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NEET AIPMT SOLVED PAPER SCREENING 2007

  • question_answer
                    A uniform rod AB of length \[l\] and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is \[\frac{m{{l}^{2}}}{3}\], the initial angular acceleration of the rod will be:                                                                                                                      

    A)                  \[\frac{2g}{3l}\]                              

    B)                  \[mg\frac{l}{2}\]             

    C)                  \[\frac{3}{2}gl\]                              

    D)                  \[\frac{3g}{2l}\]

    Correct Answer: D

    Solution :

                    The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is                 \[I=\frac{m{{l}^{2}}}{3}\]                 where m is mass of rod and \[l\] is length.                  Torque \[(\tau =I\alpha )\] acting on centre of gravity of rod is given by                 \[\tau =mg\frac{l}{2}\]                 or            \[I\alpha =mg\frac{l}{2}\]                 or            \[\frac{m{{l}^{2}}}{3}\alpha =mg\frac{l}{2}\]                 or            \[\alpha =\frac{3g}{2l}\]


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