Question

Consider a water droplet of diameter $0.2mm$ where the outside pressure is $1.5N/c{m}^2$ at $25^{\circ }C.$ The pressure inside the droplet, when the surface tension at $25^{\circ }C$ is $0.08N/m$ is

• A. $0.32N/c{m}^2$
• B. $1.18N/c{m}^2$
• C. $1.82N/c{m}^2$
• D. $1.66N/c{m}^2$

Answer 1

Answer: (D)

Given, diameter of water droplet,
$d=0.2mm$
$\therefore$ radius, $r=0.1mm=10^{-4}m$
Pressure outside droplet,
$p_0=1.5N/c{m}^2=1.5\times 10^{4}N/m^2$
Surface tension,
$T=0.08N/m$
$\therefore$ Pressure inside droplet,
$p=\frac{2T}{r}+p_0$
Putting the given values, we get
$=\frac{2\times 0.08}{10^{-4}}+1.5\times 10^4$
$=0.16\times 10^4+1.5\times 10^4$
$=1.66\times 10^{4}N/m^2$
$=1.66N/c{m}^2$