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A heavy particle is attached to one end of a light string of length l whose other end is fixed at O. The oarticle is projectyed horizontally with a velocity `v_0` from its lowest position A. When the angular displacement of the string is more than`90^(@)`, the particle leaves the circular path at B. The string again becomes taut at C such that B,O,C are collinear. Find `v_0` in terms of l and g

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The correct Answer is:
`((4+3sqrt(2))/2 gl`
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