Mensuration-8

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There is water in a cylindrical tub, the diameter of whose cross section is 8 cm. 105 iron balls
each of diameter 2 cm are dropped in. By how much will the level of water rise in the cylinder?

Solution: Volume of each lead shot = 4/3 x π x r3

= 4/3 x 22/7 x 1 x 1 x 1

88/21 c.c.

Volume of 105 lead shots = 88/21 x 105 = 440 c.c.

If the level of water in the cylinder is raised by h cm., volume of raised liquid = the volume of water displaced by the lead shots (by Archimedes principle)

V = 22/7 x 8/2 x 8/2 x h

Therefore, 22/7 x 8/2 x 8/2 x h = 440

Hence, h = 440 x 7/22 x 2/8 x 2/8 = 8.75 cm

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