Fluid Statics — Free-Surface Tilt in an Accelerating Open Tank An open tank containing a liquid starts from rest and moves with a uniform horizontal acceleration a. What is the relation between the inclination angle θ made by the free liquid surface and the horizontal, in terms of a and gravitational acceleration g?

Introduction / Context:
When a container with a free liquid surface accelerates horizontally, the surface reorients to remain perpendicular to the resultant of the real gravitational field and the inertial (body) acceleration field. This is a classic fluid statics-in-accelerating-frames problem used in marine, aerospace, and process engineering.

Given Data / Assumptions:

• Open tank with a liquid and a free surface exposed to atmosphere.
• Uniform horizontal acceleration a; gravity acts vertically downward with magnitude g.
• Neglect surface tension and sloshing transients; consider steady inclined surface after acceleration is established.

Concept / Approach:
In a non-inertial frame attached to the tank, an inertial body force per unit mass acts opposite to the acceleration, of magnitude a. The free surface aligns normal to the resultant acceleration vector formed by g downward and a backward horizontally. The slope of the surface is given by tan θ = a / g, where θ is the angle with the horizontal.

Step-by-Step Solution:

Verification / Alternative check:
If a = 0, tan θ = 0, so the surface is horizontal as expected. If a increases, θ increases smoothly; in the limit a ≫ g, the surface approaches a near-vertical orientation, consistent with intuition.

Why Other Options Are Wrong:
2 a / g and a / (2 g) insert unjustified factors; a^2 / (2 g) has wrong dimensional form inside tan and no basis in this static equilibrium model.

Common Pitfalls:
Mixing up θ with the complement angle; forgetting that the surface is orthogonal to the resultant acceleration, not parallel.

Final Answer:
tan θ = a / g.