# Surface Area and Volumes – Exercise 13.8 – Class IX

Assume π = 22/7 , unless stated otherwise.

1. Find the volume of a sphere whose radius is

(i) 7 cm

(ii) 0.63 m

Solution:

(i) Volume of sphere = 4/3 πr3 = 22/7 x 4/3 x 73 = 1437.33 m3

(ii) Volume of sphere = 4/3 πr3 = 22/7 x 4/3 x (0.63)3 = 1.05m3

1. Find the amount of water displaced by a solid spherical ball of diameter

(i) 28 cm

(ii) 0.21 m

Solution:

(i)d = 28 cm

r = d/2 = 28/2 = 14 cm

Volume of sphere = 4/3 πr3 = 22/7 x 4/3 x 143 = 11498.67 m3

(ii) d = 0.21 m

r = d/2 = 0.21/2 = 0.105 m

Volume of sphere = 4/3 πr3 = 22/7 x 4/3 x (0.105)3 = 0.004851 m3

1. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3?

Solution:

d = 4.2 cm

r = d/2 = 4.2/2 =  2.1 cm

Volume of sphere = 4/3 πr3 = 22/7 x 4/3 x (2.1)3  = 38.808 cm3

Density of metal for 1 cm3  = 8.9 g

Density of metal for 38.808 cm3  = 38.808 x 8.9 =  345.39g

1. 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?

solution:

Let diameter of earth = d  and radius of earth = d/2

Volume of earth  = 4/3 πr3 = 4/322/7 x (d/2)3

Given, the diameter of the moon is approximately one-fourth of the diameter of the earth.

i.e., diameter of moon = 1/4 x d  = d/4

Then radius of moon = d/4 x 1/2 = d/8

Volume of moon  = 4/3 πr3 = 4/322/7 x (d/8)3

Fraction of the volume of the earth is the volume of the moon = [4/322/7 x (d/2)3]/[ 4/322/7 x (d/8)3]

= (1/2)3/(1/8)3

= (1/8)/(1/512)

= 512/8 = 64

⇒Volume of the moon = 1/64 x volume of earth

Hence, volume of the moon is 1/64 of volume of earth.

1. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Solution:

d = 10.5 cm

r = d/2 = 10.5/2 = 5.25 cm

Volume of the hemisphere bowl = 2/3 πr3

= 2/3 x 22/7 x (5.25)3

= 303.1875 cm3

Therefore, the hemispherical bowl can hold 303.1875/1000 litres = 0.303 litres of milk.

1. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Solution:

Inner radius of the tank, r = 1m

Thickness of the iron sheet = 1 m = 0.01 m

⸫External radius of the tank , R = (1+ 0.01) = 1.01 m

Volume of the iron used to make the tank = 2/3 π(R3 – r3)

= 2/3 x 22/7 x (1.013 – 13)

= 0.0635m3

1. Find the volume of a sphere whose surface area is 154 cm2.

Solution:

Surface area if the given sphere = 154cm2

We know, Surface area of the given sphere = 4πr2 = 154

4x22/7x r2 = 154

r2 = 154×7/4×22 = 49/4

r  = 7/2

Volume of sphere = 4/3  πr3 = 4/3 x 22/7 x (7/2)3 = 179.67cm3

1. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square metre, find the

(i) inside surface area of the dome,

(ii) volume of the air inside the dome.

Solution:

(i) inside surface area of the dome = total cost/cost of white  washing per m2 = 498.96/2 =  249.48 m2

(ii) Let the radius of the dome be ‘r’ m

Then 2πr2 = 249.48

2 x 22/7 x r2 =  249.48

r2 = 249.48×7/2×22­ = 3969/100

r = 63/10 cm

volume of the air inside the dome = 2/3πr3 = 2/3 x 22/7 x (6.3)3 = 523.908 cm3

1. 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 S and S′.

Solution:

(i) Volume of a sphere of radius, r = 4/3 πr3

Volume of 27 such spheres = 27 x 4/3 πr3  =  36 πr3

Volume of the sphere with radius r’ = 4/3 πr3

⇒ 36 πr3  = 4/3 πr3

⇒ 27r3 = r’3

⇒ r’ = ∛(27r3)

⇒ r’ = 3r

1. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?

Solution:

d = 3.5mm

r = 3.5/2 = 1.75 mm

Volume of sphere = 4/3 πr3

= 4/3 x 22/7 x (1.75)3

= 22.46 mm3

Therefore, 22.46mm3 medicine is needed to fill this capsule.