126 + Interview Questions in Performance And Flight Mechanics in Aircraft Design Page 1 InterviewSolution

1.	Find the velocity for minimum thrust required at steady level flight if wing loading is 75N/m^2 and induced drag factor K is 0.0025. Consider CD0 as 0.02 and density as sea level.(a) 6.45m/s(b) 97m/s(c) 120m/s(d) 75m/sThe question was asked by my college professor while I was bunking the class.I need to ask this question from Flight Mechanics topic in section Performance and Flight Mechanics of Aircraft Design
Answer» Right choice is (a) 6.45m/s For EXPLANATION: GIVEN, CDO = 0.02, k = 0.0025. By solving these we GET answer as 6.45m/s.

Discussion

2.	Following diagram represents ______________(a) drag polar(b) weight polar(c) thrust polar(d) lift curve slopeThis question was addressed to me in a national level competition.Question is taken from Flight Mechanics in chapter Performance and Flight Mechanics of Aircraft Design
Answer» RIGHT option is (a) drag polar Easy explanation: Above diagram represents the typical drag polar. A drag polar is simple graphical representation of variation of drag coefficient with respect to LIFT coefficient or AOA. Drag polar can be USED to ESTIMATE drag characteristics. Lift curve slope is change in lift with respect to change in AOA.

Discussion

3.	If static pressure of flow is 1200psf and operating Mach number is 1.19 then, determine the external flow dynamic pressure.(a) 1189.52 psf(b) 12.89 psf(c) 112 psf(d) 2587 psfI got this question in an interview for internship.This interesting question is from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design
Answer» RIGHT option is (a) 1189.52 psf For explanation: DYNAMIC pressure = 0.7STATIC pressureMach number^2 = 0.712001.19^2= 1189.52 psf.

Discussion

4.	Which of the following is correct for a typical glider?(a) Drag D = Wcos (gamma)(b) Drag D = Warcsine (gamma)(c) Drag D = W(d) Drag D = Thrust TI had been asked this question in a job interview.My doubt stems from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design
Answer» Correct answer is (a) Drag D = Wcos (gamma) The EXPLANATION: Above EQUATION is used to provide information about drag of ATYPICAL glider. Drag can be given by, Drag D = Wcos (gamma) where, D is drag, W is WEIGHT of the glider and gamma is the GLIDE angle.

Discussion

5.	Consider the vertical velocity of the aircraft is 10m/s and horizontal velocity is 12 m/s. Determine the value of climb gradient.(a) 0.833(b) 1.89(c) 8(d) 2.483This question was posed to me during a job interview.This intriguing question comes from Flight Mechanics topic in chapter Performance and Flight Mechanics of Aircraft Design
Answer» Correct OPTION is (a) 0.833 The best I can explain: Climb GRADIENT = vertical VELOCITY/ horizontal velocity = 10/12 = 0.833.

Discussion

6.	All the gliders are unpowered.(a) True(b) FalseThe question was posed to me during a job interview.My question is from Flight Mechanics in chapter Performance and Flight Mechanics of Aircraft Design
Answer» CORRECT option is (a) True Easy EXPLANATION: Typically, gliders are built with zero thrust. HOWEVER, in modern days there gliders which operates with some power PLANT. Hence, statement is not correct. Not all the gliders are unpowered, some are powered as WELL.

Discussion

7.	If Total energy of an aircraft is found to be 1200 unit and kinetic energy is 1100 unit then, determine the potential energy.(a) 100 unit(b) 213 unit(c) 456 unit(d) 500 unitThe question was asked during an interview for a job.My enquiry is from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design
Answer» Right CHOICE is (a) 100 unit Explanation: Potential ENERGY = TOTAL energy – KINETIC energy = 1200 – 1100 = 100 unit.

Discussion

8.	Consider a glider is descending from 2km altitude. If it covers range of 30km then, at which aerodynamic efficiency the glider was flying?(a) 15(b) 60(c) 12(d) 21This question was posed to me in quiz.This interesting question is from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design
Answer» Correct ANSWER is (a) 15 The EXPLANATION is: Aerodynamic EFFICIENCY = range/altitude = 30/2 = 15.

Discussion

9.	Variometer is used for _____________(a) informing pilot when glider is in lift(b) increase lift(c) thrust reversal(d) increase thrust producedThe question was asked in quiz.My doubt is from Flight Mechanics topic in chapter Performance and Flight Mechanics of Aircraft Design
Answer» Correct option is (a) INFORMING pilot when glider is in lift To explain: Variometer is used to provide INFORMATION to the pilot about glider. It is used to provide data about when GLIDE is in lift. Thrust reversal are used to increase drag during LANDING and to reduce landing distance.

Discussion

10.	The level flight envelope is determined by ________(a) stall limit lines and from zero specific power(b) wing aileron size(c) spoilers area(d) tail areaI had been asked this question in an internship interview.Asked question is from Flight Mechanics topic in chapter Performance and Flight Mechanics of Aircraft Design
Answer» The CORRECT option is (a) stall limit lines and from zero SPECIFIC power Explanation: The level flight operating envelope is determined by using stall limit lines and zero specific power. The zero specific power LIT is typically shown for both maximum Thrust and for MILITARY thrust. Tail area is based on tail sizing.

Discussion

11.	Operating envelope is also known as ________(a) flight envelope(b) fly by wire(c) stall speed diagram(d) lift curveI had been asked this question in an international level competition.Enquiry is from Flight Mechanics in chapter Performance and Flight Mechanics of Aircraft Design
Answer» CORRECT choice is (a) flight envelope For EXPLANATION: OPERATING envelope is ALSO known as flight envelope. Fly by wire is a typical system used for converting pilot’s input to signal. Lift curve is used to illustrate the lift variation with respect to the angle of attack.

Discussion

12.	If flow has Mach number of 1.2 then, determine the ratio of total and external flow dynamic pressure.(a) 2.4(b) 1(c) 3.8(d) 5.6The question was asked in class test.The doubt is from Flight Mechanics in chapter Performance and Flight Mechanics of Aircraft Design
Answer» Correct option is (a) 2.4 Best explanation: GIVEN, Mach NUMBER M = 1.2 Ratio of TOTAL pressure and external flow dynamic pressure = (1.428/M^2)[1+0.2M^2]^3.5 = (1.428/1.2^2)[1+0.21.2^2]^3.5 = (1.428/1.2^2)*[1+0.288]^3.5 = 2.40.

Discussion

13.	Determine the acceleration of an aircraft which is in ground roll segment. Consider weight of the aircraft as 150kN, thrust as 110kN and drag as 10kN. Assume weight on the wheels as 60kN and rolling friction as 0.05.(a) 5.68 m/s^2(b) 100kN(c) 100 m(d) 3.5689 Kg*m/sThis question was addressed to me during an internship interview.My query is from Flight Mechanics topic in chapter Performance and Flight Mechanics of Aircraft Design
Answer» Right choice is (a) 5.68 m/s^2 Explanation: GIVEN, weight W=150KN, thrust T=100KN, drag D = 10KN, ROLLING friction f= 0.05, weight on wheels w = 60KN. Acceleration a = G(T- D – fw)/W = 9.81(100 -10-0.0560)/150 = 9.8187/150 = 5.68 m/s^2.

Discussion

14.	If stall speed of the aircraft is 10m/s then, find the approximate value of takeoff speed.(a) 11m/s(b) 1m/s(c) 9m/s(d) 7m/sThe question was asked in an interview for internship.This question is from Flight Mechanics in division Performance and Flight Mechanics of Aircraft Design
Answer» CORRECT choice is (a) 11m/s Easy explanation: TAKEOFF speed = 1.1STALL speed = 1.110 = 11m/s.

Discussion

15.	Find the required average velocity during flare for military aircraft if stall speed is 45 unit.(a) 51.75 unit(b) 60 unit(c) 45 unit(d) 100 unitI got this question during an interview.This is a very interesting question from Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design
Answer» RIGHT CHOICE is (a) 51.75 UNIT The best explanation: AVERAGE velocity during flare = 1.15stall SPEED = 1.1545 = 51.75 unit.

Discussion

16.	Find the approach speed for military aircraft. Considered stall speed is 22unit.(a) 26.4 unit(b) 20 m(c) 12 ft(d) 45 unitThis question was addressed to me in quiz.This key question is from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design
Answer» CORRECT option is (a) 26.4 unit Easy EXPLANATION: For military aircraft, APPROACH SPEED = 1.2STALL speed = 1.222 = 26.4 unit.

Discussion

17.	Find the value of lift coefficient to ensure power required minimum condition for cruise if CD0/K is given as 0.480.(a) 1.2(b) 2.1(c) 1.7(d) 7.5I have been asked this question in a job interview.The origin of the question is Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design
Answer» RIGHT OPTION is (a) 1.2 Easiest EXPLANATION: Lift coefficient = [3CD0/K]^0.5 = [30.480]^0.5 = 1.2.

Discussion

18.	Find the drag at which thrust required is minimum. Consider CD0 as 0.015 and reference area is 5m^2. Consider steady level flight with q of 15Pa.(a) 2.25N(b) 22.5N(c) 225N(d) 2250NThis question was posed to me during an internship interview.This intriguing question comes from Flight Mechanics topic in division Performance and Flight Mechanics of Aircraft Design
Answer» The CORRECT OPTION is (a) 2.25N Explanation: Given, steady level flight with CD0 of 0 015, q as 15Pa. Reference AREA S = 5m^2. Now, drag at thrust required MINIMUM is given by, D = 2CD0qS = 20.015155=2.25N.

Discussion

19.	An aircraft is operating with CL of 1.5 and has power required of 120W. If we want to reduce the power required by half then, find the value of CL at this power required condition. Consider steady level flight.(a) 2.38(b) 1.946(c) 9.45(d) 5.59The question was posed to me by my school principal while I was bunking the class.I need to ask this question from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design
Answer» Right answer is (a) 2.38 Explanation: GIVEN, P1 = 120W, CL1 = 1.5, P2 = P1/2 = 120/2 = 60W. Now, CL for new POWER required P2 is given by, CL2 = CL1[P1/P2]^(2/3) = 1.5[120/60]^(2/3) = 2.38.

Discussion

20.	Find instantaneous range of a jet propelled aircraft if lift to drag ratio is 12 and weight fraction at cruise W1/W0 is 1.9. Consider cruise speed of 120m/s and SFC of 0.00161 per second.(a) 574km(b) 574m(c) 574miles(d) 574ftI have been asked this question in unit test.Question is from Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design
Answer» The correct OPTION is (a) 574KM To explain I would SAY: Range = (LIFT to DRAG)(speed/SFC)ln (W1/W0) = 12(120/0.00161)ln (1.9) = 574Km.

Discussion

21.	Find lift to weight ratio if climb angle is 45°.(a) 0.707(b) 1(c) 1.34(d) 0.992This question was addressed to me during an interview.The above asked question is from Flight Mechanics in portion Performance and Flight Mechanics of Aircraft Design
Answer» Right OPTION is (a) 0.707 To elaborate: LIFT to WEIGHT ratio = COS(climb angle) = cos(45°) = 0.707.

Discussion

22.	Determine the value of climb angle if, excess thrust is 40 unit and weight of the aircraft is 60 unit. Consider steady climb.(a) 41.8(b) 50(c) 78(d) 12The question was asked in quiz.I need to ask this question from Flight Mechanics topic in section Performance and Flight Mechanics of Aircraft Design
Answer» Correct CHOICE is (a) 41.8 For explanation I would say: CLIMB angle = ARCSINE (excess Thrust/weight) = arcsine (40/60) = 41.8°.

Discussion

23.	A typical glider has weight of 150N and has glide angle of 2.5°. Find Lift produced by the glider.(a) 149.85N(b) 150kN(c) 322N(d) 120KnThis question was addressed to me in an online interview.The query is from Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design
Answer» Correct choice is (a) 149.85N Explanation: Lift = WEIGHT* COS (GLIDE angle) = 150*cos (2.5°) = 149.85 N.

Discussion

24.	If a sailplane has glide ratio of 45 then, find horizontal distance travelled by this sailplane. Consider altitude loss of 1000ft.(a) 45000ft(b) 1000ft(c) 3456ft(d) 987ftI had been asked this question in unit test.The question is from Flight Mechanics topic in division Performance and Flight Mechanics of Aircraft Design
Answer» The correct ANSWER is (a) 45000ft To EXPLAIN I WOULD SAY: Horizontal distance = glide RATIOaltitude loss = 451000 = 45000ft.

Discussion

25.	Find the total energy of an aircraft if kinetic energy and potential energy are 3600 and 1200 units respectively.(a) 4800 unit(b) 2800 unit(c) 1234 unit(d) 5789 unitThis question was addressed to me in an interview.My doubt stems from Flight Mechanics topic in chapter Performance and Flight Mechanics of Aircraft Design
Answer» RIGHT choice is (a) 4800 unit The best explanation: Total energy = KINETIC energy + potential energy = 3600 + 1200 = 4800 unit.

Discussion

26.	Determine the value of sink rate of glider which is operating with 4.2° of glide angle and has speed of 120m/s.(a) 8.78m/s(b) 9.8ms(c) 56.5ms(d) 8.7msThe question was posed to me at a job interview.My question is taken from Flight Mechanics topic in chapter Performance and Flight Mechanics of Aircraft Design
Answer» The CORRECT option is (a) 8.78m/s Easy EXPLANATION: Sink rate = SPEEDsine (GLIDE ANGLE) = 120sine (4.2°) = 8.78m/s.

Discussion

27.	Following diagram represents ____________(a) segments of typical takeoff(b) landing phase(c) cruise to climb mission segment(d) approachI had been asked this question in an online interview.My question is taken from Flight Mechanics in portion Performance and Flight Mechanics of Aircraft Design
Answer» The correct choice is (a) SEGMENTS of typical takeoff Explanation: Above diagram is illustrating the typical segments of a typical takeoff. As shown takeoff segment is subdivided into number of segments such as ground ROLL, transition etc. After ground roll aircraft follows a near CIRCULAR arc until it reaches CLIMB ANGLE as shown.

Discussion

28.	An aircraft weighing 200kg is in level turning operation. Find the value of vertical component of lift.(a) 1.962 KN(b) 200N(c) 2KN(d) 0.2KNThe question was posed to me during a job interview.My doubt stems from Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design
Answer» Right option is (a) 1.962 KN Easy explanation: Given, mass = 200kg WEIGHT = 200*9.81 = 1962 N. At Level TURN, Vertical component of lift is equal to weight. Hence, for given level turning FLIGHT Maneuver, vertical component of lift = weight = 1962N = 1.962KN.

Discussion

29.	During level turning flight, lift of wing is canted.(a) True(b) FalseThis question was posed to me in an online quiz.The above asked question is from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design
Answer» Right answer is (a) True The BEST I can explain: In a typical LEVEL turning FLIGHT Maneuver, the lift of the wing is canted. As a result of this, horizontal component of the lift will exert the centripetal FORCE required to turn. Typically, total lift is n times more than that of the weight of an aircraft. Hence, the above statement is true.

Discussion

30.	For a typical gliding flight, thrust will be ________(a) zero(b) equal to the lift(c) always same as drag(d) always same as weightI have been asked this question in an interview for internship.Asked question is from Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design
Answer» RIGHT answer is (a) zero To explain I would SAY: For a typical gliding flight, Thrust Will be zero. Gliding flight is often termed as unpowered flight. Lift and DRAG during gliding flight will be determined by using glide ANGLE.

Discussion

31.	Consider turning glide operation with bank angle of 1.5 and Weight of 125 unit. Find lift.(a) 125.042(b) 56(c) 120(d) 345This question was addressed to me in exam.Query is from Flight Mechanics topic in division Performance and Flight Mechanics of Aircraft Design
Answer» The correct option is (a) 125.042 For EXPLANATION I would SAY: LIFT= Weight/cosine (bank angle) = 125/cosine 1.5 = 125.042 unit.

Discussion

32.	Find the approximate value of time required to change energy height. Consider change in energy height is 250m and specific power for the operation is 100m/s.(a) 2.5s(b) 4.5min(c) 2.5hr(d) 2.59minThis question was addressed to me by my college director while I was bunking the class.This question is from Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design
Answer» Correct ANSWER is (a) 2.5s Explanation: Time REQUIRED to change energy height can be approximated as FOLLOWS, Time required = change in energy height/specific POWER = 250/100 = 2.5 s.

Discussion

33.	If Stall speed is 140fps then, find approximate radius of transition arc.(a) 4018 feet(b) 5015 km(c) 1m(d) 10cmThe question was posed to me in an interview.The query is from Flight Mechanics topic in section Performance and Flight Mechanics of Aircraft Design
Answer» Right OPTION is (a) 4018 FEET The explanation: Approximate VALUE of radius of TRANSITION arc in feet = 0.205Stall speed^2 = 0.205140*140 = 4018 feet.

Discussion

34.	Find the required amount of touchdown speed if stall speed is 100knots.(a) 115 knot(b) 120 knot(c) 100 knot(d) 99.2 knotI have been asked this question by my college director while I was bunking the class.My doubt is from Flight Mechanics in division Performance and Flight Mechanics of Aircraft Design
Answer» The CORRECT OPTION is (a) 115 knot To ELABORATE: Touchdown speed = 1.15stall speed = 1.15100 = 115 knot.

Discussion

35.	Which of the following is correct?(a) Aircraft decelerates from Approach to Touchdown speed during flare(b) Lift is always same as weight(c) Drag is always same as Thrust(d) Thrust loading is defined as the ratio of lift to dragThe question was posed to me in my homework.This intriguing question originated from Flight Mechanics topic in section Performance and Flight Mechanics of Aircraft Design
Answer» The CORRECT CHOICE is (a) Aircraft decelerates from Approach to Touchdown speed during FLARE The best explanation: During flare aircraft will decelerates from Approach speed to touchdown speed. Lift is not always same as WEIGHT. Drag is not always same as Thrust. During CRUISE, lift is equal to weight and Thrust is equal to drag.

Discussion

36.	We can design our aircraft to cruise for 3 different lift to drag ratio. Lift to drag ratio are as follows: 10, 12 and 11.25. Now, if we want to reduce thrust required to fly then, at which value of lift to drag ratio our aircraft needs to be designed so that it can operate with minimum thrust required?(a) 12(b) 10(c) 11.25(d) 10.62I have been asked this question in an online quiz.This key question is from Flight Mechanics topic in division Performance and Flight Mechanics of Aircraft Design
Answer» The correct answer is (a) 12 Explanation: Here, we’ve been asked to choose from 3 different lift to DRAG ratio so that our PRODUCT can operate with minimum thrust required. To have thrust required minimum at CRUISE we should design our aircraft with maximum value of lift to drag ratio. From given values, 12 is the highest value available. Hence, among given choices correct CHOICE WOULD be 12.

Discussion

37.	Let’s consider we are designing an aircraft which can have 0.075 rad/s as its turn rate during level turning operation. If load factor is set to be 1.1 then, at which velocity we should fly our aircraft to obtain mentioned requirement.(a) 59.8unit(b) 590unit(c) 5980unit(d) 0894unitI got this question during a job interview.I'd like to ask this question from Flight Mechanics topic in division Performance and Flight Mechanics of Aircraft Design
Answer» Correct option is (a) 59.8unit Best explanation: GIVEN, load FACTOR n=1.2, Turn rate = 0.075 rad/s Velocity V = \(\frac{G\SQRT{n^2-1}}{turn \,rate}\) = \(\frac{9.81\sqrt{1.2^2-1}}{0.075}\) = 59.8 UNIT.

Discussion

38.	In Steady level unaccelerated flight, lift is equal to weight.(a) True(b) FalseThis question was posed to me in examination.The question is from Flight Mechanics in chapter Performance and Flight Mechanics of Aircraft Design
Answer» Correct option is (a) True To elaborate: In steady LEVEL unaccelerated FLIGHT, lift is equal to weight of the aircraft. Drag is equal to THRUST of the aircraft. This is also CALLED cruise flight. For steady level unaccelerated flight, Lift = weight = QSCL.

Discussion

39.	Find the approximate value of climb angle if Thrust is 1500N, drag is 1000N and weight of the aircraft is 2500N.(a) 11.53 degree(b) 30 degree(c) 40 degree(d) 1 degreeI got this question in exam.Enquiry is from Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design
Answer» CORRECT CHOICE is (a) 11.53 DEGREE To explain: Given, steady climb, Thrust=1500N, drag D=1000N, WEIGHT W=2500N. Climb angle = arcsine [{T-D}/W] = arcsine [{1500-1000}/2500] = 11.53 degree.

Discussion

40.	Which of the following is correct value for rate of climb if an aircraft has free stream velocity of 125 m/s? Consider steady climb at climb angle of 10°.(a) 21.70 m/s(b) 12 m(c) 35 m/min(d) 1.302 min/mI got this question during an internship interview.I'd like to ask this question from Flight Mechanics topic in division Performance and Flight Mechanics of Aircraft Design
Answer» CORRECT option is (a) 21.70 m/s Explanation: RATE of CLIMB = free stream velocitysine (climb ANGLE) = 125sin (10) = 21.70 m/s.

Discussion

41.	If climb speed during takeoff is 120 unit then, find stall speed.(a) 100knots(b) 240knots(c) 1267knots(d) 215knotsI have been asked this question by my college director while I was bunking the class.My query is from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design
Answer» The correct answer is (a) 100KNOTS For explanation: STALL SPEED = CLIMB speed / 1.2 = 120/1.2 = 100knots.

Discussion

42.	Following diagram represents _________(a) specific power vs mach number(b) mach number for various propulsion systems(c) drag polar(d) lift curve slopeI had been asked this question during an interview for a job.The question is from Flight Mechanics topic in section Performance and Flight Mechanics of Aircraft Design
Answer» Right answer is (a) specific power vs mach number The best I can EXPLAIN: The above diagram is showing specific power vs Mach number relations. As shown in the diagram, specific power is INITIALLY increases with Mach number but then after certain value of Mach number it BEGINS to DECREASE.

Discussion

43.	Find the average velocity at transition if, stall speed is 15 unit.(a) 17.25 unit(b) 19 unit(c) 30 unit(d) 45 unitI have been asked this question by my college director while I was bunking the class.My question is based upon Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design
Answer» Correct choice is (a) 17.25 UNIT For explanation: AVERAGE VELOCITY = 1.15stall SPEED = 1.15 15 = 17.25 unit.

Discussion

44.	Find the approach speed if, stall speed is 12 unit.(a) 15.6 unit(b) 12 unit(c) 10 unit(d) 1 unitThe question was asked in homework.Question is from Flight Mechanics topic in chapter Performance and Flight Mechanics of Aircraft Design
Answer» RIGHT ANSWER is (a) 15.6 unit Explanation: APPROACH SPEED = 1.3* STALL speed = 1.3*12 = 15.6 unit.

Discussion

45.	Which of the following is correct?(a) Thrust required is minimum if L/D is maximum during cruise(b) Thrust required is minimum if L/D is minimum for cruise(c) Power required is minimum if L/D is maximum(d) Thrust required is maximum if L/D is maximum during steady stateThis question was addressed to me in final exam.Origin of the question is Flight Mechanics topic in division Performance and Flight Mechanics of Aircraft Design
Answer» Correct answer is (a) Thrust required is MINIMUM if L/D is MAXIMUM during cruise Easiest explanation: At steady level FLIGHT or in cruise, thrust LOADING or thrust to weight RATIO is inversely proportional lift to drag ratio. Hence thrust required will be minimum of lift to drag ratio is maximum.

Discussion

46.	If stall speed is 120 knots then, find average velocity at flare.(a) 147.6(b) 120(c) 110(d) 567I have been asked this question in unit test.This question is from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design
Answer» The correct CHOICE is (a) 147.6 Explanation: Average VELOCITY at Flare = 1.23STALL speed = 1.23120 = 147.6 KNOTS.

Discussion

47.	Following diagram represents ________(a) altitude effect on power required(b) altitude effect on thrust(c) drag polar(d) lift curveThe question was posed to me in exam.My doubt is from Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design
Answer» Right OPTION is (a) ALTITUDE effect on power required Easiest explanation: Above diagram is showing a typical power required graph. Above diagram is showing the altitude effects on the power required. Altitude effects on the thrust required will be different from the power required. Drag polar will show drag properties VARIATION with lift. Lift CURVE is used to represent lift coefficient variation with AOA.

Discussion

48.	The segment marked by ‘?’ in following diagram is showing _____(a) approach(b) climb(c) takeoff(d) cruiseThis question was posed to me in class test.The above asked question is from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design
Answer» Right option is (a) approach The explanation is: The term marked by? Is showing the Approach segment of AIRCRAFT landing phase. Climbing phase is used to INCREASE the altitude of the aircraft. Cruise segment has lift EQUAL to weight and Thrust equal to drag. Takeoff phase is DIFFERENT from the landing phase.

Discussion

49.	Let’s consider our aircraft has to generate thrust of 150KN during cruise. If at the cruise thrust loading is 0.86 then, at which value of Lift our aircraft is flying?(a) 174.4KN(b) 450KN(c) 100N(d) 174NI have been asked this question in a job interview.This is a very interesting question from Flight Mechanics in portion Performance and Flight Mechanics of Aircraft Design
Answer» RIGHT ANSWER is (a) 174.4KN Best explanation: For cruise lift is GIVEN by, L = Thrust/Thrust loading = 150/0.86 = 174.41KN.

Discussion

50.	If my glider has lift coefficient at maximum lift to drag as 1.5 and induced drag factor K is 0.007 then, find parasite drag coefficient.(a) 0.01575(b) 0.01 N(c) 0.02N(d) 0.0157NThis question was posed to me by my college director while I was bunking the class.I would like to ask this question from Flight Mechanics topic in division Performance and Flight Mechanics of Aircraft Design
Answer» The CORRECT CHOICE is (a) 0.01575 For explanation I WOULD SAY: PARASITE drag coefficient = Klift coefficient^2 = 0.0071.5*1.5 = 0.01575.

Discussion

Explore topic-wise InterviewSolutions in .

Following diagram represents ______________(a) drag polar(b) weight polar(c) thrust polar(d) lift curve slopeThis question was addressed to me in a national level competition.Question is taken from Flight Mechanics in chapter Performance and Flight Mechanics of Aircraft Design

Which of the following is correct for a typical glider?(a) Drag D = W*cos (gamma)(b) Drag D = W*arcsine (gamma)(c) Drag D = W(d) Drag D = Thrust TI had been asked this question in a job interview.My doubt stems from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design

All the gliders are unpowered.(a) True(b) FalseThe question was posed to me during a job interview.My question is from Flight Mechanics in chapter Performance and Flight Mechanics of Aircraft Design

Variometer is used for _____________(a) informing pilot when glider is in lift(b) increase lift(c) thrust reversal(d) increase thrust producedThe question was asked in quiz.My doubt is from Flight Mechanics topic in chapter Performance and Flight Mechanics of Aircraft Design

Operating envelope is also known as ________(a) flight envelope(b) fly by wire(c) stall speed diagram(d) lift curveI had been asked this question in an international level competition.Enquiry is from Flight Mechanics in chapter Performance and Flight Mechanics of Aircraft Design

If flow has Mach number of 1.2 then, determine the ratio of total and external flow dynamic pressure.(a) 2.4(b) 1(c) 3.8(d) 5.6The question was asked in class test.The doubt is from Flight Mechanics in chapter Performance and Flight Mechanics of Aircraft Design

If stall speed of the aircraft is 10m/s then, find the approximate value of takeoff speed.(a) 11m/s(b) 1m/s(c) 9m/s(d) 7m/sThe question was asked in an interview for internship.This question is from Flight Mechanics in division Performance and Flight Mechanics of Aircraft Design

Find the approach speed for military aircraft. Considered stall speed is 22unit.(a) 26.4 unit(b) 20 m(c) 12 ft(d) 45 unitThis question was addressed to me in quiz.This key question is from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design

Find lift to weight ratio if climb angle is 45°.(a) 0.707(b) 1(c) 1.34(d) 0.992This question was addressed to me during an interview.The above asked question is from Flight Mechanics in portion Performance and Flight Mechanics of Aircraft Design

A typical glider has weight of 150N and has glide angle of 2.5°. Find Lift produced by the glider.(a) 149.85N(b) 150kN(c) 322N(d) 120KnThis question was addressed to me in an online interview.The query is from Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design

Following diagram represents ____________(a) segments of typical takeoff(b) landing phase(c) cruise to climb mission segment(d) approachI had been asked this question in an online interview.My question is taken from Flight Mechanics in portion Performance and Flight Mechanics of Aircraft Design

An aircraft weighing 200kg is in level turning operation. Find the value of vertical component of lift.(a) 1.962 KN(b) 200N(c) 2KN(d) 0.2KNThe question was posed to me during a job interview.My doubt stems from Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design

During level turning flight, lift of wing is canted.(a) True(b) FalseThis question was posed to me in an online quiz.The above asked question is from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design

Consider turning glide operation with bank angle of 1.5 and Weight of 125 unit. Find lift.(a) 125.042(b) 56(c) 120(d) 345This question was addressed to me in exam.Query is from Flight Mechanics topic in division Performance and Flight Mechanics of Aircraft Design

If Stall speed is 140fps then, find approximate radius of transition arc.(a) 4018 feet(b) 5015 km(c) 1m(d) 10cmThe question was posed to me in an interview.The query is from Flight Mechanics topic in section Performance and Flight Mechanics of Aircraft Design

In Steady level unaccelerated flight, lift is equal to weight.(a) True(b) FalseThis question was posed to me in examination.The question is from Flight Mechanics in chapter Performance and Flight Mechanics of Aircraft Design

Find the approximate value of climb angle if Thrust is 1500N, drag is 1000N and weight of the aircraft is 2500N.(a) 11.53 degree(b) 30 degree(c) 40 degree(d) 1 degreeI got this question in exam.Enquiry is from Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design

If climb speed during takeoff is 120 unit then, find stall speed.(a) 100knots(b) 240knots(c) 1267knots(d) 215knotsI have been asked this question by my college director while I was bunking the class.My query is from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design

Find the approach speed if, stall speed is 12 unit.(a) 15.6 unit(b) 12 unit(c) 10 unit(d) 1 unitThe question was asked in homework.Question is from Flight Mechanics topic in chapter Performance and Flight Mechanics of Aircraft Design

If stall speed is 120 knots then, find average velocity at flare.(a) 147.6(b) 120(c) 110(d) 567I have been asked this question in unit test.This question is from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design

Following diagram represents ________(a) altitude effect on power required(b) altitude effect on thrust(c) drag polar(d) lift curveThe question was posed to me in exam.My doubt is from Flight Mechanics topic in portion Performance and Flight Mechanics of Aircraft Design

The segment marked by ‘?’ in following diagram is showing _____(a) approach(b) climb(c) takeoff(d) cruiseThis question was posed to me in class test.The above asked question is from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design

Which of the following is correct for a typical glider?(a) Drag D = Wcos (gamma)(b) Drag D = Warcsine (gamma)(c) Drag D = W(d) Drag D = Thrust TI had been asked this question in a job interview.My doubt stems from Flight Mechanics in section Performance and Flight Mechanics of Aircraft Design