A square plate of 0.1 m side moves parallel to another plate with a velocity of `0.1 ms^(-1)` both plates immersed in water. If the viscous force is 0.02 N and the coefficient of viscosity 0.01 poise, what is the distance apart?

Explanation

To solve the problem step by step, we will use the formula for viscous force and the relationship between shear stress, viscosity, and the distance between the plates.

### Step 1: Identify the given values
- Side of the square plate, \( L = 0.1 \, \text{m} \)
- Velocity of the plate, \( v = 0.1 \, \text{m/s} \)
- Viscous force, \( F = 0.02 \, \text{N} \)
- Coefficient of viscosity, \( \eta = 0.01 \, \text{poise} = 0.01 \times 0.1 \, \text{N s/m}^2 = 0.001 \, \text{N s/m}^2 \) (since \( 1 \, \text{poise} = 0.1 \, \text{N s/m}^2 \))

### Step 2: Calculate the area of the plate
The area \( A \) of the square plate can be calculated as:
\[ A = L^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \]

### Step 3: Use the formula for viscous force
The formula for viscous force \( F \) is given by:
\[ F = \eta \cdot A \cdot \frac{du}{dy} \]
where \( \frac{du}{dy} \) is the velocity gradient, and \( dy \) is the distance between the plates.

### Step 4: Rearranging the formula to find \( dy \)
We can rearrange the formula to solve for \( dy \):
\[ dy = \frac{\eta \cdot A \cdot v}{F} \]

### Step 5: Substitute the known values into the equation
Substituting the known values:
\[ dy = \frac{(0.001 \, \text{N s/m}^2) \cdot (0.01 \, \text{m}^2) \cdot (0.1 \, \text{m/s})}{0.02 \, \text{N}} \]

### Step 6: Calculate \( dy \)
Calculating the numerator:
\[ 0.001 \cdot 0.01 \cdot 0.1 = 0.000001 \, \text{N m/s} \]
Now, divide by the viscous force:
\[ dy = \frac{0.000001 \, \text{N m/s}}{0.02 \, \text{N}} = 0.00005 \, \text{m} \]

### Step 7: Convert \( dy \) to centimeters
To convert meters to centimeters:
\[ dy = 0.00005 \, \text{m} \times 100 = 0.005 \, \text{cm} \]

### Final Answer
The distance apart between the plates is \( 0.00005 \, \text{m} \) or \( 0.005 \, \text{cm} \).

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