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A bead of mass ‘m’ slides without friction on the wall of a vertical circular hoop of radius ‘R’ as shown in figure. The bead moves under the combined action of gravity and a massless spring (k) attached to the bottom of the hoop. The equilibrium length of the spring is ‘R’. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes ‘R’, would be (spring constant is ‘k’, g is acceleration due to gravity)

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$2 \sqrt{gR + \frac{{kR}^{2}}{m}}$

$\sqrt{2 Rg + \frac{4 {kR}^{2}}{m}}$

$\sqrt{2 R g + \frac{k R^{2}}{m}}$

$\sqrt{3 Rg + \frac{{kR}^{2}}{m}}$

answer is D.

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Detailed Solution

Work energy theorem

$Mg (R + Rcos 60) + \frac{1}{2} k (R^{2} - 0^{2}) = \frac{1}{2} {mv}^{2}$

$Mg \frac{3 R}{2} + \frac{{KR}^{2}}{2} = \frac{1}{2} {mv}^{2}$

$V = \sqrt{3 gR + \frac{{KR}^{2}}{m}}$

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