$A$ particle of mass $m$,attached to a string of length $l$,is moving in a vertical circle. If the particle is just looping the loop without the string slackening,and $v_A, v_B, v_D$ are the speeds at positions $A, B, D$ shown in the figure,then:

$A$ body of mass $100 \ g$ is moving in a circular path of radius $2 \ m$ on a vertical plane as shown in the figure. The velocity of the body at point $A$ is $10 \ m/s$. The ratio of its kinetic energies at point $B$ and $C$ is: (Take acceleration due to gravity as $10 \ m/s^2$)

$A$ body of mass $m \ kg$ is rotating in a vertical circle at the end of a string of length $r \ m$. What is the difference in the kinetic energy at the top and the bottom of the circle?

$A$ particle of mass $m$ is tied to one end of a string of length $l$. The particle is held horizontal with the string taut. It is then projected upward with a velocity $u$. The tension in the string is $\frac{mg}{2}$ when it is inclined at an angle $30^o$ to the horizontal. The value of $u$ is

As per the given figure,to complete the circular loop,what should be the radius if the initial height is $5 \, m$ (in $, m$)?

$A$ stone of mass $1 \ kg$ tied to a light inextensible string of length $L = \frac{5}{3} \ m$ is rotating in a circular path of radius $L$ in a vertical plane. If the ratio of maximum tension in the string to the minimum tension in the string is $3$,the speed of the stone at the highest point of the circle is ($g =$ acceleration due to gravity).