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

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
and `r=`radius of droplest
`(4)/(3)piR^3=8xx(4)/(3)pir^3`
[Since, volume remains constant]
`r=(R )/(2)`
Since, excess pressure inside drop`=(2T)/(R )`
`[T-`surface tension, `R-`radius]
Pressure difference between inner and out surface of big drop will be half of that for smaller droplet.