A) \[\frac{b}{2a}\]
B) \[\sqrt{\frac{2a}{b}}\]
C) \[\sqrt{\frac{2a}{a}}\]
D) \[\frac{a}{2a}\]
Correct Answer: B
Solution :
[b] The position of equilibrium corresponds to \[F(x)=0\] Since \[F(x)=\frac{-dU(x)}{dx}\] So \[F(x)=-\frac{d}{dx}\left( \frac{a}{{{x}^{4}}}-\frac{b}{{{x}^{2}}} \right)\]or \[F(x)=-\frac{4a}{{{x}^{5}}}-\frac{2b}{{{x}^{3}}}\] For equilibrium, \[F(x)=0\], therefore \[\frac{4a}{{{x}^{5}}}-\frac{2b}{{{x}^{3}}}=0\Rightarrow x=\pm \sqrt{\frac{2a}{b}}\] \[\frac{{{d}^{2}}U(x)}{d{{x}^{2}}}=-\frac{20a}{{{x}^{6}}}+\frac{8b}{{{x}^{4}}}\] Putting \[x=\pm \sqrt{\frac{2a}{b}}\]gives \[\frac{{{d}^{2}}U(x)}{d{{x}^{2}}}\]as negative So U is maximum. Hence, it is position of unstable equilibrium.
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