Skip to content
dronacharya.ai
← JEE Advanced 2024 · Paper 2
PhysicsInteger+4Question 30 of 51

JEE Advanced 2024 · Paper 2, Physics question 13

A spherical soap bubble inside an air chamber at pressure P0=105 PaP_0 = 10^5\,\mathrm{Pa} has a certain radius so that the excess pressure inside the bubble is ΔP=144 Pa\Delta P = 144\,\mathrm{Pa}. Now, the chamber pressure is reduced to 8P0/278P_0/27 so that the bubble radius and its excess pressure change. In this process, all the temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure ΔP\Delta P in both the cases to be much smaller than the chamber pressure. The new excess pressure ΔP\Delta P in Pa is _____.

Answer and worked solution

Official answer: 96

For a soap bubble (two surfaces), ΔP=4Tr\Delta P=\frac{4T}{r}.

Since ΔP≪P0\Delta P\ll P_0, the air inside is at about the chamber pressure. The trapped air expands isothermally:
P0r3=8P027r′3  ⟹  r′=32r.P_0r^3=\frac{8P_0}{27}r'^3\implies r'=\frac32r.

So the excess pressure falls by the same factor:
ΔP′=144⋅23=96 Pa.\Delta P'=144\cdot\frac{2}{3}=96\ \mathrm{Pa}.