To solve this problem, we need to find the ratio of translational kinetic energy $K_t$ to the total kinetic energy $(K_t + K_r)$ for a solid sphere in rolling motion.$\\$For a solid sphere, the moment of inertia $I$ about its center is $\frac{2}{5} m r^2,$ where $m$ is the mass and $r$ is the radius.$\\$The translational kinetic energy $K_t$ is given by:$\\ K_t = \frac{1}{2} m v^2 \\$The rotational kinetic energy $K_r$ is given by:$\\ K_r = \frac{1}{2} I \omega^2 \\$Since the sphere is rolling without slipping, the angular velocity $\omega$ is related to the linear velocity $v$ by $\omega = \frac{v}{r}. \\$Substitute $I = \frac{2}{5} m r^2$ and $\omega = \frac{v}{r}$ into the expression for $K_r: \\ K_r = \frac{1}{2} \left(\frac{2}{5} m r^2\right) \left(\frac{v}{r}\right)^2 \\ K_r = \frac{1}{2} \cdot \frac{2}{5} m r^2 \cdot \frac{v^2}{r^2} \\ K_r = \frac{1}{5} m v^2 \\$Now, the total kinetic energy $(K_t + K_r)$ is:$\\ K_t + K_r = \frac{1}{2} m v^2 + \frac{1}{5} m v^2 \\ K_t + K_r = \left(\frac{1}{2} + \frac{1}{5}\right) m v^2 \\ K_t + K_r = \left(\frac{5}{10} + \frac{2}{10}\right) m v^2 \\ K_t + K_r = \frac{7}{10} m v^2 \\$The ratio $K_t : (K_t + K_r)$ is:$\\ \frac{K_t}{K_t + K_r} = \frac{\frac{1}{2} m v^2}{\frac{7}{10} m v^2} \\ \frac{K_t}{K_t + K_r} = \frac{\frac{1}{2}}{\frac{7}{10}} \\ \frac{K_t}{K_t + K_r} = \frac{1}{2} \times \frac{10}{7} \\ \frac{K_t}{K_t + K_r} = \frac{10}{14} \\ \frac{K_t}{K_t + K_r} = \frac{5}{7} \\$Therefore, the ratio $K_t : (K_t + K_r)$ for the sphere is $5:7. \\$This corresponds to Option 2.