Question

Let air be at rest at the front edge of a wing and air passing over the surface of the wing at a fast speed $v.$ If the density of air is $\rho $, the highest value for $v$ in streamline flow when atmospheric pressure is ${P_{atmosphere}}$ is :
A) ${\left[ {\dfrac{{{P_{atmosphere}}}}{\rho }} \right]^{\dfrac{{ - 1}}{2}}}$
B) ${\left[ {\dfrac{{{P_{atmosphere}}}}{\rho }} \right]^2}$
C) ${\left[ {\dfrac{{{P_{atmosphere}}}}{\rho }} \right]^{\dfrac{1}{2}}}$
D) $\left[ {\dfrac{{{P_{atmosphere}}}}{\rho }} \right]$

Answer 1

Hint: Use the Bernoulli’s Principle, we get that the addition of kinetic energy, potential energy, and internal energy is constant.
 ${P_1} + \dfrac{1}{2}\rho v_1^2 + \rho g{z_1} = {P_2} + \dfrac{1}{2}\rho v_2^2 + \rho g{z_2}$
Use the expression-
${P_{atmosphere}} = \rho gh$
Then, apply the equation of motion, ${V^2} = {U^2} + 2as$ and compare both the equations.

Complete step by step answer:
This question uses the Bernoulli’s principle so, we should know what is Bernoulli’s principle
Bernoulli’s Principle states that, in steady flow, the sum of all energy associated with the fluid along a streamline is the same at all points on streamline. This means that the sum of kinetic energy, potential energy and internal energy associated with fluid remains constant. Mathematically, it can be represented as:
${P_1} + \dfrac{1}{2}\rho v_1^2 + \rho g{z_1} = {P_2} + \dfrac{1}{2}\rho v_2^2 + \rho g{z_2}$
Now, according to the question
Let the density of air be $\rho $, atmospheric pressure be ${P_{atmosphere}}$ and height be $h$.
We know that,
The pressure at some height can be given by
${P_{atmospheric}} = \rho gh \cdots (1)$
Now, let the final speed of air be $V$ and the initial speed of air be $U$
As said in question that the air is at rest at the initial time
So, the initial speed of $U = 0$
Now, we have to find the highest value of $V$
$\therefore height = \dfrac{h}{2}$
Now, using the equation of motion which is-
${V^2} = {U^2} + 2as$
Here, $a = g$ because $g$ is the acceleration due to gravity and $s = \dfrac{h}{2}$ because it is the distance of ground from the plane.
Using the equation of motion
$
\Rightarrow {V^2} = 0 + 2 \times \dfrac{h}{2} \times g \\
\Rightarrow {V^2} = gh \cdots (2) \\
 $
Now, from equation $(1)$ and $(2)$, put the value of $gh$ from equation $(2)$ in equation $(1)$
$
  \Rightarrow {P_{atmospheric}} = \rho {V^2} \\
  \Rightarrow {V^2} = \dfrac{{{P_{atmospheric}}}}{\rho } \\
  \Rightarrow V = \sqrt {\dfrac{{{P_{atmospheric}}}}{\rho }} \\
 $
It can also be written as
\[\Rightarrow V = {\left( {\dfrac{{{P_{atmosphere}}}}{\rho }} \right)^{\dfrac{1}{2}}}\]

Therefore, option (C) is the correct option.

Note:
In airplanes, the difference in airspeed is calculated by Bernoulli’s principle so that the shape of wings is such that the air passes at high speed over the upper surface than the lower surface to create pressure difference.