Current Electricity question for JEE Main (JEE Main 2019 (9 Jan, Shift 1)), with solution

Answer: (B) $1.0\%$

With volume $V = Al$ fixed, $A = V/l$, so
$$R = \frac{\rho l}{A} = \frac{\rho l^2}{V} \propto l^2$$
For a small change, $\dfrac{\Delta R}{R} = 2\dfrac{\Delta l}{l} = 2 \times 0.5\% = 1.0\%$.

Solution by Sreeraj P, M.Sc Physics