Question 80 of 118
If the ratio of lengths, radii and Young's moduli of steel and brass wires in the figure are and respectively, then the corresponding ratio of increase in their lengths is :
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Answer: (C)
According to questions,
$\begin{aligned}
& \frac{\ell_{\mathrm{s}}}{\ell_{\mathrm{b}}}=\mathrm{a}, \frac{\mathrm{r}_{\mathrm{s}}}{\mathrm{r}_{\mathrm{b}}}=\mathrm{b}, \frac{\mathrm{y}_{\mathrm{s}}}{\mathrm{y}_{\mathrm{b}}}=\mathrm{c}, \frac{\Delta \ell \mathrm{s}}{\Delta \ell_{\mathrm{b}}}=? \\
& \text { As, } \mathrm{y}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell} \Rightarrow \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{Ay}} \\
& \Delta \ell_{\mathrm{s}}=\frac{3 \mathrm{mg} \ell_{\mathrm{s}}}{\pi \mathrm{r}_{\mathrm{s}}^2 \cdot \mathrm{y}_{\mathrm{s}}}\left[\because \mathrm{F}_{\mathrm{s}}=(\mathrm{M}+2 \mathrm{M}) \mathrm{g}\right] \\
& \Delta \ell_{\mathrm{b}}=\frac{2 \mathrm{Mg}\ell_{\mathrm{b}}}{\pi \mathrm{r}_{\mathrm{b}}^2 \cdot \mathrm{y}_{\mathrm{b}}}\left[\because \mathrm{F}_{\mathrm{b}}=2 \mathrm{Mg}\right]
\end{aligned}$
$\begin{aligned}
& \frac{\ell_{\mathrm{s}}}{\ell_{\mathrm{b}}}=\mathrm{a}, \frac{\mathrm{r}_{\mathrm{s}}}{\mathrm{r}_{\mathrm{b}}}=\mathrm{b}, \frac{\mathrm{y}_{\mathrm{s}}}{\mathrm{y}_{\mathrm{b}}}=\mathrm{c}, \frac{\Delta \ell \mathrm{s}}{\Delta \ell_{\mathrm{b}}}=? \\
& \text { As, } \mathrm{y}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell} \Rightarrow \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{Ay}} \\
& \Delta \ell_{\mathrm{s}}=\frac{3 \mathrm{mg} \ell_{\mathrm{s}}}{\pi \mathrm{r}_{\mathrm{s}}^2 \cdot \mathrm{y}_{\mathrm{s}}}\left[\because \mathrm{F}_{\mathrm{s}}=(\mathrm{M}+2 \mathrm{M}) \mathrm{g}\right] \\
& \Delta \ell_{\mathrm{b}}=\frac{2 \mathrm{Mg}\ell_{\mathrm{b}}}{\pi \mathrm{r}_{\mathrm{b}}^2 \cdot \mathrm{y}_{\mathrm{b}}}\left[\because \mathrm{F}_{\mathrm{b}}=2 \mathrm{Mg}\right]
\end{aligned}$
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