An aluminium wire is stretched to make its length, 0 . 4 % larger....

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

Resistance of a wire is given by, $R = \dfrac{ρl}{A} = \dfrac{ρl}{\dfrac{V}{l}} = \dfrac{ρ{l}^{2}}{V}$

As change in length is small,

$\dfrac{∆ R}{R} \times 100$