Question

A spherical drop of water has $1mm$ radius. If the surface tension of the water is $50\times {10}^{-3}N/m$, then the difference of pressure between inside and outside the spherical drop is :

• A. $25N/m^2$
• B. $10000N/m^2$
• C. $100N/m^2$
• D. $50N/m^2$

Answer 1

Answer: (C)

$100N/m^2$

Pressure difference $P_{diff}=\frac{2T}{R}=\frac{2\times 50\times {10}^{-3}}{1\times {10}^{-3}}=100$

Where,
$T=$ Surface Tension
$R=$ Radius of spherical drop