Find the ratio of their volumes.
View Answer & Explanation
Answer: Volume ∝ r²h, so ratio = (3² × 1) : (1² × 3) = 9 : 3 = 3 : 1.
Teacher's Note:
1. The factor ⅓π is common and cancels.
2. Compare r²h for both cones.
Q4. The radii of two cones are in the ratio 2 : 1 and their volumes are equal. Find the ratio of their heights.
View Answer & Explanation
Answer: r₁²h₁ = r₂²h₂, so h₁/h₂ = (r₂/r₁)² = 1/4. The ratio of heights is 1 : 4.
Teacher's Note:
1. Equal volumes mean r²h is equal.
2. Square the inverse ratio of radii.
Q5. Find the curved surface area of a right circular cone of height 15 cm and base diameter 16 cm. (π = 22/7)
View Answer & Explanation
Answer: r = 8 cm and l = √(15² + 8²) = 17 cm. CSA = πrl = (22/7) × 8 × 17 = 136π ≈ 427.43 cm².
Teacher's Note:
1. Find the slant height with Pythagoras theorem.
2. Use CSA = πrl.
Q6. Find the volume of the largest cone that can be cut from a solid hemisphere of radius r.
View Answer & Explanation
Answer: The cone has base radius r and height r, so volume = ⅓πr²·r = \( \dfrac{1}{3} \)πr³.
Teacher's Note:
1. The biggest cone has the same base as the hemisphere.
2. Its height equals the radius.
Q7. Two cubes, each of volume 64 cm³, are joined end to end. Find the volume and the total surface area of the resulting cuboid.
View Answer & Explanation
Answer: Edge = 4 cm, so the cuboid is 8 cm × 4 cm × 4 cm. Volume = 128 cm³. Surface area = 2(32 + 16 + 32) = 160 cm².
Teacher's Note:
1. Find the edge of each cube from its volume.
2. The joined solid has length 8 cm.
Q8. The largest sphere is carved out of a cube of side 7 cm. Find the volume of the sphere. (π = 22/7)
View Answer & Explanation
Answer: Diameter = 7 cm, so r = 3.5 cm. Volume = (4/3)(22/7)(3.5)³ = 179.67 cm³.
Teacher's Note:
1. The sphere's diameter equals the cube's edge.
2. Use V = (4/3)πr³.
Part B: Combination of Solids
Q9. A circus tent is cylindrical up to a height of 3 m and conical above it. The base diameter is 105 m and the slant height of the conical part is 53 m. Find the canvas used. (π = 22/7)
View Answer & Explanation
Answer: r = 52.5 m. Cylindrical part = 2 × (22/7) × 52.5 × 3 = 990 m². Conical part = (22/7) × 52.5 × 53 = 8745 m². Total canvas = 9735 m².
Teacher's Note:
1. Canvas covers only the curved surfaces.
2. Add the cylinder and the cone.
Q10. A vessel is a hemispherical bowl surmounted by a hollow cylinder of the same diameter. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find its capacity. (π = 22/7)
View Answer & Explanation
Answer: r = 7 cm and cylinder height = 13 − 7 = 6 cm. Volume = π × 49 × 6 + ⅔π × 343 = (22/7)(294 + 228.67) = 1642.67 cm³.
Teacher's Note:
1. Subtract the hemisphere's height from the total height.
2. Add the volumes of the cylinder and the hemisphere.
Q11. A solid toy is a right circular cylinder with a hemisphere at one end and a cone at the other. The common diameter is 4.2 cm; the heights of the cylindrical and conical parts are 12 cm and 7 cm. Find the volume of the toy. (π = 22/7)
View Answer & Explanation
Answer: r = 2.1 cm. Cylinder = (22/7)(4.41)(12) = 166.32; hemisphere = ⅔(22/7)(9.261) = 19.404; cone = ⅓(22/7)(4.41)(7) = 32.34. Total = 218.064 cm³.
Teacher's Note:
1. Compute the three volumes separately.
2. Add them.
Q12. A solid is a cone standing on a hemisphere, both with radius 1 cm, and the height of the cone equals its radius. Find the volume in terms of π.
View Answer & Explanation
Answer: Volume = ⅓π(1)²(1) + ⅔π(1)³ = π/3 + 2π/3 = π cm³.
Teacher's Note:
1. The cone's height is 1 cm.
2. Add the cone and hemisphere volumes.
Q13. A hemispherical depression is cut from one face of a cubical block so that the diameter of the hemisphere equals the edge l of the cube. Find the surface area of the remaining solid.
View Answer & Explanation
Answer: Surface = 6l² − πr² + 2πr² with r = l/2, i.e. 6l² + πl²/4 = \( \dfrac{l^2}{4} \)(24 + π).
Teacher's Note:
1. The circular opening is removed and a curved surface is added.
2. Substitute r = l/2.
Q14. A glass of inner diameter 5 cm and height 10 cm has a hemispherical raised portion at the bottom. Find its apparent capacity and actual capacity. (π = 3.14)
View Answer & Explanation
Answer: Apparent capacity = 3.14 × 2.5² × 10 = 196.25 cm³. Hemisphere = ⅔ × 3.14 × 2.5³ = 32.71 cm³. Actual capacity = 196.25 − 32.71 = 163.54 cm³.
Teacher's Note:
1. Apparent capacity ignores the raised portion.
2. Subtract the hemisphere's volume for the actual capacity.
Part C: Long Answer Questions
Q15. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume is 1/27 of the volume of the given cone, at what height above the base is the section made?
View Answer & Explanation
Answer: Similar cones: volume ratio = (h/30)³ = 1/27, so h = 10 cm. The section is 30 − 10 = 20 cm above the base.
Teacher's Note:
1. Volumes of similar cones are in the cube ratio of their heights.
2. Subtract the small cone's height from 30 cm.
Q16. A solid wooden toy is a right circular cone mounted on a hemisphere of radius 4.2 cm. The total height of the toy is 10.2 cm. Find its volume. (π = 22/7)
View Answer & Explanation
Answer: Cone height = 10.2 − 4.2 = 6 cm. Volume = ⅔(22/7)(4.2)³ + ⅓(22/7)(4.2)²(6) = 155.23 + 110.88 = 266.11 cm³.
Teacher's Note:
1. Find the height of the cone part.
2. Add the two volumes.
Q17. A tent is a right circular cylinder up to a height of 3 m and then a cone, with a maximum height of 13.5 m above the ground. The radius is 14 m. Find the cost of painting the inner side at Rs. 2 per square metre. (π = 22/7)
View Answer & Explanation
Answer: Cone height = 10.5 m, slant height = √(14² + 10.5²) = 17.5 m. Area = 2π(14)(3) + π(14)(17.5) = 264 + 770 = 1034 m². Cost = 1034 × 2 = Rs. 2068.
Teacher's Note:
1. Find the slant height of the cone.
2. Add the two curved areas and multiply by the rate.
Q18. A pen stand is a cuboid 15 cm × 10 cm × 3.5 cm with four conical depressions, each of diameter 1 cm and depth 1.4 cm. Find the volume of wood in the stand. (π = 22/7)
View Answer & Explanation
Answer: Cuboid = 525 cm³. One cone = ⅓(22/7)(0.5)²(1.4) = 0.3667 cm³, so four = 1.4667 cm³. Wood = 525 − 1.47 = 523.53 cm³.
Teacher's Note:
1. Subtract the volume of the four cones.
2. Each cone has radius 0.5 cm.
Q19. Three cubes, each of edge 5 cm, are joined end to end. Find the total surface area of the resulting cuboid.
View Answer & Explanation
Answer: The cuboid is 15 cm × 5 cm × 5 cm. Surface area = 2(75 + 75 + 25) = 350 cm².
Teacher's Note:
1. Joined faces disappear, so use the dimensions of the new cuboid.
2. Apply 2(lb + bh + hl).
Free study material for Mathematics
Download CBSE Practice Material: Class 10 Mathematics Chapter 12 Surface Area and Volume
Download Practice Sheet: Chapter 12 Surface Area and Volume (Class 10 Mathematics)
Access structured practice worksheets for Chapter 12 Surface Area and Volume designed in alignment with the latest CBSE curriculum for Class 10 Mathematics. These printable problem sets help students build accuracy and prepare effectively for school tests.
Concept Clarification for Chapter 12 Surface Area and Volume
Cross-reference your completed exercises with comprehensive NCERT solutions for Class 10 Mathematics to ensure absolute clarity across all sub-topics in this chapter.
Additional Study Resources for Class 10 Mathematics
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FAQs
You can download the teacher-verified PDF for CBSE Class 10 Mathematics Surface Areas and Volumes Worksheet Set 08 from StudiesToday.com. These practice sheets for Class 10 Mathematics are designed as per the latest CBSE academic session.
Yes, our CBSE Class 10 Mathematics Surface Areas and Volumes Worksheet Set 08 includes a variety of questions like Case-based studies, Assertion-Reasoning, and MCQs as per the 50% competency-based weightage in the latest curriculum for Class 10.
Yes, we have provided detailed solutions for CBSE Class 10 Mathematics Surface Areas and Volumes Worksheet Set 08 to help Class 10 and follow the official CBSE marking scheme.
Daily practice with these Mathematics worksheets helps in identifying understanding gaps. It also improves question solving speed and ensures that Class 10 students get more marks in CBSE exams.
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