$A$ body of mass '$m$' attached at the end of a string is just completing the loop in a vertical circle. The apparent weight of the body at the lowest point in its path is ($g =$ gravitational acceleration)

$A$ particle starts sliding from the top of a sphere of diameter $42 \, m$. At what height from the bottom of the sphere will the particle lose contact with the sphere?

$A$ bullet of mass $m$ strikes a pendulum-bob of mass $M$ with velocity $u$. It passes through it and emerges out with a velocity $u/2$ from the bob. The length of the pendulum is $\ell$. What should be the minimum value of $u$,if the pendulum-bob will swing through a complete circle?

$A$ small bob tied at one end of a thin string of length $1\, m$ is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio $5: 1$. The velocity of the bob at the highest position is ............. $m/s$ (Take $g=10\, m/s^2$).

In the given figure,for an initial velocity $u = u_0/3$,find the height from the ground at which the block leaves the hemisphere,where $u_0 = \sqrt{gr}$.

$A$ stone of mass $m$ $kg$ is tied to a string of length $L$ $m$ and moved in a vertical circle of radius $49$ $cm$ in a vertical plane. If it completes $30$ revolutions per minute,the tension in the string when it is at the lowermost point is nearly. [Take $\pi^2=10$ and acceleration due to gravity,$g=10$ $m/s^2$]