A particle is projected from a point $P(\sqrt{3}R,0)$  along the axis of an annular uniform disc of mass M and inner and outer radii $R$ and $\sqrt{6}R$  respectively as shown in the figure. If the escape velocity of the particle from this point is  $2\sqrt{\frac{GM}{N\times R}}$, find $N$

Potential at point P due to elementary is equal to $dv=-\frac{Gdm}{x}$ where
$dm=\frac{M}{\pi [{\left(\sqrt{6}R\right)}^2-R^2]}2\pi rdr=\frac{2}{5}M\frac{rdr}{R^2}$

Hence $dV=\frac{-G2Mrdr}{x5R^2}$
as  $r^2+{\left(\sqrt{3}R\right)}^2=x^2$
We can write,  2rdr = 2xdx
Hence   $dV=\frac{-G2M}{5R^{2}x}xdx=\frac{-2}{5}\frac{GM}{R^2}dx$
$V=\underset{x_1}{\overset{x_2}{\int }}\frac{-2}{5}\frac{GM}{R^2}dx=-\frac{2}{5}\frac{GM}{R^2}(x_2-x_1)$
Here,  $x_2=\sqrt{{\left(\sqrt{3}R\right)}^2+{\left(\sqrt{6}R\right)}^2}=3R$
$x_1=\sqrt{{\left(\sqrt{3}R\right)}^2+R^2}=2R$
Hence $V=-\frac{2}{5}\frac{GM}{R}$
To escape a particle to infinity  $\frac{1}{2}m{v}^2\ge \frac{2}{5}\frac{GM}{R}m$
Hence   $v\ge \sqrt{\frac{4GM}{5R}}$