A wire of resistance R is stretched till its length is double of the…

Resistance R of a wire of length $l$ and radius r with specific resistance K for its material is given by

$\mathrm{R}=\rho l / \pi \mathrm{r}^2$                    (i)

As the wire is stretched, there is no change in its mass M.

But $\mathrm{M}=\pi \mathrm{r}^2 l \mathrm{~d} \quad$ or $\pi \mathrm{r}^2=\mathrm{M} / l \mathrm{~d}$

Here, $\mathrm{R}=\frac{\rho l}{\mathrm{M} / l \mathrm{~d}}=\left(\frac{\rho \mathrm{d}}{\mathrm{M}}\right) l^2$                  (ii)

Let the resistance of the stretched wire by $R^{\prime}$; then

$\mathrm{R}^{\prime}=\left(\frac{\rho \mathrm{d}}{\mathrm{M}}\right)(2 l)^2=4 \mathrm{R}$