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(D) The moment of inertia of a thin uniform rod about an axis passing through its center of mass and perpendicular to its length is $I_{cm} = \frac{ML

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$\frac{7 ML^2}{48}$

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(D) The moment of inertia of a thin uniform rod about an axis passing through its center of mass and perpendicular to its length is $I_{cm} = \frac{ML^2}{12}$.According to the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance between the center of mass and the given axis.The center of mass of the rod is at a distance $L/2$ from one end.The given axis is at a distance $L/4$ from the same end.Therefore,the distance $d$ between the center of mass and the axis is $d = |L/2 - L/4| = L/4$.Substituting these values into the parallel axis theorem:$I = \frac{ML^2}{12} + M(L/4)^2$$I = \frac{ML^2}{12} + \frac{ML^2}{16}$Taking the least common multiple of $12$ and $16$,which is $48$:$I = \frac{4ML^2 + 3ML^2}{48} = \frac{7ML^2}{48}$.

The moment of inertia of a thin uniform rod of mass $M$ and length $L$,about an axis perpendicular to the length of the rod and at a distance $L/4$ from one end is:

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