Question

The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is

• A. 4
• B. 2
• C. 1
• D. 0.125

Answer 1

The correct option is D

0.125

Let the surface tension of soap solution be S
Now for bubble 1,$\Delta P_1=\frac{4S}{r_1}$
and for bubble 2, $\Delta P_2=\frac{4S}{r_1}$
given,$\Delta P_1=2\Delta P_2.......\left(1\right)$
also, $V_1$=$n{V}_2$
$\frac{4}{3}\pi r_1^3=n\times \frac{4}{3}\pi r_2^3$
$r_1=(n{)}^{1/3}{r}_2....(2)$
from(1),
$\frac{4S}{r_1}=\frac{8S}{r_2}$
$r_2=2r_1$
utilize in (2),
$n^{1/3}=\frac{1}{2}$
$n=\frac{1}{8}=0.125$