Air flows horizontally with a speed v= 106 km/hr. A house has plane roof of area `A = 20m^(2)` The magnitude of aerodynamic lift of the roof is

Options

A. `1.1 2 7 xx 10 ^(4) N`
B. `5.0 xx 10 ^(4) N`
C. `1.127 xx 10 ^(5) N`
D. `3.127 xx 10 ^(4)N`

Answer

To find the magnitude of aerodynamic lift on the roof of the house, we will use Bernoulli's theorem and the formula for lift. Here’s a step-by-step solution:

### Step 1: Convert the speed from km/hr to m/s
Given speed \( v = 106 \, \text{km/hr} \).

To convert km/hr to m/s, we use the conversion factor:
\[ 1 \, \text{km/hr} = \frac{5}{18} \, \text{m/s} \]
So,
\[ v = 106 \times \frac{5}{18} = 29.44 \, \text{m/s} \]

### Step 2: Identify the parameters for Bernoulli's equation
We have:
- \( v_1 = 29.44 \, \text{m/s} \) (speed of air above the roof)
- \( v_2 = 0 \, \text{m/s} \) (speed of air below the roof)
- The height \( h_1 \) and \( h_2 \) are approximately equal since the roof is flat.

### Step 3: Apply Bernoulli's theorem
According to Bernoulli's theorem:
\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]
Since \( v_2 = 0 \):
\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 \]
Rearranging gives:
\[ P_1 - P_2 = \frac{1}{2} \rho v_1^2 \]
This pressure difference \( \Delta P = P_1 - P_2 \) will be used to calculate the lift.

### Step 4: Calculate the pressure difference
Assuming the density of air \( \rho \approx 1.3 \, \text{kg/m}^3 \):
\[ \Delta P = \frac{1}{2} \cdot 1.3 \cdot (29.44)^2 \]
Calculating \( (29.44)^2 \):
\[ (29.44)^2 \approx 867.78 \]
Now, substituting this value:
\[ \Delta P = \frac{1}{2} \cdot 1.3 \cdot 867.78 \approx 561.06 \, \text{Pa} \]

### Step 5: Calculate the aerodynamic lift
The aerodynamic lift \( F \) can be calculated using the formula:
\[ F = \Delta P \cdot A \]
Where \( A = 20 \, \text{m}^2 \):
\[ F = 561.06 \, \text{Pa} \cdot 20 \, \text{m}^2 = 11221.2 \, \text{N} \]

### Final Answer
The magnitude of aerodynamic lift of the roof is approximately:
\[ F \approx 11221.2 \, \text{N} \]