JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Heat & Thermodynamics

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

1 7^{\circ} C

. The volume of the bubble when it reaches the surface, where the water temperature is

2 7^{\circ} C

, is ____

cm^{3}

. (

g = 10 m / s^{2}

, density of water

= 1 0^{3} kg / m^{3}

, and 1 atm pressure is

1 0^{5} Pa

)

Answer: (D)

Solution

The gas in the bubble obeys

\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}

. At the bottom,

P_{1} = P_{a t m} + ρ g h = 1 0^{5} + 1 0^{3} \times 10 \times 5 = 1.5 \times 1 0^{5} Pa

V_{1} = 2.9 cm^{3}

and

T_{1} = 17 + 273 = 290 K

. At the surface,

P_{2} = 1 0^{5} Pa

and

T_{2} = 27 + 273 = 300 K

. Hence

V_{2} = V_{1} \frac{P _{1} T _{2}}{P _{2} T _{1}} = 2.9 \times \frac{1.5 \times 1 0 ^{5} \times 300}{1 0 ^{5} \times 290} = \frac{1.5 \times 2.9 \times 300}{290} = 4.5 cm^{3}

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option