Length of a simple pendulum is $1 \,m$. When its bob is at its lowest point its velocity is $7 \,ms^{-1}$. If the bob leaves its circular path at a height $h$ above the centre of the circle, then the value of $h$ is $(g=10 \,ms^{-2})$. (in $\,m$)

$A$ particle is tied to a $20 \ cm$ long string. It performs circular motion in a vertical plane. What is the angular speed of the particle when the tension in the string at the top is zero? (Take $g = 9.8 \ m/s^2$)

$A$ bob of mass $m$ is suspended at a point $O$ by a light string of length $l$ and is set to perform vertical circular motion as shown in the figure. Initially,by applying a horizontal velocity $v_0$ at point $A$,the string becomes slack when the bob reaches point $D$. The ratio of the kinetic energy of the bob at points $B$ and $C$ is:

$A$ small body of mass $m$ slides down from the top of a hemisphere of radius $r$. The surfaces of the block and the hemisphere are frictionless. The height at which the body loses contact with the surface of the sphere is

$A$ weightless string can support a tension up to $30 \,N$. $A$ stone of mass $0.5 \,kg$ is tied to its one end and is revolved in a circular path of radius $2 \,m$ in a vertical plane. Then the maximum angular velocity of the stone will be (acceleration due to gravity $g=10 \,m/s^2$)

$A$ 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: