Question

The excess pressure due to surface tension inside a spherical drop is $6$ units. If eight such drops combine, then the excess pressure due to surface tension inside the larger drop is

• A. 3 units
• B. 6 units
• C. 12 units
• D. 48 units

Answer 1

Answer: (A)

Let $P_1$ be the excess pressure inside the smaller drop, then,
$P_1-P_0=\frac{2\sigma }{r}=6$ units
Let $P_2$ be the excess pressure inside the larger drop.
$P_2-P_0=\frac{2\sigma }{R}=\Delta P$
Now, $8\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3\Rightarrow R=2r$
$\frac{\Delta P}{6}=\frac{r}{R}=\frac{1}{2}\Rightarrow \Delta P=\frac{6}{2}=3$ units.