A ring (R), disc (D), a solid sphere (S) and a hollow sphere with thin walls (H) al having the same mass but different radii, start together from rest at the top of an inclined plane and roll down without slipping then –
(A) All of them will reach the bottom of the inclined together
(B) The body with the maximum radius will reach the bottom first
(C) They will reach the bottom in the order S, D, H and R.
(D) All of them will have the same Kinetic energy at the bottom of the incline.

Hint: In order to solve this question, we need to know the concept of moment of Inertia, Kinetic energy and potential energy. The moment of inertia of a rigid body is a quantity that determines the torque needed for a desired angular acceleration about a rotational axis.

Complete step by step answer:
The moment of inertia depends on the body’s mass distribution and the axis chosen with larger moments requiring more torque to change the body’s rate of rotation.
The moment of inertia of ring, disc, solid sphere and hollow sphere are $m{r^2},\dfrac{1}{2}m{r^2},\dfrac{2}{5}m{r^2}{\text{ and }}\dfrac{2}{3}m{r^2}$ respectively. As the moment of inertia is the resistance to rotational motion, we see that the object with the least moment of inertia will reach the bottom first. Thus, the order of reaching the bottom is S, D, H and R.
Also, from work-energy theorem, wall forces $ = \Delta K \cdot E,{\text{ }}mgh = K \cdot E \Rightarrow all{\text{ will have same KE}}$ at the bottom. The gain in total KE is equal to loss in gravitational PE which in turn is the same for all four objects and equal to mgh, where h is the height of the inclined plane.
Also, $mgh = \dfrac{1}{2}I{w^2} + \dfrac{1}{2}m{v^2}$
$
   \Rightarrow mgh = \dfrac{1}{2}Km{r^2}{w^2} + \dfrac{1}{2}m{v^2} \\
   \Rightarrow mgh = \dfrac{1}{2}Km{v^2} + \dfrac{1}{2}m{v^2} \\
   \Rightarrow mgh = \dfrac{1}{2}\left( {1 + k} \right)m{v^2} \\
   \Rightarrow 2gh = \left( {1 + k} \right){v^2} \\
   \Rightarrow {v^2} = \dfrac{{2gh}}{{\left( {1 + k} \right)}} \\
$
Acceleration of body rolling purely on the inclined plane $a = \dfrac{{g\sin \theta }}{{1 + K}}$ where\[\;K\] is a constant.
Now, $v = at \Rightarrow t = \dfrac{{\sqrt {2gh/\left( {1 + K} \right)} }}{{g\sin \theta /\left( {1 + K} \right)}} = \sqrt {\dfrac{{2gh/\left( {1 + K} \right)}}{{{g^2}{{\sin }^2}\theta /{{\left( {1 + K} \right)}^2}}}} $
$ \Rightarrow t = \sqrt {\dfrac{{2gh}}{{\left( {1 + K} \right)}} \times \dfrac{{{{\left( {1 + K} \right)}^2}}}{{{g^2}\sin \theta }}} = \sqrt {\dfrac{{2h\left( {1 + K} \right)}}{{g\sin \theta }}} $
As, ${K_R} > {K_H} > {K_D} > {K_S} \Rightarrow {t_R} > {t_H} > {t_D} > {t_S}$
Hence, they will reach the bottom in the order S, D, H and R.
$\therefore $ options C and D are correct.

Note: The principle of work-energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the Kinetic energy of the particle. This definition can be extended to rigid bodies by defining the work of the torque and rotational KE. The work w done by the net force on a particle equals the change in particles \[KE.\]
$w = \Delta KE = \dfrac{1}{2}mv_f^2 = \dfrac{1}{2}mv_i^2$
Where \[\;{v_f}{\text{ and }}{v_i}\] are the speeds of the particle after and before the application of force and m is the mass of the particle.