Water at `20^(@)` is flowing in a pipe of radius `20.0 cm` . The viscosity of water at `20^(@)C` is `1.005` centipise. If the water's speed in the centre of the pipe is `3.00m//s`, what is water's speed:
(a) `10.0cm` from the centre of the pipe(half way between the centre and the walls)
(b) at the wall of the pipe?

Explanation

To solve the problem, we will use the concept of laminar flow in a pipe and apply the Hagen-Poiseuille equation. The velocity profile of a fluid flowing through a circular pipe is parabolic, and we can use the following relationship to find the velocity at different points in the pipe.

### Given Data:
- Radius of the pipe, \( R = 20.0 \, \text{cm} = 0.20 \, \text{m} \)
- Viscosity of water, \( \eta = 1.005 \, \text{cP} = 1.005 \times 10^{-3} \, \text{Pa.s} \) (since \( 1 \, \text{cP} = 1 \times 10^{-3} \, \text{Pa.s} \))
- Speed of water at the center of the pipe, \( V_1 = 3.00 \, \text{m/s} \)
- Distance from the center for part (a), \( r_1 = 10.0 \, \text{cm} = 0.10 \, \text{m} \)
- Distance from the center for part (b), \( r_2 = R = 20.0 \, \text{cm} = 0.20 \, \text{m} \)

### Step-by-Step Solution:

#### Part (a): Speed at 10.0 cm from the center of the pipe

1. **Use the velocity ratio equation:**
The velocity at a distance \( r \) from the center of the pipe can be related to the velocity at the center using the formula:
\[ \frac{V_1}{V_2} = \frac{R^2 - 0^2}{R^2 - r^2} \]
where \( V_2 \) is the velocity at distance \( r \) from the center.

2. **Substitute the known values:**
For \( r_1 = 10.0 \, \text{cm} = 0.10 \, \text{m} \):
\[ \frac{3.00}{V_2} = \frac{(0.20)^2 - 0^2}{(0.20)^2 - (0.10)^2} \]

3. **Calculate the squares:**
\[ (0.20)^2 = 0.04 \, \text{m}^2 \]
\[ (0.10)^2 = 0.01 \, \text{m}^2 \]

4. **Plug in the values:**
\[ \frac{3.00}{V_2} = \frac{0.04}{0.04 - 0.01} = \frac{0.04}{0.03} \]

5. **Simplify the equation:**
\[ \frac{3.00}{V_2} = \frac{4}{3} \]

6. **Cross-multiply to solve for \( V_2 \):**
\[ 3.00 \times 3 = 4 \times V_2 \implies 9.00 = 4V_2 \implies V_2 = \frac{9.00}{4} = 2.25 \, \text{m/s} \]

#### Part (b): Speed at the wall of the pipe

1. **Use the same velocity ratio equation:**
For \( r_2 = R = 20.0 \, \text{cm} = 0.20 \, \text{m} \):
\[ \frac{V_1}{V_2} = \frac{R^2 - 0^2}{R^2 - R^2} \]

2. **Notice the denominator:**
The denominator becomes zero:
\[ \frac{V_1}{V_2} = \frac{0.04}{0} \implies \text{undefined} \]

3. **Interpret the result:**
This indicates that the velocity at the wall of the pipe is zero:
\[ V_2 = 0 \, \text{m/s} \]

### Final Answers:
- (a) The speed of water at 10.0 cm from the center of the pipe is \( 2.25 \, \text{m/s} \).
- (b) The speed of water at the wall of the pipe is \( 0 \, \text{m/s} \).