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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)
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Verified Answer
The correct answer is:
$\frac{\rho_{1}}{\rho_{w}\left(\rho_{2}-\rho_{1}\right)}\left[w_{1}\left(\rho_{2}-\rho_{w}\right)-w_{2} \rho_{2}\right]$
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 $x$
$$
\begin{array}{l}
w_{1}-w_{2}=\left(v_{1}+v_{2}\right) \rho_{w} \\
w_{1}-w_{2}=v_{1} \rho_{w}+v_{2} g_{w} \\
w_{1}-w_{2}=\left(\frac{x}{\rho_{1}} \rho_{w}+\frac{w_{1}-x}{\rho_{2}} \rho_{w}\right)
w_{1}-w_{2}=\frac{x \rho_{\mathcal{P}} \rho_{w}+\left(w_{1}-x\right) \rho_{w} \rho_{1}}{\rho_{\jmath} \rho_{2}}
\end{array} \quad\left(\because v=\frac{m}{\rho}\right)
$$
$$
\begin{aligned} w_{1} \rho_{1} \rho_{2}-w_{2} \rho_{1} \rho_{2} &=x \rho_{2} \rho_{w}+w_{1} \rho_{w} \rho_{1}-x \rho_{w} \rho_{1} \\ x\left(\rho_{2}-\rho_{1}\right) \rho_{w} &=\rho_{1}\left[w_{1}\left(\rho_{2}-\rho_{w}\right)-w_{2} \rho_{2}\right] \end{aligned}\\
x=\frac{\rho_{1}}{\rho_{w}\left(\rho_{2}-\rho_{1}\right)}\left[w_{1}\left(\rho_{2}-\rho_{w}\right)-w_{z} \rho_{2}\right]
$$
$$
F=v \rho_{w} g \Rightarrow\left(w_{1}-w_{2}\right) g=v \rho_{w} g
$$
Let, the total volume be $v$ and first metal weight be $x$
$$
\begin{array}{l}
w_{1}-w_{2}=\left(v_{1}+v_{2}\right) \rho_{w} \\
w_{1}-w_{2}=v_{1} \rho_{w}+v_{2} g_{w} \\
w_{1}-w_{2}=\left(\frac{x}{\rho_{1}} \rho_{w}+\frac{w_{1}-x}{\rho_{2}} \rho_{w}\right)
w_{1}-w_{2}=\frac{x \rho_{\mathcal{P}} \rho_{w}+\left(w_{1}-x\right) \rho_{w} \rho_{1}}{\rho_{\jmath} \rho_{2}}
\end{array} \quad\left(\because v=\frac{m}{\rho}\right)
$$
$$
\begin{aligned} w_{1} \rho_{1} \rho_{2}-w_{2} \rho_{1} \rho_{2} &=x \rho_{2} \rho_{w}+w_{1} \rho_{w} \rho_{1}-x \rho_{w} \rho_{1} \\ x\left(\rho_{2}-\rho_{1}\right) \rho_{w} &=\rho_{1}\left[w_{1}\left(\rho_{2}-\rho_{w}\right)-w_{2} \rho_{2}\right] \end{aligned}\\
x=\frac{\rho_{1}}{\rho_{w}\left(\rho_{2}-\rho_{1}\right)}\left[w_{1}\left(\rho_{2}-\rho_{w}\right)-w_{z} \rho_{2}\right]
$$
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