A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is

For a solid sphere rolling without slipping, we need to find the ratio of rotational kinetic energy to total kinetic energy.

$$I = \frac{2}{5}mR^2$$

$$K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times \frac{2}{5}mR^2 \times \omega^2 = \frac{1}{5}mR^2\omega^2$$

For rolling without slipping, $$v = R\omega$$

$$K_{trans} = \frac{1}{2}mv^2 = \frac{1}{2}mR^2\omega^2$$

$$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mR^2\omega^2 + \frac{1}{5}mR^2\omega^2 = \frac{7}{10}mR^2\omega^2$$

$$\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{5}mR^2\omega^2}{\frac{7}{10}mR^2\omega^2} = \frac{\frac{1}{5}}{\frac{7}{10}} = \frac{1}{5} \times \frac{10}{7} = \frac{2}{7}$$

Hence, the correct answer is Option C.