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The ring which can slide along the rod are kept at mid-point of a smooth rod of length L. The rod is rotated with constant angular velocity $ω$ about vertical axis passing through its one end. The ring is released from mid point .The velocity of the ring , when it just leaves the rod is = $\sqrt{\frac{x}{2}} ω L$ , then the value of x is

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answer is 7.

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Detailed Solution

$α_{r} = v \frac{d v}{d x} \Rightarrow ω^{2} x d x = v d v$

$\int_{ℓ / 2}^{ℓ} ω^{2} x d x = \int_{0}^{v} v d v$

$ω^{2} {[\frac{x^{2}}{2}]}^{ℓ} = {[\frac{v^{2}}{2}]}^{v} \frac{ω^{2}}{2} [ℓ^{2} - \frac{ℓ^{2}}{4}] = \frac{v^{2}}{2} \Rightarrow v^{2} = \frac{3}{4} w^{2} ℓ^{2}$

$∴$ Along radius velocity $v = \sqrt{\frac{3}{4}} ℓ ω$

Along tangent velocity $v = ℓ ω$

$∴$ Velocity $= \sqrt{{(\sqrt{\frac{3}{4}} ℓ ω)}^{2} + {(ℓ ω)}^{2}} = \frac{\sqrt{7}}{2} ℓ ω$

$∴ x = 7$

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