flat plate is separated from a large plate by a layer of glycerine of thickness `3 xx 10^(-3)m`. If the coefficient of viscosity of glycerine is` 2 Ns//m^(2)` what is the force required to keep the plate moving with a velocity of` 6xx10^(-2) m//s`. Area of the plate is ` 4.8 xx 10 ^(-3) m `

Explanation

To solve the problem, we will use the formula for the force required to keep a plate moving through a viscous fluid. The formula is given by:

\[ F = \eta \cdot A \cdot \frac{du}{dy} \]

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

### Step 1: Identify the given values
- Coefficient of viscosity (\( \eta \)) = \( 2 \, \text{Ns/m}^2 \)
- Thickness of the glycerine layer (\( dy \)) = \( 3 \times 10^{-3} \, \text{m} \)
- Velocity of the plate (\( du \)) = \( 6 \times 10^{-2} \, \text{m/s} \)
- Area of the plate (\( A \)) = \( 4.8 \times 10^{-3} \, \text{m}^2 \)

### Step 2: Calculate the velocity gradient (\( \frac{du}{dy} \))
The velocity gradient is calculated as:

\[ \frac{du}{dy} = \frac{du}{dy} = \frac{6 \times 10^{-2}}{3 \times 10^{-3}} = 20 \, \text{s}^{-1} \]

### Step 3: Substitute the values into the formula
Now we will substitute the values into the formula for force:

\[ F = \eta \cdot A \cdot \frac{du}{dy} \]

Substituting the known values:

\[ F = 2 \cdot (4.8 \times 10^{-3}) \cdot 20 \]

### Step 4: Calculate the force
Now we perform the calculation:

\[ F = 2 \cdot 4.8 \times 10^{-3} \cdot 20 = 0.192 \, \text{N} \]

### Final Answer
The force required to keep the plate moving is:

\[ F = 0.192 \, \text{N} \]