Question

A spherical drop of water has $1\, mm$ radius. If the surface tension of water is $75 \times 10^{-3} N / m$, then difference of pressure between inside and outside of the drop is

• A. $35\, N / m ^{2}$
• B. $70\, N / m ^{2}$
• C. $140\, N / m ^{2}$
• D. $150\, N / m ^{2}$

Answer 1

Answer: (D)

Excess pressure $=\frac{2 T}{R}$
$=\frac{2 \times 75 \times 10^{-3}}{1 \times 10^{-3}}$
$=150\, N / m ^{2}$
$T=$ surface tension
$R=$ radius