Physics - W.E.P. Question with Solution | TestHub

Given, $U=\frac{M}{r^6}-\frac{N}{r^{12}}$ $\begin{array}{l}\therefore \mathrm{F}=\frac{-\mathrm{du}}{\mathrm{dr}}=\frac{-\mathrm{d}}{\mathrm{dr}}\left(\frac{\mathrm{M}}{{\mathrm{r}}^6}-\frac{\mathrm{N}}{{\mathrm{r}}^{12}}\right)\\ =-\left(\frac{-6\mathrm{M}}{{\mathrm{r}}^7}+\frac{12\mathrm{N}}{{\mathrm{r}}^{13}}\right)=\left(\frac{6\mathrm{M}}{{\mathrm{r}}^7}-\frac{12\mathrm{N}}{{\mathrm{r}}^{13}}\right)\end{array}$ For equilibrium position, $\mathrm{F}=0$ $\therefore \frac{6\mathrm{M}}{{\mathrm{r}}^7}=\frac{12\mathrm{N}}{{\mathrm{r}}^{13}}\text{ or }{\mathrm{r}}^6=\frac{2\mathrm{N}}{\mathrm{M}}$ Hence, $U=\frac{M}{(2N/M)}-\frac{N}{(2N/M{)}^2}=\frac{M^2}{4N}$