To determine the composition of a bimetallic | vedclass

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(A) Let $x$ be the weight of the first metal and $(w_1 - x)$ be the weight of the second metal in air.Let $v_1$ and $v_2$ be the volumes of the two me

Vedclass · Accepted Answer

$\frac{\rho_1}{\rho_w(\rho_2-\rho_1)}[w_1(\rho_2-\rho_w)-w_2 \rho_2]$

Vedclass · Answer

(A) Let $x$ be the weight of the first metal and $(w_1 - x)$ be the weight of the second metal in air.Let $v_1$ and $v_2$ be the volumes of the two metals respectively.$v_1 = \frac{x}{\rho_1}$ and $v_2 = \frac{w_1 - x}{\rho_2}$.The total volume of the alloy is $V = v_1 + v_2 = \frac{x}{\rho_1} + \frac{w_1 - x}{\rho_2}$.According to Archimedes' Principle, the loss in weight equals the weight of the displaced water:$w_1 - w_2 = V \rho_w = \left( \frac{x}{\rho_1} + \frac{w_1 - x}{\rho_2} \right) \rho_w$.Multiplying by $\rho_1 \rho_2$:$(w_1 - w_2) \rho_1 \rho_2 = (x \rho_2 + (w_1 - x) \rho_1) \rho_w$.$(w_1 - w_2) \rho_1 \rho_2 = (x \rho_2 + w_1 \rho_1 - x \rho_1) \rho_w$.$(w_1 - w_2) \rho_1 \rho_2 = x(\rho_2 - \rho_1) \rho_w + w_1 \rho_1 \rho_w$.$x(\rho_2 - \rho_1) \rho_w = w_1 \rho_1 \rho_2 - w_2 \rho_1 \rho_2 - w_1 \rho_1 \rho_w$.$x(\rho_2 - \rho_1) \rho_w = w_1 \rho_1(\rho_2 - \rho_w) - w_2 \rho_1 \rho_2$.$x = \frac{\rho_1}{\rho_w(\rho_2 - \rho_1)} [w_1(\rho_2 - \rho_w) - w_2 \rho_2]$.

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