Fluid Mechanics

JEE Main level, single correct, medium

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Question 5 of 192

An air bubble of volume 2.9 cm32.9 \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∘C17^{\circ} \mathrm{C}. The volume of the bubble when it reaches the surface, where the water temperature is 27∘C27^{\circ} \mathrm{C}, is ____\_\_\_\_ cm3\mathrm{cm}^{3}.
(g=10 m/s2\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right., density of water =103 kg/m3=10^{3} \mathrm{~kg} / \mathrm{m}^{3}, and 1 atm pressure is 105 Pa)\left.10^{5} \mathrm{~Pa}\right)

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Answer: (A) 4.5

An air bubble rises through water with initial volume 2.9 cm³ at the bottom. Calculating pressures at both locations:
At bottom (5 m depth): P1=Patm+ρgh=105+103(10)(5)=1.5×105P_1 = P_{atm} + \rho gh = 10^5 + 10^3(10)(5) = 1.5 \times 10^5 Pa
At surface: P2=105P_2 = 10^5 Pa
Initial temperature: T1=17°C=290T_1 = 17°C = 290 K, Final temperature: T2=27°C=300T_2 = 27°C = 300 K
Using ideal gas law: P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}
V2=V1×P1P2×T2T1=2.9×1.5×105105×300290V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 2.9 \times \frac{1.5 \times 10^5}{10^5} \times \frac{300}{290}
=2.9×1.5×3029=2.9×1.5×1.034≈4.5= 2.9 \times 1.5 \times \frac{30}{29} = 2.9 \times 1.5 \times 1.034 \approx 4.5 cm³

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