Question

A piece of solid weighs $120\mathrm{g}$ in air, $80\mathrm{g}$ in water and $60\mathrm{g}$ in a liquid. The relative density of the solid and that of the liquid are respectively:

Answer 1

Step 1. Given data

The weight of the solid in air is given as $120\mathrm{g}$,

Weight in water is given as $80\mathrm{g}$,

Weight in liquid is given as $60\mathrm{g}$.

We have to find the relative density of solid and liquid.

Step 2. Formula to be used.

The relative density of a substance is defined as the ratio of the density of the substance to the density of water. It is represented by symbol $\mathrm{R}.\mathrm{D}.$

Thus,

$\mathrm{R}.\mathrm{D}.=\frac{\mathrm{Density}\mathrm{of}\mathrm{the}\mathrm{substance}}{\mathrm{Density}\mathrm{of}\mathrm{water}}$

Step 3. Find the relative density of solid

The relative density of a substance is defined as the ratio of the density of the substance to the density of water. It is represented by symbol $\mathrm{R}.\mathrm{D}.$

Thus,

$\mathrm{R}.\mathrm{D}.=\frac{\mathrm{Density}\mathrm{of}\mathrm{the}\mathrm{substance}}{\mathrm{Density}\mathrm{of}\mathrm{water}}$

The relative density of solid is given by:

$\Rightarrow$ $\mathrm{R}.\mathrm{D}.=\frac{{\mathrm{Wt}}_{\left(\mathrm{air}\right)}}{{\mathrm{Wt}}_{\left(\mathrm{air}\right)}-{\mathrm{Wt}}_{\left(\mathrm{water}\right)}}$

$=\frac{120}{120-80}$

$=3$

Step 4. Find the relative density of liquid.

The relative density of liquid is,

$\Rightarrow$ $\mathrm{R}.\mathrm{D}.=\frac{{\mathrm{Wt}}_{\left(\mathrm{air}\right)}-W{\mathrm{t}}_{\left(\mathrm{liquid}\right)}}{{\mathrm{Wt}}_{\left(\mathrm{air}\right)}-{\mathrm{Wt}}_{\left(\mathrm{water}\right)}}$

$=\frac{120-60}{120-80}$

$=\frac{3}{2}$

Hence, the relative density of the solid and that of the liquid are $3$ and $\frac{3}{2}$ respectively.