A thin uniform annular disc( see figure) of mass M has an outer radius 4R and an inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is
 

 

Mass of an elemental ring of thickness dr and radius r is 

$\mathrm{dm}=\mathrm{\sigma }2\mathrm{\pi rdr}=\frac{\mathrm{M}\times 2\mathrm{\pi rdr}}{\mathrm{\pi }(16{\mathrm{R}}^2-9{\mathrm{R}}^2)}=\frac{2\mathrm{Mrdr}}{7R^2}$

$\mathrm{V}=-\underset{3\mathrm{R}}{\overset{4\mathrm{R}}{\int }}\frac{\mathrm{Gdm}}{\sqrt{{\mathrm{r}}^2+16{\mathrm{R}}^2}}=-\frac{2\mathrm{MG}}{7{\mathrm{R}}^2}{\int }_{3\mathrm{R}}^{4\mathrm{R}}\frac{\mathrm{rdr}}{\sqrt{{\mathrm{r}}^2+16{\mathrm{R}}^2}}$

Solving the above equation gives

$-V=\frac{2GM}{7R}(4\sqrt{2}-5)$

work required to take a unit mass from P to infinity 

$W=0-V=-V=\frac{2GM}{7R}(4\sqrt{2}-5)$