Students can Download Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 9 Maths 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) r = 7 cm, V = ?

∴ the volume of the sphere is 1437\(\frac{1}{3}\) cm^{3}

(ii) Radius of sphere = 0.63

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}.

(ii) radius of ball = \(\frac{0.21}{2}\) = 0.105 m

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:

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

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:

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

Radius of hemispherical bowl = \(\frac{2}{3}\) πr^{3}

= 0.3031875 = 0.303 litres (approximats).

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.

= 0.06348 m^{3} (approximate)

Question 7.

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

Solution:

Surface area of a 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 the 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) Let the inner redius be ‘r’

2πr^{2} = 249.48

Volume of air inside the dome =

volume of hemispherical dome

= 523.908 m^{3}

∴ Volume of air inside the dome is 523 .9 m^{3}

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

= \(\frac{4}{3}\)πr^{3}

r^{3} = 27r^{3}

r^{1} = 3r

(ii) Surface area of 1 solid iron sphere of radius r = 4πr^{2}

Surface area of iron sphere of radius r1 = 4π(r1)^{2}

4π(3r)^{2} = 36π^{2}

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: