JEE Main 2024 — Mechanical Properties of Solids Question with Solution

The formula to calculate the Young's modulus of the material of the wire, in the first case, can be written as

$Y=\frac{\frac{F}{\pi r^2}}{\frac{l}{L}} ...\left(1\right)$

From equation (1), it follows that

$F=Y\pi r^2\times \frac{l}{L} ...\left(2\right)$

For the second case, the Young's modulus is given by

$Y=\frac{\frac{F/2}{\pi r^2/4}}{\frac{l\text{'}}{L}} ...\left(3\right)$

From equation (3), it follows that

$F=Y\frac{l\text{'}}{L}\times 2\times \frac{\pi r^2}{4} ...\left(4\right)$

From equations (2) and (4), it implies that

$$\begin{gathered} Y\pi r^2\times \frac{l}{L}=\mathrm{Y}\frac{l\text{'}}{L}\times 2\times \frac{\pi {\mathrm{r}}^2}{4} \\ \Rightarrow l\text{'}=2l \end{gathered}$$