question_answer

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\[\text{AB=D}\]. The height h is equal to [NEET - 2018]

A) \[\frac{7}{5}D\]

B) \[D\]

C) \[\frac{3}{2}D\]

D) \[\frac{5}{4}D\]

Correct Answer: D

Solution :

[d]

As track is frictionless, so total mechanical energy will remain constant

\[\text{T}\text{.M}\text{.}{{\text{E}}_{\text{l}}}\text{=T}\text{.M}\text{.}{{\text{E}}_{\text{F}}}\]

\[\text{0+mgh=}\frac{1}{2}m{{v}_{L}}^{2}+0\]

\[h=\frac{v_{L}^{2}}{2g}\]

For completing the vertical circle, \[{{v}_{L}}\ge \sqrt{5gR}\]

\[h=\frac{5gR}{2g}=\frac{5}{2}R=\frac{5}{4}D\]