$A$ solid disc of radius $a$ and mass $m$ rolls down without slipping on an inclined plane making an angle $\theta$ with the horizontal. The acceleration of the disc will be $\frac{2}{b} g \sin \theta$ where $b$ is $........$. (Round off to the Nearest Integer)
($g =$ acceleration due to gravity)
($\theta =$ angle as shown in figure)

$A$ disc rolls without slipping down an inclined plane of length $L$ and inclination $\theta$. What will be its velocity at the bottom?

$A$ ring takes times $t_1$ and $t_2$ to reach the bottom of an inclined plane of length $L$ by sliding and rolling,respectively. What is the ratio of $t_1$ to $t_2$?

The linear speed of a uniform spherical shell after rolling down an inclined plane of vertical height $h$ from rest is:

$A$ uniform solid sphere of mass $m$ and radius $r$ rolls without slipping down an inclined plane,inclined at an angle $45^{\circ}$ to the horizontal. Find the magnitude of the minimum frictional coefficient at which slipping is absent.

$A$ 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 $g=10 \,m \,s^{-2}$.