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(B) When the rod rotates through an angle $	heta$,the vertical fall $h$ of the centre of gravity is given by geometry.From the figure,the distance of

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$\sqrt{\frac{6g}{L}} \sin \frac{	heta}{2}$

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(B) When the rod rotates through an angle $	heta$,the vertical fall $h$ of the centre of gravity is given by geometry.From the figure,the distance of the centre of gravity from the pivot $B$ is $L/2$. The vertical component of this distance after rotation is $(L/2) \cos 	heta$.The vertical fall $h$ of the centre of gravity is $h = L/2 - (L/2) \cos 	heta = \frac{L}{2}(1 - \cos 	heta)$.According to the law of conservation of energy,the decrease in potential energy equals the increase in rotational kinetic energy.Decrease in potential energy $= Mgh = Mg \frac{L}{2}(1 - \cos 	heta)$.Rotational kinetic energy $= \frac{1}{2} I \omega^2$,where $I = \frac{ML^2}{3}$ is the moment of inertia about the end $B$.Equating the two: $Mg \frac{L}{2}(1 - \cos 	heta) = \frac{1}{2} \left( \frac{ML^2}{3} ight) \omega^2$.$Mg \frac{L}{2}(1 - \cos 	heta) = \frac{ML^2}{6} \omega^2$.Using the trigonometric identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$:$Mg \frac{L}{2} \cdot 2 \sin^2(	heta/2) = \frac{ML^2}{6} \omega^2$.$MgL \sin^2(	heta/2) = \frac{ML^2}{6} \omega^2$.$\omega^2 = \frac{6g}{L} \sin^2(	heta/2)$.$\omega = \sqrt{\frac{6g}{L}} \sin \left( \frac{	heta}{2} ight)$.

$A$ uniform rod of length $L$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $\theta$,its angular velocity $\omega$ is given as

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