$A$ disc of mass $M$ and radius $R$ rolls on a horizontal surface and then rolls up an inclined plane as shown in the figure. If the velocity of the disc is $v,$ the height to which the disc will rise will be

An inclined plane makes an angle of $30^{\circ}$ with the horizontal. $A$ solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration ($g=$ acceleration due to gravity,$\sin 30^{\circ}=0.5$).

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

$A$ cylinder is rolling down on an inclined plane of inclination $60^{\circ}$. Its acceleration during rolling down will be $\frac{x}{\sqrt{3}} \ m/s^2$,where $x=$ . . . . . . . (use $g=10 \ m/s^2$)

$A$ spherical object of mass $M$ and radius $R$ has a moment of inertia $I$ about its axis. If this object rolls down an inclined plane of angle $\theta$ without slipping,its acceleration is given by:

$STATEMENT-1$: Two cylinders,one hollow (metal) and the other solid (wood) with the same mass and identical dimensions,are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.
$STATEMENT-2$: By the principle of conservation of energy,the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.