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J & K CET Medical J & K - CET Medical Solved Paper-2008

  • question_answer
    The moment of inertia of a circular ring of mass kg about an axis passing through its centre and perpendicular to its plane is \[4\text{ }kg-{{m}^{2}}.\] The diameter of the ring is

    A)  2 m                                       

    B)  4 m

    C)  5 m                                       

    D)  6 m

    Correct Answer: B

    Solution :

                     Moment of inertia of circular ring about an axis passing through its centre of mass and perpendicular to its plane \[I=M{{R}^{2}}\] Here,     \[I=4kg-{{m}^{2}},\] \[m=1\text{ }kg\] \[\therefore \]  \[{{R}^{2}}=\frac{4}{1}=4\] or           \[R=2\text{ }m\] Therefore, diameter of ring = 4 m.


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