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JIPMER Jipmer Medical Solved Paper-2001

  • question_answer
    Two rigid bodies A and B rotate with rotational kinetic energies \[{{E}_{A}}\] and \[{{E}_{B}}\] respectively. The moments of inertia of A and B about the axis of rotation are \[{{I}_{A}}\] and \[{{I}_{B}}\]respectively. If\[{{I}_{A}}=\frac{{{I}_{B}}}{4}\] and\[{{E}_{A}}=100\,{{E}_{B}}\] the ratio of angular momentum\[({{L}_{A}})\]of A to the angular momentum \[({{L}_{B}})\] of B is:

    A)          25              

    B)                         5/4                       

    C)  5                            

    D)         1/4

    Correct Answer: C

    Solution :

    From the formula, rotational K.E. \[{{E}_{rotational}}=\frac{1}{2}I{{\omega }^{2}}=\frac{{{L}^{2}}}{2I}\]        \[(\because L=I\omega )\] Therefore, \[{{L}^{2}}=2EI\] Hence   \[L_{A}^{2}=2{{E}_{A}}{{I}_{A}}\]                                             ?(i) and        \[L_{B}^{2}=2{{E}_{B}}{{I}_{B}}\]                                              ?(ii) From equation (i) and (ii) \[\frac{{{L}_{A}}}{{{L}_{B}}}=\sqrt{\frac{2{{E}_{A}}{{I}_{B}}}{2{{E}_{B}}{{I}_{A}}}}=\sqrt{\frac{100}{4}}=5\]    \[\left( \begin{align}   & \because \,\,\,\,\,\,\,\,\frac{{{E}_{B}}}{{{E}_{A}}}=4 \\  & \text{and}\,\,\,\frac{{{I}_{A}}}{{{I}_{B}}}=100 \\ \end{align} \right)\]


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