The moment of inertia of a unifmm rod of mass M and length L about an axis through centre and perpendicular to length L is given by (ML^(2))/12 . Now consider one such rod pivoted at its centre 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 gets embedded in one end of the rod. The angular velocity CD of the rod after collision will be

The moment of inertia of a unifmm rod of mass M and length L about an

1.	The moment of inertia of a unifmm rod of mass M and length L about an axis through centre and perpendicular to length L is given by (ML^(2))/12 . Now consider one such rod pivoted at its centre 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 gets embedded in one end of the rod. The angular velocity CD of the rod after collision will be
Answer» `V//L` `2V//L` `3V//2L` `6V//L` Solution :Angular momentum CONSERVATION gives `MVL/2=(ML^(2))/3omegarArromega=(3V)/(2L)`

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