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Question #42103

Difficulty: 8/10 | 🟢 Free

The Question

A horizontal Venturi tube has a wide section of cross-sectional area $A_1$ and a narrow throat of area $A_2$ . An ideal, incompressible fluid of density $\rho$ flows through the tube with a constant volume flow rate $Q$ . A rectangular bypass channel of width $w$ and height $d$ connects the wide section to the throat. A rigid dielectric slab of dielectric constant $K$ fits perfectly inside this channel and can slide frictionlessly. The top and bottom walls of the channel act as parallel capacitor plates connected to a constant DC voltage $V$ . The slab is partially inserted into the capacitor. The fluid pressure difference exerts a force on the slab that is perfectly balanced by the electrostatic force. Ignoring fringe effects, what is the voltage $V$ required to keep the slab in equilibrium?

Options

A.
$Q d \sqrt{ \frac{\rho}{\epsilon_0 (K-1)} \left( \frac{1}{A_2^2} - \frac{1}{A_1^2} \right) }$
B.
$Q d \sqrt{ \frac{2\rho}{\epsilon_0 (K-1)} \left( \frac{1}{A_2^2} - \frac{1}{A_1^2} \right) }$
C.
$Q \sqrt{ \frac{\rho d}{\epsilon_0 (K-1)} \left( \frac{1}{A_2^2} - \frac{1}{A_1^2} \right) }$
D.
$Q d \sqrt{ \frac{\rho}{\epsilon_0 (K-1)} \left( \frac{1}{A_2^2} + \frac{1}{A_1^2} \right) }$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Venturi Meter Logic|Dielectric Slab Displacement|Vector Calculus

Why This Question Is Tricky

Common Mistakes

This question creates a 'Lethal Collision' by coupling fluid dynamics (Bernoulli's principle) with electrodynamics (virtual work on a dielectric).

A common trap is miscalculating the electrostatic force by incorrectly differentiating the energy $U = \frac{1}{2}CV^2$ without accounting for the work done by the battery at constant voltage, leading to sign errors or missing factors of 1/2.

Furthermore, students often fail at the geometric coupling: the fluid pressure acts on the cross-sectional area $w \cdot d$ , while the capacitance gradient depends on the width $w$ and separation $d$ .

Failing to correctly identify the area over which the pressure difference acts leads to dimensional inconsistencies and incorrect dependencies on $d$ .

Quick Facts About This Question

Question Information

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