Explanation

(d) Given $l^{'} = l + 100\%\ of\ l = 2l$
Initial volume = final volume
ie, $\pi r^{2}l = \pi r^{'2}l^{'}$
$\Longrightarrow \ \ \ \ r^{'2} = \frac{r^{2}l}{l^{'}} = r^{2} \times \frac{l}{2l}$
$\Longrightarrow \ \ \ \ \ r^{'2} = \frac{r^{2}}{2}\ $
$\therefore R^{'} = \rho\frac{l^{'}}{A^{'}} = \rho\frac{2l}{\pi r^{'2}}\ \ \ \ \ \ \left( \because R = \frac{\rho l}{A} \right)$
$\ \ \ \ \ \ \ \ \ \ = \frac{\rho 4l}{\pi r^{2}}$
Thus, $\mathrm{\Delta}R = R^{'} - R = 4R - R = 3R$
$\therefore\ \ \ \%\mathrm{\Delta}R = \frac{3R}{R} \times 100\%$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 300\%$