A particle is attached with a string of length $l$ which is fixed at point O on an inclined plane. What minimum velocity should be given to the particle along the incline so that it may complete a circle on inclined plane? (The plane is smooth and initially particle was resting on the inclined plane.)

\sqrt{5 g l}

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\sqrt{\frac{5 g l}{2}}

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\sqrt{\frac{5 \sqrt{3} g l}{2}}

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\sqrt{4 g l}

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Solution

The correct option is A $\sqrt{\frac{5 g l}{2}}$
For a particle to complete vertical circle,

$\frac{1}{2} m v_{b o t t o m}^{2} = m g \times 2 l + \frac{1}{2} m v_{t o p}^{2}$

$\frac{m v_{t o p}^{2}}{l} = m g$

Solving, $v_{b o t t o m} = \sqrt{5 g l}$

In the given problem, the circle is inclined at $30^{o}$ to the horizontal which changes the effective acceleration due to gravity. Only the component of $g$ along the plane has an effect on centripetal acceleration. The component of $g$ perpendicular to the plane is balanced by the normal reaction.

$∴ g_{e f f} = g s i n θ = g s i n 30^{0} = g / 2 v_{m i n} = \sqrt{5 g_{e f f} l} = \sqrt{\frac{5 g l}{2}}$

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