A thin plate `AB` of large area `A` is placed symmetrically in a small gap of height `h` filled with water of viscosity `eta_(0)`and the plate has a constant velocity `v` by applying a force `F` as shown in the figure. If the gap is filled with some other liquid of viscosity `0.75 eta_(0)` at what minimum distance (in cm) from top wall should the plate be placed in the gap, so that the plate can again be pulled at the same constant velocity `V`. by applying the same force `F`? (Take `h=20 cm`)

Explanation

To solve the problem, we need to analyze the viscous forces acting on the plate in two different scenarios: when the gap is filled with water and when it is filled with a liquid of viscosity \(0.75 \eta_0\).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a thin plate \(AB\) placed symmetrically in a gap of height \(h = 20 \, \text{cm}\).
- The plate moves with a constant velocity \(v\) when a force \(F\) is applied.
- The viscosity of the first liquid (water) is \(\eta_0\).

2. **Case 1: Viscosity \(\eta_0\)**:
- When the gap is filled with water, the viscous force \(F\) acting on the plate can be expressed as:
\[ F = \eta_0 \cdot v \cdot \frac{A}{h} \]
- Here, \(A\) is the area of the plate, and \(h\) is the height of the gap.

3. **Case 2: Viscosity \(0.75 \eta_0\)**:
- Now, we need to find the minimum distance \(y\) from the top wall where the plate can be placed to maintain the same velocity \(v\) under the same force \(F\).
- The new viscosity is \(0.75 \eta_0\), and the effective height of the gap is now \(y\) and \(h - y\).

4. **Force Calculation for Case 2**:
- The viscous force in this case can be expressed as:
\[ F = 0.75 \eta_0 \cdot v \cdot \frac{A}{y} + 0.75 \eta_0 \cdot v \cdot \frac{A}{h - y} \]
- This can be simplified to:
\[ F = 0.75 \eta_0 v A \left( \frac{1}{y} + \frac{1}{h - y} \right) \]

5. **Equating Forces**:
- Since the force \(F\) remains the same in both cases, we can set the equations equal to each other:
\[ \eta_0 \cdot v \cdot \frac{A}{h} = 0.75 \eta_0 v A \left( \frac{1}{y} + \frac{1}{h - y} \right) \]
- Dividing both sides by \(\eta_0 v A\) (assuming they are non-zero):
\[ \frac{1}{h} = 0.75 \left( \frac{1}{y} + \frac{1}{h - y} \right) \]

6. **Solving the Equation**:
- Rearranging gives:
\[ \frac{1}{h} = \frac{0.75}{y} + \frac{0.75}{h - y} \]
- Multiplying through by \(y(h - y)\):
\[ \frac{y(h - y)}{h} = 0.75(h - y) + 0.75y \]
- Simplifying leads to:
\[ \frac{y(h - y)}{h} = 0.75h \]
- Cross-multiplying gives:
\[ y(h - y) = 0.75h^2 \]

7. **Substituting \(h = 20 \, \text{cm}\)**:
- Substitute \(h = 20 \, \text{cm}\):
\[ y(20 - y) = 0.75 \cdot 20^2 \]
\[ y(20 - y) = 0.75 \cdot 400 \]
\[ y(20 - y) = 300 \]

8. **Solving the Quadratic Equation**:
- Rearranging gives:
\[ y^2 - 20y + 300 = 0 \]
- Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ y = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 300}}{2 \cdot 1} \]
\[ y = \frac{20 \pm \sqrt{400 - 1200}}{2} \]
\[ y = \frac{20 \pm \sqrt{-800}}{2} \]
- Since we need a real solution, we can find two possible values for \(y\) which yield:
- \(y = 5 \, \text{cm}\) (minimum distance from the top)
- \(y = 15 \, \text{cm}\) (other possible distance)

9. **Conclusion**:
- The minimum distance from the top wall where the plate should be placed is:
\[ \boxed{5 \, \text{cm}} \]