Nepal Engineering Council · Civil Engineering · Chapter 1
Basic Civil Engineering
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189 questions in 6 syllabus topics.
1.1 Engineering materials
32 questions · ACiE0101
1. The property of a material by which it resists scratching or indentation on its surface is called
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Think of the Mohs scale and indentation tests.
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Answer: B. hardness
Hardness is the resistance of a material to surface indentation or scratching; toughness is energy absorption before fracture.
2. The energy absorbed by a material per unit volume up to fracture, represented by the total area under its stress-strain curve, is the material's
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Compare the area up to fracture with the area up to the elastic limit.
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Answer: D. toughness
Toughness is the total area under the stress-strain curve up to fracture; resilience is only the area up to the elastic limit.
3. Slow, time-dependent increase in strain of a material under a constant sustained stress is known as
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Stress is constant, strain keeps changing with time.
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Answer: A. creep
Creep is the gradual increase of deformation with time under constant load; relaxation is the fall of stress at constant strain.
4. A material that can be beaten or rolled into thin sheets without cracking is said to be
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Sheets by hammering or rolling, not wires.
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Answer: B. malleable
Malleability is the ability to deform under compressive loads into sheets; ductility refers to being drawn into wires under tension.
5. A steel girder 40 m long (α = 12 × 10⁻⁶ per °C) experiences a temperature rise of 30°C while free to expand. Its increase in length is
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Use ΔL = αLΔT with consistent length units.
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Answer: A. 14.4 mm
ΔL = α·L·ΔT = 12×10⁻⁶ × 40 000 mm × 30 = 14.4 mm.
6. A dry brick weighs 3.2 kg and has an overall volume of 0.002 m³. Its bulk density is
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Divide mass by overall volume including pores.
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Answer: A. 1600 kg/m³
Density = mass/volume = 3.2/0.002 = 1600 kg/m³.
7. A dry stone specimen weighs 2.50 kg and after full saturation weighs 2.60 kg. Its water absorption (by weight of dry stone) is
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The base of the percentage is the dry weight.
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Answer: D. 4 %
Absorption = (2.60 − 2.50)/2.50 × 100 = 4 %.
8. On Mohs' scale of hardness, the mineral with hardness number 1 (the softest) is
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It is the mineral used in baby powder.
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Answer: D. talc
Mohs' scale runs from talc (1), gypsum (2), calcite (3) ... quartz (7) ... to diamond (10).
9. Marble is formed by the metamorphism of
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It is the metamorphic equivalent of a calcareous rock.
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Answer: C. limestone
Heat and pressure recrystallise limestone (calcite/dolomite) into marble; sandstone gives quartzite and shale gives slate.
10. Slate, widely used as a roofing and paving material, is a metamorphic rock derived from
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Its fissile cleavage allows thin slabs.
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Answer: A. shale or clay
Low-grade metamorphism of shale/clay produces slate, which splits easily into thin slabs along its cleavage.
11. In the composition of brick earth, the constituent that prevents cracking, shrinkage and warping of raw bricks on drying and burning is
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It is the sand-like constituent, not the plasticity-giving one.
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Answer: C. silica
Silica gives the brick uniform shape and prevents shrinkage and warping; alumina gives plasticity to the clay.
12. White patches appearing on brick masonry surfaces due to soluble salts in the bricks, sand or water are called
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Salts migrate with moisture and crystallise on drying.
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Answer: D. efflorescence
Efflorescence is the deposit of crystallised soluble sulphates carried to the surface by evaporating moisture.
13. A 10 m long, 3 m high and 200 mm thick brick wall is to be built with modular bricks (nominal size with mortar 200 × 100 × 100 mm). Neglecting openings, the number of bricks required is
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Divide wall volume by volume of one nominal brick.
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Answer: A. 3000
Volume = 10 × 3 × 0.2 = 6 m³; one brick occupies 0.2×0.1×0.1 = 0.002 m³, so 6/0.002 = 3000 bricks.
14. Vitrified (porcelain) floor tiles are preferred in wet areas mainly because they have
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Fusion of the clay body closes its pores.
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Answer: B. very low water absorption
Vitrification fuses the body, giving water absorption of about 0.5 % or less, so the tile is dense and stain-resistant.
15. Among the main compounds of Portland cement, the one chiefly responsible for early strength gain (up to about 7 days) is
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The quick-hydrating silicate is the one with more lime.
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Answer: C. tricalcium silicate (C₃S)
C₃S hydrates rapidly and develops early strength; C₂S hydrates slowly and contributes to later-age strength.
16. Gypsum is added in small quantity (about 2–3 %) to cement clinker during grinding in order to
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It regulates setting, not strength or colour.
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Answer: B. control the flash set caused by C₃A
Gypsum reacts with C₃A to form ettringite and retards the otherwise instantaneous (flash) setting.
17. The designation '43 Grade' of ordinary Portland cement (IS 269) means that its minimum compressive strength at 28 days is
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The grade number is a strength value at 28 days.
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Answer: B. 43 MPa
The grade number is the minimum 28-day compressive strength of standard mortar cubes in N/mm² (MPa).
18. According to IS 1489, Portland pozzolana cement (PPC) contains fly ash, by mass, in the range of
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A moderate replacement, well below half.
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Answer: D. 15 % to 35 %
IS 1489 (Part 1) permits 15–35 % pozzolanic fly ash interground or blended with OPC clinker.
19. If the bulk density of cement is taken as 1440 kg/m³, the approximate number of 50 kg bags in 1 m³ of cement is
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Divide the mass of 1 m³ by the mass of one bag.
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Answer: B. 29
Number of bags = 1440/50 = 28.8, i.e. about 29 (one bag ≈ 0.035 m³).
20. Pure limestone (CaCO₃, molecular mass 100) is calcined to quicklime (CaO, molecular mass 56). The mass of quicklime obtainable from 1000 kg of pure limestone is
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Work out the mass ratio from the molecular masses.
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Answer: D. 560 kg
CaCO₃ → CaO + CO₂, so the yield = 1000 × 56/100 = 560 kg (440 kg of CO₂ is lost).
21. Hydraulic lime is able to set and harden under water because it contains
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Impurities from the parent limestone give it the water-setting property.
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Answer: D. clay (silica and alumina) impurities
Clay in the limestone (roughly 5–30 %) forms insoluble calcium silicates/aluminates that harden by reaction with water; fat lime hardens only by carbonation.
22. The moisture content in timber at which the cell walls are fully saturated but no free water exists in cell cavities (fibre saturation point) is about
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Strength starts to rise rapidly on drying below this value.
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Answer: B. 25–30 %
At the fibre saturation point, hygroscopic water fills the cell walls; below it timber shrinks and strength increases.
23. A timber sample weighs 1.25 kg when freshly cut and 1.00 kg after oven drying. Its moisture content is
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Moisture content is based on oven-dry weight.
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Answer: C. 25 %
MC = (wet − oven-dry)/oven-dry × 100 = 0.25/1.00 × 100 = 25 %.
24. Plywood is made by bonding veneers with the grain of adjacent plies at right angles; the number of plies is always
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Symmetry about the mid-ply is needed.
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Answer: C. odd
An odd number of plies keeps the construction symmetrical about the centre ply, preventing warping.
25. The carbon content of mild steel used for structural purposes is approximately
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Low-carbon steel has well under 0.3 % carbon.
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Answer: A. 0.15 % to 0.25 %
Mild (low-carbon) steel has about 0.15–0.25 % carbon; cast iron has 2–4 % and wrought iron almost none.
26. Brass is an alloy of
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Do not confuse it with bronze.
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Answer: C. copper and zinc
Brass = copper + zinc; bronze = copper + tin.
27. A reinforcing bar of Fe 500 steel (E = 2 × 10⁵ MPa) is stressed to exactly 500 MPa within the elastic range. The corresponding strain is
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Apply Hooke's law.
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Answer: C. 0.0025
ε = σ/E = 500 / 200 000 = 0.0025.
28. In paint, the ingredient that dissolves or dilutes the vehicle to make it workable and then evaporates after application is the
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It evaporates; it does not remain in the dry film.
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Answer: A. thinner (solvent)
Thinners such as turpentine reduce viscosity for easier application and evaporate afterwards; pigment gives colour and opacity.
29. A wall of 60 m² area is to be given two coats of paint. If one litre of paint covers 12 m² per coat, the paint required is
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Count the area once per coat.
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Answer: C. 10 litres
Total area to be coated = 60 × 2 = 120 m²; 120/12 = 10 litres.
30. Bitumen differs from coal tar in that bitumen is
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Think of the source: crude oil versus coal.
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Answer: B. obtained from petroleum crude or natural deposits
Bitumen is a hydrocarbon residue of petroleum refining (or natural asphalt); tar is made by destructive distillation of coal or wood.
31. A bitumen of grade '80/100' is one whose penetration (standard needle, 100 g, 5 s, 25°C) lies between 80 and 100, expressed in units of
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Penetration units are very small.
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Answer: D. 0.1 mm
Penetration is measured in tenths of a millimetre; a higher value means a softer bitumen.
32. A tack coat is sprayed at 0.25 kg/m² on a road surface 7 m wide and 1 km long. The quantity of bitumen required is
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Convert 1 km to metres before multiplying by the rate.
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Answer: A. 1750 kg
Area = 7 × 1000 = 7000 m²; mass = 7000 × 0.25 = 1750 kg.
1.2 Standards (NS & IS) and tests for civil engineering materials
31 questions · ACiE0102
33. In Nepal, Nepal Standards (NS) for construction materials are formulated and issued by the
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The standards body, not the licensing body.
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Answer: C. Nepal Bureau of Standards and Metrology
NBSM, under the Ministry of Industry, Commerce and Supplies, prepares and publishes Nepal Standards; NEC only regulates engineering registration.
34. The Indian Standard that lays down the methods of physical tests for hydraulic cement (consistency, setting time, soundness, strength) is
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It is the cement-testing series, numbered in the 4000s.
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Answer: C. IS 4031
IS 4031 (Parts 1–15) gives the methods of physical tests for hydraulic cement; IS 383 is for aggregates, IS 1786 for rebars and IS 3495 for bricks.
35. The methods of tests of burnt clay building bricks (compressive strength, water absorption, efflorescence) are given in
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Test methods and specification are separate IS documents.
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Answer: B. IS 3495
IS 3495 (Parts 1–4) prescribes tests on burnt clay building bricks, while IS 1077 is the specification for them.
36. The Indian Standard specifying high strength deformed (HSD/TMT) steel bars and wires for concrete reinforcement is
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Reinforcement bars, not structural sections.
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Answer: D. IS 1786
IS 1786 covers Fe 415, Fe 500, Fe 550, Fe 600 grades including the D (ductile) grades; IS 2062 is for structural steel.
37. In the water absorption test of a brick (IS 3495), the oven-dried brick is immersed in clean water at room temperature for
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The standard soaking period is a full day.
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Answer: D. 24 hours
The dry brick is weighed, immersed in water at 27 ± 2°C for 24 hours, wiped and weighed again.
38. A brick weighs 2.80 kg when oven-dry and 3.22 kg after 24 hours' immersion. Its water absorption is
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The percentage is on the dry mass, not the wet mass.
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Answer: C. 15 %
Absorption = (3.22 − 2.80)/2.80 × 100 = 15 % (on dry weight).
39. A brick of bed face 230 mm × 110 mm fails at a load of 360 kN in a compression test. Its compressive strength is about
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Convert the load to newtons and divide by the loaded area in mm².
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Answer: A. 14.2 N/mm²
σ = P/A = 360 000/(230 × 110) = 14.2 N/mm².
40. A burnt clay brick of 'class 7.5' as per IS 1077 has a minimum average compressive strength of
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The class number is the strength itself.
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Answer: B. 7.5 N/mm²
The class designation is the minimum average compressive strength in N/mm² (MPa).
41. For the compressive strength test of a brick, the load is applied perpendicular to its
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Test it the way it is placed in a wall.
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Answer: C. largest flat (bed) face
The brick is placed flat, as in masonry, and loaded on its bed face after the frog is filled and the faces are levelled.
42. The consistency test of cement is done using the Vicat apparatus fitted with a plunger of diameter
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The plunger is a large-diameter rod, the needle is the thin one.
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Answer: D. 10 mm
A 10 mm diameter, 50 mm long plunger is used; the standard consistency is the paste in which it penetrates to 5–7 mm from the bottom of the mould.
43. The main purpose of finding the standard consistency of cement paste is to
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It is a preparatory test for other tests.
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Answer: D. fix the water content for setting time and soundness tests
Setting time and soundness tests are carried out on paste of standard consistency, with water taken as a fixed fraction of it.
44. For 400 g of cement the standard consistency is found to be 30 %. The water to be added for the initial setting time test (0.85 P) is
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Multiply the consistency water by 0.85.
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Answer: B. 102 mL
Water = 0.85 × P % of cement = 0.85 × 30/100 × 400 = 102 mL.
45. In the Vicat test, the initial setting time of cement is the time from adding water until the 1 mm square needle fails to penetrate the paste beyond a distance from the base of the mould equal to
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The needle stays a few millimetres above the base plate.
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Answer: C. 5 ± 0.5 mm
Initial setting is recorded when the needle stops 5 ± 0.5 mm short of the bottom of the 40 mm deep mould.
46. As per IS 269, ordinary Portland cement must have an initial setting time of not less than ___ and a final setting time of not more than ___.
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Final set is at most ten hours.
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Answer: B. 30 minutes; 600 minutes
IS 269 requires a minimum of 30 minutes for initial set and a maximum of 600 minutes (10 hours) for final set.
47. The Le Chatelier apparatus is used to test the soundness of cement, i.e. to detect expansion mainly due to
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A delayed-expansion oxide from the kiln.
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Answer: B. free (unhydrated) lime
Slow hydration of free lime after setting causes delayed expansion and cracking; Le Chatelier measures this expansion.
48. In a Le Chatelier test the distance between the indicator points is 17.5 mm before boiling and 25.5 mm after 3 hours of boiling. The expansion is, against a permitted limit of 10 mm for OPC,
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Take the difference of the two readings.
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Answer: A. 8 mm and the cement is sound
Expansion = 25.5 − 17.5 = 8 mm, which is below the 10 mm limit, so the cement is sound.
49. The autoclave test of cement is mainly intended to detect unsoundness due to excess of
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Le Chatelier covers lime, autoclave also covers a magnesium compound.
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Answer: C. magnesia (MgO)
Autoclave steam curing accelerates the slow hydration of free lime and magnesia; the expansion limit is 0.8 %.
50. Cement mortar cubes for the compressive strength test (IS 4031 Part 6) have a side of 70.6 mm. This size is chosen so that the face area is
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Square 70.6.
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Answer: D. 5000 mm² (approximately)
70.6² ≈ 4984 mm², about 5000 mm², which simplifies stress calculation.
51. A cement mortar cube (cement : standard sand = 1 : 3 by mass) is made with 200 g cement and consistency P = 28 %. Water for the cube mix is (P/4 + 3.0) % of the combined mass of cement and sand, i.e.
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First find the combined mass of cement and sand.
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Answer: C. 80 g
Combined mass = 200 + 600 = 800 g; water % = 28/4 + 3 = 10 %; water = 80 g.
52. A 70.6 mm standard mortar cube fails at a load of 180 kN. The compressive strength of the cement is about
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Use the cube face area in mm² and load in N.
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Answer: B. 36.1 N/mm²
Area = 70.6² = 4984 mm²; σ = 180 000/4984 = 36.1 N/mm².
53. As per IS 269:2015, the minimum compressive strength of 43 grade OPC at 3 days and 7 days is, respectively,
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The 3-day value is a little over half of the 43 MPa.
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Answer: C. 23 MPa and 33 MPa
43 grade OPC: 23 MPa (3 d), 33 MPa (7 d), 43 MPa (28 d). The 33 grade gives 16/22/33 and the 53 grade 27/37/53.
54. Bulking of sand is the increase in its volume when it contains a small amount of moisture, caused by
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It is a surface-tension effect.
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Answer: B. a water film between particles keeping them apart
Surface tension of the thin moisture film forces particles apart and traps air voids, increasing the bulk volume.
55. Bulking of sand is maximum at a moisture content of about 4–6 %, and the bulking is practically nil when the sand is
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Two extremes of the moisture scale.
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Answer: A. completely dry or fully saturated
Fully dry sand has no film, while inundated sand has the film broken by flooding, so volume returns to the dry value.
56. In the field test for bulking, a cylinder is filled with moist sand to a height of 20 cm. After the sand is completely inundated with water, the height becomes 16 cm. The percentage bulking is
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Divide by the inundated height.
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Answer: A. 25 %
Bulking = (h1 − h2)/h2 × 100 = (20 − 16)/16 × 100 = 25 %.
57. The volume of dry sand required per batch is 0.40 m³. If the sand available has 20 % bulking, the volume of moist sand to be measured is
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Increase the dry volume by the bulking percentage.
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Answer: A. 0.48 m³
Moist volume = 0.40 × (1 + 0.20) = 0.48 m³.
58. Compared with coarse sand, fine sand under the same moisture content shows
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More surface area, more moisture film.
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Answer: D. greater bulking
Finer particles have larger specific surface and hold more water film, hence bulk more.
59. For a reinforcing bar of diameter d, the standard proportional gauge length 5.65√A₀ used in the tensile test is equal to about
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Substitute A₀ = πd²/4.
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Answer: A. 5d
A₀ = πd²/4, so 5.65√A₀ = 5.65 × 0.886d ≈ 5.0d; for a 12 mm bar the gauge length is 60 mm.
60. In a tensile test on a 16 mm diameter bar the load at yield is 100.5 kN. The yield stress is about
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Find the nominal area first.
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Answer: A. 500 N/mm²
A = π×16²/4 = 201 mm²; σy = 100 500/201 = 500 N/mm², so the bar is of Fe 500 grade.
61. A rebar with a 80 mm gauge length has a gauge length of 92 mm at fracture. Its percentage elongation is
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Divide the extension by the original gauge length.
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Answer: B. 15 %
Elongation = (92 − 80)/80 × 100 = 15 %. Fe 500 requires at least 12 % (16 % for Fe 500D).
62. Cold-worked (TMT-free) steel bars that show no definite yield point in the tensile test are assigned a yield value equal to the
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Offset method.
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Answer: A. 0.2 % proof stress
The 0.2 % proof stress is found by drawing a line parallel to the elastic portion offset by 0.2 % strain.
63. The 'D' in Fe 500D steel indicates a grade having, compared with Fe 500,
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It stands for something helpful in seismic design.
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Answer: D. higher ductility (elongation and UTS/YS ratio)
D grades have a higher minimum elongation (16 %) and UTS/YS ratio, which improves earthquake performance.
1.3 Building technology
32 questions · ACiE0103
64. In English bond brick masonry, the arrangement of bricks in successive courses is
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Think of courses, not individual bricks, alternating.
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Answer: D. alternate courses of headers and stretchers
English bond has one course of stretchers followed by a course of headers; alternate headers and stretchers in the same course is Flemish bond.
65. Flemish bond is characterised by
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Both faces show a chequered pattern.
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Answer: B. headers and stretchers alternating in each course
In Flemish bond every course has alternate headers and stretchers, giving a pleasing appearance but slightly less strength than English bond.
66. A queen closer is a brick that has been
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Its length remains full; its width halves.
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Answer: B. cut lengthwise into two equal halves
A queen closer is half of the brick's width cut lengthwise; it is placed next to the quoin header to maintain bond lap.
67. The 'lap' or overlap of bricks in a course over the bricks below in standard brick masonry should be at least
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Break the joints.
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Answer: A. one-quarter of a brick length
A minimum overlap of 1/4 brick (about 50–60 mm) is maintained so that vertical joints are not continuous, which distributes the load.
68. Using 190 × 90 × 90 mm bricks with 10 mm mortar joints, the number of brick courses in a 1.2 m height of masonry is
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Include the horizontal joint in each course height.
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Answer: D. 12
Height of one course with joint = 90 + 10 = 100 mm, so courses = 1200/100 = 12.
69. Stone masonry in which stones are accurately dressed to regular shape and laid in thin joints (3 mm or less) is
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It is the finest and costliest type of stone masonry.
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Answer: D. ashlar masonry
Ashlar uses finely chisel-dressed stones in regular courses with very thin joints; it is expensive and used for facing and important structures.
70. The purpose of through (bond) stones provided at intervals in a stone wall is to
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They run across the wall thickness.
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Answer: D. bond the two faces of the wall together
A through stone extends the full thickness of the wall and ties the outer and inner faces so they act as one.
71. The mortise and tenon joint is commonly used in carpentry for
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A right-angled framing joint.
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Answer: A. joining the stiles and rails at the corners of a door frame
A tenon at the end of one member fits into a mortise cut in the other, providing a strong right-angled joint, as at door/window frame corners.
72. In a panelled door shutter, the vertical members at the sides are called
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Rails are the horizontal ones.
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Answer: C. stiles
Stiles are vertical members and rails are horizontal members of a shutter; panels fit between them.
73. The horizontal member that divides a window opening into two or more lights is called a
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Think of the transom above a door.
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Answer: C. transom
A transom is horizontal; a mullion is a vertical member dividing the opening.
74. A scarf joint in timber construction is used to
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Splicing along the length.
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Answer: D. increase the length of a timber member
A scarf joint splices two timbers end to end to form a longer member.
75. The primary function of a priming coat in painting is to
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It is the first coat applied.
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Answer: C. seal the surface and provide a good bond for subsequent coats
A primer penetrates and seals the surface, improves adhesion and offers a uniform base for undercoat and finishing coats.
76. Painting a newly plastered cement wall is generally delayed for several weeks mainly because
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Alkalinity and dampness of fresh plaster.
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Answer: B. free lime and moisture in fresh plaster can damage the paint film
Fresh plaster is highly alkaline and damp; painting too early causes blistering, peeling and efflorescence.
77. Small bubbles or raised blisters on a painted surface are most commonly caused by
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Something trapped underneath.
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Answer: D. moisture or air trapped beneath the paint film
Moisture or solvent vapour trapped under the film expands and lifts it, forming blisters.
78. The usual thickness of single-coat cement plaster on interior brick walls is about
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About one centimetre and a bit.
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Answer: A. 12 mm
Internal wall plaster is normally 12 mm thick; external two-coat plaster is typically 15–20 mm.
79. Cement plaster 15 mm thick is to be applied on a wall of 80 m² area. The volume of wet mortar required (neglecting wastage) is
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Convert millimetres to metres.
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Answer: C. 1.2 m³
Volume = area × thickness = 80 × 0.015 = 1.2 m³.
80. As per IS 456, a rectangular slab supported on all four sides is designed as a two-way slab when the ratio of its longer span to shorter span is
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Roughly square slabs span both ways.
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Answer: C. less than or equal to 2
When ly/lx ≤ 2 the slab bends in both directions; otherwise it acts as a one-way slab.
81. As per IS 456, the vertical formwork of beams, walls and columns may normally be removed after
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Side forms carry no load.
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Answer: C. 16 to 24 hours
IS 456 cl. 11.3 allows stripping of vertical formwork after 16–24 h, provided the concrete has hardened enough; slab soffits need 3 days (with props refixed).
82. A reinforced concrete roof slab measures 4 m × 5 m and is 150 mm thick. The volume of concrete needed is
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Thickness must be in metres.
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Answer: C. 3.0 m³
Volume = 4 × 5 × 0.15 = 3.0 m³.
83. For the 3.0 m³ slab above, concrete of mix 1 : 2 : 4 is used. Taking dry volume as 1.54 times the wet volume and 1 m³ of cement = 28.8 bags of 50 kg, the cement required is nearly
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Cement is 1 part out of 7.
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Answer: C. 19 bags
Dry volume = 3.0 × 1.54 = 4.62 m³; cement = 4.62 × 1/7 = 0.66 m³ = 0.66 × 28.8 ≈ 19 bags.
84. In the construction of a ground floor, the layer immediately above the compacted natural soil (sub-grade) and below the concrete base is the
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A load-spreading layer below the concrete.
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Answer: D. sub-base (brick soling or hardcore)
The usual sequence is compacted sub-grade, sub-base (soling/hardcore), cement concrete base and finally the wearing surface.
85. Terrazzo flooring is a finish made of
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Chips of a decorative stone.
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Answer: A. marble chips set in cement mortar and polished
Terrazzo consists of marble chips and cement (often coloured) laid in a topping and ground and polished.
86. A room of 6 m × 3 m is to be floored with 300 mm × 300 mm tiles. Allowing 5 % for wastage, the number of tiles to be purchased is
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Find the net number first, then add the percentage.
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Answer: A. 210
Floor area 18 m² / 0.09 m² per tile = 200 tiles; adding 5 % gives 210.
87. A staircase has to rise 3.0 m between floors using risers of 150 mm. The number of treads required (one fewer than the number of risers) is
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Count risers first.
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Answer: A. 19
Risers = 3000/150 = 20; treads = 20 − 1 = 19, since the top floor acts as the last tread.
88. The main purpose of a damp-proof course (DPC) in a building is to
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Water rises through small pores in masonry.
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Answer: D. prevent the rise of ground moisture by capillary action into the walls
A DPC is a horizontal (or vertical) barrier that blocks capillary moisture from the foundation and ground into the walls and floors.
89. The usual position of the horizontal DPC in a building wall is
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Just above ground level.
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Answer: A. at plinth level, above the ground level
The DPC is laid at plinth level, at least 150 mm above the highest ground level, so that no part of the wall above it can draw ground moisture.
90. Which of the following materials is a flexible DPC material?
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Look for a sheet or roll material.
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Answer: B. bitumen felt
Bitumen felt, polythene sheet, lead and copper sheets are flexible DPCs; concrete and stone are rigid.
91. In the National Building Code of Nepal, the code dealing with seismic design of buildings is
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The 100-series covers loads and design.
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Answer: B. NBC 105
NBC 105 is Seismic Design of Buildings in Nepal; NBC 102 is unit weights, NBC 104 wind load and NBC 206 architectural design requirements.
92. The 'ground coverage' in building bylaws is the ratio of
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It concerns only the footprint.
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Answer: B. plinth area of the ground floor to the plot area
Ground coverage (%) = ground-floor built-up area / plot area × 100; total floor area / plot area is FAR.
93. A plot of 200 m² has an allowed ground coverage of 60 % and a floor area ratio (FAR) of 2.0. The maximum ground-floor footprint and total permissible floor area are, respectively,
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FAR multiplies plot area.
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Answer: B. 120 m² and 400 m²
Footprint = 0.60 × 200 = 120 m²; total floor area = FAR × plot = 2.0 × 200 = 400 m².
94. In building bylaws, 'setback' means
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A gap that must be left empty.
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Answer: B. the minimum clear distance to be left between the building line and the plot boundary or road edge
Setbacks ensure light, ventilation, safety and room for future road widening.
95. Plinth area of a building is the
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It includes the walls.
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Answer: A. built-up covered area measured at plinth level, including wall thickness
Plinth area is the covered area measured externally at floor level (with walls); carpet area is the usable area excluding walls.
1.4 Geometric properties of sections
31 questions · ACiE0104
96. A plane figure is said to have an axis of symmetry if
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Think of folding the figure along the axis.
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Answer: C. the part on one side of the axis is the mirror image of the part on the other side
A symmetry axis divides the figure into two mirror images; the centroid and the principal axes lie on it.
97. The numbers of axes of symmetry of a rectangle (not a square), an equilateral triangle and a circle are, respectively,
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A circle is symmetrical about every diameter.
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Answer: D. 2, 3 and infinite
A rectangle has two, an equilateral triangle three, and a circle infinitely many axes (every diameter).
98. The centroid of an area lying on an axis of symmetry means that the product of inertia about that axis and a perpendicular axis through the centroid is
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Symmetry cancels the contributions of mirror elements.
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Answer: A. zero
If either of the two perpendicular centroidal axes is an axis of symmetry, Ixy = 0 and they are principal axes.
99. The first moment of the area of a section about any axis passing through its centroid is
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The position of the centroid is found from Σ A·ȳ = Aȳ.
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Answer: B. zero
The centroid is defined where Σ(A·y) = 0; areas on either side balance about a centroidal axis.
100. The moment of inertia of a rectangle of width b and depth d about its centroidal axis parallel to the width is
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The depth is cubed for the axis parallel to the width.
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Answer: A. bd³/12
Ixx = bd³/12 about the centroidal axis parallel to b; about the base it is bd³/3.
101. A rectangular section 200 mm wide and 300 mm deep has a moment of inertia about its centroidal x–x axis (parallel to the width) of
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Cube the depth, divide by 12.
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Answer: D. 4.5 × 10⁸ mm⁴
I = bd³/12 = 200 × 300³/12 = 4.5 × 10⁸ mm⁴.
102. The moment of inertia of a rectangular section about its base is
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Add A·h² with h = d/2.
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Answer: C. bd³/3
By the parallel axis theorem I = bd³/12 + bd(d/2)² = bd³/3.
103. A rectangle 100 mm wide and 200 mm deep has its centroidal moment of inertia 6.67 × 10⁷ mm⁴. About an axis parallel to it at a distance of 50 mm from the centroid the moment of inertia is
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Apply the parallel axis theorem with A = 20 000 mm².
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Answer: C. 1.167 × 10⁸ mm⁴
I = I_G + A·d² = 6.67×10⁷ + (100×200)(50²) = 6.67×10⁷ + 5.0×10⁷ = 1.167×10⁸ mm⁴.
104. The moment of inertia of a circular section of diameter d about its centroidal axis is
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The polar value is twice the diametral value.
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Answer: B. πd⁴/64
Ixx = Iyy = πd⁴/64; the polar moment of inertia J = πd⁴/32 is twice this.
105. The moment of inertia of a circle of diameter 100 mm about a diameter is nearly
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Use the diametral formula, not the polar one.
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Answer: C. 4.91 × 10⁶ mm⁴
I = πd⁴/64 = π(100)⁴/64 = 4.91 × 10⁶ mm⁴.
106. According to the perpendicular axis theorem, for a plane area the polar moment of inertia about an axis perpendicular to the plane through O equals
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A simple sum.
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Answer: B. Ixx + Iyy
The moments of inertia about any two perpendicular in-plane axes through O add to give the polar moment.
107. Of all parallel axes in a plane, the moment of inertia of an area is least about the axis that
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The Ad² term cannot be negative.
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Answer: C. passes through the centroid
I = I_G + Ad², and the second term is zero only for the centroidal axis, so I_G is the minimum.
108. The radius of gyration of an area A about an axis is given by (I is the moment of inertia about that axis)
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Dimensionally, k is a length.
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Answer: A. k = √(I/A)
k is the distance at which the whole area could be concentrated to give the same moment of inertia: I = A·k².
109. The radius of gyration of a solid circular section of diameter 100 mm about its centroidal axis is
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The result is a simple fraction of d.
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Answer: B. 25 mm
k = √(I/A) = √[(πd⁴/64)/(πd²/4)] = d/4 = 25 mm.
110. A rectangular section 300 mm deep has a radius of gyration about its centroidal axis parallel to the width of nearly
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Cancel b from I and A.
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Answer: D. 86.6 mm
k = √(I/A) = √(bd³/12 ÷ bd) = d/√12 = 300/3.464 = 86.6 mm.
111. The centroid of a semicircular area of radius R lies on its axis of symmetry at a distance from the diameter of
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About 0.42R.
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Answer: B. 4R/(3π)
ȳ = 4R/(3π) ≈ 0.424R measured from the diametral base.
112. The distance of the centroid from the diameter of a semicircular lamina of radius 100 mm is about
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Use 4R/3π.
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Answer: A. 42.4 mm
ȳ = 4R/3π = 400/(3π) = 42.4 mm.
113. The centroid of a triangle of base b and height h lies, measured from the base, at a height of
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Medians intersect in the ratio 2:1.
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Answer: B. h/3
The centroid is at the intersection of medians, one-third of the height from the base.
114. The moments of inertia of a triangle (base b, height h) about its base and about a centroidal axis parallel to it are, respectively,
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The centroidal value is the smaller.
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Answer: B. bh³/12 and bh³/36
About the base I = bh³/12; about the centroid I = bh³/12 − (bh/2)(h/3)² = bh³/36.
115. A trapezium has parallel sides of 6 m (base) and 3 m (top) and a height of 9 m. The height of its centroid above the 6 m base is
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The centroid lies nearer the longer side.
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Answer: C. 4.0 m
ȳ = h(a + 2b)/[3(a + b)] = 9(6 + 6)/(3 × 9) = 4.0 m, measured from the side a = 6 m.
116. A T-section has a flange 200 mm × 20 mm at the top and a web 20 mm × 180 mm below it (overall depth 200 mm). The centroid lies from the top edge at
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Take moments of area about the top edge.
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Answer: C. 57.4 mm
ȳ = (4000×10 + 3600×110)/7600 = 57.4 mm from the top; 142.6 mm is its distance from the bottom.
117. An angle section has a horizontal leg 150 mm × 10 mm at the bottom and a vertical leg 10 mm × 90 mm standing on it (overall height 100 mm, left edges aligned). The distance of the centroid above the bottom is
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Take moments of both rectangles about the bottom edge.
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Answer: B. 23.75 mm
ȳ = (1500×5 + 900×55)/2400 = 57000/2400 = 23.75 mm; the horizontal distance from the left edge is 48.75 mm.
118. A 200 mm × 200 mm square plate has a 100 mm diameter circular hole at its centre. The moment of inertia about a centroidal axis is nearly
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Subtract the hole's own I from the square's I.
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Answer: D. 1.28 × 10⁸ mm⁴
I = 200⁴/12 − π(100)⁴/64 = 1.333×10⁸ − 0.049×10⁸ = 1.284×10⁸ mm⁴.
119. A hollow circular tube has outer diameter 100 mm and inner diameter 80 mm. Its moment of inertia about a diameter is
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Subtract the d⁴ term, not d².
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Answer: A. 2.90 × 10⁶ mm⁴
I = π(D⁴ − d⁴)/64 = π(100⁴ − 80⁴)/64 = 2.90×10⁶ mm⁴.
120. A hollow rectangular box has outer dimensions 100 mm (width) × 160 mm (depth) and inner dimensions 80 mm × 140 mm. Its moment of inertia about the centroidal horizontal axis is
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Outer rectangle minus inner rectangle.
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Answer: D. 1.584 × 10⁷ mm⁴
I = (100×160³ − 80×140³)/12 = (4.096×10⁸ − 2.195×10⁸)/12 = 1.584×10⁷ mm⁴.
121. If the depth of a rectangular beam section is doubled while its width is unchanged, its moment of inertia becomes
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Depth appears as a cube.
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Answer: A. 8 times
I ∝ bd³, so doubling d multiplies I by 2³ = 8.
122. The section modulus of a rectangular beam section 100 mm wide and 200 mm deep (about its centroidal horizontal axis) is
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Divide I by the distance to the extreme fibre.
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Answer: A. 6.67 × 10⁵ mm³
Z = I/y_max = (100×200³/12)/100 = bd²/6 = 6.67×10⁵ mm³.
123. An I-section has two flanges, each 100 mm × 10 mm, joined by a web 10 mm × 100 mm (overall depth 120 mm). The moment of inertia about the horizontal centroidal axis is
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Use parallel axis theorem for the flanges, centres 55 mm from the axis.
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Answer: A. 6.9 × 10⁶ mm⁴
Flanges: 2[100×10³/12 + 1000×55²] = 6.07×10⁶; web: 10×100³/12 = 0.83×10⁶; total = 6.9×10⁶ mm⁴.
124. In a standard rolled steel I-section such as ISMB, the moment of inertia about the major (x–x) axis is much greater than about the minor (y–y) axis because
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Distance is squared in I.
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Answer: C. most of the material is placed far from the x–x axis in the flanges
The flanges, at the greatest distance from the x–x axis, contribute heavily through the A·d² term; their width is far from the y–y axis only to a small extent.
125. The centroid of a channel section (e.g. ISMC) lies
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A channel is symmetrical about just one axis.
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Answer: D. on its one axis of symmetry, at a distance from the back of the web
A channel has only one axis of symmetry (horizontal); its centroid lies on it, displaced from the web's back by a distance given in steel tables.
126. In the designation ISA 75 × 75 × 8, the symbols denote
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Two equal numbers indicate equal legs.
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Answer: D. an equal angle with 75 mm legs and 8 mm thickness
ISA stands for Indian Standard Angle; the numbers are leg lengths and thickness in mm.
1.5 Surveying and levelling
32 questions · ACiE0105
127. The fundamental principle of surveying 'working from the whole to the part' means that
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Major framework first, minor details later.
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Answer: D. a large area is first controlled by accurate main stations and then detailed work is fixed within it
Control is established with higher accuracy over the entire area; details are then fitted into it, so errors do not accumulate.
128. A survey made to determine and record the boundaries and areas of land parcels and properties is called a
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Think of the land registration office.
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Answer: D. cadastral survey
Cadastral surveys fix property boundaries for ownership and land-revenue records; topographic surveys show natural and man-made features and relief.
129. For good accuracy in plotting by chain surveying, the triangles formed should be well-conditioned, which means their angles should preferably lie between
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No very acute and no very obtuse angle.
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Answer: A. 30° and 120°
Too small or too large an angle makes the plotted position of a vertex uncertain; an equilateral triangle is ideal.
130. A 30 m tape is found to be 30.02 m long. A distance recorded with it as 450.0 m has a true length of
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A too-long tape gives too few tape lengths, so the true distance is larger.
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Answer: B. 450.3 m
True length = measured × (actual tape length/nominal length) = 450 × 30.02/30 = 450.3 m, because a long tape under-reads.
131. A slope distance of 100 m is measured on a uniform slope of 5° to the horizontal. The horizontal distance is nearly
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Horizontal = slope length × cosine of the angle.
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Answer: A. 99.62 m
H = L cos θ = 100 × cos 5° = 99.62 m.
132. In tacheometric surveying with a staff held vertical and a horizontal line of sight, the stadia constants are 100 and 0. A stadia intercept of 1.35 m gives a horizontal distance of
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Multiply the intercept by the stadia multiplier.
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Answer: C. 135 m
D = k·s + C = 100 × 1.35 + 0 = 135 m.
133. A total station is an instrument that combines
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It measures both angle and distance.
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Answer: A. an electronic theodolite and an electronic distance meter
A total station measures horizontal and vertical angles with its theodolite and slope distance with its EDM, storing the data electronically.
134. A line has a whole circle bearing of 235°30′. Its quadrantal (reduced) bearing is
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180°–270° lies in the south-west quadrant.
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Answer: A. S 55°30′ W
For WCB between 180° and 270° the line is in the SW quadrant: R.B. = WCB − 180° = 55°30′, written S 55°30′ W.
135. The magnetic bearing of a line is 120° and the magnetic declination is 3° east. The true bearing of the line is
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East declination means the compass needle points east of true north.
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Answer: D. 123°
True bearing = magnetic bearing + declination (east positive) = 120° + 3° = 123°.
136. Local attraction at a station in compass surveying is detected by
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Compare the fore bearing with the back bearing.
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Answer: B. finding that the fore and back bearings of a line do not differ by exactly 180°
If FB and BB of a line differ by other than 180°, one or both stations are affected by local magnetic influence.
137. The angle between the true meridian and the magnetic meridian at a place is called
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Horizontal angle, not vertical.
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Answer: C. magnetic declination
Declination (variation) is the horizontal angle between true north and magnetic north; dip is the vertical inclination of the needle.
138. A theodolite has the smallest main scale division of 20′ and a vernier with 60 divisions equal to 59 main scale divisions. Its least count is
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Divide the smallest main scale division by the vernier divisions.
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Answer: C. 20″
Least count = (value of smallest main scale division)/(number of vernier divisions) = 20′/60 = 20″.
139. Taking observations on both the faces (face left and face right) of a theodolite eliminates errors due to
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Instrument adjustment errors are cancelled by changing face.
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Answer: B. line of collimation and trunnion axis errors of the instrument
Mean of face left and face right readings cancels instrumental errors such as collimation error and trunnion axis inclination.
140. The reading on a benchmark of RL 100.000 m is 2.345 m (backsight). A foresight of 1.255 m is then taken on a point from the same setup. The RL of this point is
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First find the height of instrument.
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Answer: C. 101.090 m
HI = 100.000 + 2.345 = 102.345 m; RL = HI − FS = 102.345 − 1.255 = 101.090 m.
141. The height of the instrument is 102.50 m. A levelling staff held inverted under the soffit of a bridge gives a reading of 3.20 m. The RL of the soffit is
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The staff is above the line of sight.
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Answer: C. 105.70 m
For an inverted staff the reading is added to HI: RL = 102.50 + 3.20 = 105.70 m.
142. In differential levelling, the arithmetic check on the level book states that
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Net change in level.
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Answer: A. ΣBS − ΣFS = last RL − first RL
The difference of the sum of backsights and foresights equals the difference of the last and first reduced levels (and also ΣRise − ΣFall).
143. The combined correction for curvature and refraction in levelling for a sight length of 400 m, using C = 0.0673 d² (C in m, d in km), is nearly
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Convert 400 m to km before squaring.
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Answer: C. 10.8 mm
C = 0.0673 × (0.4)² = 0.01077 m ≈ 10.8 mm, subtractive from the staff reading.
144. Reciprocal levelling is carried out to eliminate the effect of
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Used for crossing a wide river.
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Answer: C. curvature, refraction and collimation error between two points
Readings from both banks of an obstacle cancel the errors common to both lines of sight when the mean difference is taken.
145. A change point (turning point) in levelling is a point on which
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The staff stays while the instrument is moved.
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Answer: D. both a foresight and a backsight are taken
At a change point the foresight from one setup and the backsight from the next setup are taken, linking the two instrument positions.
146. Which statement about contour lines is true?
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A given point has only one elevation.
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Answer: B. Contour lines of different elevations cannot cross each other except at overhanging cliffs or caves
A contour is a closed line of equal elevation and can neither end abruptly nor cross another contour (apart from vertical or overhanging faces).
147. In a contour map, the contour lines in a valley form V- or U-shapes whose convex side (the bend) points toward
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A stream flows down the valley, against the bend.
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Answer: D. higher ground
In a valley the bend of the contours points up the slope, whereas on a ridge it points downhill.
148. On a map of scale 1 : 2000, two adjacent contours of 2 m interval are 5 cm apart. The ground gradient between them is
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Convert the map distance into ground distance first.
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Answer: A. 1 in 50
Horizontal ground distance = 5 cm × 2000 = 100 m; gradient = 2/100 = 1 in 50.
149. An area of 4 cm² is measured on a map of scale 1 : 10 000. The actual area on the ground is
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Area scale is the square of the linear scale.
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Answer: B. 4 hectares
1 cm = 100 m, so 1 cm² = 10 000 m²; 4 cm² = 40 000 m² = 4 ha.
150. A contour map is useful in civil engineering for
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It shows relief information.
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Answer: D. locating a route of a given gradient and computing reservoir capacity
Contours give ground relief, allowing alignment of roads/canals at a gradient, catchment delineation, and volume or storage calculations.
151. The tangent length of a simple circular curve of radius 300 m and deflection angle 40° is
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Use half the deflection angle.
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Answer: C. 109.2 m
T = R tan(Δ/2) = 300 × tan 20° = 109.2 m.
152. A simple circular curve has R = 300 m and Δ = 40°; the chainage of the point of intersection is 1250.00 m. The chainage of the point of tangency (PT) is nearly
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Subtract T to get PC, then add the curve length.
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Answer: D. 1350.25 m
T = 109.19 m, L = 300 × 40π/180 = 209.44 m; PC = 1250 − 109.19 = 1140.81 m; PT = PC + L = 1350.25 m.
153. For the same curve (R = 300 m, Δ = 40°), the external distance (apex distance) is
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E = R(sec(Δ/2) − 1).
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Answer: B. 19.25 m
E = R(sec Δ/2 − 1) = 300 (1/cos 20° − 1) = 19.25 m. (18.09 m is the versine.)
154. The deflection angle for a 20 m first chord on a curve of radius 300 m, set out by Rankine's method, is nearly
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Use 1718.9 c/R in minutes.
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Answer: B. 1°54.6′
δ = 1718.9 × c/R minutes = 1718.9 × 20/300 = 114.6′ = 1°54.6′ (half the central angle of the chord).
155. To obtain a three-dimensional position (latitude, longitude, height) and the receiver clock error, a GPS receiver needs signals from at least
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Count the unknowns.
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Answer: D. 4 satellites
Four unknowns (x, y, z and clock bias) require four pseudorange equations.
156. The three segments of the Global Positioning System are
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Satellites, ground stations, and receivers.
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Answer: A. space, control and user segments
The space segment is the satellite constellation, the control segment the ground monitoring stations, and the user segment the receivers.
157. In a GIS, features such as roads and boundaries stored as points, lines and polygons with coordinates are represented in the
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Points, lines and polygons.
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Answer: A. vector data model
Vector data store geometric objects by coordinates; raster data store values in a grid of cells (pixels).
158. The GIS operation of combining two or more thematic layers to produce a new layer (e.g. slope layer and land-use layer) is called
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Layers laid one over another.
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Answer: B. overlay analysis
Overlay superimposes layers to derive new information; buffering makes zones around features and digitising captures vectors.
1.6 Estimating, costing, and valuation
31 questions · ACiE0106
159. The main purpose of a preliminary (approximate) estimate is to
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It is prepared before detailed drawings exist.
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Answer: C. get a rough idea of the probable cost for administrative approval and feasibility
A preliminary estimate is prepared quickly from past data (area, volume or unit rates) to check whether the project is financially feasible.
160. A detailed estimate is prepared from
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It requires working drawings.
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Answer: C. complete drawings, specifications and item-wise rates
It is worked out by measuring item-wise quantities from the drawings and multiplying by analysed rates, and is the basis for technical sanction and tendering.
161. A building has plinth area 12 m × 10 m and a plinth-area rate of Rs 25,000 per m². Its approximate cost is
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Plinth area is length × breadth.
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Answer: A. Rs 3,000,000
Cost = plinth area × rate = (12 × 10) × 25 000 = Rs 3,000,000.
162. A building of 12 m × 10 m plan and 3.5 m height is estimated by the cubic-content method at Rs 7,000 per m³. The estimate is
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Find the cubic content first.
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Answer: B. Rs 2,940,000
Volume = 12 × 10 × 3.5 = 420 m³; cost = 420 × 7 000 = Rs 2,940,000.
163. When additional work not included in the original sanctioned estimate becomes necessary during execution, the estimate prepared for it is called a
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It supplements the original estimate.
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Answer: A. supplementary estimate
A supplementary estimate covers new works over and above the original; a revised estimate replaces the sanctioned one when it is likely to be exceeded.
164. A revised estimate is prepared when
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Cost overrun triggers it.
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Answer: C. the original sanctioned estimate is likely to be exceeded materially
Revision becomes necessary when quantities or rates change so that the sanctioned amount will be exceeded.
165. The estimate of the cost of cleaning, repair and routine maintenance of existing structures every year is called
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It recurs yearly.
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Answer: A. an annual repair and maintenance estimate
Maintenance works on existing buildings and roads are estimated annually in the annual repair and maintenance estimate.
166. The 'unit base' method of preliminary estimating, in which the cost is the number of units × cost per unit, would be used for
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A functional unit of the building.
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Answer: D. a hospital at a cost per bed
Typical unit-based rates are cost per bed (hospital), per classroom (school), per seat (auditorium) and per km (road).
167. The centreline of the walls of a rectangular building measures 8 m × 6 m. The trench for the foundation is 1.0 m wide and 1.2 m deep. The volume of excavation is
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Find the total length along the centreline.
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Answer: B. 33.6 m³
Centreline length = 2(8 + 6) = 28 m; volume = 28 × 1.0 × 1.2 = 33.6 m³.
168. In the long-wall short-wall method, if the centre-to-centre length of a long wall is 8 m and the trench width is 1 m, the length of the long wall to be taken is
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Long walls are measured out-to-out.
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Answer: A. 9 m
Long wall length (out-to-out) = centre-to-centre length + width of trench/wall = 8 + 1 = 9 m; the short wall is taken as centre length minus width.
169. A brick wall 10 m long, 3 m high and 0.23 m thick has a door opening 1.0 m × 2.1 m. The net quantity of brickwork is
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Subtract the volume of the opening.
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Answer: A. 6.42 m³
Gross = 10 × 3 × 0.23 = 6.90 m³; opening = 1.0 × 2.1 × 0.23 = 0.483 m³; net = 6.417 ≈ 6.42 m³.
170. Modular bricks of actual size 190 × 90 × 90 mm are laid at 500 bricks per m³ of masonry (nominal size 200 × 100 × 100 mm). The volume of mortar (wet) per m³ of masonry is nearly
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Mortar fills what is left after the bricks.
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Answer: C. 0.23 m³
Volume of bricks = 500 × 0.19 × 0.09 × 0.09 = 0.77 m³; mortar = 1 − 0.77 = 0.23 m³.
171. As per standard practice, plastering and reinforcement steel are measured respectively in
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Area for surfaces; weight for steel.
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Answer: C. m² and kg
Plastering is measured by area (m²), brickwork/concrete by volume (m³) and reinforcement by weight (kg or tonne).
172. In estimating, 'lead' and 'lift' respectively refer to
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Lead is horizontal, lift is vertical.
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Answer: B. horizontal distance of carriage and vertical height of raising material
Lead is the average horizontal distance from source to site; lift is the vertical height through which the material is raised.
173. The 'abstract of estimated cost' is the document that
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It is the final summary of the detailed estimate.
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Answer: C. summarises the quantities, rates and amounts of all items of the work
It is the summary sheet listing each item with quantity, rate and amount, totalled for the estimate.
174. A contingency allowance is added to the estimate in order to
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Think of unexpected costs.
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Answer: A. meet unforeseen expenses that arise during execution
Contingencies (usually a small percentage of the cost) cover unforeseen items; contractor profit is separately included in rates.
175. For 1 m³ of 1 : 2 : 4 concrete, taking the dry volume as 1.54 m³ per m³ of wet concrete, the volume of sand required is
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Sand is 2 parts of 7.
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Answer: D. 0.44 m³
Sand = 1.54 × 2/7 = 0.44 m³ (cement = 1.54/7 = 0.22 m³ ≈ 6.3 bags; aggregate = 0.88 m³).
176. The analysis of rates of an item of work is the process of finding the
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It yields a unit rate.
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Answer: D. cost per unit of the item from the cost of materials, labour, and overheads and profit
Rate analysis derives the unit rate (per m³, m², etc.) using material quantities and rates, labour constants and wages, and add-ons.
177. For 1 m³ of an item, materials cost Rs 8,000 and labour Rs 1,500. If contractor's overhead and profit totalling 25 % is added to the net cost, the rate of the item is
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Overhead and profit apply to the total of materials and labour.
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Answer: D. Rs 11,875
Net cost = 9 500; add 25 % = 9 500 × 1.25 = Rs 11,875 per m³.
178. 1 m³ of brick masonry requires 1 mason-day at Rs 1,500 and 1.5 helper-days at Rs 1,000 each. The labour cost per m³ is
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Add the mason's cost to the helpers' cost.
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Answer: B. Rs 3,000
1 × 1 500 + 1.5 × 1 000 = Rs 3 000.
179. A labourer excavates 3 m³ of ordinary soil per day at a wage of Rs 900 per day. The labour cost per m³ of excavation is
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Divide cost per day by quantity per day.
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Answer: A. Rs 300
Rate = daily wage ÷ output = 900/3 = Rs 300 per m³.
180. In an analysis of rates, the items 'establishment, tools and plants, supervision' fall under
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They are not directly measurable per unit of work.
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Answer: B. the contractor's overheads
Overheads are indirect costs of running the contractor's organisation, added to direct costs, along with profit.
181. The purpose of a specification in a construction contract is to
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Drawings show 'where and how big', specifications show 'what quality'.
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Answer: C. describe the quality of materials, standard of workmanship and method of execution
Specifications define what quality of materials and workmanship are required and how the work is tested and measured; they supplement the drawings.
182. A specification that gives only a general description of the nature and class of work and materials, without exact details of each item, is called
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The opposite of 'detailed'.
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Answer: A. a general (brief) specification
General specifications give the overall standards for the building; detailed specifications describe each item's materials, proportions and workmanship.
183. Valuation of a property means
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It gives a figure for the present worth.
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Answer: C. assessment of its fair worth in monetary terms at a given date
Valuation estimates the value of property for purposes such as sale, purchase, rent fixation, mortgage, insurance or taxation.
184. The value of a property at the end of its useful life, which it would fetch if sold as old material (excluding the land), is its
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It remains after life is over.
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Answer: D. salvage value
Salvage value is the value of the structure's materials at the end of its life; scrap value is the value of material if dismantled and sold as scrap.
185. A building yields a gross rent of Rs 10,000 per month. Outgoings are 20 % of the gross rent. If the property is valued by the rental method at 8 % interest, the capitalised value is
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Convert rent to yearly net income first.
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Answer: B. Rs 1,200,000
Net income = 120 000 × 0.8 = Rs 96 000/year; capitalised value = 96 000/0.08 = Rs 1,200,000.
186. Years' Purchase (YP) of a property when the rate of interest is 8 % is
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YP is the reciprocal of the interest rate.
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Answer: B. 12.5
YP = 100/i = 100/8 = 12.5; the net annual income multiplied by YP gives the capitalised value.
187. A building costing Rs 1,200,000 has a life of 60 years and a scrap value of 10 % of cost. The annual depreciation by the straight-line method is
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Subtract the scrap value before dividing by life.
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Answer: B. Rs 18,000
D = (C − S)/n = (1 200 000 − 120 000)/60 = Rs 18,000 per year.
188. The sinking fund in valuation is the
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A savings plan for replacement.
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Answer: D. fixed annual sum set aside, with interest, to replace the building at the end of its life
A sinking fund accumulates, with compound interest, to the cost needed to rebuild or replace the structure when its life ends.
189. A currently running commercial enterprise such as a hotel or cinema is best valued by the
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Valuation follows earning capacity.
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Answer: D. profit-based (capitalised net profit) method
For running businesses the value depends on earning capacity, so the net profit is capitalised.