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Question #43911

Difficulty: 8/10 | 🟢 Free

The Question

A long, vertical capillary tube of internal radius $r$ is partially immersed in a large reservoir of an ideal, incompressible liquid of density $\rho$ and surface tension $T$ . The angle of contact between the liquid and the tube is $0^\circ$ . The liquid rises to an equilibrium height $h_0$ inside the tube. The liquid column is then given a small vertical displacement and released, causing it to execute simple harmonic motion. The unsteady flow in the reservoir just below the tube opening adds an effective inertial mass equivalent to a liquid column of length $\Delta L = 0.6r$ (end correction). Assuming $g$ is the acceleration due to gravity, which of the following expressions represents the angular frequency $\omega$ of the small oscillations?

Options

A.
$\sqrt{\frac{\rho g^2 r}{2T + 0.6 \rho g r^2}}$
B.
$\sqrt{\frac{\rho g^2 r}{2T}}$
C.
$\sqrt{\frac{2 \rho g^2 r}{2T + 0.6 \rho g r^2}}$
D.
$\sqrt{\frac{\rho g^2 r}{2T - 0.6 \rho g r^2}}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Capillary Rise Mechanics|End Correction|Small Angle Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the transition from fluid statics to unsteady fluid dynamics.

Students commonly fall into the trap of modeling the oscillating capillary column using only the static mass of the liquid inside the tube ( $m = \rho \pi r^2 h_0$ ), which leads to the incorrect angular frequency $\omega = \sqrt{g/h_0}$ (Option B).

This naive approach ignores the kinetic energy and inertia of the fluid in the reservoir converging into the tube.

By applying the unsteady Bernoulli principle, the 'end correction' emerges naturally as an added mass term ( $\Delta L = 0.6r$ ).

The lethal collision occurs when students misapply static Jurin's Law to a dynamic system, failing to account for the momentum of the surrounding fluid and missing the $0.6 \rho g r^2$ inertial term in the denominator.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Mechanical Properties of Fluids - Class 11 Part 2
Concept	Capillary Rise Mechanics
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free