Describe an experiment, using Archimedes' principle, to find the relative density of a liquid.

By definition, relative density of a liquid is given as —

$\text{R.D.} = \dfrac{\text{Wt of given volume of liq}}{\text{Wt of same volume of water}}$

By Archimedes' Principle if a solid is immersed in a liquid or water, it displaces the liquid or water equal to it's volume. Therefore, above equation takes the form —

$\text{R.D.} = \dfrac{\text{Wt of liq displaced by body}}{\text{Wt of water displaced by body}} \\[0.5em] = \dfrac{(\text{Wt of body})_\text{air}-(\text{Wt of body})_\text{liq}}{(\text{Wt of body})_\text{air}-(\text{Wt of body})_\text{water}}$

Thus, to find the relative density of a liquid using Archimedes' Principle, we take a body which is heavier than both the given liquid and water and also insoluble in both. The body is first weighed in air, then in liquid and then after washing it with water and drying, it is weighed in water. If the weight of the body in air is W₁ gf, in liquid is W₂ gf and in water is W₃ gf, then from above we get,

$\text{R.D. of liquid} = \dfrac{W$1 - W2}{W1 - W3}