A rod of mass m, length l is as shown. A bullet of mass m moving with a speed V strikes as shown and get embedded in to the rod. Then the angular velocity of the rod $ω_{M}$ achieved for $x = x_{M};$ then

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$\frac{3 V_{x}}{L^{2} + 3 x^{2}} = ω$

$\frac{12 V_{x}}{L^{2} + 12 x^{2}} = ω$

$x_{M} = \frac{L}{\sqrt{3}}$

$ω_{M} = \frac{\sqrt{3} V}{2 l}$

answer is A, D.

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Detailed Solution

$mVx = (\frac{{mL}^{2}}{3} + {mx}^{2}) ω$

$ω = \frac{3 νx}{L^{2} + 3 x^{2}}$

For maximum $\frac{dω}{dx} = 0 \Rightarrow x = \frac{L}{\sqrt{3}} \Rightarrow ω = \frac{\sqrt{3} V}{2 l}$

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