$A$ heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when:

$A$ bob of mass $M$ is suspended by a massless string of length $L$. The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$. The angle $\theta$ at which the speed of the bob is half of that at $A$, satisfies:

$A$ bob of mass $m$ is tied to a string of length $l$. What is the minimum kinetic energy that must be provided to the bob at the lowest point so that it completes a full vertical circle (in $;mgl$)?

$A$ 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

In the figure shown,a particle is released from the position $A$ on a smooth track. When the particle reaches at $B$,then the normal reaction on it by the track is .........

$A$ bucket tied at the end of a $1.6\, m$ long string is whirled in a vertical circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill,when the bucket is at the highest position? (Take $g = 10\, m/s^2$)