Q222 / 225 · JEE Main 2003 · Physics

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

  • A
    $200 \%$
  • B
    $100 \%$
  • C
    $50 \%$
  • Correct answer
    D
    $300 \%$

Solution

$\mathrm{L}_1=2 \mathrm{l}$ 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}}$$;$ $\mathrm{R}=\rho \frac{1}{\pi^2}$ $\mathrm{R}_{\text {new }}=(\rho) \frac{21}{(\pi)(\mathrm{r} / \sqrt{2})^2}=\frac{(\rho) 4 \mathrm{l}}{(\pi) \mathrm{r}^2}=4 \times \mathrm{R}$ $\therefore \Delta \mathrm{R}=4 \mathrm{R}-\mathrm{R}=3 \mathrm{R}$ $\frac{\Delta \mathrm{R}}{\mathrm{R}} \%=\frac{3 \mathrm{R}}{\mathrm{R}} \times 100=300 \%$

Hence, the correct answer is option: 'D'.

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