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 constituents 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. $\frac{{\rho }_1}{{\rho }_w({\rho }_2-{\rho }_1)}[w_1({\rho }_2-{\rho }_w)-w_2{\rho }_2]$
• B. $\frac{{\rho }_1}{{\rho }_w({\rho }_2+{\rho }_1)}[w_1({\rho }_2-{\rho }_w)+w_2{\rho }_2]$
• C. $\frac{{\rho }_1}{{\rho }_w({\rho }_2-{\rho }_1)}[w_1({\rho }_2+{\rho }_w)-w_2{\rho }_1]$
• D. $\frac{{\rho }_1}{{\rho }_w({\rho }_2-{\rho }_1)}[w_1({\rho }_1-{\rho }_w)-w_2{\rho }_1]$

Answer 1

The correct option is A $\frac{{\rho }_1}{{\rho }_w({\rho }_2-{\rho }_1)}[w_1({\rho }_2-{\rho }_w)-w_2{\rho }_2]$

Let $V$ = volume of alloy and $W_{\alpha }$ and $W_{\beta }$ be the weights of the first and second metals in air respecticely.

We have:

$Vg=\frac{W_{\alpha }}{{\rho }_1}+\frac{W_{\beta }}{{\rho }_2}{W}_{\alpha }+W_{\beta }=w_1...\left(1\right)w_2=w_1-V{\rho }_{w}g=w_1-(\frac{W_{\alpha }}{{\rho }_1}+\frac{W_{\beta }}{{\rho }_2}){\rho }_w=W_{\alpha }(1-\frac{{\rho }_w}{{\rho }_1})+W_{\beta }(1-\frac{{\rho }_w}{{\rho }_2})\Rightarrow w_2=W_{\alpha }(1-\frac{{\rho }_w}{{\rho }_1})+W_{\beta }(1-\frac{{\rho }_w}{{\rho }_2})......\left(2\right)$

Solving equations (1) and (2) we get:

$W_{\alpha }=\frac{{\rho }_1\left(w_1\right({\rho }_2-{\rho }_w)-w_2{\rho }_2)}{{\rho }_w({\rho }_2-{\rho }_1)}$