The length of a metallic wire is increased by 20 % and its area of...

The length of a metallic wire is increased by $20 %$ and its area of cross-section is reduced by $4 % .$ The percentage change in resistance of the metallic wire is $________ .$

Correct answer

Step-by-step solution

The given data is
$l ' = 1 . 2lA ' = 0 . 96A$
Using the formula $R = \dfrac{ρl}{A}$ , the new resistance is
${R}^{'} = \dfrac{1 . 2ρl}{0 . 96A}$
$⇒ {R}^{'} = 1 . 25R$
$⇒ \DeltaR = 1 . 25R - R = 0 . 25R$
Hence, the percentage increase is
$= \dfrac{\DeltaR}{R} \times 100 = \dfrac{0 . 25R}{R} \times 100 = 25 %$

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