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GATE_2016_CE
Q. 1 – Q. 5 carry one mark each.
01. 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
01. Ans:(c)
Sol: In if clause (type2) ‘were’ is in the past tense so the so main clause should be in the conditional
clause. Therefore ‘C’ is the best answer
02. He turned a deaf ear to my request. What does the underlined phrasal verb mean?
(a) Ignored
(b) appreciated
(c) twisted
(d) returned
02. Ans:(a)
Sol: ‘turned a deaf ear’ means ignored
03. 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
03. Ans:(c)
Sol: Where there is a will there is a way. It is a quotation
04. (x% of y) + (y% of x) is equivalent to _____
(a) 2% of xy
(b) 2% of (xy/100)
(c) xy% of 100
(d) 100% of xy
04. Ans: (a)
Sol: (x% of y) + (y% of x) =
x
y
y
x
100
100

xy  xy
100

2xy
100
= 2% of xy
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:3:
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SET_2_Forenoon Session
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GATE_2016_CE
05. 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
05. Ans: (a)
Sol: 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.
Option ‘b’:
original number = 57
After reversing, the number is formed = 75
The difference between these two numbers = 75 – 57 = 18
 It is not.
06. 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.
7
6.5
9
8
9.5
8
10 9
8
8
7.5
P
Q
6
6.5
4
2000 2001 2002 2003 2004 2005 2006
If the amounts invested in the companies, P and Q, in 2006 are in the ratio 8:9, then the amounts
received after one year as interests from companies P and Q would be in the ratio:
(a) 2:3
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(b) 3:4
(c) 6:7
(d) 4:3
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SET_2_Forenoon Session
06. Ans: (d)
Sol: 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
6% of 8
6
8
100
4
Q
:
:
:
4% of 9
4
9
100
3
07. 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 reminder of their greatness.
(d) A king’s greatness, as we know him today, is interpreted by historians.
07. Ans:(d)
Sol: ‘Today, historians correlate greatness of a king at his time with the availability of evidence.’ This
statement leads to the best inference option ‘d’
08. Fact 1: Humans are mammals.
Fact 2: Some humans are engineers.
Fact 3: Engineers build houses
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
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(b) III only
(c) I, II and III
(d) I only
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GATE_2016_CE
08. Ans: (b)
Sol: From given facts, the following venn diagram is possible.
H = Humans
BH
M = Mammals
M
E = Engineers
H
E
BH = Build houses
 From above diagram, statement III is true.
09. 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?
(a) x2
(b) 0.75 x2
09. Ans: (D)
Sol: Base perimeter of square pyramid = x = p
x
Slant height   l
2
Lateral surface area of pyramid
1
  Base perimeter (p)  slant height (l)
2
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(c) 0.50 x2
(d) 0.25 x2
x/2
x/4
x/4
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x
2
SET_2_Forenoon Session
1
 pl
2
1
x
 x
2
2
2
x

4
= 0.25x2

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 read by Ananth, twice
that to be read by Bharath? Assume Ananth and Bharath read all the pages with constant pace.
(a) 1
(b) 2
(c) 3
(d) 4
10. Ans: (c)
Sol: Ananth takes 6 hours to read a book
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’
1 hrs
Ananth
Read
2 pages
Bharath
Not read
10 pages
Read
3 pages
Not read
9 pages
10 : 9
 it is not
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Option ‘b’
2 hrs
Ananth
Read
4 pages
Bharath
Not read
8 pages
Read
6 pages
Not read
6 pages
8 : 6
4: 3
 it is not
Options ‘c’
3 hrs
Ananth
Read
6 pages
Bharath
not Read
6 pages
Read
9 pages
not Read
3 pages
2 :1
 After 3 hours is the number of pages to be read by Ananth, twice that to be read by Bharath.
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SET_2_Forenoon Session
Q1 – Q. 25 carry One Mark each
01. 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)
01. Ans: 54.5
Sol: Increasing order of spot speeds
32, 45, 49, 51, 53, 56, 60, 62, 66, 79
Median = middle values average 
53  56
= 54.5 kmph
2
02. The optimum value of the function f(x) = x2  4x + 2 is
(a) 2 (maximum)
(c)  2 (maximum)
(b) 2 (minimum)
(d) 2 (minimum)
02. Ans: (d)
Sol: f(x) = x2 – 4x +2
f(x) = 2x – 4 = 0  x = 2
f(x) = 2 > 0 minimum at x = 2
 min value is f(2) = 4 – 8 +2 = –2
03. The fourier series of the function,
f(x) = 0, – < x  0
= –x, 0 < x < 
In the interval [–,] is
f x  
  sin x sin 2 x sin 3x
 2  cos x cos 3x

 

 .....  


 ....
4   12
2
3

32
  1
The convergence of the above Fourier series at x = 0 gives
1
2


2
6
n 1 n

(a)
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
(b)

n 1
 1n 1
n
2

2
12

(c)
1
 2n  1
n 1
2

2
8

(d)
 1n 1
 2n  1
n 1


4
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GATE_2016_CE
03. Ans: (c)
Sol: f 0  
 2 1 1 1

  2  2  2  ... 
4  1 3 5

   
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
1  2  2  ... 
8
3
5


1
 2n  1
n 1
2
2

8
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SET_2_Forenoon Session
04. X and Y are two random independent events. It is known that P(X) = 0.40 and P(XYC) = 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
04. Ans: (a)
Sol: P (X) = 0.40 P(XUYC) = 0.7
P (Y) = ?
P(XUYC) = P (X) + P(YC) P(X) P(YC)
= P(X) + P(YC) (1  P(X))
0.7 = 0.4 + P(YC) (0.6)
0.7  0.4 = P(YC)
0 .3
 PY C   P(YC) = 0.5
0 .6
P(Y) = 0.5
P(XUY) = P(X) + (Y) P(X) P(Y)
= 0.4 + 0.5  0.4  0.5 = 0.9  0.2 = 0.7
xy
?
x 0 x  y2
y0
05. What is the value of lim
2
(b) 1
(a) 1
(c) 0
(d) Limit does not exist
05. Ans: (d)
xy
x 0 x  y 2
Sol: lt
2
It is in the form
0
0
Put y = mx
As y 0, mx 0
x 2m
m

2
2
x 0 x 1  m
1  m2
lt


For different values of m we get different limits it is not unique therefore limit does not exist.
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GATE_2016_CE
06. The kinematic indeterminacy of the plane truss shown in the figure is
(a) 11
(b) 8
(c) 3
(d) 0
06. Ans: (a)
Sol: Dk = 2j –r ; j = 7; r = 3
Dk = 2 × 7 – 3 = 11
07. As per IS 456-2000 for the design of reinforced concrete beam, the maximum allowable shear stress
(c max) depends on the
(a) grade of concrete and grade of soil
(b) grade of concrete only
(c) grade of steel only
(d) grade of concrete and percentage of reinforcement
07. Ans: (b)
Sol: max depends on grade of concrete as per IS: 456-2000
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SET_2_Forenoon Session
08. 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.
A
D
P
L
Rope
L
Rigid arm
B
L
C
L
Under the action of a concentrated load P at C as shown, the magnitude of tension developed in the
rope is
(a)
3P
(b)
2
P
(c)
2
3P
8
(d)
2P
08. Ans: (b)
VA
Sol: MA = 0
P ×l – TX × l – TY × l = 0
2×T×
T
1
2
A
HA
TY =T sin45o
T
P
l
P
45o
P 2
P

2
2
B
l
C
TX = Tcos45o
09. 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) 7.5 MPa
(c) 5.0 MPa
(d) 3.5 MPa
09. Ans: (d)
Sol: Minimum compressive strength of burnt clay brick = 3.5 MPa
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GATE_2016_CE
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.
Activity
Duration
Activity
Duration
(days)
(days)
A
Inauguration
1
G
Flooring
25
B
Foundation work
7
H
Electrification
7
C
Structural construction-1
30
I
Plumbing
7
D
Structural construction-2
30
J
Wood work
7
E
Brick masonry work
25
K
Coloring
3
F
Plastering
7
L
Handing over function
1
A
1
E
4
C
F
6
8
J
B
2
3
11 L
10 K
D
G
5
H
7
12
I
9
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
10. Ans: (c)
Sol:
38
63
38
E(25)
4
6
63
F(7)
70
70
8
J(7)
C(30)
A(1)
1
B(7)
2
10
(0)
3
D)30)
0
0
1
1
8
8
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11
L(1)
12
I(7)
G(25)
5
38
K(3)
38
7
63
H(7)
63
77
9
70
77
80
80
81
81
70
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Path
SET_2_Forenoon Session
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)
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
(b) becomes half
(c) becomes four-times
(d) remains the same
11. Ans: (d)
Sol: For pure clays, the bearing capacity is independent of the footing width.
12. The relationship between the specific gravity of sand (G) and the hydraulic gradient (i) to initiate
quick condition in the sand layer having porosity of 30% is
(a) G = 0.7i+1
(b) G = 1.43 i –1
(c) G = 1.43 i + 1
(d) G = 0.7 i –1
12. Ans: (c)
Sol: At quick condition, i = ic
i = (G 1) (1 n)
= (G 1) (1 0.3)
= ( G 1) (0.7)
 G = 1.43 i + 1
13. The results of a consolidation test on an undistributed soil, sampled at a depth of 10 m below the
ground level are as follows:
Saturated unit weight
: 16 kN/m3
Pre-consolidation pressure : 90 kPa
The water table was encountered at the ground level. Assuming the unit weight of water as 10
kN/m3, the over-consolidation ratio of the soil is:
(a) 0.67
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(b) 1.50
(c) 1.77
(d) 2.00
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GATE_2016_CE
13. Ans: (b)
Sol: At 10 m depth,  = 10
= [16  10]  10 = 60 kPa
 c 90
OCR =

 1.50
 60
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).
1m
Gate
4m
3m
Weir
Floor
P
10 m
5m
25 m
S
R
5m
5m
Figure-I
Q
40 m
Figure-II
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) Data not sufficient
14. Ans: (b)
Sol: hR = h – hP
2.75 = 4 – hP
hP = 4 – 2.75 = 1.25 m
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SET_2_Forenoon Session
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
km3) drained out from the area is ______
15. Ans: 0.2
Sol: Sy = n – Sr = 25 – 5 = 20% = 0.2
Amount of water drained = Sy  Thickness of aquifer  Surface area
= 0.2 
100
1000 = 0.2 Km3
100 1000
16. Group I contains the types of fluids while Group II contains the shear stress-rate of shear relationship
Yield stress shear stress
of different types of fluids, as shown in the figure
5
4
23
1
Rate of shear
Group-I
Group-II
P. Newtonian fluid
1.
Curve 1
Q. Pseudo plastic fluid
2.
Curve 2
R. Plastic fluid
3.
Curve 3
S. Dilatant fluid
4.
Curve 4
5.
Curve 5
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
16. Ans: (c)
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GATE_2016_CE
17. The atmospheric layer closest to the earth surface is
(a) the mesosphere
(b) the stratosphere
(c) the thermosphere (d) the troposphere
17. Ans: (d)
18. A water supply board is responsible for treating 1500 m3/day of water. A settling column analysis
indicates that an overflow rate of 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_______
18. Ans: 6.9
Sol: Q = 1500 m3/day
v0 = 20 m/day
Total surface area required =
Q
1500
=
= 75 m2
20
vo
Surface area of each setting tank =
75
= 37.5 m2
2
 2
d = 37.5
4
 d = 6.9 m
Diameter of setting tank d = 6.9 m
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 CaCO 3 before supplying
to 4 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_______
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19. Ans: 385.714
Q
75 mg/l
(Q – QB) 4540
Sol: Cin = 420 mg/l
Cout = 75 mg/l
QB
Q = 4  540 = 2160 lit/day
C  Cout
420  75
 = in
100
100 =
420
Cin
= 82.142%
CB= C
= 420
Q
 = 100%
C mix 
75 
SET_2_Forenoon Session
Cout = 0
 = 100%
Cout = 0
(Q – QB)
Cin = 420
(Q  Q B )C out  Q B C B
Q
(Q  Q B )  0  Q B  420
 QB = 385.714 lit/day
2160
Flow that can be by passed = 385.714 Lit/day
20. The sound pressure (expressed in µPa) of the faintest sound that a normal healthy individual can
hear is
(a) 0.2
(b) 2
(c) 20
(d) 55
20. Ans: (C)
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GATE_2016_CE
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
21. Ans: (b)
1
 Eh 3
4
Sol: Radius of relative stiffness,   
2 
12k (1   ) 
Statement -1: False
Directly proportional to modulus of elasticity and also 
( As  increases l decreases)
Statement-2: False
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
22. Ans: (a)
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
23. Ans : (d)
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SET_2_Forenoon Session
24. The reduced bearing of a 10 m long line is N30oE. The departure of the line is
(a) 10.00 m
(b) 8.66 m
(c) 7.52 m
(d) 5.00 m
24. Ans: (d)
Sol: Departure = D = l sin = 10  sin30o = 5.0 m
25. A circular curve of radius R connects two straights with a deflection angle of 60. The tangent length
is
(a) 0.577 R
(b) 1.155 R
(c) 1.732 R
(d) 3.464 R
25. Ans: (a)
Sol: Tangent = T = R tan

 R tan 30 o  0.577 R
2
Q.26 – Q.55 carry Two Mark each
26. Consider the following linear system
x+ 2y –3z = a
2x + 3y + 3z = b
5x + 9y – 6z = c
This system is consistent if a,b and c satisfy the equation
(a) 7a – b –c = 0
(b) 3a + b – c = 0
(c) 3a – b + c = 0
(d) 7a – b + c = 0
26. Ans: (b)
1 2  3 a 


Sol: (AB)   2 3 3 b 
5 9  6 c


(R2 – 2R1); (R3 – 5R1)
a 
1 2  3


  0  1 9 b  2a 
 0  1 9 c  5a 


(R3 – R2)
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a
1 2  3



  0 1 9
b  2a 
0 0
0 (c  b  3a ) 

(c –b–3a) = 0
3a + b – c = 0
27. If f(x) and g(x) are two probability density functions,
x
: a  x  0
a 1

 x
f x      1 : 0  x  a
 a
: otherwise
 0


 x
 a :  a  x  0

x
g x   
: 0xa
a
: otherwise
0


Which of the following statements 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.
27. Ans: (b)
Sol: Ef x  

 xf x dx

a
 x2

  x2

 x dx
=    x dx   
a
a


 a
0
0
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SET_2_Forenoon Session
a
 x3 x2 
  x3 x2 
 
=     
2 0
 3a 2  a  3a
  a2 a2    a2 a2 
   
 
=  
3
2
3
2

 
=0
Egx  
 x2
x2
dx

 a
0 a dx
a
0
a
0
a
 x3 
 x3 
=      
 3a   a  3a 0
 a2  a2
=    
 3 3
=0
 
Vf x   E X 2  EX 
2
a
 x3
  x3

2


E X     x dx   
 x 2 dx
a
a


a 
0
 
0
2
0
a
 x 4 x3 
 x 4 x3 
 
=      
 4a 3   a  4a 3  0
 a3 a3   a3 a3 
=         
4 3  4 3
 
a3
EX 
6
2
Vf x  
a3
6
 
Vg x   E X 2  EX 
 
0
E X2   
a
2
a
x3
x3
dx   dx
a
a
0
0
a
 x4 
 x4 
=      
 4a   a  4a  0
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 a3  a3
= 0      
 4 4
=
2a 3 a 3

4
2
a3
2
Vgx  
28. The angle of intersection of the curves
x2 = 4y and y2 = 4x at point (0,0) is
(a) 0o
(b) 30o
(c) 45o
(d) 90o
28. Ans: (d)
Sol:
Angle between the curves is angle between the tangents at the point of intersection.
29. The area between the parabola x2 = 8y and the straight line y = 8 is ___________
29. Ans: 85.33
Sol:
Y
y=8
(-8, 8)
(8, 8)
2
x = 8y
X
8
Area
= 2  8 y dy
0
8

y 
=2 8
 85.33
3 
2  0
3
2
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30. The quadratic approximation of f(x) = x3 – 3x2 –5 at the point x = 0 is
(a) 3x2 – 6x –5
(b) –3x2 –5
(c) –3x2 + 6x – 5
(d) 3x2 – 5
30. Ans: (b)
Sol: Quadratic Approximation
2

x  0
= f 0   x  0  f ' 0  
f ' ' 0 
2!
f x   x 3  3x 2  5  f 0   5
f ' x   3x 2  6 x  f ' 0   0
f = 6x – 6  f (x) = –6

x2
 6
Equation is =  5  x 0  
2
 3 x 2  5
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?
y
x
z
(a) 0.00
(b) 0.25
(c) 0.50
(d) 1.00
31. Ans: (c)
Sol: For hydrostatic state  = 0.5
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32. For the stress state (in MPa) shown in the figure, the major principal stress is 10 MPa.
5
5
5

The shear stress  is
5
(a) 10.0 MPa
(b) 5.0 MPa
(c) 2.5 MPa
(d) 0.0 MPa
32. Ans: (b)
Sol:
Major principal stress
10  1 
x  y
2
 x  y 
   2xy
 
 2 
2
55
55
2

 
   xy
2
 2 
2
 xy = 5 MPa
33. The portal frame shown in figure is subjected to a uniformly distributed vertical load w (per unit
length)
w
Q
R
L/2
S
P
L
The bending moment in the beam at the joint ‘Q’ is
(a) Zero
ACE Engineering Academy
(b)
wL2
hogging 
24
(c)
wL2
hogging 
12
(d)
wL2
sagging 
8
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33. Ans: (a)
VQ
Sol: DS = 0 Determinate structure
MQ = 0
Q MQP
MQP = 0
l/2
MQ = 0
P
HP =0
VP
FBD of member QP
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 1 m, 0.05 m2 and 30 × 106 N/m2, respectively, and
W be 100 kN.
A,E
45o
Q
45o
A,E
90o
2A,E
S
90o
A,EL
45o
45o
R
A,E
W
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/m2, Shortening = 0.0118 m
(b) Compressive force = 14.14 kN; stress = 141.4 kN/m2, Extension = 0.0118 m
(c) Compressive force = 100 kN; stress =
1000 kN/m2, Shortening = 0.0417 m
(d) Compressive force = 100 kN; stress =
1000 kN/m2, Extension = 0.0417 m
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34. Ans: (c)
FQS
F
W

 SR
sin 90 sin 45 sin 45
Sol:
FQS 
W
FQS
 FSR
2
45o
45o
FSR
45o
W
FQS
90o
45o
FSR
Lami’s triangle
FQS
sin 45
FQR 

FQR
sin 90

FQP
FQP
B
FQR
sin 45
Q
o
o
45
W
1

W
2 sin 45
45
FQS
90o
FQP
Lami’s triangle
FQR = 100 kN (Comp.); stress =
100
2000

kN / m 2
0.05  2
2
 1000 kN / m 2
Shortening 
W  2 100  10 3  1  2 1000  10 3


= 0.0417 m
AE
2  0.05  30  10 6
30  10 6
35. A haunched (varying depth) reinforced concreted 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 _____
5m
X
10 kN/m
400 m
600 m
X
effective span = 20 m
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35. Ans: 65
wx
 w 
Sol: BM@ section XX = M = 
 (x) 
2
 2 

2
10  20
(5)  10  5  2.5
2
= 375 kN-m
SF @ X =
10 20
5  10 = 50 kN
2
(SF)X = Vu 
= 50 
Mu
tan()
d
0.2 m
10 m

375  0.2 


0.5  10 
Design SF, (SF)x = 50 + 15 = 65 kN
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 _________
11.25 kN
P
Q
150 mm
36. Ans: 3
11.25 kN
S
R
150 mm
150 mm
Sol:
f cr 
M 11.25  10 3  150

 3MPa
Z
 150  150 2 


6


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37. Two bolted plates under tension with alternative arrangement of bolt holes are shown in figure 1
and 2. The hole diameter, pitch, and gauge length are d, p and g respectively.
P
d
P
Figure 1
g
P
P
P
Figure 2
Which one of the following conditions must be ensured to have higher net tensile capacity of
configuration shown in figure 2 than that shown in figure 1?
(a) p 2  2gd
(b) p 2  4gd
(c) p 2  4gd
(d) p  4gd
37. Ans: (c)
Sol: The net tensile capacity of the plate Tdn = 0.9 An fu/γm1
The net tensile capacity of the plate configuration -1, Tdn1 = 0.9 An1 fu/γm1
The net tensile capacity of the plate configuration -1, Tdn2 = 0.9 An2 fu/γm1
For same material and size of the plate, the net tensile capacity of the plate depends on Net
effective sectional area of plate (An).
Net effective sectional area of plate configuration 1 An1 = (B×t-d×t)
Net effective sectional area of plate configuration 2 An2 = (B×t-2×d×t +p2t/4g = B×t- d×t - d×t +p2t/4g)
[Assume width of plate = B, Thickness of plate = t; Diameter of bolt hole = d; staggered pitch
distance = p and Gauge distance = g]
As per Question Tdn2 > Tdn1 => An2 > An1
=> An2 > An1
=> (B×t-d×t -d×t + p2t/4g) > (B×t-d×t)
=> -d×t + p2t/4g > 0
=> p2t/4g > d × t
=> p2 > 4gd
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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 (MP, 2Mp) as shown.
P
2L/3
2MP
Mp
L
L
The minimum value of load (P) at which the beam would collapse (ultimate load) is
(a) 7.5 Mp /L
(b) 5.0 Mp /L
(c) 4.5 Mp /L
(d) 2.5 Mp /L
38. Ans: (a)
Sol:  
2
4
  1
3
3
1 
2l/3 P

2
4l/3

 
MP
2MP

P    2M P   2M P   2M P 1  M P 1
1
2MP
2MP
Mechanism-I
2

P    4M P   3M P  
3
2
= 5.5 Mp
P
8.25M P

2l/3 P

2
1  
3
2
  5M P 
3
P
7 .5 M P

ACE Engineering Academy
1


2MP

P×1 = 2 Mp + Mp + Mp + Mp
P
4l/3
MP
MP

MP
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39. The activity-on-arrow network of activities for a construction project is shown in figure. The
durations (expressed in days) of the activities are mentioned below the arrows.
Q
3
P
2
10
R
4
20
T
5
30
5
3
X 90
2
80
U
3
50
40
W
3
70
V
2
60
The critical duration for this construction project is
(a) 13 days
(b) 14 days
39. Ans: (c)
T(5)
30
Sol:
(c) 15 days
Q(3)
70
(0)
W(3)
U(3)
10 P(2) 20 R(4) 40 (0)
(d) 16 days
80 X(2) 90
50
(0)
60 V(2)
S(3)
Path
Duration
P–Q–T–W–X
:
2 + 3 + 5 + 3 + 2 = 15
P–Q–Dummy–U–W–X
:
2+3+0+3+3+2 = 13
P–R–Dummy–U–W–X
:
2 + 4 + 0 +3+3+2 = 14
P–R–Dummy–V–X
:
2 + 4 + 0 +2+2 = 10
P–S–V–X
:
2+3+2+2=9
Critical path duration = 15 days
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40. The seepage occurring through an earthen dam is represented by a flownet 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 cm 3/s per m length of the dam) through
the earthen dam is_______.
40. Ans: 500
Sol: Nf = 20, Nd = 10,
hf = 5m
K = 3 mm/min

3
m / sec
1000  60
Q  Kh f .

Nf
Nd
3
20
= 5 10–4 m3/sec/m
5
1000  60
10
= 500 cm3/sec/m
41. The soil profile at a site consists of a 5 m thick sand layer underlain 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 _______
bulk=16.5 kN/m3
1m
Sand
4m
3m
sat = 19 kN/m3, w = 9.81 kN/m3,=32o
C– Soil
sat = 18.5 kN/m3, w = 9.81 kN/m3,
c= 25 kN/m2 , =24o
A
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41. Ans: 69.67
Sol: Taking 2 = 24 (at point A)
K a2 
1  sin 
 0.422
1  sin 
At point A,  = 1 + 4 + 3
 v = 1  16.5 + 4 (19  9.81) + 3(18.5  9.81)
= 79.33 kPa
p a  K a 2  v  2C 2 K a 2   w h
 0.422  79.33  2  25 0.422  9.81  7
= 69.67 kPa
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
42. Ans: (b)
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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 ______
600 mm
300 mm
P
43. Ans: 0.0455
Soil 1
Soil 2
150 mm
150 mm
Q
Sol: 1 ml = 1 cm3
Q
200
cm 3 / sec
15  60
A = 80 cm2,
h = 600  300 = 300 mm
= 30 cm
L = 150 + 150 = 300 mm
= 30 cm
Q = K. i. A
200
h
 K .A
15  60
L
K
30
 80
30
K = 2.778 103 cm/sec
Where,
K = Equivalent or Average permeability
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K = 2.778 103 cm/sec
K = 2.778  102 mm/sec
K
Z1  Z 2
Z1 Z 2

K1 K 2
2.778  102 =
7500 
150  150
150 150

0.02 K 2
150
= 107. 99  102
K2
K2 = 0.0455 mm/sec
44. A 4m 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,  = 28.63o, sat = 19 kN/m3, bulk = 17 kN/m3 and w = 9.81 kN/m3. The values of
bearing capacity factors for different  are given below:

Nc
Nq
N
15o
12.9
4.4
2.5
20o
17.7
7.4
5.0
25o
25.1
12.7
9.7
30o
37.2
22.5
19.7
1.5 m
5.5 m
4m
Using Terzaghi’s bearing capacity equation and a factor of safety F s = 2.5, the net safe bearing
capacity (expressed in kN/m2) for local shear failure of the soil is _______
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44. Ans: 298.50
Sol: For local shear failure, use Cm and m to compute bearing capacity
2
2
C m  .C   35  23.33 kN / m 2
3
3
tan  m 
2
tan 
3
tan  m 
2
 tan 28.63
3
m = 19.998 say 20
For m = 20, N C  17.7
N q  7.4
N   5 (from table given)
q ns 




1
C m N C  D N q  1  0.5BN 
F

1
23.33  17.7  17  1.57.4  1  0.5  17  4  5
2 .5
qns = 298.50 kN/m2
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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 30o 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 ________
.
Water jet
30o
Plate
20 mm
200 mm
F
45. Ans: 7.85
Sol:
Hinge
O
Plate
100 mm
FH
200 mm
F
F.B.D
Moment of forces about hinge point O, for equilibrium in vertical position
F  200 = FH  100
F
FH AV 2 .sin 30 o

2
2

1
 1000  (0.02) 2  (10) 2  sin 30 o 
4
2
= 7.85 N
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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 _______.
46. Ans: 60
Sol: Given 1hr UHG
2 hr SHG ord @ 1 PM = ?
By method of super position 1 hr UHG  2 hr UHG
Time
Ist storm
IInd storm
1hr UHG ord
1hr delayed
2hr DRH ord
2hr UHG ord
2hr DRH/2
1hr UHG ord
10 AM
0
–
11 AM
3
0
12 PM
12
3
1 PM
8
12
20
20/2 = 10
2hr UHG ord @ 1 PM = 10 m3/sec
Runoff R (excess rainfall) = 4 cm
2hr DRH ord @ 1 PM = 10 4 = 40 m3/sec
(R = 4 cm)
2hr SHG ord @ 1 PM = 40 + 20 = 60 m3/sec
Flow of river@1 PM = 60 m3/sec
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___________
47. Ans: 0.203
b=3m
Q = 6 m3/s
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yp = 0.5 m
A1 = 3  0.5 = 1.5 m2
V1 
Q
6

 4 m/s
A 1 1 .5
E1 = E2 + (z)Max
E 1  y1 
V12
42
 0 .5 
2g
2  9.81
= 1.315 m
1/ 3
 Q2 
y c 2   2 
 gb 
E2 = Ec =
 62 

2
 9.81 3 
1/ 3
= 0.74 m
3
 0.74 = 1.112 m
2
1.315 = 1.112 + (Z)max
(Z)max = 1.315  1.112
= 0.203 m
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/m 3 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 ___________.
48. Ans: 6714.6
Sol:
Velocity at the jet = V  2gH
 2  10  500  100 m/s
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V2
Velocity head at the jet 
= 500 m
2g
Head loss due to friction 
5
 500 = 25 m
100
Net head after losses = 500  25 = 475 m
Discharge = AV 

(0.15)2  100 = 1.767 m3/s
4
Power developed = o. . Q. Hnet
= 0.8  10  1.767  475= 6714.6 kW
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
49. Ans: (d)
Sol: actual velocity va =
but va =
100
distance
=
= 1 m/day
time
100
v
 v = va  

v = 1  0.15 = 0.15 m/day
but v = Ki
K=
;
i=
3
100
0.15
0.15 100
v
=
=
= 5 m/day
i
3 /100
3
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50. A sample of water has been analyzed for common ions and results are presented in the form of a bar
diagram as shown.
meq/L
0
6.35
2+
Ca
Na+
Mg
HCO 3
meq/L 0
4.10
2.65
2+
SO 24 
3.30
6.85
+
K
Cl–
3.90
6.75
The non-carbonate hardness (expressed in mg/L as CaCO3) of the sample is
(a) 40
(b) 165
(c) 195
(d) 205
50. Ans: (a)
Sol: Ca = 2.65 m.eq/lit
Mg = 1.45 m.eq/lit
TH = 2.65  50 + 1.45  50 = 205 mg/l as CaCO3
HCO3 = 3.3 m. eq/lit
TA = 3.3  50 = 165 mg/l as CaCO3
TH > TA
CH = TA = 165 mg/l
NCH = TH – CH = 205 – 165 = 40 mg/l as CaCO3
Non carbonacreous hardness = 40 mg/l as CaCO3
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 _________
51. Ans: 67.979
Sol: L60 = L30 – 20 log10
L60 = 74 – 20 log10
60
30
60
= 67.979 dB
30
Noise reading at a distance 60 m = 67.979 dB
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52. For a wastewater sample, the three-day biochemical oxygen demand at incubation temperature of


20oC BOD 3day , 20o 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 20oC BOD 5day , 20o c would be _________
52. Ans: 276.15
Sol: y 320
o
C
 L 0 [1  e  K 20 3 ]
200 = L 0 1  e 0.223   L0 = 413.95 mg/l
y 520
o
C
 L 0 [1  e  K 20 5 ]
= 413.95[1–e–0.22 5] = 276.15 mg/l
5 day BOD@ 200C = 276.15 mg/l
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 10s. Pedestrain crossings at this 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
53. Ans: (a)
Sol:
y = y1 + y2 + y3
= 0.3 + 0.25 + 0.25 = 0.8
Lost time, L = 10 sec
Cycle time, C o 
1.5L  5 1.5  10  5
= 100s

1 y
1  0 .8
Green times:
y1
(C o  L )
y
0 .3

(100  10)  33.75  34
0 .8
0.25
G2 
(100  10)  28.125  28
0 .8
G3 = G2 = 28
G1 
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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________
54. Ans: 92.14
Sol: Downward gradient, N =  3%
f = 0.35
Sb 
But
(vi ) 2  (v f ) 2
2g ( f  N )
v i  100 
v f  50 
Sb 
5
 27.77 m/s
18
5
= 13.88 m/s
18
27.77 2  13.88 2
3 

2  9.81 0.35 

100 

Sb = 92.14 m
55. A tall tower was photographed from an elevation of 700 m above the datum. The radial distance 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 __________
55. Ans: 120.4 m
Sol: Relief Displacement = d = r2 – ro = 112.50 – 82.40
= 30.10 m
h2 
d (H  h 1 ) 30.1(700  250)

r2
112.50
= 120.40 m
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