JEE Main 2024 — 6 April, Shift 1 — Question 55

We know $\mathrm{R}=\frac{\rho l}{\mathrm{~A}}, \mathrm{R} \propto \frac{l}{\mathrm{r}^{2}}$

As we starch the wire, its length will increase but its radius will decrease keeping the volume constant

$$\begin{aligned} & V_{i}=V_{f} & \pi \mathrm{r}^{2} l=\pi \frac{\mathrm{r}^{2}}{4} l_{\mathrm{f}} & l_{\mathrm{f}}=4 l & \frac{\mathrm{R}_{\text {new }}}{\mathrm{R}_{\text {old }}}=\left(\frac{4 l}{\frac{\mathrm{r}^{2}}{4}}\right) \frac{\mathrm{r}^{2}}{l}=16 & R_{\text {new }}=16 R & \therefore \quad \mathrm{x}=16 \end{aligned}$$