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100+ Free Aragón PAU Technology and Engineering II Practice Questions
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2026 Statistics
Key Facts: Aragón PAU Technology and Engineering II Exam
90 min
Time Limit
PAU Organising Commission / UNIZAR
0–10
Grading Scale
Gobierno de Aragón
4.0
Min. Access Phase Score
UNIZAR PAU Guidelines
EUR 75.00
Ordinary Registration Fee
UNIZAR PAU inscription page 2026
6 Blocks
Curriculum Content Areas
2º Bachillerato Technology & Engineering II Syllabus
The Aragón PAU Technology and Engineering II exam (Tecnología e Ingeniería II) is administered by the PAU Organising Commission of Aragón and Universidad de Zaragoza (UNIZAR) for students completing 2nd Bachillerato. The exam lasts 90 minutes and is graded on a 0–10 scale (minimum 4.0 required in the Access Phase). Note that local questions on this platform are an English-language MCQ study adaptation created to help students master the underlying 2nd Bachillerato curriculum.
Sample Aragón PAU Technology and Engineering II Practice Questions
Try these sample questions to test your Aragón PAU Technology and 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 bar with an initial cross-sectional area of A0 = 100 mm² is subjected to an axial tensile force of F = 50 kN. What is the engineering stress (σ) experienced by the bar?
A.500 kPa
B.500 MPa
C.50 MPa
D.5 GPa
Explanation: Engineering stress is defined as force divided by initial area: σ = F / A0. Converting units: F = 50,000 N and A0 = 100 × 10^-6 m². Therefore, σ = 50,000 N / (100 × 10^-6 m²) = 500,000,000 Pa = 500 MPa.
2A tensile specimen with a gauge length of L0 = 200 mm elongates to a length of L = 200.4 mm under an applied load. What is the engineering strain (ε)?
A.0.02 (2.0%)
B.0.004 (0.4%)
C.0.002 (0.2%)
D.0.0002 (0.02%)
Explanation: Engineering strain is the ratio of elongation to original length: ε = ΔL / L0. Elongation ΔL = 200.4 - 200 = 0.4 mm. Thus, ε = 0.4 mm / 200 mm = 0.002, or 0.2%.
3A structural steel rod of diameter d = 10 mm and original length L0 = 0.5 m is subjected to a tensile force F = 16.5 kN within its elastic limit. Given Young's modulus E = 210 GPa for steel, calculate the total elongation (ΔL) of the rod.
A.1.00 mm
B.2.00 mm
C.0.50 mm
D.0.25 mm
Explanation: Cross-sectional area A0 = π d² / 4 = π (0.01 m)² / 4 = 7.854 × 10^-5 m². Stress σ = F / A0 = 16,500 N / (7.854 × 10^-5 m²) = 210.08 MPa. Using Hooke's law (ε = σ / E): strain ε = 210.08 × 10^6 Pa / (210 × 10^9 Pa) = 1.00 × 10^-3. Elongation ΔL = L0 × ε = 0.5 m × 1.00 × 10^-3 = 5.0 × 10^-4 m = 0.50 mm.
4In a Brinell hardness test on a metallic alloy, a hardened steel ball of diameter D = 10 mm is applied under a load F = 3000 kgf. The measured indentation diameter is d = 4.0 mm. Calculate the Brinell Hardness Number (HB). [Formula: HB = 2F / (π D (D - √(D² - d²)))]
A.150 HB
B.229 HB
C.310 HB
D.450 HB
Explanation: Compute the term √(D² - d²) = √(10² - 4²) = √84 ≈ 9.16515 mm. Then (D - √(D² - d²)) = 10 - 9.16515 = 0.83485 mm. Indentation surface area A = π × 10 × 0.83485 ≈ 26.227 mm². HB = 2 × 3000 / 26.227 = 6000 / 26.227 ≈ 228.8 HB, which rounds to 229 HB.
5In a Charpy impact test, a heavy pendulum striker of mass m = 20 kg is released from a height h1 = 1.5 m. After fracturing the notched test specimen, the pendulum swings to a maximum height h2 = 0.6 m on the opposite side. Assuming g = 9.8 m/s², what is the impact energy absorbed by the specimen?
A.117.6 J
B.176.4 J
C.411.6 J
D.294.0 J
Explanation: Impact energy absorbed equals the difference in potential energy before and after fracture: ΔE = m g (h1 - h2). ΔE = 20 kg × 9.8 m/s² × (1.5 m - 0.6 m) = 196 × 0.9 = 176.4 J.
6Which type of indenter is used in the Vickers hardness test?
A.A cylindrical carbide pin with a flat tip
B.A square-based diamond pyramid with an angle of 136° between opposite faces
C.A hardened steel or tungsten carbide ball of 10 mm diameter
D.A diamond cone with an included angle of 120°
Explanation: The Vickers hardness test utilizes a square-based diamond pyramid indenter with a 136° angle between opposite faces. Indentation diagonals are measured optically.
7What is the primary microstructural goal and effect of quenching (temple) heat treatment applied to carbon steel?
A.To transform austenite rapidly into martensite, maximizing hardness and tensile strength
B.To transform martensite into coarse pearlite to maximize ductility
C.To remove all carbon from the iron lattice via surface decarburization
D.To produce a pure ferrite matrix with low yield strength
Explanation: Quenching involves heating steel above its critical upper transformation temperature to form austenite, followed by rapid cooling in water or oil. This prevents diffusion and traps carbon in a supersaturated body-centered tetragonal structure known as martensite, conferring high hardness and strength.
8Why is tempering (revenido) mandatory immediately following quenching in the heat treatment of tool steel?
A.To melt grain boundaries and weld microscopic internal microcracks
B.To increase the hardness even further beyond the quenched state
C.To relieve internal quenching stresses and increase toughness while reducing extreme brittleness
D.To convert martensite back into 100% untransformed austenite
Explanation: As-quenched martensite is extremely hard but very brittle and contains severe internal stresses. Tempering (reheating below A1 temperature) allows partial carbon diffusion to form tempered martensite, relieving internal stresses and restoring impact toughness and ductility.
9Full annealing (recocido de regeneración) of steel involves heating above the critical temperature followed by:
A.Very slow cooling inside the shut-down furnace to obtain maximum softness and machinability
B.Rapid quenching in cold brine to freeze the crystal structure
C.Forced air cooling using high-velocity industrial fans
D.Immediate mechanical forging while red-hot
Explanation: Full annealing requires heating steel above A3/A1 to form austenite, followed by slow cooling within the furnace (often < 20°C/hour). This produces coarse pearlite and ferrite, ensuring minimum hardness, maximum ductility, and optimal machinability.
10A binary Cu-Ni phase diagram shows complete liquid and solid solubility. An alloy containing 40 wt% Ni is held at 1200°C in a two-phase (Liquid + α) region. The liquid phase contains wL = 32 wt% Ni and the solid α phase contains wα = 50 wt% Ni. Using the lever rule, determine the mass fraction of the liquid phase (WL).
A.36.0%
B.44.4%
C.64.0%
D.55.6%
Explanation: By the lever rule, the fraction of liquid WL = (wα - w0) / (wα - wL). Here w0 = 40, wα = 50, and wL = 32. WL = (50 - 40) / (50 - 32) = 10 / 18 = 0.5556 = 55.6%.
About the Aragón PAU Technology and Engineering II Practice Questions
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