Question

An ornament weighing $36g$ in air, weighs only $34g$ in water. Assuming that some copper is mixed with gold to prepare the ornament, find the amount of copper in it. specific gravity of gold is $19.3$ and that of copper is $8.9$.

Answer 1

Here, $m_o=m_{av}+m_{cu}=36g------1$
Let $V$ be the volume of the ornament lie $c{m}^2$
so, $V\times P_w\times g=2\times g$
$\Rightarrow (v_{au}+v_{cu})\times P_w\times g=2\times g$

$\Rightarrow (\frac{m}{{\rho }_{au}}+\frac{m}{{\rho }_{cu}})P_w\times g=2\times g$

$\Rightarrow (\frac{m_{av}}{19.3}+\frac{m_{cu}}{8.9})\times 1=2$

$\Rightarrow 8.9m_{au}+19.3m_{cu}=2\times 19.3\times 8.9$

$=343.54---------\left(2\right)$
from eqn(1) and (2)

$\Rightarrow 8.9(m_{au}+m_{cu})=8.9\times 36$

$\Rightarrow m_{au}=2.225g$

$\Rightarrow m_{au}=2.2g$.