Question

A spherical soap bubble of radius 1 cm is formed inside another bubble of radius 3 cm The radius of a single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is

• A. 0.75 cm
• B. 0.75 m
• C. 7.5 cm
• D. 7.5 m

Answer 1

Answer: (A)

Pressure outside the bigger drop $=P_{1}$
Pressure inside the bigger drop $=P_{2}$
Radius of bigger drop, $r_{1}=3 \,cm$
Excess pressure $=P_{2}-P_{1}=\frac{4 S}{r_{1}}=\frac{4 S}{3}$
Pressure inside the smaller drop $=P_{3}$
Excess pressure $=P_{3}-P_{2}=\frac{4 S}{r_{2}}=\frac{4 S}{1}$
Pressure difference between inner side of smaller drop and outer side of bigger drop
$=P_{3}-P_{1}=P_{3}-P_{2}+P_{2}-P_{1}=\frac{4 S}{1}+\frac{4 S}{3}=\frac{16 S}{3}$
This pressure difference should exist in a single drop of radius $r$.
$\therefore \frac{4 S}{r}=\frac{16 S}{3} $
or $r=\frac{3}{4}\, cm$
$ =0.75 \,cm$