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

track from height h into vertical circle AB

1. $\frac32 D$

2. $\frac75 D$

3. D

4. $\frac54 D$

Doubtly Asked 30 Oct 2026 NEET UG 2018 Practice test neet Save WhatsApp Report

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Answer: 4. $\frac54 D$

To just complete the vertical circle of radius $R=D/2$, the speed at the top B must satisfy $v_B^2=gR$.

Energy conservation from the start to B (height $2R$): $mgh=mg(2R)+\tfrac12mv_B^2=2mgR+\tfrac12mgR$.

$$h=\frac52R=\frac52\cdot\frac D2=\frac54D$$

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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