Question

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

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

Answer 1

The correct option is C $3units$

Given: The excess pressure due to surface tension inside a spherical drop $=6units$
When eight drops combine, then a big drop will made and the volume of both can be written as
$8\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3;$
$R=2r$
As the pressure excess for a single surface is given by $P=2T/R$
Or $R=\frac{2T}{P}$
And for two surfaces $p=2\times 2T/R$
$R=2\times \frac{2T}{p}$
Now,
$\frac{2T}{p}=\frac{2\times 2T}{p}$
Or $P=\frac{p}{2}=\frac{6}{2}units=3units$
Final answer: $\left(a\right)$