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The potential energy of a system of two particles is given by $\mathrm{v=-\frac{m}{r^{2}}+\frac{n}{r^{8}}}$ , where $\mathrm{r}$ between the particle and $\mathrm{m}$ and $\mathrm{n}$ are positive constant. If the system is in equilibrium at $\mathrm{r=r_{0}}$ then $\mathrm{r_{0}}$ in terms of $\mathrm{m}$ and $\mathrm{n}$ .
Option: 1
$r_{0}^{3}=2 \sqrt{\frac{n}{m}}$

Option: 2
$r_{0}^{3}=2 \frac{n}{m}$

Option: 3
$r_{0}^{3}=\frac{4 n}{m}$

Option: 4
$r_{0}^{3}=8 \frac{n}{m}$

Answers (1)

best_answer

For equilibrium -

$\left.\frac{d V}{d r}\right|_{r=r_{0}}=0 \\$

$\mathrm{\frac{d}{d r}\left[-m r^{-2}+n r-8\right]_{r=r_{0}}=0}$
$\mathrm{{\left[2 m r^{-3}-8 n r^{-9}\right]_{\gamma=r_{0}}=0}}$
$\mathrm{{2 m r_{0}^{-3}=8 n r_{0}^{-9}}$
$\mathrm{\frac{r_{0}^{-3}}{r_{0}-9}=\frac{8 n}{2 m}}$
$\mathrm{r_{0}^{-3+9}=\frac{4 n}{m} }$
$\mathrm{r_{0}^{6}=\frac{4 n}{m} \Rightarrow r_{0}^{3}=\sqrt[2]{\frac{n}{m}} }$ Ans.

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SANGALDEEP SINGH

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