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An inclined track ends in a circular loop of diameter ‘D’. From what height on the track a particle should be released so that it completes that loop in the vertical plane ?

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$\frac{5 D}{2}$

$\frac{2 D}{5}$

$\frac{5 D}{4}$

$\frac{4 D}{5}$

answer is C.

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Detailed Solution

For completion of circular path

minimum velocity = $\sqrt{5 g l}$

By conservation of energy,

$\frac{1}{2} m v^{2} = m g h v = \sqrt{5 g l} \frac{5}{2} m g l = m g h h = \frac{5}{2} l$

Also,

$l = \frac{D}{2} h = \frac{5}{2} \times \frac{D}{2} = \frac{5 D}{4}$

Hence the correct answer is $\frac{5 D}{4} .$

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