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Question 1Engineering Fundamentals and Applied Science

A pipeline inspection ROV is rated to a depth of 300 m in seawater of density 1025 kg/m³. Taking g as 9.81 m/s², what is the hydrostatic pressure at that depth, ignoring atmospheric pressure?

  1. About 1.01 MPa
  2. About 0.302 MPa
  3. About 3.02 MPa Correct answer.
  4. About 30.2 MPa

Explanation

Hydrostatic pressure is p = ρgh, so 1025 × 9.81 × 300 ≈ 3.02 × 10⁶ Pa, that is roughly 3.02 MPa. This gauge-style value ignores the atmospheric pressure acting at the free surface.

Question 2Engineering Fundamentals and Applied Science

In a simple ideal vapour-compression refrigeration cycle, which process occurs across the expansion device?

  1. Constant-pressure heat absorption from the refrigerated space
  2. Isenthalpic throttling from condenser pressure to evaporator pressure Correct answer.
  3. Isentropic compression from evaporator pressure to condenser pressure
  4. Constant-pressure heat rejection to the surroundings

Explanation

The expansion valve or capillary tube throttles the refrigerant with no work and negligible heat transfer, so enthalpy stays essentially constant while pressure and temperature drop. Compression, condensation and evaporation occur in the other three components.

Question 3Engineering Fundamentals and Applied Science

A 400 V three-phase induction motor draws a line current of 20 A at a power factor of 0.85. What is the input active power, assuming a balanced load?

  1. About 11.8 kW Correct answer.
  2. About 6.8 kW
  3. About 13.6 kW
  4. About 20.4 kW

Explanation

For a balanced three-phase load, P = √3 × V_L × I_L × cos φ = 1.732 × 400 × 20 × 0.85 ≈ 11.8 kW. The √3 factor arises from combining the three phase powers of a star or delta load.

Question 4Engineering Fundamentals and Applied Science

A simply supported beam of span 6 m carries a single point load of 12 kN at midspan. What is the maximum bending moment?

  1. 72 kN·m
  2. 18 kN·m Correct answer.
  3. 9 kN·m
  4. 36 kN·m

Explanation

For a central point load on a simply supported beam, the maximum moment is PL/4 = 12 × 6 / 4 = 18 kN·m, occurring under the load. The reactions are each 6 kN, and the moment at midspan equals 6 kN × 3 m.

Question 5Engineering Fundamentals and Applied Science

Which material property best characterises a metal's resistance to local plastic deformation by indentation?

  1. Hardness Correct answer.
  2. Stiffness
  3. Toughness
  4. Ductility

Explanation

Hardness quantifies resistance to indentation or scratching and is commonly measured by Brinell, Vickers or Rockwell tests. It correlates loosely with tensile strength but is a distinct surface-mechanical property.

Question 6Engineering Fundamentals and Applied Science

A control valve has a rangeability of 50:1 and a maximum controllable flow of 100 m³/h. What is the minimum controllable flow at the bottom of its useful range?

  1. 2 m³/h Correct answer.
  2. 50 m³/h
  3. 0.5 m³/h
  4. 5 m³/h

Explanation

Rangeability is the ratio of maximum to minimum controllable flow, so the minimum is 100 / 50 = 2 m³/h. Below this value the valve characteristic is no longer predictable.

Question 7Engineering Fundamentals and Applied Science

In a four-stroke diesel engine, during which stroke is the intake valve open and the piston moving toward bottom dead centre?

  1. Exhaust stroke
  2. Intake stroke Correct answer.
  3. Compression stroke
  4. Power stroke

Explanation

The intake stroke draws fresh charge into the cylinder as the piston descends from top dead centre to bottom dead centre with the intake valve open. Compression, power and exhaust follow in that order.

Question 8Engineering Fundamentals and Applied Science

A strain gauge with a gauge factor of 2.0 is bonded to a steel member and experiences a strain of 500 microstrain. What fractional change in resistance does it produce?

  1. 0.002
  2. 0.0005
  3. 0.001 Correct answer.
  4. 0.01

Explanation

Gauge factor GF = (ΔR/R)/ε, so ΔR/R = GF × ε = 2.0 × 500 × 10⁻⁶ = 0.001. This small fractional change is why bridge circuits and amplification are used.

Question 9Engineering Fundamentals and Applied Science

Which heat transfer mode requires no physical medium and can occur through a vacuum?

  1. Natural convection
  2. Forced convection
  3. Radiation Correct answer.
  4. Conduction

Explanation

Thermal radiation is transported by electromagnetic waves and needs no intervening material, which is why the Sun's heat reaches Earth through space. Conduction and convection both require matter.

Question 10Engineering Fundamentals and Applied Science

A gear train has a driving gear with 20 teeth and a driven gear with 60 teeth. If the driving gear rotates at 900 rpm, what is the driven gear speed?

  1. 300 rpm Correct answer.
  2. 2700 rpm
  3. 450 rpm
  4. 1800 rpm

Explanation

For meshing gears, speed ratio is inversely proportional to tooth count: N_driven = 900 × 20/60 = 300 rpm. The torque is multiplied by the same 3:1 ratio in the opposite sense.

Question 11Engineering Fundamentals and Applied Science

A DC shunt motor is running at steady speed with a constant load torque. If the field flux is reduced while the armature voltage is held constant, what happens to the speed?

  1. It increases until the load balance is restored Correct answer.
  2. It decreases proportionally to the flux
  3. It falls to zero immediately
  4. It remains unchanged because the load torque is constant

Explanation

Back emf E = kΦω, and with fixed terminal voltage a reduced flux requires a higher speed to generate the same back emf. The motor accelerates until the new speed satisfies the load torque, which is why field weakening gives speed control above base speed.

Question 12Engineering Fundamentals and Applied Science

A steel shaft transmits 15 kW at 1500 rpm. What is the approximate torque on the shaft?

  1. 955 N·m
  2. 9.55 N·m
  3. 95.5 N·m Correct answer.
  4. 150 N·m

Explanation

Power equals torque times angular velocity, so T = P/ω with ω = 2π × 1500/60 ≈ 157 rad/s. Thus T = 15000/157 ≈ 95.5 N·m.

Question 13Engineering Fundamentals and Applied Science

In a closed-loop control system, what is the primary effect of adding integral action to a proportional controller?

  1. It guarantees a faster settling time in every process
  2. It removes the need for a feedback sensor
  3. It increases the steady-state error for a step input
  4. It eliminates steady-state offset for a constant setpoint Correct answer.

Explanation

Integral action accumulates the error over time and keeps adjusting the output until the error reaches zero, removing offset. The trade-off is reduced phase margin and possible oscillation if the gain is too high.

Question 14Engineering Fundamentals and Applied Science

Which dimensionless group compares inertial forces to viscous forces in a flowing fluid?

  1. Froude number
  2. Mach number
  3. Reynolds number Correct answer.
  4. Prandtl number

Explanation

The Reynolds number Re = ρvD/μ expresses the ratio of inertial to viscous forces and determines whether flow is laminar or turbulent. Each of the other groups compares a different pair of effects.

Question 15Engineering Fundamentals and Applied Science

A 10 kΩ resistor and a 2.2 kΩ resistor are connected in parallel. What is the equivalent resistance?

  1. About 0.82 kΩ
  2. About 12.2 kΩ
  3. About 5.0 kΩ
  4. About 1.80 kΩ Correct answer.

Explanation

For two parallel resistors, R = R1R2/(R1+R2) = (10 × 2.2)/(12.2) ≈ 1.80 kΩ. The result must be smaller than the smallest resistor, which is a useful sanity check.

Question 16Engineering Fundamentals and Applied Science

A cylindrical pressure vessel with closed ends has an internal diameter of 1 m and a wall thickness of 10 mm under 2 MPa internal pressure. What is the approximate hoop stress?

  1. 200 MPa
  2. 100 MPa Correct answer.
  3. 10 MPa
  4. 50 MPa

Explanation

For a thin-walled cylinder, hoop stress σ = pr/t with r the internal radius, so σ = 2 × 0.5 / 0.010 = 100 MPa. The longitudinal stress would be half this value.

Question 17Engineering Fundamentals and Applied Science

Which failure mode is characterised by progressive crack growth under repeated cyclic loading at stress levels below the yield strength?

  1. Fatigue Correct answer.
  2. Corrosion
  3. Creep
  4. Brittle fracture

Explanation

Fatigue is cumulative damage from cyclic stress, often initiating at stress concentrations and propagating until final fracture. It can occur well below yield, which is why endurance limits and S-N curves are used in design.

Question 18Engineering Fundamentals and Applied Science

A 24 V DC supply feeds a load through a cable with a total loop resistance of 0.5 Ω. If the load draws 4 A, what is the voltage at the load terminals?

  1. 24 V
  2. 22 V Correct answer.
  3. 26 V
  4. 20 V

Explanation

The cable voltage drop is I × R = 4 × 0.5 = 2 V, so the load sees 24 − 2 = 22 V. Cable resistance is often specified as a loop value covering both conductors.

Question 19Engineering Fundamentals and Applied Science

A pump delivers 0.02 m³/s against a total head of 30 m. Taking water density as 1000 kg/m³ and g as 9.81 m/s², what is the hydraulic power delivered to the fluid?

  1. About 59 kW
  2. About 0.59 kW
  3. About 2.9 kW
  4. About 5.9 kW Correct answer.

Explanation

Hydraulic power is P = ρgQH = 1000 × 9.81 × 0.02 × 30 ≈ 5.9 kW. Shaft power would be higher, divided by the pump efficiency.

Question 20Engineering Fundamentals and Applied Science

A thermocouple is used to measure the temperature of a gas stream in a duct whose walls are much colder. Which error is most significant, and how is it reduced?

  1. Conduction error, reduced by lengthening the thermocouple wires
  2. Cold-junction error, reduced by removing the reference junction
  3. Radiation error, reduced by a radiation shield around the junction Correct answer.
  4. Seebeck error, reduced by reversing the polarity

Explanation

A hot junction facing cold walls loses heat by radiation, so it reads lower than the true gas temperature. A concentric radiation shield reduces the view factor to the walls and brings the reading closer to the gas temperature.

Question 21Engineering Fundamentals and Applied Science

What is the principal advantage of a star-delta starter for a three-phase squirrel-cage induction motor?

  1. It increases the starting torque above the direct-on-line value
  2. It reduces the starting current to about one third of the direct-on-line value Correct answer.
  3. It eliminates the need for overload protection
  4. It allows continuous speed control during running

Explanation

During starting the winding is connected in star, so each phase receives 1/√3 of the line voltage and the line current falls to about one third of the direct-on-line value. The penalty is that starting torque also drops to about one third.

Question 22Engineering Fundamentals and Applied Science

A ball is thrown vertically upward at 20 m/s. Neglecting air resistance and taking g as 9.81 m/s², what maximum height does it reach?

  1. About 20.4 m Correct answer.
  2. About 2.0 m
  3. About 40.8 m
  4. About 10.2 m

Explanation

Using v² = u² − 2gh with v = 0 at the top gives h = u²/(2g) = 400/19.62 ≈ 20.4 m. The time to reach the top is about 2.04 s.

Question 23Engineering Fundamentals and Applied Science

A steel component in a marine environment is coupled to a graphite composite part. Which corrosion mechanism is most likely to be accelerated on the steel?

  1. Galvanic corrosion, with the steel acting as the anode Correct answer.
  2. Uniform atmospheric corrosion only
  3. Passivation of the steel by the graphite
  4. Stress-corrosion cracking from the graphite

Explanation

Graphite is strongly cathodic relative to steel, so in the presence of seawater the steel becomes the anode and dissolves at an accelerated rate. The larger the cathode-to-anode area ratio, the more severe the attack.

Question 24Engineering Fundamentals and Applied Science

A signal of 0–10 V is digitised by a 12-bit analogue-to-digital converter with a 0–10 V range. What is the approximate resolution?

  1. 1.2 mV
  2. 10 mV
  3. 2.4 mV Correct answer.
  4. 4.9 mV

Explanation

A 12-bit converter has 2¹² = 4096 steps, so the resolution is 10 V / 4096 ≈ 2.4 mV. This is the smallest input change that produces a distinct output code.

Question 25Engineering Fundamentals and Applied Science

In a vapour-compression refrigeration cycle, the coefficient of performance is defined as the cooling effect divided by the compressor work input. If the evaporator absorbs 12 kW and the compressor consumes 4 kW, what is the COP?

  1. 3.0 Correct answer.
  2. 0.33
  3. 4.0
  4. 2.0

Explanation

COP = cooling effect / work input = 12 / 4 = 3.0. The condenser must then reject 16 kW, the sum of the cooling effect and the compressor work.

Question 26Engineering Fundamentals and Applied Science

A simply supported beam of span 4 m carries a central point load of 12 kN. Neglecting the self-weight of the beam, what is the magnitude of the bending moment at mid-span?

  1. 24 kN·m
  2. 6 kN·m
  3. 48 kN·m
  4. 12 kN·m Correct answer.

Explanation

For a simply supported beam with a central point load W over span L, each reaction equals W/2, so the maximum bending moment occurs at mid-span and equals WL/4. Substituting W = 12 kN and L = 4 m gives 12 × 4 / 4 = 12 kN·m. The moment diagram is triangular, peaking under the load.

Question 27Engineering Fundamentals and Applied Science

In a first-law analysis of a steady-flow adiabatic nozzle with negligible change in potential energy, which quantity is conserved between inlet and outlet?

  1. Total enthalpy Correct answer.
  2. Static pressure
  3. Entropy
  4. Mass flow rate multiplied by velocity

Explanation

For an adiabatic steady-flow device with no shaft work and negligible potential energy change, the steady-flow energy equation reduces to h₁ + V₁²/2 = h₂ + V₂²/2, so stagnation (total) enthalpy is constant. Static enthalpy falls as velocity rises, which is why the fluid accelerates and pressure drops. Mass flow rate is also conserved, but the quantity that links the thermodynamic and kinetic terms is total enthalpy.

Question 28Engineering Fundamentals and Applied Science

A 400 V, 50 Hz three-phase induction motor has 4 poles and runs at 1440 rpm at full load. What is the approximate slip?

  1. 2.4 %
  2. 6.0 %
  3. 4.0 % Correct answer.
  4. 10.0 %

Explanation

Synchronous speed is ns = 120f/p = 120 × 50 / 4 = 1500 rpm. Slip is s = (ns − n)/ns = (1500 − 1440)/1500 = 0.04, or 4 %. Slip is normally a few percent at full load for a standard cage motor.

Question 29Engineering Fundamentals and Applied Science

A platinum resistance thermometer has a resistance of 100.0 Ω at 0 °C and a temperature coefficient of resistance of 0.00385 per °C. Assuming a linear characteristic, what is its resistance at 200 °C?

  1. 138.5 Ω
  2. 154.0 Ω
  3. 123.1 Ω
  4. 177.0 Ω Correct answer.

Explanation

Using the linear approximation R = R₀(1 + αΔT), the fractional change is 0.00385 × 200 = 0.770. Multiplying by 100.0 Ω gives 177.0 Ω. The linear model is a standard first-order treatment of a PRT over a moderate range, though real platinum sensors show slight curvature described by a Callendar–Van Dusen polynomial.

Question 30Engineering Fundamentals and Applied Science

A plant has an open-loop transfer function G(s)H(s) = K/[s(s+2)(s+8)]. Applying the Routh–Hurwitz criterion, what is the maximum value of K for which the closed-loop system remains stable?

  1. 160
  2. 20
  3. 80 Correct answer.
  4. 16

Explanation

The closed-loop characteristic equation is s³ + 10s² + 16s + K = 0. The Routh array gives the s¹ row entry as (10×16 − K)/10, which must be positive, so K < 160. The s⁰ row requires K > 0, so the stability boundary is K = 160 and the maximum stable gain is 160.

Question 31Engineering Design and Analysis

In a design failure mode and effects analysis (DFMEA), the risk priority number (RPN) is traditionally calculated as the product of which three factors?

  1. Detection, tolerance stack-up, and process capability
  2. Severity, occurrence, and detection Correct answer.
  3. Severity, probability of failure, and cost of repair
  4. Occurrence, mean time between failures, and test coverage

Explanation

Traditional DFMEA ranks each failure mode by multiplying severity, occurrence, and detection ratings, each usually scored on a 1–10 scale. The product produces an RPN used to prioritize which failure modes need mitigation first. Cost, MTBF, and tolerance stack-up are not part of the classic RPN formula.

Question 32Engineering Design and Analysis

A tolerance stack-up analysis is being performed on a linear assembly of five dimensioned parts. Which method accounts for the statistical distribution of dimensions rather than assuming every part is at its worst-case limit?

  1. Worst-case (arithmetic) stack-up
  2. Datum shift analysis
  3. Monte Carlo sensitivity ranking only
  4. Root-sum-square (statistical) stack-up Correct answer.

Explanation

Root-sum-square stack-up combines individual tolerances as the square root of the sum of squares, reflecting the low probability that all parts sit simultaneously at their extremes. Worst-case stack-up simply sums tolerances and is conservative. Datum shift and sensitivity ranking are separate concepts and do not by themselves model distributions.

Question 33Engineering Design and Analysis

An engineering team needs to estimate the fatigue life of a rotating shaft subjected to fully reversed bending with a known alternating stress well below the yield strength. Which design approach is most appropriate?

  1. Stress-life (S-N) fatigue analysis with a fatigue strength reduction factor Correct answer.
  2. Linear elastic fracture mechanics applied to an assumed initial crack
  3. Yield-strength comparison against the peak stress
  4. Ultimate-strength comparison against the mean stress

Explanation

For high-cycle fatigue with fully reversed loading and stresses below yield, stress-life (S-N) methods predict cycles to failure and incorporate surface, size, and notch factors. Comparing to yield or ultimate strength ignores cyclic damage accumulation. Fracture mechanics is used when a crack is already assumed or detected.

Question 34Engineering Design and Analysis

A cantilever beam of rectangular cross section is loaded at its free end. If the beam's height is doubled while its width and length remain unchanged, how does the maximum bending deflection at the tip change?

  1. It decreases by a factor of 4
  2. It decreases by a factor of 2
  3. It remains unchanged because length is constant
  4. It decreases by a factor of 8 Correct answer.

Explanation

Tip deflection of an end-loaded cantilever is proportional to 1/I, and the area moment of inertia for a rectangle scales with height cubed. Doubling the height increases I by 2^3 = 8, so deflection falls to one-eighth. Width and length do not change, so no other factor applies.

Question 35Engineering Design and Analysis

During concept selection, a design team scores alternatives against weighted criteria using a decision matrix. What is the primary purpose of normalizing the raw scores before weighting?

  1. To eliminate the need for subjective weighting factors
  2. To ensure criteria with larger numeric ranges do not dominate the weighted total Correct answer.
  3. To convert all criteria into equivalent monetary units
  4. To guarantee that the highest total score is always the cheapest concept

Explanation

Normalization rescales each criterion to a comparable range so that a criterion measured on a large numeric scale does not overwhelm others when weighted. Weighting still expresses relative importance and remains subjective. Normalization does not convert criteria to money or guarantee cost outcomes.

Question 36Engineering Design and Analysis

In a design, a critical feature's dimension is specified as 25.00 ± 0.10 mm. Process data show the mean is 25.03 mm with a standard deviation of 0.02 mm. What is the process capability index Cpk for this feature?

  1. 1.67
  2. 0.83
  3. 1.50
  4. 1.17 Correct answer.

Explanation

Cpk is the minimum of (USL − mean)/(3σ) and (mean − LSL)/(3σ). Here USL = 25.10 and LSL = 24.90, so the upper side is (25.10 − 25.03)/(0.06) = 1.167 and the lower side is (25.03 − 24.90)/0.06 = 2.167; the minimum is about 1.17. The off-center mean limits capability on the upper side.

Question 37Engineering Design and Analysis

A product is designed to survive a specified shock load. Which design change most directly increases the strain-energy absorption capacity of a structural member made from the same material?

  1. Increasing the member's volume in the highly stressed region Correct answer.
  2. Increasing the material's yield strength without changing geometry
  3. Increasing the member's stiffness while keeping the load path identical
  4. Reducing the cross-sectional area to lower mass

Explanation

Toughness is energy absorbed per unit volume, so total strain-energy capacity depends on both material toughness and the volume of material undergoing strain. Adding volume in the highly stressed region raises the total energy the member can absorb. Stiffness and reduced area do not increase absorbed energy.

Question 38Engineering Design and Analysis

In a kinematic analysis of a four-bar linkage, Grashof's criterion is used to determine which characteristic?

  1. The number of degrees of freedom of the mechanism
  2. The velocity ratio between input and output links
  3. The maximum transmission angle achievable at the coupler
  4. Whether at least one link can make a full rotation relative to the others Correct answer.

Explanation

Grashof's inequality (shortest plus longest link sum less than or equal to the sum of the other two) indicates whether the linkage is a crank-rocker, double-crank, or double-rocker type, meaning whether a link can fully rotate. Degrees of freedom come from the Grübler/Kutzbach criterion, and transmission angle and velocity ratio are separate analyses.

Question 39Engineering Design and Analysis

A rotating machine exhibits vibration at exactly the shaft rotational frequency. Which fault is most consistent with this signature?

  1. Rolling-element bearing outer-race defect
  2. Mass unbalance of the rotor Correct answer.
  3. Looseness in the mounting feet
  4. Gear tooth meshing error

Explanation

Mass unbalance produces a dominant once-per-revolution (1X) vibration component because the centrifugal force rotates with the shaft. Bearing defects appear at characteristic defect frequencies, gear errors at mesh frequency and sidebands, and looseness often shows harmonics or subharmonics.

Question 40Engineering Design and Analysis

A design team is choosing between two materials for a beam that must minimize mass while resisting bending. The beam geometry and applied load are fixed, and failure is by yielding. Which material index should guide the selection?

  1. Square root of strength divided by density Correct answer.
  2. Density divided by strength
  3. Strength squared divided by density
  4. Strength divided by density

Explanation

For a beam of fixed length and stiffness or strength requirement, minimizing mass leads to a material index of the square root of strength divided by density. The exponent arises because the section dimensions can be adjusted, so the mass scales with strength^(1/2)/density. Using strength/density or its inverse gives the wrong scaling.

Question 41Engineering Design and Analysis

Which document in a design control process formally records the design inputs, outputs, reviews, and verification activities for a medical device?

  1. Bill of materials
  2. Process flow diagram
  3. Design history file Correct answer.
  4. Supplier quality agreement

Explanation

The design history file compiles records demonstrating that the design was developed according to the approved design plan, including inputs, outputs, reviews, and verification. A bill of materials lists parts, a process flow diagram shows manufacturing steps, and a supplier agreement governs procurement.

Question 42Engineering Design and Analysis

A pressure vessel cylinder has an internal radius r and wall thickness t with t much smaller than r. According to thin-wall theory, the hoop stress is approximately:

  1. p/(rt)
  2. pr/(2t)
  3. pr/t Correct answer.
  4. 2pr/t

Explanation

Thin-wall cylinder theory gives hoop stress as pr/t and longitudinal stress as pr/(2t). The hoop stress is twice the longitudinal stress for a closed-end cylinder. The other expressions misplace or alter the factor of two.

Question 43Engineering Design and Analysis

A design requires a bolted joint to maintain clamp load under cyclic service. Which change best reduces the alternating load carried by the bolt?

  1. Reducing the bolt diameter to lower its stiffness
  2. Increasing the stiffness of the clamped members relative to the bolt Correct answer.
  3. Increasing bolt preload while keeping joint stiffness constant
  4. Removing the washers to shorten the grip length

Explanation

In a preloaded joint, the fraction of external load taken by the bolt depends on the relative stiffness of bolt and clamped members. Making the members stiffer relative to the bolt diverts more of the cyclic load into the members and reduces the bolt's alternating stress. Preload affects mean stress, and reducing bolt diameter generally raises stress.

Question 44Engineering Design and Analysis

In geometric dimensioning and tolerancing, a position tolerance applied at maximum material condition (MMC) provides what benefit?

  1. It requires the feature to be perfectly centered at MMC
  2. It eliminates the need for a datum reference frame
  3. It allows a bonus tolerance as the feature departs from MMC Correct answer.
  4. It fixes the tolerance zone regardless of feature size

Explanation

The MMC modifier grants bonus tolerance equal to the amount the actual feature size departs from maximum material condition toward least material condition. This can lower manufacturing cost while protecting assembly. It does not fix the zone, demand perfect centering, or remove datums.

Question 45Engineering Design and Analysis

A heat sink must dissipate a fixed power while keeping the component junction temperature below a limit. Which combination of thermal resistances determines the required heat-sink resistance?

  1. The sum of all resistances multiplied by the dissipated power
  2. The heat-sink resistance alone, independent of other interfaces
  3. Only the junction-to-case resistance
  4. Junction-to-case plus case-to-sink resistances subtracted from the total allowed resistance Correct answer.

Explanation

The total allowed thermal resistance from junction to ambient equals (Tj,max − Tambient)/power. The heat sink's allowable resistance is that total minus the junction-to-case and case-to-sink (interface) resistances. Ignoring interface resistances or using only one term gives an over-optimistic design.

Question 46Engineering Design and Analysis

A team is performing a design of experiments to study three factors at two levels each. Which design requires the fewest runs while still allowing estimation of all main effects and all two-factor interactions?

  1. A 2^(3-1) fractional factorial with 4 runs
  2. Full 2^3 factorial with 8 runs Correct answer.
  3. One-factor-at-a-time with 6 runs
  4. A Plackett-Burman design with 8 runs

Explanation

A full 2^3 factorial uses 8 runs and cleanly estimates three main effects and three two-factor interactions. A 2^(3-1) fractional design with 4 runs confounds interactions with main effects, and one-factor-at-a-time cannot estimate interactions at all. Plackett-Burman designs also alias interactions.

Question 47Engineering Design and Analysis

Which analysis method is most appropriate for predicting the temperature distribution in a complex electronic enclosure with convection and radiation boundary conditions?

  1. Root-locus stability analysis
  2. Hand calculation using a single lumped thermal resistance
  3. Finite element or finite volume thermal simulation Correct answer.
  4. Mohr's circle construction

Explanation

Complex geometries with combined convection and radiation are best handled by numerical thermal simulation such as finite element or finite volume methods. A single lumped resistance cannot capture spatial gradients, and Mohr's circle and root locus belong to stress and control analysis respectively.

Question 48Engineering Design and Analysis

A shaft is loaded in combined steady torsion and fully reversed bending. Which failure criterion is commonly applied to estimate fatigue failure under this combined loading?

  1. Coulomb friction criterion
  2. Distortion energy (von Mises) combined with a fatigue damage model Correct answer.
  3. Maximum shear stress compared only to ultimate strength
  4. Maximum principal stress compared to yield strength

Explanation

For combined loading under fatigue, the distortion energy (von Mises) equivalent stress is commonly combined with an S-N or damage model to estimate life. Comparing to yield or ultimate strength addresses static failure, not cyclic damage. Coulomb friction is unrelated to fatigue strength.

Question 49Engineering Design and Analysis

During a design review, a fault tree analysis shows a top event with two independent basic events combined through an AND gate. If each basic event has a probability of 0.05, what is the approximate probability of the top event?

  1. 0.0025 Correct answer.
  2. 0.95
  3. 0.05
  4. 0.10

Explanation

An AND gate requires all input events to occur, so for independent events the probabilities multiply: 0.05 × 0.05 = 0.0025. An OR gate would sum the probabilities approximately. The other values do not follow from the AND logic.

Question 50Engineering Design and Analysis

A product must meet a reliability target of 0.90 at 1,000 hours. Assuming a constant failure rate (exponential distribution), what approximate failure rate is required?

  1. About 1.0 × 10^-3 per hour
  2. About 1.0 × 10^-5 per hour
  3. About 9.0 × 10^-4 per hour
  4. About 1.05 × 10^-4 per hour Correct answer.

Explanation

For the exponential model, reliability R = exp(−λt). Solving λ = −ln(0.90)/1000 gives approximately 1.05 × 10^-4 per hour. The other values either misplace the logarithm or invert the calculation.