Assume π = 22/7 , unless stated otherwise.
- Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m
Solution:
(i) Volume of sphere = 4/3 πr3 = 22/7 x 4/3 x 73 = 1437.33 m3
(ii) Volume of sphere = 4/3 πr3 = 22/7 x 4/3 x (0.63)3 = 1.05m3
- Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm
(ii) 0.21 m
Solution:
(i)d = 28 cm
r = d/2 = 28/2 = 14 cm
Volume of sphere = 4/3 πr3 = 22/7 x 4/3 x 143 = 11498.67 m3
(ii) d = 0.21 m
r = d/2 = 0.21/2 = 0.105 m
Volume of sphere = 4/3 πr3 = 22/7 x 4/3 x (0.105)3 = 0.004851 m3
- 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:
d = 4.2 cm
r = d/2 = 4.2/2 = 2.1 cm
Volume of sphere = 4/3 πr3 = 22/7 x 4/3 x (2.1)3 = 38.808 cm3
Density of metal for 1 cm3 = 8.9 g
Density of metal for 38.808 cm3 = 38.808 x 8.9 = 345.39g
- 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 diameter of earth = d and radius of earth = d/2
Volume of earth = 4/3 πr3 = 4/3 x 22/7 x (d/2)3
Given, the diameter of the moon is approximately one-fourth of the diameter of the earth.
i.e., diameter of moon = 1/4 x d = d/4
Then radius of moon = d/4 x 1/2 = d/8
Volume of moon = 4/3 πr3 = 4/3 x 22/7 x (d/8)3
Fraction of the volume of the earth is the volume of the moon = [4/3 x 22/7 x (d/2)3]/[ 4/3 x 22/7 x (d/8)3]
= (1/2)3/(1/8)3
= (1/8)/(1/512)
= 512/8 = 64
⇒Volume of the moon = 1/64 x volume of earth
Hence, volume of the moon is 1/64 of volume of earth.
- How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Solution:
d = 10.5 cm
r = d/2 = 10.5/2 = 5.25 cm
Volume of the hemisphere bowl = 2/3 πr3
= 2/3 x 22/7 x (5.25)3
= 303.1875 cm3
Therefore, the hemispherical bowl can hold 303.1875/1000 litres = 0.303 litres of milk.
- 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 of the tank, r = 1m
Thickness of the iron sheet = 1 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 = 2/3 π(R3 – r3)
= 2/3 x 22/7 x (1.013 – 13)
= 0.0635m3
- Find the volume of a sphere whose surface area is 154 cm2.
Solution:
Surface area if the given sphere = 154cm2
We know, Surface area of the given sphere = 4πr2 = 154
4x22/7x r2 = 154
r2 = 154×7/4×22 = 49/4
r = 7/2
Volume of sphere = 4/3 πr3 = 4/3 x 22/7 x (7/2)3 = 179.67cm3
- 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) inside surface area of the dome = total cost/cost of white washing per m2 = 498.96/2 = 249.48 m2
(ii) Let the radius of the dome be ‘r’ m
Then 2πr2 = 249.48
2 x 22/7 x r2 = 249.48
r2 = 249.48×7/2×22 = 3969/100
r = 63/10 cm
volume of the air inside the dome = 2/3πr3 = 2/3 x 22/7 x (6.3)3 = 523.908 cm3
- 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) Volume of a sphere of radius, r = 4/3 πr3
Volume of 27 such spheres = 27 x 4/3 πr3 = 36 πr3
Volume of the sphere with radius r’ = 4/3 πr3
⇒ 36 πr3 = 4/3 πr3
⇒ 27r3 = r’3
⇒ r’ = ∛(27r3)
⇒ r’ = 3r
- 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:
d = 3.5mm
r = 3.5/2 = 1.75 mm
Volume of sphere = 4/3 πr3
= 4/3 x 22/7 x (1.75)3
= 22.46 mm3
Therefore, 22.46mm3 medicine is needed to fill this capsule.
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