Question

An aeroplane of mass \( 4.5 \times 10^4 \) kg and total wing area of \( 600 \, \text{m}^2 \) is travelling at a constant height. The pressure difference between the upper and lower surfaces of its wings is (Acceleration due to gravity \( = 10 \, \text{m s}^{-2} \))

• A. \( 500 \, \text{N m}^{-2} \)
• B. \( 825 \, \text{N m}^{-2} \)
• C. \( 600 \, \text{N m}^{-2} \)
• D. \( 750 \, \text{N m}^{-2} \)

Hint:

For an aeroplane flying at a constant height (level flight): Lift Force = Weight of the aeroplane. Lift Force = Pressure Difference \( \times \) Wing Area (\( F_{lift} = \Delta P \cdot A \)). So, \( \Delta P \cdot A = mg \implies \Delta P = \frac{mg}{A} \). Pressure is Force/Area, so units are \( \text{N/m}^2 \) or Pascals (Pa).

Answer 1

The Correct Option is D

The aeroplane is travelling at a constant height.
This means the vertical forces are balanced.
The upward lift force \( F_{lift} \) generated by the wings must be equal to the weight \( mg \) of the aeroplane.
Mass \( m = 4.
5 \times 10^4 \) kg.
Acceleration due to gravity \( g = 10 \, \text{m s}^{-2} \).
Weight \( W = mg = (4.
5 \times 10^4 \text{ kg}) \times (10 \, \text{m s}^{-2}) = 4.
5 \times 10^5 \) N.
So, \( F_{lift} = 4.
5 \times 10^5 \) N.
The lift force is generated due to the pressure difference \( \Delta P \) between the lower and upper surfaces of the wings.
\( F_{lift} = \Delta P \times A_{wing} \), where \( A_{wing} \) is the total wing area.
Given \( A_{wing} = 600 \, \text{m}^2 \).
So, \( \Delta P = \frac{F_{lift}}{A_{wing}} \).
\[ \Delta P = \frac{4.
5 \times 10^5 \, \text{N}}{600 \, \text{m}^2} = \frac{450000}{600} \, \text{N m}^{-2} \] \[ \Delta P = \frac{4500}{6} \, \text{N m}^{-2} \] \[ \Delta P = \frac{1500}{2} = 750 \, \text{N m}^{-2} \] This matches option (4).