Difficulty: Easy
Correct Answer: Newton·second per square metre (N·s/m² = Pa·s)
Explanation:
Introduction / Context:
Dynamic viscosity quantifies a fluid’s resistance to shear deformation and appears in Newton’s law of viscosity, τ = μ (du/dy). Correct units prevent errors in calculations of laminar flow, boundary layers, and Reynolds numbers.
Given Data / Assumptions:
Concept / Approach:
From τ = μ (du/dy), rearrange to μ = τ / (du/dy). Substituting dimensions gives μ units = (N/m²) / (s^-1) = N·s/m². Since N/m² is a Pascal (Pa), μ is equivalently expressed as Pa·s. This is distinct from kinematic viscosity ν = μ/ρ with units m²/s.
Step-by-Step Solution:
Verification / Alternative check:
For water at 20°C, μ ≈ 1.002 × 10^-3 Pa·s. Converting to base SI gives ≈ 1.002 × 10^-3 kg/(m·s), consistent with the derivation.
Why Other Options Are Wrong:
(a) kg·s/m² is not dimensionally equivalent to kg/(m·s); it inverts time. (c) N·s²/m³ mismatches stress-rate dimensions. (d) m²/s is kinematic viscosity ν, not μ. (e) kg/m³ is density, another property entirely.
Common Pitfalls:
Confusing Pa·s with m²/s; mixing SI with CGS (poise, centipoise: 1 cP = 1 mPa·s).
Final Answer:
Newton·second per square metre (N·s/m² = Pa·s)
Discussion & Comments