$U (r) = \frac{A}{r^{2}} - \frac{B}{r}$ where r is the distance from the centre of the force and A and B are positive constants.

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The work required move the particle from equilibrium distance to infinity is

\frac{B^{2}}{4A}

Only (1)

The equilibrium distance is given by

2 A/B

The work required move the particle from equilibrium distance to infinity is

\frac{B}{4A}

answer is A, B.

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Detailed Solution

F = \frac{du}{dr} = 0

$\frac{- 2A}{r^{3}} + \frac{B}{r^{2}} = 0$

$\frac{2A}{r^{3}} = \frac{B}{r^{2}} \Rightarrow r = \frac{2A}{B}$

$W = \frac{{AB}^{2}}{{4A}^{2}} - \frac{B^{2}}{2A}$

$W = \frac{B^{2}}{4A}$

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