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Question #42380

Difficulty: 8/10 | 🟢 Free

The Question

A U-tube of uniform cross-section has a horizontal base of length 2L2L and two vertical arms. The horizontal base is completely filled with a non-homogeneous liquid whose density varies linearly from ρ0\rho_0 at the left end to 3ρ03\rho_0 at the right end. The vertical arms contain a homogeneous liquid of density 2ρ02\rho_0. Massless, freely moving pistons separate the two liquids at the junctions of the base and the arms. The entire U-tube is rotated with a constant angular velocity ω\omega about a vertical axis located at a distance dd to the left of the left end of the horizontal base. In the steady state, what is the difference in height Δh\Delta h between the liquid levels in the right and left vertical arms?

Options

  • A.

    ω2Lg(2d+73L)\frac{\omega^2 L}{g} \left(2d + \frac{7}{3} L\right)

  • B.

    ω2Lg(2d+2L)\frac{\omega^2 L}{g} (2d + 2L)

  • C.

    ω2Lg(d+76L)\frac{\omega^2 L}{g} \left(d + \frac{7}{6} L\right)

  • D.

    ω2Lg(4d+143L)\frac{\omega^2 L}{g} \left(4d + \frac{14}{3} L\right)

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Parallel and Perpendicular Axes|Accelerated Frame Hydrostatics|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze hydrostatics in an accelerated frame with a variable density mass distribution.

The common trap is attempting a brute-force integration of the pressure differential, which is algebraically tedious and prone to errors.

The elegant, advanced approach recognizes that the work done by the pressure difference equals the change in the fluid's centrifugal potential energy.

By applying the definition of the center of mass (analogous to the Parallel Axis Theorem for first moments), the effective inertial force depends only on the total mass and the position of the center of mass.

Students who assume a uniform average density will incorrectly calculate the CM position and choose the trap option.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterSystem of Particles and Rotational Motion - Class 11 Part 1
ConceptParallel and Perpendicular Axes
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree