A spherical soap bubble inside an air chamber at pressure $P_0 = 10^5$ Pa has a certain radius so that the excess pressure inside the bubble is $\Delta P = 144$ Pa. Now, the chamber pressure is reduced to $8P_0/27$ so that the bubble radius and its excess pressure change. In this process, all the te...

The correct numerical answer is 96. Click to view the complete step-by-step calculation.

JEE Advanced 2024 Paper 2 Physics Question 13 Solution

Let $P_1$ and $P_2$ be the chamber pressures. $P_1 = P_0$ , $P_2 = \frac{8}{27}P_0$ . Let $r_1, r_2$ be radii.
Excess pressure $\Delta P = \frac{4T}{r}$ . Given $\Delta P_1 = 144$ Pa.
Pressure inside bubble $P_{in} = P_{chamber} + \Delta P$ . Since $\Delta P \ll P_{chamber}$ , $P_{in} \approx P_{chamber}$ .
For constant temperature and constant moles of air inside: $P_{in1} V_1 = P_{in2} V_2 \Rightarrow P_1 r_1^3 = P_2 r_2^3$ .
$P_0 r_1^3 = \left( \frac{8}{27} P_0 \right) r_2^3 \Rightarrow \frac{r_2^3}{r_1^3} = \frac{27}{8} \Rightarrow \frac{r_2}{r_1} = \frac{3}{2}$ .
The new excess pressure $\Delta P_2 = \frac{4T}{r_2}$ .
$\frac{\Delta P_2}{\Delta P_1} = \frac{r_1}{r_2} = \frac{2}{3}$ .
$\Delta P_2 = \frac{2}{3} \times 144 = 2 \times 48 = 96$ Pa.