A) \[\frac{{{A}^{2}}}{4B}\]
B) \[\frac{{{B}^{2}}}{4A}\]
C) \[\frac{2B}{A}\]
D) \[-\frac{{{B}^{2}}}{4A}\]
Correct Answer: A
Solution :
As it is quite clear that at equilibrium \[F=0\] \[f=\frac{dU}{dr}\] So, for equilibrium \[\frac{dU}{dr}=0\] Here, \[U=\frac{A}{{{r}^{6}}}-\frac{B}{{{r}^{12}}}\] \[\frac{dU}{dr}=\frac{d(A{{r}^{-6}})}{dr}-\frac{d(B{{r}^{-12}})}{dr}=0\] \[\frac{-6A}{{{r}^{7}}}+\frac{12B}{{{r}^{13}}}=0\] \[-A+\frac{2B}{{{r}^{6}}}=0\] or \[{{r}^{6}}=\frac{2B}{A}\] So, at \[{{r}^{6}}=\frac{2B}{A}\] \[{{U}_{0}}=\frac{A}{(2B/A)}-\frac{B}{(4{{B}^{2}}/{{A}^{2}})}=\frac{{{A}^{2}}}{2B}-\frac{{{A}^{2}}}{4B}=\frac{{{A}^{2}}}{4B}\]
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