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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 x100 100
xy xy
100
2xy
100
= 2% of xy
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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.
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 (b) 3:4 (c) 6:7 (d) 4:3
2000 2001 2002 2003 2004 2005 2006
6.58 8 8
7.56.5
4
6
89109.59
7 P Q
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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 Q
6% of 8 : 4% of 96
8100 :
49
100
4 : 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 (b) III only (c) I, II and III (d) I only
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08. Ans: (b)
Sol: 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.
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 (c) 0.50 x2 (d) 0.25 x2
09. Ans: (D)Sol: Base perimeter of square pyramid = x = p
Slant height 2x
l
Lateral surface area of pyramid
21 Base perimeter (p) slant height (l)
BH
E
M
H
x/2
x/4x/4
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2x
p21
l
2x
x21
4x2
= 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
6 = 2 pages
If Bharath read 12 number of pages in 4 hrs
1 hr =12
4 = 3 pages
Option ‘a’
it is not
1 hrs
Ananth Bharath
Read2 pages
Not read10 pages
Read3 pages
Not read9 pages
10 : 9
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Option ‘b’
it is not
Options ‘c’
After 3 hours is the number of pages to be read by Ananth, twice that to be read by Bharath.
3 hrs
Ananth Bharath
Read6 pages
not Read6 pages
Read9 pages
not Read3 pages
2 : 1
2 hrs
Ananth Bharath
Read4 pages
Not read8 pages
Read6 pages
Not read6 pages
8 : 6
4 : 3
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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 average2
5653 = 54.5 kmph
02. The optimum value of the function f(x) = x2 4x + 2 is
(a) 2 (maximum) (b) 2 (minimum) (c) 2 (maximum) (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
....
3
x3sin
2
x2sin
1
xsin.....
23
x3cos21
xcos2
4xf
The convergence of the above Fourier series at x = 0 gives
(a)
1n
2
2 6n
1(b)
12n
1 2
1n2
1n
(c)
1n
2
2 81n2
1(d)
1n
1n
41n2
1
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03. Ans: (c)
Sol:
...
51
31
112
40f 222
....5
1
3
11
2
42
0f0f22
.....5
1
3
11
2
42 22
8...
5
1
3
11
2
22
1n
2
2 81n2
1
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04. X and Y are two random independent events. It is known that P(X) = 0.40 and P(XYC) = 0.7.
Which one of the following is the value of P(XY) ?
(a) 0.7 (b) 0.5 (c) 0.4 (d) 0.3
04. Ans: (a)
Sol: P (X) = 0.40 P(XUYC) = 0.7P (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)
CYP6.0
3.0 P(YC) = 0.5
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
05. What is the value of ?yx
xylim
220x0y
(a) 1 (b) 1 (c) 0 (d) Limit does not exist
05. Ans: (d)
Sol:220x yx
xylt
It is in the form00
Put y = mx
As y0, mx 0
222
2
0x m1m
m1xmx
lt
For different values of m we get different limits it is not unique therefore limit does not exist.
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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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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.
Under the action of a concentrated load P at C as shown, the magnitude of tension developed in the
rope is
(a)2
P3(b)
2
P(c)
8
P3(d) P2
08. Ans: (b)
Sol:MA = 0
P ×l – TX × l – TY × l = 0
2 × T × P2
1
2
P
2
2PT
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
L
CL L
Rigid arm
RopeL
A D
B
P
lB C
45o
TX = Tcos45o
T
TY =T sin45o
Pl
HA
VA
A
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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
(days)
Activity Duration
(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
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:
4 6 8
5 7 9
10 11 121 2 3
A BC
E F
DG H
I
J
K L
4 6 8
5 7 9
10 11 121 2 3A(1)
C(30)
E(25) F(7)
D)30)
G(25) H(7)I(7)
J(7)
K(3) L(1)
0 0 1 1 8 8
38 38 63 63 70 70
77 77 80 80 81 81
70 7063 6338 38
B(7)(0)
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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)
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 (b) 1.50 (c) 1.77 (d) 2.00
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13. Ans: (b)
Sol: At 10 m depth, = 10
= [16 10] 10 = 60 kPa
OCR = 50.160
90c
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).
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
1 m
3 m
10 m 5 m 25 mP 5 m
Gate
Weir
Floor
Figure-I
40 m
R5 m
Q
4 m
S
Figure-II
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15. Water table of an aquifer drops by 100 cm over an area of 1000 km2. 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
=100
0.2 1000100 1000
= 0.2 Km3
16. Group I contains the types of fluids 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. 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)
5
423
1Yie
ld s
tres
s sh
ear
stre
ss
Rate of shear
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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 =ov
Q =
1500
20 = 75 m2
Surface area of each setting tank =75
2 = 37.5 m2
2d4
= 37.5
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 CaCO3 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
Sol: Cin = 420 mg/l
Cout = 75 mg/l
Q = 4 540 = 2160 lit/day
= in out
in
C C100
C
=
420 75100
420
= 82.142%
= 100%
Q
CQC)QQ(C BBoutB
mix
2160
420Q0)QQ(75 BB QB = 385.714 lit/day
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)
Q(Q – QB)
(Q – QB)
Q 75 mg/l4540
QB
CB= C = 420 = 100%
Cout = 0
Cout = 0
Cin = 420
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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)
Sol: Radius of relative stiffness,4
1
2
3
)1(k12
Eh
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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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 = R577.030tanR2
tanR o
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)
Sol:
c695
b332
a321
)AB(
(R2 – 2R1); (R3 – 5R1)
a5c910
a2b910
a321
(R3 – R2)
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)a3bc(000
a2b910
a321
(c –b–3a) = 0
3a + b – c = 0
27. If f(x) and g(x) are two probability density functions,
otherwise:0
ax0:1a
x
0xa:1a
x
xf
otherwise:0
ax0:a
x
0xa:a
x
xg
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:
dxxxfxfE
=
0
a
a
0
22
dxxax
dxxax
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=a
0
230
a
23
2
x
a3
x
2
x
a3
x
=
2a
3a
2a
3a 2222
= 0
a
0
20
a
2
dxax
dxax
xgE
=a
0
30
a
3
a3x
a3x
=3a
3a 22
= 0
22 XEXExfV
0
a
a
0
23
23
2 dxxa
xdxx
a
xXE
=a
0
340
a
34
3x
a4x
3x
a4x
=
3a
4a
3a
4a 3333
6
aXE
32
6
axfV
3
22 XEXExgV
dxa
xdx
a
xXE
a
0
30
a
32
=a
0
40
a
4
a4x
a4x
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= 4a
4a
033
=2a
4a2 33
2a
xgV3
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:
Area = 8
0
dyy82
= 33.85
2
3y
82
8
0
2
3
(-8, 8) (8, 8)
x2 = 8yX
Y
y = 8
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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
= 0''f!2
0x0'f0x0f
2
50f5x3xxf 23
00'fx6x3x'f 2
f = 6x – 6 f (x) = –6
Equation is = 62
x0x5
2
5x3 2
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 (b) 0.25 (c) 0.50 (d) 1.00
31. Ans: (c)
Sol: For hydrostatic state = 0.5
y
x
z
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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 (b) 5.0 MPa (c) 2.5 MPa (d) 0.0 MPa
32. Ans: (b)
Sol:
Major principal stress
2xy
2
yxyx1 22
10
2xy
2
2
55
2
55
xy = 5 MPa
33. The portal frame shown in figure is subjected to a uniformly distributed vertical load w (per unit
length)
The bending moment in the beam at the joint ‘Q’ is
(a) Zero (b) hogging24
wL2
(c) hogging12
wL2
(d) sagging8
wL2
5
5
5
5
Q R
SP
L
L/2
w
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33. Ans: (a)
Sol: DS = 0 Determinate structure
MQ = 0
MQP = 0
MQ = 0
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.
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
45o
45o
90oS
A,EA,E
R45o
45o
A,E A,EL
Q
W
90o
2A,E
HP =0 VP
MQP
P
VQ
l/2
Q
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34. Ans: (c)
Sol:45sin
F
45sin
F
90sin
W SRQS
SRQS F2
WF
45sin
F
90sin
F
45sin
F QPQRQS
W45sin
1
2
WFQR
FQR = 100 kN (Comp.); stress = 2m/kN2
2000
205.0
100
2m/kN1000
Shortening6
3
6
3
1030
101000
103005.02
2110100
AE
2W
= 0.0417 m
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 _____
effective span = 20 m
400 m
600 m
5m 10 kN/mX
X
45o
FQS
B
45o
90o
Q
FQP
FQP
FQR
Lami’s triangle
90o
45o
45o
45o45o
W
Lami’s triangle
FQS
FSR
FQS FSR
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35. Ans: 65
Sol: BM@ section XX = M =2
wx)x(
2
w 2
5.2510)5(2
2010
= 375 kN-m
SF @ X =2
20105 10 = 50 kN
(SF)X = )tan(d
MV u
u
=
10
2.0
5.0
37550
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 _________
36. Ans: 3
Sol:
6
150150
1501025.11
Z
Mf
2
3
cr 3MPa
150 mm 150 mm 150 mm
SPRQ
11.25 kN 11.25 kN
0.2 m
10 m
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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.
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) gd2p 2 (b) gd4p2 (c) gd4p 2 (d) gd4p
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
P Pd
Figure 1
P
g PP
Figure 2
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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.
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: 13
4
3
2
21
1P1PPP MM2M2M2P
2M3M4
3
2P PP
= 5.5 Mp
PM25.8
P
3
21
P×1 = 2 Mp + Mp + Mp + Mp
PM53
2P
PM5.7
P
2MP
LL
Mp
2L/3
P
1
2MP
2MP
2MP
4l/32l/3 P
MP
Mechanism-I
MP
2MP
MP
4l/32l/3 P
1
MP
Mechanism-II
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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.
The critical duration for this construction project is
(a) 13 days (b) 14 days (c) 15 days (d) 16 days
39. Ans: (c)
Sol:
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
10 20 40 50
30 70
80 90
60
T5
Q3
P2
R4
U3
W3
X2
V2
53
10 20 40 50
30 70
80 90
60
T(5)
Q(3)
P(2) R(4)
U(3)
W(3)
X(2)
V(2)S(3)
(0)
(0)
(0)
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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 cm3/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
sec/m601000
3
d
ff N
N.KhQ
10
205
601000
3
= 5 10–4 m3/sec/m
= 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 _______
1m
4m
3m
A
Sand
sat = 19 kN/m3, w = 9.81 kN/m3,=32o
sat = 18.5 kN/m3, w = 9.81 kN/m3,c= 25 kN/m2 , =24o
bulk=16.5 kN/m3
C– Soil
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41. Ans: 69.67
Sol: Taking 2 = 24 (at point A)
422.0sin1
sin1K
2a
At point A, = 1 + 4 + 3v = 1 16.5 + 4 (19 9.81) + 3(18.5 9.81)
= 79.33 kPa
hKC2Kp wa2vaa 22
781.9422.025233.79422.0
= 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 ______
43. Ans: 0.0455
Sol: 1 ml = 1 cm3
sec/cm6015
200Q 3
A = 80 cm2,
h = 600 300 = 300 mm
= 30 cm
L = 150 + 150 = 300 mm
= 30 cm
Q = K. i. A
A.L
hK
6015
200
8030
30K
K = 2.778 103 cm/sec
Where,
K = Equivalent or Average permeability
300 mm
QP
600 mm
Soil 1 Soil 2
150 mm 150 mm
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K = 2.778 103 cm/sec
K = 2.778 102 mm/sec
2
2
1
1
21
K
Z
K
ZZZ
K
2.778 102 =
2K
150
02.0
150150150
2K
1507500 = 107. 99 102
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
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 _______
5.5 m
4 m
1.5 m
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44. Ans: 298.50
Sol: For local shear failure, use Cm and m to compute bearing capacity
2m m/kN33.2335
3
2C.
3
2C
tan tan3
2m
tan 63.28tan3
2m
m = 19.998 say 20
For m = 20, 7.17NC
4.7N q
5N (from table given)
BN5.01NDNCF
1q qCmns
54175.014.75.1177.1733.235.2
1
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 ________
.
45. Ans: 7.85
Sol:
Moment of forces about hinge point O, for equilibrium in vertical position
F 200 = FH 100
2
30sin.AV
2
FF
o2H
2
130sin)10()02.0(
41000 o22
= 7.85 N
200 mm
Plate
20 mm
30o
Water jet
F
200 mm
Plate
F
100 mm
FH
Hinge
O
F.B.D
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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
1hr UHG ord
IInd storm
1hr delayed
1hr UHG ord
2hr DRH ord 2hr UHG ord
2hr DRH/2
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 = 3 m
Q = 6 m3/s
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yp = 0.5 m
A1 = 3 0.5 = 1.5 m2
5.1
6
A
QV
11 4 m/s
E1 = E2 + (z)Max
81.92
45.0
g2
VyE
221
11
= 1.315 m
3/1
2
23/1
2
2
2c 381.9
6
gb
Qy
= 0.74 m
E2 = Ec =2
3 0.74 = 1.112 m
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/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 ___________.
48. Ans: 6714.6
Sol:
Velocity at the jet = gH2V
500102 100 m/s
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Velocity head at the jetg2
V2
= 500 m
Head loss due to friction 500100
5 = 25 m
Net head after losses = 500 25 = 475 m
Discharge =4
AV (0.15)2 100 = 1.767 m3/s
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 =time
cetandis=
100
100 = 1 m/day
but va =v v = va
v = 1 0.15 = 0.15 m/day
but v = Ki ; i =3
100
K =i
v =
0.15
3 /100 =
0.15 100
3
= 5 m/day
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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.
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 – 10
6020log
30
L60 = 74 – 10
6020log
30 = 67.979 dB
Noise reading at a distance 60 m = 67.979 dB
0 2.65 4.10 6.35 6.85
K+Na+Mg2+Ca2+
3HCO 2
4SO Cl–
3.30 3.90 6.750meq/L
meq/L
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52. For a wastewater sample, the three-day biochemical oxygen demand at incubation temperature of
20oC c20,day3 oBOD 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 c20,day5 oBOD would be _________
52. Ans: 276.15
Sol: ]e1[Ly 3K0
C203
20o
200 = 0.22 30L 1 e L0 = 413.95 mg/l
]e1[Ly 5K0
C205
20o
= 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,8.01
5105.1
y1
5L5.1Co
= 100s
Green times:
)LC(y
yG o
11
3475.33)10100(8.0
3.0
28125.28)10100(8.0
25.0G 2
G3 = G2 = 28
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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
)Nf(g2
)v()v(S
2f
2i
b
But 18
5100v i 27.77 m/s
18
550v f = 13.88 m/s
100
335.081.92
88.1377.27S
22
b
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
50.112
)250700(1.30
r
)hH(dh
2
12
= 120.40 m