Explanation

(c) Resistance of wire given by
$R = \rho\frac{l}{A}$
Also, volume (V)=Length$(l)$×Area $(A)$
Where, $A = \pi r^{2}$ (r is radius)
When the wire is stretched its volume (V) remains constant.
Hence,
$R = \frac{\rho V}{\pi^{2}r^{4}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots.(i)$
When radius is halved
$R^{'} = \frac{pV}{\pi r^{2}\left( \frac{r}{2} \right)^{4}}\ \ \ldots\ldots\ldots\ldots\ldots(ii)$
$\therefore\frac{R^{'}}{R} = \frac{16\rho V}{\pi^{2}r^{4}} \times \frac{\pi^{2}r^{4}}{\rho V} = 16$
$\Longrightarrow R^{'} = 16R$
Hence, new resistance increases to sixteen times its original value.