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The potential clergy of a system of two particles is given by $\mathrm{v=-\frac{m}{r^2}+\frac{n}{r^8}}$ , where $\mathrm{r}$ is distance between the particle and $\mathrm{m}$ and $\mathrm{n}$ are positive constant. If the system is in equilibrium at $r=\mathrm{r_0}$ , then $\mathrm{r_0}$ in terms of $\mathrm{m}$ and $\mathrm{n}$ -
Option: 1
$\mathrm{r_0^3=2 \sqrt{\frac{n}{m}}}$

Option: 2
$\mathrm{r_0^3=2 \frac{n}{m}}$

Option: 3
$\mathrm{r_0^3= 4\frac{ n}{m}}$

Option: 4
$\mathrm{r_0^3=8 \frac{n}{m}}$

Answers (1)

best_answer

For equilibrium -

$\mathrm{\left.\frac{d V}{d r}\right|_{r=r_0}=0 }$

$\mathrm{\frac{d}{d r}\left[-m r^{-2}+n r^{-8}\right]_{r=r_0}=0}$

$\mathrm{ {\left[2 m r^{-3}-8 n r^{-9}\right]_{r=r_0}=0}}$

$\mathrm{2 m r_0^{-3}=8 n r_0^{-9}}$

$\mathrm{\frac{r_0^{-3}}{r_0^{-9}}=\frac{8 n}{2 m}}$

$\mathrm{r_0^{-3+9}=\frac{4 n}{m}}$

$\mathrm{ r_0^6=\frac{4 n}{m}}$

$\mathrm{ \Rightarrow r_0^3=2 \sqrt{\frac{n}{m}}}$

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