$A$ disc is given an angular velocity $\omega_0$ as shown in the figure and is kept on an inclined plane of inclination $\theta$ having a coefficient of friction $\mu = \tan \theta$. If the initial acceleration of the centre of mass is $a$ and the angular acceleration is $\alpha$,then:

Two uniform solid spheres having unequal masses and unequal radii are released from rest from the same height on a rough incline. If the spheres roll without slipping,

$A$ uniform solid spherical ball is rolling down a smooth inclined plane from a height $h$. The velocity attained by the ball when it reaches the bottom of the inclined plane is $v$. If the ball is now thrown vertically upwards with the same velocity $v$, the maximum height to which the ball will rise is

$A$ solid sphere rolls down two different inclined planes of the same heights but different angles of inclination.
$(a)$ Will it reach the bottom with the same speed in each case?
$(b)$ Will it take longer to roll down one plane than the other?
$(c)$ If so,which one and why?

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

$A$ tennis ball (treated as a hollow spherical shell) starting from $O$ rolls down a hill. At point $A$,the ball becomes airborne,leaving at an angle of $30^\circ$ with the horizontal. The ball strikes the ground at $B$. What is the value of the distance $AB$ (in $m$)? (Moment of inertia of a spherical shell of mass $m$ and radius $R$ about its diameter is $I = \frac{2}{3}mR^2$).