By Bernuolli's equation we know that
[tex]\frac{1}{2}\rho v_1^2 + \rho gh_1 + P_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2[/tex]
now since we know that height is almost same for two sides of wing
now we will have
[tex]\frac{1}{2}\rho v_1^2 + P_1 = \frac{1}{2}\rho v_2^2 + P_2[/tex]
now pressure on lower side of wing must be more than the pressure on the upper side
[tex]P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2)[/tex]
so speed on the top is more than the speed at the bottom
so correct answer will be
If air speed is greater along the top surface of a bird's wings, pressure of the moving air there is REDUCED.
