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NEET AIPMT SOLVED PAPER SCREENING 2012

  • question_answer
    The potential energy of a particle in a force field is \[U=\frac{A}{{{r}^{2}}}-\frac{A}{r},\] where A and B are positive constants and r is the distance of particle from the centre of the field. For  stable equilibrium, the distance of the particle is

    A) B/2A                                     

    B) 2A/B     

    C)         A/B                      

    D)        B/A

    Correct Answer: B

    Solution :

    Given, the potential energy of a particle in a force field \[U=\frac{A}{{{r}^{2}}}-\frac{B}{{{r}^{1}}}\] For stable equilibrium, \[F=-\frac{dU}{dr}=0\] \[0=-\frac{2A}{{{r}^{3}}}+\frac{B}{{{r}^{2}}}\]or\[\frac{2A}{r}=B\] The distance of particle from the centre of the field        \[r=\frac{2A}{B}\]


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