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NEET AIPMT SOLVED PAPER 2003

  • question_answer
                    A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity \[\omega \]. Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be:                                                                                                       

    A)                 \[\frac{(M+4m)\,\omega }{M}\]                              

    B)                 \[\frac{(M-4m)\,\omega }{M+4m}\]

    C)                 \[\frac{M\,\omega }{4m}\]

    D)                 \[\frac{M\,\omega }{M+4m}\]

    Correct Answer: D

    Solution :

                    Key Idea: If external torque acting on the system is zero, hence angular momentum remains conserved.                 \[{{\tau }_{\text{ext}}}=0\] \[or\frac{dL}{dt}=0\] or            L = constant or            \[I\omega \] = constant \[\therefore \]  \[{{I}_{1}}\,{{\omega }_{1}}={{I}_{2}}\,{{\omega }_{2}}\]                                               ?(i) Here,     \[{{I}_{1}}=M{{r}^{2}},\,\,{{\omega }_{1}}=\omega ,\,\,{{I}_{2}}=M{{r}^{2}}+4m{{r}^{2}}\]                 Hence, Eq. (i) can be written as                 \[M{{r}^{2}}\omega =(M{{r}^{2}}+4m{{r}^{2}})\,{{\omega }_{2}}\]                 \[\therefore {{\omega }_{2}}=\frac{M\omega }{M+4m}\]


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