Q 11-05-078JEE MainJEE Main 2024 (29 Jan, Shift 2)Medium
A bob of mass $m$ is suspended by a light string of length $L$. It is imparted a minimum horizontal velocity at the lowest point $A$ such that it just completes a half circle, reaching the topmost position $B$. The ratio of kinetic energies $\dfrac{(K.E.)_A}{(K.E.)_B}$ is:
Answer: (B) $5:1$
To just reach the top with the string taut, the tension at B is zero:
$$mg = \frac{mv_B^2}{L} \;\Rightarrow\; v_B^2 = gL$$
Energy conservation from A to B (rise $2L$):
$$v_A^2 = v_B^2 + 4gL = 5gL$$
$$\frac{(K.E.)_A}{(K.E.)_B} = \frac{v_A^2}{v_B^2} = \frac51$$
Solution by Sreeraj P, M.Sc Physics