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A particle of mass $m$ is suspended by a string of length $l$ from a fixed rigid support. A sufficient horizontal velocity $v_{0} = \sqrt{3 gl}$ is imparted to it suddenly. Calculate the angle made by the string with the vertical when the acceleration of the particle is inclined to the string by $45 °$ .

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$\frac{π}{6}$

$θ = 60 °$

$\frac{π}{2}$

$θ = 80 °$

answer is D.

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Detailed Solution

$h = l (1 - cosθ)$

$v^{2} = v_{0}^{2} - 2 gh = 3 gl - 2 gl (1 - cosθ)$

$= gl (1 + 2 cosθ)$

At $45 °$ means radial and tangential components of acceleration are equal.

$\frac{v^{2}}{l} = gsinθ$

$1 + 2 cosθ = sinθ$

Solving the equation, we get $θ = 90 °$ or $\frac{π}{2}$

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