Physics - Fluid Question with Solution | TestHub

Excess pressure inside a soap bubble is given by $\Delta P=\frac{4T}{R}$

Let the atmospheric pressure be $P_0=1$ atm.

For bubble A, excess pressure $\Delta P_A=1.01-1=0.01$ atm.

For bubble B, excess pressure $\Delta P_B=1.02-1=0.02$ atm.

Since $\Delta P\propto \frac{1}{R}$, we have $R\propto \frac{1}{\Delta P}$

The volume of a bubble is $V=\frac{4}{3}\pi R^3$, so $V\propto \frac{1}{(\Delta P{)}^3}$

The ratio of volumes is $\frac{V_A}{V_B}={\left(\frac{\Delta P_B}{\Delta P_A}\right)}^3={\left(\frac{0.02}{0.01}\right)}^3=2^3=8$

Thus, the ratio of volume of A to that of B is $8:1$.

Answer: $8:1$