Fluid mechanics — dimensional analysis refresher What is the correct fundamental dimensional formula for absolute (dynamic) viscosity μ in terms of mass (M), length (L), and time (T)?

Introduction / Context:
Absolute (dynamic) viscosity μ links shear stress to velocity gradient in Newtonian fluids via Newton's law of viscosity. Knowing its dimensions is essential for checking equations, converting units (Pa·s, N·s/m^2, kg·m^-1·s^-1), and building dimensionless groups such as Reynolds number.

Given Data / Assumptions:

• We are using fundamental dimensions: mass (M), length (L), and time (T).
• Newton's law of viscosity: τ = μ * (du/dy).
• Shear stress τ has units of force/area.

Concept / Approach:
Start from τ with dimensions of pressure: M L^-1 T^-2. The velocity gradient du/dy has dimensions T^-1. Solving μ = τ / (du/dy) gives the dimensional formula for viscosity. This approach prevents memorization errors and is reliable for cross-checking.

Step-by-Step Solution:
Write τ dimensions: τ = Force/Area = (M L T^-2)/L^2 = M L^-1 T^-2.Velocity gradient du/dy = (L T^-1)/L = T^-1.Compute μ: μ = τ / (du/dy) = (M L^-1 T^-2)/(T^-1) = M L^-1 T^-1.

Verification / Alternative check:
Use SI units: 1 Pa·s = 1 N·s/m^2 = (kg·m·s^-2)·s·m^-2 = kg·m^-1·s^-1, which maps to M L^-1 T^-1.

Why Other Options Are Wrong:

• MLT^-1, ML^-1T, MLT: do not match the derived expression; they misplace the exponents or omit the negative exponent on time or length.

Common Pitfalls:

• Confusing dynamic viscosity with kinematic viscosity ν, which has dimensions L^2 T^-1.
• Dropping the negative sign on the length exponent when expanding pressure units.

Final Answer:
ML-1T-1