For the arrangement shown in the figure find the net gravitational field at point ‘O’ is

Gravitational field due to arc  $E_1=\frac{2G\left(2M\right)\sin \left(\pi /6\right)}{\frac{\pi }{3}{R}^2}$
$E_1=\frac{6GM}{\pi R^2}towardsleft$
Gravitational field due to a rod at the axial point 
$E_2=\frac{GM}{R}\left[\frac{1}{R}-\frac{1}{2R}\right]$ 
$E_2=\frac{GM}{2R^2}towardsright$ 
Net gravitational field
$E=E_1-E_2$   towards left  $\left[\because E_1>E_2\right]$
$=\frac{6GM}{\pi R^2}-\frac{GM}{2R^2}$

$=\frac{GM}{R^2}\left[\frac{6}{\pi }-\frac{1}{2}\right]$

$E=\frac{GM}{R^2}\left[\frac{12-\pi }{2\pi }\right]towardsleft$