To solve this problem, we need to find the ratio of accelerations for two scenarios:$\\$1. A solid sphere rolling down an incline without slipping.$\\$2. A solid sphere slipping down the incline without rolling.$\\$Let's analyze each case:$\\$• For the sphere rolling without slipping:$\\$The forces acting on the sphere are gravitational force, normal force, and friction.$\\$The frictional force provides the torque necessary for rolling.$\\$Using Newton's second law for translation:$\\ m g \sin \theta - f = m a_1 \quad$(1)$\\$Using the rotational motion equation:$\\ f R = I \alpha \quad$(2)$\\$For a solid sphere, the moment of inertia $I = \frac{2}{5} m R^2$ and $\alpha = \frac{a_1}{R}. \\$Substitute $I$ and $\alpha$ in equation (2):$\\ f R = \frac{2}{5} m R^2 \frac{a_1}{R} \\ f = \frac{2}{5} m a_1 \quad$(3)$\\$Substitute equation (3) into equation (1):$\\ m g \sin \theta - \frac{2}{5} m a_1 = m a_1 \\ m g \sin \theta = m a_1 + \frac{2}{5} m a_1 \\ m g \sin \theta = \frac{7}{5} m a_1 \\ a_1 = \frac{5}{7} g \sin \theta \\$• For the sphere slipping without rolling:$\\$There is no frictional force providing torque, so the sphere slides down.$\\$Using Newton's second law for translation:$\\ m g \sin \theta = m a_2 \\ a_2 = g \sin \theta \\$Now, find the ratio of the accelerations:$\\ \frac{a_1}{a_2} = \frac{\frac{5}{7} g \sin \theta}{g \sin \theta} \\ \frac{a_1}{a_2} = \frac{5}{7} \\$Therefore, the ratio of the acceleration for a solid sphere rolling down an incline without slipping$\\$to the acceleration of the sphere slipping down the incline without rolling is $5:7. \\$This corresponds to Option 1.