# NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.8

Question 1.
Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m
Solution:
(i) We have radius of sphere = 7 cm

∴ Volume of sphere = $$\frac{4}{3} \pi r^{3}$$
= $$\frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7$$
= 1437 $$\frac {1}{3}$$ cm3

(ii) We have
Radius of sphere = 0.63 m

∴ Volume of sphere = $$\frac{4}{3} \pi r^{3}$$
= $$\frac{4}{3} \times \frac{22}{7} \times 0.63 \times 0.63 \times 0.63$$
= 1.0478 m3
= 1.05 m3 (approx)

Question 2.
Find the amount of water displaced by a solid spherical ball of diameter.
(i) 28 cm
(ii) 0.21 m
Solution:
(i) We have given that

Diameter of spherical ball = 28 cm
∴ Radius of spherical ball = $$\frac {28}{2}$$ = 14 cm
Volume of spherical ball = $$\frac{4}{3} \pi r^{3}$$
= $$\frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14$$
= 11498$$\frac {2}{3}$$ cm3
Therefore, amount of water displaced by solid sphere is 11498$$\frac {2}{3}$$ cm3.

(ii) Diameter of spherical ball = 0.21 m

∴ Radius of spherical a ball = $$\frac {0.21}{2}$$ = 0.105 m
So, volume of spherical ball = $$\frac{4}{3} \pi r^{3}$$
= $$\frac{4}{3} \times \frac{22}{7}$$ × 0.105 × 0.105 × 0.105
= 0.004851 m3
Therefore, the amount of water displaced by the solid sphere is 0.004851 m3

Question 3.
The diameter of a metallic ball is 4.2. What is the mass of the ball, if the density of the metal is 8.9 g per cm3?
Solution:

We have given that diameter of metallic ball = 4.2 cm
Radius of metallic ball = $$\frac {4.2}{2}$$ = 2.1 cm
volume of metallic ball = $$\frac{4}{3} \pi r^{3}$$
= $$\frac{4}{3} \times \frac{22}{7}$$ × 2.1 × 2.1 × 2.1
= 38.808 cm3
The density of the metal is 8.9 g per cm3.
Mass of the metallic ball = 38.808 × 8.9 = 345.39 g (approx)

Question 4.
The diameter of the moon is approximately one-fourth the diameter of the earth. What fraction of the volume of the earth is the volume of the moon.
Solution:
Let the diameter of earth = x

∴ Radius of earth = $$\frac{x}{2}$$
Volume of earth = $$\frac{4}{3} \pi r^{3}$$
= $$\frac{4}{3} \pi \times \frac{x}{2} \times \frac{x}{2} \times \frac{x}{2}$$
= $$\frac{\pi x^{3}}{6}$$
Again, the diameter of the moon is one-fourth of the diameter of the earth.

Diameter of moon = $$\frac{x}{4}$$
Radius of moon = $$\frac{x}{8}$$
∴ Volume of moon = $$\frac{4}{3} \pi r^{3}$$
= $$\frac{4}{3} \pi \times \frac{x}{8} \times \frac{x}{8} \times \frac{x}{8}$$
= $$\frac{x^{3} \pi}{384}$$
Therefore, fraction of the volume of the moon and the volume of earth is = $$\frac{\frac{\pi x^{3}}{384}}{\frac{\pi x^{3}}{6}}=\frac{6}{384}=\frac{1}{64}$$

Question 5.
How many liters of milk can a hemispherical bowl of diameter 10.5 cm hold.
Solution:
We have a diameter of hemispherical bowl = 10.5 cm

∴ Radius of hemispherical = $$\frac{10.5}{2}$$ = 5.25 cm
∴ Volume of hemisphere = $$\frac{2}{3} \pi r^{3}$$
= $$\frac{2}{3} \times \frac{22}{7}$$ × 5.25 × 5.25 × 5.25
= 303.187 cm3
We know that 1000 cm3 = 1 litre
∴ 303.18 cm3 = 0.303 litre (approx)
Therefore, a hemispherical bowl of diameter 10.2 cm holds 0.303 liters of milk.

Question 6.
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:
We have given that
The inner radius of hemispherical tank = 1 m
and thickness of iron sheet = 1 cm = 0.01 m
The outer radius of hemispherical tank = 1.01 m

So, volume of iron used to make the tank = Volume of out spherical tank – Volume of inner spherical tank
= $$\left.\frac{2}{3} \pi(1.01)^{3}\right)-\frac{2}{3} \pi(1)^{3}$$
= $$\frac{2}{3}$$ × π × 1.030301 – $$\frac{2}{3}$$ × π × 1
= $$\frac{2}{3}$$ × π × (1.030301 – 1)
= $$\frac{2}{3} \times \frac{22}{7} \times 0.030301$$
= 0.06348 m3 (approx)
Therefore, the volume of the iron used to make the tank is 0.06348 m3 (approx)

Question 7.
Find the volume of a sphere whose surface area is 154 cm2.
Solution:
We have given that
The surface area of a sphere is 154 cm2
but, the surface area of sphere = 4πr2

Volume of sphere = $$\frac{4}{3} \pi r^{3}$$
= $$\frac{4}{3} \times \frac{22}{7}$$ × 3.5 × 3.5 × 3.5
= 179$$\frac{2}{3}$$ cm3

Question 8.
A dome of a building is in the form of a hemisphere. From inside it was while washed at the cost of Rs. 498.96. If the cost of whitewashing is Rs. 2.00 per square meter find the:
(i) Inside the surface area of the dome.
(ii) Volume of the air inside the dome.
Solution:
(i) We have given that Rs. 2 is the rate of whitewashing of 1 m2 area
Rs. 1 is the rate of white-washing $$\frac{1}{2}$$ m2 area

Rs. 498.96 is the rate of white-washing = $$\frac{1}{2}$$ × 498.96 = 249.48 m2 area.
Therefore, according to question, inside surface area of dome = 249.48 m2

(ii) We have,
surface area of hemispherical dome = 249.48 m2
but surface area of hemispherical dome = 2πr2
⇒ 2πr2 = 249.48
⇒ r2 = $$\frac{249.48 \times 7}{2 \times 22}$$
⇒ r2 = 39.69
⇒ r = √39.69 = 6.3 m
Therefore,
Volume of hemispherical dome = $$\frac{2}{3} \pi r^{3}$$
= $$\frac{2}{3} \times \frac{22}{7}$$ × 6.3 × 6.3 × 6.3
= 523.908 m3
Therefore, the volume of the air inside the dome is 523.9 m3 (approx).

Question 9.
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:
We have given that

The radius of sphere = r
Surface area(s) = 4πr2
and volume of sphere = $$\frac{4}{3} \pi r^{3}$$
Volume of 27 such speres = $$\frac{4}{3} \times \pi r^{3} \times 27$$ = 36πr3

According to question Volume of 27 small sphere = volume of big sphere
or, 36πr3 = $$\frac{4}{3} \pi\left(r^{\prime}\right)^{3}$$ (∵ radius = r’)
⇒ (r’)3 = $$\frac{36 \pi r^{3} \times 3}{4 \times \pi}$$
(∴ volume of big sphere = volume of 27 small sphere)
⇒ (r’)3 = 27r3
⇒ r’ = 3r
Therefore, radius r’ of new sphere is equal to 3r.

(ii) Surface area of new-sphere (s’) =4π(r’)2
and surface area of small sphere = 4πr2
∴ Ratio between them = $$\left(\frac{4 \pi r^{2}}{4 \pi\left(r^{\prime}\right)^{2}}\right)$$
= $$\left(\frac{4 \pi r^{2}}{4 \pi \times 3 r \times 3 r}\right)$$ (∴ r’ = 3r)
= $$\frac{1}{9}$$
= 1 : 9

Question 10.
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:
We have given

Diameter of spherical capsule = 3.5 mm
∴ Radius of spherical capsule = $$\frac{3.5}{2}$$ = 1.75 mm
∴ Volume of spherical capsule = $$\frac{4}{3} \pi r^{3}$$
= $$\frac{4}{3} \times \frac{22}{7}$$ × 1.75 × 1.75 × 1.75
= 22.4583 mm3 (approx)
Therefore, medicine is needed to fill the capsule is 22.4583 mm3 (approx)

error: Content is protected !!