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 $8.9$.

• A. 2.2 g
• B. 4.4 g
• C. 1.1 g
• D. 3.6 g

Answer 1

Answer: (A)

Given that $m_{\text{real }}=36g,m_{\text{app }}=34g$.
Density of gold ${\rho }_{Au}=19.3g/cc$
Density of copper ${\rho }_{Cu}=8.9g/cc$
We know that loss of weight $=$ weight of displaced water
$=36-34=2g=$ Buoyant force $=B$
Here, $m_{\text{real }}=m_{Aw}+m_{Cu}=36g$ ...(i)
Let $v$ be the volume of the ornament in centimetres. Then
$B=v\times {\rho }_w\times g=2\times g$
$\Rightarrow \left(\frac{m_{Au}}{{\rho }_{Au}}+\frac{m_{Cu}}{{\rho }_{Cu}}\right){\rho }_w\times g=2\times g$
$m_{Au}{\rho }_{Cu}+m_{Cu}{\rho }_{Au}=2{\rho }_{Acc}{\rho }_{Cu}$
$8.9m_{Au}+19.3m_{Cu}=2\times 19.3\times 8.9=343.54$ ...(ii)
From Eqs. (i) and (ii), $8.9m_{Au}+19.3m_{Cu}=343.54$
$\Rightarrow 8.9\left(m_{Au}+m_{Cu}\right)+10.4m_{Cu}=343.54$
$\Rightarrow 8.9\times 36+10.4m_{Cu}=343.54$
$m_{Cu}=2.225g$
So the amount of copper in the ornament is $2.2g$.