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Fluid Mechanics

Section: Physics

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Question : 5 of 18

Marks: +1, -0

An aeroplane of mass

4.5 \times 10^4 \text{ 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}

[AP EAPCET 22 May 2025 S2]

$500 \text{ N m}^{-2}$
$825 \text{ N m}^{-2}$
$600 \text{ N m}^{-2}$
$750 \text{ N m}^{-2}$

Solution:

Mass of the aeroplane,

m=4.5 \times 10^4 \text{ kg}

Total wing area,

A=600 \text{ m}^2

Acceleration due to gravity,

g=10 \text{ m/s}^2

The aeroplane is travelling at a constant height, meaning lift = weight.

Step 1. Lift

=

Pressure difference × Area

L=\Delta P \times A

Step 2. Lift = Weight

L=m g

So,

\Delta P \times A=m g

Step 3. Solve for pressure difference

\Delta P = \frac{m g}{A}

Substitute the known values:

\Delta P = \frac{4.5 \times 10^4 \times 10}{600}

\Delta P = \frac{4.5 \times 10^5}{600}

\Delta P = 750 \text{ N/m}^2

Answer:

750 \text{ N/m}^2

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