BCECE Medical BCECE Medical Solved Papers-2008

  • question_answer
    Moment of inertia of ring about its diameter is Then, moment of inertia about an axis passing through centre perpendicular to its plane is

    A)  \[2I\]

    B)  \[\frac{I}{2}\]

    C)  \[\frac{3}{2}I\]

    D)  \[I\]

    Correct Answer: A

    Solution :

    By the theorem of perpendicular axes, the moment of inertia about the central axis \[{{I}_{C}}\], will be equal to the sum of its moments of inertia about two mutually perpendicular diameters lying in its plane. Thus,                 \[{{I}_{d}}=I=\frac{1}{2}M{{R}^{2}}\] \[\therefore \] \[{{I}_{C}}=I+I=\frac{1}{2}M{{R}^{2}}+\frac{1}{2}M{{R}^{2}}\] \[=I+I=2I\]


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