Rotating liquid in a cylinder: The rise of liquid at the wall relative to the initial level is ________ the depression at the axis for solid-body rotation.

Introduction / Context:
When a liquid in a cylinder rotates as a solid body, the free surface forms a paraboloid described by z = (ω^2 * r^2) / (2 * g) + constant. Volume conservation around the free surface implies that what rises at the walls must be balanced by a matching depression at the axis.

Given Data / Assumptions:

• Open cylinder with a free surface.
• Steady, solid-body rotation with angular speed ω.
• Negligible evaporation and no inflow/outflow during rotation.

Concept / Approach:
The parabolic surface has its minimum at the axis and maximum at the wall. Integrating the free-surface elevation relative to the initial flat level over the cross-section yields zero net change in volume; thus, the peak rise at the wall equals the central depression in magnitude for the same reference level.

Step-by-Step Solution:

Verification / Alternative check:
Compute mean of Δz over the radius; it is zero for the chosen constant. Therefore, positive volume at the rim equals negative volume at the center, confirming equality of magnitudes.

Why Other Options Are Wrong:

• Less than / more than: Would violate volume conservation for the steady parabolic surface.

Common Pitfalls:
Confusing instantaneous sloshing (transient) with steady solid-body rotation; forgetting that the constant is chosen to conserve volume.

Final Answer:
same as