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

As body is at rest initially, i.e., speed=0

At point A, speed=v As track is frictionless, so total mechanical energy will remain constant.

$\therefore (\mathrm{T}.\mathrm{M}.\mathrm{E}{)}_i=(\mathrm{T}.\mathrm{M}.\mathrm{E}{)}_f$

$0+mgh=\frac{1}{2}m{v}^2+0\text{ or } h=\frac{v^2}{2g}$

$\text{ For completing the vertical circle, }v\ge \sqrt{5gR}$

$\therefore h=\frac{5gR}{2g}=\frac{5}{2}R=\frac{5}{4}D$