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

$A$ car of mass $m$ is crossing a convex bridge of radius of curvature $R$ with speed $V$. At the highest point, the thrust (normal force) is ($g =$ gravitational acceleration).

$A$ particle rotates in a horizontal circle of radius $R$ in a conical funnel with constant speed $V$. The inner surface of the funnel is smooth. The height of the plane of the circle from the vertex of the funnel is (where $g$ is the acceleration due to gravity):

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

Two particles of mass $m$ are constrained to move along two horizontal frictionless rails that make an angle $2\theta$ with respect to each other. They are connected by a spring with spring constant $k$. The angular frequency of small oscillations for the motion where the two masses always stay at the same distance from the meeting point of the rails is:

$A$ 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? $\left(g=10 \,m/s^{2}\right)$