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JEE Main & Advanced Physics Rotational Motion Question Bank Self Evaluation Test - System of Particles Rotational Motion

  • question_answer
    A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotational energy will be

    A) \[\frac{{{K}^{2}}}{{{R}^{2}}}\]

    B) \[\frac{{{K}^{2}}}{{{K}^{2}}+{{R}^{2}}}\]

    C) \[\frac{{{R}^{2}}}{{{K}^{2}}+{{R}^{2}}}\]

    D) \[\frac{{{K}^{2}}+{{R}^{2}}}{{{R}^{2}}}\]

    Correct Answer: B

    Solution :

    [b] \[\frac{Rotational\,\,KE}{Total\,\,KE}=\frac{\frac{1}{2}mv\left( \frac{{{K}^{2}}}{{{R}^{2}}} \right)}{\frac{1}{2}m{{v}^{2}}\left( 1+\frac{{{K}^{2}}}{{{R}^{2}}} \right)}~~\] \[=\frac{{{K}^{2}}}{{{K}^{2}}+{{R}^{2}}}\]


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