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MGIMS WARDHA MGIMS WARDHA Solved Paper-2010

  • question_answer
    The moment of inertia of a ring about one of its diameter is\[I.\]What will be its moment of inertia about a tangent parallel to the diameter?

    A)  4\[I\]                                   

    B)  \[2\,I\]

    C)  \[\frac{3}{2}I\]                                

    D)  \[3\,I\]

    Correct Answer: D

    Solution :

                     \[I=\frac{1}{2}M{{R}^{2}}\] According to theorem of parallel axes, \[\therefore \]  \[I'=\frac{1}{2}M{{R}^{2}}+M{{R}^{2}}=\frac{3}{2}M{{R}^{2}}=3I\]


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