An air bubble of radius 1.0 mm is observed at a depth 20 cm below the free surface of a liquid having surface tension 0.095 J/m² and density 10³ kg/m³. The difference between pressure inside the bubble and atmospheric pressure is __________ N/m². (Take g = 10 m/s²) Numerical answer type. Work it out, then open the answer.

Answer: 2190. Air bubble in liquid (one surface): Pin - P₀ = hρ g + 2T/r. hρ g = 0.2 × 10³ × 10 = 2000 Pa 2T/r = 2 × 0.09510⁻³ = 190 Pa Difference = 2000 + 190 = 2190 N/m²

An air bubble of radius 1.0 mm is observed at a depth 20 cm below the… | JEE Main 2025 (23 Jan, Shift 2)

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Excess Pressure: Drops & Bubbles

An air bubble of radius 1.0 mm is observed at a depth 20 cm below the free surface of a liquid having surface tension $0.095\text{ J/m}^2$ and density $10^3\text{ kg/m}^3$ . The difference between pressure inside the bubble and atmospheric pressure is __________ $\text{N/m}^2$ . (Take $g = 10\text{ m/s}^2$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 2190

Air bubble in liquid (one surface): $P_{in} - P_0 = h\rho g + \frac{2T}{r}$ .
$h\rho g = 0.2 \times 10^3 \times 10 = 2000\text{ Pa}$
$\frac{2T}{r} = \frac{2 \times 0.095}{10^{-3}} = 190\text{ Pa}$
Difference $= 2000 + 190 = 2190\text{ N/m}^2$

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