Q 12-03-165JEE MainJEE Main 2022 (26 Jun, Shift 1)Easy
An aluminium wire is stretched to make its length $0.4\%$ larger. The percentage change in resistance is
Answer: (C) $0.8\%$
With volume constant, $R\propto l^2$, so $\dfrac{\Delta R}{R} = 2\dfrac{\Delta l}{l} = 2\times0.4\% = 0.8\%$.
Solution by Sreeraj P, M.Sc Physics