Aircraft Design MCQ – Aerodynamics – Lift

1 - Question

 Following diagram represents ___________

a) uncambered wing lift curve
b) cambered airfoil lift curve
c) cambered wing lift curve slope
d) uncambered airfoil
View Answer Answer: a
Explanation: A typical diagram of lift curve is illustrated in above figure. The diagram is applicable to uncambered wing. Airfoil lift curve will be different from wing. Typically, lift coefficient of airfoil is greater than the wing.

2 - Question

Following diagram represents _______________

a) cambered wing lift curve
b) drag polar
c) thrust loading
d) power loading
View Answer Answer: a
Explanation: A typical, lift curve is shown for Cambered airfoil. As shown in the diagram, camber wing with cambered airfoils will produce lift at small angle. Drag polar is used to provide information about drag characteristics. Thrust loading is defined as ratio of weight to thrust.

3 - Question

At same value of maximum lift AOA, lift coefficient of Cambered wing is _________
a) higher than uncambered wing
b) lower than uncambered wing
c) same as cambered airfoil always
d) not dependent on camber
View Answer Answer: a
Explanation: Lift coefficient at maximum lift AOA for a cambered wing is higher than that of the uncambered wing. At maximum lift AOA, value of lift coefficient will be maximum as well. Further increase in angle will reduce the amount of lift generated by wing.

4 - Question

 Following diagram represents _______

a) lift curve slope vs mach number
b) drag polar
c) thrust required minimum
d) wing loading chart
View Answer Answer: a
Explanation: The above diagram is showing a typical variation between lift curve and Mach number. Drag polar is nothing but the drag variation with lift or angle of attack. It is used to estimate drag properties of airfoil and wing. Wing loading is defined as the ratio of weight to the reference area.

5 - Question

A 2D airfoil is operating at 0.6M. Find the approximate value of lift curve slope in per rad.
a) 7.85
b) 1.9
c) 2.3
d) 3.87
View Answer Answer: a
Explanation: Given, mach number M = 0.6 2D lift curve slope = 2*π / (1-M2)0.5 = 2*π / (1-0.62)0.5 = 7.85 per rad.

6 - Question

A supersonic 2D airfoil has Mach number of 1.5. Find Lift curve slope of the airfoil.
a) 0.062 per degree
b) 0.062 per rad
c) 24 per degree
d) 12.56 per degree
View Answer Answer: a
Explanation: Given, mach number M = 1.5 2D lift curve slope = 4 / (M2 – 1)0.5 = 4 / (1.52 – 1)0.5 = 3.57 per rad = 3.57/57.3 = 0.062 per degree.

7 - Question

If an airfoil is designed to provide lift curve slope as 7.8, then find airfoil efficiency. The actual lift curve slope of the airfoil is 7.4.
a) 94.87%
b) 45%
c) 15.6%
d) 98.7%
View Answer Answer: a
Explanation: Airfoil efficiency = Actual lift curve slope / theoretical or designed lift curve slope = 7.4/7.8 = 0.9487 = 94.87%.

8 - Question

A subsonic aircraft wing has aspect ratio of 8. Now we have been asked to provide winglet to this wing. Evaluate approximate value of effective aspect ratio of winglet is considered.
a) 9.6
b) 23
c) 32
d) 4.89
View Answer Answer: a
Explanation: Approximate effective aspect ratio = 1.2*Wing aspect ratio = 1.2*8 = 9.6.

9 - Question

A subsonic aircraft has unswept wing which has maximum lift coefficient of 1.4. Now, consider wing is provided sweep of 25° at quarter chord point. Now, determine the change in maximum lift coefficient due to the provision of sweep.
a) Maximum Lift coefficient will be reduced by 0.2580 amount
b) Increment by 1.245
c) Decreases by 0.0476 amount
d) Maximum lift coefficient will increase by 56%
View Answer Answer: a
Explanation: Given, unswept maximum lift coefficient CL1 = 1.4 Now, sweep of 25° is provided at quarter chord point. Hence sweep angle at quarter chord s = 25° Maximum lift coefficient CL2 = CL1*0.9*cos (a) = 1.4*0.9*cos (25°) = 1.1419. Now, change in maximum lift coefficient = CL2 – CL1 = 1.1419 – 1.4 = -0.2580. Hence, by implementing the sweep of 25° maximum lift coefficient is reduced by 0.2508 unit.

10 - Question

 Following diagram represents __________


a) plain flap
b) aileron
c) rudder
d) spoiler
View Answer Answer: a
Explanation: Flap is a high lift device. Flap is used to increase lift produced by wing during landing and takeoff. Aileron is used to bank the aircraft. Rudder provides yawing motion to the aircraft. Spoiler is used to provide additional drag during landing.

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