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

The correct option is A $2.2$ g

Given $m_g+m_c=36\times {10}^{-3}kg........\left(I\right)$
Applying the Archimedes principle,
loss in weight of the ornament in water = the weight of the water displaced
$\Rightarrow$ 2g of water is displaced.
Since volume of water displaced = Volume of the ornament
$\Rightarrow$ Volume of the ornament $=\frac{2\times {10}^{-3}}{1000}=2\times {10}^{-6}{m}^3$

$\because$ Volume of the ornament = Volume of gold in ornament + Volume of copper in the ornament
$\Rightarrow 2\times {10}^{-6}{m}^3=\frac{m_g}{19.3\times {10}^3}+\frac{m_c}{8.9\times {10}^3}\Rightarrow 8.9m_g+19.3m_c=2\times 19.3\times 8.9\times {10}^{-3}\Rightarrow 8.9m_g+19.3m_c=343.54\times {10}^{-3}kg........\left(II\right)$

Solving $\left(I\right)$ and $\left(II\right)$, we have
$m_g=33.775\times {10}^{-3}kg\text{and}{m}_c=2.225\times {10}^{-3}kg$

So mass of copper in the ornament $=2.225\times {10}^{-3}kg\approx 2.2gm$