$A$ body of mass $M = 100 \; g$ is rotating with an angular frequency of $2/\pi \; \text{rev/sec}$ as shown in the figure. What is the angle made by the string with the vertical? $(g = 10 \; m/s^2)$

Two stones of masses $m$ and $3m$ are whirled in horizontal circles,the heavier one in a radius of $(r/3)$ and the lighter one in a radius of $r$. The tangential speed of the lighter stone is $n$ times that of the heavier stone when they experience the same centripetal force. The value of $n$ 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 $m/s^2$):

$A$ conical pendulum of length $1\,m$ makes an angle $\theta = 45^\circ$ with respect to the $Z$-axis and moves in a circle in the $XY$ plane. The radius of the circle is $0.4\,m$ and its centre is vertically below $O$. The speed of the pendulum,in its circular path,will be ..... $m/s$ (Take $g = 10\,ms^{-2}$)

$A$ 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 centripetal force $F$ is kept constant but the angular velocity is doubled,the new radius of the path (original radius $R$) will be:

If the radius of curvature of the path of two particles of same masses are in the ratio $1 : 2$,then in order to have constant centripetal force,their velocity,should be in the ratio of