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NEET generalGeneralHard

Question

A body initially at rest and sliding along a frictionless track from a height hh (as shown in the figure) just completes a vertical circle of diameter AB=DAB = D. The height hh is equal to

1

75D\frac{7}{5}D

2

D

3

32D\frac{3}{2}D

4

54D\frac{5}{4}D

Step-by-Step Solution

For a body to complete a vertical circle, the minimum velocity at the bottom of the circle is 5gR\sqrt{5gR}. Using conservation of energy, mgh=12mv2=12m(5gR)=52mgRmgh = \frac{1}{2}mv^2 = \frac{1}{2}m(5gR) = \frac{5}{2}mgR. Since D=2RD = 2R, h=54Dh = \frac{5}{4}D.

Exam Context & Concepts Covered

This question aligns with the NEET general syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

generalinitiallyslidingfrictionlessheightfigure

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