$A$ small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial position. The object is

$A$ uniform spherical object of mass $M$ and radius $R$ has a moment of inertia $I$. It rolls down an inclined plane of angle $\theta$ without slipping. What is its acceleration?

One solid sphere and one disc of the same radius are rolling down an inclined plane without slipping. One reaches the bottom earlier than the other due to:

$A$ solid sphere,a disc,and a solid cylinder,all of the same mass and made of the same material,are allowed to roll down from rest on an inclined plane. Then,which of the following is true?

$A$ small disc of mass $500 \ g$ and radius $5 \ cm$ rolls down an inclined plane without slipping. The speed of its center of mass when it reaches the bottom of the inclined plane depends on:

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