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, $8\left(\frac{4}{3} \pi r^{3}\right)=\frac{4}{3} \pi R^{3} \Rightarrow R=2 r$
$\frac{\Delta P}{6}=\frac{r}{R}=\frac{1}{2} \Rightarrow \Delta P=\frac{6}{2}=3$ units.