 Solution

Define ${\gamma }_{SV}$, ${\gamma }_{SL}$ and ${\gamma }_{LV}$ as the solid-vapor, solid-liquid and liquid-vapor interfacial free energies per actual area. For the corresponding planar material, Young's contact-angle equation isConsider advancing the macroscopic liquid footprint over projected area $A$, while retaining the specified wetting only to each sphere's equator. The newly wetted upper hemispheres have actual area $2\varphi A$. The gaps occupy projected area $(1-\varphi )A$, across which a liquid-vapor interface remains instead of solid-liquid contact.

The microscopic free-energy change, apart from the outer macroscopic droplet surface, isEquating this with $-{\gamma }_{LV}\cos {\theta }^{\ast }$, the macroscopic wetting contribution, yieldsThe factor $2$ comes from the curved hemisphere area; the negative gap term is the trapped liquid-vapor interface. This combines Wenzel roughness on the wetted solid fraction with Cassie-type air gaps. It differs from both homogeneous Wenzel wetting and a Cassie formula that treats the solid contacts as flat.

For $\varphi =0.9069$ and $\theta =68^{\circ }$, this gives $\cos {\theta }^{\ast }\simeq 0.5864$ and ${\theta }^{\ast }\simeq 54.1^{\circ }$. The hybrid state is less wettable than the fully penetrated hemispherical model, but still not hydrophobic. The result is conditional on the assumed equatorial meniscus and retained air: pinning or metastability may be required, and allowing further penetration can change the equilibrium geometry.