Question

A goldsmith desires to test the purity of a gold ornament suspected to the mixed with copper. The ornament weights $0.25kg$ in air and is observe to displace $0.015$ litre of water when immersed in it. Densities of gold and copper with respect to water are, respectively, $19.3$ and $8.9$. The approximate percentage of copper in the ornament is

• A. $5\%$
• B. $10\%$
• C. $15\%$
• D. $25\%$

Answer 1

The correct option is D $25\%$

Let volume of gold in ornament $V_1$ and that of copper $=V_2$ and density of gold $=\rho g$ and that of copper$={\rho }_c$

$\Rightarrow \rho g{V}_1+{\rho }_{c}V-2=0.25\to \left(1\right)$

Volume of ornament $=$Volume of water displaced

$\Rightarrow V_1+V-2=0.15\times {10}^{-3}ms\Rightarrow V_2=(0.015\times {10}^{-3}-V_1)$

According to question-

$\frac{{\rho }_g}{{\rho }_w}=19.3$ and $\frac{{\rho }_c}{{\rho }_w}=8.9{\rho }_g=19.3\times {10}^{3}kg/m^3$

${\rho }_c=8.9\times {10}^{3}kg/m^3$

Putting these value in equation $\left(1\right)$

$(19.3\times {10}^3)V_1+(1.9\times {10}^3)(0.0015\times {10}^{-3}-V_1)0.25\left[\right(19.3\times {10}^3)-(1.9\times {10}^3\left)\right]V_1+(8.9\times 0.015)=0.2510.4\times {10}^3{V}_1=0.12V_1=\frac{0.12}{10.4\times {10}^3}=0.0115\times {10}^{-3}=1.15\times {10}^{-5}{m}^3$

$V_2=(0.015\times {10}^3-0.0115\times {10}^{-3})=0.0035\times {10}^{-3}\%ofcopper=\left(\frac{V_2}{V_1+V_2}\right)\times 100=\left[\frac{0.0035\times {10}^{-3}}{(0.0115+0.0035)\times {10}^{-3}}\right]\times 100$

$\approx 23.34\%approx25\%$