$A$ body of mass $m$ slides down along a frictionless inclined plane from height $h$ and just completes motion in a vertical circle of radius $R = 2 \ m$ after reaching the bottom. What is the value of $h$? [Use $g = 10 \ m/s^2$]

Length of a simple pendulum is $1 \,m$. When its bob is at its lowest point its velocity is $7 \,ms^{-1}$. If the bob leaves its circular path at a height $h$ above the centre of the circle, then the value of $h$ is $(g=10 \,ms^{-2})$. (in $\,m$)

$A$ can filled with water is revolved in a vertical circle of radius $r$ with constant speed such that the water just does not fall out. The time period of revolution is ($g=$ acceleration due to gravity).

$A$ particle is moving along a vertical circle of radius $R$. At point $P$,what will be the velocity of the particle? (Assume the critical condition at the highest point $C$).

$A$ particle of mass $m$ is tied to one end of a string of length $l$. The particle is held horizontal with the string taut. It is then projected upward with a velocity $u$. The tension in the string is $\frac{mg}{2}$ when it is inclined at an angle $30^o$ to the horizontal. The value of $u$ is

$A$ particle is rotated in a vertical circle by connecting it to a light rod of length $l$ and keeping the other end of the rod fixed. The minimum speed of the particle when the light rod is horizontal for which the particle will complete the circle is