A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water.
Calculate the rise in water level, assuming that no water overflows.
Solution:
When a solid sphere is totally submerged in water, the volume of the water that it displaces is equal to its own volume.
Given internal diameter of cylindrical can = 21 cm
Radius of the cylindrical can, R = 21/2 cm
Diameter of sphere = 10.5 cm
Radius of the sphere, r = 10.5/2 = 21/4 cm
Let the rise in water level be h.
Rise in volume of water = Volume of sphere immersed
π R2h = (4/3)π r3
π (21/2)2h = (4/3)π ×(21/4)3
(21/2) ×(21/2)×h = (4/3)× (21/4)×(21/4)× (21/4)
h = 21/12
h = 7/4
h = 1.75 cm
Hence the rise in water level is 1.75 cm.
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