In the S.I. system, identify the physical quantity that uses m^2/s as its unit.

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:

  • Kinematic viscosity is denoted by nu and defined as dynamic viscosity mu divided by density rho.
  • Dynamic viscosity mu in S.I. is Pa·s (or N·s/m^2), and density rho is kg/m^3.
  • Other listed quantities have different standard S.I. units (e.g., pressure → N/m^2).


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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