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  • The potential energy of a system of two particles is given by <img alt="v=-\frac{m}{r^2}+\frac{n}{r^8}" src="https://entrancecorner.oncodecogs.com/gif.latex?v%3D-%5Cfrac%7Bm%7D%7Br%5E2%7D&plus;

The potential energy of a system of two particles is given by v=-\frac{m}{r^2}+\frac{n}{r^8}, where r is distance between the particle and m and n are positive constant. If the system is in equilibrium at r=r_0 then r_0 in terms of m and 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 -
\mathrm{\begin{aligned} & \left.\frac{d V}{d r}\right|_{r=r_0}=0 \\ & \frac{d}{d r}\left[-m r^{-2}+n r-8\right]_{r=r_0}=0 \\ & {\left[2 m r^{-3}-8 n r^{-9}\right]_{r=r_0=0}} \\ & 2 m r_0^{-3}=8 n r_0^{-9} \\ & \frac{r_0^{-3}}{r_0^{-9}}=\frac{8 n}{2 m} \\ & r_0^{-3+9}=\frac{4 n}{m} \\ & r_0^6=\frac{4 n}{m} \Rightarrow r_0^3=\sqrt[2]{\frac{n}{m}} \text { Ans. } \end{aligned} }

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