When an object moves through a fluid, as when a ball falls through air or a glass sphere falls through water te fluid exerts a viscous foce F on the object this force tends to slow the object for a small sphere of radius `r` moving is given by stoke's law, `F_(w)=6pietarv`. in this formula `eta` in the coefficient of viscosity of the fluid which is the proportionality constant that determines how much tangential force is required to move a fluid layer at a constant speed v, when the layer has an area A and is located a perpendicular distance z from and immobile surface. the magnitude of the force is given by `F=etaAv//z`. For a viscous fluid to move from location 2 to location 1 along 2 must exceed that at location 1, poiseuilles's law given the volumes flow rate Q that results from such a pressure difference `P_(2)-P_(1)`. The flow rate of expressed by the formula `Q=(piR^(4)(P_(2)-P_(1)))/(8etaL)` poiseuille's law remains valid as long as the fluid flow is laminar. For a sfficiently high speed however the flow becomes turbulent flow is laminar as long as the reynolds number is less than approximately 2000. This number is given by the formula `R_(e)=(2overline(v)rhoR)/(eta)` In which `overline(v)` is the average speed `rho` is the density `eta` is the coefficient of viscosity of the fluid and R is the radius of the pipe. Take the density of water to be `rho=1000kg//m^(3)`
Q. Calculate the highest average speed that blood `(rho~~1000kg//m^(3)`) could have and still remain in laminar flow when it flows through the arorta `(R=8xx10^(-3)m`) Take the coeffiicient of viscosity of blood to be `4xx10^(-3)Pa-s`

Options

A. `0.5m//s`
B. `1.0m//s`
C. 1.5m/s
D. 2.0m/s

Answer

To solve the problem of finding the highest average speed that blood can have while still remaining in laminar flow through the aorta, we will use the Reynolds number formula and the given parameters.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Density of blood, \( \rho = 1000 \, \text{kg/m}^3 \)
- Radius of the aorta, \( R = 8 \times 10^{-3} \, \text{m} \)
- Coefficient of viscosity of blood, \( \eta = 4 \times 10^{-3} \, \text{Pa.s} \)
- The maximum Reynolds number for laminar flow, \( R_e = 2000 \)

2. **Write the Formula for Reynolds Number:**
The Reynolds number \( R_e \) is given by the formula:
\[ R_e = \frac{2 \overline{v} \rho R}{\eta} \]
where \( \overline{v} \) is the average speed of the fluid.

3. **Rearrange the Formula to Solve for Average Speed:**
To find the average speed \( \overline{v} \), we rearrange the formula:
\[ \overline{v} = \frac{R_e \cdot \eta}{2 \cdot \rho \cdot R} \]

4. **Substitute the Known Values into the Formula:**
Now we substitute the known values into the rearranged formula:
\[ \overline{v} = \frac{2000 \cdot (4 \times 10^{-3})}{2 \cdot (1000) \cdot (8 \times 10^{-3})} \]

5. **Calculate the Numerator:**
Calculate the numerator:
\[ 2000 \cdot (4 \times 10^{-3}) = 8 \, \text{(unit: Pa.s)} \]

6. **Calculate the Denominator:**
Calculate the denominator:
\[ 2 \cdot (1000) \cdot (8 \times 10^{-3}) = 16 \, \text{(unit: kg/m.s)} \]

7. **Calculate the Average Speed:**
Now, we can calculate \( \overline{v} \):
\[ \overline{v} = \frac{8}{16} = 0.5 \, \text{m/s} \]

8. **Conclusion:**
The highest average speed that blood could have while still remaining in laminar flow through the aorta is:
\[ \overline{v} = 0.5 \, \text{m/s} \]