A river `10 m` deep is flowing at `5ms^(-1)`. The shearing stress between horizontal layers of the rivers is (`eta=10^-(3) SI` units)

Options

A. `10^(-3)N//m^(2)`
B. `0.8xx10^(-3)N//m^(2)`
C. `0.5xx10^(-3)N//m^(2)`
D. `1 N//m^(2)`

Explanation

To find the shearing stress between the horizontal layers of the river, we can follow these steps:

### Step 1: Understand the formula for shearing stress
The shearing stress (σ) can be defined as:
\[ \sigma = \eta \frac{dv}{dx} \]
where:
- \( \sigma \) is the shearing stress,
- \( \eta \) is the dynamic viscosity of the fluid,
- \( dv \) is the change in velocity,
- \( dx \) is the change in depth.

### Step 2: Identify the given values
From the problem statement, we have:
- Depth of the river, \( dx = 10 \, m \)
- Velocity of the river, \( dv = 5 \, m/s \)
- Dynamic viscosity, \( \eta = 10^{-3} \, \text{SI units} \)

### Step 3: Substitute the values into the formula
Now we can substitute the known values into the shearing stress formula:
\[ \sigma = \eta \frac{dv}{dx} = 10^{-3} \times \frac{5}{10} \]

### Step 4: Calculate the shearing stress
Now we perform the calculation:
\[ \sigma = 10^{-3} \times 0.5 = 0.5 \times 10^{-3} \, \text{N/m}^2 \]

### Step 5: Write the final answer
Thus, the shearing stress between the horizontal layers of the river is:
\[ \sigma = 0.5 \times 10^{-3} \, \text{N/m}^2 \]