Accelerated Tanks – Free Surface Inclination An open tank of liquid is accelerated up an inclined plane. The inclination of the liquid free surface varies with the tank acceleration according to which relation?

Introduction:
When a container of liquid undergoes linear acceleration, the free surface tilts until it becomes perpendicular to the resultant of gravitational acceleration and the imposed acceleration. This question asks how the inclination depends on the acceleration magnitude.

Given Data / Assumptions:

• Open tank with a free surface.
• Tank accelerated along an incline with constant acceleration a.
• Gravity g acts vertically downward.

Concept / Approach:

The free surface aligns perpendicular to the resultant acceleration vector formed by g and the applied acceleration a. The tilt angle theta measured from the horizontal satisfies tan(theta) = a_parallel / g if a acts along the tank axis. Hence theta increases as a increases; for small angles, theta ≈ a / g, showing direct proportionality.

Step-by-Step Solution:

Verification / Alternative check:

Plotting theta vs a from tan(theta) = a / g shows monotonic increase; for a much less than g, theta ≈ a / g in radians, confirming proportionality.

Why Other Options Are Wrong:

Equal to the acceleration: Dimensionally meaningless; angle is not in acceleration units.Inversely proportional: Contradicts tan(theta) = a / g.Independent or proportional to acceleration squared: Both inconsistent with the tan relation.

Common Pitfalls:

Forgetting that the free surface is always perpendicular to the resultant acceleration and misinterpreting “inclination” as a length rather than an angle.

Final Answer:

directly proportional