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(C) This is a case of a physical pendulum.The formula for the time period of a physical pendulum is given by $T = 2 \pi \sqrt{\frac{I}{m g d}}$,where

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$2 \pi \sqrt{\frac{2 l}{3 g}}$

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(C) This is a case of a physical pendulum.The formula for the time period of a physical pendulum is given by $T = 2 \pi \sqrt{\frac{I}{m g d}}$,where $I$ is the moment of inertia about the pivot point,$m$ is the mass,and $d$ is the distance from the pivot to the center of mass.For a uniform rod of mass $m$ and length $l$ suspended about its end:$1$. The moment of inertia about the end is $I = \frac{m l^2}{3}$.$2$. The distance from the pivot to the center of mass is $d = \frac{l}{2}$.Substituting these values into the formula:$T = 2 \pi \sqrt{\frac{\frac{m l^2}{3}}{m g (\frac{l}{2})}}$$T = 2 \pi \sqrt{\frac{m l^2}{3} \cdot \frac{2}{m g l}}$$T = 2 \pi \sqrt{\frac{2 l}{3 g}}$Thus,the correct option is $C$.

$A$ uniform rod of mass $m$ and length $l$ is suspended about its end. The time period of small angular oscillations is ..........

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