JEE Main & Advanced JEE Main Paper (Held on 7 May 2012)

  • question_answer
    A solid sphere having mass m and radius r rolls down an inclined plane. Then its kinetic energy is   JEE Main Online Paper (Held On 07 May 2012)

    A) \[\frac{5}{7}\]rotational and \[\frac{2}{7}\]translational

    B)                        \[\frac{2}{7}\]rotational and \[\frac{5}{7}\]translational

    C)                        \[\frac{2}{5}\]rotational and \[\frac{3}{5}\]translational

    D)                        \[\frac{1}{2}\]rotational and \[\frac{1}{2}\]translational

    Correct Answer: B

    Solution :

                    \[K.{{E}_{rotational}}=\frac{1}{2}I{{\omega }^{2}}\] \[=\frac{1}{2}\frac{2}{5}\omega {{r}^{2}}{{d}^{2}}\left( \because {{I}_{Solid\,sphere}}=\frac{2}{5}m{{r}^{2}} \right)\] \[K.{{E}_{translational}}=\frac{1}{2}m{{v}^{2}}\] \[\therefore \]\[\frac{K.{{E}_{rotational}}}{K.{{E}_{translational}}}=\frac{2}{5}\] Hence option (b) is correct


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