Ce gate-2022-forenoon-solution

Solution
12-02-2022 | FORENOON SESSION
Detailed
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Ph:
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APTITUDE
1. You should _________ when to say _________.
(a) no / no (b) no / know
(c) know / know (d) know / no
Sol: (d)
2 Two straight lines pass through the origin (x0, y0)
= (0,0). One of them passes through the point
(x1, y1) = (1, 3) and the other passes through the
point (x2, y2) = (1,2).
What is the area enclosed between the straight
lines in the interval [0, 1] on the x-axis?
(a) 0.5 (b) 1.0
(c) 1.5 (d) 2.0
Sol: (a)
Equation of first straight line passing through (0,
0) and (1, 3)
y – y1 =
 


 

 
2 1
1
2 1
y y
(x x )
x x
y – 0 =



3 0
(x 0)
1 0
y = 3x
Equation of second stragith line passing through
(0, 0) and (1, 2)
y – y1 =



2 1
1
2 1
y y
(x x )
x x
y – 0 =



2 0
(x 0)
1 0
y = 2x
x
y = 2x
y = 3x
x = 1
y
Area = 

1
0
(3x 2x) dx
=
 
 
 
 
 
1
2
2
0
3x 1
x
2 2
= 0.5
3. If
p : q = 1 : 2
q : r = 4 : 3
r : s = 4 : 5
and u is 50% more than s, what is the ratio
p : u ?
(a) 2 : 15 (b) 16 : 15
(c) 1 : 5 (d) 16 : 45
Sol: (d)
p : q = 1 : 2 ...(i)
q : r = 4 : 3 ...(ii)
r : s = 4 : 5 ...(iii)
Multiply (i) by 8
(ii) by 4
(iii) by 3
p : q = 8 : 16 ...(a)
q : r = 16 : 12 ...(b)
r : s = 12 : 15 ...(c)
From (a) (b) and (c)
p : q : r : s = 8 : 16 : 12 : 15
‘ Let s = 15
So u = 1.5 × 15 = 22.5
Now
p
u
=
8 80 16
= =
22.5 225 45
4. Given the statements:
 P is the sister of Q.
 Q is the husband of R.
 R is the mother of S.
 T is the husband of P.
Based on the above information, T is _________
of S.
(a) the grandfather (b) an uncle
(c) the father (d) a brother
Sol: (b)
male
female
husband
brother/sister
mother/father

Solution
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Now the family tree for given question
T P Q R
S
From the family tree, its clear that T is an
uncle-in-law of S
5. In the following diagram, the point R is the center
of the circle. The lines PQ and ZV are tangential
to the circle. The relation among the areas of the
squares, PXWR, RUVZ and SPQT is
S P
T Q R
U
V
Z
W
X
(a) Area of SPQT = Area of RUVZ = Area of
PXWR
(b) Area of SPQT = Area of PXWR – Area of
RUVZ
(c) Area of PXWR = Area of SPQT – Area of
RUVZ
(d) Area of PXWR = Area of RUVZ – Area of
SPQT
Sol: (b)
From PQR by Pythagoras theorem,
PR2 = PQ2 + QR2
Also, QR = RZ,
 PR2 = PQ2 + RZ2
 Area of PXWR = Area of SPQT + Area of
RUVZ
 Area of SPQT = Area of PXWR – Area of
RUVZ
6. Healthy eating is a critical component of healthy
aging. When should one start eating healthy? It
turns out that it is never too early. For example,
babies who start eating healthy in the first year
are more likely to have better overall health as
they get older.
Which one of the following is the CORRECT logical
inference based on the information in the above
passage?
(a) Healthy eating is important for those with
good health conditions, but not for others
(b) Eating healthy can be started at any age,
earlier the better
(c) Eating healthy and better overall health are
more correlated at a young age, but not
older age
(d) Healthy eating is more important for adults
than kids
Sol: (b)
7. P invested Rs 5000 per month for 6 months of a
year and Q invested Rs x per month for 8 months
of the year in a partnership business. The profit
is shared in proportion to the total investment
made in that year.
If at the end of that investment year, Q receives
4
9
of the total profit, what is the value of x (in
Rs)?
(a) 2500 (b) 3000
(c) 4687 (d) 8437
Sol: (b)
Let total profit = K
So, profit by Q =
4
K
9
Profit by P =  
4 5K
K K
9 9
Profit by P
Profit by Q
=
 
 
 
5
4
5
4
=
P P
Q Q
C T
C T
5
4
=


5000 6
x 8
x =
 

5000 6 4
8 5
x = 3000

Solution
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8.
11
9
7
14
3
2
4
3 4 5 6 7 8 9
0
2
4
6
8
10
12
14
16
Marks
Frequency
The above frequency chart shows the frequency
distribution of marks obtained by a set of students
in an exam.
From the data presented above, which one of the
following is CORRECT?
(a) mean > mode > median
(b) mode > median > mean
(c) mode > mean > median
(d) median > mode > mean
Sol: (b)
1 1 2 2 7 7
1 2 3 7
f x + f x .....+ f x
Mean =
f + f + f .....+ f
3×3 + 9× 4 +11×5 + 7× 6 +14×7 + 2×8 + 4×9
=
3 + 9 +11+ 7 +14 + 2+ 4
292
= = 5.84
50
Median =
   
   
   
th th
n n
+ +1
2 2
2
=
   
   
   
th th
50 50
+ +1
2 2
2
=
th th
25 + 26
2
=
6 + 6
= 6
2
Mode = frequently occured observation (data with
high frequency)
= 7
Mode > Medium > Mean
9. In the square grid shown on the left, a person
standing at P2 position is required to move to P5
position.
The only movement allowed for a step involves,
“two moves along one direction followed by one
move in a perpendicular direction”. The permissible
directions for movement are shown as dotted
arrows in the right.
For example, a person at a given position Y can
move only to the positions marked X on the right.
Without occupying any of the shaded squares at
the end of each step, the minimum number of
steps required to go from P2 to P5 is
1 2 3 4 5
P
Q
R
S
T
X X
X
X
X
X
Y
X X
Example: Allowed steps for a person at Y
(a) 4 (b) 5
(c) 6 (d) 7
Sol: (b)
Minimum number of steps required are:
P2  Q4  S3  T5  R4  P5
10.
Consider a cube made by folding a single sheet
of paper of appropriate shape.

Solution
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The interior faces of the cube are all blank.
However, the exterior faces that are not visible in
the above view may not be blank.
Which one of the following represents a possible
unfolding of the cube?
(a) (b)
(c) (d)
Sol: (d)
Black edge is perpendicular to given line and white
shaded edge is parallel to given line so only
possibility is option (d).
TECHNICAL
11. Consider the following expression:
z = sin(y + it) + cos(y – it)
where z, y, and t are variables, and i = –1 is
a complex number. The partial differential equation
derived from the above expression is
(a)
 
 
2 2
2 2
z z
+ = 0
t y
(b)
 
 
2 2
2 2
z z
– = 0
t y
(c)
 
 
z z
– i = 0
t y
(d)
 
 
z z
+i = 0
t y
Sol: (a)
z = sin(y + it) + cos(y – it)


z
y
= cos(y + it) – sin(y – it)


2
2 2
z
y
= –sin(y + it) – cos(y – it)

 

2
2 2
z
z
y
…(i)


z
t
= icos(y + it) + isin(y – it)


2
2 2
z
t
= +sin(y + it) + cos(y – it) = z …(ii)
Adding (i) and (ii)
 
 
 
2 2
2 2 2 2
z z
0
y t
12. For the equation
 
 
 
3/2
3
2
3
d y dy
+ x + x y = 0
dx
dx
the correct description is
(a) an ordinary differential equation of order 3
and degree 2.
(b) an ordinary differential equation of order 3
and degree 3.
(c) an ordinary differential equation of order 2
and degree 3.
(d) an ordinary differential equation of order 3
and degree 3/2.
Sol: (a)
 
 
 
3/2
3
2
3
d y dy
+ x + x y = 0
dx
dx
Power of
 
 
 
dy
dx
is fractional, make it integer.
.
 
 
 
3/2
3
2
3
d y dy
+ x y = –x
dx
dx
   
   
 
 
2 3
3
2 2
3
d y dy
+ x y = x
dx
dx
Now order = 3 and
degree = 2
13. The hoop stress at a point on the surface of a
thin cylindrical pressure vessel is computed to be
30.0 MPa. The value of maximum shear stress at
this point is
(a) 7.5 MPa (b) 15.0 MPa
(c) 30.0 MPa (d) 22.5 MPa
Sol: (b)
Given,
 

absmax
pd
4 t

Solution
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Hoop stress   
h
pd
( ) 30 MPa
2t

 
       
 
 
absmax h
pd 1 1 30
15 MPa
2t 2 2 2
14. In the context of elastic theory of reinforced
concrete, the modular ratio is defined as the ratio
of
(a) Young’s modulus of elasticity of reinforce-
ment material to Young’s modulus of elas-
ticity of concrete.
(b) Young’s modulus of elasticity of concrete to
Young’s modulus of elasticity of reinforce-
ment material.
(c) Shear modulus of reinforcement material to
the shear modulus of concrete.
(d) Young’s modulus of elasticity of reinforce-
ment material to the shear modulus of con-
crete.
Sol: (a)
This is a question of working stress method i.e.
elastic theory.
Modular ratio
=
s
c
E Young's modulus of steel
=
E Young's modulusof concrete
Hence, option (a) is correct.
15. Which of the following equations is correct for the
Pozzolanic reaction?
(a) Ca(OH)2 + Reactive Superplasticiser + H2O
 C-S-H
(b) Ca(OH)2 + Reactive Silicon dioxide + H2O
 C-S-H
(c) Ca(OH)2 + Reactive Sulphates + H2O
 C-S-H
(d) Ca(OH)2 + Reactive Sulphur + H2O
 C-S-H
Sol: (b)
• Pozzolanic materials have no cementing
properties itself but have the property of
combining with lime to produce stable
compound.
• Pozzolana is considered as siliceous and
aluminous materials and when added in
cement it have SiO2 and Al2O3 form.
• So, pozzolanic reaction :
Ca(OH)2 + Reactive slilica-di-oxide +
H2O  C-S-H gel or tobermonite gel
16. Consider the cross-section of a beam made up of
thin uniform elements having thickness t (t << a)
shown in the figure. The (x, y) coordinates of the
points along the center-line of the cross-section
are given in the figure.
y
K (a, 3a)
a
2a
(–2a, 3a)
H
t
t
x
3a
J (0, 0)
The coordinates of the shear center of this cross-
section are:
(a) x = 0, y = 3a (b) x = 2a, y = 2a
(c) x = – a, y = 2a (d) x = – 2a, y = a
Sol: (a)
Shear centre of section consisting of two
intersecting narrow rectangles always lies at the
intersection of centrelines of two rectangles.
Example:
G
S
S, G
S = Shear centre
G = Centre of gravity
K(a, 3a)
H
t
(–2a, 3a)
t
J(0, 0)
3a
Intersection point
2a a
y
x
Coordinate of shear centre (0, 3a).

Solution
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17. Four different soils are classified as CH, ML, SP,
and SW, as per the Unified Soil Classification
System. Which one of the following options
correctly represents their arrangement in the
decreasing order of hydraulic conductivity?
(a) SW, SP, ML, CH (b) CH, ML, SP, SW
(c) SP, SW, CH, ML (d) ML, SP, CH, SW
Sol: (a)
Hydraulic conductivity Order.
Gravel > Sand > silt > lay
18. Let 
v and 
h denote the effective vertical stress
and effective horizontal stress, respectively. Which
one of the following conditions must be satisfied
for a soil element to reach the failure state under
Rankine’s passive earth pressure condition?
(a)  
  
v h (b)  
  
v h
(c)  
 
v h
= (d)  
 
v h
+ = 0
Sol: (a)
We know,  
 
h v
= K
For passive earth pressure,
K = 
P
K 1





h
v
= 
P
K 1
 
h > 
v
19. With respect to fluid flow, match the following in
Column X with Column Y:
Column X Column Y
P. Viscosity I. Mach number
Q. Gravity II. Reynolds number
R. Compressibility III. Euler number
S. Pressure IV. Froude number
Which one of the following combinations is
correct?
(a) (P) – (II), (Q) – (IV), (R) – (I), (S) – (III)
(b) (P) – (III), (Q) – (IV), (R) – (I), (S) – (II)
(c) (P) – (IV), (Q) – (II), (R) – (I), (S) – (III)
(d) (P) – (II), (Q) – (IV), (R) – (III), (S) – (I)
Sol: (a)
Reynold’s number is defined when apart from
inertial force, viscous forces are dominant.
Reynold’s number (Re)
=
 
 
   
 
 
2 2
Inertial force L V VL
Viscous force VL
Froude’s number (Fe): It is used when in addition
to inertial force, gravity forces are important.
Fe =

 

2 2
3
Inertial force V L V
Gravity force L g gL
Euler number (Eu): It is used when apart from
inertial force, only pressure forces are dominant.
Eu =

 
 
 

 
2 2
2
Inertial force V L V
Pressure force PL P
Mach number (M): It is used when in addition to
inertial force, compressibility forces are dominant.
M =
  
    
 
 
 

 
2 2
2
Inertial force L V V V
Elastic force C
KL K
where, C = Velocity of sound in the medium.
20. Let  represent soil suction head and K represent
hydraulic conductivity of the soil. If the soil
moisture content  increases, which one of the
following statements is TRUE?
(a)  decreases and K increases
(b)  increases and K decreases
(c) Both  and K decrease
(d) Both  and K increase
Sol: (a)
21. A rectangular channel with Gradually Varied Flow
(GVF) has a changing bed slope. If the change is
from a steeper slope to a steep slope, the resulting
GVF profile is
(a) S3
(b) S1
(c) S2
(d) either S1 or S2, depending on the magnitude
of the slopes
Sol: (a)

Solution
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Steep
CDL
NDL
S3
Steeper
NDL
CDL
22. The total hardness in raw water is 500 milligram
per liter as CaCO3. The total hardness of this raw
water, expressed in milligram equivalent per liter,
is
(Consider the atomic weights of Ca, C, and O as
40 g/mol, 12 g/mol, and 16 g/mol, respectively.)
(a) 10 (b) 100
(c) 1 (d) 5
Sol: (a)
TH = 3
mg
500 as CaCO
L
=
 
500 meq
100 2 L
= 10 meq/L
23. An aerial photograph is taken from a flight at a
height of 3.5 km above mean sea level, using a
camera of focal length 152 mm. If the average
ground elevation is 460 m above mean sea level,
then the scale of the photograph is
(a) 1 : 20000 (b) 1 : 20
(c) 1 : 100000 (d) 1 : 2800
Sol: (a)
H = 3.5 Km = 3500 m
f = 152 mm
havg = 460 m
Scale =

–3
avg
f 152×10 1
= =
H h 3500 – 460 20000
24. A line between stations P and Q laid on a slope
of 1 in 5 was measured as 350 m using a 50 m
tape. The tape is known to be short by 0.1 m.
The corrected horizontal length (in m) of the line
PQ will be
(a) 342.52 (b) 349.30
(c) 356.20 (d) 350.70
Sol: (a)
350m
P
Q
5
1 
1
tan =
5
Horizontal distance of line PQ, = 
350cos
=
350×5
26
= 343.20 m
Tape is 0.1 m short.
Nominal length of tape, l = 50 m
Actual length of tape, 
l = 50 – 0.1 = 49.9 m
Corrected horizontal length of line PQ
=

 
 
 
×343.2
l
l
=
 
 
 
49.9
×343.2
50
= 342.517 m
 342.52 m
25. The matrix M is defined as
M =
 
 
 
1 3
4 2
and has eigenvalues 5 and –2. The matrix Q is
formed as
Q = M3 – 4 M2 – 2 M
Which of the following is/are the eigenvalue(s) of
matrix Q ?
(a) 15 (b) 25
(c) –20 (d) –30
Sol: (a,c)
Eigen values of M are  5, –2 so eigen value
of Q = M3 – 4M2 – 2M are
are
53 – 4 × 52 – 2 × 5 = 15,
(–2)3 – 4 × (–2)2–2 × –2 = – 20
26. For wastewater coming from a wood pulping
industry, Chemical Oxygen Demand (COD) and
5-day Biochemical Oxygen Demand (BOD5) were
determined. For this wastewater, which of the
following statement(s) is/are correct?

Solution
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(a) COD > BOD5 (b) COD  BOD5
(c) COD < BOD5 (d) COD = BOD5
Sol: (a,b)
We know COD is the oxygen demand for both
biodegradable and non-biodegradable organic
matter. Hence COD > BOD5 also COD  BOD5.
27. Which of the following process(es) can be used
for conversion of salt water into fresh water?
(a) Microfiltration (b) Electrodialysis
(c) Ultrafiltration (d) Reverse osmosis
Sol: (b,d)
Desalination can be done by
• Electrodialysis
• Reverse osmosis
28. A horizontal curve is to be designed in a region
with limited space. Which of the following
measure(s) can be used to decrease the radius
of curvature?
(a) Decrease the design speed.
(b) Increase the superelevation.
(c) Increase the design speed.
(d) Restrict vehicles with higher weight from
using the facility.
Sol: (d)
e + f =
2
V
127R
• From above equations, the value of super-
elevation needed increases with increase in
speed with decrease in radius of curve for a
constant value of f.
• But from practical point of view, it will be
necessary to limit the maximum allowable
superelevation to avoid very high value of e.
• In this case, heavily loaded vehicles are not
safe because of slow speed, therefore we
restrict the vehicles with higher weight from
using the facility
29. Consider the following recursive iteration scheme
for different values of variable P with the initial
guess x1 = 1:
 
 
 
n+1 n
n
1 P
x = x + , n = 1, 2, 3, 4, 5
2 x
For P = 2, x5 is obtained to be 1.414, rounded-
off to three decimal places. For P = 3, x5 is
obtained to be 1.732, rounded-off to three decimal
places.
If P = 10, the numerical value of x5 is ________.
(round off to three decimal places)
Sol: (3.1 to 3.2)
xn+1 =
 
 
 
n
n
1 P
x +
2 x
Converges when 
n+1 n
x = x =
 =
 

 

 
1 P
+
2

2
=

P
2
2 = P
 = P
Given recursive relation is finding square root of P
When P = 2, 5
x = 2 = 1.4124 2135
When P = 3, 5
x = 3 = 1.732 05080
Similarily when P = 10,
5
x = 10 = 3.162 277
Ans. = 3.162
30. The Fourier cosine series of a function is given
by:
 

 n
n=0
f x = f cosnx
For f(x) = cos4x, the numerical value of (f4 + f5)
is _________. (round off to three decimal places)
Sol: (0.125)
cos4x = cos2x cos2x
=
 
  
  
  
1 cos 2x 1 cos2x
2 2
=   2
1
[1 2cos2x cos 2x]
4
=

 
 
 
 
1 (1 cos4x)
1 2cos2x
4 2
=   
1 cos2x 1 cos4x
4 2 8 8

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
=  
3 cos2x cos4x
8 2 8
f4 =
1
,
8
f5 = 0
 f4 + f5 =  
1
0 0.125
8
31. An uncompacted heap of soil has a volume of
10000 m3 and void ratio of 1. If the soil is
compacted to a volume of 7500 m3, then the
corresponding void ratio of the compacted soil is
__________. (round off to one decimal place)
Sol: (0.5)
Uncompacted
air
W
S
V1
VS
Compacted
V2
VS
air
W
S
1
V =  
1 s
1+ e V
Vs =
1
1
V
1+ e
V2 =  
2 S
1+e V
=  
 
 
 
1
2
1
V
1+ e
1+ e
2
1
V
V =
2
1
1+ e
1+ e
7500
10000
= 2
1+ e
1+1
3
4
= 2
1+ e
2
e2 =
3 1
– 1= = 0.5
2 z
32. A concentrated vertical load of 3000 kN is applied
on a horizontal ground surface. Points P and Q
are at depths 1 m and 2 m below the ground,
respectively, along the line of application of the
load. Considering the ground to be a linearly
elastic, isotropic, semi-infinite medium, the ratio
of the increase in vertical stress at P to the
increase in vertical stress at Q is ________. (in
integer)
Sol: (4)
Q=300 KN
1m
2m
P
Q
Increase in vertical stress,
z = 2
KQ
z
z  2
1
z


P
Q
=
   
 
   
 
2 2
Q
P
Z 2
= = 4
Z 1
33. At a site, Static Cone Penetration Test was carried
out. The measured point (tip) resistance qc was
1000 kPa at a certain depth. The friction ratio (fr)
was estimated as 1 % at the same depth.
The value of sleeve (side) friction (in kPa) at that
depth was _____________ . (in integer)
Sol: (10)
f
R =
s
c
f
f
Rf = Friction ratio
fs = Sleeve (side) friction
qc = Cone tip resistance
1
100
= s
f
1000
fs = 10 KPa
34. During a particular stage of the growth of a crop,
the consumptive use of water is 2.8 mm/day. The
amount of water available in the soil is 50% of the
maximum depth of available water in the root zone.
Consider the maximum root zone depth of the
crop as 80 mm and the irrigation efficiency as
70%.
The interval between irrigation (in days) will be
_________. (round off to the nearest integer)
Sol: (*)
Data insufficient

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
35. The bearing of a survey line is N31°17'W. Its
azimuth observed from north is ______ deg.
(round off to two decimal places)
Sol: (328.72)
W E
A
S
31°17’
N
Azimuth of line (A) = 360° – 31°17'
= 328°43'
=
43
328 +
60
= 328.717°
= 328.72°
36. The Cartesian coordinates of a point P in a right-
handed coordinate system are (1, 1, 1). The
transformed coordinates of P due to a 45°
clockwise rotation of the coordinate system about
the positive x-axis are
(a)  
1
, 0, 2 (b)  
1, 0, – 2
(c)  
–1, 0, 2 (d)  
–1
, 0, – 2
Sol: (a)
Rotation matrix in R3 around the x axis is given by
 
 
 
 
 
 
1 0 0
0 cos – sin
0 sin cos
so the transfrmed coordinates of P due to a 45°
clockwise rotation
 
 
 
 
x
y
z
=
   
   
   
   
   
 
 
 
1 0 0 1
1 1
0 – 1
2 2
1 1 1
0
2 2
=
 
 
 
 
 
1
0
2
37. A semi-circular bar of radius R m, in a vertical
plane, is fixed at the end G, as shown in figure.
A horizontal load of magnitude P kN is applied at
the end H. The magnitude of the axial force, shear
force, and bending moment at point Q for  
= 45 ,
respectively, are
Q
R
H
P O
G

(a)
P P PR
kN, kN, and kNm
2 2 2
(b)
P P
kN, kN, and 0 kNm
2 2
(c)
P PR
0kN, kN, and kNm
2 2
(d)
P PR
kN, 0 kN, and kNm
2 2
Sol: (a)
R
Q
H O
G

P
M = PRsin45°
Q
P
P
Rcos45°
45°
Rsin45°
BM at Q = PR sin 45° =
PR
2
kNm
Q
P
c
o
s
4
5
°
P
Psin45° 45°

Shear force = Pcos45° =
P
2
kN
Axial force = Psin45° =
P
2
kN
38. A weld is used for joining an angle section ISA
100 mm × 100 mm × 10 mm to a gusset plate
of thickness 15 mm to transmit a tensile load.
The permissible stress in the angle is 150 MPa
and the permissible shear stress on the section
through the throat of the fillet weld is 108 MPa.

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
The location of the centroid of the angle is
represented by Cyy in the figure, where Cyy =
28.4 mm. The area of cross-section of the angle
is 1903 mm2. Assuming the effective throat
thickness of the weld to be 0.7 times the given
weld size, the lengths L1 and L2 (rounded off to
the nearest integer) of the weld required to transmit
a load equal to the full strength of the tension
member are, respectively
L1
5 mm weld
5 mm weld
100 mm
Cyy
L2
15 mm thick gusset plate
(a) 541 mm and 214 mm
(b) 214 mm and 541 mm
(c) 380 mm and 151 mm
(d) 151 mm and 380 mm
Sol: (a)
The question is based on WSM principle as
permissible stresses are given. Here, it is required
to calculate L1 and L2. So as to design a moment
free connection.
L1
L2
5mm weld
Cyy
Given data,
ISA 100 mm × 100 mm × 10 mm
Thickness of gueset plate = 15 MPa
Permissible shear stress in the weld = 108 MPa
Cyy = 28.4 mm
Area of angle section = 1903 mm2
Full strength of the tension member assuming
that the section is uniformly stressed i.e.
neglecting the effect of shear lag.
T = Area × Permissible stress
= 1903 × 150 × 10–3 = 285.45 kN
Assuming force developed for weld length L1 = T1
For weld length L2 = T2
 T1 + T2 = T
Considering moment about ‘T2’
T(100 – 28.4) = T1 × 100

 
1
285.45× 100 – 28.4
T = = 204.38kN
100
 T2 = k
285.45 – 204.38 = 81.07 N
T1 = 0.7 × Weld size × L1 × Permissible
stress in weld
 L1 =
3
204.38×10
0.7×5×108
= 540.69mm = 541mm
L2 =
3
81.07×10
0.7×5×108
= 214.47mm = 214mm
39. The project activities are given in the following
table along with the duration and dependency.


P 10
Q 12
R 5 P
S 10 Q
T 10 P, Q
Activities Duration (days) Depends on
Which one of the following combinations is
correct?
(a) Total duration of the project = 22 days, Critical
path is 
Q S
(b) Total duration of the project = 20 days, Critical
path is 
Q T
(c) Total duration of the project = 22 days, Critical
path is 
P T
(d) Total duration of the project = 2 days, Critical
path is 
P R
Sol: (a)
2
1
3
4 5
R, 5 days
T, 10 days
P, 10 days
Q, 12 days
S, 10 days
(i) P R
(ii) P T
(iii) Q T
(iv) Q S




10 + 5 = 15 days
10 + 10 = 20 days
12 + 10 = 22 days
12 + 10 = 22 days
Paths Project duration

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
 Total duration of the project,
= maximum of [(i), (ii), (iii) (iv)] = 22 days
Here, critical paths are Q T

Q S

40. The correct match between the physical states of
the soils given in Group I and the governing
conditions given in Group II is

1. normally consolidated
P. sensitivity 16
soil
2. quick day Q. dilation angle = 0
3. sand in critical state R. liquid limit > 50
S. over consolidation
4. clay of high plasticity
ratio = 1
Group I Group II
(a) 1-S, 2-P, 3-Q, 4-R
(b) 1-Q, 2-S, 3-P, 4-R
(c) 1-Q, 2-P, 3-R, 4-S
(d) 1-S, 2-Q, 3-P, 4-R
Sol: (a)
41. As per Rankine’s theory of earth pressure, the
inclination of failure planes is


 
 
 
45 +
2
with
respect to the direction of the minor principal
stress.
The above statement is correct for which one of
the following options?
(a) Only the active state and not the passive
state
(b) Only the passive state and not the active
state
(c) Both active as well as passive states
(d) Neither active nor passive state
Sol: (a)
In case of active state minor principal plane is
vertical plane and minor principal stress direction
is horizontal. Hence failure plane makes

45 +
2
angle with respect to direction of minor principal
stress.
42. Henry’s law constant for transferring O2 from air
into water, at room temperature, is

mmol
1.3
liter – atm
Given that the partial pressure of
O2 in the atmosphere is 0.21 atm, the
concentration of dissolved oxygen (mg/liter) in
water in equilibrium with the atmosphere at room
temperature is
(Consider the molecular weight of O2 as
32 g/mol)
(a) 8.7 (b) 0.8
(c) 198.1 (d) 0.2
Sol: (a)
Henry’s law
Cg = KP
Cg = Concentration of dissolved gas
K = Henry’s law constant
P = Partial pressure of gas
Cg =
m mol
1.3 ×0.21atm
lit – atm
= 0.273 m mol/lit
= 0.273 × 32 mg/lit
= 8.736 mg/lit
43. In a water sample, the concentrations of
2+ 2+ –
3
Ca , Mg and HCO are 100 mg/L, 36 mg/L
and 122 mg/L, respectively. The atomic masses
of various elements are: Ca = 40, Mg = 24, H =
1, C = 12, O = 16
The total hardness and the temporary hardness
in the water sample (in mg/L as CaCO3) will be
(a) 400 and 100, respectively
(b) 400 and 300, respectively
(c) 500 and 100, respectively
(d) 800 and 200, respectively
Sol: (a)
2+
Ca = 100 mg/L
2+
Mg = 36 mg/L
–
3
HCO = 122 mg/L
Total hardness,
TH =    
 
2+ 2+
Ca + Mg
=
   
   
   
100 36
+ meq L
40 2 24 2
= 8 meq/L
= 3
8×50mg L as CaCO

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
= 3
400 mg L as CaCO
Alkalinity=  
 
–
3
HCO
=
 
 
 
122
meq L
611
= 2 meq/L
= 3
2×50 mg L as CaCO
= 100 mg/L as CaCO3
CH = Temporary hardness
= Min (TH, alkalinity)
= min (400, 100)
= 100 mg/L as CaCO3
TH = 400, CH = 100
44. Consider the four points P, Q, R and S shown in
the Greenshields fundamental speed-flow diagram.
Denote their corresponding traffic densities by kP,
kQ, kR, and kS, respectively. the correct order of
these densities is
R
S
Q
Speed
P
Flow
(a) kP > kQ > kR > kS
(b) kS > kR > kQ > kP
(c) kQ > kR > kS > kP
(d) kQ > kR > kP > kS
Sol: (a)
• When the speed of traffic flow decreases and
becomes zero, the density attains the
maximum value whereas volume becomes
zero, it means at point p there is maximum
density
KP - Jam density
• For increasing value of speeds, density
decreases, whereas volume increase upto a
certain limit (upto f
V 2 )
• At high speeds the volume starts decreasing
and density keeps on further reducing.
Space
mean
speed
Vf
Space
mean speed
V /2
f
qmax
K /2
J
Density KJ
Volume (q)
Volume (q)
Slope = Vf
Density (K)
K /2
J KJ
Hence,   
P Q R S
K K K K
45. Let max {a, b} denote the maximum of two real
number a and b.
Which of the following statement(s) is/are TRUE
about the function f(x) = max {3 – x, x–1}?
(a) It is continuous on its domain.
(b) It has a local minimum at x = 2.
(c) It has a local maximum at x = 2.
(d) It is differentiable on its domain.
Sol: (a,b)
f(x) = max {3 – x, x – 1}
y
y=3-x
3
x
y
y=x-1
1 x
both intersecting at
3 – x = x – 1
2x = 4
x = 2
y = max {3 –x, x–1}
y
y = 3–x
0 1 2
corner so not
differentiate at x=2
x
y = x–1

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
46. A horizontal force of P kN is applied to a
homogeneous body of weight 25 kN, as shown in
the figure. The coefficient of friction between the
body and the floor is 0.3. Which of the following
statement(s) is/are correct?
1 m
P
2 m
(a) The motion of the body will occur by over-
turning.
(b) Sliding of the body never occurs.
(c) No motion occurs for P  6 kN.
(d) The motion of the body will occur by sliding
only.
Sol: (a,c)
2m
1m
P
W
0.5m
CG
x
y
O
N
Fs
Motion of block will occur in two ways, the block
may slide to the left or it may tip about edge O.
Case-I: Checking motion of block in sliding
 
x
F 0
P – Fs = 0
P = Fs
and
 
y
F 0
N – W = 0
N = W = 25 kN
So, force due to friction (Fs) = N
= 0.3 × 25 = 7.5 kN
Hence sliding will occur at 7.5 kN.
Case-II: Checking motion of block against
tipping (over turning)
 
O
M 0 
–P × 2 + W × 0.5 = 0
2 × P = 25 × 0.5
P =


25 0.5
6.25 kN
2
Block will overturn at P = 6.25 kN
Motion of block will occur by overturning before
sliding at P = 7.5 kN
For larger force, both sliding and overturning will
take place.
Hence correction options are (a) and (c).
47. In the context of cross-drainage structures, the
correct statement(s) regarding the relative positions
of a natural drain (stream/river) and an irrigation
canal, is/are
(a) In an aqueduct, natural drain water goes un-
der the irrigation canal, whereas in a super-
passage, natural drain water goes over the
irrigation canal.
(b) In a level crossing, natural drain water goes
through the irrigation canal.
(c) In an aqueduct, natural drain water goes over
the irrigation canal, whereas in a super-pas-
sage, natural drain water goes under the
irrigation canal.
(d) In a canal syphon, natural drain water goes
through the irrigation canal.
Sol: (a,b)
In Aqueduct and Syphon aqueduct under irrigation
canal or canal passes over natural drain.
Canal
Road
Drain
Aqueduct
Canal
Syphon aqueduct

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
In Super passage and Canal syphon or simply
syphon, natural drain passes over irrigation canal
or canal goes under natural drain.
Drain
Canal
Super passage
Canal
Canal
Drain
Canal Syphon
In level crossing, natural drain water goes through
the irrigation canal.
Incoming canal
regulator (sometimes
provided)
Canal
C
a
n
a
l
Outgoing canal
regulator
(Cross regulator)
Drainage
Drainage
Regulator/A
barrage
Note:
Cross drainage
work
Aqueduct
Typical cross-section Used when
- HFL of drain is sufficiently below
the bottomof the canal
- Drainage water water flows
- HFL of drain is higher than canal
bed
- Water passes through aqueduct
barrels under syphonic action
- FSL of canal is sufficiently below
the bottomof the draintrough
- Canal water flows freely under
gravity
- FSL of canal is sufficiently above
the bed level of drainagetrough
- Canal flows under syphonic
actionunder thetrough
- Generally provided when a large
canal and a huge drainage (such
as a stream or river) approach
each other practically at the same
level.
Syphon
aqueduct
Super
passage
Canal
Road
Drain
Canal
Drain
Canal
Canal
Canal
Drain
Canal syphon
or syphon
Level
crossing
Incoming canal
regulator (sometimes
provided)
Canal
C
a
n
a
l
Outgoing canal
regulator
(Cross regulator)
Drainage
Drainage
Regulator/A
barrage
48. Consider the differential equation
 
dy
= 4 x + 2 – y
dx
For the initial condition y = 3 at x = 1, the value
of y at x = 1.4 obtained using Euler’s method
with a step-size of 0.2 is _________. (round off to
one decimal place)
Sol: (6.4 to 6.4)
x0 =  0
1 y = 3
x1 =  1
1.2 y = ?
x2 =  2
1.4 y = ?
Using Euler’s forward method
yn+1 = yn + hf(xn, yn)
y1 = y0 + hf (x0, y0)
= 3 + 0.2 × (4(1 + 2)–3)
= 4.8
y2 = y1 + hf(x1, y1)
= 4.8 + 0.2 (4(1.2 + 2)–4.8)
= 6.4
49. A set of observations of independent variable (x)
and the corresponding dependent variable (y) is
given below.
x 5 2 4 3
y 16 10 13 12
Based on the data, the coefficient a of the linear
regression model
y = a + bx
is estimated as 6.1.
The coefficient b is ______________ . (round off
to one decimal place)
Sol: (1.9)
x y
5 16
2 10
4 13
3 12
x =
  
 
5 2 4 3 14
3.5
4 4
y =
  

16 10 13 12
12.75
4
y = a + bx

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
Given a = 6.1
y = 
a bx
 12.75 = 6.1 + 6 × (3.5)
 
b 1.9
50. The plane truss shown in the figure is subjected
to an external force P. It is given that P = 70 kN,
a = 2 m, and b = 3m.
P
C a D a E a G a I
b
J
H
F
A
a
a
B
The magnitude (absolute value) of force (in kN) in
member EF is ____. (round off to the nearest
integer)
Sol: (30 kN)
Various zero force members in the structure are
shown below.
A
F H
J HJ
HA
B
C D E G
I
RA
RJ
P
a
a
a a a a
b
0 0
0
0
0
0
0
0
0
After removing zero force members the structure
becomes.
A
F
J HJ
HA
E G
RA
RJ
P
a
1
1
Taking sectional plane (1)-(1)
Consider portion on left side
A
RA
4m
HA
FEF
FEG
70
4m
E
x
y
 
E
M 0 
RA × 4 = HA × 4
 RA = HA
 
y
F 0
RA + FEF = 70 …(i)
Now,  
J
M 0 
RA × 8 – HA × 1 – 70 × 4 = 0
or RA × 8 – RA × 1 = 280
RA = 40 kN
Putting in equation (i)
So, FEF = 70 – 40 = 30 kN
51. Consider the linearly elastic plane frame shown
in the figure. Members HF, FK and FG are welded
together at joint F. Joints K, G and H are fixed
supports. A counter-clockwise moment M is
applied at joint F. Consider flexural rigidity EI =
105 kN-m2 for each member and neglect axial
deformations.
H
EI
5 m
M
F
EI EI
K
3 m 3 m
4 m
G
If the magnitude (absolute value) of the support
moment at H is 10 kN-m, the magnitude (absolute
value) of the applied moment M (in kN-m) to
maintain static equilibrium is ___________. (round
off to the nearest integer)

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
Sol: (60_60)
H
EI
5 m
M
F
EI EI
K
3 m 3 m
4 m
G
FK = FG =  
2 2
(4) (3) 5 m
Distribution factor (DF)
   
   
   
     
     
     
   
   
   
Relative Joint
Joint Member D.F.
stiffness stiffness
4EI 1
FG
5 3
4EI 12EI 1
F FH
5 5 3
4EI 1
FK
5 3
MFH =
 

 
 
1
M
3
MHF = 10 kNm (given)
As far end fixed, and no loading in span FH,
 MHF =  FH
1
M
2
=
   
   
   
   
1 1 1
M M
2 3 6

1
M
6
= MHF = 10 kNm
 
M 60 kNm
52. Consider a simply supported beam PQ as shown
in the figure. A truck having 100 kN on the front
axle and 200 kN on the rear axle, moves from left
to right. The spacing between the axles is 3 m.
The maximum bending moment at point R is
_________ kNm. (in integer)
P R Q
4 m
1 m
Sol: (180_180)
Q
P
3m
R
1m = a 4m = b
h = 4/5
 
 
 
1
5
ILD for BM at R
Height of the ILD = h =
 

 

 
a b
(a b)
=
 

 

 
4 1
(4 1)
=
 
 
 
4
5
units
For the B.M. at R to be maximum the heavier
load, i.e., the 200 kN load (rear load) should be
placed at R. The front axle load, i.e. 100 kN load
should be placed 3m on the right side of R.
Maximum BM at R =
 
  
 
 
4 1
200 100
5 5
= 180 kNm
53. A reinforced concrete beam with rectangular cross
section (width = 300 mm, effective depth = 580
mm) is made of M30 grade concrete. It has 1%
longitudinal tension reinforcement of Fe 415 grade
steel. The design shear strength for this beam is
0.66 N/mm2. The beam has to resist a factored
shear force of 440 kN. The spacing of two-legged,
10 mm diameter vertical stirrups of Fe 415 grade
steel is __________mm. (round off to the nearest
integer)
Sol: (101 mm)
This is a question of vertical shear reinforcement
design.
Data given,
b = 300 mm
d = 580 mm
Grade of concrete = M30
Grade of steel = Fe415
C = 0.66 N/mm2

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
Vu = 440 kN
Shear stress = 2 legged 10 mm dia.
Nominal Shear Stress,
 u
v
V
=
bd
=
3
440×10
= 2.53MPa
300×580
Spacing ‘SV’ can be calculated from the following
expression  
  
y SV V C
V
d
0.87f A – bd
S
 V
S   
 
y SV
V C
0.87f A
– b

 
 2
0.87× 415×2× ×10
4
2.53 – 0.66 ×300
= 101.09 mm
Spacing SV from minimum shear reinforcement
criteria.
SV 
y SV
0.87f A
0.4b

 2
0.87× 415×2× ×10
4
0.4×300
= 472.61mm
As per IS code SV  min (300 mm , 0.75d)
 min (300 mm, 435 mm)
So, SV
 101.09 mm
SV = 101 mm, rounding off the nearest integer
54. A square concrete pile of 10 m length is driven
into a deep layer of uniform homogeneous clay.
Average unconfined compressive strength of the
clay, determined through laboratory tests on
undisturbed samples extracted from the clay layer,
is 100 kPa. If the ultimate compressive load
capacity of the driven pile is 632 kN, the required
width of the pile is _______ mm. (in integer)
(Bearing capacity factor Nc = 9; adhesion factor
 = 0.7)
Sol: (400)
Given, L = 10 m, qunconfined = 100 kPa, Qu = 632
N, NC =9,  = 0.7
We know,
Qu = up b s s
A
q . + f A
qup =
 
 
 
ub C
100
C N = ×9 = 450kPa
2
Ab = a2 [a = width of pile]
fs =  
u
100
C = 0.7× 35KPa
2
fs = 4a×L = 4×10×a = 40a
632 = 450a2 + 35 × 40a
2
a + 3.11a – 1.404 = 0
a = 0.4, –3.51
Required width of pile = 0.4 m
= 0.4 × 1000 mm
= 400 mm
55. A raft foundation of 30 m × 25 m is proposed to
be constructed at a depth of 8 m in a sand layer.
A 25 m thick saturated clay layer exists 2 m
below the base of the raft foundation. Below the
clay layer, a dense sand layer exists at the site.
A 25 mm thick undisturbed sample was collected
from the mid-depth of the clay layer and tested in
a laboratory oedometer under double drainage
condition. It was found that the soil sample had
undergone 50 % consolidation settlement in 10
minutes.
The time (in days) required for 25 % consolidation
settlement of the raft foundation will be
___________. (round off to the nearest integer)
Sol: (1739)
Sand
30m × 25m
2m
25 m Saturated clay
Dense sand
For undisturbed sample
H1 = 25 mm, U1 = 50%, t1 = 10 min
U1 < 60%
1
V
T =
 
2 2
1
U = ×0.5 = 0.196
4 4
1
V
T =
V 1
2
1
C t
H
Double drainage condition,
0.196 =
 
 
 
V
2
–3
C ×10
25
×10
2

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
CV = –6 2
3.0625×10 m min
For raft foundation,
U2 = 25%,

2
2
V
T = ×0.25 = 0.0491
4
, H2 = 25 m
0.491 =
 
 
 
–6
2
2
3.0625×10 × t
25
2
t2 = 6
2.504×10 min
=
6
2.504×10
days
60×24
= 1739.21 days = 1739 days
56. A two-hour duration storm event with uniform
excess rainfall of 3 cm occurred on a watershed.
The ordinates of streamflow hydrograph resulting
from this event are given in the table.
3
0 1 2 3 4 5 6 7
( )
10 16 34 40 31 25 16 10
( /s)
Time
hours
Streamflow
m
Considering a constant baseflow of 10 m3/s, the
peak flow ordinate (in m3/s) of one-hour unit
hydrograph for the watershed is ________ . (in
integer)
Sol: (12)

Time Flood Hydrograph BaseFlow 2hrs.3cm DRH 2hr.UH = DRH 3
0 10 10 0 0
1 16 10 6 2
2 34 10 24 8
3 40 10 30 10
4 31 10 2.1 7
5 25 10 15 5
6 16 10 6 2
7 10 10 0 0
 
 
 
A B
A B A B
S – S
Time 2hr.UH S S S – S 1hrUH =
1
2
0 0 0 0 0
1 2 2 0 2 4
2 8 8 2 6 12
3 10 12 8 4 8
4 7 15 12 3 6
5 5 17 15 2 4
6 2 17 17 0 0
7 0 17 17 0
Max. ordinate of 1 Hr UH = 12 m3/sec
57. Two reservoirs are connected by two parallel pipes
of equal length and of diameters 20 cm and 10
cm, as shown in the figure (not drawn to scale).
When the difference in the water levels of the
reservoirs is 5 m, the ratio of discharge in the
larger diameter pipe to the discharge in the smaller
diameter pipe is ____________. (round off to two
decimal places)
(Consider only loss due to friction and neglect all
other losses. Assume the friction factor to be the
same for both the pipes)
Reservoir
20 cm
5 m
Reservoir
10 cm
Sol: (5.6 to 5.7)
Reservoir
20 cm
5 m
Reservoir
10 cm
For parallel connection of pipes, head loss remains
constant through each pipe.
As per question, we have to consider loss due to
friction only and neglect all other losses.
Assuming friction factor to be same for both the
pipes, we have
(hf)larger dia = (hf)smaller dia
 
 
2 2
larger dia smaller dia
5 5
f (Q ) f (Q )
12.1D 12.1d
[Darcy-Weisbach equation]

     
  
     
   
 
5 5/2
larger dia
smaller dia
Q D 20
5.6568
Q d 10
58. Depth of water flowing in a 3 m wide rectangular
channel is 2 m. The channel carries a discharge
of 12 m3/s. Take g = 9.8 m/s2. The bed width (in
m) at contraction, which just causes the critical
flow, is _________ without changing the upstream
water level. (round off to two decimal places)

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
Sol: (2.1 to 2.2)
Given: B = 3m, y = 2m, Q = 12 m3/sec
Velocity (v) =
   
 
   

   
Q 12
2 m/sec
A 2 3
B
y
Cross-section 1
B Bmin
(Plan)
1
2
2
Specific energy at section (1-1) = E =
 

 
 
 
2
v
y
2g
 E =
 
   
  
2
(2) 108
2 m
2 9.8 49
When channel section is contracted to minimum
width and for constant discharge Q, the flow over
contracted section will be critical flow and under
the assumption that no energy loss has taken
place.
E = Ec =
 
 
 
108
9
We have that, Ec =
3
2
yc (for rectangular cross-
section)
 yc =
   
  
   
   
c
2 2 108 72
E
3 3 49 49
For critical flow condition, 
2
3
Q T
1
gA
 

2
min
3
min c
Q B
1
g(B y )
 (Bmin)2 =
 
 
 
 
2
3
c
Q
gy
 Bmin =
 
 
  
 
 
 
 
 
 
 
1/2
2
3
(12)
2.152 m
72
9.8
49
59. A wastewater sample contains two nitrogen
species, namely ammonia and nitrate. Consider
the atomic weight of N, H, and O as 14 g/mol, 1
g/mol, and 16 g/mol, respectively. In this
wastewater, the concentration of ammonia is 34
mg NH3/liter and that of nitrate is 6.2 mg 3
–
3
NO /liter. The total nitrogen concentration in this
wastewater is _________ milligram nitrogen per
liter. (round off to one decimal place)
Sol: (58.8)
 
3
NH = 34 mg/L
 
 
–
3
NO = 6.2 mg/L
Total nitrogen concentration in waste water
=    
34 6.2 meq
+
14 + 3 14 +16×3 L
=
34 6.2
+ meq L
17 62
= 2 + 0.1meq L
= 2.1 meq/L
=   2
2.1× 14×2 mg L as N
= 2
58.8 mg L as N
60. A 2 % sewage sample (in distilled water) was
incubated for 3 days at 27 °C temperature. After
incubation, a dissolved oxygen depletion of 10
mg/L was recorded. The biochemical oxygen
demand (BOD) rate constant at 27 °C was found
to be 0.23 day-1 (at base e).
The ultimate BOD (in mg/L) of the sewage will
be_____________. (round off to the nearest
integer)
Sol: (1003)
Depletion of dissolved oxygen after 3 days
incubation = 10 mg/L
Dilution factor =
Vol. of dilutedsample
Vol.of undiluted sewage sample
=
100
= 50
2
BOD3 = DO consumed × dilution factor
= 10×50 = 500mg L
BOD3 =  
–kt
u
BOD 1– e
500 =  
–0.23×3
u
BOD 1– e
BODu = 1003.162 mg L
= 1003 mg/L
61. A water treatment plant has a sedimentation basin
of depth 3 m, width 5 m, and length 40 m. The
water inflow rate is 500 m3/h. The removal fraction

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
of particles having a settling velocity of 1.0 m/h
is_______. (round off to one decimal place)
(Consider the particle density as 2650 kg/m3 and
liquid density as 991 kg/m3)
Sol: (0.4)
H = 3 m, W = 5m, L = 40 m
Q = 500 m3/h
P = 2650 Kg/m3
L = 991 Kg/m3
Overflow rate, VS =
Q 500
= m/h
WL 5× 40
= 2.5 m/h
Removal fraction of particles having setting velocity
of 1 m/h
=
1
= 0.4
2.5
62. A two-phase signalized intersection is designed
with a cycle time of 100 s. The amber and red
times for each phase are 4 s and 50 s,
respectively. If the total lost time per phase due
to start-up and clearance is 2 s, the effective
green time of each phase is ______s. (in integer)
Sol: (48)
RA
GB
GA
RB
C0
AB
AA
Two phase signalized intersections,
Cycle time = 100 secs.
Amber time = 4 secs. [each phase]
Red time = 50 secs. [each phase]
Total lost time = 2 sec. [each phase]
Effective green time = ? [each phase]
Total Effective green = Cycle time – Lost time
= 100 – 2 × 2 = 96 sec.
As there is symmetry in phases so effective green
time per phase
=
96
= 48 sec.
2
63. At a traffic intersection, cars and buses arrive
randomly according to independent Poisson
processes at an average rate of 4 vehicles per
hour and 2 vehicles per hour, respectively. The
probability of observing at least 2 vehicles in 30
minutes is ______. (round off to two decimal
places)
Sol: (80-82%)
1 = 
4 1
veh/ sec
3600 900
2 = 
2 1
veh/sec
3600 1800
 =   
1 2 =  
1 1 1
veh/sec
900 1800 600
t = 30 min = 30 × 60 = 1800 sec

P(n 2) =  
1 [P(0) P(1)]
P(n) =


t n
e ( t)
n!

P(n 2) =
 
  
  
 
 






1 0
1800
600 1
e 1800
600
1
0!
  
 

 
 
  





1 1
1800
600 1
e 1800
600
1!

P(n 2) = 1 – (e–3 + 3e–3) = 1 – 0.1991
 
 
P(n 2) 0.80085
P(n 2) 80.08%
64. The vehicle count obtained in every 10 minute
interval of a traffic volume survey done in peak
one hour is given below:






0 10 10
10 20 11
20 30 12
30 40 15
40 50 13
50 60 11
Time interval
Vehicle Count
(inminutes)
The peak hour factor (PHF) for 10 minute sub-
interval is __________. (round off to one decimal
place)
Sol: (0.75-0.85)
Peak hour factor

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
=
Peak flow during 1hour
6×Peak flow during 10 minutes
=
 
10 +11+12 +15 +13 +11
6×15
=
72
6×15
PHF = 0.8
65. For the dual-wheel carrying assembly shown in
the figure, P is the load on each wheel, a is the
radius of the contact area of the wheel, s is the
spacing between the wheels, and d is the clear
distance between the wheels. Assuming that the
ground is an elastic, homogeneous, and isotropic
half space, the ratio of Equivalent Single Wheel
Load (ESWL) at depth z = d/2 to the ESWL at
depth z = 2s is ___________. (round off to one
decimal place)
(Consider the influence angle to be 45° for the
linear dispersion of stress with depth)
P
2a
s
d
2a
P
z
Sol: (0.5)
P S
d
P
d/2
2s
2P
P
d
z =
2
z = 2S
Z(log scale)
ESWL
(log scale)
At z =
d
2
, equivalent single wheel load = P
(Because of no stress overlap)
at z=2S, ESWL = 2P [Because of stress overlap]
 
 
d 2
2S
ESWL
ESWL
=
P 1
= = 0.5
2P 2

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