$A$ stone of mass $m$ $kg$ is tied to a string of length $L$ $m$ and moved in a vertical circle of radius $49$ $cm$ in a vertical plane. If it completes $30$ revolutions per minute,the tension in the string when it is at the lowermost point is nearly. [Take $\pi^2=10$ and acceleration due to gravity,$g=10$ $m/s^2$]

$A$ stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is

$A$ particle attached to a string of length $r$ is moving in a vertical circular path. If the speed of the particle at the highest point is $\sqrt{7gr}$,then the ratio of the tension in the string at the highest point to the tension at the lowest point is ($g=$ acceleration due to gravity).

$A$ 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?

$A$ stone of mass $1 \ kg$ is tied to one end of a light string and is whirled in a vertical circle as shown in the figure. The tension in the string when the string becomes horizontal is ......... $N$.

$A$ 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