A piece of solid weight 120 g in air 80 g in water and 60g in liquid. The relative density at the solid and that at the liquid are respectively

Given weight of solid in air$w_a=120\text{g}$

Weight of solid in water $w_w=80\text{g}$

Weight of solid in liquid $w_L=60\text{g}$

The relative density of an object is equal to the ratio of its weight in air and the difference of its weight in air and its weight when the object is submerge in liquid

For relative density of solid ${\rho }_s=\frac{120\text{g}}{120\text{g-80}\text{g}}=\frac{120}{40}=3$

Hence relative density of solid ${\rho }_s=3$

How relative density of liquid can be calculated as the ratio of loss of weight of solid in liquid to the loss f weight in water

e.g., relative density of liquid ${\rho }_L=\frac{120-60}{120-80}$

${\rho }_L=\frac{60}{40}=1.5$