There is a 1mm thick layer of glycerine between a flat plate of area `100cm^(2)` and and a big plate. If te coefficient of viscosity of glycerine is `1.0kg//m-sec`, then how much force is required to move the plate with a velocity of 7 cm/sec.

Options

A. 3.5 N
B. 0.7 N
C. 1.4 N
D. None of these

Answer

To solve the problem, we will use the formula for viscous force, which is given by:

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

Where:
- \( F \) is the viscous force,
- \( \mu \) 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:**
- Coefficient of viscosity, \( \mu = 1 \, \text{kg/m·s} \)
- Area of the plate, \( A = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 0.01 \, \text{m}^2 \)
- Velocity, \( v = 7 \, \text{cm/s} = 7 \times 10^{-2} \, \text{m/s} \)
- Thickness of the glycerin layer, \( dx = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)

2. **Calculate the velocity gradient \( \frac{dv}{dx} \):**
\[ \frac{dv}{dx} = \frac{v}{dx} = \frac{7 \times 10^{-2} \, \text{m/s}}{1 \times 10^{-3} \, \text{m}} = 70 \, \text{s}^{-1} \]

3. **Substitute the values into the viscous force formula:**
\[ F = \mu \cdot A \cdot \frac{dv}{dx} \]
\[ F = 1 \, \text{kg/m·s} \cdot 0.01 \, \text{m}^2 \cdot 70 \, \text{s}^{-1} \]

4. **Calculate the force:**
\[ F = 1 \cdot 0.01 \cdot 70 = 0.7 \, \text{N} \]

Thus, the force required to move the plate with a velocity of 7 cm/s is **0.7 N**.