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Consider a rod of mass M and length L pivoted at its centre is free to rotate in a vertical plane. The rod is at rest in the vertical position. A bullet of mass M moving horizontally at a speed v strikes and embedded in one end of the rod. The angular velocity of the rod just after the collision will be

a
vL
b
2vL
c
3v2L
d
6vL

detailed solution

Correct option is C

Apply conservation of angular momentum about the pivot point0+MVL/2=Iω HERE I=ML212+ML22⇒  ω = 3v2L

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