# 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 5000kgm3 weighs 0.5kgf in air. if it floats in a liquid of density 8000kgm3, 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/m3, 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/m3

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/m3

This is the density of paraffin..

Density of the solid is = 5000 kg/m3

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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