Identify the correct dimension (M, L, T) for dynamic viscosity (absolute viscosity) in SI base dimensions.

Introduction / Context:
Dimensional analysis is fundamental in fluid mechanics. Dynamic viscosity μ relates shear stress to shear rate in Newton’s law of viscosity, and its correct dimensional form is essential for deriving non-dimensional groups and checking equations for consistency.

Given Data / Assumptions:

• Newtonian constitutive equation: τ = μ * (du/dy).
• Shear stress τ has dimension of force/area; shear rate has dimension of 1/time.
• Base dimensions: M (mass), L (length), T (time).

Concept / Approach:

From τ = μ * (du/dy), rearrange to μ = τ / (du/dy). Determine the dimensions of τ and du/dy, then form μ. This ensures the result is independent of the unit system (SI, CGS) as long as base dimensions are used consistently.

Step-by-Step Solution:

Verification / Alternative check:

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

Why Other Options Are Wrong:

L^2 T^-1, L T^-2: Lack mass dimension and do not match μ definitions. M L^-1 T^-2: That is stress, not viscosity.

Common Pitfalls:

Confusing dynamic viscosity with kinematic viscosity (ν), which has dimensions L^2 T^-1.

Final Answer:

M L^-1 T^-1