Current Electricity

JEE Main level, single correct, medium

132 of 380

Question 132 of 380

An aluminium wire is stretched to make its length, 0.40.4% larger. The percentage change in resistance is

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Answer: (C) 0.80.8%

Volume of wire having cross-sectional area AA and length ll is, V=AlV=Al
Resistance of a wire is given by, R=ρlA=ρlVl=ρl2VR=\dfrac{\rho l}{A}=\dfrac{\rho l}{\dfrac{V}{l}}=\dfrac{\rho {l}^{2}}{V}
As change in length is small,
∆RR×100\dfrac{∆R}{R}\times 100%=\dfrac{2∆l}{l}\times 100%=2\times 0.4%=0.8%

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