:2:
GATE_2016_CE
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.
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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
04.
(B) tagafer
(C) tagazen
(D) relffer
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 __________.
05.
(A) 56
(B) 64
(C) 72
(D) 96
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.
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SET_1_Forenoon Session
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\
Elegance
Smooth
Soft
Executive
Q1
27300
20009
17602
9999
Q2
25222
19392
18445
8942
Q3
28976
22429
19544
10234
Q4
21012
18229
16595
10109
Product
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. 78
Rs. 173
Rs. 48
Rs. 63
Total Revenue
102510
80059
72186
39284
 48
 63
 78
 173
4920480
5043717
5630508
6796132
More revenue is on executive
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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
08.
(B) (ii) only
(C) (i) and (ii)
(D) neither (i) nor (ii)
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
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SET_1_Forenoon Session
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
7
5
5
5
f    2   3   5
2
2
2
= – 2 – 3 – 5 = –10
 This is not a factor of f(x)
Option ‘B’ is true
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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
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SET_1_Forenoon Session
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: 0.998
Sol: f(x) = 3x –ex + sinx =0
xo = 0.333
f(x) = 3 – ex – cosx
x1  x o 
f x o 
 0.998
f x o 
  0.390 
 0.333  

 0.604 
= 0.333 + 0.645
= 0.998
02. The type of partial differential equation
 2p  2p
 2p
p p


3
2 
= 0 is
2
2
x
y
xy
x y
(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
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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 equation
(A) C coskt [C 1 e 
  C e 
2
k/ x


k/ x

u
2u
  2 is of the form
t
x
 ]
(B) C e kt [C1e 

  C e 
2
k/ x


 ]
k/ x


(D) C sin kt [C1 cos k /  x  C 2 sin  k /  x ]
(C) C e kt [C1 cos k /  x  C 2 sin  k /  x ]
05. Ans: (C)
Sol:
y
2y
 2
t
x
The solution of heat equation is





u(xy) = C e kt C1 cos k /  x  C 2 sin  k /  x
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SET_1_Forenoon Session
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
A
L
B
60
60
E
L
G
L
P
F
60
60
D
L
C
L
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
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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
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SET_1_Forenoon Session
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)
4c
 Kp
(B)
2c K p

(C)
4c K a

(D)
4c
 Ka
13. Ans: (D)
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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________
1.2
1
Discharge (m3/s)
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
6
7
Time (hours)
14. Ans: 21.6
Sol: Area of catchment =
1/ 2 1 6  60  60 1
 4 ha
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)
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SET_1_Forenoon Session
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

y2 1
  1  1  8Fr21
y1 2



1
 1  1  8  10 2
2

= 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
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(B) 29
(C) 31
(D) 32
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GATE_2016_CE
19. Ans: (B)
Sol: v =
2
m / sec
60
Total surface area of bags required =
Q
v
10
= 300 m2
2 / 60
Surface area of each bag = dH
=  0.45  7.5
Number of bags required =
300
= 28.29 ≃ 29
 0.45  7.5
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
(A) 3
4
2
1
(C) 3
2
4
1
P
Q
R
S
(B)
2
3
4
1
(D)
1
4
2
3
20. Ans: (A)
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SET_1_Forenoon Session
21. During a forensic investigation of pavement failure, an engineer reconstructed the graphs P, Q, R
Flow
P
Bitumen Content
R
Q
Bitumen Content
S
Air Voids
Voids filled
With bitumen
Marshall Stability
and S, using partial and damaged old reports.
Bitumen Content
Bitumen Content
Theoretically plausible correct graphs according to the 'Marshall mixture design output' are
(A) P, Q, R
(B) P, Q, S
21.
Ans: (B)
Sol:
The graph wrong among the given is ‘R’
(C) Q, R, S
(D) R, S, P
Voids filled
with Bitumen
The correct graph should be
Bitumen Content 
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, q 
3600
Ht
q
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3600
 1200 vehicles/hr
3
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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:
Bottom of beam
2.96 m
0.645 m
Floor reduced level = 40.5 m
RL of bottom of beam = 40.5 + 0.645 + 2.96
= 44.105 m
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SET_1_Forenoon Session
Q. 26 – Q. 55 carry two marks each
26. Probability density function of a random variable X is given below
0.25
f x   
0
if 1  x  5
otherwise
P(x  4) is
(A)
3
4
(B)
1
2
(C)
1
4
(D)
1
8
26. Ans: (A)
Sol:
if 1  x  5
otherwise
0.25
f x   
0
4
1
4


1
Px  4   f x dx    f x dx    f x dx 
4
=
1
1
 4 dx  4 x 
1
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4
1

3
4
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27. The value of
(A)


0
GATE_2016_CE
 sin x
1
dx  
dx is
2
0
1 x
x

2
(B) 
(C)
3
2
(D) 1
Ans: (B)
27.
Sol:


0
 sin x
1
dx
dx  
2
0
x
1 x


= tan 1 x  0 
=


2
 
 
2 2
28. The area of the region bounded by the parabola y = x2 + 1 and the straight line
x + y = 3 is
(A)
59
6
(B)
9
2
(C)
28.
Ans: (B)
Sol:
y = x2 + 1; x + y = 3 area bounded by parabola
10
3
(D)
7
6
x+y=3
 x + x2 + 1 = 3
 x2 + x – 2 = 0
 (x + 2) (x – 1) = 0
= x = –2, 1
A=
 dydx
1 3 x
=
  dydx
 2 x 2 1
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1
=
 y
3 x
x 2 1
SET_1_Forenoon Session
dx
2
1
=
 3  x  x
2
 1dx
2
1
=
 2  x  x dx
2
2
1

x2 x3 
 
= 2x 
2
3  2

1 1 
8

=  2     4  2  
2 3 
3

 12  3  2    10 
=


6

  3 
=
7 10 27 9
 

6 3
6 2
29. The magnitudes of vectors P, Q and R are 100 kN, 250 kN and 150 kN, respectively as shown in the
figure.
Q
P
90o
45o
R
60o
X
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°
ACE Engineering Academy
(B) 368.1 and 94.7°
(C) 330.4 and 118.9°
(D) 400.1 and 113.5°
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29.
Ans: (C)
Sol:
RX = FX = 100 cos60o –250 sin15 –150 cos 15
GATE_2016_CE
= –159.59 kN
RY = FY = 100 sin 60 + 250 cos15 – 150 sin15
= 289.261 kN
R=
R 2X  R 2Y
 159.59 2  289.2612
= 330.36 kN
tan  
RY
RX

289.261
 1.815
159.59
 = 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
d4y
d2y

3
 108x 2 are
4
2
dx
dx
(A) [c1  c 2 x  c3 sin 3x  c 4 cos 3x ] and [3x 4  12x 2  c]
(B) [c 2 x  c3 sin 3x  c 4 cos 3x ] and [5x 4  12x 2  c]
(C) [c1  c3 sin 3x  c 4 cos 3x ] and [3x 4  12x 2  c]
(D) [c1  c 2 x  c3 sin 3x  c 4 cos 3x ] and [5x 4  12x 2  c]
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30.
Ans: (A)
Sol:
d2y
d4y

 108x 2
3
4
2
dx
dx
SET_1_Forenoon Session
A = m4 + 3m2 = 0
 m = 0, 0, 
3i
C.F = c1  c 2 x  c 3 sin 3x  c 4 cos 3x
1


2
PI   4
108x
2
 D  3D 
=
1

D 
3D 2 1 
2 
 3D 
4
1
=
3D 2
=
1
3D 2
=
1
3D 2
108x 2
1
 D2 
2
1  3  108 x


 D2

 ... 108 x 2
1 
3



x2 
2

= 3x4 – 12x2
108
x
72


2

y(x) = c 1  c 2 x  c 3 sin 3x  c 4 cos 3x  3x 4  12 x 2
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.
w = 20 kN/m
P
Q
EI = 30  10 N.m
6
1m
1m
2
1m
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
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(B) 0.943 × 10-7
(C) 4.310 × 10-7
(D) 5.910 × 10-7
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31.
GATE_2016_CE
Ans: NO ANSWER
Sol:
20 kN/m
P
1m
1m
1m
10 kN
10 kN
Parabolic
Straight
12.5
10
Q
10
Straight
BMD
Max. BM at the centre of beam = 10  1.5 – 20  0.5  0.25 = 12.5 kN-m
2 .5
EI
10
EI
10
EI
Q
P
S
12.5
EI
Conjugate Beam with M diagram
EI
Maximum slope in real beam occurs at supports P and Q (both)
 Max. slope in real beam = shear force
1  10 
 2.5 
 10  2
Or reaction at supports on conjugate beam  (1)   0.5     (0.5)

2  EI 
 EI  3
 EI 

10.833 10.833

 3.6110 4 radian
3
EI
30  10
Another method for Question 31.
Using Macaulay’s double integration method:
X
P
20 kN/m
Q
1m
1m
1m
10 kN
10 kN
X
x
Moment at section X-X is
M x  R Q (x) 
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20
20
( x  1) 2  ( x  2) 2
2
2
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SET_1_Forenoon Session
d2y
EI 2  M x
dx
dy 10 x 2 10
10
EI

 ( x  1) 3  ( x  2) 3  C1
dx
2
3
3
5x 3 10
10
EI( y) 
 ( x  1) 4  ( x  2) 4  C1 x  C 2
3 12
12
@ 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
Q 
10.83
10.83

= 3.61  10–4 (radian)
6
EI
(3  10 / 1000)
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.
w
Q
P
L
Figure I
w
w
Hinge
Y
X
L
Z
L
Figure II
Which one of the following relationships is true?
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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.
20 kN
10 kN
J
10 kN
H
G
1m
K
U
1m
L
V
M
F
E
T
S
2m
2m
R
2m
Q
2m
P
2m
1m
N 1m
2m
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
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SET_1_Forenoon Session
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.
B
2m
P
C
6m
A
1.5 m
1.5 m
Which one of the following sets gives the correct magnitudes of RAv, RBh and RAh ?
1
2
(A) R AV  0; R Bh  P; and R Ah  P
3
3
(B) R AV  0; R Bh 
2
1
P; and R Ah  P
3
3
3
1.5
1.5
1.5
P (D) R AV  P; R Bh 
P; and R Ah 
P
(C) R AV  P; R Bh  P; and R Ah 
8
8
8
8
34.
Ans: (D)
Sol:
FY = 0
RAV = P
MB = 0
RAV × 1.5 – P × 3 + RAh × 8 = 0
R Ah 
1.5P
8
FX = 0
RBh = RAh =
1.5P
8
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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
35.
Ans: (B)
Sol:
Min % Steel
(B) 205 and 4000
(C) 270 and 2000
(D) 300 and 2500
(A st ) min 0.85

b.d
fy
(A st ) min 
0.85
 250  400
415
= 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.
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SET_1_Forenoon Session
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,
E ce 
Ec
5000 25

1 
1  1.5
= 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)
8M p
L
(B)
6M p
L
(C)
4M p
L
(D)
2M p
L
37.Ans: (B)
Sol:
W
WC    3M P 
A

 
2

WC     3M P 
2
WC 
6M p

ACE Engineering Academy
A
l/2
l/2
l/2
l/2



MP
B


MP
B
C
MP
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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).
P
100 mm
150 mm
P
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)
P  Pdw  L w t t
fu
3 mw
270  103  L w  7 
410
410
)
 (2   j  7 
3  1.25
3  1.25
lj = 101.83 mm ≈ 105mm
[Refer Problem No: 06 & Page-19, Class Room Practice Questions, Chapter-3 (welded connections)
of Course material]
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SET_1_Forenoon Session
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.
E
8-10-14
F
6-8-11
5-7-10
G
H
7-12-18
I
The expected completion time (in days) of the project is _________
39.
Ans: 38
Sol:
E
8-10-14
F
6-8-11
G
Activity
5-7-10
H
7-12-18
I
Exp. Duration
tE 
t o  4t m  t p
6
E–F
8  4(10)  14
 10.33
6
F–G
6  4(8)  11
 8.16
6
G–H
5  4(7)  10
 7.16
6
H–I
7  4(12)  18
 12.16
6
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 _________
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40.
Ans: 42
Sol:
n = 0.7, S = 40%, V = 100 m3
n
GATE_2016_CE
VV
; VV  n.V  0.7  100  70 m 3
V
S
Vw
; Vw  S.Vv  0.4  70  28 m 3
Vv
 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.
Cohesionless
backfill
Gravity
Retaining
Wall
6m
P
4m
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
41.
(B) 180
(C) 240
(D) 360
Ans: (A)
Sol:
0
6m
Fa
2m
P
40 kPa
2m
W
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SET_1_Forenoon Session
1
Total active force, Fa=  40  6  120 kN
2
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.
GL
1m
Sand
GWT
1m
Clay
Hard Stratum
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: e 
wG
 0.25  2.70  0.675
S
G  e
 2.70  0.675 
 sat   w 
 10

 1 e 
 1  0.675 
= 20.15 kN/m3
At centre of clay,
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GATE_2016_CE
 1  17  0.520.15  10
= 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,
f  o  
= 22.075 + 50
= 72.075 kN/m2


As the preconsolidaiton stress c  60 kPa lies between o and f , the clay undergoes
consolidation partly in OC state and partly in NC state
The ultimate consolidation settlement, Sf
Sf  H.
 1
  
  
CR
C
log10  c   H. C log10  f 
1  eo
1  eo
 c 
 o 
0.05
0.5
 60 
 72.075 
log10 
log10 
 1


1  0.675
1  0.675
 22.075 
 60 
= 0.03677 m
= 36.77 mm
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SET_1_Forenoon Session
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 ________
3m
1m
Soil
sat = 18 kN/m3
5m
X
X
1m
2m
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
=
3
 5  2.5 m
6
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
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GATE_2016_CE
1st Method:
  z   w .h
3 

 18  9.815  9.81 3   1
6 

= 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
3  3  u
= 200  150
= 50 kPa
 
 


1  3 tan 2 45    2C tan 45  
2
2


22 
22 


 50 tan 2 45    2  15 tan 45  
2
2 


= 154.377 kPa
1  1  u
  3   d  u 
154.377 = (200 + d  150)
 d = 104.37 kPa
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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 _________
65 m
5m
Drain holes
Q
P
10 m
40 m
45.
Sol:
H =65 m
5m=H
Q
P
10 m
40 m
①
wH = 650
②
wH = 50
③
④
(w = 10 kN/m3)
H  H' 

 250
 w H '
3 

Area of uplift pressure distribution diagram gives uplift force
Total Area = (1) + (2) + (3) + (4)
(1) 10  250 = 2500 kN/m
(2)
1
 10  400  2000 kN/m
2
(3) 40  50 = 2000 kN/m
(4)
1
 40  200  4000 kN/m
2
Total uplift force on base = 10,500 kN/m
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46. Seepage is occurring through a porous media shown in the figure. The hydraulic conductivity values
(k1, k2, k3) are in m/day.
Impervious
15 m
10 m
P
3m
k1 = 2
k2 = 3
20 m
10 m
k3 = 1
3m
Q
20 m
10 m
The seepage discharge (m3 /day per m) through the porous media at section PQ is
(A)
7
12
(B)
1
2
46.
Ans: 0.5
Sol:
Equivalent hydraulic conductivity,
k
(C)
9
16
(D)
3
4
Z1  Z 2  Z 3
Z1 Z 2 Z 3


K1 K 2 K 3

20  30  10 60

 2 m / day
20 30 10 30


2
3
1
Q = K i. A
=K
h
.A
L
= 2
15  10  3  1  1
60
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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: (D)
Sol:
b = 4 m,
So = 0.001
Q = 16 m3/s, R 
A  4y n

P  4  2 y n



n = 0.012
g = 10 m/s2
 Q2
y c   2
 gb



1/ 3
 16 2
 
2
 10  4



1/ 3
= 1.86 m
Q = AV
1
16  (4 y n )   R 2 / 3 S10/ 2
n
1  4y n

16  4 y n 
0.012  4  2 y n
 4y n
y n 
 4  2y n



2/3




2/3
(0.001) 1 / 2
4  0.012
0.001
Trial and error
Assume yn = 1.86 m
L.H.S = 1.79
 yc > yn
or yn < yc
So > Sc
Steep slope
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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 __________
5m
Sector gate
30
30
Spillway
48. Ans: 127.04
Sol:
A
5m
Sector gate
C
D
30
30
O
5m
B
Vertical height of the sector gate (AB)
AB = 5 m (Equilateral le)
Area of segment = Area of sector  Area of triangle
1
1

A  (  5 2 )    AB  OD 
6
2

   52   1

    5  5 cos 30 o 
A  

 6  2
A = 13.09  10.82 = 2.27 m2
Horizontal component of hydrostatic force on the sector gate
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GATE_2016_CE
FH = . g. h . Aprojected vertical plane
5
= 1000  10   (5  1) N/m
2
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 =
FH2  FV2
 125 2  22.7 2
= 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 = 2 y 1  m 2
b  2
1
1
y  2y 1 
3
3
 b = 1.154  y
b = 1.154  2
b = 2.3 m
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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 =
1014
= 3.154  10–5 mol/lit
3.169 1010
(PH)B = log10
1
1
= log10
= 4.5

3.154 105
H 
B
(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– e
 AW
Q
when  = 96%
A = 5600 m2
 = 97%
A = 6100 m2
 = 99%
A=?
0.96 = 1 – e
5600 w
185
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 W = 0.1063 m/sec
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0.99 = 1 – e
 A 0.1063
185
GATE_2016_CE
 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–K2]
100 = L0[1–e–K2] ………. I
y4 = L0[1–e–K4]
155 = L0[1–e–K4] ………...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
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53.
Ans: (B)
Sol:
No. of lanes, n = 2
SET_1_Forenoon Session
R = 50 m
l=5m
For V = 60 kmph
Wm 

Wp 

For V = 80 kmph
n 2
2R
Wm 
2(5) 2
 0.5
2  50

V
9.5 R
Wp 
60
= 0.89 m
9.5 50
n 2
2R
(2)(5) 2
 0.5
2  50
V
9.5 R

80
= 1.19 m
9.5 50
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
20 kmph
V
q = 50 veh/hr
but q = K. (Vrelative)
50 = K(V–20)
50 = KV – 20 K  (1)
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b)
GATE_2016_CE
Moving observer opposite to the direction of motion
30 kmph
V
q = 200 veh/hr
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
_________
T
10.5
2.555
0.555
Q
P
BM 450.000
16.5
60 m
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
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