Question

An aeroplane has a mass of \[1600\,{\text{kg}}\]. Area of each wing is \[50\,{{\text{m}}^2}\]. At certain altitude where density is \[1\,{\text{kg}} \cdot {{\text{m}}^{ - 3}}\] velocity of the air just above and below the wing are respectively, \[70\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] and \[50\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\]. The vertical acceleration of the aeroplane is [Assume speed of aeroplane is constant and all the lift is provided by the wings]
A. \[37.5\,{\text{m/}}{{\text{s}}^2}\] upward
B. \[4.6\,{\text{kg}} \cdot {{\text{m}}^{ - 3}}\] downward
C. \[2.3\,{\text{kg}} \cdot {{\text{m}}^{ - 3}}\] upward
D. \[2.3\,{\text{kg}} \cdot {{\text{m}}^{ - 3}}\] down

Answer 1

Hint: We can use the expression for Bernoulli’s principle. Using this principle, we can calculate the airlift on the aeroplane by considering the potential energy term for the upper and lower surface of the wings of aeroplane is the same. Then determine the acceleration of the aeroplane using the relation between the force and acceleration of an object.

Formula used:
The expression for Bernoulli’s principle is given by
\[{P_0} + \dfrac{1}{2}\rho {v^2} + \rho gh = {\text{constant}}\] …… (1)
Here, \[{P_0}\] is the atmospheric pressure, \[\rho \] is density of the fluid, \[v\] is velocity of the fluid, \[g\] is acceleration due to gravity and \[h\] is height of the fluid from the ground.

Complete step by step answer:
We have given that the mass of the aeroplane is \[1600\,{\text{kg}}\] and area of each wing of the aeroplane is \[50\,{{\text{m}}^2}\].
\[m = 1600\,{\text{kg}}\]
\[\Rightarrow A = 50\,{{\text{m}}^2}\]
The velocities of the air just above and below the wing are \[70\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] and \[50\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] respectively.
\[{v_{top}} = 70\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\]
\[\Rightarrow{v_{bottom}} = 50\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\]
We are asked to calculate the vertical acceleration of the aeroplane.We can use Bernoulli’s theorem to determine the air lift on the aeroplane due to its wings.Since the height difference between the upper and lower surface of the wings of aeroplane is very negligible, we can neglect the potential energy term in the Bernoulli’s theorem.

Let us rewrite equation (1) for the upper and lower surface of the aeroplane.
\[{P_{top}} + \dfrac{1}{2}\rho v_{top}^2 = {P_{bottom}} + \dfrac{1}{2}\rho v_{bottom}^2\]
Here, \[{P_{top}}\] and \[{P_{bottom}}\] are the atmospheric pressures on the upper and lower surfaces of the wings respectively.
\[ \Rightarrow {P_{bottom}} - {P_{top}} = \dfrac{1}{2}\rho v_{top}^2 - \dfrac{1}{2}\rho v_{bottom}^2\]
\[ \Rightarrow \Delta P = \dfrac{1}{2}\rho \left( {v_{top}^2 - v_{bottom}^2} \right)\]
\[ \Rightarrow \dfrac{F}{A} = \dfrac{1}{2}\rho \left( {v_{top}^2 - v_{bottom}^2} \right)\]
\[ \Rightarrow F = \dfrac{1}{2}\rho \left( {v_{top}^2 - v_{bottom}^2} \right)A\]
Here, \[F\] is the air lift on the wings of the plane.
Substitute \[1\,{\text{kg}} \cdot {{\text{m}}^{ - 3}}\] for \[\rho \], \[70\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] for \[{v_{top}}\], \[50\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] for \[{v_{bottom}}\] and \[50\,{{\text{m}}^2}\] for \[A\] in the above equation.
\[ \Rightarrow F = \dfrac{1}{2}\left( {1\,{\text{kg}} \cdot {{\text{m}}^{ - 3}}} \right)\left( {{{\left( {70\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)}^2} - {{\left( {50\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)}^2}} \right)\left( {50\,{{\text{m}}^2}} \right)\]
\[ \Rightarrow F = 25\left( {4900 - 2500} \right)\]
\[ \Rightarrow F = 60000\,{\text{N}}\]
Hence, the air lift on the aeroplane is \[60000\,{\text{N}}\].

Let us now calculate the vertical acceleration of the aeroplane.
The air lift on the wings of the plane is balanced by weight of the aeroplane.
\[F = ma\]
\[ \Rightarrow a = \dfrac{F}{m}\]
Substitute \[60000\,{\text{N}}\] for \[F\] and \[1600\,{\text{kg}}\] for \[m\] in the above equation.
\[ \Rightarrow a = \dfrac{{60000\,{\text{N}}}}{{1600\,{\text{kg}}}}\]
\[ \therefore a = 37.5\,{\text{m/}}{{\text{s}}^2}\]
Therefore, the vertical acceleration of the aeroplane is \[37.5\,{\text{m/}}{{\text{s}}^2}\] in the upward direction.

Hence, the correct option is A.

Note:The students may think how we determined the direction of the vertical acceleration of the aeroplane is in the upward direction. From the final answer, we can see that the value of the vertical acceleration of the aeroplane is positive which indicates that the direction of the acceleration of the aeroplane must be in the upward direction.