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A uniform rod OP of mass $m$ and length $l$ is hinged at the point O. The rod rotates about the vertical axis $O O'$ . It makes a constant angle $θ$ with $O O'$ .The moment of inertia of the rod about the axis of rotation is given by

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$(m l^{2} / 12) \sin θ$

$(m l^{2} / 3) \cos^{2} θ$

$m l^{2} / 3$

$(m l^{2} / 3) \sin^{2} θ$

answer is B.

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Detailed Solution

$\begin{array}{l} \int dI = \int_{0}^{1} \frac{m}{1} \sin θ xdx \\ I = \frac{m}{1} \sin θ \int_{0}^{1} x d x \\ I = \frac{1}{3} {ml}^{2} \sin θ \end{array}$

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