According to questions, $\begin{aligned}
& \frac{\ell_{\mathrm{s}}}{\ell_{\mathrm{b}}}=\mathrm{a}, \frac{\mathrm{r}{\mathrm{s}}}{\mathrm{r}{\mathrm{b}}}=\mathrm{b}, \frac{\mathrm{y}{\mathrm{s}}}{\mathrm{y}{\mathrm{b}}}=\mathrm{c}, \frac{\Delta \ell \mathrm{s}}{\Delta \ell_{\mathrm{b}}}=? \
& \text { As, } \mathrm{y}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell} \Rightarrow \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{Ay}} \
& \Delta \ell_{\mathrm{s}}=\frac{3 \mathrm{mg} \ell_{\mathrm{s}}}{\pi \mathrm{r}{\mathrm{s}}^2 \cdot \mathrm{y}{\mathrm{s}}}\left[\because \mathrm{F}{\mathrm{s}}=(\mathrm{M}+2 \mathrm{M}) \mathrm{g}\right] \
& \Delta \ell{\mathrm{b}}=\frac{2 \mathrm{Mg}\ell_{\mathrm{b}}}{\pi \mathrm{r}{\mathrm{b}}^2 \cdot \mathrm{y}{\mathrm{b}}}\left[\because \mathrm{F}{\mathrm{b}}=2 \mathrm{Mg}\right]
\end{aligned}\therefore \frac{\Delta \ell{\mathrm{s}}}{\Delta \ell_{\mathrm{b}}}=\frac{\frac{3 \mathrm{Mg} \ell_{\mathrm{s}}}{\pi \mathrm{r}{\mathrm{s}}^2 \cdot \mathrm{y}{\mathrm{s}}}}{\frac{2 \mathrm{Mg} \cdot \ell_{\mathrm{b}}}{\pi \mathrm{r}{\mathrm{b}}^2 \cdot \mathrm{y}{\mathrm{b}}}}=\frac{3 \mathrm{a}}{2 \mathrm{b}^2 \mathrm{C}}$
JEE Main 2013 — Physics Mechanics
2013mcqhard
If the ratio of lengths, radii and Young's moduli of steel and brass wires in the figure are a,b and c respectively, then the corresponding ratio of increase in their lengths is : 
Official previous-year question
Held on 9 Apr 2013 · Verified 6 Jul 2026.
Options
- A
2ab23c
- B
b2a2c
- C
2b2c3a
- D
b22ac
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