91Ó°ÊÓ

The mass is \(m = 1.7 \times {10^6}\;{\rm{kg}}\).

The speed of air near the lower surface is \(v = 95\;{\rm{m/s}}\).

The surface area is \(A = 1200\;{{\rm{m}}^2}\).

In this problem, Bernoulli’s equation will be applied for calculating the speed of the air flow. Firstly, determine the difference in pressure at the lower and upper of the wings.

The relation from Bernoulli’s equation is given by,

\({v_0} = \sqrt {\frac{{2\left( {{P_{\rm{L}}} - {P_{\rm{U}}}} \right)}}{\rho } + {v^2}} \)…… (i)

Here, \({P_{\rm{L}}}\) and \({P_{\rm{U}}}\) are the pressure at lower and upper of the wings respectively and \(\rho \) is the density of air.

The relation of pressure difference is given by,

\(\begin{array}{l}\left( {{P_{\rm{L}}} - {P_{\rm{U}}}} \right) = \frac{F}{A}\\\left( {{P_{\rm{L}}} - {P_{\rm{U}}}} \right) = \frac{{mg}}{A}\end{array}\)……. (ii)

Here, \(g\) is the gravitational acceleration and \(F\) is the weight of the airplane.

On plugging the values of equation (ii) in the equation (i),

\(\begin{array}{l}{v_0} = \sqrt {\frac{{2mg}}{{\rho A}} + {v^2}} \\{v_0} = \sqrt {\frac{{2\left( {1.7 \times {{10}^6}\;{\rm{kg}}} \right)\left( {9.80\;{\rm{m/}}{{\rm{s}}^2}} \right)}}{{\left( {1.29\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {1200\;{{\rm{m}}^2}} \right)}} + {{\left( {95\;{\rm{m/s}}} \right)}^2}} \\{v_0} = 174.8\;{\rm{m/s}}\end{array}\)

Thus, the speed of the air flow is \(174.8\;{\rm{m/s}}\).