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(B) The moment of inertia of a thin uniform rod about an axis passing through its center of mass $(CM)$ and perpendicular to its length is given by $I

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$\frac{M L^{2}}{9}$

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(B) The moment of inertia of a thin uniform rod about an axis passing through its center of mass $(CM)$ and perpendicular to its length is given by $I_{CM} = \frac{M L^{2}}{12}$.The axis of rotation is at a distance $x$ from the center of mass. Since the axis is at a distance $L/3$ from one end,and the center of mass is at $L/2$ from the same end,the distance $x$ between the axis and the center of mass is $x = |L/2 - L/3| = L/6$.Using the parallel axis theorem,$I = I_{CM} + M x^{2}$.Substituting the values,we get $I = \frac{M L^{2}}{12} + M \left( \frac{L}{6} \right)^{2}$.$I = \frac{M L^{2}}{12} + \frac{M L^{2}}{36}$.Taking the least common multiple,$I = \frac{3 M L^{2} + M L^{2}}{36} = \frac{4 M L^{2}}{36} = \frac{M L^{2}}{9}$.

Column-$I$	Column-$II$
$1. m = -2$	$a$. Convex mirror
$2. m = -\frac{1}{2}$	$b$. Concave mirror
$3. m = +2$	$c$. Real Image
$4. m = +\frac{1}{2}$	$d$. Virtual image

The moment of inertia of a thin uniform 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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