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Q.1Q.1Q.1Q.1Q.1 Sum of two positive number is 100. After substracting 5 from each number product ofresulting number is 0. One of the original number is(a) 95 (b) 92(c) 85 (d) 90
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)Let the two positive numbers be x and y.∴ x + y = 100 ...(i)
(x – 5) (y – 5) = 0 ...(ii)⇒ x = 5 or y = 5If x = 5 then y = 95If y = 5 then x = 95∴ One of the number is 95 since 5 is not in any of the options.
End of Solution
Q.2Q.2Q.2Q.2Q.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.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)0,
O
1,
P
2,
Q
3,
R
4,
S
5,
T
6,
U
7,
V
8,
W
9
X∴ 45 is coded as ‘ST ’.
End of Solution
Q.3Q.3Q.3Q.3Q.3 Unit place in 26591749110016 is(a) 6 (b) 1(c) 3 (d) 9
Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)2, a, (a + d), (a + 2d), ...... (a + 6d), 34∴ Total number of terms of AP(n) = 9Let sum of seven inserted numbers = S
∴ S =7
[ ( 6 )] 7[ 3 ]2
a a d a d+ + = +
Tn = 34Also, a – 2 = (a + d – a)⇒ a – d = 2Similarly 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
End of Solution
Q.5Q.5Q.5Q.5Q.5 Five friends P, Q, R, S and T went to a picnic. They sleep in a row P, Q, T don’t wantto sleep near R as he snore loud. P and S ignoring Q as he hug legs in the sleep. Whatis the correct sequence of their position of sleeps.(a) RSTPQ (b) PTQSR(c) RSPTQ (d) PQRST
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)Option (c) satisfies the given conditions in the paragraph.
End of Solution
Q.6Q.6Q.6Q.6Q.6 The total montly expenditure of a family is shown in a pie-chart below. The money spenton different expenditures is shown below. What is the extra money spent on educationsas compared to transport in percent will be
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)Let total monhly earining = Rs. 100
Monthly spent on education =15
100 Rs.15100
× =
% money extra spent on education as compard to transportation
=15 10
100 50%10−
× =
End of Solution
General English
Q.7Q.7Q.7Q.7Q.7 It is a common critism that most of the academicians live in their ________, so theyare not aware of the real life challenge.(a) Ivory tower (b) homes(c) Glass palaces (d) Big flats
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)
End of Solution
Q.8Q.8Q.8Q.8Q.8 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.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)
End of Solution
Q.9Q.9Q.9Q.9Q.9 If Fuse : Fusion :: Use : ___________.(a) Usage (b) Usion(c) Uses (d) User
Q.2Q.2Q.2Q.2Q.2 A cantilever beam PQ of uniform flexure rigidity is subjected to concentrated momentM at R
L2
RQ
M
Deflection at free end Q.
(a)23
4MLEI
(b)23
8MLEI
(c)2
4ML
EI(d)
2
6ML
EI
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)
θ1
C1
C2
θ1B1
BA
L/2 L/2
δ1
M
C
δc = δ1 = CC1 + C1C2
= BB1 + C1C2
=
2
2 22 2
L LM M
LEI EI
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎛ ⎞+ ⎜ ⎟⎝ ⎠
δc =23
8MLEI
End of Solution
Q.3Q.3Q.3Q.3Q.3 The initial condition is shown in figure. Axial stiffness of each of the bar 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 ________.
2 mm
Rigid
W
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= –56.25 kNm or 56.25 kNm (hogging)ILD for BM at the centre of span ‘QS’
P Q S T
50 kN/m
1.5 m
ILD MC
0.750.75
1.25
(–) (–)
(+)
2.5 m2.5 m1.5 m C
MC =1
1.5 0.75 50 22
⎡ ⎤− × × × ×⎢ ⎥⎣ ⎦
= –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, magnitude of maximum hogging momentremains same throughout span QS.
Q.7Q.7Q.7Q.7Q.7 For a certain region:Pan evaporation = 100 mmEffective rainfall = 20 mmCrop coefficient = 0.4Irrigation efficiency = 0.5What will be the amount of water required for irrigation?
Ans.Ans.Ans.Ans.Ans. (40 mm)(40 mm)(40 mm)(40 mm)(40 mm)Water required by crop = 100 × 0.4 mm = 40 mmEffective rainfall = 20 mmAdditional water requried = 20 mm
Amount of water required after accounting irrigation efficiency = 20
40 mm0.5
=
End of Solution
Geotechnical Engineering
Q.8Q.8Q.8Q.8Q.8 Water flows in upward direction in a tank through 2.5 m thick sand layer. The e andG of sand are 0.58 and 2.7. Sand is fully saturated. γw is 10 kN/m3. Effective stressat point A, located 1 m above the base of the tank is _____ kN/m2.
Q.9Q.9Q.9Q.9Q.9 SPT was conducted at survey 1.5 m internal upto 30 m depth. At 3 m depth, the observedno. of hammer blows for 3 successive 150 mm penetration were 8, 6 and 9. SPT, Nvalue at 3 m depth is(a) 15 (b) 23(c) 17 (d) 14
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)No. of blows for each 150 mm penetration 8, 6 and 9.We will not consider first 150 mm number of blows.Hence, for last 300 mm, number of blows are 15.Hence, observed SPT number = 15.
End of Solution
Q.10Q.10Q.10Q.10Q.10 Velocity of flow is proportional to the first power of hydraulic gradient in Darcy’s law, thelaw 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.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)Darcy’s law is valid for laminar flow condition.
Q.11Q.11Q.11Q.11Q.11 A drained triaxial test is conducted on a sand sample. If σd = 150 kPa and consolidationstress is 50 kPa the calculated the value of angle of internal friction (φ)?
Q.12Q.12Q.12Q.12Q.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 ______.
Q.13Q.13Q.13Q.13Q.13 A vertical retaining wall of 5 m height has to support soil of γ = 18 kN/m3, effectivecohesion 12 kN/m2 and effective friction angle 30°. As per Rankine, assuming that tensioncrack has occurred the lateral active thrust on wall per meter length is _____.
Q.Q.Q.Q.Q.1515151515 A fill of 2 m thick sand with unit weight of 20 kN/m3 is placed above the clay layerto 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____________.
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.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)Velocity in x-depth, ux = u sinθVelocity in y-depth, uy = u cosθ
∂φ−
∂x= ux
Integrating it φ = –uxx + f (x) + c= –(u sinθ) x + f (y) + c ...(i)
y∂φ
−∂ = yu
Integrating it φ = –uy y + f (x) + c= –(u cosθ) y + f (x) + c ...(ii)
Q.17Q.17Q.17Q.17Q.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.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)For stability of floating body M lies above G
GM > 0
End of Solution
Q.18Q.18Q.18Q.18Q.18 A circular water tank of 2 m dia has circular orifice of dia 1 m at bottom. Water is enteringthe tank at 20 l/s and escaping through orifice. Take CD for orifice = 0.8. Neglect anyfriction losses. g = 9.81 m/s2. Height of water level in the tank at steady rate is _____.
Ans.Ans.Ans.Ans.Ans. (0.5164 m)(0.5164 m)(0.5164 m)(0.5164 m)(0.5164 m)
H
Q = 20 /sl
dc
= 0.1 m = 0.8d
Assume H is the level of weter in the tank in steady condition.For steady water level in the tank
Discharge through orifice = Water enters in the tank
Q.19Q.19Q.19Q.19Q.19 For a 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 theenergy loss in jump?(a) 114.2 MW (b) 114.2 kW(c) 141.2 J/s (d) 141.2 HP
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)Assuming channle bed to be horizontal and frictionless.
q =126
= 2 m3/s/m
Y1 = 0.3 m
Y2
Initial Froude No. (Fr) =1/22
31
qgY
⎛ ⎞⎜ ⎟⎝ ⎠
=
1/22
3
2
9.81 0.3
⎛ ⎞⎜ ⎟×⎝ ⎠
= 3.88
From Belenger’s Momentum equation for a rectangular channel
Q.20Q.20Q.20Q.20Q.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____________.
q = 1 – P = 0.98∴ Probability of non-occurance of an event is given by,
Assurance = q n
= (0.98)25
= 0.603
End of Solution
Environmental Engineering
Q.22Q.22Q.22Q.22Q.22 In a homogenous unconfined aquifer of area 3 km2, water table elevation is 102 m. Afternatural 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 volumeof ground water pumped after the natural recharge is __________ Mm3.
Q.23Q.23Q.23Q.23Q.23 35.67 mg HCl is added to distilled water and the total solution is made 1 l. The atomicweight 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.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)HCl→ H+ + Cl–
1 mole of HCL gives 1 mole H+ ions36.5 gm of HCl gives 1 gm of H+ ions
35.67 mg = +135.67 0.977 mg of H
36.5× =
=3
4 +0.977 109.77 10 moles of H
1
−−×
= ×
pH = –log10[H+] = –log10[9.77 × 10–4]
= –log109.77 + 4 log1010= 4 – 0.989 = 3.01
End of Solution
Q.24Q.24Q.24Q.24Q.24 As chlorine reacts rapidly with water to form Cl–, HOCl and H+. The most activedisinfectant in the chlorination process is
2(g) 2C H O HOC C H− ++ + +l l l���⇀↽���
(a) HOCl (b) H+
(c) H2O (d) Cl–
Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)
End of Solution
Q.25Q.25Q.25Q.25Q.25 A gaseous chemical has concentration of 41.6 μmole/m3 in air at 1 atm pressure andtemp 293 k gas constant R is 82.05 × 10–6 m3/atm/molK. Assume ideal gas low is valid,the concentration of gaseous chemical will be __________ ppm.
Q.26Q.26Q.26Q.26Q.26 For a river the discharge is 1000 MLD. BODs for the river is 5 mg/l and dissolved oxygenis 8 mg/l before receiving waste water discharge at a location. For existing environmentcondition dissolved oxygen is 10 mg/l. Waste water discharge of 100 MLD drains intothe river. BOD ultimate of waste water is 200 mg/l and dissolved oxygen is 2 mg/l fallsat a location. Assume complete mixing. Immediate do deficit is ______ mg/l.
Q.27Q.27Q.27Q.27Q.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 with80% of efficiency will be _______ μm.
Q.28Q.28Q.28Q.28Q.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 ofstream 40 m2 which remain constant. BOD exertion rate constant is 0.3/day (loge). BODremaining at 3 km downstream from mixing location is _________.
Q.29Q.29Q.29Q.29Q.29 A water supply scheme transport 10 MLD water through a 450 mm diameter pipe fora distance 2.5 km. A chlorine dose of 3.5 mg/L is applied at the starting point. It isdecided to increase the flow from 10 MLD to 13 MLD in the pipeline. Assume exponentfor concentration n = 0.36 with this increased flow in order to attain the same level ofdisinfection. The chlorine dose to be applied at the starting point.(a) 3.95 (b) 4.4(c) 5.55 (d) 4.75
Q.30Q.30Q.30Q.30Q.30 The los angles test for stone aggregate is used to examine(a) crushing strength (b) abrasion(c) toughness (d) none of these
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)Los angles test of stone is used for Abrasian resistance.
End of Solution
Q.31Q.31Q.31Q.31Q.31 During the process of hydration of cement, due to increase in C2S content in cementclinker, the heat of hydration(a) does not change (b) initially decrease the increase(c) decrease (d) increase
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)
End of Solution
Geometics Engineering
Q.32Q.32Q.32Q.32Q.32 An open traverse PQRST is surveyed using theodolite
If the independent coordinated (Northing and Easting) of station P are (400, 200 m) theindependent 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
Q.35Q.35Q.35Q.35Q.35 As per IRC, the minimum width of median in urban areas is ____________.
Ans.Ans.Ans.Ans.Ans. (1.2 m)(1.2 m)(1.2 m)(1.2 m)(1.2 m)Desirable width of median in urban roads = 5 mAnd As per IRC : 86-1983 min. width = 1.2 m
End of Solution
Q.36Q.36Q.36Q.36Q.36 For the contraction joint, which is correct position of dowel bar
(a)d/2 d d/3- /4
(b)d/2
(c)d/2
(d)d/2 d d/3- /4
Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)
End of Solution
RCC Structures and Prestressed Concrete
Q.37Q.37Q.37Q.37Q.37 Calculate the design bending moment of a singly reinforced sectionIf width = 300 mmEffective depth = 450 mmArea of steel in tension = 942 mm2
Q.38Q.38Q.38Q.38Q.38 A simply supported prismatic concrete beam of rectangular cross-section of span 8 mis prestressed with effective prestress force of 600 kN. Eccentricity is zero at supportand varies linearly to value ‘e’ at mid span. Required value of e ________.
To support a point load applied at mid span (W)= 12 kN
Balancing load = Point load2P sin θ = W
2/ 2e
PL
⎛ ⎞⎜ ⎟⎝ ⎠ = 2
2 2PeL
×= W
4PeL
= W
e =12000 N × 8000 mm
4 4 600 1000 NWL
P=
× ×= 40 mm
End of Solution
Q.39Q.39Q.39Q.39Q.39 The variation of bond stress varies as
(a) (b)
(c) (d)
Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)
End of Solution
Design of Steel Structures
Q.40Q.40Q.40Q.40Q.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 __________.
Q.45Q.45Q.45Q.45Q.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 d z dy y dz+ + + + +∫ x x x is
Ans.Ans.Ans.Ans.Ans. (3)(3)(3)(3)(3)
I = [( ) ( ) ( )]C
yd dy zd dz zdy ydz+ + + + +∫ x x x x
= ( )(1,1,1)(0,0,0)[ ( ) ( ) ( )]C
d y d z d yz y yz z+ + = + +∫ x x x x
= (1 + 1 + 1) – (0 + 0 + 0) = 3
End of Solution
Q.46Q.46Q.46Q.46Q.46 If the true value of ln2 is 0.69. If value of ln2 is also calculated by linear regressionbetween ln1 and ln6 then calculate the percentage error in the value of ln2.
Ans.Ans.Ans.Ans.Ans. (48.11%)(48.11%)(48.11%)(48.11%)(48.11%)True value In2 = 0.69 = T
Q.47Q.47Q.47Q.47Q.47 θ = f (t, z) and D = f(θ), k = f(θ) then 2
2
( )( ) 0
kD
z tz
∂ θ ∂ θ ∂θθ + − =∂ ∂∂
is _______.
(a) Second order linear equation (b) Second order non-linear equation(c) Second degree linear equation (d) Second degree non-linear equation
Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)∵ 1st term of given D. Equation contains product of dependent variable with it’sderivative, so it is non-linear and also we have 2nd order derivative so it’s order is twoi.e., 2nd order is non linear.