A copper wire is stretched to make it 0.5 % longer. The percentage change in its electrical resistance if…

Volume, $V=A.l$

In differential form,
$$\begin{gathered} \frac{dV}{V}=\frac{dA}{A}+\frac{dl}{l} \\ 0=\frac{dA}{A}+\frac{dl}{l} \end{gathered}$$

$\Rightarrow \frac{dA}{A}=-\frac{dl}{l}$

Also resistance of wire is as follows $R=\rho \frac{l}{A}$
diffrentiation formula for resistance is as follows

 $$\begin{gathered} \frac{dR}{R}=\frac{dl}{l}-\frac{dA}{A} \\ \frac{dR}{R}=\frac{dl}{l}-(-\frac{dl}{l}) \end{gathered}$$
$\frac{dR}{R}=2·\frac{dl}{l}$

$$\begin{gathered} \frac{dR}{R}\times 100=2·\frac{dl}{l}\times 100 \\ \frac{dR}{R}\%=2\times 0.5\% \\ \frac{dR}{R}\%=1\% \end{gathered}$$

$\therefore \%$ change in $R=2\times 0.5\%=1\%$