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Question #42241

Difficulty: 8/10 | 🟢 Free

The Question

A horizontal Venturi meter is mounted on a cart that moves with a constant horizontal acceleration aa in the direction of the fluid flow. The fluid has density ρ\rho. The cross-sectional areas of the wide and narrow sections of the meter are AA and a0a_0, respectively. A U-tube manometer containing a liquid of density ρm\rho_m (ρm>ρ\rho_m > \rho) is connected between the wide and narrow sections. The horizontal distance between the two tapping points is LL. Assume the flow rate is large enough such that the liquid level in the manometer arm connected to the wide section is lower than that in the other arm by a height hh. Which of the following expressions gives the volume flow rate QQ of the fluid?

Options

  • A.

    Aa02(ρm−ρ)(gh+aL)ρ(A2−a02)A a_0 \sqrt{ \frac{2 (\rho_m - \rho) (g h + a L)}{\rho (A^2 - a_0^2)} }

  • B.

    Aa02(ρm−ρ)ghρ(A2−a02)A a_0 \sqrt{ \frac{2 (\rho_m - \rho) g h}{\rho (A^2 - a_0^2)} }

  • C.

    Aa02[(ρm−ρ)gh−ρaL]ρ(A2−a02)A a_0 \sqrt{ \frac{2 [(\rho_m - \rho) g h - \rho a L]}{\rho (A^2 - a_0^2)} }

  • D.

    Aa02[(ρm−ρ)gh+ρmaL]ρ(A2−a02)A a_0 \sqrt{ \frac{2 [(\rho_m - \rho) g h + \rho_m a L]}{\rho (A^2 - a_0^2)} }

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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

Accelerated Frame Hydrostatics|Venturi Meter Logic|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the application of fluid dynamics in non-inertial reference frames, specifically the modification of Bernoulli's principle and manometry due to a pseudo-force field.

Students commonly fall into the trap of either completely ignoring the horizontal acceleration (treating it as a standard Venturi meter problem) or applying the pseudo-force gradient only to the main flowing fluid while forgetting that the static manometer fluid is also subject to the same inertial frame shift.

By requiring the simultaneous evaluation of the effective potential gradient ρ(a⃗⋅Δr⃗)\rho(\vec{a} \cdot \Delta\vec{r}) in both the dynamic pipe flow and the static U-tube, the problem isolates the misconception that pressure is solely a function of depth, reinforcing that it depends on the total effective gravity vector g⃗eff=g⃗−a⃗\vec{g}_{eff} = \vec{g} - \vec{a}.

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Question Information

SubjectPhysics
ChapterMechanical Properties of Fluids - Class 11 Part 2
ConceptAccelerated Frame Hydrostatics
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree