$\begin{aligned}
& \mathrm{KE}{(T)}=\frac{1}{2} m v^2 \
& \mathrm{KE}{(R)}=\frac{1}{2} \cdot \frac{2}{5} m R^2 \cdot \frac{v^2}{R^2}=\frac{1}{2} m v^2\left(\frac{2}{5}\right)
\end{aligned}So,\frac{\mathrm{KE}{(T)}}{\mathrm{KE}{(R)}}=\frac{5}{2}$
A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :
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