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BVP Medical BVP Medical Solved Paper-2001

  • question_answer
    A circular ring of mass M and radius R is rotating about it axis at an angular frequency\[\omega \]. Two blocks each of mass mare gently placed on the opposite ends of a diameter of the ring. The angular frequency becomes\[\omega \]. Then the ratio of\[\frac{\omega }{\omega }\]will be:

    A) \[\frac{M}{M+2m}\]                      

    B)  \[\frac{m}{2M}\]

    C) \[\frac{m}{2M+M}\]                      

    D) \[\frac{2M}{(2m+M)}\]

    Correct Answer: A

    Solution :

                                          Applying the law of conservation of angular momentum, we have \[I\omega =I\omega \]                                .....(i) Here,      \[I=M{{R}^{2}}\] \[I=(M+2m){{R}^{2}}\] Hence, putting the values of \[I\] and \[I\] in equation (i) we obtain \[M{{R}^{2}}\omega =(M+2m){{R}^{2}}\omega \] So,  \[\frac{\omega }{\omega }=\frac{M}{M+2m}\]     


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