Surface Areas and Volumes — Class 10 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Surface Areas and Volumes" — 9 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Mathematics chapter "Surface Areas and Volumes" — 9 important questions with detailed answers for CBSE…
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NCERT Solutions for Class 10 Mathematics — Surface Areas and Volumes. Step-by-step answers to all 9 textbook questions from this chapter, written for CBSE board preparation and free to use.
Surface Areas and Volumes — All 9 Questions Solved
Q1. A solid is formed by combining a cylinder and a hemisphere at one end. The cylinder has radius 3.5 cm and height 10 cm. The hemisphere has the same radius. Find the total surface area. (Use π = 22/7)
Step 1: Total surface area = Curved surface area of cylinder + Base area of cylinder + Curved surface area of hemisphere
Step 2: Curved surface area of cylinder = 2πrh = 2 × (22/7) × 3.5 × 10
Step 3: = 2 × (22/7) × 35 = 2 × 22 × 5 = 220 cm²
Step 4: Base area of cylinder = πr² = (22/7) × (3.5)²
Step 5: = (22/7) × 12.25 = 22 × 1.75 = 38.5 cm²
Step 6: Curved surface area of hemisphere = 2πr² = 2 × (22/7) × (3.5)²
Step 7: = 2 × (22/7) × 12.25 = 2 × 38.5 = 77 cm²
Step 8: Total surface area = 220 + 38.5 + 77 = 335.5 cm²
Final Answer: 335.5 cm²
Q2. A cone is placed on top of a cylinder. The cylinder has radius 5 cm and height 8 cm. The cone has the same radius and slant height 13 cm. Find the total volume and total curved surface area.
Step 1: Volume of cylinder = πr²h = (22/7) × 5² × 8
Step 2: = (22/7) × 25 × 8 = (22/7) × 200
Step 3: = 4400/7 ≈ 628.57 cm³
Step 4: For cone, slant height l = 13 cm, radius r = 5 cm
Step 5: Height of cone: h² = l² − r² = 13² − 5² = 169 − 25 = 144
Step 6: h = 12 cm
Step 7: Volume of cone = (1/3)πr²h = (1/3) × (22/7) × 25 × 12
Step 8: = (1/3) × (22/7) × 300 = (22 × 100)/7 = 2200/7 ≈ 314.29 cm³
Step 9: Total volume = 4400/7 + 2200/7 = 6600/7 ≈ 942.86 cm³
Step 10: Curved surface area of cylinder = 2πrh = 2 × (22/7) × 5 × 8 = (22/7) × 80 = 1760/7 cm²
Step 11: Curved surface area of cone = πrl = (22/7) × 5 × 13 = 1430/7 cm²
Step 12: Total curved surface area = 1760/7 + 1430/7 = 3190/7 ≈ 455.71 cm²
Final Answer: Total volume ≈ 942.86 cm³, Total curved surface area ≈ 455.71 cm²
Q3. Find the volume of a frustum of a cone if the radii of the two circular ends are 5 cm and 3 cm, and the height is 6 cm. (Use π = 22/7)
Step 1: Volume of frustum = (1/3)πh(R² + r² + Rr)
Step 2: Here, R = 5 cm (larger radius), r = 3 cm (smaller radius), h = 6 cm
Step 3: Volume = (1/3) × (22/7) × 6 × (5² + 3² + 5×3)
Step 4: Volume = (1/3) × (22/7) × 6 × (25 + 9 + 15)
Step 5: Volume = (1/3) × (22/7) × 6 × 49
Step 6: Volume = (22/7) × 2 × 49
Step 7: Volume = (22/7) × 98
Step 8: Volume = 22 × 14 = 308 cm³
Final Answer: 308 cm³
Q4. A solid is made by combining a hemisphere and a cone. The hemisphere has radius 4 cm. The cone has the same base radius and height 3 cm. Find the total surface area of the solid. (Use π = 3.14)
Step 1: Total surface area = Curved surface area of hemisphere + Curved surface area of cone
Step 2: Curved surface area of hemisphere = 2πr² = 2 × 3.14 × 4²
Step 3: = 2 × 3.14 × 16 = 100.48 cm²
Step 4: For cone, r = 4 cm, h = 3 cm
Step 5: Slant height l = √(r² + h²) = √(16 + 9) = √25 = 5 cm
Step 6: Curved surface area of cone = πrl = 3.14 × 4 × 5 = 62.8 cm²
Step 7: Total surface area = 100.48 + 62.8 = 163.28 cm²
Final Answer: 163.28 cm²
Q5. Find the volume of a composite solid formed by joining a cylinder (radius 6 cm, height 10 cm) and a cone (same radius, height 8 cm) at the cylinder's top.
Step 1: Volume of cylinder = πr²h = (22/7) × 6² × 10
Step 2: = (22/7) × 36 × 10 = (22/7) × 360
Step 3: = 7920/7 ≈ 1131.43 cm³
Step 4: Volume of cone = (1/3)πr²h = (1/3) × (22/7) × 6² × 8
Step 5: = (1/3) × (22/7) × 36 × 8
Step 6: = (1/3) × (22/7) × 288
Step 7: = (22 × 96)/7 = 2112/7 ≈ 301.71 cm³
Step 8: Total volume = 7920/7 + 2112/7 = 10032/7 ≈ 1433.14 cm³
Final Answer: 10032/7 cm³ or approximately 1433.14 cm³
Q6. A frustum of a cone has slant height 10 cm, and the radii of the two ends are 6 cm and 4 cm. Find the curved surface area of the frustum. (Use π = 22/7)
Step 1: Curved surface area of frustum = π(R + r)l
Step 2: Here, R = 6 cm (larger radius), r = 4 cm (smaller radius), l = 10 cm (slant height)
Step 3: Curved surface area = (22/7) × (6 + 4) × 10
Step 4: = (22/7) × 10 × 10
Step 5: = (22/7) × 100
Step 6: = 2200/7 ≈ 314.29 cm²
Final Answer: 2200/7 cm² or approximately 314.29 cm²
Q7. Find the total surface area of a hemisphere with radius 7 cm (including the flat base). (Use π = 22/7)
Step 1: Total surface area = Curved surface area + Base area
Step 2: Curved surface area of hemisphere = 2πr²
Step 3: = 2 × (22/7) × 7² = 2 × (22/7) × 49
Step 4: = 2 × 22 × 7 = 308 cm²
Step 5: Base area = πr² = (22/7) × 7² = 22 × 7 = 154 cm²
Step 6: Total surface area = 308 + 154 = 462 cm²
Final Answer: 462 cm²
Q8. A cone has height 24 cm and base radius 7 cm. Find its volume and lateral surface area. (Use π = 22/7)
Step 1: Volume of cone = (1/3)πr²h
Step 2: = (1/3) × (22/7) × 7² × 24
Step 3: = (1/3) × (22/7) × 49 × 24
Step 4: = (1/3) × 22 × 7 × 24
Step 5: = (1/3) × 3696 = 1232 cm³
Step 6: Slant height l = √(h² + r²) = √(24² + 7²) = √(576 + 49) = √625 = 25 cm
Step 7: Lateral surface area = πrl = (22/7) × 7 × 25
Step 8: = 22 × 25 = 550 cm²
Final Answer: Volume = 1232 cm³, Lateral surface area = 550 cm²
Q9. Find the volume of a frustum with height 7 cm, upper radius 2 cm, and lower radius 4 cm. (Use π = 22/7)
Step 1: Volume of frustum = (1/3)πh(R² + r² + Rr)
Step 2: Here, h = 7 cm, R = 4 cm (lower radius), r = 2 cm (upper radius)
Step 3: Volume = (1/3) × (22/7) × 7 × (4² + 2² + 4×2)
Step 4: Volume = (1/3) × (22/7) × 7 × (16 + 4 + 8)
Step 5: Volume = (1/3) × (22/7) × 7 × 28
Step 6: Volume = (1/3) × 22 × 28
Step 7: Volume = (616)/3 ≈ 205.33 cm³
Final Answer: 616/3 cm³ or approximately 205.33 cm³
Surface Areas and Volumes — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. A solid metallic sphere of radius 10 cm is melted and recast into small cones, each of radius 5 cm and height 10 cm. How many cones are formed?
- A. 8
- B. 16
- C. 32
- D. 64
Answer: B. 16 — The volume of the sphere is (4/3)π(10)³ and the volume of one cone is (1/3)π(5)²(10). The number of cones formed is the ratio of the sphere's volume to the cone's volume, which simplifies to 16.
MCQ 2. The curved surface area of a cylinder with radius 7 cm and height 10 cm is:
- A. 220 cm²
- B. 308 cm²
- C. 440 cm²
- D. 770 cm²
Answer: C. 440 cm² — Curved surface area of a cylinder = 2πrh. Substituting r=7 cm and h=10 cm, we get 2 * (22/7) * 7 * 10 = 440 cm².
MCQ 3. If the radii of two spheres are in the ratio 2:3, what is the ratio of their volumes?
- A. 2:3
- B. 4:9
- C. 8:27
- D. 16:81
Answer: C. 8:27 — The volume of a sphere is proportional to the cube of its radius (V = (4/3)πr³). If radii are r₁:r₂ = 2:3, then volumes V₁:V₂ = r₁³:r₂³ = 2³:3³ = 8:27.
MCQ 4. A toy is in the form of a cone mounted on a hemisphere. The diameter of the base of the cone and hemisphere is 6 cm and the height of the cone is 4 cm. Find the total surface area of the toy. (Use π = 3.14)
- A. 94.2 cm²
- B. 103.62 cm²
- C. 122.46 cm²
- D. 188.4 cm²
Answer: B. 103.62 cm² — The total surface area is the sum of the curved surface area of the cone (πrl) and the curved surface area of the hemisphere (2πr²). First, calculate the slant height of the cone. Then substitute the values to find the sum.
MCQ 5. A cylindrical bucket, 32 cm high and with a radius of base 18 cm, is full of sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius of the base of the heap.
- A. 24 cm
- B. 30 cm
- C. 36 cm
- D. 48 cm
Answer: C. 36 cm — The volume of sand in the cylindrical bucket is equal to the volume of the conical heap. Equating the volume of the cylinder (πr²h) to the volume of the cone ((1/3)πR²H) allows us to solve for the radius of the conical heap.
MCQ 6. The radii of the top and bottom of a bucket of slant height 35 cm are 28 cm and 7 cm respectively. Find the capacity of the bucket. (Use π = 22/7)
- A. 30184 cm³
- B. 32704 cm³
- C. 33612 cm³
- D. 34588 cm³
Answer: A. 30184 cm³ — First, use the slant height and radii to find the height of the frustum. Then, apply the formula for the volume of a frustum: (1/3)πh(R² + r² + Rr), substituting the calculated height and given radii.
MCQ 7. The total surface area of a cube whose side is 5 cm is:
- A. 25 cm²
- B. 125 cm²
- C. 150 cm²
- D. 250 cm²
Answer: C. 150 cm² — The total surface area of a cube is 6a², where 'a' is the side length. For a = 5 cm, TSA = 6 * 5² = 150 cm².
MCQ 8. 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.
- A. π cm³
- B. 2π cm³
- C. 3π cm³
- D. (2/3)π cm³
Answer: A. π cm³ — The total volume is the sum of the volume of the cone ((1/3)πr²h) and the volume of the hemisphere ((2/3)πr³). Given that r=1 cm and h=1 cm, the total volume is (1/3)π(1)²(1) + (2/3)π(1)³ = π cm³.
MCQ 9. The volume of a hemisphere is 19404 cm³. The radius of the hemisphere is: (Use π = 22/7)
- A. 7 cm
- B. 14 cm
- C. 21 cm
- D. 28 cm
Answer: C. 21 cm — Use the formula for the volume of a hemisphere, V = (2/3)πr³. Substitute the given volume and solve for r by isolating r³ and finding its cube root.
MCQ 10. The curved surface area of a cone with radius 3 cm and slant height 5 cm is:
- A. 15π cm²
- B. 20π cm²
- C. 24π cm²
- D. 30π cm²
Answer: A. 15π cm² — The curved surface area of a cone is given by the formula πrl, where r is the radius and l is the slant height. Substituting r=3 cm and l=5 cm gives 15π cm².
MCQ 11. How many spherical bullets can be made from a solid cuboid of lead with dimensions 66 cm x 42 cm x 21 cm, each bullet having a diameter of 4.2 cm? (Use π = 22/7)
- A. 1000
- B. 1250
- C. 1500
- D. 1750
Answer: C. 1500 — First, calculate the volume of the cuboid (length × width × height). Then, calculate the volume of one spherical bullet (4/3)πr³, where r is half the diameter. Finally, divide the volume of the cuboid by the volume of one bullet to find the number of bullets.
MCQ 12. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. (Use π = 22/7)
- A. 200 mm²
- B. 220 mm²
- C. 250 mm²
- D. 280 mm²
Answer: B. 220 mm² — The surface area of the capsule is the sum of the curved surface area of the cylindrical part and the curved surface area of the two hemispherical ends. First, find the height of the cylindrical part by subtracting twice the radius from the total length. Then apply the combined surface area formula 2πr(h + 2r).
MCQ 13. The volume of a right circular cylinder with radius 7 cm and height 10 cm is: (Use π = 22/7)
- A. 440 cm³
- B. 770 cm³
- C. 1540 cm³
- D. 2200 cm³
Answer: C. 1540 cm³ — The volume of a cylinder is given by the formula πr²h. Substitute the given radius r=7 cm and height h=10 cm to calculate the volume.
MCQ 14. The diameter of a metallic hemisphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire. (Use π = 22/7)
- A. 12 m
- B. 18 m
- C. 24 m
- D. 36 m
Answer: B. 18 m — The volume of the hemisphere is equal to the volume of the cylindrical wire. Calculate the volume of the hemisphere using (2/3)πR³. Then, calculate the volume of the wire as πr²L, where r is the wire's radius and L is its length. Equate the two volumes to solve for L.
MCQ 15. A container, open from the top and made of metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of the metal sheet used, if it costs Rs 8 per 100 cm². (Use π = 3.14)
- A. Rs 125.60
- B. Rs 156.75
- C. Rs 178.20
- D. Rs 190.50
Answer: B. Rs 156.75 — First, calculate the slant height of the frustum using the height and radii. Then, find the curved surface area of the frustum using π(R+r)l. Add the area of the bottom circular base (πr²) as the container is open at the top. Finally, calculate the total cost based on the given rate.
MCQ 16. If the height of a cylinder is halved and its radius is doubled, then the curved surface area of the new cylinder compared to the original cylinder will be:
- A. Half
- B. Same
- C. Double
- D. Four times
Answer: B. Same — The curved surface area of a cylinder is 2πrh. When the height is halved (h' = h/2) and the radius is doubled (r' = 2r), the new curved surface area is 2π(2r)(h/2) = 2πrh, which is the same as the original.
MCQ 17. A cone has a base radius of 7 cm and a height of 9 cm. What is its volume? (Use π = 22/7)
- A. 154 cm³
- B. 308 cm³
- C. 462 cm³
- D. 616 cm³
Answer: C. 462 cm³ — The volume of a cone is given by the formula (1/3)πr²h. Substitute the given radius r=7 cm and height h=9 cm to calculate the volume.
MCQ 18. The surface area of a sphere is 616 cm². Find its radius. (Use π = 22/7)
- A. 7 cm
- B. 14 cm
- C. 21 cm
- D. 28 cm
Answer: A. 7 cm — Use the formula for the surface area of a sphere, SA = 4πr². Substitute the given surface area and solve for r by isolating r² and finding its square root.
MCQ 19. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use π = 3.14)
- A. 892.26 kg
- B. 921.64 kg
- C. 987.52 kg
- D. 1024.16 kg
Answer: A. 892.26 kg — Calculate the volume of each cylindrical part separately using the formula πr²h. Add the two volumes to get the total volume of the pole. Finally, multiply the total volume by the given mass per cm³ and convert to kilograms.
MCQ 20. If the dimensions of a cuboid are doubled, then its volume increases by:
- A. 2 times
- B. 4 times
- C. 6 times
- D. 8 times
Answer: D. 8 times — The volume of a cuboid is lwh. If each dimension (length, width, height) is doubled, the new volume becomes (2l)(2w)(2h) = 8lwh, which is 8 times the original volume.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.