How is the relative density of a solid determined by using Archimedes principle? Derive the formula used.

Relative density of a substance = $\frac{\mathrm{Density}\mathrm{of}\mathrm{the}\mathrm{substance}}{\mathrm{Density}\mathrm{of}\mathrm{water}}$

We have, Density of a substance = $\frac{\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{substance}}{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{substance}}$

Therefore, Relative Density = $\frac{{\displaystyle \raisebox{1ex}{$\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{substance}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{substance}$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{Mass}\mathrm{of}\mathrm{water}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Volume}\mathrm{of}\mathrm{water}$}\right.}}$

• Case I - When the immersed solid is denser than water and insoluble in it

From Archimedes' principle, loss in the weight of a solid when immersed in water is equal to the weight of the water displaced by the solid when placed in it.

Relative density of the solid = $\frac{\mathrm{Weight}\mathrm{of}\mathrm{solid}\mathrm{in}\mathrm{air}}{\mathrm{Loss}\mathrm{in}\mathrm{weight}\mathrm{of}\mathrm{solid}\mathrm{in}\mathrm{water}}$ = $\frac{W_1}{W_1-W2}$

• Case II - When immersed solid is lighter than that of water

Relative density of the solid = $\frac{\mathrm{Weight}\mathrm{of}\mathrm{solid}\mathrm{in}\mathrm{air}}{\mathrm{Loss}\mathrm{in}\mathrm{the}\mathrm{weight}\mathrm{of}\mathrm{cork}\mathrm{when}\mathrm{immersed}\mathrm{in}\mathrm{water}}$ = $\frac{W_2-W_1}{W_2-W_3}$