It is desired to design an aeroplane with a lift of $1000\,N{m^{ - 2}}$ of wing area. If the velocity of streamline flow part of the lower wing surface is $100\,m{s^{ - 1}}$, then the required velocity over the upper surface is (given density of air is $1.3\,kg{m^{ - 3}}$ ):
A. $57\,m{s^{ - 1}}$
B. $77.4\,m{s^{ - 1}}$
C. $107.4\,m{s^{ - 1}}$
D. $127.2\,m{s^{ - 1}}$
Answer
594.6k+ views
Hint: Here we have to use Bernoulli's principle to get the answer.
The cumulative mechanical energy of the flowing fluid, consisting of the gravitational potential elevation energy, the fluid pressure energy and the fluid motion kinetic energy, remains constant. Here we have taken Bernoulli’s equation and equated it on the left hand side and right hand side as the values for the lower wing surface are given and we have to find the values for the upper wing surface.
Complete step by step answer:
Bernoulli 's equation formula is the relationship between a fluid in a container's pressure, kinetic energy, and gravitational potential energy.
Bernoulli’s formula is given by:
$p + \dfrac{1}{2}\rho {v^2} + \rho gh = {\text{constant}}$
Where
$p$ is the pressure exerted
$v$ is the velocity
$\rho $ is the density of the liquid
$h$ is the height of the container.
Bernoulli’s equation is given by:
${p_1} + \dfrac{1}
{2}\rho {v_1}^2 + \rho g{y_1} = {p_2} + \dfrac{1}
{2}\rho {v_2}^2 + \rho g{y_2}$
Given,
Pressure difference, ${P_1} - {P_2} = 1000\,N{m^{ - 2}}$
Density, $\rho = 1.3\,kg{m^{ - 3}}$
Velocity of the streamline flow, ${v_1}^2 = 100\,m{s^{ - 1}}$
Using Bernoulli’s theorem, we get:
$
{P_1} + \dfrac{1}
{2}\rho {v_1}^2 = {P_2} + \dfrac{1}
{2}\rho {v_2}^2 \\
\Rightarrow \dfrac{1}
{2}\rho \left( {{v_2}^2 - {v_1}^2} \right) = {P_1} - {P_2} \\
\Rightarrow {v_2}^2 = \dfrac{{2\left( {{P_1} - {P_2}} \right)}}
{\rho } + {v_1}^2 \\
\Rightarrow {v_2}^2 = \dfrac{{2000}}
{{1.3}} + {\left( {100} \right)^2} \\
\Rightarrow {v_2} = 107.4\,m{s^{ - 1}} \\
$
So, the correct answer is “Option C”.
Additional Information:
- The definition of the theory of Bernoulli is the idea that an increase in the speed of a liquid causes a decrease in pressure and a decrease in the speed of a liquid creates an increase in pressure.
- The theory of Bernoulli, the physical principle proposed by Daniel Bernoulli, states that the friction inside the fluid reduces as the speed of a flowing fluid (liquid or gas) increases. Because in the narrower pipe, the speed is greater, the kinetic energy of the volume is greater.
- The theory of Bernoulli can be generalized to multiple conditions in real life. This theory, for instance, explains why aeroplane wings are bent around the top and why ships have to navigate when they move away from each other. The pressure above the wing is lower than below it, supplying the wing with lift from below.
Note:
Here we have to use the pressure difference and cannot use only one pressure. If only one value of pressure was given then we don’t have to use the pressure difference. Also, we have to see whether the units are in standard form or not. If the unit of one quantity is in standard form and the unit of another quantity in C.G.S then the answer would differ by a factor of ten or hundred.
The cumulative mechanical energy of the flowing fluid, consisting of the gravitational potential elevation energy, the fluid pressure energy and the fluid motion kinetic energy, remains constant. Here we have taken Bernoulli’s equation and equated it on the left hand side and right hand side as the values for the lower wing surface are given and we have to find the values for the upper wing surface.
Complete step by step answer:
Bernoulli 's equation formula is the relationship between a fluid in a container's pressure, kinetic energy, and gravitational potential energy.
Bernoulli’s formula is given by:
$p + \dfrac{1}{2}\rho {v^2} + \rho gh = {\text{constant}}$
Where
$p$ is the pressure exerted
$v$ is the velocity
$\rho $ is the density of the liquid
$h$ is the height of the container.
Bernoulli’s equation is given by:
${p_1} + \dfrac{1}
{2}\rho {v_1}^2 + \rho g{y_1} = {p_2} + \dfrac{1}
{2}\rho {v_2}^2 + \rho g{y_2}$
Given,
Pressure difference, ${P_1} - {P_2} = 1000\,N{m^{ - 2}}$
Density, $\rho = 1.3\,kg{m^{ - 3}}$
Velocity of the streamline flow, ${v_1}^2 = 100\,m{s^{ - 1}}$
Using Bernoulli’s theorem, we get:
$
{P_1} + \dfrac{1}
{2}\rho {v_1}^2 = {P_2} + \dfrac{1}
{2}\rho {v_2}^2 \\
\Rightarrow \dfrac{1}
{2}\rho \left( {{v_2}^2 - {v_1}^2} \right) = {P_1} - {P_2} \\
\Rightarrow {v_2}^2 = \dfrac{{2\left( {{P_1} - {P_2}} \right)}}
{\rho } + {v_1}^2 \\
\Rightarrow {v_2}^2 = \dfrac{{2000}}
{{1.3}} + {\left( {100} \right)^2} \\
\Rightarrow {v_2} = 107.4\,m{s^{ - 1}} \\
$
So, the correct answer is “Option C”.
Additional Information:
- The definition of the theory of Bernoulli is the idea that an increase in the speed of a liquid causes a decrease in pressure and a decrease in the speed of a liquid creates an increase in pressure.
- The theory of Bernoulli, the physical principle proposed by Daniel Bernoulli, states that the friction inside the fluid reduces as the speed of a flowing fluid (liquid or gas) increases. Because in the narrower pipe, the speed is greater, the kinetic energy of the volume is greater.
- The theory of Bernoulli can be generalized to multiple conditions in real life. This theory, for instance, explains why aeroplane wings are bent around the top and why ships have to navigate when they move away from each other. The pressure above the wing is lower than below it, supplying the wing with lift from below.
Note:
Here we have to use the pressure difference and cannot use only one pressure. If only one value of pressure was given then we don’t have to use the pressure difference. Also, we have to see whether the units are in standard form or not. If the unit of one quantity is in standard form and the unit of another quantity in C.G.S then the answer would differ by a factor of ten or hundred.
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