An aeroplane is flying at a velocity of `"900 km h"^(-1)` loops a vertical circular loop. If the maximum force pressing the pilot against the seat is five times his weight, what would be the diameter (in m) of the loop? `[g=10ms^(-2)]`

Answer

To solve the problem, we need to determine the diameter of a vertical circular loop that an aeroplane is flying through, given that the maximum force pressing the pilot against the seat is five times his weight.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Velocity of the aeroplane, \( v = 900 \, \text{km/h} \)
- Maximum force pressing the pilot against the seat = \( 5 \times \text{weight of the pilot} \)
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)

2. **Convert Velocity to m/s:**
\[ v = 900 \, \text{km/h} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = 250 \, \text{m/s} \]

3. **Understanding Forces at the Bottom of the Loop:**
- At the bottom of the loop, the forces acting on the pilot are the normal force \( N \) (upwards) and the weight \( mg \) (downwards).
- According to the problem, the maximum normal force is \( N = 5mg \).

4. **Apply Newton's Second Law for Circular Motion:**
- The net force towards the center of the circular path provides the centripetal force:
\[ N - mg = \frac{mv^2}{r} \]
- Substituting \( N = 5mg \) into the equation:
\[ 5mg - mg = \frac{mv^2}{r} \]
- This simplifies to:
\[ 4mg = \frac{mv^2}{r} \]

5. **Canceling Mass \( m \):**
- Since \( m \) appears in all terms, we can cancel it out (assuming \( m \neq 0 \)):
\[ 4g = \frac{v^2}{r} \]

6. **Rearranging for Radius \( r \):**
\[ r = \frac{v^2}{4g} \]

7. **Substituting Known Values:**
- Substitute \( v = 250 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ r = \frac{(250)^2}{4 \times 10} = \frac{62500}{40} = 1562.5 \, \text{m} \]

8. **Calculating Diameter \( d \):**
\[ d = 2r = 2 \times 1562.5 = 3125 \, \text{m} \]

### Final Answer:
The diameter of the loop is \( 3125 \, \text{m} \).

---