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All 99 questions from the WBPSC AE Civil 2014 paper (Assistant Engineer — 2014 (Civil)), with the correct answer and a short explanation under each one. Tap Show answer after you have tried a question. Want exam conditions? Take this paper as a timed mock test — scored +2 / −2/3 like the real exam.
| Paper | Assistant Engineer — 2014 (Civil) |
| Questions | 99 |
| Marking (current pattern) | +2 for a right answer, −2/3 for a wrong one |
| Answers on this page | All 99 |
Q1.Flow through a syphon pipe stops if the summit is ________ from the inlet.
Answer: (c) greater than 7.5m
A siphon carries water over a summit that stands above the hydraulic grade line, so the pressure at the summit is below atmospheric. Applying Bernoulli between the upstream water surface and the summit: ps/γ = − (hs + V²/2g + hf) Water cannot sustain an absolute pressure lower than its vapour pressure — about 2.5 m of water (absolute) at ordinary temperatures, taking dissolved-air release into account.
Q2.Mechanical stabilization of a soil is done with the help of
Answer: (d) proper grading
Mechanical stabilisation improves a soil purely by altering its grading — blending a deficient soil with borrow material so that the mixture has a well-graded particle-size distribution — and then compacting it. No chemical reaction is involved; strength comes from grain interlock (internal friction) supplied by the coarse fraction plus cohesion supplied by a controlled amount of binder fines.
Q3.If the R.L. of a B.M. is 100m, the back-sight is 1.215m and the foresight is 1.870m, the R.L. of the forward station is
Answer: (a) 99.345m
Levelling is bookkeeping about a single horizontal line of sight. Every staff reading is measured down from that line, so a station with a larger staff reading is lower. H.I. = R.L.BM + B.S. R.L.forward = H.I. − F.S. Equivalently, the rise-and-fall check: B.S. − F.S. = 1.215 − 1.870 = −0.655 m, i.e. a fall of 0.655 m, so the forward station lies below the bench mark.
Q4.The flow is laminar if Reynolds number is
Answer: (b) less than 2000
Reynolds number is the ratio of inertia force to viscous force, Re = ρVD/μ = VD/ν. When viscous forces dominate the disturbances are damped out and the flow stays in orderly layers.
Q5.The deflection at the mid point of a simply supported beam of length L under a concentrated load W at centre is
Answer: (b) WL³/48EI
For a simply supported span L with a central point load W, integrating EI d²y/dx² = M (or applying the unit-load method) gives the central deflection δmax = WL³/48EI Two checks that settle the answer without memorising it: Dimensions. Deflection is a length, so the numerator must carry L³ — any option with L² is dimensionally impossible. Comparison.
Q6.(continued) A normally consolidated clay settles 1 cm when effective stress was increased from 100 kN/m² to 200 kN/m². If the effective stress is further increased from 200 kN/m² to 400 kN/m² then settlement of the same clay is
Answer: (a) 1 cm
For a normally consolidated clay the compression curve is a straight line on a semi-logarithmic plot of void ratio against effective stress, so consolidation settlement depends on the ratio of the stresses, not their difference: ΔH = (Cc H)/(1 + e0) · log10(σ′2/σ′1) The first increment doubles the stress (100 → 200) and so does the second (200 → 400). Each is one and the same log-cycle step, log 2 = 0.301, so each produces the same settlement of 1 cm.
Q7.In a flow-net diagram the length of the flow line in the square is 2m, the total head loss is 18m and the number of potential drop is 12. The value of exit gradient
Answer: (b) 0.75
In a flow net the total head loss H is divided equally among the potential drops, so each field loses Δh = H / Nd and the hydraulic gradient in any field is that drop divided by the length of the flow path across the field, i = Δh / l. The exit gradient is simply this quantity evaluated in the last field at the downstream exit face, where l is smallest and i therefore largest — which is exactly why piping starts there.
Q8.The minimum centre to centre distance between the piles of a pile group in clay should be equal to the
Answer: (c) 3 times the diameter of pile
Piles driven close together in clay interfere in two ways: the stress bulbs overlap so the group acts as a single block, and driving remoulds and heaves the surrounding clay. IS 2911 (Part 1) therefore fixes minimum centre-to-centre spacings: Friction piles (the normal case in clay): 3d, where d is the shaft diameter. End-bearing piles passing through relatively compressible strata: 2.5d. End-bearing piles resting on hard rock: 2d may be permitted.
Q9.A rectangular beam is to be cut from a circular log of wood of diameter D. The depth of the strongest section is
Answer: (d) D√(2/3)
The strongest rectangular beam is the one with the largest section modulus Z = bd²/6, cut from a log of diameter D so that b² + d² = D². Substituting b² = D² − d²: Z ∝ d(D² − d²)½·d → maximise b(D² − b²) Differentiating with respect to b gives D² − 3b² = 0, so b = D/√3 and d = √(D² − D²/3) = D√(2/3) = 0.816 D. The strongest section has the proportion d : b = √2 : 1.
Q10.Chezy's formula for discharge through an open channel is Q = AC√(mi) where m is
Answer: (c) hydraulic mean depth
In Q = AC√(mi), the symbol m is the hydraulic mean depth (also called hydraulic radius, R): m = A / P = cross-sectional area of flow ÷ wetted perimeter Chezy's equation comes from balancing the gravity component of the water weight along the channel against the boundary shear: τ0 = γ m i, and with τ0 = kρV² this gives V = C√(mi). Here C is Chezy's constant (dimensions L½T−1, typically 40–70) and i is the bed slope.
Q11.A triangular DRH due to a storm has a time base of 40 hours and peak flow of 20 m³/sec occurring at 20 hrs. from the start. If the catchment area is 14.4 sq. km., the rainfall excess in the storm was
Answer: (b) 10 cm
The area under a direct runoff hydrograph is the volume of direct runoff; dividing it by the catchment area converts it to a depth — the rainfall excess. For a triangular DRH the area is simply ½ × base × peak, so the peak time does not enter the volume calculation at all (it only fixes the shape).
Q12.If a 4 hr. unit hydrograph on a certain basin has a peak ordinate of 80 m³/s, the peak ordinate of a 2-hr. unit hydrograph for the same basin will be
Answer: (b) greater than 80 m³/s
A unit hydrograph always represents 1 cm of rainfall excess, whatever its duration. Compressing the same 1 cm into 2 hours instead of 4 hours doubles the intensity of the effective rainfall, so the runoff arrives faster and more concentrated: the peak is higher, the time to peak shorter, and the time base slightly shorter. The general rule: as the unit duration decreases, the peak of the unit hydrograph increases.
Q13.A cantilever beam of length L carries a point load W at the free end. The strain energy stored is
Answer: (b) W²L³/6EI
Strain energy stored in bending is U = ∫0L M² dx / (2EI) For a cantilever with a load W at the free end, measuring x from the free end, M = Wx, so U = ∫0L W²x² dx /(2EI) = W²L³/(6EI). Cross-check with Castigliano's theorem: δ = ∂U/∂W = 2WL³/6EI = WL³/3EI, the standard tip deflection — proof that the coefficient 1/6 is right. Also note that energy must be quadratic in the load (it equals ½Wδ), so any option linear in W is wrong at a glance.
Q14.Orifice meter is used to measure
Answer: (a) discharge in a pipe
An orifice meter is a discharge-measuring device for a pipe running full. A thin plate with a concentric sharp-edged hole (typically d/D = 0.5) is bolted between flanges; the contraction of the jet raises the velocity and drops the pressure, and the differential head read on a manometer gives the flow: Q = Cd · a0a1/√(a1²−a0²) · √(2gh) with Cd ≈ 0.62–0.65.
Q15.Minimum compressive strength of burnt clay bricks shall not be less than
Answer: (b) 10 MPa
Burnt clay bricks are graded by their average compressive strength of a set of five bricks tested in accordance with IS 3495, the class designation being the strength itself. For bricks accepted for ordinary load-bearing brickwork the strength demanded is 10 N/mm² (≈ 105 kg/cm²) — this is the figure quoted for first-class bricks in building specifications, and CPWD/PWD schedules write it as 105 kg/cm².
Q16.Rumble strips are preferred on main roads as they
Answer: (b) produce noise and physical sensation on the steering
Rumble strips are shallow raised or grooved bands laid across the carriageway on the approach to a hazard — a junction, a toll plaza, a school zone, a sharp curve. Their purpose is to alert the driver: the tyres passing over them generate a distinct noise and a vibration felt through the steering wheel and the seat, which is exactly the sensory warning that makes the driver look up and ease off.
Q17.A principal plane contains
Answer: (b) no shear stress
A principal plane is defined by the absence of shear stress on it; the normal stress acting on it is then a principal stress. Everything else follows from that definition, it is not an extra property to be memorised: Because τ = 0, the resultant stress on the plane is purely normal. The two principal stresses in a plane problem turn out to be the algebraic maximum and minimum normal stresses, but each acts on its own plane — no single plane carries both.
Q18.For a hypersonic flow, match number is
Answer: (d) more than 6
Mach number M = V/c, where c = √(kRT) is the speed of sound, measures the compressibility of the flow. The standard regimes are: Incompressible M < 0.3 (density change under 5%) Subsonic 0.3 < M < 1 Transonic M ≈ 1 Supersonic M > 1 Hypersonic M > 6 (some texts use 5) — here shock layers are thin, air dissociates and aerodynamic heating dominates.
Q19.Coefficient of discharge in a nozzle is
Answer: (a) actual discharge/theoretical discharge
The three coefficients of a flow-measuring outlet, with their standard definitions: Coefficient of discharge Cd = Qactual / Qtheoretical Coefficient of velocity Cv = Vactual / Vtheoretical (loss of energy at entry) Coefficient of contraction Cc = avena contracta / aorifice They are linked by Cd = Cv × Cc. All three are less than unity, so every coefficient must be written with the actual quantity on top.
Q20.Number of bricks required for one cubic metre of brickwork are
Answer: (c) 500
Take the modular brick of IS 1077, 190 × 90 × 90 mm. In brickwork each brick carries a 10 mm mortar joint on one face in each direction, so the volume occupied by one brick plus its share of mortar is the nominal size: 200 × 100 × 100 mm = 0.002 m³ Hence the number in 1 m³ = 1/0.002 = 500 bricks. Working from the brick alone (0.19 × 0.09 × 0.09 = 0.001539 m³) would give 650 and would ignore the mortar, which occupies roughly 23% of the volume of the masonry.
Q21.The cable carrying a load of wkN/m of horizontal span is stretched between supports l meter apart. The supports are at same level and central dip is h m. The maximum tension in cable is
Answer: (d) (wl/2)√(1+(l/4h)²)
A cable carrying a load uniform per unit horizontal span hangs in a parabola. Two forces act at a support: The horizontal pull, constant along the cable, obtained by taking moments about the lowest point: H = wl²/8h. The vertical reaction, half the total load: V = wl/2.
Q22.In the quadrilateral bearing system a whole circle bearing of 294°30' can be expressed as
Answer: (b) N 65°30' W
Converting a whole-circle bearing (WCB) to a quadrantal or reduced bearing (RB) needs one rule per quadrant, and the letter pair follows the quadrant the line falls in: 0°–90° (NE): RB = WCB, written N…E 90°–180° (SE): RB = 180° − WCB, written S…E 180°–270° (SW): RB = WCB − 180°, written S…W 270°–360° (NW): RB = 360° − WCB, written N…W Here WCB = 294°30' lies in the fourth (NW) quadrant, so RB = 360° − 294°30' = 65°30', i.e. N 65°30' W.
Q23.The intensity of irrigation means
Answer: (a) percentage of culturable commanded area to be irrigated annually
Take the definitions in order, because the marks hang on which area is in the denominator: GCA — gross commanded area: all land the canal can command topographically, including villages, roads and ponds. CCA — culturable commanded area: the part of the GCA that is fit for cultivation. Intensity of irrigation — the percentage of the CCA that is actually proposed to be irrigated in a year (or in a given crop season).
Q24.If L is the length of transition curve and R is the radius of circular curve, then the shift of the curve is directly proportional to
Answer: (b) 1/R and L²
Inserting a transition curve of length L between the straight and a circular curve of radius R requires the circular curve to be moved bodily inwards by the shift S = L² / (24R) So the shift varies directly as L² and inversely as R. The result comes from the cubic parabola/clothoid: the offset of a transition at its end is L²/6R, and the circular curve produces an offset L²/8R over the same half-length; the difference L²/6R − L²/8R = L²/24R is the shift.
Q25.For a single reinforced beam if concrete reaches its maximum allowable stress earlier than steel, the beam is known as
Answer: (b) over reinforced
The classification depends on which material reaches its permissible stress first, and that in turn depends on where the neutral axis sits relative to the critical (balanced) position xc: Over-reinforced (steel more than balanced, x > xc): concrete reaches its maximum permissible stress first. Failure is by crushing of the concrete — sudden, brittle, without warning. Not permitted by design codes.
Q26.If the time required for 50% consolidation of a remolded sample of clay with single drainage is one minute, then the time required to consolidate the same sample of clay with same degree of consolidation with double drainage in
Answer: (a) 1/4 minute
The whole of consolidation theory is contained in the time factor Tv = cv t / d² where d is the length of the longest drainage path. For a given degree of consolidation, Tv is a fixed number (T50 = 0.197), so t ∝ d². Opening the second face halves the drainage path: with single drainage d = H, with double drainage d = H/2. The time therefore falls to one quarter — 15 seconds, i.e. ¼ minute.
Q27.The phreatic line in an earth dam may be
Answer: (c) parabolic
The phreatic line (top flow line, seepage line) is the uppermost streamline, on which the pressure is atmospheric.
Q28.In a liquid limit test, the moisture content at 10 blows was 70% and that at 100 blows was 20%. Then the liquid limit of the soil, is
Answer: (b) 50%
Casagrande's flow curve is the plot of water content against the logarithm of the number of blows; it is a straight line: w = − IF log10 N + C The liquid limit is the water content corresponding to N = 25 blows, read off this line. Here the two given points are exactly one log-cycle apart, so the flow index is the plain difference of water contents, IF = 70 − 20 = 50%.
Q29.For a 'best' symmetrical trapezoidal section of an open channel with a given area of section and side slopes, which one of the the following statements holds true:
Answer: (a) Half the top width is equal to one of the side slopes
The most efficient (best) section is the one that passes a given discharge with the smallest wetted perimeter, i.e. the greatest hydraulic mean depth. For a trapezoid with fixed area and fixed side slope the conditions come out as: Each sloping side equals half the top width — the section is a half-hexagon when the side slope is also free to vary (60° sides). R = A/P = y/2: the hydraulic mean depth is half the water depth.
Q30.A hydrograph is a plot of
Answer: (c) steam flow against time
A hydrograph is a plot of the discharge (stream flow) of a river against time at a particular gauging section. It shows the complete response: a rising limb, a crest segment, a recession limb, and the base flow beneath it all. The other 'graphs' in hydrology are worth fixing in the same breath: Hyetograph — rainfall intensity vs time (a bar chart). Mass curve — cumulative rainfall or runoff vs time.
Q31.If an upgrade of 1.5% followed by a down grade of 0.5% and rate of change of grade is 0.2% per 20 m chain, then the length of vertical curve is
Answer: (b) 200 m
On a vertical curve the grade changes at a uniform rate r, so the curve length is simply the total change of grade divided by the permitted rate of change: L = N / r The total change of grade N is the algebraic difference. An up-grade is positive and a down-grade negative, so N = (+1.5) − (−0.5) = 2%. At 0.2% per 20 m chain the curve needs 2/0.2 = 10 chains, that is 10 × 20 = 200 m. Because the grades change sign the curve is a summit (crest) curve.
Q32.Radius of equivalent circular contact area of a dual wheel for a 80 KN single axle load with contact pressure of 552 KPa is
Answer: (a) 152 mm
The pavement design load is applied through the tyres of a dual wheel, and for analysis the pair is replaced by a single circular contact area of the same total load and the same tyre pressure.
Q33.Minimum value of Marshall stability for DBM mix
Answer: (d) 820 kg
The Marshall method fixes the bitumen content of a dense mix by testing 101.6 mm diameter, 63.5 mm high compacted briquettes at 60 °C in a loading head at 50 mm/min. Two readings are taken: stability (the maximum load in kg or kN, a measure of resistance to shoving) and flow (the deformation at that load, in units of 0.25 mm).
Q34.Which of the following types of sleepers is preferred?
Answer: (d) Concrete sleeper
Indian Railways has standardised on the prestressed concrete (PSC) sleeper for all main-line track, and it is the sleeper of choice wherever the traffic justifies the cost: Long life, 40–60 years, against 12–15 for treated wood. Heavy — about 270 kg — which gives excellent lateral and longitudinal resistance, essential for long welded rail. Immune to termites, fire, corrosion and moisture; no creep; maintains gauge well.
Q35.Minimum frequency of field density tests for compaction control related to mass earthwork is
Answer: (a) 1 in 5000 m³
Field density testing (sand-replacement or core-cutter, IS 2720 Part 28/29, or a nuclear gauge) is the acceptance test for compaction. The frequency is graded according to how much the layer matters to the pavement: Mass earthwork in embankment: one test per 5000 m³ — a large volume, because the bulk fill deep in the embankment is less critical.
Q36.An isochrone is a line on the basin map
Answer: (c) joining points having equal time of travel of surface runoff to the catchment outlet
An isochrone is a contour of travel time: it joins all points in the catchment from which surface runoff takes the same time to reach the outlet. Isochrones divide the basin into sub-areas A₁, A₂, A₃ …, each contributing during successive one-unit time intervals; that is the basis of the time–area method and of the time of concentration tc, which is the travel time from the hydraulically most remote point.
Q37.The maximum value of centrifugal ratio on roads and railway respectively are taken as
Answer: (c) 1/4 and 1/8
The centrifugal ratio is the ratio of the centrifugal force to the weight of the vehicle: P/W = V²/(gR) Its permitted maximum is fixed by overturning and skidding, and the two modes of transport use different limits: Roads: 1/4 — a road vehicle has a wide wheel base relative to its centre of gravity height and a high tyre–road friction, so a larger ratio is tolerable.
Q38.If the coefficient of permeability and that of percolation of a soil are 1×10⁻³ cm/sec and 2×10⁻³ cm/sec respectively, its void ratio is
Answer: (b) 1.0
Darcy's coefficient of permeability k is defined on the superficial (discharge) velocity — the flow divided by the total area of the soil. Water actually moves only through the voids, so the true seepage velocity is larger in the ratio of total area to void area, i.e. 1/n. The coefficient of percolation is defined on that seepage velocity: kp = k / n Hence n = k/kp = 0.5, and the void ratio follows from e = n/(1 − n).
Q39.The angle of internal friction of clays as per UU tests is usually
Answer: (a) 0° to 5°
An unconsolidated-undrained (UU or Q) triaxial test is run with the drainage valve shut throughout. On a saturated clay the whole of any increase in cell pressure is carried by the pore water (Skempton's B = 1), so the effective stress on the specimen never changes, whatever the confining pressure.
Q40.In an unconfined compression test on a clay specimen of initial volume V and length L, the area of cross-section at failure is taken as
Answer: (a) V/(L-ΔL)
The unconfined compression test is a quick undrained test on a saturated clay. No drainage occurs, so the specimen deforms at constant volume: it shortens by ΔL and bulges outwards to keep V unchanged.
Q41.If the focal length of the object glass is 25 cm and the distance from object glass to the transition axis 15 cm, the additive constant is
Answer: (b) 0.4
In the stadia system with an external-focusing telescope the staff intercept s gives the distance as D = (f/i)·s + (f + d) where f/i is the multiplying constant (fixed at 100 by choosing the stadia interval) and (f + d) is the additive constant: the focal length of the object glass plus the distance from the object glass to the vertical (trunnion) axis of the instrument, because distances are reckoned from that axis.
Q42.According to Terzaghi's theory, the ultimate bearing capacity at ground surface for a purely cohesive soil and for a smooth base of strip footing is (C = Cohesion of Soil)
Answer: (b) 5.14 C
Terzaghi's bearing capacity equation for a strip footing is qu = cNc + γDfNq + 0.5γBNγ For a purely cohesive soil φ = 0, so Nq = 1 and Nγ = 0; and at the ground surface Df = 0, so the surcharge term vanishes too. The whole capacity reduces to qu = cNc, and the value of Nc depends entirely on the roughness assumed for the footing base: Smooth base (the Prandtl–Reissner solution, φ = 0): Nc = π + 2 = 5.14 → qu = 5.14 c.
Q43.The porosity and specific gravity of solid of a sand lying below a masonary dam is 40% and 2.67 respectively. The maximum permissible upward gradient with a factor of safety 4 is
Answer: (a) 0.25
Upward seepage reduces the effective stress in the sand; when the exit gradient reaches the critical hydraulic gradient the effective stress becomes zero and the sand boils (quick condition, piping): ic = (G − 1)/(1 + e) = γ′/γw The void ratio comes from the porosity, e = n/(1 − n) = 0.4/0.6 = 0.667. The permissible gradient is the critical value divided by the factor of safety.
Q44.The specific yield of soil depends upon
Answer: (d) all of the above
Specific yield Sy is the volume of water an unconfined aquifer releases by gravity drainage per unit volume of aquifer, expressed as a percentage.
Q45.A hyetograph is a graph representing
Answer: (b) rainfall intensity with time
A hyetograph is a bar chart of rainfall intensity (mm/h) against time for a single storm at a station. Its ordinate is intensity, its abscissa time, and the area under it equals the total rainfall depth of the storm — which is why it is the standard input for computing rainfall excess by the φ- index or infiltration methods and for constructing design storms.
Q46.Relation between transmissibility (T) and permeability (K) for an aquifer of depth (d), is
Answer: (b) T = K.d
Transmissibility (transmissivity) is the rate at which water passes through the entire saturated thickness of an aquifer under unit hydraulic gradient, per unit width: T = K b (written here as T = K d) where K is the coefficient of permeability (m/day) and d the saturated depth or thickness (m).
Q47.The probability of a 10 cm. rain in 1 hour occuring at a station B is found to be 1/60. What is the probability that a 1 hour rain of magnitude 10 cm. or larger will occur in station B once in 30 successive years is
Answer: (b) 0.307
Occurrences of a rare rainfall in successive years are independent Bernoulli trials with p = 1/60 each year. For exactly one occurrence in n = 30 years the binomial law gives P1 = nC1 p (1 − p)n−1 = n p qn−1 Note the exponent is n − 1 = 29, not 30: one of the thirty years is the year of occurrence. Contrast this with the risk of at least one occurrence, R = 1 − qn = 1 − (59/60)30 = 0.396, which is the assurance formula used in design.
Q48.The disinfection efficiency of chlorine in water treatment
Answer: (d) is reduced by increased pH value
Chlorine added to water hydrolyses to hypochlorous acid, which then dissociates: Cl2 + H2O → HOCl + H+ + Cl−; HOCl ⇌ H+ + OCl− (pKa ≈ 7.5) HOCl is the effective disinfectant — some 80 times more germicidal than the hypochlorite ion OCl−, because the neutral molecule penetrates the negatively charged cell wall. As the pH rises the equilibrium shifts to OCl−: at pH 6 about 97% is HOCl, at pH 7.5 half and half, at pH 9 only about 3%.
Q49.Eutrophication in water bodies are caused due to presence of high levels of
Answer: (a) nitrogen and phosphorous
Eutrophication is the over-enrichment of a lake or slow-moving water body with the plant nutrients nitrogen and phosphorus, coming from agricultural fertiliser runoff, detergents and inadequately treated sewage. Phosphorus is normally the limiting nutrient in fresh water, so a concentration above roughly 0.01–0.02 mg/l of P is enough to trigger it.
Q50.Froude number is
Answer: (b) inertia force/gravity force
Froude number is the similarity parameter for flows in which gravity is the dominant force — free- surface flows, spillways, waves, ship models: Fr = inertia force / gravity force = V/√(gL) Written with the forces themselves, inertia ∝ ρL²V² and gravity ∝ ρL³g, so their ratio is V²/(gL) and the Froude number is conventionally defined as its square root.
Q51.Percentage of alumina in a good brick earth lies between
Answer: (b) 20 to 30%
Good brick earth is a three-component mix, and each component has a job. The standard proportions asked for again and again are: Silica (sand) 50–60% — prevents shrinkage, warping and cracking, and gives the burnt brick its uniform shape. Too much and the brick becomes brittle. Alumina (clay) 20–30% — the plastic fraction. It lets the wet earth be moulded and holds the mass together until vitrification takes over. Too much and the brick shrinks, warps and cracks on drying.
Q52.A canal was designed to supply irrigation needs of 120 hectares of land growing rice of 140 days base period having a delta of 134 cm. In this canal is used to irrigate wheat of base period 120 days having a delta of 52 cm., the area (in hectares) that can be irrigated is
Answer: (a) 2650
This is a duty–delta problem. Delta (Δ) is the total depth of water a crop needs over its whole base period; base period (B) is the number of days that water is required; duty (D) is the area a unit of continuous discharge will mature. They are locked together by Δ = 8.64 B / D ⟹ D = 8.64 B / Δ (Δ in metres, B in days, D in hectares per cumec) The canal's discharge capacity is fixed by the rice design.
Q53.A circular sewer 2 m diameter has to carry a discharge of 2 m³/s when flowing nearly full. What is the minimum required slope to initiate the flow? (Assume Manning's n 0.015)
Answer: (a) 0.00023
For any circular section running full, the hydraulic radius is a fixed fraction of the diameter — this is the single fact the question turns on: R = A/P = (πD²/4) ÷ (πD) = D/4 So for a 2 m sewer, R = 0.5 m — not 1.0 m.
Q54.A rainfall is considered acid rain if the pH of rainwater is
Answer: (b) less than 5.6
Rainfall is never neutral. Atmospheric CO2 dissolves in the falling drop and forms carbonic acid: CO2 + H2O ⇌ H2CO3 ⇌ H+ + HCO3– At the pre-industrial partial pressure of CO2 this equilibrium fixes the pH of clean rain at about 5.6. That, and not 7.0, is therefore the datum.
Q55.Bayd and Foster's method for determination of ESWL based on
Answer: (b) equal vertical stress criterion
An equivalent single wheel load (ESWL) is the load on a single wheel that produces the same effect at a given depth as the dual wheel set actually applied. Which "effect" is matched defines the method. Boyd and Foster chose equal vertical stress and turned it into a straight-line construction on a log–log plot of ESWL against depth: At depth z ≤ d/2 (d = clear gap between the tyres) the stress bulbs have not yet merged, so each wheel acts alone: ESWL = P.
Q56.Quantity of bitumen/10 m² area required for first coat of surface dressing is
Answer: (a) 14 kg
Surface dressing is a thin sprayed-binder-and-chippings seal. In a two-coat dressing the first coat carries the larger chippings (about 19 mm nominal) and therefore needs the heavier spray of binder; the second coat uses smaller chippings and less binder. The MoRTH/IRC spray rates for surface dressing are of the order of 1.0–1.4 kg/m², that is 10–14 kg per 10 m². The first coat of a two-coat dressing is specified at 14 kg per 10 m².
Q57.Early attainment of strength in rapid hardening cement is mainly due to
Answer: (c) tricalcium silicate
Strength in Portland cement comes almost entirely from the two calcium silicates, and they differ only in rate: C3S (tricalcium silicate, alite) — hydrates quickly and supplies practically all the strength up to 28 days. C2S (dicalcium silicate, belite) — hydrates slowly and supplies the later strength beyond 28 days. C3A — flash set and heat, negligible strength. C4AF — colour, negligible strength.
Q58.Minimum time of diverstion of traffic after repairing of a road is
Answer: (c) 24 hours
After a bituminous repair — pothole patching, a premix carpet or a surface dressing — the binder must cool and the aggregate must key into it before it can take channelised wheel loads. Standard maintenance practice is to keep traffic off the repaired stretch for at least 24 hours, and where a seal coat has been applied, to admit traffic afterwards at controlled speed for a further day so that the chippings are rolled in rather than whipped off.
Q59.Percentage of more cement should be used while concreting under water thatn that required for same mix placed in the dry
Answer: (b) 10%
Concrete placed under water is deposited through a tremie or a bottom-opening bucket and can never be compacted by vibration. It must therefore be highly cohesive and self-flowing: slump of the order of 150–180 mm, a rich mix, and generous fines so it does not segregate or wash out as it leaves the tremie. IS 456:2000, clause 14.2, meets this by requiring that concrete placed under water contain at least 10% more cement than the same mix placed in the dry.
Q60.If the magnetic bearing of the sun at a place at noon in southern hemisphere is 167°, the magnetic of declination at that place is
Answer: (c) 13° E
At local apparent noon the sun lies exactly on the observer's meridian, so its true bearing is 180° (due south) for a northern-hemisphere station and 0°/360° (due north) for a southern- hemisphere station.
Q61.A 30 cm square bearing plate settles by 1.5 cm in a plate loading test on a cohesive soil when the intensity of loading is 2 kg/cm². The settlement of a prototype shallow footing 1.0 m square under the same intensity of loading is
Answer: (c) 5 cm
The plate load test measures settlement at a scale far smaller than the real footing, so the result must be extrapolated. The two extrapolation laws are completely different, and the soil type decides which one to use: Clay (cohesive): the modulus is essentially independent of confining stress, so settlement is directly proportional to the loaded width — Sf = Sp (Bf/Bp).
Q62.The fluid generally used for grouting is
Answer: (d) all of the above
Grouting injects a fluid into voids, fissures or loose ground so that on setting it seals the seepage path or stiffens the mass. The grout is selected by the size of the opening it must enter: Cement–water grout (with or without sand or bentonite) — the workhorse for rock fissures, gravels and contact grouting under dam foundations. It cannot penetrate fine sands, since the cement grains block the pores.
Q63.If a cohesive soil specimen is subjected to a vertical compressive load, the inclination of the cracks to the horizontal is
Answer: (b) 45°
Failure in a soil specimen occurs on the plane where the shear stress first reaches the Mohr– Coulomb envelope, and from the geometry of the Mohr circle that plane is inclined to the plane of the major principal stress by θ = 45° + φ/2 In a vertical compression test the major principal stress is vertical, so its plane (the plane it acts on) is horizontal, and θ is measured from the horizontal.
Q64.When bitumen is added to soil mainly it
Answer: (c) acts as water proofing agent
Bituminous stabilisation works by two mechanisms, and which one dominates depends on the soil: In cohesionless soils (sand–bitumen) the binder supplies the cohesion that the sand lacks. In cohesive soils (soil–bitumen) the clay already has cohesion; what destroys it is water. Here the bitumen coats and seals the particles, so the soil cannot take up moisture and lose strength.
Q65.If pan evaporation is denoted by Ep and the actual evaporation by E, then
Answer: (a) Ep > E
Pan evaporation Ep is measured from a small, shallow, isolated body of water — usually a US Class A pan, 1.21 m diameter and 255 mm deep, standing on a wooden frame. Such a pan: heats up faster than a lake because of its small volume and heat gain through the sides and bottom; sits in air that has not been humidified by upwind evaporation, so the vapour pressure deficit driving evaporation is larger.
Q66.A 6 h storm had 6 cm. of rainfall and the resulting runoff was 3 cm. if the φ-index remains at the same value, the runoff due to 12 cm. rainfall in 9 h in the catchment is
Answer: (d) 7.5 cm
The φ-index is the constant rate of loss that, subtracted from the hyetograph, leaves exactly the observed runoff. It is a lumped index — it absorbs infiltration, interception, depression storage and evaporation into one average rate — and it has units of depth per unit time, so it must always be multiplied by the storm duration, never applied as a fixed depth or a fixed percentage. That is the whole trap in this question.
Q67.A trickling filter is primarily a
Answer: (b) biological oxidation process to remove BOD from sewage
A trickling filter does not filter. Sewage is sprayed by a rotary distributor over a bed of 25–75 mm stone or plastic media 1.8–3 m deep; the media are simply a support on which a zoogleal biofilm of bacteria, fungi and protozoa grows. As the sewage trickles down in a thin film, the organisms adsorb and oxidise the dissolved and colloidal organic matter, taking oxygen from air that ventilates naturally through the underdrains and the voids.
Q68.According to Bharat Stage IV norms, air pollutants which are monitored for petrol- driven vehicles are
Answer: (d) CO, HC
Spark-ignition (petrol) and compression-ignition (diesel) engines fail in opposite directions, and the emission norms follow that split: Petrol engines run near stoichiometric with a fuel-rich transient behaviour, so incomplete combustion dominates: carbon monoxide and unburnt hydrocarbons are the characteristic pollutants, and they are the two measured in the routine Pollution-Under-Control check on a petrol vehicle.
Q69.Equivalent axle load factor (EALF) defines
Answer: (c) damage per pass to a pavement by the axle in question relative to the damage per pass of a standard axle
pass of a standard axle Pavement design cannot add axle loads arithmetically because damage is grossly non-linear in load.
Q70.Geosynthetics is used in pavement construction for
Answer: (d) all of the above
Geosynthetics earn their place in a pavement by performing more than one function at once, and all four of the classical functions apply in road construction: Separation — a geotextile between a soft subgrade and the granular sub-base stops the aggregate from punching into the soil and the soil from pumping up into the aggregate, preserving the designed layer thickness.
Q71.Vee-bee time for low workability as per IS standard is
Answer: (b) 5-10 secs
The Vee-Bee consistometer measures workability by time: a slump cone of concrete is placed in a cylindrical container on a vibrating table, a transparent disc is lowered on top, and the vibrator is started. The Vee-Bee degree is the number of seconds for the concrete to remould completely and just cover the underside of the disc.
Q72.Minimum thickness of lean concrete layer below foundation concrete should be
Answer: (a) 100 mm
Before any structural footing is cast, a levelling course of lean concrete — commonly M10 or a 1:4:8 nominal mix — is laid on the prepared bed. Its minimum thickness is 100 mm.
Q73.If G is specific gravity of sand particles, e is void ratio, the critically hydraulic gradient is
Answer: (c) (G-1)/(1+e)
When water seeps upwards through a soil, the seepage force acts against gravity and reduces the effective stress.
Q74.Higher efficiency for removal of SPM less than 1 μm can be achieved with
Answer: (b) Bag Filter
Every particulate control device has a fractional efficiency curve, and the question is which device holds up in the sub-micron region. Fabric filter (bag house) — collection occurs on the dust cake, not the cloth, by interception, inertial impaction and, crucially for fine particles, Brownian diffusion. Diffusion becomes more effective as the particle gets smaller, so the bag filter holds efficiencies above 99% right down to about 0.01 μm.
Q75.Ca++ = 40 mg/L; Mg++ = 12 mg/L. What is the total hardness?
Answer: (a) 150 mg/L as CaCO3
Hardness is always reported on a common basis — mg/L as CaCO3 — because the ions responsible have different atomic weights and different valencies. The conversion is by chemical equivalents: Hardness as CaCO3 = concentration × (50 ÷ equivalent weight of the ion) Equivalent weights: Ca2+ = 40/2 = 20; Mg2+ = 24.3/2 = 12.15; CaCO3 = 100/2 = 50.
Q76.Two sound sources are emitting sound simultaneously with sound levels of 60 dB and 70 dB. What will be the resultant sound level?
Answer: (a) 70.4 dB
Decibels are logarithmic, so sound levels can never be added arithmetically. Convert each level back to intensity, add the intensities, and convert back: Ltotal = 10 log10(10L₁/10 + 10L₂/10) The result is governed by a rule worth memorising: two equal sources add 3 dB; sources 10 dB apart add only 0.4 dB.
Q77.Diameter of holes for bolts/rivets less than 25 mm diameter should be larger by
Answer: (a) 1.5 mm
A rivet is driven hot and upset to fill its hole, so the hole must be slightly larger than the shank. IS 800 fixes the allowance by size: Nominal diameter up to and including 25 mm: hole = nominal + 1.5 mm. Nominal diameter over 25 mm: hole = nominal + 2.0 mm. The enlarged size is the gross diameter, and all rivet strength calculations — shear capacity, bearing capacity, net area of the connected plate — are performed on the gross diameter, not the nominal one.
Q78.Wet mix macadam is used for construction of
Answer: (b) subbase/base
Wet mix macadam (WMM) is a well-graded crushed aggregate, 0–53 mm, mixed with water in a pug mill or batching plant at optimum moisture content, laid by paver and rolled to a dense, tightly keyed layer 75–200 mm thick. It has no binder — it derives its strength from mechanical interlock and dense grading.
Q79.Constituent(s) of road marking meterial is/are
Answer: (d) all of the above
A thermoplastic road marking is a formulated composite, and all three items listed are genuine constituents: Binder — the thermoplastic resin (usually a hydrocarbon or alkyd resin) that melts at about 180–200 °C, wets the aggregate and pigment and bonds the mark to the road. Glass beads — intermixed through the material and also drop-applied on the hot surface. They retro-reflect headlamp light back to the driver, which is what makes the marking visible at night.
Q80.In RCC column with spiral reinforcement minimum number of longitudinal bars in contact with the spiral shall less than
Answer: (d) 6
IS 456:2000, clause 26.5.3.1(c), sets the minimum number of longitudinal bars in a compression member: Four in a rectangular column — one in each corner of the tie. Six in a circular column with helical (spiral) reinforcement.
Q81.Battens shall be designed to carry bending moment and shear force arising from transverse shear force V which is ________ % of total axial load on column.
Answer: (b) 2.5
A battened column behaves as a Vierendeel frame: the two main components act as chords and the battens as rigid transverse members. Even a nominally axially loaded column is never perfectly straight, so a transverse shear develops. IS 800:2007 covers this by requiring battens (and lacing, clause 7.6.6.1) to be designed for a transverse shear Vt = 2.5% of the total axial force on the column, shared equally between the battened faces.
Q82.A tank of 5m long, 2m wide and 2m depth is completely filled with oil. If it is accelerated along the length at the rate of 1.5 m/sec². the amount of oil is in (m³)
Answer: (c) 15.29
An open tank under constant horizontal acceleration reaches relative equilibrium: the liquid moves as a rigid body and the free surface becomes a plane inclined at tan θ = a / g Because the tank starts completely full, the surface cannot rise at the rear — the brim is already at the top. Instead the whole tilted surface hinges about the rear top corner and the surplus liquid spills out over the front, leaving a wedge of air at the front end.
Q83.In an open channel under critical depth
Answer: (b) specific energy is minimum
Specific energy is the energy head measured from the channel bed: E = y + V²/2g = y + q²/(2gy²) (for a rectangular channel of unit discharge q) Plotted against depth for a fixed q, this is a curve with two limbs: as y → 0 the velocity head blows up, and as y → ∞ the curve approaches the 45° line E = y. Between them lies a minimum.
Q84.NO2 concentration in air is 470 μg/m³ at ambient condition with temperature of 25°C. The concentration in ppm will be
Answer: (c) 0.25 ppm
Converting a mass concentration to a volume fraction needs the molar volume of the gas. At 25 °C and 1 atm one mole occupies 24.45 litres (at 0 °C it is 22.4 L), so ppm = (μg/m³ × 24.45) ÷ (1000 × M) with M the molecular weight in g/mol. For NO2, M = 14 + 32 = 46.
Q85.Maximum area of compression reinforcement, in doubly reinforced beam of width b and depth D is
Answer: (d) 0.04bD
IS 456:2000 clause 26.5.1 governs steel limits in beams, and both the compression and the tension steel are capped at the same value: Maximum compression reinforcement: 0.04 bD (clause 26.5.1.2) Maximum tension reinforcement: 0.04 bD (clause 26.5.1.1(b)) Minimum tension reinforcement: As/bd = 0.85/fy, which is 0.205% for Fe 415. Note that the limits are written on the overall depth D, not the effective depth d, whereas the minimum is written on d.
Q86.If the total distrubuted load on purlin is W including wind load, the maximun bending moment is
Answer: (d) Wl/10
Purlins carry the roof sheeting between trusses. They are continuous over several trusses rather than simply supported, so a purely simple-beam moment would be conservative. IS 800 therefore permits angle and channel purlins conforming to the empirical rules to be designed for M = W L / 10 where W is the total distributed load on the purlin including wind and L is the span (the centre-to-centre spacing of the trusses).
Q87.A stream function ψ = 4x³ - y². The magnitude of velocity at a point (1,2) is
Answer: (b) 12.65
The stream function is defined so that it automatically satisfies continuity. The velocity components follow from it as u = ∂ψ/∂y and v = −∂ψ/∂x Learn the sign pattern: the y-derivative gives the x-velocity with a plus sign, the x-derivative gives the y-velocity with a minus sign. Once both components are found the resultant is the vector sum, never the arithmetic sum.
Q88.The ratio of average velocity to maximum velocity for steady laminar flow in a circular pipe is
Answer: (c) 0.5
For steady laminar (Hagen–Poiseuille) flow in a circular pipe the velocity profile is a paraboloid of revolution: u = umax [1 − (r/R)²], with umax = (−dp/dx) R²/4μ Integrating over the section, the mean velocity is V̄ = (1/πR²)∫0R u 2πr dr = umax/2 so V̄/umax = 0.5. This is a purely geometric consequence of the paraboloid: the mean ordinate of a paraboloid of revolution is exactly half its peak.
Q89.Main function of alumina in brick earth is
Answer: (a) to impart plasticity
Alumina is the clay fraction of brick earth, and its defining property is plasticity. Wetted alumina lets the earth be pugged, moulded and pressed into shape, and holds the green brick together until burning vitrifies the mass. Without sufficient alumina the earth crumbles and cannot be moulded at all. The limit of 20–30% exists because plasticity is bought at a price: alumina holds a great deal of water, so an excess causes heavy drying shrinkage, warping and cracking.
Q90.In case of pretensioning, tensioning force should be applied
Answer: (a) gradually and uniformly
In pretensioning the wires are stressed against abutments before the concrete is cast, and the force is transferred to the hardened concrete by bond alone when the wires are released.
Q91.Sag roads in a roof truss are used for connecting
Answer: (b) Purlin
Purlins on a sloping roof carry loads that do not act along a principal axis: the gravity load resolves into a component normal to the roof and a component parallel to the slope. The parallel component bends the purlin about its weak axis and makes it sag down the slope. Sag rods are light tie rods, usually 16–22 mm diameter, run up the slope from purlin to purlin at mid-span (or at the third points for long spans), hanging ultimately from the ridge.
Q92.The slenderness ratio of single angle discontinuous strut shall not exceed
Answer: (b) 180
A discontinuous strut is one that is interrupted at its ends and connected to a gusset — the typical web member of a roof truss. Because a single angle is connected through one leg only, the load is eccentric and the member has a very low minor-axis radius of gyration, so a strict slenderness limit is imposed. IS 800:2007, clause 7.5 and Table 3, caps the effective slenderness ratio of such a member at 180.
Q93.In which of the following types of flow, the losses are maximum?
Answer: (b) Turbulant flow
In laminar flow the fluid moves in orderly layers, momentum is exchanged only by molecular viscosity, and the head loss is proportional to the first power of velocity: hf = 32 μ V L / (ρ g D²) ⟹ hf ∝ V, with f = 64/Re In turbulent flow the eddies transport momentum bodily across the flow.
Q94.Tranquil flow must always occur
Answer: (b) above critical depth
Open channel flow is classified by the Froude number F = V/√(gD), where D is the hydraulic depth: F < 1 — subcritical, tranquil or streaming flow. Depth greater than critical, velocity less than critical. Disturbances travel both upstream and downstream, so control is exercised from the downstream end. F = 1 — critical flow, at the critical depth. F > 1 — supercritical, rapid or shooting flow.
Q95.Compressive strength of any structural concrete sample should not be less than specified characteristic compressive strength minus
Answer: (b) 3 MPa
Acceptance of concrete is judged on two counts, and the individual-result rule is the one this question asks for. The strength of any individual sample must not fall below fck − 3 N/mm² while the mean of any four non-overlapping consecutive samples must not fall below fck + 0.825σ or fck + 3, whichever is greater.
Q96.In aerobic environment, nitrosomonas covert
Answer: (a) NH3 to NO2⁻
Nitrification is a two-stage aerobic, autotrophic oxidation carried out by two distinct genera: NH4+ → NO2– by Nitrosomonas, then NO2– → NO3– by Nitrobacter Both organisms are strict aerobes, slow growing, and highly sensitive to pH (optimum 7.5–8.5) and to temperature — which is why nitrification is the first process to fail in a cold or overloaded plant.
Q97.Sewage treatment in an oxidation pond is done by
Answer: (c) algal-bacterial symbiosis
An oxidation (stabilisation) pond is a shallow basin, 1–1.5 m deep, in which sewage is treated by a self-sustaining partnership rather than by machinery: Aerobic bacteria oxidise the organic matter, consuming O2 and releasing CO2, ammonia and phosphate. Algae use that CO2, ammonia and phosphate, together with sunlight, to photosynthesise — and release O2 back into the water.
Q98.Combined flakiness and elongation indices of coarse aggregates for WBM used in subbase/base courses should be
Answer: (a) ≤ 30%
Flakiness index is the percentage by weight of particles whose least dimension (thickness) is less than 0.6 of their mean sieve size — measured on a thickness gauge. Elongation index is the percentage whose greatest dimension (length) exceeds 1.8 times the mean sieve size — measured on a length gauge. Both are determined to IS 2386 (Part I). Flaky and elongated particles break under the roller, do not interlock, and produce a layer with high voids and poor stability.
Q99.Photochemical smog is caused due to presence of
Answer: (c) NO, Hydrocarbon and sunlight
Photochemical (Los Angeles-type, oxidising) smog is not emitted — it is manufactured in the atmosphere. Three ingredients are indispensable: oxides of nitrogen, hydrocarbons (VOCs) and sunlight. The chain begins with the photolysis of nitrogen dioxide: NO2 + hν (λ < 430 nm) → NO + O O + O2 + M → O3 Hydrocarbon radicals then convert NO back to NO2 without consuming ozone, so O3 accumulates, and further reaction gives peroxyacetyl nitrate (PAN), aldehydes and formaldehyde.
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