Question

Co-efficient of viscosity of a fluid divided by its mass density is known as

• A. Apparent viscosity
• B. Surface tension
• C. Kinematic viscosity
• D. Specific gravity

Hint:

Viscosity Definitions:
KINEMATIC VISCOSITY ($\nu$) = $\frac{\text{Dynamic Viscosity } (\mu)}{\text{Density } (\rho)}$. (Units: $\text{m}^2/\text{s}$ or Stokes/cSt).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Fluid property definitions: Kinematic Viscosity ($ u$) is defined as the ratio of dynamic (absolute) viscosity ($\mu$) to fluid mass density ($ ho$).
Key Formula or Approach:
\[ \mathbf{Kinematic \text{ } Viscosity \text{ } (\nu)} = \frac{\text{Dynamic Viscosity } (\mu)}{\text{Mass Density } (\rho)} \quad [\text{SI Unit: m}^2/\text{s}, \; \text{CGS Unit: Stoke (St) / Centistoke (cSt)}] \]

Step 2: Detailed Explanation:
In fluid mechanics, rheology, and dairy hydraulics:
1. Dynamic Viscosity ($\mu$ Absolute Viscosity): The internal molecular resistance of a fluid to shear deformation (measured in $\text{Pa}\cdot\text{s}$ or Poise).
2. Kinematic Viscosity ($\nu$) (C): Defined mathematically as the dynamic coefficient of viscosity divided by fluid mass density:
\[ \nu = \frac{\mu}{\rho} \]
- In the SI system, it has units of $\text{m}^2/\text{s}$; in CGS units, $1\text{ Stoke} = 1\text{ cm}^2/\text{s} = 10^{-4}\text{ m}^2/\text{s}$ ($1\text{ cSt} = 1\text{ mm}^2/\text{s}$).
- It represents the rate at which momentum diffuses through fluid shear layers, utilized directly in Reynolds number calculations ($Re = \frac{v D}{\nu}$).

Step 3: Final Answer:
Therefore, dynamic viscosity divided by mass density is Kinematic viscosity, corresponding to option (C).