Question

A soap bubble of radius $1.0cm$ is formed inside another soap bubble of radius $2.0cm$. The radius of an another soap bubble which has the same pressure difference as that between the inside of the smaller and outside of large soap bubble, in meters is

• A. $6.67\times {10}^{-3}$
• B. $3.34\times {10}^{-3}$
• C. $2.23\times {10}^{-3}$
• D. $4.5\times {10}^{-3}$

Answer 1

Answer: (A)

$6.67\times {10}^{-3}$

Given : $a=1cm$ $b=2cm$
Let surface tension of soap be $T$.
For bubble of radius $a$ :
$P_2-P_3=\frac{4T}{a}$ .....(1)
For bubble of radius $b$ :
$P_1-P_2=\frac{4T}{b}$ .....(2)
Adding (1) and (2), we get $P_1-P_3=\frac{4T}{a}+\frac{4T}{b}$
$\therefore$ $\frac{4T}{R}=\frac{4T}{a}+\frac{4T}{b}$
Or $\frac{1}{R}=\frac{1}{a}+\frac{1}{b}$
$\therefore$ $\frac{1}{R}=\frac{1}{2}+\frac{1}{1}$
We get $R=\frac{2}{3}cm=6.67\times {10}^{-3}m$