$A$ cylinder is rolling down an inclined plane of inclination $30^\circ$ without slipping. Its linear acceleration along the inclined plane will be:

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 object of uniform density rolls up a curved path with an initial velocity $v_0$ as shown in the figure. If the maximum height attained by the object is $\frac{7v_0^2}{10g}$ ($g =$ acceleration due to gravity), the object is a . . . . . . .

$A$ solid cylinder is rolling down on an inclined plane of angle $\theta$. The coefficient of static friction between the plane and the cylinder is $\mu_s$. The condition for the cylinder not to slip is

$A$ disc of mass $m$ and radius $r$ rolls down an inclined plane of height $h$. When it reaches the bottom of the plane,its rotational kinetic energy is ($g=$ acceleration due to gravity).

$A$ thin uniform circular ring is rolling down an inclined plane of inclination $30^o$ without slipping. Its linear acceleration along the inclined plane will be