UPSC Civil Engineering 2026 — Paper II

(a) 'Explain with a neat sketch the essential features' — the answer must both draw a labelled sketch (plan of two successive courses plus an elevation of a one-brick wall) and list the defining rules of English bond: alternate header and stretcher courses, the queen closer next to the quoin header, the quarter-brick lap and the avoidance of continuous vertical joints. (b) For each of the five named cements the answer must state the site conditions/works for which it is the best choice AND the reason (the compositional or physical property that makes it suitable). All five sub-parts (i)–(v) must be attempted; 2 marks each. (c) Using the declining/diminishing balance (constant percentage) method of depreciation with P = Rs 80,000, scrap value S = Rs 20,000 and life n = 10 years, calculate (i) the annual percentage rate of depreciation and (ii) the depreciation fund accumulated after 2 years. (d) Two parts: (a) state the purpose/functions of a marshalling yard in railway operation, and (b) discuss the various types — flat (level) yard, gravity yard and hump yard — with how each works and where each is suitable. (e) Two linked demands: (a) set out the key steps of flexible-overlay design over an existing flexible pavement by the Benkelman Beam Deflection method (IRC:81), and (b) explain why, when and how the corrections — leg/beam movement, pavement temperature and seasonal (subgrade moisture) variation — are applied, ending in the characteristic deflection and overlay thickness.

• Labelled sketch: elevation and plans of two successive courses of a one-brick wall
• Alternate courses of headers and stretchers
• Queen closer next to the quoin header of the header course
• Quarter-brick lap between courses, with no continuous vertical joints
• PPC: hydraulic, marine and mass works or aggressive exposure; the pozzolana takes up free lime, giving lower heat, permeability and better sulphate resistance
• Rapid hardening: early strength (precast, early formwork removal, repair, cold weather); more C₃S and finer grinding
• Quick setting: underwater or running water, grouting; less gypsum and added aluminate, with set in about 5 and 30 minutes
• Low heat: mass concrete such as dams; less C₃S and C₃A, more C₂S, so slow heat release
• Air-entraining: freeze-thaw and de-icing exposure; tiny bubbles relieve freezing pressure
• S = P(1 − r)ⁿ, so (1 − r)¹⁰ = 0.25
• r = 12.94% per year
• Book value after 2 years = 80,000 × 0.8706² ≈ ₹60,629
• Depreciation fund = ₹19,371
• Purpose: receive, sort and make up wagons into new trains by destination; other functions such as storage and inspection
• Flat yard: level tracks, shunting locomotive pushes wagons; simple, slow, suits small traffic
• Gravity yard: falling gradient, wagons roll by gravity under brake control; moderate traffic
• Hump yard: locomotive pushes rake to an artificial hump, wagons roll down to sorting sidings with retarders; fastest, for heavy traffic
• Steps: design traffic (msa), condition survey and section division, deflection test (8170 kg, 5.6 kg/cm², at most 50 m apart)
• Leg correction: X_a = 2(d₀ − d_f), X_t = X_a + 2.91Y, and the rebound deflection is twice X_t
• Temperature: 0.01 mm per °C about 35 °C for bituminous layers of 40 mm or more; seasonal: chart factor by soil, rainfall and moisture
• Characteristic deflection mean + 2σ (NH and SH) or mean + σ, and the overlay thickness from the chart against traffic

(a) A descending 0.5% gradient meets a descending 3.0% gradient, so the deviation angle N = 0.005 - 0.03 in magnitude 0.025 and the curve is a SUMMIT curve. Compute the curve length required for (i) stopping sight distance and (ii) overtaking sight distance at a design speed of 65 km/h, with f = 0.35 and acceleration 3.6 km/h/s, stating all assumed data. (b) Reduce the fly-levelling readings (4 backsights, 3 foresights) from a benchmark of RL 381.605 to get the height of instrument at the last setup, then compute the staff readings needed to set the tops of four pegs at 50 m intervals on a 1 in 100 falling gradient starting from a first-peg RL of 380.500. (c) Four deliverables from the 11-activity table: (1) draw the activity-on-arrow network for events 1 to 8, (2) do the forward and backward passes, (3) identify the critical path and project duration, and (4) tabulate the activity times (EST, EFT, LST, LFT) and total float of every activity.

• N = 0.025 and a summit curve; assumed t = 2.5 s (SSD), 2.0 s (OSD), H = 1.2 m, h = 0.15 m (SSD) and 1.2 m (OSD)
• SSD = 0.278Vt + V²/254f = 92.7 m
• OSD = d₁ + d₂ + d₃ ≈ 308 m, with v_b = V − 16, s = 0.7v_b + 6 and T = √(4s/a)
• Curve lengths by testing L > S then L < S: about 9.5 m for SSD (minimum practical length used) and 232 m for OSD
• HI by set-up: 383.820, 383.825, 382.130 and 381.400; change-point RLs 382.795, 380.535 and 379.745
• Arithmetic check ΣBS − ΣFS = −1.860 matching the RL difference
• Peg RLs 380.500, 380.000, 379.500 and 379.000
• Staff readings 0.900, 1.400, 1.900 and 2.400 m from HI 381.400
• Correct network with the 11 activities and their durations
• Forward pass E: 0, 10, 12, 18, 26, 34, 40, 52 and backward pass L: 0, 18, 12, 18, 26, 34, 40, 52
• Critical path 1-3-4-5-6-7-8, project duration 52 days
• Activity table with total floats: 1-2: 8, 2-5: 8, 2-7: 18, 3-6: 17, 5-7: 4, and zero on the critical activities

(a) Three demands: define the Gantt/bar chart in the context of construction project management, illustrate it with a neat labelled sketch, and discuss its limitations with suitable construction examples (the limitations carry the larger weight). (b) Explain, with a neat sketch of the strength versus water-cement ratio curve, how the degree of compaction achieved by vibration modifies the inverse strength-w/c relationship (Abrams' law holds only for fully compacted concrete), and then state the advantages of vibrating fresh concrete. (c(i)) For a BG curve of R = 1000 m with cant provided for an equilibrium speed of 90 km/h, compute the equilibrium cant and then the maximum permissible speed (cant + cant deficiency of 75 mm) and the minimum permissible speed (cant - cant excess of 75 mm). (c(ii)) Two parts in about 75 words: give a precise definition of remote sensing, and then bring out how it differs from photogrammetry (purpose, spectral range, sensors and output).

• Definition: horizontal bars against a time scale, showing start, finish, duration and progress
• Neat labelled sketch with an activity list, time axis and progress or today line
• Limitations: no dependencies, no critical path or float, and poor for large complex projects
• Further limitations with a construction example: hard to update, no uncertainty or resource information
• Abrams' law: strength falls as w/c rises, valid only for fully compacted concrete
• Sketch of strength against w/c: vibrated curve keeps rising at low w/c, hand-compacted curve turns down because of voids
• Vibration makes stiff, low w/c mixes compactable; 1% voids costs about 5–6% strength
• Advantages: denser and stronger, lower w/c, better bond, finish and economy, fewer voids
• C = GV²/127R with G = 1750 mm gives 111.6 mm
• Maximum: cant plus deficiency 186.6 mm, V_max ≈ 116.4 km/h
• Minimum: cant minus excess 36.6 mm, V_min ≈ 51.5 km/h
• Definition: information without contact, by recording reflected or emitted electromagnetic energy
• Photogrammetry: accurate measurements and maps from overlapping photographs
• Differences in purpose, spectral range (visible only against multispectral, thermal and microwave), sensors and output

(a) Three demands: define the effective CBR of a subgrade as used in the latest IRC flexible pavement design guidelines, discuss why it matters for pavement design, and describe how it is determined (two-layer subgrade, chart, and conversion to resilient modulus). (b(i)) 'List the various applications' — an enumerated answer is wanted, covering the planning, design, terminal, modelling and traffic-management uses to which origin-destination data are put; a one-line mention of the survey methods is enough by way of context. (b(ii)) Two demands: explain what orientation of a plane table means and why it is necessary, and then discuss each method of orientation — by magnetic needle, by backsighting and by resection (three-point and two-point problems) — bringing out the suitability, advantages and disadvantages of each. (c(i)) Two demands: briefly differentiate accelerating admixtures from retarding admixtures (what each does to setting and strength, with named chemicals), and explain how each modifies the properties of fresh and hardened concrete, with the situations in which each is used. (c(ii)) Two demands: define creep of hardened concrete precisely, and draw the labelled time-versus-deformation diagram showing elastic strain on loading, creep under sustained load, instantaneous recovery and creep recovery on unloading, and the residual permanent strain.

• Definition: single CBR (or modulus) equivalent to the layered subgrade, giving the same surface deflection
• Significance: the subgrade strength governs pavement thickness and the subgrade strain criterion; a weak embankment below a good top layer is otherwise ignored
• Calculation: 4-day soaked CBR of each layer converted to M_RS = 10 CBR (≤ 5%) or 17.6 CBR^0.64 (> 5%)
• Two-layer (500 mm over semi-infinite) IITPAVE deflection for 40 kN and 0.56 MPa, then effective modulus, capped at 100 MPa, and back to effective CBR
• Planning: desire lines, trip generation and attraction, forecasting, new links and bypasses
• Design: location of bypasses, intersection and flyover types, lanes and capacity
• Terminals and facilities: bus, truck and rail terminals, parking, toll plazas, public transport routes
• Modelling and traffic management: four-stage models, one-way schemes, diversions, signal coordination
• Orientation: table set so that the sheet lines are parallel to the ground lines; needed for consistent plotting from different stations
• Magnetic needle: quick, small scale, affected by local attraction
• Backsighting: accurate, standard for traversing, needs a plotted line and visibility, error propagates
• Resection: three-point (Bessel, Lehmann, mechanical; danger circle) and two-point problems
• Accelerators shorten the setting time and raise early strength (CaCl₂, calcium formate, triethanolamine)
• Retarders delay the setting time (gypsum, sugar, lignosulphonate, tartaric acid)
• Effects on fresh concrete (setting time, workability) and hardened concrete (early and later strength, shrinkage)
• Uses: cold weather, repair, shotcrete for accelerators; hot weather, long haul, large pours for retarders
• Definition: increase in strain with time under sustained constant stress, over and above the instantaneous elastic strain
• Diagram: elastic strain on loading, creep strain rising at a decreasing rate to a limit
• On unloading: instantaneous recovery and creep recovery, both labelled
• Residual permanent strain shown, and axes labelled with t₀ and t₁

(a) Three demands: distinguish a flood hydrograph from a mass curve, construct and plot the flood hydrograph from the twelve monthly discharges given for the dam site, and read off the minimum and maximum discharges in the river. (b) From the six annual rainfall values, compute (i) the standard error (percentage error) in the estimate of mean rainfall from the existing six gauges, and (ii) the optimum number of gauges for a 10% error, using N = (Cv/e)^2. (c) From a steady-state pumping test on an unconfined aquifer (Q = 5400 l/min, initial saturated thickness 25 m, drawdowns 1.11 m at 30 m and 0.53 m at 90 m), find the coefficient of permeability and hence the transmissivity of the aquifer. (d) Two demands: state which variation in demand governs the design of each of the five named units of a water supply system and why, and then work out the actual design values for a town of 15000 persons at 270 litres per capita per day. (e) Write and balance the three oxidation steps for glycine — carbon to CO2 with nitrogen released as ammonia, ammonia to nitrite, nitrite to nitrate — sum the oxygen demanded in all three and express the theoretical oxygen demand per mole and per gram of glycine.

• Hydrograph: discharge against time; mass curve: cumulative volume against time (its integral, slope = discharge)
• Hydrograph plotted with axes labelled and the points joined
• Maximum 210 m³/s in August and minimum 100 m³/s in December
• Mean 118.1 cm, standard deviation 35.4 cm (n − 1 divisor), C_v = 29.9%
• Standard error ε = C_v/√N = 12.2%
• N = (C_v/e)² = 8.97, so about 9 gauges (3 more)
• Thiem equation for an unconfined aquifer Q = πK(h₂² − h₁²)/ln(r₂/r₁)
• h₁ = 23.89 m and h₂ = 24.47 m (thickness 25 m less drawdown), Q = 0.09 m³/s
• K ≈ 1.12 × 10⁻³ m/s (about 97 m/day)
• T = K b ≈ 0.028 m²/s (about 2420 m²/day)
• Average demand 4050 m³/day; maximum daily 1.8 × average = 7290 m³/day; peak hourly 2.7 × average hourly = 455.6 m³/h
• Filters, sedimentation tank and pumps: maximum daily demand, with reasons
• Distribution mains: maximum hourly demand of the maximum day
• Clear water reservoir: hourly variation (balancing) and reserve, about one-third of the daily demand
• Step 1: C₂H₅NO₂ + 1.5 O₂ → 2 CO₂ + H₂O + NH₃
• Steps 2 and 3: NH₃ + 1.5 O₂ → NO₂⁻ + H⁺ + H₂O, and NO₂⁻ + 0.5 O₂ → NO₃⁻
• Total 3.5 mol O₂ per mol of glycine = 112 g
• ThOD = 112/75 = 1.49 g O₂/g glycine

(a) Using Brune's trap-efficiency versus capacity-inflow relation supplied in the table, find the probable useful life of a 30 Mcum reservoir (annual inflow 60 Mcum, sediment inflow 360000 tonnes per year at 1200 kg/cum) up to the point at which 80% of the initial capacity is silted, by the incremental (step-by-step) method because trap efficiency falls as the capacity shrinks. (b) For a 75 cm deep uniform sand bed (d = 0.55 mm, G = 2.67, porosity 0.40) expanded by 50% during backwashing, compute (i) the backwash (fluidisation) rate, using Stokes' law because the drag coefficient is prescribed as CD = 24/R, and (ii) the head loss through the expanded bed. (c(i)) Explain how the velocity in a circular sewer changes as the depth of flow (and hence the discharge) varies — through the hydraulic mean depth in Manning's equation — and state the design consequences (self-cleansing velocity at minimum flow, non-scouring velocity at maximum flow). (c(ii)) Three demands in about 105 words: compute the threshold odour number for 20 mL of sample made up to 200 mL, comment on whether such water is acceptable to consumers, and state the correct procedure for an odour test given that individual sensitivity to odour varies.

• Sediment volume 360 000 × 1000/1200 = 0.30 Mm³/year; initial capacity-inflow ratio 0.5
• Life ends at 24 Mm³ lost, capacity 6 Mm³ (ratio 0.1)
• Steps of 6 Mm³ using mean trap efficiency 95.75%, 95.25%, 94% and 90% from the table
• Times 20.9, 21.0, 21.3 and 22.2 years, total about 85 years
• Stokes settling velocity v_s = g(G − 1)d²/18ν = 21.2 cm/s
• Expanded-bed porosity 0.60 from (1 − ε_e)L_e = (1 − ε)L
• Backwash velocity v_b = v_s ε_e^4.5 ≈ 2.13 cm/s (about 77 m/h)
• Head loss (1 − ε)(G − 1)L = 0.75 m
• Manning v = (1/n)R^(2/3)S^(1/2); velocity depends on R = A/P at fixed n and S
• Same velocity at full and half depth (R = D/4); maximum v ≈ 1.14 V at 0.81D and maximum Q ≈ 1.07 Q at 0.95D
• Velocity falls below half depth (about 0.7 V at 0.25D)
• Design: self-cleansing velocity at minimum flow and non-scouring velocity at maximum flow
• TON = (A + B)/A = 200/20 = 10
• Comment: not acceptable for drinking (limit about 3), needs treatment
• Procedure: panel of at least five persons, odour-free room and dilution water, blanks, increasing concentrations, geometric mean of the thresholds

(a) For a 5 MLD sedimentation tank treating sewage at 250 ppm SS with 60% removal, compute the daily sludge production by weight and by volume at 97% and at 96% moisture content (wet sludge specific gravity 1.02), and then the percentage reduction in sludge volume caused by the 1% reduction in moisture content. (b) Two demands: draw a neat labelled sketch (plan) of a diversion headwork showing all its components, and then explain briefly the function of each of the five named items — silt excluder, cross regulator, canal fall, aqueduct and escape. (c(i)) Three quantities are asked for a mass concrete dam with river bed width 100 m, 10 m depth of water upstream, deepest foundation at RL 100.0 m and spillway crest at RL 160.0 m: (I) the hydraulic height, (II) the total (structural) height, and (III) the total hydrostatic force on the upstream face. The definitions of the two heights must be stated because the data fix them only through an assumption. (c(ii)) Classify the five named hydraulic structures by their mode of operation: dams as storage works, weir/barrage as diversion (river) works, aqueduct and super passage as cross-drainage works, and falls as canal drop/energy-dissipation works — with the distinguishing feature of each class.

• Dry solids 5 × 10⁶ l × 250 mg/l × 0.60 = 750 kg/day
• At 97% moisture: 750/0.03 = 25 000 kg/day and volume 24.51 m³/day
• At 96% moisture: 750/0.04 = 18 750 kg/day and volume 18.38 m³/day
• Reduction in volume (24.51 − 18.38)/24.51 = 25%
• Labelled plan: weir or barrage, undersluices, divide wall, head regulator, silt excluder, fish ladder, guide banks and marginal bunds
• Silt excluder: tunnels in the undersluice pocket that pass bottom silt downstream (not a silt ejector)
• Cross regulator: gate across the parent canal below an offtake to head up water and control discharge; canal fall: drop and energy dissipation on steep ground
• Aqueduct: canal carried over a drainage; escape: releases surplus canal water to a natural drain
• Definitions: hydraulic height is the highest controlled water level above the lowest original bed; total height is the top of dam above the lowest foundation
• Assumptions stated: reservoir at crest RL 160 with bed at RL 150, top of dam at crest
• Hydraulic height 10.0 m and total height 60.0 m
• F = ½γh²b = ½ × 9.81 × 10² × 100 = 49 050 kN, acting at h/3 above the bed
• Dams: storage works
• Barrage or weir: diversion (river) works
• Aqueduct and super passage: cross-drainage works
• Falls: canal drop or energy-dissipation works, with the distinguishing feature of each class

(a(i)) Three demands: explain Lacey's silt factor and state what f = 1.0 signifies in terms of the bed material, then design the canal section (bed width, depth, velocity, area, perimeter, hydraulic radius) and find the discharge for S = 1/5000, f = 0.90 and side slopes of 1/2 H : 1 V. (a(ii)) In about 75 words, enumerate the effects of waterlogging — on soil aeration and crop yield, on salinity/alkalinity, and the ancillary agricultural, health and structural consequences. (b) For a 1000 MW coal-fired plant at 70% load factor and 42% efficiency (coal CV 30 MJ/kg, ash 7% of which 77% is fly ash, carbon 78%, sulphur 1%), find the annual coal consumption and the annual emissions of particulates, CO2 and SO2 with no control device, and then the particulate emission with a 99%-efficient electrostatic precipitator. (c) 'Enumerate the key additions ... to emphasize that it is a paradigm shift' — the answer must first establish what the Solid Waste Management Rules, 2016 already provided, then list the additions made by the 2026 Rules, and argue why the additions amount to a change of paradigm rather than an incremental amendment.

• f = 1.76√d (d in mm); f = 1.0 is standard medium silt, d ≈ 0.323 mm
• Lacey's equations: V = (Qf²/140)^(1/6), R = 5V²/2f, P = 4.75√Q, S = f^(5/3)/(3340Q^(1/6))
• Q ≈ 3.9 m³/s, V ≈ 0.53 m/s, A ≈ 7.37 m², P ≈ 9.4 m, R ≈ 0.79 m
• Trapezoid solved with P = B + 2.236D and A = BD + 0.5D²: D ≈ 0.95 m, B ≈ 7.3 m
• Rise of water table into the root zone, poor aeration and reduced crop yield
• Salinity and alkalinity from capillary rise and evaporation
• Ancillary effects: delayed tillage and sowing, mosquito breeding and disease
• Damage to foundations, roads and canal linings
• Average output 700 MW; energy 6.132 × 10⁶ MWh/yr; heat input at 42% efficiency
• Coal ≈ 1.75 × 10⁶ t/yr (about 55.6 kg/s)
• Fly ash ≈ 9.4 × 10⁴ t/yr with no control; ≈ 944 t/yr with a 99% ESP
• CO₂ ≈ 5.0 Mt/yr using 44/12 of the carbon mass; SO₂ ≈ 35 000 t/yr using 64/32 of the sulphur mass
• Notified 28 January 2026, in force 1 April 2026, superseding the 2016 Rules
• Four-stream source segregation (wet, dry, sanitary, special care) against three in 2016
• Polluter-pays environmental compensation and extended bulk waste generator responsibility (20 000 m², 40 000 l/day or 100 kg/day)
• Central online portal, landfill restriction and audits, legacy dumpsite biomining, RDF targets of 5% to 15%
• Paradigm shift: from local-body collection and disposal to generator responsibility, circular economy and enforcement by liability and digital tracking