A Solid Of Density 5000 Kg M3 Weight 0.5 Kgf In Air. It Is Completely Immersed In Water Of Density 1000 Kg M3. Calculate The Apparent Weight Of The Solid In Water.

Density,\( \rho = 5000kg/m^{3}\)

Weight = 0.5 kg f in air

mass M = 0.5 kg

Volume V = M /\( \rho\)

= 0.5 / 5000 = \( 10^{-4}m^{3}\)

mass of water displaced = V * density of water

= \( 10^{-4} * 1000 = 0.1 kg\)

Apparent weight = wt in the air – wt of water displaced.

= 0.5 kg f – 0.1 kg f

= 0.4 kg f

Therefore, the apparent weight of solid in water is 0.4 kg f

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