A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB : D. The height h is equal to

For completing the vertical circle the minimum velocity of the body at .B should be $\sqrt{5\mathrm{gR}}$, where R = $\frac{\mathrm{D}}{2}$
Hence, ${\mathrm{v}}_{\mathrm{A}} = \sqrt{\frac{5}{2}\mathrm{gD}}$
Now applying C.O.M.E. between O and A,

$∆\mathrm{K} + ∆\mathrm{U} = 0 \Rightarrow (\frac{1}{2}{{\mathrm{mv}}_{\mathrm{A}}}^2-0)-\mathrm{mgh} = 0$

$\frac{1}{2}\mathrm{m}.\frac{5}{2}\mathrm{gD} = \mathrm{mgh} \mathrm{or} \mathrm{h} = \frac{5}{4}\mathrm{D}$