$A$ 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)

$A$ body of mass $1\,kg$ is rotating in a vertical circle of radius $1\,m$. What will be the difference in its kinetic energy at the top and bottom of the circle? $(g = 10\,m/s^2)$

$A$ 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. If $\theta$ is the instantaneous angle which the line joining the particle and the centre of the loop makes with the vertical,then identify the correct statement$(s)$ related to the normal reaction $N$ between the block and the surface.

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

The maximum value attained by the tension in the string of a swinging pendulum is four times the minimum value it attains. There is no slack in the string. The angular amplitude of the pendulum is (in $^{\circ}$)

$A$ $2 \, kg$ stone at the end of a string $1 \, m$ long is whirled in a vertical circle at a constant speed. The speed of the stone is $4 \, m/s$. The tension in the string will be $52 \, N$ when the stone is: