$A$ solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is:

$A$ ring of radius $3a$ is fixed rigidly on a table. $A$ small ring of mass $m$ and radius $a$ rolls inside it without slipping,as shown in the figure. The small ring is released from position $A$ (the horizontal position). What will be the speed of the center of the small ring when it reaches the lowest point?

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

$A$ ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle $60^{\circ}$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $(2-\sqrt{3}) / \sqrt{10} \,s$, then the height of the top of the inclined plane, in metres, is. . . . . . Take $g=10 \,m \,s^{-2}$.

$A$ sphere rolls down on an inclined plane of inclination $\theta$. What is the acceleration as the sphere reaches the bottom?

$A$ solid sphere of mass $m$ rolls down an inclined plane without slipping,starting from rest at the top of an inclined plane. The linear speed of the sphere at the bottom of the inclined plane is $v$. The kinetic energy of the sphere at the bottom is