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2026 Statistics

Key Facts: Extremadura PAU Technology and Engineering II (Tecnología e Ingeniería II - UEx 2026) Exam

90 min

Duración oficial del examen escrito de PAU Tecnología e Ingeniería II en Extremadura

Comisión Organizadora de la PAU de Extremadura / UEx

EUR 78.26

Tasa ordinaria de inscripción en la PAU de Extremadura

Universidad de Extremadura (UEx)

0–10

Escala de calificación (mínimo 4.0 en Fase de Acceso para hacer media)

Normativa PAU Extremadura

5 Bloques

Materiales, Termodinámica, Mecánica/Fluídica, Electrónica Digital y Control Automático

Decreto de Currículo de 2º Bachillerato de Extremadura

100

Preguntas cuantitativas y conceptuales de práctica en este banco

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Prepare for the UEx 2026 PAU Technology and Engineering II exam with 100 high-quality practice questions covering materials testing, thermodynamics, fluid power, digital electronics, and control systems.

Sample Extremadura PAU Technology and Engineering II (Tecnología e Ingeniería II - UEx 2026) Practice Questions

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1A cylindrical steel test specimen with an initial diameter of 10 mm is subjected to a static axial tensile load of 15 kN. What is the normal tensile stress (σ) induced in the cross-section of the specimen?
A.190.99 MPa
B.47.75 MPa
C.150.00 MPa
D.477.46 MPa
Explanation: Normal tensile stress is given by σ = F / A0. The initial cross-sectional area is A0 = π · d² / 4 = π · (10 mm)² / 4 = 78.54 mm². Converting 15 kN to 15,000 N: σ = 15,000 N / 78.54 mm² = 190.99 N/mm² = 190.99 MPa.
2A metal structural rod with an original gauge length of 200 mm elongates by 0.4 mm under an axial tensile load within its elastic region. What is the engineering unit strain (ε) of the rod?
A.0.02 (2.0%)
B.0.002 (0.2%)
C.0.0004 (0.04%)
D.0.004 (0.4%)
Explanation: Engineering unit strain is defined as ε = ΔL / L0. Dividing elongation by original length: ε = 0.4 mm / 200 mm = 0.002, which equals 0.2%.
3During a tensile test within the elastic limit, a metallic alloy exhibits a tensile stress of 210 MPa and an engineering strain of 0.001. What is the Young's modulus (modulus of elasticity, E) of the alloy?
A.21 GPa
B.105 GPa
C.210 GPa
D.420 GPa
Explanation: According to Hooke's Law (σ = E · ε), Young's modulus is E = σ / ε = 210 MPa / 0.001 = 210,000 MPa = 210 GPa.
4In a Brinell hardness test, a hardened steel ball with diameter D = 10 mm is pressed into a metal sample under a load P = 3000 kgf, producing an indentation diameter d = 4.0 mm. What is the Brinell Hardness Number (HB)?
A.114.4 HB
B.150.0 HB
C.300.0 HB
D.228.8 HB
Explanation: Brinell hardness is HB = (2 · P) / [π · D · (D - √(D² - d²))]. Substituting values: √(100 - 16) = √84 = 9.16515 mm. Denominator = π · 10 · (10 - 9.16515) = 31.4159 · 0.83485 = 26.227 mm². HB = 6000 kgf / 26.227 mm² = 228.77 ≈ 228.8 HB.
5A standard Charpy impact test uses a pendulum hammer with an initial potential energy of 300 J. After fracturing a notched specimen of cross-sectional area 0.8 cm², the hammer rises to a residual energy height of 180 J. What is the impact toughness (resilience, KCU) of the material?
A.150 J/cm²
B.375 J/cm²
C.225 J/cm²
D.120 J/cm²
Explanation: Absorbed energy ΔE = E_initial - E_residual = 300 J - 180 J = 120 J. The impact toughness KCU is ΔE / A = 120 J / 0.8 cm² = 150 J/cm².
6Which heat treatment process involves heating steel above its critical transformation temperature (austenite range), holding it, and rapidly cooling (quenching) it in water or oil to achieve high hardness?
A.Annealing (Recocido)
B.Quenching (Temple)
C.Tempering (Revenido)
D.Normalizing (Normalizado)
Explanation: Quenching (temple) rapidly cools austenite to prevent equilibrium diffusion, trapping carbon in a supersaturated body-centered tetragonal structure known as martensite, which gives maximum hardness.
7A steel bar of diameter 12 mm and length 300 mm is pulled by a tensile force F = 24 kN. Given Young's modulus E = 200 GPa, what is the total elongation (ΔL) of the bar?
A.1.273 mm
B.0.159 mm
C.0.318 mm
D.0.637 mm
Explanation: Area A0 = π · (12)² / 4 = 113.10 mm². Normal stress σ = 24,000 N / 113.10 mm² = 212.21 MPa. Strain ε = σ / E = 212.21 MPa / 200,000 MPa = 0.001061. Elongation ΔL = ε · L0 = 0.001061 · 300 mm = 0.318 mm.
8During an axial tensile test, a cylindrical metal rod experiences a longitudinal strain ε_z = +0.0015 and a transverse diametral strain ε_x = -0.00045. What is the Poisson's ratio (ν) of the material?
A.0.45
B.0.33
C.0.25
D.0.30
Explanation: Poisson's ratio is defined as the negative ratio of transverse strain to axial strain: ν = - ε_transverse / ε_axial = - (-0.00045) / 0.0015 = 0.30.
9A steel railway track segment of initial length L0 = 20 m with coefficient of thermal expansion α = 12 × 10⁻⁶ /°C is fixed rigidly between unyielding abutments. If the temperature increases by ΔT = 50°C, what thermal compressive stress (σ) develops in the steel (E = 200 GPa)?
A.120 MPa
B.240 MPa
C.60 MPa
D.12 MPa
Explanation: Thermal strain prevented by rigid constraints is ε_th = α · ΔT = (12 × 10⁻⁶ /°C) · 50°C = 0.0006. Thermal stress is σ = E · ε_th = 200,000 MPa · 0.0006 = 120 MPa.
10An isotropic metal alloy has a Young's modulus E = 208 GPa and Poisson's ratio ν = 0.30. Using the elastic relationship G = E / [2(1 + ν)], what is the shear modulus (modulus of rigidity, G) of the alloy?
A.104 GPa
B.80 GPa
C.160 GPa
D.95 GPa
Explanation: Shear modulus G = E / [2(1 + ν)] = 208 GPa / [2 · (1 + 0.30)] = 208 / 2.60 = 80 GPa.

About the Extremadura PAU Technology and Engineering II (Tecnología e Ingeniería II - UEx 2026) Practice Questions

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