JEE Main 2013 — Mechanical Properties of Solids Question with Solution
JEE Main 2013 (09 Apr Online)
Question
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 :
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
According to questions,
$\begin{aligned}
& \frac{\ell_{\mathrm{s}}}{\ell_{\mathrm{b}}}=\mathrm{a}, \frac{\mathrm{r}_{\mathrm{s}}}{\mathrm{r}_{\mathrm{b}}}=\mathrm{b}, \frac{\mathrm{y}_{\mathrm{s}}}{\mathrm{y}_{\mathrm{b}}}=\mathrm{c}, \frac{\Delta \ell \mathrm{s}}{\Delta \ell_{\mathrm{b}}}=? \\
& \text { As, } \mathrm{y}=\frac{\mathrm{F} \ell}{\mathrm{A} \Delta \ell} \Rightarrow \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{Ay}} \\
& \Delta \ell_{\mathrm{s}}=\frac{3 \mathrm{mg} \ell_{\mathrm{s}}}{\pi \mathrm{r}_{\mathrm{s}}^2 \cdot \mathrm{y}_{\mathrm{s}}}\left[\because \mathrm{F}_{\mathrm{s}}=(\mathrm{M}+2 \mathrm{M}) \mathrm{g}\right] \\
& \Delta \ell_{\mathrm{b}}=\frac{2 \mathrm{Mg}\ell_{\mathrm{b}}}{\pi \mathrm{r}_{\mathrm{b}}^2 \cdot \mathrm{y}_{\mathrm{b}}}\left[\because \mathrm{F}_{\mathrm{b}}=2 \mathrm{Mg}\right]
\end{aligned}$
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This is a previous-year question from JEE Main 2013, covering the Mechanical Properties of Solids chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.