GATE 2016 : Civil Engineering
Detailed Solutions :
Date : 05-02-2016, Forenoon Session
Set - 1
General Aptitude :
Q.1 – Q.5 carry one mark each.
Q.1
Ans.
Sol.
Q.2
Ans.
Sol.
Q.3
Ans.
Sol.
Q.4
Ans.
Sol.
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 minister does not meet me.
(B) I will leave the place until the minster does not 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.
(D)
‘until’ itself is negative so it can’t take one more negative i.e., ‘does not’. Hence, Option (D) is the right
answer
A rewording of something written or spoken is a __________.
(A) Paraphrase
(B) Paradox
(C) Paradigm
(D) Paraffin
(A)
‘paraphrase’ means a restatement of a text, passage or a rewording of something written or spoken.
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
(A)
‘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.
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
(C)
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.
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GATE 2016 [CE] Set - 1
Q.5
Ans.
Sol.
2
A cube is built using 64 cubic blocks of side and 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
(D)
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.
Q.6 – Q.10 Carry two marks each.
Q.6 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.
Ans.
Sol.
Quarter/ Product Elegance
Smooth
Soft
Executive
27300
20009
17602
9999
Q1
25222
19392
18445
8942
Q2
28976
22429
19544
10234
Q3
18229
16595
10109
21012
Q4
Which product contributes the greatest fraction to the revenue of the company in that year?
(A) Elegance
(B) Executive
(C) Smooth
(D) Soft
(B)
Elegance
27300
25222
28976
21012
102510
Rs. 48
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Smooth
20009
19392
22429
18229
80059
Rs. 63
Soft
17602
18445
19544
16595
72186
Rs. 78
Executive
9999
8942
10234
10109
39284
Rs. 173
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GATE 2016 [CE] Set - 1
3
Total Revenue
Q.7
Ans.
Sol.
Q.8
Ans.
Sol.
Q.9
102510
80059
72186
39284
48
63
78
173
4920480
5043717
5630508
6796132
More revenue is on executive
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.
(D)
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’
Consider the following statements relating of 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)
(A)
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
If f ( x)  2 x 7  3x  5 , which of the following is a factor of f ( x) ?
(A) ( x 3  8)
Ans.
Sol.
(B) ( x  1)
(C) (2 x  5)
(D) ( x  1)
(B)
Option (A)
x 2  8  0, x3   8, x   2
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GATE 2016 [CE] Set - 1
4
For x   2, in f ( x)  2 x 7  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
For x  1, in f ( x)  2 x 7  3x  5
f (1)  2(1)7  3(1)  5  5  5  0
 ( x  1) is a factor of f ( x)
Hence, Option ‘B’ is true
Q.10 In a process, the number of cycles of failure decreases exponentially with an increase in load. At a load
of 80 units, it takes 100 cycles of 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
Ans. (B)
Sol.
Load
Failure for cycles
80
100
40
10000
There is one failure 5000 cycles load must be between 80 and 40.
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GATE 2016 [CE] Set - 1
5
Technical :
Q.1 - Q. 25 Carry one mark each
Q.1
Newton-Raphson method is to be used to find root of equation 3x  e x  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)
Ans.
0.36
f ( x)  3x  e x  sin x
Sol.
f '( x)  3  e x  cos x
f (0.333)  3  0.333  e0.333  sin (0.333)  use in radians
  0.06926
f '(0.333)  3  e0.333  cos (0.333)  use in radians
 2.549
x1  0.333 
Q.2
(  0.06926)
 0.357  0.36
2.549
2 P 2 P
2 P
P P
 2 3
2

 0 is
The type of partial differential equation
2
x
y
xy
x y
(A) Elliptic
(B) Parabolic
(C) Hyperbolic
Ans.
(C)
Sol.
The given partial differential equation is of the form,
A
where,
2 P
2 P
2 P

B

C

x 2
xy
y 2
(D) None of these

P P 
f  P , x, y , ,   0
x y 

A  1, B  3, C  1
B2  4ac  9  4
50

Q.3
Partial differential equation is hyperbolic
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
Ans.
(D)
Q.4
Type II error in hypothesis testing is
(C) 2
(D) 1
(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
Ans.
(A)
Sol.
Type II error means acceptance of the null hypothesis when it is false and should be rejected
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GATE 2016 [CE] Set - 1
Q.5
6
The solution of the partial differential equation
(A) C cos( kt ) C1e (

k / ) x
 C2 e  (
k / ) x
u
 2u
  2 is of the form
t
x


(C) Ce kt C1 cos( k /  ) x  C2 sin( k /  ) x 
Ans.
Sol.
(B) Ce kt C1e (

k / ) x
 C2 e  (
k / ) x


(D) C sin(kt ) C1 cos( k /  ) x  C2 sin( k /  ) x 
(C)
The given partial differential equation is
y
2 y
 2
t
x
This is heat equation and its solution is given by

 k

k 
u ( x, y )  Ce kt C1 cos 
 x  C2 sin  
 x 
 
   

Q.6
Consider the plan truss with load P as shown in the figure. Let the horizontal and vertical reactions at the
joint B be H B 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 , H B and VC ?
(A) VB  0, H B  0, VC  P
(C) VB 
Ans.
Sol.
P
P
, H B  P(sin 600 ), VC 
2
2
(B) VB 
P
P
, H B  0, VC 
2
2
(D) VB  P, H B  P(cos 600 ), VC  0
(A)
The force P is passing through support at C

VC  P
And other two reaction become zero
VB  0, H B  0
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7
Q.7
GATE 2016 [CE] Set - 1
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 c max is required to take care of
Ans.
Sol.
Q.8
Ans.
Sol.
(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.
(D)
If v  c max , diagonal compression failure occurs in concrete
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 _________.
100
Elastic section modulus of the section Z e  500 cm3  500 103 mm3
Plastic section modulus of the section Z p  650 cm3  650 103 mm3
Design bending compressive stress f ed  200 Mpa
The bending strength of laterally unsupported beam is given by
M b  b , Z p . f ed
b  1.0 (For plastic and compact sections)

Ze
(For semi compact sections)
Zp
Ze  Elastic section modulus of steel beam
The design flexural capacity of the section ( M d )
M d  Z e . f ed  500 103  200  100 106 N-mm  100 kN-m
Q.9
Bull’s trench kiln is used in the manufacturing of
(A) Lime
(B) Cement
(C) Bricks
(D) None of these
Ans. (C)
Sol. Bulls Trench kiln is used for manufacturing bricks
Q.10 The compound which is largely responsible for initial setting and early strength gain of Ordinary
Portland Cement is
(A) C3 A
(B) C3 S
(C) C2 S
(D) C4 AF
Ans.
Sol.
(B)
The compound responsible for initial setting and earth strength gain in OPC is C3 S .
Q.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
Ans. (B)
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GATE 2016 [CE] Set - 1
8
Q.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
Ans.
(D)
Sol.
Plastic limit, wp  26%
(C) CL-ML
(D) CI
Liquid limit, wL  48%
Since wL lies in the range of 35 – 50%, it is intermediate compressible. Hence soils is CI.
Q.13 A vertical cut is to be made in a soil mass having cohesion c, angle of internal friction  and unit weight
 . Consider K a and K p as the coefficients of active and passive earth pressures, respectively, the
maximum depth of unsupported excavation is
(A)
Ans.
4c
 Kp
(B)
2c K p

(C)
4c K a

(D)
4c
 Ka
(D)
Maximum depth of unsupported excavation is given by,
4c
 Ka
where, c  cohesion of soil,   unit weight, K a  coefficients of active earth pressure.
Q.14 The direction runoff hydrograph is response to 5 cm rainfall excess in a catchment is shown in the
figure. The area of the catchment (expressed in hectares) is ______
Ans.
21.6
Sol.
d (cm )  Area of catchment  Area of triangular portion
1
 1 6  60  60
1
 4 ha
Area of catchment  2
5
10
100
A  21.6 ha
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GATE 2016 [CE] Set - 1
Q.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 and 3
(B) P-1, Q-1 and 2
(C) P-1 and 2, Q-1
(D) P-1 and 2, Q-1 and 2
Ans. (B)
Q.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 ________.
Ans. 13.651
Sol. Pre-jump froude number, Fr1  10
y2 1 
 1  1  8Fr21 

y1 2 
1
  1  1  8 102 

2
 13.651
Q.17 Pre-cursors to photochemical oxidants are
(A) NOX ,VOCS and sunlight
(B) SO2 , CO2 and sunlight
(C) H 2 S , CO and sunlight
(D) SO2 , NH 3 and sunlight
Ans.
(A)
Photochemical smog is formed in the atmosphere when pre-cursor pollutants including Nitrogen Oxide
(NO x ) and Volatile Organic Compound (VOCs ) undergo reaction in sunlight to form smog, of which
ozone is the principle component.
Q.18 Crown corrosion in a reinforced concrete sewer is caused by:
(B) CO2
(C) CH 4
(A) H 2 S
Ans.
(D) NH 3
(A)
Bacteria in the slime under flowing sewage convert sulphate in the sewage into sulphides. Sulphides in
the liquid make their way to the surface of the sewage and reduced into the sewer atmosphere or
Hydrogen Sulphide ( H 2S ) gas. Bacterial action converts atmospheric H 2S gas to Sulphuric Acid which
causes corrosion in the crown of the pipe and this corrosion is called crown corrosion.
Q.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
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GATE 2016 [CE] Set - 1
Ans.
Sol.
10
(B)
Given : Diameter of bag = 0.45 m,
Length of bag = 7.5 m,
Discharge = 10 m3 /s.
2
Velocity of air, v  m / sec
60
Total surface area of bags required 
Q
v
10
 300 m 2
2
60
Surface area of each bag  dH
  0.45  7.5
300
 28.29  29
Number of bags required 
 0.45  7.5
Q.20 Match the items is 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
(A) P-3, Q-4, R-2, S-1
(B) P-2, Q-3, R-4, S-1
(C) P-3, Q-2, R-4, S-1
(D) P-1, Q-4, R-2, S-3
Ans. (A)
Q.21 During a forensic investigation of pavement failure, an engineer reconstructed the graphs P, Q, R and S,
using partial and damaged old reports.
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GATE 2016 [CE] Set - 1
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
Ans.
Sol.
(B)
The graph between VFB and Bitumen content is
Q.22 In a one-lane homogeneous traffic stream, the observed average headway is 3.0 s. The flow (expressed
in vehicles /hr) in this traffic stream is _______.
Ans. 1200
Sol.
Average time head way, H t  3s
The flow of traffic stream, q 
3600
Ht
3600
 1200 vehicles/hr
3
Q.23 The minimum number of satellites needed for a GPS to determine its position precisely is
(A) 2
(B) 3
(C) 4
(D) 24
Ans. (C)
At a minimum, four satellite must be in view of the receiver for it to compute four unknown quantity
(three position coordinate and one for clock deviation from satellite time).
Q.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+
Ans. (C)
The sun provide a very convenient source of energy for remote sensing. The Sun's energy is either
reflected, as it is for visible wavelengths, or absorbed and then re-emitted, as it is for thermal infrared
wavelengths. Remote sensing system which measure energy i.e., naturally available are called as passive
remote sensing.
Q.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
Ans. (D)
q
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GATE 2016 [CE] Set - 1
12
Sol.
Floor reduced level  40.5 m
RL of bottom of beam  40.5  0.645  2.96
 44.105 m
Q.26 – Q. 55 Carry two marks each
Q.26 Probability density function of a random variable X is given below
0.25 if 1  x  5
f (x)  
otherwise
 0
P( X  4) is
(A)
Ans.
Sol.
3
4
1
2
(B)
(C)
1
4
(D)
1
8
3
2
(D) 1
(A)
0.25 if 1 x  5
f ( x)  
otherwise
 0
4
P( x  4) 

1
f ( x)d ( x) 



4
f ( x) d ( x)   f ( x) d ( x )
1
4
1
1
3
  dx  ( x)14 
4
4
4
1
Q.27 The value of
(A)
Ans.
(B)
Sol.


0


0
 sin x
1

dx
0 x dx is
1  x2

2
(B) 
 sin x
1
dx  
dx
2
0
1 x
x


  tan 1 ( x)  
0
2

(C)

  sin x

 0 x dx  2 by using laplace transformation 
 
 
2 2
Q.28 The area of the region bounded by the parabola y  x 2  1 and the straight line x  y  3 is
(A)
59
6
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(B)
9
2
(C)
10
3
(D)
7
6
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GATE 2016 [CE] Set - 1
13
Ans.
(B)
Sol.
y  x 2  1; x  y  3 are 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 3 x
  dydx
 2 x 1
1
1
2
2
  [ y ]3x2 x1dx   (2  x  x 2 )dx
1

1 1 
8
x 2 x3 

  2 x      2      4  2  
2 3 
3
2 3  2 

 12  3  2   10  7 10 27 9



  
6
6 2

  3  6 3
Q.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
Ans.
(A) 290.9 and 96.00
(C)
(B) 368.1 and 94.70
Sol.
RX   FX  100 cos 600  250 sin15  150 cos15
(C) 330.4 and 118.90
(D) 400.1 and 113.50
 159.59 kN
RY   FY  100sin 600  250 cos15  150sin15
 280.261 kN
R  RX2  RY2
 159.592  289.2612  330.36 kN
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GATE 2016 [CE] Set - 1
14
tan  
RY 289.261

 1.815
RX
159.59
  61.10
Angle with respect to X-axis
( )
 118.90
Q.30 The respective expressions for complimentary function and particular integral part of the solution of the
differential equation
d4y
d2y

3
 108 x 2 are
4
2
dx
dx
(A) c1  c2 x  c3 sin 3 x  c4 cos 3 x  and 3 x 4  12 x 2  c 
(B)  c2 x  c3 sin 3 x  c4 cos 3 x  and 5 x 4  12 x 2  c 
(C) c1  c3 sin 3x  c4 cos 3x  and 3 x 4  12 x 2  c 
Ans.
Sol.
(D) c1  c2 x  c3 sin 3 x  c4 cos 3 x  and 5 x 4  12 x 2  c 
(A)
d4y
d2y

3
 108 x 2
dx 4
dx 2
A  m4  3m2  0
 m  0, 0  3 i
C.F.  c1  c2 x  c3 sin 3x  c4 cos 3x
1


108 x 2 
PI   4
2 
D
3
D



1

D4 
3D 2 1 
2
 3D 

1
3D 2
 D2 
1
2
1  3  108 x  3D 2



1
3D 2

x2 
2
4
2

108
x
72

  3 x  12 x
2


108 x 2
 D2

2
1  3  ... 108 x


y ( x)  c1c2 x  c3 sin 3x  c1 cos 3x  3x ' 12 x 2
Q.31 A 3 m long simple 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
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(B) 0.943 107
(C) 4.310 107
(D) 5.910 107
© Copyright
15
Ans.
Sol.
GATE 2016 [CE] Set - 1
No answer
Maximum 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)
 Maximum slope in real beam = shear force
1  10 
 10  2
 2.5 
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:

Moment at section X-X is
20
20
M x  RQ ( x)  ( x  1)2  ( x  2)2
2
2
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GATE 2016 [CE] Set - 1
EI
16
d2y
 Mx
dx 
dy 10 x 2 10
10
EI

 ( x  1) 2  ( x  2)3  C1
dx
2
3
3
EI ( y ) 
5 x 2 10
10
 ( x  1) 2  ( x  2) 4  C1 x  C2
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 
Q.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 as Y) are shown in the Figures I and II respectively. The spans of PQ
and XZ and 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
max1 and max1 , respectively, in the case of beam PQ and the corresponding quantities for the beam XZ
be max 2 and max 2 , respectively.
Which one of the following relationships is true?
(A) max1  max 2 and max1  max 2
(B) max1  max 2 and max1  max 2
(C) max1  max 2 and max1  max 2
(D) max1  max 2 and max1  max 2
Ans.
(D)
Sol.
Figure: I is the free body diagram of left part of figure II

max1  max 2
ymax1  ymax 2
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17
Q.33 A plane truss with applied loads is shown in the figure.
Ans.
Sol.
The members which do not carry any force are
(A) FT, TG, HU, MP, PL
(C) FT, GS, HU, MP, QL
(A)
Taking FBD of jt F  FFT  0
(B) ET, GS, UR, VR, QL
(D) MP, PL, HU, FT, UR
Taking FBD of jtT  FTG  0
Taking FBD of jtH  FHU  0
Taking FBD of jtM  FMP  0
Taking FBD of jtP  FPL  0
Q.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 reaction at
supports A and B, respectively and RAv be the vertical reaction at support A. Self-weight of the member
may be ignored.
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GATE 2016 [CE] Set - 1
18
Which one of the following sets given the correct magnitudes of RAv , RBh and RAh ?
Ans.
Sol.
1
2
(A) RAv  0, RBh  P and RAh  P
3
3
3
1.5
P
(C) RAv  P, RBh  P and RAh 
8
8
(D)
 FY  0
RAV  P
2
1
P and RAh  P
3
3
1.5
1.5
P and RAh 
P
(D) RAv  P, RBh 
8
8
(B) RAv  0, RBh 
 MB  0
RAV 1.5  P  3  RAh  8  0
.5P
8
 FX  0
RAh 
.5P
8
Q.35 A reinforced concrete (RC) beam with width of 250 mm and effective depth of 400 mm is reinforced
with Fe 415 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
Ans. (B)
Sol. Min % Steel
( Ast ) min 0.85

b.d
fy
RBh  RAh 
( Ast )min 
0.85
 250  400
415
 205 mm 2
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 (A st ) max as per options given.
Q.36 For M 25 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 _________
Ans. 10000
Sol. Long term modulus of concrete,
Ee
5000 25
 10000MPa

1 
1  1.5
Q.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 M p , the magnitude of the collapse load is
Ece 
(A)
8M p
L
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(B)
6M p
L
(C)
4M p
L
(D)
2M p
L
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19
Ans.
(B)
Sol.
WC    3 M P 
GATE 2016 [CE] Set - 1

 
2

WC     3 M P 
2
6M P
WC 

Q.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 (  mw  1.25 ).
Ans.
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
(B)
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GATE 2016 [CE] Set - 1
Sol.
20
fu  410 N / mm2 ; f y  250 N / mm2 ;  mw  1.25; S  10 mm ; Factored load P  270 kN
Assume length of each side load (P) to the design shear strength of fillet weld (Pdw )
P  Pdw  Lwtt
fu
3 mw
270 103  Lw  7 
410 
410 
 2 i  7 

3 1.25 
3 1.25 
l j  101.83 mm  105 mm
Q.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.
Ans.
Sol.
The expected completion time (in days) of the project is ________
38
Activity
Exp. Duration
t0  4tm  t p
tE 
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(2)  18
 12.16
6
Expected completion time of the project  10.33  8.16  7.16  12.16  37.81 days.
Q.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 ________
Ans. 42
Sol. Given : Porosity n  0.7,
Degree of saturation S  40%,
Volume of the soil V  100 m3 .
VV
;VV  n, V  0.7 100  70 m3
V
V
S  W ; VW  S .VV  0.4  70  28 m3
VV
n

Vol. of air, Va  VV  VW  70  28  42 m3
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21
Q.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.
Ans.
Sol.
The minimum weight of the wall (expressed in kN per m length) required to prevent it from overturning
about it toe (Point P) is
(A) 120
(B) 180
(C) 240
(D) 360
(A)
Given : Active earth pressure at the bottom of the wall = 40 kPa,
Width of wall = 4 m
1
Total active force, Fa   40  6  120 kN
2
Taking moments about ‘P’
Fa  2  W  2

W  Fa  120 kN
Q.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
© Copyright
:
17 kN / m3
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GATE 2016 [CE] Set - 1
22
A compacted fill of 2.5 m height with unit weight of 20 kN / m 3
Assuming unit weight of water as 10 kN / m3 , the ultimate consolidation settlement (expressed in mm)
of the clay layer is _________.
Ans.
36.77
Sol.
For clay : void ratio, e 
wG
 0.25  2.70  0.675
S
G  e 
 2.70  0.675 
 sat   w 

10
 1  0.675 
 1  e 
 20.15 kN / m3
At centre of clay,
 117  0.5[20.15  10]
 22.075 kN / m 2
Applied load intensity, q  2.5  20
 50 kN / m 2 (  ')

Final effective stress at end of consolidation at centre of day,
 f '  0'   '
 22.075  50
 72.075 kN / m2
As the preconsolidation stress (c  60 kPa) lies between  0' and  f ' , the clay undergoes consolidation
partly in OC state and partly in NC state
The ultimate consolidation settlement, S f
S f  H.
 1
 ' 
  f' 
C
CR
log10  c   H . C log10 

1  e0
1  e0
  '0 
  c' 
0.05
0.5
 60 
 72.075 
log10 
log10 
 1
 0.03677 m

1  0.675
1  0.675
 22.075 
 60 
 36.77 mm
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GATE 2016 [CE] Set - 1
Q.43 A seepage flow condition is shown in the figure. The saturated unit weight of the soil  sat  18 kN / m3 .
Using unit weight of water,  w  9.81 kN / m3 , the effective vertical stress (expressed in kN / m2 ) on
plane X-X is _________.
Ans.
Sol.
65.475
1st Method:
Effective stress,  '   ' z   w .h
 3 
 (18  9.81)5  9.81 3  1  65.475 kPa
 6 
2nd Method:
Total loss of head, h f  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   w 4
 18  5  9.81 4  129.24 kPa
Neutral stress, u   w hp  9.81 6.5
 63.765 kPa

 '    u  129.24  63.765  65.475 kPa
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GATE 2016 [CE] Set - 1
24
Q.44 A drained triaxial compression test on a saturated clay yielded the effective shear strength parameters as
c '  15 kPa and  '  220 . 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 _______.
Ans.
104.37
Sol.
Given : Effective shear strength parameters C '  15 kPa, '= 220
In Consolidated Undrained (CU) test :
3  200 kPa ,
Pore water pressure, u  150 kPa .
To find deviator stress,  d
3'  3  u  200  150
'
' 


1'  3' tan 2  45    2C ' tan  45  
2
2


22 
22 


 50' tan 2  45    2 15 tan  45    154.377 kPa
2
2 


1'  1  u  (3  d  u )
154.377  (200  d  150)

deviator stress, d  104.37 kPa
Q.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 )
is _______.
Ans.
10500
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GATE 2016 [CE] Set - 1
25
Sol.
Area of uplift pressure distribution diagram given uplift force
Total Area  (1)  (2)  (3)  (4)
(1) 10  250  2500 kN / m
(3) 40  50  2000 kN / m
1
10  400  2000 kN / m
2
1
 40  200  4000 kN / m
(4)
2
(2)
Total uplift force on base  10500 kN / m
Q.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
7
1
9
3
(A)
(B)
(C)
(D)
12
2
16
4
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GATE 2016 [CE] Set - 1
26
Ans.
Sol.
0.5
Equivalent hydraulic conductivity,
20  30  10 60
Z  Z 2  Z3


 2 m / day
k 1
20 30 10 30
Z1 Z 2 Z 3
 


2 3 1
K1 K 2 K 3
h
(15  10)
1
 3 1 
Discharge,
Q  kiA  K A  2 
L
60
2
Q.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 / s 2 , the category of the channel slope is
(A) Horizontal
(B) Mild
(C) Critical
(D) Steep
Ans. (B)
1
Sol.
Q  AR 2/3 S 1/2
n
For wide rectangular channel,
By
By
Hydraulic mean depth,
R

y
B  2y B
Area,
A  By
1

16 
 By  ( y ) 2/3  (0.001)1/2
0.012
Put B  4 m , we get
yn  1.284 m
1/3
Also, critical depth,
 q2 
yc   
 g 
yn  yc
1/3
 42 


 9.81 
 1.177 m
As,
Hence, slope is mild.
Q.48 A sector gate is provided on a spillway as shown in the figure. Assuming g  10 m / s 2 , the resultant
force per meter length (expressed in kN/m) on the gate will be ________.
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27
Ans.
Sol.
GATE 2016 [CE] Set - 1
127.04
Vertical height of the sector gate (AB)
AB  2  5  sin (300 )  5 m
Area of segment  Area of sector – Area of triangle
1
1

A  ( 52 )    AB  OD 
6
2

  52   1
0
A
    5  5 cos 30 

 6  2
A  13.09  10.82  2.27 m 2
Horizontal component of hydrostatic force on the sector gate
FH  .g .h. Aproject vertical plate
5
 1000 10   (5 1) N/ m
2
FH  125 kN / meter length
Vertical component of hydrostatic force on the section 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
 1252  22.7 2  127.04 kN / m
Q.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 _______.
Ans. 2.3
Sol. For hydraulically efficient trapezoidal channel,
lateral side = horizontal side
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GATE 2016 [CE] Set - 1
28
From figure,

B sin 600  d
d
B
 2.31m
sin 600
Q.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 ______.
Ans. 4.5
Sol.
It is known that pH   log10 [H  ]

(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
10 14
 3.154  10 5 mol / lit
3.169  10 10
1
1
 4.5
(pH) B  log10   log10
(H ) B
3.154 105
(H  ) B 
(pH) of effluent from industry B, (pH) B  4.5
Q.51 An electrostatic precipitator (ESP) with 5600 m2 of collector plate area is 956 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 _____.
Ans. 8012.38
Sol. Efficiency () of electrostatic precipitator (ESP) is given by,
  1 e
where,
 AW
Q
A = Plate area of collector,
W  Migration velocity, cm/s
Q  Gas flow through the precipitator, m 3 /s .
When   96%
A  5600 m 2
  97%
  99%
A  6100 m 2
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A?
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GATE 2016 [CE] Set - 1
29
0.96  1  e
0.99  1  e
5600W
185
 W  0.1063 m / sec
 A0.1063
185
 A  8012.38 m 2
Plate area of collector A  8012.38 m 2
Q.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 ________.
Ans.
0.3 d 1
Sol.
BOD of sewage sample after 2 days, y2  100 mg / l
y2  L0 [1  e K 2 ]
100  L0 [1  e K 2 ]
… (i)
BOD of sewage sample after 4 days, y4  L0 [1  e K 4 ]
155  L0 [1  e K 4 ]
… (ii)
By solving (i) and (ii),
K  0.3 d 1 (base e) (or) 0.13 d 1 (base 10)
Q.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
Ans.
Sol.
(A) 0.89 m, 0.50 m, 1.19 m and 0.50 m
(C) 0.50 m, 1.19 m, 0.50 and 0.89 m
(B)
No. of lanes, n  2
Radius of curve, R  50 m
(B) 0.50 m, 0.89 m, 0.50 m and 1.19 m
(D) 1.19 m, 0.50 m, 0.89 m and 0.50 m
Wheelbase, l  5 m
(i) For V  60 kmph
n 2 2(5) 2

 0.5
2 R 2  50
V
60
Psychological widening, W p 

 0.89 m
9.5 R 9.5 50
Mechanical widening,
Wm 
(ii) For V  80 kmph
n 2 2(5) 2

 0.5
2 R 2  50
V
80
Psychological widening, W p 

 1.19 m
9.5 R 9.5 50
Mechanical widening,
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Wm 
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GATE 2016 [CE] Set - 1
30
Q.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 _______.
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  20K
… (i)
(b) Moving observer opposite to the direction of motion
Q  K .V
200  K (V  30)
200  KV  30 K
… (ii)
Solving equations (i) and (ii)
Density of traffic stream, K  3 veh / km
Q.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.
Reading 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 ______.
Ans.
476.911 m
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GATE 2016 [CE] Set - 1
Sol.
x
y
In BET ,
tan16.50 

y  3.3759 x
In AOT ,
tan10.50 
x2
60  y
….. (i)
….. (ii)
Put value of y in equation (ii),
0.1853 
x2
60  3.3759 x
11.1203  0.6255 x  x  2
x  24.3568 m
So, the reduced level of tower,
T  450.000  2.555  24.356
 476.911m

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GATE 2016 : Civil Engineering
Detailed Solutions :
Date : 07-02-2016, Forenoon Session
Set - 2
General Aptitude :
Q.1 – Q.5 carry one mark each.
Q.1
Ans.
Sol.
Q.2
Ans.
Q.3
Ans.
Sol.
Q.4
If I were you, I ______ that laptop. It’s much too expensive.
(A) Won’t buy
(B) Shan’t buy
(C) Wouldn’t buy
(D) Would buy
(C)
In if clause (type 2) ‘were’ is in the past tense so the main clause should be in the conditional clause.
Therefore ‘C’ is the best answer.
He turned a deaf ear to my request.
What does the underlined phrasal verb mean?
(A) Ignored
(B) Appreciated
(C) Twisted
(D) Returned
(A)
Choose the most appropriate set of words from the options given below to complete the following
sentence.
_________ _________ is a will, ________ is a way.
(A) Wear, there, their
(B) Were, their, there
(C) Where, there, there
(D) Where, their, their
(C)
Where there is a will there is a way. it is a quotation.
( x % of y ) + ( y % of x ) is equivalent to ____________
 xy 
(B) 2% of 

 100 
(A) 2% of xy
Ans.
(A)
Sol.
( x % of y )  ( y % of x) 
(C) xy % of 100
(D) 100% of xy
x
y
y
x
100
100
xy  xy 2 xy

100
100
 2 % of xy

Q.5
Ans.
The sum of the digits of a two-digit number is 12. If the new number formed by reversing the digits is
greater than the original number by 54, find the original number.
(A) 39
(B) 57
(C) 66
(D) 93
(A)
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GATE 2016 [CE] Set - 2
Sol.
2
The new number formed by reversing the digits is greater than the original number is possible in options
A and B only.
Options ‘A’ :
Sum of two digits in the number =3 + 9 = 12
After reversing the two digits number = 93
The difference between the new number formed and original number = 93 – 39 = 54
 Option ‘A’ is correct.
Q.6 – Q.10 carry two marks each.
Q.6
Two finance companies, P and Q , declared fixed annual rates of interest on the amounts invested with
them. The rates of interest offered by these companies may differ from year to year. Year-wise annual
rates of interest offered by these companies are shown by the line graph provided below.
If the amounts invested in the companies, P and Q , in 2006 are in the ratio 8 : 9, then the amount s
received after one year as interests from companies P and Q would be in the ratio :
Ans.
Sol.
Q.7
Ans.
(A) 2 : 3
(B) 3 : 4
(C) 6 : 7
(D) 4 : 3
(D)
The amounts invested in the companies of, P and Q in 2006 = 8 : 9
The rate of interest of company ‘P’ in 2006 = 6%
The rate of interest of company ‘Q’ in 2006 = 4%
The amounts received after one year by P and Q companies in 2006 year
P
:
Q
6 % of 8 :
4 % of 9
6
4
:
8
9
100
100
4
:
3
Today we consider, Ashoka as a great ruler because of the copious evidence he left behind in the form of
stone carved edicts. Historians tend to correlate greatness of a king at his time with the availability of
evidence today.
Which of the following can be logically inferred from the above sentences?
(A) Emperors who do not leave significant sculpted evidence are completely forgotten.
(B) Ashoka produced stone-carved edicts to ensure that later historians will respect him.
(C) Statues of kings are a reminder of their greatness.
(D) A king's greatness, as we know him today, is interpreted by historians.
(D)
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GATE 2016 [CE] Set - 2
3
Q.8
Ans.
Sol.
Q.9
Fact 1 : Humans are mammals.
Fact 2 : Some humans are engineers.
Fact 3 : Engineers build hours.
If the above statements are facts, which of the following can be logically inferred?
I. All mammals build houses.
II. Engineers are mammals.
III. Some humans are not engineers.
(A) II only
(B) III only
(C) I, II and III
(D) I only
(B)
From given facts, the following venn diagram is possible.
H = Humans
M = Mammals
E = Engineers
BH = Build houses
 From above diagram, statement III is true.
A square pyramid has a base perimeter x and the slant height is half of the perimeter. What is the lateral
surface area of the pyramid?
(B) 0.75 x 2
(A) x 2
Ans.
Sol.
(C) 0.50 x 2
(D) 0.25 x 2
(D)
Base perimeter of square pyramid  x  p
Slant height 
x
l
2
1
Lateral surface area of pyramid   Base perimeter (p)  slant height (l)
2
x 1
  pl
2 2
1
x
  x
2
2

x2
4
 0.25 x2
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GATE 2016 [CE] Set - 2
4
Q.10 Ananth takes 6 hours and Bharath takes 4 hours to read a book. Both started reading copies of the book
at the same time. After how many hours is the number of pages to be rad by Ananth, twice that to be
read by Bharath? Assume Ananth and Bharath read all the pages with constant pace.
(A) 1
(B) 2
Ans.
(C)
Sol.
Ananth takes 6 hours to read a book
(C) 3
(D) 4
Bharath takes 4 hours to read a book
L.C.M = 12
The number of pages read by Ananth and Bharath must be 12 (or) multiple of 12 only.
If Ananth read 12 number of pages in 6 hrs

1 hr 
12
 2 pages
6
If Bharath read 12 number of pages in 4 hrs
1 hr 
12
 3 pages
4
Option ‘A’

It is not
Option ‘B’

It is not
Option ‘C’

After 3 hours is the number of pages to be read by Ananth, twice that to be read by Bharath.
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GATE 2016 [CE] Set - 2
Technical :
Q.1 – Q.25 carry one mark each.
Q.1 The spot speeds (expressed in km/hr) observed at a road section are 66, 62, 45, 79, 32, 51, 56, 60, 53,
and 49. The median speed (expressed in km/hr) is _____________
(Note : Answer with one decimal accuracy)
Ans. (54.5)
Sol. Median = middle value of observed data
Increasing order of spot speeds
32, 45, 49, 51, 53, 56, 60, 62, 66, 79
When two values in middle take average of these two
53  56
= 54.5 krnph
Median = middle values average =
2
Q.2 The optimum value of the function f ( x)  x 2  4 x  2 is
(A) 2 (maximum)
(B) 2 (minimum)
(C) – 2 (maximum)
(D) – 2 (minimum)
Ans. (D)
f ( x)  x 2  4 x  2
Sol.
f '( x )  2 x  4  0
Q.3
x2
f "( x )  2  0 minimum at x  2
 min value is f (2)  4  8  2   2
The Fourier series of the function,
f ( x )  0,
  x  0
f ( x )    x,
0 x
in the interval [ , ] is
 2  cos x cos 3 x
  sin x sin 2 x
  2 
 ....  

2
4  1
3
2
  1
The convergence of the above Fourier series at x  0 gives



1 2
(1) n 1 2
1
2

(A)  2 
(B) 
(C)


2
6
12
n2
8
n 1
n 1 n
n 1 (2n  1)
(C)
Put x  0 in the given series,
 2 1 1 1

f (0)    2  2  2  ... 
4  1 3 5

f ( x) 
Ans.
Sol.
 f "( x)  0, Minimum 


 f "( x)  0, Maximum 

sin 3 x

 ....
3

( 1) n 1 


4
n 1 (2n  1)

(D)
cos (0)  1


sin (0)  0 
f (0 )  f (0 )  2 
1 1

  1  2  2  ... 
2
4  3 5

  2
1 1

  1  2  2  ... 
2 4  3 5

1 1
2


...

32 52
8
2

1



2
8
n 1 (2n  1)
1

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GATE 2016 [CE] Set - 2
Q.4
Ans.
Sol.
6
X and Y are two random independent events. It is known that P ( X )  0.40 and P( X  Y C )  0.7.
Which one of the following is the value of P ( X  Y ) ?
(A) 0.7
(B) 0.5
(C) 0.4
(D) 0.3
(A)
Given P ( x )  0.40
P( X  Y C )  0.7

Hence,
(Y  X )  0.3
P ( X  Y )  0.3  0.4  0.7
Q.5
What is the value of lim
Ans.
(A) 1
(D)
x 0
y 0
Sol.
(B) – 1
lim
x 0
y 0
It is in the form
Put,
As
xy
?
x  y2
2
(C) 0
(D) Limit does not exist
xy
x  y2
2
0
0
y  mx
y  0, mx  0
x2m
m

2
2
x 0 x (1  m )
1  m2
For different values of m we get different limits it is not unique therefore limit does not exist.
The kinematic indeterminacy of the plane truss shown in the figure is
lim
Q.6
(A) 11
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(B) 8
(C) 3
(D) 0
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GATE 2016 [CE] Set - 2
7
Ans.
Sol.
(A)
Kinematic indeterminacy,
Dk  2 j  re
Where
j  no. of joins
re  no. of equilibrium equations
Q.7
Ans.
Sol.
Q.8
Ans.
Sol.
 2  7  3  11
As per IS 456-2000 for the design of reinforced concrete beam, the maximum allowable shear stress
( c max ) depends on the
(A)
(B)
(C)
(D)
(B)
(i)
Grade of concrete and grade of steel
Grade of concrete only
Grade of steel only
Grade of concrete and percentage of reinforcement
max depends on grade of concrete as per IS : 456-2000.
(ii) Percentage of steel.
An assembly made of a rigid arm A-B-C hinged at end A and supported by an elastic rope C-D at end C
is shown in the figure. The members may be assumed to be weightless and the lengths of the respective
members are as shown in the figure
Under the action of a concentrated load P at C as shown, the magnitude of tension developed in the rope
is
3P
P
3P
(B)
(C)
(D) 2P
(A)
8
2
2
(B)
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GATE 2016 [CE] Set - 2
8
Take moment about A,
As A is hinged,
Hence,
M A  0

RD  2L  P  L  0

RD 
P
2
At joint D :
Fy  0

T cos 450 

T
Fy  verticle forces
P
2
P
2
Q.9
As per Indian standards for bricks, minimum acceptable compressive strength of any class of burnt clay
bricks in dry state is
(A) 10.0 MPa
(B) 0.75 MPa
(C) 5.0 MPa
(D) 3.5 MPa
Ans. (D)
Q.10 A construction project consists of twelve activities. The estimated duration (in days) required to
complete each of the activities along with the corresponding network diagram is shown below.
A
B
C
D
E
F
Activity
Inauguration
Foundation work
Structural construction-1
Structural construction-2
Brick masonry work
Plastering
Duration (days)
1
7
30
30
25
7
G
H
I
J
K
L
Activity
Flooring
Electrification
Plumbing
Wood work
Coloring
Handing over function
Duration (days)
25
7
7
7
3
1
Total floats (in days) for the activities 5-7 and 11-12 for the project are, respectively,
(A) 25 and 1
(B) 1 and 1
(C) 0 and 0
(D) 81 and 0
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GATE 2016 [CE] Set - 2
9
Ans.
Sol.
(C)
Path
Duration
A-B-C-E-F-J-K-L
81
A-B-C-Dummy-G-H-I-K-L
81
A-B-D-G-H-I-K-L
81
All are critical activities, hence total floats of the activities 5 – 7 and 11 – 12 are 0 and 0 (default).
Q.11 A strip footing is resting on the surface of a purely clayey soil deposit. If the width of the footing is
doubled, the ultimate bearing capacity of the soil
(A) Becomes double
(C) Becomes half
(C) Becomes four-times (D) Remains the same
Ans. (D)
qu  CNC
Sol.
Where, qu  Ultimate load
C  Cohesive strength
NC  Bearing factor for cohesion
In case of clay ultimate bearing capacity is independent of width of footing.
Q.12 The results of a consolidation test on an undisturbed soil, sampled at a depth of 10 m below the ground
level are as follows:
(A) G  0.7i  1
Ans.
Sol.
(B) G  1.43 i  1
(C) G  1.43 i  1
(D) G  0.7 i  1
(C)
ic 
G 1
 (G  1)(1  n)
1 e

Where,
 1

 1 n 

 1 e

ic  Hydraulic gradient at critical condition
G  Specific gravity of soil
e  Void ratio of soil
n  Porosity of soil
ic  (G  1)(1  0.3)  (G  1)  0.7
G
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ic
 1  1.43ic  1
0.7
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GATE 2016 [CE] Set - 2
10
Q.13 The results of a consolidation test on an undisturbed soil, sampled at a depth of 10 m below the ground
level are as follows:
Saturated unit weight
:
16 kN / m 3
Pre-consolidation pressure
:
90 kPa
The water table was encountered at the ground level. Assuming the unit weight of water as 10 kN / m 3 ,
the over-consolidation ratio of the soil is
(A) 0.67
(B) 1.50
(C) 1.77
(D) 2.00
Ans. (B)
Sol. Given data : Saturated unit weight (  sat ) = 16 kN / m 3
Pre-consolidation pressure = 90 kPa
Unit weight of water (  w ) = 10 kN / m 3
Depth ( h ) = 10 m
Over-Consolidation Ratio (OCR) =
Maximum effective stress in past
Maximum effective stress in present
Maximum effective stress in present  h sat  h w
 10 sat  10 w  10 16  10 10  60 kN/m 2
90
 1.5
60
Q.14 Profile of a weir on permeable foundation is shown in figure I and an elementary profile of ‘upstream
pile only case’ according to Khosla’s theory is shown in figure II. The uplift pressure heads at key points
Q, R and S are 3.14 m, 2.75 m and 0 m, respectively (refer figure II).

OCR 
What is the uplift pressure head at point P downstream of the weir (junction of floor and pile as shown
in the figure I)?
(A) 2.75 m
(B) 1.25 m
(C) 0.8 m
(D) Date not sufficient
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11
Ans.
(B)
Sol.
Given :
GATE 2016 [CE] Set - 2
hR  2.75m
hQ  3.14 m
hS  0 m
Total head
H  3 1  4 m
hR
 100
H
2.75
R 
 100  68.75%
4
R 
P  100  R  31.25%
Now,
P 
Pressure head at point P
100
Total head
31.25 

hP
 100
4
hP  1.25 m
Q.15 Water table of an aquifer drops by 100 cm over an area of 1000 km 2 . The porosity and specific retention
of the aquifer material are 25% and 5% respectively. The amount of water (expressed in km 2 ) drained
out from the area is _________
Ans. (0.2)
Sol. Given :
Porosity (n) = 25%
Specific retention ( Sr )  5%
Surface area  1000 km 2
specific retention + specific yield = porosity
Sr  S y  n
S y  n  S r  25  5  20 %  0.2
Amount of water drained  Sr  Thickness of aquifer  Surface area
100
 1000  0.2 km3
100  1000
Q.16 Group-I contains the types of fluid while Group-II contains the shear stress –rate of shear relationship
of different types of fluids, as shown in the figure.
Group-I
Group-II
P. Newtonian fluid
1. Curve 1
Q. Pseudo plastic fluid 2. Curve2
R. Plastic fluid
3. Curve 3
S. Dilatant fluid
4. Curve 4
5. Curve 5
 0.2 
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GATE 2016 [CE] Set - 2
12
The correct match between Group I and Group II is
(A) P-2, Q-4, R-1, S-5
(B) P-2, Q-5, R-4, S-1
(C) P-2, Q-4, R-5, S-3
(D) P-2, Q-1, R-3, S-4
Ans. (C)
Q.17 The atmospheric layer closest to the earth surface is
(A) The mesosphere
(B) The stratosphere
(C) The thermosphere
Ans. (D)
Sol.
(D) The troposphere
Q.18 A water supply board is responsible for treating 1500 m 3 /day of water. A settling column analysis
indicates that an overflow rate 20 m/day will produce satisfactory removal for a depth of 3.1 m. It is
decided to have two circular settling tanks in parallel. The required diameter (expressed in m ) of the
settling tanks is _______
Ans. (6.19)
Sol.
Given :
Discharge (Q) = 1500 m3 /day
Overflow rate (OFR) = 20 m/day
 Tanks are in parallel.
Discharge to be treated by one tank,
1500
Q
 750 m3 /day
2
Q
750
Surface area, A 

 37.5 m 2
OFR 20
 tank is circular


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d2
4
d  6.91m
37.5  
where, d  dia. Of circular tank
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13
GATE 2016 [CE] Set - 2
Q.19 The hardness of a ground water sample was found to be 420 mg/L as CaCO3 . A softener containing ion
exchange resins was installed to reduce the total hardness to 75 mg/L as CaCO3 before supplying to 4
Ans.
Sol.
households. Each household gets treated water at a rate of 540 L/day. If the efficiency of the softener is
100%, the bypass flow rate (expressed in L/day) is ______
(385.7)
Total water to be treated = rate of treated water per household  number of households
 540  4  2160 l/day
Resultant hardness required
 75 mg/l
Let bypass rate be x l/day having hardness of 420 mg/l
Resultant hardness required
 75 mg/l
CH mix 
CH1Q1  CH 2Q2
Q
x  420  (2160  x)0
2160
2160  75
x
 385.7 l/day
420
Q.20 The sound pressure (expressed in Pa ) of the faintest sound that a normal healthy individual can hear is
75 
Ans.
Sol.
(A) 0.2
(B) 2
(C) 20
(D) 55
(C)
The sound pressure of the faintest sound that a normal healthy individual can hear is 20 Pa . It is taken
as reference sound pressure level.
A 20 Pa pressure is 0 dB on the sound pressure level scale.
Q.21 In the context of the IRC 58-2011 guidelines for rigid pavement design, consider the following pair of
statements.
I: Radius of relative stiffness is directly related to modulus of elasticity of concrete and inversely
related to Poisson’s ratio
II: Radius of relative stiffness is directly related to thickness of slab and modulus of subgrade reaction.
Which one of the following combinations is correct?
(A) I: True; II: True
(B) I: False; II: False
(C) I: True; II: False
(D) I: False; II: True
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GATE 2016 [CE] Set - 2
Ans.
Sol.
14
(C)
Radius of relative stiffness,
1

4
Eh3
l
2 
12k(1   ) 
Where, l  Radius of relative stiffness, E  Modulus of elasticity
  Poisson’s ratio, K  Modulus of subgrade reaction
l  E,
l
1

 Statement 1 is true,
1
l
k
Statement 2 is false
 Option (C) is correct.
Q.22 If the total number of commercial vehicles per day ranges from 3000 to 6000, the minimum percentage
of commercial traffic to be surveyed for axle load is
(A) 15
(B) 20
(C) 25
(D) 30
Ans. (A)
Q.23 Optimal flight planning for a photogrammetric survey should be carried out considering
(A) Only side-lap
(B) Only end-lap
(C) Either side-lap or end-lap
(D) Both side-lap as well as end-lap
Ans. (D)
Q.24 The reduced bearing of a 10 m long line is N 30 0 E . The departure of the line is
(A) 10.00 m
(B) 8.66 m
(C) 7.52 m
(D) 5.00 m
Ans. (D)
Sol. The departure of the line,
D  l sin   10 sin 300
10

5 m
2
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15
GATE 2016 [CE] Set - 2
Q.25 A circular curve of radius R connects two straights with a deflection angle of 60 0 . The tangent length is
(A) 0.557 R
(B) 1.155 R
(C) 1.732 R
(D) 3.464 R
Ans. (A)
Sol.
Tangent length, VT1  R tan

2
 R tan 300  0.577 R
Q.26 – Q.55 carry two mark each.
Q.26 Consider the following linear system
x  2 y  3z  a
2 x  3 y  3z  b
5x  9 y  6z  c
This system is consistent if a , b and c satisfy the equation
(A) 7 a  b  c  0
(B) 3a  b  c  0
(C) 3a  b  c  0
Ans. (B)
Sol. For consistent solution
Rank of ( A : B) = Rank of A
Hence, ( A : B)  ( A)
(D) 7 a  b  c  0
 1 2 3 a 


( A : B)   2 3 3 b 
5 9 6 c 


( R2  2 R1 ), ( R3  5R1 )
a 
 1 2 3


  0 1 9 b  2a 
 0 1 9 c  5a 


( R3  R2 )
a
 1 2 3



  0 1 9
b  2a 
 0 0 0 (c  b  3a ) 


(c  b  3a )  0
3a  b  c  0
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GATE 2016 [CE] Set - 2
16
Q.27 If f ( x ) and g ( x ) are two probability density functions,
 x
 x
 a 1 :  a  x  0
 a :  a  x  0


 x
 x
f ( x)    1 : 0  x  a
g ( x)  
: 0 xa
a
a


: otherwise
 0
 0 : otherwise




Which one of the following statement is true?
(A) Mean of f ( x ) and g ( x ) are same ; Variance of f ( x ) and g ( x ) are same
(B) Mean of f ( x ) and g ( x ) are same ; Variance of f ( x ) and g ( x ) are different
(C) Mean of f ( x ) and g ( x ) are different ; Variance of f ( x ) and g ( x ) are same
(D) Mean of f ( x ) and g ( x ) are different ; Variance of f ( x ) and g ( x ) are different
Ans.
(B)

Sol.
Mean of f ( x ) 
E f ( x) 
 xf ( x) dx

a
 x2

  x2



x
dx
 a  a  0  a  x  dx
0

0
a
 x3 x 2 
  x3 x 2 
    
 
2 0
 3a 2  a  3a
  a2 a2    a2 a2 
 
 
 
2  3
2 
 3
0
 x2
x2
E g ( x)  
dx   dx
a
a
0
a
0
Mean of g ( x) 
a
0
a
 x3 
 x3 
 a2  a2
0
         
 3a  a  3a 0
 3  3
Variance of f ( x) 
Vf ( x)  E ( X 2 )  ( E ( X ))2
a
 x2
  x3

2
 x 2  dx
E ( X )     x  dx   
a
a


a 
0
0
2
0
a
 x 4 x3 
  x 4 x3 
 a3 a3    a3 a3 
 
   
       
3 0
3
 4a 3   a  4a
 4 3  4
E( X 2 ) 
Vf ( x) 
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a3
6
a3
6
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GATE 2016 [CE] Set - 2
17
Vg ( x)  E ( X 2 )  ( E ( X )) 2
Variance of g ( x) 
 x3
x3
E( X )  
dx   dx
a
a
0
a
a
0
2
0
a
 x4 
 x4 
 a3  a3
        (0)     
 4a   a  4a  0
 4 4
2a 3 a 3


4
2
3
a
Vg ( x) 
2
Q.28 The angle of intersection of the curves x 2  4 y and y 2  4 x at point (0, 0) is
Ans.
Sol.
(A) 0 0
(D)
(B) 300
(C) 450
(D) 900
Angle between the curves is angle between the tangents at the point of intersection.
Q.29 The area between the parabola x 2  8 y and the straight line y  8 is ______.
Ans. (85.33)
Sol.
y
Area  2  x dy
x2  8 y  x  8 y
0
8
 2  8y dy
0
8
 3
 y2 
 2 8    85.33
3
 
 2 0
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GATE 2016 [CE] Set - 2
18
Q.30 The quadratic approximation of f ( x)  x3  3x 2  5 at the point x  0 is
(A) 3x 2  6 x  5
(B)  3x 2  5
(C)  3x 2  6 x  5
Ans.
(B)
Sol.
Quadratic Approximation  f (0)  ( x  0) f '(0) 
( x  0) 2
f "(0)
2!
f ( x)  x3  3x 2  5

f (0)  5
f '( x)  3x 2  6 x

f '(0)  0
f "( x)  6 x  6

f "( x)  6
 Equation is  5  x(0) 
(D) 3 x 2  5
x2
( 6)
2
 3 x 2  5
Q.31 An elastic isotropic body is in a hydrostatic state of stress as shown in the figure. For no change in the
volume to occur, what should be its Poisson’s ratio?
(A) 0.00
Ans.
(C)
Sol.
Volumetric strain,
(B) 0.25
    y  z
V   x
3

V 
(C) 0.50

 (1  2)

Where,
(D) 1.00
  Poisson’s ratio
Change in volume V

Total volume
V

V   x   y   z

3
V 
As
V  0

Enter  x   y   z  0 or 1  2  0

1  2  0

 (1  2)

  0.5
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GATE 2016 [CE] Set - 2
19
Q.32 For the stress state (in MPa) shown in the figure, the major principal stress is 10 MPa.
The shear stress  is
(A) 10.0 MPa
Ans.
Sol.
(B) 5.0 MPa
(C) 2.5 MPa
(D) 0.0 MPa
(B)
Major/minor principal stress
1/2 
x   y
2
 x   y 
2
 
   xy
 2 
2
55
 55
2
 
 Major principal stress 1 
   xy
2
2


2
 10  5   xy

 xy  5 MPa
Q.33 The portal frame shown in the figure is subjected to a uniformly distributed vertical load w (per unit
length).
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GATE 2016 [CE] Set - 2
20
The bending moment in the beam at the joint ' Q ' is
(A) Zero
Ans.
Sol.
(B)
wL2
(hogging)
24
(C)
wL2
(hogging)
12
(D)
wL2
(sagging)
8
(A)
As there is no horizontal force,
HP  HS  0
Hence
Moment about point Q,
M Q  H P 
L
0
2
 BM at Q  0
Q.34 Consider the structural system shown in the figure under the action of weight W . All the joints are
hinged. The properties of the members in terms of length ( L ), area ( A) and the modulus of elasticity
( E ) are also given in the figure. Let L, A and E be 1m, 0.05 m 2 and 30 106 N/m 2 , respectively, and
W be 100 kN.
Which one of the following sets gives the correct values of the force, stress and change in length of the
horizontal member QR ?
(A) Compressive force = 25 kN ; Stress  250 kN/m 2 ; Shortening = 0.0118 m
(B) Compressive force = 14.14 kN ; Stress  141.4 kN/m 2 ; Extension = 0.0118 m
(C) Compressive force = 100 kN ; Stress  1000 kN/m 2 ; Shortening = 0.0417 m
(D) Compressive force = 100 kN ; Stress  1000 kN/m 2 ; Extension = 0.0417 m
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GATE 2016 [CE] Set - 2
21
Ans.
Sol.
(C)
Given data :
L  1m , A  0.05 m 2 , E  30 106 N/m 2
Consider joint ‘S’
FSQ  FSR
2 FSQ cos 450  W
W
W
 2
2
2
W
FSQ 

2
As the truss is symmetrical
W
FQP  FPR 

2
Now consider joint ‘ Q ’

FSQ 
FQP  FQS 
(Tensile)
W
2
Fx  0
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GATE 2016 [CE] Set - 2
22

FQP

FQR  W

FQR  100 kN (Compressive)
2

FQS
2
 FQR  0
(Compressive)
Stress in member QR ,
QR 
FQR
2
100
100 100

 1000 kN/m 2
2 1.05
25

QR 

QR  1000 kN/m 2
As the member QR consist compressive tone so it will go under shortening
 Shortening,  
P(Length) FQR LQR

2 AE
2 AE
LQR  L2  L2  2 L

100 103  2  1 100  103  2
2



6
6
2  0.05  30  10
0.1 30  10
30
  0.0471

Q.35 A haunched (varying depth) reinforced concrete beam is simply supported at both ends, as shown in the
figure. The beam is subjected to a uniformly distributed factored load of intensity 10 kN/m. The design
shear force (expressed in kN) at the section X -X of the beam is ______.
Ans.
Sol.
(65)
Shear force at section X-X,
Vu  100  5 10  50 kN
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23
GATE 2016 [CE] Set - 2
Depth at section X-X,
d  400 
200
 5  500 mm  0.5 m
10
Moment at section X-X,
M u  100  5  10  2.5  5  375 kNm
Design shear force at section X-X,
Vu ,design  Vu 
Mu
375 200
tan   50 

 65 kN
d
0.5 10000
Q.36 A 450 mm long plain concrete prism is subjected to the concentrated vertical loads as shown in the
figure cross section of the prism is given as 150 mm  150 mm . Considering linear stress distribution
across the cross-section, the modulus of rupture (expressed in MPa) is ______.
Ans.
(3)
Sol.
Bending moment at Q
BM Q  11.25  150  1.6875  106 N-mm
Bending equation

 MQ

y
I
where,
  Modulus of rupture or bending stress
y  Height of neutral axis from base
y
d 150

 75 mm
2
2
I  Moment of inertia
I

© Copyright
bd 3 (150) 4

 mm 4
12
12
1.6875  106  75

 3MPa
(150) 4
12
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GATE 2016 [CE] Set - 2
24
Q.37 Two bolted plates under tension with alternative arrangement of bolt holes are shown in figures 1 and 2.
The hole diameter, pitch and gauge length are d , p and g , respectively.
Which one of the following must be ensured to have higher net tensile capacity of configuration shown
in figure 2 than that shown in figure 1?
(A) p 2  2 gd
Ans.
Sol.
(B) p 2  4 gd
(C) p 2  4 gd
(D) p  4 gd
(C)
Tensile strength of plate in arrangement (2) will be greater than in arrangement (1)
*As per IS code 800 : 2007 close 6.3
f up  
fup 

 0.9 Anet
   0.9 Anet

 m1  2 
 m1 1

( Anet ) 2  ( Anet )1

p2  
 B  2 d 
 t    ( B  d )t 1
4g  2

where,
t  Thickness of plate
B  Width of plate
d  Depth of plate
p2
 Bd
B  2d 
4g
p2
d
4g
p 2  4 gd
Q.38 A fixed-end beam is subjected to a concentrated load ( P ) as shown in the figure. The beam has two
different segments having different plastic moment capacities ( M p , 2 M p ) as shown.
The minimum value of load ( P ) at which the beam would collapse (ultimate load) is
(A)
7.5 M p
L
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(B)
5.0 M p
L
(C)
4.5 M p
L
(D)
2.5 M p
L
© Copyright
25
Ans.
Sol.
GATE 2016 [CE] Set - 2
(A)
DS  2
 Number of plastic hinge required for complete collapse = No. of static indeterminacy + 1
 DS  1
 2 1  3
Mechanism 1 :
Take hinges at the point of load act and the fixed ends.
2L 4L

3
3
  2

For principal of virtual work done,






 2L 
 2M P   2M P   2M P   M P   P 
  0
 3 
2 PL
4M P   3M P  

3
4 PL
8 M P  3M P  

3
4P L
11M P  u
3
33 M P
Pu 
4 L
M
Pu  8.25 P
L
Mechanism 2 :
Take hinges at the point of change of plastic moment, and fixed ends
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GATE 2016 [CE] Set - 2
26
 2L 
2M P  M P  M P  M P  P 

 3 
2 PL
5M P 

3
15M P
Pu 

2L
M

Pu  7.5 P
L
Q.39 The activity-on-arrow network of activities for a construction project is shown in the figure. The
durations (expressed in days) of the activities are mentioned below the arrows.
Ans.
Sol.
The critical duration for this construction project is
(A) 13 days
(B) 14 days
(C) 15 days
(C)
The critical duration of construction is the longest duration of network.
Path
P-Q-T-W-X
P-Q-Dummy-U-W-X
P-R-Dummy-U-W-X
P-R-Dummy-V-X
P-S-V-X
The critical duration is 15 days.
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(D) 16 days
Duration
2 + 3 + 5 + 3 + 2 = 15
2 + 3 + 0 + 3 + 3 + 2 = 13
2 + 4 + 0 + 3 + 3 + 2 = 14
2 + 4 + 0 + 2 + 2 – 10
2+3+2+2=9
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GATE 2016 [CE] Set - 2
27
Q.40 The seepage occurring through an earthen dam is represented by a flow net comprising of 10
equipotential drops and 20 flow channels. The coefficient of permeability of the soil is 3 mm/min and
the head loss is 5 m. The rate of seepage (expressed in cm3/s per m length of the dam) through the
earthen dam is ______.
Ans.
(500)
KH  N f
q
Sol.
Nd
Where, q  rate of seepage through soil in m3 /sec
K  Permeability in m/sec
N f  No. of flow channels
N d  No. of equipotential lines
H  Head loss in m
K
3 mm 3  10 3 m

m/s
min
60
H  5 m, N f  20, N d  10
q
3  10 3
20 3
 5
m /sec per m length of dam
60
10
3  10 3  5  20
 106 cm 3 /sec per m length of dam
q
60  10
q  500 cm3 /sec per m length of dam
Q.41 The soil profile at a site consists of a 5 m thick sand layer underline by a c   soil as shown in figure.
The water table is found 1 m below the ground level. The entire soil mass is retained by a concrete
retaining wall and is in the active state. The back of the wall is smooth and vertical. The total active
earth pressure (expressed in kN/m2) at point A as per Rankine's theory is ______.
Ans.
Sol.
(69.65)
In c- soil
Earth pressure
pa  ka v  2c ka
Where,
ka  Coefficient of active earth pressure
v  Effective pressure
c  Cohesive strength
ka 
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1  sin  1  sin 24

 0.4217
1  sin  1  sin 24
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GATE 2016 [CE] Set - 2
28
Note : Below water table, water pressure will not be multiplied by ka at point A
Effective pressure v  1  b  4 sat  3 sat
Where,
 b   bulk ,  '   sat   w
v  1  b  (4   ' 4   w )  (3 ' 3 w )
v  (1  b  4 ' 3 ')  (4 w  3 w )
Earth pressure at ' A ' ,
pa  ka v  2c ka
pa  ka [  b  4 ' 3 ']  4W  3W  2  c ' ka
pa  0.4217[16.5  4 (19  9.81)  3(18.5  9.81)] 7  9.81  2  25  0.4217
pa  69.65 kN/m 2
Q.42 OMC-SP and MDD-SP denote the optimum moisture content and maximum dry density obtained from
standard Proctor compaction test, respectively. OMC-MP and MDD-MP denote the optimum moisture
content and maximum dry density obtained from the modified Proctor compaction test, respectively.
Which one of the following is correct?
(A) OMC-SP < OMC-MP and MDD-SP < MDD-MP
(B) OMC-SP > OMC-MP and MDD-SP < MDD-MP
(C) OMC-SP < OMC-MP and MDD-SP > MDD-MP
(D) OMC-SP > OMC-MP and MDD-SP > MDD-MP
Ans. (B)
Sol.
Q.43 Water flows from P to Q through two soil samples, Soil 1 and Soil 2, having cross sectional area of 80
cm2 as shown in the figure. Over a period of 15 minutes, 200 ml of water was observed to pass through
any cross section. The flow conditions can be assumed to be steady state. If the coefficient of
permeability of Soil 1 is 0.02 mm/s, the coefficient of permeability of Soil 2 (expressed in mm/s) would
be ______
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GATE 2016 [CE] Set - 2
29
Ans.
(0.045)
Sol.
Discharge 
200 ml
200 cm3
200 103 mm3



15min 15  60 sec
900
sec
As per Darcy
q  k avg iA
kavg 
 Zi
Z
 i
ki
( Flow is normal to bedding plane)
150  150
150 150

0.02 k
Head difference 600  300
i

1
length
300
kavg 
A  80 cm2  80 102 mm2


 150  150 
200  103
2
q
1
80
10




150 150 
900



 0.02 k 
k  0.045 mm/sec
Q.44 A 4 m wide strip footing is founded at a depth of 1.5 m below the ground surface in a c   soil as
shown in the figure. The water table is at a depth of 5.5 m below ground surface. The soil properties are
: c '  35 kN/m2 ,  '  28.630 ,  sat  19 kN/m 3 ,  bulk  17 kN/m 3 and  w  9.81 kN/m 3 . the values of
bearing capacity factors for different  ' are given below.
'
Nc
Nq
N
150
12.9
17.7
25.1
37.2
4.4
7.4
12.7
22.5
2.5
5.0
9.7
19.7
20 0
250
300
Using Terzaghi's bearing capacity equation and a factor of safety Fs  2.5, the net safe bearing capacity
(expressed in kN/m2) for local shear failure of the soil is ______.
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GATE 2016 [CE] Set - 2
30
Ans.
(298.48)
Sol.
Local shear failure is occurring hence modified c and  should be used
cm 
2
2
c   35  23.33 kN/m 2
3
3
tan m 
2
2
tan   tan 28.63
3
3
 m  20 0
for  m  20 0 , N c  17.7, N q  7.4, N   5.0
 The water table is at a depth of 5.5 m below ground surface
D f  B  1.5  4 m hence no effect on bearing capacity
 As per Terzaghi for strip footing
qu  cN c  D f N q  0.5 B  N 
2
qu   35 17.7  17 1.5  7.4  0.5  4  17  5
3
qu  771.7 kN/m 2
qnu  qu    qu  D f  771.7  17  1.5
qnu  746.2 kN/m 2
Net safe bearing capacity
qns 
qnu 746.2

 298.48 kN/m 2
2.5
F
where,
F  Factor of safety
Q.45 A square plate is suspended vertically from one of its edges using a hinge support as shown in figure. A
water jet of 20 mm diameter having a velocity of 10 m/s strikes the plate at its mid-point, at an angle of
300 with the vertical. Consider g as 9.81 m/s2 and neglect the self-weight of the plate. The force F
(expressed in N) required to keep the plate in its vertical position is ______.
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GATE 2016 [CE] Set - 2
31
Ans.
Sol.
(7.85)
Force exerted by jet in x-direction
Fx  m [v sin  0]
Fx   q  v sin   (103 )

(0.02) 2  10 10 sin 300
4
Fx  15.7079 N
Taking moment about hinge
0.2
 F  0.2
2
F  7.85 N
Fx 
Q.46 The ordinates of a one-hour unit hydrograph at sixty minute interval are 0, 3, 12, 8, 6, 3 and 0 m3/s. A
two-hour storm of 4 cm excess rainfall occurred in the basin from 10 AM. Considering constant base
flow of 20 m3/s, the flow of the river (expressed in m3/s) at 1 PM is ______.
Ans. (60)
Sol.
Ordinate of 1 Offset ordinate Ordinate of 2
Ordinate of
Ordinate of
Time (1)
3
3
3
hr UH (m /s)
hr DRH (m /s) DRH of 4 cm
flood
(m /s)
rainfall
excess
hydrograph
(2)
(4)
=
(2)
+
(3)
(3)
(m3/s)
(m3/s)
(6) = (5) + 20
4
(5) = (4) 
2
10 : 00 am
0
–
0
0
20
11 : 00 am
3
0
3
6
26
12 : 00 pm
12
3
15
30
50
01 : 00 pm
8
12
20
40
60
02 : 00 pm
6
8
14
28
48
03 : 00 pm
3
6
9
18
38
05 : 00 pm
0
3
3
6
26
0
0
20
Ordinate of 01 : 00 pm = 60 m3/s.
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GATE 2016 [CE] Set - 2
32
Q.47 A 3 m wide rectangular channel carries a flow of 6 m3/s. The depth of flow at a section P is 0.5 m. A
flat-topped hump is to be placed at the downstream of the section P. Assume negligible energy loss
between section P and hump, and consider g as 9.81 m/s2. The maximum height of the hump (expressed
in m) which will not change the depth of flow at section P is ______.
Ans. (0.20)
Sol.
B  3m
Q  6 m3 /s
y p  0.5 m
A1  3  0.5  1.5 m 2
V1 
Q
6

 4 m/s
A1 1.5
Q  AV
1 1
According to energy equation
E1  E2  (z )max
E1  y1 
V12
42
 1.315 m
 0.5 
2g
2  9.81
1
3
1
3
 Q   6

yc   2   
 0.74 m
2 
 gB   9.81 3 
2
2
3
E2  Ec   0.74  1.112 m
2
3 

 Ec  yc 
2 

1.315  1.112  (Z )max
(Z )max  1.315  1.112
 0.203m
Q.48 A penstock of 1 m diameter and 5 km length is used to supply water from a reservoir to an impulse
turbine. A nozzle of 15 cm diameter is fixed at the end of the penstock. The elevation difference
between the turbine and water level in the reservoir is 500 m. Consider the head loss due to friction as
5% of the velocity head available at the jet. Assume unit weight of water = 10 kN/m3 and acceleration
due to gravity (g) = 10 m/s2. If the overall efficiency is 80%, power generated (expressed in kW and
rounded to nearest integer) is ______.
Ans. (6570)
Head loss ( hL ) = 5%
Sol. Given :
Efficiency (0 ) = 80%
Head difference ( H ) = 500 m
Apply energy equation at the free surface of reservoir and exit of nozzle
H = hL 
v2
2g
v12
500  Head loss 
2g
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33
GATE 2016 [CE] Set - 2
v12 v12
500  0.05

2g 2g
v12 v12
500  0.05

2g 2g
v1  97.59 m/sec
Water power (W.P.) 
1 2
mv1
2
1 3 
(10 )  (0.15) 2 (97.59)  8212.178 kW
2
4
Shaft power (S.P.)
Now,
0 
W.P
S.P.
0.8 
8212.178
S.P.  6569.74 kW
S.P.  6570 kW
Q.49 A tracer takes 100 days to travel from Well-1 to Well-2, which are 100 m apart. The elevation of water
surface in Well-2 is 3 m below that in Well-1. Assuming porosity equal to 15%, the coefficient of
permeability (expressed in m/day) is
(A) 0.30
(B) 0.45
(C) 1.00
(D) 5.00
Ans. (D)
Sol. Seepage velocity
v distance
Where, v  Discharge velocity
Vs  
time
n
n  Porosity = 0.15 (Given)
k  Permeability
v  ki
As per Darcy
ki distance
100 m


n
time
100 days

head difference
3

lenght
100
k 3
100

m / day
0.15  100 100
k  5 m / day
Q.50 A sample of water has been analyzed for common ions and results are presented in the form of a bar
diagram as shown.
Hydraulic gradient, i 
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GATE 2016 [CE] Set - 2
34
The non-carbonate hardness (expressed in mg/L as CaCO3) of the sample is
(A) 40
Ans.
(B) 165
(C) 195
(D) 205
(A)
Ca 2  2.65 meq / l , Mg 2  1.45 meq / l
Sol.
Na 2  2.25 meq / l , K 2  0.5 meq / l
HCO3  3.3 meq / l , SO42  0.6 meq / l
Cl   2.85 meq / l
Hardness is due to multivalent metallic cations, i.e. Ca 2  and Mg 2
Total hardness ( mg / l as CaCO3 )  (Total meq / l )  (eq. weight of CaCO3 in mg)
 (2.65  1.45)  50 mg/ l as CaCO3
 205 mg / l as CaCO3
Alkalinity is due to the presence of HCO3 in this case
Alkalinity ( mg / l as CaCO3 )  3.3  50 mg / l as CaCO3
Now, Carbonate Hardness  Min {Total Hardness, Alkalinity}
 165 mg / l
 Non-carbonate hardness = Total hardness – Carbonate Hardness
 205  165  40 mg / l as CaCO3
Q.51 A noise meter located at a distance of 30 m from a point source recorded 74 dB. The reading at a
distance of 60 m from the point source would be ________ dB.
Ans.
(67.98)
Sol.
Given :
Initial distance ( R1 ) = 60 m
Final distance ( R2 ) = 30 m
Initial reading ( L1 ) = 74 dB
Required reading ( L2 ) = ?
R 
L2  L1  20 log  2 
 R1 
 74  20 log 0.5
 67.98 dB
Q.52 For a wastewater sample, the three-day biochemical oxygen demand at incubation temperature of 20 0 C
(BOD3day , 200 C ) is estimated as 200 mg/L. taking the value of the first order BOD reaction rate constant
as 0.22 day 1 , the five-day BOD (expressed in mg/L) of the wastewater at incubation temperature of
200 C (BOD5 day , 200 C ) would be ______.
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GATE 2016 [CE] Set - 2
35
Ans.
(276.19)
Sol.
Given :
Rate constant ( k ) = 0.22 day 1
BOD3  200 mg/L
BOD5  ?
Now,
BODt  BODU (1  e  kt )
Where, BODt  BOD at time t
BODU  Ultimate BOD
BOD5  L0 (1  e 0.225 )
… (i)
BOD3  L0 (1  e 0.223 )
… (ii)
From equation (i) and (ii), we get
BOD5 (1  e0.225 )

BOD3 (1  e0.223 )
0.667
0.483
 200  1.38  276.19 mg / l
BOD5  200 
Q.53 The critical flow ratios for a three-phase signal are found to be 0.30, 0.25, and 0.25. The total time lost
in the cycle is 10 s. Pedestrian crossings at the junction are not significant. The respective Green times
(expressed in seconds and rounded off to the nearest integer) for the three phases are
(A) 34, 28 and 28
(B) 40, 25, and 25
(C) 40, 30 and 30
(D) 50, 25, and 25
Ans. (A)
Y  y1  y2  y3  0.30  0.25  0.25  0.80
Sol.
Now Total lost time,
L  10 sec (given)
 Optimum cycle time,
1.5 L  5
C0 
1 Y
(1.5  10)  5 15  5 20



 100 sec
C0 
1  0.80
0.20 0.2
Now green times are calculated by,
y
0.30
G1  1 (C0  L) 
(100  10)  33.75  34 sec
y
0.80
G2 
y2
0.25
(C0  L) 
(100  10)  28.11  28 sec
y
0.80
G3 
y3
0.25
(C0  L) 
(100  10)  28.11  28 sec
y
0.80
Q.54 A motorist traveling at 100 km/h on a highway needs to take the next exit, which has a speed limit of 50
km/h. The section of the roadway before the ramp entry has a downgrade of 3% and coefficient of
friction ( f ) is 0.35. In order to enter the ramp at the maximum allowable speed limit, the braking
distance (expressed in m) from the exit ramp is ______.
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GATE 2016 [CE] Set - 2
Ans.
Sol.
36
(92.32)
Total energy lost between point (1) and (2)  Work done by frictional force
1 2 1 2
mv1  mv2  mg  0.03s  f  (mg)  s
2
2
(0.278 100)2 (0.278  50)2

 9.81 0.03  s  0.35  9.81 s
2
2
289.815  3.1392s
s  92.32 m

Q.55 A tall tower was photographed from an elevation of 700 m above the datum. The radial distances of the
top and bottom of the tower from the principal points are 112.50 mm and 82.40 mm, respectively. If the
bottom of the tower is at an elevation 250 m above the datum, then the height (expressed in m) of the
tower is ______.
Ans. (120.4)
Sol.
Given : Flying height H  700 m , havg  250
Relief distance, d  r1  r  112.5  82.40  30.1 mm
hr1
H  havg

d

30.1 

[where h is height of tower]
h  112.5
700  250
h  120.4 mm

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