Identify the correct dimension (M, L, T) for dynamic viscosity (absolute viscosity) in SI base dimensions.
-
AM L^-1 T^-1
-
BL^2 T^-1
-
CL T^-2
-
DM L^-1 T^-2
Answer
Correct Answer: M L^-1 T^-1
Explanation
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:
Shear stress τ → force/area = (M L T^-2) / L^2 = M L^-1 T^-2.Shear rate du/dy → velocity gradient = (L T^-1) / L = T^-1.Dynamic viscosity μ = τ / (du/dy) = (M L^-1 T^-2) / (T^-1) = M L^-1 T^-1.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