A copper wire is stretched to make it 0.5% longer. The percentage...

Volume, $V = A . l$

In differential form,
$\dfrac{dV}{V} = \dfrac{dA}{A} + \dfrac{dl}{l}0 = \dfrac{dA}{A} + \dfrac{dl}{l}$

$⇒ \dfrac{dA}{A} = - \dfrac{dl}{l}$

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

$\dfrac{dR}{R} = \dfrac{dl}{l} - \dfrac{dA}{A}\dfrac{dR}{R} = \dfrac{dl}{l} - ( - \dfrac{dl}{l} )$
$\dfrac{dR}{R} = 2 \cdot \dfrac{dl}{l}$

$\dfrac{dR}{R} \times 100 = 2 \cdot \dfrac{dl}{l} \times 100\dfrac{dR}{R}$

$∴$ change in $R = 2 \times 0.5$