$A$ string of length '$L$' fixed at one end carries a mass '$m$' at the other end. The string makes $\frac{3}{\pi}$ r.p.s. around the vertical axis through the fixed end. The tension in the string is (in $mL$)

$A$ ball of mass $m$ is attached to the free end of a string of length $l$. The ball is moving in a horizontal circular path about the vertical axis as shown in the diagram. The angular velocity $\omega$ of the ball will be ($T =$ Tension in the string).

$A$ particle moves in a horizontal circle on the smooth inner surface of a hemispherical bowl of radius $R$. The plane of motion is at a depth $d$ below the centre of the hemisphere. The speed of the particle is:

The $x$ and $y$ displacements of a particle are given as $x(t) = a \sin \omega t$ and $y(t) = a \sin 2 \omega t$. Its trajectory will look like:

$A$ particle of mass $m$ is moving in a circular path of radius $r$ in a plane. Its angular momentum is $L$. The value of the centripetal force acting on the particle is:

$A$ string of length $2 \ m$ can withstand a maximum tension of $16000 \ N$. $A$ particle of mass $2 \ kg$ is tied to one end of the string and rotated in a horizontal circle. What is the maximum number of rotations per second the particle can make so that the string does not break? (Take $\pi^2 = g = 10 \ m/s^2$)