A uniform thin rod of length l and mass m is hinged at a distance l/4 from one of the end and released from horizontal position as shown in Fig. The angular velocity of the rod as it passes the vertical position is

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$\sqrt{\frac{3 g}{7 l}}$

$2 \sqrt{\frac{5 g}{7 l}}$

$2 \sqrt{\frac{6 g}{7 l}}$

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

answer is B.

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Detailed Solution

$m g (\frac{l}{4}) = \frac{1}{2} [\frac{m l^{2}}{12} + m {(\frac{l}{4})}^{2}] ω^{2}$ $\Rightarrow ω = \sqrt{\frac{24 g}{7 l}} = 2 \sqrt{\frac{6 g}{7 l}}$

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