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

Suppose, R = radius of water dropand r = radius of droplets∴ 43πR3=8×43πr3(Since volume remains constant)∴ r=R2Excess pressure inside big drop=2TRExcess pressure inside small drop=2Tr=2TR2=4TR( T-surface tension, R-radius)Therefore, pressure difference between inner and outer surface of big drop will be half of that for smaller droplet.