NCERT Solutions for Class 9 Maths - Surface Areas and Volumes (Exercise 11.4)

Q1. Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m (Assume π = 22 / 7)
Ans: (i) Radius of sphere, r = 7 cm
Using, Volume of sphere = (4 / 3) πr³
= (4 / 3) × (22 / 7) × 7³ = 4312 / 3
Hence, volume of the sphere is 4312 / 3 cm³
(ii) Radius of sphere, r = 0.63 m
Using, volume of sphere = (4 / 3) πr³
= (4 / 3) × (22 / 7) × 0.63³ = 1.0478
Hence, volume of the sphere is 1.05 m³ (approx).

Q2. Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m (Assume π = 22 / 7)
Ans: (i) Diameter = 28 cm
Radius, r = 28 / 2 cm = 14cm
Volume of the solid spherical ball = (4 / 3) πr³
Volume of the ball = (4 / 3) × (22 / 7) × 14³ = 34496 / 3
Hence, volume of the ball is 34496 / 3 cm³
(ii) Diameter = 0.21 m
Radius of the ball = 0.21 / 2 m = 0.105 m
Volume of the ball = (4 / 3)πr³
Volume of the ball = (4 / 3) × (22 / 7) × 0.105³
Hence, volume of the ball = 0.004851 m³

Q3. The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³? (Assume π = 22 / 7)
Ans:  Given,
Diameter of a metallic ball = 4.2 cm
Radius(r) of the metallic ball, r = 4.2 / 2 cm = 2.1 cm
Volume formula = 4 / 3 πr³
Volume of the metallic ball = (4 / 3) × (22 / 7) × 2.1³
Volume of the metallic ball = 38.808 cm³
Now, using relationship between, density, mass and volume,
Density = Mass / Volume
Mass = Density × volume
= (8.9 × 38.808) g
= 345.3912 g
Mass of the ball is 345.39 g (approx).

Q4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Ans: Let the diameter of earth be “d”. Therefore, the radius of earth will be will be d/2
Diameter of moon will be d / 4 and the radius of moon will be d/8
Find the volume of the moon:
Volume of the moon = (4 / 3) πr³ = (4 / 3) π (d / 8)³ = 4 / 3π(d³ / 512)
Find the volume of the earth:
Volume of the earth = (4 / 3) πr³ = (4 / 3) π (d / 2)³ = 4 / 3π(d³ / 8)
Fraction of the volume of the earth is the volume of the moon

Volume of moon is of the 1 / 64 volume of earth.

Q5. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22 / 7)
Ans: Diameter of hemispherical bowl = 10.5 cm
Radius of hemispherical bowl, r = 10.5 / 2 cm = 5.25 cm
Formula for volume of the hemispherical bowl = (2 / 3) πr³
Volume of the hemispherical bowl = (2 / 3) × (22 / 7) × 5.25³ = 303.1875
Volume of the hemispherical bowl is 303.1875 cm³
Capacity of the bowl = (303.1875) / 1000 L = 0.303 litres(approx.)
Therefore, hemispherical bowl can hold 0.303 litres of milk.

Q6. A hemi spherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. (Assume π = 22/7)
Ans: Inner Radius of the tank, (r) = 1m
Outer Radius (R) = 1.01m
Volume of the iron used in the tank = (2 / 3) π (R³ – r³)
Put values,
Volume of the iron used in the hemispherical tank = (2 / 3) × (22 / 7) × (1.01³ – 1³) = 0.06348
So, volume of the iron used in the hemispherical tank is 0.06348 m³.

Q7. Find the volume of a sphere whose surface area is 154 cm². (Assume π = 22 / 7)
Ans: Let r be the radius of a sphere.
Surface area of sphere = 4πr²
4πr² = 154 cm² (given)
r² = (154 × 7) / (4 × 22) r = 7 / 2
Radius is 7 / 2 cm
Now, Volume of the sphere = (4 / 3) πr³
Volume of the sphere = (4 / 3) x (22 / 7) x (7 / 2)³ = 179 x 2 / 3
Volume of the sphere is 179 x 2 / 3 cm³

Q8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing isRs20 per square meter, find the
(i) Inside surface area of the dome
(ii) volume of the air inside the dome (Assume π = 22 / 7)
Ans: (i) Cost of white-washing the dome from inside = Rs 4989.60
Cost of white-washing 1m² area = Rs 20
CSA of the inner side of dome = 498.96 / 2 m²  = 249.48 m²
(ii) Let the inner radius of the hemispherical dome be r.
CSA of inner side of dome = 249.48 m² (from (i))
Formula to find CSA of a hemi sphere = 2πr²
2πr² = 249.48
2 × (22 / 7) × r² = 249.48
r² = (249.48 × 7) / (2 × 22)
r² = 39.69
r = 6.3
So, radius is 6.3 m
Volume of air inside the dome = Volume of hemispherical dome
Using formula, volume of the hemisphere
= 2 / 3 πr³ = (2 / 3) × (22 / 7) × 6.3 × 6.3 × 6.3
= 523.908
= 523.9(approx.)
Volume of air inside the dome is 523.9 m³.

Q9. Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the
(i) radius r’ of the new sphere,
(ii) ratio of Sand S’.
Ans: Volume of the solid sphere = (4 / 3)πr³
Volume of twenty seven solid sphere = 27 × (4 / 3)πr³ = 36 πr³
(i) New solid iron sphere radius = r’
Volume of this new sphere = (4/3)π(r’)³
(4 / 3)π(r’)³ = 36 πr³
(r’)³ = 27r³
r’= 3r
Radius of new sphere will be 3r (thrice the radius of original sphere)
(ii) Surface area of iron sphere of radius r, S = 4πr²
Surface area of iron sphere of radius r’= 4π (r’)²
Now
S / S’ = (4πr²) / ( 4π (r’)²)
S / S’ = r² / (3r’)² = 1 / 9
The ratio of S and S’ is 1: 9.

Q10. A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm³) is needed to fill this capsule? (Assume π = 22 / 7)
Ans: Diameter of capsule = 3.5 mm
Radius of capsule, say r = diameter / 2 = (3.5 / 2) mm = 1.75mm
Volume of spherical capsule = 4 / 3 πr³
Volume of spherical capsule = (4 / 3) × (22 / 7) × (1.75)³ = 22.458
The volume of the spherical capsule is 22.46 mm³.