Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.8 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.8

Assume π = \(\frac{22}{7}\), unless stated otherwise.

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 (r) = 7 cm

∴ Volume of sphere = \(\frac{4}{3}\)πr^{3}

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

= \(\frac{88 \times 49}{3}\)

= \(\frac{4312}{3}\)

= 4312\(\frac{1}{3}\) cm^{3}.

Hence,volume of the spher = 4312\(\frac{1}{3}\) cm^{3}.

(ii) We have,

Radius of the sphere (r) = 0.63 m

∴ Volume of the sphere = \(\frac{4}{3}\)πr^{3}

= \(\frac{4}{3} \times \frac{22}{7} \times(0.63)^3\)

= 88 × 0.03 × 0.3969

= 1.0478 m^{3} = 1.05 m^{3}

Hence,volume of the sphere = 1.05 m^{3}.

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,

Diameter of spherical ball = 28 cm

∴ Radius of the spherical ball (r) = \(\frac{28}{2}\)

= 14 cm

Amount of water displaced by spherical ball = Volume of spherical ball

= \(\frac{4}{3} \pi r^3=\frac{4}{3} \times \frac{22}{7} \times(14)^3\)

= \(\frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14\)

= \(\frac{88 \times 2 \times 196}{3}\)

= \(\frac{34496}{3}\)

= 11498\(\frac{2}{3}\) cm^{3}

Amount of water displaced by spherical ball

=11498\(\frac{2}{3}\) cm^{3}

(ii) We have,

Diameter of spherical ball = 0.21 m

∴ Radius of spherical ball (r) = \(\frac{0.21}{2}\)

= 0.105 m

Amount of water displaced by spherical ball = Volume of spherical ball

= \(\frac{4}{3}\)πr^{3}

= \(\frac{4}{3} \times \frac{22}{7} \times(0.105)^3\)

= \(\frac{4}{3} \times \frac{22}{7}\) × 0.105 × 0.105 × 0.105

= 88 × 0·005 × 0.011025

= 0.004851 m^{3}

Hence, amount of water displaced by spherical ball = 0.004851 m^{3}.

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 cm^{3}?

Solution:

We have,

Diameter of ball = 4.2 cm

∴ Radius of the ball (r) = \(\frac{4.2}{2}\) = 2.1 cm

Since, ball is in the shape of sphere, therefore we need to calculate the volume of sphere for its mass.

∴ Volume of spherical ball = \(\frac{4}{3}\)πr^{3}

= \(\frac{4}{3} \times \frac{22}{7} \times(2.1)^3\)

= \(\frac{88}{21}\) × 2.1 × 2.1 × 2.1

= 8.8 × 4.41 = 38.808 cm^{3}

mass of the ball = volume × density

= 38.808 × 8.9

= 345.39 grams (approx.)

Hence,mass of the ball = 345.39 grams (approx.)

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:

Let the diameter of the earth be 2x.

∴ Radius of the earth (R) = \(\frac{2 x}{2}\) = x

According to question,

Diameter of the moon = \(\frac{1}{4}\) of diameter of the earth = \(\frac{1}{4}\) × 2x = \(\frac{x}{2}\)

∴ Radius of the moon (r) = \(\frac{x}{4}\)

Volume of the earth = \(\frac{4}{3}\)πr^{3} = \(\frac{4}{3}\)πx^{3}

Volume of the moon = \(\frac{4}{3}\)πr^{3} = \(\frac{4}{3} \pi\left(\frac{x}{4}\right)^3\)

= \(\frac{4}{3} \pi \frac{x^3}{64}\)

⇒ Volume of the moon = \(\frac{1}{4}\) × volume of the earth

Hence, \(\frac{1}{64}\) fraction of volume of earth is volume of the moon.

Question 5.

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

Solution:

We have,

Diameter of hemispherical bowl = 10.5 cm

∴ Radius of hemispherical bowl (r) = \(\frac{10.5}{2}\)

= 5.25 cm

∴ Volume of the hemispherical bowl

= \(\frac{2}{3} \pi r^3\)

= \(\frac{2}{3} \times \frac{22}{7} \times(5.25)^3\)

= \(\frac{2}{3} \times \frac{22}{7}\) × 5.25 × 5.25 × 5.25

= 44 × 0.25 × 27.5625

= 303.1875 cm^{3}

= 0.3031875 litres

= 0.303 litres (approx.)

Hence, hemispherical bowl, can hold 0.303 litres (approx.) 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,

Inner radius of the hemispherical tank (r) = 1 m

Thickness of the iron sheet = 1 cm

= 0.01 m

∴ Outer radius of the hemispherical tank

(R) = 1 + 0.01 = 1.01 m

Volume of the iron used to make the tank = External volume – Internal volume

= \(\frac{4}{3}\)πR^{3} – \(\frac{2}{3}\)πr^{3}

= \(\frac{2}{3}\)π[R^{3} – r^{3}]

= \(\frac{2}{3} \times \frac{22}{7}\)[(1.01)^{3} – (1)^{3}]

= \(\frac{44}{21}\)(1.030301 – 1)

= \(\frac{44}{21}\) × 0.030301

= 0·06348 m^{3} (approx.)

Hence, volume of the iron used to make the tank = 0·06348 m^{3} (approx.)

Question 7.

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

Solution:

Let the radius of sphere be r сm.

Surface area of the sphere = 154 cm^{2} (given)

Hence,volume of the sphere = 179\(\frac{2}{3}\) cm^{3}.

Question 8.

A dome of a building is in the form of a hemisphere. From inside, it was whitewashed at the cost of Rs. 498.96. If the cost of the 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:

Let inner radius of dome be r m.

We have, Rate of white washed = Rs. 2.00 per m^{2}

Total cost of white washed = Rs. 498.96

(i) Inside curved surface area of the dome

= \(\frac{\text { Total cost }}{1 m^2 \cos t}=\frac{498.96}{2}\)

(ii) We have,

Inside C.S.A. of the dome = 249.48

2πr^{2} = 249.48

2 × \(\frac{22}{7}\) × r^{2} = 249.48

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

r^{2} = 39.69

r = \(\sqrt{39.69}\)

r = 63 m.

Volume of air inside the dome = \(\frac{2}{3}\)πr^{3}

= \(\frac{2}{3} \times \frac{22}{7} \times(6.3)^3\)

= \(\frac{2 \times 22}{21}\) × 6.3 × 6.3 × 6.3

= 44 × 0.3 × 39.69

= 523.908 m^{3}

= 523.9 m^{3} (approx.)

Hence, (i) Curved surface area of the dome = 249.48 m^{2}

(ii) Volume of air inside the dome = 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) radio of S and S’.

Solution:

We have,

Twenty seven spheres, each of radius r and surface area S are melted to form a sphere with surface area S’.

(i) Then S = 4πr^{2} …..(i)

Volume of one sphere = \(\frac{2}{3}\)πr^{3}

Volume of 27 spheres = 27 × \(\frac{2}{3}\)πr^{3}

Volume of new sphere = Volume of 27 spheres

⇒ \(\frac{4}{3} \pi r^{\prime} 3=27 \times \frac{4}{3} \pi r^3\)

⇒ \(r^{\prime 3}=\frac{27 \times \frac{4}{3} \pi r^3}{\frac{4}{3} \pi}\)

⇒ r’^{3} = 27r^{3} = (3r)^{3}

⇒ r’= 3r

(ii) Surface area of new sphere (S’) = 4πr’^{2}

= 4π × (3r)^{2} = 4π × 9r^{2}

= 36πr^{2}

Now, \(\frac{S}{S^{\prime}}=\frac{4 \pi r^2}{36 \pi r^2}\)

⇒ \(\frac{S}{S^{\prime}}=\frac{1}{9}\)

⇒ S : S’ = 1 : 9

Hence (i)Radius of new sphere (r’) = 3r

(ii) S : S’ = 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,

Diameter of spherical capsule = 3.5 mm

∴ Radius of spherical capsule (r) = \(\frac{3.5}{2}\)

= 1.75 mm

Volume of the capsule = \(\frac{4}{3}\)πr^{3}

= \(\frac{4}{3} \times \frac{22}{7} \times(1.75)^3\)

= \(\frac{4}{3} \times \frac{22}{7}\) × 1.75 × 1.75 × 1.75

= \(\frac{88 \times 0.25 \times 3.0625}{3}\)

= \(\frac{67.375}{3}\)

= 22.46 mm^{3} (approx.)

Hence, volume of medicine in the capsule = 22.46 mm^{3} (approx.)