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100+ Free Balearic PAU Technology & Engineering II Practice Questions
Balearic Islands PAU Technology and Engineering II Examination — UIB (Tecnologia i Enginyeria II 2026) practice questions are available now; exam metadata is being verified.
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Sample Balearic PAU Technology & Engineering II Practice Questions
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1A cylindrical steel bar with a diameter of $d = 10\text{ mm}$ is subjected to an axial tensile force of $F = 50\text{ kN}$. What is the normal tensile stress $\sigma$ induced in the bar?
A.636.62 MPa
B.159.15 MPa
C.500.00 MPa
D.318.31 MPa
Explanation: The cross-sectional area of the cylindrical bar is $A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (10\text{ mm})^2 = 78.54\text{ mm}^2$. Tensile stress is calculated as $\sigma = \frac{F}{A} = \frac{50000\text{ N}}{78.54\text{ mm}^2} = 636.62\text{ N/mm}^2 = 636.62\text{ MPa}$.
2A metal wire of initial length $L_0 = 2.0\text{ m}$ undergoes an elongation of $\Delta L = 1.2\text{ mm}$ under load. What is the unit engineering strain $\varepsilon$?
A.0.0600 (6.00%)
B.0.0006 (0.06%)
C.0.0060 (0.60%)
D.0.0012 (0.12%)
Explanation: Engineering strain is defined as the ratio of elongation to initial length: $\varepsilon = \frac{\Delta L}{L_0} = \frac{1.2\text{ mm}}{2000\text{ mm}} = 0.0006$, which corresponds to $0.06\%$.
3A structural steel alloy has a Young's modulus of elasticity $E = 210\text{ GPa}$. If it experiences an elastic strain of $\varepsilon = 0.001$, what is the elastic stress $\sigma$?
A.2100 MPa
B.2.1 GPa
C.210 MPa
D.21 MPa
Explanation: According to Hooke's Law within the elastic regime, $\sigma = E \cdot \varepsilon = 210 \times 10^9\text{ Pa} \times 0.001 = 210 \times 10^6\text{ Pa} = 210\text{ MPa}$.
4Which type of penetrator (indenter) is standardly used in the Brinell hardness test (HB)?
A.A 120° diamond pyramid
B.A 136° diamond square-based pyramid
C.A conical steel needle
D.A hardened steel or tungsten carbide ball
Explanation: The Brinell hardness test uses a spherical indenter made of hardened steel (or tungsten carbide for very hard materials) of standardized diameter (typically 10 mm, 5 mm, or 2.5 mm).
5What is the primary microstructural objective of quenching (temple) a medium-carbon steel from its austenitizing temperature?
A.To transform austenite rapidly into martensite, achieving high hardness
B.To form a soft, ductile coarse pearlite structure
C.To relieve internal residual stresses without changing phase
D.To promote graphite flake precipitation
Explanation: Quenching involves rapid cooling (in water, oil, or forced air) from the austenitic region to suppress diffusion, causing diffusionless shear transformation of austenite into hard, brittle martensite.
6Why is a tempering heat treatment (revenido) almost always performed immediately after quenching steel?
A.To prevent surface oxidation and decarburization during storage
B.To reduce excessive brittleness and internal stresses while restoring acceptable toughness
C.To further increase the hardness of martensite to maximum theoretical limits
D.To transform martensite back into 100% pure austenite
Explanation: As-quenched martensite is extremely hard but excessively brittle and highly stressed. Tempering (reheating below $A_{c1}$) relieves internal stresses and precipitates fine carbides, increasing toughness and ductility at a slight expense of hardness.
7How do thermoplastic polymers differ fundamentally from thermosetting polymers regarding heat response?
A.Thermoplastics contain cross-linked 3D network polymer chains.
B.Thermosets consist solely of linear or branched chains with weak van der Waals intermolecular bonds.
C.Thermoplastics soften and melt upon heating and can be repeatedly re-molded, whereas thermosets decompose permanently when heated.
D.Thermosets can be melted and re-molded repeatedly, whereas thermoplastics decompose upon heating.
Explanation: Thermoplastics consist of linear or branched polymer chains held together by weak secondary bonds (van der Waals), allowing them to soften/melt reversibly upon heating. Thermosets contain permanent covalent cross-links, forming a 3D rigid network that decomposes/chars rather than melting.
8In a fiber-reinforced composite material (e.g., carbon-fiber reinforced epoxy), what is the main mechanical function of the matrix material?
A.To carry almost 100% of the tensile load along the longitudinal fiber direction
B.To increase the electrical conductivity of the composite to metallic levels
C.To reduce the overall density of the composite below that of air
D.To bind the fibers together, transfer external loads to the fibers, and protect them from environmental damage
Explanation: The fibers provide high strength and stiffness, while the matrix binds the reinforcement fibers, distributes and transfers shear stresses between fibers, and protects them from abrasion and corrosion.
9A cylindrical titanium test specimen with diameter $d_0 = 12.00\text{ mm}$ undergoes axial tension. Under load, its axial strain is measured as $\varepsilon_x = +0.0015$ and its diameter decreases by $\Delta d = -0.0054\text{ mm}$. What is Poisson's ratio $\nu$ for this alloy?
A.0.30
B.0.45
C.0.25
D.0.36
Explanation: Lateral strain is $\varepsilon_y = \frac{\Delta d}{d_0} = \frac{-0.0054\text{ mm}}{12.00\text{ mm}} = -0.00045$. Poisson's ratio is $\nu = -\frac{\varepsilon_y}{\varepsilon_x} = -\frac{-0.00045}{0.0015} = 0.30$.
10A Brinell hardness test is conducted using a ball indenter of diameter $D = 10\text{ mm}$ under a test load $P = 3000\text{ kgf}$. The resulting impression diameter is measured as $d = 4.0\text{ mm}$. What is the Brinell Hardness Number ($HB$)?
A.254.65 HB
B.228.88 HB
C.238.73 HB
D.198.94 HB
Explanation: Brinell hardness formula: $HB = \frac{2P}{\pi D \left(D - \sqrt{D^2 - d^2}\right)}$. Here $D = 10$, $d = 4$, $P = 3000$. $\sqrt{100 - 16} = \sqrt{84} = 9.16515$. $D - \sqrt{84} = 10 - 9.16515 = 0.83485$. Area $A_{imp} = \frac{\pi \cdot 10 \cdot 0.83485}{2} = 13.1132\text{ mm}^2$. $HB = \frac{3000}{13.1132} = 228.88\text{ kgf/mm}^2$.
About the Balearic PAU Technology & Engineering II Practice Questions
Verified exam format metadata for Balearic Islands PAU Technology and Engineering II Examination — UIB (Tecnologia i Enginyeria II 2026) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.