Verified Solution • 2025–26 Session moderate

Q9.20: NCERT Physics - Part 2 MECHANICAL PROPERTIES OF FLUIDS 9.20 Solution: What is the excess pressure inside a bubble of soap solution of …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often apply the single‑interface formula ΔP=2γ/r to a soap bubble, forgetting it has two curved surfaces; the correct excess pressure is ΔP=4γ/r. Remember to use 4γ/r for a soap bubble and 2γ/r only when a single liquid–gas interface is present, and always add the ambient (atmospheric) pressure to the hydrostatic term.

What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 �C) is 2.50×10−22.50 \times 10^{-2} N m__? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01×1051.01 \times 10^5 Pa).

Step-by-Step Verified Solution

**Answer:** Excess pressure = 20 Pa; Inside pressure = 1.06×1051.06 \times 10^5 Pa. **Step-by-Step Explanation:** Excess pressure inside soap bubble (two surfaces):
ΔP=4Sr=4×0.0250.005=20 Pa\Delta P = \frac{4S}{r} = \frac{4 \times 0.025}{0.005} = 20 \text{ Pa}
Pressure at depth 40 cm in soap solution:
Pliquid=P0+ρgh=1.01×105+1.20×103×9.8×0.40=1.01×105+4704=1.057×105 PaP_{liquid} = P_0 + \rho g h = 1.01 \times 10^5 + 1.20 \times 10^3 \times 9.8 \times 0.40 = 1.01 \times 10^5 + 4704 = 1.057 \times 10^5 \text{ Pa}
Inside pressure of air bubble:
P=Pliquid+2Sr=1.057×105+2×0.0250.005=1.057×105+10=1.06×105 PaP = P_{liquid} + \frac{2S}{r} = 1.057 \times 10^5 + \frac{2 \times 0.025}{0.005} = 1.057 \times 10^5 + 10 = 1.06 \times 10^5 \text{ Pa}

All Questions in this Chapter

Q9.1 Q9.2 Q9.3 Q9.4 Q9.5 Q9.6 Q9.7 Q9.8 Q9.9 Q9.10 Q9.11 Q9.12 Q9.13 Q9.14 Q9.15 Q9.16 Q9.17 Q9.18 Q9.19 Q9.20

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