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(B) The moment of inertia of a thin uniform rod of length $L$ and mass $M$ about an axis passing through its center of mass $(CM)$ and perpendicular t

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

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(B) The moment of inertia of a thin uniform rod of length $L$ and mass $M$ about an axis passing through its center of mass $(CM)$ and perpendicular to the rod is $I_{CM} = \frac{M L^2}{12}$.The distance of the given axis from the center of mass is $x = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}$.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,$I = \frac{3 M L^2 + M L^2}{36} = \frac{4 M L^2}{36} = \frac{M L^2}{9}$.

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 $\frac{L}{3}$ from one of its ends and perpendicular to the rod is

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