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Aeronautical Engineering practice questions

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Question 1Mathematics and Basic Engineering Sciences

A wind-tunnel data reduction uses the lift coefficient CL = L / (q∞ S), where L is lift in newtons and S is wing area in square metres. For CL to be dimensionless, which SI unit must the dynamic pressure q∞ carry?

  1. kg·s⁻²
  2. kg·m·s⁻²
  3. kg·m⁻¹·s⁻² Correct answer.
  4. kg·m⁻²·s⁻³

Explanation

Lift is a force, N = kg·m·s⁻², and area is m², so q∞ must be N/m². Dividing kg·m·s⁻² by m² gives kg·m⁻¹·s⁻², which is the pascal. This also matches q∞ = ½ρV², with ρ in kg·m⁻³ and V² in m²·s⁻².

Question 2Mathematics and Basic Engineering Sciences

A linearised longitudinal model of an aircraft has eigenvalues −0.8 ± 2.1i for the short-period mode and −1.5 for the phugoid. What does the short-period eigenvalue pair indicate?

  1. Stable non-oscillatory mode with time constant 1.5 s
  2. Neutral oscillation at 0.8 rad·s⁻¹ with no damping
  3. Stable oscillation with decay rate 0.8 s⁻¹ and frequency 2.1 rad·s⁻¹ Correct answer.
  4. Stable oscillation with growth rate 0.8 s⁻¹ and frequency 2.1 rad·s⁻¹

Explanation

A complex pair s = σ ± iωd gives oscillation at ωd = 2.1 rad·s⁻¹ and an envelope e^(σt). A negative real part σ = −0.8 s⁻¹ means the oscillation decays, not grows. The real eigenvalue −1.5 belongs to a separate non-oscillatory mode and does not set the short-period frequency.

Question 3Mathematics and Basic Engineering Sciences

A 180 kg engine is mounted 1.5 m forward of a wing spar. Taking g = 9.81 m·s⁻², what is the magnitude of the static moment of the engine weight about the spar?

  1. 1 766 N·m
  2. 265 N·m
  3. 3 531 N·m
  4. 2 649 N·m Correct answer.

Explanation

The engine weight is W = mg = 180 × 9.81 = 1 766 N. The moment arm is 1.5 m, so the moment magnitude is M = Wd = 1 766 × 1.5 ≈ 2 649 N·m. The calculation uses the perpendicular distance from the spar to the line of action of the weight.

Question 4Mathematics and Basic Engineering Sciences

A first-order lag element G(s) = K/(τs + 1) with τ = 2.0 s is subjected to a unit step input. At what time after the step does the output reach 63.2% of its final value?

  1. 1.0 s
  2. 2.0 s Correct answer.
  3. 4.0 s
  4. 0.5 s

Explanation

For a first-order lag, the step response is y(t) = K(1 − e^(−t/τ)). The 63.2% point corresponds to one time constant because 1 − e^(−1) ≈ 0.632. With τ = 2.0 s, that time is 2.0 s.

Question 5Mathematics and Basic Engineering Sciences

A pitot-static probe on a light aircraft measures a stagnation pressure of 101.8 kPa and a static pressure of 100.0 kPa. If the local air density is 1.20 kg·m⁻³ and the flow is incompressible, what is the true airspeed?

  1. 30.0 m·s⁻¹
  2. 77.5 m·s⁻¹
  3. 38.7 m·s⁻¹
  4. 54.8 m·s⁻¹ Correct answer.

Explanation

For incompressible flow, Bernoulli's relation gives q∞ = p0 − p = ½ρV², so V = √(2Δp/ρ). Here Δp = 1.8 kPa = 1 800 Pa, giving V = √(2 × 1 800 / 1.20) = √3 000 ≈ 54.8 m·s⁻¹. Compressibility corrections are small at this speed, so the incompressible result is acceptable.

Question 6Mathematics and Basic Engineering Sciences

For a thin-walled circular tube carrying torque T, the shear stress is approximately τ = T/(2πr²t), where r is the mean radius and t is the wall thickness. If the mean radius is doubled and the wall thickness is halved while T is unchanged, by what factor does τ change?

  1. It remains unchanged.
  2. It becomes one quarter of its original value.
  3. It doubles.
  4. It becomes one half of its original value. Correct answer.

Explanation

The denominator depends on r²t. Doubling r multiplies r² by 4, while halving t multiplies t by 0.5, so the product r²t becomes 4 × 0.5 = 2 times larger. Because τ is inversely proportional to r²t, the stress falls to one half of its original value.

Question 7Mathematics and Basic Engineering Sciences

The time to failure of a hydraulic actuator is modelled by an exponential distribution with constant failure rate λ. What is the probability that the actuator survives beyond its mean time to failure, 1/λ?

  1. 0.500
  2. 0.135
  3. 0.368 Correct answer.
  4. 0.632

Explanation

The exponential survival function is R(t) = e^(−λt). Evaluating at the mean time t = 1/λ gives R = e^(−1) ≈ 0.368. The value 0.632 is the corresponding probability of failure by the mean time, not survival beyond it.

Question 8Mathematics and Basic Engineering Sciences

An ideal Brayton cycle operates with a compressor pressure ratio of 16 and air with γ = 1.4. Using the cold-air-standard relation η = 1 − rp^((1−γ)/γ), what is the cycle thermal efficiency?

  1. 0.35
  2. 0.45
  3. 0.55 Correct answer.
  4. 0.65

Explanation

For γ = 1.4, the exponent is (1 − γ)/γ = −0.2857. With rp = 16, rp^(−0.2857) ≈ 0.453, so η = 1 − 0.453 ≈ 0.547, or about 0.55. This is the ideal-cycle efficiency before accounting for component losses.

Question 9Mathematics and Basic Engineering Sciences

A two-dimensional incompressible velocity field is given by u = 2x and v = −2y, with coordinates in metres and velocity in m·s⁻¹. What is the convective acceleration vector at the point (x, y) = (1, 2) m?

  1. (4, −8) m·s⁻²
  2. (4, 8) m·s⁻² Correct answer.
  3. (2, −4) m·s⁻²
  4. (−4, 8) m·s⁻²

Explanation

Convective acceleration is (V·∇)V. Here ∂u/∂x = 2, ∂u/∂y = 0, ∂v/∂x = 0, and ∂v/∂y = −2. At (1, 2), u = 2 and v = −4, so ax = u(2) + v(0) = 4 and ay = u(0) + v(−2) = 8, giving (4, 8) m·s⁻².

Question 10Mathematics and Basic Engineering Sciences

A feedback control system has the characteristic equation s³ + 4s² + (K + 1)s + 2K = 0, where K is a positive gain. Using the Routh–Hurwitz criterion, which range of K guarantees closed-loop stability?

  1. K > 0 Correct answer.
  2. K < −2
  3. 0 < K < 4
  4. K > 2

Explanation

The Routh array first column contains 1, 4, (K + 2)/2, and 2K. For stability all first-column entries must be positive, which requires K > −2 and K > 0. Since K is stated to be positive, the binding condition is simply K > 0.

Question 11Mathematics and Basic Engineering Sciences

The Reynolds number Re = ρVL/μ is used to classify boundary-layer regimes on aircraft surfaces. What is the SI unit of Re?

  1. kg·m⁻³ (density)
  2. m·s⁻¹ (velocity)
  3. Dimensionless Correct answer.
  4. m²·s⁻¹ (kinematic viscosity)

Explanation

The grouping ρVL/μ has units (kg·m⁻³)(m·s⁻¹)(m)/(kg·m⁻¹·s⁻¹). The numerator reduces to kg·m⁻¹·s⁻¹, which cancels exactly with the denominator, leaving no unit. Reynolds number is therefore dimensionless, so transition criteria such as Re ≈ 5×10^5 are independent of the unit system.

Question 12Mathematics and Basic Engineering Sciences

In the International Standard Atmosphere at sea level, T = 288.15 K. Taking γ = 1.4 and R = 287 J·kg⁻¹·K⁻¹ for air, the speed of sound a = √(γRT) is closest to:

  1. 295 m/s
  2. 380 m/s
  3. 340 m/s Correct answer.
  4. 420 m/s

Explanation

For an ideal gas the speed of sound is a = √(γRT). Substituting γ = 1.4, R = 287 J·kg⁻¹·K⁻¹ and T = 288.15 K gives γRT ≈ 1.158×10^5 m²·s⁻², so a ≈ 340 m/s. The other values correspond to temperatures well away from ISA sea level.

Question 13Mathematics and Basic Engineering Sciences

A pitch-rate response of an aircraft is approximated by a second-order transfer function with denominator s² + 2ζω_n s + ω_n². If ζ = 0.7, the response is best described as:

  1. Undamped
  2. Underdamped Correct answer.
  3. Overdamped
  4. Critically damped

Explanation

The denominator s² + 2ζω_n s + ω_n² has roots with negative real part and nonzero imaginary part when 0 < ζ < 1. That is the underdamped case, where the pitch-rate response overshoots before settling. Critical damping occurs at ζ = 1, overdamping at ζ > 1, and undamped oscillation at ζ = 0.

Question 14Mathematics and Basic Engineering Sciences

A two-dimensional incompressible flow is described by the stream function ψ = 3xy, with coordinates in metres and ψ in m²·s⁻¹. What is the horizontal velocity component u at the point (x, y) = (2 m, 1 m)?

  1. 3 m/s
  2. 1.5 m/s
  3. 6 m/s Correct answer.
  4. −6 m/s

Explanation

For a two-dimensional incompressible flow, the velocity components follow u = ∂ψ/∂y and v = −∂ψ/∂x. With ψ = 3xy, u = 3x, so at x = 2 m the horizontal component is 6 m/s. The vertical component there is v = −3y = −3 m/s.

Question 15Mathematics and Basic Engineering Sciences

A nonlinear equation f(x) = x³ − 2x − 5 = 0 is to be solved by the Newton–Raphson method. Starting from x₀ = 2, what is the first iterate x₁?

  1. 2.1 Correct answer.
  2. 2.05
  3. 2.2
  4. 1.9

Explanation

Newton–Raphson uses x₁ = x₀ − f(x₀)/f′(x₀). Here f(2) = 8 − 4 − 5 = −1 and f′(x) = 3x² − 2, so f′(2) = 10. Therefore x₁ = 2 − (−1)/10 = 2.1.

Question 16Mathematics and Basic Engineering Sciences

A solid circular shaft of diameter 50 mm transmits a torque of 2.0 kN·m. Using τ_max = 16T/(πd³), what is the maximum shear stress?

  1. 163 MPa
  2. 81.5 MPa Correct answer.
  3. 20.4 MPa
  4. 40.7 MPa

Explanation

The maximum shear stress in a solid circular shaft is τ_max = 16T/(πd³). With T = 2.0×10³ N·m and d = 0.050 m, τ_max = 32,000/(π × 1.25×10⁻⁴) ≈ 81.5×10⁶ Pa, or 81.5 MPa. Using the radius instead of the diameter in this formula gives an incorrect factor.

Question 17Mathematics and Basic Engineering Sciences

For laminar flow over a flat plate, the boundary-layer thickness is estimated by δ ≈ 5x/√Re_x. At a station where x = 1.0 m and Re_x = 5.0×10^5, what is δ?

  1. 14.1 mm
  2. 2.2 mm
  3. 7.1 mm Correct answer.
  4. 5.0 mm

Explanation

The laminar flat-plate correlation gives δ ≈ 5x/√Re_x. Substituting x = 1.0 m and Re_x = 5.0×10^5 gives δ ≈ 5/(707.1) m = 0.00707 m, or about 7.1 mm. The boundary layer therefore remains thin at this station.

Question 18Mathematics and Basic Engineering Sciences

A unity-feedback flight-control loop has open-loop transfer function G(s) = K/[s(s+2)]. The closed-loop characteristic equation is s² + 2s + K = 0. What value of K gives a damping ratio ζ = 0.5?

  1. 8
  2. 4 Correct answer.
  3. 1
  4. 2

Explanation

Comparing s² + 2s + K = 0 with s² + 2ζω_n s + ω_n² gives ω_n = √K and 2ζω_n = 2. Hence ζ = 1/√K, and setting ζ = 0.5 requires √K = 2, so K = 4. Lower gains give heavier damping, while higher gains reduce damping.

Question 19Mathematics and Basic Engineering Sciences

The characteristic polynomial of a feedback control system is s⁴ + 2s³ + 3s² + 4s + K = 0, where K is a positive gain. Using the Routh–Hurwitz criterion, what is the range of K for closed-loop stability?

  1. K > 0
  2. 0 < K < 2 Correct answer.
  3. 0 < K < 4
  4. K > 2

Explanation

The Routh array for s⁴ + 2s³ + 3s² + 4s + K has first-column entries 1, 2, 1, 4 − 2K and K. Stability requires every first-column entry to be positive, so K > 0 and 4 − 2K > 0. Combining these inequalities gives 0 < K < 2.

Question 20Mathematics and Basic Engineering Sciences

A flight-control computer uses three identical gyros in a 2-out-of-3 majority-vote configuration. Each gyro operates independently with reliability 0.9. What is the reliability of the majority-vote set?

  1. 0.729
  2. 0.900
  3. 0.972 Correct answer.
  4. 0.990

Explanation

A 2-out-of-3 majority set works if exactly two gyros or all three gyros operate. Its reliability is C(3,2)(0.9)²(0.1) + (0.9)³ = 0.243 + 0.729 = 0.972. This is higher than a single gyro but below the ideal of 1.0 because two simultaneous failures still cause loss of the set.

Question 21Mathematics and Basic Engineering Sciences

A rectangular wing has a span of 12 m and a constant chord of 1.5 m. Using AR = b²/S, what is its aspect ratio?

  1. 9.0
  2. 8.0 Correct answer.
  3. 6.0
  4. 12.0

Explanation

Wing area is S = b × c = 12 × 1.5 = 18 m², so AR = b²/S = 144/18 = 8.0. Aspect ratio is dimensionless because both b² and S have units of area.

Question 22Mathematics and Basic Engineering Sciences

In Laplace-domain flight-control analysis, what is the Laplace transform of a constant command a for t ≥ 0, with zero initial conditions?

  1. a / s Correct answer.
  2. a / (s + 1)
  3. a / s²
  4. a s

Explanation

For a constant a defined for t ≥ 0, the Laplace transform is a∫₀^∞ e^(−st) dt = a/s for Re(s) > 0. The ramp a t transforms to a/s², not to a/s.

Question 23Mathematics and Basic Engineering Sciences

A wing fuel tank contains kerosene of density 800 kg/m³. Taking g = 9.81 m/s², what is the gauge pressure at the tank bottom when the fuel depth is 2.5 m?

  1. 19.6 kPa Correct answer.
  2. 9.81 kPa
  3. 196 kPa
  4. 24.5 kPa

Explanation

For an incompressible fluid at rest, gauge pressure increases with depth according to Δp = ρgh. Substituting 800 kg/m³ × 9.81 m/s² × 2.5 m gives 19,620 Pa, or about 19.6 kPa.

Question 24Mathematics and Basic Engineering Sciences

A wing spar is idealised as a simply supported beam of span 6.0 m carrying a uniformly distributed lift load of 4.0 kN/m. What is the maximum bending moment?

  1. 18.0 kN·m Correct answer.
  2. 9.0 kN·m
  3. 36.0 kN·m
  4. 12.0 kN·m

Explanation

For a simply supported beam with a uniformly distributed load w over span L, the maximum bending moment is wL²/8 at midspan. Here w = 4.0 kN/m and L = 6.0 m, giving 4.0 × 36 / 8 = 18 kN·m.

Question 25Mathematics and Basic Engineering Sciences

Fuel-burn rate measurements are 0 kg/s at t = 0, 4 kg/s at t = 10 s, and 6 kg/s at t = 20 s. Using the trapezoidal rule, what is the estimated total fuel used over 0–20 s?

  1. 100 kg
  2. 120 kg
  3. 70 kg Correct answer.
  4. 80 kg

Explanation

The trapezoidal rule applies (y_i + y_{i+1})h/2 to each interval and sums the results. With h = 10 s, the two intervals contribute 20 kg and 50 kg, so the total is 70 kg.

Question 26Mathematics and Basic Engineering Sciences

A static-pressure sensor error is normally distributed with mean 0 Pa and standard deviation 2 Pa. What is the probability that a reading has an error between −2 Pa and +2 Pa?

  1. 99.7%
  2. 50.0%
  3. 95.4%
  4. 68.3% Correct answer.

Explanation

For a normal distribution, the interval mean ± one standard deviation contains about 68.3% of the probability. Here the limits are 0 ± 2 Pa, exactly one standard deviation from the mean.

Question 27Mathematics and Basic Engineering Sciences

An avionics unit of mass 0.50 kg and specific heat 900 J kg⁻¹ K⁻¹ has an exposed area 0.050 m² and convective coefficient 20 W m⁻² K⁻¹. Using the lumped-capacitance time constant τ = mc/(hA), what is τ?

  1. 90 s
  2. 450 s Correct answer.
  3. 900 s
  4. 45 s

Explanation

The lumped-capacitance time constant is τ = mc/(hA). The thermal capacity mc is 0.50 × 900 = 450 J/K and hA is 20 × 0.050 = 1.0 W/K, so τ = 450 s.

Question 28Mathematics and Basic Engineering Sciences

An aluminium strut of length 1.2 m and cross-sectional area 200 mm² carries an axial tensile load of 14 kN. Taking E = 70 GPa, what is the elastic elongation? Use δ = FL/(AE).

  1. 2.4 mm
  2. 0.12 mm
  3. 0.60 mm
  4. 1.2 mm Correct answer.

Explanation

Convert the area to 200 × 10⁻⁶ m² and use δ = FL/(AE). The product AE is 200 × 10⁻⁶ × 70 × 10⁹ = 14 × 10⁶ N, so δ = (14,000 × 1.2)/(14 × 10⁶) = 0.0012 m = 1.2 mm.

Question 29Mathematics and Basic Engineering Sciences

A wing-skin element is in plane stress with σx = 80 MPa, σy = 20 MPa and τxy = 40 MPa. What are the principal stresses?

  1. 100 MPa and 0 MPa Correct answer.
  2. 90 MPa and 10 MPa
  3. 80 MPa and 20 MPa
  4. 120 MPa and −20 MPa

Explanation

The principal stresses are σ₁,₂ = (σx + σy)/2 ± sqrt[((σx − σy)/2)² + τxy²]. The centre is 50 MPa and the radius is sqrt(30² + 40²) = 50 MPa, giving 100 MPa and 0 MPa.

Question 30Mathematics and Basic Engineering Sciences

A cantilever wing spar has tip deflection δ = PL³/(3EI) under an end load P. If the span L is doubled and the second moment of area I is halved while P and E remain constant, by what factor does the tip deflection change?

  1. 16 Correct answer.
  2. 4
  3. 32
  4. 8

Explanation

Tip deflection scales as L³/I when P and E are fixed. Doubling L multiplies deflection by 8, and halving I multiplies it by a further 2, so the combined factor is 16.

Question 31Mathematics and Basic Engineering Sciences

For an incompressible flow over a wing, which condition must the velocity field V satisfy at every point where the flow is continuous?

  1. ∇·V = 0 Correct answer.
  2. ∇p = 0
  3. ∇×V = 0
  4. ∂V/∂t = 0

Explanation

For a constant-density (incompressible) flow, conservation of mass reduces to the continuity condition ∇·V = 0. This must hold at every point where the velocity field is continuous and differentiable.

Question 32Mathematics and Basic Engineering Sciences

An aerodynamic force F = (3i + 4j) N acts at a point on an aileron, and the hinge-line direction is the unit vector u = 0.6i + 0.8j. What is the component of the force along the hinge line?

  1. 2.5 N
  2. 7.0 N
  3. 5.0 N Correct answer.
  4. 4.0 N

Explanation

The component along the hinge line is the dot product of the force with the unit hinge-line vector: F·u = (3)(0.6) + (4)(0.8) = 1.8 + 3.2 = 5.0 N. Because u is a unit vector, no further division by its length is needed.

Question 33Mathematics and Basic Engineering Sciences

An aircraft climbs at a constant speed of 120 m/s along a straight path inclined at 6° above the horizontal. Taking sin 6° ≈ 0.1045 and cos 6° ≈ 0.9945, what is the vertical speed?

  1. 2.1 m/s
  2. 12.5 m/s Correct answer.
  3. 119.3 m/s
  4. 20.9 m/s

Explanation

The vertical speed is the component of the flight-path velocity perpendicular to the horizontal, so V_vertical = V sin θ. With V = 120 m/s and sin 6° ≈ 0.1045, this gives 12.5 m/s.

Question 34Mathematics and Basic Engineering Sciences

Air flows over a wing section at 120 m/s where the static pressure is 85.0 kPa and density is 1.00 kg/m³. Neglecting compressibility and height changes, what is the total pressure at that point?

  1. 79.3 kPa
  2. 101.3 kPa
  3. 85.7 kPa
  4. 92.2 kPa Correct answer.

Explanation

For steady, incompressible, frictionless flow along a streamline, total pressure is p0 = p + 0.5ρV². Substituting gives 85,000 + 0.5(1.00)(120)² = 92,200 Pa, or 92.2 kPa.

Question 35Mathematics and Basic Engineering Sciences

A circular fuselage test section is idealized as a thin-walled cylinder of diameter 1.20 m and wall thickness 3.0 mm. If the internal pressure is 0.80 MPa, what is the hoop stress?

  1. 160 MPa Correct answer.
  2. 80 MPa
  3. 320 MPa
  4. 40 MPa

Explanation

For a thin-walled cylindrical pressure vessel, the hoop stress is σ_h = p r / t = p d / (2t). Using p = 0.80 MPa, d = 1.20 m, and t = 0.003 m gives 160 MPa.

Question 36Mathematics and Basic Engineering Sciences

A wing spar is modelled as a beam with a rectangular cross-section 100 mm wide and 200 mm deep. At a section where the bending moment is 50 kN·m, what is the maximum bending stress? Use I = b h³/12 and y = ±h/2.

  1. 37.5 MPa
  2. 300 MPa
  3. 75 MPa Correct answer.
  4. 150 MPa

Explanation

The second moment of area is I = b h³/12 = 0.100(0.200)³/12 = 6.667×10⁻⁵ m⁴. Maximum bending stress occurs at y = h/2 = 0.100 m, so σ = M y / I = (50,000)(0.100)/(6.667×10⁻⁵) = 75 MPa.

Question 37Mathematics and Basic Engineering Sciences

Air enters an aircraft heat exchanger at 50 kg/s and is heated by 20 K at essentially constant pressure. Taking cp = 1005 J kg⁻¹ K⁻¹, what heat-transfer rate is required?

  1. 2.01 MW
  2. 20.1 kW
  3. 1.01 MW Correct answer.
  4. 0.50 MW

Explanation

At essentially constant pressure, the heat rate is Q = m cp ΔT. Substituting 50 kg/s, 1005 J kg⁻¹ K⁻¹, and 20 K gives 1,005,000 W, approximately 1.01 MW.

Question 38Mathematics and Basic Engineering Sciences

A simplified fuel-flow transient is modelled by dy/dt = −2y with y(0) = 1. Using the explicit Euler method with step size 0.10 s, what is the estimate of y after two steps?

  1. 0.60
  2. 0.64 Correct answer.
  3. 0.80
  4. 0.81

Explanation

Explicit Euler advances y_{n+1} = y_n + h f(y_n), with f(y) = −2y and h = 0.10. First step: y1 = 1 + 0.10(−2) = 0.80; second step: y2 = 0.80 + 0.10(−1.60) = 0.64.

Question 39Mathematics and Basic Engineering Sciences

The skin temperature over a wing panel is modelled by T(x,y) = x²y + y³, with x and y in metres and T in kelvin. At the point (1,2), what is the maximum rate of increase of temperature per metre?

  1. 12.8 K/m
  2. 4.0 K/m
  3. 13.6 K/m Correct answer.
  4. 17.0 K/m

Explanation

The greatest rate of increase at a point is the magnitude of the gradient. Here ∂T/∂x = 2xy = 4 and ∂T/∂y = x² + 3y² = 13 at (1,2), so |∇T| = √(4² + 13²) = √185 ≈ 13.6 K/m.

Question 40Mathematics and Basic Engineering Sciences

A single-cell wing torsion box has enclosed area A = 0.80 m², constant skin thickness t = 2.0 mm, perimeter s = 5.0 m, shear modulus G = 28 GPa, and length L = 3.0 m. It carries torque T = 20 kN·m. What is the angle of twist? Use θ = T L s / (4 A² G t) for constant thickness.

  1. 1.05 × 10⁻³ rad
  2. 2.09 × 10⁻¹ rad
  3. 2.09 × 10⁻³ rad Correct answer.
  4. 4.19 × 10⁻³ rad

Explanation

For a constant-thickness single-cell section, the twist is θ = T L s / (4 A² G t). Substituting T = 20,000 N·m, L = 3.0 m, s = 5.0 m, A = 0.80 m², G = 28×10⁹ Pa, and t = 0.002 m gives θ ≈ 2.09×10⁻³ rad.

Question 41Mathematics and Basic Engineering Sciences

An air-data pressure transducer on a test wing reports a differential pressure of 2.0 kPa. Expressed in coherent SI base units, this pressure is:

  1. 2000 kg·m·s⁻²
  2. 2000 kg·m⁻¹·s⁻² Correct answer.
  3. 2000 kg·m⁻¹·s⁻¹
  4. 2000 kg·m⁻²·s⁻²

Explanation

The pascal is defined as one newton per square metre, and a newton is kg·m·s⁻². Dividing by m² gives kg·m⁻¹·s⁻², so 2.0 kPa = 2000 Pa = 2000 kg·m⁻¹·s⁻².

Question 42Mathematics and Basic Engineering Sciences

A 1.5 m long wing-fairing control rod stretches by 0.45 mm under a taxi-braking load. What is the average axial strain in the rod?

  1. 6.75×10⁻³
  2. 6.75×10⁻⁴
  3. 3.0×10⁻⁴ Correct answer.
  4. 3.0×10⁻³

Explanation

Average axial strain is change in length divided by original length. Convert 0.45 mm to 4.5×10⁻⁴ m and divide by 1.5 m, giving 3.0×10⁻⁴.

Question 43Mathematics and Basic Engineering Sciences

A rectangular spar cross-section is 60 mm wide and 120 mm deep. What is its second moment of area about the centroidal axis parallel to the width?

  1. 8.64×10⁶ mm⁴ Correct answer.
  2. 2.16×10⁶ mm⁴
  3. 4.32×10⁶ mm⁴
  4. 1.08×10⁷ mm⁴

Explanation

For a rectangle, I = b h³/12 about the centroidal axis parallel to side b. Taking b = 60 mm and h = 120 mm gives 60 × 120³ / 12 = 8.64×10⁶ mm⁴.

Question 44Mathematics and Basic Engineering Sciences

Air with density 1.10 kg/m³ and dynamic viscosity 1.80×10⁻⁵ Pa·s flows at 60 m/s over a wing chord of 1.20 m. What is the chord Reynolds number?

  1. 4.40×10⁶ Correct answer.
  2. 8.80×10⁶
  3. 2.20×10⁶
  4. 3.67×10⁵

Explanation

The chord Reynolds number is Re = ρ V c / μ. Substituting 1.10 × 60 × 1.20 / (1.80×10⁻⁵) gives 4.40×10⁶.

Question 45Mathematics and Basic Engineering Sciences

Air enters a converging engine inlet at 1.00 kg/m³, 80 m/s and 0.50 m². At the exit the density is 1.10 kg/m³ and area is 0.35 m². What is the exit velocity?

  1. 125.7 m/s
  2. 91.4 m/s
  3. 103.9 m/s Correct answer.
  4. 114.3 m/s

Explanation

Mass flow is conserved, so ρ₁A₁V₁ = ρ₂A₂V₂. The inlet mass flow is 1.00×0.50×80 = 40 kg/s; dividing by (1.10×0.35) gives about 103.9 m/s.

Question 46Mathematics and Basic Engineering Sciences

A cabin wall panel of area 2.0 m² and thickness 20 mm has thermal conductivity 0.040 W m⁻¹ K⁻¹. In steady state the temperature difference across it is 30 K. What is the heat-transfer rate through the panel?

  1. 2.4 kW
  2. 60 W
  3. 120 W Correct answer.
  4. 240 W

Explanation

For one-dimensional steady conduction, Q = k A ΔT / L. Substituting 0.040×2.0×30/0.020 gives 120 W.

Question 47Mathematics and Basic Engineering Sciences

A first-order actuator response to a unit step is Y(s) = K/[s(τs+1)]. Using the final-value theorem, what is the steady-state output y(∞)?

  1. K/τ
  2. Infinite
  3. K Correct answer.
  4. 0

Explanation

The final-value theorem gives y(∞) = lim_{s→0} sY(s). Here sY(s) = K/(τs+1), whose limit as s tends to zero is K.

Question 48Mathematics and Basic Engineering Sciences

A wing access panel is fastened by two vertical bolts at x=0 and x=1.0 m. A vertical load of 12 kN acts at x=0.40 m. Treating the panel as a rigid body in static equilibrium, what is the vertical reaction at the bolt at x=0?

  1. 12.0 kN
  2. 7.2 kN Correct answer.
  3. 4.8 kN
  4. 6.0 kN

Explanation

For static equilibrium, take moments about the bolt at x = 1.0 m. The 12 kN load has a moment arm of 0.60 m, so R₀ × 1.0 = 12 × 0.60, giving R₀ = 7.2 kN.

Question 49Mathematics and Basic Engineering Sciences

In a wind-tunnel study, drag D of a geometrically similar wing is assumed to depend on density ρ, speed V, dynamic viscosity μ, speed of sound a, and chord c. How many independent dimensionless groups does dimensional analysis produce for these six variables?

  1. 2
  2. 3 Correct answer.
  3. 4
  4. 5

Explanation

There are six variables and three independent dimensions, M, L and T. Dimensional analysis therefore yields n − m = 6 − 3 = 3 independent dimensionless groups, commonly drag coefficient, Reynolds number and Mach number.

Question 50Mathematics and Basic Engineering Sciences

A simplified lateral-directional model has system matrix A = [[-0.5, 1.0], [-2.0, -0.5]]. What best describes its eigenvalues?

  1. A complex pair with negative real part Correct answer.
  2. A complex pair with positive real part
  3. Two distinct negative real roots
  4. A repeated negative real root

Explanation

The trace is −1 and the determinant is (−0.5)(−0.5) − (1.0)(−2.0) = 2.25. The characteristic equation λ² + λ + 2.25 = 0 has discriminant 1 − 9 < 0, so the eigenvalues are a complex pair with real part −0.5.