If a copper wire is stretched to increase its length by 20 % . The percentage increase in resistance of the…

Before heating, resistance of wire is $R_0=\left(\frac{\rho l_0}{A_0}\right)$

After heating, resistance is $R=\frac{\rho l}{A}$

As volume = constant

Therefore,

$A_0{l}_0=1.2l_{0}A$

$\Rightarrow A=\left(\frac{A_0}{1.2}\right)$

So,

$$\begin{gathered} R=\frac{\rho l}{A} \\ =\frac{\rho \times \left(1.2l_0\right)}{\left({\displaystyle \frac{A_0}{1.2}}\right)} \\ =\frac{\rho l_0}{A_0}\times 1.44 \\ =1.44 R_0 \end{gathered}$$

Percentage increase in $R$ $=\frac{R-R_0}{R_0}\times 100=44\%$