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100+ Free Madrid PAU Tech & Engineering II Practice Questions
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2026 Statistics
Key Facts: Madrid PAU Tech & Engineering II Exam
PAU Organising Commission / UCM
Exam Body
Community of Madrid University Admissions
90 min
Duration
Madrid PAU Regulations
0–10
Scoring
PAU Spain Marking Scheme
2º Bach.
Target Level
Madrid Education Department
€93.02
Registration Fee
UCM PAU Fee Schedule
Sample Madrid PAU Tech & Engineering II Practice Questions
Try these sample questions to test your Madrid 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 diameter of 10 mm is subjected to an axial tensile force of 50 kN. Assuming pure elastic behavior, what is the engineering stress induced in the specimen?
A.636.6 MPa
B.159.2 MPa
C.318.3 MPa
D.795.8 MPa
Explanation: Engineering stress is calculated as sigma = F / A_0. The initial cross-sectional area A_0 = pi * (d/2)^2 = pi * (0.005 m)^2 = 7.854 * 10^-5 m^2. Dividing the force of 50,000 N by 7.854 * 10^-5 m^2 yields 636.6 * 10^6 Pa = 636.6 MPa.
2During a tensile test on a metallic bar with an initial gauge length of 50 mm, an applied stress of 210 MPa produces an elastic elongation of 0.05 mm. What is the Young's Modulus (modulus of elasticity) of the material?
A.210 GPa
B.105 GPa
C.420 GPa
D.21 GPa
Explanation: Engineering strain is epsilon = delta_L / L_0 = 0.05 mm / 50 mm = 0.001. According to Hooke's Law, Young's Modulus E = sigma / epsilon = 210 MPa / 0.001 = 210,000 MPa = 210 GPa.
3In a mechanical stress-strain diagram obtained from a standard tensile test, what does the yield strength (límite elástico) represent?
A.The maximum stress the material can sustain before permanent plastic deformation occurs
B.The maximum stress achieved right at the point of physical fracture
C.The slope of the stress-strain curve in the plastic deformation region
D.The total energy absorbed per unit volume up to catastrophic failure
Explanation: Yield strength defines the transition point between elastic (reversible) deformation and plastic (permanent) deformation. Up to the yield strength, the specimen returns to its original dimensions upon unloading.
4A structural steel alloy has a yield strength of 400 MPa and a Young's Modulus of 200 GPa. What is its modulus of resilience (elastic strain energy storage capacity per unit volume)?
A.400 kJ/m³
B.800 kJ/m³
C.200 kJ/m³
D.1600 kJ/m³
Explanation: The modulus of resilience U_r is the area under the elastic region of the stress-strain curve: U_r = sigma_y^2 / (2 * E). Substituting values yields (400 * 10^6)^2 / (2 * 200 * 10^9) = 1.6 * 10^17 / 4 * 10^11 = 400,000 J/m³ = 400 kJ/m³.
5A material specimen experiences an engineering stress of 300 MPa under a tensile load with a corresponding engineering strain of 0.10. Assuming uniform deformation without necking, what is the true stress in the specimen?
A.330 MPa
B.270 MPa
C.300 MPa
D.360 MPa
Explanation: Assuming constant volume during uniform plastic deformation, true stress sigma_true = sigma_eng * (1 + epsilon_eng). Here, sigma_true = 300 * (1 + 0.10) = 330 MPa.
6In a Brinell hardness test using a steel ball indenter of diameter D = 10 mm and a standard test load P = 3000 kgf, an indentation diameter d = 4.0 mm is measured. Calculate the Brinell Hardness Number (HB).
A.228.7 HB
B.238.7 HB
C.218.7 HB
D.248.7 HB
Explanation: Brinell hardness formula is HB = (2 * P) / [pi * D * (D - sqrt(D^2 - d^2))]. Here D^2 - d^2 = 100 - 16 = 84, sqrt(84) = 9.16515 mm. Thus (D - 9.16515) = 0.83485 mm. Denominator = pi * 10 * 0.83485 = 26.227. HB = 6000 / 26.227 = 228.7 HB.
7What is the key geometric specification of the diamond pyramid indenter used in the Vickers hardness test (HV)?
A.A square-based pyramid with an apex angle of 136° between opposite faces
B.A spherical diamond ball with a 1.588 mm diameter
C.A conical diamond point with a 120° apex angle
D.A cylindrical tungsten carbide pin with a flat tip
Explanation: The Vickers hardness test uses a square-based diamond pyramid indenter with an angle of 136° between opposite faces, providing a single continuous hardness scale across soft and hard metals.
8How does the Rockwell hardness test differ fundamentally from the Brinell and Vickers hardness testing methods?
A.It directly measures the permanent depth of indentation rather than optical surface area
B.It measures the rebound kinetic energy of a falling hammer
C.It relies on ultrasonic frequency damping upon surface contact
D.It determines the scratching resistance using Mohs mineral standards
Explanation: Rockwell hardness testing measures the permanent depth of indentation produced by a minor and major load sequence, allowing direct digital or dial reading without optical measurement of impression diagonals or diameters.
9In a standard Charpy pendulum impact test, a hammer of mass m = 20 kg is released from an initial height h = 1.5 m. After fracturing the notched specimen, the hammer swings up to a final height h' = 0.6 m. Taking g = 9.81 m/s², calculate the impact energy absorbed by the specimen.
A.176.6 J
B.294.3 J
C.117.7 J
D.58.9 J
Explanation: The impact energy absorbed is equal to the change in gravitational potential energy of the pendulum: E = m * g * (h - h') = 20 kg * 9.81 m/s² * (1.5 m - 0.6 m) = 196.2 * 0.9 = 176.58 J (176.6 J).
10A binary Cu-Ni phase diagram displays complete solid solubility. An alloy containing 40 wt% Ni is slowly cooled to 1200°C. At this temperature, the liquid phase L contains 30 wt% Ni and the solid phase alpha contains 50 wt% Ni. Using the lever rule, calculate the mass fraction of the liquid phase.
A.0.50 (50%)
B.0.25 (25%)
C.0.75 (75%)
D.0.40 (40%)
Explanation: According to the lever rule, the mass fraction of liquid phase W_L = (C_alpha - C_overall) / (C_alpha - C_L) = (50 - 40) / (50 - 30) = 10 / 20 = 0.50 or 50%.
About the Madrid PAU Tech & Engineering II Practice Questions
Verified exam format metadata for Madrid PAU Technology and Engineering II (Tecnología e Ingeniería II - UCM 2026) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.