A metal plate of area `10^(-2)m^(2)` is placed on a liquid layer of thickness `2xx10^(-3)m`. If the liquid has coefficient of viscosity 2 S.I. units the force required to move the plate with a velocity of `3(cm)/(s)` is

Options

A. 0.3 N
B. 0.03 N
C. 3 N
D. 30 N

Explanation

To solve the problem, we will use the formula for viscous force given by Newton's law of viscosity:

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

Where:
- \( F \) is the force required to move the plate,
- \( \eta \) is the coefficient of viscosity,
- \( A \) is the area of the plate,
- \( \frac{dv}{dx} \) is the velocity gradient.

### Step-by-Step Solution:

1. **Identify the given values:**
- Area of the plate, \( A = 10^{-2} \, m^2 \)
- Thickness of the liquid layer, \( dx = 2 \times 10^{-3} \, m \)
- Coefficient of viscosity, \( \eta = 2 \, \text{Pascal seconds (Pa·s)} \)
- Velocity of the plate, \( v = 3 \, \text{cm/s} = 3 \times 10^{-2} \, m/s \)

2. **Calculate the velocity gradient \( \frac{dv}{dx} \):**
- Since the plate is moving with a velocity \( v \) and the velocity at the bottom (in contact with the liquid) is zero, the change in velocity \( dv \) is equal to \( v \).
- Thus, \( dv = 3 \times 10^{-2} \, m/s \).
- The velocity gradient \( \frac{dv}{dx} \) is given by:
\[ \frac{dv}{dx} = \frac{dv}{dx} = \frac{3 \times 10^{-2}}{2 \times 10^{-3}} = \frac{3 \times 10^{-2}}{2 \times 10^{-3}} = \frac{3}{2} \times 10^{1} = 1.5 \times 10^{1} \, s^{-1} \]

3. **Substitute the values into the viscous force formula:**
\[ F = \eta \cdot A \cdot \frac{dv}{dx} \]
\[ F = 2 \cdot (10^{-2}) \cdot (1.5 \times 10^{1}) \]

4. **Calculate the force:**
\[ F = 2 \cdot 10^{-2} \cdot 1.5 \times 10^{1} = 3 \times 10^{-1} \, N \]
\[ F = 0.3 \, N \]

### Final Answer:
The force required to move the plate with a velocity of \( 3 \, \text{cm/s} \) is \( 0.3 \, N \).

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