$A$ bucket containing water is revolved in a vertical circle of radius $r$. To prevent the water from falling down,the minimum frequency of revolution required is ($g =$ acceleration due to gravity).

$A$ particle of mass $m$ is tied to a string of length $l$ and whirled in a vertical circle. The difference in tension and kinetic energy at the lowest and highest positions of the circular path will be:

$A$ body of mass $m \ kg$ slides from rest along the curve of a vertical circle from point $A$ to $B$ on a frictionless path. The velocity of the body at $B$ is: (Given: $R = 14 \ m$,$g = 10 \ m/s^2$,and $\sqrt{2} = 1.4$) (in $m/s$)

$A$ body tied to a string of length $L$ is revolved in a vertical circle with minimum velocity. When the body reaches the uppermost point,the string breaks and the body moves under the influence of the gravitational field of the Earth along a parabolic path. The horizontal range $AC$ of the body will be

$A$ particle originally at rest at the highest point of a smooth vertical circle (radius $R = 1.8 \ m$) is slightly displaced. It will leave the circle at a vertical distance $h$ from the bottom. The value of $h$ will be ..... $m$.

$A$ particle with charge $Q \text{ C}$, tied at the end of an inextensible string of length $R \text{ m}$, revolves in a vertical plane. At the centre of the circular trajectory, there is a fixed charge of magnitude $Q \text{ C}$. 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 is