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The correct choice is (a) AIRCRAFT which is statically stable may or may not be dynamically stable

The explanation is: If an aircraft has initial TENDENCY to return to its ORIGINAL equilibrium condition after being disturbed then, it is said that the aircraft is statically stable. However, it does not mean that the aircraft is dynamically stable as well. Dynamic stability is DEFINED by certain finite time PERIOD. Lift is not always same as weight.

The CORRECT choice is (a) True

To explain I would say: Thrust is a propulsive force produced by the ENGINE of an AIRCRAFT. Thrust will affect the stability of the aircraft. The DIRECT moment of the thrust, inlet normal force due to turning of air etc. will INFLUENCE on the aircraft stability.

The CORRECT answer is (a) Aircraft 1

The best explanation: Pitching moment coefficient curve is shown for 3 different CONFIGURATIONS. To trim at positive AOA, value of zero LIFT pitching moment coefficient Cm0 should be positive. As can be seen in the diagram aircraft 1 has positive value of Cm0. Hence, aircraft 1 will result in trim at positive AOA.

The correct choice is (a) \(\frac{DCM}{d\delta e}\)

To explain: Elevator control power is defined as the CHANGE in moment COEFFICIENT to the change in elevator angle. Elevator control power is given by, Cmδe = \(\frac{dCm}{d\delta e}\) Where, Cmδe is known as elevator control power. Larger VALUE INDICATES more effectiveness of control power.

The correct choice is (a) it has initial tendency to come back to its original equilibrium condition after being disturbed

To explain I WOULD SAY: An AIRCRAFT or an object is said to be statically stable if and only if it has initial tendency to return to its original equilibrium position after being disturbed. Lift is not ALWAYS greater than the weight. At cruise it will be equal to weight of the aircraft.

The correct CHOICE is (a) -0.0096 PER degree

Explanation: Given, LIFT CURVE slope C= 0.032 per degree, Xcg = 0.3

Neutral point is located at 60% of chord i.e. Xnp = 0.6.

Pitching MOMENT derivative = -C*[Xnp-Xcg] = -0.0032*[0.6-0.2] = -0.0096 per degree.

The correct ANSWER is (a) -0.107

The best EXPLANATION: Given, wing alone ARRANGEMENT, CMac = -0.126, CL0 = 0.38.

Now, CM0 is given by,

Cm0 = CMac + CL0*[Xcg-Xac]

= -0.126 + 0.38*[0.3-0.25] = -1.07.

The CORRECT choice is (a) Cm0 > 0

Easy explanation: To trim an aircraft at positive AOA, it is required that aircraft has positive value of ZERO lift MOMENT coefficient. Hence, to trim aircraft at a positive angle of attack, aircraft should be designed with Cm0 positive i.e., Cm0>0. Negative value of Cm0 will not allow the aircraft to trim at positive AOA.

The correct CHOICE is (a) 0.9

To explain I would say: Given, ELEVATOR float E = 1.8°. CHδe = 0.003, CHɑt=-0.006.

Now, TAIL angle of ATTACK = – e* CHδe / CHɑt = -1.8*0.003/-0.006 = 0.9°.

The correct choice is (a) Slope of yawing moment CURVE POSITIVE

The best I can explain: An AIRCRAFT is said to be in directional stability if the yawing moment curve slope is positive. Negative pitching moment COEFFICIENT curve slope is minimum criteria for longitudinal static stability. Positive value of zero lift pitching moment coefficient will be used to DESIGN an aircraft to trim at positive AOA.

Right answer is (a) HINGE moment coefficient is function of tail angle of attack

Explanation: Hinge moment coefficient is function of number of factors such as tail angle of attack, tail setting angle, ETC. Due to downwash the angle of attack at tail will be more. Lifting PROPERTY of an airfoil and sphere will be DIFFERENT. Airfoil is 2d SHAPE and sphere is 3d shape.

The CORRECT choice is (a) damping coefficient divided by critical damping coefficient

The explanation is: Damping ratio is DEFINED as ratio of damping coefficient to the critical damping coefficient. Lift to drag ratio is USED to provide information about aerodynamics of the AIRCRAFT. It is often termed as aerodynamic EFFICIENCY. Thrust to weight ratio is called Thrust loading.

The correct option is (a) Phugoid

To explain I WOULD say: Phugoid MODE is an example of longitudinal mode. DUTCH and lateral mode are not example of longitudinal mode. Aileron is known as primary control SURFACE. Aileron is used to control the aircraft rolling MOMENT.

The correct option is (a) statically stable

For explanation I would say: A typical illustration of STABILITY criteria is shown in the diagram. As shown in the figure, if ball is DISTURBED from equilibrium POSITION it will return to its original equilibrium position. It has an initial tendency to return to the equilibrium position. Hence, the above diagram is illustrating the CONCEPT of statically stable object.

The correct OPTION is (a) WN = \(\sqrt{\FRAC{k}{m}}\)

Explanation: A typical RELATION of natural frequency is illustrated in the above question. Natural frequency in the terms of spring COEFFICIENT and mass can be given as follows Wn = \(\sqrt{\frac{k}{m}}\) where, Wn = natural frequency, k is spring coefficient and m is mass of the system.

The correct CHOICE is (a) 50 cm

To explain I WOULD say: For rectangular wing, LOCATION of neutral POINT = chord/4 = 2/4 = 0.5m = 50cm.

The correct ANSWER is (a) directional stability

The best explanation: Directional stability of the AIRCRAFT is defined as the stability about the yawing axis. In typical aircraft yaw and rolling both will be PRODUCED by DEFLECTING the rudder. Stability about rolling moment is called lateral stability.

Correct answer is (a) 14.14

For EXPLANATION: GIVEN damping RATIO d = 0.05

Lift to DRAG = 0.707/d = 0.707/0.05 = 14.14.

Right CHOICE is (a) typical roll STABILITY concept

For EXPLANATION: Above diagram is illustrating the concept of roll stability. Typical lateral stability criteria can be observed in the diagram. Longitudinal stability is represented by using pitching moment curve. LIFT curve will be used to PROVIDE relationship between lift and angle of attack. Drag polar will correlate drag and lift.

Correct OPTION is (a) Dutch roll is not considered as a longitudinal mode

To elaborate: Dutch roll is not longitudinal mode. Typical longitudinal modes INCLUDE phugoid mode and SHORT period mode. Lift curve slope is defined as ratio of CHANGE in lift coefficient to the change in AOA. STABILITY and controllability are inverse of each other.

Correct choice is (a) -0.156

For explanation I would say: Moment COEFFICIENT about CG = Moment coefficient about AERODYNAMIC CENTRE of + (wing LIFT coefficient*[Xcg-Xac])

= -0.216 + (1.2[0.3-0.25]) = -0.156.

Correct answer is (a) RIGHT wing goes back

To explain I would SAY: Yawing moment is said to be positive by sign convention if right wing goes back and VICE versa. If nose pitches up then it is a positive pitching moment and similarly if nose pitches down it is CALLED negative pitching moment as per the sign convention is considered.

The correct option is (a) lateral stability

The best I can explain: Stability about the rolling is termed as lateral stability. Lateral and directional stability are CLOSELY coupled. When we deflect only aileron it will generate rolling as WELL as it will ALSO cause the AIRCRAFT to yaw. Longitudinal stability is Stability about pitching MOMENT. Elevator is used for such purposes.

The correct option is (a) elevator to float

For explanation I would say: In stick free, we allow elevator to float as ANGLE of attack or relative WIND changes. In stick fixed we will deflect elevator at some angle and then fix it at that angle. Stick free stability depends UPON number of FACTORS such as HINGE moment etc.

Correct OPTION is (a) cg position influence on MOMENT coefficient curve

For explanation I would SAY: The above diagram is illustrating the influence of CG position on moment coefficient curve. It can be used to illustrate the concept of neutral point. Lift curve is used to show the VARIATION of lift with respect to the aircraft ANGLE of attack. Drag polar is representation of drag characteristics.

Right choice is (a) aircraft NUMBER 1

To explain: As shown in the diagram typical pitching moment coefficient curve is represented. Aircraft number 1 will be STATICALLY STABLE. As can be observed in the diagram if aircraft number 1 is disturbed by some forces; let’s consider some UPWARD direction gust is encountered. At such conditions it will generate negative pitching moment to oppose the upward deflection and hence, it has initial TENDENCY to return to the original equilibrium position. Therefore it is statically stable.

The CORRECT OPTION is (a) 2.08 deg

To explain: Given, tail AOA ɑ = 2°, CHδe = -0.005, CHɑt=-0.0052

Elevator float = -(CHɑt / CHδe)*ɑ = -(-0.0052/-0.005)*2 = -2.08°.

Considering only MAGNITUDE. Hence, 2.08°.

The CORRECT choice is (a) 0 unit

For explanation: Given, aircraft is in cruise, lift = 100N.

At cruise CONDITION, net rolling MOMENT is ZERO. As at cruise all the forces and moments are in equilibrium or in balanced.

Right option is (a) pitching MOMENT COEFFICIENT diagram of unstable aircraft

The explanation: As shown in the diagram typical pitching moment coefficient curve is shown. As shown it is USED to provide relationship between pitching moment coefficient and angle of attack. Given diagram is illustrating the concept of STATICALLY unstable aircraft. Lift curve is RELATED to lift and AOA.

The correct choice is (a) -4.72°

Easy EXPLANATION: Given, NORTH VELOCITY n = 60.5m/s, EAST to west velocity e = 5m/s.

For given cross wind condition sideslip angle β is given by,

β = -ARCTAN (e/n) = -arctan (5/60.5) = -4.72°.

Correct answer is (a) 0.324

For explanation: Location of NEUTRAL point = XCG – moment COEFFICIENT SLOPE/lift curve slope

= 0.3 – 0.116/4.76 = 0.3-0.024 = 0.324.

The correct option is (a) False

Best explanation: Canards are typically used to provide longitudinal instability. Canards are located at the FORWARD SECTION of the aircraft. They are located ahead of the CG of aircraft typically. Hence, if some disturbance is GIVEN then, it will not have initial tendency to RETURN to its original equilibrium position. Hence, it will not provide static STABILITY.

Correct option is (a) 0.32

For EXPLANATION: Given, SM = 0.12, Xcg = 0.2 (20% of CHORD = 0.2*chord)

Neutral POINT position = SM + Xcg = 0.12 + 0.2 = 0.32.

Right CHOICE is (a) 1.43

Easiest explanation: GIVEN, v = 2, V = [u^2+v^2+w^2] ^0.5 = [80*80+2*2+4.5*4.5] ^0.5 = 80.1521.

Sideslip ANGLE = ARCSINE (v/V) = arcsine (2/80.1521) = 1.43°.

The correct option is (a) hinge MOMENT

To explain: Stick FORCE is function of hinge moment. Lofting is a mathematical modelling used to define different cross section or geometry. Fuel air ratio is related to aircraft engine. PRESSURE ratio of INLET will be dependent of the TYPE of inlet.

Correct choice is (a) 0.1619 rad

Easy explanation: Trim lift coefficient C = 0.75, lift CURVE slope c = 4.5 per rad, ELEVATOR deflection E = 1.056 per rad, and trim elevator ANGLE E = 0.020 rad.

Trim angle = [C-e*E]/c

= [0.75-1.056*0.020]/4.5

= [0.75-0.02112]/4.5

= 0.728/4.5 = 0.1619 rad.

Correct option is (a) 8.9%

For EXPLANATION I would SAY: Static Margin = – (moment COEFFICIENT SLOPE/lift curve slope)

= -(-0.58/6.5) = 0.089 = 0.089*100% = 8.9%.

Right answer is (a) True

The best I can explain: Yes, it is true that aircraft can suffer from the adverse yaw phenomenon DUE to rolling MOTION. When an aircraft is banked to execute turning operation, the AILERON MAY produce a yawing motion that OPPOSES the turn. This is called adverse yaw.

The correct OPTION is (a) 0.39

To elaborate: Typically, STICK FREE neutral point is located at 2-5% ahead of stick fixed neutral point.

Stick free neutral point = 2 to 5 % ahead of stick fixed = 0.02*0.4 to 0.05*0.4 ahead of stick fixed = 0.008 to 0.1 ahead of stick fixed neutral point

Hence, neutral point location of stick free = 0.4-(0.008 to 0.1) = 0.392 to 0.39.

Hence, Correct answer from the GIVEN OPTIONS Will be 0.39.

Correct option is (a) -1.1172 per rad

For explanation: Given, tail efficiency e = 0.95, tail volume ratio of HORIZONTAL tail v = 0.7, lift curve slope of tail c = 4.2 per rad, downwash DERIVATIVE d = 0.6

Pitching moment COEFFICIENT slope for the tail = -e*v*c*(1-d)

= -0.95*0.7*4.2*(1-0.6) = -1.1172 per rad.