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(D) For a particle to just complete a vertical circle without the string slackening,the minimum speed at the highest point $A$ must be $v_A = \sqrt{gl

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All of the above

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(D) For a particle to just complete a vertical circle without the string slackening,the minimum speed at the highest point $A$ must be $v_A = \sqrt{gl}$.Using the principle of conservation of energy between $A$ and $B$ (lowest point):$\frac{1}{2}mv_A^2 + mg(2l) = \frac{1}{2}mv_B^2$$\frac{1}{2}m(gl) + 2mgl = \frac{1}{2}mv_B^2$$v_B^2 = gl + 4gl = 5gl \implies v_B = \sqrt{5gl}$.Using conservation of energy between $A$ and $D$ (horizontal position):$\frac{1}{2}mv_A^2 + mgl = \frac{1}{2}mv_D^2$$\frac{1}{2}m(gl) + mgl = \frac{1}{2}mv_D^2$$v_D^2 = gl + 2gl = 3gl \implies v_D = \sqrt{3gl}$.Comparing the speeds: $v_B = \sqrt{5gl}$,$v_D = \sqrt{3gl}$,$v_A = \sqrt{gl}$. Thus,$v_B > v_D > v_A$.The tension at any point making an angle $	heta$ with the vertical downward direction is given by $T = \frac{mv^2}{l} + mg\cos	heta$. At point $D$,the angle with the vertical is $90^{\circ}$.$T_D = \frac{m(3gl)}{l} + mg\cos(90^{\circ}) = 3mg + 0 = 3mg$.Since all statements are correct,the correct option is $D$.

$A$ particle of mass $m$,attached to a string of length $l$,is moving in a vertical circle. If the particle is just looping the loop without the string slackening,and $v_A, v_B, v_D$ are the speeds at positions $A, B, D$ shown in the figure,then:

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