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

Answer: (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 $x$
$w_1-w_2=\left(v_1+v_2\right){\rho }_w$
$w_1-w_2=v_1{\rho }_w+v_2{\rho }_w\left(\because v=\frac{m}{\rho }\right)$
$w_1-w_2=\left(\frac{x}{{\rho }_1}{\rho }_w+\frac{w_1-x}{{\rho }_2}{\rho }_w\right)$
$w_1-w_2=\frac{x{\rho }_2{\rho }_w+\left(w_1-x\right){\rho }_w{\rho }_1}{{\rho }_1{\rho }_2}$
$w_1{\rho }_1{\rho }_2-w_2{\rho }_1{\rho }_2=x{\rho }_2{\rho }_w+w_1{\rho }_w{\rho }_1-x{\rho }_w{\rho }_1$
$x\left({\rho }_2-{\rho }_1\right){\rho }_w={\rho }_1\left[w_1\left({\rho }_2-{\rho }_w\right)-w_2{\rho }_2\right]$
$x=\frac{{\rho }_1}{{\rho }_w\left({\rho }_2-{\rho }_1\right)}\left[w_1\left({\rho }_2-{\rho }_w\right)-w_2{\rho }_2\right]$