There is a 1 mm thick layer of oil between a flat plate of area `10^(-2)m^(2)` and a big plate. How much force is required to move the plate with a velocity of `1.5 cm//s^`. The coefficient of viscosity of oil is 1 poise

Options

A. `1.5xx10^(-3)N`
B. `1.3xx10^(-5)N`
C. `1.5xx10^(-2)N`
D. `1.5xx10^(2)N`

Answer

To solve the problem step by step, we will use the principles of fluid mechanics, particularly Newton's law of viscosity.

### Step 1: Write down the given data
- Area of the plate, \( A = 10^{-2} \, \text{m}^2 \)
- Thickness of the oil layer, \( \Delta x = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Velocity of the upper plate, \( V = 1.5 \, \text{cm/s} = 1.5 \times 10^{-2} \, \text{m/s} \)
- Coefficient of viscosity of oil, \( \eta = 1 \, \text{poise} = 0.1 \, \text{Pa.s} \)

### Step 2: Calculate the velocity gradient
The velocity gradient (\( \frac{du}{dy} \)) can be calculated using the formula:
\[ \frac{du}{dy} = \frac{V - 0}{\Delta x} = \frac{1.5 \times 10^{-2} \, \text{m/s}}{1 \times 10^{-3} \, \text{m}} = 15 \, \text{s}^{-1} \]

### Step 3: Apply Newton's law of viscosity
According to Newton's law of viscosity, the shear stress (\( \tau \)) is given by:
\[ \tau = \eta \frac{du}{dy} \]
Substituting the values:
\[ \tau = 0.1 \, \text{Pa.s} \times 15 \, \text{s}^{-1} = 1.5 \, \text{Pa} \]

### Step 4: Calculate the force required
The shear stress is also defined as the force per unit area:
\[ \tau = \frac{F}{A} \]
Rearranging this gives:
\[ F = \tau \times A \]
Substituting the values:
\[ F = 1.5 \, \text{Pa} \times 10^{-2} \, \text{m}^2 = 1.5 \times 10^{-2} \, \text{N} \]

### Final Answer
The force required to move the plate with a velocity of \( 1.5 \, \text{cm/s} \) is:
\[ F = 1.5 \times 10^{-2} \, \text{N} \]

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