A body weighs W₁ gf in air and when immersed in a liquid, it weighs W₂ gf, while it weighs W₃ gf on immersing it in water. Find: (i) volume of the body (ii) upthrust due to liquid (iii) relative density of the solid and (iv) relative density of the liquid.

Given,

Weight of the body in air = W₁ gf

Weight of the body in liquid = W₂ gf

Weight of the body in water = W₃ gf

(i) Let V be the volume of the body.

Upthrust due to water = loss in weight when immersed in water = (W₁ - W₃) gf    …[Eq 1]

Weight of water displaced = Volume of water displaced x density of water x g

Volume of water displaced = Volume of the body = V

Density of water = 1 g cm^-3

∴ Weight of water displaced = V x 1 x g = V gf     …[Eq 2]

But weight of water displaced is equal to upthrust due to water

∴ From eqns 1 & 2,

V = (W₁ - W₃) cm³

∴ Volume of the body = V = (W₁ - W₃) cm³

(ii) Upthrust due to liquid = loss in weight when immersed in liquid = (W₁ - W₂) gf

(iii) Relative density of the solid

$\text{R.D. of solid} \\[0.5em] = \dfrac{(\text{Wt of body})$\text{air}}{(\text{Wt of body})\text{air}-(\text{Wt of body})\text{water}} \\[0.5em] = \dfrac{W1}{W1 - W3}

(iv) Relative density of the liquid

$\text{R.D. of liquid} \\[0.5em] = \dfrac{(\text{Wt of body})$\text{air}-(\text{Wt of body})\text{liquid}}{(\text{Wt of body})\text{air}-(\text{Wt of body})\text{water}} \\[0.5em] = \dfrac{W1 - W2}{W1 - W3}