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(B) In a vertical circular motion,the tension $T$ in the string at any point is given by $T = \frac{mv^2}{r} + mg \cos 		heta$,where $	heta$ is the

Vedclass · Accepted Answer

the mass is at the lowest point of the circle.

Vedclass · Answer

(B) In a vertical circular motion,the tension $T$ in the string at any point is given by $T = \frac{mv^2}{r} + mg \cos 		heta$,where $	heta$ is the angle with the downward vertical.At the lowest point,$	heta = 0^{\circ}$,so $\cos 0^{\circ} = 1$. The tension is $T_{low} = \frac{mv^2}{r} + mg$.At the highest point,$	heta = 180^{\circ}$,so $\cos 180^{\circ} = -1$. The tension is $T_{high} = \frac{mv^2}{r} - mg$.Since the tension is maximum at the lowest point of the circle,the probability of the wire breaking is maximum at this position.Therefore,option $B$ is correct.

$A$ heavy mass is attached at one end of a thin wire and whirled in a vertical circle. The chances of breaking the wire are maximum when:

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