An air bubble of radius 0.1 cm is in a liquid having surface tension 0.06 N/m and density 103 kg/m3. The pressure inside the bubble is 1100 Nm ⁻² greater than the atmospheric pressure. At what depth is the bubble below the surface of the liquid? (g = 9.8 ms ⁻²) (a) 0.1 m (b) 0.15 m (c) 0.20 m (d) 0.25 m

Answer: (a). An air bubble in a liquid has one surface: Pin - P₀ = hρ g + 2T/r. 2T/r = 2 × 0.0610⁻³ = 120 Pa hρ g = 1100 - 120 = 980 Pa h = (980)/(10³ × 9.8) = 0.1 m So the answer is (a) 0.1 m.

An air bubble of radius 0.1 cm is in a liquid having surface tension… | JEE Main 2014 (11 Apr, Shift 1)

Q58JEE Main 2014 · 11 APR · SHIFT 1Medium

Excess Pressure: Drops & Bubbles

An air bubble of radius 0.1 cm is in a liquid having surface tension 0.06 N/m and density 103 kg/m3. The pressure inside the bubble is 1100 Nm ${}^{-2}$ greater than the atmospheric pressure. At what depth is the bubble below the surface of the liquid? (g = 9.8 ms ${}^{-2}$ )

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Answer: (a)

An air bubble in a liquid has one surface: $P_{in} - P_0 = h\rho g + \frac{2T}{r}$ .
$\frac{2T}{r} = \frac{2 \times 0.06}{10^{-3}} = 120\text{ Pa}$
$h\rho g = 1100 - 120 = 980\text{ Pa}$
$h = \frac{980}{10^3 \times 9.8} = 0.1\text{ m}$
So the answer is (a) $0.1\text{ m}$ .

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