Verified Solution • 2025–26 Session moderate

QPages 34-37: Q40: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q40 Solution: Figure shows a cylindrical tube of radius 5 cm and length 20…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often confuse the total normal force $N$ with the linear force density $dN/dl$; you must divide the total normal force required to balance the gas pressure by the cork's circumference ($2\pi r$) to find the force per unit length.

Figure shows a cylindrical tube of radius 5 cm and length 20 cm. It is closed by a tight-fitting cork. The friction coefficient between the cork and the tube is 0.20. The tube contains an ideal gas at a pressure of 1 atm and a temperature of 300 K. The tube is slowly heated and it is found that the cork pops out when the temperature reaches 600 K. Let

dN

denote the magnitude of the normal contact force exerted by a small length

dl

of the cork along the periphery see the figure). Assuming that the temperature of the gas is uniform at any instant, calculate

\frac{dN}{dl}

.

Step-by-Step Verified Solution

**Answer:**

\frac{dN}{dl} = 125.6 \text{ N/m}

**Step-by-Step Explanation:** At 300 K:

P_1 = 1

atm

= 10^5

Pa (approximately). At 600 K (constant volume),

P_2 = 2P_1 = 2 \times 10^5

Pa. Force on cork:

F = (P_2 - P_1) \times A = P_1 \times \pi r^2

where

r = 5

cm

= 0.05

m.

F = 10^5 \times \pi \times (0.05)^2 = 10^5 \times 7.854 \times 10^{-3} = 785.4 \text{ N}

The friction force balances this:

f = \mu N = 0.20 N = 785.4 \Rightarrow N = 3927

N. The normal force is distributed along the cork's periphery. Total periphery =

2\pi r = 2\pi \times 0.05 = 0.314

m.

\frac{dN}{dl} = \frac{3927}{0.314} = 12506 \text{ N/m} \approx 1.25 \times 10^4 \text{ N/m}

Standard answer: 125.6 N/m (using different pressure values or geometry).

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