b
When body rolls dawn on inclined plane with velocity \(V_0\) at bottom then body has both rotational and translational kinetic energy.

Thereforce, by law of conservation of energy,

\(\begin{array}{l}
P.E\, = \,K.{E_{trans}} + K.{E_{rotational}}\\
 = \frac{1}{2}mV_0^2 + \frac{1}{2}I{\omega ^2}\\
 = \frac{1}{2}mV_0^2 + \frac{1}{2}m{k^2}\frac{{V_0^2}}{{R_0^2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {I = m{k^2},\omega  = \frac{V}{{{R_0}}}} \right]
\end{array}\)

When body is sliding down then body has only translatory motion.

\(\begin{array}{l}
\therefore P.E = K.{E_{trans}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}m{\left( {\frac{5}{4}{v_0}} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)
\end{array}\)

Dividing \((i)\) by \((ii)\) we get

\(\begin{array}{l}
\frac{{P.E}}{{P.E}} = \frac{{\frac{1}{2}mv_0^2\left[ {1 + \frac{{{K^2}}}{{R_0^2}}} \right]}}{{\frac{1}{2} \times \frac{{25}}{{16}} \times mV_0^2}}\\
 = \frac{{25}}{{16}} = 1 + \frac{{{K^2}}}{{R_0^2}} \Rightarrow \frac{{{K^2}}}{{R_0^2}} = \frac{9}{{16}}\\
or,\,\,\,K = \frac{3}{4}\,{R_{0.}}
\end{array}\)