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100+ Free Pancyprian Appl. Mech. Sci. III Practice Questions

Cyprus Pancyprian Access Examination Technical School Applied Mechanical Science III (Theoretical Direction) (ΕΦΑΡΜΟΣΜΕΝΗ ΜΗΧΑΝΙΚΗ ΕΠΙΣΤΗΜΗ ΙΙΙ Τ.Σ. (Θ.Κ.), code 414) practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: Pancyprian Appl. Mech. Sci. III Exam

2.5 hours

Written paper duration per the 2026 Pancyprian timetable (08:00–10:30)

Cyprus Examinations Service

Code 414

Subject code in the official 2026 exam guide (ΕΦΑΡΜΟΣΜΕΝΗ ΜΗΧΑΝΙΚΗ ΕΠΙΣΤΗΜΗ ΙΙΙ Τ.Σ. (Θ.Κ.))

Cyprus Examinations Service

0–20

Marking scale; feeds the access/ranking grade with no pass/fail mark

Cyprus Examinations Service

EUR 25

2026 fee per access subject (not code 032)

Cyprus Examinations Service

10 June 2026

2026 exam date for this subject per the official timetable

Cyprus Examinations Service timetable

100

Original practice questions in this study bank

OpenExamPrep

100 free MCQs covering Pancyprian Applied Mechanical Science III (code 414) with real stress, strain, thermo, fluid and mechanism calculations. This is an English-language study adaptation — the official exam is a 2.5-hour Greek written paper with constructed-response problem-solving, so practise full written solutions separately.

Sample Pancyprian Appl. Mech. Sci. III Practice Questions

Try these sample questions to test your Pancyprian Appl. Mech. Sci. III exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A steel rod carries an axial tensile force of 48 kN. The rod has a circular cross-section of diameter 20 mm. What is the average normal stress in the rod?
A.76.4 MPa
B.152.8 MPa
C.38.2 MPa
D.240 MPa
Explanation: σ = F/A. A = πd²/4 = π(0.020)²/4 = 3.142×10⁻⁴ m². With F = 48 000 N, σ = 48 000 / 3.142×10⁻⁴ = 1.528×10⁸ Pa = 152.8 MPa.
2A bolted joint carries a shear force of 18 kN on a single bolt of diameter 12 mm. Assuming single shear and uniform stress, the average shear stress in the bolt is closest to:
A.79.6 MPa
B.159.2 MPa
C.39.8 MPa
D.127 MPa
Explanation: τ = F/A with A = πd²/4 = π(0.012)²/4 = 1.131×10⁻⁴ m². τ = 18 000 / 1.131×10⁻⁴ = 1.591×10⁸ Pa ≈ 159.2 MPa.
3A rectangular bar 40 mm × 10 mm carries an axial compressive load of 20 kN. The compressive stress is:
A.50 MPa
B.5 MPa
C.500 MPa
D.20 MPa
Explanation: A = 0.040 × 0.010 = 4.0×10⁻⁴ m². σ = F/A = 20 000 / 4.0×10⁻⁴ = 5.0×10⁷ Pa = 50 MPa.
4A hollow circular shaft has outer diameter Do = 60 mm and inner diameter Di = 40 mm. It transmits a torque T = 1.0 kN·m. The maximum shear stress due to torsion is closest to:
A.29.5 MPa
B.51.5 MPa
C.15.0 MPa
D.73.0 MPa
Explanation: J = (π/32)(Do⁴ − Di⁴) = (π/32)(0.060⁴ − 0.040⁴) = (π/32)(1.296×10⁻⁵ − 2.56×10⁻⁶) = 1.021×10⁻⁶ m⁴. τ_max = T c / J = 1000 × 0.030 / 1.021×10⁻⁶ ≈ 2.94×10⁷ Pa ≈ 29.4 MPa, closest to 29.5 MPa.
5A simply supported beam of length L = 2.0 m carries a central concentrated load P = 4 kN. The maximum bending moment is:
A.1.0 kN·m
B.2.0 kN·m
C.4.0 kN·m
D.8.0 kN·m
Explanation: For a central point load on a simply supported beam, M_max = PL/4 = (4 kN)(2.0 m)/4 = 2.0 kN·m at mid-span.
6A rectangular beam has width b = 20 mm and depth h = 40 mm. The second moment of area about the neutral axis parallel to the width is:
A.1.067×10⁻⁷ m⁴
B.2.133×10⁻⁷ m⁴
C.5.333×10⁻⁸ m⁴
D.3.2×10⁻⁶ m⁴
Explanation: I = bh³/12 = (0.020)(0.040)³/12 = (0.020)(6.4×10⁻⁵)/12 = 1.067×10⁻⁷ m⁴.
7A beam cross-section has I = 2.0×10⁻⁶ m⁴. At a section where M = 1.5 kN·m, the bending stress at y = 40 mm from the neutral axis is:
A.30 MPa
B.7.5 MPa
C.75 MPa
D.3.0 MPa
Explanation: σ = My/I = (1500 N·m)(0.040 m) / 2.0×10⁻⁶ m⁴ = 3.0×10⁷ Pa = 30 MPa.
8A material has ultimate tensile strength σ_u = 400 MPa. For a design tensile stress of 100 MPa, the factor of safety based on ultimate strength is:
A.2
B.4
C.0.25
D.5
Explanation: Factor of safety n = σ_u / σ_allow = 400 / 100 = 4.
9A thin-walled cylindrical pressure vessel has internal diameter 200 mm and wall thickness 5 mm. For internal pressure p = 2.0 MPa, the hoop (circumferential) stress is:
A.20 MPa
B.40 MPa
C.10 MPa
D.80 MPa
Explanation: For a thin cylinder, σ_h = pD/(2t) = (2.0 MPa)(200 mm)/(2×5 mm) = 40 MPa.
10For the thin cylinder with p = 2.0 MPa, D = 200 mm and t = 5 mm, the longitudinal stress is:
A.40 MPa
B.20 MPa
C.10 MPa
D.80 MPa
Explanation: σ_l = pD/(4t) = (2.0)(200)/(4×5) = 20 MPa, half the hoop stress for a closed thin cylinder.

About the Pancyprian Appl. Mech. Sci. III Practice Questions

Verified exam format metadata for Cyprus Pancyprian Access Examination Technical School Applied Mechanical Science III (Theoretical Direction) (ΕΦΑΡΜΟΣΜΕΝΗ ΜΗΧΑΝΙΚΗ ΕΠΙΣΤΗΜΗ ΙΙΙ Τ.Σ. (Θ.Κ.), code 414) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.