When a sphere with moment of inertia $I$ rolls down an inclined plane,what percentage of its total energy is rotational kinetic energy?

Two uniform similar discs roll down two inclined planes of length $S$ and $2S$ respectively,as shown in the figure. The velocities of the two discs at the points $A$ and $B$ at the bottom of the inclined planes are related as:

$A$ sphere,a disc,a ring,and a hollow sphere of the same radius are rolled down from the same height on an inclined plane simultaneously. The order in which these objects reach the bottom is:

$A$ solid cylinder of mass $m$ and radius $R$ rolls down an inclined plane of height $h$ without slipping. The speed of its centre of mass when it reaches the bottom is

$A$ solid cylinder rolls down an inclined plane from a height $h$. At any moment,the ratio of rotational kinetic energy to the total kinetic energy is:

$A$ solid sphere rolls down without slipping from the top of an inclined plane of height $28 \text{ m}$ and angle of inclination $30^{\circ}$. The velocity of the sphere, when it reaches the bottom of the plane is (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)