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:

$A$ body attached to one end of a string performs motion along a vertical circle. Its centripetal acceleration,when the string is horizontal,will be [ $g$ = 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 highest and lowest positions of the circular path will be:

$A$ small body slides down a smooth uneven surface from a height $H$,which eventually emerges into a circular loop of radius $R (< H)$. What should be the value of $H$,so that the force on the body at $A$ is $\sqrt{2}$ times its weight?

$A$ 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).

$A$ hollow sphere has a radius of $6.4 \, m$. The minimum velocity required by a motorcyclist at the bottom to complete the vertical circle is ........... $m/s$.