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Canary Islands PAU Technology and Engineering II - independent engineering foundations study materials
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Canary Islands PAU Technology and Engineering II - independent engineering foundations practice questions

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Question 1Materials, testing and selection

A cylindrical steel specimen with an initial cross-sectional area of 100 mm² is subjected to an axial tensile force of 50 kN. What is the normal tensile stress induced in the specimen?

  1. 500 MPa Correct answer.
  2. 50 MPa
  3. 5 MPa
  4. 5000 MPa

Explanation

Normal tensile stress is calculated as σ = F / A0. Converting 50 kN to 50,000 N and dividing by 100 mm² gives σ = 50,000 N / 100 mm² = 500 N/mm² = 500 MPa.

Question 2Materials, testing and selection

A metal bar with an initial length of 200 mm elongates by 0.4 mm under a tensile stress of 400 MPa within its elastic limit. What is the Young's modulus (modulus of elasticity) of the material?

  1. 400 GPa
  2. 20 GPa
  3. 100 GPa
  4. 200 GPa Correct answer.

Explanation

Engineering strain is ε = ΔL / L0 = 0.4 mm / 200 mm = 0.002. According to Hooke's Law (σ = E · ε), Young's modulus is E = σ / ε = 400 MPa / 0.002 = 200,000 MPa = 200 GPa.

Question 3Materials, testing and selection

A cylindrical alloy test bar has a diameter of 10 mm and an ultimate tensile strength (UTS) of 600 MPa. What maximum tensile force can the bar support before fracture?

  1. 188.50 kN
  2. 60.00 kN
  3. 47.12 kN Correct answer.
  4. 18.85 kN

Explanation

The initial cross-sectional area is A0 = π · d² / 4 = π · (10 mm)² / 4 ≈ 78.54 mm². Maximum force is F_max = UTS · A0 = 600 N/mm² · 78.54 mm² = 47,124 N ≈ 47.12 kN.

Question 4Materials, testing and selection

In a Charpy impact test, a pendulum with an initial potential energy of 300 J breaks a notched specimen and retains 120 J of residual energy. If the cross-sectional area at the notch is 0.8 cm², what is the resilience (KCV impact toughness) of the material?

  1. 375 J/cm²
  2. 180 J/cm²
  3. 225 J/cm² Correct answer.
  4. 150 J/cm²

Explanation

Absorbed energy is ΔE = E1 - E2 = 300 J - 120 J = 180 J. Impact resilience KCV is the absorbed energy per unit area: KCV = ΔE / S = 180 J / 0.8 cm² = 225 J/cm².

Question 5Materials, testing and selection

Which indenter and measurement principle are used in the standard Brinell hardness test (HB)?

  1. A square-based diamond pyramid with a 136° face angle, measuring indentation diagonals
  2. A steel cone pressed under load, measuring the diameter of the elastic recovery zone
  3. A hardened steel or tungsten carbide sphere pressed under load, measuring the surface area of the spherical indentation Correct answer.
  4. A 120° diamond pyramid pressed under load, measuring the depth of penetration

Explanation

Brinell hardness testing (HB) uses a hardened steel or tungsten carbide ball (typically 10 mm diameter) pressed into the material under a specified load F. Hardness is derived from load divided by the surface area of the spherical impression.

Question 6Materials, testing and selection

A structural steel has a yield strength σ_y = 300 MPa and a Young's modulus E = 200 GPa. What is the strain energy density stored at the elastic limit (modulus of resilience U_e)?

  1. 0.450 MJ/m³
  2. 0.075 MJ/m³
  3. 0.225 MJ/m³ Correct answer.
  4. 1.500 MJ/m³

Explanation

The modulus of resilience is U_e = σ_y² / (2 E). Converting E = 200 GPa = 200,000 MPa = 200,000 × 10⁶ Pa, U_e = (300 × 10⁶)² / (2 · 200 × 10⁹) = (9 × 10¹⁴) / (4 × 10¹¹) = 2.25 × 10⁵ J/m³ = 0.225 MJ/m³.

Question 7Materials, testing and selection

What is the primary objective of quenching (templado) in heat-treating medium-to-high carbon steels?

  1. Rapid cooling from austenite to transform the micro-structure into hard, brittle martensite Correct answer.
  2. Slow cooling in a furnace to produce soft, ductile coarse pearlite
  3. Reheating hardened steel below A1 to relieve internal stresses and increase toughness
  4. Diffusing nitrogen into the steel surface to form hard iron nitrides

Explanation

Quenching involves heating steel above its critical temperature into the austenitic range and then cooling it rapidly (in water, oil, or air) to trap carbon in a body-centered tetragonal lattice, forming martensite.

Question 8Materials, testing and selection

Why is tempering (revenido) almost always performed immediately after quenching steel?

  1. To increase carbon content in the core of the steel component
  2. To reduce extreme brittleness and relieve internal quenching stresses while retaining acceptable hardness Correct answer.
  3. To cause rapid grain growth and maximize electrical conductivity
  4. To transform remaining ferrite into austenite

Explanation

As-quenched martensite is extremely hard and brittle with high internal stresses. Tempering reheats the component below A1 (200 °C - 650 °C), allowing martensite to decompose slightly into tempered martensite, restoring toughness and impact resistance.

Question 9Materials, testing and selection

Which heat treatment process involves heating steel above its critical transformation temperature, holding it, and then cooling it very slowly inside the furnace?

  1. Normalizing (normalizado)
  2. Quenching (templado)
  3. Carburizing (cementación)
  4. Annealing (recocido) Correct answer.

Explanation

Full annealing (recocido de regeneración) heats steel into the austenitic region followed by slow furnace cooling. This produces a soft, strain-free structure with maximum ductility and minimum hardness.

Question 10Materials, testing and selection

What is the purpose of carburizing (cementación) as a surface engineering treatment for low-carbon steel parts?

  1. To eliminate residual stresses throughout the entire cross-section of heavy shafts
  2. To enrich the outer surface layer with carbon, producing a wear-resistant hard case after quenching while maintaining a tough core Correct answer.
  3. To coat the surface with zinc to prevent oxidation
  4. To remove carbon from the surface layer to improve corrosion resistance

Explanation

Carburizing diffuses carbon into the surface of low-carbon steel at elevated temperatures (900-950 °C). Subsequent quenching hardens the high-carbon case to withstand surface wear, while the low-carbon core remains ductile and shock-resistant.

Question 11Materials, testing and selection

During a tensile test on a cylindrical specimen, initial length L0 = 50 mm and diameter D0 = 10 mm. At fracture, final length Lf = 62 mm and neck diameter Df = 8 mm. Calculate the percentage elongation (% EL) and percentage reduction of area (% RA).

  1. % EL = 24.0%, % RA = 36.0% Correct answer.
  2. % EL = 12.0%, % RA = 20.0%
  3. % EL = 24.0%, % RA = 20.0%
  4. % EL = 12.0%, % RA = 36.0%

Explanation

% EL = [(Lf - L0) / L0] · 100 = [(62 - 50) / 50] · 100 = (12 / 50) · 100 = 24.0%. Initial area A0 = π·10²/4 = 25π; final area Af = π·8²/4 = 16π. % RA = [(A0 - Af) / A0] · 100 = [(25π - 16π) / 25π] · 100 = (9 / 25) · 100 = 36.0%.

Question 12Materials, testing and selection

What mechanical degradation mechanism involves slow, continuous, time-dependent plastic deformation of a component under constant load at elevated temperatures (typically T > 0.4 T_melting in Kelvin)?

  1. Creep (fluencia a alta temperatura) Correct answer.
  2. Fatigue (fatiga)
  3. Brittle fracture (fractura frágil)
  4. Elastic spring-back (recuperación elástica)

Explanation

Creep is the time-dependent permanent deformation that occurs when materials are subjected to persistent mechanical stresses at elevated temperatures relative to their absolute melting point.

Question 13Materials, testing and selection

A steel clevis pin with a diameter of 20 mm connects two structural rods in double shear. If a total shear force V = 60 kN is applied to the joint, calculate the average shear stress τ in the pin.

  1. 47.7 MPa
  2. 300.0 MPa
  3. 191.0 MPa
  4. 95.5 MPa Correct answer.

Explanation

In double shear, the load is resisted across two circular cross-sections. Area of one cross-section is A1 = π · (20)² / 4 = 314.16 mm². Total shear area A = 2 · A1 = 628.32 mm². Average shear stress τ = V / A = 60,000 N / 628.32 mm² ≈ 95.5 N/mm² = 95.5 MPa.

Question 14Materials, testing and selection

On the iron-carbon (Fe-Fe3C) phase diagram, what micro-structural transformation occurs at the eutectoid composition (0.77 wt% Carbon) at 727 °C during slow cooling?

  1. Austenite transforms completely into pearlite (an interleaved mixture of ferrite and cementite) Correct answer.
  2. Liquid phase transforms directly into austenite and cementite
  3. Ferrite transforms into martensite
  4. Cementite decomposes into graphite and iron

Explanation

At the eutectoid point (0.77% C, 727 °C), solid austenite (γ) decomposes simultaneously into ferrite (α, ~0.022% C) and cementite (Fe3C, 6.67% C), creating the lamellar pearlite micro-structure.

Question 15Materials, testing and selection

Which combination of indenter and major load is used for the standard Rockwell C hardness scale (HRC)?

  1. 10 mm tungsten carbide sphere with a major load of 3000 kgf
  2. 120° diamond cone (Brale indenter) with a major load of 150 kgf Correct answer.
  3. 1/16-inch steel ball with a major load of 100 kgf
  4. 136° diamond pyramid with a major load of 30 kgf

Explanation

Rockwell C (HRC) is used for hard materials (e.g. quenched steels) and specifies a 120° diamond cone (Brale indenter) with a total major load of 150 kgf (pre-load 10 kgf + main load 140 kgf).

Question 16Materials, testing and selection

A steel tie rod (E = 200 GPa, coefficient of thermal expansion α = 12 × 10⁻⁶ K⁻¹) is rigidly fixed between two unyielding supports at 20 °C. If the temperature rises to 70 °C, what thermal compressive stress is induced in the rod?

  1. 120 MPa Correct answer.
  2. 240 MPa
  3. 60 MPa
  4. 12 MPa

Explanation

The temperature change is ΔT = 70 °C - 20 °C = 50 K. Thermal strain prevented by supports is ε_th = α · ΔT = (12 × 10⁻⁶ K⁻¹) · 50 K = 600 × 10⁻⁶ = 0.0006. Thermal stress is σ = E · ε_th = 200,000 MPa · 0.0006 = 120 MPa.

Question 17Statics, strength and mechanical transmission

A simple gear pair consists of a driving pinion with Z1 = 20 teeth rotating at N1 = 1200 rpm and a driven gear with Z2 = 60 teeth. What is the rotational speed N2 of the driven gear?

  1. 400 rpm Correct answer.
  2. 3600 rpm
  3. 600 rpm
  4. 1800 rpm

Explanation

The velocity ratio for gears is N1 / N2 = Z2 / Z1. Rearranging for N2 gives N2 = N1 · (Z1 / Z2) = 1200 rpm · (20 / 60) = 1200 / 3 = 400 rpm.

Question 18Statics, strength and mechanical transmission

A compound gear train consists of input gear Z1 = 15 teeth, intermediate gears Z2 = 45 teeth (meshing with Z1) and Z3 = 20 teeth (mounted on the same shaft as Z2), and output gear Z4 = 80 teeth (meshing with Z3). If input shaft rotates at 1800 rpm, what is the output speed?

  1. 150 rpm Correct answer.
  2. 300 rpm
  3. 600 rpm
  4. 75 rpm

Explanation

Overall gear ratio is i = (Z1 · Z3) / (Z2 · Z4) = (15 · 20) / (45 · 80) = 300 / 3600 = 1 / 12. Output rotational speed N_out = N_in · i = 1800 rpm / 12 = 150 rpm.

Question 19Statics, strength and mechanical transmission

An electric motor delivers a steady output torque of 50 N·m at a rotational speed of 1500 rpm. What mechanical power is delivered by the motor shaft?

  1. 15.71 kW
  2. 4.71 kW
  3. 75.00 kW
  4. 7.85 kW Correct answer.

Explanation

Angular velocity is ω = 2π · N / 60 = 2π · 1500 / 60 = 157.08 rad/s. Mechanical power P = T · ω = 50 N·m · 157.08 rad/s = 7854 W = 7.854 kW ≈ 7.85 kW.

Question 20Statics, strength and mechanical transmission

A belt drive system uses a driver pulley of diameter d1 = 100 mm rotating at n1 = 1440 rpm to drive an output pulley of diameter d2 = 360 mm. Assuming no belt slippage, calculate the linear belt speed v.

  1. 15.08 m/s
  2. 452.4 m/s
  3. 2.51 m/s
  4. 7.54 m/s Correct answer.

Explanation

Linear speed of the belt is v = ω1 · r1 = (2π · 1440 / 60 rad/s) · (0.05 m) = 150.8 rad/s · 0.05 m ≈ 7.54 m/s. Alternatively v = π · d1 · (n1/60) = π · 0.1 m · 24 s⁻¹ = 7.54 m/s.

Question 21Statics, strength and mechanical transmission

A industrial gearbox consists of two reduction stages with individual mechanical efficiencies of η1 = 95% and η2 = 90%. If the input power supplied is 10 kW, what output mechanical power is available at the final shaft?

  1. 8.05 kW
  2. 9.50 kW
  3. 9.25 kW
  4. 8.55 kW Correct answer.

Explanation

Total efficiency of cascaded mechanical stages is the product of individual efficiencies: η_total = η1 · η2 = 0.95 · 0.90 = 0.855 (85.5%). Output power P_out = P_in · η_total = 10 kW · 0.855 = 8.55 kW.

Question 22Statics, strength and mechanical transmission

Two standard spur gears mesh together with Z1 = 20 teeth and Z2 = 60 teeth. If both gears have a module m = 2.5 mm, what is the center distance (a) between their parallel shafts?

  1. 100 mm Correct answer.
  2. 200 mm
  3. 80 mm
  4. 150 mm

Explanation

Pitch diameters are d1 = m · Z1 = 2.5 · 20 = 50 mm, and d2 = m · Z2 = 2.5 · 60 = 150 mm. Center distance for external meshing gears is a = (d1 + d2) / 2 = (50 + 150) / 2 = 100 mm. Alternatively a = m · (Z1 + Z2) / 2 = 2.5 · (80) / 2 = 100 mm.

Question 23Statics, strength and mechanical transmission

A single-thread worm (Z1 = 1) rotating at 1500 rpm drives a worm gear (corona) with Z2 = 50 teeth. What is the rotational speed of the worm gear?

  1. 75000 rpm
  2. 150 rpm
  3. 30 rpm Correct answer.
  4. 50 rpm

Explanation

The velocity reduction ratio for a worm-gear set is i = Z1 / Z2 = 1 / 50. Output speed N2 = N1 · (Z1 / Z2) = 1500 rpm / 50 = 30 rpm.

Question 24Statics, strength and mechanical transmission

An electric motor rates P = 15 kW output at a rated speed n = 2850 rpm. Calculate the output torque produced at the motor shaft.

  1. 5.26 N·m
  2. 50.26 N·m Correct answer.
  3. 100.52 N·m
  4. 314.16 N·m

Explanation

Angular velocity is ω = 2π · n / 60 = 2π · 2850 / 60 = 298.45 rad/s. Torque T = P / ω = 15,000 W / 298.45 rad/s ≈ 50.26 N·m.

Question 25Statics, strength and mechanical transmission

A differential pulley system (polipasto) with a theoretical mechanical advantage of MA = 4 and a mechanical efficiency of η = 80% is used to hoist a load of 4000 N. What effort force F must be applied to lift the load?

  1. 1000 N
  2. 1600 N
  3. 1250 N Correct answer.
  4. 800 N

Explanation

Actual mechanical advantage is AMA = MA · η = 4 · 0.80 = 3.2. Effort force required is F = Load / AMA = 4000 N / 3.2 = 1250 N.

Question 26Statics, strength and mechanical transmission

A rack and pinion mechanism features a pinion with module m = 2 mm and Z = 25 teeth, rotating at n = 300 rpm. What is the linear displacement speed of the driven rack?

  1. 47.12 m/s
  2. 0.250 m/s
  3. 0.785 m/s Correct answer.
  4. 1.571 m/s

Explanation

Pinion pitch diameter d = m · Z = 2 · 25 = 50 mm = 0.05 m. Linear rack velocity v = ω · r = (2π · 300 / 60 rad/s) · (0.025 m) = 31.416 rad/s · 0.025 m ≈ 0.7854 m/s.

Question 27Statics, strength and mechanical transmission

In a planetary gear set, the sun gear has Z_s = 20 teeth (rotating clockwise at 1200 rpm), the ring gear has Z_r = 80 teeth (fixed, 0 rpm), and planet gears are held by planet carrier C. Calculate the rotational speed and direction of planet carrier C using Willis' equation.

  1. 300 rpm counter-clockwise
  2. 240 rpm clockwise Correct answer.
  3. 120 rpm clockwise
  4. 480 rpm clockwise

Explanation

Willis equation: (N_s - N_c) / (N_r - N_c) = -Z_r / Z_s. Substituting N_r = 0: (1200 - N_c) / (0 - N_c) = -80 / 20 = -4. Simplifying: 1200 - N_c = 4 N_c => 5 N_c = 1200 => N_c = 240 rpm (clockwise).

Question 28Statics, strength and mechanical transmission

An industrial transmission gearbox receives P_in = 20 kW with an overall mechanical efficiency of η = 92%. How much heat power P_loss is continuously dissipated by internal friction inside the housing?

  1. 3.2 kW
  2. 0.8 kW
  3. 18.4 kW
  4. 1.6 kW Correct answer.

Explanation

Power lost to mechanical friction is P_loss = P_in · (1 - η) = 20 kW · (1 - 0.92) = 20 kW · 0.08 = 1.6 kW (1600 W).

Question 29Statics, strength and mechanical transmission

Compared to straight spur gears, what is a distinct operational characteristic of helical gear transmissions?

  1. Ability to connect non-parallel intersecting shafts at 90 degrees without bevel geometry
  2. Guaranteed 100% mechanical efficiency due to rolling point contact
  3. Lower manufacturing cost and zero axial thrust forces under all operating speeds
  4. Smoother and quieter tooth engagement with higher load capacity, but generating axial thrust force along shafts Correct answer.

Explanation

Helical gear teeth meet gradually along a helix, providing smoother, quieter operation and higher load capability than spur gears. However, the angled teeth create axial thrust components that require thrust bearings.

Question 30Statics, strength and mechanical transmission

A Class 1 lever has an effort arm of L_e = 1.5 m and a load arm of L_l = 0.3 m. What minimum effort force F_e is required to balance a mass of 200 kg placed at the load arm? (Use g = 9.81 m/s²)

  1. 654.0 N
  2. 981.0 N
  3. 1962.0 N
  4. 392.4 N Correct answer.

Explanation

Load force is F_l = m · g = 200 kg · 9.81 m/s² = 1962 N. By the principle of moments (F_e · L_e = F_l · L_l): F_e = 1962 N · (0.3 m / 1.5 m) = 1962 · 0.2 = 392.4 N.

Question 31Statics, strength and mechanical transmission

A heavy flywheel with a mass moment of inertia I = 2.5 kg·m² is acted upon by a net accelerating torque T = 15 N·m. Calculate the resulting angular acceleration α of the flywheel.

  1. 6.0 rad/s² Correct answer.
  2. 37.5 rad/s²
  3. 0.16 rad/s²
  4. 12.0 rad/s²

Explanation

Newton's second law for rotation states T = I · α. Solving for angular acceleration gives α = T / I = 15 N·m / 2.5 kg·m² = 6.0 rad/s².

Question 32Statics, strength and mechanical transmission

In flat belt drive design, what physical parameters determine the maximum torque that can be transmitted without slipping according to the belt friction equation (T1 / T2 = e^(μ·α))?

  1. Coefficient of friction μ between belt and pulley, and wrap angle α in radians Correct answer.
  2. Pulley shaft diameter and belt thickness alone
  3. Ambient humidity and ambient air density alone
  4. Gear module and pitch circle teeth count

Explanation

The maximum tension ratio before slip is given by Euler-Eytelwein equation T1 / T2 = e^(μ·α), where μ is the coefficient of friction and α is the contact wrap angle in radians.

Question 33Statics, strength and mechanical transmission

A power screw (husillo-tuerca) mechanism has a single-thread lead pitch p = 5 mm per revolution. If the screw turns at n = 120 rpm, what is the linear translation velocity of the nut?

  1. 10.0 mm/s Correct answer.
  2. 600.0 mm/s
  3. 24.0 mm/s
  4. 2.0 mm/s

Explanation

Linear velocity is v = p · (n / 60) = 5 mm · (120 rpm / 60 s) = 5 mm · 2 s⁻¹ = 10.0 mm/s (or 600 mm/min).

Question 34Electrical circuits, machines and digital logic

Applying De Morgan's Laws, what is the simplified Boolean expression equivalent to F = NOT( A · B + C )?

  1. F = (NOT A · NOT B) + NOT C
  2. F = (A + B) · NOT C
  3. F = (NOT A + NOT B) · NOT C Correct answer.
  4. F = NOT A · NOT B · NOT C

Explanation

Applying De Morgan's theorem NOT(X + Y) = NOT X · NOT Y: F = NOT( A · B ) · NOT C. Applying De Morgan's theorem to the first term NOT(A · B) = (NOT A + NOT B) yields F = (NOT A + NOT B) · NOT C.

Question 35Electrical circuits, machines and digital logic

A 3-variable logic function F(A, B, C) has minterms m(1, 3, 5, 7) = A'B'C + A'BC + AB'C + ABC. What is the simplified expression for F obtained using a Karnaugh map?

  1. F = A + C
  2. F = C Correct answer.
  3. F = A' + C
  4. F = B'C

Explanation

Group all four minterms: m(1,3,5,7) = C · (A'B' + A'B + AB' + AB) = C · 1 = C. On a 3-variable K-map, these four adjacent minterms form a quad group covering the entire C = 1 column/row.

Question 36Electrical circuits, machines and digital logic

A 4-variable Boolean function F(A, B, C, D) is defined by minterms Σm(0, 2, 8, 10). What minimal Sum of Products (SOP) expression is obtained by grouping these corners on a Karnaugh map?

  1. F = B' · D' Correct answer.
  2. F = A' · C'
  3. F = B · D
  4. F = A' · D'

Explanation

The four corner minterms on a 4-variable K-map (0000, 0010, 1000, 1010) are adjacent across boundaries. In all four minterms, B = 0 and D = 0. Grouping all four corners yields the single product term F = B' · D'.

Question 37Electrical circuits, machines and digital logic

What is the truth table definition and Boolean equation for a 2-input XOR (Exclusive-OR) logic gate?

  1. F = A + B; output is 0 only when both inputs are 0
  2. F = A' · B'; output is 1 when both inputs are 0
  3. F = A ⊕ B = A · B' + A' · B; output is 1 when inputs are different Correct answer.
  4. F = A · B; output is 1 only when both inputs are 1

Explanation

The XOR gate evaluates to 1 when the inputs differ (A=0, B=1 or A=1, B=0), represented algebraically by F = A ⊕ B = A · B' + A' · B.

Question 38Electrical circuits, machines and digital logic

What are the boolean logic equations for the Sum (S) and Carry-out (C_out) of a 1-bit Full Adder with inputs A, B, and C_in?

  1. S = A + B + C_in ; C_out = A · B · C_in
  2. S = A · B · C_in ; C_out = A ⊕ B ⊕ C_in
  3. S = A ⊕ B ; C_out = A · B
  4. S = A ⊕ B ⊕ C_in ; C_out = A · B + C_in · (A ⊕ B) Correct answer.

Explanation

A full adder computes the arithmetic sum bit S = A ⊕ B ⊕ C_in and carry output C_out = A · B + C_in · (A ⊕ B) = A·B + A·C_in + B·C_in.

Question 39Electrical circuits, machines and digital logic

A 4-to-1 Multiplexer has data inputs I0, I1, I2, I3 and select lines S1 (MSB), S0 (LSB). What data input appears at the output Y when S1 = 1 and S0 = 0?

  1. I1
  2. I2 Correct answer.
  3. I3
  4. I0

Explanation

Select inputs S1 S0 = 1 0 represent binary 2 (decimal 2). Therefore, data input line I2 is selected and routed to output Y.

Question 40Electrical circuits, machines and digital logic

A 2-to-4 decoder with active-LOW outputs (Y0', Y1', Y2', Y3') receives input address A1 = 0, A0 = 1. Assuming the decoder is enabled, what state will the output lines assume?

  1. Y1' = 1 (HIGH), while Y0' = Y2' = Y3' = 0 (LOW)
  2. Y3' = 0 (LOW), while Y0' = Y1' = Y2' = 1 (HIGH)
  3. Y1' = 0 (LOW), while Y0' = Y2' = Y3' = 1 (HIGH) Correct answer.
  4. Y0' = 0 (LOW), while Y1' = Y2' = Y3' = 1 (HIGH)

Explanation

Address inputs A1 A0 = 0 1 select line 1. For active-LOW outputs, the selected line goes LOW (0) while all unselected lines remain HIGH (1). Thus Y1' = 0 and Y0' = Y2' = Y3' = 1.

Question 41Electrical circuits, machines and digital logic

How many 2-input NAND gates are required to construct a 2-input OR gate (F = A + B)?

  1. 1 gate
  2. 2 gates
  3. 4 gates
  4. 3 gates Correct answer.

Explanation

By De Morgan's law, A + B = NOT( NOT A · NOT B ). Connect two NAND gates as inverters to produce NOT A and NOT B, then feed these into a third NAND gate: NOT( NOT A · NOT B ) = A + B. Total = 3 NAND gates.

Question 42Electrical circuits, machines and digital logic

In a basic active-HIGH SR latch built with NOR gates, what occurs when inputs S = 1 and R = 1 are applied simultaneously?

  1. Set state: output Q is forced to 1 and Q' to 0
  2. Normal toggle mode: output Q flips to Q'
  3. No change: output retains previous state
  4. Invalid (forbidden) state: both outputs Q and Q' are forced to 0, causing unpredictable state when inputs return to 0 0 Correct answer.

Explanation

For a NOR-gate SR latch, applying S=1 and R=1 forces both outputs Q and Q' to LOW (0), violating the complementary condition Q' = NOT Q. If S and R drop to 0 simultaneously, race conditions lead to an unpredictable final state.

Question 43Electrical circuits, machines and digital logic

What is the output transition of a edge-triggered JK Flip-Flop when control inputs are set to J = 1 and K = 1 upon arrival of an active clock pulse?

  1. Set mode: output is set to 1 (Q_next = 1)
  2. Hold mode: output retains its previous state (Q_next = Q)
  3. Reset mode: output is cleared to 0 (Q_next = 0)
  4. Toggle mode: the output Q_next becomes the complement of the previous state (Q_next = Q') Correct answer.

Explanation

When J=1 and K=1, the JK flip-flop operates in toggle mode, inverting its current state Q on every active clock edge.

Question 44Electrical circuits, machines and digital logic

What is the characteristic equation and functional behavior of a D (Data) Flip-Flop?

  1. Q_next = NOT D; output assumes the complement of D
  2. Q_next = D; output assumes the value of D at the active clock edge Correct answer.
  3. Q_next = D ⊕ Q; output toggles whenever D = 1
  4. Q_next = D · Q; output goes HIGH only if D = 1 and Q = 1

Explanation

The D flip-flop transfers the logic level present at its D input directly to output Q upon the active edge of the clock signal, characterized by Q_next = D.

Question 45Electrical circuits, machines and digital logic

A 4-bit synchronous binary counter consists of 4 flip-flops. How many distinct counting states (modulo) can this counter sequence through?

  1. 16 states (Modulo-16, 0 to 15) Correct answer.
  2. 8 states (Modulo-8, 0 to 7)
  3. 32 states (Modulo-32, 0 to 31)
  4. 4 states (Modulo-4, 0 to 3)

Explanation

A counter with n flip-flops has 2ⁿ distinct binary states. For n = 4, the modulo is 2⁴ = 16 states (0000_2 to 1111_2).

Question 46Electrical circuits, machines and digital logic

Convert the 12-bit binary integer 110101101011_2 into its equivalent Hexadecimal representation.

  1. E5B_16
  2. C6A_16
  3. D6B_16 Correct answer.
  4. D5C_16

Explanation

Group binary digits into nibbles from right to left: (1101) (0110) (1011). Convert each nibble: 1101_2 = 13 = D; 0110_2 = 6; 1011_2 = 11 = B. Thus hex equivalent is D6B_16.

Question 47Electrical circuits, machines and digital logic

Represent the decimal value -45 as an 8-bit signed binary integer in Two's Complement format.

  1. 00101101_2
  2. 11010011_2 Correct answer.
  3. 11010010_2
  4. 10101101_2

Explanation

+45 in 8-bit binary is 00101101_2. Invert all bits (One's complement): 11010010_2. Add 1 to LSB (Two's complement): 11010010 + 1 = 11010011_2.

Question 48Electrical circuits, machines and digital logic

A 4-to-2 Priority Encoder has inputs D3, D2, D1, D0 where D3 has the highest priority and D0 the lowest. If input combination D3 D2 D1 D0 = 0 1 1 0 is applied, what binary code appears on outputs Y1 Y0?

  1. Y1 Y0 = 11_2 (decimal 3)
  2. Y1 Y0 = 10_2 (decimal 2) Correct answer.
  3. Y1 Y0 = 00_2 (decimal 0)
  4. Y1 Y0 = 01_2 (decimal 1)

Explanation

Priority encoder outputs the binary code of the highest-order active input. Here D3=0, but D2=1. Since D2 is active and higher priority than D1=1, the encoder outputs code 2, which is Y1 Y0 = 10_2.

Question 49Electrical circuits, machines and digital logic

What is the characteristic equation for a T (Toggle) Flip-Flop relating next state Q_next to input T and current state Q?

  1. Q_next = T · Q
  2. Q_next = T ⊕ Q = T · Q' + T' · Q Correct answer.
  3. Q_next = T + Q
  4. Q_next = NOT T

Explanation

A T flip-flop holds its state when T = 0 (Q_next = Q) and toggles its state when T = 1 (Q_next = Q'). This behavior is expressed by XOR logic: Q_next = T ⊕ Q = T · Q' + T' · Q.

Question 50Electrical circuits, machines and digital logic

In combinational circuit design, what causes a static 1-hazard, and how can it be eliminated in a Karnaugh map implementation?

  1. It is caused by high fan-out loading; eliminated by inserting an inverting buffer
  2. It is caused by ground bounce; eliminated by reducing supply voltage
  3. It is caused by unequal propagation delays when a single variable changes, creating a temporary 0 glitch on an output that should remain 1; eliminated by adding redundant consensus prime implicant terms Correct answer.
  4. It is caused by clock jitter in sequential flip-flops; eliminated by adding a decoupling capacitor

Explanation

A static 1-hazard occurs when an input transition moves between adjacent K-map groupings, causing propagation delay differences to produce a temporary 0 glitch. Adding redundant overlap terms (consensus terms) covers the transition seamlessly.