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Punjab Medical Punjab - MET Solved Paper-2009

  • question_answer
    Radius of gyration of disc of mass \[50g\] and radius \[2.5cm\] about an axis passing through its centre of gravity and perpendicular to the plane is

    A) \[6.54cm\]                         

    B) \[3.64cm\]

    C) \[1.77cm\]                         

    D) \[0.88cm\]

    Correct Answer: C

    Solution :

    Here mass\[M=50g\]and radius\[R=-2.5cm\]. Required moment of inertia of the disc is given by              \[I=\frac{M{{R}^{2}}}{2}=M{{K}^{2}}\] So,     \[{{K}^{2}}=\frac{{{R}^{2}}}{2}\] or        \[K=\frac{R}{\sqrt{2}}=\frac{2.5}{\sqrt{2}}=\frac{2.5\sqrt{2}}{2}\]


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