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

  $W=\Delta U=U_f-U_i=U_{\infty }-U_P$  

$=-U_P=m{V}_P=-V_P$    $(asm=1)$
Upon integration, potential at point P will be obtained as given below Let dM be the mass of small ring as shown
 $dM=\frac{M}{\pi {\left(4R\right)}^2-\pi {\left(3R\right)}^2}\left(2\pi r\right)dr=\frac{2Mrdr}{7R^2}$
$d{V}_p=-\frac{GdM}{\sqrt{16R^2+r^2}}$
 

$=-\frac{2GM}{7R^2}\underset{3R}{\overset{4R}{\int }}\frac{r}{\sqrt{16R^2+r^2}}dr$

  $\therefore W=+\frac{2GM}{7R}\left(4\sqrt{2-}5\right)$