Question

Dynamic viscosity of a fluid is 2.2 poise and specific gravity is 0.7. Then kinematic viscosity in SI units is:

• A. \( 3.14 \times 10^{-4} \, \text{m}^2/\text{s} \)
• B. \( 3.14 \times 10^{-3} \, \text{m}^2/\text{s} \)
• C. \( 1.5 \times 10^{-4} \, \text{stokes} \)
• D. \( 1.5 \times 10^{-3} \, \text{stokes} \)

Hint:

To calculate kinematic viscosity, convert dynamic viscosity from poise to pascal-seconds and use the formula \( \nu = \frac{\mu}{\rho} \), where \( \rho \) is the fluid's density.

Answer 1

The Correct Option is A

Step 1: Formula for Kinematic Viscosity.
The kinematic viscosity \( \nu \) is given by the formula: \[ \nu = \frac{\mu}{\rho} \] Where: - \( \nu \) is the kinematic viscosity (in \( \text{m}^2/\text{s} \)) - \( \mu \) is the dynamic viscosity (in \( \text{Pa} \cdot \text{s} \) or poise) - \( \rho \) is the density (in \( \text{kg/m}^3 \))
Step 2: Converting units.
- The given dynamic viscosity \( \mu = 2.2 \) poise. - \( 1 \, \text{poise} = 0.1 \, \text{Pa} \cdot \text{s} \), so: \[ \mu = 2.2 \times 0.1 = 0.22 \, \text{Pa} \cdot \text{s} \] - The specific gravity \( SG = 0.7 \), and density \( \rho = SG \times 1000 = 0.7 \times 1000 = 700 \, \text{kg/m}^3 \).
Step 3: Calculating Kinematic Viscosity.
Using the formula for kinematic viscosity: \[ \nu = \frac{0.22}{700} = 3.14 \times 10^{-4} \, \text{m}^2/\text{s} \]
Step 4: Conclusion.
Therefore, the correct kinematic viscosity is \( 3.14 \times 10^{-4} \, \text{m}^2/\text{s} \), and the correct answer is (1).