A uniform hemispherical shell placed on an inclined plane of inclination $\theta$ is at rest. The angle made by the circular face of the shell with horizontal is $\varphi ={\sin }^{-1}\left(n\sin \theta \right).$ The value of $n$ is

 

For rotational equilibrium net torque about any point due to normal reaction, weight and friction should be zero. For this weight should pass through point of contact. Therefore centre of mass of the shell should be above the point of contact on the vertical line passing through the point of contact. Centre of mass of the hemispherical shall is at a distance $\frac{R}{2}$ from the centre of the shell

$From\Delta ACG, \frac{CG}{\sin \theta }=\frac{CA}{\sin (180-\varphi )}\Rightarrow \frac{ }{\sin \theta }=\frac{R}{\sin \varphi }\Rightarrow \sin \varphi =2\sin \theta \Rightarrow \varphi ={\sin }^{-1}\left(2\sin \theta \right)$