Between a plate of area `100 cm^(2)` and another plate of area `100 m^(2)` there is a `1 mm`, thick layer of water, if the coefficient of viscosity of water is `0.01` poise, then the force required to move the smaller plate with a velocity `10 cms^(-1)` with reference to large plate is

Options

A. `100 dyn`
B. `10^(4)dyn`
C. `10^(6) dyn`
D. `10^(9)dyn`

Explanation

To solve the problem, we will use the formula for the force required to move a plate through a viscous fluid, which is derived from Newton's law of viscosity. The formula is given by:

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

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

### Step 1: Convert the given values to consistent units
- The area of the smaller plate is given as \( 100 \, \text{cm}^2 \). We will convert this to \( \text{m}^2 \):
\[ A = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 0.01 \, \text{m}^2 \]

- The coefficient of viscosity \( \eta \) is given as \( 0.01 \, \text{poise} \). We will convert this to SI units (Pascal-seconds):
\[ 1 \, \text{poise} = 0.1 \, \text{Pa} \cdot \text{s} \implies \eta = 0.01 \, \text{poise} = 0.01 \times 0.1 \, \text{Pa} \cdot \text{s} = 0.001 \, \text{Pa} \cdot \text{s} \]

- The velocity \( v \) is given as \( 10 \, \text{cm/s} \):
\[ v = 10 \, \text{cm/s} = 0.1 \, \text{m/s} \]

- The thickness of the water layer \( dy \) is given as \( 1 \, \text{mm} \):
\[ dy = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \]

### Step 2: Calculate the velocity gradient \( \frac{dv}{dy} \)
The velocity gradient \( \frac{dv}{dy} \) is calculated as:
\[ \frac{dv}{dy} = \frac{v}{dy} = \frac{0.1 \, \text{m/s}}{1 \times 10^{-3} \, \text{m}} = 100 \, \text{s}^{-1} \]

### Step 3: Substitute the values into the formula for force
Now we can substitute the values into the formula:
\[ F = \eta \cdot A \cdot \frac{dv}{dy} \]
Substituting the values we have:
\[ F = 0.001 \, \text{Pa} \cdot \text{s} \cdot 0.01 \, \text{m}^2 \cdot 100 \, \text{s}^{-1} \]
\[ F = 0.001 \cdot 0.01 \cdot 100 = 0.001 \, \text{N} \]

### Final Answer
The force required to move the smaller plate with a velocity of \( 10 \, \text{cm/s} \) is \( 0.001 \, \text{N} \).

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