A uniform rod of length 2.0 m is suspended through an end & oscillate with small amplitude. The time period is -

1.60s

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1.80s

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2.00s

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2.40s

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Solution

The correct option is D 2.40s
This is the special case of physical pendulum and in this case
$T = 2 π \sqrt{\frac{2 l}{3 g}}$

$\Rightarrow T = 2 \times 3.14 \sqrt{\frac{2 \times 2}{3 \times 9.8}}$

$= 2.31 s e c \approx 2.4 s e c$

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