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A thin rod of length l and mass m is turned at mid- point 0 at an angle of $60^{\circ}$ . The moment of inertia of the rod about an axis passing through 0 and perpendicular to the plane of the rod will be


\[\begin {array} {1 1} (a)\;\frac{ml^2}{3} & \quad (b)\;\frac{ml^2}{6} \\ (c)\;\frac{Ml^2}{8} & \quad  (d)\;\frac{Ml^2}{12} \end {array}\]
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Since axis of rotation is perpendicular to the length of the rod, this is not different from the M.I of a uniform rod about a perpendicular axis passing through its centre.
answered Nov 26, 2013 by meena.p

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