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Question

# Moment of inertia of a uniform rod of mass M and length L about and axis through its centre and perpendicular to its length is given by ML212. Now, consider one such rod pivoted at its centre free to rotate in vertical plane. The rod is at rest in vertical position. A bullet of mass M moving horizontally a speed v strikes and is embedded in one end of the rod. The angular velocity of the rod just after collision will be

A
vL
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B
2vL
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C
3v2L
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D
6vL
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Solution

## The correct option is C 3v2LAngular Momentum Conservation will take place.Initial Angular Momentum =MvL2Final Moment of Inertia, I=ML212+M(L2)2=ML23Final Angular Momentum, L=Iω=ML23ω∴MvL2=ML23ωω=3v2L

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