A metal plate ofarea 100 sq. cm, rests on a layer of oil 2mm thick. A force of 0.1 N applied parallel to the plate horizontally keeps it moving with uniform speed of 1cm/ s. What is the coefficient of viscosity of oil ?

Options

A. `0.5Ns//m^2`
B. `1Ns//m^2`
C. `1.5Ns//m^2`
D. `2Ns//m^2`

Explanation

To find the coefficient of viscosity of the oil, we can follow these steps:

### Step 1: Identify the given values
- Area of the plate (A) = 100 sq. cm = \(100 \times 10^{-4}\) sq. m = \(0.01\) m²
- Thickness of the oil layer (dy) = 2 mm = \(2 \times 10^{-3}\) m
- Force applied (F) = 0.1 N
- Velocity of the plate (du) = 1 cm/s = \(1 \times 10^{-2}\) m/s

### Step 2: Calculate the shear stress (τ)
Shear stress (τ) is given by the formula:
\[ \tau = \frac{F}{A} \]
Substituting the values:
\[ \tau = \frac{0.1 \, \text{N}}{0.01 \, \text{m}^2} = 10 \, \text{N/m}^2 \]

### Step 3: Calculate the velocity gradient (du/dy)
The velocity gradient (du/dy) is given by:
\[ \frac{du}{dy} = \frac{du}{dy} = \frac{1 \times 10^{-2} \, \text{m/s}}{2 \times 10^{-3} \, \text{m}} = 5 \, \text{s}^{-1} \]

### Step 4: Use Newton's law of viscosity
According to Newton's law of viscosity:
\[ \tau = \mu \frac{du}{dy} \]
Where:
- τ = shear stress
- μ = coefficient of viscosity
- \(\frac{du}{dy}\) = velocity gradient

### Step 5: Rearrange to find the coefficient of viscosity (μ)
Rearranging the equation to solve for μ:
\[ \mu = \frac{\tau}{\frac{du}{dy}} \]
Substituting the values:
\[ \mu = \frac{10 \, \text{N/m}^2}{5 \, \text{s}^{-1}} = 2 \, \text{N s/m}^2 \]

### Final Answer
The coefficient of viscosity of the oil is:
\[ \mu = 2 \, \text{N s/m}^2 \]

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