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NEET Sample Paper NEET Sample Test Paper-73

  • question_answer
    A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to\[-K/{{r}^{2}}\], where K is a constant. The total energy of the particle is

    A) \[\frac{K}{2r}\]                        

    B) \[-\frac{K}{2r}\]

    C) \[-\frac{K}{r}\]                          

    D) \[\frac{K}{r}\]

    Correct Answer: B

    Solution :

    Here \[\frac{m{{\nu }^{2}}}{r}\,\,=\,\,\frac{K}{{{r}^{2}}}\,\,\,\Rightarrow \,\,K.E.=\,\,\frac{1}{2}\,m{{\nu }^{2}}=\frac{K}{2r}\] \[U=-\int_{\infty }^{r}{F.dr=-\int_{\infty }^{r}{\left( -\frac{K}{{{r}^{2}}} \right)}\,dr=-\frac{K}{r}}\] Total energy \[\operatorname{E}=K.E. + P.E.=\frac{K}{2r}-\frac{K}{r}=-\frac{K}{2r}\]


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