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(C) Let the length of the string be $l$. When the mass $m$ is released from point $A$,it moves in a circular path of radius $l$.By the law of conserva

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$3mg$

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(C) Let the length of the string be $l$. When the mass $m$ is released from point $A$,it moves in a circular path of radius $l$.By the law of conservation of energy between point $A$ and point $B$:$PE_A + KE_A = PE_B + KE_B$Taking the potential energy at $B$ as zero,$PE_A = mgl$ and $KE_A = 0$.At point $B$,$PE_B = 0$ and $KE_B = \frac{1}{2}mv^2$.So,$mgl = \frac{1}{2}mv^2 \implies v^2 = 2gl$.At point $B$,the forces acting on the mass are tension $T$ (upwards) and weight $mg$ (downwards).The net centripetal force is provided by the tension and gravity:$T - mg = \frac{mv^2}{l}$Substituting $v^2 = 2gl$:$T - mg = \frac{m(2gl)}{l} = 2mg$$T = 2mg + mg = 3mg$.Therefore,the tension at point $B$ is $3mg$.

$A$ mass $m$ is released from point $A$ as shown in the figure. The tension in the string at point $B$ will be:

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