A flat plate of 20 cm squre moves over another similar plate with a thin layer of 0.4 cm of a liquid between them. If a force of one kg. wt. moves one of the plates uniformly with a velocity with a velcoity of `1 ms^(-1)` calculate the strain rate, sheraing stres and coefficient of viscosity.

Explanation

To solve the problem step by step, we will calculate the strain rate, shearing stress, and coefficient of viscosity for the given scenario.

### Step 1: Calculate the area of the square plate
The side of the plate is given as 20 cm. To find the area in square meters:
\[ \text{Area} = \text{side}^2 = (20 \, \text{cm})^2 = 400 \, \text{cm}^2 \]
Converting cm² to m²:
\[ \text{Area} = 400 \, \text{cm}^2 \times \left( \frac{1 \, \text{m}}{100 \, \text{cm}} \right)^2 = 400 \times 10^{-4} \, \text{m}^2 = 0.04 \, \text{m}^2 \]

### Step 2: Calculate the force acting on the plate
The force is given as 1 kg weight. To convert this to Newtons:
\[ \text{Force} = 1 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 9.8 \, \text{N} \]

### Step 3: Convert the distance between the plates to meters
The distance between the plates is given as 0.4 cm. Converting this to meters:
\[ \text{Distance} = 0.4 \, \text{cm} = 0.4 \times 10^{-2} \, \text{m} = 0.004 \, \text{m} \]

### Step 4: Calculate the strain rate
The strain rate is defined as the change in velocity (dV) divided by the distance (dx):
\[ \text{Strain Rate} = \frac{dV}{dx} = \frac{1 \, \text{m/s}}{0.004 \, \text{m}} = 250 \, \text{s}^{-1} \]

### Step 5: Calculate the shearing stress
Shearing stress is calculated using the formula:
\[ \text{Shearing Stress} = \frac{\text{Force}}{\text{Area}} = \frac{9.8 \, \text{N}}{0.04 \, \text{m}^2} = 245 \, \text{N/m}^2 \]

### Step 6: Calculate the coefficient of viscosity
The coefficient of viscosity (η) can be calculated using the formula:
\[ \eta = \frac{\text{Shearing Stress}}{\text{Strain Rate}} = \frac{245 \, \text{N/m}^2}{250 \, \text{s}^{-1}} = 0.98 \, \text{Pa.s} = 0.98 \, \text{poise} \]

### Final Answers:
- Strain Rate: \( 250 \, \text{s}^{-1} \)
- Shearing Stress: \( 245 \, \text{N/m}^2 \)
- Coefficient of Viscosity: \( 0.98 \, \text{poise} \)