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The potential energy of a particle in a ...

The potential energy of a particle in a conservative field is `U = frac{a}{r^3}-frac{b}{r^2}` , where a and b are positive constants and r is the distance of particle from the centre of field. For equilibrium, the value of r is

A

`frac{2a}{b}`

B

`frac{3a}{2b}`

C

`frac{a}{2b}`

D

`frac{a}{b}`

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The correct Answer is:
b
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