Question

A spherical drop of water has radius $r\text{m}$. If surface tension of water is $70\times {10}^{-3}\text{N/m}$ and the pressure difference between inside and outside of the spherical drop is $280{\text{N/m}}^2$. Find the value of $r$.

• A. $1\text{mm}$
• B. $0.5\text{mm}$
• C. $0.4\text{mm}$
• D. $0.2\text{mm}$

Answer 1

The correct option is B $0.5\text{mm}$

We know,
For spherical drop of water,
Pressure difference, $\Delta P=\frac{2T}{r}$
where,
$T=$ Surface tension of water
$r=$ radius of spherical drop
$\therefore 280=\frac{2\times 70\times {10}^{-3}}{r}$
$\Rightarrow r=\frac{2\times 70\times {10}^{-3}}{280}\text{m}=0.5\text{mm}$