The velocity of water in river is 180 km `h^(-1)` near the surface .If the river is 5 m deep, then the shearing stress between the surface layer and the bottom layer is ( coefficient of viscosity of water `eta =10^(-3)` Pa s)

Options

A. `10^(-2)N m^(-2)`
B. `10^(-3)N m^(-2)`
C. `10^(-4)N m^(-2)`
D. `10^(-5)N m^(-2)`

Explanation

To solve the problem of calculating the shearing stress between the surface layer and the bottom layer of the river, we can follow these steps:

### Step 1: Convert the velocity from km/h to m/s
The given velocity of water is 180 km/h. We need to convert this to meters per second (m/s) using the conversion factor \( \frac{5}{18} \).

\[ \text{Velocity} = 180 \, \text{km/h} \times \frac{5}{18} = 50 \, \text{m/s} \]

### Step 2: Identify the change in velocity (dv)
The change in velocity (dv) is the difference between the velocity at the surface and the velocity at the bottom layer. The velocity at the bottom layer is 0 m/s (since it is at rest), and the velocity at the surface is 50 m/s.

\[ dv = 50 \, \text{m/s} - 0 \, \text{m/s} = 50 \, \text{m/s} \]

### Step 3: Determine the depth (dx)
The depth of the river is given as 5 m. This will be our dx.

\[ dx = 5 \, \text{m} \]

### Step 4: Calculate the velocity gradient (dv/dx)
Now we can calculate the velocity gradient (dv/dx), which is the change in velocity per unit depth.

\[ \frac{dv}{dx} = \frac{50 \, \text{m/s}}{5 \, \text{m}} = 10 \, \text{s}^{-1} \]

### Step 5: Use the formula for shearing stress
The formula for shearing stress (\( \tau \)) in terms of viscosity (\( \eta \)) and the velocity gradient (\( \frac{dv}{dx} \)) is given by:

\[ \tau = \eta \frac{dv}{dx} \]

Given that \( \eta = 10^{-3} \, \text{Pa s} \), we can substitute the values into the equation.

\[ \tau = 10^{-3} \, \text{Pa s} \times 10 \, \text{s}^{-1} = 10^{-2} \, \text{Pa} \]

### Final Answer
The shearing stress between the surface layer and the bottom layer of the river is:

\[ \tau = 0.01 \, \text{Pa} \]

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