An ornament weighing 36 g in air , weighs only 34 g in water. Assuming that some copper is mixed with gold to prepare an ornament the amount of copper in it is

(specific gravity of gold is 19.3 & copper is 8.9)

Let m is the mass of copper present in the ornament

Then mass of gold present in it $=36-\text{m}$

Let V is the volume of ornament

$\text{V}={\text{V}}_{\text{copper}}+{\text{V}}_{\text{Gold}}$                                                                           $\because {\rho }_{copper}=8.9$

$\text{V}=\frac{\text{m}}{8.9}+\frac{36-\text{m}}{19.3}$                                                                           ${\rho }_{Gold}=19.3$

We know weight lost by the ornament up on immersing in water

${\text{W}}_{\text{air}}-{\text{W}}_{\text{in}\text{water}}={\text{V}}_{\text{ornament}}\times {\rho }_w\times g$

$\Rightarrow 36\times \text{g}-34\times \text{g}=\left(\frac{\text{m}}{8.9}+\frac{36-\text{m}}{19.3}\right)\times 1\text{g/cc}\times \text{g}$

$\Rightarrow 2=\frac{\text{m}}{8.9}+\frac{36-\text{m}}{19.3}$

$2\times 8.9\times 19.3=19.3\text{m}+36\times 8.9-8.9\text{m}$

$\therefore 10.4\text{m}=2\times 8.9\times 19.3-36\times 8.9$

$\therefore \text{m}=2.2\text{g}$