JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2026
- Hard
- MCQ
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 \(17^{\circ} C\). The volume of the bubble when it reaches the surface, where the water temperature is \(27^{\circ} C\), is ___________ \(cm ^3\). ( \(g =10 m / s ^2\), density of water \(=10^3 kg / m ^3\), and 1 atm pressure is \(10^5 Pa\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 4.5
Step-by-step Solution
Detailed explanation
For an air bubble rising in water, the no. of moles remain constant
\(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)
\(\frac{\left( P _{ atm }+\rho gh \right) 2.9 cm^3}{290 K}=\frac{\left( P _{ atm }\right) V _2}{300}\)
\(V _2=4.5 cm^3\)
\(\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}\)
\(\frac{\left( P _{ atm }+\rho gh \right) 2.9 cm^3}{290 K}=\frac{\left( P _{ atm }\right) V _2}{300}\)
\(V _2=4.5 cm^3\)
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