## Karnataka Board Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 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) r = 7 cm, V = ?

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 solid spherical ball, diameter, d = 28 cm,

∴ radius, r = \(\frac{28}{2}\) = 14 cm.

Volume of solid spherical ball, V = \(\frac{4}{3}\) πr^{3}.

Amount of water displaced by the ball = 0.00485 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:

Diameter of metallic ball, d = 4.2 cm. density = 8.9 gm/cm^{3}

mass, m = ? d = 4.2 cm.

∴ V = 38.808 cm^{3}.

∴ Mass = Density × Volume

= 8.9 × 38.808

= 345.40 gm.

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 earth be ‘d’ unit.

∴ Volume of Moon = \(\frac{1}{64}\) × Volume of Earth

Question 5.

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

Solution:

Diameter of a hemispherical bowl, d = 10.5 cm

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:

In a hemispherical tank,

Inner radius, r_{1} = 1 m

Outer radius, r_{2} = 1 + 0.01 = 1.01 m (∵ 1 cm = 0.01)

∴ Volume of hemispherical tank, V

= Volume of outer diameter – Volume of inner diameter.

Question 7.

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

Solution:

Surface area of sphere, 4πr^{2} =154 cm^{2}.

Volume of Sphere, V = ?

A = 4πr^{2} = 154

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) Cost of white-washing dome is Rs. 498.96

Cost of white-washing is Rs. 2 per sq. metre.

∴ Surface Area = \(\frac{498.96}{2}\) = 249.48 m^{2}.

(ii) Surface Area of hemisphere, V = 2πr^{2}

∴ Inner volume of hemisphere dome, V

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_{1}. Find the

(i) radius r of the new sphere,

(ii) ratio of S and S_{1}

Solution:

(i) Volume of 1 sphere, V = \(\frac{4}{3}\)πr^{3}

Volume of 27 solid sphere

= 27 × \(\frac{4}{3}\)πr^{3}

Let r_{1}is the radius of the new sphere.

Volume of new sphere = Volume of 27 solid sphere

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 capsule of medicine, d

d = 3.5 mm = \(\frac{7}{2}\) mm.

##### KSEEB Solutions for Class 9 Maths