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JEE Main & Advanced JEE Main Paper (Held On 11 April 2015)

  • question_answer
    A particle is moving in a circle of radius r under the action of a force \[F=\alpha {{r}^{2}}\]which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy=0 for r=0): [JEE Main Online Paper (Held On 11 April 2015)]  

    A) \[\alpha {{r}^{3}}\]

    B) \[\frac{1}{2}\alpha {{r}^{3}}\]

    C) \[\frac{4}{3}\alpha {{r}^{3}}\]

    D) \[\frac{5}{6}\alpha {{r}^{3}}\]

    Correct Answer: D

    Solution :

    \[\frac{m{{V}^{2}}}{r}=\alpha {{r}^{2}}\] \[\therefore \]\[K.E.=\frac{\alpha {{r}^{3}}}{2}\] \[\Delta P.E=\int\limits_{0}^{r}{\alpha {{r}^{2}}.dr}\] \[P.E=\frac{\alpha {{r}^{3}}}{3}\] \[T.E=\frac{\alpha {{r}^{3}}}{2}+\frac{\alpha {{r}^{3}}}{3}\] \[\] \[T.E=\frac{5}{6}\alpha {{r}^{3}}\]


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