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A rough L-shaped rod of circular cross section is located in a horizontal plane and a sleeve of mass m is inserted onto the rod. The rod is rotated with a constant angular velocity $ω$ in the horizontal plane. The lengths $l_{1}$ and $l_{2}$ are shown in figure. When the sleeve is about to slip on rod, if N is the normal reaction, f is the frictional force acting on the sleeve by the rod and $ω_{0}$ is the angular velocity, then ( $μ =$ co-efficient of friction between rod and sleeve).

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$N = m ω^{2} l_{1}$

$ω_{0}^{4} = \frac{μ^{2} g^{2}}{l_{2}^{2} - μ^{2} l_{1}^{2}}$

$N = m \sqrt{g^{2} + ω_{0}^{4} l_{1}^{2}}$

$f = μ N$

answer is B, C, D.

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Detailed Solution

$N_{1}^{2} + N_{2}^{2} = N^{2}, N_{^{2}} = m g, m ω^{2} x \sin ϕ = f f = m ω^{2} l_{2}, N_{1} = m ω^{2} l_{1}$

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