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J & K CET Engineering J and K - CET Engineering Solved Paper-2008

  • question_answer
    The moment of inertia of a circular ring of mass \[1\text{ }kg\]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, \[I=4\,kg-{{m}^{2}},\] \[m=1kg\] \[\therefore \] \[{{R}^{2}}=\frac{4}{1}=4\] or \[R=2m\] Therefore, diameter of ring \[=4\text{ }m\].


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