$A$ body is placed as shown in the figure. To complete one full vertical circular loop,the height $h$ must be:

$A$ body of mass $m \ kg$ is rotating in a vertical circle at the end of a string of length $r \ m$. What is the difference in the kinetic energy at the top and the bottom of the circle?

$A$ stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is

$A$ simple pendulum with a bob of mass $m = 1 \ kg$, charge $q = 5 \ \mu C$ and string length $l = 1 \ m$ is given a horizontal velocity $u$ in a uniform electric field $E = 2 \times 10^6 \ V/m$ at its bottom-most point $A$, as shown in the figure. It is given a speed $u$ such that the particle leaves the circular path at its topmost point $C$. Find the speed $u$. (Take $g = 10 \ m/s^2$)

$A$ ball attached to one end of a string swings in a vertical plane such that its acceleration at point $A$ (extreme position) is equal to its acceleration at point $B$ (mean position). The angle $\theta$ is :-

$A$ stone is tied to a string of length $L$ and is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is