Solved: null Question for NEET

NEET Hard

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

$\frac{7}{5}D$

$\frac{3}{2}D$

$\frac{5}{4}D$

Step-by-Step Solution

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

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