Q87 / 90 · JEE Main 2024 (06 Apr Shift 1) · Physics
A wire of resistance $R$ and radius $r$ is stretched till its radius became $r / 2$. If new resistance of the stretched wire is $x R$, then value of $x$ is _______
Correct answer: 16
Solution
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
$\begin{aligned}
& \mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}} \\
& \pi \mathrm{r}^2 l=\pi \frac{\mathrm{r}^2}{4} l_{\mathrm{f}} \\
& l_{\mathrm{f}}=4 l \\
& \frac{\mathrm{R}_{\text {new }}}{\mathrm{R}_{\text {old }}}=\left(\frac{4 l}{\frac{\mathrm{r}^2}{4}}\right) \frac{\mathrm{r}^2}{l}=16 \\
& \mathrm{R}_{\text {new }}=16 \mathrm{R} \\
\therefore & \mathrm{x}=16
\end{aligned}$
Hence, the correct answer is 16.