a piece of iron weighs 44.5gf in air, 39.5gf in water 40.3gf in parrafin oil. find density of iron density of paraffin oil.

a solid of density 5000kgm^{3 }weighs 0.5kgf in air. if it floats in a liquid of density 8000kgm^{3}, what is its apparent weight?

Let the volume of iron be V.

Its weight in air is = 44.5 gf = 0.445g N

Its weight in water is = 39.5 gf = 0.395g N

Its weight in paraffin is 40.3 gf = 0.403g N

If the volume of the iron piece is V and its density is d then,

Vdg = 0.445 g

=> Vd = 0.445 ……….(1)

We know density of water is 1000 kg/m^{3}, so,

V(1000)g = 0.445g - 0.395g

=> 1000V = 0.05

=> V = 0.00005 …….(2)

(1) and (2) => d = 0.445/0.00005 = 8900 kg/m^{3}

This is the density of iron.

If q is the density of paraffin then,

Vqg = 0.445g - 0.403g

=> 0.00005q = 0.042

=> q = 840 kg/m^{3}

This is the density of paraffin..

Density of the solid is = 5000 kg/m^{3}

Its weight in air is, W = 0.5 kgf = 0.5g N

The buoyant force (B) by the liquid is equal to its weight since it floats in the liquid.

B = 0.5g N

Thus, the apparent weight in the liquid is = W – B = 0

When a body floats, its apparent weight is zero.

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