The blood speed in a normal segment of a horizontal artery is 0.11 m/s . An abnormal segment of the artery is narrowed down by plaque to one-fourth the normal cross-sectional area . What is the difference in blood pressure between the normal and constricted segments of the artery ? (Take density `rho` of blood as 1060 `kg //m^(3)`)

Options

A. 96 Pa
B. 100 Pa
C. 91 Pa
D. 87 Pa

Answer

To find the difference in blood pressure between the normal and constricted segments of the artery, we can use Bernoulli's equation along with the principle of continuity. Here’s a step-by-step solution:

### Step 1: Understand the problem
We have a horizontal artery where the blood speed in the normal segment is given as \( V_1 = 0.11 \, \text{m/s} \). The cross-sectional area of the abnormal segment is one-fourth that of the normal segment, which implies that the velocity in the constricted segment \( V_2 \) will be greater than \( V_1 \).

### Step 2: Use the continuity equation
According to the continuity equation, the mass flow rate must remain constant. This can be expressed as:
\[ A_1 V_1 = A_2 V_2 \]
Given \( A_2 = \frac{1}{4} A_1 \), we can write:
\[ A_1 V_1 = \frac{1}{4} A_1 V_2 \]
Dividing both sides by \( A_1 \) (assuming \( A_1 \neq 0 \)):
\[ V_1 = \frac{1}{4} V_2 \]
Rearranging gives:
\[ V_2 = 4 V_1 \]

### Step 3: Substitute the known values
Now substituting \( V_1 = 0.11 \, \text{m/s} \):
\[ V_2 = 4 \times 0.11 = 0.44 \, \text{m/s} \]

### Step 4: Apply Bernoulli's equation
Bernoulli's equation states that for a horizontal flow:
\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]
Rearranging gives:
\[ P_1 - P_2 = \frac{1}{2} \rho V_2^2 - \frac{1}{2} \rho V_1^2 \]
Factoring out \( \frac{1}{2} \rho \):
\[ P_1 - P_2 = \frac{1}{2} \rho (V_2^2 - V_1^2) \]

### Step 5: Substitute the density and velocities
Given the density of blood \( \rho = 1060 \, \text{kg/m}^3 \):
\[ P_1 - P_2 = \frac{1}{2} \times 1060 \times (0.44^2 - 0.11^2) \]

### Step 6: Calculate \( V_2^2 \) and \( V_1^2 \)
Calculating the squares:
\[ 0.44^2 = 0.1936 \]
\[ 0.11^2 = 0.0121 \]
Now substituting these values:
\[ P_1 - P_2 = \frac{1}{2} \times 1060 \times (0.1936 - 0.0121) \]
\[ = \frac{1}{2} \times 1060 \times 0.1815 \]

### Step 7: Final calculation
Calculating the pressure difference:
\[ P_1 - P_2 = 530 \times 0.1815 \approx 96.2 \, \text{Pa} \]

Thus, the difference in blood pressure between the normal and constricted segments of the artery is approximately **96 Pa**.

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