As we starch the wire, its length will increase but its radius will decrease keeping the volume constant
$\mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}}$
$\pi^2 l=\pi \frac{\mathrm{r}^2}{4} l_{\mathrm{f}}$
$l_{\mathrm{f}}=4 l$
$\frac{\mathrm{R}_{\text {nelv }}}{\mathrm{R}_{\text {old }}}=\left(\frac{4 l}{\frac{\mathrm{r}^2}{4}}\right) \frac{\mathrm{r}^2}{l}=16$
$\mathrm{R}_{\text {neer }}=16 \mathrm{R}$
$\therefore \mathrm{x}=16$
The current in resistance $R _2$ would be zero if
$(A)$ $V_1=V_2$ and $R_1=R_2=R_3$
$(B)$ $V_1=V_2$ and $R_1=2 R_2=R_3$
$(C)$ $V_1=2 V_2$ and $2 R_1=2 R_2=R_3$
$(D)$ $2 V _1= V _2$ and $2 R _1= R _2= R _3$





