If a copper wire is stretched to make its radius decrease by 0.1%, then percentage increase in resistance is approximately 

In case of stretching volume of wire is assumed to remain unchanged. 

$\therefore lA=$constant and $L=\frac{c}{A}$

$\therefore R=\rho .\frac{l}{A}=\frac{\rho .C}{A^2}=\frac{\rho .C}{{\left(A{r}^2\right)}^2}=R\propto \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r^4$}\right.$

$\therefore \frac{\Delta R}{R}=-4\frac{\Delta r}{r}\times 100\%$

$=(-4)\times (-0.1\%)=+0.4\%$