Hyderabad | Delhi | Ahmedabad | Pune | Bhubaneswar | Bengaluru ...€¦ · So, 2013 & 2019 have same calendar 09. The ratio of ‘the sum of the odd positive integers from 1 to 100’

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Questions with Detailed Solutions

GATE-2020

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CIVIL ENGINEERING(Afternoon Session)

ACE Engineering Publications Hyderabad Delhi Pune Bhubaneswar Bengaluru Lucknow Chennai Vijayawada Vizag Tirupati Kolkata Ahmedabad

SUBJECTWISE WEIGHTAGE

S. No. NAME OF THE SUBJECT 1 MARK QUESTIONS

2 MARKS QUESTIONS

01 Engineering Mechanics - 1

02 Strength of Material 2 2

03 Structural Analysis 1 1

04 Construction Materials and Management 1 1

05 Reinforced Cement Concrete 1 2

06 Design of Steel Structures - 1

07 Geotechnical Engineering 4 5

08 Fluid Mechanics & Hydraulic Machines 2 2

09 Hydrology 2 -

10 Irrigation Engineering 1 2

11 Environmental Engineering 3 4

12 Transportation Engineering 3 3

13 Geomatics Engineering (Surveying) - 2

14 Engineering Mathematics 5 4

15 Verbal Ability 2 1

16 Numerical Ability 3 4

Total No. of Questions 30 35

GATE - 2020CIVIL ENGINEERING

Question with Detailed Solutions

09/02/20

AfternoonSession


2 Civil Engineering

Section : General Aptitude

01. If f x x2=] g for each , ,x then

f x

f f f xd 3 3-^ ^ ^

]]h gghh

is equal to _____

(a) f(x) (b) (f(x))4

(c) (f(x))2 (d) (f(x))3

01. Ans: (d)Sol: f(x) = x2

f(f(x)) = f(x2) = x4

f(f(f(x))) = f(x4) = x8

So, f x

f f f x

x

xx f x2

86 3= = =^ ^

]] ^ ]gghh gh

02. Nominal interest rate is defined as the amount paid by the borrower to the lender for using the borrowed amount for a specific period of time. Real interest rate calculated on the basis of actual value (inflation-adjusted), is approximately equal to the difference between nominal rate and expected rate of inflation in the economy.(a) Under low inflation, real interest rate is high

and borrowers get benefited.(b) Under low inflation, real interest rate is low

and borrowers get benefited.(c) Under high inflation, real interest rate is low

and borrowers get benefited.(d) Under high inflation, real interest rate is low

lenders get benefited.02. Ans: (c)Sol: The term ‘inflation - adjusted’ suggests option (c).

03. Select the most appropriate word that can replace the underlined word without changing the meaning of the sentence:

Now - a - days, most children have a tendency to belittle the legitimate concerns of their parents.

(a) applaud (b) disparage (c) begrudge (d) reduce03. Ans: (b)Sol: Belittle means disparage(to describe someone as

unimportant).

04. The monthly distribution of 9 Watt LED bulbs sold by two firms X and Y from January to June 2018 is shown in the pie-chart and the corresponding table. If the total number of LED bulbs sold by two firms during April-June 2018 is 50000, then the number of LED bulbs sold by the firm Y during April-June 2018 is _____.

15%January (15%)

February (20%)10%

10%

15%20%

30%

March (30%)

April (15%)

May (10%)

June (10%)

Month Ratio of LED bulbs sold by two firms (X : Y)

January 7 : 8February 2 : 3March 2 : 1April 3 : 2May 1 : 4June 9 : 11

(a) 9750 (b) 8250 (c) 11250 (d) 875004. Ans: (*)Sol: Question wrong Question change required : April - June 2018 → Jan - June 2018 Total bulbs = 50000 Answer: 9750 April - June 2018 total bulbs sold = 50000

05. Rescue teams deployed ______disaster hit areas combat_____ a lot of difficulties to save the people.

(a) with, with (b) to, to (c) with, at (d) in, with


3 GATE_2020_Questions with Solutions


4 Civil Engineering

05. Ans: (d)Sol: Deployed in : combat with

06. After the inauguration of the new building the Head of the Department (HoD) collated faculty preferences for office space. P wanted a room adjacent to the lab. Q wanted to be close to the lift. R wanted a view of the playground and S wanted a corner office.

Assuming that everyone was satisfied, which among the following shows a possible allocation

LiftP

S R HOD Q

Lab

Play Ground

(a)

(c) (d)

(b)

Roa

dR

oad

Roa

dR

oad

LiftP

S R HOD Q

Lab

Play Ground

LiftP

S R P HOD

Lab

Play Ground

LiftP

HOD Q R S

Lab

Play Ground

06. Ans: (a)Sol: 1. P wanted a room adjacent to lab option ‘c’ eliminated 2. S wanted a corner office option ‘d’ eliminated 3. Q want to be close to the shift option ‘b’ eliminated Best possible option A, satisfying faculty

preferences

07. Select the word that fits the analogy: Partial : Impartial : : Popular: ___ (a) Impopular (b) Mispopular (c) Unpopular (d) Dispopular07. Ans: (c)Sol: Antonyms analogy

08. For the year 2019, which of the previous year’s calendar can be used?

(a) 2014 (b) 2012 (c) 2011 (d) 201308. Ans: (d)Sol: Non - leap year repeats after 6 (or) 11 years

(Note: Above statement is valid only if leap year is repeated every 4 years)

2013 + 6 = 2019 2013 & 2019 calender is sameMethod - 2

:odd days

odd days

2013

1

2014

1

2015

1

2016

2

2017

1

2018

1

2019

7 0= =1 2 34444444444444444444444444 4444444444444444444444444

So, 2013 & 2019 have same calendar

09. The ratio of ‘the sum of the odd positive integers from 1 to 100’ to ‘the sum of the even positive integers from 150 to 200’ is _______

(a) 1 : 2 (b) 45 : 95 (c) 1 : 1 (d) 50 : 9109. Ans: (d)Sol: Sum of odd positive integers from 1 to 100 50

2

1 992500� � �^ h

Sum of even positive integers from 150 to 200

262

150 20013 350#� � �^ h

Ratio = 13 350

2500

91

50

#=



10. In a school of 1000 students, 300 students play chess and 600 students play football. If 50 students play both chess and football, the number of students who play neither is ____

(a) 50 (b) 200 (c) 100 (d) 15010. Ans:(d)Sol:

Chess(300) Football (600)

1000

neither

300+600-50=850150

50

150 Students play neither chess not football

Section : Civil Engineering

01. Muskingum method is used in (a) hydrologic channel routing (b) hydraulic reservoir routing (c) hydraulic channel routing (d) hydrologic reservoir routing01. Ans: (d)Sol: Hydrologic channel routing - Muskingum method

02. The ordinary differential equation

sindx

d ux u x2 02

22� � � is

(a) nonlinear and homogeneous (b) linear and non homogeneous (c) non linear and non homogeneous (d) linear and homogeneous02. Ans: (b)

Sol: sindx

d ux u x2 02

22� � �

It is a linear non-homogeneous equation

03. Velocity distribution in a boundary layer is given by

sinU

u y

2

��

�3

c m , where u is the velocity at vertical

co-ordinate y, U∞ is the free stream velocity and d is the boundary layer thickness. The values of U∞ and

d are 0.3 m/s and 1.0 m, respectively. The velocity

gradient y

u

22c m (in s-1, round to two decimal places)

at y = 0, is 03. Ans: 0.47Sol: Given: sin

U

U y

2

��

�3

c m

U∞ = 0.3 m/s ẟ = 1 m

To find: @dy

duy 0=

. . cosdy

duU

y

2 2��

�� 3 c m

@ y = 0, cos y

21

��

�c m

.

dy

du

2 1

0 3& #

# ��

= 0.471 s-1

04. As per IS 456:2000, the pH value of water for concrete mix shall NOT be less than

(a) 6.0 (b) 5.5 (c) 4.5 (d) 5.004. Ans: (a)Sol: As per clause 5.4.2 of IS 456:2000, the pH value of

water shall not be less than 6.

05. The traffic starts discharging from an approach at an intersection with the signal turning green. The constant headway considered from the fourth or fifth headway position is referred to as

(a) intersection headway (b) saturation headway (c) discharge headway (d) effective headway05. Ans:(b)Sol: Saturation headway starts after fourth or fifth

headway


6 Civil Engineering

06. Super passage is a canal cross-drainage structure in which(a) canal water flows under pressure below a

natural stream(b) natural stream water flows with free surface

below a canal (c) natural stream water flows under pressure

below a canal(d) canal water flows with free surface below a

natural stream06. Ans: (b)Sol: In super passage, canal runs below drain. FSL of

canal is lower than bed of natural stream canal water runs under gravity.

07. 24-h traffic count at a road section was observed to be 1000 vehicles on a Tuesday in the month of July. If daily adjustment factor for Tuesday is 1.121 and monthly adjustment factor for July is 0.913, the Annual Average Daily Traffic (in veh/day, round off to the nearest integer) is _____

07. Ans: 1023Sol: ADT = Daily Adj factor × 24 hr traffic ADT = 1.21 × 1000 = 1121 v/day AADT = Monthly exp factor × ADT = 0.913 × 1121 = 1023.5 veh/day

08. For an axle load of 15 tonne on a road, the Vehicle Damage Factor (round off to two decimal places), in terms of the standard axle load of 8 tonne, is_______

08. Ans: 12.36

Sol: VDFstandard load

actual load4

= b l

. .VDF8

1512 359 12 36

4

.= =b l

09. The following partial differential equation is defined for u:u (x, y)

; ;

y

u

x

uy x x x02

2

1 222

22

$ # #=

The set of auxiliary conditions necessary to solve the equation uniquely, is (a) one initial condition and two boundary conditions(b) three initial conditions(c) two initial conditions and one boundary condition(d) three boundary conditions

09. Ans: (a)

Sol: Given: ; ;y

u

x

uy x x x02

2

1 222

22

$ # #=

It needs one initial and two boundary conditions.

10. The state of stress represented by Mohr’s circle shown in the figure is

Normal Stress

Shear Stress

(0,0)

(a) biaxial tension of equal magnitude (b) pure shear (c) hydrostatic stress (d) uniaxial tension

10. Ans: (b)Sol: Under pure shear condition, centre of Mohr’s circle

coincides with origin.




8 Civil Engineering

11. A weightless cantilever beam of span L is loaded as shown in the figure. For the entire span of the beam, the material properties are identical and the cross-section is rectangular with constant width.

From the flexure -critical perspective, the most economical longitudinal profile of the beam to carry the given loads amongst the options given below, is

P

PLL

(b)

(d)

(a)

(c)

11. Ans: (d)Sol:

A B M= PL

MA = 0

RA = P

M M

SFD

BMD

+veM

BA

+ve

P

F 0u-5 �� RA – P = 0 RA = P M 0A^ �� P(L) – PL – MA = 0 MA = 0 For most economical (Beam of uniform strength)

non prismatic beam should be used (i.e. c/s dimensions should be changed as per the B.M along length of beam).

Maximum cross section is provided where the bending moment is maximum

i.e.

Max. BM Occurs

12. A gas contains two types of suspended particles having average sizes of 2 mm and 50 mm. Amongst the options given below, the most suitable pollution control strategy for removal of these particles is (a) electrostatic precipitator followed by cyclonic

separator (b) electrostatic precipitator followed by venturi

scrubber(c) settling chamber followed by bag filter (d) bag filter followed by electrostatic precipitator

12. Ans: (a)Sol: Particles of 50 m separated by gravity settling

chamber and 2 m are separated by Fabric filters [bag filters]

13. The relationship between oxygen consumption and equivalent biodegradable organic removal (i.e., BOD) in a closed container with respect to time is shown in the figure.

BOD Exerted

L Remaining

0 2 4 6Time, Days

Oxy

gen

cons

umpt

ion

and

Equi

vale

nt o

rgan

ic re

mov

al

8 10 12

L0

Assume that the rate of oxygen consumption is directly proportional to the amount of degradable organic matter and is expressed as ,

dt

dLkL

tt��



where, L1 (in mg/litre) is the oxygen equivalent of the organics remaining at time t and k(in d-1) is the degradation rate constant. Lo is the oxygen equivalent of organic matter at time, t = 0.

In the above context, the correct expression is

(a) Lt = Lo (1 - e-kt) (b) BODt = Lo - Lt

(c) BOD5 = L5 (d) Lo = Lte-kt

13. Ans: (d)Sol: As per BOD Kinetics: Lo = Initial organic matter concentration Lt = Remaining organic matter concentration at any

time ‘t’

dt

dLLt` ��

Consider exponential decline, .dt

dLK Lt��

dt

dLKdt� �##

ln (Lt) = –k(t) + C Applying Boundary Conditions: At t = 0, Lt = Lo

ln (Lo) = C ∴ ln (Lt) = – Kt + ln (Lo)

L

Le

o

t kt" = -

Organic matter remaining at time t, Lt = Lo.e

–Kt

yt = BOD consumed during ‘t’ days = Organic matter consumed yt = Lo – Lt

.y L L et o okt` � � -

= Lo [1 – e–kt]

14. A one -dimensional consolidation test is carried out on a standard 19 mm thick clay sample. The oedometer’s deflection gauge indicates a reading of 2.1 mm. just before removal of the load, without allowing any swelling. The void ratio is 0.62 at this stage. The initial void ratio (round off to two decimal places) of the standard specimen is ________.

14. Ans: 0.82Sol: Initial After loading H = 19 mm ΔH = 2.1 mm eo = ? ef = 0.62

H

H

e

e

1 o

� ��

. .

e

e

19

2 1

1

0 62

o

o� ��

..

e

e0 11

1

0 62

o

o� ��

0.11 + 0.11 eo = eo - 0.62 eo = 0.820 eo = 0.82

15. A triangular direct runoff hydrograph due to a storm has a time base of 90 hours. The peak flow of 60 m3/s occurs at 20 hours from the start of the storm. The area of catchment is 300 km2. The rainfall excess of the storm (in cm), is __

(a) 5.40 (b) 3.24 (c) 2.00 (d) 6.4815. Ans: (b)Sol: Triangular DRH Time base 90 hrs = TB Qp = 60 m3/s C.A = 300 km2

R = ?

60 m3/s

90 Hr

20 Hr

Qp = Q

t

Area of DRH = C.A × R R

2

190 3600 60 300 10

6# # # # #=

R = 0.0324 m R = 3.24 cm


10 Civil Engineering

16. Soil deposit formed due to transportation by wind is termed as

(a) alluvial deposit (b) aeolian deposit (c) lacustrine deposit (d) estuarine deposit16. Ans: (b)Sol: Soil deposited by wind is known as Aeolian soil Soil deposited by water on river bed is known as

Alluvial soil Soil deposited by water lake bed is known as

Lacustrine soil.

17. A fair (unbiased) coin is tossed 15 times. The probability of getting exactly 8 Heads (round off to three decimal places), is ______

17. Ans: 0.196Sol: Given: n = 15, ,P q

2

1

2

1= =

Let x = Number of heads ( )P x C p q8

nxx n x= = -

C2

1

2

1158

8 7= b bl l

( )

C

215

158= = 0.196

18. The integral

x x x dx5 4 3 23 2

0

1

+ + +^ h#

is estimated numerically using three alternative methods namely the rectangular, trapezoidal and Simpson’s rules with a common step size. In this context, which one of the following statements is TRUE?(a) Simpson’s rule as well as rectangular rule of

estimation will give NON-zero error.(b) Simpson’s rule, rectangular rule as well as

trapezoidal rule of estimation will give NON-zero error.

(c) Only the rectangular rule of estimation will give zero error.

(d) Only Simpson’s rule of estimation will give zero error.

18. Ans: (d)Sol: The integrand is f(x) = 5x3 + 4x2 + 3x + 2 It is a polynomial of degree 3 Simpson’s rule gives the exact value of the integral

19. A sample of 500 g dry sand, when poured into a 2 litre capacity cylinder which is partially filled with water, displaces 188 cm3 of water. The density of water is 1 g/cm3. The specific gravity of the sand is

(a) 2.52 (b) 2.66 (c) 2.72 (d) 2.5519. Ans: (b)Sol: Given: Weight of solid particles, Ws = 500 g The volume of water replaced by dry soil, is equal

to the volume of the soil solid particles. ∴ VS = 188 cm3

. /V

Wg cc

188

5002 66s

S

S��

The specific gravity of soil (soil solids) ,

.G

1

2 66S

w

s

��

∴ Gs = 2.66

20. Two identically sized primary settling tanks receive water for Type -I settling (discrete particles in dilute suspension) under laminar flow conditions. The Surface Overflow Rate (SOR) maintained in the two tanks are 30 m3/m2. d and 15 m3/m2.d The lowest diameters of the particles, which shall be settled out completely under SORs of 30 m3/m2.d and 15 m3/m2.d are designated as d30 and d15, respectively. The ratio, d

d15

30 (round off to two decimal places), is ______

20. Ans: 1.414Sol: For PST, Vo = Vs

and in such a case, Vs ∝ d2

( )

( )

d

d

V

V

s

s

152302

15

30` =

d

d

15

302

15

30` = =





21. The ratio of the plastic moment capacity of a beam section to its yield moment capacity is termed as

(a) shape factor (b) slenderness ratio (c) load factor (d) aspect ratio21. Ans: (a)

Sol: Shape factor = M

M

Yield Moment Capacity

Plastic Moment Capacity

y

p =

22. The velocity components in the x and y directions for an incompressible flow are given as u = ( - 5 + 6x) and v = -(9+6y), respectively. The equation of the streamline is

(a) y

x

9 6

5 6constant�

� � �

(b) (-5+6x) - (9+6y) = constant

(c) (-5 + 6x) (9 + 6y) = constant

(d) x

y

5 6

9 6

� ��

= constant

22. Ans: (c)Sol: u = (-5 + 6x) v = -(9 + 6y)

dx

dy

x

y

5 6

9 6� � �� ^_ i

h

x

dx

y

dy

5 6 9 6& � � �

��

^ _h i

By integration

ln x ln y ln C6

15 6

6

19 6� � �� ^ _h i

⇒ (-5 + 6x) (9 + 6y) = constant

23. A soil had dry unit weight of 15.5 kN/m3, specific gravity of 2.65 and degree of saturation of 72%. Considering the unit weight of water as 10 kN/m3, the water content of the soil (in %, round off to two decimal places) is ___

23. Ans: 19.28Sol: γ = 15.5 k N/m3 G = 2.65 Sr = 72% γw = 10 kN/m3

w = ?

e

G

1d

w��

.

.e

G1

15 5

2 65 101

d

w #��

∴ e = 0.7096∴ e.sr = w.G

.

. .w

2 65

0 7096 0 72#= = 0.19281= 19.28%

24. The value of Limx

x

7

9 2020

x

2

++

"3 is

(a) indeterminable (b) 1

(c) 9

7 (d) 3

24. Ans: (d)Sol:

lim limx

x

x

xx

7

9 2020

12

7

92020

x x

22

2

��

�

�

" "\ 3

cb l

m

lim

x

x

17

92020

x

2�

�

�

"3 b l

9 3= =

25. The maximum applied load on a cylindrical concrete specimen of diameter 150 mm and length 300 mm tested as per the split tensile strength test guidlines of IS 5816:1999 is 157 kN. The split tensile strength (in MPa, round of to one decimal place) of the specimen is________

25. Ans: 2.2Sol: Split tensile strength of concrete

. /fDL

PN mm

2

150 300

2 157 102 221s

32

# ## #

� ��

26. Alkalinity of water, in equivalent/litre (eq/litre), is given by

{HCO3–} + 2{CO3

2–} + {OH–} – {H+} Where, {} represents concentration in mol/litre. For

a water sample, the concentrations of HCO3– = 2

×10–3 mol/litre, CO32– = 3.04 ×10–4 mol/litre and pH



of water = 9.0. The atomic weights are: Ca = 40; C = 12; and O = 16. If the concentration of OH– and H+ are NEGLECTED, the alkalinity of water sample (in mg/litre as CaCO3), is

(a) 130.4 (b) 50.0 (c) 100.0 (d) 65.2

26. Ans: (a)Sol: CO3

–2 = 3.04 ×10–4 mole/L = 3.04 ×10–4 ×60 ×103

= 18.24 mg/L HCO3

– = 2 ×10–3 mole/L = 2 ×10–3 ×61 ×103

= 122 mg/L ∴Total Alkalinity = [ ] [ ]CO HCO

30

50

61

5032

3# #+- -

.TA 18 2430

50122

61

50` # #� �

= 30.4 + 100 = 130.4 mg/L as CaCO3

27. Joints I, J, K, L, Q and M of the frame shown in the figure (not drawn to the scale) are pins. Continuous members IQ and LJ are connected through a pin at N. Continuous members JM and KQ are connected through a pin at P. The frame has hinge supports at joints R and S. The loads acting at joints I, J and K are along the negative Y direction and the loads acting at joints L and M are along the positive X direction.

10 kNY

X

10 kN

1m 1m 1m 1m 1m 1m

10 kN

10 kN

QL

I J

NR P S

K

M

10 kN

1m1m

The magnitude of the horizontal component of reaction (in kN) at S, is

(a) 10 (b) 20 (c) 5 (d) 15

27. Ans: (a)Sol:

450

900

450

FPK

FPK

Fks

10 kN

10 kN

Adopting Lami’s Theorem

sin

F

sin45 90

10KS =

F kN comp2

10KS = ^ h

900

450 450

M

FPMFPM

FMS

10 kN

Adopting Lami’s theorem

sin

F

sin45 90

10MS =

F comp2

10MS = ^ h

450

450

FKS

FMSsin450

FKSsin450

FMScos450

cos450

FMS

FKS

Vs

HS

S

Vs=0

The magnitude of the horizontal comp of reaction at S

HS = FKS cos 45 + FMS cos 45

2

10

2

1

2

10

2

1# #� � = 10 kN



28. The theodolite was set up at a station P. The angle of depression to a vane 2 m above the foot of a staff held at another station Q was 45o. The horizontal distance between stations P and Q is 20 m. The staff reading at a benchmark S of RL 433.050 m is 2.905 m. Neglecting the errors due to curvature and refraction, the RL of the station Q (in m), is

(a) 435.955 (b) 413.050 (c) 413.955 (d) 431.05028. Ans: (c)Sol:

2 m

20 m

S433.05 m

2905 m

P

Q

45°x

Tanx

x m4520

20o &= =

RL of Q = 433.05 + 2.905 - 20 - 2 = 413.955 m

29. A cast iron pipe of diameter 600 mm and length 400 m carries water from a tank and discharges freely into air at a point 4.5 m below the water surface in the tank. The friction factor of the pipe is 0.018. Consider acceleration due to gravity as 9.81 m/s2. The velocity of the flow in pipe (in m/s, round off to two decimal places) is ____________.

29. Ans: 2.6Sol: Given: L = 400 m D = 0.6 m f = 0.018 g = 9.81 m/s2

1

2

H = 4.5 m

To find: V = ? By B.Eqn b/w (1) & (2)

q

P

g

VZ

q

P

g

VZ h

2 2

1 12

12 2

2

2� � � � � �

.g

V

D

fL

g

V4 5

2 2

22

22

� �

.. .

.V4 5

2 9 81

11

0 6

0 018 40022& #

#� �b l

⇒ V2 = 2.606 m/s

30. A theodolite is set up at station A. The RL of instrument axis is 212.250 m. The angle of elevation to the top of a 4 m long staff, held vertical at station B, is 7o. The horizontal distance between stations A and B is 400 m. Neglecting the errors due to curvature of earth and refraction, the RL (in m, round off to three decimal places) of station B is ____________.

30. Ans: 257.363 Sol:

4 m

400 m

x

A

B7°

(4 + x) = 400 Tan 7o ⇒ x = 45.113 m RL of B = 212.25 + 45.113 = 257.363 m



31. The flow-density relationship of traffic on a highway is shown in the figure

Flow

Density

The correct representaion of speed-density relationship of the traffic on this highway is

Spee

d

Density

1.

3.

Spee

d

Density

4.

Spee

d

Density

2.

Spee

d

Density

31. Ans: (d)Sol: Initially uniform & hyperbola is required in the

second part.

Flow

Density

Spee

d

Density

Daganzo model

32. A footing of size 2 m × 2 m transferring a pressure of 200 kN/m2, is placed at a depth of 1.5 m below the ground as shown in the figure (not drawn to the scale). The clay stratum is normally consolidated. The clay has specific gravity of 2.65 and compression index of 0.3.

200 kN/m2

Silty Sand GWTγd = 15 kN/m3

γd = 18 kN/m3

γd = 17 kN/m3Clay

Dense Sand

1.5

m1.

5 m

1m0.5 m

Considering 2:1 (vertical to horizontal) method of load distribution and γw = 10 kN/m3, the primary consolidation settlement (in mm, round off to two decimal places) of the clay stratum is ___________.

32. Ans: 74.23Sol: Footing size = 2 m × 2 m Clay: .

e

G e

1sat

o

ow� ��

�

.

e

e

10

17

1

2 65

o

o� ��

1.7 + 1.7 eo = 2.65 + eo eo = 1.357 Final consolidation settlement

'

' 'S H

e

Clog

1f o

o

c

o

o10 �

� �� ; E

Ho = 1.5 m, Cc = 0.3, eo = 1.357 At centre of clay σo' = (1.5 + 0.5 )γd + 0.5 γ'1 + 0.7 γ'2 σo' = 2 × 15 + 0.5 × 8 + 0.75 × 7 σo' = 39.25 kPa Increase in stress at centre of clay layer due to

footing load, '

B Z

Q2��

�] g --- for 2 V to 1 H load dispersion

'.

. kPa2 1 75

80056 892��

��^ h

.

..

.

. .S log

1 1 357

0 31 5

39 25

39 25 56 89f 10# #� �

�b l

Sf = 0.7427 m = 74.27 mm



33. A 10 m high slope of dry clay soil (unit weight = 20 kN/m3), with a slope angle of 45o and the circular slip surface, is shown in the figure (not drawn to the scale). The weight of the slip wedge is denoted by W. The undrained unit cohesion (cu) is 60 kPa.

Unit weight = 20 kN/m3

= 60 kPa

10 m

10 m

W

4.48 m

45°

The factor of safety of the slope against slip failure, is

(a) 0.58 (b) 1.57 (c) 1.67 (d) 1.8433. Ans: (*)Sol:

Unit weight = 20 kN/m3

= 60 kPa

10 m

10 m

A

B C

W

4.48 m

45°

θ

The triangle ABC shown is a right angled triangle ∴ θ = 90o

L R180

o# #� ��

. m10 90180

15 708# # ��

Area of the triangle ABC = m2

110 10 50

2# # =

Area of sector ABC = area of circle 360

o

o

# �� R

360

2 #� ��

. m10360

9078 54

2 2#�� Area of the wedge = area of the sector - area of

triangle a = 78.54 - 50

a = 28.54 m2

weight of the wedge W = γ × volume of wedge = γ × (a × 1) = 20 × 28.54 × 1 = 570.8 kN/m

.FOS

W x

C IRu=

. .

.FOS

570 8 4 48

60 15 708 10

## #=

F = 3.69 (check)

34. The planar structure RST shown in the figure is roller-supported at S and pin-supported at R. Members RS and ST have uniform flexural rigidity (EI) and S is a rigid joint. Consider only bending deformation and neglect effects of self-weight and axial stiffening.

T

SR

L

L/2

P

When the structure is subjected to a concentrated horizontal load P at the end T, the magnitude of rotation at the support R, is

(a) EI

PL

12

2

(b) EI

PL

6

(c) EI

PL

6

2

(d) EI

PL

12

3

34. Ans: (d)Sol: P

M PL2=

θR

QEI

ML

EI

PLL

EI

PL

6 6

2

12R

2#= = =

b l





35. For the hottest month of the year at the proposed airport site, the monthly mean of the average daily temperature is 39oC. The monthly mean of the maximum daily temperature is 48oC for the same month of the year. From the given information, the calculated Airport Reference Temperature (in o C), is

(a) 36 (b) 42 (c) 39 (d) 48

35. Ans: (b)

Sol: ART TT T

3a

m a� � �b l

C393

48 3942o� � � �b l

36. Permeability tests were carried out on the samples collected from two different layers as shown int he figure (not drawn to the scale). The relevant horizontal (kh) and vertical (kv) coefficients of permeability are indicated for each layer

Ground Level

Layer 1

Layer 2

kh1 = 4.4 × 10−3 m/s

kr1 = 4 × 10−3 m/s

4 m

3 m

kh2 = 6 × 10−1 m/s

kr2 = 5.5 × 10−1 m/s

The ratio of the equivalent horizontal to vertical coefficients of permeability, is

(a) 80.20 (b) 37.29 (c) 0.03 (d) 68.2536. Ans: (b)Sol:

z1

z2

kh1

kh2

Kh : Kv:

. /K m s4 4 10h3

1 #= - /k m s4 10v

31 #= -

/K m s6 10h1

2 #= - .K 5 5 10v1

2 #= -

Equivalent Horizontal permeability,

KZ Z

K Z K Zh e

h h

1 2

1 21 2� ��] g

.

3 4

4 4 10 3 6 10 43 1# # # #� �

�- -

= 0.3447 m/s Equivalent vertical permeability,

z1

z2

kv1

kv2

K

K

Z

K

Z

Z ZV e

v v

1 2

1 2

1 2

��] g

.

..K

4 10

3

5 5 10

4

3 4

757 272

79 2437 10v e

3 1

3

# #

#��

- -

-] g

.

..

K

K

9 2437 10

0 344737 29

v

h

e3#

= =-b l

37. The diameter and height of a right circular cylinder are 3 cm and 4 cm, respectively. The absolute error in each of these two measurements is 0.2 cm. The absolute error in the computed volume (in cm3, round off to three decimal places), is _________.

37. Ans: 5.184Sol: V d h

4

2��

Vdh d d h

4

2

4

2

� � � � ��

given ẟd = 0.2 cm, ẟ h = 0.2 cm d = 3 cm, h = 4 cm

. .V

4

2 3 4 0 2

4

3 0 22# # # # #� � ��

ẟ V = 1.65 π ẟ V = 5.1836 cm3 = 5.184 cm3



38. The cross-section of the reinforced concrete beam having an effective depth of 500 mm is shown in the figure (not drawn to the scale ). The grades of concrete and steel used are M35 and Fe550, respectively. The area of tension reinforcement is 400 mm2. It is given that corresponding to 0.2% proof stress, the material safety factor is 1.15 and the yield strain of Fe550 steel is 0.0044.

500 mm

500

mm

100 mm

200 mm

AH = 400 mmt

As per IS 456:2000, the limiting depth (in mm, round off to the nearest integer) of the neutral axis measured from the extreme compression fiber, is _________.

38. Ans: 220 mmSol: Limiting depth of N.A

. .

..X d d

0 0035 0 0044

0 00350 44maxu � � �; E

= 0.44 × 500 = 220 mm Xu max > 100 mm ∴ N.A lies in web

39. Two steel plates are lap jointed in a workshop using 6 mm thick fillet weld as shown in the figure (not drawn to the scale). The ultimate strength of the weld is 410 MPa.

120 mm

200 mm

Filletweld

As per Limit State Design of IS 800:2007, the design capacity (in kN, round off to three decimal places) of the welded connection, is __________.

39. Ans: 413.586

Sol: Design Shear capacity of fillet weld (Pdw) = P

P PL t f

3dw

mw

w t u

## #

��

Lw = 2 × 200 +120 = 520 mm tt = K × S = 0.7 × 6 = 4.2 mm γmw = 1.25( for workshop welding) = 1.50 (for site welding)

.

..P P N

3 1 25

520 4 2 410413586 02dw

## #= = =

= 413.586 kN

40. A hydraulic jump occurs in a triangular (V-shaped) channel with side slopes 1:1 (vertical to horizontal). The sequent depths are 0.5 m and 1.5 m. The flow rate (in m3/s, round off to two decimal places) in the channel is ________.

40. Ans: 1.73Sol:

1

190°45° 45°

Given: Side slopes = 1 : 1 depths y1 = 0.5 m y2 = 1.5 m To find: Flow rate (m3/s) Specific force equation:

. .g A

Qz

g A

Qz

1

2

12

2

2� � �

For triangle channel z1 = y

3

1 , zy

32

2=

,A y A y1 12

2 22` = =

g

Q

y yy y

1 1

3

12

12

22 2

313& � � �< 7F A



. . .

. .Q

9 81 0 5

1

1 5

1

3

11 5 0 5

2

2 23 3& � � �] ] ] ]g g g g; 6E @

.

. .Q

9 813 556

3

13 25

2

& =5 5? ?

.

. .Q

3 3 556

3 25 9 812& ##=

⇒ Q2 = 2.9886 .Q 2 9886& = . .Q 1 7287 1 73& -= m3/s

41. The ion product of water (pKw) is 14. If a rain water sample has a pH of 5.6, the concentration of OH– in the sample (in 10–9 mol/litre, round off to one decimal place), is ________.

41. Ans: 3.98Sol: pH = 5.6 pOH = 14 – 5.6 = 8.4 [OH–] = 10–8.4 mole/L = 3.98 ×10–9 mole/L

42. The plane truss has hinge supports at P and W and is subjected to the horizontal forces as shown in the figure (not drawn to the scale)

10 kN

10 kN

10 kN

2 mS V Z

Y

X

W

R

Q

P

2 m

4 m

4 m

4 m

Representing the tensile force with ‘+’ sign and the compressive force with ‘–’ sign, the force in member XW (in kN, round off to the nearest integer), is ___.

42. Ans:30 kN (compression)Sol:

10 kN

10 kN

10 kN FBC

FAB

FCP FPQ

x

PB C

4 m

4 m

2 m 2 m

x

Adopting method of sections and taking section X - X as shown in above figure and selecting the upper part of section cut :

Σ MB = 0 ⇒ 10 × 4 + 10 × 8 - FPQ × 4 = 0

F kN compression4

10 4 10 830PQ

# #� � � ^ h

43. An ordinary differential equation is given below

dx

d y

dx

dyy6 02

2

� � �

The general solution of the above equation (with constant C1 and C2), is

(a) ( )y x C e C xex x

1 3 2 2� �-

(b) ( )y x C xe C ex x

1 3 2 2� �-

(c) ( )y x C e C ex x

1 3 2 2� �-

(d) ( )y x C e C ex x

1 3 2 2� � -

43. Ans: (d)Sol: Given:

dx

d y

dx

dyy6 02

2

� � �

The auxiliary equation is 6D2 + D – 1 = 0 ⇒ 6D2 + 3D – 2D – 1 = 0 ⇒ 3D (2D + 1) – (2D + 1) = 0 ⇒ (3D–1) (2D+1) = 0 ⇒ ,D

3

1

2

1� � ∴The general solution is y C e C e

x x1 3 2 2� � -





44. A waste to energy plant burns dry solid waste of composition: Carbon = 35%, Oxygen = 26%, Hydrogen = 10%, Sulphur = 6%, Nitrogen = 3% and Inerts = 20%. Burning rate is 1000 tonnes/d. Oxygen in air by weight is 23%. Assume complete conversion of Carbon to CO2, Hydrogen to H2O, Sulphur to SO2 and nitrogen to NO2.

Given Atomic weights: H = 1, C = 12, N = 14, O = 16, S = 32

The stoichiometric (theoretical) amount of air (in tonnes/d, round off to the nearest integer) required for complete burning of this waste, is _____.

44. Ans: 6957Sol: So, the MSW quantity = 1000 tonnes/D Excluding inert gases = 200 tonnes/D ---------------- 800 tonnes/D

80 gm MSW needs 160 g of Oxygen

800 tonnes MSW needs ?

/O ton day80

800 16016002

#= =

Air contains only 0.23 [23%] of O2,

So, Air .

. /ton D0 23

16006956 5= =

45. A constant-head permeability test was conducted on a soil specimen under a hydraulic gradient of 2.5. The soil specimen has specific gravity of 2.65 and saturated water content of 20%. If the coefficient of permeability of the soil is 0.1 cm/s, the seepage velocity (in cm/s, round off to two decimal places) through the soil specimen is _________.

45. Ans: 0.72Sol: Constant Head test: i = 2.5 G = 2.65 w = 20% S = 100% k = 0.1 cm/s Vs = ?

Discharge velocity, v = ki = 0.1 × 2.5 = 0.25 cm/s ∴ . .e s w Gr = e = 0.2 × 2.65 e = 0.53 .n

e

e

10 3464� � �

Seepage velocity, vn

vs =

∴ .

.. /V cm s

0 3464

0 250 7216s = =

Vs = 0.72 cm/s

46. A 5 m high vertical wall has a saturated clay backfill. The saturated unit weight and cohesion of clay are 18 kN/m3 and 20 kPa, respectively. The angle of internal friction of clay is zero. In order to prevent development of tension zone, the height of the wall is required to be increased. Dry sand is used as backfill above the clay for the increased portion of the wall. The unit weight and angle of internal friction of sand are 16 kN/m3 and 30o, respectively. Assume that the back of the wall is smooth and top of the backfill is horizontal. To prevent the development of tension zone, the minimum height (in m, round off to one decimal place) by which the wall has to be raised, is _______.

46. Ans: 25 Sol:

X

5 m

Dry sandφ1=300, γ1=16 kΝ/m3,C1=0

φ2=00,

γ2=18 kΝ/m3,C2=20 kPa

Clay



K

sin

sin

sin

sin

1

1

1 30

1 30

3

1a

1

11 �

��

� ��

Ksin

sin

1

11a

2

22 �

��

�

For no tension crack condition, the active pressure at the top of clay shall be zero.

At a depth of ‘x’ from top, as shown in figure, the active pressure, Pa just below the sand is:

P K C K2a a v a22 2��

x0 1 2 20 11# #��

x0 1 16 2 20 1# #� � x = 2.5 m

47. The Fourier series to represent x – x2 for –p ≤ x ≤ p is given by

cos sinx xa

a nx b nx2

on n

nn

2

11

� � � �33

==

//

The value of ao (round off to two decimal places), is _____.

47. Ans: -6.57Sol: Let ( ) ,f x x x x

2 # #� ��

( )a f x dx1

o ��r

r

-

#

( )x x dx1 2

�� r

r

-

#

xdx x dx

1 2

�� r

r

r

r

--

* 4##

x dx10 2

2

0��

r

> H#

x123 o

3

�� rb l; E

1

3

23

�� : D

3

2 2�� = –6.57

48. Crops are grown in a field having soil, which has field capacity of 30% and permanent wilting point of 13%. The effective depth of root zone is 80 cm. Irrigation water is supplied when the average soil moisture drops to 20%. Consider density of the soil as 1500 kg/m3 and density of water as 1000 kg/m3. If the daily consumptive use of water for the crops is 2 mm, the frequency of irrigating the crops (in days), is

(a) 7 (b) 10 (c) 13 (d) 1148. Ans: (*)Sol: dw = sd(FC - OMC)

1000

150080

100

30 20� �^ h

= 12 cm = 120 mm cu = 2 mm/day

fc

dwdays

2

12060

u= = =

49. A concrete beam of span 15 m, 150 mm wide and 350 mm deep is prestressed with a parabolic cable as shown in the figure (not drawn to scale). Coefficient of friction for the cable is 0.35, and coefficient of wave effect is 0.0015 per meter.

50 mm

Cable

Centoidal axis

50 m

m

350

mm

15 m

7.5 m

70mm

If the cable is tensioned from one end only, the percentage loss (round off to one decimal place) in the cable force due to friction, is _______.

49. Ans: 4.5%Sol: Loss of stress due to friction ∆σ = σo [mα + KL] % Loss of stress due to friction when cable is

tensioned from one end



⇒ h = 50 + 70 mm = 120 mm = 0.12 m

⇒ ..

L

h8

15

8 0 120 064

#��

⇒ KL

100 100o o

o# #T

��

��

��8 B

= [0.35 × 0.064 + 0.0015 × 15] × 100 = 4.49% ≈ 4.5%

50. A prismatic linearly elastic bar of length L, cross-sectional area A, and made up of a material with Young’s modulus E, is subjected to axial tensile force as shown in the figures. When the bar is subjected to axial tensile forces P1 and P2, the strain energies stored in the bar are U1 and U2, respectively.

P1

P2

(P1+P2)

If U is the strain energy stored int he same bar when subjected to an axial tensile force (P1 + P2), the correct relationship is

(a) U = U1– U2 (b) U > U1 + U2

(c) U = U1 + U2 (d) U < U1 + U2

50. Ans: (b)

Sol: UAE

P L

2Axial load

2

=

U AEP L2112

=

U AEP L2222

=

( )U AE

P P L21 2

2

=+

P1

P2

(P2+P2)

as (P1 + P2)2 > P1

2 + P22

U > U1 + U2

51. Group-I gives a list of test methods for evaluating properties of aggregates. Group-II gives the list of properties to be evaluated.

Group-I Test Methods P. Soundness test Q. Crushing test R. Los Angeles abrasion test S. Stripping value test Group-II Properties 1. Strength 2. Resistance to weathering 3. Adhesion 4. Hardness The correct match of test methods under Group-I to

properties under Group-II, is (a) P-2; Q-4; R-3; S-1 (b) P-2; Q-1; R-4; S-3 (c) P-3; Q-4; R-1; S-2 (d) P-4; Q-1; R-2; S-351. Ans: (b)Sol:

1. Soundness Test of aggregates is used to determine their resistance against weathering affects.

2. Crushing test is used to determine the strength of the aggregates.

3. Los Angeles Abrasion test is used to determine the harness of aggregates.

4. Stripping value test is used to determine the adhesion better aggregates and bitumen.

52. The design speed of a two-lane two-way road is 60 km/h and the longitudinal coefficient of friction is 0.36. The reaction time of a driver is 2.5 seconds. Consider acceleration due to gravity as 9.8 m/s2. The intermediate sight distance (in m, round off to the nearest integer) required for the load is ______.

52. Ans: 162Sol: V = 60 kmph (16.67 m/s), f = 0.36 ; t = 2.5 sec g = 9.8 m/s ISD = 2(SSD) = vt

gf

v2

2

2

+< F

. .. .

.2 16 67 2 5

2 9 8 0 36

16 672

# # #� �; E

= 162 m







53. A sample of water contains an organic compound C8H16O8 at a concentration of 10–3 mol/litre. Given that the atomic weight of C = 12 g/mol, H = 1 g/mol, and O = 16 g/mol, the theoretical oxygen demand of water (in g of O2 per litre, round off to two decimal places), is ________.

53. Ans: 0.256Sol: C8H16O8 = 10–3 mole/L Molecular weight of C8H16O8 = 240 gm/mole ∴ C8H16O8 = 240 ×10–3 = 0.24 g/L C8H16O8 + 8 O2 →8CO2 + 8 H2 O Mol. weight : 240 gm 8 ×32 gm weight : 0.24 g/L ?

Theoretical oxygen demand .0 24240

8 32##=

= 0.256 g/L

54. A concrete dam holds 10 m of static water as shown in the figure (not drawn to the scale). The uplift is assumed to vary linearly from full hydrostatic head at the heel, to zero at the toe of dam. The coefficient of friction between the dam and foundation soil is 0.45. Specific weights of concrete and water are 24 kN/m3 and 9.81 kN/m3, respectively.

Dam

B

Soil

10.0

m

For NO sliding condition, the required minimum base width base width B (in m, round off to two decimal places) is _________.

54. Ans: (15.36)Sol:

..

bS C

H

0 459 81

241

10

��

�^ bh l

. .

. m0 45 1 446

1015 36= =] g

bsafe = 15.36 m i.e., = 16 m

55. A 4 × 4 matrix [P] is given below

[ ]P

0

2

0

0

1

3

0

0

3

0

6

1

0

4

1

6

��

R

T

SSSSSSSSSSS

V

X

WWWWWWWWWWW

The eigen values of [P] are (a) 0, 3, 6, 6 (b) 1, 2, 5, 7 (c) 3, 4, 5, 7 (d) 1, 2, 3, 4

55. Ans: (b)Sol: Given:

P

0

2

0

0

1

3

0

0

3

0

6

1

0

4

1

6

��

R

T

SSSSSSSSSSS

V

X

WWWWWWWWWWW

P

0

2

0

0

1

3

0

0

3

0

6

1

0

4

1

6

��

Expanding along the first column

( ( ( )2

1

0

0

3

6

1

0

1

6

2 36 1 3 0 0� � � �# -

= 70 We know that det (P) = Product of eigen values Option (b) is the only possible answer





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