Class 10 Mathematics
Chapter 12: Surface Areas and Volumes
Exercise 12.2 – Stepwise Solutions
Instruction: Unless stated otherwise, take π = 22/7.
Q1. A solid is in the shape of a cone standing on a hemisphere with both radii = 1 cm and height of cone = radius.
Step 1: Radius r = 1 cm
Height of cone h = 1 cm
Volume of hemisphere
V = (2/3)πr³
= (2/3)π(1³)
= 2π/3
Volume of cone
V = (1/3)πr²h
= (1/3)π(1²)(1)
= π/3
Total volume
= 2π/3 + π/3
= π
Answer: π cm³
Q2. A model is shaped like a cylinder with two cones attached at its ends.
Diameter = 3 cm
Radius r = 1.5 cm
Total length = 12 cm
Each cone height = 2 cm
Step 1: Height of cylinder
= 12 − 2 − 2
= 8 cm
Volume of cylinder
= πr²h
= π(1.5)²(8)
= π(2.25)(8)
= 18π
Volume of one cone
= (1/3)πr²h
= (1/3)π(1.5²)(2)
= (1/3)π(2.25)(2)
= 1.5π
Two cones volume = 3π
Total volume
= 18π + 3π
= 21π
Answer: 21π cm³
Q3. Gulab jamun problem
Length = 5 cm
Diameter = 2.8 cm
Radius r = 1.4 cm
Step 1: Height of cylindrical part
Total length = cylinder height + diameter
5 = h + 2.8
h = 2.2 cm
Volume of cylinder
= πr²h
= (22/7)(1.4²)(2.2)
1.4² = 1.96
= (22/7)(1.96)(2.2)
= 13.552 cm³
Volume of sphere
= (4/3)πr³
= (4/3)(22/7)(1.4³)
1.4³ = 2.744
= 11.49 cm³
Total volume of one gulab jamun
= 13.552 + 11.49
= 25.04 cm³
Volume of 45 gulab jamuns
= 45 × 25.04
= 1126.8 cm³
Syrup = 30
= 0.30 × 1126.8
≈ 338 cm³
Q4. Pen stand with 4 conical depressions
Cuboid dimensions = 15 × 10 × 3.5 cm
Volume of cuboid
= l × b × h
= 15 × 10 × 3.5
= 525 cm³
Volume of one cone
= (1/3)πr²h
= (1/3)(22/7)(0.5²)(1.4)
0.5² = 0.25
= (1/3)(22/7)(0.35)
= 0.366 cm³
4 cones volume
= 4 × 0.366
= 1.464 cm³
Volume of wood
= 525 − 1.464
= 523.54 cm³
Q5. Inverted cone filled with water
Radius r = 5 cm
Height h = 8 cm
Volume of cone
= (1/3)πr²h
= (1/3)(22/7)(25)(8)
= 209.52 cm³
Water overflow = 1/4
= 52.38 cm³
Volume of one lead shot
Radius = 0.5 cm
= (4/3)πr³
= (4/3)(22/7)(0.125)
= 0.524 cm³
Number of lead shots
= 52.38 ÷ 0.524
≈ 100 shots
Q6. Iron pole with two cylinders
First cylinder
Radius = 12 cm
Height = 220 cm
Volume
= πr²h
= 3.14 × 144 × 220
= 99475.2 cm³
Second cylinder
Radius = 8 cm
Height = 60 cm
Volume
= 3.14 × 64 × 60
= 12057.6 cm³
Total volume
= 111532.8 cm³
Mass
= 111532.8 × 8
= 892262 g
= 892.26 kg
Q7. Cone on hemisphere placed in cylinder
Radius = 60 cm
Cylinder height = 180 cm
Volume cylinder
= πr²h
= π(60²)(180)
= π(3600)(180)
= 648000π
Volume hemisphere
= (2/3)πr³
= (2/3)π(216000)
= 144000π
Volume cone
= (1/3)πr²h
= (1/3)π(3600)(120)
= 144000π
Total solid volume
= 288000π
Water left
= 648000π − 288000π
= 360000π cm³
Q8. Spherical vessel with cylindrical neck
Sphere radius = 8.5/2 = 4.25 cm
Volume sphere
= (4/3)πr³
= (4/3)(3.14)(4.25³)
= 321.4 cm³
Volume cylinder
Radius = 1 cm
Height = 8 cm
= πr²h
= 3.14 × 1 × 8
= 25.12 cm³
Total volume
= 321.4 + 25.12
= 346.52 cm³
Measured volume = 345 cm³
Conclusion: The child’s answer is approximately correct.
Final Answers Summary
- Q1 → π cm³
- Q2 → 21π cm³
- Q3 → ≈ 338 cm³ syrup
- Q4 → 523.54 cm³
- Q5 → ≈ 100 lead shots
- Q6 → ≈ 892 kg
- Q7 → 360000π cm³
- Q8 → Child is correct