$A$ steel wire can withstand a load up to $2940 \ N$. $A$ load of $150 \ kg$ is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position,so that the wire does not break when the load passes through the position of equilibrium,is (in $^{\circ}$)

$A$ 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

$A$ stone of mass $m$ tied to the end of a string revolves in a vertical circle of radius $R$. The net forces at the lowest and highest points of the circle directed vertically downwards are:

Lowest Point	Highest Point
$(a) \ mg - T_1$	$mg + T_2$
$(b) \ mg + T_1$	$mg - T_2$
$(c) \ mg + T_1 - \frac{mv_1^2}{R}$	$mg - T_2 + \frac{mv_2^2}{R}$
$(d) \ mg - T_1 - \frac{mv_1^2}{R}$	$mg + T_2 + \frac{mv_2^2}{R}$

$T_1$ and $v_1$ denote the tension and speed at the lowest point. $T_2$ and $v_2$ denote corresponding values at the highest point.

$A$ 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? $(...\, m)$

$A$ ball of mass $2 \,g$ released from the top of an inclined plane describes a circular motion of radius $20 \,cm$ in the vertical plane upon reaching the bottom. The minimum height of the inclined plane is (in $\,cm$)

$A$ body of mass $100 \ g$ is moving in a circular path of radius $2 \ m$ on a vertical plane as shown in the figure. The velocity of the body at point $A$ is $10 \ m/s$. The ratio of its kinetic energies at point $B$ and $C$ is: (Take acceleration due to gravity as $10 \ m/s^2$)