BCECE Engineering BCECE Engineering Solved Paper-2008

  • question_answer
    Moment of inertia of ring about its diameter is \[I.\] 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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