One end of string of length 1m is tied to a body of mass 0.5 kg. It is whirled in a vertical circle as shown in figure. If the angular frequency of the body is 4rad/sec.
 

List 1	List 2
I) Tension at the lower most point B	P)  12.5N
II) Tension at the top most point A	Q) 13N
III) Tension at point C	R) 8N
IV) Tension at point D	S) 3N
	T) 10N

For the particle to just complete to circular motion $V_B=\sqrt{5gR}$

For its $mgh=\frac{1}{2}m{V}_B^2\Rightarrow h=\frac{5}{2}R=2.5R$

Velocity at C will be $\sqrt{3gR}$

Force extorted by the particle on the track at point C is

$F=\frac{m{V}^2}{R}=\frac{m{\left(\sqrt{3g}R\right)}^2}{R}=3mg$

At A, F =0, for the force at A to be more than

its will be $mg+N=\frac{m{V}_A^2}{R}$

$N=\frac{m{V}_A^2}{R}-mg>mg\therefore V_A>\sqrt{2gR}$

Applying L.C of energy between 0 and A we get $mg\left(2R\right)+\frac{1}{2}m{V}_A^2$

if $V_A>\sqrt{2gR}$then $h>3R.$