$A$ simple pendulum has a length of $2 \,m$ and a bob of mass $100 \,g$. It is whirled in a horizontal plane. If the string breaks under a tension of $10 \,N$, what is the angle made by the string with the vertical? $\left(g=10 \,m/s^{2}\right)$

$A$ particle is describing circular motion in a horizontal plane in contact with the smooth inside surface of a fixed right circular cone with its axis vertical and vertex down. The height of the plane of motion above the vertex is $h$ and the semivertical angle of the cone is $\alpha$. The period of revolution of the particle

$A$ stone of mass $2 \,kg$ is tied at one end of a string of length $2 \,m$ and is whirled in a horizontal circle. If the string can withstand a maximum tension of $64 \,N$, then the permissible maximum number of rotations per minute is

$A$ 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 $m = 0.5 \ kg$ is attached to the end of a string having length $L = 0.5 \ m$. The ball is rotated on a horizontal circular path about a vertical axis. The maximum tension that the string can bear is $324 \ N$. The maximum possible value of angular velocity of the ball (in $rad/s$) is

$A$ man carrying a monkey on his shoulder is cycling smoothly on a circular track of radius $9 \,m$ and completes $120$ revolutions in $3$ minutes. The magnitude of the centripetal acceleration of the monkey is (in $m/s^2$):