Dimensional analysis: what are the base dimensions of dynamic viscosity μ in M–L–T terms?

Introduction / Context:
Dynamic viscosity μ quantifies a fluid’s resistance to shear. Correct base dimensions are essential for checking equations, converting units, and building dimensionless groups like Reynolds number.

Given Data / Assumptions:

• Shear stress τ = μ * (du/dy).
• Shear stress has dimensions of force/area.
• Velocity gradient du/dy has dimensions of T^-1.

Concept / Approach:

Start by writing τ in dimensions. Force is M L T^-2; area is L^2; hence τ has dimensions M L^-1 T^-2. Since τ = μ * (du/dy) and du/dy has dimensions T^-1, solving for μ yields μ = τ / (du/dy) with dimensions M L^-1 T^-1.

Step-by-Step Solution:

Verification / Alternative check:

In SI, μ units are Pa·s = (N/m^2)·s = (kg·m^-1·s^-2)·s = kg·m^-1·s^-1, matching M L^-1 T^-1.

Why Other Options Are Wrong:

(a) corresponds to impulse per area; (b) and (c) are dimensionally inconsistent for μ. (e) not needed.

Common Pitfalls:

Confusing dynamic viscosity with kinematic viscosity ν whose dimensions are L^2 T^-1.

Final Answer:

M L^-1 T^-1