A cylinder of mass M$$_C$$ and sphere of mass M$$_S$$ are placed at points A and B of two inclines, respectively. (See figure). If they roll on the incline without slipping such that their accelerations are the same, then the ratio $$\frac{\sin\theta_C}{\sin\theta_S}$$ is:

The linear acceleration $$a$$ of a rigid body rolling down an inclined plane of angle $$\theta$$ without slipping is given by:

$$a = \frac{g \sin \theta}{1 + \frac{I}{M R^2}}$$

Assuming a standard solid cylinder, its moment of inertia is $$I = \frac{1}{2} M_C R^2$$,

$$a_C = \frac{g \sin \theta_C}{1 + \frac{1}{2}} = \frac{g \sin \theta_C}{\frac{3}{2}} = \frac{2}{3} g \sin \theta_C$$

Assuming a standard solid sphere, its moment of inertia is $$I = \frac{2}{5} M_S R^2$$,

$$a_S = \frac{g \sin \theta_S}{1 + \frac{2}{5}} = \frac{g \sin \theta_S}{\frac{7}{5}} = \frac{5}{7} g \sin \theta_S$$

$$\frac{2}{3} g \sin \theta_C = \frac{5}{7} g \sin \theta_S$$

$$\frac{\sin \theta_C}{\sin \theta_S} = \frac{\frac{5}{7}}{\frac{2}{3}} = \frac{5}{7} \times \frac{3}{2} = \frac{15}{14}$$