A ring of mass m can slide over a smooth vertical rod as shown in figure. The ring is connected to a spring of force constant $k = 4 mg/R$ , where 2R is the natural length of the spring. The other end of spring is fixed to the ground at a horizontal distance 2R from the base of the rod. If the mass is released at a height 1.5R, then the velocity of the ring as it reaches the ground is

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2 \sqrt{gR}

\sqrt{2gR}

\sqrt{gR}

\sqrt{3gR}

answer is B.

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Detailed Solution

mg (\frac{3}{2} x) + \frac{1}{2} k x^{2} = \frac{1}{2} {mv}^{2}

(∵ x = \frac{R}{2})

$3 \frac{mgR}{2} + \frac{1}{2} \frac{4mg}{R} . \frac{R^{2}}{4} = \frac{1}{2} {mv}^{2}$

$3gR + \frac{{4gR}^{2}}{4R} = v^{2}$

$4 gR = v^{2}$

$v = 2 \sqrt{gR}$

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