The diameter of a wire is reduced to one-fifth of its original value by stretching it. It its initial resistance is $R$ , what would be its resistance after reduction of the diameter?

\frac{R}{625}

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\frac{R}{25}

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25 R

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625 R

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Solution

The correct option is D $625 R$
Initial volume $=$ Final volume
$A_{1} L_{1} = A_{2} L_{2}$
$π r_{1}^{2} l_{1} = π r_{2}^{2} l_{2}$
$\frac{r_{1}^{2}}{r_{2}^{2}} = \frac{l_{2}}{l_{1}}$
$\frac{R_{1}}{R_{2}} = \frac{ρ l_{1}}{A_{1}} \times \frac{A_{2}}{ρ l_{2}}$
$\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} \times \frac{π r_{1}^{2}}{π r_{2}^{2}} \Rightarrow \frac{R_{1}}{R_{2}} = {(\frac{r_{2}}{r_{1}})}^{4}$
$D_{2} = \frac{D_{1}}{5} r_{2} = \frac{r_{1}}{5}$
$\frac{R_{1}}{R_{2}} = \frac{1}{625}$

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