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A cylindrical metallic wire is stretched to increase its length if the resistance of the wire is increased by 4% then the percentage increase in its length is

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answer is D.

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Detailed Solution

Area x length = constant

$\Rightarrow \frac{d A}{A} + \frac{d l}{l} = 0$

$\frac{d A}{A} = - \frac{d l}{l}$

$R = \frac{ρ L}{A}$

$\Rightarrow \frac{d R}{R} \times 100 = [\frac{d ρ}{ρ} + \frac{d l}{L} - \frac{d A}{A}] \times 100$

$= [0 + \frac{d l}{l} + \frac{d l}{l}] \times 100$

$= \frac{2 d l}{l}$

$4 = 2 \frac{d l}{l} \Rightarrow \frac{d l}{l} = 2 %$

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