Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A body initially at rest and sliding along a frictionless track from a height $h$ just completes a vertical circle of diameter $AB = D$ . The height $h$ is equal to:

\frac{3}{2}D

\frac{7}{4}D

\frac{5}{4}D

Step-by-Step Solution

Identify Condition for Vertical Circle: For a body to just complete a vertical circle, the minimum velocity ( $v$ ) required at the lowest point of the circle is given by $v = \sqrt{5gR}$ , where $R$ is the radius [Class 11 Physics, Ch 5, Example 5.7].
Relate Radius to Diameter: Given the diameter $D$ , the radius $R = \frac{D}{2}$ . Thus, $v = \sqrt{5g\frac{D}{2}}$ .
Apply Conservation of Mechanical Energy: The body starts from rest at height $h$ . The potential energy lost is converted into kinetic energy at the bottom of the circle. $PE_{initial} = KE_{final}$ $mgh = \frac{1}{2}mv^2$ [Class 11 Physics, Ch 5, Sec 5.8]
Substitute and Solve: $mgh = \frac{1}{2}m \left( \sqrt{\frac{5gD}{2}} \right)^2$ $gh = \frac{1}{2} \left( \frac{5gD}{2} \right)$ $h = \frac{5}{4}D$

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