Question

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

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

Answer 1

The correct option is D $150{\text{N/m}}^2$

Given,
Radius of spherical drop of water, $R=1\text{mm}={10}^{-3}\text{m}$
Surface Tension, $T=75\times {10}^{-3}\text{N/m}$
To find: excess pressure inside water droplet, $\Delta P=?$
We know that, excess pressure inside a water droplet is given by
$\Delta P=\frac{2T}{R}$
On putting all the data,
$\Delta P=\frac{2\times 75\times {10}^{-3}}{{10}^{-3}}=150{\text{N/m}}^2$