$A$ 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 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

$A$ block follows the path as shown in the figure from height $h$. If the radius of the circular path is $r$,then the relation that holds good to complete a full circle is

$A$ bucket full of water is revolved in a vertical circle of radius $2\,m$. What should be the maximum time-period of revolution so that the water does not fall off the bucket? (Take $g = 10\,m/s^2$) ......... $\sec$.

$A$ body of mass '$m$' attached at the end of a string is just completing the loop in a vertical circle. The apparent weight of the body at the lowest point in its path is ($g =$ gravitational acceleration)

$A$ stone of mass $m$,tied to a string,is being whirled in a vertical circle with a uniform speed. The tension in the string is