NEET Sample Paper NEET Sample Test Paper-74

  • question_answer
    If a sphere rolls down from an inclined plane without slipping, then the ratio of translational energy to its total energy is

    A)  \[\frac{7}{2}\]                         

    B)  \[\frac{2}{7}\]

    C)  \[\frac{5}{7}\]                         

    D)  \[\frac{7}{5}\]

    Correct Answer: B

    Solution :

    \[{{(KE)}_{rotation}}=\frac{1}{2}/{{\omega }^{2}}=\frac{1}{2}\left( \frac{2}{5}M{{R}^{2}} \right){{\omega }^{2}}\] \[=\,\,\,\frac{1}{5}M{{R}^{2}}\,{{\omega }^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \because \,\,\,l=\frac{2}{5}M{{R}^{2}} \right)\] \[{{(KE)}_{translation}}=\frac{1}{2}M{{v}^{2}}=\frac{1}{2}M{{R}^{2}}\,{{\omega }^{2}}\] \[{{(KE)}_{Total}}=\,\,\frac{7}{10}M{{R}^{2}}\,{{\omega }^{2}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{{(KE)}_{rotation}}}{{{(KE)}_{Total}}}=\frac{2}{7}\]


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