A neckless weighing 50 gm in air, but it weighs 46 g in water. Assume copper is mixed with gold to prepare the neckless. Find how much copper is present in it. (Specific gravity of gold is 20 and that of copper is 10).

Let m be the mass of copper in neckless.

Mass of gold = (50-m)

Volume of copper = ${\mathrm{V}}_1 = \frac{\mathrm{m}}{10}$

Volume of gold = ${\text{V}}_2 = \frac{50-\mathrm{m}}{20}$

when immersed in water ${\mathrm{\rho }}_{\mathrm{W}} = 1 \mathrm{gm}/{\mathrm{cm}}^3$

Decrease in weight = upthrust

$(50-46)\mathrm{g} = ({\mathrm{V}}_1+{\mathrm{V}}_2){\mathrm{\rho }}_{\mathrm{w}}\mathrm{g}$

$\Rightarrow 4 = \frac{\mathrm{m}}{10}+\frac{50-\mathrm{m}}{20} \Rightarrow \mathrm{m} = 30\mathrm{g}$