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MGIMS WARDHA MGIMS WARDHA Solved Paper-2011

  • question_answer
    The moments of inertia of two freely rotating bodies A and B are \[{{I}_{A}}\] and \[{{I}_{B}}\] respectively \[{{I}_{A}}>{{I}_{B}}\]and their angular momenta are equal. If \[{{K}_{A}}\] and \[{{K}_{B}}\] are their kinetic energies then

    A)  \[{{K}_{A}}={{K}_{B}}\]                

    B)  \[{{K}_{A}}>{{K}_{B}}\]

    C)  \[{{K}_{A}}<{{K}_{B}}\]                                

    D)  \[{{K}_{A}}=2{{K}_{B}}\]

    Correct Answer: C

    Solution :

                     Kinetic energy \[K=\frac{{{L}^{2}}}{2I}\] If angular momenta are equal, then \[K\propto \frac{1}{I}\] i.e.,              \[IK=\]constant \[\Rightarrow \]               \[{{I}_{A}}{{K}_{A}}={{I}_{B}}{{K}_{B}}\] \[\Rightarrow \]               \[\frac{{{I}_{A}}}{{{I}_{B}}}=\frac{{{K}_{B}}}{{{K}_{A}}}\]                 \[{{I}_{A}}>{{I}_{B}}\] \[\therefore \]  \[{{K}_{B}}>{{K}_{A}}\]


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