Physics - Circular Motion Question with Solution | TestHub

The acceleration is horizontal (towards the centre of the circle), not along the funnel surface. Resolving: $N\cos \theta =mg$
$$\text{ and }N\sin \theta =\frac{m{v}^2}{R}\text{, }$$
so $N=mg\sec \theta \ne mg\cos \theta$ (Assertion true).
The Reason is true and is exactly why the usual $mg\cos \theta$ result (which needs equilibrium along the normal) fails here.