Bernoulli's Theorem PYQ | JEE Main Physics 2025 | Aircraft Lift Force

 

❓ Question

A fully loaded Boeing aircraft has

$$m=5.4\times10^5\ \text{kg},\qquad A=500\ \text{m}^2,\qquad g=10\ \text{m/s}^2.$$

It flies horizontally with average air speed

$$v_{\text{avg}}=1080\ \text{km/h}.$$

Given that the density of air is

$$\rho=1.2\ \text{kg/m}^3,$$

find the percentage fractional increase in the speed of air over the upper surface of the wing relative to the lower surface.

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✍️ Solution

For level flight, the lift force balances the weight of the aircraft.

$$F_L=mg.$$

Hence the pressure difference between the lower and upper surfaces is

$$\Delta P=\frac{F_L}{A} =\frac{mg}{A}.$$

Substituting the given values,

$$\Delta P = \frac{5.4\times10^5\times10}{500} = 1.08\times10^4\ \text{Pa}.$$

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Apply Bernoulli's Theorem

Since both wing surfaces are at nearly the same height,

$$P_1+\frac12\rho v_1^2 = P_2+\frac12\rho v_2^2.$$

Therefore,

$$2(P_1-P_2) = \rho\left(v_2^2-v_1^2\right).$$

Using

$$v_2^2-v_1^2=(v_2-v_1)(v_2+v_1),$$

we get

$$2\Delta P = \rho(v_2-v_1)(v_2+v_1).$$

Since

$$v_{\text{avg}} =\frac{v_1+v_2}{2},$$
$$v_1+v_2 = 2v_{\text{avg}}.$$

Hence,

$$v_2-v_1 = \frac{\Delta P}{\rho v_{\text{avg}}}.$$

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Calculate the Speed Difference

Convert the average speed:

$$1080\ \text{km/h} = 300\ \text{m/s}.$$

Therefore,

$$v_2-v_1 = \frac{1.08\times10^4}{1.2\times300} = 30\ \text{m/s}.$$

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Fractional Increase in Speed

$$\frac{v_2-v_1}{v_{\text{avg}}}\times100 = \frac{30}{300}\times100 = 10\%.$$

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✅ Final Answer

$$\boxed{10\%}$$