Question

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