When a disc is rotating with angular velocity $\omega$,a particle situated at a distance of $4 \ cm$ just begins to slip. If the angular velocity is doubled,at what distance will the particle start to slip (in $cm$)?

$A$ single wire $ACB$ passes through a smooth ring at $C$ which revolves at a constant speed in a horizontal circle of radius $r$ as shown in the figure. The speed of revolution is

$A$ 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

$A$ particle starts from rest and performs circular motion of constant radius with speed given by $v = \alpha \sqrt{x}$,where $\alpha$ is a constant and $x$ is the distance covered. The correct graph of the magnitude of its tangential acceleration $(a_t)$ and centripetal acceleration $(a_c)$ versus $t$ will be:

$A$ particle moves with constant angular velocity in a circular path of a certain radius and is acted upon by a certain centripetal force $F$. If the angular velocity is kept the same but the radius of the path is halved,the new force will be

$A$ ball of mass $0.5 \ kg$ is attached to a string of length $50 \ cm$. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is $400 \ N$. The maximum possible value of angular velocity of the ball in $rad/s$ is: