GATE 2020 Civil Engineering Previous Year Paper
Civil Engineering
General Aptitude
Q.1 Sum of two positive numbers is 100. After subtracting 5 from each number product of the resulting number is 0. One of the original number is
(a) 95
(b) 92
(c) 85
(d) 90
Ans. (a)
Let the two positive numbers be x and y.
∴ x + y = 100 …(i)
(x – 5) (y – 5) = 0 …(ii)
⇒ x = 5 or y = 5
If x = 5 then y = 95
If y = 5 then x = 95
∴ One of the numbers is 95 since 5 is not in any of the options.
Q.2 If 0, 1, 2, …… 8, 9 are coded as O, P, Q, ….. W, X then 45 is coded as
(a) TS
(b) SS
(c) ST
(d) SU
Ans. (c)
∴ 45 is coded as ‘ST’.
Q.3 Unit place in 26591749110016 is
(a) 6
(b) 1
(c) 3
(d) 9
Ans. (b)
≃ 26591749(110016)
≃ Unit place of 9° = 1
∴ Cycle of equation is (9, 1) (9, 1), (9, 1)
So the answer will be 1.
Q.4 Insert seven numbers between 2 and 34 such is an AP. The sum of those seven inserted seven number is
(a) 124
(b) 130
(c) 120
(d) 126
Ans. (d)
2, a, (a + d), (a + 2d), …… (a + 6d), 34
∴ Total number of terms of AP(n) = 9
Let sum of seven inserted numbers = S
∴ S = 7/2[a + (a + 6d )] = 7[a + 3d ]
Tn = 34
Also, a – 2 = (a + d – a)
⇒ a – d = 2
Similarly a – 2 = 34 – (a + 6d)
⇒ a – 2 = 34 – a – 6d
⇒ 2a = 36 – 6d = 36 – 6(a – 2)
⇒ 2a = 36 – 6a + 12
⇒ 8a = 48
⇒ a = 6
∴ d = a – 2 = 6 – 2 = 4
∴ S = 7(a + 3d) = 7(6 + 3 × 4) = 126
General English
Q.5 It is a common criticism that most of the academicians live in their ________, so they are not aware of the real life challenge.
(a) Ivory tower
(b) homes
(c) Glass palaces
(d) Big flats
Ans. (a)
Q.6 His hunger for reading is insatiable. He reads indiscriminately. He is most certainly a/an _________ reader.
(a) all round
(b) voracious
(c) wise
(d) precarious
Ans. (b)
Q.7 If Fuse : Fusion :: Use : ___________.
(a) Usage
(b) Usion
(c) Uses
(d) User
Ans. (a)
Solid Mechanics
Q.1 In a 2-D stress analysis, the state of stress at point P is 
(a) (σxx – σyy)2 + 4τ2xy = 0
(b) τxy
(c) σxx + σyy = 0
(d) σxxσyy – τ2xy = 0
Ans. (c)
In pure shear condition
σx = 0, σy = 0, τxy = τ
For this condition (c) is correct
σxx + σyy = 0
Q.2 A cantilever beam PQ of uniform flexural rigidity is subjected to concentrated moment M at R
Deflection at free end Q.
(a) 3ML2/8EI
(b) ML2/6EI
(c) 3ML2/4EI
(d) ML2/6EI
Ans. (b)
(d) M
δc = δ1 = CC1 + C1C2
= BB1 + C1C2
= ML222EI+ML2EIL2
δc = 3ML28EI
Q.3 The initial condition is shown in figure. Axial stiffness of each of the bars is 10 kN/mm. A load W is applied such that the system remains horizontal and total deflection is 5 mm. The load W will be ________.
Ans. (130 kN)
Structural Analysis
Q.4 How many zero force members are there in the above truss
(a) 6
(b) 7
(c) 8
(d) 9
Ans. (c)
As ΔAB = 0, hence FAB = 0
Total number of zero force member = 8
Q.5 A planer elastic structure is subjected to UDL as shown in figure. Where maximum BM any in the structure will be ________.
Ans. (0)
Mx = 0 of all points
Q.6 Distributed load (continuous or in patches) of 50 kN/m may occupy any position on the grid the maximum –ve BM at R will be
(a) 56.25 kN-m
(b) 150 kN-m
(c) 93.75 kN-m
(d) 22.5 kN-m
Ans. (a)
ILD for MS
∴ Mmax at support = − [½ × 1.5 × 1.5 × 50]
= –56.25 kNm or 56.25 kNm (hogging)
ILD for BM at the centre of span ‘QS’
MC = − [½ × 1.5 × 0.75 × 50] × 2
= –56.25 kNm or 56.25 kNm (hogging)
So, it can be concluded that for maximum hogging moment, overhang span (PQ & ST) can be loaded. And because of symmetry, the magnitude of maximum hogging moment remains the same throughout span QS.
Irrigation Engineering
Q.7 For a certain region:
Pan evaporation = 100 mm
Effective rainfall = 20 mm
Crop coefficient = 0.4
Irrigation efficiency = 0.5
What will be the amount of water required for irrigation?
Ans. (40 mm)
Water required by crop = 100 × 0.4 mm = 40 mm
Effective rainfall = 20 mm
Additional water required = 20 mm
Amount of water required after accounting irrigation efficiency = 20/0.5 = 40mm
Geotechnical Engineering
Q.8 Water flows in upward direction in a tank through 2.5 m thick sand layer. The e and G of sand are 0.58 and 2.7. Sand is fully saturated. γw is 10 kN/m3. Effective stress at point A, located 1 m above the base of the tank is _____ kN/m2.
Ans. (8.94 kN/m 2)
Q.9 SPT was conducted at a survey 1.5 m internal upto 30 m depth. At 3 m depth, the observed no. of hammer blows for 3 successive 150 mm penetration were 8, 6 and 9. SPT, N value at 3 m depth is
(a) 15
(b) 23
(c) 17
(d) 14
Ans. (a)
No. of blows for each 150 mm penetration 8, 6 and 9.
We will not consider the first 150 mm number of blows.
Hence, for the last 300 mm, the number of blows is 15.
Hence, observed SPT number = 15.
Q.10 Velocity of flow is proportional to the first power of hydraulic gradient in Darcy’s law, the law is applicable to
(a) turbulent flow in porous media
(b) transitional flow in porous media
(c) laminar flow in porous media
(d) laminar as well as turbulent flow in porous media
Ans. (c)
Darcy’s law is valid for laminar flow conditions.
Q.11 A drained triaxial test is conducted on a sand sample. If σd = 150 kPa and consolidation stress is 50 kPa the calculated the value of angle of internal friction (φ)?
Ans. (36.87°)
Q.12 A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. γsat and effective angle of internal friction of sand are 18 kN/m3 and 38°. Assume γw = 10 kN/m3. Seepage is parallel to the slope. Factor of safety against shear failure is ______.
Ans. (0.601)
Q.13 A vertical retaining wall of 5 m height has to support soil of γ = 18 kN/m3, effective cohesion 12 kN/m2 and effective friction angle 30°. As per Rankine, assuming that tension crack has occurred the lateral active thrust on wall per meter length is _____.
Ans. (21.714 kN/m)
After tension crack
Pa =½ × 16.144(5 – 2.309)
= 21.714 kN/m
Q.14 The water table is at a depth of 3 m below ground level, γw = 9.81 kN/m3. Surcharge is applied instantaneously Immediately after pre-loading, the effective stress (kN/m2) at point P and Q will be
(a) 54 and 95
(b) 36 and 90
(c) 36 and 126
(d) 124 and 204
Ans. (a)
( σ )P =q + 3γ = 3 × 18 = 54 Surcharge is applied suddenly.
∴ Excess pore water pressure developed.
( σ )Q = ( q + 3 γ b + 4 γ sat ) − (4 γ w + 70)
= 70 + 3 × 18 + 4 × 20 – (4 × 9.81 + 70)
= 94.76 kN/m2
Q.15 A fill of 2 m thick sand with unit weight of 20 kN/m3 is placed above the clay layer to accelerate the rate of consolidation of clay, cv = 9 × 10–2 m2/year. mv = 2.2 × 10–4 m2/kN. The settlement of clay layer 10 year after the construction is ____________.
Ans. (18.832)
Fluid Mechanics
Q.16 Uniform flow with velocity u makes angle Q with axis. The velocity potential φ is
(a) ±u(x sinθ – y cosθ)
(b) ±u(y sinθ + x cosθ)
(c) ±u(x sinθ + y cosθ)
(d) ±u(y sinθ – x cosθ)
Ans. (c)
Velocity in x-depth, ux = usinθ
Velocity in y-depth, uy = ucosθ
− ∂φ/∂x = ux
Integrating it φ = –uxx + f(x) + c
= –(usinθ)x + f(y) + c …(i)
− ∂φ/∂y= u y
Integrating it φ = –uy y + f(x) + c
= –(ucosθ)y + f(x) + c …(ii)
By equation (i) and (ii),
φ = − u ( x sin θ + y cos θ )
If we take ∂φ/∂x = ux and ∂φ/∂y = uy
Then φ = u( x sin θ + y cos θ )
So, φ = ± u (xsinθ + ycosθ)
Q.17 A floating body in a liquid is in a stable condition if
(a) metacentre lies above centre of gravity
(b) metacentre lies below centre of gravity
(c) centre of buoyancy lies below the centre of gravity
(d) centre of buoyancy lies above the centre of gravity
Ans. (a)
For stability of floating body M lies above G
GM > 0
Q.18 A circular water tank of 2 m dia has circular orifice of dia 1 m at bottom. Water is entering the tank at 20 l/s and escaping through orifice. Take CD for orifice = 0.8. Neglect any friction losses. g = 9.81 m/s2. Height of the water level in the tank at a steady rate is _____.
Ans. (0.5164 m)
Open Channel Flow
Q.19 For an open channel flow discharge is 12 m3/s and width is 6 m. Hydraulic jump is formed. Depth at upstream was 30 cm. Take g = 9.81 m/s2 and ρw = 100 kg/m3. Calculate the energy loss in jump?
(a) 114.2 MW
(b) 114.2 kW
(c) 141.2 J/s
(d) 141.2 HP
Ans. (a)
Q.20 For a rectangular open channel flow, the width of section is 4 m. The discharge is 6 m3/s. The Manning’s roughness coefficient is 0.02. The critical velocity for the channel will be ____________.
Ans. (2.45 m/s)
Critical depth (YC) = q2g1/3
=1.529.811/3=0.612 cm
Critical velocity (VC) = gYc = 9.81 × 0.612 = 2.45 m/s
Engineering Hydrology
Q.21 The probability that a 50 year flood may not occur at all during the 25 year life of a project is _________.
Ans. (0.603)
P = 1/T = 1/50 = 0.02
q = 1 – P = 0.98
∴ Probability of non-occurrence of an event is given by,
Assurance = qn
= (0.98)25
= 0.603
Environmental Engineering
Q.22 In a homogenous unconfined aquifer of area 3 km2, water table elevation is 102 m. After natural recharge of volume 0.9 million cubic meter (Mm3) the water table rose to 103.2 m. After this recharge, pumping took place and WT dropped down to 101.20 m. The volume of groundwater pumped after the natural recharge is __________ Mm3.
Ans. (1.5 Mm3)
VR = 0.9 Mm3
V = 3 × (103.2 – 102)
= 3 × 1.2 = 3.6 Mm3
ys or yR = VRV = 0.9/3.6
Now, 1/S = VD/V
VD = 0.9/3.6 [ 3 × ( 103.2 − 101.2 ) ]
VD = 1.5 Mm3
Q.23 35.67 mg HCl is added to distilled water and the total solution is made 1 l. The atomic weight of H and Cl is 1 and 35.5. Neglecting the dissociation of H2O, the pH of solution is
(a) 3.5
(b) 2.5
(c) 2.01
(d) 3.01
Ans. (d)
HCl →H+ + Cl–
1 mole of HCL gives 1 mole H+ ions
36.5 gm of HCl gives 1 gm of H+ ions
35.67 mg = 1/36.5 × 35.67 = 0.977 mg of H +
= 0.977 × 10− 3/1 = 9.77 × 10 − 4 moles of H+
pH = –log10[H+] = –log10[9.77 × 10–4]
= –log109.77 + 4log1010
= 4 – 0.989 = 3.01
Q.24 As chlorine reacts rapidly with water to form Cl–, HOCl and H+. The most active disinfectant in the chlorination process is
Cl2(g) + H2O ⇌ HOC l + Cl− + H+
(a) HOCl
(b) H+
(c) H2O
(d) Cl–
Ans. (a)
Q.25 A gaseous chemical has a concentration of 41.6 μmole/m3 in air at 1 atm pressure and temp 293 k gas constant R is 82.05 × 10–6 m3/atm/molK. Assume ideal gas law is valid, the concentration of gaseous chemical will be __________ ppm.
Ans. (1)
Q.26 For a river the discharge is 1000 MLD. BODs for the river are 5 mg/l and dissolved oxygen is 8 mg/l before receiving wastewater discharge at a location. For existing environment conditions dissolved oxygen is 10 mg/l. Waste water discharge of 100 MLD drains into the river. BOD ultimate of waste water is 200 mg/l and dissolved oxygen is 2 mg/l falls at a location. Assume complete mixing. Immediate do deficit is ______ mg/l.
Ans. (2.545)
Q.27 Surface overloading rate of primary settling tank (discrete) is 200 l/m2 per day. ν = 1.01 × 10–2 cm2/s. G = 2.64, the minimum dia of particle that will be removed with 80% of efficiency will be _______ μm.
Ans. (14.53)
Q.28 Stream with a flow rate of 5 m3/sec, BODU = 30 mg/L. Waste water discharge 0.2 m3/s, BOD5 = 500 mg/l joins the stream at a location and mix up instantaneously. Area of stream 40 m2 which remains constant. BOD exertion rate constant is 0.3/day (loge). BOD remaining 3 km downstream from the mixing location is _________.
Ans. (49.57)
Q.29 A water supply scheme transport 10 MLD water through a 450 mm diameter pipe for a distance 2.5 km. A chlorine dose of 3.5 mg/L is applied at the starting point. It is decided to increase the flow from 10 MLD to 13 MLD in the pipeline. Assume exponent for concentration n = 0.36 with this increased flow in order to attain the same level of disinfection. The chlorine is applied at the starting point.
(a) 3.95
(b) 4.4
(c) 5.55
(d) 4.75
Ans. (d)
Construction Materials
Q.30 The los angeles test for stone aggregate is used to examine
(a) crushing strength
(b) abrasion
(c) toughness
(d) none of these
Ans. (b)
Los Angeles test of stone is used for Abrasion resistance.
Q.31 During the process of hydration of cement, due to increase in C2S content in cement ,clinker, the heat of hydration
(a) does not change
(b) initially decrease the increase
(c) decrease
(d) increase
Ans. (c)
Geomatics Engineering
Q.32 An open traverse PQRST is surveyed using theodolite
If the independent coordinated (Northing and Easting) of station P are (400, 200 m) the independent coordinates of station T are
(a) 194.7, 370.1
(b) 405.3, 229.9
(c) 205.3, 429.9
(d) 394.7, 170.1
Ans. (d)
ΔL = –5.3
ΔD = –29.9
T, Northing{400 + (–5.3) = 394.7
T, Easting{200 + (–29.9) = 170.1
T [394.7 m, 170.1 m]
Q.33 The length of line segment SP is __________.
Ans. (44.784 m)
ΔL = 40cos80° + 50cos10° + 30cos210°
= 30.20
ΔD = 40sin80° + 50sin10° + 30sin210°
= 33.07
Length, SP = √ΔL2 + ΔD2
= 44.784 m
Highway Engineering
Q.34 Gradient of a road is 4.5%, radius is 100 m the compensated as per IRC will be _____.
Ans. (4%)
Gradient = 4.5%, R = 100 m
Grade compensation = (30 + R/R) ⊁ (75/R)% = 30 + 100/100 ⊁ 75/100 = 1.3% ⊁ 0.75} G.C = 0.75
Compensated Gradient = Gradient G.C = 4.5% – 0.75 = 3.75 ⊀ 4%
Hence C.G = 4%
Q.35 As per IRC, the minimum width of median in urban areas is ____________.
Ans. (1.2 m)
Desirable width of median in urban roads = 5m
And As per IRC : 86-1983 min. width = 1.2m
Q.36 For the contraction joint, which is correct position of dowel bar
Ans. (b)
RCC Structures and Prestressed Concrete
Q.37 Calculate the design bending moment of a singly reinforced section
If width = 300 mm
Effective depth = 450 mm
Area of steel in tension = 942 mm2
Grade of concrete is M25 and Fe500 Steel
Ans. (158.28 kN-m)
Q.38 A simply supported prismatic concrete beam of rectangular cross-section of span 8 m is prestressed with effective prestress force of 600 kN. Eccentricity is zero at support and varies linearly to value ‘e’ at mid span. Required value of e ________.
Ans. (40 mm)
P = 600 kN
Simply supported span = L = 8 m
To support a point load applied at mid span (W)
= 12 kN
Balancing load = Point load
2P sin θ = W
2P (E/L/2) = 2
2Pe × 2/L = W
4Pe/L = W
e = WL/4P = 12000N × 8000mm/ 4 × 600 × 1000N
= 40mm
Q.39 The variation of bond stress varies as
Ans. (c)
Design of Steel Structures
Q.40 For a given beam shown below the permissible stress in the weld is 20 N/mm2,
Ixx = 7.73 × 106 mm4. The allowable shear force in kN will be __________.
Ans. (323.527 kN)
Engineering Mathematics
Q.41 The area of an ellipse represented by an equation 
(a) πab/4
(b) 4πab/3
(c) πab
(d) πab/2
Ans. (c)
Q.42 The value of
(a) 1/2
(b) 1/4
(c) 0
(d) 1
Ans. (b)
Q.43 for the ordinary differential equation d2x/dt2 -5dx/dt+6x=0,intial condition x(0) =0 and dx/dt (0) =10 ,the solution of the given equation
(a) -10e2t+10e3t
(b)10e2t+10e3t
(c) 5e2t+6e3t
(d) -5e2t+6e3t
Ans.(a)
Q.44 
Ans.(3)
Q.45 If C represents, a line segment between (0, 0, 0) and (1, 1, 1) in Cartesian coordinate system, the value of line integral ∫C[ ( y + z ) dx + ( x + z )dy + ( x + y )dz] is
Ans. (3)
Q.46 If the true value of ln2 is 0.69. If value of ln2 is also calculated by linear regression between ln1 and ln6 then calculate the percentage error in the value of ln2.
Ans. (48.11%)
Q.47 θ = f(t, z) and D = f(θ), k = f(θ) then D ( θ ) ∂ ∂ z2 θ 2 + ∂ k ∂ ( z θ ) − ∂θ ∂ t = 0 is _______.
(a) Second order linear equation
(b) Second order nonlinear equation
(c) Second degree linear equation
(d) Second degree non-linear equation
Ans. (d)
∵ 1st term of given D. Equation contains a product of dependent variable with its derivative, so it is non-linear and also we have 2nd order derivative so it’s order is two i.e., 2nd order is non linear.
