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: 2 : GATE_2016_CE

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Q. 1 – Q. 5 carry one mark each.

01. Out of the following four sentences, select the most suitable sentence with respect to grammar and

usage.

(A) I will not leave the place until the minister does not meet me.

(B) I will not leave the place until the minister doesn’t meet me.

(C) I will not leave the place until the minister meet me.

(D) I will not leave the place until the minister meets me.

01 Ans: (D)

Sol: ‘until’ itself is negative so it can’t take one more negative i.e., ‘does not’. Hence, Option (D) is the

right answer

02. A rewording of something written or spoken is a ______________.

(A) paraphrase (B) paradox

(C) paradigm (D) paraffin

02. Ans: (A)

Sol: ‘paraphrase’ means a restatement of a text, passage or a rewording of something written or spoken.

03. Archimedes said, “Give me a lever long enough and a fulcrum on which to place it, and I will move

the world.” The sentence above is an example of a ___________ statement.

(A) figurative (B) collateral

(C) literal (D) figurine

03. Ans: (A)

Sol: ‘figurative’ means representing by a figure or resemblance or expressing one thing in terms

normally denoting another with which it may be regarded as analogous.

: 3 : SET_1_Forenoon Session


04. If ‘relftaga’ means carefree, ‘otaga’ means careful and ‘fertaga’ means careless, which of the

following could mean ‘aftercare’?

(A) zentaga (B) tagafer (C) tagazen (D) relffer

04. Ans: (C)

Sol: From given codes

relftaga carefree

Otaga careful

Fertaga careless

From these codes, clearly known that “care” means “taga,” from given alternatives, option ‘C’ is

correct.

05. A cube is built using 64 cubic blocks of side one unit. After it is built, one cubic block is removed

from every corner of the cube. The resulting surface area of the body (in square units) after the

removal is __________.

(A) 56 (B) 64

(C) 72 (D) 96

05. Ans: (D)

Sol: From given data, 64 cubic blocks of one unit

Sizes are formed

No of faces of the

Cube is ‘6’

No of corners of the

Cube is ‘8’

After removing one

Cubic block from

Each corner,

The resulting surface area of the body

= 6 (4) = 96 sq. Units.

: 4 : GATE_2016_CE




Q. 6 – Q. 10 carry two marks each.

06. A shaving set company sells 4 different types of razors, Elegance, Smooth, Soft and Executive.

Elegance sells at Rs. 48, Smooth at Rs. 63, Soft at Rs. 78 and Executive at Rs. 173 per piece. The

table below shows the numbers of each razor sold in each quarter of a year.

Quarter\

ProductElegance Smooth Soft Executive

Q1 27300 20009 17602 9999

Q2 25222 19392 18445 8942

Q3 28976 22429 19544 10234

Q4 21012 18229 16595 10109

Which product contributes the greatest fraction to the revenue of the company in that year?

(A) Elegance (B) Executive

(C) Smooth (D) Soft

06. Ans: (B)

Elegance Smooth Soft Executive

27300 20009 17602 9999

25222 19392 18445 8942

28976 22429 19544 10234

21012 18229 16595 10109

102510 80059 72186 39284

Rs. 48 Rs. 63 Rs. 78 Rs. 173

Total Revenue

102510 80059 72186 39284

48 63 78 173

4920480 5043717 5630508 6796132

More revenue is on executive

: 6 : GATE_2016_CE


07. Indian currency notes show the denomination indicated in at least seventeen languages. If this is not

an indication of the nation’s diversity, nothing else is.

Which of the following can be logically inferred from the above sentences?

(A) India is a country of exactly seventeen languages.

(B) Linguistic pluralism is the only indicator of a nation’s diversity.

(C) Indian currency notes have sufficient space for all the Indian languages.

(D) Linguistic pluralism is strong evidence of India’s diversity.

07. Ans: (D)

Sol: If seventeen languages were not an indication of the nation’s diversity, nothing else is. If nothing

else is so the best inference is option ‘D’

08. Consider the following statements relating to the level of poker play of four players P, Q, R and S.

I. P always beats Q

II. R always beats S

III. S loses to P only sometimes

IV. R always loses to Q

Which of the following can be logically inferred from the above statements?

(i) P is likely to beat all the three other players

(ii) S is the absolute worst player in the set

(A) (i) only (B) (ii) only (C) (i) and (ii) (D) neither (i) nor (ii)

08. Ans: (A)

Sol: From the given data

Player P > Player Q

Player R > Player S

Player S < Player P (sometimes only)

Player R < Player Q

Player ‘P’ > Player ‘Q’ > Player ‘R’ > Player ‘S’

Logically, the statement (i) is definitely true, but statement (ii) is not

Option ‘A’ is true



09. If f(x) = 2x7 + 3x 5, which of the following is a factor of f(x)?

(A) (x 3 +8) (B) (x–1) (C) (2x–5) (D) (x+1)

09. Ans: (B)Sol: Option (A)

x3 + 8 = 0, x3 = –8, x = –2

Sub x = –2, in f(x) = 2x7 + 3x –5

F(–2) = 2 (–2)7 + 3(–2) –5

= –256 – 6 – 5 = – 267

This is not a factor of f(x)

Option (B)

X – 1 = 0, x = 1

Sub @ x = 1, in f(x) = 2 x7 + 3x –5

f(1) = 2(1)7 + 3(1) –5 = 5–5=0

This is a factor of f(x)

Option (C)

2x – f = 0, x=5/2

Sub @ x = 5/2 in f(x) = 2x7 + 3x–5

525

325

225

f7

= – 2 – 3 – 5 = –10

This is not a factor of f(x)

Option ‘B’ is true

: 8 : GATE_2016_CE


10. In a process, the number of cycles to failure decreases exponentially with an increase in load. At a

load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for

failure. The load for which the failure will happen in 5000 cycles is ________.

(A) 40.00 (B) 46.02 (C) 60.01 (D) 92.02

10. Ans: (B)

load failure for cycles

80 100

40 10000

There is one failure 5000 cycles load must be between 80 and 40

46.02



Q. 1 – Q. 25 carry one mark each

01. Newton-Raphson method is to be used to find root of equation 3x – ex + sin x = 0. If the initial trial

value for the root is taken as 0.333, the next approximation for the root would be _________ (note:

answer up to three decimal)

01. Ans: 0.3601

Sol: o

o01 xf

xfxx

333.0f

333.0f333.0x1

f (0.333) = 3 (0.333) e0.333 + sin (0.333) = 0.999 1.3951 + 0.3268 = 0.0693

f (0.333) = 3 e0.333+ cos (0.333) = 3 1.3951 +0.9450 = 2.5499

x1 = 0.333 +5499.2

0693.0

= 0.333 + 0.02717 = 0.3601

02. The type of partial differential equation

yp

xp

2yx

p3

yp

xp 2

2

2

2

2

= 0 is

(A) elliptic (B) parabolic (C) hyperbolic (D) none of these

02. Ans: (C)

Sol: B2 – 4ac = 9 – 4

= 5 > 0

PDE is Hyperbolic

03. If the entries in each column of a square matrix M add up to 1, then an eigenvalue of M is

(A) 4 (B) 3 (C) 2 (D) 1

03. Ans: All Options are correct

: 10 : GATE_2016_CE


04. Type II error in hypothesis testing is

(A)acceptance of the null hypothesis when it is false and should be rejected

(B) rejection of the null hypothesis when it is true and should be accepted

(C) rejection of the null hypothesis when it is false and should be rejected

(D)acceptance of the null hypothesis when it is true and should be accepted

04. Ans: (A)

Sol: Type II error means acceptance of the null hypothesis when it is false and should be

rejected

05. The solution of the partial differential equation2

2

x

u

t

u

is of the form

(A) ]eCeC[ktcosC x/k2

x/k1

(B) ]eCeC[eC x/k2

x/k1

kt

(C) ]x/ksinCx/kcosC[eC 21kt (D) ]x/ksinCx/kcosC[ktsinC 21

05. Ans: (C)

Sol:2

2

x

y

t

y

The solution of heat equation is

u(xy) = x/ksinCx/kcosCeC 21kt



06. Consider the plane truss with load P as shown in the figure. Let the horizontal and vertical reactions

at the joint B be HB and VB, respectively and VC be the vertical reaction at the joint C

Which one of the following sets gives the correct values of VB, HB and VC?

(A) VB = 0; HB = 0; VC = P (B) VB = P/2; HB = 0; VC = P/2

(C) VB = P/2; HB = P (sin 60°); VC = P/2 (D) VB = P; HB = P (cos 60°); VC = 0

06. Ans: (A)

Sol: The force P is passing through support at C

VC = P

and other two reactions become zero

VB = 0, HB = 0

07. In shear design of an RC beam, other than the allowable shear strength of concrete (c), there is also

an additional check suggested in IS 456-2000 with respect to the maximum permissible shear stress

(c max). The check for τc max is required to take care of

(A)additional shear resistance from reinforcing steel

(B) additional shear stress that comes from accidental loading

(C) possibility of failure of concrete by diagonal tension

(D)possibility of crushing of concrete by diagonal compression

07. Ans: (D)

Sol: If v > c max , diagonal compression failure occurs in concrete

L L

L

L

L

P

D

E

A

F

6060

60 60

G

CB

: 12 : GATE_2016_CE


08. The semi-compact section of a laterally unsupported steel beam has an elastic section modulus,

plastic section modulus and design bending compressive stress of 500 cm3, 650 cm3 and 200 MPa,

respectively. The design flexural capacity (expressed in kNm) of the section is _________

08. Ans: 100

Sol: Elastic section modulus of the section Ze=500 cm3 = 500 × 103 mm3

Plastic section modulus of the section Zp=650 cm3 =650 × 103 mm3

Design bending compressive stress fcd = 200 Mpa

The bending strength of laterally unsupported beam is given by

Md = b.Zp.fcd

b = 1.0 (For plastic and compact sections)

= Ze/Zp (for semi compact sections)

Ze = Elastic section modulus of steel beam

The design flexural capacity of the section (Md)

Md = Ze.fcd = 500 × 103 × 200 = 100 × 106 N-mm = 100 kN-m

09. Bull's trench kiln is used in the manufacturing of

(A) lime (B) cement (C) bricks (D) none of these

09. Ans: (C)

Sol: Bulls Trench kiln is used for manufacturing bricks

10. The compound which is largely responsible for initial setting and early strength gain of Ordinary

Portland Cement is

(A) C3A (B) C3S (C) C2S (D) C4AF

10. Ans: (B)

Sol: The compound responsible for initial setting and early strength gain in OPC is C3S



11. In the consolidated undrained triaxial test on a saturated soil sample, the pore water pressure is zero

(A) during shearing stage only

(B) at the end of consolidation stage only

(C) both at the end of consolidation and during shearing stages

(D) under none of the above conditions

11. Ans: (B)

12. A fine grained soil is found to be plastic in the water content range of 26-48%. As per Indian

Standard Classification System, the soil is classified as

(A) CL (B) CH (C) CL-ML (D) CI

12. Ans: (D)

Sol: wP = 26%,

wL = 48%

Since wL lies in the range of 35 50%, it is intermediate compressible. Hence soils is CI.

13. A vertical cut is to be made in a soil mass having cohesion c, angle of internal friction , and unit

weight γ. Considering Ka and Kp as the coefficients of active and passive earth pressures,

respectively, the maximum depth of unsupported excavation is

(A)pK

c4

(B)

pKc2

(C)

aKc4(D)

aK

c4

13. Ans: (D)

: 14 : GATE_2016_CE


14. The direct runoff hydrograph in response to 5 cm rainfall excess in a catchment is shown in the

figure. The area of the catchment (expressed in hectares) is________

14. Ans: 21.6

Sol: Area of catchment =4

1/ 2 1 6 60 60 1ha

5 /100 10

A = 21.6 ha

15. The type of flood routing (Group I) and the equation(s) used for the purpose (Group II) are given

below

Group – I Group II

P. Hydrologic flood routing 1. Continuity equation

Q. Hydraulic flood routing 2. Momentum equation

3. Energy equation

The correct match is

(A) P-1; Q-1, 2 & 3 (B) P-1; Q-1 & 2 (C) P-1&2; Q-1 (D) P-1&2; Q- 1&2

15. Ans: (B)

0 1 2 3 4 5 6 70

0.2

0.4

0.6

0.8

1

1.2

Time (hours)

Dis

char

ge(m

3 /s)



16. The pre-jump Froude Number for a particular flow in a horizontal rectangular channel is 10. The

ratio of sequent depths (i.e., post-jump depth to pre-jump depth) is ________

16. Ans: 13.651

Sol:

Fr1 = 10

21r

1

2 F8112

1

y

y

2108112

1

= 13.651

17. Pre-cursors to photochemical oxidants are

(A) NOX, VOCs and sunlight (B) SO2, CO2 and sunlight

(C) H2S, CO and sunlight (D) SO2, NH3 and sunlight

17. Ans: (A)

18. Crown corrosion in a reinforced concrete sewer is caused by:

(A) H2S (B) CO2 (C) CH4 (D) NH3

18. Ans: (A)

19. It was decided to construct a fabric filter, using bags of 0.45 m diameter and 7.5 m long, for

removing industrial stack gas containing particulates. The expected rate of airflow into the filter is

10 m3/s. If the filtering velocity is 2.0 m/min, the minimum number of bags (rounded to nearest

higher integer) required for continuous cleaning operation is

(A) 27 (B) 29 (C) 31 (D) 32

: 16 : GATE_2016_CE


19. Ans: (B)

Sol: v =2

m / sec60

Total surface area of bags required =v

Q

10

2 / 60 = 300 m2

Surface area of each bag = dH

= 0.45 7.5

Number of bags required =300

0.45 7.5 = 28.29 ≃ 29

20. Match the items in Group – I with those in Group – II and choose the right combination.

Group - I Group - II

P. Activated sludge process 1. Nitrifiers and denitrifiers

Q. Rising of sludge 2. Autotrophic bacteria

R. Conventional nitrification 3. Heterotrophic bacteria

S. Biological nitrogen removal 4. Denitrifiers

P Q R S P Q R S

(A)3 4 2 1 (B) 2 3 4 1

(C) 3 2 4 1 (D) 1 4 2 3

20. Ans: (A)



21. During a forensic investigation of pavement failure, an engineer reconstructed the graphs P, Q, R

and S, using partial and damaged old reports.

Theoretically plausible correct graphs according to the 'Marshall mixture design output' are

(A) P, Q, R (B) P, Q, S (C) Q, R, S (D) R, S, P

21. Ans: (B)

Sol: The graph wrong among the given is ‘R’

The correct graph should be

22. In a one-lane one-way homogeneous traffic stream, the observed average headway is

3.0 s. The flow (expressed in vehicles/hr) in this traffic stream is________

22. Ans: 1200

Sol: Average time head way, Ht = 3s

The flow of traffic stream,tH

3600q

12003

3600q vehicles/hr

Bitumen Content

Mar

shal

l Sta

bilit

y

Bitumen Content Bitumen Content

Bitumen Content

Voi

ds f

illed

With

bitu

men

Air

Voi

dsFl

ow

P Q

R S

Bitumen Content

Voids filledwith Bitumen

: 18 : GATE_2016_CE


23. The minimum number of satellites needed for a GPS to determine its position precisely is

(A) 2 (B) 3 (C) 4 (D) 24

23. Ans: (C)

24. The system that uses the Sun as a source of electromagnetic energy and records the naturally

radiated and reflected energy from the object is called

(A) Geographical Information System (B) Global Positioning System

(C) Passive Remote Sensing (D) Active Remote Sensing

24. Ans: (C)

25. The staff reading taken on a workshop floor using a level is 0.645 m. The inverted staff reading

taken to the bottom of a beam is 2.960 m. The reduced level of the floor is 40.500 m. The reduced

level (expressed in m) of the bottom of the beam is

(A) 44.105 (B) 43.460 (C) 42.815 (D) 41.145

25. Ans: (D)

Sol:

Floor reduced level = 40.5 m

RL of bottom of beam = 40.5 + 0.645 + 2.96

= 44.105 m

0.645 m

2.96 m

Bottom of beam



Q. 26 – Q. 55 carry two marks each

26. Probability density function of a random variable X is given below

otherwise0

5x1if25.0xf

P(x 4) is

(A)4

3(B)

2

1(C)

4

1(D)

8

1

26. Ans: (A)

Sol:

otherwise0

5x1if25.0xf

4 1 4

1

xdxfxdxfxdxf4xP

= 4

3x

4

1dx

4

1 41

4

1

: 20 : GATE_2016_CE


27. The value of

00 2 dx

xxsin

dxx1

1 is

(A)2

(B) (C)2

3(D) 1

27. Ans: (B)

Sol:

00 2 dx

xxsin

dxx1

1

= 2

xtan 01

=

22

28. The area of the region bounded by the parabola y = x2 + 1 and the straight line

x + y = 3 is

(A)6

59(B)

29

(C)3

10(D)

67

28. Ans: (B)

Sol: y = x2 + 1; x + y = 3 area bounded by parabola

x + y = 3

x + x2 + 1 = 3

x2 + x – 2 = 0

(x + 2) (x – 1) = 0

= x = –2, 1

A = dydx

=

1

2

x3

1x2

dydx



=

1

2

x31x dxy 2

= dx1xx31

2

2

= dxxx21

2

2

=1

2

32

3

x

2

xx2

=

3

824

3

1

2

12

=

3

10

6

2312

=2

9

6

27

3

10

6

7

29. The magnitudes of vectors P, Q and R are 100 kN, 250 kN and 150 kN, respectively as shown in the

figure.

The respective values of the magnitude (in kN) and the direction (with respect to the x-axis) of the

resultant vector are

(A) 290.9 and 96.0° (B) 368.1 and 94.7° (C) 330.4 and 118.9° (D) 400.1 and 113.5°

R X

Q

P

60o45o

90o

: 22 : GATE_2016_CE


29. Ans: (C)

Sol: RX = FX = 100 cos60o –250 sin15 –150 cos 15

= –159.59 kN

RY = FY = 100 sin 60 + 250 cos15 – 150 sin15

= 289.261 kN

R = 2Y

2X RR

22 261.28959.159

= 330.36 kN

X

Y

R

Rtan

815.159.159

261.289

= 61.1o

Angle with respect to X-axis ( )

= 118.9o

30. The respective expressions for complimentary function and particular integral part of the solution of

the differential equation 22

2

4

4

x108dx

yd3

dxyd

are

(A) ]x3coscx3sincxcc[ 4321 and ]cx12x3[ 24

(B) ]x3coscx3sincxc[ 432 and ]cx12x5[ 24

(C) ]x3coscx3sincc[ 431 and ]cx12x3[ 24

(D) ]x3coscx3sincxcc[ 4321 and ]cx12x5[ 24



30. Ans: (A)

Sol: 22

2

4

4

x108dx

yd3

dxyd

A = m4 + 3m2 = 0

m = 0, 0, 3 i

C.F = x3coscx3sincxcc 4321

2

24x108

D3D

1PI

= 2

2

42

x108

D3

D1D3

1

= 2

12

2x108

3

D1

D3

1

= 22

2x108...

3

D1

D3

1

=

2

x72x108

D3

1 22

2= 3x4 – 12x2

y(x) = 244321 x12x3x3coscx3sincxcc

31. A 3 m long simply supported beam of uniform cross section is subjected to a uniformly distributed

load of w = 20 kN/m in the central 1 m as shown in the figure.

If the flexural rigidity (EI) of the beam is 30 106 N-m2, the maximum slope (expressed in radians)

of the deformed beam is

(A) 0.681 × 10-7 (B) 0.943 × 10-7 (C) 4.310 × 10-7 (D) 5.910 × 10-7

w = 20 kN/m

EI = 30 106 N.m2

1 m 1 m 1 m

P Q

: 24 : GATE_2016_CE


31. Ans: NO ANSWER

Sol:

Max. BM at the centre of beam = 10 1.5 – 20 0.5 0.25 = 12.5 kN-m

Maximum slope in real beam occurs at supports P and Q (both)

Max. slope in real beam = shear force

Or reaction at supports on conjugate beam

EI

5.2)5.0(

3

2

EI

105.0

EI

10)1(

2

1

radian1061.31030

833.10

EI

833.10 43

Another method for Question 31.

Using Macaulay’s double integration method:

Moment at section X-X is

22Qx )2x(

2

20)1x(

2

20)x(RM

10 kN 10 kN1 m

1 m 1 m

P Q20 kN/m

x

X

X

Straight Straight

Parabolic

10 12.5 10

BMD

10 kN 10 kN

1 m 1 m 1 mP Q

20 kN/m

P Q

EI

10 EI

5.2

EI

5.12

EI

10

S

Conjugate Beam withEI

M diagram



x2

2

Mdx

ydEI

133

2

C)2x(3

10)1x(

3

10

2

x10

dx

dyEI

2144

3

CxC)2x(12

10)1x(

12

10

3

x5)y(EI

@ x = 0; y = 0, C2 = 0

@ x = 3m; y = 0, C1 = –10.83

Slope at any support (here slope at support Q)

Use x = 0 in slope equation

EI (Q) = C1

)1000/103(83.10

EI83.10

6Q = 3.61 10–4 (radian)

32. Two beams PQ (fixed at P and with a roller support at Q, as shown in Figure I, which allows vertical

movement) and XZ (with a hinge at Y) are shown in the Figures I and II respectively. The spans of

PQ and XZ are L and 2L respectively. Both the beams are under the action of uniformly distributed

load (W) and have the same flexural stiffness, EI (where, E and I respectively denote modulus of

elasticity and moment of inertia about axis of bending). Let the maximum deflection and maximum

rotation be δmax1 and θmax1, respectively, in the case of beam PQ and the corresponding quantities for

the beam XZ be δmax2 and θmax2, respectively.

Which one of the following relationships is true?

w

P Q

L

Figure Iw

X Hinge

L

Figure II

w

Z

L

Y

: 26 : GATE_2016_CE


(A) δmax1 ≠ δmax2 and θmax1 ≠ θmax2 (B) δmax1 = δmax2 and θmax1 ≠ θmax2

(C) δmax1 ≠ δmax2 and θmax1 = θmax2 (D) δmax1 = δmax2 and θmax1 = θmax2

32. Ans: (D)

Sol: Figure: I is the free body diagram of left part of figure II

max1 = max 2

ymax 1 = ymax 2

33. A plane truss with applied loads is shown in the figure.

The members which do not carry any force are

(A) FT, TG, HU, MP, PL (B) ET, GS, UR, VR, QL

(C) FT, GS, HU, MP, QL (D) MP, PL, HU, FT, UR

33. Ans: (A)

Sol: Taking FBD of jtF = FFT = 0

Taking FBD of jtT = FTG = 0

Taking FBD of jtH = FHU = 0

Taking FBD of jtM = FMP = 0

Taking FBD of jtP = FPL = 0

10 kN

10 kN

EF

G

HJ

KL

M

NPQRST

1 m

1 m

1 m

1 m

U V

2 m 2 m 2 m 2 m 2 m 2 m

20 kN



34. A rigid member ACB is shown in the figure. The member is supported at A and B by pinned and

guided roller supports, respectively. A force P acts at C as shown. Let RAh and RBh be the horizontal

reactions at supports A and B, respectively, and RAv be the vertical reaction at support A. Self weight

of the member may be ignored.

Which one of the following sets gives the correct magnitudes of RAv, RBh and RAh ?

(A) P32

Rand;P31

R;0R AhBhAV (B) P31

Rand;P32

R;0R AhBhAV

(C) P85.1

Rand;P83

R;PR AhBhAV (D) P85.1

Rand;P85.1

R;PR AhBhAV

34. Ans: (D)

Sol: FY = 0 RAV = P MB = 0

RAV × 1.5 – P × 3 + RAh × 8 = 0

8

P5.1R Ah

FX = 0

RBh = RAh =8

P5.1

6 m

2 m

A

1.5 m 1.5 m

B

C

P

: 28 : GATE_2016_CE


35. A reinforced concrete (RC) beam with width of 250 mm and effective depth of 400 mm is reinforced

with Fe415 steel. As per the provisions of IS 456-2000, the minimum and maximum amount of

tensile reinforcement (expressed in mm2) for the section are, respectively

(A) 250 and 3500 (B) 205 and 4000 (C) 270 and 2000 (D) 300 and 2500

35. Ans: (B)

Sol: Min % Steel

y

minst

f

85.0

d.b

)A(

400250415

85.0)A( minst

= 205 mm2

Maximum reinforcement cannot be calculated as total depth is not given

Ast max = 4% Ag = 4% (b D)

Infact, there is no need to calculate (Ast)max as per options given.



36. For M25 concrete with creep coefficient of 1.5, the long-term static modulus of elasticity (expressed

in MPa) as per the provisions of IS:456-2000 is ________

36. Ans: 10,000

Sol: Long term modulus of concrete,

5.11

255000

1

EE c

ce

= 10,000 MPa

37. A propped cantilever of span L carries a vertical concentrated load at the mid-span. If the plastic

moment capacity of the section is Mp, the magnitude of the collapse load is

(A)L

M8 p (B)L

M6 p (C)L

M4 p (D)L

M2 p

37.Ans: (B)

Sol:

PC M3W

2

PC M32

W

pC

M6W

l/2

MP

MP

MP

A B

C

l/2

l/2l/2

W

A B

: 30 : GATE_2016_CE


38. Two plates are connected by fillet welds of size 10 mm and subjected to tension, as shown in the

figure. The thickness of each plate is 12 mm. The yield stress and the ultimate tensile stress of steel

are 250 MPa and 410 MPa, respectively. The welding is done in the workshop (γmw = 1.25).

As per the Limit State Method of IS 800: 2007, the minimum length (rounded off to the nearest

higher multiple of 5 mm) of each weld to transmit a force P equal to 270 kN (factored) is

(A) 90 mm (B) 105 mm (C) 110 mm (D) 115 mm

38. Ans: (B)

Sol: fu=410 N/mm2; fy=250 N/mm2; γmw=1.25; S = 10mm; Factored load P = 270 kN

Assume length of each side weld lj and Effective length of fillet Weld Lw = 2 lj

Effective throat thickness tt= K× S = 0.7 × 10 = 7 mm

Equating Design axial load (P) to the design shear strength of fillet weld (Pdw)

mw

utwdw

3

ftLPP

25.13

41072(

25.13

4107L10270 jw

3

)

lj = 101.83 mm ≈ 105mm

[Refer Problem No: 06 & Page-19, Class Room Practice Questions, Chapter-3 (welded connections)

of Course material]

P

P

150 mm

100 mm



39. The Optimistic Time (O), Most likely Time (M) and Pessimistic Time (P) (in days) of the activities

in the critical path are given below in the format O-M-P.

The expected completion time (in days) of the project is _________

39. Ans: 38

Sol:

Activity Exp. Duration

6

tt4tt pmo

E

E – F33.10

614)10(48

F – G16.8

611)8(46

G – H16.7

610)7(45

H – I16.12

6

18)12(47

Expected completion time of the project = 10.33 + 8.16 + 7.16 + 12.16

= 37.81 days

40. The porosity (n) and the degree of saturation (S) of a soil sample are 0.7 and 40%, respectively. In a

100 m3 volume of the soil, the volume (expressed in m3) of air is _________

E F G H I8-10-14 6-8-11 5-7-10 7-12-18

E F G H I8-10-14 6-8-11 5-7-10 7-12-18

: 32 : GATE_2016_CE


40. Ans: 42

Sol: n = 0.7, S = 40%, V = 100 m3

3V

V m701007.0V.nV;V

Vn

3vw

v

w m28704.0V.SV;VV

S

Vol. of air, Va = VV VW = 70 28 = 42 m3

41. A homogeneous gravity retaining wall supporting a cohesionless backfill is shown in the figure. The

lateral active earth pressure at the bottom of the wall is 40 kPa.

The minimum weight of the wall (expressed in kN per m length) required to prevent it from

overturning about its toe (Point P) is

(A) 120 (B) 180 (C) 240 (D) 360

41. Ans: (A)

Sol:

P4 m

6 m

Cohesionlessbackfill

GravityRetaining

Wall

6 m

2 m

2 m

40 kPa

Fa

0

P

W



Total active force, Fa= kN1206402

1

Taking moments about ‘P’

Fa 2 = W 2

W = Fa = 120 kN

42. An undisturbed soil sample was taken from the middle of a clay layer (i.e., 1.5 m below GL), as

shown in figure. The water table was at the top of clay layer. Laboratory test results are as follows:

Natural water content of clay : 25%

Preconsolidation pressure of clay : 60 kPa

Compression index of clay : 0.50

Recompression index of clay : 0.05

Specific gravity of clay : 2.70

Bulk unit weight of sand : 17 kN/m3

A compacted fill of 2.5 m height with unit weight of 20 kN/m3 is placed at the ground level.

Assuming unit weight of water as10 kN/m3, the ultimate consolidation settlement (expressed in mm)

of the clay layer is ____________

42. Ans: 36.77

Sol: For clay: 675.070.225.0S

wGe

675.01

675.070.210

e1

eGwsat

= 20.15 kN/m3

At centre of clay,

Sand

Clay

GL

GWT

Hard Stratum

1 m

1 m

: 34 : GATE_2016_CE


1015.205.0171

= 22.075 kN/m2

Applied load intensity, q = 2.5 20

= 50 kN/m2 ( = )

Final effective stress at end of consolidation at centre of day,

of

= 22.075 + 50

= 72.075 kN/m2

As the preconsolidaiton stress kPa60c lies between o and f , the clay undergoes

consolidation partly in OC state and partly in NC state

The ultimate consolidation settlement, Sf

c

f10

o

C

o

c10

o

Rf log

e1

C.Hlog

e1

C.HS

60

075.72log

675.01

5.01

075.22

60log

675.01

05.01 1010 = 0.03677 m

= 36.77 mm



43. A seepage flow condition is shown in the figure. The saturated unit weight of the soil γsat = 18

kN/m3. Using unit weight of water, γw = 9.81 kN/m3, the effective vertical stress (expressed in

kN/m2 ) on plane X-X is ________

43. Ans: 65.475

Sol: 2nd method:

Total loss of head, hf =3

Total length of seepage, L = 6 m

Loss of head for 5 m seepage length

= m5.256

3

Pressure head at X – X is :

hp = 9 2.5 = 6.5 m

At plane XX:

Total stress, = sat 5 + w4

= 18 5 + 9.81 4

= 129.24 kPa

Neutral stress, u = whp = 9.81 6.5

= 63.765 kPa

= u = 129.24 63.765 = 65.475 kPa

3 m

1m

5m

1m

2m

Soilsat = 18 kN/m3

XX

: 36 : GATE_2016_CE


1st Method:

h.z w

1

6

3381.9581.918

= 65.475 kPa

44. A drained triaxial compression test on a saturated clay yielded the effective shear strength

parameters as c' = 15 kPa and ' = 22 . Consolidated Undrained triaxial test on an identical sample

of this clay at a cell pressure of 200 kPa developed a pore water pressure of 150 kPa at failure. The

deviator stress (expressed in kPa) at failure is _________

44. Ans: 104.37

Sol: C = 15 kPa, = 22

In CU test :

3 = 200 kPa,

u = 150 kPa. To find d

u33

= 200 150

= 50 kPa

245tanC2

245tan2

31

2

2245tan152

2

2245tan50 2

= 154.377 kPa

u11

ud3

154.377 = (200 + d 150)

d = 104.37 kPa



10 m40 m

45. A concrete gravity dam section is shown in the figure. Assuming unit weight of water as 10 kN/m3

and unit weight of concrete as 24 kN/m3, the uplift force per unit length of the dam (expressed in

kN/m) at PQ is _________

45.

Sol:

Area of uplift pressure distribution diagram gives uplift force

Total Area = (1) + (2) + (3) + (4)

(1) 10 250 = 2500 kN/m (2) 2000400102

1 kN/m

(3) 40 50 = 2000 kN/m (4) 4000200402

1 kN/m

Total uplift force on base = 10,500 kN/m

40 m10 m

P Q

5 mDrain holes

65 m

wH = 650

wH = 50 (w = 10 kN/m3)

2503

'HH'Hw

①

②

③④

P Q 5 m = H

H =65 m

: 38 : GATE_2016_CE


46. Seepage is occurring through a porous media shown in the figure. The hydraulic conductivity values

(k1, k2, k3) are in m/day.

The seepage discharge (m3 /day per m) through the porous media at section PQ is

(A)12

7(B)

21

(C)169

(D)43

46. Ans: 0.5

Sol: Equivalent hydraulic conductivity,

3

3

2

2

1

1

321

K

Z

K

Z

K

ZZZZ

k

day/m230

60

1

10

3

30

2

20103020

Q = K i. A

= A.L

hK

=

2

113

60

10152

20 m 10 m 20 m 10 m

k1 = 2 k2 = 3 k3 = 1 3 m10 m

Q

P

Impervious

3 m

15 m

: 40 : GATE_2016_CE


47. A 4 m wide rectangular channel, having bed slope of 0.001 carries a discharge of

16 m3 /s. Considering Manning's roughness coefficient = 0.012 and g = 10 m/s2 , the category of the

channel slope is

(A) horizontal (B) mild (C) critical (D) steep

47. Ans: (B)

Sol: b = 4 m, So = 0.001

Q = 16 m3/s,

n

n

y24

y4

P

AR

n = 0.012

g = 10 m/s2

3/1

2

23/1

2

2

c410

16

gb

Qy

= 1.16 m

Q = AV

2/10

3/2n SR

n

1)y4(16

2/1

3/2

n

nn )001.0(

y24

y4

012.0

1y416

001.0

012.04

y24

y4y

3/2

n

nn

= 1.55

Trial and error

Assume yn = 1.16 m

L.H.S = 1.79

yn > yc

So < Sc

Mild slope



48. A sector gate is provided on a spillway as shown in the figure. Assuming g = 10 m/s2 , the resultant

force per meter length (expressed in kN/m) on the gate will be __________

48. Ans: 127.04

Sol:

Vertical height of the sector gate (AB)

AB = 5 m (Equilateral le)

Area of segment = Area of sector Area of triangle

ODAB

2

1)5(

6

1A 2

o

2

30cos552

1

6

5A

A = 13.09 10.82 = 2.27 m2

Horizontal component of hydrostatic force on the sector gate

FH = . g. h . Aprojected vertical plane

Sector gate 5 m

3030

Spillway

Sector gate 5 m

3030

5 m

OCD

B

A

: 42 : GATE_2016_CE


= )15(2

5101000 N/m

FH = 125 kN/meter length

Vertical component of hydrostatic force on the sector gate,

FV = weight of the segmental block of water ACBD

= . g. (volume of fluid displaced)

= . g. A. L

= 1000 10 2.27 1

FV = 22.7 kN/meter length

Resultant hydrostatic force = 2V

2H FF

22 7.22125

= 127.04 kN/m

49. A hydraulically efficient trapezoidal channel section has a uniform flow depth of 2 m. The bed width

(expressed in m) of the channel is __________

49. Ans: 2.3

Sol:

T = 2I

b + 2my = 2m1y2

3

11y2y

3

12b

b = 1.154 y

b = 1.154 2

b = 2.3 m



50. Effluent from an industry 'A' has a pH of 4.2. The effluent from another industry 'B' has double the

hydroxyl (OH-) ion concentration than the effluent from industry 'A'. pH of effluent from the

industry 'B' will be __________

50. Ans: 4.5

Sol: (PH)A = 4.2 (H+)A = 10–4.2

(OH–)A = 10–9.8 mol/lit

(OH–)B = 2 (OH–)A = 2 10–9.8 mol/lit

= 3.169 10 –10 mol/lit

(H+)B =14

10

10

3.169 10

= 3.154 10–5 mol/lit

(PH)B = 10

B

1log

H= 10 5

1log

3.154 10 = 4.5

(PH) of effluent from industry B (PH)B = 4.5

51. An electrostatic precipitator (ESP) with 5600 m2 of collector plate area is 96 percent efficient in

treating 185 m3/s of flue gas from a 200 MW thermal power plant. It was found that in order to

achieve 97 percent efficiency, the collector plate area should be 6100 m2 . In order to increase the

efficiency to 99 percent, the ESP collector plate area (expressed in m2 ) would be ______

51. Ans: 8012.38

Sol: of ESP is given by

= 1–AW

Qe

when = 96% A = 5600 m2

= 97% A = 6100 m2

= 99% A = ?

0.96 = 1 –5600 w

185e

W = 0.1063 m/sec

: 44 : GATE_2016_CE


0.99 = 1 –A 0.1063

185e

A = 8012.38 m2

plate area of collector A = 8012.38 m2

52. The 2-day and 4-day BOD values of a sewage sample are 100 mg/L and 155 mg/L, respectively. The

value of BOD rate constant (expressed in per day) is _________

52. Ans: 0.3 d–1

Sol: y2 = 100 mg/l

y4 = 155 mg/l

y2 = L0[1–e–K2]

100 = L0[1–e–K2] ………. I

y4 = L0[1–e–K4]

155 = L0[1–e–K4] ………...II

By solving I and II

K = 0.3 d–1 (bass e) (or) 0.13 d–1 (base 10)

53. A two lane, one-way road with radius of 50 m is predominantly carrying lorries with wheelbase of 5

m. The speed of lorries is restricted to be between 60 kmph and 80 kmph. The mechanical widening

and psychological widening required at 60 kmph are designated as Wme,60 and Wps,60, respectively.

The mechanical widening and psychological widening required at 80 kmph are designated as Wme,80

and Wps,80, respectively. The correct values of Wme,60, Wps,60, Wme,80, Wps,80, respectively are

(A) 0.89 m, 0.50 m, 1.19 m, and 0.50 m

(B) 0.50 m, 0.89 m, 0.50 m, and 1.19 m

(C) 0.50 m, 1.19 m, 0.50 m, and 0.89 m

(D) 1.19 m, 0.50 m, 0.89 m, and 0.50 m



53. Ans: (B)

Sol: No. of lanes, n = 2

R = 50 m

l = 5 m

For V = 60 kmph For V = 80 kmph

R2

nW

2

m

R2

nW

2

m

5.0502

)5(2 2

5.0502

)5)(2( 2

R5.9

VWp

R5.9

VWp

505.9

60 = 0.89 m

505.9

80 = 1.19 m

54. While traveling along and against the traffic stream, a moving observer measured the relative flows

as 50 vehicles/hr and 200 vehicles/hr, respectively. The average speeds of the moving observer while

traveling along and against the stream are 20 km/hr and 30 km/hr, respectively. The density of the

traffic stream (expressed in vehicles/km) is _________

54. Ans: 3 veh/km

Sol: Assume speed of traffic = V

a) Moving observer along the direction of traffic

but q = K. (Vrelative)

50 = K(V–20)

50 = KV – 20 K (1)

20 kmph

V

q = 50 veh/hr

: 46 : GATE_2016_CE


b) Moving observer opposite to the direction of motion

Q = K.V

200 = K(V + 30)

200 = KV + 30 K (2)

Solving equations (1) and (2)

Density of traffic stream, K = 3 veh/km

55. The vertical angles subtended by the top of a tower T at two instrument stations set up at P and Q,

are shown in the figure. The two stations are in line with the tower and spaced at a distance of 60 m.

Readings taken from these two stations on a leveling staff placed at the benchmark (BM = 450.000

m) are also shown in the figure. The reduced level of the top of the tower T (expressed in m) is

_________

55. Ans: 482.2498 m

Sol: V2 = D tan 16.5 = 0.2962D

V1 = (D + 60) tan 10.5 = (D + 60) = 0.1853

(V1 – V2) = 2.555 – 0.555 = 2.0

D = 100.2525 m

V2 = 29.6948 m

RL of T = 450.000 + 2.555 + 29.6948 = 482.2498 m

60 mP

Q

10.5 16.52.5550.555

BM 450.000

T

30 kmph

V

q = 200 veh/hr

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