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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.
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.
CBSE | Class 10 | Exercise 1 | maths | Maths | Mensuration | ML Aggarwal
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;

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

R2h = (4/3)r3

×(21/2)2h = (4/3)××(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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