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# Moment of inertia of a thin rod of mass M and length l about an axis as shown .

$\begin {array} {1 1} (a)\;\frac{Ml^2 \sin \theta}{12} & \quad (b)\;\frac{Ml^2 \cos \theta}{12} \\ (c)\;\frac{Ml^2}{12} & \quad (d)\;\frac{Ml^2}{3} \end {array}$

Man per unit length $= \large\frac{M}{l}$
Consider an element of width dx at a distance x from 0
mass of element $=dm= \large\frac{M}{2}$$dx$
M.I= man $\times$ perpendicular distance from axis of rotation
Hence $\large\frac{Ml^2 \sin \theta}{12}$ is a correct answer.
edited Jun 18, 2014