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.

Given,

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

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

$\text{R.D. of iron} = \dfrac{\text {Density of iron in kg m}^{-3}}{\text{1000 kg cm}^{-3}} \\[0.5em] = \dfrac{8.9 \times 10^3}{1000} \\[0.5em] = 8.9 \\[1.5em] \text{From relation, R.D.} = \dfrac{W${1}}{W{1} − W{2}} \\[0.5em] \Rightarrow 8.9 = \dfrac{44.5}{44.5 - W{2}} \\[0.5em] \Rightarrow 8.9 (44.5 - W{2}) = 44.5 \\[0.5em] \Rightarrow (8.9 \times 44.5) - (8.9W{2} ) = 44.5 \\[0.5em] \Rightarrow 396.05 - 8.9W{2} = 44.5 \\[0.5em] \Rightarrow W{2} = \dfrac{396.05 - 44.5}{8.9} \\[0.5em] \Rightarrow W{2} = \dfrac{351.55}{8.9} \\[0.5em] \Rightarrow W{2} = 39.5 \text{ gm} \\[0.5em]

Hence,

Weight of iron piece when immersed in water = 39.5 gm