A ball of mass $m$ is released from rest at $\text{A}$ as shown in the figure. Find the minimum height $h$, so that ball will just complete the circular motion on the track (all surfaces are smooth).

The correct option is D $h=\frac{5}{2}R$
For just completing the circular motion on the track, normal reaction at $C$
$N=0$


Applying equation of dynamics towards centre at point $\text{C}$
$N+mg=\frac{m{v}_C^2}{R}$
$N=0\Rightarrow mg=\frac{m{v}_C^2}{R}$
$\Rightarrow v_C=\sqrt{gR}$
$v_C$ is minimum speed the particle needs to have at $\text{C}$, else it will leave the track.

Applying energy conservation between points $\text{A}$ and $\text{C}$ and taking reference of $PE=0$ at bottom:
$P{E}_A+K{E}_A=P{E}_C+K{E}_C$
$\Rightarrow mgh+0=mg\left(2R\right)+\frac{1}{2}m{v}_C^2$
$\Rightarrow mgh=2mgR+\frac{1}{2}mgR$
$\Rightarrow mgh=\frac{5}{2}mgR$
$\therefore h=\frac{5}{2}R$ is the minimum height for the ball to just complete the circular motion on track.