$A$ body of mass $m \ kg$ slides from rest along the curve of a vertical circle from point $A$ to $B$ on a frictionless path. The velocity of the body at $B$ is: (Given: $R = 14 \ m$,$g = 10 \ m/s^2$,and $\sqrt{2} = 1.4$) (in $m/s$)

$A$ particle with charge $Q$ coulomb, tied at the end of an inextensible string of length $R$ metre, revolves in a vertical plane. At the centre of the circular trajectory, there is a fixed charge of magnitude $Q$ coulomb. The mass of the moving charge $M$ is such that $Mg = \frac{Q^2}{4 \pi \epsilon_0 R^2}$. If at the highest position of the particle, the tension of the string just vanishes, the horizontal velocity at the lowest point has to be

$A$ body of mass $m$ hangs at one end of a string of length $l$,the other end of which is fixed. It is given a horizontal velocity such that the string just reaches a position where it makes an angle of $60^{\circ}$ with the vertical. The tension in the string at the mean position is:

$A$ 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

The minimum velocity $v$ required for a simple pendulum to complete one full rotation in a vertical plane is given. If the length of the string is reduced to one-fourth,what will be the new required velocity?

$A$ bob of mass $m$ is tied to a string of length $l$. What is the minimum kinetic energy that must be provided to the bob at the lowest point so that it completes a full vertical circle (in $;mgl$)?