$A$ mass is performing vertical circular motion (see figure). If the average velocity of the particle is increased,then at which point will the string break?

$A$ 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?

$A$ stone of mass $900 \,g$ is tied to a string and moved in a vertical circle of radius $1 \,m$ making $10 \,rpm$. The tension in the string, when the stone is at the lowest point is (if $\pi^2=9.8$ and $g=9.8 \,m/s^2$): (in $\,N$)

The maximum value attained by the tension in the string of a swinging pendulum is four times the minimum value it attains. There is no slack in the string. The angular amplitude of the pendulum is (in $^{\circ}$)

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 .........

$A$ 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).