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

Physics · Mechanical Properties of Matter

JEE Main 2025 — 23 January, Evening Shift — Question 67

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

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

Answer: 2190

Numerical answer — enter this value.

Step-by-step solution

IMAGES

$\Delta \mathrm{P}=\mathrm{P}_{\text {in }}-\mathrm{P}_{0}$

$=\rho g h+\frac{2 \mathrm{~T}}{\mathrm{R}}=\frac{1000 \times 10 \times 20}{100}+\frac{2 \times 0.095}{10^{-3}}$

$=2000+190$

$=2190$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Physics
Chapter: Mechanical Properties of Matter
Topic: Surface Tension and Surface Energy