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(C) Let the mass of the rod be $m$. The distance of the center of mass of the rod from the pivot $O$ is $r_{cm} = L + L/2 = 3L/2$.Linear impulse $P =

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$18$

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(C) Let the mass of the rod be $m$. The distance of the center of mass of the rod from the pivot $O$ is $r_{cm} = L + L/2 = 3L/2$.Linear impulse $P = \Delta p = m v_{cm} = m (\omega r_{cm}) = m \omega (3L/2)$.Thus, $P = \frac{3}{2} m \omega L$ --- $(i)$Angular impulse about pivot $O$ is $J_{	heta} = \int 	au dt = \Delta L_{ang}$.The impulse $P$ is applied at a distance $r = L + L/2 - x = 3L/2 - x$ from the pivot $O$ (based on the figure, the impulse is at $x$ from the midpoint towards the pivot end).$P (3L/2 - x) = I_O \omega$, where $I_O$ is the moment of inertia about $O$.$I_O = I_{cm} + m(r_{cm})^2 = \frac{mL^2}{12} + m(3L/2)^2 = mL^2 (\frac{1}{12} + \frac{9}{4}) = mL^2 (\frac{1+27}{12}) = \frac{28}{12} mL^2 = \frac{7}{3} mL^2$.So, $P (3L/2 - x) = (\frac{7}{3} mL^2) \omega$ --- $(ii)$Dividing $(ii)$ by $(i)$:$\frac{P(3L/2 - x)}{P} = \frac{(7/3) mL^2 \omega}{(3/2) m \omega L}$$3L/2 - x = \frac{7}{3} \cdot \frac{2}{3} L = \frac{14}{9} L$$x = \frac{3}{2} L - \frac{14}{9} L = \frac{27-28}{18} L$. Since $x$ is a distance, we take the magnitude: $x = L/18$.Comparing $x = L/n$ with $x = L/18$, we get $n = 18$.

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