NCERT solutions for class 9 Maths chapter 13 Surface Areas and Volumes (Exercise

NCERT solutions for class 9 Maths

Surface Areas and Volumes

Chapter 13 Exercise 13.8

EX 13.8 QUESTION 1.

Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 cm
Solution:
(i) Radius (r) = 7cm
Volume of sphere = $\frac { 4 }{ 3 }$ πr = $\frac { 4 }{ 3 }$ x $\frac { 22 }{ 7 }$ x 7 x 7 x 7 = 4312 / 3 cm³ = 1437 $\frac { 1 }{ 3 }$ cm³
Thus, the volume of sphere = 1437 $\frac { 1 }{ 3 }$ cm³

(ii) Radius (r) = 0.63 m

Volume of sphere = $\frac { 4 }{ 3 }$ πr = $\frac { 4 }{ 3 }$ x $\frac { 22 }{ 7 }$ x 0.63 x 0.63 x 0.63 = 1.05 m³ (approx.)

Thus, the volume of sphere = 1.05 m³ (approx.)

Ex 13.8 Question 2.

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

(i) 28 cm
(ii) 0.21 m
Solution:
(i) Radius of the ball (r) cm $\frac { 28 }{ 2 }$ cm = 14cm
Volume of the spherical ball = $\frac { 4 }{ 3 }$ πr³

(ii) Radius(r) = 0.21/2 m = 0.105m
Volume of sphere = $\frac { 4 }{ 3 }$ πr³ = $\frac { 4 }{ 3 }$ x $\frac { 22 }{ 7 }$ x 0.105 x 0.105 x 0.105 = 0.004861 m³
Thus, the amount of water displayed = 0.004861 m³.

Ex 13.8 Question 3.

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:
Radius (r) = $\frac { 4.2 }{ 2 }$ cm = 2.1cm
Volume of sphere = $\frac { 4 }{ 3 }$ πr³ = $\frac { 4 }{ 3 }$ x $\frac { 22 }{ 7 }$ x 2.1 x 2.1 x 2.1 = 38.808 cm³
Mass of 1 cm³ = 8.9 g

∴ Mass of 38.808 cm = 8.9 x 38.808 = 345.39 g (approx.)
Thus, the mass of ball = 345.39 g (approx.)

Ex 13.8 Question 4.

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:

Ex 13.8 Question 5.

How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Solution:
Radius of the hemispherical bowl (r) = $\frac { 10.5 }{ 2 }$ cm

Thus, the capacity of the bowl = 0.303 litres (approx.)

Ex 13.8 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:

Ex 13.8 Question 7.

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

Ex 13.8 Question 8.

A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of ₹498.96. If the cost of white washing is ₹2.00 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
Solution:

Ex 13.8 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:
(i) Let the radius of a sphere = r and the radius of new sphere = r’.
Volume of solid sphere = $\frac { 4 }{ 3 }$ πr³
Volume of 27 solid spheres = 27 x [ $\frac { 4 }{ 3 }$ πr³]
∴ Volume of the new sphere = Volume of 27 solid spheres

Hence, the radius of a new sphere is 3r.

(ii) Surface area of a sphere S = 4πr²
Surface area of a sphere S’ = 4π (r’) = 4π(3r)² = 36πr²

Thus, S : S’ = 1 : 9

Ex 13.8 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:
Radius of the spherical capsule (r) = $\frac { 3.5 }{ 2 }$ mm = 1.75mm
∴ Volume of the spherical capsule = $\frac { 4 }{ 3 }$ πr³
= $\frac { 4 }{ 3 }$ x $\frac { 22 }{ 7 }$ x 1.75 x 1.75 x 1.75
= 22.46 mm³ (approx.)

NCERT solutions for class 9 Maths chapter 13 Surface Areas and Volumes (Exercise – 13.8)