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 )

Physics · Mechanical Properties of Matter

JEE Main 2025 — 24 January, Morning Shift — Question 49

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 $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )

Option A:
0.02
Option B:
0.1
Option C:
0.25
Option D:
0.05
Correct

Answer: D

Step-by-step solution

T is surface tension

$P$ in air bubble $=P_{0}+\rho g h+\frac{2 T}{R}$

$\mathrm{P}_{\text {in }}-\mathrm{P}_{0}=\rho g h+\frac{2 \mathrm{~T}}{\mathrm{R}}=2100$

$\frac{2 T}{R}=2100-\rho g h$

$\mathrm{T}=\frac{\mathrm{R}}{2}\left(2100-10^{3} \times 10 \times 0.2\right)$

$=\frac{1}{20}(2100-2000) \times 10^{-2}$

$=0.05$

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Exam: JEE Main 2025
Paper: 24 January, Morning Shift
Subject: Physics
Chapter: Mechanical Properties of Matter
Topic: Surface Tension and Surface Energy

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Step-by-step solution

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