Difficulty: Easy
Correct Answer: Kinematic viscosity
Explanation:
Introduction / Context:Recognizing units is a powerful shortcut in engineering exams. The unit m^2/s commonly appears in fluid mechanics when describing how momentum diffuses through a fluid, a property central to laminar versus turbulent behavior and to boundary-layer growth on surfaces.
Given Data / Assumptions:
Concept / Approach:Kinematic viscosity nu = mu / rho characterizes diffusivity of momentum. It determines Reynolds number Re = V * L / nu, which separates flow regimes and strongly influences drag, head loss, and heat/mass transfer coefficients. Because it is a ratio of viscosity to density, its S.I. unit simplifies to m^2/s, highlighting its “diffusivity-like” nature similar to thermal diffusivity (alpha) with units m^2/s.
Step-by-Step Solution:
Start with nu = mu / rho.Replace mu by N·s/m^2 and rho by kg/m^3: nu units = (N·s/m^2) / (kg/m^3).Use 1 N = 1 kg·m/s^2 to simplify: (kg·m/s^2 * s / m^2) / (kg/m^3) → (kg·m/(s·m^2)) / (kg/m^3) → (m/(s·m^2)) * (m^3/kg) * (kg) = m^2/s.Verification / Alternative check:
Compare with water at 20°C: nu ≈ 1.0 × 10^-6 m^2/s (1 cSt), which matches the known magnitude and unit.Why Other Options Are Wrong:
Surface tension: N/m.Pressure: N/m^2 (Pa).Power: watt (W = J/s = N·m/s).Specific weight: N/m^3.Common Pitfalls:
Mixing dynamic viscosity (Pa·s) with kinematic viscosity (m^2/s).Confusing m^2/s with m/s (velocity) or m^2 (area).Final Answer:
Kinematic viscosity
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