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

NCERT Solutions for Class 9 Maths Chapter 13 Surface
Areas and Volumes Ex 13.8 are the part
of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Chapter 13 Surface
Areas and Volumes Ex 13.8.

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

**Ex
13.8 Class 9 Maths Question 1.
**Find the volume of a sphere whose radius is:

(i) 7 cm

(ii) 0.63 cm

**Solution:
**(i) Here, radius (r) = 7 cm

Thus, the required volume = 143713 cm

^{3}

(ii) Here, radius (r) = 0.63 m

Thus, the required volume is 1.05 m^{3} (approx.)

**Ex
13.8 Class 9 Maths 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 ball = 28 cm

Radius of the ball (r) = 28/2 cm = 14 cm

(ii)
Diameter of the ball = 0.21 m

⇒ Radius (r) = 0.21/2 m = 21/200 m

Thus, the amount of water displaced = 0.004851 m^{3}.

**Ex 13.8 Class 9 Maths 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:
**Diameter of the metallic ball = 4.2 cm

⇒ Radius of the metallic ball (r) = 4.2/2 cm = 2.1 cm

Density of the metal = 8.9 g per cm

^{3}

∴ Mass of the ball = 8.9 × [Volume of the ball]

Thus, the mass of ball is 345.39 g (approx.)

**Ex
13.8 Class 9 Maths 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 2r.

⇒ Radius of the earth = 2r/2 = r

Since, diameter of the moon = ¼ (Diameter of the earth)

⇒ Radius of the moon = ¼ (Radius of the earth)

Radius of the moon = ¼ (r) = r/4

∴ Volume of the earth = 4/3 Ï€r

^{3}and

**Ex
13.8 Class 9 Maths Question 5.**

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

**Solution:
**Diameter of the hemispherical bowl = 10.5 cm

⇒ Radius of the hemispherical bowl (r) = 10.5/2 cm = 105/20 cm

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

**Ex
13.8 Class 9 Maths 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:
**Inner radius (r) = 1 m

∵ Thickness = 1 cm = 1/100 m = 0 .01 m

Thus, the required volume of the iron used = 0.06348
m^{3} (approx.)

**Ex 13.8 Class 9 Maths Question 7.
**Find the volume of a sphere whose surface area is
154 cm

^{2}.

**Solution:
**Let ‘r’ be the radius of the sphere.

∴ Surface area of the sphere = 4Ï€r

^{2}

4Ï€r

^{2}= 154 [Given]

**Ex
13.8 Class 9 Maths 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:
**(i) Total cost of white-washing = ₹ 498.96

Cost of 1 m² of white-washing = ₹ 2

∴ Area = Total/Cost/Cost of 1 m

^{2}area = 498.96/2 = 249.48 m

^{2}

Thus, the required surface area of the dome is 249.48 m

^{2}.

(ii) Let ‘r’ be the radius of the hemispherical
dome.

∴ Inside surface area of the dome = 2Ï€r^{2}

Now, volume of air inside the dome = Volume of a hemisphere

^{3}(approx.).

**Ex
13.8 Class 9 Maths 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) The radius of the small sphere is r.

∴ Its volume = 4/3 Ï€r

^{3}

Volume of 27 small spheres 27 × [4/3 Ï€r

^{3}]

Let the radius of the new sphere be r’.

∴ Volume of the new sphere = 4/3 Ï€(r’)

^{3}

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

(ii) Surface area of a sphere = 4Ï€r^{2}

S = 4Ï€r^{2} and S’ = 4Ï€ (3r)^{2} [∵ r’ = 3r]

Thus, S : S’ = 1 : 9

**Ex
13.8 Class 9 Maths 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:
**Diameter of the spherical capsule = 3.5 mm

⇒ Radius of the spherical capsule (r) = 3.5/2 mm

= 22.45833 mm

^{3}

^{3}(approx.)

Thus, the required quantity of medicine = 22.46
mm^{3 }(approx.)

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