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If a wire is stretched, so that its length is 20% more than its initial length, the percentage increase in the resistance of the wire is :-

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25%

40%

44%

10%

answer is C.

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Detailed Solution

l₁=100
L₂=120
$\begin{array}{l} \frac{R_{1}}{R_{2}} = (\frac{l_{1}}{l_{2}}) 2 \\ \frac{R_{1} - R_{2}}{R_{2}} \times 100 \end{array}$

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