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(D) Let the velocity at the top be $v_t = n\sqrt{gR}$. The kinetic energy at the top is $K_t = \frac{1}{2}mv_t^2 = \frac{1}{2}m(n^2gR)$.Using the prin

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$\frac{n^2+4}{n^2}$

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(D) Let the velocity at the top be $v_t = n\sqrt{gR}$. The kinetic energy at the top is $K_t = \frac{1}{2}mv_t^2 = \frac{1}{2}m(n^2gR)$.Using the principle of conservation of mechanical energy between the top and the bottom points: $K_t + U_t = K_b + U_b$.Taking the potential energy at the bottom as $0$,the potential energy at the top is $mg(2R)$.So,$\frac{1}{2}mn^2gR + 2mgR = \frac{1}{2}mv_b^2$.Multiplying by $\frac{2}{m}$,we get $n^2gR + 4gR = v_b^2$,so $v_b^2 = (n^2+4)gR$.The kinetic energy at the bottom is $K_b = \frac{1}{2}mv_b^2 = \frac{1}{2}m(n^2+4)gR$.The ratio of kinetic energy at the bottom to that at the top is $\frac{K_b}{K_t} = \frac{\frac{1}{2}m(n^2+4)gR}{\frac{1}{2}mn^2gR} = \frac{n^2+4}{n^2}$.

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