Consider a water droplet of diameter 0.2 mm where the outside pressure is `1.5N//cm^(2)` at `25^(@)C`. The pressure inside the droplet, when the surface tension at `25^(@)C` is 0.08 N/m is

Options

A. `0.32N//cm^(2)`
B. `1.18N//cm^(2)`
C. `1.82N//cm^(2)`
D. `1.66N//cm^(2)`

Answer

D

Explanation

Given, diameter of water droplet,
d=0.2 mm
`:.` radius, `r=0.1mm=10^(-4)m`
Pressure outside droplet,
`p_(0)=1.5N//cm^(2)=1.5xx10^(4)N//m^(2)`
Surface tension,
`T=0.08N//m`
`:.` Pressure inside droplet,
`p=(2T)/(r)+p_(0)`
Putting the given values, we get
`=(2xx0.08)/(10^(-4))+1.5xx10^(4)`
`=0.16xx10^(4)+1.5xx10^(4)`
`=1.66xx10^(4)N//m^(2)`
`=1.66N//cm^(2)`