$A$ solid sphere is rolling down an inclined plane. What is the ratio of its translational kinetic energy to its rotational kinetic energy?

$A$ body is rolling down an inclined plane. If the kinetic energy of rotation is $40\%$ of the translational kinetic energy,then the body is a

$A$ solid cylinder of mass $M$ and radius $R$ rolls down an inclined plane of height $h$. When it reaches the foot of the plane,its rotational kinetic energy is ($g=$ acceleration due to gravity).

Derive expressions for the kinetic energy and velocity of a body rolling without sliding down an inclined plane of inclination $\theta$.

$A$ 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 $= 10 \text{ ms}^{-2}$)

$A$ solid spherical ball is rolled up an inclined plane of angle of inclination $30^{\circ}$ with an initial speed of $4 \ m/s$ at the bottom of the inclination. How far will the ball go up the plane (in $cm$)? (Use $g=10 \ m/s^2$)