$A$ particle starts sliding from the top of a sphere of diameter $42 \, m$. At what height from the bottom of the sphere will the particle lose contact with the sphere?

$A$ bob of mass $m$ is suspended by a light string of length $L$. It is imparted a minimum horizontal velocity at the lowest point $A$ such that it just completes a full vertical circle,reaching the topmost position $B$. The ratio of kinetic energies $\frac{(\text{K.E.})_A}{(\text{K.E.})_B}$ 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

$A$ 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 :-

An object is at the top of a smooth sphere which is kept fixed. As the object slides down after being given a negligible side push,what is the behavior of the magnitude of the acceleration of the object during its motion until it reaches the ground?

$A$ stone of mass $900 \,g$ is tied to a string and moved in a vertical circle of radius $1 \,m$ making $10 \,rpm$. The tension in the string, when the stone is at the lowest point is (if $\pi^2=9.8$ and $g=9.8 \,m/s^2$): (in $\,N$)