Explanation

(c) When hollow cylinder slides with out rolling, it possess only translational kinetic energy, $K_{\tau} = \frac{1}{2}mv^{2}$
When it rolls without slipping, it possess both types of kinetic energy,
$K_{N} = \frac{1}{2}mv^{2}\left( 1 + \frac{K^{2}}{R^{2}} \right)$
$\therefore\frac{K_{T}}{K_{N}} = \frac{1}{\left( 1 + \frac{K^{2}}{R^{2}} \right)} = \frac{1}{2}$ [For hollow cylinder $\frac{K^{2}}{R^{2}} = 1$]