An aluminium wire is stretched to make its length, 0 . 4 % larger. The percentage change in resistance is

Volume of wire having cross-sectional area $A$ and length $l$ is, $V=Al$

Resistance of a wire is given by, $R=\frac{\rho l}{A}=\frac{\rho l}{{\displaystyle \frac{V}{l}}}=\frac{\rho l^2}{V}$

As change in length is small,

$\frac{∆R}{R}\times 100\%=\frac{2∆l}{l}\times 100\%=2\times 0.4\%=0.8\%$