A water drop is divided into 8 equal droplets. The pressure difference between inner and outer sides of the big drop

Suppose, R = radius of water drop and r = radius of droplets43πR3 = 8×43πr3[Since, volume remains constant]r = R2Since, excess pressure inside drop = 2TR[T-surface tension, R-radius]Pressure difference between inner and outer surface of big drop will be half of that for smaller droplet.