To solve the problem of finding the shearing stress between the horizontal layers of water in a river, we will follow these steps:
### Step 1: Convert the velocity from km/h to m/s
The velocity of water near the surface is given as \( 18 \, \text{km/h} \). We need to convert this to meters per second (m/s).
\[
\text{Velocity} = 18 \, \text{km/h} = \frac{18 \times 1000 \, \text{m}}{3600 \, \text{s}} = 5 \, \text{m/s}
\]
### Step 2: Calculate the velocity gradient (dv/dy)
The depth of the river is given as \( 5 \, \text{m} \). The velocity gradient (change in velocity with respect to depth) can be calculated using the formula:
\[
\frac{dv}{dy} = \frac{\Delta v}{\Delta y}
\]
Here, \( \Delta v \) is the change in velocity, which is \( 5 \, \text{m/s} \) (from the bottom where velocity is 0 to the surface where it is 5 m/s), and \( \Delta y \) is the depth of the river, which is \( 5 \, \text{m} \).
\[
\frac{dv}{dy} = \frac{5 \, \text{m/s}}{5 \, \text{m}} = 1 \, \text{s}^{-1}
\]
### Step 3: Use the formula for shear stress
The formula for shear stress (\( \tau \)) in a fluid is given by:
\[
\tau = \eta \cdot \frac{dv}{dy}
\]
Where:
- \( \tau \) is the shear stress,
- \( \eta \) is the coefficient of viscosity,
- \( \frac{dv}{dy} \) is the velocity gradient.
Given that the coefficient of viscosity of water \( \eta = 10^{-2} \, \text{poise} \), we need to convert this to SI units (Pascal-seconds):
\[
1 \, \text{poise} = 0.1 \, \text{Pa} \cdot \text{s} \Rightarrow 10^{-2} \, \text{poise} = 10^{-2} \times 0.1 \, \text{Pa} \cdot \text{s} = 10^{-3} \, \text{Pa} \cdot \text{s}
\]
### Step 4: Calculate the shear stress
Now we can substitute the values into the shear stress formula:
\[
\tau = 10^{-3} \, \text{Pa} \cdot \text{s} \cdot 1 \, \text{s}^{-1} = 10^{-3} \, \text{Pa}
\]
### Final Answer
The shearing stress between the horizontal layers of water is:
\[
\tau = 10^{-3} \, \text{N/m}^2
\]
