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(D) The lift force $F_L$ must balance the weight of the aircraft: $F_L = mg = (5.4 	imes 10^5 \, kg) 	imes (10 \, m/s^2) = 5.4 	imes 10^6 \, N$.The

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$10$

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(D) The lift force $F_L$ must balance the weight of the aircraft: $F_L = mg = (5.4 	imes 10^5 \, kg) 	imes (10 \, m/s^2) = 5.4 	imes 10^6 \, N$.The pressure difference $\Delta P = P_2 - P_1$ between the lower and upper surfaces is given by $\Delta P = \frac{F_L}{A} = \frac{5.4 	imes 10^6 \, N}{500 \, m^2} = 1.08 	imes 10^4 \, Pa$.Using Bernoulli's equation: $P_1 + \frac{1}{2} ho v_1^2 = P_2 + \frac{1}{2} ho v_2^2$,where $v_1$ is the speed on the lower surface and $v_2$ is the speed on the upper surface.$P_2 - P_1 = \frac{1}{2} ho (v_1^2 - v_2^2) = \frac{1}{2} ho (v_1 - v_2)(v_1 + v_2)$.The speed of the aircraft is $v = 1080 \, km/h = 1080 	imes \frac{5}{18} \, m/s = 300 \, m/s$. Assuming $v_1 + v_2 \approx 2v = 600 \, m/s$.$1.08 	imes 10^4 = \frac{1}{2} 	imes 1.2 	imes (v_2 - v_1) 	imes 600$.$1.08 	imes 10^4 = 360 	imes (v_2 - v_1) \implies v_2 - v_1 = \frac{10800}{360} = 30 \, m/s$.The fractional increase is $\frac{v_2 - v_1}{v} 	imes 100 = \frac{30}{300} 	imes 100 = 10 \%$.

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