Question

An alloy of gold and copper weighs $0.2\,kg$ in air and $0.188\,kg$ in water. The densities of gold and copper are $19.3 \times {10^3}\,kg{m^{ - 3}}$ and $8.93 \times {10^3}\,kg{m^{ - 3}}$ respectively. The amount of gold in block is nearly:
A) $12 \times {10^{ - 3}}\,kg$
B) $17 \times {10^{ - 3}}\,kg$
C) $0.173\,kg$
D) $0.388\,kg$

Answer 1

Hint: In this problem, the alloy’s weight in air and weight in water are given, and the density of the gold and copper also given. By finding the amount of alloy in air and the amount of alloy in water, after finding the amount of alloy in air and water, and by equating the amount of alloy in air and amount of alloy in water, then the amount of gold can be determined.

Useful formula:
Amount of alloy in air is the sum of amount of gold and amount of copper,
${V_1}{\rho _1}g + {V_2}{\rho _2}g$
Where, ${V_1}$ is the volume of the gold, ${\rho _1}$ is the density of the gold, ${V_2}$ is the volume of the copper, ${\rho _2}$ is the density of the copper and $g$ is the acceleration due to gravity.
Amount of alloy in water,
${V_1}\left( {{\rho _1} - \rho } \right)g + {V_2}\left( {{\rho _2} - \rho } \right)g$
Here, the density of water is subtracted with the density of gold and density of copper.

Complete step by step solution:
Given that,
An alloy of gold and copper weighs $0.2\,kg$ in air,
An alloy of gold and copper weighs $0.188\,kg$ in water.
The density of gold, ${\rho _1} = 19.3 \times {10^3}\,kg{m^{ - 3}}$,
The density copper, ${\rho _2} = 8.93 \times {10^3}\,kg{m^{ - 3}}$.
Now,
The amount of alloy in air,
${V_1}{\rho _1}g + {V_2}{\rho _2}g = 0.2\,...................\left( 1 \right)$
On substituting the density values and the acceleration due to gravity in the equation (1), then
${V_1}\left( {19.3 \times {{10}^3} \times 9.81} \right) + {V_2}\left( {8.93 \times {{10}^3} \times 9.81} \right) = 0.2$
By multiplying the terms, then
$189333{V_1} + 87603.3{V_2} = 0.2\,................\left( 2 \right)$
The amount of alloy in water,
${V_1}\left( {{\rho _1} - \rho } \right)g + {V_2}\left( {{\rho _2} - \rho } \right)g = 0.188\,.................\left( 3 \right)$
On substituting the density values and the acceleration due to gravity in the equation (1), then ${V_1}\left( {19.3 \times {{10}^3} - 1000} \right)9.81 + {V_2}\left( {8.93 \times {{10}^3} - 1000} \right)9.81 = 0.188$
By solving the above equation, then
$179523{V_1} + 77793.3{V_2} = 0.188\,................\left( 4 \right)$
On subtracting the equation (2) and equation (4), then the value of ${V_1}$ and ${V_2}$ are,
${V_1} = 9 \times {10^{ - 7}}\,{m^3}$ and ${V_2} = 3 \times {10^{ - 7}}\,{m^3}$
The amount of gold is,
 $ \Rightarrow {V_1} \times {\rho _1}$
On substituting the volume and density valuer in the above equation, then
$ \Rightarrow 9 \times {10^{ - 7}} \times 19.3 \times {10^3}$
On multiplying,
The amount of gold is $0.017\,kg$ is also equal to $17 \times {10^{ - 3}}\,kg$

Hence, the option (B) is correct.

Note: In equation (3), the density of the gold and the density of the copper is subtracted by the density of water because the alloy is in water. Then by subtracting the equation (2) and the equation (4), the volume of the gold and copper is determined and the volume is multiplied with the density, the weight of the gold is determined.