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Question #42220

Difficulty: 8/10 | 🟢 Free

The Question

A two-dimensional open container has a parabolic cross-section given by $y = c x^2$ , where $y$ is the vertical axis and $x$ is the horizontal axis. The container is filled with an incompressible liquid of density $\rho$ up to a height $H$ (when at rest). The container is then given a constant horizontal acceleration $a$ in the $+x$ direction. Assuming the liquid does not spill, what is the steady-state gauge pressure at the lowest point of the container (the origin)?

Options

A.
$\rho g H$
B.
$\rho g H - \frac{\rho a^2}{4 c g}$
C.
$\rho g H + \frac{\rho a^2}{4 c g}$
D.
$\rho g H - \frac{\rho a^2}{2 c g}$

Check Your Answer

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

Hydrostatic Paradox|Accelerated Frame Hydrostatics|Frame of Reference Inertial vs Non-Inertial

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of fluid statics in non-inertial frames with integral calculus for volume conservation.

The primary trap lies in the assumption that the free surface of an accelerated liquid simply rotates about its initial geometric center (which would imply the y-intercept $y_0$ remains $H$ and the base pressure remains $\rho g H$ ).

This heuristic is only true for containers with uniform rectangular cross-sections.

For a parabolic (non-uniform) container, the condition of constant liquid volume forces the pivot point to shift downwards, fundamentally altering the hydrostatic pressure at the base.

Quick Facts About This Question

Question Information

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