Critical Speed & Looping question from JEE Main 2014 (19 Apr, Shift 1)

Answer: (c)

1. Height of X above O is $R\cos\theta$; height of Y is $R\sin\phi$.
2. Energy: $v^2 = 2gR(\cos\theta - \sin\phi)$ at Y.
3. Normal reaction vanishes when the radial part of gravity alone gives the centripetal force: $mg\sin\phi = \frac{mv^2}{R}$.
4. $\sin\phi = 2(\cos\theta - \sin\phi) \Rightarrow 3\sin\phi = 2\cos\theta$
5. So the answer is (c) $\sin\phi = \frac{2}{3}\cos\theta$.