$A$ bullet of mass $m$ moving with velocity $v$ passes through a pendulum bob of mass $M$ and emerges with a velocity $v/2$. What is the minimum value of $v$ so that the pendulum bob completes a full vertical circle of length $\ell$?

$A$ bucket full of water is revolved in a vertical circle of radius $2\,m$. What should be the maximum time-period of revolution so that the water does not fall off the bucket? (Take $g = 10\,m/s^2$) ......... $\sec$.

$A$ simple pendulum of length $L$ carries a bob of mass $m$. When the bob is at its lowest position, it is given the minimum horizontal speed necessary for it to move in a vertical circle about the point of suspension. When the string is horizontal, the net force on the bob is

$A$ particle of mass $m$ is released from a height $H$ on a smooth curved surface which ends into a vertical loop of radius $R$,as shown. If $\theta$ is the instantaneous angle which the line joining the particle and the centre of the loop makes with the vertical,then identify the correct statement$(s)$ related to the normal reaction $N$ between the block and the surface.

$A$ block of mass $m$ at the end of a string is whirled round in a vertical circle of radius $R$. The critical speed of the block at the top of its swing below which the string would slacken before the block reaches the top is

$A$ 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).