At Q ;

The equation of motion of the ball is

$$ \begin{aligned} & \frac{m v^{2}}{R}=N-m g \sin \alpha \\\\ & N=\frac{m v^{2}}{R}+m g \sin \alpha .......(i) \end{aligned} $$

From figure, $$h = R\sin \alpha $$

From law of conservation of energy,

$$ (mg)R \sin \alpha=\frac{1}{2} m v^{2} $$

$\therefore \quad m v^{2} / R=2 m g \sin \alpha$

So, Eq. (i) becomes,

$N=2 m g \sin \alpha+m g \sin \alpha=3 m g \sin \alpha$

$\therefore$ Ratio, $\frac{m v^{2}}{R N}=\frac{2}{3}=$ constant

$A=2 / 3$, which is independent of $\alpha$.