An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 kg/m³. If the pressure inside the bubble is 2100 N/m² greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use g = 10 m/s²) (a) 0.02 (b) 0.05 (c) 0.25 (d) 0.1

Answer: (b). An air bubble inside a liquid has one surface: Pin = P₀ + ρ g h + 2T/r. ρ g h = 1000 × 10 × 0.2 = 2000 N/m ², so 2T/r = 2100 - 2000 = 100 N/m ². T = 100 × 10⁻³2 = 0.05 N/m So the answer is (b) 0.05.

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the… | JEE Main 2025 (24 Jan, Shift 1)

Q18JEE Main 2025 · 24 JAN · SHIFT 1Medium

Excess Pressure: Drops & Bubbles

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density $1000\text{ kg/m}^3$ . If the pressure inside the bubble is $2100\text{ N/m}^2$ greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use $g = 10\text{ m/s}^2$ )

Show answer and solution

Answer: (b)

An air bubble inside a liquid has one surface: $P_{in} = P_0 + \rho g h + \frac{2T}{r}$ .
$\rho g h = 1000 \times 10 \times 0.2 = 2000$ N/m $^2$ , so $\frac{2T}{r} = 2100 - 2000 = 100$ N/m $^2$ .
$T = \frac{100 \times 10^{-3}}{2} = 0.05$ N/m
So the answer is (b) 0.05.

Report a mistake

Excess Pressure: Drops & Bubbles question from JEE Main 2025 (24 Jan, Shift 1)

More questions on Excess Pressure: Drops & Bubbles