Two metal plates of area `2 xx10^(-4) m^(2)` each, are kept in water and one plate is moved over the other with a certain velocity. The distance between the plates is `2 xx 10^(-4) m`. If the horizontal force applied to move the plate is `10^(-3)N`, calculate the velocity of the plate. Given `eta` of water is` 10^(-3)` decapoise.

Explanation

To solve the problem, we will use the formula for viscous force in a fluid, which is given by:

\[ F = \eta \cdot A \cdot \frac{dv}{dx} \]

Where:
- \( F \) is the horizontal force applied (in Newtons),
- \( \eta \) is the dynamic viscosity of the fluid (in Pascal-seconds or N·s/m²),
- \( A \) is the area of the plates (in m²),
- \( \frac{dv}{dx} \) is the velocity gradient (in m/s per m).

### Step-by-step Solution:

1. **Identify the given values:**
- Area of each plate, \( A = 2 \times 10^{-4} \, m^2 \)
- Distance between the plates, \( dx = 2 \times 10^{-4} \, m \)
- Horizontal force applied, \( F = 10^{-3} \, N \)
- Viscosity of water, \( \eta = 10^{-3} \, \text{decapoise} = 10^{-3} \times 0.1 \, \text{N·s/m²} = 10^{-4} \, \text{N·s/m²} \)

2. **Rearranging the viscous force formula:**
We need to find the velocity \( dv \). Rearranging the formula gives us:
\[ dv = \frac{F \cdot dx}{\eta \cdot A} \]

3. **Substituting the values into the equation:**
\[ dv = \frac{(10^{-3} \, N) \cdot (2 \times 10^{-4} \, m)}{(10^{-4} \, \text{N·s/m²}) \cdot (2 \times 10^{-4} \, m^2)} \]

4. **Calculating the numerator:**
\[ \text{Numerator} = 10^{-3} \cdot 2 \times 10^{-4} = 2 \times 10^{-7} \, N \cdot m \]

5. **Calculating the denominator:**
\[ \text{Denominator} = 10^{-4} \cdot 2 \times 10^{-4} = 2 \times 10^{-8} \, \text{N·s} \]

6. **Dividing the numerator by the denominator:**
\[ dv = \frac{2 \times 10^{-7}}{2 \times 10^{-8}} = 10^{1} = 1 \, m/s \]

### Final Answer:
The velocity of the plate is \( 1 \, m/s \).