Question

A piece of iron weighs 44.5 gf in air. If the density of iron is 8.9 x 103 kg m-3, find the weight of the iron piece when immersed in water.

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Answer

Weight of the solid in air (W1) = 44.5 gf

Density of the iron = 8.9 x 103 kg m-3

RD of iron = 8.9 × 1000\over 1000

= 8.9

RD = W_1\over W_1 - W_2

⇒ 8.9 = 44.5\over 44.5 - W_2

⇒ 8.9 (44.5 – W2) = 44.5

⇒ 396.05 – 8.9 W2 = 44.5

⇒ 8.9 W2 =  396.05 – 44.5

⇒ W2 = 351.55\over 8.9 = 39.5 gm

Hence,

Weight of iron piece when immersed in water = 39.5 gm

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