Water is flowing on a horizontal fixed surface, such that its flow velocity varies with y (vertical direction) as `v=k((2y^(2))/(a^(2)) -(y^(3))/(a^(3)))`. If coefficient of viscosity for water is `eta`, what will be shear stress between layers of water at y =a.

Options

A. `(eta k)/(a)`
B. `(eta)/(ka)`
C. `(eta a)/( k)`
D. None of these

Answer

To find the shear stress between layers of water at \( y = a \), we will follow these steps:

### Step 1: Understand the velocity profile
The velocity of water flow is given by:
\[ v = k\left(\frac{2y^2}{a^2} - \frac{y^3}{a^3}\right) \]
This equation indicates that the velocity varies with the vertical position \( y \).

### Step 2: Apply Newton's law of viscosity
According to Newton's law of viscosity, the shear stress \( \tau \) is related to the viscosity \( \eta \) and the velocity gradient \( \frac{dv}{dy} \) as follows:
\[ \tau = \eta \frac{dv}{dy} \]

### Step 3: Differentiate the velocity equation
We need to differentiate the velocity \( v \) with respect to \( y \):
\[ \frac{dv}{dy} = \frac{d}{dy}\left(k\left(\frac{2y^2}{a^2} - \frac{y^3}{a^3}\right)\right) \]
Using the power rule for differentiation:
\[ \frac{dv}{dy} = k\left(\frac{d}{dy}\left(\frac{2y^2}{a^2}\right) - \frac{d}{dy}\left(\frac{y^3}{a^3}\right)\right) \]
Calculating the derivatives:
\[ \frac{d}{dy}\left(\frac{2y^2}{a^2}\right) = \frac{4y}{a^2}, \quad \frac{d}{dy}\left(\frac{y^3}{a^3}\right) = \frac{3y^2}{a^3} \]
Thus,
\[ \frac{dv}{dy} = k\left(\frac{4y}{a^2} - \frac{3y^2}{a^3}\right) \]

### Step 4: Evaluate \( \frac{dv}{dy} \) at \( y = a \)
Now, we substitute \( y = a \) into the expression for \( \frac{dv}{dy} \):
\[ \frac{dv}{dy}\bigg|_{y=a} = k\left(\frac{4a}{a^2} - \frac{3a^2}{a^3}\right) \]
Simplifying this gives:
\[ \frac{dv}{dy}\bigg|_{y=a} = k\left(\frac{4}{a} - \frac{3}{a}\right) = k\left(\frac{1}{a}\right) = \frac{k}{a} \]

### Step 5: Calculate the shear stress
Now we can substitute \( \frac{dv}{dy} \) back into the equation for shear stress:
\[ \tau = \eta \frac{dv}{dy} = \eta \left(\frac{k}{a}\right) \]

### Final Result
Thus, the shear stress between layers of water at \( y = a \) is:
\[ \tau = \frac{\eta k}{a} \]