Q 11-05-064JEE MainJEE Main 2025 (29 Jan, Shift 1)Medium
A body of mass $m$ connected to a massless and unstretchable string goes in a vertical circle of radius $R$ under gravity $g$. The other end of the string is fixed at the centre of the circle. If the velocity at the top of the circular path is $n\sqrt{gR}$, where $n \ge 1$, then the ratio of the kinetic energy of the body at the bottom to that at the top of the circle is
Answer: (B) $\dfrac{n^2 + 4}{n^2}$
Energy conservation from top to bottom (drop $2R$):
$$v_b^2 = v_t^2 + 4gR = n^2gR + 4gR$$
$$\frac{K_b}{K_t} = \frac{v_b^2}{v_t^2} = \frac{n^2 + 4}{n^2}$$
Solution by Sreeraj P, M.Sc Physics