The ratios of lengths, areas of cross-section and Young's modulii of steel to that of brass wires shown in the figure are $a, b$ and $c$ respectively. The ratio of increase in the lengths of brass to that of steel wires is [Assume that the masses of steel and brass wires are negligible]

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Aicharya · Accepted Answer

Now, $ T^{\prime}=T+3 g \Rightarrow T=4 g $ So, $T^{\prime}=7 \mathrm{~g}$ So, $\frac{F_S}{F_B}=\frac{T^{\prime}}{T}=\frac{7}{4}$ We know that, $ \begin{aligned} & Y=\frac{F / A}{\Delta L / L} \Rightarrow \Delta L=\frac{F L}{A Y} \\ & \frac{\Delta L_S}{\Delta L_B}=\left(\frac{F_S}{F_B}\right)\left(\frac{L_S}{L_B}\right)\left(\frac{A_B}{A_S}\right)\left(\frac{Y_B}{Y_S}\right) \\ & =\frac{7}{4} \times(a)\left(\frac{1}{b}\right) \times\left(\frac{1}{c}\right) \\ & \text { or, } \frac{\Delta L_S}{\Delta L_B}=\frac{7 a}{4 b c} \text { or, } \frac{\Delta L_B}{\Delta L_S}=\frac{4 b c}{7 a} \end{aligned} $

The ratios of lengths, areas of cross-section and Young's modulii of steel to that of brass wires shown in…

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