Question

To determine the composition of a bimetallic alloy, a sample is first weighed in air and then in water. These weights are found to be $w_1$ and $w_2$ respectively. If the densities of the two constituent metals are $\rho_1$ and $\rho_2$ respectively, then the weight of the first metal in the sample is (where $\rho_w$ is the density of water)

• A. $\frac{\rho_{1}}{\rho_{w}\left(\rho_{2}-\rho_{1}\right)}\left[w_{1}\left(\rho_{2}-\rho_{w}\right)-w_{2}\rho_{2}\right]$
• B. $\frac{\rho_{1}}{\rho_{w}\left(\rho_{2}+\rho_{1}\right)}\left[w_{1}\left(\rho_{2}-\rho_{w}\right)+w_{2}\rho_{2}\right]$
• C. $\frac{\rho_{1}}{\rho_{w}\left(\rho_{2}-\rho_{1}\right)}\left[w_{1}\left(\rho_{2}+\rho_{w}\right)-w_{2}\rho_{1}\right]$
• D. $\frac{\rho_{1}}{\rho_{w}\left(\rho_{2}-\rho_{1}\right)}\left[w_{1}\left(\rho_{1}+\rho_{w}\right)-w_{2}\rho_{1}\right]$

Answer 1

The Correct Option is A

By Archimedes' Principle
$F=v \rho_{w} g$
$\Rightarrow\left(w_{1}-w_{2}\right) g=v \rho_{w} g$
Let, the total volume be $v$ and first metal weight be $x$
$w_{1}-w_{2}=\left(v_{1}+v_{2}\right) \rho_{w}$
$w_{1}-w_{2}=v_{1} \rho_{w}+v_{2} \rho_{w} \,\,\,\left(\because v=\frac{m}{\rho}\right)$
$w_{1}-w_{2}=\left(\frac{x}{\rho_{1}} \rho_{w}+\frac{w_{1}-x}{\rho_{2}} \rho_{w}\right)$
$w_{1}-w_{2}=\frac{x \rho_{2} \rho_{w}+\left(w_{1}-x\right) \rho_{w} \rho_{1}}{\rho_{1} \rho_{2}}$
$w_{1}\, \rho_{1} \,\rho_{2}-w_{2} \,\rho_{1}\, \rho_{2}=x \rho_{2} \,\rho_{w}+w_{1} \,\rho_{w} \rho_{1}-x \rho_{w}\, \rho_{1}$
$x\left(\rho_{2}-\rho_{1}\right) \rho_{w}=\rho_{1}\left[w_{1}\left(\rho_{2}-\rho_{w}\right)-w_{2}\, \rho_{2}\right]$
$x=\frac{\rho_{1}}{\rho_{w}\left(\rho_{2}-\rho_{1}\right)}\left[w_{1}\left(\rho_{2}-\rho_{w}\right)-w_{2} \,\rho_{2}\right]$