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

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Fluid Mechanics

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 / m^{3}

. If the pressure inside the bubble is

2100 N / m^{2}

greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use

g = 10 m / s^{2}

)

Answer: (D)

For an air bubble of radius

R

at depth

h

inside a liquid, the excess pressure inside the bubble over the surrounding liquid pressure comes from a single curved surface:

P_{in} - P_{l i q u i d a t d e pt h} = \frac{2 T}{R}

The liquid pressure at depth

h

exceeds atmospheric pressure by

ρ g h

, so

P_{in} - P_{0} = ρ g h + \frac{2 T}{R}

With

P_{in} - P_{0} = 2100

N/m²,

ρ = 1000

kg/m³,

g = 10

m/s²,

h = 0.20

R = 0.1

= 1 0^{- 3}

ρ g h = 1000 \times 10 \times 0.2 = 2000

N/m²

\frac{2 T}{R} = 2100 - 2000 = 100

T = \frac{100 \times R}{2} = \frac{100 \times 1 0 ^{- 3}}{2} = 0.05

N/m

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option