Bihar Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.8 Textbook Questions and Answers.

## BSEB Bihar Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.8

Question 1.

Find the volume of a sphere whose radius is

(i) 7 cm

(ii) 0.63 m

Solution:

(i) We have : r = radius of the sphere = 7 cm.

(ii) We have : r = radius of the sphere = 0.63 m

Question 2.

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

(i) 28 cm

(ii) 0.21 m

Solution:

(i) Diameter of the spherical ball = 28 cm

∴ Radius = \(\frac { 28 }{ 2 }\) cm = 14 cm

Amount of water displaced by the spherical ball

= Its volume = \(\frac { 4 }{ 3 }\) πr³

= (\(\frac { 4 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x 14 x 14 x 14) cm³

= \(\frac { 34496 }{ 3 }\) cm³

= 11498\(\frac { 2 }{ 3 }\) cm³

= 11498\(\frac { 2 }{ 3 }\) cm³.

(ii) Diameter of the spherical ball = 0.21 m

∴ Radius = (\(\frac { 0.21 }{ 2 }\)) = 0.105 m

Amount of water displaced by the spherical ball

= Its volume = \(\frac { 4 }{ 3 }\) πr³

= (\(\frac { 4 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x 0.105 x 0105 x 0105) m³

= 0.004851 m³.

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:

Diameter of the ball = 4.2 cm

Density of the metal is 8.9 g per cm³

Mass of the ball = (38.808 x 8.9) g = 345.3912 g.

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 the moon be r. Then, the radius of the moon = \(\frac { r }{ 2 }\)

According to the question, diameter of the earth is 4r, so 4 r its radius = \(\frac { 4r }{ 2 }\) = 2r

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

Question 5.

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

Solution:

Diameter of a hemispherical bowl = 10.5 cm.

Hence, the hemispherical bowl can hold 303 l (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:

Let R cm and r cm be respectively the external and internal radii of the hemispherical vessel. Then,

R = 1.01 (as thickness = 1 cm = .01 m)

and r = 1 m.

Volume of iron used = External volume – Internal volume

Question 7.

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

Solution:

Let r cm be the radius of the sphere.

So, surface area = 154 cm²

Question 8.

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

(ii) Let r be the radius of the dome.

∴ Surface area = 2πr²

Volume of the air inside the dome = Volume of the dome

= \(\frac { 2 }{ 3 }\) πr³

= \(\frac { 2 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x 6.3 x 6.3 x 6.3 m³

= 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:

(i) Volume of 27 solid sphere of radius

r = 27 x \(\frac { 4 }{ 3 }\)πr³ … (1)

Volume of the new sphere of radius r’ = \(\frac { 4 }{ 3 }\) πr’³ … (2)

According to the problem, we have

\(\frac { 4 }{ 3 }\)πr’³ = 27 x \(\frac { 4 }{ 3 }\)πr³

⇒ r’3 = 27r³ = (3r)³

∴ r’ = 3r.

(ii) Required ratio = \(\frac { S }{ S’ }\) = \(\frac{4 \pi r^{2}}{4 \pi r^{22}}\) = \(\frac{r^{2}}{(3 r)^{2}}\)

= \(\frac{r^{2}}{9 r^{2}}\) = \(\frac { 1 }{ 9 }\) = 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:

Diameter of the spherical capsule = 3.5 mm

Radius = \(\frac { 3.5 }{ 3 }\) mm

= 1.75 mm.

Medicine needed for its filling = Volume of spherical capsule

= \(\frac { 4 }{ 3 }\)πr³

= (\(\frac { 4 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x 1.75 x 1.75 x 1.75) m³

= 22.46 m³ (approx).