Question

The excess pressure inside a soap bubble a is twice that in another soap bubble b the ratio of volumes of a and b is

Answer 1

Answer:

Excess pressure inside the soap bubble = 4 T/ R

Pa = 2 Pb

4T/R = 8T/r

4/8 = R/r

R/r = 1/2

volume = 4/3 π r ^3

$volume \: \alpha \: {r}^{3}$

$\frac{v1}{v2} = \: \frac{1}{ \sqrt[3]{2} }$