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Memory Based Questions of GATE 2020

Civil Engineering | Forenoon Session

NON-TECHNICAL SECTION

General Aptitude

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. (b)(b)(b)(b)(b)� 26591749(110016)

� Unit place of 9° = 1∴ Cycle of equatio is (9, 1) (9, 1), (9, 1)So answer will be 1.

End of Solution

Q.4Q.4Q.4Q.4Q.4 Insert seven number between 2 and 34 such is an AP. The sum of those seven insertedseven number is(a) 124 (b) 130(c) 120 (d) 126




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

General English

Q.5Q.5Q.5Q.5Q.5 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.6Q.6Q.6Q.6Q.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.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)

End of Solution

Q.7Q.7Q.7Q.7Q.7 If Fuse : Fusion :: Use : ___________.(a) Usage (b) Usion(c) Uses (d) User


End of Solution




TECHNICAL SECTION

Solid Mechanics

Q.1Q.1Q.1Q.1Q.1 In a 2-D stress analysis, the state of stress at point P is [σ] = y

y y

σ τ⎡ ⎤⎢ ⎥τ σ⎢ ⎥⎣ ⎦

xx x

x x. The necessary

and sufficient condition for existence of the state of pure shear at point P(a) (σxx – σyy)

2 + 4τ2xy = 0 (b) τxy

(c) σxx + σyy = 0 (d) σxxσyy – τ2xy = 0

Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)

ττ

In pure shear conditionσx = 0, σy = 0, τxy = τ

σ σxx = –σ στ σyy

xy

= = –

σ σxx = σ στyy

xy

= – = 0

σ

σ

σσ

σ

σ

σσ or

τ-Axis

σ-AxisCO

For this condition(c) is correct

σxx + σyy = 0

End of Solution




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


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




Ans.Ans.Ans.Ans.Ans. (130 kN)(130 kN)(130 kN)(130 kN)(130 kN)

(2)(1) (3)

P2 P3P1

P1P2 P3

W

P1 + P1 + P2 = W ...(i)

δ1 = 15 mm =P LAE

10 kN/mmAEL

=

δ2 = 23 mm =P LAE

So, P1 = 10 × 5 = 50 kNP2 = 10 × 3 = 30 kNW = 2(50) + 30 = 130 kN

End of Solution

Structural Analysis

Q.4Q.4Q.4Q.4Q.4

P

How many zero force members are there in the above truss(a) 6 (b) 7(c) 8 (d) 9





P

B

A

L L L3

2

4

1

Sequence 1 — 2 3 4 AB— — —

As ΔAB = 0, hence FAB = 0Total number of zero force member = 8

End of Solution

Q.5Q.5Q.5Q.5Q.5 A planer elastic structure is subjected to UDL as shown in figure. Where maximum BMany in the structure will be ________.

8 m

12 kN/m

Parabolic profile2.4 m

Ans.Ans.Ans.Ans.Ans. (0)(0)(0)(0)(0)

8 m

Parabolic profile2.4 m

Mx = 0 of all points

End of Solution

Q.6Q.6Q.6Q.6Q.6 Distributed load (continuous or in patches) of 50 kN/m may occupy any position on thegrid the maximum –ve BM at R will be

1.5 m 3 m 2 m 1.5 m

P Q R S T

(a) 56.25 kN-m (b) 150 kN-m(c) 93.75 kN-m (d) 22.5 kN-m

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Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)ILD for MS

P Q S T

50 kN/m

1.5 mILD MS

1.5 m

θ = 1 unit

∴ Mmax at support =1

1.5 1.5 502

⎡ ⎤− × × ×⎢ ⎥⎣ ⎦

= –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.

End of Solution




Irrigation Engineering

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.

1 m

2.5 m

Water

Outward flow

1.2 m

1 m

Inward Flow

Filter Media

A




Ans.Ans.Ans.Ans.Ans. (8.94 kN/m(8.94 kN/m(8.94 kN/m(8.94 kN/m(8.94 kN/m22222)))))

1 m

2.5 m

1.5 m

1 m

1.2 m

A

γa =1 2.7 1

101 1 0.58G

e ω− −⎛ ⎞ ⎛ ⎞γ = ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+ + = 10.759

σ = zγ – izγω

= 1.21.5( ) 1.5

2.5 ω⎛ ⎞γ − × × γ⎜ ⎟⎝ ⎠

= 8.939 kN/m2

End of Solution

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.

End of Solution




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)?

Ans.Ans.Ans.Ans.Ans. (36.87°)(36.87°)(36.87°)(36.87°)(36.87°)Sand (C = 0); σd = 150; σ3 = 50; σ1 = 200

σ1 = 23 tan 45 2 tan 45

2 2c

φ φ⎛ ⎞ ⎛ ⎞σ + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

200 = 250 tan 452φ⎛ ⎞+⎜ ⎟⎝ ⎠

φ = 36.87°

End of Solution

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 ______.

Ans.Ans.Ans.Ans.Ans. (0.601)(0.601)(0.601)(0.601)(0.601)

F.O.S. =tan 18 10 tan38tan 18 tan30sat

γ ′ φ − °⎛ ⎞⋅ = ⎜ ⎟⎝ ⎠γ β °

= 0.601

End of Solution

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 _____.

Ans.Ans.Ans.Ans.Ans. (21.714 kN/m)(21.714 kN/m)(21.714 kN/m)(21.714 kN/m)(21.714 kN/m)

5 m C

K

= 12 = 30° = 18 kN/m

= 1/3

φγ 3

a

–2C K√ a

2Cγ√Ka

= = 2.309×

×

2 121183

γ − =2 16.144a aK Z C K




After tension crack

2.309

16.144

Pa =1

16.144(5 2.309)2

× −

= 21.714 kN/m

End of Solution

Q.14Q.14Q.14Q.14Q.14 The water table is at a depth of 3 m below ground level, γw = 9.81 kN/m3. Surchargeis applied instantaneously

3 m

4 m

3 m

Sand, = 18 kN/mγB3

Silty clay ( = 20 kN/mγsat3)

Clay ( = 21 kN/mγsat3)

Boulder clay

q = 70 kN/m2

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





3 m

4 m

3 m

γ = 18

γ = 20

γ = 20

q = 70 kN/m2

P

Q

( )Pσ = q + 3γ = 3 × 18 = 54

Surcharge is applied suddenly.

∴ Excess pore water pressure developed.

( )Qσ = sat( 3 4 ) (4 70)b wq + γ + γ − γ +

= 70 + 3 × 18 + 4 × 20 – (4 × 9.81 + 70)

= 94.76 kN/m2

End of Solution

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____________.

3 m

10 m Clay

Sand

Impervious strata





Δσ = 2 × 20 = 40 kN/m2

ΔH = Vm HΔσ

= 2.2 × 10–4 × 40 × 10 × 103 mm= 88 mm

Tv =2

2 2

9 10 100.036

5

C t

H

−× × ×= =

Tv = 2

4U

π

U =0.036 4

0.214× =

πΔh after 10 years = 0.214 × 88 = 18.832 mm

End of Solution

Fluid Mechanics

Q.16Q.16Q.16Q.16Q.16 u

y

θ

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)




By equation (i) and (ii),

φ = ( sin cos )u y− θ + θx

If we take∂φ∂x

= ux and yu∂φ=

∂y

Then φ = ( sin cos )u yθ + θx

So, φ = ( sin cos )u y± θ + θx

End of Solution

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

· · 2dc a gH = 20 × 10–3

20.8 (0.1) 24

gHπ

× = 0.02

H = 0.5164 m

End of Solution




Open Channel Flow

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

2

1

YY

= ( )21

11 1 8

2F− + +

= ( )211 1 8 3.88

2− + + ×

= 5.018∴ Y2 = 5.018 × 0.3 = 1.505 m

Head loss in the jump (hL) =3

2 1

1 2

( )4

Y YYY−

=3(1.505 0.3)

4 1.505 0.3−

× ×

= 0.968 mPower lost in the jump = γwQhL

= (9.81 × 12 × 0.968) kW= 114.04 kW

End of Solution




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____________.

Ans.Ans.Ans.Ans.Ans. (2.45 m/s)(2.45 m/s)(2.45 m/s)(2.45 m/s)(2.45 m/s)

Critical depth (YC) =1/32q

g

⎛ ⎞⎜ ⎟⎝ ⎠

=

1/321.59.81

⎛ ⎞⎜ ⎟⎝ ⎠

= 0.612 m

Critical velocity (VC) = 9.81 0.612CgY = × = 2.45 m/s

End of Solution

Engineering Hydrology

Q.21Q.21Q.21Q.21Q.21 The probability that a 50 year flood may not occur at all during 25 year life of a projectis _________.


P =1 1

0.0250T

= =

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.

Ans.Ans.Ans.Ans.Ans. (1.5 Mm(1.5 Mm(1.5 Mm(1.5 Mm(1.5 Mm33333)))))

103.2 m

102 m

101.20 m

VR = 0.9 Mm3

V = 3 × (103.2 – 102)




= 3 × 1.2 = 3.6 Mm3

ys or yR =0.93.6

VRV

=

Now,1S

= DVV

VD = ( )0.93 103.2 101.2

3.6⎡ ⎤× −⎣ ⎦

VD = 1.5 Mm3

End of Solution

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–


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.




Ans.Ans.Ans.Ans.Ans. (1)(1)(1)(1)(1)PV = nRT

V =nRT

P

=6 6

6 341.6 10 32.05 10 293 10 m1

− −−× × ×

× =

41.6 μ mole of gas volume of 10–6 m3

So, 1 ppm =3

6 6 31 part of gas 1 m of gas

10 parts of air 10 m of air=

1 ppm =6

6 3

41.6 10 moles

10 m

× μ

So, 41.6 μ moles/m3 = 1 ppm

End of Solution

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.


DOmix =· · 2 100 8 1000

100 1000s S R R

s R

DO Q DO QQ Q

+ × + ×=

+ +

= 7.45 mg/lDO = DOsat – DOmix = 10 – 7.45 = 2.545 mg/l

End of Solution

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.


η = 80 = 100s

s

uv

×

us =3

40.8 2000 101.85 10 m/sec

86400

−−× ×

= ×

us =( ) 21

18

G wdP

−




1.85 × 10–4 =2

2 3 4

(2.64 1) 9.81

18 1.02 10 10 10

d− −

− × ×× × × ×

d = 1.453 × 10–5 m= 14.53 μm

End of Solution

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 _________.


t =2d

; where, v = 0.2 5

0.13 m/sec40

S RQ QA+ +

= =

t =33 10

0.26 days0.13 86400

×=

×BOD5 = BODu(1 – e–k × 5)

BODu = 0.3 5

500

(1 )e− +− = 643.66 mg/l

DOmix =BOD BODR u S u

S R

Q QQ Q

+ ⋅+

=5 30 0.2 643.66

53.6 mg/5 0.2

× + ×=

+l

Ct = Co e–k × t

= 53.6e–0.3 × 0.26

= 49.57 mg/l

End of Solution

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

Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)Waterson law, tch = Constant

1 1nt c = 2 2

nt c




11

1

hd cv =

22

2

nd cv

0.86(3.5)10

=0.86

2( )13

c

c2 =1/0.8613

3.5 4.74710

⎛ ⎞ × =⎜ ⎟⎝ ⎠

= 4.75 mg/l

End of Solution

Construction Materials

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


End of Solution

Geometics Engineering

Q.32Q.32Q.32Q.32Q.32 An open traverse PQRST is surveyed using theodolite

Line Northing (m) Southing (m) Easting (m) Westing (m)110.2 – 45.5 –80.6 – – 60.1

– 90.7 – 70.8– 105.4 55.5 –

PQQRRSST

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




Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)

Northing

400 m, 200 m

EastingWesting

Southing

ΔL = –5.3ΔD = –29.9

T, Northing{400 + (–5.3) = 394.7T, Easting{200 + (–29.9) = 170.1

T [394.7 m, 170.1 m]

End of Solution

Q.33Q.33Q.33Q.33Q.33 The length of line segment SP is __________.

Segment

PQ

QR

RS

Length

40

50

30

Bearing

80°

10°

210°

Ans.Ans.Ans.Ans.Ans. (44.784 m)(44.784 m)(44.784 m)(44.784 m)(44.784 m)ΔL = 40cos80° + 50cos10° + 30cos210°

= 30.20

ΔD = 40sin80° + 50sin10° + 30sin210°

= 33.07

Length, SP = 2 2L DΔ + Δ= 44.784 m

End of Solution




Highway Engineering

Q.34Q.34Q.34Q.34Q.34 Gradient of a road is 4.5%, radius is 100 m the compensated as per IRC will be _____.

Ans.Ans.Ans.Ans.Ans. (4%)(4%)(4%)(4%)(4%)Gradient = 4.5%, R = 100 m

Grade compensation = 30 75 30 100 75

% 1.3% 0.75 . 0.75/ / /100 100

RGC

R R⎫+ +⎛ ⎞ ⎛ ⎞> = > = > =⎬⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎭

Compansated Gradient = Gradient G.C = 4.5% – 0.75 = 3.75 /< 4%

Hence C.G = 4%

End of Solution

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


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

Grade of concrete is M25 and Fe500 Steel




Ans.Ans.Ans.Ans.Ans. (158.28 kN-m)(158.28 kN-m)(158.28 kN-m)(158.28 kN-m)(158.28 kN-m)

300 mm

450 mm

M25 concreteFe500 steel

B = 300 mm

d = 450 mm

Ast = 942 mm2

Mu = ?

(i) xulim = 0.46 × d = 0.46 × 450 = 207 mm

(ii) xu =0.87 0.87 500 942

151.77 mm0.36 0.36 25 300

y st

ck

f A

f B

⋅ ⋅ × ×= =⋅ ⋅ × ×

(iii) xu < xu lim It is an under reinforcement section.

(iv) Mu = 0.36 ⋅ fck ⋅ B ⋅ xu ⋅ (d – 0.42xu)

= 0.36 × 25 × 300 × 151.77 × (450 – 0.42 × 151.77)/106

= 158.28 kN-m

End of Solution

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 ________.

Ans.Ans.Ans.Ans.Ans. (40 mm)(40 mm)(40 mm)(40 mm)(40 mm)

P L/2e

12 kN

600 kN600 kN




P = 600 kNSimply supported span = L = 8 m

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)


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 __________.

10 m

m12

0 m

m10

mm

x x

100 mm




Ans.Ans.Ans.Ans.Ans. (323.527 kN)(323.527 kN)(323.527 kN)(323.527 kN)(323.527 kN)

q =FAy

b×I

b = 4t = (4 × 7 × 10)

100 =( )

( )3

100 10 55

773 10 4 7

F × × ×× × ×

F = 323.527 kN

End of Solution

Engineering Mathematics

Q.41Q.41Q.41Q.41Q.41 The area of an ellipse represented by an equation 2 2

2 2 1y

a b+ =x

is

(a)4abπ

(b)4

3abπ

(c) abπ (d)2abπ


Area =

2 2

2 2

(1) (1)

b aa a

ba y x aa

dyd dyd+ − +

=− =− − +

=∫ ∫ ∫ ∫x

x

x x

=

2 2

0 0

4 (1)

ba

a a

y

dy d

−

= =∫ ∫

x

x

x

=

2 2

0 0

4

ba

a a

y

d

−

= =∫ ∫

x

x

x

= πab

End of Solution




Q.42Q.42Q.42Q.42Q.42 The value of 2

2

5 4lim

4 2→∞

− ++x

x xx x

is

(a)12

(b)14

(c) 0 (d) 1


It is in ∞⎛ ⎞

⎜ ⎟⎝ ⎠∞ from so by L-Hospital Rule

=2 5

lim8 2x→∞

− ∞⎛ ⎞ =⎜ ⎟⎝ ⎠+ ∞xx

=2 1

lim8 4x→∞

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

End of Solution

Q.43Q.43Q.43Q.43Q.43 For the ordinary differential equation 2

25 6 0

d ddtdt

− + =x x x , initial condition x(0) = 0 and

(0)ddt

x = 10, the solution of the given equation

(a) –10e2t + 10e3t (b) 10e2t + 10e3t

(c) 5e2t + 6e3t (d) –5e2t + 6e3t

Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)A.E. is m2 – 5m + 6 = 0 ⇒ m = 2, 3 so Cf = C1e

2t + C2e3t.

PI = 0 and G. Solution is x = CF + PI = C1e2t + C2e

3t and 2 31 22 3t td

C e C edt

= +x

.

Now, using initial conditions we get C1 = –10, C2 = 10.x = –10e2t + 10e3t

End of Solution

Q.44Q.44Q.44Q.44Q.44

1

2

3

1 3 2 1

2 2 3 1

4 4 6 2

2 5 2 1

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥− ⎢ ⎥⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥− ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

xxx

Value of x3 will be ___________.


[A : B] = Converting intoan Echelon Form

1 3 2 1 1 3 2 1

2 2 3 1 0 1 2 1

4 4 6 2 0 0 1 3

3 5 2 1 0 0 0 0

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥− − − −⎢ ⎥ ⎢ ⎥⎯⎯⎯⎯⎯⎯⎯→⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

� �

� �

� �

� �




⇒1

2

3

1 3 2

0 1 2

0 0 1

0 0 0

⎡ ⎤⎡ ⎤⎢ ⎥− − ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦

⎣ ⎦

xxx

=

1

1

3

0

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

⇒ x3 = 3

End of Solution

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


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

x y = ln xx0 = 1 0x1 = 6 1.79

Divided differentiation

1.79 06 1

−−

= [ ]0 10.358 ,f= x x

Approx:ln2 = f [x0] + (x – x0) f [x0, x1]

= 0 + (2 – 1) 0.358= 0.358 = A

% error = 100 48.11%T A

T−

× =

End of Solution




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.

End of Solution

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