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

$\begin{array}{l} R = \frac{{δl}^{2}}{V} \Rightarrow {Rαl}^{2}; V is the volume of wire . \\ \frac{R_{2}}{R_{1}} = {(\frac{l_{2}}{l_{1}})}^{2} = {(\frac{120}{100})}^{2} = \frac{144}{100} \end{array}$

$\Rightarrow (\frac{R_{2}}{R_{1}} - 1) \times 100 = (\frac{144}{100} - 1) \times 100 = 44 %$

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