An inclined track ends in a circular loop of diameter ‘D’. From what height on the track a particle should be released so that it completes that loop in the vertical plane?

To follow complete circular path, Minimum velocity at lowermost point of the vertical path is,$\mathrm{v}=\sqrt{5\mathrm{gR}} =\sqrt{\frac{5\mathrm{gD}}{2}}$

using conservation of mechanical energy,
$\frac{1}{2}{\mathrm{mv}}^2=\mathrm{mgh}$

$$\begin{gathered} \Rightarrow \frac{1}{2}\mathrm{m}\frac{5\mathrm{gD}}{2}=\mathrm{mgh} \\ \Rightarrow \mathrm{h}=\frac{5\mathrm{D}}{4} \end{gathered}$$