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(D) Let the rod be $AB$ of length $L$ and mass $M$. The center of mass of the rod is at point $O$,which is at a distance of $L/2$ from either end.The

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$\frac{ML^2}{9}$

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(D) Let the rod be $AB$ of length $L$ and mass $M$. The center of mass of the rod is at point $O$,which is at a distance of $L/2$ from either end.The axis of rotation passes through point $C$,which is at a distance of $L/3$ from end $B$. Therefore,the distance between the center of mass $O$ and the axis of rotation $C$ is $d = OC = L/2 - L/3 = L/6$.According to the parallel axis theorem,the moment of inertia $I$ about an axis passing through $C$ is given by $I = I_{CM} + Md^2$,where $I_{CM} = ML^2/12$ is the moment of inertia about the center of mass.Substituting the values: $I = \frac{ML^2}{12} + M\left(\frac{L}{6}\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{36} = \frac{3ML^2 + ML^2}{36} = \frac{4ML^2}{36} = \frac{ML^2}{9}$.

The moment of inertia of a uniform thin rod of length $L$ and mass $M$ about an axis passing through a point at a distance of $L/3$ from one of its ends and perpendicular to the rod is:

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