$A$ particle originally at rest at the highest point of a smooth vertical circle (radius $R = 1.8 \ m$) is slightly displaced. It will leave the circle at a vertical distance $h$ from the bottom. The value of $h$ will be ..... $m$.

$A$ point mass '$m$' is moved in a vertical circle of radius '$r$' with the help of a string. The velocity of the mass is $\sqrt{7gr}$ at the lowest point. The tension in the string at the lowest point is .......... $mg$.

The minimum velocity $v$ required for a simple pendulum to complete one full rotation in a vertical plane is given. If the length of the string is reduced to one-fourth,what will be the new required velocity?

$A$ bob of mass $m$ is suspended by a light string of length $L$. It is imparted a minimum horizontal velocity at the lowest point $A$ such that it just completes a full vertical circle,reaching the topmost position $B$. The ratio of kinetic energies $\frac{(\text{K.E.})_A}{(\text{K.E.})_B}$ is:

$A$ particle of mass $200 \,g$ is moving in a circle of radius $2 \,m$. The particle is just 'looping the loop'. What are the speed of the particle and the tension in the string at the highest point of the circular path? (Take $g = 10 \,m/s^2$)

$A$ 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 $g = 10 \ m/s^2$)