An air bubble of volume 2.9 cm³ rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is 17. The volume of the bubble when it reaches the surface, where the water temperature is 27, is cm³. ( g = 10 m/s², density of water = 10³ kg/m³, and 1 atm pressure is 10⁵ Pa) (a) 2.0 (b) 4.2 (c) 3.0 (d) 4.5

Answer: (d). Pressure at bottom P₁ = P₀ + ρ g h = 10⁵ + 10³ × 10 × 5 = 1.5 × 10⁵ Pa; at surface P₂ = 10⁵ Pa. (P₁V₁)/(T₁) = (P₂V₂)/(T₂) with T₁ = 290 K, T₂ = 300 K. V₂ = 2.9 × 1.5 × (300)/(290) = 4.5 cm³. So the answer is (d) 4.5.

An air bubble of volume 2.9 cm³ rises from the bottom of a swimming… | JEE Main 2026 (23 Jan, Shift 2)

Q14JEE Main 2026 · 23 JAN · SHIFT 2Medium

Gas Laws & Ideal Gas Equation

An air bubble of volume $2.9\text{ cm}^{3}$ rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is $17^\circ\text{C}$ . The volume of the bubble when it reaches the surface, where the water temperature is $27^\circ\text{C}$ , is $\underline{\qquad}\ \text{cm}^{3}$ .

( $g = 10\text{ m/s}^{2}$ , density of water $= 10^{3}\text{ kg/m}^{3}$ , and 1 atm pressure is $10^{5}\text{ Pa}$ )

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Answer: (d)

Pressure at bottom $P_1 = P_0 + \rho g h = 10^5 + 10^3 \times 10 \times 5 = 1.5 \times 10^5\text{ Pa}$ ; at surface $P_2 = 10^5\text{ Pa}$ .
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ with $T_1 = 290\text{ K}$ , $T_2 = 300\text{ K}$ .
$V_2 = 2.9 \times 1.5 \times \frac{300}{290} = 4.5\text{ cm}^3$ .
So the answer is (d) 4.5.

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