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

Answer

To solve the problem of finding the shearing stress between the horizontal layers of a river, we can follow these steps:

### Step 1: Understand the given data
- Depth of the river (h) = 10 m
- Velocity of the river (v) = 5 m/s
- Viscosity (η) = \(10^{-3}\) SI units (Pa.s)

### Step 2: Identify the formula for shearing stress
The shearing stress (τ) between horizontal layers of a fluid can be expressed as:
\[ \tau = \eta \frac{dv}{dx} \]
where:
- τ is the shearing stress,
- η is the dynamic viscosity,
- \(\frac{dv}{dx}\) is the velocity gradient.

### Step 3: Calculate the velocity gradient
The velocity gradient \(\frac{dv}{dx}\) can be defined as the change in velocity (Δv) over the change in distance (Δx). Here, we can consider:
- Δv = 5 m/s (the velocity of the top layer),
- Δx = 10 m (the depth of the river).

Thus, the velocity gradient is:
\[ \frac{dv}{dx} = \frac{5 \, \text{m/s}}{10 \, \text{m}} = 0.5 \, \text{s}^{-1} \]

### Step 4: Substitute values into the shearing stress formula
Now, substituting the values of η and \(\frac{dv}{dx}\) into the shearing stress formula:
\[ \tau = 10^{-3} \, \text{Pa.s} \times 0.5 \, \text{s}^{-1} \]
\[ \tau = 0.5 \times 10^{-3} \, \text{Pa} = 5 \times 10^{-4} \, \text{Pa} \]

### Step 5: Convert the units if necessary
Since 1 Pa = 1 N/m², we can express the shearing stress as:
\[ \tau = 5 \times 10^{-4} \, \text{N/m}^2 \]

### Conclusion
The shearing stress between the horizontal layers of the river is:
\[ \tau = 5 \times 10^{-4} \, \text{N/m}^2 \text{ or } 0.5 \times 10^{-3} \, \text{N/m}^2 \]

### Final Answer
Thus, the correct option is:
\[ \text{Option 3: } 0.5 \times 10^{-3} \, \text{N/m}^2 \]

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