A plate of area `2m^(2)` is made to move horizontally with a speed of `2m//s` by applying a horizontal tangential force over the free surface of a liquid. If the depth of the liquid is 1m and the liquid in contact with the bed is stationary. Coefficient of viscosity of liquid is `0.01` poise. Find the tangential force needed to move the plate.

Explanation

To find the tangential force needed to move the plate, we will use the formula for viscous force in a fluid, which is given by:

\[ F = \eta \cdot A \cdot \frac{\Delta v}{\Delta y} \]

Where:
- \( F \) is the tangential force,
- \( \eta \) is the coefficient of viscosity,
- \( A \) is the area of the plate,
- \( \Delta v \) is the change in velocity,
- \( \Delta y \) is the depth of the liquid.

### Step-by-step Solution:

1. **Identify the given values:**
- Area of the plate, \( A = 2 \, \text{m}^2 \)
- Speed of the plate, \( v = 2 \, \text{m/s} \)
- Coefficient of viscosity, \( \eta = 0.01 \, \text{poise} = 0.01 \times 10^{-1} \, \text{N s/m}^2 = 0.001 \, \text{N s/m}^2 \)
- Depth of the liquid, \( \Delta y = 1 \, \text{m} \)
- The velocity of the liquid in contact with the bed is stationary, so \( v_{stationary} = 0 \, \text{m/s} \).

2. **Calculate the change in velocity (\( \Delta v \)):**
\[ \Delta v = v - v_{stationary} = 2 \, \text{m/s} - 0 \, \text{m/s} = 2 \, \text{m/s} \]

3. **Substitute the values into the formula:**
\[ F = \eta \cdot A \cdot \frac{\Delta v}{\Delta y} \]
\[ F = 0.001 \, \text{N s/m}^2 \cdot 2 \, \text{m}^2 \cdot \frac{2 \, \text{m/s}}{1 \, \text{m}} \]

4. **Calculate the force:**
\[ F = 0.001 \cdot 2 \cdot 2 = 0.004 \, \text{N} \]

5. **Convert to standard form:**
\[ F = 4 \times 10^{-3} \, \text{N} \]

### Final Answer:
The tangential force needed to move the plate is \( 4 \times 10^{-3} \, \text{N} \).