An ornament weighing 369 m in air weighs only 349 m in water. Assuming that some copper is mixed with gold to prepare the ornament. Find out the amount of copper in it. Specific gravity of gold is 19.3 and that of copper is 8.9.

Data:- $m_{air}=36 gm$

$m_{water}=34 gm$

sp. gravity $S_{gold}=19.3$

sp. gravity $S_{copper}=8.9$

To Find:- $W_{copper}=?$

Solution:-

$m_{air}=36 gm$

$\therefore \text{ weight of ornament in air is } W_{air}=36\times 10^{-3}\times 9.81=0.35316 N$

$m_{water}=34 gm$

$\therefore \text{ weight of ornament in water is } W_{water}=34\times 10^{-3}\times 9.81=0.33354 N$

$\text{Buoyant Force }(F_B)=W_{air}-W_{water}=0.01962 N$

$\text{Weight of fluid displaced }(F_B)=W_{water}\times \text{ Vol. of ornament}$

$0.01962=9810\times \text{ Vol. of ornament}$

$\therefore \text{ Vol. of ornament}=2\times 10^{-6} m^3$

But, ornament contains Gold and Copper

$\therefore \text{ Volume of ornament (V) }=V_{gold}+V_{copper}$............(1)

$V_{gold}\frac{\text{Weight of gold}}{W_{gold}}$

$V_{gold}\frac{\text{Weight of gold}}{19.3\times 9810}$

Similarly,

$V_{copper}\frac{\text{Weight of copper}}{W_{copper}}$

$V_{copper}\frac{\text{Weight of copper}}{8.9\times 9810}$

$\therefore$ Equation (1) becomes

$2\times 10^{-6}=V_{gold}+V_{copper}$

$2\times 10^{-6}=\frac{W_{gold}}{19.3\times 9810}+\frac{W_{copper}}{8.9\times 9810}$..........(2)

$\text{Also, weight of ornament}=W_{gold}+W_{copper}$

$0.35316=W_{gold}+W_{copper}$...........(3)

Solving equation (2) and equation (3)

$W_{gold}=0.33133N$ and $W_{copper}=0.0218N$