$A$ 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 $g = 10 \ m/s^2$]

$A$ particle of mass $m$ is released from a height $H$ on a smooth curved surface which ends into a vertical loop of radius $R$,as shown. The minimum value of $H$ required so that the particle makes a complete vertical circle is given by (in $R$)

$A$ 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 :-

Consider a particle moving with a minimum speed $v$ at the highest point of a vertical circle of radius $R$. If the radius of the circle is doubled, the corresponding minimum speed will be:

$A$ mass $m$ is released from point $A$ as shown in the figure. The tension in the string at point $B$ will be:

$A$ smooth inclined plane ends in a vertical circular loop, as shown in the figure. $A$ small body is released from height $h$ as shown. If the body exerts a force of three times its weight on the plane at the highest point of the circle, then the height $h = \alpha R$. The value of $\alpha$ is . . . . . . .