1. Find the volume of a sphere whose radius is
(i) 7 cm (ii) 0.63 m
(Assume π =22/7)
Solution:
(i) Radius of sphere, r = 7 cm
Using, Volume of sphere = (4/3) πr3
= (4/3)×(22/7)×73
= 4312/3
Hence, volume of the sphere is 4312/3 cm3
(ii) Radius of sphere, r = 0.63 m
Using, volume of sphere = (4/3) πr3
= (4/3)×(22/7)×0.633
= 1.0478
Hence, volume of the sphere is 1.05 m3 (approx).
2. Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm (ii) 0.21 m
(Assume π =22/7)
Solution:
(i) Diameter = 28 cm
Radius, r = 28/2 cm = 14cm
Volume of the solid spherical ball = (4/3) πr3
Volume of the ball = (4/3)×(22/7)×143 = 34496/3
Hence, volume of the ball is 34496/3 cm3
(ii) Diameter = 0.21 m
Radius of the ball =0.21/2 m= 0.105 m
Volume of the ball = (4/3 )πr3
Volume of the ball = (4/3)× (22/7)×0.1053 m3
Hence, volume of the ball = 0.004851 m3
3.The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3? (Assume π=22/7)
Solution:
Given,
Diameter of a metallic ball = 4.2 cm
Radius(r) of the metallic ball, r = 4.2/2 cm = 2.1 cm
Volume formula = 4/3 πr3
Volume of the metallic ball = (4/3)×(22/7)×2.1 cm3
Volume of the metallic ball = 38.808 cm3
Now, using relationship between, density, mass and volume,
Density = Mass/Volume
Mass = Density × volume
= (8.9×38.808) g
= 345.3912 g
Mass of the ball is 345.39 g (approx).
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”. Therefore, the radius of earth will be will be d/2
Diameter of moon will be d/4 and the radius of moon will be d/8
Find the volume of the moon :
Volume of the moon = (4/3) πr3 = (4/3) π (d/8)3 = 4/3π(d3/512)
Find the volume of the earth :
Volume of the earth = (4/3) πr3= (4/3) π (d/2)3 = 4/3π(d3/8)
Fraction of the volume of the earth is the volume of the moon
Answer: Volume of moon is of the 1/64 volume of earth.
5. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22/7)
Solution:
Diameter of hemispherical bowl = 10.5 cm
Radius of hemispherical bowl, r = 10.5/2 cm = 5.25 cm
Formula for volume of the hemispherical bowl = (2/3) πr3
Volume of the hemispherical bowl = (2/3)×(22/7)×5.253 = 303.1875
Volume of the hemispherical bowl is 303.1875 cm3
Capacity of the bowl = (303.1875)/1000 L = 0.303 litres(approx.)
Therefore, hemispherical bowl can hold 0.303 litres of milk.
6. A hemi spherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. (Assume π = 22/7)
Solution:
Inner Radius of the tank, (r ) = 1m
Outer Radius (R ) = 1.01m
Volume of the iron used in the tank = (2/3) π(R3– r3)
Put values,
Volume of the iron used in the hemispherical tank = (2/3)×(22/7)×(1.013– 13) = 0.06348
So, volume of the iron used in the hemispherical tank is 0.06348 m3.
7. Find the volume of a sphere whose surface area is 154 cm2. (Assume π = 22/7)
Solution:
Let r be the radius of a sphere.
Surface area of sphere = 4πr2
4πr2 = 154 cm2 (given)
r2 = (154×7)/(4 ×22)
r = 7/2
Radius is 7/2 cm
Now,
Volume of the sphere = (4/3) πr3
8. A dome of a building is in the form of a hemi sphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing isRs20 per square meter, find the
(i) inside surface area of the dome (ii) volume of the air inside the dome
(Assume π = 22/7)
Solution:
(i) Cost of white-washing the dome from inside = Rs 4989.60
Cost of white-washing 1m2 area = Rs 20
CSA of the inner side of dome = 498.96/2 m2 = 249.48 m2
(ii) Let the inner radius of the hemispherical dome be r.
CSA of inner side of dome = 249.48 m2 (from (i))
Formula to find CSA of a hemi sphere = 2πr2
2πr2 = 249.48
2×(22/7)×r2 = 249.48
r2 = (249.48×7)/(2×22)
r2 = 39.69
r = 6.3
So, radius is 6.3 m
Volume of air inside the dome = Volume of hemispherical dome
Using formula, volume of the hemisphere = 2/3 πr3
= (2/3)×(22/7)×6.3×6.3×6.3
= 523.908
= 523.9(approx.)
Answer: Volume of air inside the dome is 523.9 m3.
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 Sand S’.
Solution:
Volume of the solid sphere = (4/3)πr3
Volume of twenty seven solid sphere = 27×(4/3)πr3 = 36 π r3
(i) New solid iron sphere radius = r’
Volume of this new sphere = (4/3)π(r’)3
(4/3)π(r’)3 = 36 π r3
(r’)3 = 27r3
r’= 3r
Radius of new sphere will be 3r (thrice the radius of original sphere)
(ii) Surface area of iron sphere of radius r, S =4πr2
Surface area of iron sphere of radius r’= 4π (r’)2
Now
S/S’ = (4πr2)/( 4π (r’)2)
S/S’ = r2/(3r’)2 = 1/9
The ratio of S and S’ is 1: 9.
10. A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm3) is needed to fill this capsule? (Assume π = 22/7)
Solution:
Diameter of capsule = 3.5 mm
Radius of capsule, say r = diameter/ 2 = (3.5/2) mm = 1.75mm
Volume of spherical capsule = 4/3 πr3
Volume of spherical capsule = (4/3)×(22/7)×(1.75)3 = 22.458
Answer: The volume of the spherical capsule is 22.46 mm3.