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(B) When the rod is released,its potential energy is converted into rotational kinetic energy about the pivot point $P$.The initial potential energy o

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$\sqrt{3gl}$

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(B) When the rod is released,its potential energy is converted into rotational kinetic energy about the pivot point $P$.The initial potential energy of the rod is $U = mg(\frac{l}{2})$,because the center of gravity is at a distance of $\frac{l}{2}$ from the pivot point $P$.The rotational kinetic energy of the rod is $K = \frac{1}{2}I\omega^2$,where $I$ is the moment of inertia of the rod about the end $P$,given by $I = \frac{ml^2}{3}$.By the law of conservation of energy,the initial potential energy equals the final rotational kinetic energy:$mg(\frac{l}{2}) = \frac{1}{2}I\omega^2$$mg(\frac{l}{2}) = \frac{1}{2}(\frac{ml^2}{3})\omega^2$$g = \frac{l}{3}\omega^2$$\omega^2 = \frac{3g}{l} \implies \omega = \sqrt{\frac{3g}{l}}$The linear velocity $v$ of the upper end of the rod is given by $v = \omega l$.Substituting the value of $\omega$:$v = \sqrt{\frac{3g}{l}} 	imes l = \sqrt{3gl}$

$A$ rod of length $l$ is held vertically stationary with its lower end located at a point $P$ on a horizontal plane. When the rod is released to topple about $P$,the velocity of the upper end of the rod with which it hits the ground is

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