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\(\frac{1}{3}\) cm3
(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 = \(\frac{4}{3}\) πr3.
(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 cm3 ?
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 = \(\frac{2}{3}\) πr3
= 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, r1 = 1 m
Outer radius, r2 = 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 m3 (approximate)
Question 7.
Find the volume of a sphere whose surface area is 154 cm2.
Solution:
Surface area of a sphere, 4πr2 =154 cm2.
Volume of Sphere, V =?
A = 4πr2 = 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 m2.
(ii) Let the inner redius be ‘r’
2πr2 = 249.48
Volume of air inside the dome =
volume of hemispherical dome
= 523.908 m3
∴ Volume of air inside the dome is 523 .9 m3
Question 9.
Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S1. Find the
(i) radius r of the new sphere,
(ii) ratio of S and S1
Solution:
(i) Volume of 1 sphere, V = \(\frac{4}{3}\)πr3
Volume of 27 solid sphere
= 27 × \(\frac{4}{3}\)πr3
Let r1is the radius of the new sphere.
Volume of new sphere = Volume of 27 solid sphere
= \(\frac{4}{3}\)πr3
r3 = 27r3
r1 = 3r
(ii) Surface area of 1 solid iron sphere of radius r = 4πr2
Surface area of iron sphere of radius r1 = 4π(r1)2
4π(3r)2 = 36π2
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:
