As shown in the figure, a mass $m$ and another block of mass $10m$ are connected with a string of length $L$. Friction is sufficient to prevent the slipping of the $10m$ block. Mass $m$ is given a velocity $u$ in the vertical direction. For the complete circular motion of mass $m$:

$A$ body crosses the topmost point of a vertical circle with critical speed. Its centripetal acceleration,when the string is horizontal,will be

In the case of vertical circular motion of a particle attached to a thread of length $r$, if the tension in the thread is zero at an angle of $30^{\circ}$ with the horizontal as shown in the figure, the velocity at the bottom point $(A)$ of the circular path is ($g =$ gravitational acceleration).

$A$ particle is kept at rest at the top of a sphere of diameter $42\, m$. When disturbed slightly,it slides down. At what height $H$ from the bottom will the particle leave the sphere? $(...\, m)$

$A$ particle with charge $Q$ coulomb, tied at the end of an inextensible string of length $R$ metre, revolves in a vertical plane. At the centre of the circular trajectory, there is a fixed charge of magnitude $Q$ coulomb. The mass of the moving charge $M$ is such that $Mg = \frac{Q^2}{4 \pi \epsilon_0 R^2}$. If at the highest position of the particle, the tension of the string just vanishes, the horizontal velocity at the lowest point has to be

$A$ particle of mass $m$ is released from a height $H$ on a smooth curved surface which ends into a vertical loop of radius $R$,as shown. The minimum value of $H$ required so that the particle makes a complete vertical circle is given by (in $R$)