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Question

# A uniform thin rod of mass M and length L is hinged by a frictionless pivot at its end O, as shown in the diagram. A bullet of mass m moving horizontally with a velocity v, strikes the free end of the rod and gets embedded in it. The angular velocity of the system about O, just after the collision is,

A
mvL(M+m)
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B
2mvL(M+2m)
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C
3mvL(M+3m)
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D
mvLM
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Solution

## The correct option is C 3mvL(M+3m)Let ω be the angular velocity acquired by the system (rod + bullet) immediately after the collision. Since no external torque acts, the angular momentum of the system is conserved, thus mvL=Iω −(1) Where I is the moment of inertia of the system about an axis passing throug O and perpendicular to the rod. ∴I=MOI of rod about O+MOI of bullet stuck at its lower end about O =13ML2+mL2=13(M+3m)L2 −(2) Using Eq. (1) in Eq. (2), we have, mvL=13(M+3m)L2ω ∴ω=3mvL(M+3m) Hence, (C) is the correct answer.

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