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Time period of a simple pendulum of length L is $T_{1}$ and time period of a uniform rod of the same length L pivoted about one end and oscillating in a vertical plane is $T_{2} .$ Amplitude of oscillation in both the case is small. Then $\frac{T_{1}}{T_{2}}$ is

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$\sqrt{\frac{1}{3}}$

$\sqrt{\frac{4}{3}}$

$\sqrt{\frac{3}{2}}$

answer is C.

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Detailed Solution

$T_{1} = 2 π \sqrt{\frac{L}{g}}$ and $T_{2} = 2 π \sqrt{\frac{I}{M g l}}$ (physical pendulum)

$\Rightarrow T_{2} = 2 π \sqrt{\frac{M L^{2} / 3}{M g (L / 2)}} = 2 π \sqrt{\frac{2 L}{3 g}} \Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{3}{2}}$

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