The diagram shows a small bead of mass m carrying charge + q at P. The bead can freely move on a smooth fixed ring placed on a smooth horizontal plane. In the same plane, a charge + Q has also been fixed as shown. The potential at the point P due to + Q is $V$ . The speed with which the bead should be projected from the point P so that it can complete a circle along the ring, should be greater than

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$\sqrt{\frac{6 q V}{m}}$

$\sqrt{\frac{q V}{m}}$

$\sqrt{\frac{3 q V}{m}}$

$\sqrt{\frac{q V}{6 m}}$

answer is A.

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Detailed Solution

As the bead moves form P to M, its distance from fixed charge Q decreases; therefore, its potential energy increases. It has smallest speed at M. For completing the circle, KE at M must be > 0.
$∴ K E_{M} + k \frac{q Q}{a} = \frac{1}{2} m u^{2} + k \frac{Q q}{4 a}$
It is given that $k \frac{Q q}{4 a} = V$
$∴ K E_{M} + 4 q V = \frac{1}{2} m u^{2} + q V$
$K E_{M} = \frac{1}{2} m u^{2} - 3 q V$
$K E_{M} > 0$
$\Rightarrow \frac{1}{2} m u^{2} > 3 q V$
$\Rightarrow u > \sqrt{\frac{6 q V}{m}}$
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