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MGIMS WARDHA MGIMS WARDHA Solved Paper-2014

  • question_answer
    The moment of inertia of a uniform ring of mass\[m\]and radius r about an axis,\[AA\] touching the ring tangentially and lying in the plane of the ring only, as shown in figure , is

    A)  \[\frac{5}{4}m{{r}^{2}}\]                             

    B)  \[\frac{5}{2}m{{r}^{2}}\]  

    C)  \[\frac{1}{4}m{{r}^{2}}\]                             

    D)  \[\frac{1}{2}m{{r}^{2}}\]

    Correct Answer: A

    Solution :

                    By parallel axis theorem, \[I={{I}_{cm}}+m{{h}^{2}}\] \[=\frac{m{{r}^{2}}}{4}+m{{r}^{2}}\] \[=\frac{5}{4}m{{r}^{2}}\]           


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