Nepal Engineering Council · Civil Engineering · Chapter 9
Transportation
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187 questions in 6 syllabus topics.
9.1 Highway planning and survey
31 questions · ACiE0901
1. Which mode of transport offers the greatest flexibility for door-to-door movement of passengers and goods?
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Think about which mode needs no terminal change at either end.
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Answer: A. Roadways
Roads can reach almost every origin and destination and allow vehicles to start, stop and change route at will, which no other mode can do.
2. For moving very heavy bulk cargo over long distances at the lowest cost per tonne-km, the most economical mode is
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Low speed is acceptable for bulk goods.
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Answer: D. waterways
Water transport has the lowest energy and cost per tonne-km, though it is slow and tied to navigable routes.
3. The Tribhuvan Rajpath, Nepal's first highway connecting the Kathmandu Valley to the plains, links Kathmandu with
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It heads south, not north or west.
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Answer: C. the Indian border at Raxaul via Birgunj
The Tribhuvan Rajpath runs from Kathmandu through Bhimphedi and Hetauda to Birgunj and the Raxaul border.
4. The Araniko Highway in Nepal connects Kathmandu with
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It is named after a Nepali architect who worked in China.
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Answer: B. Kodari on the Tibet (China) border
The Araniko Highway runs north-east from Kathmandu through Banepa and Dhulikhel to Kodari at the Nepal-Tibet border.
5. The Jayakar Committee (1927) in India is remembered for recommending
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Think of funding rather than design.
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Answer: B. a road development fund built from an extra tax on petrol
The Jayakar Committee led to the Central Road Fund (1929) from a petrol tax and later to the Indian Roads Congress (1934).
6. The Nagpur Road Plan (1943) classified roads into National Highways, State Highways, District Roads and Village Roads and proposed
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The pattern is named after a star and a mesh.
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Answer: A. a star and grid pattern of road network
The Nagpur Plan (a 20-year plan) adopted a star and grid road pattern, with no village more than a fixed distance from a major road.
7. In Nepal, the Strategic Road Network (SRN) consists of
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The strategic network carries the longest-distance, most important traffic.
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Answer: D. national highways and feeder roads
The Nepal Road Standard places national highways and feeder roads in the Strategic Road Network; district, urban and village roads form the local road network.
8. A road of national importance linking the capital with regional centres and international border crossings is classified in Nepal as a
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It is the top class of the Nepal road classification.
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Answer: C. national highway
National highways form the main arteries of the country, connecting the capital, regions and border points.
9. Under the IRC classification, roads serving rural areas of production and giving them an outlet to market centres or to higher-order roads are called
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The class sits between Major District Roads and Village Roads.
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Answer: C. Other District Roads (ODR)
ODRs serve rural production areas and link them to markets, MDRs, SHs or NHs; Village Roads link villages to ODRs.
10. A road that has cross-drainage structures (bridges and culverts) so that traffic is not interrupted during monsoon is called
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The name describes when it stays open.
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Answer: B. an all-weather road
All-weather roads have permanent cross-drainage works at streams; fair-weather roads have no or only partial ones and may be cut during rains.
11. The correct sequence of engineering surveys for a new highway is
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Detail increases at each stage.
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Answer: A. map study, reconnaissance, preliminary survey, final location survey
The study goes from general to detailed: map study, field reconnaissance, preliminary (detailed) survey and finally location of the chosen centre line on the ground.
12. Which instrument is mainly used for quick measurement of slopes and elevations during reconnaissance in hilly terrain?
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Light, portable and approximate.
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Answer: D. Abney level or clinometer with aneroid barometer
Handheld clinometers, Abney levels and aneroid barometers give rapid, rough slopes and heights suitable for reconnaissance.
13. The main purpose of the preliminary survey of a highway is to
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It comes before the alignment is finalised.
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Answer: D. collect detailed topographic data along trial lines and compare alternatives with cost estimates
The preliminary survey runs a primary traverse, levelling and topography along the shortlisted routes so that the best alignment can be chosen on the basis of cost.
14. Staking of the centre line, detailed levelling, cross-sections and setting out of curves are carried out in the
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It is the last survey stage.
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Answer: C. final location survey
The final location survey transfers the selected alignment to the ground and gathers the detailed data needed for design and drawings.
15. The four basic requirements of an ideal highway alignment are that it should be
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Four words, all about length, ease, safety and cost.
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Answer: B. short, easy, safe and economical
An ideal alignment is short, easy (gentle gradients, curves), safe (adequate sight distance, stability) and economical in construction and maintenance.
16. Which of the following is a positive control (obligatory point) for highway alignment?
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Controls that attract the alignment, not repel it.
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Answer: A. A mountain pass or saddle
A saddle or pass, a good bridge site or an important town must be included, so they attract the alignment; religious places, forests and expensive land are avoided.
17. From the viewpoint of economy, a highway should preferably cross a river
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A shorter bridge costs less.
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Answer: A. at right angles at a narrow, stable reach
A right-angle crossing at a stable narrow reach minimises bridge length and foundation problems.
18. For a hill road in a snowy region, which slope aspect is generally preferred for the alignment?
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Think of where the sun is in Nepal's sky.
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Answer: D. South-facing slopes
South-facing slopes receive more sun in the northern hemisphere, so snow and ice melt early and the slopes are drier and more stable.
19. Which of the following is the best practice with respect to geometric features of an alignment?
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Driver expectancy matters.
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Answer: C. Avoid a sharp curve at the end of a long straight stretch
Sharp curves after a long straight surprise drivers travelling at high speed; the curve radii should be generous and consistent.
20. The annual cost method of comparing alternative alignments includes
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Include cost to build, upkeep and use.
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Answer: B. annualised capital cost, maintenance cost and vehicle operating cost
The total yearly cost of each alignment (interest and depreciation on capital, maintenance, plus road user costs) is compared; the lowest is preferred.
21. When comparing mutually exclusive alignments by benefit-cost analysis, the alternative with the extra investment is justified if its incremental BCR is
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Benefits must beat costs.
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Answer: B. greater than 1
An increment of investment is justified when its extra benefits exceed its extra costs, i.e. incremental BCR > 1.
22. The Nepal Road Standard (NRS) is the official standard for the design of roads in Nepal and is issued by the
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Think of the agency that builds strategic roads.
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Answer: A. Department of Roads
The Department of Roads (Ministry of Physical Infrastructure and Transport) issues the Nepal Road Standard for the strategic road network.
23. The Nepal Rural Road Standard mainly applies to
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It is for the local road network.
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Answer: D. district and village roads
Rural roads (district and village roads) are designed to the separate rural road standard, with lower design speeds and widths than the strategic network.
24. On a hill road, the ruling gradient is 6%. How much road length is needed to gain 90 m of elevation on a continuous grade?
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Length = rise divided by the gradient as a fraction.
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Answer: C. 1500 m
Length = rise / gradient = 90 / 0.06 = 1500 m.
25. A district of 2,500 km² contains 475 km of roads. The road density per 100 km² of area is
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Road length divided by area, scaled to 100 km².
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Answer: C. 19.0 km
Density = 475 / 2500 × 100 = 19.0 km per 100 km².
26. Alignment B is 2 km shorter than A but needs Rs 6 million more annual capital charge. Traffic is 1500 vehicles/day and vehicle operating cost is Rs 12 per vehicle-km. The net annual saving of B over A is about
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Subtract extra capital charge from the operating-cost saving.
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Answer: B. Rs 7.14 million
Saving = 2 × 1500 × 365 × 12 = Rs 13.14 million; net = 13.14 − 6.00 = Rs 7.14 million.
27. On a topographic map of scale 1:5000, a proposed road section measures 3.2 cm. The actual length on the ground is
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Multiply by the scale denominator, then convert cm to m.
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Answer: A. 160 m
Ground length = 3.2 cm × 5000 = 16,000 cm = 160 m.
28. Two points on a proposed alignment are at elevations of 1240 m and 1300 m, with 1200 m horizontal distance between them. The average gradient is
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Rise over run, expressed as a percentage.
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Answer: D. 5%
Gradient = 60 / 1200 = 0.05 = 5%.
29. Under IRC terrain classification, a stretch where the general cross slope of the country is 18% is classified as
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Check which cross slope band 18% falls in.
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Answer: D. rolling terrain
IRC: plain 0-10%, rolling 10-25%, mountainous 25-60%, steep above 60%. 18% is rolling.
30. A project costs Rs 30 million with a capital recovery charge of 12% per year and yearly maintenance of Rs 0.9 million. Yearly benefits are Rs 5.4 million. The benefit-cost ratio is
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Annualise the capital before dividing.
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Answer: C. 1.2
Annual cost = 0.12 × 30 + 0.9 = 4.5 million; BCR = 5.4 / 4.5 = 1.2.
31. A road gradient of 1 in 25 is equal to
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Convert the ratio to a percentage.
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Answer: B. 4%
Gradient = 1/25 = 0.04 = 4%.
9.2 Geometric design of highway
31 questions · ACiE0902
32. The design speed of a highway is best defined as
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It is a design choice, not a measured average.
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Answer: A. the speed chosen to fix the geometric features that control vehicle operation on the road
Design speed is the selected speed used to determine all speed-dependent geometric elements such as curves, sight distances and superelevation.
33. The design hourly volume used for geometric design is commonly taken as the
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It is neither the very highest nor the average.
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Answer: D. 30th highest hourly volume of the year
The 30th highest hour avoids designing for rare peaks while not being overloaded too often; it lies near the knee of the hourly-volume curve.
34. Camber is provided on a straight road mainly to
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Think of rain, not curves.
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Answer: C. drain rainwater quickly from the carriageway surface
Camber is the transverse slope from the crown to the edges; it sheds surface water and so protects the pavement.
35. A two-lane carriageway 7.0 m wide has a camber of 2.5% (crown at centre). The crown is higher than the pavement edge by
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Use half the width, not the full width.
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Answer: B. 8.75 cm
Each half width is 3.5 m; rise = 3.5 × 0.025 = 0.0875 m = 8.75 cm.
36. The IRC-recommended carriageway width for a two-lane road without kerbs is
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Two lanes of about 3.5 m each.
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Answer: B. 7.0 m
IRC recommends 3.75 m for a single lane and 7.0 m for a two-lane carriageway.
37. The maximum superelevation generally permitted on plain and rolling terrain roads (IRC) is
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It is smaller than the hill-road limit.
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Answer: A. 7%
IRC limits superelevation to 7% in plain and rolling terrain; 10% is allowed on hill roads not bound by snow, and 4% in urban areas.
38. As per IRC, superelevation is calculated for
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Slow vehicles must not slide inwards.
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Answer: D. 75% of the design speed, neglecting lateral friction
The required superelevation is e = V²/(225R) with V taken as 0.75 × design speed, and then checked at full speed for friction.
39. A highway curve is designed for a design speed of 80 km/h. For R = 250 m, the superelevation e = V²/(225R), with V = 0.75 × design speed, is
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Take 75% of the speed first.
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Answer: C. 6.4%
V = 60 km/h; e = 60² / (225 × 250) = 0.064 = 6.4%.
40. The maximum coefficient of lateral friction adopted by IRC for horizontal curve design is
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It is a comfort limit, not a slipping limit.
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Answer: C. 0.15
Lateral friction between tyre and pavement is limited to 0.15 for comfort and safety; 0.35-0.40 values relate to longitudinal friction.
41. The minimum radius of a horizontal curve for a design speed of 80 km/h with e = 0.07 and f = 0.15 is about
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Use e + f in the denominator.
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Answer: B. 229 m
R = V² / [127 (e + f)] = 80² / (127 × 0.22) = 229 m.
42. A curve of radius 100 m has e = 0.07 and the allowable f = 0.15. The maximum safe speed is about
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Use e + f = 0.22.
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Answer: A. 53 km/h
V = √[127 R (e + f)] = √(127 × 100 × 0.22) = 52.9 km/h.
43. A transition curve between a straight and a circular curve is introduced mainly to
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Think of what changes along the transition.
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Answer: D. allow gradual change of curvature, centrifugal acceleration and superelevation
The transition gives a smooth entry: curvature and superelevation grow gradually, and centrifugal acceleration builds at a comfortable rate.
44. A spiral (clothoid) transition curve has the property that its radius of curvature
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Radius is infinity at the straight and R at the circle.
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Answer: D. varies inversely with the distance along the curve
For a clothoid, R × L = constant, so curvature increases linearly with length, giving a uniform rate of change of centrifugal acceleration.
45. The IRC empirical formula for minimum length of transition curve in plain and rolling terrain is Ls = 2.7 V²/R. For V = 50 km/h and R = 150 m, Ls is
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Square the speed before dividing.
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Answer: C. 45 m
Ls = 2.7 × 50² / 150 = 2.7 × 2500 / 150 = 45 m.
46. The limiting gradient on a highway is
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It sits between ruling and exceptional gradient.
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Answer: B. a gradient steeper than the ruling gradient, adopted where the ruling gradient would cause heavy cost, for short lengths
Ruling gradient is the design maximum; limiting gradient may be used for short stretches in difficult terrain, and exceptional gradient only in extreme cases.
47. For a horizontal curve of radius 60 m on a hill road with a ruling gradient, grade compensation is the lower of (30 + R)/R % and 75/R %. The compensation is
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Compute both and take the smaller.
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Answer: A. 1.25%
(30 + 60)/60 = 1.5%; 75/60 = 1.25%; the lower value 1.25% is adopted, subject to the maximum.
48. The two components of stopping sight distance are
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One part occurs before the brakes are applied.
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Answer: A. lag distance (reaction time) and braking distance
SSD = v t + v² / [2g (f ± n)], i.e. distance travelled during reaction time plus braking distance.
49. A vehicle at 60 km/h, with reaction time 2.5 s and f = 0.36 on level ground. The stopping sight distance is about
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Add the lag distance to the braking distance.
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Answer: D. 81 m
v = 16.67 m/s; lag = 41.7 m; braking = 16.67² / (2 × 9.81 × 0.36) = 39.3 m; SSD = 81 m.
50. At 50 km/h, with t = 2.5 s and f = 0.37, the SSD on a 3% downgrade is about
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A downgrade lengthens braking distance.
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Answer: C. 64 m
On a downgrade the braking term uses (f − n): 34.7 + 13.89² / [2 × 9.81 × (0.37 − 0.03)] = 34.7 + 28.9 = 63.6 m. Level road gives 61 m.
51. As per IRC, the intermediate sight distance is
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The name hints that it lies between SSD and OSD.
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Answer: B. twice the stopping sight distance
ISD = 2 × SSD; it is the minimum where OSD cannot be provided throughout, and gives limited overtaking opportunity.
52. A summit curve joins grades +3% and −2%. The length (assuming L > SSD) for SSD = 100 m, with eye height 1.2 m and object height 0.15 m, L = N S² / 4.4 is
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Algebraic difference of grades is 0.05.
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Answer: B. 114 m
N = 0.03 − (−0.02) = 0.05; L = 0.05 × 100² / 4.4 = 113.6 m.
53. A valley curve with N = 0.07 and headlight sight distance S = 80 m, using L = N S² / (1.5 + 0.035 S) with L > S, has length
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The denominator is 1.5 + 2.8.
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Answer: A. 104 m
L = 0.07 × 6400 / (1.5 + 0.035 × 80) = 448 / 4.3 = 104.2 m.
54. The design of a valley curve is governed by
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Night driving is critical.
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Answer: D. headlight sight distance at night and comfort against centrifugal force
Valley curves are checked for headlight sight distance and for rider comfort at the bottom, where vertical centrifugal force adds to weight.
55. A two-lane pavement (n = 2, wheel base l = 6.1 m) on a curve of R = 150 m, V = 60 km/h needs total extra widening of about
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Add mechanical and psychological parts.
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Answer: C. 0.76 m
Mechanical = n l² / (2R) = 2 × 37.2 / 300 = 0.248 m; psychological = V / (9.5 √R) = 60 / (9.5 × 12.25) = 0.516 m; total 0.76 m.
56. Psychological widening on a curve is provided to
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The cause is driver behaviour.
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Answer: C. give extra clearance because drivers drift from the lane edge at higher speed
Mechanical widening is due to the off-tracking of rear wheels; psychological widening gives extra space for the driver's lateral variability at speed and curvature.
57. On a horizontal curve of R = 300 m, SSD S = 150 m, the clear setback from the centre line of the inside lane for a single-lane road (m = R − R cos(S/2R)) is about
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Use the angle in radians.
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Answer: B. 9.3 m
α/2 = S / (2R) = 0.25 rad; m = 300 (1 − cos 0.25) = 300 × 0.0311 = 9.33 m.
58. By the rational method, Q = C i A / 360 (A in ha, i in mm/h). For C = 0.5, i = 80 mm/h and A = 20 ha, the peak runoff is
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Check unit consistency: ha and mm/h.
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Answer: A. 2.22 m³/s
Q = 0.5 × 80 × 20 / 360 = 2.22 m³/s.
59. A rectangular lined side drain 0.6 m wide, flowing 0.3 m deep, slope 0.01, n = 0.015. Discharge by Manning's formula is about
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Hydraulic radius = area / wetted perimeter.
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Answer: D. 0.34 m³/s
A = 0.18 m²; P = 1.2 m; R = 0.15 m; V = (1/0.015) × 0.15^(2/3) × 0.01^(1/2) = 1.88 m/s; Q = 0.34 m³/s.
60. On a hill road, a catch water drain is provided
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Where does the extra water come from?
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Answer: D. above the top of the cut slope to intercept surface water from the hillside
A catch-water drain on the uphill side stops runoff from flowing over the cut slope and into the road, preventing erosion and slips.
61. A hairpin bend on a hill road should have
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Think about trucks turning tightly.
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Answer: C. a gentle gradient (not steeper than about 2.5%) on the curve
Hairpins are given a flat gradient (about 1 in 40 or 2.5%) through the bend since vehicles need extra traction at tight radii.
62. On the hill side of a hill road, the wall provided to support the cut face of the slope is called a
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The hill face is like a chest.
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Answer: B. breast wall
A breast wall supports a cut slope above the road; a retaining wall supports fill on the valley side.
9.3 Highway materials
32 questions · ACiE0903
63. The ability of road aggregate to resist fracture under sudden shock or repeated blows is measured by the
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Hammer blows are used.
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Answer: A. aggregate impact value test
The impact test drops a 13.5-14 kg hammer 15 times on aggregate and measures the fines produced, showing toughness against shock.
64. In the Los Angeles abrasion test, the abrasion value is the percentage of the original sample that passes which IS sieve after the test?
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It is a fine sieve, much smaller than 4.75 mm.
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Answer: D. 1.70 mm
The sample is tumbled with steel balls in a drum for 500 revolutions (grades A-D); the loss is the portion passing the 1.70 mm sieve.
65. The soundness test on road aggregate assesses its
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It is a durability test.
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Answer: C. resistance to weathering, using cycles of sodium or magnesium sulphate
Aggregate is soaked in sulphate solution and oven-dried for five cycles; the weight loss measures its durability against freeze-thaw or salt action.
66. In the shape test, an aggregate particle is classed as flaky if its least dimension is less than
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Flaky means thin.
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Answer: B. 0.6 times its mean sieve size
IS 2386 defines flaky particles as having thickness less than 0.6 × mean dimension; elongated ones have length greater than 1.8 × mean dimension.
67. The stripping value test (static immersion) determines the
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It involves coating and soaking.
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Answer: B. affinity of aggregate for bitumen and its resistance to water displacement
Aggregate coated with bitumen is immersed in water for 24 h at 40°C; the percentage of stripped area indicates adhesion.
68. Skid resistance of the wearing course aggregate is judged by its
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Think of wet, polished surfaces.
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Answer: A. polished stone value (PSV)
High PSV means the aggregate resists polishing by traffic and keeps surface texture, hence good wet-weather skid resistance.
69. A gradation with a wide range of particle sizes, giving minimum voids and a high density, is called
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Opposite of single-size.
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Answer: D. dense (well) graded
Dense graded aggregate has a continuous distribution from coarse to fine particles, with voids filled by smaller particles.
70. Bitumen differs from tar mainly because bitumen
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Look at the source of each.
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Answer: C. is obtained as residue from crude petroleum distillation
Bitumen is a petroleum-derived binder; tar is obtained by destructive distillation of coal or wood and is more temperature susceptible.
71. In the standard penetration test of bitumen, a penetration value of 65 means that the needle penetrated
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The unit is a tenth of a millimetre.
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Answer: C. 6.5 mm
The needle (100 g, 5 s, 25°C) penetration is measured in units of 0.1 mm, so 65 = 6.5 mm. Harder bitumen has lower penetration.
72. The softening point of bitumen is determined by the
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A ball and a ring.
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Answer: B. ring and ball test, heating in water at 5°C per minute
A steel ball placed on a bitumen disc in a ring sinks as bitumen softens; the temperature when it touches the plate 25 mm below is the softening point.
73. The ductility test of bitumen is carried out at 27°C with the briquette pulled at
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Roughly 5 cm each minute.
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Answer: A. 50 mm per minute
Ductility is the length in cm to which a bitumen thread stretches before breaking, at 27°C and 50 mm/min; it indicates cohesion and flexibility.
74. The fire point of bitumen is
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Burning requires more heat than a flash.
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Answer: D. higher than the flash point, when the sample burns for at least 5 seconds
Flash point: vapours momentarily flash when a flame passes; fire point: vapours ignite and keep burning for at least 5 s. Fire point is always higher.
75. The grade VG-30 bitumen indicates an absolute viscosity of about
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The number is roughly the viscosity in hundreds of poise.
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Answer: D. 3000 poise at 60°C
Viscosity grades are based on absolute viscosity at 60°C: VG-10 about 1000, VG-20 about 2000, VG-30 about 3000, VG-40 about 4000 poise.
76. Medium curing (MC) cutback bitumen is bitumen that has been diluted with
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It is a middle volatility solvent.
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Answer: C. kerosene
RC uses gasoline or naphtha, MC uses kerosene, SC uses heavy oils; the thinner evaporates and leaves the bitumen binder.
77. A cationic bitumen emulsion is preferred with siliceous aggregates because
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Opposites attract.
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Answer: B. positively charged bitumen droplets are attracted to the negatively charged aggregate surface
Siliceous aggregates are electro-negative; cationic droplets are attracted and also work on damp surfaces.
78. A standard Marshall specimen has dimensions of
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It is 4 inches by 2.5 inches.
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Answer: A. 101.6 mm diameter and 63.5 mm height
The Marshall mould is 4 in diameter (101.6 mm) and 2.5 in tall (63.5 mm), compacted with a 4.54 kg hammer dropped from 457 mm (75 blows each face for heavy traffic).
79. In Marshall design, a very low air-void content in the compacted asphalt mixture mainly leads to
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Bitumen needs room.
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Answer: A. bleeding and rutting in service
Too few voids leave no room for bitumen expansion and further traffic compaction, so the mix flushes and ruts; high voids cause permeability and ageing.
80. The loading rate used for testing a Marshall specimen at 60°C is
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About 2 inches per minute.
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Answer: D. 50.8 mm per minute
The Marshall test head is loaded at 50.8 mm (2 in) per minute after 30-40 minutes in the 60°C water bath; max load is stability and deformation is flow.
81. In the CBR test the plunger has a diameter of 50 mm and penetrates the soil at
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Slow, controlled penetration.
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Answer: C. 1.25 mm per minute
The CBR plunger is 50 mm diameter and is pushed at 1.25 mm/min; loads at 2.5 mm and 5.0 mm penetration are compared to standard crushed stone loads.
82. In the AASHTO soil classification, the best group for subgrade use is
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Smallest numbers are the best.
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Answer: B. A-1-a
A-1 soils (stone fragments, gravel and sand) are excellent subgrades; A-7 clays rate poorest.
83. Using Fuller's maximum density curve P = 100 (d/D)^0.5, for a maximum aggregate size D = 20 mm, the percentage passing the 10 mm sieve is
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Take the square root of 0.5.
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Answer: B. 70.7%
P = 100 × (10/20)^0.5 = 100 × 0.707 = 70.7%.
84. A 2000 g aggregate sample has 480 g of flaky particles. The flakiness index is
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Weight of flaky divided by total weight.
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Answer: A. 24%
Flakiness index = 480 / 2000 × 100 = 24%.
85. In an aggregate crushing value test, 2.50 kg of aggregate is crushed and 0.70 kg passes the 2.36 mm IS sieve. The ACV is
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Passing weight over original weight.
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Answer: D. 28%
ACV = 0.70 / 2.50 × 100 = 28%.
86. Oven-dried aggregate weighs 2000 g and the saturated surface-dry weight is 2020 g. The water absorption is
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Divide by the dry weight.
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Answer: C. 1.0%
Absorption = (2020 − 2000) / 2000 × 100 = 1.0%.
87. Marshall tests give bitumen contents of 4.9% at maximum stability, 5.3% at maximum bulk density and 5.1% at 4% air voids. The optimum bitumen content is
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Average the three.
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Answer: C. 5.1%
OBC is the average of the three contents: (4.9 + 5.3 + 5.1)/3 = 5.1%.
88. A mix has 5% bitumen (Gb = 1.02) and 95% aggregate (G = 2.65) by weight. The theoretical maximum specific gravity of the mix is
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Use 100 / sum of (weight / specific gravity).
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Answer: B. 2.45
Gt = 100 / (5/1.02 + 95/2.65) = 100 / (4.90 + 35.85) = 2.454.
89. The mix above (Gt = 2.454) has a bulk specific gravity Gm = 2.36. The air-void content is about
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Compare compacted and theoretical densities.
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Answer: A. 3.8%
Vv = (Gt − Gm)/Gt × 100 = (2.454 − 2.36)/2.454 × 100 = 3.8%.
90. The mix above also has Vb = 11.6% (volume of bitumen). With Vv = 3.8%, the voids in mineral aggregate (VMA) are about
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VMA is the sum of air and bitumen volumes.
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Answer: D. 15.4%
VMA = Vv + Vb = 3.8 + 11.6 = 15.4%.
91. In a CBR test, the load at 2.5 mm penetration is 0.85 kN and at 5.0 mm is 1.10 kN (standard loads 13.44 kN and 20.16 kN). The CBR value is
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Compute both and take the higher unless re-test.
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Answer: D. 6.3%
2.5 mm: 0.85/13.44 = 6.3%; 5 mm: 1.10/20.16 = 5.5%; the 2.5 mm value (higher) governs.
92. As per IRC:37, the resilient modulus of a subgrade with a CBR of 8% is, using MR = 17.6 (CBR)^0.64, about
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The power formula applies above 5%.
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Answer: C. 67 MPa
17.6 × 8^0.64 = 17.6 × 3.78 = 66.6 MPa. (For CBR ≤ 5%, MR = 10 CBR.)
93. A 75 cm plate-load test gives a pressure of 0.07 N/mm² at a deflection of 1.25 mm. The modulus of subgrade reaction k is
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Pressure divided by deflection.
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Answer: B. 0.056 N/mm³
k = 0.07 / 1.25 = 0.056 N/mm³ (56 MPa/m).
94. A soil has 60% passing the 0.075 mm sieve, LL = 45% and PI = 15%. Using GI = 0.2a + 0.005ac + 0.01bd (a = F−35, b = F−15, c = LL−40, d = PI−10), the group index is
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Compute a, b, c, d first.
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Answer: A. 8
a = 25, b = 45, c = 5, d = 5; GI = 5 + 0.625 + 2.25 = 7.9, i.e. 8.
9.4 Traffic engineering and safety
32 questions · ACiE0904
95. PIEV time of a driver stands for
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Four mental stages from seeing to acting.
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Answer: A. perception, intellection, emotion and volition
The PIEV time is the interval between seeing an obstruction and starting to apply the brakes; a value of 2.5 s is used in design.
96. The Passenger Car Unit (PCU) is used in traffic studies to
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The car is the reference vehicle.
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Answer: D. convert a mixed traffic stream into equivalent passenger cars
Each vehicle type is assigned a factor relative to a car (1.0), so volumes and capacities can be expressed in a single unit.
97. Which shape is normally used for regulatory traffic signs such as speed limit or no-entry signs?
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Compulsory signs use a round shape.
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Answer: C. Circular with red border
Regulatory signs are circular (the stop sign is octagonal and give-way inverted triangle); warning signs are triangular and informatory signs are rectangular.
98. A continuous double line along the centre of the road means that
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A solid line is a barrier.
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Answer: B. overtaking is prohibited for traffic in both directions
A solid centre line prohibits overtaking or crossing it; broken lines allow overtaking when it is safe.
99. One-way streets increase the capacity of an urban network mainly because they
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Think of conflict points.
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Answer: B. remove opposing and crossing conflicts at intersections
With traffic in only one direction, right-turn (opposing) conflicts vanish and signal coordination is easier, so capacity rises.
100. The speed at which 85% of vehicles travel at or below is used as
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It is a percentile, not an average.
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Answer: A. the upper limit for speed regulation
The 85th percentile speed is a standard basis for setting upper speed limits; the 15th percentile is used for lower limits.
101. The peak hour factor (PHF) is defined as
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It cannot exceed 1.
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Answer: D. peak-hour volume divided by 4 times the maximum 15-minute volume
PHF = V / (4 × V15max). It is 1.0 for perfectly uniform flow and falls as peaking within the hour increases.
102. In an origin and destination survey, the roadside interview method
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It involves stopping vehicles.
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Answer: C. asks drivers of stopped vehicles at a selected point about their origin and destination
Drivers are stopped briefly and interviewed; other methods are home interview, licence plate matching, tags and postcards.
103. The maximum number of vehicles that can pass a section in one hour under ideal roadway and traffic conditions is the
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It is the theoretical maximum.
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Answer: C. basic capacity
Basic capacity applies to ideal conditions; possible capacity is under prevailing conditions; practical capacity is the volume at the design level of service.
104. Level of service A on a highway describes
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The letter A is the best grade.
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Answer: B. free flow with low volumes and high speeds
LOS A is the best condition, with drivers free to choose speed; LOS F is breakdown flow.
105. Among different parking arrangements along a kerb, the arrangement that accommodates the maximum number of vehicles per unit length of kerb is
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Cars side by side, nose-in.
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Answer: A. right-angle (90°) parking
90° parking uses less kerb length per car but needs the greatest depth of road and makes manoeuvring harder.
106. Parking turnover is defined as
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It measures how often bays change vehicles.
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Answer: D. the number of vehicles parked per bay over a given period
Turnover = number of vehicles parked / (number of bays × duration). Occupancy is the share of bays used; accumulation is the number parked at an instant.
107. A collision diagram in accident analysis shows
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It is about movements in the crash.
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Answer: D. the type and location pattern of accidents at a spot
Collision diagrams use arrows to show how vehicles collided; condition diagrams show the physical features (kerbs, signs, road width) of the site.
108. The four E's of road safety are
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Law and teaching are two of them.
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Answer: C. engineering, education, enforcement and emergency care
Road safety programmes combine engineering measures, education, enforcement of traffic laws and emergency medical care.
109. The number of conflict points at an uncontrolled four-leg intersection with two-way traffic is
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More legs means many more conflicts.
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Answer: B. 32
A four-leg two-way intersection has 32 conflicts (8 diverging, 8 merging and 16 crossing); a T-junction has 9.
110. A rotary intersection mainly converts crossing conflicts into
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Vehicles circulate in one direction.
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Answer: A. weaving (merging and diverging) conflicts
All vehicles move in one direction around the island, so crossing movements become weaving movements, which are less severe.
111. In Webster's method, the optimum signal cycle length is Co = (1.5 L + 5)/(1 − Y), where L is the
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It is time that is wasted.
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Answer: A. total lost time per cycle
L is the sum of lost times (start-up plus clearance) over all phases in a cycle; Y is the sum of critical flow ratios.
112. Which of the following improves night-time visibility of a road?
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They bounce light back to its source.
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Answer: D. Retro-reflective road markings and cat's-eye studs
Retro-reflective materials return headlight beams to the driver; studs and delineators give guidance in the dark and rain.
113. The main purpose of an anti-glare screen on a median is to
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It is about opposing headlights.
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Answer: C. reduce dazzle from the headlights of opposing vehicles
The screen blocks high-beam glare from opposing traffic at night, preserving the driver's vision.
114. Crash barriers (guard rails) are mainly installed on
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Think of where a vehicle might fall.
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Answer: B. embankments, sharp curves and bridge approaches to prevent vehicles leaving the road
Barriers redirect errant vehicles and reduce the severity of run-off-road crashes at high-risk locations.
115. The peak hour volume at a section is 1200 veh/h and the maximum 15-minute volume in that hour is 340 vehicles. The peak hour factor is
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Multiply the 15-minute count by 4.
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Answer: B. 0.88
PHF = 1200 / (4 × 340) = 0.882.
116. If the centre-to-centre spacing of vehicles is 25 m, the basic capacity of a lane at 40 km/h is
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Use capacity = 1000 V / S.
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Answer: A. 1600 vehicles per hour
Capacity = 1000 V / S = 1000 × 40 / 25 = 1600 veh/h.
117. A traffic stream has a density of 40 veh/km and a space mean speed of 55 km/h. The flow is
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Flow = density × speed.
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Answer: D. 2200 veh/h
q = k × u = 40 × 55 = 2200 veh/h.
118. In Greenshields' model, u = uf (1 − k/kj), with free flow speed 80 km/h and jam density 120 veh/km. The maximum flow is
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Maximum occurs at half the jam density.
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Answer: C. 2400 veh/h
qmax = uf kj / 4 = 80 × 120 / 4 = 2400 veh/h, at k = 60 veh/km and u = 40 km/h.
119. Spot speeds of four vehicles are 20, 40, 50 and 60 km/h. The space mean speed is about
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Use the harmonic mean.
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Answer: C. 35.8 km/h
Harmonic mean = 4 / (1/20 + 1/40 + 1/50 + 1/60) = 4 / 0.1117 = 35.8 km/h. The arithmetic mean is 42.5.
120. A road section of 10 km with an AADT of 8000 vehicles recorded 12 accidents in a year. The accident rate per 100 million vehicle-km is about
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Vehicle-km per year = AADT × 365 × length.
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Answer: B. 41
Rate = 12 × 10⁸ / (8000 × 365 × 10) = 41.1.
121. A rotary has weaving width w = 10 m, average entry width e = 7 m, weaving length l = 40 m, and proportion of weaving traffic p = 0.4. The practical weaving capacity Qp = 280 w (1 + e/w)(1 − p/3)/(1 + w/l) is
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Compute each bracket separately.
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Answer: A. 3300 PCU/h
Qp = 280 × 10 × 1.7 × 0.8667 / 1.25 = 3300 PCU/h.
122. A 3-phase signal has lost time of 12 s per cycle and sum of critical flow ratios Y = 0.55. Webster's optimum cycle length is
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Use Webster's formula.
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Answer: D. 51 s
Co = (1.5 × 12 + 5)/(1 − 0.55) = 23/0.45 = 51.1 s.
123. From the same signal (Co = 51 s, L = 12 s, Y = 0.55) a phase has y = 0.30. Its effective green time is about
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Distribute the effective green in proportion to y.
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Answer: D. 21 s
Total effective green = Co − L = 39 s; g = (0.30/0.55) × 39 = 21.3 s.
124. A pedestrian crossing is 14 m wide. Using 7 s plus 1.2 m/s walking speed, the minimum pedestrian green time is about
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Walking time is distance divided by speed.
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Answer: C. 18.7 s
Gp = 7 + d / 1.2 = 7 + 14 / 1.2 = 18.7 s.
125. A peak-hour stream has 600 cars, 100 trucks (3.0 PCU each) and 200 two-wheelers (0.5 PCU each). The total flow in PCU/h is
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Weight each class by its PCU.
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Answer: B. 1000
600 × 1 + 100 × 3 + 200 × 0.5 = 600 + 300 + 100 = 1000 PCU/h.
126. 120 vehicles used 20 parking bays over an 8-hour period. The parking turnover is
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Divide by bays and hours.
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Answer: A. 0.75 vehicles per bay per hour
Turnover = 120 / (20 × 8) = 0.75 veh/bay/h.
9.5 Road pavement
30 questions · ACiE0905
127. In a flexible pavement, loads are transferred to the subgrade mainly by
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It is a layered, granular mechanism.
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Answer: A. grain-to-grain contact, spreading the load over a wider area with depth
Flexible pavements have little flexural rigidity; the wheel load is dissipated through the layers by particle contact, so stress at the subgrade is low.
128. A rigid pavement distributes wheel loads over a large subgrade area mainly through its
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Think of a stiff plate.
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Answer: D. flexural strength (slab action)
A concrete slab is stiff and acts as a beam or plate on an elastic foundation, so design is governed by flexural stresses.
129. The correct order of layers in a flexible pavement from top to bottom is
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Strongest material is at the top.
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Answer: C. surface course, base course, sub-base course, subgrade
Layer quality decreases with depth: the best materials are near the top where stresses are highest.
130. A prime coat is applied
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It is applied on an unbound granular surface.
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Answer: B. on a granular base before laying a bituminous layer, to seal it and bond with the layer
Prime coat is a low-viscosity bitumen sprayed on the base to penetrate it, plug capillary voids and give a bond to the next bituminous layer. A tack coat bonds new bituminous layer to an existing bituminous or concrete surface.
131. The standard axle adopted in IRC and Nepal DoR flexible pavement design is a single axle of
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About 8.2 tonnes.
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Answer: B. 80 kN (8160 kg) with dual wheels
Traffic is converted into the number of equivalent 80 kN single-axle passes (msa) using the vehicle damage factor.
132. According to the fourth-power law, a doubling of the axle load increases pavement damage by a factor of about
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Raise 2 to the fourth power.
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Answer: A. 16
Relative damage ∝ (P/Ps)⁴, so doubling load gives 2⁴ = 16 times the damage.
133. Vehicle damage factor (VDF) is the
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It converts vehicles to standard axles.
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Answer: D. number of standard axle passes equivalent in damage to one pass of the vehicle
Each commercial vehicle is converted to equivalent standard axles by its VDF (damage per vehicle), so mixed traffic can be summed.
134. For a dual-wheel assembly with centre-to-centre spacing S, the equivalent single wheel load (ESWL) at depths greater than 2S is
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The wheels behave together at depth.
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Answer: C. twice the load on one wheel
At great depth the stress bulbs overlap completely so the two wheels act as a single wheel carrying the total load; near the surface ESWL equals a single wheel load.
135. The tyre contact area of a wheel is approximated as a circle; its contact pressure is approximately
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Think of a balloon pressing on a floor.
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Answer: C. equal to the tyre inflation pressure
Contact pressure is generally taken as equal to or slightly above the inflation pressure; contact area = load / pressure.
136. The subgrade strength parameter used in flexible pavement design by the CBR method is the
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It is a penetration test.
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Answer: B. soaked California Bearing Ratio
Flexible design charts (IRC 37 and the Nepal DoR guideline) use the design CBR of the subgrade; rigid pavement design uses the modulus of subgrade reaction k.
137. The assumption of Boussinesq's theory for stress distribution in a pavement is that the soil is
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The medium extends infinitely in depth.
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Answer: A. a homogeneous, isotropic, linearly elastic half-space
Boussinesq considers a point load on the surface of a semi-infinite, homogeneous, isotropic elastic medium; Burmister extended this to layered systems.
138. The three critical load positions considered in Westergaard's analysis of a rigid pavement slab are
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Positions on the plan of a slab.
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Answer: D. interior, edge and corner
Westergaard derived stress and deflection formulas for loads at the interior, the edge and the corner of a slab.
139. The radius of relative stiffness l of a concrete slab on a subgrade increases with
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l is proportional to h to the power 3/4.
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Answer: D. an increase in the slab thickness
l = [E h³ / (12 (1 − μ²) k)]^(1/4): it grows with E and h and decreases with k.
140. During the daytime, when the top of a concrete slab is warmer than the bottom, warping causes
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The hot top wants to expand.
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Answer: C. tension at the bottom of the slab (interior) restrained by its weight
The top expands more than the bottom, so the slab tends to curl down at the edges; the self weight restrains this, giving tension at the bottom and compression at the top.
141. Dowel bars in transverse joints of a concrete pavement are provided to
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They allow slabs to slide along their length.
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Answer: B. transfer load across the joint while permitting movement
Dowels are smooth bars placed along the traffic direction to give load transfer; tie bars at longitudinal joints are deformed bars that keep slabs from separating.
142. Tie bars in a longitudinal joint of a concrete pavement are used to
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They tie, not slide.
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Answer: A. keep the adjacent slabs from drifting apart
Tie bars are deformed steel bars across longitudinal joints; they hold slabs together without providing much load transfer.
143. Frictional restraint of the subgrade against shrinkage of a concrete slab is considered in the design of
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Longer slabs have more restraint.
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Answer: A. steel reinforcement for contraction between joints
In reinforced slabs the steel resists the friction pull as the slab shortens: As = f W L / (2 s).
144. Pumping in rigid pavements is the
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It is related to water and joints.
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Answer: D. ejection of water and fine subgrade material through joints under traffic loads
Pumping occurs when water collects beneath slab edges and is forced out with fines by moving loads, leaving voids and leading to faulting.
145. Compared with flexible pavements, rigid pavements generally have
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Pay now or pay later.
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Answer: C. higher initial cost, lower maintenance and longer design life
Concrete pavements cost more at the start but last about 30 years or more with little maintenance.
146. Rutting in a flexible pavement is mainly caused by
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Look at wheel tracks.
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Answer: B. permanent deformation in the wheel paths of the subgrade or bituminous layers
Repeated loading causes cumulative permanent deformation (shear flow in mix or subgrade compression), forming longitudinal depressions.
147. An axle load of 120 kN is applied to a pavement. Relative to an 80 kN standard axle, its damage factor by the fourth-power law is
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Raise the load ratio to the 4th power.
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Answer: B. 5.06
VDF = (120/80)⁴ = 1.5⁴ = 5.06.
148. A road has initial commercial traffic A = 800 CV/day, lane distribution factor D = 0.75, VDF F = 3.5, growth rate r = 7.5% and design life n = 15 years. Using N = 365 [(1+r)ⁿ − 1]/r × A × D × F, the design traffic is approximately
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Compute the growth factor first.
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Answer: A. 20 msa
Growth factor = (1.075¹⁵ − 1)/0.075 = 26.12; N = 365 × 26.12 × 800 × 0.75 × 3.5 = 20.0 × 10⁶ standard axles.
149. The traffic growth factor [(1+r)ⁿ − 1]/r for r = 5% and a design period of 10 years is
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Compute 1.05 to the 10th power.
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Answer: D. 12.58
(1.05¹⁰ − 1)/0.05 = (1.6289 − 1)/0.05 = 12.58.
150. A concrete slab has E = 30,000 MPa, h = 250 mm, μ = 0.15 and k = 0.05 N/mm³. The radius of relative stiffness l = [E h³ / (12 (1 − μ²) k)]^(1/4) is about
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Take the fourth root of the whole ratio.
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Answer: C. 946 mm
E h³ = 30000 × 1.5625×10⁷ = 4.69×10¹¹; denominator = 12 × 0.9775 × 0.05 = 0.5865; ratio = 7.99×10¹¹; fourth root = 946 mm.
151. A circular load of intensity 0.6 MPa and radius 150 mm acts on the surface of a homogeneous half-space. The vertical stress under its centre at 300 mm depth (σz = p [1 − z³/(a² + z²)^1.5]) is about
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Compute the bracket first.
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Answer: C. 0.17 MPa
z³ = 2.7×10⁷; (a² + z²)^1.5 = (112500)^1.5 = 3.773×10⁷; ratio = 0.716; σz = 0.6 × 0.284 = 0.171 MPa.
152. A tyre load of 20 kN acts at a contact pressure of 0.7 MPa. Assuming a circular area, the contact radius is about
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Area = load / pressure, then A = πa².
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Answer: B. 95 mm
Area = 20,000 / 0.7 = 28,571 mm²; a = √(A/π) = √9095 = 95.4 mm.
153. The Westergaard edge warping stress is σ = E α t Cx / 2. For E = 30,000 MPa, α = 10×10⁻⁶ per °C, temperature difference t = 15°C, and Cx = 0.8, the stress is
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Don't forget the factor of 2.
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Answer: A. 1.8 MPa
σ = 30000 × 10×10⁻⁶ × 15 × 0.8 / 2 = 1.8 MPa.
154. A fully restrained concrete slab (E = 30,000 MPa, α = 10×10⁻⁶ per °C) is subjected to a uniform temperature drop of 20°C. The tensile stress developed is
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Full restraint stress equals E times strain.
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Answer: D. 6.0 MPa
σ = E α ΔT = 30000 × 10×10⁻⁶ × 20 = 6.0 MPa.
155. A 200 mm concrete slab (unit weight 24 kN/m³ ≈ 2400 kg/m³) with contraction joints 15 m apart, friction coefficient 1.5 and allowable steel stress 1400 kg/cm². The steel required per metre width (As = f W L/2s) is about
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Find W in kg/m² first.
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Answer: D. 3.86 cm²/m
W = 0.2 × 2400 = 480 kg/m²; As = 1.5 × 480 × 15 / (2 × 1400) = 3.86 cm²/m.
156. The Westergaard corner stress σc = (3P/h²)[1 − (a√2/l)^0.6] for P = 40 kN, h = 250 mm, a = 150 mm and l = 946 mm is about
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Evaluate the bracket separately.
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Answer: C. 1.14 MPa
3P/h² = 120000/62500 = 1.92; (150 × 1.414/946)^0.6 = 0.224^0.6 = 0.41; σ = 1.92 × 0.59 = 1.14 MPa.
9.6 Road construction & maintenance
31 questions · ACiE0906
157. A motor grader is mainly used in road construction for
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It has a long adjustable blade.
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Answer: A. spreading, shaping and fine-grading of materials
The grader's blade spreads and trims embankment, sub-base and base layers to the required levels and camber.
158. Sheepsfoot rollers are best suited to the compaction of
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Feet penetrate and knead.
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Answer: D. cohesive (clayey) soils
The projecting feet knead and compact clay in layers, giving high pressure on a small area; granular soils respond better to vibratory rollers.
159. A bituminous paver is used to
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It places the mix on the road.
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Answer: C. lay the hot mix to uniform thickness and level before compaction
The paver receives hot mix in a hopper, spreads it with augers and screed to a controlled thickness and gives initial compaction.
160. The first step of road subgrade preparation is typically
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Weeds and roots must go first.
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Answer: B. clearing and grubbing, removing vegetation and topsoil
The site is cleared of trees, roots and organic topsoil before excavation or filling; the subgrade is then shaped and compacted.
161. Compaction of soil in the field is controlled by comparing the field dry density with the maximum dry density obtained from the
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OMC and MDD come from it.
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Answer: B. Proctor compaction test
The Proctor test gives the OMC and MDD; field density is expressed as a percentage of MDD (relative compaction).
162. The field dry density of a fine-grained subgrade can be determined by the
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A steel cylinder is driven into the ground.
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Answer: A. core cutter method
The core cutter drives a steel cylinder of known volume into the soil to get density; the sand replacement method is used for granular soils with gravel.
163. Lime stabilisation is most effective for
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Lime reacts with clay minerals.
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Answer: D. plastic clay soils
Lime reduces plasticity index, improves workability and gains strength through slow pozzolanic reactions with clay minerals; cement is more suitable for sands.
164. Cement stabilised soil layers should be compacted
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Time is of the essence once water is added.
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Answer: C. soon after mixing and before the cement begins to set
Delay after adding water and cement breaks the cementation bonds that start forming, reducing strength.
165. In construction of water bound macadam (WBM), the screenings (binding material) are applied
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The fine material goes in after the large stones.
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Answer: C. after the coarse aggregate is rolled dry, then watered and rolled to form slurry
The coarse aggregate is first rolled dry; screenings are then spread and the layer is wetted and rolled to fill voids with slurry.
166. In penetration macadam, the bitumen is
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The binder flows into voids from above.
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Answer: B. sprayed hot over compacted open-graded coarse aggregate, followed by key aggregate
A grout of hot bitumen penetrates the open-graded stone skeleton; the key aggregate is then spread and rolled.
167. Bituminous macadam (BM) differs from dense bituminous macadam (DBM) mainly because BM
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Open grading means more voids.
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Answer: A. uses open-graded aggregate with a low bitumen content
BM is open-graded premixed aggregate with 3.3 to 3.5% bitumen; DBM is dense graded with higher binder content and lower voids.
168. The correct sequence of rolling in compacting an asphalt concrete layer is
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Think of heavy first, smooth last.
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Answer: D. breakdown (initial), intermediate, finish rolling
A heavy steel roller does the breakdown rolling while the mix is hot; a pneumatic roller kneads in the intermediate pass; a smooth steel roller removes marks in the finish pass.
169. A tack coat is applied before laying a new bituminous layer to
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Think of glue between layers.
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Answer: D. bond the new layer to the existing surface
A thin film of bitumen emulsion or cutback is sprayed on the existing bituminous or concrete surface to create the bond.
170. In concrete pavement (PQC) construction, a polythene separation membrane is placed over the sub-base to
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Slab should slide a little.
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Answer: C. reduce friction between the slab and the sub-base
The sheet lets the slab move with temperature and shrinkage, reducing restraint stresses and cracking, and prevents loss of water into the sub-base.
171. Transverse contraction joints in a concrete pavement are normally sawn to a depth of about
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Deep enough to induce a crack, not to cut through.
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Answer: B. one-third of the slab thickness
The sawn groove creates a plane of weakness (about D/4 to D/3) so shrinkage cracks form at the joint rather than randomly.
172. Curing of a newly laid cement concrete pavement is mainly done to
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Hydration requires water.
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Answer: A. retain moisture so that cement can hydrate and reduce shrinkage cracking
Keeping the slab moist (wet burlap, membranes, ponding) allows hydration and prevents plastic shrinkage cracks.
173. Filling of potholes and sealing of cracks in a road are classified as
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Done every year without major works.
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Answer: A. routine maintenance
Routine maintenance covers small, regularly needed works (patching, crack sealing, drain cleaning, shoulder repair); periodic maintenance includes resurfacing.
174. Clearing of landslide debris and repair of flood-damaged sections of a hill road is a case of
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It cannot be planned in advance.
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Answer: D. emergency (urgent) maintenance
Emergency maintenance restores traffic after unforeseen events such as landslides, floods and washouts.
175. Bleeding of a bituminous surface is usually corrected by
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Add a material that soaks up extra bitumen.
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Answer: C. spreading hot coarse sand or fine stone chips and rolling
Excess bitumen is blotted by spreading and rolling in chips or coarse sand, restoring surface friction.
176. In a proper pothole repair, the damaged area is
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Cut neat sides before filling.
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Answer: B. cut to a regular rectangle with vertical sides, cleaned, tacked, filled and compacted
Removing all loose material and giving vertical sides makes a bonded patch; the new mix is placed and compacted in layers.
177. The Benkelman beam is used to measure
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It measures how far the pavement moves.
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Answer: B. the rebound deflection of a pavement under a standard wheel load
Rebound deflection measurements are used to compute characteristic deflection and design the overlay for rehabilitation.
178. A roller of effective width 2.0 m moves at 3 km/h, compacts a 0.15 m lift in 8 passes, working efficiency 0.8. The output in compacted m³ per hour is
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Convert km/h to m/h.
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Answer: A. 90
Q = W × V × t × E / N = 2.0 × 3000 × 0.15 × 0.8 / 8 = 90 m³/h.
179. Tack coat is applied at 0.25 kg/m² on a 7 m wide, 1000 m long section. The quantity of binder required is
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Area multiplied by spray rate.
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Answer: D. 1750 kg
Area = 7 × 1000 = 7000 m²; quantity = 7000 × 0.25 = 1750 kg.
180. A 75 mm thick dense bituminous macadam layer is laid over 7.0 m × 1000 m with compacted density 2.35 t/m³. The quantity of mix needed is
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Volume times density.
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Answer: C. 1234 t
Volume = 7 × 1000 × 0.075 = 525 m³; mass = 525 × 2.35 = 1233.75 t.
181. A concrete pavement 7.5 m wide, 1000 m long and 250 mm thick requires concrete volume of
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Width times length times thickness.
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Answer: C. 1875 m³
Volume = 7.5 × 1000 × 0.25 = 1875 m³.
182. The zero-air-void dry unit weight for G = 2.7 at a water content of 15% is (γw = 9.81 kN/m³)
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Use 1 + wG in the denominator.
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Answer: B. 18.85 kN/m³
γd = G γw / (1 + w G) = 2.7 × 9.81 / (1 + 0.15 × 2.7) = 26.49 / 1.405 = 18.85 kN/m³.
183. A field soil has a wet unit weight of 20.4 kN/m³ at 12% water content. Its MDD is 19.0 kN/m³. The relative compaction is about
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Find dry unit weight first.
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Answer: A. 95.9%
γd = 20.4 / 1.12 = 18.21 kN/m³; relative compaction = 18.21/19.0 × 100 = 95.9%.
184. In a sand replacement test, the sand filling the hole weighs 1.62 kg (sand bulk density 1.45 g/cm³). The excavated soil weighs 2150 g with 10% water content. The field dry density is about
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Divide the bulk density by 1 + w.
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Answer: D. 1.75 g/cm³
Volume = 1620/1.45 = 1117 cm³; bulk density = 2150/1117 = 1.92 g/cm³; dry density = 1.92/1.10 = 1.75 g/cm³.
185. To raise the water content of soil at a density of 18 kN/m³ (dry) from 9% to the OMC of 14%, the water to be added per m³ of compacted soil is about
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Water difference of 5% on the dry mass.
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Answer: D. 92 litres
Dry mass = 18/9.81 × 1000 = 1835 kg/m³; water = 0.05 × 1835 = 91.7 kg = about 92 L.
186. A 150 mm thick, 7.0 m wide, 1000 m long layer of soil with dry density 1.8 t/m³ is stabilised with 6% cement by dry weight. The cement required is
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Find the dry soil mass first.
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Answer: C. 113.4 t
Dry soil = 7 × 1000 × 0.15 × 1.8 = 1890 t; cement = 0.06 × 1890 = 113.4 t.
187. A Benkelman beam survey gives rebound deflections with a mean of 1.1 mm and a standard deviation of 0.2 mm. The characteristic deflection (mean + 2σ) used for a national highway overlay design is
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Add two standard deviations.
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Answer: B. 1.5 mm
Characteristic deflection = 1.1 + 2 × 0.2 = 1.5 mm.