Nepal Engineering Council · Mechanical Engineering · Chapter 1
Basic Mechanical Engineering Concept
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183 questions in 6 syllabus topics.
1.1 Mechanical drawing
31 questions · AMeE0101
1. A hole is dimensioned 50 +0.025/0 mm and a shaft 50 −0.025/−0.050 mm. The maximum clearance between them is
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Use the extreme limits that give the largest gap.
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Answer: A. 0.075 mm
Maximum clearance = largest hole − smallest shaft = 50.025 − 49.950 = 0.075 mm.
2. The upper and lower limits of a shaft are 30.021 mm and 30.000 mm. Its tolerance is
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Tolerance is the permitted variation in size.
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Answer: C. 0.021 mm
Tolerance = upper limit − lower limit = 30.021 − 30.000 = 0.021 mm.
3. A single-riveted lap joint has pitch 50 mm and rivet hole diameter 20 mm. The tearing efficiency of the plate is
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Compare the plate strength between holes with the solid plate strength per pitch.
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Answer: B. 60%
η_t = (p − d)/p = (50 − 20)/50 = 0.6 = 60%.
4. A rivet of 20 mm diameter in single shear has permissible shear stress 60 MPa. Its shear strength is approximately
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Shear area of a rivet is the cross-sectional area of its shank.
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Answer: D. 18.85 kN
P_s = (π/4)d²τ = (π/4)(20²)(60) ≈ 18 850 N = 18.85 kN.
5. A 16 mm rivet in double shear has τ = 80 MPa. The load it can carry in shear is nearly
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Two shear planes double the resisting area.
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Answer: A. 32.2 kN
P = 2 × (π/4)(16²)(80) ≈ 32 170 N ≈ 32.2 kN.
6. A fillet weld of leg size 8 mm and length 100 mm is loaded in shear. If the permissible shear stress is 80 MPa, the strength of the weld is about
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Shear acts on the throat section, not the leg.
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Answer: C. 45.2 kN
Throat = 0.707 s = 5.66 mm. P = 0.707 × 8 × 100 × 80 ≈ 45 250 N ≈ 45.2 kN.
7. The throat thickness of an equal-leg fillet weld with leg size 10 mm is
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The throat is the shortest distance from the root to the face.
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Answer: B. 7.07 mm
Throat t = s·cos45° = 0.707 × 10 = 7.07 mm.
8. An M20 bolt has a tensile stress area of 245 mm². Under an axial load of 49 kN the tensile stress is
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Stress = load divided by area, using N and mm².
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Answer: D. 200 MPa
σ = P/A = 49 000/245 = 200 MPa.
9. A 10 mm × 8 mm × 50 mm (b × h × l) sunk key transmits 200 N·m on a 40 mm diameter shaft. The shear stress in the key is
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Tangential force at the shaft surface acts over b × l.
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Answer: A. 20 MPa
F = T/(d/2) = 200 000/20 = 10 000 N; τ = F/(b·l) = 10 000/(10 × 50) = 20 MPa.
10. A double-start thread has a pitch of 3 mm. The distance the nut advances in one revolution (lead) is
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Lead and pitch differ in multi-start threads.
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Answer: C. 6 mm
Lead = number of starts × pitch = 2 × 3 = 6 mm.
11. A component is drawn to a scale of 1:5. A line measuring 40 mm on the drawing represents an actual length of
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1:5 means the drawing is smaller than the object.
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Answer: B. 200 mm
Actual = drawing length × 5 = 40 × 5 = 200 mm (reduction scale).
12. A hole is 40 +0.025/0 mm and a shaft is 40 +0.059/+0.043 mm. The minimum interference is
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Use the smallest shaft and the largest hole.
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Answer: D. 0.018 mm
Minimum interference = smallest shaft − largest hole = 40.043 − 40.025 = 0.018 mm.
13. Deviations of a surface profile from its mean line at four points are +2, −3, +4 and −1 µm. The arithmetic mean roughness Ra is
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Ignore signs before averaging.
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Answer: A. 2.5 µm
Ra is the mean of absolute deviations: (2 + 3 + 4 + 1)/4 = 2.5 µm.
14. A taper key of 1 in 100 taper is 200 mm long. The difference in its depth between the two ends is
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Multiply length by the taper ratio.
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Answer: C. 2 mm
Taper 1:100 over 200 mm gives 200/100 = 2 mm change in thickness.
15. Which of the following is NOT a type of fit between a hole and a shaft?
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Recall the three standard fit families.
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Answer: B. Transference fit
The three recognised fits are clearance, transition and interference; 'transference' does not exist.
16. In the hole-basis system of fits, the hole
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The letter H designates the hole.
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Answer: D. has zero lower deviation (fundamental deviation H)
In a hole-basis system the lower limit of the hole equals the basic size (H hole) and shaft deviations vary to give the fit.
17. Which of the following fits is a clearance fit?
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Shaft letters near 'a' give the largest clearance, those after 'n' give interference.
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Answer: A. H7/g6
H7/g6 gives a small positive clearance (sliding fit); p, s and u shafts give interference fits.
18. A fit in which the tolerance zones of hole and shaft overlap so that either clearance or interference may result is called
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Think of 'in between' the other two fits.
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Answer: C. a transition fit
Overlapping tolerance zones give a transition fit, e.g. H7/k6.
19. In first-angle projection, the plan (top view) is placed
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The object lies between the observer and the plane.
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Answer: B. below the front view
In first angle method the object lies in the first quadrant; the top view is drawn below the elevation and the left view to the right.
20. In engineering drawings, hidden edges are shown by
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Edges that cannot be seen are broken.
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Answer: D. thin dashed lines
Hidden outlines are drawn with short dashes (type E/F); chain lines are for centres and cutting planes.
21. Section lines (hatching) in a sectional view are normally drawn at
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Same hatch direction for the same part.
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Answer: A. 45° to the main outline, thin and equally spaced
Standard practice is thin continuous lines, equally spaced, inclined at 45° to the principal outline or axis.
22. The thread designation M10 × 1.5 indicates
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Letter M then diameter then pitch.
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Answer: C. ISO metric thread of 10 mm nominal diameter and 1.5 mm pitch
M stands for ISO metric; 10 is the nominal (major) diameter and 1.5 mm the pitch.
23. The included angle of an ISO metric screw thread profile is
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Whitworth is the one with 55°.
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Answer: B. 60°
ISO metric and Unified threads have a 60° thread angle; Whitworth is 55° and Acme 29°.
24. In riveting, 'caulking' is done to
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Think of pressure vessels.
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Answer: D. make the joint leak-proof (fluid tight)
Caulking and fullering upset the plate edge against the adjoining plate to make boiler joints steam/fluid tight.
25. The distance between the centres of adjacent rivets in the same row is called
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Back pitch is between rows.
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Answer: A. pitch
Pitch is the centre-to-centre distance of neighbouring rivets in a row; margin is the distance from the hole centre to plate edge.
26. In a standard welding symbol, a triangle on the arrow side of the reference line indicates
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The triangle resembles the weld cross-section.
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Answer: C. a fillet weld on the arrow side of the joint
A right-angled triangle symbol denotes a fillet weld; its position (above/below the line) shows which side.
27. Which welded joint is formed by placing two plates in the same plane edge to edge and welding?
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The plates 'butt' against each other.
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Answer: B. Butt joint
Plates in the same plane welded along their abutting edges form a butt joint.
28. A key that is semicircular in shape, fitted in a circular keyseat milled in the shaft, is called
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Named after its inventor.
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Answer: D. a Woodruff key
The Woodruff key is a segment of a disc; it is common in machine tools and automobile shafts.
29. A feather key differs from a sunk taper key mainly because it
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It permits relative axial movement.
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Answer: A. is parallel and allows axial sliding of the hub
A feather key is a parallel key fixed to the shaft (or hub) that lets the mating part slide along the shaft.
30. The surface finish value Ra is defined as
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The A in Ra stands for 'arithmetic'.
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Answer: C. the arithmetic average of absolute profile deviations from the mean line
Ra = (1/L)∫|y|dx, the arithmetic mean of absolute deviations from the mean line over the sampling length.
31. ISO roughness grade number N7 corresponds to Ra equal to
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Each step doubles the Ra value.
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Answer: B. 1.6 µm
ISO grades: N6 = 0.8, N7 = 1.6, N8 = 3.2, N9 = 6.3 µm.
1.2 Engineering materials
32 questions · AMeE0102
32. A 10 mm diameter steel bar carries a tensile load of 20 kN. The engineering stress in the bar is nearly
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Area of a circle is πd²/4; keep N and mm².
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Answer: A. 254.6 MPa
σ = P/A = 20 000/[(π/4)(10²)] = 20 000/78.54 ≈ 254.6 MPa.
33. A bar of original gauge length 50 mm extends by 0.15 mm under load. The axial strain is
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Strain is change in length divided by original length.
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Answer: C. 0.003
ε = ΔL/L = 0.15/50 = 0.003 (dimensionless).
34. In the elastic range a steel specimen shows a stress of 200 MPa at a strain of 0.001. Young's modulus is
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E is the slope of the elastic portion of the stress-strain curve.
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Answer: B. 200 GPa
E = σ/ε = 200 MPa/0.001 = 200 000 MPa = 200 GPa.
35. A tensile specimen of gauge length 50 mm has a gauge length of 62 mm after fracture. The percentage elongation is
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Divide the increase in gauge length by the original gauge length.
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Answer: D. 24%
Elongation = (62 − 50)/50 × 100 = 24%.
36. A round tensile specimen of initial diameter 12.5 mm has a diameter of 8 mm at the fracture neck. The percentage reduction in area is about
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Areas scale with the square of the diameter.
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Answer: D. 59%
Reduction = 1 − (8/12.5)² = 1 − 0.4096 = 0.590, i.e. about 59%.
37. A bar under tension has an axial strain of 0.001 and a lateral strain of −0.0003. Poisson's ratio is
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Ratio of lateral to axial strain in magnitude.
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Answer: B. 0.3
ν = −(lateral strain)/(axial strain) = 0.0003/0.001 = 0.3.
38. A mild steel member has a yield strength of 250 MPa and a working stress of 100 MPa. The factor of safety based on yield is
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Strength divided by allowable stress.
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Answer: C. 2.5
FOS = yield strength/working stress = 250/100 = 2.5.
39. For steel, the endurance limit of a polished rotating-beam specimen is approximately half of the ultimate tensile strength. For σ_ut = 600 MPa the endurance limit is about
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Use the standard 0.5 rule.
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Answer: A. 300 MPa
S_e' ≈ 0.5 σ_ut = 0.5 × 600 = 300 MPa (rule of thumb for steels).
40. A Brinell test uses a 10 mm ball and a 3000 kgf load, producing an indentation of 4 mm diameter. The Brinell hardness number is approximately
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Use the spherical-cap area formula.
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Answer: B. 229
HB = 2P/[πD(D − √(D² − d²))] = 6000/[π·10·(10 − 9.165)] ≈ 229 kgf/mm².
41. A steel has σ_y = 250 MPa and E = 200 GPa. Its modulus of resilience σ_y²/2E is
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Energy per unit volume stored at the elastic limit.
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Answer: D. 0.156 MJ/m³
σ_y²/(2E) = (250²)/(2 × 200 000) = 0.15625 MPa = 0.156 MJ/m³.
42. A steel rod (E = 200 GPa, α = 12 × 10⁻⁶ /°C) is rigidly fixed at both ends and heated by 50 °C. The thermal stress developed is
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Fully restrained expansion gives σ = EαΔT.
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Answer: A. 120 MPa
σ = EαΔT = 200 000 × 12×10⁻⁶ × 50 = 120 MPa (compressive).
43. In a fatigue cycle the maximum stress is +300 MPa and the minimum is −100 MPa. The stress amplitude is
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Amplitude is half the stress range.
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Answer: C. 200 MPa
σ_a = (σ_max − σ_min)/2 = (300 + 100)/2 = 200 MPa (mean stress is 100 MPa).
44. A plate with a hole has a theoretical stress concentration factor K_t = 2.5. If the nominal stress is 80 MPa, the maximum stress at the hole is
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Multiply nominal stress by the concentration factor.
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Answer: A. 200 MPa
σ_max = K_t σ_nom = 2.5 × 80 = 200 MPa.
45. The property of a material by which it can be drawn into thin wires is called
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Wire drawing is tensile plastic flow.
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Answer: C. ductility
Ductility is the ability to undergo large plastic deformation in tension, e.g. wire drawing; malleability refers to compression (rolling, hammering).
46. Toughness of a material is best represented by
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It is an energy measure.
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Answer: B. the total area under the stress-strain curve
Toughness is energy absorbed per unit volume up to fracture, i.e. area under the σ–ε curve.
47. Which pair of tests measures the impact toughness of a material?
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Both use a swinging pendulum.
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Answer: D. Charpy and Izod
Charpy (simply supported) and Izod (cantilever) notched-bar tests measure energy absorbed in impact.
48. The Rockwell C scale hardness test uses
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C scale for hard steels.
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Answer: D. a diamond cone indenter with 150 kgf major load
HRC uses a 120° diamond spheroconical (Brale) indenter, 10 kgf minor and 150 kgf major load.
49. For a material with no well-defined yield point (e.g. aluminium alloy), yield strength is usually specified by
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Offset method.
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Answer: B. the 0.2% offset (proof) stress
A line parallel to the elastic slope offset by 0.2% strain locates the 0.2% proof stress.
50. Which of the following indicates a material of high stiffness?
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Stiffness and strength are different properties.
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Answer: C. High modulus of elasticity
Stiffness is resistance to elastic deformation, measured by Young's modulus.
51. Brass is an alloy mainly of
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Bronze has the tin.
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Answer: A. copper and zinc
Brass is Cu–Zn; bronze is Cu–Sn.
52. The minimum chromium content that makes steel 'stainless' by forming a passive oxide film is about
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The film is chromium oxide.
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Answer: B. 10.5–12%
Steels with at least about 10.5–12% Cr develop a self-healing Cr₂O₃ film and resist rusting.
53. Duralumin is an age-hardenable alloy of aluminium containing mainly
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Age hardening by CuAl₂ precipitates.
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Answer: D. copper (about 4%) with small amounts of Mg and Mn
Duralumin is Al + ~4% Cu, ~0.5% Mg, ~0.5% Mn; widely used in aircraft structures.
54. Fatigue failure of a component is characterised by
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Think of a paper clip bent back and forth.
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Answer: A. failure under repeated stresses below the static ultimate strength
Fatigue is progressive crack growth under cyclic loading that can fail the part at stresses well below σ_ut or even σ_y.
55. In an S–N curve for steel, the endurance limit is the stress
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The curve flattens out.
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Answer: C. below which the material can sustain infinite cycles without failure
Steels show a horizontal knee near 10⁶–10⁷ cycles; stress below the limit gives infinite life.
56. Beach marks on a fractured surface are a characteristic sign of
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Look at the shape of ripples on a shore.
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Answer: A. fatigue failure
Concentric beach (clamshell) marks record crack-front arrests during progressive fatigue crack growth.
57. Creep is the
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It is time dependent.
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Answer: C. slow time-dependent plastic deformation under constant load at elevated temperature
Creep occurs under steady stress, typically above about 0.4 T_m (absolute melting temperature).
58. Which of the following is the correct order of the creep stages in a creep curve?
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The steady-state stage is in the middle.
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Answer: B. primary, secondary (steady state), tertiary
Decreasing strain rate (primary), constant minimum rate (secondary), then accelerating rate to rupture (tertiary).
59. Galvanic corrosion occurs when
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A galvanic cell is formed.
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Answer: D. two dissimilar metals are in electrical contact in the presence of an electrolyte
The more anodic metal corrodes when coupled with a more noble metal through an electrolyte.
60. Sacrificial anodes such as zinc or magnesium protect steel structures because they
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The protecting metal gets consumed.
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Answer: D. are more anodic and corrode in preference to steel
Zinc/magnesium are lower in the galvanic series, so they supply electrons and become the corroding anode (cathodic protection).
61. Galvanising of steel sheets means coating with
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Used on roofing sheets and pipes.
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Answer: B. zinc
Galvanising applies zinc by hot dipping or electroplating; tin coating is called tinning.
62. Which of the following is NOT a method of corrosion control?
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Look for the method that promotes the reaction.
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Answer: C. Increasing the temperature of the environment
Raising the temperature generally speeds up corrosion; coatings, cathodic protection and inhibitors reduce it.
63. Stress corrosion cracking requires
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All three conditions must act together.
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Answer: A. a tensile stress, a susceptible material and a specific corrosive environment together
SCC arises from combined sustained tensile stress and a specific environment (e.g. brass in ammonia, stainless steel in chlorides).
1.3 Material science
31 questions · AMeE0103
64. Copper is FCC with atomic radius 0.128 nm. Its lattice parameter a = 4R/√2 is about
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Face diagonal holds 4 radii.
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Answer: A. 0.362 nm
For FCC atoms touch along the face diagonal: √2 a = 4R, so a = 4(0.128)/1.414 = 0.362 nm.
65. Iron (BCC) has an atomic radius of 0.125 nm. Its lattice parameter a = 4R/√3 is about
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Body diagonal holds 4 radii.
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Answer: C. 0.289 nm
In BCC atoms touch along the body diagonal: √3 a = 4R, so a = 4(0.125)/1.732 = 0.289 nm.
66. Copper (FCC, a = 0.362 nm, atomic mass 63.5 g/mol) has a theoretical density of about
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Four atoms per FCC cell.
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Answer: B. 8.9 g/cm³
ρ = nA/(a³N_A) = 4 × 63.5/[(0.362×10⁻⁷ cm)³ × 6.022×10²³] ≈ 8.9 g/cm³.
67. A plane cuts the crystal axes at a/2, b and ∞. Its Miller indices are
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Take reciprocals of the intercepts.
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Answer: D. (210)
Intercepts in terms of a, b, c are 1/2, 1, ∞; reciprocals are 2, 1, 0, giving (210).
68. A single crystal under 50 MPa tension has its slip plane normal at 45° and slip direction at 45° to the axis. The resolved shear stress is
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Product of two cosines.
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Answer: A. 25 MPa
τ = σ cosφ cosλ = 50 × 0.707 × 0.707 = 25 MPa (Schmid's law).
69. A metal sheet is cold rolled from 10 mm to 7.5 mm thickness. The percentage cold work is
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Reduction in thickness relative to the original.
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Answer: C. 25%
%CW = (A0 − Ad)/A0 × 100 = (10 − 7.5)/10 × 100 = 25%.
70. Yield strength follows σ_y = σ₀ + k·d^(−1/2) with σ₀ = 70 MPa and k = 15 MPa·mm^½. For grain size d = 0.25 mm, σ_y is
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Finer grains give higher strength.
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Answer: B. 100 MPa
d^(−1/2) = 1/0.5 = 2, so σ_y = 70 + 15 × 2 = 100 MPa (Hall–Petch).
71. A 0.4% carbon steel is slowly cooled to just below 727 °C. Taking ferrite at 0.022% C and eutectoid at 0.77% C, the weight fraction of proeutectoid ferrite is about
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Take the opposite lever arm over the full tie-line.
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Answer: D. 49%
Lever rule: W_α' = (0.77 − 0.40)/(0.77 − 0.022) = 0.37/0.748 ≈ 0.49.
72. Using 0.022%, 0.77% and 6.67% C for ferrite, eutectoid and cementite, the fraction of cementite in pearlite is about
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Pearlite is mostly ferrite.
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Answer: A. 11%
W_Fe₃C = (0.77 − 0.022)/(6.67 − 0.022) = 0.1125 ≈ 11%.
73. At a temperature in a two-phase field of an A–B alloy (40 wt% B), the liquid is 30 wt% B and the solid is 50 wt% B. The fraction of liquid is
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Lever rule on the tie line.
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Answer: C. 50%
W_L = (C_S − C₀)/(C_S − C_L) = (50 − 40)/(50 − 30) = 0.5.
74. Polyethylene has a mer mass of 28 g/mol. A sample with number-average molecular mass 28 000 g/mol has a degree of polymerisation of
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Chain mass divided by repeat unit mass.
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Answer: B. 1000
DP = M_n / mer mass = 28 000/28 = 1000.
75. A composite has 30% by volume of fibres (E = 200 GPa) in a matrix (E = 4 GPa). By the iso-strain rule of mixtures, the longitudinal modulus is
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Weight each modulus by its volume fraction.
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Answer: D. 62.8 GPa
E_c = V_f E_f + V_m E_m = 0.3 × 200 + 0.7 × 4 = 62.8 GPa.
76. The number of atoms per unit cell of an FCC crystal is
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Count shared corner and face atoms.
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Answer: A. 4
8 corners × 1/8 + 6 faces × 1/2 = 4 atoms.
77. The atomic packing factor of FCC and HCP (ideal) structures is
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They are the densest sphere packings.
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Answer: C. 0.74
Both are close-packed with APF = π/(3√2) = 0.74; BCC is 0.68 and simple cubic is 0.52.
78. Which of the following has a coordination number of 8?
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Count nearest neighbours of the body centre atom.
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Answer: B. BCC
In BCC the central atom touches 8 corner atoms. FCC and HCP have 12; simple cubic has 6.
79. Plastic deformation of crystalline metals by slip takes place by the movement of
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A line defect.
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Answer: D. dislocations
Slip occurs through dislocation motion on close-packed planes along close-packed directions.
80. Increasing the amount of cold work in a metal generally
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Work hardening.
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Answer: A. raises strength and hardness and lowers ductility
Strain hardening multiplies dislocations that hinder each other's motion.
81. The correct sequence of stages when a cold-worked metal is annealed is
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Think of rising temperature.
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Answer: C. recovery, recrystallisation, grain growth
Heating first relieves stresses (recovery), then new strain-free grains form (recrystallisation), then they coarsen.
82. Hot working of a metal is carried out
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It depends on recrystallisation, not on a fixed temperature.
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Answer: B. above the recrystallisation temperature
Hot working is performed above the recrystallisation temperature so strain hardening is continuously removed.
83. Strengthening by dissolving a different-sized atom in a solvent metal lattice is called
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The atoms are dissolved, not precipitated.
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Answer: D. solid solution strengthening
Solute atoms distort the lattice and impede dislocation motion.
84. Age (precipitation) hardening depends on
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Needs a sloping solvus line.
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Answer: A. decreasing solid solubility with falling temperature
Solution treatment, quenching and ageing produce fine coherent precipitates which obstruct dislocations.
85. In the iron–carbon diagram, the eutectoid reaction occurs at about
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The one with solid-to-solid change.
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Answer: C. 727 °C and 0.77% C
Austenite → ferrite + cementite (pearlite) at 727 °C and 0.77% C; the eutectic is at 1147 °C, 4.3% C.
86. Cementite (Fe₃C) contains how much carbon by weight?
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It is a hard, brittle compound.
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Answer: B. 6.67%
Fe₃C: 12/(3 × 55.85 + 12) = 6.67% C.
87. Pearlite is a mixture of
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Lamellar structure.
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Answer: D. alternate lamellae of ferrite and cementite
Pearlite is the eutectoid product of austenite, formed of alternating plates of α-ferrite and Fe₃C.
88. Steel is distinguished from cast iron by the carbon content, which is
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Max solubility in austenite.
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Answer: A. below about 2.1% for steel
Steels have up to ~2.1% C (max solubility in austenite); cast irons have above that.
89. In grey cast iron the graphite is present mainly as
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Cause of brittleness and damping.
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Answer: C. flakes
Grey iron has graphite flakes in a pearlitic or ferritic matrix; nodular iron has spheroidal graphite.
90. Nodular (SG) cast iron is produced by adding a small amount of
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A light alkaline-earth metal.
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Answer: B. magnesium (or cerium) to the molten iron
Mg/Ce treatment changes graphite shape from flakes to spheroids, greatly increasing ductility.
91. White cast iron has its carbon mostly in the form of
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Fracture surface is white.
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Answer: D. cementite
White iron is rapidly solidified so carbon stays as Fe₃C; it is hard and brittle.
92. Thermosetting polymers, once cured,
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Think of an egg being boiled.
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Answer: A. cannot be remelted because of cross-linking
Cross-linked network structure decomposes on heating instead of melting (e.g. bakelite, epoxy).
93. The glass transition temperature T_g of a polymer is the temperature
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Applies to the amorphous region.
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Answer: C. at which an amorphous polymer changes from a hard, glassy to a rubbery state
Below T_g chain segments are frozen; above it the polymer is leathery or rubbery.
94. In a fibre-reinforced composite the main function of the matrix is to
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Fibres are the strong stiff part.
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Answer: B. bind the fibres, transfer load to them and protect them
The matrix holds fibres in place, transmits stress between them and shields them from damage.
1.4 Basic electrical and electronics
30 questions · AMeE0104
95. A 100 W, 230 V incandescent lamp is switched on at its rated voltage. Its hot resistance is
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Use P = V²/R.
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Answer: A. 529 Ω
R = V²/P = 230²/100 = 529 Ω.
96. Resistors of 6 Ω and 3 Ω are connected in parallel. The equivalent resistance is
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Product over sum for two resistors.
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Answer: C. 2 Ω
R = (6 × 3)/(6 + 3) = 2 Ω.
97. At a junction currents of 5 A and 3 A flow in, and 2 A flows out through one branch. The current in the remaining outgoing branch is
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Current in equals current out.
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Answer: B. 6 A
KCL: 5 + 3 = 2 + I, so I = 6 A.
98. A current of 2 A flows for 5 minutes. The charge transferred is
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Convert minutes to seconds first.
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Answer: D. 600 C
Q = It = 2 × (5 × 60) = 600 C.
99. A 2 kW heater is used for 5 hours. The energy consumed is
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Energy is power × time; check the unit.
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Answer: A. 10 kWh
E = P × t = 2 kW × 5 h = 10 kWh (units).
100. A sinusoidal supply has a peak value of 311 V. Its r.m.s. value is nearly
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Divide the peak by √2.
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Answer: C. 220 V
V_rms = V_m/√2 = 311/1.414 ≈ 220 V.
101. A series R–L circuit has R = 30 Ω and X_L = 40 Ω and is connected to a 100 V a.c. supply. The current is
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Combine R and X as the hypotenuse.
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Answer: B. 2 A
Z = √(30² + 40²) = 50 Ω; I = 100/50 = 2 A.
102. A load draws 4 kW real power and 5 kVA apparent power. The power factor is
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Real power divided by apparent power.
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Answer: D. 0.8
PF = P/S = 4/5 = 0.8.
103. An ideal transformer has 1000 primary turns and is rated 2200 V / 220 V. The secondary has
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Turns ratio equals voltage ratio.
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Answer: A. 100 turns
N₂/N₁ = V₂/V₁ = 220/2200, so N₂ = 100.
104. A 4-pole, 50 Hz induction motor runs at 1440 rpm. Its slip is
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Compute synchronous speed first.
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Answer: C. 4%
N_s = 120f/P = 1500 rpm; slip = (1500 − 1440)/1500 = 0.04 = 4%.
105. A 220 V d.c. shunt motor draws an armature current of 20 A and has armature resistance 0.5 Ω. The back e.m.f. is
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Subtract the armature drop from the supply.
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Answer: B. 210 V
E_b = V − I_aR_a = 220 − 20 × 0.5 = 210 V.
106. An inverting op-amp has R_f = 100 kΩ and R_in = 10 kΩ. The voltage gain is
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The sign shows phase inversion.
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Answer: D. −10
A_v = −R_f/R_in = −100/10 = −10.
107. A transistor has collector current 2 mA for a base current of 20 µA. The current gain β is
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Collector over base current.
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Answer: A. 100
β = I_C/I_B = 2 mA/20 µA = 100.
108. A series circuit has L = 1 mH and C = 1 µF. The resonant frequency is about
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Resonance when X_L = X_C.
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Answer: C. 5.03 kHz
f₀ = 1/(2π√LC) = 1/(2π√(10⁻⁹)) ≈ 5033 Hz.
109. Kirchhoff's voltage law states that
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Closed loop.
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Answer: B. the algebraic sum of voltages around any closed loop is zero
KVL is conservation of energy around a closed path.
110. Ohm's law is valid for a conductor when
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Temperature changes resistance.
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Answer: D. temperature and other physical conditions are constant
V ∝ I only for ohmic materials at constant temperature.
111. A resistor with colour bands brown, black, red (and gold tolerance) has a value of
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Third band is the multiplier.
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Answer: A. 1 kΩ
Brown = 1, black = 0, red = ×10²: 10 × 100 = 1 kΩ.
112. A capacitor in a steady d.c. circuit behaves as
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Final steady state.
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Answer: C. an open circuit after it is fully charged
Once charged no current flows, so it blocks d.c.
113. In a pure inductor the current
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Inductors resist change in current.
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Answer: B. lags the voltage by 90°
v = L di/dt, so current lags voltage by 90° and the power factor is zero.
114. In a series R–L–C circuit at resonance
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Reactances cancel.
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Answer: D. the impedance is minimum and equals R
At resonance X_L = X_C, so Z = R and current is maximum at unity power factor.
115. A transformer cannot be operated on d.c. supply because
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Faraday's law.
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Answer: A. a constant flux induces no e.m.f. and the primary may burn out
Induction needs changing flux; with d.c. the primary current is limited only by winding resistance.
116. The rotor of an induction motor always runs
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If speeds were equal, no torque would exist.
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Answer: C. at a speed slightly less than synchronous speed
Relative motion between field and rotor is needed to induce rotor current.
117. Which d.c. motor has the highest starting torque?
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Field is in series with the armature.
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Answer: B. Series motor
In a series motor torque ∝ I_a², so a large starting current gives a high starting torque.
118. A low-pass filter
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Look at the name.
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Answer: D. passes low frequencies and attenuates high frequencies
An RC low-pass filter has gain falling above f_c = 1/(2πRC).
119. A relay is basically
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Low power controls high power.
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Answer: A. an electromagnetically operated switch
A small current in the coil moves an armature to open or close contacts in another circuit.
120. The typical forward voltage drop of a silicon p–n junction diode is about
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Compare with germanium.
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Answer: C. 0.7 V
Silicon diodes start conducting near 0.6–0.7 V; germanium near 0.3 V.
121. A Zener diode is used as a voltage regulator when operated in
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Reverse bias.
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Answer: B. reverse breakdown region
Voltage across a Zener stays nearly constant in breakdown over a wide current range.
122. For a transistor to work as a linear amplifier the base–emitter and base–collector junctions are
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Saturation is both forward.
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Answer: D. forward biased and reverse biased respectively
The active region has B–E forward, B–C reverse.
123. A common-emitter amplifier produces a phase shift between input and output of
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Inverting amplifier.
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Answer: A. 180°
A rise in base voltage raises collector current and lowers collector voltage, inverting the signal.
124. For sustained oscillations in an oscillator, the loop gain Aβ must be
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It needs positive feedback.
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Answer: C. unity with zero net phase shift (positive feedback)
Barkhausen criterion: |Aβ| = 1 and total phase shift 0° or 360°.
1.5 Mechanical workshop
29 questions · AMeE0105
125. A job of 50 mm diameter is turned at 400 rpm. The cutting speed is
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Use D in mm and N in rpm.
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Answer: A. 62.8 m/min
V = πDN/1000 = π × 50 × 400/1000 = 62.8 m/min.
126. A 40 mm diameter bar is to be turned at a cutting speed of 30 m/min. The required spindle speed is nearly
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Rearrange V = πDN/1000.
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Answer: C. 239 rpm
N = 1000V/(πD) = 1000 × 30/(π × 40) ≈ 239 rpm.
127. A 150 mm long cut is taken on a lathe at a feed of 0.2 mm/rev and 500 rpm. The machining time is
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Feed rate per minute = f × N.
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Answer: B. 1.5 min
T = L/(f N) = 150/(0.2 × 500) = 1.5 min.
128. In turning, V = 60 m/min, feed f = 0.25 mm/rev and depth of cut d = 2 mm. The material removal rate is
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Convert V to mm/min.
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Answer: D. 30 000 mm³/min
MRR = V f d = 60 000 mm/min × 0.25 × 2 = 30 000 mm³/min.
129. A job 200 mm long is to be turned with a taper from 40 mm to 30 mm over a taper length of 100 mm by the tailstock set-over method. The set-over is
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The total job length appears in the numerator.
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Answer: A. 10 mm
Set-over = L(D − d)/(2l) = 200 × 10/(2 × 100) = 10 mm.
130. A taper from 50 mm to 40 mm diameter over a length of 50 mm is turned using the compound rest. The compound rest must be set at about
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Use the half-taper angle.
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Answer: C. 5.7°
tan α = (D − d)/(2l) = 10/100 = 0.1, so α = 5.71° (half angle of taper).
131. A hole 40 mm deep is drilled at a feed of 0.2 mm/rev at 500 rpm. The drilling time (ignoring approach) is
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Penetration rate = f × N.
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Answer: B. 0.4 min
t = depth/(f N) = 40/(0.2 × 500) = 0.4 min.
132. A shaper makes 30 double strokes per minute with a stroke length of 200 mm. The cutting stroke takes 2/3 of the cycle time. The average cutting speed is
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Only the cutting-stroke time counts.
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Answer: D. 9 m/min
Cycle = 1/30 min; cutting time = (2/3)/30 = 0.0222 min; V = 0.2 m/0.0222 min = 9 m/min.
133. In simple indexing on a milling machine (40:1 worm gear ratio), the crank movement required for cutting 8 divisions is
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40 divided by number of divisions.
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Answer: A. 5 full turns
Index crank turns = 40/N = 40/8 = 5 complete turns.
134. A grinding wheel of diameter 300 mm runs at 1900 rpm. Its peripheral speed is about
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Convert diameter to metres and rpm to rev/s.
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Answer: C. 29.8 m/s
V = πDN/60 = π × 0.3 × 1900/60 ≈ 29.8 m/s.
135. The tap drill diameter for an M12 × 1.75 internal thread is about
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Subtract the pitch.
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Answer: B. 10.25 mm
Tap drill = major diameter − pitch = 12 − 1.75 = 10.25 mm.
136. In arc welding the arc voltage is 25 V, current 200 A, travel speed 5 mm/s and heat transfer efficiency 0.8. The heat input per unit length is
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Power divided by speed, times efficiency.
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Answer: D. 800 J/mm
H = ηVI/v = 0.8 × 25 × 200/5 = 800 J/mm.
137. To cut a 1.5 mm pitch thread on a lathe with a lead screw of 6 mm pitch, the gear ratio (driver : driven) between spindle and lead screw should be
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Lead screw moves 4 times more per turn than needed.
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Answer: A. 1 : 4
Ratio = pitch of job/pitch of lead screw = 1.5/6 = 1/4.
138. Which of the following is the most suitable fire extinguisher for an electrical fire in a workshop?
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Avoid conductive agents.
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Answer: C. Carbon dioxide (CO₂) extinguisher
CO₂ is non-conducting and leaves no residue; water and foam conduct electricity.
139. Safety goggles in a workshop are mainly used to protect against
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Eyes.
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Answer: B. flying chips and particles
Eye protection is mandatory in machining and grinding operations.
140. Long loose hair and loose clothing should be avoided near a rotating lathe because they
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Rotating parts.
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Answer: D. can be caught by the rotating parts
Entanglement in a rotating chuck or job is a major cause of serious accidents.
141. A hacksaw blade is mounted so that its teeth point
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Forward stroke cuts.
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Answer: A. away from the handle (cutting on the forward stroke)
A hacksaw cuts on the forward push stroke; pressure is released on the return stroke.
142. The movement of the lathe carriage and the cross-slide for facing is controlled by
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Hangs in front of the carriage.
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Answer: C. the apron mechanism
The apron houses the feed gears, half-nut and clutches that move the carriage and cross-slide.
143. A three-jaw chuck is best suited for holding
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Self-centring.
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Answer: B. round and hexagonal stock quickly and concentrically
Its jaws move together via a scroll, giving self-centring.
144. The quick-return mechanism in a shaper makes
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Idle stroke.
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Answer: D. the return stroke faster than the cutting stroke
Quick return reduces idle time per cycle.
145. In up (conventional) milling, the cutter rotation is
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Chip thickness grows from zero.
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Answer: A. opposite to the direction of feed of the workpiece
In up-milling chip thickness starts at zero and increases; in down milling the rotation and feed directions coincide.
146. Dressing a grinding wheel is done to
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Resharpening the surface.
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Answer: C. remove clogged or dulled grains and expose sharp new cutting grains
Dressing sharpens the wheel; truing restores its shape.
147. The included point angle of a standard general-purpose twist drill is
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Slightly less than 120°.
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Answer: B. 118°
General purpose drills have 118° point angle; harder materials use up to 140°.
148. Reaming is carried out to
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Finishing operation.
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Answer: D. finish a drilled hole to accurate size and good surface finish
A reamer removes a small allowance, improving diameter accuracy and surface finish.
149. In arc welding, the flux coating on the electrode
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Shields against oxygen and nitrogen.
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Answer: A. produces shielding gas and slag that protect the molten pool
Coating decomposes to shielding gas, stabilises the arc and forms slag.
150. A neutral oxy-acetylene flame is obtained when
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Neither carburising nor oxidising.
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Answer: C. the oxygen to acetylene ratio is about 1 : 1
Neutral flame (O₂:C₂H₂ ≈ 1:1) is used for welding mild steel.
151. Which of the following is a non-consumable electrode welding process?
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Tungsten.
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Answer: B. TIG (GTAW)
TIG uses a tungsten electrode that is not melted; filler is added separately.
152. Brazing differs from welding in that
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Base metal does not melt.
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Answer: D. the filler metal melts above 450 °C but below the melting point of the base metal
In brazing the base metal does not melt; filler metals melt above 450 °C (soldering is below 450 °C).
153. Spot welding is an example of
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Heat from current through contact resistance.
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Answer: A. resistance welding
Heat is generated by I²R at the contact between overlapping sheets under pressure.
1.6 Oorganization management
30 questions · AMeE0106
154. A firm has fixed costs of Rs 200 000, selling price Rs 50 per unit and variable cost Rs 30 per unit. The break-even quantity is
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Divide fixed cost by contribution per unit.
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Answer: A. 10 000 units
BEP = FC/(P − V) = 200 000/(50 − 30) = 10 000 units.
155. A project costing Rs 250 000 yields an annual profit of Rs 50 000. The return on investment is
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Profit divided by investment.
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Answer: C. 20%
ROI = 50 000/250 000 × 100 = 20%.
156. An investment of Rs 300 000 returns Rs 75 000 net cash per year. The payback period is
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Years needed to recover the investment.
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Answer: B. 4 years
Payback = investment/annual inflow = 300 000/75 000 = 4 years.
157. Annual demand is 1200 units, ordering cost Rs 100 per order and holding cost Rs 6 per unit per year. The economic order quantity is
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Use the square-root EOQ formula.
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Answer: D. 200 units
EOQ = √(2DS/H) = √(2 × 1200 × 100/6) = √40 000 = 200 units.
158. A PERT activity has optimistic 4 days, most likely 6 days and pessimistic 14 days. The expected time is
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Most likely time carries weight 4.
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Answer: A. 7 days
t_e = (a + 4m + b)/6 = (4 + 24 + 14)/6 = 7 days.
159. Equipment costing Rs 100 000 with salvage value Rs 10 000 and life of 10 years is depreciated by the straight-line method. Annual depreciation is
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Depreciable amount over life.
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Answer: C. Rs 9 000
(Cost − salvage)/life = 90 000/10 = Rs 9 000.
160. A company with an average of 120 employees has 12 separations in a year. The labour turnover rate is
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Separations over average headcount.
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Answer: B. 10%
Turnover = separations/average workforce × 100 = 12/120 × 100 = 10%.
161. A worker produces 240 units in an 8-hour shift. The labour productivity is
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Output per unit of input.
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Answer: D. 30 units per hour
Productivity = output/input = 240/8 = 30 units per labour-hour.
162. An organization in which authority flows vertically from top to bottom through a single chain of command is a
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Simplest structure.
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Answer: A. line organization
Line organization has direct vertical lines of authority and unity of command.
163. The main drawback of a functional organization (Taylor) is that
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Think of 'one man, one boss'.
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Answer: C. an employee reports to several specialist bosses, violating unity of command
Functional foremen each control a specific function, so workers receive orders from multiple supervisors.
164. A matrix organization combines
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Dual reporting.
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Answer: B. functional departments with project teams
Employees report to both a functional manager and a project manager.
165. Which form of business ownership has unlimited liability and a single owner?
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One person.
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Answer: D. Sole proprietorship
A sole proprietor owns, manages and bears all risk personally with unlimited liability.
166. Scientific management is associated with
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Principles of Scientific Management, 1911.
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Answer: A. F. W. Taylor
Taylor developed time and motion study and scientific selection and training.
167. The 14 principles of management and the functions of planning, organising, commanding, coordinating and controlling were proposed by
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French mining engineer.
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Answer: C. Henri Fayol
Fayol, in General and Industrial Management, set out the administrative principles.
168. The Hawthorne experiments, which highlighted the importance of social and human factors at work, were conducted by
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Human relations school.
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Answer: B. Elton Mayo
Mayo's studies at Western Electric's Hawthorne plant founded the human relations school.
169. Max Weber is best known for his theory of
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Rule-based administration.
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Answer: D. bureaucracy
Weber proposed an ideal bureaucracy based on hierarchy, rules and impersonal procedures.
170. In Maslow's hierarchy of needs, the highest-level need is
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Top of the pyramid.
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Answer: A. self-actualisation
Order: physiological, safety, social, esteem, self-actualisation.
171. Herzberg's two-factor theory classifies factors such as salary and working conditions as
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They avoid dissatisfaction.
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Answer: C. hygiene factors
Hygiene factors prevent dissatisfaction but do not motivate; achievement and recognition are motivators.
172. McGregor's Theory X assumes that workers
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Pessimistic view of workers.
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Answer: B. dislike work and must be controlled and directed
Theory X is the traditional authoritarian view; Theory Y assumes the opposite.
173. A leader who involves subordinates in decision-making is called
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Consultation.
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Answer: D. democratic (participative)
Participative leaders consult the group, improving morale and acceptance.
174. Which of the following is a barrier to effective communication?
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Barrier means obstacle.
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Answer: A. Differences in language and perception
Language, semantics, perception and noise distort messages; feedback and listening reduce distortion.
175. The informal communication channel in an organization is commonly known as
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Name suggests rumours.
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Answer: C. the grapevine
Grapevine is a spontaneous, unofficial network spreading information quickly.
176. An entrepreneur is best described as a person who
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Innovation and risk bearing.
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Answer: B. organises resources and takes risks to start a new enterprise
Entrepreneurs identify opportunities, mobilise resources and bear business risk.
177. A business plan generally includes
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It outlines the whole venture.
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Answer: D. executive summary, market analysis, operations and financial plan
A plan covers the idea, marketing, production/operations, organisation and projected finances.
178. Recruitment in human resource management refers to
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It comes before selection.
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Answer: A. searching for and attracting potential candidates for vacant posts
Recruitment generates a pool of applicants; selection then chooses among them.
179. Performance appraisal is used mainly to
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Feedback on employees.
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Answer: C. evaluate employee performance for feedback, promotion and training needs
Appraisal compares actual performance against standards.
180. A Management Information System (MIS) primarily provides
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Support for decision makers.
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Answer: B. timely and relevant information to support managerial decisions
MIS collects, processes and reports information to planning and control decisions.
181. In a SWOT analysis the letters W and T stand for
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Internal and external factors.
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Answer: D. weaknesses and threats
SWOT = strengths, weaknesses (internal) and opportunities, threats (external).
182. Management by Objectives (MBO) was popularised by
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Modern management guru.
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Answer: A. Peter Drucker
Drucker introduced MBO in The Practice of Management (1954).
183. Technology management mainly deals with
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Strategy for technology.
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Answer: C. planning, acquiring and using technology to gain competitive advantage
It links technology strategy with business strategy through acquisition, development and transfer.