Question

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

• A. $0.32 \,N / cm ^{2}$
• B. $1.18 \,N / cm ^{2}$
• C. $1.82\, N / cm ^{2}$
• D. $1.66 \,N / cm ^{2}$

Answer 1

Answer: (D)

Given, diameter of water droplet,
$d=0.2 \,mm$
$\therefore $ radius, $r=0.1 \,mm =10^{-4} \, m$
Pressure outside droplet,
$p_{0}=1.5 \,N / cm ^{2}=1.5 \times 10^{4} \,N / m ^{2}$
Surface tension,
$T=0.08 \,N / m$
$\therefore $ Pressure inside droplet,
$p=\frac{2 T}{r}+p_{0}$
Putting the given values, we get
$=\frac{2 \times 0.08}{10^{-4}}+1.5 \times 10^{4}$
$=0.16 \times 10^{4}+1.5 \times 10^{4}$
$=1.66 \times 10^{4} \, N / m ^{2}$
$=1.66 \,N / cm ^{2}$