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Question #76668

Difficulty: 5/10 | 🟢 Free

The Question

A closed cubical container of side length $L$ is initially half-filled with an incompressible liquid of density $\rho$ . The remaining space is filled with air at atmospheric pressure $P_0$ . The container is given a constant horizontal acceleration $a = 2g$ in the positive $x$ -direction (parallel to one of its horizontal edges). Assuming the air pressure inside remains $P_0$ , what is the magnitude of the net upward force exerted on the top wall of the container?

(Note: Ignore the mass of the top wall itself. Consider only the forces due to the pressure difference between the inside and the outside of the top wall, where the outside is at atmospheric pressure $P_0$ .)

Options

A.
$\frac{1}{16} \rho g L^3$
B.
$\frac{1}{8} \rho g L^3$
C.
$\frac{9}{16} \rho g L^3$
D.
$\frac{1}{2} \rho g L^3$

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

Frame of Reference Inertial vs Non-Inertial|Hydrostatic Paradox|Vector Geometry

Why This Question Is Tricky

Examiner Levers

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This question tests the application of the Equivalence Principle and fluid statics in a non-inertial frame, specifically targeting the 'Hydrostatic Paradox' misconception.

Students often erroneously assume that isobaric surfaces must remain horizontal or fail to correctly map the inclined free surface ( $\tan\theta = a/g$ ).

A lethal trap in this geometry is assuming the free surface passes through the container's corners, which violates volume conservation.

Furthermore, students may incorrectly integrate pressure over the entire top wall, ignoring that the liquid detaches and cannot exert negative gauge pressure.

The correct framework requires applying volume conservation to find the exact free surface equation, determining the partially wetted length on the top wall, and integrating the effective pressure gradient.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Laws of Motion - Class 11 Part 1
Concept	Frame of Reference Inertial vs Non-Inertial
Difficulty	5
Cognitive Load	6
Target Solve Time	140 sec
Access	Free