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(C) Let the length of the rod be $\ell$ and its mass be $m$. The rod rotates about the fixed end on the ground.Initially,the center of mass $(CM)$ of

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$\sqrt{3g\ell}$

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(C) Let the length of the rod be $\ell$ and its mass be $m$. The rod rotates about the fixed end on the ground.Initially,the center of mass $(CM)$ of the rod is at a height $h = \frac{\ell}{2}$ from the ground.When the rod hits the ground,the potential energy lost by the rod is converted into rotational kinetic energy.Loss in potential energy = $mg \Delta h = mg \frac{\ell}{2}$.Rotational kinetic energy = $\frac{1}{2} I \omega^2$,where $I$ is the moment of inertia about the end,$I = \frac{m\ell^2}{3}$.Equating the two: $mg \frac{\ell}{2} = \frac{1}{2} \left( \frac{m\ell^2}{3} ight) \omega^2$.Simplifying,$g\ell = \frac{\ell^2}{3} \omega^2$,so $\omega^2 = \frac{3g}{\ell}$,which gives $\omega = \sqrt{\frac{3g}{\ell}}$.The velocity $v$ of the upper end is $v = \omega \ell$.Substituting $\omega$,we get $v = \sqrt{\frac{3g}{\ell}} 	imes \ell = \sqrt{3g\ell}$.

Column-$I$	Column-$II$
$(1)$ $W = F \Delta x$	$(a)$ $P = \tau \omega$
$(2)$ $P = Fv$	$(b)$ $W = \tau \Delta \theta$
	$(c)$ $L = I \omega$

$A$ thin meter rod is placed with one end on the ground. It is allowed to fall such that the contact point remains fixed. Find the velocity of the upper end when it hits the ground.

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Column-$I$ Column-$II$

$(1)$ $W = F \Delta x$ $(a)$ $P = \tau \omega$

$(2)$ $P = Fv$ $(b)$ $W = \tau \Delta \theta$

$(c)$ $L = I \omega$

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