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# 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:

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Solution

## Step 1. Given dataThe 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 solidThe 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:$⇒$ $\mathrm{R}.\mathrm{D}.=\frac{{\mathrm{Wt}}_{\left(\mathrm{air}\right)}}{{\mathrm{Wt}}_{\left(\mathrm{air}\right)}-{\mathrm{Wt}}_{\left(\mathrm{water}\right)}}$ $=\frac{120}{120-80}$ $=3$Step 4. Find the relative density of liquid.The relative density of liquid is,$⇒$ $\mathrm{R}.\mathrm{D}.=\frac{{\mathrm{Wt}}_{\left(\mathrm{air}\right)}-W{\mathrm{t}}_{\left(\mathrm{liquid}\right)}}{{\mathrm{Wt}}_{\left(\mathrm{air}\right)}-{\mathrm{Wt}}_{\left(\mathrm{water}\right)}}$ $=\frac{120-60}{120-80}$ $=\frac{3}{2}$Hence, the relative density of the solid and that of the liquid are $3$ and $\frac{3}{2}$ respectively.

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