$A$ 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:

$A$ cylinder of mass $m$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$. What will be the velocity of its center of mass when the cylinder reaches the bottom?

$A$ body starts rolling down an inclined plane of length $L$ and height $h$. This body reaches the bottom of the plane in time $t$. The relation between $L$ and $t$ is:

$A$ solid cylinder rolls up an inclined plane of angle of inclination $30^{\circ}$. At the bottom of the inclined plane,the centre of mass of the cylinder has a speed of $5 \; m/s$.
$(a)$ How far will the cylinder go up the plane?
$(b)$ How long will it take to return to the bottom?

$A$ sphere of radius $R$ is rolling down an inclined plane whose angle of inclination is $\theta$. Its acceleration would be

$A$ solid sphere rolls down from the top of an inclined plane,$7 \ m$ high,without slipping. Its linear speed at the foot of the plane is $\left(g = 10 \ m/s^2\right)$