An aircraft executes a horizontal loop with a speed of 150 m/s with its, wings banked at an angle of `12^(@)`. The radius of the loop is `(g=10m//s^(2))`

Options

A. `10.6 km`
B. 9.6 km
C. 7.4 km
D. 5.8 km

Answer

To find the radius of the loop executed by the aircraft, we can use the formula for the banking angle in circular motion. The formula is given by:

\[ \tan(\theta) = \frac{V^2}{Rg} \]

Where:
- \( \theta \) is the banking angle,
- \( V \) is the speed of the aircraft,
- \( R \) is the radius of the loop,
- \( g \) is the acceleration due to gravity.

Given:
- \( V = 150 \, \text{m/s} \)
- \( \theta = 12^\circ \)
- \( g = 10 \, \text{m/s}^2 \)

We need to find the radius \( R \).

### Step 1: Convert the angle to radians (if necessary)
In this case, we can directly use the angle in degrees since we will be using the tangent function.

### Step 2: Calculate \( \tan(12^\circ) \)
Using a calculator or trigonometric tables, we find:

\[ \tan(12^\circ) \approx 0.2126 \]

### Step 3: Substitute the values into the formula
We can rearrange the formula to solve for \( R \):

\[ R = \frac{V^2}{g \cdot \tan(\theta)} \]

Substituting the known values:

\[ R = \frac{(150)^2}{10 \cdot 0.2126} \]

### Step 4: Calculate \( R \)
Calculating the numerator:

\[ (150)^2 = 22500 \]

Calculating the denominator:

\[ 10 \cdot 0.2126 = 2.126 \]

Now substituting these values back into the equation for \( R \):

\[ R = \frac{22500}{2.126} \approx 10563.5 \, \text{m} \]

### Final Answer
The radius of the loop is approximately \( 10563.5 \, \text{m} \).

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