A metal plate of area `0.02m^(2) ` is lying on a liquid layer of thickness `10^(-3)` m and coeffcient of viscosity 120 poise .Calcualate the horizontal force requirred to move the plate with a speed of `0.025ms^(-1)`

Answer

To solve the problem, we need to calculate the horizontal force required to move a metal plate lying on a liquid layer. We will use the formula for viscous force, which is given by:

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

Where:
- \( F \) is the horizontal force,
- \( \eta \) is the coefficient of viscosity,
- \( A \) is the area of the plate,
- \( v \) is the velocity of the plate,
- \( d \) is the thickness of the liquid layer.

### Step-by-Step Solution:

**Step 1: Convert the coefficient of viscosity from poise to SI units.**
- Given: \( \eta = 120 \) poise.
- 1 poise = 0.1 Pa·s, so:
\[ \eta = 120 \times 0.1 = 12 \, \text{Pa·s} \]

**Step 2: Convert the area from \( m^2 \) to \( cm^2 \).**
- Given: \( A = 0.02 \, m^2 \).
- To convert to \( cm^2 \):
\[ A = 0.02 \, m^2 \times (10000 \, cm^2/m^2) = 200 \, cm^2 \]

**Step 3: Convert the thickness from \( m \) to \( cm \).**
- Given: \( d = 10^{-3} \, m \).
- To convert to \( cm \):
\[ d = 10^{-3} \, m \times 100 = 0.1 \, cm \]

**Step 4: Use the formula to calculate the force.**
- Given: \( v = 0.025 \, m/s \) (which is already in SI units).
- Now substituting the values into the formula:
\[ F = \eta \cdot A \cdot \frac{v}{d} \]
\[ F = 12 \, \text{Pa·s} \cdot 0.02 \, m^2 \cdot \frac{0.025 \, m/s}{0.001 \, m} \]

**Step 5: Calculate the force.**
- First, calculate \( \frac{v}{d} \):
\[ \frac{0.025}{0.001} = 25 \, s^{-1} \]
- Now substituting back:
\[ F = 12 \cdot 0.02 \cdot 25 \]
\[ F = 12 \cdot 0.5 = 6 \, N \]

### Final Answer:
The horizontal force required to move the plate with a speed of \( 0.025 \, m/s \) is \( 6 \, N \).

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