An air bubble of volume 2.9 cm 3 rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is 17 ° C . The volume of the bubble when it reaches the surface, where the water temperature is 27 ° C , is \ \ \ \ cm 3 . ( g =10 m / s 2 , density of water =10 3 kg / m 3 , and 1 atm pressure is 10 5 Pa )

Physics · Fluid Mechanics

JEE Main 2026 — 23 January, Evening Shift — Question 34

An air bubble of volume $2.9 \mathrm{~cm}^{3}$ rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is $17^{\circ} \mathrm{C}$ . The volume of the bubble when it reaches the surface, where the water temperature is $27^{\circ} \mathrm{C}$ , is $\_\_\_\_$ $\mathrm{cm}^{3}$ . ( $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ , density of water $=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ , and 1 atm pressure is $10^{5} \mathrm{~Pa}$ )

Option A:
4.2
Option B:
2
Option C:
3
Option D:
4.5
Correct

Answer: D

Step-by-step solution

For an air bubble rising in water, the no. of moles remain constant $\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}$ $\frac{\left(\mathrm{P}_{\mathrm{atm}}+\rho \mathrm{gh}\right) 2.9 \mathrm{~cm}^{3}}{290 \mathrm{~K}}=\frac{\left(\mathrm{P}_{\mathrm{atm}}\right) \mathrm{V}_{2}}{300}$ $\mathrm{V}_{2}=4.5 \mathrm{~cm}^{3}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Evening Shift
Subject: Physics
Chapter: Fluid Mechanics
Topic: Buoyancy and Archimedes' Principle