When wire is stretched twice, its new length be $l$. Then

$l^{\prime}=2 l$

On stretching volume of the wire remains constant.

$\therefore l A=l^{\prime} A^{\prime}$ where $A^{\prime}$ is the new cross-sectional area

or $\quad A^{\prime}=\frac{l}{l^{\prime}} A=\frac{l}{2 l} A=\frac{A}{2}$

$\therefore \quad$ Resistance of the stretched wire is

$R^{\prime} =\rho \frac{l^{\prime}}{A^{\prime}} =\rho \frac{2 l}{(A / 2)}=4 \rho \frac{l}{A}$

$=4(4 \,\Omega) =16\, \Omega (\text { Using }(\mathrm{i}))$