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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2007

  • question_answer
    Moment of inertia of ring about its diameter is\[I\]. Then. moment of inertia about an axis passing through centre and perpendicular to its plane is

    A)  \[2I\]             

    B)  \[\frac{I}{2}\]

    C)  \[\frac{3}{2}I\]            

    D)  \[I\]

    Correct Answer: A

    Solution :

     \[{{I}_{d}}=I=\frac{1}{2}M{{R}^{2}}\] \[{{I}_{c}}={{I}_{d}}+{{I}_{d}}{{R}^{2}}\] [According to per- pendicular axis theorem] \[=I+I\] \[{{I}_{c}}=2I\]


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