A particle located in a one-dimensional potential field has its potential energy function…
Physics · JEE Advanced · NTA Exams — Work, Energy and Power
A particle located in a one-dimensional potential field has its potential energy function as U(x) =\(\frac{a}{x^{4}}-\frac{b}{x^{2}}\)where a and b are positive constants. The position of equilibrium x-corresponds to
- \(\frac{b}{2 a}\)
- \(\sqrt{\frac{2 a}{b}}\)
- \(\sqrt{\frac{2 b}{a}}\)
- \(\frac{a}{2 b}\)
Answer
(B) 2 a b
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