2. Solved Papers
  3. NEET

NEET AIPMT SOLVED PAPER SCREENING 2006

  • question_answer
                    A uniform rod of length \[l\] and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is:                                               (Moment of inertia of rod about A is \[\frac{m{{l}^{2}}}{3}\])

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

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

    C)                  \[\frac{3g}{2{{l}^{2}}}\]                

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

    Correct Answer: A

    Solution :

                    The moment of inertia of the uniform rod about an axis through one end and perpendicular to length is                 \[I=\frac{m{{l}^{2}}}{3}\]                 where m is mass of rod and \[l\] its 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            \[\frac{m{{l}^{2}}}{3}\alpha =mg\frac{l}{2}\]                 \[\therefore \]  \[\alpha =\frac{3g}{2l}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner