If a copper wire is stretched to increase its length by 20 % . The...

Before heating, resistance of wire is ${R}_{0} = \left(\dfrac{ρ{l}_{0}}{{A}_{0}}\right)$

After heating, resistance is $R = \dfrac{ρl}{A}$

As volume = constant

Therefore,

${A}_{0}{l}_{0} = 1 . 2{l}_{0}A$

$⇒ A = \left(\dfrac{{A}_{0}}{1 . 2}\right)$

So,

$R = \dfrac{ρl}{A} = \dfrac{ρ \times \left(1 . 2{l}_{0}\right)}{\left(\dfrac{{A}_{0}}{1 . 2}\right)} = \dfrac{ρ{l}_{0}}{{A}_{0}} \times 1 . 44 = 1 . 44{R}_{0}$

Percentage increase in $R$ $= \dfrac{R - {R}_{0}}{{R}_{0}} \times 100 = 44$