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.
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Asame as
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Bless than
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Cmore than
Answer
Correct Answer: same as
Explanation
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:
Free-surface shape: Δz(r) = (ω^2 * r^2) / (2 * g) − constant.Choose constant so the average elevation change over the disk is zero.Then Δz at r = 0 equals −Δz at r = R, making rise at wall = depression at axis.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