A typical riverborne silt particle has a radius of `20(mu)m` and a density of `2xx10^(3) kg//m^(3)`. The viscosity of water is `1.0 mPI`. Find the terminal speed with which such a particle will settle to the bottom of a motionless volume of water.

Explanation

To find the terminal speed with which a silt particle settles in water, we can use the formula for terminal velocity derived from Stokes' law:

\[ V_t = \frac{2}{9} \cdot \frac{r^2 \cdot g \cdot (\rho_s - \rho_w)}{\eta} \]

Where:
- \( V_t \) = terminal velocity (m/s)
- \( r \) = radius of the particle (m)
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))
- \( \rho_s \) = density of the particle (kg/m³)
- \( \rho_w \) = density of the fluid (kg/m³)
- \( \eta \) = viscosity of the fluid (Pa·s)

### Step 1: Convert Units
- Given radius \( r = 20 \, \mu m = 20 \times 10^{-6} \, m \)
- Given density of the particle \( \rho_s = 2 \times 10^3 \, kg/m^3 \)
- Given density of water \( \rho_w = 1000 \, kg/m^3 \)
- Given viscosity \( \eta = 1.0 \, mPa \cdot s = 1.0 \times 10^{-3} \, Pa \cdot s \)

### Step 2: Substitute Values into the Formula
Now, we can substitute these values into the terminal velocity formula:

\[ V_t = \frac{2}{9} \cdot \frac{(20 \times 10^{-6})^2 \cdot 9.8 \cdot (2000 - 1000)}{1.0 \times 10^{-3}} \]

### Step 3: Calculate the Values
1. Calculate \( (20 \times 10^{-6})^2 \):
\[ (20 \times 10^{-6})^2 = 400 \times 10^{-12} \, m^2 \]

2. Calculate the difference in density \( \rho_s - \rho_w \):
\[ 2000 - 1000 = 1000 \, kg/m^3 \]

3. Substitute these values back into the equation:
\[ V_t = \frac{2}{9} \cdot \frac{400 \times 10^{-12} \cdot 9.8 \cdot 1000}{1.0 \times 10^{-3}} \]

4. Simplify the fraction:
\[ V_t = \frac{2}{9} \cdot \frac{400 \cdot 9.8 \cdot 1000 \times 10^{-12}}{10^{-3}} \]
\[ = \frac{2}{9} \cdot (400 \cdot 9.8 \cdot 1000 \times 10^{-9}) \]
\[ = \frac{2 \cdot 3920 \times 10^{-9}}{9} \]
\[ = \frac{7840 \times 10^{-9}}{9} \]

5. Calculate \( \frac{7840}{9} \):
\[ \approx 871.11 \times 10^{-9} \, m/s \]

### Step 4: Convert to mm/s
Convert \( V_t \) from m/s to mm/s:
\[ V_t \approx 0.871 \, mm/s \]

### Final Answer
The terminal speed with which the silt particle will settle to the bottom of the water is approximately:
\[ \text{Terminal speed} \approx 0.871 \, mm/s \]