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

Let us consider a circular elemental area of radius $x$and thickness $dx.$The area of the shaded portion = $2\pi xdx.$ Let $dm$be the mass of the shaded portion.

Therefore     $\frac{Mass}{area}=\frac{M}{\pi \left(4R^2\right)-\pi \left(3R^2\right)}=\frac{dm}{2\pi xdx}$

                    $dm =\frac{2M}{7R^2}xdx$

The gravitational potential of the mass $dm$ at $P$is

$dV =\frac{-G dm}{\sqrt{{\left(4R\right)}^2+x^2}}=-\frac{G}{\sqrt{16R^2+x^2}}\times \frac{2M}{7R^2}xdx$

$= \frac{-2GM}{7R^2}\frac{xdx}{\sqrt{16R^2+x^2} } \left(1\right)$

Suppose $16 R^2 +x^2=t^2$

               $$\begin{gathered} \Rightarrow 2xdx =2tdt \\ \Rightarrow xdx =tdt \end{gathered}$$

Also for $x =3R, t=5R$

and for  $x = 4R, t =4\sqrt{2}R$

On integrating equation (1), taking the above limits, we get

$$\begin{gathered} V = -{\int }_{5R}^{4\sqrt{2}R}\frac{2GM}{7R^2}dt = \frac{-2GM}{7R^2}{\left[t\right]}_{5R}^{4\sqrt{2}R} \\ = \frac{-2GM}{7R^2}\left[4\sqrt{2}R-5R\right] = V = \frac{-2GM}{7R}\left(4\sqrt{2}-5\right) \end{gathered}$$

Now $\frac{W_{P\infty }}{1}=V_{\infty }-V_P=-V_P$

Therefore    $W_{P\infty }=\frac{2GM}{7R}(4\sqrt{2}-5)$