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 \(\frac {1}{3}\) cm^{3}

(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 m^{3}

= 1.05 m^{3} (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\(\frac {2}{3}\) cm^{3}

Therefore, amount of water displaced by solid sphere is 11498\(\frac {2}{3}\) cm^{3}.

(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 m^{3}

Therefore, the amount of water displaced by the solid sphere is 0.004851 m^{3}

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 cm^{3}?

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 cm^{3}

The density of the metal is 8.9 g per cm^{3}.

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 cm^{3}

We know that 1000 cm^{3} = 1 litre

∴ 303.18 cm^{3} = 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 m^{3} (approx)

Therefore, the volume of the iron used to make the tank is 0.06348 m^{3} (approx)

Question 7.

Find the volume of a sphere whose surface area is 154 cm^{2}.

Solution:

We have given that

The surface area of a sphere is 154 cm^{2}

but, the surface area of sphere = 4πr^{2}

Volume of sphere = \(\frac{4}{3} \pi r^{3}\)

= \(\frac{4}{3} \times \frac{22}{7}\) × 3.5 × 3.5 × 3.5

= 179\(\frac{2}{3}\) cm^{3}

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 m^{2} area

Rs. 1 is the rate of white-washing \(\frac{1}{2}\) m^{2} area

Rs. 498.96 is the rate of white-washing = \(\frac{1}{2}\) × 498.96 = 249.48 m^{2} area.

Therefore, according to question, inside surface area of dome = 249.48 m^{2}

(ii) We have,

surface area of hemispherical dome = 249.48 m^{2}

but surface area of hemispherical dome = 2πr^{2}

⇒ 2πr^{2} = 249.48

⇒ r^{2} = \(\frac{249.48 \times 7}{2 \times 22}\)

⇒ r^{2} = 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 m^{3}

Therefore, the volume of the air inside the dome is 523.9 m^{3} (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) = 4πr^{2}

and volume of sphere = \(\frac{4}{3} \pi r^{3}\)

Volume of 27 such speres = \(\frac{4}{3} \times \pi r^{3} \times 27\) = 36πr^{3}

According to question Volume of 27 small sphere = volume of big sphere

or, 36πr^{3} = \(\frac{4}{3} \pi\left(r^{\prime}\right)^{3}\) (∵ 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} = 27r^{3}

⇒ 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 = 4πr^{2}

∴ 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 mm^{3}) 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 mm^{3} (approx)

Therefore, medicine is needed to fill the capsule is 22.4583 mm^{3} (approx)