Question

An ornament weighing $36g$ in air weighs only $34g$ in water. Assuming that some coppers 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. $2.2g$
• B. $4.4g$
• C. $1.1g$
• D. $3.6g$

Answer 1

The correct option is A $2.2g$

Given :

Weight of ornament in air, $W_{air}=36g$

Weight of ornament in water, $W_{water}=34g$

Specific gravity of gold, ${\rho }_g=19.3g/cc$

Specific gravity of copper, ${\rho }_c=8.9g/cc$

Let $V_c$ be the volume of copper and $V_g$ be the volume $0$ gols

Loss of weight $=$ weight of displacement water (buoyance) $=36-34=2g$

$W=V_g\times {\rho }_g\times g+V_c\times {\rho }_c\times g$

$36g=19.3V_g\times g+8.9V_c\times g$

$\Rightarrow m_g+m_c=36⟶\left(1\right)$

Now, $(V_g+V_c){\rho }_w\times g=2\times g$

$\Rightarrow (\frac{m_g}{{\rho }_g}+\frac{m_c}{{\rho }_c}){\rho }_w\times g=2\times g$

$\Rightarrow (\frac{m_g}{19.3}+\frac{m_c}{8.9})=2⟶\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$

mass of gold in orrnament, $m_g=33.75$

mass of copper in ornament, $m_c=2.225$

Since, the goldsmith argues that he has not mixed copper or any other material with gold, rather some cavities might have been left inside the ornament.

Now, $\Rightarrow (\frac{m_g}{{\rho }_g}+V_{cavity}){\rho }_w\times g=2\times g$

$\Rightarrow (\frac{33.75}{19.3}+V_{cavity})=2\Rightarrow V_{cavity}=0.251$