Class 9 Mathematics
Chapter 11: Surface Areas and Volumes
Exercise 11.4 – Stepwise Solutions
Important Formula Used
Volume of sphere = (4/3)πr³
Volume of hemisphere = (2/3)πr³
Surface area of sphere = 4πr²
Note: Assume π = 22/7 unless stated otherwise.
Q1. Find the volume of a sphere
(i) r = 7 cm
Volume = (4/3) × (22/7) × 7³
= (4/3) × (22/7) × 343
= 1437.33 cm³
(ii) r = 0.63 m
= (4/3) × (22/7) × (0.63)³
= 1.047 m³ (approx.)
Q2. Find the amount of water displaced by spherical ball
(i) Diameter = 28 cm
r = 14 cm
Volume = (4/3) × (22/7) × 14³
= 11494.67 cm³
(ii) Diameter = 0.21 m
r = 0.105 m
Volume = (4/3) × (22/7) × (0.105)³
= 0.00485 m³
Q3. Diameter = 4.2 cm, density = 8.9 g/cm³
r = 2.1 cm
Volume = (4/3) × (22/7) × (2.1)³
= 38.808 cm³
Mass = density × volume
= 8.9 × 38.808
= 345.39 g
Answer: 345.39 g
Q4. Diameter ratio = 1 : 4
Volume ∝ (diameter)³
Ratio = (1/4)³ : 1³
= 1/64 : 1
= 1 : 64
Q5. Diameter = 10.5 cm
r = 5.25 cm
Volume of hemisphere = (2/3)πr³
= (2/3) × (22/7) × (5.25)³
= 303.19 cm³
1 litre = 1000 cm³
Capacity = 0.303 L
Answer: 0.303 litres
Q6. Thickness = 1 cm, inner radius = 1 m
Outer radius = 1 + 0.01 = 1.01 m
Volume of iron = volume of outer hemisphere − inner hemisphere
= (2/3)π(1.01³ − 1³)
= (2/3) × (22/7) × (1.0303 − 1)
= 0.063 m³ (approx.)
Answer: 0.063 m³
Q7. Surface area = 154 cm²
4πr² = 154
r = 3.5 cm
Volume = (4/3)πr³
= (4/3) × (22/7) × (3.5)³
= 179.67 cm³
Answer: 179.67 cm³
Q8. Cost = ₹4989.60, rate = ₹20/m²
Area = 4989.60 / 20 = 249.48 m²
Inside surface area = 2πr²
2 × (22/7) × r² = 249.48
r² = 39.69
r = 6.3 m
(i) Inside surface area = 249.48 m²
(ii) Volume
= (2/3)πr³
= (2/3) × (22/7) × (6.3)³
= 523.91 m³
Q9. 27 spheres melted
Volume conserved:
27 × (4/3)πr³ = (4/3)πr′³
r′³ = 27r³
r′ = 3r
(i) Radius = 3r
(ii) Ratio of surface areas
S ∝ r², S′ ∝ (3r)² = 9r²
Ratio = 1 : 9
Q10. Diameter = 3.5 mm
r = 1.75 mm
Volume = (4/3) × (22/7) × (1.75)³
= 22.45 mm³
Answer: 22.45 mm³