Question

At a certain point in oil, the shear stress and velocity gradient are 0.2 N/m$^2$ and 0.1 s$^{-1}$ respectively, then the dynamic viscosity is

• A. 2 N-s/m$^2$
• B. 50 N-s/m$^2$
• C. 0.5 N-s/m$^2$
• D. 0.02 N-s/m$^2$

Hint:

Newton's law of viscosity, $\tau = \mu (du/dy)$, is a cornerstone of fluid mechanics.
Remember that dynamic viscosity $\mu$ is the proportionality constant between shear stress and the rate of shear strain.
Ensure your units are consistent (e.g., Pascals for stress, s$^{-1}$ for gradient) to get viscosity in Pa·s.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to calculate the dynamic viscosity of an oil, given the shear stress and the velocity gradient at a point.

Step 2: Key Formula or Approach:
For a Newtonian fluid (which oils are generally assumed to be), the relationship between shear stress ($\tau$), dynamic viscosity ($\mu$), and velocity gradient (rate of shear strain, $du/dy$) is given by Newton's law of viscosity:
\[ \tau = \mu \frac{du}{dy} \]

Step 3: Detailed Explanation:
We are given:
- Shear stress ($\tau$) = 0.2 N/m$^2$
- Velocity gradient ($du/dy$) = 0.1 s$^{-1}$
We need to find the dynamic viscosity, $\mu$. Rearrange the formula:
\[ \mu = \frac{\tau}{du/dy} \] Substitute the given values:
\[ \mu = \frac{0.2 \text{ N/m}^2}{0.1 \text{ s}^{-1}} \] \[ \mu = 2 \frac{\text{N}}{\text{m}^2} \cdot \text{s} \quad \text{or} \quad 2 \text{ N-s/m}^2 \] The unit N-s/m$^2$ is also known as a Pascal-second (Pa·s).

Step 4: Final Answer:
The dynamic viscosity is 2 N-s/m$^2$.