Class 9 - Mathematics
Surface Areas and Volumes - Exercise 13.8
Top Block 1
Exercise 13.8
Question :1 : Find the volume of a sphere whose radius is
(i) 7 cm (ii) 0.63 m
Answer :
(i) Here, radius (r) = 7 cm
Volume of the sphere = 4πr3 /3
= (4/3) * (22/7) * 7 * 7 * 7
= (4/3) * 22 * 7 * 7
= 4312/3 cm3
(ii) Here, radius (r) = 0.63 m
Volume of the sphere = 4πr3 /3
= (4/3) * (22/7) * 0.63 * 0.63 * 0.63
= (4/3) * 0.09 * 0.63 * 0.63
= 4 * 0.09 * 0.21 * 0.63
= 1.05 m3 (approx.)
Question : 2: Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm (ii) 0.21 m
Answer :
(i) Diameter of the ball = 28 cm
Radius of the ball = 28/2 = 14 cm
Now, volume of the spherical ball = 4πr3 /3
= (4/3) * (22/7) * 14 * 14 * 14
= (4/3) * 22 * 14 * 14 * 2
= 34496/3 cm3
(ii) Diameter of the ball = 0.21 m
Radius of the ball = 0.21/2 m
Now, volume of the spherical ball = 4πr3 /3
= (4/3) * (22/7) * (0.21/2) * (0.21/2) * (0.21/2)
= (4/3) * 22 * (0.03/2) * (0.21/2) * (0.21/2)
= 11 * 0.03 * 0.07 * 0.21
= 0.004851cm3
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?
Answer :
Diameter of the metallic ball = 4.2 cm
Radius (r) of the metallic ball = 4.2/2 = 2.1 cm
Now, volume of the ball = 4πr3 /3
= (4/3) * (22/7) * 2.1 * 2.1 * 2.1
= 4 * 22 * 0.7 * 0.3 * 2.1
= 38.808 cm3
Density of the metal = 8.9 g/cm3
So, mass of the ball = 8.9 * 38.808 = 345.39 g
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?
Answer :
Let diameter of the earth be 2r.
Then radius of the earth = 2r/2 = r
So, the diameter of the moon = 2r/4 = r/2
Radius of the moon = (r/2)/2 = r/4
Volume of the earth = 4πr3 /3 ………..1
Volume of the moon = {4π(r/4)3 }/3 ………..2
Now, Volume of the earth/ Volume of the moon = {4πr3 /3}/{4π(r/4)3 }/3
= r3 /(r3 /64)
= 64
⇒ Volume of the moon = (1/64) * Volume of the earth
Hence, volume of the moon is 1/164 of volume of the earth.
Question : 5: How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Answer :
Diameter of the hemisphere = 10.5 cm
Radius (r) of the hemisphere = 10.5/2 = 5.25 cm
Volume of the hemisphere = 2πr3 /3
= (2/3) * (22/7) * 5.25 * 5.25 * 5.25
= 2 * 22 * 1.75 * 5.25 * 0.75
= 303 cm3
= 303/1000 l
= 0.303 l
Hence, the hemisphere bowl can hold 0.303 litres 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.
Answer :
Inner radius (r) of the tank = 1m
Thickness of the iron sheet = 1 cm = 1/100 m = 0.01 m
External radius of the tank (R) = 1 + 0.01 = 1.01 m
Volume of the iron used to make the tank = 2π(R3 – r3)/3
= (2/3) * (22/7) * [(1.01)3 – 13]
= (2/3) * (22/7) * 0.030301
= 0.06348 m3
Question : 7: Find the volume of a sphere whose surface area is 154 cm2.
Answer :
Let r be the radius of the sphere.
Its surface area = 4πr2
⇒ 4πr2 = 154
⇒ r2 = 154/4π
⇒ r2 = 154/(4 * 22/7)
⇒ r2 = (154 * 7)/(4 * 22)
⇒ r2 = (7 * 7)/4
⇒ r2 = 49/4
⇒ r = √(49/4)
⇒ r = 7/2
⇒ r = 3.5
Now, volume of the sphere = 4πr3 /3
= (4/3) * (22/7) * 3.5 * 3.5 * 3.5
= (4/3) * 22 * 0.5 * 3.5 * 3.5
= 539/3 cm3
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.
Answer :
(i) Inner surface of the dome = Total cost/Cost of white per m2
= 498.96/2
= 249.48 m2
(ii) Let radius of the dome be r m.
Then, 2πr2 = 249.48
⇒ r2 = 249.48/2π
⇒ r2 = 249.48/(2 * 22/7)
⇒ r2 = (249.48 * 7)/(2 * 22)
⇒ r2 = (11.34 * 7)/2
⇒ r2 = 5.67 * 7
⇒ r2 = 39.69
⇒ r = √39.69
⇒ r = 6.3 cm
So, volume of the air inside the dome = 2πr3 /3
= (2/3) * (22/7) * 6.3 * 6.3 * 6.3
= 2 * 22 * 0.9 * 2.1 * 6.3
= 523.9 cm3
Mddle block 1
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′.
Answer :
(i) Volume of a sphere of radius r = 4πr3 /3
Volume of 27 such sphere = 27 * 4πr3 /3
= 9 * 4πr3
= 36πr3
Volume of a sphere of radius r’ = 4π(r’)3 /3
⇒ 4π(r’)3 /3 = 36πr3
⇒ (r’)3 /3 = 9r3
⇒ (r’)3 = 9 * 3 * r3
⇒ (r’)3 = 27r3
⇒ (r’)3 = (3r)3
⇒ r’ = 3r
(ii) Surface are (S) of the sphere with radius r = 4πr2
Surface are (S’) of the sphere with radius r = 4π(r’)2
= 4π(3r)2
= 4π * 9r2
= 36πr2
Now, S/S’ = 4πr2 /36πr2
⇒ S/S’ = 1/9
⇒ S : S’ = 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?
Answer :
Given, diameter of sphere = 3.5 mm
Radius of the sphere = 3.5/2 = 1.75 mm
Now, volume of the capsule = 4πr3 /3
= (4/3) * (22/7) * 1.75 * 1.75 * 1.75
= (4/3) * 22 * 0.25 * 1.75 * 1.75
= 67.375/3
= 22.46 mm3 (approx.)
Hence, 22.46 mm3 of medicine is needed to fill the capsule.
