Find the maximum magnitude of the attraction force.

F_{m a x} = \frac{b^{3}}{27 a^{2}}

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F_{m a x} = - \frac{b^{3}}{27 a^{2}}

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F_{m a x} = - 3 \frac{b^{3}}{27 a^{2}}

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F_{m a x} = 3 \frac{b^{3}}{27 a^{2}}

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Solution

The correct option is B $F_{m a x} = \frac{b^{3}}{27 a^{2}}$
We have,
$F = \frac{d U}{d r}$

$F = - 2 a r^{- 3} + b r^{- 2}$ .........(1)

$F = b r^{- 2} - 2 a r^{- 3}$ ..........(2)

$F = \frac{b}{r^{2}} - \frac{2 a}{r^{3}}$

Now,

$F^{'} (r) = 0$
$(- 2 a r^{- 3} + b r^{- 2})^{^{'}} = 6 a r^{- 4} - 2 b r^{- 3} = \frac{6 a}{r^{4}} - \frac{2 b}{r^{3}}$ .......from(1)

$\Rightarrow \frac{6 a}{r^{4}} = \frac{2 b}{r^{3}}$

$∴ r = \frac{3 a}{b}$

Hence the maximum force is
$F_{m a x} = b (\frac{3 a}{b})^{- 2} - 2 a (\frac{3 a}{b})^{- 3}$ ....from(2)
$F_{m a x} = \frac{b^{3}}{9 a^{2}} - 2 \frac{b^{3}}{27 a^{2}} = \frac{b^{3}}{27 a^{2}}$

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