Consider a situation in which a ring, a solid cylinder and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and having identical diameter. The correct statement for this situation is

When a rigid body rolls down an inclined plane without slipping, we use energy conservation. Starting from rest at height $$h$$, all potential energy converts to translational and rotational kinetic energy:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Using the rolling condition $$\omega = v/R$$ and writing $$I = kmR^2$$ where $$k$$ is a constant depending on the body's geometry:

$$mgh = \frac{1}{2}mv^2(1 + k)$$ $$v = \sqrt{\frac{2gh}{1+k}}$$

The moment of inertia factors are: for a ring, $$I = mR^2$$ so $$k = 1$$; for a solid cylinder, $$I = \frac{1}{2}mR^2$$ so $$k = \frac{1}{2}$$; for a solid sphere, $$I = \frac{2}{5}mR^2$$ so $$k = \frac{2}{5}$$.

The velocity at the bottom is: $$v_{\text{ring}} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$$ $$v_{\text{cylinder}} = \sqrt{\frac{2gh}{1+\frac{1}{2}}} = \sqrt{\frac{4gh}{3}}$$ $$v_{\text{sphere}} = \sqrt{\frac{2gh}{1+\frac{2}{5}}} = \sqrt{\frac{10gh}{7}}$$

Comparing: $$\frac{10}{7} > \frac{4}{3} > 1$$, so $$v_{\text{sphere}} > v_{\text{cylinder}} > v_{\text{ring}}$$. The sphere has the greatest and the ring has the least velocity at the bottom.