What is the force due to viscosity acting on a layer of water of area `4xx10^(-2)m^2`, if the relative velocity between the two layers of water, separated by 0.4 mm is 5 cm/s ? The coefficient of viscosity of water = 0.01 poise.[1 poise=0.1 `Ns//m^2`]

Options

A. `5xx10^(-3)` N
B. `8xx10^(-4)` N
C. `12xx10^(-3)` N
D. `5xx10^(-5)` N

Explanation

To find the force due to viscosity acting on a layer of water, we can use Newton's law of viscosity, which states:

\[ F = \mu A \frac{du}{dy} \]

Where:
- \( F \) is the force due to viscosity,
- \( \mu \) is the coefficient of viscosity,
- \( A \) is the area,
- \( \frac{du}{dy} \) is the velocity gradient.

### Step-by-Step Solution:

1. **Identify the given values:**
- Area \( A = 4 \times 10^{-2} \, m^2 \)
- Relative velocity \( du = 5 \, cm/s = 5 \times 10^{-2} \, m/s \) (convert cm/s to m/s)
- Separation distance \( dy = 0.4 \, mm = 0.4 \times 10^{-3} \, m \) (convert mm to m)
- Coefficient of viscosity \( \mu = 0.01 \, \text{poise} = 0.01 \times 0.1 \, \text{Ns/m}^2 = 0.001 \, \text{Ns/m}^2 \) (convert poise to Ns/m²)

2. **Calculate the velocity gradient \( \frac{du}{dy} \):**
\[ \frac{du}{dy} = \frac{5 \times 10^{-2} \, m/s}{0.4 \times 10^{-3} \, m} = \frac{5 \times 10^{-2}}{0.4 \times 10^{-3}} = \frac{5}{0.4} \times 10^{1} = 12.5 \, s^{-1} \]

3. **Substitute the values into the formula:**
\[ F = \mu A \frac{du}{dy} \]
\[ F = (0.001 \, \text{Ns/m}^2) \times (4 \times 10^{-2} \, m^2) \times (12.5 \, s^{-1}) \]

4. **Calculate the force \( F \):**
\[ F = 0.001 \times 4 \times 10^{-2} \times 12.5 \]
\[ F = 0.0005 \times 12.5 = 0.00625 \, N \]

5. **Final Result:**
The force due to viscosity acting on the layer of water is:
\[ F = 0.00625 \, N \]