$A$ body of mass $m$ is tied to a string of length $l$ and rotated in a vertical circle. What is the difference in tension between the lowest point and the highest point (in $,mg$)?

$A$ body slides down a frictionless track which ends in a circular loop of diameter $D$. The minimum height $h$ of the body in terms of $D$ so that it may just complete the loop 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 $g = 10 \ m/s^2$)

$A$ particle of mass $m$,attached to a string of length $l$,is moving in a vertical circle. If the particle is just looping the loop without the string slackening,and $v_A, v_B, v_D$ are the speeds at positions $A, B, D$ shown in the figure,then:

$A$ body of mass $m$ is suspended from a string of length $l$. What is the minimum horizontal velocity that should be given to the body in its lowest position so that it may complete one full revolution in a vertical plane with the point of suspension as the center of the circle?

$A$ small block slides down from the top of a hemisphere of radius $R=3\, m$ as shown in the figure. The height $h$ at which the block will lose contact with the surface of the sphere is $............. \; m$. (Assume there is no friction between the block and the hemisphere)