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(B) For completing the vertical circular motion,the minimum velocity at the uppermost point is $v = \sqrt{gL}$.When the string breaks at the top,the b

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$x = 2L$

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(B) For completing the vertical circular motion,the minimum velocity at the uppermost point is $v = \sqrt{gL}$.When the string breaks at the top,the body acts as a projectile launched horizontally from a height $h = 2L$ (the diameter of the circle).The time taken by the body to reach the ground is given by $t = \sqrt{\frac{2h}{g}}$.Substituting $h = 2L$,we get $t = \sqrt{\frac{2(2L)}{g}} = \sqrt{\frac{4L}{g}} = 2\sqrt{\frac{L}{g}}$.The horizontal range $x$ is given by the product of horizontal velocity and time: $x = v 	imes t$.Substituting the values,$x = \sqrt{gL} 	imes 2\sqrt{\frac{L}{g}} = 2\sqrt{gL 	imes \frac{L}{g}} = 2L$.

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