JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Fluid Mechanics

The half-immersed droplet is supported by two upward effects, the buoyant thrust on its submerged half and the surface-tension pull around the contact circle, balancing its full weight. Writing the weight as $d\cdot \frac{4}{3}\pi r^{3}g$, the buoyancy on the lower half as $\rho \cdot \frac{2}{3}\pi r^{3}g$, and the surface-tension force as $T\cdot 2\pi r$, equilibrium requires $d\cdot \frac{4}{3}\pi r^{3}g=\rho \cdot \frac{2}{3}\pi r^{3}g+2\pi rT$. Collecting the volume terms gives $\frac{2}{3}\pi r^{3}g(2d-\rho )=2\pi rT$, so $r=\sqrt{\frac{3T}{(2d-\rho )g}}$.