A uniform rod of mass m, length l rests on a smooth horizontal surface. Rod is given a sharp horizontal impulse P perpendicular to the rod at a distance $\frac{l}{4}$ from the centre. The angular velocity of rod will be

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$\frac{P}{ml}$

$\frac{P}{2 ml}$

$\frac{2 P}{ml}$

$\frac{3 P}{ml}$

answer is D.

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Detailed Solution

$P \times \frac{l}{4} = \frac{{ml}^{2}}{12} \times ω \Rightarrow ω = \frac{3 P}{ml}$

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