A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is  $\frac{2\mathrm{GM}}{\mathrm{xR}}(4\sqrt{2}-5)$  Find the value of x.

$\sigma =\frac{M}{A}; \sigma =\frac{M}{16\pi R^2-9\pi R^2} ; \sigma =\frac{M}{7\pi R^2}$

Potential due to small ring of mass dm at p is

$\mathrm{dV}=\frac{-\mathrm{Gdm}}{\mathrm{r}}=\frac{-\mathrm{G}2\pi \mathrm{xdx}\cdot \sigma }{\mathrm{r}}$

$\mathrm{dV}=-2\pi \mathrm{G}\sigma \frac{\mathrm{xdx}}{\sqrt{16{\mathrm{R}}^2+{\mathrm{x}}^2}}$

$\therefore$  Total potential at P,

$V_p={\int }_{3R}^{4R}dV; V_P=-2\pi G\sigma {\int }_{3R}^{4R}\frac{xdx}{\sqrt{16R^2+x^2}}$

$V_p=-2\pi \frac{GM}{7\pi R^2}[\sqrt{16R^2+16R^2}-\sqrt{16R^2+9R^2}]$

${\mathrm{V}}_{\mathrm{P}}=\frac{-2\mathrm{GM}}{7\mathrm{R}}[4\sqrt{2}-5]; {\mathrm{V}}_{\propto }=0\mathrm{W}=\mathrm{\Delta Vm}$

$\therefore \mathrm{W}=[{\mathrm{V}}_{\mathrm{\infty }}-{\mathrm{V}}_{\mathrm{P}}]$.......(1)

$\mathrm{W}=\frac{2\mathrm{GM}}{7\mathrm{R}}[4\sqrt{2}-5]$