A metal square plate of `10cm` side rests on `2mm` thick caster oil layer. Calculate the horizontal force needed to move the plate with speed `3cm s^(-1) : (` Coefficient of viscosity of caster oil is 15 poise. )

Options

A. `2.25xx10^(-2)N`
B. `2.25xx10^(-1)N`
C. `2.25xx10^(-3)N`
D. `2.25xx10^(-4)N`

Answer

To solve the problem, we need to calculate the horizontal force required to move a metal square plate resting on a layer of castor oil. We will use the formula for viscous force:

\[ F = \eta \cdot A \cdot \frac{\Delta V}{\Delta X} \]

Where:
- \( F \) is the force,
- \( \eta \) is the coefficient of viscosity,
- \( A \) is the area of the plate,
- \( \Delta V \) is the velocity,
- \( \Delta X \) is the thickness of the oil layer.

### Step 1: Convert the dimensions to SI units
- The side of the square plate is given as \( 10 \, \text{cm} \). Convert this to meters:
\[ 10 \, \text{cm} = 10 \times 10^{-2} \, \text{m} = 0.1 \, \text{m} \]

- The thickness of the castor oil layer is \( 2 \, \text{mm} \). Convert this to meters:
\[ 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \]

- The speed is given as \( 3 \, \text{cm/s} \). Convert this to meters per second:
\[ 3 \, \text{cm/s} = 3 \times 10^{-2} \, \text{m/s} \]

### Step 2: Calculate the area of the square plate
The area \( A \) of the square plate can be calculated using the formula:
\[ A = \text{side}^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \]

### Step 3: Identify the coefficient of viscosity
The coefficient of viscosity \( \eta \) is given as \( 15 \, \text{poise} \). We need to convert this to SI units (Pascal-seconds):
\[ 1 \, \text{poise} = 0.1 \, \text{Pa} \cdot \text{s} \implies 15 \, \text{poise} = 15 \times 0.1 = 1.5 \, \text{Pa} \cdot \text{s} \]

### Step 4: Substitute the values into the formula
Now we can substitute the values into the formula for force:
\[ F = \eta \cdot A \cdot \frac{\Delta V}{\Delta X} \]
Substituting the values:
\[ F = (1.5 \, \text{Pa} \cdot \text{s}) \cdot (0.01 \, \text{m}^2) \cdot \frac{(3 \times 10^{-2} \, \text{m/s})}{(2 \times 10^{-3} \, \text{m})} \]

### Step 5: Calculate the force
Calculating the fraction:
\[ \frac{3 \times 10^{-2}}{2 \times 10^{-3}} = 15 \]
Now substituting this back into the equation:
\[ F = 1.5 \cdot 0.01 \cdot 15 \]
\[ F = 1.5 \cdot 0.15 = 0.225 \, \text{N} \]

### Final Answer
The horizontal force needed to move the plate is:
\[ F = 0.225 \, \text{N} \quad \text{or} \quad 2.25 \times 10^{-1} \, \text{N} \]