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The length of a given cylindrical wire is increased by $100 \%$. Due to the consequent decrease in diameter the change in the resistance of the wire will be
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$300 \%$
$300 \%$
$\mathrm{L}_1=2 \mathrm{l}$ or $\left(\pi \mathrm{r}^2 \mathrm{l}\right)=\left(\pi \mathrm{r}_2^2\right)(2 \mathrm{l})$
$\Rightarrow \mathrm{r}_2=\frac{\mathrm{r}}{\sqrt{2}}$$;$ $\mathrm{R}=\rho \frac{1}{\pi^2}$
$\mathrm{R}_{\text {new }}=(\rho) \frac{21}{(\pi)(\mathrm{r} / \sqrt{2})^2}=\frac{(\rho) 4 \mathrm{l}}{(\pi) \mathrm{r}^2}=4 \times \mathrm{R}$
$\therefore \Delta \mathrm{R}=4 \mathrm{R}-\mathrm{R}=3 \mathrm{R}$
$\frac{\Delta \mathrm{R}}{\mathrm{R}} \%=\frac{3 \mathrm{R}}{\mathrm{R}} \times 100=300 \%$
$\Rightarrow \mathrm{r}_2=\frac{\mathrm{r}}{\sqrt{2}}$$;$ $\mathrm{R}=\rho \frac{1}{\pi^2}$
$\mathrm{R}_{\text {new }}=(\rho) \frac{21}{(\pi)(\mathrm{r} / \sqrt{2})^2}=\frac{(\rho) 4 \mathrm{l}}{(\pi) \mathrm{r}^2}=4 \times \mathrm{R}$
$\therefore \Delta \mathrm{R}=4 \mathrm{R}-\mathrm{R}=3 \mathrm{R}$
$\frac{\Delta \mathrm{R}}{\mathrm{R}} \%=\frac{3 \mathrm{R}}{\mathrm{R}} \times 100=300 \%$
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