Verified Solution • 2025–26 Session moderate

QPages 34-37: Q25: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q25 Solution: An air bubble of radius 2.0 mm is formed at the bottom of a 3.3 …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often apply $P_1V_1=P_2V_2$ but then set $r_2 = r_1\,P_1/P_2$, forgetting that $V\propto r^3$. The correct relation is $r_2 = r_1\,(P_1/P_2)^{1/3}$; always take the cube root when converting volume changes to radius changes.

An air bubble of radius 2.0 mm is formed at the bottom of a 3.3 m deep river. Calculate the radius of the bubble as it comes to the surface. Atmospheric pressure =

1.0 \times 10^5 \text{ Pa}

and density of water =

1000 \text{ kg m}^{-3}

.

Step-by-Step Verified Solution

**Answer:**

r = 2.2 \text{ mm}

**Step-by-Step Explanation:** At bottom:

P_1 = P_{atm} + \rho g h = 10^5 + 1000 \times 10 \times 3.3 = 1.33 \times 10^5

Pa.

V_1 = \frac{4}{3}\pi r_1^3

. At surface:

P_2 = 10^5

Pa.

V_2 = \frac{4}{3}\pi r_2^3

.

P_1 V_1 = P_2 V_2 \Rightarrow r_2 = r_1 \left(\frac{P_1}{P_2}\right)^{1/3} = 2.0 \times \left(\frac{1.33}{1.0}\right)^{1/3} = 2.2 \text{ mm}

All Questions in this Chapter

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