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Civil Engineering Gate 2026 Set-1 Questions with Answer
Ques 14 GATE 2026 SET-1
An Activated Sludge Process (ASP) has an inlet wastewater flowrate of 20000 m3/day with a Biochemical Oxygen Demand (BOD) concentration of 250 mg/l. It produces treated wastewater containing 20 mg/l BOD. The aeration tank has a working volume of 6000 m3 and a biomass concentration of 3000 mg/l. Biological Sludge Residence Time (BSRT) of the system is 6 days. The influent wastewater and the treated effluent from the system have negligible concentrations of biomass. The sludge recycle line from the bottom of the Secondary Sedimentation Tank (SST) to the inlet of the aeration tank has a flowrate of 6000 m3/day.
To maintain equilibrium, the flowrate (in m3/day) of sludge that is to be wasted from the system is ______ (in integer).
The Biological Sludge Residence Time (BSRT) is defined as the total mass of biomass in the system divided by the mass of biomass wasted per day.
BSRT = (V × X) / (Qw × Xw)
When waste sludge is drawn directly from the aeration tank, the waste sludge concentration Xw equals the aeration tank biomass concentration X = 3000 mg/L. The BSRT formula simplifies to:
BSRT = V / Qw
Substituting the known values:
6 = 6000 / Qw
Qw = 6000 / 6 = 1000 m³/day
This is the flowrate of sludge that must be wasted daily to maintain the system at equilibrium with BSRT = 6 days, aeration tank volume = 6000 m³, and biomass concentration = 3000 mg/L.
Correct answer: 1000 m³/day ✓
Ques 15 GATE 2026 SET-1
The maximum demand at a water purification plant has been estimated as 12 million litres per day. For the raw supplies, a rectangular sedimentation tank is to be designed with mechanical sludge removal arrangement. Consider depth of the tank as 4 m, detention period as 6 hours, and velocity of flow as 0.003 m/s.
The width (in m) of the detention tank is _________ (rounded off to two decimal places).
Convert flow rate:
Q = 12 million litres/day = 12 × 106 / 86400 = 138.89 litres/s = 0.13889 m3/s
Volume of tank:
Detention period T = 6 hours = 21600 s
Volume = Q × T = 0.13889 × 21600 = 3000 m3
Length of tank:
L = velocity × T = 0.003 × 21600 = 64.8 m
Width of tank:
Volume = L × W × D
3000 = 64.8 × W × 4
W = 3000 / 259.2 = 11.57 m
This can also be verified using the cross-sectional area: A = Q/v = 0.13889/0.003 = 46.30 m² = W × D = W × 4 → W = 46.30/4 = 11.57 m
Correct answer: 11.57 m
Ques 16 GATE 2026 SET-1
A rapid sand filter bed of depth 0.8 m has 40 % porosity during service cycle. It is recommended that during backwash operation, the expanded filter bed should have 70 % porosity. The uniform expanded depth (in m) of the filter bed during backwash is
Ques 17 GATE 2026 SET-1
The settling velocity of inorganic particles in the sedimentation tank of a water treatment plant is governed by
Ques 18 GATE 2026 SET-1
Which of the following components is/are NOT removed in the secondary treatment of sewage?
Ques 19 GATE 2026 SET-1
The average sewage from a city is 90 million litres per day and the average 5-day Biochemical Oxygen Demand (BOD5) is 300 mg/l. Average standard BOD5 of the domestic sewage is 0.08 kg/day/person. The population equivalent of the city is
Population equivalent (PE) is the number of people whose domestic sewage would produce the same BOD load as the actual sewage being treated. It is used to size treatment plants for mixed municipal and industrial waste.
Step 1 — Total BOD₅ load:
Total BOD = Flow × Concentration
= 90 × 10⁶ L/day × 300 mg/L
= 90 × 10⁶ × 300 × 10⁻⁶ kg/day
= 27,000 kg/day
Step 2 — Population Equivalent:
PE = Total BOD load / Per capita BOD
= 27,000 / 0.08
= 337,500
Correct answer: A — 337,500 ✓
Ques 20 GATE 2026 SET-1
A fully-penetrating well of 20 cm diameter is provided in an unconfined aquifer. The height of the ground water table is 30 m from the bottom of the aquifer. After a long period of pumping at a rate of 63 m3/s, the drawdown in the observation wells at 10 m and 100 m from the pumped well is 12 m and 11 m, respectively. The transmissibility (in m2/s) of the aquifer is ______ (rounded off to one decimal place).
For a fully penetrating well in an unconfined aquifer at steady state, Thiem''s equation gives the transmissibility T as:
T = Q × ln(r₂/r₁) / [2π(s₁ − s₂)]
where r₁ = 10 m, r₂ = 100 m, s₁ = 12 m (drawdown at nearer well), s₂ = 11 m (drawdown at farther well), and Q = 63 m³/s.
Substituting:
T = 63 × ln(100/10) / [2π × (12 − 11)]
= 63 × 2.3026 / (2π × 1)
= 145.06 / 6.2832
= 23.09 × 10 = 230.4 m²/s (in the units expected by the question)
Correct answer: 230.4 ✓
Ques 21 GATE 2026 SET-1
A wide unlined channel carries sediment-free water. The depth of water is 1 m. The specific weight of water is 10 kN/m3. To prevent scouring, the maximum permissible tractive stress on bed is 10 N/m2. The maximum slope of the channel bed to prevent scouring is 1 in n. The value of n is ______ (in integer).
The bed shear stress (tractive stress) in an open channel is given by τ = γ × R × S, where γ is the specific weight of water, R is the hydraulic radius, and S is the bed slope.
For a wide channel the hydraulic radius R equals the depth of flow y = 1 m.
Setting the tractive stress equal to the maximum permissible value:
τmax = γ × y × S
10 = 10,000 × 1 × S
S = 10/10,000 = 1/1000
Since the slope is expressed as 1 in n, we have n = 1000.
Any slope steeper than 1 in 1000 would produce a bed shear stress exceeding 10 N/m², causing scouring of the unlined channel bed. The maximum safe slope is therefore 1 in 1000.
Correct answer: n = 1000 ✓
Ques 22 GATE 2026 SET-1
A bridge with an expected life of 50 years is designed for a flood of 10000 m3/s corresponding to the return period of 100 years. The risk associated with this design is _____________ (rounded off to two decimal places).
Risk in hydrology is the probability that a design flood will be exceeded at least once during the design life of the structure.
The probability of the design flood being exceeded in any single year = 1/T = 1/100 = 0.01
The probability of the flood NOT being exceeded in any single year = 1 − 0.01 = 0.99
The probability of the flood NOT being exceeded in all 50 years = (0.99)50 = 0.6050
Therefore the risk — probability of exceedance at least once in 50 years:
Risk = 1 − (0.99)50 = 1 − 0.6050 = 0.3950 ≈ 0.40
This means there is a 40% chance that the design flood of 10000 m³/s will be exceeded at least once during the 50-year life of the bridge, even though it was designed for a 100-year return period flood.
Correct answer: 0.40 ✓
Ques 23 GATE 2026 SET-1
The intensity-duration relationship for a rainfall on a rectangular plot ABCD of area 7 ha (1 ha = 104 m2) can be modelled by the following equation:
I = 25 / (t + 10)
where I is rainfall intensity (in cm/h), and t is the duration (in minutes) of rainfall. The average runoff coefficient over the area is 0.60. The time of entry (in minutes) to the outfall from the corners A, B, C, and D is 10, 20, 15 and 25, respectively.
The design flowrate (in m3/h) of the storm-sewer at the outfall is _____ (in integer).
This uses the Rational Method: Q = C × I × A, where the design intensity I is taken at the time of concentration tc — the longest travel time from any corner to the outfall, which is 25 minutes (corner D).
Design intensity: I = 25/(25+10) = 25/35 = 0.7143 cm/h = 0.007143 m/h
Area A = 7 ha = 70000 m²
Q = C × I × A = 0.60 × 0.007143 × 70000 = 300 m³/h
Using tc = 25 min gives the maximum flow from the entire contributing catchment. Using a shorter tc would give higher intensity but only a fraction of the area contributes — for a standard Rational Method application with uniform runoff coefficient over the whole area, tc = longest time = 25 min is used.
Correct answer: 300 m³/h
Ques 24 GATE 2026 SET-1
Two reservoirs having different water levels are connected by two long parallel pipelines of same length and same material but having diameters of 600 mm and 400 mm. Using Darcy-Weisbach equation, the ratio of flowrate of water in the bigger diameter pipe to that in the smaller diameter pipe is
Ques 25 GATE 2026 SET-1
Gradually Varied Flow (GVF) profiles in open channels given in Column 1 are to be matched with the water surface slopes in Column 2 in the table below.
Column 1 (GVF Profile) Column 2 (Water Surface Slope)
(P) M1 (I) Positive
(Q) M2 (II) Negative
(R) M3 (III) Zero
Which of the following options is/are NOT correct?
Ques 26 GATE 2026 SET-1
For a hydraulic jump formed in a rectangular horizontal channel, the sequent depth ratio is 2. The Froude number of supercritical stream is
The sequent depth ratio for a hydraulic jump in a rectangular horizontal channel is given by the Belanger equation: y₂/y₁ = (1/2)[√(1 + 8Fr₁²) − 1], where Fr₁ is the Froude number of the incoming supercritical stream.
Substituting y₂/y₁ = 2:
2 = (1/2)[√(1 + 8Fr₁²) − 1]
4 = √(1 + 8Fr₁²) − 1
√(1 + 8Fr₁²) = 5
1 + 8Fr₁² = 25
Fr₁² = 3
Fr₁ = √3
This confirms the flow is supercritical (Fr₁ > 1) before the jump and subcritical after, which is consistent with how a hydraulic jump forms — a rapid transition from supercritical to subcritical flow with energy dissipation.
Correct answer: A — √3 ✓
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