An aluminium wire is stretched to make its length, $$0.4\%$$ larger. The percentage change in resistance is

An aluminium wire is stretched to make its length $$0.4\%$$ larger and we wish to determine the resulting percentage change in its resistance.

The resistance of the wire is given by $$R = \frac{\rho l}{A}$$ where $$\rho$$ is the resistivity, $$l$$ is the length, and $$A$$ is the cross-sectional area.

Since the volume of the wire remains constant during stretching, we have $$V = A\,l = \text{constant}$$ which implies $$A = \frac{V}{l}\,.$$

Substituting this expression for $$A$$ into the resistance formula yields $$R = \frac{\rho\,l}{A} = \frac{\rho\,l^2}{V}$$ so that $$R\propto l^2$$ when $$\rho$$ and $$V$$ are constant.

Taking differentials then gives $$\frac{\Delta R}{R} = 2\,\frac{\Delta l}{l}\,. $$ Since the length increases by $$0.4\%$$, we obtain $$\frac{\Delta R}{R} = 2 \times 0.4\% = 0.8\% \,. $$

Therefore, the percentage change in resistance is $$0.8\%$$, and the correct answer is Option C.