A piece of wire is redrawn by pulling it until its length is doubled. Compare the new resistance with the original value.

Given that, length of the wire is doubled.

But, volume of the wire remain in same both cases.

Volume = Area of cross-section x Length 

             V = A'L' = AL 

or          $\frac{\mathrm{A}\text{'}}{\mathrm{A}} = \frac{\mathrm{L}}{\mathrm{L}\text{'}} = \frac{\mathrm{L}}{2\mathrm{L}} = \frac{1}{2}$         [$\because$ L' = 2L]

$\therefore$ Ratio of the resistances are,  

$\frac{\mathrm{R}\text{'}}{\mathrm{R}} = \frac{\mathrm{\rho }{\displaystyle \frac{\mathrm{L}\text{'}}{\mathrm{A}\text{'}}}}{\mathrm{\rho }{\displaystyle \frac{\mathrm{L}}{\mathrm{A}}}} = \frac{\mathrm{L}\text{'}}{\mathrm{L}}\times \frac{\mathrm{A}}{\mathrm{A}\text{'}} = \frac{2}{1}\times \frac{2}{1} = 4$

i.e.,                    $\mathrm{R}\text{'} = 4\mathrm{R}$ 

Thus, the new resistance is four times the original resistance.