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100+ Free Galicia PAU Technology and Engineering II Practice Questions

Galicia PAU Technology and Engineering II (Tecnología e Ingeniería II 2º Bachillerato) practice questions are available now; exam metadata is being verified.

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Key Facts: Galicia PAU Technology and Engineering II Exam

90 min

Exam Time Limit

Comisión Interuniversitaria de Galicia

0–10

Grading Scale

CIUG Galicia

4.0

Min. Access Phase Score

Comisión Interuniversitaria de Galicia

EUR 63.67

Ordinary Registration Fee

CIUG Galicia

5 Blocks

Curriculum Content Areas

2º Bachillerato Technology & Engineering II LOMLOE Syllabus

The Galicia PAU Technology and Engineering II exam (Tecnología e Ingeniería II) is administered by the Comisión Interuniversitaria de Galicia (CIUG) 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 Galicia PAU Technology and Engineering II Practice Questions

Try these sample questions to test your Galicia 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 80 mm² is subjected to an axial tensile load of 40 kN. What normal tensile stress is 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 40 kN to 40,000 N and dividing by 80 mm² yields σ = 40,000 N / 80 mm² = 500 N/mm² = 500 MPa.
2A metallic test specimen with an initial gauge length of 200 mm elongates to 200.4 mm under elastic tensile loading. What unitless strain (ε) does the specimen experience?
A.0.002
B.0.02
C.0.0002
D.0.2
Explanation: Unitless strain is defined as ε = ΔL / L₀. The elongation is ΔL = 200.4 mm - 200 mm = 0.4 mm. Dividing by initial length L₀ = 200 mm gives ε = 0.4 / 200 = 0.002 (or 0.2%).
3A structural steel alloy exhibits a normal stress of 420 MPa at an elastic strain of 0.002. What is its Young's modulus of elasticity (E)?
A.210 GPa
B.21 GPa
C.2100 GPa
D.420 GPa
Explanation: According to Hooke's law, Young's modulus is E = σ / ε. Substituting σ = 420 MPa and ε = 0.002 gives E = 420 / 0.002 = 210,000 MPa = 210 GPa.
4A Charpy impact test specimen with a V-notch cross-sectional area of 0.8 cm² absorbs 60 J of energy upon fracture. What is the material's impact resilience (K)?
A.75 J/cm²
B.48 J/cm²
C.7.5 J/cm²
D.750 J/cm²
Explanation: Impact resilience is calculated as ρ = K / A, where K is absorbed energy in Joules and A is cross-sectional area at the notch in cm². Thus, ρ = 60 J / 0.8 cm² = 75 J/cm².
5Which thermal heat treatment involves heating steel above its critical temperature into the austenite range followed by rapid quenching in water or oil to form martensite?
A.Quenching (Temple)
B.Tempering (Revenido)
C.Annealing (Recocido)
D.Normalizing (Normalizado)
Explanation: Quenching (temple) consists of rapid cooling from the austenitic phase to freeze carbon in a supersaturated body-centered tetragonal lattice called martensite, maximizing hardness and strength.
6In the iron-carbon (Fe-Fe₃C) equilibrium phase diagram, what carbon content defines the exact eutectoid steel composition that transforms entirely into pearlite at 727 °C?
A.0.77 wt% C
B.2.11 wt% C
C.4.30 wt% C
D.0.022 wt% C
Explanation: The eutectoid point in the Fe-Fe₃C phase diagram occurs at 0.77 wt% C (often rounded to 0.8%), where solid austenite transforms upon cooling into a lamellar mixture of ferrite and cementite known as pearlite.
7A Brinell hardness test is conducted using a hardened steel ball of diameter D = 10 mm and a standard test load F = 3000 kgf. If the resulting indentation diameter is d = 4 mm, what is the Brinell Hardness Number (HB)?
A.229 HB
B.180 HB
C.315 HB
D.145 HB
Explanation: Brinell hardness formula is HB = 2F / [π D (D - √(D² - d²))]. Substituting F = 3000 kgf, D = 10 mm, and d = 4 mm gives HB = 6000 / [π × 10 × (10 - √(100 - 16))] = 6000 / [31.416 × (10 - 9.165)] = 6000 / 26.227 = 228.8 ≈ 229 HB.
8A hypoeutectoid steel containing 0.40 wt% C is slowly cooled to just below 727 °C. Assuming maximum ferrite carbon solubility is 0.022 wt% C and eutectoid carbon content is 0.77 wt% C, what is the mass fraction of proeutectoid alpha-ferrite?
A.49.5%
B.52.0%
C.40.0%
D.65.3%
Explanation: By the lever rule, the mass fraction of proeutectoid ferrite just below 727 °C is W_α = (C_eutectoid - C₀) / (C_eutectoid - C_α) = (0.77 - 0.40) / (0.77 - 0.022) = 0.37 / 0.748 = 0.4946 ≈ 49.5%.
9A steel component with Young's modulus E = 200 GPa is stressed to 400 MPa, exceeding its proportional limit. If total measured strain under load is ε_total = 0.006, what is the permanent plastic strain (ε_p) remaining after complete unloading?
A.0.004
B.0.002
C.0.006
D.0.001
Explanation: Elastic strain recovery upon unloading follows Hooke's law: ε_e = σ / E = 400 MPa / 200,000 MPa = 0.002. Permanent plastic strain is ε_p = ε_total - ε_e = 0.006 - 0.002 = 0.004.
10In a Charpy impact pendulum test, a 20 kg hammer is released from an initial height h₁ = 1.5 m. After fracturing the specimen, the hammer swings up to a maximum height h₂ = 0.6 m. Taking g = 9.8 m/s², how much energy (K) was absorbed by the test piece?
A.176.4 J
B.294.0 J
C.117.6 J
D.150.0 J
Explanation: Absorbed impact energy equals the loss of gravitational potential energy: K = m · g · (h₁ - h₂) = 20 kg × 9.8 m/s² × (1.5 m - 0.6 m) = 196 × 0.9 = 176.4 J.

About the Galicia PAU Technology and Engineering II Practice Questions

Verified exam format metadata for Galicia PAU Technology and Engineering II (Tecnología e Ingeniería II 2º Bachillerato) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.