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CMC Medical CMC-Medical VELLORE Solved Paper-2009

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

    A)  6.54 cm               

    B)  3.64 cm

    C)  1.77 cm                               

    D)  0.88 cm

    Correct Answer: C

    Solution :

                     Here mass M = 50 g and radius R = 2.5 cm. 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            \[K=\frac{R}{\sqrt{2}}=\frac{2.5}{\sqrt{2}}=\frac{2.5\sqrt{2}}{2}\] \[=1.767\,cm\] \[=1.77\,cm\]


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