QJEEM24Phaeb2: JEE Main 2024 Physics - Class 11 JEEM24Phaeb2 Solution: A plane is in level flight at constant speed and each of its…

To solve this problem, we need to employ Bernoulli's equation, which is applied to describe the behavior of fluid flow. For a fluid in steady flow, the principle states that the sum of the pressure potential energy density, kinetic energy density, and the gravitational potential energy density has the same value at all points along a streamline. Since the plane is in level flight, we can ignore changes in gravitational potential energy. Bernoulli's equation can be written as:

$P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2,$

where,

• $P_1$ and $P_2$ are the static pressures on the lower and upper surfaces of the wing respectively,
• $v_1$ is the airspeed over the lower surface of the wing,
• $v_2$ is the airspeed over the upper surface of the wing,
• $\rho$ is the density of air,
• P_2 > P_1 because v_2 > v_1 .

First, let's convert the airspeeds to $\mathrm{m/s}$ :

$v_1 = 180 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{3600 \mathrm{s}} = 50 \mathrm{m/s},$

$v_2 = 252 \frac{\mathrm{km}}{\mathrm{h}} \times \frac{1000 \mathrm{m}}{3600 \mathrm{s}} = 70 \mathrm{m/s}.$

The pressure difference between the lower and upper wing surfaces can thus be calculated using Bernoulli's equation:

$\Delta P = P_1 - P_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2.$

Substitute the values (with $\rho = 1 \mathrm{~kg/m}^3$ ):

$\Delta P = \frac{1}{2} (1 \mathrm{~kg/m}^3) (70 \mathrm{m/s})^2 - \frac{1}{2} (1 \mathrm{~kg/m}^3) (50 \mathrm{m/s})^2,$

$\Delta P = \frac{1}{2} (4900 \mathrm{~kg/m \cdot s}^2) - \frac{1}{2} (2500 \mathrm{~kg/m \cdot s}^2),$

$\Delta P = \frac{1}{2} (2400 \mathrm{~kg/m \cdot s}^2),$

$\Delta P = 1200 \mathrm{~N/m}^2.$

The lift force generated by the pressure difference over one wing is $\Delta P \times \text{wing area}$ , and since there are two wings, we must double the lift force generated by one wing to find the total lift force sustaining the plane. The weight of the plane is effectively the lift force when in level flight at constant speed. So:

$\text{Lift} = 2 \times \Delta P \times \text{wing area},$

$\text{Lift} = 2 \times 1200 \mathrm{~N/m}^2 \times 40 \mathrm{~m}^2,$

$\text{Lift} = 2 \times 48000 \mathrm{N},$

$\text{Lift} = 96000 \mathrm{N}.$

To find the mass $m$ of the plane, we use Newton's second law, where lift force is equal to the weight ( $mg$ , where $g$ is the acceleration due to gravity):

$m g = \text{Lift},$

$m = \frac{\text{Lift}}{g},$

$m = \frac{96000 \mathrm{N}}{10 \mathrm{m/s}^2},$

$m = 9600 \mathrm{kg}.$

Therefore, the mass of the plane is 9600 kg.