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Memory Based Questions of GATE 2020

Civil Engineering | Afternoon Session

NON-TECHNICAL SECTION

General English

Q.1Q.1Q.1Q.1Q.1 Now a days most children have a tendency to belittle the legitimate concern of theirparents?Which one of the following word can replace the underlined word without changing thesense of sentence?(a) Reduce (b) Disparage(c) (d)

Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)

End of Solution

Q.2Q.2Q.2Q.2Q.2 Partiality : Impartiality : : Popular : ?(a) Impopular (b) Dispopular(c) Unpopular (d) Mispopular

Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)Partiality : Impartiality : : Popular : Unpopular

End of Solution

Q.3Q.3Q.3Q.3Q.3 Rescue teams deployed ___________ disaster hit areas, combat ________ a lot ofdifficulties to save the people.(a) to, with (b) to, to(c) with, with (d) in, with

Ans.Ans.Ans.Ans.Ans. (a)(a)(a)(a)(a)

End of Solutions

General Aptitude

Q.4Q.4Q.4Q.4Q.4 Percentage LED bulbs sold by two firms X and Y from January to June is given below.

JanuaryFebruaryMarchAprilMayJune

15%20%30%15%10%10%

10

30

20

10

15

15

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Ratio of LED bulbs sold by X and Y are as shown below

JanuaryFebruaryMarchAprilMayJune

7 : 82 : 32 : 13 : 21 : 49 : 11

X : Y

If total bulbs sold by X and Y in April-June is 50000 then the number of bulbs soldby firm Y during April to June is ___________(a) 8250 (b) 8750(c) 9750 (d) 11250

Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)x : y

April = 15% → 3 : 2May = 10% → 1 : 4June = 10% → 9 : 11

y = 6 + 8 + 5.5 = 19.5Total LED bulbs = 50,000

y = 19.5% of 50,000

y =19.5

50,000 9750100

× =

End of Solution

Q.5Q.5Q.5Q.5Q.5 Out of 1000 students, 300 play chess 600 play football and 50 play both games. Howmany students play none of the sports?

Ans.Ans.Ans.Ans.Ans. (150)(150)(150)(150)(150)

T = 1000

250 50 550

600 Football300 Chess

Total number of students playing sports = 850Total number of students not playing sports = 1000 – 850 = 150

End of Solution

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Q.6Q.6Q.6Q.6Q.6 When was the calendar of 2019 repeated?(a) 2011 (b) 2012(c) 2013 (d) 2014

Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)YYYYYearearearearear Number of odd daysNumber of odd daysNumber of odd daysNumber of odd daysNumber of odd days2013 12014 12015 12016 22017 12018 1TTTTTotalotalotalotalotal 77777Number of odd days in 2019 = 1, so 2013 calendar is same as 2019.

End of Solution

Q.7Q.7Q.7Q.7Q.7 Nominal interest rate is defined as the amount paid by the borrower to the lender forusing the borrowed amount for a specific period of time. Real interest rate is calculatedon the basis of actual rate (inflation-adjusted) and is approximately equal to the differencebetween nominal rate and expected rate of inflation in the economy.(a) Under low inflation, real interest rate is low and borrowers get benefitted(b) Under high inflation, real interest rate is low and lender gets benefitted(c) Under low inflation, real interest rate is high and borrowers benefitted(d) Under high inflation, real interest rate is low and borrowers benefitted

Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)Real interest rate = Nominal rate – Inflation

End of Solution

Q.8Q.8Q.8Q.8Q.8 HOD of a college wants to rearrange four faculty rooms based on their requirements.P needs room near to lab. Q needs room near to lift. R needs visibility to the playground,and S need a corner room. Which of the following arrangement will be satisfying therequirements?

(a)

Play Ground

Lab

Roa

d

P Lift

HOD QRS

(b)

Play Ground

Lab

Roa

d

P Lift

R QSHOD




(c)

Play Ground

Lab

Roa

d

Q Lift

P HODRS

(d)

Play Ground

Lab

Roa

d

P Lift

R SQHOD


End of Solution

Q.9Q.9Q.9Q.9Q.9 Ratio of sum of odd positive integers from 1 to 100 to the sum of even positive integersfrom 150 to 200 will be(a) 45 : 95 (b) 50 : 91(c) 1 : 1 (d) 1 : 2


Sum of old numbers from 1 to 100Sum of even numbers from 150 to 200

= 250 50 50 2500

175 26 175 26 4550×

= =× ×

Ratio = 50 : 91From 1 to 100 = 50 odd numberFrom 150 to 200 = 26 even number

End of Solution

Q.10Q.10Q.10Q.10Q.10 An unbiased fair coin was tossed 15 times. What is the probability of getting exactly8 heads?

Ans.Ans.Ans.Ans.Ans. (0.196)(0.196)(0.196)(0.196)(0.196)

P(H) =12

P(T) =12

Probability of getting exactly 8 heads out of 15 trial = 8 15 8

158

1 12 2

C−⎡ ⎤ ⎡ ⎤×⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= 0.196

End of Solution




TECHNICAL SECTION

Engineering Mathematics

Q.1Q.1Q.1Q.1Q.12

22

2 sin 0d u

ud

− + =x xx

(a) linear, homogenous (b) non-linear, homogeneous(c) linear, non-homogeneous (d) non-linear, non-homogeneous

Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)Its solution is of the type u = f(x), i.e., dependent variable is u.Hence, given equation is Linear & Non-Homogeneous.

End of Solution

Q.2Q.2Q.2Q.2Q.229 2020

lim7→ ∞

++x

xx

Ans.Ans.Ans.Ans.Ans. (3)(3)(3)(3)(3)

2

20203 1lim

71x→∞

+

⎛ ⎞+⎜ ⎟⎝ ⎠

xx

xx

= 3

End of Solution

Q.3Q.3Q.3Q.3Q.3 The diameter and length of a cylinder is 3 m and 4 m respectively. There is an absoluteerror of 0.2 m in each of these measurements. Find the absolute error in the volume.

Ans.Ans.Ans.Ans.Ans. (7.35(7.35(7.35(7.35(7.35πππππ)))))Let diameter = x and height = y

then V =2 2

2 4yy π⎛ ⎞π =⎜ ⎟⎝ ⎠

x x...(i)

∵ V = ( , ), so = v v

f x y dv d dyy

⎛ ⎞∂ ∂⎛ ⎞ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∂ ∂x

x

dv =21

2 4xy d dy

⎛ ⎞π⎛ ⎞π +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠xx ...(ii)

So,dvv

=

2

222 4

4

y dx dy d dyyy

π π+= +

π

x x xxx

So, % error in volume = 100%dvv

×




= 2 100%d dy

y⎡ ⎤

+ ×⎢ ⎥⎣ ⎦

xx

=( 0.2) ( 0.2)

2 100%3 4

± ±⎡ ⎤× + ×⎢ ⎥⎣ ⎦

= ( )405

3⎛ ⎞± + ±⎜ ⎟⎝ ⎠

= (±13.3) + (±5) � 18.3%

Now exact volume is V =2 2(3) 4

94 4

yπ π ×= = π

x

Approximate value of V =2

2 4dv y d dyπ π

= xx x +

= ( ) 2(3)(4) 0.2 (3) ( 0.2)2 4π π

± + ±

= ±1.65 πSo absolute error in volume = |v – dv | = 7.35π

End of Solution

Q.4Q.4Q.4Q.4Q.4 For the given 4 × 4 matrix. What are the eigen values?

0 1 3 0

2 3 0 4

0 0 6 1

0 0 1 6

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(a) (1, 0, 3, 4) (b) (1, 2, 5, 7)(c) (3, 4, 5, 7) (d) (6, 3, 6, 6)

Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)|P | = 70 and Trace (P) = 15

So, only option, i.e., (b) (1, 2, 5, 7) satisfies.

End of Solution

Q.5Q.5Q.5Q.5Q.5 Expansion of x-x2 as per Fourier series is

0

1 1cos sin

2 n nn n

a a n b n∞ ∞

= =+ +∑ ∑x x

Find the value of a0 = __________.

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Ans.Ans.Ans.Ans.Ans. (–6.58)(–6.58)(–6.58)(–6.58)(–6.58)

a0 =1

( )f dπ

−ππ ∫ x x = 21( )d

π

−π

−π ∫ x x x = 2

0

1 dπ−

π ∫ 2x x

=3

0

1 23

π⎛ ⎞− ⎜ ⎟π ⎝ ⎠

x =

232 2

3 3− π⎡ ⎤− π =⎣ ⎦π

= –6.58

End of Solution

Q.6Q.6Q.6Q.6Q.6 If f (x) = x2

then ( ( ( ))

( )f f f

fxx

(a) f(x) (b) (f(x))2

(c) (f(x))3 (d) (f(x))4

Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)f(x) = x2

f(f(f(x))) = f(f(x2)) = f(x4) = (x4)2 = x8

So,( ( ( )))

( )f f f

fxx

=8

6 2 3 32

( ) ( ( ))f= = =x x x xx

End of Solution

Q.7Q.7Q.7Q.7Q.7 An ordinary differential equation is given below2

2

60

d y dyy

dd+ − =

xxThe general solution of the above equation is

(a) 3 21 2( )y C e C e

−= +

x x

x x (b) 3 21 2( )y C e C e

−= +

x x

x

(c) 3 21 2( )y C e C e= +x x

x (d) 3 21 2( )y C e C e

− −= +

x x

x x


2

2

6d y dyy

dd+ −

xx= 0

(6D2 + D – 1)y = 06D2 + 3D – 2D – 1 = 0

3D(2D + 1) – 1(2D + 1) = 0(2D + 1)(3D – 1) = 0

D =1

2−

, D = 13

y = /3 /21 2C e C e−+x x

End of Solution




Q.8Q.8Q.8Q.8Q.8 The integral ( )1

3 2

0

5 4 3 2x x x dx+ + +∫ is estimate numerically using 3 alternative methods,

namely the rectangle, trapezoidal and Simpson’s rule with a common step size. In thiscontext which one of the following statements is true?(a) Simpson’s rule, rectangle rule as well as trapezoidal rule of estimation will give non-

zero error.(b) Only Simpson rule will give zero error(c) Simpson and rectangle give non-zero error(d) Only rectangle will give zero error

Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)Because integral is a polynomial of 3rd degree.

End of Solution

Q.9Q.9Q.9Q.9Q.92

2

du d udt dx

= can be solved by

(a) one boundary, 2 initial condition (b) 2 boundary, 1 initial condition(c) 0 boundary, 2 initial condition (d) 2 boundary, 0 initial condition

Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)Two boundaries and one initial condition i.e., u(x, 0); u(0, t); u(l, t)

End of Solution

Solid Mechanics

Q.10Q.10Q.10Q.10Q.10

P

PL

What will be the most economical section of beam

(a) (b)

(c) (d)





P

BPL

L

MA

V = A P

(–PL) + (PL) + (–MA) = 0MA = 0

A B

PL

Bending moment diagram

For most economical,Maximum cross-section is given where maximum bending moment occurs.

So, option (a) is correct.

End of Solution

Q.11Q.11Q.11Q.11Q.11 The following Mohr’s circle corresponds to which of the following state of stress

Shear stress

Normal stress

σv

–σh σh

–σv

(a) Uniaxial tension (b) Hydrostatic loading(c) Biaxial tension (d) Pure shear

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

Shear stress

Normal stress

σv

–σh σh

–σv

In pure shear condition, Mohr’s circle has its center at origin.

End of Solution




Q.12Q.12Q.12Q.12Q.12 Strain energy for following figure as given

P1, U1

(P P1 2+ )

P2 , U2

The strain energy in bar when both (P1 + P2) is applied is U. Which of following canbe inferred?(a) U > U1 + U2 (b) U = U1 + U2

(c) U < U1 + U2 (d) None of these


P1, U1 U1 = 2

1

2P LAE

P2 , U2 U2 = 22

2P LAE

(P P1 2+ ) U = ( )2

1 2

2

P P LAE

+

( )21 2P P+ > 2 2

1 2P P+

U > U1 + U2

End of Solution

Q.13Q.13Q.13Q.13Q.13 Considering constant EI of the given frame, what will be the slope at R?

P T

S

L/2

L

(a)2

6PLEI

(b)2

12PL

EI

(c)3

6PLEI

(d)3

12PL

EI





P T

SLR

L/2

PSLR

PL2

θR θS

θR =( /2)

6PL L

EI

θR =2

12PL

EI

End of Solution

Construction Materials

Q.14Q.14Q.14Q.14Q.14 Match the following :List IList IList IList IList I List IIList IIList IIList IIList II

PPPPP..... Soundness test 1.1.1.1.1. StrengthQ.Q.Q.Q.Q. Crushing test 2.2.2.2.2. Resistance to weatheringR.R.R.R.R. Los angeles abrasian test 3.3.3.3.3. AdhesionS.S.S.S.S. Stripping value test 4.4.4.4.4. HardnessWhich of the following is correct?(a) P-2, Q-1, R-4, S-3 (b) P-2, Q-4, R-1, S-3(d) P-1, Q-2, R-3, S-4 (d) P-4, Q-3, R-2, S-1


End of Solution

Q.15Q.15Q.15Q.15Q.15 The maximum applied load on a cylindrical concrete specimen of diameter 150 mm andlength 300 mm tested as per the split tensile strength test guidelines of IS 5816-1999is 157 kN. The split tensile strength (in MPa) __________ of the specimen is





P

D

L

P = 157 kND = 150 mmL = 300 mm

In split tensile strength test, split tensile strength of concrete

fet =2PDLπ

= 2 157000

150 300×

π × ×= 2.22 N/mm2

End of Solution

Geotechnical Engineering

Q.16Q.16Q.16Q.16Q.16 Find the FOS of the slope.

10 m

10 m

4.48m

W

Cu = 60 kPa = 20 kN/mγ 3


10 m

10 m

4.48m

W

Cu = 60 kPa = 20 kN/mγ 3

C r × ( )θ45°

Cu = 60Consider unit length of slope




Circular arc area = 2 Area of 360

rθ

× π − Δ

= 290 110 10 10 28.54

360 2× π × − × × =

Height of wedge = 28.54γ = 570.8 kN/m

FOS =[ ] 60 10 10( ) 4

570.8 4.48c r r

W x

π× × ×× θ ×=

× ×

= 1.84

End of Solution

Q.17Q.17Q.17Q.17Q.17 In an oedometer test initial depth of the soil sample is 19 mm and the reduction in depthis 2.1 mm after application of load. The final value of void ratio is 0.62. The initial voidratio is

Ans.Ans.Ans.Ans.Ans. (0.8213)(0.8213)(0.8213)(0.8213)(0.8213)Oedometer reading = 2.1 mm

16 mm thick, e = 0.62

0

HHΔ

= 0

0 01 1fe e e

e eΔ −

=+ +

2.1 mm19 mm

= 0

0

0.621

ee

−+

e0 = 0.8213

End of Solution

Q.18Q.18Q.18Q.18Q.18 500 gm of dry soil was poured into a 2 L cylinder. The cylinder is partially filled withwater. The soil replaces 188 cm3 of water. What is the specific gravity of the soil?

Ans.Ans.Ans.Ans.Ans. (2.66)(2.66)(2.66)(2.66)(2.66)Ws = 500 gmVs = 188 cc

γs =s

s

WV

Gs =500 / 188

2.661

s

w

γ= =

γ

End of Solution

Q.19Q.19Q.19Q.19Q.19 A saturated clay, c = 20 kN/m2 and γ = 18 kN/m3 is retained by 5 m retaining wall withφ = 0°. Considering backfill to be smooth, what height of retained wall to be raised bydry sand with φ = 30°, γ = 16 kN/m3 so that no tension cracks are developed? ____m





q = γ xx

5 m

Cu = 20 kPa = 0φ

To prevent tension crack,

q =2 2 20

401a

c

k

×= =

q = 40dγ =x

x =40

2.5 m16

=

End of Solution

Q.20Q.20Q.20Q.20Q.20 Soil deposited by wind is known as(a) Lacustrine soil (b) Aeoline soil(c) Black cotton soil (d) Alluvial soil

Ans.Ans.Ans.Ans.Ans. (b)(b)(b)(b)(b)Soil deposited by wind is Aeoline soil.

End of Solution

Q.21Q.21Q.21Q.21Q.21 The dry density of soil is 15.5 kN/m3. What is the percentage water content of the soilif it is 72% saturated? [Given G = 2.65]

Ans.Ans.Ans.Ans.Ans. (19.28)(19.28)(19.28)(19.28)(19.28)γd = 15.5 kN/m3, G = 2.65, S = 72%

γd =2.65 10

15.51 1

wGe e

γ ×= =

+ +

e = 0.7096

w =0.72 0.7096

0.19282.65

SeG

×= =

w = 19.28%

End of Solution




Q.22Q.22Q.22Q.22Q.22 200 kN/m2 pressure load is being applied at foundation level. The settlement in the claylayer of thickness 1.5 m, will be ________ mm.[Given CC = 0.3, Gclay = 2.65, γw = 10 kN/m3]

1.5 m

1.5 m

0.5 m

0.5 m

200 kN/m2

γ = 15 kN/m3

2 m

φ = 30°Sand

z = 1.75 mC ClayC

γsat = 17 kN/m3

Dense sand

2V1H


1.5 m

1.5 m

0.5 m

0.5 m

200 kN/m2

γ = 15 kN/m3

2 m

φ = 30°Sand

z = 1.75 mC ClayC

γsat = 17 kN/m3

Dense sand

2V1H

γsat =32.65

10 17 kN/m1 1wG e e

e e+ +⎛ ⎞ ⎛ ⎞γ = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+ +

e0 = 1.3571. H0 = 1.5 m

2. 0 at c/c of compressible layerσ

0σ = ( )sat2 0.5 0.75 1.25d sat wc⎡ ⎤γ + γ + γ − γ⎣ ⎦

= 2 × 15 + 0.5 × 18 + 0.75 × 17 – 1.25 × 10= 39.25 kN/m2

3. Δσ =( )

2

2ForceArea 2

qBB nZ

=+

= 22

200 2 256.89 kN/m

12 2 1.75

2

× × =⎛ ⎞+ × ×⎜ ⎟⎝ ⎠




4. ΔH = 0 0

0 0log

1CH C

e⎛ ⎞σ + Δσ⎜ ⎟+ σ⎝ ⎠

=1.5 0.3 39.25 56.89

log1 1.357 39.25

× +⎛ ⎞⎜ ⎟⎝ ⎠+

= 74.28 mm

End of Solution

Engineering Hydrology

Q.23Q.23Q.23Q.23Q.23 In a storm with rainfall duration of 90 hour, peak of 60 m3/sec occurs at 20 hours. Areaof catchment is 300 km2. What will be the rainfall excess depth?

Ans.Ans.Ans.Ans.Ans. (3.24 cm)(3.24 cm)(3.24 cm)(3.24 cm)(3.24 cm)

20 hr 70 hr

60 m /sec3

Time (hr)

Dis

char

ge (m

/sec

)3 Ac = 300 km2

⇒

3

6 2

1 60m /s 90 3600 s2 100 cm

300 10 m

⎡ ⎤× × ×⎢ ⎥×⎢ ⎥

×⎢ ⎥⎣ ⎦

= Rainfall excess

∴ Rainfall excess = 3.24 cm

End of Solution

Environmental Engineering

Q.24Q.24Q.24Q.24Q.24 A solid waste having following composition C = 35%, O = 26%, H = 10%, N = 3%,S = 6% and inert gases = 20%. 1000 tonnes/day of solid waste is combusted and itcompletely transforms into CO2, SO2, NO2 and H2O. The air has 23% of oxygen. Whatis the amount of air required (in tonnes/day) to complete the combustionstoichiometrically?




Ans.Ans.Ans.Ans.Ans. (6956.5)(6956.5)(6956.5)(6956.5)(6956.5)Let MSW be 100 gmThe proportion of C = 35 gm = 2.91 moles

O = 26 gm = 1.625 molesH = 10 gm = 10 molesN = 3 gm = 0.214 molesS = 6 gm = 0.1875 moles

C2.91 O1.625 H10 N0.214 S0.1875 + 5O2 → 2.9CO2 + 0.1875SO2 + 0.214NO2 + 5H2O

From reaction1 mole MSW requires 5 mole O2 for combustion80 gm of MSW compound reacts with 160 gm of O2 for combustion

800 tonne/day –––––––– 160

800 1600 t/d80

× =

Air requires 1600

6956.5 t/d0.23

=

End of Solution

Q.25Q.25Q.25Q.25Q.25 C8 H16 O8 of concentration 10–3 moles/litre require O2.MW of c = 12, H = 1 and 0 = 16. What will be the theoretical oxygen demand _________(in gm/litre).

Ans.Ans.Ans.Ans.Ans. (0.256)(0.256)(0.256)(0.256)(0.256)C8H16O8 of conc. 10–3 moles/lt required O2 (in gm/lt)

C8H16O8 + 8O2 → 8CO2 + 8H2O 1 mole 8 mole

1 mole of C8H16O2 requires 8 moles of O2 for its decomposition240 gm = 128 gm

or 10–3 moles = 8 × 10–3 moles= 8 × 10–3 × 32= 0.256 gm/lt

End of Solution

Q.26Q.26Q.26Q.26Q.26 A gas contains two types of suspended particles having average size 2 μm and 50 μm.Most suitable strategy will be(a) Settling chamber followed by bag filter(b) Bag filter followed by ESP(c) ESP followed by cyclonic separator(d) ESP followed by venturi scrubber


End of Solutio




Q.27Q.27Q.27Q.27Q.27 For a solution of pH 5.6 the concentration of [OH–] will be X × 10–9 mol/litre? The valueof X will be

Ans.Ans.Ans.Ans.Ans. (3.98)(3.98)(3.98)(3.98)(3.98)pH + pOH = 14

pOH = 14 – 5.6 = 8.4–log [OH–] = 8.4

[OH–] = 10–8.4 moles/lt= 10–8.4+9 × 10–9 moles/lt= 3.98 × 10–9 moles/lt

End of Solution

Q.28Q.28Q.28Q.28Q.28 In a constant head permeability test, the hydraulic gradient is 2.5. The specific gravityof the medium is 2.65 with water content of 20%. The coefficient of permeability,k = 0.1 cm/s. The seepage velocity (in cm/s) will be


e =0.2 2.65

0.531

wGs

×= =

n = 0.34641

ee

=+

Vs =v Kn n

=i

=0.1 × 2.50.3464

= 0.721 cm/sec

End of Solution

Q.29Q.29Q.29Q.29Q.29 Two primary sedimentation tank have OFR as 30 m3/m2/day and 15 m3/m2/day. Thediameter of particle removed from these primary sedimentation tank for OFR of 30 m3/m2/dayand 15 m3/m2/day are d30 and d15 respectively. What is the ratio of d30 to d15?

Ans.Ans.Ans.Ans.Ans. (1.4142)(1.4142)(1.4142)(1.4142)(1.4142)For type-I setting, Stokes law is applicable.

Vs ∝ d2

230215

d

d=

3015

= 2

30

15

dd = 2 = 1.4142

End of Solution




Q.30Q.30Q.30Q.30Q.30 Alkalinity of water in equivalent/litre is given by

{ } { } { } { }23 3HCO 2 CO OH H {}− − − ++ + − represents concentration in mol/litre. For a water

sample, the concentration of 3HCO− = 2 × 10–3 mol/litre, 2CO − = 3.04 × 10–4 mol/litre

and pH of water = 9.0. The atomic weight are Ca = 40, C = 12 and O = 16. If theconcentration of OH– and H+ are neglected the alkalinity of water sample (in mg/litre)as CaCO3 is _________.(a) 50 (b) 65.2(c) 100 (d) 130.4


End of Solution

Q.31Q.31Q.31Q.31Q.31 The relationship between oxygen consumption and equivalent biodegradable organicremoval (i.e. BOD) in a closed container with respect to time is shown in figure.

L (remaining)

BOD exertedL0

days2 4 6 8 10 12

Assume that the rate of oxygen consumption is directly proportional to the amount of

degradable organic matter and is expressed as ,tdLkL

dt= − where Lt is the oxygen

equivalent of the organic matter remaining at time ‘t ’. k (d –1) represent the degradationrate constant, L0 is the oxygen equivalent of organic matter at time t = 0Correct expression(a) L0 = Lte

–kt (b) BOD5 = L5

(c) Lt = L0(1 – e–kt) (d) BODt = L0 – Lt


End of Solution




Irrigation Engineering

Q.32Q.32Q.32Q.32Q.32 Which of the following condition is super passage?(a) Canal FSL over natural stream bed level(b) Canal FSL below natural stream bed level(c) Natural stream FSL over canal bed level(d) Natural stream FSL below canal bed level


End of Solution

Q.33Q.33Q.33Q.33Q.33 Muskingham method is used for(a) Hydraulic reservoir routing (b) Hydraulic channel routing(c) Hydrologic reservoir routing (d) Hydrologic channel routing


End of Solution

Q.34Q.34Q.34Q.34Q.34 The field capacity of soil is 30% and the permanent wilting point is 13%. Irrigation isdone when the moisture content falls to 20%. The density of soil is 1500 kg/m3 anddensity of water is 1000 kg/m3. Consumptive use is 2 mm per day. The effective depthof root zone is 80 cm. What is the frequency of irrigation?(a) 7 days (b) 11 days(c) 9 days (d) 13 days

Ans.Ans.Ans.Ans.Ans. (*)(*)(*)(*)(*)FC = 30%

PWP = 13%

FC = 30%

OMC = 20%

PWP = 13%

dw = ( )·d

wd FC OMCγ

× −γ

= ( )150080 0.3 0.2

1000× −

= 12 cm or 120 mmConsumptive use = 2 mm/day

So, frequency of irrigation = 120

60 days2

=

End of Solution




Q.35Q.35Q.35Q.35Q.35 The uplift pressure varies in triangular shape being zero at the toe. What is the safewidth ‘B’ such that there will be no sliding?

10 m

B

γconcrete = 24 kN/m3, γw = 9.81 kN/m3, coefficient of friction = 0.45

Ans.Ans.Ans.Ans.Ans. (15.873 m)(15.873 m)(15.873 m)(15.873 m)(15.873 m)

10 m

B

10γω

μ = 0.45γconc. = 24 kN/m3

Bmin sliding =10

0.45(2.4 1)−

= 15.873 m

End of Solution

Highway Engineering

Q.36Q.36Q.36Q.36Q.36 The design speed of a two lane two way road is 60 km/hr and the longitudinal coefficientof friction is 0.36. The reaction time of the driven is 2.5 sec. Consider acceleration dueto gravity as 9.8 m/sec2. The intermediate sight distance (in m) required for the roadis ________.




Ans.Ans.Ans.Ans.Ans. (162)(162)(162)(162)(162)Given : f = 0.36; v = 60 km; g = 9.8 m/s2; tR = 2.5s

SSD =2

0.278254RV

Vtf

⎛ ⎞+⎜ ⎟⎝ ⎠

=260

0.278 60 2.5254 0.36

× × +×

= 41.7 + 39.37 = 81 mISD = 2 × SSD = 81 × 2 = 162 m

End of Solution

Q.37Q.37Q.37Q.37Q.37 Traffic starts discharging from an approach at an intersection with the signal turninggreen. The constant head was considered from the fourth or fifth headway position isreferred as _______.(a) Intersection headway (b) Saturation headway(c) Effective headway (d) Discharge headway


End of Solution

Q.38Q.38Q.38Q.38Q.38 The average daily temperature of an airport is 39°C. The average maximum temperatureis noted to be 48°C. What will be the airport reference temperature?(a) 36 (b) 39(c) 42 (d) 46

Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)Ta = 39°CTm = 48°C

ATR =3

m aa

T TT

−⎛ ⎞+ ⎜ ⎟⎝ ⎠

=48 39

39 42 C3−⎛ ⎞+ = °⎜ ⎟⎝ ⎠

End of Solution

Q.39Q.39Q.39Q.39Q.39 What is the value of VDF for axle load 15 tonnes considering standard axle load as 8tonnes?

Ans.Ans.Ans.Ans.Ans. (12.35)(12.35)(12.35)(12.35)(12.35)Axle load = 15 T

Standard axle load = 8 T

VDF =415

8⎡ ⎤⎢ ⎥⎣ ⎦

= 12.35

End of Solution




Q.40Q.40Q.40Q.40Q.40 24 hr traffic count at a road section was observed to be 1000 vehicles on a Tuesdayin the month of July. If daily adjustment factor for Tuesday is 1.121 and monthlyadjustment factor for July is 0.913, the annual average daily traffic is ______ veh/day.

Ans.Ans.Ans.Ans.Ans. (1023.473)(1023.473)(1023.473)(1023.473)(1023.473)Tweek = DEF × T24

= 1.121 × 1000 = 1121AADT = MEF × ADT

= 0.913 × 1121= 1023.473

End of Solution

Q.41Q.41Q.41Q.41Q.41 The variation between q and v is given as

q

v

The variation between v vs k will be

(a)

v

k

(b)

v

k

(c)

v

k

(d)

v

k

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

End of Solution




Geometics Engineering

Q.42Q.42Q.42Q.42Q.42 A theodolite is set up at station A. The R.L. of instrument axis is 212.250 m. The angleof elevation to the top of a 4 m long staff, held vertical at station B is 7°. The horizontaldistance between stations A and B is 400 m. Neglecting the errors due to curvatureof earth and refraction, the RL of station B is __________ m.


x

4 m

400 m

7°

V = 400 tan 7° = 49.113x = (49.113 – 4) = 45.113

RLB = 212.25 + 45.113 = 257.363 m

End of Solution

Q.43Q.43Q.43Q.43Q.43 Theodolite is set at station P. Angle of depression to a vane 2 m above the foot of staffat Q is 45°. Distance between P and Q = 20 m (horizontal).Staff reading at bench mark S = 2.905 mR.L. of bench mark = 433.050 mR.L. of station Q is ________ m.


x

2 m

P

2.905 m

SBench mark = 433.05 mRL

Staff 20 m

Q

45°

20x

= tan 45°

x = 20 mRL of Q = 433.05 + 2.905 – x – 2

= 433.05 + 2.905 – 20 – 2= 413.955 m

End of Solution




FLUID MECHANICS

Q.44Q.44Q.44Q.44Q.44 Velocity component in x and y direction of incompressible flow are u = –5 + 6x, v = –(9 + 6y)respectively. Equation of stream line is _______

(a)(9 6 )

constant5 6

yx

+=

− +(b) ( 5 6 ) (9 6 ) constantx y− + − + =

(c)5 6

constant9 6

xy

− +=

+ (d) (–5 + 6x) (9 + 6y) = constant

Ans.Ans.Ans.Ans.Ans. (d)(d)(d)(d)(d)Given :

u = –5 + 6xv = –(9 + 6y)

Equation of streamline

dux

=dyv

5 6d

− +xx

=(9 6 )

dyy− +

Integrating it,

ln(–5 + 6x)6

= –ln(9 + 6y) ln6 6

C+

1ln( 5 6 ) (9 6 )

6y− + ⋅ +x =

1ln

6C

Take antilog,(–5 + 6x)(9 + 6y) = constant

u.v = constant

End of Solution

Q.45Q.45Q.45Q.45Q.45 A hydraulic jump occurs in a triangular channel with side slopes 1 : 1 (V : H). The sequentdepths are 0.5 m and 1.5 m respectively. The flow rate is _________ m3/s.

Ans.Ans.Ans.Ans.Ans. (1.728 m(1.728 m(1.728 m(1.728 m(1.728 m33333/s)/s)/s)/s)/s)

Y

T Y = 2

Y = Y3 1

1




A =1

22

Y Y× × = Y2

Y =3Y

For a horizontal and frictionless channel

Specific Force (F) =2Q

AYAg

+ = Constant

⇒2

223 ( )

Y QYY g

⎛ ⎞ +⎜ ⎟⎝ ⎠= Constant

⇒3 2

23Y Q

gY+ = Constant

If Y1 and Y2 are conjugate depths

13 2

213

Y Q

gY+ =

23 2

23

Y QgY

+

⇒3 2

20.5

3 0.5Q

g+

×=

3 2

21.5

3 1.5Q

g+

×

⇒3 31.5 0.5

3 3− =

2

2 2

1 1

0.5 1.5

Qg

⎛ ⎞−⎜ ⎟⎝ ⎠Q = 1.728 m3/sec

End of Solution

Q.46Q.46Q.46Q.46Q.46 A cast iron pipe of diameter 600 mm and length 400 m is discharging into open air.The level of discharging pipe is 4.5 m below the water surface in the tank. The frictionfactor of the pipe is 0.018. What is the velocity (in m/s) of flow in the pipe?


(1)

(2)

4.5 m

gDfL

= 9.81 m/s = 0.6 m

= 0.018 = 400 m

2




Apply energy equation between (1) and (2)

21 1

12P V

zg g

+ +ρ

=2

2 222 f

P Vz h

g g+ + +

ρ

4.5 =2 2 2

0.52 2 2

f L V V VgD g g

⋅ ⋅ + +

4.5 =2 2(0.018)(400) 1.5

2(9.81)(0.6) 2V V

g⋅ +

4.5 =2 212 1.5

2 2V Vg g

+

V2 = 6.54V = 2.557 m/s

End of Solution

Q.47Q.47Q.47Q.47Q.47 Velocity profile

sin2

yU U∞

π⎛ ⎞= ×⎜ ⎟⎝ ⎠δ

Given, U∞ = 0.3 m/s and δ = 1 m. What is dudy

at y = 0?

Ans.Ans.Ans.Ans.Ans. (0.47)(0.47)(0.47)(0.47)(0.47)Given :

uu∞

= sin2

yπ⎛ ⎞⋅⎜ ⎟⎝ ⎠δu∞ = 0.3 m/s

δ = 1 m

dudy

= sin2

d yu

dy ∞π⎛ ⎞⋅ ⋅⎜ ⎟⎝ ⎠δ

= cos2 2

u y∞ ⋅ π π⎛ ⎞⋅⎜ ⎟⎝ ⎠δ δAt y = 0 and δ = 1

0y

dudy =

=0.3 0

cos2(1) 2 1

π π⎛ ⎞⋅⎜ ⎟⎝ ⎠

= 0.47 m/s/m

End of Solution




RCC Structures and Prestressed Concrete

Q.48Q.48Q.48Q.48Q.48500 mm

100 mm

A st = 400 mm2

500

mm

Yield strain of steel = 0.0044, M35, Fe550The limiting depth of N.A. ___________ mm.

Ans.Ans.Ans.Ans.Ans. (221.52)(221.52)(221.52)(221.52)(221.52)For a RCC T-Beam

500

d =500 mm

xu,lim

d – xu,lim

0.0035

0.0044

(For limiting depth of neutral axis)Considering d = 500 mm

, lim

0.0035

ux=

, lim

0.0044

ud − x

d – xu,lim = ,lim0.00440.0035 u× x

35 × 500 = 35xu,lim + 44xu,lim

= 79xu,lim

xu,lim =35 500

79×

= 221.52 mm

Limiting depth of neutral axisx0,lim = 221.52 mm

End of Solution

Q.49Q.49Q.49Q.49Q.49 Minimum pH of water allowed for concrete mix design?(a) 4.5 (b) 5.5(c) 6 (d) 5




Ans.Ans.Ans.Ans.Ans. (c)(c)(c)(c)(c)1. Minimum pH value of water for concrete = 6.0As per IS code provision no. 5.4.2, the pH value of water shall not less than 6.0.

End of Solution

Q.50Q.50Q.50Q.50Q.50 Consider the PSC beam as shown in the figure below.

50 mm

70 mm

7.5 m 7.5 m

What will be the percentage frictional loss in the PSC (in %) __________ consideringjacking to be done from one end.[Given, μ = 0.35, k = 0.0015/m]

Ans.Ans.Ans.Ans.Ans. (4.49%)(4.49%)(4.49%)(4.49%)(4.49%)

50 mm

70 mm

7.5 m 7.5 m

Jacking from one endx = L = 15 m

Wobble correction factor, K = 0.0015Coefficient of friction = 0.35 = μ

P = Not givenp0 = Unknown

Change of gradient, α = tan α = 8hL

= 8 12015000

× = 0.064

% loss of stress in steel due to friction

=0

0

( )100

p Kp

+ μα×

x

= (0.0015 × 15 + 0.35 × 0.064) × 100= 4.49%

End of Solution




Structural Analysis

Q.51Q.51Q.51Q.51Q.51 The force at member PQ will be. Point A and Q are hinged.

4 m

4 m

4 m

Q

P

A

10 kN

10 kN

10 kN

2 m 2 m

Ans.Ans.Ans.Ans.Ans. (30 kN)(30 kN)(30 kN)(30 kN)(30 kN)Force in PQ

10 kN

10 kN

10 kNR

4 m

4 m

4 m

P

Q2m 2m

(1) (1)

Considering the section above (1) – (1)Taking moment about ‘R’

10 kN

10 kN

10 kNR

4 m

4 m

FPQ




ΣMRL = 0(10 × 4) + (10 × 8) + FPQ × 4 = 0

FPQ =120

4− = –30 kN = 30 kN (C)

End of Solution

Design of Steel Structures

Q.52Q.52Q.52Q.52Q.52 What will be the design capacity for a fillet weld as shown in figure?

P = ?120 mm

200 mm

Ultimate strength of steel is 410 MPa, thickness of weld = 6 mm.

Ans.Ans.Ans.Ans.Ans. (413.6 kN)(413.6 kN)(413.6 kN)(413.6 kN)(413.6 kN)

P = ?120 mm

200 mm

P =410

0.7 6 5203 1.25

× × ××

=716352 413598 N

3=

= 413.6 kN

End of Solution

Q.53Q.53Q.53Q.53Q.53 The ratio of plastic moment capacity to yield moment capacity is called(a) shape factor (b) load factor(c) modulus of risilience (d) rigidity modulus


Ratio of p

y

MM = Shape factor

End of Solution

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