Answer

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**Hint:**As a first step, you could recall that differentiating the potential energy with respect to distance will give the force at that point. We could thus find the force and equate it zero as at equilibrium position the force will be zero. Check whether the second derivative is positive to see whether that particular position is stable.

**Formula used:**

Force,

$F=-\dfrac{dU}{dr}$

**Complete Step by step solution:**

We are given the potential energy of a particle as,

$U=\left( \dfrac{a}{{{r}^{2}}} \right)-\left( \dfrac{b}{r} \right)$

Where, a and b are constants, r is the distance from the centre of the field.

We know that force is related to potential energy as,

$F=-\dfrac{dU}{dr}$

$\Rightarrow F=-\dfrac{dU}{dr}=2a{{r}^{-3}}-b{{r}^{-2}}$

$\therefore F=-\dfrac{b}{{{r}^{2}}}+\dfrac{2a}{{{r}^{3}}}$

We know that at equilibrium position of the particle, F=0, so,

$F=0$

$\Rightarrow F=-\dfrac{b}{{{r}^{2}}}+\dfrac{2a}{{{r}^{3}}}=0$

Let at equilibrium position,$r={{r}_{0}}$, then,

$F=-\dfrac{b}{{{r}_{0}}^{2}}+\dfrac{2a}{{{r}_{0}}^{3}}=0$

$\Rightarrow b{{r}_{0}}=2a$

$\therefore {{r}_{0}}=\dfrac{2a}{b}$

For a position to be stable, $\dfrac{{{d}^{2}}U}{d{{r}^{2}}}\rangle 0$

$\dfrac{{{d}^{2}}U}{d{{r}^{2}}}=\dfrac{6a}{{{r}^{4}}}-\dfrac{2b}{{{r}^{3}}}$

But, $r={{r}_{0}}$

$\dfrac{{{d}^{2}}U}{d{{r}^{2}}}=\dfrac{6a}{{{r}_{0}}^{4}}-\dfrac{2b}{{{r}_{0}}^{3}}=\dfrac{1}{{{r}_{0}}^{3}}\left( \dfrac{6ab}{2a}-2b \right)$

$\therefore \dfrac{{{d}^{2}}U}{d{{r}^{2}}}=\dfrac{b}{{{r}_{0}}^{3}}$

As ${{r}_{0}}$ and b are two positive constants, we could say that

$\dfrac{{{d}^{2}}U}{d{{r}^{2}}}\rangle 0$

Therefore, we found that the value of ${{r}_{0}}$ will be ${{r}_{0}}=\dfrac{2a}{b}$at equilibrium position and this position is also found to be steady.

**Hence, option C is the correct answer.**

**Note:**

From the definition of equilibrium position we could actually conclude that the position is steady or in other words stable. Also, force at a point is zero would directly imply that that particular position is stable. Since, ${{r}_{0}}=\dfrac{2a}{b}$ where both a and b are positive constants, we could conclude that ${{r}_{0}}$is also positive.

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