Question

If difference of pressure between inside and outside of the spherical drop is $70{\text{N/m}}^2$, then what is the radius of spherical drop ?
Given: Surface tension of water is $70\times {10}^{-3}\text{N/m}$.

• A. $1\text{mm}$
• B. $2\text{mm}$
• C. $0.2\text{mm}$
• D. $0.1\text{mm}$

Answer 1

The correct option is B $2\text{mm}$

Pressure difference between inside and outside of the spherical drop is,
$\Delta P=\frac{2T}{r}$
$\Delta P=70{\text{N/m}}^2$
$T=70\times {10}^{-3}\text{N/m}$
$r=$ Radius of drop
$\Rightarrow r=\frac{2T}{\Delta P}=\frac{2\times 70\times {10}^{-3}}{70}=2\times {10}^{-3}\text{m}$
$\therefore r=2\text{mm}$