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Question

A rod of length L and mass m pivoted about one end is oscillating in vertical plane. The period of oscillation for small amplitudes is

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A
T = 2(L/3g)
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B
T = 2(3L/2g)
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C
T = 2(2L/3g)
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D
T = 2(L/6g)
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Solution

The correct option is C T = 2(2L/3g)

The torque equation about the point of suspension is I = Mg(PQ) = Mg(L/2) sin . ]

For small oscillations sin = and moment of inertia about the point of suspension is I=ML23. Substituting the values, we get d2dt2=(3g/2L).

Thus, time period T = 2(2L/3g)


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