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(A) Let $N$ be the normal reaction of the weighing machine on the boy,which represents the reading of the machine.At any point in the vertical circle,

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the reading at lowermost point $L$ is greater than the reading at highest point $H$.

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(A) Let $N$ be the normal reaction of the weighing machine on the boy,which represents the reading of the machine.At any point in the vertical circle,the forces acting on the boy are the normal reaction $N$ (upward) and the gravitational force $mg$ (downward).The net centripetal force required for circular motion is $F_c = \frac{mv^2}{R}$.Since the speed $v = \sqrt{gR}$ is constant,the centripetal acceleration is $a_c = \frac{v^2}{R} = \frac{gR}{R} = g$.At the highest point $H$,the net force is $mg - N_H = ma_c = mg$. Thus,$N_H = mg - mg = 0$.At the lowest point $L$,the net force is $N_L - mg = ma_c = mg$. Thus,$N_L = mg + mg = 2mg$.Comparing the readings,$N_L = 2mg$ and $N_H = 0$. Therefore,the reading at the lowermost point $L$ is greater than the reading at the highest point $H$.Ans: $(A)$

$A$ cage revolves around a vertical circle of radius $R$ with constant linear speed $\sqrt{gR}$. The cage is connected to the revolving arm in such a manner that a boy of mass $m$ remains always vertical while standing on a weighing machine kept inside the cage. It is found that

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