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(D) To find the moment of inertia about the given axis,we use the parallel axis theorem: $I = I_{	ext{cm}} + M d^2$.Here,$I_{	ext{cm}}$ is the momen

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

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(D) To find the moment of inertia about the given axis,we use the parallel axis theorem: $I = I_{	ext{cm}} + M d^2$.Here,$I_{	ext{cm}}$ is the moment of inertia about the center of mass,which is $\frac{ML^2}{12}$.The distance $d$ between the center of mass and the axis is the difference between the distance from the end to the center of mass $(L/2)$ and the distance from the end to the axis $(L/4)$:$d = \frac{L}{2} - \frac{L}{4} = \frac{L}{4}$.Substituting these values into the theorem:$I = \frac{ML^2}{12} + M \left(\frac{L}{4}ight)^2$$I = \frac{ML^2}{12} + \frac{ML^2}{16}$Taking the least common multiple of $12$ and $16$,which is $48$:$I = \frac{4ML^2 + 3ML^2}{48} = \frac{7ML^2}{48}$.

Column-$I$	Column-$II$
$(1)$ Perpendicular Axis Theorem	$(a)$ $I = I_C + Md^2$
$(2)$ Parallel Axis Theorem	$(b)$ $I_z = I_x + I_y$

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 $\frac{L}{4}$ from one of its ends and perpendicular to the rod is:

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Match Column-$I$ with Column-$II$:
Column-$I$ Column-$II$
$(1)$ Perpendicular Axis Theorem $(a)$ $I = I_C + Md^2$
$(2)$ Parallel Axis Theorem $(b)$ $I_z = I_x + I_y$

Where, $d =$ distance between two parallel axes.

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