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

Options

A. will be the same as for smaller droplet
B. will be half of that for smaller droplet
C. will be one-fourth of that for smaller droplet
D. will be twice of that for smaller droplet

Answer

B

Explanation

Suppose `R=`radius of water drop
`:. 4/3piR^(3)=8x4/3pir^(3)`
(since volume remains constant)
`:. r=R/2`
since excess pressure inside drop `=(2T)/R`
(T- surface tension, `R` -radius)
Therefore, pressure difference between inner and outer surface of bid drop will be half of that for smaller droplet.