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 $\,m$)

$A$ 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?

As shown in the figure, a mass $m$ and another block of mass $10m$ are connected with a string of length $L$. Friction is sufficient to prevent the slipping of the $10m$ block. Mass $m$ is given a velocity $u$ in the vertical direction. For the complete circular motion of mass $m$:

$A$ particle originally at rest at the highest point of a smooth vertical circle is slightly displaced. It will leave the circle at a vertical distance $h$ below the highest point,such that

$A$ can filled with water is revolved in a vertical circle of radius $r$ with constant speed such that the water just does not fall out. The time period of revolution is ($g=$ 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: