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Q.

A thin uniform rod of mass m and length l is hinged at the upper end and free to rotate about its upper end in vertical plane . When it is at rest, it receives an impulse J at its lowest point, normal to its length. Immediately after impact,

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a

the angular momentum of the rod is JI

b

the angular velocity of the rod is 3J/ml

c

the kinetic energy of the rod is 3J2/2m

d

the linear velocity of the midpoint of the rod is 3J/2m

answer is A.

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Detailed Solution

Angular momentum = linear momentum x perpendicular distance from the point of rotation  During the course of collision angular momentum of the rod about the axis of rotation remains conserved .                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       Therefore  L=JI Also, I=ml23        ω=LI=JIml2/3=3Jml Kinetic energy =L22l=J2l22ml23=3J22m        vc=ωl2=3J2m
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