$A$ plane is inclined at an angle of $30^{\circ}$ with the horizontal. If a sphere rolls down this plane without slipping,what will be the linear acceleration of the sphere?

The moment of inertia of a solid cylinder about its axis is $I$. It is allowed to roll down an incline without slipping. If its angular velocity at the bottom is $\omega$,then the total kinetic energy $(K.E.)$ of the cylinder will be:

An inclined plane makes an angle of $30^\circ$ with the horizontal. $A$ solid sphere starts rolling down from rest without slipping. Its linear acceleration will be:

$A$ solid cylinder of diameter $30 \ cm$ is rolled down an inclined plane from a height of $2 \ m$. If there is no energy loss due to friction,the angular velocity at the base of the plane is ....... $rad/s$. (Take $g = 10 \ m/s^2$)

$A$ 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 $g=10 \ m/s^2$)

The rotational kinetic energy of a solid sphere of mass $3 \; kg$ and radius $0.2 \; m$ rolling down an inclined plane of height $7 \; m$ is (in $; J$)