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All 97 questions from the WBPSC AE Civil 2016 paper (Assistant Engineer — 2016 (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 — 2016 (Civil) |
| Questions | 97 |
| Marking (current pattern) | +2 for a right answer, −2/3 for a wrong one |
| Answers on this page | All 97 |
Q1.If the total hardness of water is greater than its total alkalinity, the carbonate hardness will be equal to
Answer: (a) total alkalinity
Hardness is caused by Ca2+ and Mg2+. Alkalinity is caused by the anions HCO3–, CO32– and OH–. Carbonate (temporary) hardness is that part of the hardness for which a carbonate or bicarbonate ion is available as the partner — so it can only be as large as whichever of the two quantities runs out first. The whole topic reduces to a two-line rule: If Total hardness > Total alkalinity → Carbonate hardness = Total alkalinity, and Non- carbonate hardness = TH − TA.
Q2.Maximum amount of water that a soil can hold by capillary action is called
Answer: (c) field capacity
Field capacity is the moisture content held by a soil against gravity after free drainage from saturation has essentially ceased — usually 2–3 days after irrigation. This water is retained in the capillary pores by surface tension, so it is exactly “the maximum amount of water a soil can hold by capillary action”. It corresponds to a soil suction of roughly 1/3 atmosphere (about 33 kPa).
Q3.What is the resultant noise level generated simultaneously from two sources with same intensity of 90 dB respectively?
Answer: (c) 93 dB
Sound levels in decibels are logarithmic, so they cannot simply be added. Intensities add; levels combine as Ltotal = 10 log10 ( 10L₁/10 + 10L₂/10 ) For two equal sources the intensity exactly doubles, and 10 log 2 = 3.01 dB. So the answer for any pair of equal sources is simply “+3 dB”, regardless of the level.
Q4.If there is a rain between 8:00 a.m. to 10:30 a.m. and the entire basin just starts contributing water at 8:30 a.m. to the outlet, then the time of concentration will be
Answer: (c) 30 minutes
The time of concentration (tc) is the time taken by runoff to travel from the hydraulically most remote point of the catchment to the outlet. The whole basin is contributing to the outlet only after this interval has elapsed from the beginning of rainfall. Rain begins at 8:00 a.m. The entire basin “just starts contributing” at 8:30 a.m. — that is the instant the most distant droplet's contribution arrives. Hence tc = 8:30 − 8:00 = 30 minutes.
Q5.First stage of water treatment is
Answer: (a) plain sedimentation
The conventional order of a surface-water treatment train is: Screening → Plain sedimentation (grit/plain settling) → Coagulation–flocculation → Clarification → Filtration → Disinfection Of the four options offered, plain sedimentation comes first. It is settling under gravity alone, with no chemicals added, and its job is to strip the coarse, readily-settleable silt so that the later chemical stages are not overloaded.
Q6.To prevent the sloughing of an earthen dam, it is necessary to provide
Answer: (c) rock toe
Sloughing is the slipping-away of the downstream face of an earth dam. It happens when the phreatic (seepage) line emerges on the downstream slope: the emerging water raises pore pressure, reduces effective stress, softens the toe, and a shallow slice slides off. The cure is therefore not to armour the face but to keep the seepage line inside the body of the dam and give the seepage a controlled exit. A rock toe (with an inverted filter behind it) does exactly this.
Q7.Alum acts as
Answer: (a) coagulant
Alum, Al2(SO4)3·18H2O, is the standard coagulant. Added to water it hydrolyses and reacts with the natural alkalinity: Al2(SO4)3 + 3Ca(HCO3)2 → 2Al(OH)3↓ + 3CaSO4 + 6CO2 The trivalent Al3+ ion neutralises the negative surface charge on the colloidal clay particles, destroying the electrostatic repulsion that keeps them dispersed — this charge neutralisation is coagulation.
Q8.Sloughing phenomena is observed in
Answer: (c) trickling filter
Sloughing (also called unloading) is the periodic shearing-off of the biological slime layer from the media of a trickling filter. As the zoogloeal film grows thicker, oxygen and food cannot penetrate to its base; the innermost layer turns anaerobic, loses its grip on the stone, and a sheet of biomass washes off with the effluent. The film then regrows.
Q9.Y-axis of hydrograph indicates
Answer: (c) discharge
A hydrograph is a plot of discharge against time at a specified section of a stream. The ordinate is therefore discharge, normally in m³/s (cumec), and the abscissa is time. Keep the family of related plots straight: Hydrograph — discharge vs time. Hyetograph — rainfall intensity vs time (usually drawn as a bar chart, inverted at the top of the same sheet). Mass curve — cumulative rainfall or runoff vs time.
Q10.Drip irrigation is used in ________ areas.
Answer: (d) arid/semi arid
Drip (trickle) irrigation delivers water drop by drop directly to the root zone through emitters on a low-pressure lateral network. Its whole economic case is water economy: field application efficiency reaches 85–95 %, against 60–70 % for surface methods, because there is virtually no conveyance loss, no deep percolation between rows and no evaporation from a wetted soil surface.
Q11.In earthen dam the phreatic line is
Answer: (b) parabolic
The phreatic line (seepage line, or top flow line) is the uppermost streamline of the seepage flow through the body of an earth dam. Along it the pressure is atmospheric. Casagrande showed that this line is very closely a parabola — specifically a base parabola with its focus at the downstream toe (or at the upstream end of the horizontal filter, if one is provided), with small corrections at the entry and exit.
Q12.Recommended per capita water demand for domestic purposes in Indian Cities may be taken as
Answer: (c) 135 lpcd
135 lpcd is the figure recommended by IS 1172 and by the CPHEEO Manual as the domestic requirement for Indian cities provided with full flushing systems.
Q13.River training works primarily to protect river bank is
Answer: (c) groynes
Groynes (also called spurs) are river-training structures built out from the bank into the stream. They work by pushing the high-velocity filament of the current away from the bank towards mid- stream, so that the erosive attack falls on the nose of the groyne rather than on the bank itself, while silt deposits in the still pockets between successive groynes and builds the bank out.
Q14.Dead storage of a reservoir is provided to
Answer: (c) accommodate silt entry
Reservoir storage is divided by elevation into: Dead storage — below the lowest outlet / sill level. It is not available for supply and is deliberately provided as a sacrificial volume for the silt that the river brings in over the design life of the reservoir (typically 100 years). Sizing it requires the sediment yield of the catchment and the trap efficiency of the reservoir.
Q15.The process by which the rain water enters the surface of the earth is called
Answer: (b) infiltration
Infiltration is the entry of water through the ground surface into the soil. It is the process asked about. Once inside, the downward movement of that water through the unsaturated zone to the water table is percolation. The two words are routinely interchanged in loose speech, and the PSC likes to test exactly that distinction.
Q16.Corrosion in concrete sewer is caused by
Answer: (a) H2S
This is the classic crown corrosion of concrete sewers, and hydrogen sulphide is the agent. The mechanism runs in two stages and both must be quoted for full marks: Below the water line — when the sewage turns septic (low DO, long detention, flat gradients, warm climate), anaerobic sulphate-reducing bacteria reduce sulphate to sulphide. H2S escapes into the air space above the flow.
Q17.Imhoff tanks are used for removing
Answer: (d) organics
An Imhoff tank is a two-storey tank: a sedimentation chamber on top and a sludge digestion chamber underneath, separated by a sloping floor with a slot. Settleable solids drop through the slot into the lower chamber and digest anaerobically there; the gas produced is vented through separate gas vents so that rising bubbles never re-mix the settling zone above.
Q18.Isohyets are
Answer: (b) line of equal precipitation
An isohyet is a line joining points of equal rainfall depth over a given period. The prefix hyet- comes from the Greek for rain, the same root as in hyetograph. The isohyetal method is also the most accurate of the three ways of computing mean areal rainfall over a catchment, because it lets the engineer draw in the effect of topography by hand: P̄ = Σ [ ai × (Pi + Pi+1)/2 ] / Σ ai where ai is the area between two successive isohyets.
Q19.Standard BOD is taken for
Answer: (c) 5 days
The standard test is BOD5 at 20 °C — five days' incubation in the dark at 20 °C. Two practical reasons fixed the choice: five days is roughly the travel time of British rivers to the sea (where the test originated), and it captures about 68 % of the ultimate carbonaceous BOD while stopping short of the nitrification stage that begins around the eighth to tenth day and would otherwise inflate the reading.
Q20.In India, the maximum permissible limit of iron in drinking water is
Answer: (a) 0.3 mg/l
IS 10500 sets the acceptable (desirable) limit for iron in drinking water at 0.3 mg/l. Above it the water stains laundry and sanitary ware a rusty brown, tastes metallic and bitter, encourages iron bacteria (Crenothrix) growth in the mains, and discolours on standing as ferrous iron oxidises to ferric. These are aesthetic and operational objections rather than a toxic hazard, which is why the limit sits at the level where the nuisance begins.
Q21.BOD is measure of
Answer: (c) biodegradable organics
BOD measures the oxygen consumed by micro-organisms while they stabilise organic matter. Only the fraction that the organisms can actually metabolise registers — that is, the biodegradable organic fraction. Refractory organics (many synthetic compounds, lignin, some pesticides) exert little or no BOD even though they are unquestionably organic and polluting.
Q22.Cyclone separators are used for
Answer: (c) SPM
A cyclone separator removes SPM — suspended particulate matter. Dust-laden gas is admitted tangentially at the top of a cylinder-on-cone body and spun into a vortex. The centrifugal force on a particle is many times gravity, so particles are flung to the wall, lose momentum against it, and spiral down the cone into the hopper, while the cleaned gas reverses direction and leaves through a central outlet tube at the top.
Q23.If x is the depth of water at any section, then the mean velocity is
Answer: (d) 0.6x
In an open channel the velocity is zero at the bed, rises rapidly through the boundary layer, and reaches its maximum a little below the free surface (at about 0.05–0.25 of the depth down, because of air resistance and secondary currents). Integrating that logarithmic profile shows that the mean velocity of the vertical occurs at 0.6 of the depth measured from the water surface — i.e. at 0.4 depth above the bed. This is the basis of stream gauging with a current meter.
Q24.Premature aging of plants occurs due to
Answer: (d) PAN
PAN — peroxyacetyl nitrate — is a secondary pollutant formed in photochemical smog from the reaction of hydrocarbons with NO2 in sunlight. It is an extremely potent phytotoxicant: it attacks the spongy mesophyll cells of young leaves, producing the characteristic glazing, silvering or bronzing of the underside, and causes premature ageing (early senescence) and leaf drop.
Q25.The total depth of water required by a crop during the entire period the crop is in the field, is known as
Answer: (a) Delta
Delta (Δ) is the total depth of water, in metres or cm, required by a crop over its entire period in the field. It is the crop's total water requirement expressed as a depth spread uniformly over the irrigated area. The three quantities are tied together by one relation that the PSC asks in some form nearly every year: Δ (metres) = 8.64 B / D where B = base period in days and D = duty in hectares per cumec.
Q26.Ductility of a flexural member (Reinforced concrete) can be increased by
Answer: (b) decreasing percentage of tensile reinforcement
Ductility is the ability of a member to undergo large inelastic deformation without a significant drop in load — it is what gives warning before collapse and what allows moment redistribution and energy dissipation in an earthquake. In a flexural RC member the ductility depends on how far the neutral axis lies from the compression face.
Q27.The shear stress distribution in viscous fluid through a circular pipe is
Answer: (b) maximum at the inside of surface
For steady laminar flow in a circular pipe, force balance on a cylindrical fluid element of radius r gives τ = (∂p/∂x) · r/2 Shear stress therefore varies linearly with radius: it is zero on the axis (r = 0, where the velocity gradient du/dr = 0) and maximum at the pipe wall — the inside surface — where τ0 = (∂p/ ∂x)·R/2 = ρ g R S/2.
Q28.The capillary rise or depression in a small diameter tube is
Answer: (c) inversely proportional to diameter
Equating the surface-tension force lifting the column to the weight of the raised liquid: σ · (π d) cos θ = ρ g · (π d²/4) · h which rearranges to h = 4 σ cos θ / (ρ g d) So capillary rise (or depression, when θ > 90° as for mercury on glass) is inversely proportional to the diameter of the tube. Halve the bore and the rise doubles. This is why fine-grained soils show far higher capillary rise than coarse sands — the pore channels act as fine tubes.
Q29.Which of the section should be used preferably at places where torsion occurs?
Answer: (c) Box section
A box (closed hollow) section is the right choice wherever torsion is significant. A closed thin- walled tube resists torsion by a continuous shear flow around the perimeter — Bredt–Batho theory: q = T/(2 Am), J = 4 Am² / ∮ (ds/t) where Am is the area enclosed by the mid-line of the wall. Because Am is squared, a closed section is enormously stiffer in torsion than an open one of identical material and wall thickness — commonly by a factor of hundreds.
Q30.Side face reinforcement in a reinforced concrete beam is to be provided for
Answer: (d) Depth of web of the beam > 750mm
IS 456:2000, Clause 26.5.1.3, states that where the depth of the web of a beam exceeds 750 mm, side face reinforcement shall be provided along the two faces. The requirement is Area not less than 0.1 % of the web area (b × dweb); Distributed equally on the two faces; Spacing not exceeding 300 mm or the web thickness, whichever is less. The purpose is crack control, not strength.
Q31.The closure of valve is considered rapid when (L = pipe length, C = celerity)
Answer: (d) the duration of closure is less than 2L/C
Water hammer theory classifies valve closure by comparing the closure time T with the time 2L/C that the pressure wave takes to travel to the reservoir and return (L = pipe length, C = celerity of the pressure wave, about 1400 m/s in water in a rigid pipe). Rapid (sudden) closure: T < 2L/C. The valve is fully shut before the relief wave gets back, so the full Joukowsky pressure rise develops: Δp = ρ C ΔV, i.e. head rise h = C V / g. Slow (gradual) closure: T > 2L/C.
Q32.The critical depth in a open channel flow is the depth of flow at which
Answer: (b) the specific energy is minimum
Specific energy is the energy per unit weight measured from the channel bed: E = y + V²/2g = y + q²/(2 g y²) (for constant unit discharge q) Plotting E against y gives the specific-energy curve, a two-limbed hyperbola-like curve with a nose pointing left. Setting dE/dy = 0: 1 − q²/(g y³) = 0 ⇒ yc = (q²/g)1/3 At this depth, called the critical depth, the specific energy is a minimum.
Q33.A beam curved in plan is designed for
Answer: (d) bending moment, shear force and torsion
A beam curved in plan (a circular balcony beam, a helical stair beam, a curved bridge girder, a ring beam of a water tank) is loaded vertically while its axis runs horizontally along a curve. Because the load and the axis are not in one vertical plane, the internal actions at any section are all three: bending moment, shear force and torsional moment. The reason is geometric.
Q34.A pitot tube is an instrument for measuring
Answer: (c) velocity of flow
A pitot tube measures velocity of flow at a point. Its open mouth faces upstream, so the fluid is brought to rest at the tip — the stagnation point. Applying Bernoulli's equation along the streamline from a point in the undisturbed stream to the stagnation point: p0/ρg = p/ρg + V²/2g ⇒ V = √(2g Δh) where Δh is the difference between the stagnation (total) head and the static head.
Q35.The percentage of HYSD bars in one-way slabs to cater for shrinkage or temperature, as per IS:456 is
Answer: (d) 0.12%
IS 456:2000, Clause 26.5.2.1: the minimum reinforcement in either direction of a slab, to control shrinkage and temperature effects, shall be 0.15 % of the gross cross-sectional area for mild steel (Fe 250); 0.12 % of the gross cross-sectional area for high-strength deformed bars (HYSD, Fe 415 and above) and welded wire fabric. The question specifies HYSD bars, so the answer is 0.12 %.
Q36.In a rectangular channel of constant width, water flows at subcritical state. At a place where there is a local rise (hump) in the channel floor, the depth of water, over the hump will be
Answer: (c) less than upstream depth
Take the specific-energy curve and think in two steps. Over a small hump of height Δz the specific energy falls: E₂ = E₁ − Δz. Moving left on the specific- energy curve means moving towards the nose. The flow is subcritical, so it sits on the upper limb, where depth decreases as E decreases. Therefore the depth over the hump is less than the upstream depth.
Q37.For two-way R.C. slab, the deflection is guided by
Answer: (a) shorter span
For a two-way slab the deflection, and hence the span/depth check for serviceability, is governed by the shorter span. IS 456 Clause 24.1 makes this explicit: for two-way slabs of shorter span up to 3.5 m with mild steel, the span/depth ratios given (35 for simply supported, 40 for continuous) are applied to the shorter span. The physical reason is that the slab distributes load to the two directions in inverse proportion to the fourth power of span.
Q38.An increase in temperature on the top fibres of a simply supported beam will cause
Answer: (a) upward deflection
Heating the top fibres makes them expand relative to the bottom. Since the two faces are monolithic, the beam must curve so that the longer (hotter) fibre lies on the outside of the curve — i.e. the beam becomes convex upwards, or hogs, lifting its mid-span. That is an upward deflection.
Q39.The eccentricity in a hollow circular column of external diameter 25 cm, internal diameter 15 cm for a eccentric load of 1000KN for no-tension condition is
Answer: (d) 4.25 cm
For an eccentric axial load on a section, the no-tension condition requires the load to fall inside the kern (core) of the section. Derive the kern radius of a hollow circle from first principles: Z = I/y = [π(D⁴ − d⁴)/64] ÷ (D/2) = π(D⁴ − d⁴)/(32D) A = π(D² − d²)/4 e = Z/A = (D⁴ − d⁴)/[8D(D² − d²)] = (D² + d²)/(8D) Note that the magnitude of the load — 1000 kN here — never enters the calculation.
Q40.Which test is not used to measure workability of concrete?
Answer: (c) Airpermeability test
Workability is the ease with which concrete can be mixed, placed, compacted and finished without segregation. The standard tests for it are: Slump test — the site test, IS 1199; good for medium to high workability. Compacting factor test — ratio of partially-compacted to fully-compacted weight; best for low-workability, dry mixes where slump is meaningless. Vee-Bee consistometer test — seconds of vibration to remould a slump cone into a cylinder; suited to very dry mixes.
Q41.The stagnation pressure is the sum of
Answer: (b) static pressure and dynamic pressure
When a moving fluid is brought to rest isentropically, its kinetic energy is converted into pressure. The resulting pressure is the stagnation (total) pressure: pstagnation = pstatic + ½ ρ V² The term ½ ρ V² is the dynamic pressure. Hence stagnation pressure is the sum of static and dynamic pressures. In head form, dividing by ρ g: Hstagnation = p/ρg + V²/2g This is exactly the principle the pitot tube exploits (see Q34).
Q42.The slab without any beam is called as
Answer: (d) Flat slab
A flat slab is a reinforced concrete slab supported directly on columns, with no intermediate beams. Load travels from the slab straight into the columns, which is why the critical design issue is punching shear around the column head. To deal with it the slab may be provided with: a drop panel — a local thickening of the slab over the column; a column head / capital — a flared enlargement at the top of the column; both, or neither (a true flat plate).
Q43.Cavitation in fluid flow occurs when
Answer: (a) the pressure of flow decreases to a valve close to vapour pressure
Cavitation is the formation and violent collapse of vapour bubbles in a flowing liquid. Bubbles nucleate wherever the local absolute pressure falls to (or below) the vapour pressure of the liquid at the prevailing temperature — at the suction eye of a pump, on the low-pressure face of a turbine runner blade or propeller, at the throat of a venturi, or downstream of a spillway irregularity.
Q44.The dimension of dynamic viscosity of fluid is
Answer: (b) ML^-1T^-1
Dynamic viscosity is defined by Newton's law of viscosity, τ = μ (du/dy), hence μ = τ/(du/dy). In the M-L-T system: τ is a stress, [ML−1T−2]; the velocity gradient is [T−1]. Dividing, [μ] = ML−1T−2 ÷ T−1 = ML−1T−1 Its SI unit is Pa·s (= N·s/m² = kg/m·s); the CGS unit is the poise, 1 Pa·s = 10 poise.
Q45.The head loss in a sudden expansion in a pipeline is given by (v1 & v2 are the velocity of fluid)
Answer: (a) (v1-v2)^2 / 2g
At a sudden enlargement the jet leaving the smaller pipe cannot follow the abrupt widening. It separates, and the resulting eddies in the corner dissipate energy.
Q46.The most essential condition for a hydraulic jump to form is
Answer: (c) the existence of supercritical flow before the jump
A hydraulic jump is the abrupt transition from supercritical to subcritical flow, with a large loss of energy in the roller. It cannot occur unless the approaching flow is supercritical, i.e. Fr₁ > 1. The reverse transition (subcritical to supercritical) happens smoothly and is not a jump.
Q47.A cantilever beam is subjected to a uniformly distributed load W (Total downward load) over its whole length L and a concentrated load W (upward) at free end. The deflection at the free end is
Answer: (c) 5WL³/24EI (upward)
Superpose the two standard cantilever cases. The wording is decisive: W is the total downward UDL, so the intensity is w = W/L. UDL of intensity w over the whole span: δ = wL⁴/8EI, downward. Point load P at the free end: δ = PL³/3EI.
Q48.The total energy line is always higher than hydraulic grade line, the vertical distance between them represents
Answer: (a) the velocity head
By definition: Hydraulic grade line (HGL) = z + p/ρg — the piezometric head; it is the level to which water would rise in a piezometer inserted at that section. Total energy line (TEL) = z + p/ρg + V²/2g — the total head. Subtracting, the vertical gap between the TEL and the HGL is exactly the velocity head V²/2g. It is therefore always positive (hence TEL is always above HGL) and it widens wherever the pipe narrows and the velocity rises.
Q49.The shear centre in case of equal angle section is
Answer: (c) at the junction of two legs
The shear centre is the point through which a transverse load must pass if the section is to bend without twisting. For any section built up from thin rectangles whose centre-lines all meet at a single point — an angle, a tee, a cruciform — the shear flow in each leg produces a force whose line of action passes through that common point. Their resultant therefore also passes through it, and the shear centre lies at the junction of the two legs.
Q50.The most essential feature of a turbulent flow in pipe is
Answer: (d) Velocity and pressure at a point exhibit irregular fluctuation
The defining feature of turbulence is the presence of random, three-dimensional fluctuations of velocity and pressure superposed on the mean motion. At a fixed point the instantaneous velocity is written u = ū + u′ where ū is the time-averaged value and u′ the fluctuation, whose time average is zero but whose mean square is not.
Q51.If a sample of dry sand tested in direct shear test gives failure shear stress as 100kN/m² at a normal stress of 200kN/m², then the angle of internal friction is given by
Answer: (d) tan⁻¹(1/2)
Dry sand is a cohesionless soil, so c = 0 and the Mohr–Coulomb failure criterion reduces to τf = σ tan φ The failure envelope is a straight line through the origin, and its slope is directly the tangent of the angle of internal friction.
Q52.A soil has a bulk density of 1.80gm/cm³ at a water content of 5%. If the void ratio remains constant, then its bulk density for a water content of 10% will be
Answer: (b) 1.88 gm/cm³
The trick here is that the void ratio is unchanged, so the dry density is unchanged. Bulk density and dry density are related by γ = γd (1 + w) Find γd from the first condition and re-apply it at the second water content.
Q53.Height of the instrument may be defined as
Answer: (b) the elevation of the plane of collimation, when the instrument is accurately levelled
In levelling, the height of instrument (HI) is not the height of the telescope above the tripod or the ground. It is the reduced level of the line of collimation — the elevation, above datum, of the horizontal line of sight when the instrument has been properly levelled.
Q54.The bearing capacity of a strip footing resting on a saturated clay is 120kN/m². The bearing capacity of a circular footing with diameter same as the width of strip footing resting on same soil will be
Answer: (a) more than 120kN/m²
Q55.Right deflection angle may be directly obtained by setting the instrument to read
Answer: (a) zero on back station
A deflection angle is the angle between the prolongation of the previous line and the next line — the amount by which the traverse 'deflects' from a straight course. To read it directly for a right deflection: Set the horizontal circle to read zero and sight the back station. Transit (plunge) the telescope, which now points along the prolongation of the back line while still reading zero. Swing the telescope to the forward station.
Q56.Two specimens of clay A and B are tested in a consolidation apparatus. If (mv)A = 3.6×10⁻⁴ m²/kN, (mv)B = 1.8×10⁻⁴ m²/kN, (cv)A = 3.8×10⁻⁴ cm²/sec, (cv)B = 1.9×10⁻⁴ cm²/sec then the ratio kA/kB is equal to
Answer: (d) 4.0
The coefficient of consolidation, coefficient of volume change and permeability are tied together by cv = k / (mv γw) ⇒ k = cv mv γw Since γw is common to both soils, the ratio of permeabilities is simply the product of the two ratios.
Q57.Closed contours, with higher value inwards, represent a
Answer: (b) hillock
On a contour map the elevation values are written on the contour lines themselves. When a set of closed contours has higher values towards the inside, the ground rises as you move inward — that is a hillock (a hill, knoll or summit).
Q58.If the R.L. of a B.M. is 100.00m, the backsight is 1.215m and the foresight is 1.870m, the R.L. of the forward station is
Answer: (a) 99.345m
Q59.If the foresight of a line AB is 35° and that of line BC is 15°, then the included angle between the lines is
Answer: (c) 160°
Read the question carefully: it asks for the included angle between the lines at station B, not for the deflection angle. The included angle is measured between the ray BA (looking back) and the ray BC (looking forward). The bearing given for AB is the direction A→B.
Q60.Taylor's stability number is given by
Answer: (c) C/FcrH
Taylor defined the dimensionless stability number of a slope as Sn = cm/(γ H) = c / (Fc γ H) where c is the cohesion, cm = c/Fc the mobilised cohesion, Fc the factor of safety with respect to cohesion, γ the bulk unit weight and H the height of the slope. Option (C), C/FcγH, is this expression. Check the dimensions, which settles it without recalling the formula: c is kN/m², γH is kN/m³ × m = kN/m², and Fc is dimensionless.
Q61.While doing levelling in undulating terrain, it is preferable to set the level on
Answer: (c) any side of the slope
The instrument should be set on the side of the slope, at an intermediate elevation, rather than on the summit or in the valley bottom. The reason is entirely practical: the staff must be readable at both the back station and the forward station, and the sight distances should be roughly balanced. From a summit, both stations lie below; the line of sight strikes low on the staff or misses it, and the sights are very unequal.
Q62.Terzaghi's bearing capacity factors are function of
Answer: (b) only φ
Terzaghi's bearing capacity equation is qu = c Nc + γ Df Nq + 0.5 γ B Nγ The three factors Nc, Nq and Nγ are functions of the angle of internal friction φ alone.
Q63.The minimum centre to centre distance between the piles of a pile group in clays should be equal to the
Answer: (c) 3 times the diameter of the pile
IS 2911 requires a minimum centre-to-centre spacing of 3 times the pile diameter for friction piles in clay (and 2.5 d for end-bearing piles resting on rock, where group action is not an issue). The governing consideration in clay is group efficiency.
Q64.ABCD is a regular parallelogram plot of land whose angle BAD is 60°. If the bearing of the line BC is 30° then the bearing of the line CD is
Answer: (d) 270°
Work through the parallelogram side by side. In parallelogram ABCD, angle BAD = 60°, so the adjacent angle ABC = 120° (co-interior angles of the parallel sides AD and BC). Given the bearing of BC = 30°, walk round the figure. In a parallelogram, side CD is anti-parallel to side AB, and side AD is anti-parallel to BC. Going around the closed figure ABCDA, each successive side turns by the exterior angle.
Q65.If Δ is the angle of deflection of a curve, T1 and T2 are the points of tangencies, the angle between the tangent at T1 and long chord T1T2 will be
Answer: (c) Δ/2
For a simple circular curve of deflection angle Δ, the tangent at the point of commencement T₁ makes an angle with the long chord T₁T₂ equal to Δ/2. The proof is the tangent–chord (alternate segment) theorem of circle geometry: the angle between a tangent and a chord drawn from the point of contact equals the inscribed angle in the alternate segment. Since the arc T₁T₂ subtends the central angle Δ, the inscribed angle is Δ/2, and so is the tangent–chord angle.
Q66.The critical hydraulic gradient of a soil is given by
Answer: (a) (G-1)/(1+e)
Quick sand (boiling) occurs when upward seepage reduces the effective stress to zero. At that moment the seepage force per unit volume equals the submerged unit weight: ic γw = γ′ = (G − 1) γw/(1 + e) so the critical hydraulic gradient is ic = (G − 1)/(1 + e) For typical values G = 2.65 and e = 0.65, ic = 1.65/1.65 = 1.0 — which is why the critical gradient is so often quoted as approximately unity.
Q67.A backsight reading on B.M. = 100m was 3.250m. The inverted staff reading to the bottom of a girder was 1.250m. The R.L. of the bottom of girder is
Answer: (c) 104.5
An inverted staff reading is used when the point of interest is above the line of collimation — the underside of a girder, a soffit, a ceiling. The staff is held upside down with its foot on the point, so the reading must be added to the height of instrument instead of subtracted. Conventionally the reading is entered in the level book with a negative sign, which makes the ordinary formula work unchanged.
Q68.A soil sample having a void ratio of 1.3, water content of 50% and a specific gravity of 2.60, is in a state of
Answer: (b) full saturation
Use the fundamental phase relation S e = w G. Compute the degree of saturation and compare with 100 %.
Q69.A slope is to be constructed at an angle of 30° to the horizontal for a soil with C = 15kN/m², φ = 22.5°, r = 19kN/m³. Taylor's stability number is 0.046. If a factor of safety (with respect to cohesion) of 1.5 is required, then the safe height of the slope will be
Answer: (d) 11.5m
Taylor's stability number is defined as Sn = cm/(γ H), where the mobilised cohesion is cm = c/Fc. Rearranging for the safe height: H = c / (Fc γ Sn)
Q70.Underload, the void ratio of a submerged saturated clay decreases from 1.00 to 0.92. What will be the settlement of a 2m thick clay of same type due to consolidation?
Answer: (c) 80mm
Consolidation settlement from a change in void ratio is Δ H = H × Δe/(1 + e₀) where e₀ is the initial void ratio and H the thickness of the compressible layer. The formula follows directly from the phase diagram: the solids do not change volume, so the entire settlement is the reduction of the void volume, and the void volume per unit total volume is e/(1+e).
Q71.If the cross sectional area of an embankment at 30m intervals are 20, 40, 60, 50 and 30 m² respectively, then the volume of the embankment on the basis of prismoidal rule is
Answer: (a) 5300 m³
The prismoidal (Simpson's) rule for volumes from cross-sectional areas at equal intervals d is V = (d/3) [ A₁ + An + 4(A₂ + A₄ + …) + 2(A₃ + A₅ + …) ] It needs an odd number of sections — an even number of intervals — which the five given sections satisfy.
Q72.For a sandy soil, the angle of internal friction is 30°. If the major principal stress at failure is 50kN/m², then the corresponding minor principal stress (in kN/m²) is
Answer: (b) 16.66
For a cohesionless soil at failure, the principal stress ratio is fixed by the friction angle: σ₁/σ₃ = tan²(45° + φ/2) = (1 + sin φ)/(1 − sin φ) = Nφ With φ = 30°, sin φ = 0.5, so Nφ = 1.5/0.5 = 3.
Q73.As per IS classification, a fine grained soil is highly compressible if its liquid limit is greater than
Answer: (c) 50%
The IS soil classification (IS 1498) divides fine-grained soils by liquid limit into three compressibility bands: Low compressibility (L) — wL < 35 %: ML, CL, OL Intermediate compressibility (I) — 35 % ≤ wL ≤ 50 %: MI, CI, OI High compressibility (H) — wL > 50 %: MH, CH, OH So a fine-grained soil is highly compressible when its liquid limit exceeds 50 %. Note the difference from the Unified (USCS) system, which has only two bands, L and H, split at wL = 50 %.
Q74.The critical height of an unsupported vertical cut in a cohesive soil is given by (c = unit cohesion, φ = angle of internal friction, r = unit weight of soil)
Answer: (a) (4c/r)tan(45°+φ/2)
The active earth pressure at depth z in a c–φ soil is pa = γ z Ka − 2c√Ka Tension exists from the surface down to the depth z₀ = 2c/(γ√Ka), where the pressure crosses zero. An unsupported vertical cut stands as long as the total net thrust is zero, i.e. down to twice that depth: Hc = 2 z₀ = 4c/(γ√Ka) = (4c/γ) tan(45° + φ/2) using 1/√Ka = √Nφ = tan(45° + φ/2).
Q75.California Bearing Ratio method of designing flexible pavements is more accurate as it involves
Answer: (a) characteristics of soils
The California Bearing Ratio is a penetration test performed on the subgrade soil (and on the pavement materials, in the laboratory or in situ). It is defined as CBR = (load resisted by the specimen / load resisted by standard crushed stone) × 100 measured at 2.5 mm and 5.0 mm penetration, the higher of the two being reported.
Q76.Bottom most component of a flexible pavement, is
Answer: (a) sub-grade
The layers of a flexible pavement, from the bottom up, are: Subgrade — the compacted natural soil on which everything rests. It is the bottom-most component. Sub-base course — granular material (GSB), spreading the load and providing drainage. Base course — the main load-spreading layer (WBM, WMM, or bituminous macadam). Surface course — binder course plus wearing course (bituminous concrete, etc.).
Q77.The maximum number of passenger car that can pass a given point on a roadway during one hour under the most ideal roadway and traffic conditions, is called
Answer: (b) Basic capacity
Basic capacity is the maximum number of vehicles that can pass a point in one hour under the most ideal roadway and traffic conditions that can possibly be attained. It is a theoretical ceiling and is never achieved on a real road. The three capacity terms form a descending sequence, and the PSC tests the distinction repeatedly: Basic capacity — ideal conditions; the theoretical maximum. Possible capacity — the maximum under the prevailing roadway and traffic conditions.
Q78.Which one of the following influences the geometric design of the highways?
Answer: (d) All of the above
Geometric design of a highway covers every visible dimension of the road: the elements that determine how it looks, how it feels to drive, and how safely traffic can move on it. All three items listed are geometric design elements: Radii of highway curves — horizontal alignment, governed by design speed, superelevation and side friction: R = V²/(127(e + f)). Width of pavement — cross-section design: 3.75 m for a single lane, 7.0 m for two lanes, plus shoulders.
Q79.If V is speed in Km/hour and R is radius of the curve, the super elevation e is equal to
Answer: (b) V²/225R
On a horizontal curve, equilibrium of the vehicle gives e + f = V²/(127 R) (V in km/h, R in metres) IRC practice is to design the superelevation for 75 % of the design speed, neglecting friction entirely — a deliberately conservative step that keeps slow-moving vehicles from sliding inward on a steeply banked curve. Setting f = 0 and V → 0.75V: e = (0.75V)²/(127R) = 0.5625 V²/(127R) = V²/(225.7 R) ≈ V²/(225 R)
Q80.The extra widening of pavements recommended by Indian Roads Congress for hill roads having radius 150 meters, is
Answer: (d) 0.0m
This item is a deliberate trap, and the trap is the phrase hill roads. Extra widening is required only on sharp curves; on hill roads the IRC table stops well short of 150 m radius: R below 20 m — 0.9 m R 21 to 40 m — 0.6 m R 41 to 60 m — 0.6 m R above 60 m — nil At R = 150 m the curve is flat enough that neither the mechanical nor the psychological component is significant, so the recommended extra widening is 0.0 m. Confirm it from the formula.
Q81.Sandstone is a (i) Sedimentary rock (ii) Aqueous rock (iii) Silicious rock. The correct answer is
Answer: (d) All of the above
Sandstone qualifies under all three headings, so the answer is 'All of the above': (i) Sedimentary — by mode of formation: it is formed by the deposition, burial and cementation of sand grains. (ii) Aqueous — by the agent of deposition: the sediment was laid down by water.
Q82.Which of the following cements contains maximum percentage of dicalcium silicate?
Answer: (b) Low heat cement
Low heat cement contains the maximum proportion of dicalcium silicate (C2S) — typically about 40–50 %, against roughly 25 % in ordinary Portland cement. The reason lies in the heats of hydration of the Bogue compounds: C3S (tricalcium silicate) — hydrates fast, gives early strength, evolves about 500 J/g. C2S (dicalcium silicate) — hydrates slowly, gives late strength (beyond 28 days), evolves only about 260 J/g.
Q83.Critical path
Answer: (a) is always longest
In a project network the critical path is the longest path, measured in time, from the start event to the end event. It is longest precisely because it is the chain of activities that has no spare time: every activity on it has zero total float, and the project duration equals the length of this path.
Q84.In waterbound macadam roads, binding material, is
Answer: (b) stone dust
In water-bound macadam (WBM) construction, coarse aggregate is spread and rolled dry, and the voids between the stones are then filled with a screening or binding material. The screenings are stone dust or fine crushed aggregate of the same rock, applied dry in three or more thin layers with rolling and brooming, after which water is sprayed and the surface rolled until a grout of stone dust and water forms and sets.
Q85.Pug mill is used for
Answer: (a) preparation of clay
A pug mill is used for the preparation (tempering) of clay in brick manufacture. It is a conical iron tub with a vertical shaft carrying horizontal knives; clay, water and any additives are fed in at the top, and as the shaft rotates the knives cut, knead and homogenise the mass. The tempered clay is extruded at the bottom in a plastic, uniform condition fit for moulding.
Q86.In PERT analysis, the time estimates of activities and probability of their occurrence follow
Answer: (c) β-distribution curve
PERT assumes each activity's duration follows a β-distribution. The choice is deliberate: the beta distribution is defined on a finite interval (so it has a definite optimistic and pessimistic limit, unlike the normal distribution which extends to infinity in both directions), and it can be skewed either way, which matches real construction activities where overruns are more likely than underruns.
Q87.Newly constructed pavement with ordinary Portland cement attains its working strength after
Answer: (d) 28 days
Ordinary Portland cement concrete attains its characteristic (working) strength at 28 days. IS 456 defines the characteristic compressive strength fck as the 28-day cube strength below which not more than 5 % of test results are expected to fall, and all design is referred to that age.
Q88.The initial setting time for ordinary Portland cement as per IS specifications should not be less than
Answer: (b) 30 minutes
IS 269 / IS 4031 lay down for ordinary Portland cement: Initial setting time — not less than 30 minutes. This is the time available for mixing, transporting, placing and compacting the concrete before the paste begins to stiffen. Too short a time and the concrete cannot be placed properly. Final setting time — not more than 600 minutes (10 hours).
Q89.For earthquake resistant masonry buildings, the vertical distance between opening one above the other in a load bearing wall shall not be less than
Answer: (b) 60cm
IS 4326 (earthquake-resistant design and construction of buildings) requires that in a load bearing wall the vertical distance between two openings one above the other shall not be less than 60 cm, and in any case not less than half the width of the smaller opening. The reason is the behaviour of the masonry pier between openings.
Q90.Maximum super elevation on hill roads should not exceed
Answer: (d) 10%
IRC practice on superelevation limits: Plain and rolling terrain — 7 % maximum; Hill roads not bound by snow — 10 % maximum; Snow-bound hill roads — 7 %; Urban roads with frequent intersections — 4 %. The hill-road allowance is higher because the radii available are severely constrained by the terrain, and steeper banking is the only way to keep the design speed reasonable on tight curves.
Q91.The cost slope is defined as
Answer: (d) (crash cost - normal cost)/(normal time - crash time)
Cost slope is the rate at which the direct cost of an activity increases per unit of time saved by crashing it: Cost slope = (Crash cost − Normal cost) / (Normal time − Crash time) Both the numerator and the denominator must be differences. The numerator is the extra money spent; the denominator is the time bought with it.
Q92.Number of bricks required for one cubic meter of brick masonry is
Answer: (c) 500
Take the standard modular brick with mortar, i.e. the nominal size 20 × 10 × 10 cm (the actual brick being 19 × 9 × 9 cm, the difference being the 10 mm mortar joint).
Q93.To compensate the loss of tractive force of vehicles along curves of radius R, the percentage reduction of gradient is
Answer: (b) 75/R
A vehicle rounding a curve loses tractive effort to curve resistance. If the curve also lies on a rising gradient, the combined demand can exceed what the vehicle can deliver, so the gradient is eased through the curve. This easing is grade compensation. IRC specifies Grade compensation = (30 + R)/R per cent, subject to a maximum of 75/R per cent Of the four options only 75/R appears in the code, and it is the value the question is after.
Q94.Percentage of carbon content in mild steel is
Answer: (a) less than 0.25
Q95.Crushing strength of good building stone should be more than
Answer: (b) 100 MPa
IS specifications require the crushing (compressive) strength of a good building stone to be not less than 100 MPa (100 N/mm²). The test is on a 40 mm cube, saturated, loaded at a specified rate, with the strength reported as the average of three specimens.
Q96.Latest start of an activity is always
Answer: (a) greater than or equal to latest event time of preceding node
By definition, the latest start time of an activity is LST = LFT − duration and the latest finish time is the latest event time of the succeeding node. Working backwards through the network, the latest start of an activity must be at least as late as the latest allowable occurrence of the event that precedes it — otherwise the activity would be required to begin before its preceding event could occur, which is impossible.
Q97.The ideal shape of a transition curve is
Answer: (a) Clothoid
The ideal transition curve is the clothoid, also called the spiral or Euler's spiral. Its defining property is exactly what a transition must provide: L r = constant, i.e. the radius varies inversely with the length travelled along the curve Since the centrifugal force is mV²/r, an inverse-linear radius means the centrifugal force grows linearly with distance along the curve.
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