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Q.

If a copper wire is stretched to increase its length by 20% the percentage increase in resistance of the wire is _____ %

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answer is 44.

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Detailed Solution

$R α l^{2}$

$\frac{R}{R_{2}} = {(\frac{L 100}{120 L})}^{2} = {(\frac{10}{12})}^{2} R_{2} = \frac{144}{100} R$

$\frac{R_{2} - R_{1}}{R_{1}} \times 100 = \frac{44}{100} \times 100 = 44$

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