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 : 
Answer & explanation
Correct answer: option 3
As track is frictionless, so total mechanical energy will remain constant
mgh = \(\frac{1}{2}mv_L^2\)
h = \(\frac{v_L^2}{2g}\)
For completing the vertical circle, vL \(\geq \) \(\sqrt{5gR}\)
h = \(\frac{5gR}{2g}\) = \(\frac{5}{4}\) D