A fighter plane is pulling out for a dive at a speed of `900 km//h`. Assuming its path to be a vertical circle of radius `2000 m` and its mass to be `16000 kg`, find the force exerted by the air on it at the lowest point. Take `g=9.8 m//s^(2)`

Options

A. `3.28xx10^(5)N`
B. `6.56xx10^(5)N`
C. `9.28xx10^(5)N`
D. `12.56xx10^(5)N`

Explanation

To solve the problem of finding the force exerted by the air on a fighter plane at the lowest point of its circular path, we can follow these steps:

### Step 1: Convert the speed from km/h to m/s
The speed of the plane is given as 900 km/h. We need to convert this to meters per second (m/s) using the conversion factor \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\).

\[ \text{Speed in m/s} = 900 \times \frac{1}{3.6} = 250 \text{ m/s} \]

### Step 2: Calculate the centripetal force required
The centripetal force (\(F_c\)) required to keep the plane moving in a circular path is given by the formula:

\[ F_c = \frac{mv^2}{r} \]

Where:
- \(m = 16000 \text{ kg}\) (mass of the plane)
- \(v = 250 \text{ m/s}\) (speed of the plane)
- \(r = 2000 \text{ m}\) (radius of the circular path)

Substituting the values into the formula:

\[ F_c = \frac{16000 \times (250)^2}{2000} \]

Calculating this gives:

\[ F_c = \frac{16000 \times 62500}{2000} = \frac{1000000000}{2000} = 500000 \text{ N} = 5 \times 10^5 \text{ N} \]

### Step 3: Calculate the gravitational force acting on the plane
The gravitational force (\(F_g\)) acting on the plane can be calculated using:

\[ F_g = mg \]

Where:
- \(g = 9.8 \text{ m/s}^2\)

Substituting the values:

\[ F_g = 16000 \times 9.8 = 156800 \text{ N} = 1.568 \times 10^5 \text{ N} \]

### Step 4: Calculate the net force exerted by the air on the plane at the lowest point
At the lowest point of the circular path, the net force (\(F_{net}\)) exerted by the air on the plane is the centripetal force minus the gravitational force (since both forces act in opposite directions).

\[ F_{net} = F_c - F_g \]

Substituting the values:

\[ F_{net} = 5 \times 10^5 - 1.568 \times 10^5 \]

Calculating this gives:

\[ F_{net} = (5 - 1.568) \times 10^5 = 3.432 \times 10^5 \text{ N} \]

### Final Answer
The force exerted by the air on the plane at the lowest point is approximately:

\[ F_{net} \approx 3.432 \times 10^5 \text{ N} \]

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