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NCERT Class 10 Mathematics Exercise 12.2 Solutions: Surface Areas and Volumes

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of Exercise 12.2 in NCERT's Class 10 Mathematics book, chapter “Surface Areas and Volumes”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

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Ex 12.2 Q1
A solid is a cone on a hemisphere, both of radius 1 cm, with the cone's height equal to its radius. Find its volume in terms of π.
1. The solid consists of a cone and a hemisphere joined at their bases.
2. The radius of both the cone and the hemisphere is 1 cm.
3. The height of the cone is equal to its radius, so h = 1 cm.
4. Volume of the cone is (1/3) × π × r² × h = (1/3) × π × 1² × 1 = π/3 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
5. Volume of the hemisphere is (2/3) × π × r³ = (2/3) × π × 1³ = 2π/3 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
6. Total volume is the sum of the cone and hemisphere volumes: π/3 + 2π/3 = 3π/3 = π cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
Answer: π cm³
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Ex 12.2 Q2
A model is a cylinder with a cone at each end: diameter 3 cm, total length 12 cm, each cone 2 cm high. Find the volume of air inside (π = 22/7).
1. The model consists of a central cylinder with a cone attached to each end.
2. The diameter of the cylinder and cones is 3 cm, so the radius r is 3 ÷ 2 = 1.5 cm.
3. The total length of the model is 12 cm, and the height of each cone is 2 cm.
4. The length of the cylindrical part is the total length minus the heights of the two cones: 12 − 2 − 2 = 8 cm.
5. The volume of the cylinder is calculated as π × r² × h = (22/7) × (1.5)² × 8.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
6. The volume of the cylinder is (22/7) × 2.25 × 8 = 56.57 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
7. The volume of one cone is calculated as (1/3) × π × r² × h_cone = (1/3) × (22/7) × (1.5)² × 2.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
8. The volume of one cone is (1/3) × (22/7) × 2.25 × 2 = 4.71 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
9. The total volume of the two cones is 2 × 4.71 = 9.43 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
10. The total volume of air inside is the sum of the cylinder and the two cones: 56.57 + 9.43 = 66 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
Answer: 66 cm³
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Ex 12.2 Q3
A gulab jamun is a cylinder with two hemispherical ends, length 5 cm and diameter 2.8 cm, and is about 30% syrup by volume. Find the syrup in 45 gulab jamuns (π = 22/7).
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. The diameter of the gulab jamun is 2.8 cm, so the radius r is 1.4 cm.
2. The total length is 5 cm, which includes the cylinder and two hemispheres (forming a full sphere).
3. The length of the cylindrical part is 5 cm minus the diameter of the sphere, so h = 5 − 2.8 = 2.2 cm.
4. Volume of the cylindrical part is πr²h = (22/7) × (1.4)² × 2.2 = 13.552 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
5. Volume of the two hemispherical ends (one full sphere) is (4/3)πr³ = (4/3) × (22/7) × (1.4)³ = 11.4987 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
6. Total volume of one gulab jamun is 13.552 + 11.4987 = 25.0507 cm³.
7. The syrup content is 30% of the total volume, so syrup in one gulab jamun is 0.30 × 25.0507 = 7.5152 cm³.
  why?
  Per cent means 'out of 100', so x% of a quantity = (x / 100) × the quantity.
8. Syrup in 45 gulab jamuns is 45 × 7.5152 = 338.184 cm³.
9. Rounding to the nearest whole number gives 338 cm³.
Answer: 338 cm³
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Ex 12.2 Q4
A wooden pen stand is a 15 cm × 10 cm × 3.5 cm cuboid with four conical holes of radius 0.5 cm and depth 1.4 cm. Find the volume of wood (π = 22/7).
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Volume of the cuboid is calculated as length × breadth × height, which is 15 × 10 × 3.5 = 525 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
2. Volume of one conical hole is calculated using the formula (1/3)πr²h, with r = 0.5 cm and h = 1.4 cm.
  why?
  Area of a circle = πr² (a semicircle is half of it, a quadrant a quarter). Take π = 22/7 or 3.14 as the question says.
3. Substituting the values, the volume of one cone is (1/3) × (22/7) × (0.5)² × 1.4 = (1/3) × (22/7) × 0.25 × 1.4 = 0.3667 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
4. Total volume of the four conical holes is 4 × 0.3667 = 1.4668 cm³.
5. Volume of wood is the volume of the cuboid minus the total volume of the four holes, which is 525 − 1.4668 = 523.5332 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
6. Rounding to two decimal places, the volume of wood is 523.53 cm³.
Answer: 523.53 cm³
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Ex 12.2 Q5
An inverted cone of height 8 cm and top radius 5 cm is full of water. Lead shots of radius 0.5 cm are dropped in until one-fourth of the water flows out. How many shots were dropped? (π = 22/7)
1. Volume of water in the cone = (1/3) × π × r² × h = (1/3) × (22/7) × 5² × 8 = 4400/21 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
2. Volume of water that flows out = (1/4) × 4400/21 = 1100/21 cm³.
3. Volume of one lead shot = (4/3) × π × r³ = (4/3) × (22/7) × (0.5)³ = 11/42 cm³.
4. Number of shots = Volume of water flowed out ÷ Volume of one shot = (1100/21) ÷ (11/42) = (1100/21) × (42/11) = 100.
Answer: 100 lead shots
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Ex 12.2 Q6
An iron pole is a cylinder of height 220 cm and base diameter 24 cm topped by a cylinder of height 60 cm and radius 8 cm. Find its mass if 1 cm³ of iron is about 8 g (π = 3.14).
1. The radius of the lower cylinder is half of its diameter, so r = 24 ÷ 2 = 12 cm.
2. The height of the lower cylinder is h = 220 cm.
3. The volume of the lower cylinder is V₁ = πr²h = 3.14 × 12² × 220 = 3.14 × 144 × 220 = 99475.2 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
4. The radius of the upper cylinder is r = 8 cm.
5. The height of the upper cylinder is h = 60 cm.
6. The volume of the upper cylinder is V₂ = πr²h = 3.14 × 8² × 60 = 3.14 × 64 × 60 = 12057.6 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
7. The total volume of the iron pole is V = V₁ + V₂ = 99475.2 + 12057.6 = 111532.8 cm³.
8. The mass of the pole is calculated as Volume × Density = 111532.8 × 8 g = 892262.4 g.
9. Converting the mass to kilograms, we divide by 1000: 892262.4 ÷ 1000 = 892.2624 kg.
10. Rounding to two decimal places, the mass is 892.26 kg.
Answer: 892.26 kg
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Ex 12.2 Q7
A solid of a cone (height 120 cm, radius 60 cm) on a hemisphere (radius 60 cm) stands in a cylinder (radius 60 cm, height 180 cm) full of water, touching the bottom. Find the water left (π = 22/7).
1. Volume of the cylinder = πr²h = (22/7) × 60² × 180 = (22/7) × 648000 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
2. Volume of the cone = (1/3)πr²h = (1/3) × (22/7) × 60² × 120 = (22/7) × 144000 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
3. Volume of the hemisphere = (2/3)πr³ = (2/3) × (22/7) × 60³ = (22/7) × 144000 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
4. The solid pushes out water equal to its volume: (22/7) × (144000 + 144000) = (22/7) × 288000 cm³.
5. Water left = (22/7) × (648000 − 288000) = (22/7) × 360000 ≈ 1131428.57 cm³.
6. In cubic metres (1 m³ = 1000000 cm³): about 1.131 m³.
Answer: About 1131428.57 cm³ = 1.131 m³ of water is left.
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Ex 12.2 Q8
A spherical glass vessel (inside diameter 8.5 cm) has a cylindrical neck 8 cm long and 2 cm in diameter. A child says it holds 345 cm³. Check her answer (π = 3.14).
1. The vessel is a sphere of radius 8.5 ÷ 2 = 4.25 cm with a cylindrical neck of radius 1 cm and length 8 cm.
2. Volume of the spherical part = (4/3) × 3.14 × 4.25³ = (4/3) × 3.14 × 76.765625 ≈ 321.39 cm³.
  why?
  Volumes: cylinder πr²h, cone ⅓πr²h, sphere ⁴⁄₃πr³, hemisphere ⅔πr³, cuboid l × b × h. The volume of a solid made of parts is the sum of the parts' volumes; melting or recasting keeps the volume the same.
3. Volume of the neck = 3.14 × 1² × 8 = 25.12 cm³.
4. Total volume = 321.39 + 25.12 = 346.51 cm³.
5. The child's 345 cm³ is not equal to 346.51 cm³, so her answer is not correct (the true capacity is about 346.51 cm³).
Answer: Capacity ≈ 346.51 cm³, so she is not correct.
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Other exercises in this chapter

Exercise 12.1 — 9 questions

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NCERT Class 10 Mathematics Exercise 12.2 Solutions: Surface Areas and Volumes

Ex 12.2 Q1

Ex 12.2 Q2

Ex 12.2 Q3

Ex 12.2 Q4

Ex 12.2 Q5

Ex 12.2 Q6

Ex 12.2 Q7

Ex 12.2 Q8