$A$ 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

$A$ bob of mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in the figure. If the string becomes slack at some point $P$ making an angle $\theta$ with the horizontal,the ratio of the speed $v_p$ of the bob at point $P$ to its initial speed $v_0$ is:

You may have seen in a circus a motorcyclist driving in vertical loops inside a 'death well' (a hollow spherical chamber with holes,so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point,with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is $25 \; m$?

$A$ 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 $m$ tied to the end of a string revolves in a vertical circle of radius $R$. The net forces at the lowest and highest points of the circle directed vertically downwards are:

Lowest Point	Highest Point
$(a) \ mg - T_1$	$mg + T_2$
$(b) \ mg + T_1$	$mg - T_2$
$(c) \ mg + T_1 - \frac{mv_1^2}{R}$	$mg - T_2 + \frac{mv_2^2}{R}$
$(d) \ mg - T_1 - \frac{mv_1^2}{R}$	$mg + T_2 + \frac{mv_2^2}{R}$

$T_1$ and $v_1$ denote the tension and speed at the lowest point. $T_2$ and $v_2$ denote corresponding values at the highest point.

The maximum and minimum tension in the string whirling in a circle of radius $2.5 \, m$ with constant velocity are in the ratio $5 : 3$. Then its velocity is: