The length of a metallic wire is increased by 20 % and its area of cross-section is reduced by 4 % . The…

The given data is 

$$\begin{gathered} l\text{'}=1.2l \\ A\text{'}=0.96A \end{gathered}$$

Using the formula $R=\frac{\rho l}{A}$, the new resistance is 

$R^{\text{'}}=\frac{1.2\rho l}{0.96A}$

$\Rightarrow R^{\text{'}}=1.25R$

$\Rightarrow \Delta R=1.25R-R=0.25R$

Hence, the percentage increase is 

$$\begin{gathered} =\frac{\Delta R}{R}\times 100 \\ =\frac{0.25R}{R}\times 100=25\% \end{gathered}$$