JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 15)
Under the same load, wire A having length $5.0 \mathrm{~m}$ and cross section $2.5 \times 10^{-5} \mathrm{~m}^{2}$ stretches
uniformly by the same amount as another wire B of length $6.0 \mathrm{~m}$ and a cross section of $3.0 \times 10^{-5}$
$\mathrm{m}^{2}$ stretches. The ratio of the Young's modulus of wire A to that of wire $B$ will be :
$1: 2$
$1: 4$
$1: 1$
$1: 10$
Explanation
$\Delta \ell=\frac{F \ell}{S Y}$
$F$ is same for both wire and $\Delta \ell$ is also same
$$ \begin{aligned} & \frac{\Delta \ell}{F}=\frac{\ell}{S Y} \\\\ & \Rightarrow \frac{\ell_{A}}{S_{A} Y_{A}}=\frac{\ell_{B}}{S_{B} Y_{B}} \\\\ & \Rightarrow \frac{5}{2.5 \times Y_{A}}=\frac{6}{3 \times Y_{B}} \\\\ & \Rightarrow \frac{Y_{A}}{Y_{B}}=1 \end{aligned} $$
$F$ is same for both wire and $\Delta \ell$ is also same
$$ \begin{aligned} & \frac{\Delta \ell}{F}=\frac{\ell}{S Y} \\\\ & \Rightarrow \frac{\ell_{A}}{S_{A} Y_{A}}=\frac{\ell_{B}}{S_{B} Y_{B}} \\\\ & \Rightarrow \frac{5}{2.5 \times Y_{A}}=\frac{6}{3 \times Y_{B}} \\\\ & \Rightarrow \frac{Y_{A}}{Y_{B}}=1 \end{aligned} $$
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