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 $\qquad$
JEE MAIN 2024, Diffcult
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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
$\mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}}$
$\pi^2 l=\pi \frac{\mathrm{r}^2}{4} l_{\mathrm{f}}$
$l_{\mathrm{f}}=4 l$
$\frac{\mathrm{R}_{\text {nelv }}}{\mathrm{R}_{\text {old }}}=\left(\frac{4 l}{\frac{\mathrm{r}^2}{4}}\right) \frac{\mathrm{r}^2}{l}=16$
$\mathrm{R}_{\text {neer }}=16 \mathrm{R}$
$\therefore \mathrm{x}=16$
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