A particle is suspended vertically from a point O by an inextensible massless string of length L. A vertical line AB is at a distance of L/8 from O as shown. The object is given a horizontal velocity u. At some point, its motion ceases to be circular and eventually the object passed through the line AB. At the instant of crossing AB, its velocity is horizontal. Then $u = \sqrt{\frac{1}{2} (x + 3 \sqrt{3}) g L},$ then the value of $x$ is

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Detailed Solution

Now, we have following equations

$T_{Q} = 0$ Therefore, $m g \sin θ = \frac{m v^{2}}{L} ....... (1)$

$v^{2} = u^{2} - 2 g h = u^{2} - 2 g L (1 + \sin θ) ......... (2)$

$Q D = \frac{1}{2} (r a n g e)____(3)$

$u = \sqrt{g L (2 + \frac{3 \sqrt{3}}{2})}$

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