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Aircraft Design - Performance And Flight Mechanics Quiz(MCQ)
1 .
Find climb angle of climb gradient is 0.6. Assume non accelerated flight.
A)
1.2
B)
4.5rad
C)
31°
D)
9.94°

Correct Answer :   31°

Explanation : Given, climb gradient G = 0.6.
Now, climb angle = arctan (G) = arctan (0.6) = 30.9 ∼ 31°.

2 .
Which of the following is correct for steady level flight?
A)
T=D
B)
T>D
C)
D<T
D)
L>W

Correct Answer :   T=D

Explaination : In steady level flight, flight path angle or climb angle is zero. Hence, conventional equation of motion reduces to the thrust T = Drag D. Steady level flight is unaccelerated and hence all the forces should give sum in respective direction as zero.

3 .
Find the drag at which thrust required is minimum. Consider CD0 as 0.015 and reference area is 5m2. Consider steady level flight with q of 15Pa.
A)
2.25N
B)
22.5N
C)
225N
D)
2250N

Correct Answer :   2.25N

Explaination : Given, steady level flight with CD0 of 0
015, q as 15Pa. Reference area S = 5m2.
Now, drag at thrust required minimum is given by,
D = 2*CD0*q*S = 2*0.015*15*5=2.25N.

4 .
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)
10
B)
10.62
C)
11.25
D)
12

Correct Answer :   12

Explaination :  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.

5 .
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)
100N
B)
174N
C)
174.4KN
D)
450KN

Correct Answer :   174.4KN

Explaination : For cruise lift is given by,
L = Thrust/Thrust loading = 150/0.86 = 174.41KN.

6 .
If an aircraft is in steady level flight and weighs 2500 kg then, find lift at the given flight condition.
A)
24.525KN
B)
24525kg
C)
2500 N
D)
35000kg

Correct Answer :   24.525KN

Explanation : Given, a steady level flight and mass m = 2500kg
Lift = m*g = 2500*9.81=24525N = 24.525 KN.

7 .
Following diagram represents ________

A)
thrust polar
B)
drag polar
C)
weight polar
D)
lift curve slope

Correct Answer :   drag polar

Explaination : 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.

8 .
An aircraft is designed with lift coefficient of 1.1 during steady level flight. If dynamic pressure is set to be 25.65 Pa then find weight of aircraft during this flight condition.
A)
28.215 N per unit area
B)
30 kg
C)
59 N
D)
125.25 N

Correct Answer :   28.215 N per unit area

Explaination : Given, steady level flight.
Lift coefficient CL=1.1, q=25.65Pa.
Since, area is not mentioned we will find weight as per unit area.
Weight can be given by,
W = q*S*CL = 25.65*1.1 = 28.215 N per area.

9 .
Following diagram represents _______

A)
lift curve
B)
lift polar
C)
drag polar
D)
thrust required curve

Correct Answer :   thrust required curve

Explaination : Above diagram is showing a typical Thrust required curve. Variation of Thrust required can be observed in the diagram. Lift curve is showing the relationship between lift and angle of attack. Drag polar is showing the graphical representation of the aircraft drag characteristics.

10 .
Following diagram represents _______

A)
lift curve
B)
drag polar
C)
lift induced thrust required
D)
zero lift thrust required

Correct Answer :   lift induced thrust required

Explaination : Para below it Lift induced thrust required is shown in the diagram. Lift induced thrust required and zero lift thrust required are not equal, they are two different properties. Lift curve and drag polar is used to provide information of the lift variation and drag variation respectively.

11 .
Calculate how much power will require to maintain cruise speed of 120 m/s if thrust required is 1KN. Consider steady level flight.
A)
120KW
B)
110N
C)
120N
D)
129W

Correct Answer :   120KW

Explanation : Power = thrust*velocity = 120*1 = 120KNm/s = 120KW.

12 .
Following diagram represents _____

A)
drag polar
B)
lift curve
C)
altitude effect on thrust
D)
altitude effect on power required

Correct Answer :   altitude effect on power required

Explaination : 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.

13 .
Determine the value of drag at which we can have minimum power if given reference area is 10m2 and zero lift drag coefficient is 0.5. Consider q as 120Pa.
A)
1.25
B)
9.85
C)
2400N
D)
4500 KN

Correct Answer :   2400N

Explaination : Drag at minimum power = q*area*[4*zero lift drag coefficient] = 120*10*4*0.5 = 2400N.

14 .
If power required at sea level altitude is 2005W then, determine power required at altitude at which density ratio is 0.9.
A)
2.113KW
B)
500W
C)
2354W
D)
211115W

Correct Answer :   2.113KW

Explanation : Power at altitude = Sea-level power*[1/density ratio]0.5
= 2005*[1/0.9]0.5 = 2113.45W = 2.113KW.

15 .
A prop driven aircraft has propeller efficiency of 0.92 and C as 0.000745 1/m. If lift to drag ratio is 14 and weight fraction as 1.145 then, find range of the propeller driven aircraft.
A)
2340km
B)
2340.95m
C)
2340.5miles
D)
2500miles

Correct Answer :   2340.95m

Explaination : Prop range = (lift to drag)*[propeller efficiency/C]*ln (weight fraction)
= 14*[0.92/0.000745]*ln (1.145) = 2340.95m.

16 .
What do you mean by rate of climb?
A)
Drag polar
B)
Thrust required
C)
Lift of an aircraft
D)
Vertical velocity of an aircraft

Correct Answer :   Vertical velocity of an aircraft

Explanation : Rate of climb is nothing but the vertical velocity of aircraft. It is the climbing rate of an aircraft. Lift is an aerodynamic force which holds the aircraft in the air. Weight is the force acting due to gravity. Thrust is propulsive force.

17 .
Ratio of vertical distance travelled to the horizontal distance travelled is known as _______
A)
thrust loss
B)
lift curve slope
C)
climb gradient
D)
power required

Correct Answer :   climb gradient

Explanation : Ratio of vertical distance travelled to the horizontal distance travelled is known as Climb gradient. Lift curve slope is change lift coefficient to the change in angle of attack. Power required is product of Thrust required and velocity of the aircraft.

18 .
Which of the following is correct in terms of fate of climb (R/C)?
A)
R/C = lift/drag
B)
R/C = excess power/weight
C)
R/C = excess lift/drag
D)
R/C = excess power/Thrust required

Correct Answer :   R/C = excess power/weight

Explanation : Rate of climb is nothing but the vertical velocity of the aircraft. Rate of climb can be defined as the ratio of excess power of the aircraft to the aircraft weight. Hence, Correct relation between rate of climb and power is as follows: R/C = excess power/weight. Where, R/C = rate of climb.

19 .
Find lift to weight ratio if climb angle is 45°.
A)
0.707
B)
0.992
C)
1
D)
1.34

Correct Answer :   0.707

Explanation : Lift to weight ratio = cos(climb angle) = cos(45°) = 0.707.

20 .
Find the approximate value of climb angle if Thrust is 1500N, drag is 1000N and weight of the aircraft is 2500N.
A)
1 degree
B)
11.53 degree
C)
30 degree
D)
40 degree

Correct Answer :   11.53 degree

Explanation : Given, steady climb, Thrust=1500N, drag D=1000N, weight W=2500N.
Climb angle = arcsine [{T-D}/W] = arcsine [{1500-1000}/2500] = 11.53 degree.

21 .
Consider steady climb from an altitude of 10km to 15km. If rate of climb is 20 m/s then, determine the time to climb.
A)
0.07 min
B)
4.16 hr
C)
25 min
D)
250s

Correct Answer :   250s

Explanation : Time to climb = change in altitude/rate of climb = (15-10)*1000/20 = 250s.

22 .
To operate at R/C of 7.08 m/s, determine the value of excess power. Consider weight of the aircraft as 13127.5 N.
A)
1.312KN
B)
13127 N
C)
92.942 kW
D)
1000W

Correct Answer :   92.942 kW

Explanation : Excess power = (R/C) * Weight = 7.08*13127.5 = 92.942 kW.

23 .
Corner speed is defined as _______
A)
maximum lift velocity
B)
maximum aerodynamic efficiency
C)
minimum instantaneous turn speed
D)
velocity for maximum instantaneous turn

Correct Answer :   velocity for maximum instantaneous turn

Explanation : Velocity for maximum instantaneous turn is termed as corner speed. Corner speed is defined by intersection of the stall limit and the Structural limit. For a typical fighter, corner speed can vary between 300 to 350 knots.

24 .
If load factor of level turning flight is found to be 1.2 and flight speed is 120m/s then, determine turn rate.
A)
0.03 rad per second
B)
1.2
C)
1.2 rad/s
D)
3.107 degree/s

Correct Answer :   3.107 degree/s

25 .
Following diagram represents __________

A)
level turn
B)
drag polar
C)
lift curve slope
D)
mach number limit

Correct Answer :   level turn

Explaination : A typical level turn is shown in the diagram. Lift curve slope is defined as the ratio of the change in lift coefficient and the change in AOA. Mach number limit is concerned with propulsion. Drag polar is graphical representation of drag.

26 .
During level turning operation my aircraft has load factor of 2. Find the required bank angle for this level turning operation.
A)
60°
B)
60rad
C)
2.1rad
D)
120rad

Correct Answer :   60°

Explaination : Bank angle = arccosine (1/load factor) = arccosine (1/2) = arccosine 0.5 = 60°.

27 .
If an aircraft can slow down during turn then, it is termed as __________
A)
straight flight
B)
instantaneous turn
C)
cruise segment
D)
non instantaneous turn

Correct Answer :   instantaneous turn

Explanation : If an aircraft can slow down during turn then, it is termed as Instantaneous turn. Here, load factor will be limited by maximum lift coefficient or structural strength of the aircraft. Cruise segment is part of mission profile of an aircraft. At cruise all forces and moments are balanced.

28 .
The aircraft _____________ during a sustained turn.
A)
can be slowed down
B)
can lower its velocity
C)
cannot be slowed down
D)
can increase lofting process effort

Correct Answer :   cannot be slowed down

Explanation : In sustained turn, aircraft is not allowed to slow down. During sustained turn aircraft cannot be slowed down or lose altitude during turn. In sustained turn thrust has to be equal to the drag. Lift is load factor times weight.

29 .
Term marked by ’?’ in the following diagram is called ______

A)
dive speed
B)
cruise speed
C)
free stream velocity
D)
corner speed

Correct Answer :   corner speed

Explaination : Above diagram is showing a typical relationship between turn rate and velocity. Above diagram is illustrating the concept of corner speed. Corner speed is velocity required for maximum instantaneous turn. Cruise speed of an aircraft is speed required to attain cruise condition or steady level flight.

30 .
Turn rate is defined as _______
A)
drag
B)
ratio of lift to drag
C)
product of thrust and velocity
D)
ratio of radial acceleration and the velocity

Correct Answer :   ratio of radial acceleration and the velocity

Explanation : Turn rate is defined as the ratio of radial acceleration Divided by the velocity. Typically, turn rate is denoted by degree per second. Lift to drag ratio is called aerodynamic efficiency of the aircraft. Thrust into velocity will result in power.

31 .
An aircraft weighing 200kg is in level turning operation. Find the value of vertical component of lift.
A)
0.2KN
B)
1.962 KN
C)
2KN
D)
200N

Correct Answer :   1.962 KN

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.

32 .
For a typical gliding flight, thrust will be ________
A)
zero
B)
equal to the lift
C)
always same as drag
D)
always same as weight

Correct Answer :   zero

Explanation : 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.

33 .
The tangent of glide angle is ________
A)
equal to thrust
B)
inverse of thrust
C)
lift to drag ratio
D)
equal to drag to lift ratio

Correct Answer :   equal to drag to lift ratio

Explanation : The tangent of glide angle is equal to the drag to lift ratio. Lift to drag ratio is called aerodynamic efficiency of the aircraft. Tangent of Glide angle is inverse of the Aerodynamic efficiency or inverse of the lift to drag ratio.

34 .
A typical glider has weight of 150N and has glide angle of 2.5°. Find Lift produced by the glider.
A)
120Kn
B)
322N
C)
149.85N
D)
150kN

Correct Answer :   149.85N

Explanation : Lift = weight* cos (glide angle) = 150*cos (2.5°) = 149.85 N.

35 .
Glider with glide angle of 4.2° is gliding with gliding ratio of _____
A)
12.9
B)
13.6
C)
15
D)
25.4

Correct Answer :   13.6

Explanation : Glide ratio = 1/tan (glide angle) = 1 / tan (4.2°) = 13.6.

36 .
Typically, a high performance unpowered aircraft is called ______
A)
jet
B)
glider
C)
sailplane
D)
twin engine

Correct Answer :   sailplane

Explanation : Typically, a high performance unpowered aircraft is termed as sailplane. A glider is crude, low performance unpowered aircraft. This is based on typical sailplane terminology.

37 .
Which of the following is correct for a typical glider?
A)
Drag D = W*cos (gamma)
B)
Drag D = W
C)
Drag D = Thrust T
D)
Drag D = W*arcsine (gamma)

Correct Answer :   Drag D = W*cos (gamma)

Explanation : Above equation is used to provide information about drag of atypical glider. Drag can be given by, Drag D = W*cos (gamma) where, D is drag, W is weight of the glider and gamma is the glide angle.

38 .
Determine the value of sink rate of glider which is operating with 4.2° of glide angle and has speed of 120m/s.
A)
8.7ms
B)
8.78m/s
C)
9.8ms
D)
56.5ms

Correct Answer :   8.78m/s

Explanation : Sink rate = speed*sine (glide angle) = 120*sine (4.2°) = 8.78m/s.

39 .
Sink rate can be used to __________
A)
estimate only wing loading
B)
calculate weight to thrust ratio
C)
determine and estimate takeoff gross weight
D)
determine the time a glider may remain in the air

Correct Answer :   determine the time a glider may remain in the air

Explanation : The time a glider may remain in the air can be determined by using the sink rate. Sink rate is the product of aircraft velocity and sine of the glide angle. It is the vertical velocity.

40 .
If a sailplane has glide ratio of 45 then, find horizontal distance travelled by this sailplane. Consider altitude loss of 1000ft.
A)
987ft
B)
1000ft
C)
3456ft
D)
45000ft

Correct Answer :   45000ft

Explanation : Horizontal distance = glide ratio*altitude loss = 45*1000 = 45000ft.

41 .
Graphical representation of sink rate of sailplane is called __________
A)
drag polar wing
B)
lift curve of wing
C)
typical speed polar
D)
propulsion systems

Correct Answer :   typical speed polar

Explanation : The graphical representation of sink rate for sailplane is called speed polar. Speed polar is also known as hodograph. Drag polar is graphical representation of drag characteristics. Lift curve is used to provide relationship between lift and angle of attack.

42 .
Velocity for the best glide ratio is 100ft/s. Determine velocity for minimum sink rate.
A)
76ft/s
B)
85ft/s
C)
124m/s
D)
219m/s

Correct Answer :   76ft/s

Explaination : Minimum sink rate velocity = 0.76*best glide ratio velocity = 0.76*100 = 76ft/s.

43 .
Variometer is used for _______
A)
increase lift
B)
thrust reversal
C)
increase thrust produced
D)
informing pilot when glider is in lift

Correct Answer :   informing pilot when glider is in lift

Explanation : 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.

44 .
Determine the value of best glide ratio velocity. Consider velocity for minimum sink rate is 10 fps.
A)
11 fps
B)
13.15 fps
C)
20 fps
D)
34 fps

Correct Answer :   13.15 fps

Explanation : Given, velocity for minimum sink rate v = 10 fps
Velocity for best glide ratio = v/0.76 = 10/0.76 = 13.15 fps.

45 .
Consider turning glide operation with bank angle of 1.5 and Weight of 125 unit. Find lift.
A)
56
B)
120
C)
125.042
D)
345

Correct Answer :   125.042

Explanation : Lift= Weight/cosine (bank angle) = 125/cosine 1.5 = 125.042 unit.

46 .
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)
21
C)
30
D)
49

Correct Answer :   15

Explanation : Aerodynamic efficiency = range/altitude = 30/2 = 15.

47 .
Find the range of glider which has L/D of 13.6 and is descending from 5km.
A)
12 m
B)
12 inch
C)
68 m
D)
68km

Correct Answer :   68km

Explanation : Range = altitude*L/D = 5*13.6 = 68km.

48 .
Total energy of an aircraft is defined as _______
A)
total kinetic energy only
B)
total potential energy only
C)
sum of the potential and kinetic energy
D)
product of kinetic and potential energy

Correct Answer :   sum of the potential and kinetic energy

Explanation : At any given point in time, the total energy of the aircraft or energy state of an aircraft is the sum of the potential energy and kinetic energy. Kinetic energy is equal to half of product of mass and square of velocity.

49 .
Total energy divided by the aircraft weight is term as _________
A)
KE
B)
PE
C)
Total energy
D)
Specific energy

Correct Answer :   Specific energy

Explanation : When total energy is divided by the aircraft weight then, it is termed as specific energy. Total energy of an aircraft is summation of kinetic and potential energy. It is often known as energy state of the aircraft at a given point in time.

50 .
If Total energy of an aircraft is found to be 1200 unit and kinetic energy is 1100 unit then, determine the potential energy.
A)
10 unit
B)
100 unit
C)
213 unit
D)
456 unit

Correct Answer :   100 unit

Explanation : Potential energy = Total energy – kinetic energy = 1200 – 1100 = 100 unit.

51 .
Find the total energy of an aircraft if kinetic energy and potential energy are 3600 and 1200 units respectively.
A)
2800 unit
B)
4800 unit
C)
5789 unit
D)
8789 unit

Correct Answer :   4800 unit

Explanation : Total energy = kinetic energy + potential energy
= 3600 + 1200 = 4800 unit.

52 .
Unit of specific energy is ______
A)
meter
B)
newton
C)
unit less
D)
meter per second

Correct Answer :   meter

Explanation : Specific energy is defined as the ratio of total energy to the weight of the aircraft. Specific energy is defined by distance unit. It can be meter, feet etc. Meter per second is the unit of velocity or speed. Newton is standard unit of force.

53 .
Following diagram represents _______

A)
zero lift
B)
zero drag
C)
drag at zero lift
D)
zero specific power Contour

Correct Answer :   zero specific power Contour

Explaination : Above diagram is illustrating the concept of zero specific power Contour. As shown in the diagram, zero specific power Contours are plotted for different values of load factor. This chart is one of the major tools for evaluation of new fighters.

54 .
Following diagram represents ________

A)
turn radius only
B)
power loading vs height
C)
turn rate and specific power relations
D)
altitude effects on thrust

Correct Answer :   turn rate and specific power relations

Explaination : Typical turn rate and specific power relations can be observed in the diagram. Power loading is defined as the ratio of weight of the aircraft to the aircraft power. It is used to provide information about performance of the aircraft.

55 .
If a Maneuver requires specific power of 1200 unit then, find the approximate value of fuel specific energy. Given thrust as 1000 unit and tsfc of 0.00042 unit.
A)
1568.45 unit
B)
2857.14 unit
C)
4500 unit
D)
6789 unit

Correct Answer :   2857.14 unit

Explanation : Fuel specific energy = specific power / (thrust*tsfc)
= 1200/1000*0.00042 = 2857.14 unit.

56 .
Falling diagram represents ________

A)
operating envelope
B)
lift curve
C)
tire sizing
D)
empty weight diagram

Correct Answer :   operating envelope

Explaination : Operating envelope is shown in the above diagram. Above diagram is typically used for fighter aircraft. Fighter operating envelope is one of the complex type of diagram. Lift curve is used to provide relationship between lift and angle of attack.

57 .
The level flight envelope is determined by ______
A)
tail area
B)
spoilers area
C)
wing aileron size
D)
stall limit lines and from zero specific power

Correct Answer :   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.

58 .
The highest altitude where Ps is zero is called ________
A)
drag polar
B)
geometric height
C)
absolute ceiling
D)
geometric altitude

Correct Answer :   absolute ceiling

Explanation : Absolute ceiling is defined as the highest altitude at where Ps is zero. Geometric height or geometric altitude is based on height measured. Drag polar is graphical representation of drag properties.

59 .
Operating envelope is also known as ________
A)
lift curve
B)
fly by wire
C)
flight envelope
D)
stall speed diagram

Correct Answer :   flight envelope

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.

60 .
If static pressure of flow is 1200psf and operating Mach number is 1.19 then, determine the external flow dynamic pressure.
A)
12.89 psf
B)
112 psf
C)
1189.52 psf
D)
2587 psf

Correct Answer :   1189.52 psf

Explanation : Dynamic pressure = 0.7*static pressure*Mach number2
= 0.7*1200*1.192 = 1189.52 psf.

61 .
Determine total pressure if static pressure is 1600psf and Mach number is 1.8. Consider flow of air.
A)
1000 psf
B)
3450 psf
C)
7892 psf
D)
9193.2 psf

Correct Answer :   9193.2 psf

Explanation : Total pressure = static pressure*[1+0.2*Mach number*Mach number]3.5
= 1600[1+0.2*1.8*1.8]3.5
= 1600*1.6483.5 = 9193.2 psf.

62 .
Following diagram represents _______
A)
approach
B)
landing phase
C)
segments of typical takeoff
D)
cruise to climb mission segment

Correct Answer :   segments of typical takeoff

Explaination : 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.

63 .
Segment marked by ‘?’ in the diagram is called _______
A)
cruise path
B)
flight path angle
C)
free stream velocity
D)
typical ground roll

Correct Answer :   typical ground roll

Explaination : A typical ground roll is shown in the diagram. Ground roll is further subdivided as level ground roll and the ground roll during rotation to AOA for liftoff. During ground roll, aircraft may experience drag, Thrust, friction on wheels etc.

64 .
Which of the following is not part of takeoff?
A)
Climb
B)
Ground roll
C)
Transition
D)
Descending

Correct Answer :   Ground roll

Explanation : Takeoff segment is subdivided into number individual segments or sections. These segments are ground roll, transition, climb etc. Descending is not the part of takeoff.

65 .
If stall speed of the aircraft is 10m/s then, find the approximate value of takeoff speed.
A)
11m/s
B)
9m/s
C)
7m/s
D)
1m/s

Correct Answer :   11m/s

Explanation : Takeoff speed = 1.1*stall speed = 1.1*10 = 11m/s.

66 .
If our design has takeoff velocity of 120 knots then, at which value of stall speed we should design our aircraft?
A)
109.08knots
B)
110 knots
C)
125 knots
D)
150 knots

Correct Answer :   109.08knots

Explanation : Stall speed = 0.909*Takeoff speed = 0.909*120 = 109.08 knots.

67 .
If climb speed during takeoff is 120 unit then, find stall speed.
A)
100knots
B)
215knots
C)
240knots
D)
1267knots

Correct Answer :   100knots

Explanation : Stall speed = climb speed / 1.2 = 120/1.2 = 100knots.

68 .
Final velocity at the end of the landing phase will be _______
A)
touchdown speed
B)
same as climb rate
C)
same as Approach speed
D)
zero

Correct Answer :   zero

Explanation : Final velocity at the end of the aircraft landing phase will be zero. It is not equal to approach speed. Touchdown speed is achieved during flare segment. Rate of climb is vertical velocity of the aircraft.

69 .
Find the required average velocity during flare for military aircraft if stall speed is 45 unit.
A)
100 unit
B)
60 unit
C)
51.75 unit
D)
32.57 unit

Correct Answer :   51.75 unit

Explanation : Average velocity during flare = 1.15*stall speed = 1.15*45 = 51.75 unit.

70 .
Following diagram represents _______

A)
climbing
B)
takeoff phase
C)
typical landing analysis segments
D)
takeoff ground roll

Correct Answer :   typical landing analysis segments

Explaination : Above diagram is showing typical landing phase of an aircraft. As shown in the diagram landing phase is subdivided into number of segments such as flare distance, Approach distance etc. Takeoff ground roll is a segment of takeoff phase. Takeoff phase includes ground roll, transition, climb etc.

71 .
The segment marked by ‘?’ in following diagram is showing _____
A)
climb
B)
approach
C)
cruise
D)
takeoff

Correct Answer :   approach

Explaination : 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.

72 .
Find the average velocity at transition if, stall speed is 15 unit.
A)
12.57 unit
B)
17.25 unit
C)
30 unit
D)
49 unit

Correct Answer :   17.25 unit

Explanation : Average velocity = 1.15*stall speed = 1.15* 15 = 17.25 unit.

73 .
Find the approach speed if, stall speed is 12 unit.
A)
15.6 unit
B)
12 unit
C)
10 unit
D)
1 unit

Correct Answer :   15.6 unit

Explanation : Approach speed = 1.3* stall speed = 1.3*12 = 15.6 unit.

74 .
For military aircraft touchdown speed is 100 knots then, approximate value of stall speed.
A)
90.90
B)
100
C)
200
D)
223

Correct Answer :   90.90

Explanation : Stall speed = touchdown speed/1.1 = 100/1.1 = 90.90.

75 .
Find the approach speed for military aircraft. Considered stall speed is 22unit.
A)
12 ft
B)
20 m
C)
26.4 unit
D)
45 unit

Correct Answer :   26.4 unit

Explanation : For military aircraft, Approach speed = 1.2*stall speed = 1.2*22 = 26.4 unit.

76 .
If stall speed is 120 knots then, find average velocity at flare.
A)
110.25
B)
125.4
C)
137.5
D)
147.6

Correct Answer :   147.6

Explanation : Average velocity at Flare = 1.23*stall speed = 1.23*120 = 147.6 knots.