Question

A plane is in level flight at constant speed and each of its two wings has an area of $25\,{{\text{m}}^2}$. If the speed of the air is $180\,{\text{km}}\,{{\text{h}}^{ - 1}}$ over the lower wing and $234\,{\text{km}}\,{{\text{h}}^{ - 1}}$ over the upper wing surface, determine the plane’s mass. (Take air density to be ${\text{1}}\,{\text{km}}\,{{\text{m}}^{ - 3}}$ )

Answer 1

We use Bernoulli's equation and given information to find the difference of the pressure. We know that force is the product pressure and area. We substitute the required values to find the force. After that we can easily find the mass of the plane using the acceleration due to gravity.

Formula used:
We use Bernoulli’s equation to solve this solution,
${P_1} - {P_2} = \dfrac{1}{2}\rho \left( {{V_2}^2 - {V_1}^2} \right)$
Where,
${P_1}$is pressure of air over the lower wing
${P_2}$ is pressure of air over the upper wing
${V_1}$is speed of air over the lower wing
${V_2}$ is speed of air over the upper wing
$\rho$density of air.

Complete step by step answer:
Area of the plane wings,
A = 2 \times 25 \\\ \Rightarrow A = 50\,{{\text{m}}^{\text{2}}} \\\
Speed of air on the lower wing,
{V_1} = 180\,{\text{km}}\,{{\text{h}}^{ - 1}} \\\ \Rightarrow {V_1} = \dfrac{{180 \times 1000}}{{3600}} \\\ \Rightarrow {V_1} = 50\,{\text{m}}\,{{\text{s}}^{ - 1}} \\\
Speed of air on the upper wing,
{V_2} = 234\,{\text{km}}\,{{\text{h}}^{ - 1}} \\\ \Rightarrow {V_2} = \dfrac{{234 \times 1000}}{{3600}} \\\ \Rightarrow {V_2} = 65\,{\text{m}}\,{{\text{s}}^{ - 1}} \\\
Density of air is,
$\rho = 1\,{\text{kg}}\,{{\text{m}}^{ - {\text{3}}}}$
Let,
The pressure of air on the lower wing is ${P_1}$
The pressure of air on the upper wing is ${P_2}$
So,
Using Bernoulli’s equation, we can obtain the upward force on the plane,

${P_1} - {P_2} = \dfrac{1}{2}\rho \left( {{V_2}^2 - {V_1}^2} \right)$
\therefore F = \left( {{P_1} - {P_2}} \right)A = \dfrac{1}{2}\rho \left( {{V_2}^2 - {V_1}^2} \right)A \\\ \Rightarrow F = \left( {{P_1} - {P_2}} \right)A = \dfrac{1}{2} \times 1 \times \left( {{{65}^2} - {{50}^2}} \right) \times 50 \\\ \Rightarrow F = 43125\,{\text{N}} \\\
Therefore,
m = \dfrac{F}{g} \\\ \Rightarrow m = \dfrac{{43125}}{{9.8}} \\\ \Rightarrow m = 4400\,{\text{kg}} \\\
Hence, the required answer is $4400\,{\text{kg}}$.

So, the correct answer is “Option A”.

Additional Information:
Bernoulli’s equation: A statement of the conservation theory of energy suitable for moving fluids can be considered to be the Bernoulli Equation. In regions where the flow velocity is increased, the qualitative action that is typically branded with the word 'Bernoulli effect' is the lowering of fluid pressure. This reduction of pressure in a flow path constriction can seem counterintuitive, but when you consider pressure to be energy density, it seems less so. Kinetic energy must increase at the cost of pressure energy during the high velocity flow through the constriction.

Note:
Remember to change to unit of speed of air on upper and lower wing from ${\text{km}}\,{{\text{h}}^{ - 1}}$ to ${\text{m}}\,{{\text{s}}^{ - 1}}$. Also convert density of air ${\text{1}}\,{\text{km}}\,{{\text{m}}^{ - 3}}$ to $1\,{\text{kg}}\,{{\text{m}}^{ - {\text{3}}}}$. A significant expression relating to the pressure, height and velocity of a fluid at a point in its flow is the Bernoulli equation. In an idealized system, the relation between these fluid conditions along a streamline often equates to the same constant along that streamline.