The velocity of water in a river is 18 km/h near the upper surface. The river 5 m deep. What is the shearing stress between the horizontal layers of water ? (Coefficient of viscosity of water= `10^(- 2)` Poise)

Options

A. `10^(-1)N//m^2`
B. `10^(-2)N//m^2`
C. `10^(-3)N//m^2`
D. `10^(-4)N//m^2`

Explanation

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

### Step-by-Step Solution:

1. **Convert the velocity from km/h to m/s**:
\[ \text{Velocity} = 18 \, \text{km/h} = \frac{18 \times 1000 \, \text{m}}{3600 \, \text{s}} = 5 \, \text{m/s} \]

2. **Identify the parameters**:
- Coefficient of viscosity, \( \mu = 10^{-2} \, \text{Poise} = 0.01 \, \text{Poise} \)
- Convert Poise to SI units (Pascal-seconds):
\[ 1 \, \text{Poise} = 0.1 \, \text{Pa} \cdot \text{s} \implies 0.01 \, \text{Poise} = 0.01 \times 0.1 \, \text{Pa} \cdot \text{s} = 0.001 \, \text{Pa} \cdot \text{s} \]

3. **Determine the velocity difference (\( du \))**:
- The velocity at the upper layer is \( 5 \, \text{m/s} \) and at the bottom layer is \( 0 \, \text{m/s} \):
\[ du = 5 \, \text{m/s} - 0 \, \text{m/s} = 5 \, \text{m/s} \]

4. **Calculate the distance between the layers (\( dy \))**:
- The distance between the upper layer and the bottom layer is given as \( 5 \, \text{m} \):
\[ dy = 5 \, \text{m} \]

5. **Apply Newton's law of viscosity**:
- The shear stress (\( \tau \)) is given by:
\[ \tau = \mu \frac{du}{dy} \]
- Substituting the values:
\[ \tau = 0.001 \, \text{Pa} \cdot \text{s} \cdot \frac{5 \, \text{m/s}}{5 \, \text{m}} = 0.001 \, \text{Pa} \cdot \text{s} \cdot 1 = 0.001 \, \text{Pa} \]

6. **Final Result**:
- The shearing stress between the horizontal layers of water is:
\[ \tau = 10^{-3} \, \text{N/m}^2 \]