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. Ie., `10mm^(2)` . The narrower part has an area `5 mm^(2)`. The pressure drop in the artery is `22 Pa`. Density of the blood is 1.06x10 (3)kg m^(-3). The speed of the blood in the artery is

Options

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

Explanation

To solve the problem of finding the speed of blood in the artery using the given data, we will follow these steps:

### Step 1: Identify the Given Data
- Cross-sectional area of the wider part (A1) = 10 mm² = \(10 \times 10^{-6}\) m²
- Cross-sectional area of the narrower part (A2) = 5 mm² = \(5 \times 10^{-6}\) m²
- Pressure drop (ΔP) = 22 Pa
- Density of blood (ρ) = \(1.06 \times 10^3\) kg/m³

### Step 2: Use the Continuity Equation
According to the continuity equation, the product of the cross-sectional area and the velocity at two points in a fluid flow is constant:
\[ A_1 V_1 = A_2 V_2 \]
Substituting the values:
\[ 10 \times 10^{-6} \cdot V_1 = 5 \times 10^{-6} \cdot V_2 \]
From this, we can express \(V_2\) in terms of \(V_1\):
\[ V_2 = 2 V_1 \quad \text{(Equation 1)} \]

### Step 3: Apply Bernoulli's Equation
For horizontal flow, Bernoulli's equation states:
\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]
Rearranging gives:
\[ P_1 - P_2 = \frac{1}{2} \rho V_2^2 - \frac{1}{2} \rho V_1^2 \]
This can be simplified to:
\[ \Delta P = \frac{1}{2} \rho (V_2^2 - V_1^2) \quad \text{(Equation 2)} \]

### Step 4: Substitute the Pressure Drop
From the problem, we know:
\[ \Delta P = 22 \, \text{Pa} \]
Substituting this into Equation 2:
\[ 22 = \frac{1}{2} \cdot (1.06 \times 10^3) \cdot (V_2^2 - V_1^2) \]

### Step 5: Substitute \(V_2\) from Equation 1
Using \(V_2 = 2V_1\) in the equation:
\[ 22 = \frac{1}{2} \cdot (1.06 \times 10^3) \cdot ((2V_1)^2 - V_1^2) \]
This simplifies to:
\[ 22 = \frac{1}{2} \cdot (1.06 \times 10^3) \cdot (4V_1^2 - V_1^2) \]
\[ 22 = \frac{1}{2} \cdot (1.06 \times 10^3) \cdot (3V_1^2) \]

### Step 6: Solve for \(V_1\)
Multiplying both sides by 2:
\[ 44 = (1.06 \times 10^3) \cdot (3V_1^2) \]
Dividing both sides by \(1.06 \times 10^3\):
\[ V_1^2 = \frac{44}{1.06 \times 10^3 \cdot 3} \]
Calculating:
\[ V_1^2 = \frac{44}{3180} \approx 0.01384 \]
Taking the square root:
\[ V_1 \approx \sqrt{0.01384} \approx 0.1177 \, \text{m/s} \approx 0.12 \, \text{m/s} \]

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

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