A small soap bubble of radius 4 cm is trapped inside another bubble of radius 6 cm without any contact. Let P2 be the pressure inside the inner bubble and P0, the pressure outside the outer bubble. Radius of another bubble with pressure difference P2 - P0 between its inside and outside would be :

Question

A small soap bubble of radius 4 cm is trapped inside another bubble of radius 6 cm without any contact. Let P2 be the pressure inside the inner bubble and P0, the pressure outside the outer bubble. Radius of another bubble with pressure difference P2 - P0 between its inside and outside would be : (A) 12 cm (B) 2.4 cm (C) 6 cm (D) 4.8 cm

Tardigrade Admin · Accepted Answer

We know that excess pressure inside a soap bubble is

\frac{4 T}{r}

,
where

T

is surface tensions of soap bubble and

r

is radius of bubble.
Given:

r_{1} = 4 c m, r_{2} = 6 c m

. Now,

P_{2} = P_{0} + \frac{4 T}{r _{1}} + \frac{4 T}{r _{2}}

\Rightarrow P_{2} = P_{0} + 4 T [\frac{1}{r _{1}} + \frac{1}{r _{2}}]

\Rightarrow (P_{2} - P_{0}) = 4 T [\frac{1}{4} + \frac{1}{6}]

\Rightarrow (P_{2} - P_{0}) = 4 T \frac{( 3 + 2 )}{12}

\Rightarrow (P_{2} - P_{0}) = 4 T \times \frac{5}{12}

\Rightarrow \frac{4 T}{( \frac{12}{5} )}

Let a single bubble to have this excess pressure has radius

r

.
Therefore, on comparison

\frac{4 T}{r} = \frac{4 T}{\frac{12}{6}}

\Rightarrow r = \frac{12}{5} = 2.4 c m