The velocity of upper layer of water in a river is 36 km`h^( -1)`. Shearing stress between horizontal layers of water is `10^(-3) Nm^(-2)` Depth of the river is __m. (Co-efficient of viscosity of water is `10^(-2)` Pa.s)

Explanation

To solve the problem, we need to find the depth of the river using the given parameters: the velocity of the upper layer of water, the shearing stress, and the coefficient of viscosity.

### Step-by-Step Solution:

**Step 1: Convert the velocity from km/h to m/s.**
- Given velocity of the upper layer of water = 36 km/h.
- To convert km/h to m/s, we use the conversion factor: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\).
- Therefore,
\[ V = 36 \text{ km/h} = \frac{36}{3.6} \text{ m/s} = 10 \text{ m/s}. \]

**Step 2: Write the formula for shear stress.**
- The shear stress (\(\tau\)) in a fluid is given by the formula:
\[ \tau = \eta \cdot \frac{\Delta V}{\Delta Y}, \]
where \(\eta\) is the coefficient of viscosity, \(\Delta V\) is the velocity difference, and \(\Delta Y\) is the depth of the fluid.

**Step 3: Substitute the known values into the shear stress formula.**
- Given:
- Shearing stress (\(\tau\)) = \(10^{-3} \text{ Nm}^{-2}\)
- Coefficient of viscosity (\(\eta\)) = \(10^{-2} \text{ Pa.s}\)
- Velocity of the upper layer (\(V\)) = \(10 \text{ m/s}\)
- We assume the velocity of the bottom layer is 0 (since it is stationary), thus \(\Delta V = V - 0 = 10 \text{ m/s}\).
- Substituting these values into the shear stress formula:
\[ 10^{-3} = 10^{-2} \cdot \frac{10}{D}. \]

**Step 4: Solve for the depth \(D\).**
- Rearranging the equation gives:
\[ D = 10^{-2} \cdot \frac{10}{10^{-3}}. \]
- Simplifying further:
\[ D = 10^{-2} \cdot 10 \cdot 10^{3} = 10^{-2} \cdot 10^{4} = 10^{2} = 100 \text{ m}. \]

### Final Answer:
The depth of the river is **100 m**.