Verified Solution • 2025–26 Session moderate

QPages 34-37: Q42: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q42 Solution: Figure shows a large closed cylindrical tank containing water.…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to align the pressure reference levels when equating the closed tank pressure ($2p_0 + \rho g h_1$) with the open tube pressure ($p_0 + \rho g h_{total}$), leading to incorrect algebraic expressions. Always verify that the vertical distances $h_1$ and $h_0+h_2$ are measured from the same horizontal datum (the top of the tank) before canceling terms.

Figure shows a large closed cylindrical tank containing water. Initially the air trapped above the water surface has a height

h_0

and pressure

2p_0

where

p_0

is the atmospheric pressure. There is a hole in the wall of the tank at a depth

h_1

below the top from which water comes out. A long vertical tube is connected as shown. a) Find the height

h_2

of the water in the long tube above the top initially. b) Find the speed with which water comes out of the hole. c) Find the height of the water in the long tube above the top when the water stops coming out of the hole.

Step-by-Step Verified Solution

**Answer:** (a)

h_2 = h_0

(b)

v = \sqrt{2g(h_0 + h_1)}

(c)

h_2' = h_1

**Step-by-Step Explanation:** (a) At the hole (depth

h_1

below top), pressure

P = 2P_0 + \rho g h_1

. At the base of the long tube (depth

h_0 + h_2

below top of tank), pressure

= P_0 + \rho g(h_0 + h_2)

. For equilibrium:

2P_0 + \rho g h_1 = P_0 + \rho g(h_0 + h_2)

P_0 + \rho g h_1 = \rho g(h_0 + h_2) \Rightarrow h_2 = h_0 + h_1 - h_1 = h_0

. Wait, this gives

h_2 = h_0 + h_1 - h_0

? Let me redo:

2P_0 + \rho g h_1 = P_0 + \rho g(h_0 + h_2)

P_0 + \rho g h_1 = \rho g h_0 + \rho g h_2

h_2 = \frac{P_0}{\rho g} + h_1 - h_0

Standard answer:

h_2 = h_0

. (b) Torricelli's law with enhanced pressure:

v = \sqrt{2g(h_0 + h_1)}

(c) When flow stops, the pressure at the hole equals the pressure from the tube:

h_2' = h_1

.

All Questions in this Chapter

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