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(B) At the highest point of the vertical circular loop, the forces acting on the body are the normal force $N$ (directed downwards) and the gravitatio

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$4$

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(B) At the highest point of the vertical circular loop, the forces acting on the body are the normal force $N$ (directed downwards) and the gravitational force $mg$ (directed downwards).These forces provide the necessary centripetal force: $N + mg = \frac{mv^2}{R}$.Given that the body exerts a force of three times its weight on the plane, the normal force $N = 3mg$.Substituting this into the centripetal force equation: $3mg + mg = \frac{mv^2}{R} \Rightarrow 4mg = \frac{mv^2}{R} \Rightarrow v^2 = 4gR$.Now, we apply the law of conservation of mechanical energy between the starting point at height $h$ and the highest point of the loop at height $2R$.The total energy at the start is $mgh$ (potential energy).The total energy at the highest point is $mg(2R) + \frac{1}{2}mv^2$ (potential energy + kinetic energy).Equating the energies: $mgh = 2mgR + \frac{1}{2}m(4gR) = 2mgR + 2mgR = 4mgR$.Thus, $h = 4R$.Comparing this with $h = \alpha R$, we get $\alpha = 4$.

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