All Practice Exams
100+ Free Canary PAU Tech & Engineering II Practice Questions
Canary Islands PAU Technology and Engineering II Examination — COPAU / ULPGC & ULL (Tecnología e Ingeniería II 2º Bachillerato 2026) practice questions are available now; exam metadata is being verified.
✓ No registration✓ No credit card✓ No hidden fees✓ Start practicing immediately
100+ Questions
100% Free
Loading practice questions...
Same family resources
Explore More Spain Canary Islands PAU (Prueba de Acceso a la Universidad)
Continue into nearby exams from the same family. Each card keeps practice questions, study guides, flashcards, videos, and articles in one place.
2026 Statistics
Key Facts: Canary PAU Tech & Engineering II Exam
90 Minutes
Exam duration
COPAU / ULPGC & ULL
EUR 76.12
Base PAU registration fee
Gobierno de Canarias / COPAU
0–10 Scale
Grading scale for Bachillerato and PAU
Spanish Ministry of Education & COPAU
6 Core Blocks
Materials, Mechanisms, Thermodynamics, Electronics, Control, Pneumatics/Hydraulics
2º Bachillerato Tecnología e Ingeniería II Syllabus
100 Questions
Practice bank size in OpenExamPrep
OpenExamPrep
Canary Islands PAU Technology and Engineering II (COPAU 2026) is a 90-minute university entrance exam assessing 2º Bachillerato Technology & Engineering II across 6 core technical domains.
Sample Canary PAU Tech & Engineering II Practice Questions
Try these sample questions to test your Canary PAU Tech & Engineering II exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1A 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?
A.500 MPa
B.50 MPa
C.5 MPa
D.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.
2A 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?
A.400 GPa
B.20 GPa
C.100 GPa
D.200 GPa
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.
3A 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?
A.188.50 kN
B.60.00 kN
C.47.12 kN
D.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.
4In 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?
A.375 J/cm²
B.180 J/cm²
C.225 J/cm²
D.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.
5Which indenter and measurement principle are used in the standard Brinell hardness test (HB)?
A.A square-based diamond pyramid with a 136° face angle, measuring indentation diagonals
B.A steel cone pressed under load, measuring the diameter of the elastic recovery zone
C.A hardened steel or tungsten carbide sphere pressed under load, measuring the surface area of the spherical indentation
D.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.
6A Vickers hardness test performed with a load of 30 kgf produces a square indentation with an average diagonal length of 0.4 mm. Calculate the Vickers hardness number (HV).
A.225.0 HV
B.347.7 HV
C.695.4 HV
D.173.8 HV
Explanation: Vickers hardness is calculated as HV = 1.8544 · F / d², where F = 30 kgf and d = 0.4 mm. HV = 1.8544 · 30 / (0.4)² = 55.632 / 0.16 = 347.7 HV.
7A 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)?
A.0.450 MJ/m³
B.0.075 MJ/m³
C.0.225 MJ/m³
D.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.
8What is the primary objective of quenching (templado) in heat-treating medium-to-high carbon steels?
A.Rapid cooling from austenite to transform the micro-structure into hard, brittle martensite
B.Slow cooling in a furnace to produce soft, ductile coarse pearlite
C.Reheating hardened steel below A1 to relieve internal stresses and increase toughness
D.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.
9Why is tempering (revenido) almost always performed immediately after quenching steel?
A.To increase carbon content in the core of the steel component
B.To reduce extreme brittleness and relieve internal quenching stresses while retaining acceptable hardness
C.To cause rapid grain growth and maximize electrical conductivity
D.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.
10Which heat treatment process involves heating steel above its critical transformation temperature, holding it, and then cooling it very slowly inside the furnace?
A.Normalizing (normalizado)
B.Quenching (templado)
C.Carburizing (cementación)
D.Annealing (recocido)
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.
About the Canary PAU Tech & Engineering II Practice Questions
Verified exam format metadata for Canary Islands PAU Technology and Engineering II Examination — COPAU / ULPGC & ULL (Tecnología e Ingeniería II 2º Bachillerato 2026) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.