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A long uniform rod of length $L$ and mass $M$ is free to rotate in a horizontal plane about a vertical axis through its one end $O$ . A spring of force constant $k$ is connected horizontally between one end of the rod and a fixed wall. When the rod is in equilibrium it is parallel to the wall .What is the period of small oscillations resulting when the rod is rotated slightly and released.

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$T = 2 \sqrt{\frac{M}{3 k}}$

$T = 2 π \sqrt{\frac{1}{3 k}}$

$T = 2 π \sqrt{\frac{M}{k}}$

$T = 2 π \sqrt{\frac{M}{3 k}}$

answer is D.

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Detailed Solution

If the rod is displaced through small angle $θ$ ,then extension in the spring will be $L θ$ .The force in the spring $F = k (L θ)$

Thus restoring torque $τ = - F (L \cos θ) = - (- k L θ) (L \cos θ)$

For small $θ, \cos θ = 1$

$∴ τ = k L^{2} (- θ)$

Angular acceleration

$α = \frac{τ}{I} = \frac{k L^{2}}{I} (- θ) ∴ T = 2 π \sqrt{\frac{I}{{kL}^{2}}} = 2 π \sqrt{\frac{{ML}^{2}}{3 {kL}^{2}}} = 2 π \sqrt{\frac{M}{3 k}}$

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