Critical Speed & Looping question from JEE Main 2022 (26 Jun, Shift 1)

Answer: (c)

1. At angle $\alpha$ below the horizontal through O, the ball has fallen $R\sin\alpha$, so $v^2 = 2gR\sin\alpha$.
2. Centripetal force $F_c = \frac{mv^2}{R} = 2mg\sin\alpha$.
3. Radial equation: $N - mg\sin\alpha = \frac{mv^2}{R} \Rightarrow N = 3mg\sin\alpha$.
4. $A = \frac{F_c}{N} = \frac{2}{3}$, independent of $\alpha$.
5. So the graph is a horizontal straight line: the answer is (c).