Consider a thin square plate floating on a viscous liquid in a large tank. The height h of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity ` u_0`. Which of the following statement is (are) true ?

Options

A. The resistive force of liquid on the plate is inversely proportional to h
B. The resistive force of liquid on the plate is independent of the area of the plate
C. The tangential (shear) stress on the floor of the tank increases with `u_0`
D. the tangential (shear) stress on the plate varies linearly with the viscosity ` eta ` of the liquid

Explanation

To solve the problem, we need to analyze the situation of a thin square plate floating on a viscous liquid and being pulled horizontally with a constant velocity \( u_0 \). We will evaluate the statements provided in the question based on the principles of fluid mechanics, particularly Newton's law of viscosity.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a tank filled with a viscous liquid, and the height \( h \) of the liquid is much smaller than the width of the tank.
- A thin square plate is floating on the surface of this liquid and is being pulled horizontally with a constant velocity \( u_0 \).

2. **Applying Newton's Law of Viscosity**:
- According to Newton's law of viscosity, the viscous force \( F \) acting on the plate can be expressed as:
\[ F = \eta A \frac{du}{dy} \]
- Where \( \eta \) is the dynamic viscosity of the liquid, \( A \) is the area of the plate, and \( \frac{du}{dy} \) is the velocity gradient perpendicular to the direction of flow.

3. **Determining the Velocity Gradient**:
- The velocity profile across the height of the liquid can be assumed linear due to the viscous drag. At the surface of the liquid (where the plate is), the velocity is \( u_0 \), and at the bottom of the liquid (at height \( h \)), the velocity is \( 0 \).
- Thus, the velocity gradient \( \frac{du}{dy} \) can be calculated as:
\[ \frac{du}{dy} = \frac{u_0 - 0}{h} = \frac{u_0}{h} \]

4. **Calculating the Viscous Force**:
- Substituting the velocity gradient back into the equation for viscous force:
\[ F = \eta A \frac{u_0}{h} \]
- This shows that the viscous force is directly proportional to the area \( A \) of the plate and inversely proportional to the height \( h \).

5. **Evaluating the Statements**:
- **Statement A**: The resistive force of the liquid is inversely proportional to \( h \).
- **True**: We derived that \( F \propto \frac{1}{h} \).
- **Statement B**: The resistive force on the plate is independent of the area.
- **False**: The resistive force is directly proportional to the area \( A \).
- **Statement C**: The shear stress on the floor of the tank will increase with \( u_0 \).
- **True**: As \( u_0 \) increases, the viscous force increases, leading to higher shear stress.
- **Statement D**: The tangential shear on the plate varies linearly with viscosity.
- **True**: The viscous force is directly proportional to the viscosity \( \eta \).

6. **Conclusion**:
- The correct statements are A, C, and D.

### Summary of Correct Statements:
- A: True
- B: False
- C: True
- D: True