1. Interpret the situation

A solid sphere has two possible kinds of kinetic energy:

• Translational kinetic energy: $K_t = \frac12 Mv^2$
• Rotational kinetic energy: $K_r = \frac12 I\omega^2$

For a rolling sphere, the rolling condition is $v = R\omega$ where $R$ is the radius of the sphere.

2. Moment of inertia of a solid sphere

For a solid sphere about its center, $I = \frac{2}{5}MR^2$

3. Compute rotational kinetic energy

$K_r = \frac12 I\omega^2 = \frac12 \cdot \frac{2}{5}MR^2\omega^2$ $K_r = \frac{1}{5}MR^2\omega^2$

Using $v = R\omega$, we get $R^2\omega^2 = v^2$. Hence, $K_r = \frac{1}{5}Mv^2$

4. Compute total kinetic energy

$K_{\text{total}} = K_t + K_r = \frac12 Mv^2 + \frac15 Mv^2$

Taking LCM $10$, $K_{\text{total}} = \left(\frac{5}{10} + \frac{2}{10}\right)Mv^2 = \frac{7}{10}Mv^2$

5. Find the required ratio

$\frac{K_r}{K_{\text{total}}} = \frac{\frac15 Mv^2}{\frac{7}{10}Mv^2}$

Cancel $Mv^2$: $\frac{K_r}{K_{\text{total}}} = \frac{1/5}{7/10} = \frac{1}{5}\cdot\frac{10}{7} = \frac{2}{7}$

6. Match with options

$\frac{K_r}{K_{\text{total}}} = \frac{2}{7}$ So the correct option is B.

Note: Although the statement says the sphere is on a frictionless plane, pure rolling cannot be maintained by a frictionless surface unless somehow already constrained. But since the question asks for the standard rolling-energy ratio, we use $v=R\omega$.