Drops of liquid of density d are floating half immersed in a liquid of density p. If the surface tension of liquid is T, then radius of the drop will be

The weight of the drop is equal to the sum of force due to surface tension on drop and force of buoyancy. Thus   43πr3dg=2πrT+12×43πr3ρg or  2r2dg=3T+r2ρg r=3Tg(2d−ρ)