$A$ $1 \,kg$ stone at the end of a $1 \,m$ long string is whirled in a vertical circle at a constant speed of $4 \,m/s$. The tension in the string is $6 \,N$ when the stone is at:

$A$ simple pendulum with a bob of mass $m = 1 \ kg$, charge $q = 5 \ \mu C$ and string length $l = 1 \ m$ is given a horizontal velocity $u$ in a uniform electric field $E = 2 \times 10^6 \ V/m$ at its bottom-most point $A$, as shown in the figure. It is given a speed $u$ such that the particle leaves the circular path at its topmost point $C$. Find the speed $u$. (Take $g = 10 \ m/s^2$)

$A$ small block slides down from the top of a hemisphere of radius $R=3\, m$ as shown in the figure. The height $h$ at which the block will lose contact with the surface of the sphere is $............. \; m$. (Assume there is no friction between the block and the hemisphere)

$A$ 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).

In the case of vertical circular motion of a particle attached to a thread of length $r$, if the tension in the thread is zero at an angle of $30^{\circ}$ with the horizontal as shown in the figure, the velocity at the bottom point $(A)$ of the circular path is ($g =$ gravitational acceleration).

$A$ small cube with mass $M$ starts at rest at point $1$ at a height $4R$,where $R$ is the radius of the circular part of the track. The cube slides down the frictionless track and around the loop. The force that the track exerts on the cube at point $2$ is nearly ......... times the cube's weight $mg$.