If 25% part of the length of a wire is stretched by 25%, then the percentage change in the resistance of the wire will be about

Solution:

Resistance of the wire after stretching, 
$\begin{array}{l}{R}^{\prime \prime }=R_1+R_2=\frac{3R}{4}+(1.25{)}^2\frac{R}{4}\\ =\frac{3R}{4}+\frac{25}{16}\frac{R}{4}=\frac{73R}{64}\\ \therefore \mathrm{\%}\text{ change }=\frac{R^{\prime \prime }-R^{\mathrm{\prime }}}{R^{\mathrm{\prime }}}\times 100=\frac{\frac{73R}{64}-R}{R}\times 100\\ =\frac{900}{64}=14.06\end{array}$