A wire is stretched to increase its length by 2%, then the percentage change in its resistance is _________.

(D)
\(R =\frac{\rho l}{A}=\frac{\rho l^2}{A l}=\frac{\rho l^2}{V}\)
Volume V remains constant on stretching
\(R \propto l^2\)
\(l_1=l\) and \(l_2=l+2 \% l=1.02 l\)
Now \(\frac{ R _2}{ R _1}=\left(\frac{l_2}{l_1}\right)^2\)
\(\begin{aligned} \therefore \frac{ R _2}{ R _1} =\left[\frac{1.02 l}{l}\right]^2 \\ \therefore R _2 =1.0404 R _1\end{aligned}\)
\(\begin{aligned} \text { Change } =\frac{R_2-R_1}{R_1} \times 100 \% \\ = \frac{1.0404 R_1-R_1}{R_1} \times 100 \% \\ = 4.04 \% \approx 4 \%\end{aligned}\)
Second Method :
\(\begin{aligned} R =\frac{\rho l^2}{V} \\ d R =\frac{\rho}{ V }(2 l) d l \\ \frac{d R }{ R } =\frac{\frac{\rho}{ V }(2 l) d l}{\frac{\rho l^2}{V}} \\ \frac{d R }{ R } =2 \frac{d l}{l} \\ =2 \times 2 \% \\ =4 \%\end{aligned}\)