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. Blood vessel is 0.10 m in length and has a radius of `1.5xx10^(-3)`m blood flows at rate of `10^(-7)m^(3)//s` through this vessel. The pressure difference that must be maintained in this flow between the two ends of the vessel is 20 Pa what is the viscosity sufficient of blood?

Options

A. `2xx10^(-3)Pa-s`
B. `1xx10^(-3)Pa-s`
C. `4xx10^(-3)Pa-s`
D. `5xx10^(-4)Pa-s`

Answer

To find the viscosity coefficient of blood, we will use Poiseuille's law, which relates the flow rate \( Q \) of a fluid through a cylindrical vessel to the pressure difference \( P_2 - P_1 \), the radius \( R \) of the vessel, the length \( L \) of the vessel, and the viscosity \( \eta \) of the fluid.

### Step-by-Step Solution:

1. **Identify the Formula**:
The formula for the flow rate \( Q \) is given by:
\[ Q = \frac{\pi R^4 (P_2 - P_1)}{8 \eta L} \]
We need to rearrange this formula to solve for the viscosity \( \eta \).

2. **Rearranging the Formula**:
Rearranging the formula to isolate \( \eta \):
\[ \eta = \frac{\pi R^4 (P_2 - P_1)}{8 Q L} \]

3. **Substituting the Given Values**:
We have the following values from the problem:
- Length of the blood vessel \( L = 0.10 \, \text{m} \)
- Radius of the blood vessel \( R = 1.5 \times 10^{-3} \, \text{m} \)
- Flow rate \( Q = 10^{-7} \, \text{m}^3/\text{s} \)
- Pressure difference \( P_2 - P_1 = 20 \, \text{Pa} \)

Now, substituting these values into the rearranged formula:
\[ \eta = \frac{\pi (1.5 \times 10^{-3})^4 (20)}{8 (10^{-7}) (0.10)} \]

4. **Calculating \( R^4 \)**:
First, calculate \( (1.5 \times 10^{-3})^4 \):
\[ (1.5 \times 10^{-3})^4 = 5.0625 \times 10^{-12} \, \text{m}^4 \]

5. **Calculating the Numerator**:
Now calculate the numerator:
\[ \text{Numerator} = \pi (5.0625 \times 10^{-12}) (20) \approx 3.14 \times 5.0625 \times 10^{-12} \times 20 \approx 3.14 \times 1.0125 \times 10^{-10} \approx 3.18 \times 10^{-10} \]

6. **Calculating the Denominator**:
Now calculate the denominator:
\[ \text{Denominator} = 8 (10^{-7}) (0.10) = 8 \times 10^{-8} \]

7. **Final Calculation**:
Now, divide the numerator by the denominator to find \( \eta \):
\[ \eta = \frac{3.18 \times 10^{-10}}{8 \times 10^{-8}} = \frac{3.18}{8} \times 10^{-2} = 0.3975 \times 10^{-2} \approx 3.975 \times 10^{-3} \, \text{Pa.s} \]

8. **Rounding Off**:
Rounding off gives us:
\[ \eta \approx 4 \times 10^{-3} \, \text{Pa.s} \]

### Final Answer:
The viscosity coefficient of blood is approximately \( 4 \times 10^{-3} \, \text{Pa.s} \).