A piece of wire is redrawn, without change in volume so that its radius is halved. Compare the new resistance with the original resistance.

Let length of original wire is $l$ and radius is $r$.

So $R=\rho \frac{l}{\pi r^{2}}$

Original volume $V _{1}=\pi r^{2} l$

When its radius halved it becomes $\frac{r}{2}$ and let length of redrawn wire becomes $l^{\prime}$

Then $R^{\prime}=4 \rho \frac{l^{\prime}}{\pi\left(\frac{r}{2}\right)^{2}}$

and volume $V _{2}=\pi\left(\frac{r}{2}\right)^{2} l^{\prime}$

Volume is same or $V _{1}= V _{2}$ or $. \pi r^{2} l=\pi\left(\frac{r}{2}\right)^{2} l^{\prime}$

or $\quad l^{\prime}=4 l$

Substitute value in $R ^{\prime}=\rho \frac{l^{\prime}}{\pi\left(\frac{r}{2}\right)^{2}}$

$R ^{\prime}=\rho \frac{4 l \times 4}{\pi r^{2}}$

$R ^{\prime}=\rho \frac{l \times 16}{\pi r^{2}}$

$R ^{\prime}=16 \times R$

So resistance increased by $16$ times.