An aircraft of mass $$4$$×$$105$$ kg with total wing $$arem2$$ in level flight at a speed of $$720$$ km $$h-1$$. The density of air at its height is $$1.2$$ kg $$m-3$$. The fractional increase in the speed of the air on the upper surface of its wings relative to the lower surface is (Take$$ g $$=$$ $$10$$ $$ms-2)

Question

An aircraft of mass $$4$$×$$105$$ kg with total wing $$arem2$$ in level flight at a speed of $$720$$ km $$h-1$$. The density of air at its height is $$1.2$$ kg $$m-3$$. The fractional increase in the speed of the air on the upper surface of its wings relative to the lower surface is (Take$$ g $$=$$ $$10$$ $$ms-2)

Infinity Learn · Accepted Answer

The weight of the aircraft is balanced by the upward force due to the Pressure difference.i.e.,  ∆P $$=$$ mgA $$=$$ (4$$×$$105kg)(10$$ $$ms-2)500$$ $$m2$$ $$=$$ $$45$$×$$104N$$ $$m-2$$      $$=$$ $$8$$×$$103Nm-2Let$$ $$v1$$, $$v2$$ are the speed of air on the lower and upper surface of the wings of the aircraft and $$P1$$, $$P2$$ are the pressure there.Using Bernoulli's theorem, we get           $$P1+12$$ρ$$v12$$  $$=$$ $$P2+$$ $$12$$ρ$$v22$$          $$P1-P2$$ $$=$$ $$12($$ρ$$v22-$$ρ$$v12)            ∆P $$=$$ ρ$$2(v2+v1)(v2-v1)$$        or  $$v2-v1$$ $$=$$ ∆PρvavHere, vav $$=$$ $$v1+v22$$ $$=$$ $$720$$ km $$h-1$$                $$=$$ $$720$$× $$518$$ $$ms-1$$ $$=$$ $$200$$ $$ms-1v2-v1vav$$ $$=$$ ∆Pρ$$v2av=$$ $$45$$×$$1041.2$$×$$(200)2$$                $$=$$  $$4$$×$$1045$$×$$1.2$$×$$4$$×$$104$$ $$=$$ $$0.17$$

An aircraft of mass 4×105 kg with total wing arem2 in level flight at a speed of 720 km h-1. The density of air at its height is 1.2 kg m-3. The fractional increase in the speed of the air on the upper surface of its wings relative to the lower surface is (Take g = 10 ms-2)

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