NEET Sample Paper NEET Sample Test Paper-39

  • question_answer The potential energy of a particle in a force field is:\[U=\frac{A}{{{r}^{2}}}-\frac{B}{r},\] 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/A          

    B)  B/2A

    C)  2A/B         

    D)  A/B

    Correct Answer: C

    Solution :

      Force experienced by the particle in field, \[F=-\left( \frac{dU}{dr} \right)=-\left( \frac{-2A}{{{r}^{3}}}+\frac{B}{{{r}^{2}}} \right)\] At equilibrium \[F=0\,F=-\left( \frac{-2A}{{{r}^{3}}}+\frac{B}{{{r}^{2}}} \right)=0\] \[\Rightarrow \]\[r=\frac{2A}{B}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner