A solid of density 5000 kg m^-3weighs 0.5 kgf in air. It is completely immersed in water of density 1000 kg m^-3 .
Calculate the apparent weight of the solid in water.

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Solution

We have,

Density of solid, d = 5000 kg/m3

Weight in air, W = 0.5 kgf = 0.5g N

Density of water, q = 1000 kg/m3

Let, V be the volume of the solid.

So, W = Vdg

=> 0.5g = V × 5000 × g

=> V = 10-4 m-3

So, Buoyant force is, B = Vqg = 10-4 × 1000 × g = 0.1 g N = 0.1 kgf

Thus, apparent weight = W – B = 0.5 – 0.1 = 0.4 kgf

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A solid of density 5000 kg m-3 weighs 0.5 kgf in air. It is completely immersed in water of density 1000 kg m-3 . Calculate the apparent weight of the solid in water.

A solid of density 5000 kg m^-3weighs 0.5 kgf in air. It is completely immersed in water of density 1000 kg m^-3 .
Calculate the apparent weight of the solid in water.