Pick the wrong statements about fluids and viscosity: Which option correctly identifies the incorrect assertions regarding equilibrium shear stress and non-Newtonian viscosity?

Difficulty: Easy

Correct Answer: both (b) and (c)

Explanation:


Introduction / Context:
Distinguishing correct from incorrect statements about viscous behavior is essential in rheology and fluid statics. This item targets understanding of shear stress in equilibrium states and how viscosity depends on shear for non-Newtonian fluids.


Given Data / Assumptions:

  • Equilibrium refers to fluid at rest with no velocity gradients.
  • Non-Newtonian fluids show apparent viscosity that varies with shear rate and sometimes time.
  • Temperature dependence applies to all fluids but is not the only dependency for non-Newtonian types.


Concept / Approach:
In a static fluid (equilibrium), shear stress is zero; only normal (pressure) stresses act. For non-Newtonian fluids, apparent viscosity depends on shear rate (and possibly time), not just temperature. Momentum transfer by viscosity indeed arises from velocity gradients (Newton’s law of viscosity) in laminar flow.


Step-by-Step Solution:

Evaluate (a): True — shear stress τ = μ du/dy in laminar flow.Evaluate (b): False — at equilibrium (no motion), τ = 0; fluids cannot sustain shear at rest.Evaluate (c): False — non-Newtonian viscosity depends on shear rate (and sometimes time), not solely on temperature.Thus, both (b) and (c) are wrong statements.


Verification / Alternative check:
Rheograms of shear-thinning/thickening fluids (e.g., polymer melts, paints) show viscosity as a function of shear rate; static fluids exhibit hydrostatic pressure with zero shear.


Why Other Options Are Wrong:

  • (a) is correct and should not be selected as wrong.
  • Any single selection misses that two statements are wrong here.


Common Pitfalls:
Confusing non-Newtonian behavior with temperature-only effects; also, assuming fluids at rest can support shear like solids—incorrect by definition.


Final Answer:
both (b) and (c)

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