The flow of blood in a large artery of a anesthetized dog is diverted through a venturimeter. The wider part of the meter has a cross-sectional area equal to that of the artery A = `10mm^(2)`. The narrower part has an area a = `5 mm^(2)`. The pressure drop in the artery is `22 Pa`. Density of the blood is `1.06xx10^(3)kgm^(-3)`. What is the speed of blood in the artery?

Options

A. `0.12 ms^(-1)`
B. `0.62 ms^(-1)`
C. `0.24 ms^(-1)`
D. `0.42 ms^(-1)`

Explanation

To find the speed of blood in the artery, we can use the principles of fluid dynamics, specifically the Bernoulli's equation and the continuity equation. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
- Cross-sectional area of the artery, \( A = 10 \, \text{mm}^2 = 10 \times 10^{-6} \, \text{m}^2 \)
- Cross-sectional area of the narrower part of the venturimeter, \( a = 5 \, \text{mm}^2 = 5 \times 10^{-6} \, \text{m}^2 \)
- Pressure drop in the artery, \( P_1 - P_2 = 22 \, \text{Pa} \)
- Density of blood, \( \rho = 1.06 \times 10^3 \, \text{kg/m}^3 \)

### Step 2: Apply the Continuity Equation
According to the continuity equation, the flow rate must be constant. Therefore, we have:
\[ A \cdot V_1 = a \cdot V_2 \]
Where:
- \( V_1 \) is the speed of blood in the artery.
- \( V_2 \) is the speed of blood in the narrower part of the venturimeter.

From this, we can express \( V_2 \) in terms of \( V_1 \):
\[ V_2 = \frac{A}{a} \cdot V_1 \]

### Step 3: Apply Bernoulli's Equation
Using Bernoulli's principle, we can write:
\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]
Rearranging gives:
\[ P_1 - P_2 = \frac{1}{2} \rho V_2^2 - \frac{1}{2} \rho V_1^2 \]
Substituting the pressure drop:
\[ 22 = \frac{1}{2} \rho (V_2^2 - V_1^2) \]

### Step 4: Substitute \( V_2 \)
Substituting \( V_2 \) from the continuity equation into the Bernoulli equation:
\[ 22 = \frac{1}{2} \rho \left( \left(\frac{A}{a} V_1\right)^2 - V_1^2 \right) \]
\[ 22 = \frac{1}{2} \rho \left( \left(\frac{A^2}{a^2} V_1^2\right) - V_1^2 \right) \]
\[ 22 = \frac{1}{2} \rho V_1^2 \left( \frac{A^2}{a^2} - 1 \right) \]

### Step 5: Solve for \( V_1^2 \)
Rearranging gives:
\[ V_1^2 = \frac{2 \cdot 22}{\rho \left( \frac{A^2}{a^2} - 1 \right)} \]

### Step 6: Calculate \( \frac{A^2}{a^2} \)
Calculating \( \frac{A^2}{a^2} \):
\[ \frac{A^2}{a^2} = \frac{(10 \times 10^{-6})^2}{(5 \times 10^{-6})^2} = \frac{100 \times 10^{-12}}{25 \times 10^{-12}} = 4 \]
Thus, \( \frac{A^2}{a^2} - 1 = 4 - 1 = 3 \).

### Step 7: Substitute Values
Now substituting back into the equation for \( V_1^2 \):
\[ V_1^2 = \frac{2 \cdot 22}{1.06 \times 10^3 \cdot 3} \]
\[ V_1^2 = \frac{44}{3180} \approx 0.013846 \]
Taking the square root gives:
\[ V_1 \approx \sqrt{0.013846} \approx 0.1177 \, \text{m/s} \]

### Final Answer
The speed of blood in the artery is approximately \( V_1 \approx 0.12 \, \text{m/s} \).

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