Afternoon Aptitude and Technical Questions

Solution
12-02-2022 | AFTERNOON SESSION
Detailed
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Ph:
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APTITUDE
1. The movie was funny and I _____.
(a) could help laughing
(b) couldn’t help laughed
(c) couldn’t help laughing
(d) could helped laughed
Sol: (c)
2.
1 1 1
x : y : z = : :
2 3 4
What is the value of
x + z – y
y
?
(a) 0.75 (b) 1.25
(c) 2.25 (d) 3.25
Sol: (b)
Let x =
K
2
y =
K
3
z =
K
4
So, the value of
x + z – y
y
=
K K K
+ –
2 4 3
K
3
=
5K 3 5
× =
12 K 4
3. Both the numerator and the denominator of 3/4
are increased by a positive integer, x, and those
of 15/17 are decreased by the same integer. This
operation results in the same value for both the
fractions.
What is the value of x ?
(a) 1 (b) 2
(c) 3 (d) 4
Sol: (c)
 
 
 
3 + x
4 + x
=
 
 
 
15 – x
17 – x
Put x = 1, 2, 3, 4
Only x = 3 is satisfying.
4. A survey of 450 students about their subjects of
interest resulted in the following outcome.
• 150 students are interested in Mathematics.
• 200 students are interested in Physics.
• 175 students are interested in Chemistry.
• 50 students are interested in Mathematics and
Physics.
• 60 students are interested in Physics and
Chemistry.
• 40 students are interested in Mathematics and
Chemistry.
• 30 students are interested in Mathematics,
Physics and Chemistry.
• Remaining students are interested in
Humanities.
Based on the above information, the number of
students interested in Humanities is
(a) 10 (b) 30
(c) 40 (d) 45
Sol: (d)
         
     
  
   
n M P C = n M +n P + n C – n M P
–n P C – n C M +n M P C
 
 
n M P C =150+200+175 – 50 – 60 – 40+30
 
 
n M P C = 405
n(Humanities) = 450 – 405
= 45
5.
For the picture shown above, which one of the
following is the correct picture representing
reflection with respect to the mirror shown as the
dotted line ?
(a)

Solution
Detailed
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(b)
(c)
(d)
Sol: (a)
6. In the last few years, several new shopping malls
were opened in the city. The total number of visitors
in the malls is impressive. However, the total
revenue generated through sales in the shops in
these malls is generally low.
Which one of the following is the CORRECT logical
inference based on the information in the above
passage ?
(a) Fewer people are visiting the malls but
spending more
(b) More people are visiting the malls but not
spending enough
(c) More people are visiting the malls and spending
more
(d) Fewer people are visiting the malls and not
spending enough
Sol: (b)
More people visiting are the walls but not spending
more. This can be conferred from the low revenue
generation through sales in the shops in these
malls.
7. In a partnership business the monthly investment
by three friends for the first six months is in the
ratio 3: 4: 5. After six months, they had to increase
their monthly investments by 10%, 15% and 20%,
respectively, of their initial monthly investment.
The new investment ratio was kept constant for
the next six months.
What is the ratio of their shares in the total profit
(in the same order) at the end of the year such
that the share is proportional to their individual
total investment over the year?
(a) 22 : 23 : 24 (b) 22 : 33 : 50
(c) 33 : 46 : 60 (d) 63 : 86 : 110
Sol: (d)
Let capitals invested by three friends for 6 months
are
C1 = 3x
C2 = 4x
C3 = 5x
P1 : P2 : P3 =  
 
 
3x×6 +1.1×3x× 6
: 4x ×6 +1.15× 4x× 6
: 5x× 6 +1.2×5x×6
= 6.3 : 8.6 :11
= 63 : 86 :110
8. Consider the following equations of straight lines:
Line L1: 2x – 3y = 5
Line L2: 3x + 2y = 8
Line L3: 4x – 6y = 5
Line L4: 6x – 9y = 6
Which one among the following is the correct
statement ?
(a) L1 is parallel to L2 and L1 is perpendicular to
L3
(b) L2 is parallel to L4 and L2 is perpendicular to
L1
(c) L3 is perpendicular to L4 and L3 is parallel to
L2
(d) L4 is perpendicular to L2 and L4 is parallel to
L3

Solution
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Sol: (d)
 
  
 
1
2 5 2
L y = + x – slope
3 3 3
 
  
 
2
3
3
L y = – x + 4 slope –
2 2
 
  
 
3
2x 5 2
L y = + + slope –
3 6 3
 
  
 
4
2x 2 2
L y = + – slope
3 3 3
L4 is perpendicular to L2.

2
–3
×
2 3
= –1
L4 is parallel to L3
 Slopes are equal.
9. Given below are two statements and four
conclusions drawn based on the statements.
Statement 1: Some soaps are clean.
Statement 2: All clean objects are wet.
Conclusion I: Some clean objects are soaps.
Conclusion II: No clean object is a soap.
Conclusion III: Some wet objects are soaps.
Conclusion IV: All wet objects are soaps.
Which one of the following options can be logically
inferred?
(a) Only conclusion I is correct
(b) Either conclusion I or conclusion II is correct
(c) Either conclusion III or conclusion IV is correct
(d) Only conclusion I and conclusion III are correct
Sol: (d)
Probability (1)
Soaps
Clean
Wet
Probability (2)
Soaps clean
Wet
Conclusion (I) followed in both possibilities.
Conclusion (II) not followed.
Conclusion (III) followed in both possibilities.
Conclusion (IV) not followed.
10. An ant walks in a straight line on a plane leaving
behind a trace of its movement. The initial position
of the ant is at point P facing east.
The ant first turns 72° anticlockwise at P, and
then does the following two steps in sequence
exactly FIVE times before halting.
East
North
1. moves forward for 10 cm.
2. turns 144° clockwise.
The pattern made by the trace left behind by the
ant is
(a)
S
T R
P Q
PQ = QR = RS = ST = TP =10 cm
(b)
T S
U R
P Q

PQ = QR = RS ST = TU = UP =10cm
(c)
S
T R
P Q
SQ = QT = TR = RP = PS = 10cm
(d)
U T
W R
P
Q
S
SW = WR =RP =PT = TQ = QU=US =10 cm
Sol: (d)
Road map:

Solution
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72°
144°
72°
x axis
R
36°
144°
y axis
S
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TECHNICAL
11. The function f(x, y) satisfies the Laplace equation
 

2
f x,y = 0
on a circular domain of radius r = 1 with its
center at point P with coordinates x = 0, y = 0.
The value of this function on the circular boundary
of this domain is equal to 3.
The numerical value f(0, 0) is :
(a) 0 (b) 2
(c) 3 (d) 1
Sol: (c)
12.
 
 
 

2 3 4
x x x
x – + – +.... dx
2 3 4
is equal to
(a)
1
+ Constant
1+ x
(b) 2
1
+ Constant
1+ x
(c)
1
– + Constant
1– x
(d) 2
1
– + Constant
1– x
Sol: (*)
I =
 

 
 

2 3 4
x x x
x – + – +.... dx
2 3 4
I =
2 3 4 5
x x x x
– + – +....
2 6 12 20
Option (a)
1
1+ x
=   
–1 2 3
1+ x =1– x + x – x .....
So, its incorrect.
Option (b)
2
1
1+ x
= (1 + x2)–1 = 
2 4 6
1– x + x – x +...
So, its incorrect.
Similarly option (c) and (d) both are incorrect.
No-correct choice given.
13. For a linear elastic and isotropic material, the
correct relationship among Young’s modulus of
elasticity (E), Poisson’s ratio (v), and shear
modulus (G) is
(a)  
E
G =
2 1+ v
(b)  
E
G =
2 1+ 2v
(c)  
G
E =
2 1+ v
(d)  
G
E =
2 1+ 2v
Sol: (a)
14. Read the following statements relating to flexure
of reinforced concrete beams:
I. In over-reinforced sections, the failure strain in
concrete reaches earlier than the yield strain
in steel.
II. In under-reinforced sections, steel reaches
yielding at a load lower than the load at which
the concrete reaches failure strain.
III. Over-reinforced beams are recommended in
practice as compared to the under-reinforced
beams.
IV. In balanced sections, the concrete reaches
failure strain earlier than the yield strain in
tensile steel.
Each of the above statements is either True or
False.
Which one of the following combinations is
correct ?
(a) I (True), II (True), III (False), IV (False)
(b) I (True), II (True), III (False), IV (True)
(c) I (False), II (False), III (True), IV (False)
(d) I (False), II (True), III (True), IV (False)
Sol: (a)
The question is based on LSM design principle
as it is describing different conditions related to
strain.
Depending on amount of reinforcement in a cross-
section, here ca be three types of sections viz.
balanced, under reinforced and over reinforced.
Balanced section is a section that is expected to
result in a balanced failure. It means at the ultimate
limit state in flexure, the concrete will attain a
limiting compressive strain of 0.0035 and steel
will attain minimum specified tensile strain of

Solution
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
y
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0.87f
0.002
E
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Under reinforced section is a section in which
steel yield before collapse.
Over reinforced section is a section in which
crushing of concrete in compression i.e. attainment
of compressive strain of 0.0035 occurs prior to
yielding of steel.
In case of over reinforced section the deflection,
crack width remain relatively low and failure occurs
without any sign of warning and hence over
reinforced flexural members are not recommended
by IS code.
Based on the above information:
Statement I is true.
Statement II is true.
Statement III is false.
Statement IV is false.
15. Match all the possible combinations between
Column X (Cement compounds) and Column Y
(Cement properties) :
Column X Column Y
(i) C3S (P) Early age strength
(ii) C2S (Q) Later age strength
(iii) C3A (R) Flash setting
(S) Highest heat of hydration
(T) Lowest heat of hydration
correct?
(a) (i) - (P), (ii) - (Q) and (T), (iii) - (R) and (S)
(b) (i) - (Q) and (T), (ii) - (P) and (S), (iii) - (R)
(c) (i) - (P), (ii) - (Q) and (R), (iii) - (T)
(d) (i) - (T), (ii) - (S), (iii) - (P) and (Q)
Sol: (a)
C3S - Responsible for early age strength
C2S - Responsible for later age strength and
lowest heat of hydration
C3A - Flash setting and highest heat of hydration.
(i) - (P)
(ii) - Q & T
(iii) - R & S
16. Consider a beam PQ fixed at P, hinged at Q, and
subjected to a load F as shown in figure (not
drawn to scale). The static and kinematic degrees
of indeterminacy, respectively, are
P
F
Q

(a) 2 and 1 (b) 2 and 0
(c) 1 and 2 (d) 2 and 2
Sol: (a)

F
P
Q
MP
RP
HP
RQ
HQ
Unknown reactions = 5
Equilibrium equation = 3
Ds = 5 – 3
s
D = 2
k
D = 1 i.e., rotation (  ) at Q.
17. Read the following statements:
(P) While designing a shallow footing in sandy
soil, monsoon season is considered for critical
design in terms of bearing capacity.
(Q) For slope stability of an earthen dam, sudden
drawdown is never a critical condition.
(R) In a sandy sea beach, quicksand condition
can arise only if the critical hydraulic gradient
exceeds the existing hydraulic gradient.
(S) The active earth thrust on a rigid retaining wall
supporting homogeneous cohesionless backfill
will reduce with the lowering of water table in
the backfill.
correct?
(a) (P)-True, (Q)-False, (R)-False, (S)-False
(b) (P)-False, (Q)-True, (R)-True, (S)-True
(c) (P)-True, (Q)-False, (R)-True, (S)-True
(d) (P)-False, (Q)-True, (R)-False, (S)-False

Solution
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Sol: (c)
18. Stresses acting on an infinitesimal soil element
are shown in the figure (with   
z x ). The major
and minor principal stresses are 1 and 3 ,
respectively. Considering the compressive stresses
as positive, which one of the following expressions
correctly represents the angle between the major
principal stress plane and the horizontal plane ?
z
zx
zx
x x
zx
zx
z
(a)

 
 
 
 
–1 zx
1 x
tan
–
(b)

 
 
 
 
–1 zx
3 x
tan
–
(c)

 
 
 
 
–1 zx
1 x
tan
+
(d)

 
 
 
 
–1 zx
1 3
tan
+
Sol: (b)
z
x x
zx
z
Mohr’s circle by considering compressive stresses
positive and shear giving anticlockwise rotation
as positive.
 
3, 0
A
C O
B
180 – 2p
 
1, 0
 
 
z xz
Q ,
p
p
 
 
x xz
P , –
2p
From triangle ACP
A
C
P
P
xy
x 3
– 
P
tan =

 
xy
x 3
–
Since, nothing is mentioned about sign convention
of shear stress, so one can adopt negative sign
for given shear stress. In that case
P
tan =

 
xz
3 x
–
19. Match Column X with Column Y:
Column X Column Y
(P) Horton equation (I) Design of alluvial channel
(Q) Penman method (II) Maximum flood discharge
(R) Chezy’s formula (III) Evapotranspiration
(S) Lacey’s theory (IV) Infiltration
(T) Dicken’s formula (V) Flow velocity
correct ?
(a) (P)-(IV), (Q)-(III), (R)-(V), (S)-(I), (T)-(II)
(b) (P)-(III), (Q)-(IV), (R)-(V), (S)-(I), (T)-(II)
(c) (P)-(IV), (Q)-(III), (R)-(II), (S)-(I), (T)-(V)
(d) (P)-(III), (Q)-(IV), (R)-(I), (S)-(V), (T)-(II)
Sol: (a)
20. In a certain month, the reference crop
evapotranspiration at a location is 6 mm/day. If
the crop coefficient and soil coefficient are 1.2
and 0.8, respectively, the actual evapotranspiration
in mm/day is
(a) 5.76 (b) 7.20
(c) 6.80 (d) 8.00
Sol: (a)
Actual evapotranspiration (ETC)
= KS × KC × Reference evapotranspiration (ET0)
= 0.8 × 1.2 × 6
= 5.76 mm

Solution
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21. The dimension of dynamic viscosity is :
(a) –1 –1
ML T (b) –1 –2
ML T
(c) –2 –2
ML T (d) 0 –1
ML T
Sol: (a)
For Newtonian fluid,
The shear stress (  ) is experssed as
 =
 
 
 
du
, where
dy
du
dy
= shear strain rate, and
 = dynamic (or absolute)
viscosity of the fluid.
  =
 
 
  
     
 
   
 
   
Force
Area
=
du Length 1
×
dy Time Length
=
 
 
 
 
 
 
 
 
2
2
Force
Length Force×Time
=
1 Length
Time
[Force] = Mass × Acceleration
=
Velocity
Mass×
Time
=
 
 
 
Length
Time
Mass×
Time
[Force] = –2
MLT
Dimension of dynamic viscosity =  

=
 
 
 
–2
–1 –1
2
MLT ×T
= ML T
L
22. A process equipment emits 5 kg/h of volatile
organic compounds (VOCs). If a hood placed over
the process equipment captures 95% of the
VOCs, then the fugitive emission in kg/h is
(a) 0.25 (b) 4.75
(c) 2.50 (d) 0.48
Sol: (a)
VOC emission = 5 Kg/h
Capturing Efficiency = 95%
Fugitive emission =
5
×5 Kg h
100
= 0.25 Kg/h
23. Match the following attributes of a city with the
appropriate scale of measurements.
Attribute Scale of
measurement
(P) Average temperature (I) Interval
(°C) of a city
(Q) Name of a city (II) Ordinal
(R) Population density (III) Nominal
of a city
(S) Ranking of a city (IV) Ratio
based on ease of
business
correct?
(a) (P)-(I), (Q)-(III), (R)-(IV), (S)-(II)
(b) (P)-(II), (Q)-(I), (R)-(IV), (S)-(III)
(c) (P)-(II), (Q)-(III), (R)-(IV), (S)-(I)
(d) (P)-(I), (Q)-(II), (R)-(III), (S)-(IV)
Sol: (b)
24. If the magnetic bearing of the Sun at a place at
noon is S 2° E, the magnetic declination (in
degrees) at that place is
(a) 2°E (b) 2°W
(c) 4°E (d) 4°W
Sol: (a)
MB =    
S2 E = 180 – 2 = 178
TB = 180°
Declination,  = TB – MB
= 180 – 178
= 2° or 2°E
25. P and Q are two square matrices of the same
order. Which of the following statement(s) is/are
correct ?
(a) If P and Q are invertible, then  –1 –1 –1
PQ = Q P
(b) If P and Q are invertible, then  –1 –1 –1
QP = P Q
(c) If P and Q are invertible, then  –1 –1 –1
PQ = P Q
(d) If P and Q are not invertible, then
 –1 –1 –1
PQ = Q P
Sol: (a, b)
If P and Q are invertible then (PQ)–1 = Q–1P–1 is
correct.
Let, PQ = C

Solution
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–1
P PQ = –1
P C
Q = –1
P C
–1
Q Q = –1 –1
Q P C
I = –1 –1
Q P C
IC–1 = –1 –1 –1
Q P CC
C–1 = –1 –1
Q P
 –1
PQ = –1 –1
Q P
Hence, proved.
Similarly, we can prove if P, Q are invertible then
 –1
QP = –1 –1
P Q
26. In a solid waste handling facility, the moisture
contents (MC) of food waste, paper waste, and
glass waste were found to be MCf, MCp, and
MCg, respectively. Similarly, the energy contents
(EC) of plastic waste, food waste, and glass waste
were found to be ECpp, ECf, and ECg,
respectively. Which of the following statement(s)
is/are correct ?
(a) MCf > MCp > MCg
(b) ECpp > ECf > ECg
(c) MCf < MCp < MCg
(d) ECpp < ECf < ECg
Sol: (a, b)
Typical specific weight and moisture content data
for residential, commercial, industrial, and
agricultural wastes.
Type of waste
Specifif weight, lb/yd
3
Moisture content,
% by weight
Range Typical Typical
Range
Residential (uncomopacted)
*Food wastes (mixed)
*Paper
Cardboard
Plastics
Textiles
Rubber
Leather
yard wastes
Wood
*Glass
Tin cans
Aluminum
Other metals
Dirt, ashes, etc.
Ashes
Rubbish
Residential yard wastes
Leaves (loose and dry)
Green grass (loose and moist)
Green grass (wet and compacted)
yard waste (shredded)
Yard waste (composted)
Municipal
In compactor truck
In landfill
Normally compacted
Well compacted
Commercial
Food wastes (wet)
Appliances
220-810
70-220
70-135
70-220
70-170
170-340
170-440
100-380
220-540
270-810
85-270
110-405
220-1940
540-1685
1095-1400
150-305
50-250
350-500
1000-1400
450-600
450-650
300-760
610-840
995-1250
800-1600
250-340
490
150
85
110
110
220
270
170
400
330
150
270
540
810
1255
220
100
400
1000
500
550
500
760
1010
910
305
50-80
4-10
4-8
1-4
6-15
1-4
8-12
30-80
15-40
1-4
2-4
2-4
2-4
6-12
6-12
5-20
20-40
40-80
50-90
20-70
40-60
70
6
5
2
10
2
10
60
20
2
3
2
3
8
6
15
30
60
80
50
50
20
25
25
25
70
1
15-40
15-40
15-40
50-80
0-2
Moisture content
Mcf > Mcp > Mcg
Typical values for short residue and energy conent
of residentail MSW.
Component
Inert residue percent
a
Range Typical
Energy , Btu/lb
b
Range Typical
*Food wastes
*Paper
Cardboard
Plastics
Textiles
Rubber
Leather
Yard wastes
Wood
Misc. organics
Organic
Inorganic
*Glass
Tin cans
Aluminum
Other metal
Dirt, ashes, etc.
Municipal solid wastes
2-8
4-8
3-6
6-20
2-4
8-20
8-20
2-6
0.6-2
–
96-99+
96-99+
90-99+
94-99+
60-80
5.0
6.0
5.0
10.0
2.5
10.0
10.0
4.5
1.5
–
98.0
98.0
96.0
98.0
70.0
1,500-3,000
5,000-8,000
6,000-7,500
12,000-16,000
6,500-8,000
9,000-12,000
6,500-8,500
1,000-8,000
7,500-8,500
–
50-100c
100-500
c
–
100–500
c
1,000-5,000
4,000-6,000
2,000
7,200
7,000
14,000
7,500
10,000
7,500
2,800
8,000
–
60
300
–
300
3,000
5,000
After complete combustion
As discarded basis
Energy content is from coatings, labels and attached materials.
a
b
c
Note : Btu/lb × 2.326 = kJ/kg
Energy content
ECpp > ECf > ECg
27. To design an optimum municipal solid waste
collection route, which of the following is/are NOT
desired :
(a) Collection vehicle should not travel twice down
the same street in a day.
(b) Waste collection on congested roads should
not occur during rush hours in morning or
evening.
(c) Collection should occur in the uphill direction.
(d) The last collection point on a route should be
as close as possible to the waste disposal
facility.
Sol: (c)
Some heuristic guidelines that should be taken
into consideration when laying out routes are as
follows :
1. Existing policies and regulations related to
such items as the point of collection and
frequency of collection must be identified.
2. Existing system characteristics such as crew
size and vehicle types must be coordinated.
3. Wherever possible, routes should be laid out
so that they begin and end near arterial
streets, using topographical and physical
barriers as route boundaries.
4. In hilly area, routes should start at the top of
the grade and proceed downhill as the vehicle
becomes loaded.

Solution
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5. Routes should be laid out so that the last
container to be collected on the route is
located nearest to the disposal site.
6. Waste generated at traffic-congested locations
should be collected as early in the day as
possible.
7. Sources at which extremely large quantities
of waste are generated should be serviced
during the first part of the days.
8. Scattered pickup points (where small quantities
of solid waste are generated) that receive the
same collection frequency should, if possible,
be serviced during one trip or on the same
day.
28. For a traffic stream, v is the space mean speed,
k is the density, q is the flow, vf is the free flow
speed, and kj is the jam density. Assume that
the speed decreases linearly with density.
Which of the following relation(s) is/are correct ?
(a)
 
 
 
j 2
j
f
k
q = k k – k
v
(b)
 
 
 
2
f
f
j
v
q = v k – k
k
(c)
 
 
 
2
f
f
j
v
q = v v – v
k
(d)
 
 
 
j 2
j
f
k
q = k v – v
v
Sol: (b)
As per Greenshield’s
V =
 
 
 
f
j
K
V 1–
K
and q = K × V
=
 
 
 
f
j
K
K 1– V
K
q =
 
 
 
2
f
f
j
V
V K – K
K
29. The error in measuring the radius of a 5 cm circular
rod was 0.2%. If the cross-sectional area of the
rod was calculated using this measurement, then
the resulting absolute percentage error in the
computed area is______.
(round off to two decimal places)
Sol: (0.40)
r = 5 cm
er =
0.2
×5cm = 0.01cm
100
A =  2
r
eA =  r
2 r.e
Absolute perecentage error in computed area
= A
e
×100
A
=


r
2
2 r.e
×100
r
=
 
 
 
r
e
2× ×100
r
= 2×0.2% = 0.4%
30. The components of pure shear strain in a sheared
material are given in the matrix form :
 
  
 
1 1
=
1 –1
Here, Trace() = 0. Given, P= Trace (8 ) and Q
= Trace (
11 ).
The numerical value of (P+Q) is ________. (in
integer)
Sol: (32)

A – I = 0


1– 1
1 –1–
= 0
  
  
1– 1– –1 = 0
 
2
– 1 – 1 = 0
 =  2
Eigen values of  are 2 and – 2
Eigen values 8
are  
8
2 and  
8
–2
Eigen values of 11
are  
11
2 and  
11
– 2 .
P = Trace 8
( ) = Sum of Eigen values
=    
8 8
2 + – 2
= 16 + 16 = 32
Q = Trace 11
( ) = Sum of Eigen values
=    
11 11
2 + – 2
= 0
P + Q = 32 + 0
= 32
31. The inside diameter of a sampler tube is 50 mm.
The inside diameter of the cutting edge is kept

Solution
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such that the Inside Clearance Ratio (ICR) is 1.0
% to minimize the friction on the sample as the
sampler tube enters into the soil.
The inside diameter (in mm) of the cutting edge
is _________________. (round off to two decimal
places)
Sol: (25)
D4
D3
D1
D2
Sampling tube
Cutting edge
D3 = 50 mm, ICR = 1%
ICR =
3 1
1
D – D
×100
D
1 =
1
1
50 – D
D
D1 = 25 mm
32. A concentrically loaded isolated square footing of
size 2 m × 2 m carries a concentrated vertical
load of 1000 kN. Considering Boussinesq’s theory
of stress distribution, the maximum depth (in m)
of the pressure bulb corresponding to 10 % of the
vertical load intensity will be ___________. (round
off to two decimal places)
Sol: (4.37)
Considering Boussinesq’s theory of stress
distribution
x =
 
 
  
 
 
 
 
 
5/2
2 2
3Q 1
2 z r
1+
z
For r = 0, 

z 2
3Q
=
2 z
z = 2 2
Q 0.1
0.1q = 0.1× = Q
B 2
1
Q
40
=
 2
3Q
2 z
z2 = 

3× 40
z = 4.37m
2
33. In a triaxial unconsolidated undrained (UU) test
on a saturated clay sample, the cell pressure
was 100 kPa. If the deviatoric stress at failure
was 150 kPa, then the undrained shear strength
of the soil is _________ kPa. (in integer)
Sol: (75)
3 = c =100kPa
d =  
1 3
– = 150Kpa
For UU test
Cuu
3 1
 
1 3
–
Undrained shear strength
=
 
 1 3
uu f
–
C = =
2
=
150
2
= 75 KPa
34. A flood control structure having an expected life
of n years is designed by considering a flood of
return period T years. When T = n, and  
n ,
the structure’s hydrologic risk of failure in
percentage is ____.
(round off to one decimal place)
Sol: (0.6)
Risk of failure = n
1– q
=  n
1– 1 – p
=
 
 
 
n
1
1– 1–
T
For T =  
n
Risk of failure =
1
1– = 0.632
e
35. The base length of the runway at the mean sea
level (MSL) is 1500 m. If the runway is located at
an altitude of 300 m above the MSL, the actual
length (in m) of the runway to be provided is
____________. (round off to the nearest integer)
Sol: (1604 to 1606)

Solution
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Correction for elevation = It should increase at a
rate of 7% per 300 m rise in elevation from MSL.
Given that,
Basic runway length at MSL = 1500 m
Elevation = 300 m
Correction =
7 300
× ×1500
100× 300
= 105 m
The actual length of runway
= 1500 + 105 = 1605 m
36. Consider the polynomial f(x) = x3 – 6x2 + 11x –
6 on the domain S given by  
1 x 3 . The first
and second derivatives are  
f (x) and f (x).
Consider the following statements:
I. The given polynomial is zero at the boundary
points x = 1 and x = 3.
II. There exists one local maxima of f(x) within
the domain S.
III. The second derivative  
f (x) 0 throughout the
domain S.
IV. There exists one local minima of f(x) within
the domain S.
The correct option is:
(a) Only statements I, II and III are correct
(b) Only statements I, II and IV are correct
(c) Only statements I and IV are correct
(d) Only statements II and IV are correct
Sol: (b)
 
f x = 3 2
x – 6x +11x – 6
f(1) = 1– 6 +11– 6 = 0
f(3) = 3 2
3 – 6×3 +11×3 – 6 = 0
Statement (I) is correct.
 

f x = 2
3x – 12x +11
 

f x =  
1
0 x = 2
3
Point of maxima
Point of minima
+ +

– 
1 2 3
1
2 –
3
1
2+
3
–
f(x) has local maxima at
1
x = 2 –
3
Statement (II) is also true.
Now,  

f x = 6x – 12
 

f x > 0
6x – 12 > 0
x > 2
Statement (III) is incorrect statement (IV) is also
correct.
 x =
1
2 +
3
is point of minima.
37. An undamped spring-mass system with mass m
and spring stiffness k is shown in the figure. The
natural frequency and natural period of this system
are  rad/s and T s, respectively. If the stiffness
of the spring is doubled and the mass is halved,
then the natural frequency and the natural period
of the modified system, respectively are
m
k
(a) 
2 rad/s and T/2 s
(b) /2 rad/s and 2T s
(c) 
4 rad/s and T/4 s
(d)  rad/s and T s
Sol: (a)
m
k
This is question of undamped free vibration of
sinle degree of freedom system.
For the above figure it is given that natural
frequency and natural period are w rad/s and T s,
respectively.
The equation for soof system is
 =
k
m
and T =


2

Solution
Detailed
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For the modified system, stiffness of the spring is
doubled and the mass is halved.
Let us assume k' = 2k and m' =
m
,
2
here k' and
m' represent the stiffness and mass of the
modified system respectively.
 
 =

 
 
 
k 2k 4k
= =
m
m m
2
=
 
 
 
 
 
k k
2 = 2 =
m m

and T' =
  
 
 

  
 
2 2 1 2 T
= = =
2 2 2

 
 

 
2
T =

 
 and T are natural frequency and natural period
of the modified system.
Hence, the natural frequency of the modified
system is doubled and natural period is halved
compared to the original given system. Hence
option (a) is correct.
38. For the square steel beam cross-section shown
in the figure, the shape factor about Z–Z axis is
S and the plastic moment capacity is MP. Consider
yield stress fy = 250 MPa and a = 100 mm.
K
a a
a
a
z z
J
H
I
The values of S and MP (rounded-off to one decimal
place) are
(a) S = 2.0, MP = 58.9 kN-m
(b) S = 2.0, MP = 100.2 kN-m
(c) S = 1.5, MP = 58.9 kN-m
(d) S = 1.5, MP = 100.2 kN-m
Sol: (a)
K
a a
a
a
z z
J
H
I
Shape factor for diamond shaped section = 2
S =
y P
P
y y e
f Z
M
=
M f Z
or, MP = S.My = S.fy.Ze
K
a a
a
a
z z
J
H
I
2a
2a
2
IZZ =
4
a
12
Zzz =
4 3
3
a × 2 2a
= mm
12×a 12
So, MP =
 
 
 
 
3
–6
2× 100
2×250× ×10 kNm
12
= 58.93 kNm
39. A post-tensioned concrete member of span 15 m
and cross-section of 450 mm × 450 mm is
prestressed with three steel tendons, each of
cross-sectional area 200 mm2. The tendons are
tensioned one after another to a stress of 1500
MPa. All the tendons are straight and located at
125 mm from the bottom of the member. Assume
the prestress to be the same in all tendons and
the modular ratio to be 6. The average loss of
prestress, due to elastic deformation of concrete,
considering all three tendons is
(a) 14.16 MPa (b) 7.08 MPa
(c) 28.32 MPa (d) 42.48 MPa
Sol: (a)
This is a question of calculation of elastic
shortening loss in post tensioned member.
Given data,
length = 15 m
b × D = 450 mm × 450 mm
Numberof tendons = 3
Cross-section area of each tendon (As)
= 200 mm2.
Prestress = 1500 MPa
Modular ratio (m) = 6

Solution
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15 m
(L-section) (Cross-section)
125 mm
e
C
L
450 mm
450
mm
From the given data, eccentricity (e) =
450
– 125 = 100 mm
2
Force in each cable (P) = 1500 × 200 × 10–3 =
300 kN
The tendons are tensioned one after another, and
hence when tendon (1) is pulled no loss in tenson
(1)
When tendon (2) is pulled, loss in tendon (1) but
no loss in tendon (2).
When tendon (3) is pulled, loss in tendon (1) and
(2), but no loss in tendon (3).
Hence, there will be 2 times losses in tendon (1),
time loss in tendon (1) and no loss in tendon (3).
While calculating elastic shortening loss, self
weight of the structure is neglected to be on the
conservative side.
Consider tensioning of tendon-1
No loss in tendon (1)
Consider tensioning of tendon-2
e
C
L
P
P
Stress in concrete at the level of prestressing
tendon
fc =
P Pe
+
A I
e =
 
 
 
 
3 3
3
300×10 300×10 ×100×100
+
450× 450 450× 450
12
= 2.36 MPa
e
P
C
L
I = moment of inertial of the section about the
centroidal axis.
As the tendons are horizontal and at the same
level fc,avg = fc.
Loss due to elastic deformation = mfc
= 6 × 2.36 = 14.16 MPa.
Considering tensioning of tendon 3
Loss due to elastic deformation in (1)
= mfc = 6 × 2.36 = 14.16 MPa
Loss due to elastic deformation in (2)
= mfc = 6 × 2.36 = 14.16 MPa
 Total loss in tendon (1) = 2 × 14.16
= 28.32 MPa
 Total loss in tendon (1)
= 2 × 14.16 = 28.32 MPa
In tendon (2) = 14.16 MPa
In tendon (3) = 0
 Average loss of pre-stress, considering all
three tendons is
=
28.32 +14.16 + 0
3
= 14.16 MPa
40. Match the following in Column X with Column Y:
Column X
(P) In a triaxial compression test, with increase
of axial strain in loose sand under drained
shear condition, the volumetric strain
(Q) In a triaxial compression test, with increase
of axial strain in loose sand under undrained
shear condition, the excess pore water
pressure
(R) In a triaxial compression test, the pore
pressure parameter “B” for a saturated soil
(S) For shallow strip footing in pure saturated
clay, Terzaghi’s bearing capacity factor (Nq)
due to surcharge
Column Y
(I) Decreases
(II) Increases
(III) Remains same
(IV) is always 0.0
(V) is always 1.0
(VI) is always 0.5

Solution
Detailed
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correct?
(a) (P)-(I), (Q)-(II), (R)-(V), (S)-(V)
(b) (P)-(II), (Q)-(I), (R)-(IV), (S)-(V)
(c) (P)-(I), (Q)-(III), (R)-(VI), (S)-(IV)
(d) (P)-(I), (Q)-(II), (R)-(V), (S)-(VI)
Sol: (a)
(P) In a triaxial compression test, with increase
of axial strain in loose sand under drained
shear condition, the volumetric strain
decreases. As the sample is compressing.
Hence, volume is decreasing.
Volume
change
+
–
Dense sand
Loose sand
(Q) In a triaxial compression test, with increase
of axial strain in loose sand under undrained
shear condition, the excess pore water
pressure increases.
(R) In a triaxial compression test, the pore pressure
parameter “B” for a saturated soil is 1.
(S) For shallow strip footing in pure saturated
clay, Terzaghi’s bearing capacity factor (Nq)
due to surcharge is 1.
41. A soil sample is underlying a water column of
height h1, as shown in the figure. The vertical
effective stresses at points A, B and C are
  
  
A B C
, and , respectively. Let 
 
sat and be
the saturated and submerged unit weights of the
soil sample, respectively, and w be the unit
weight of water. Which one of the following
expressions correctly represents the sum
  
   
A B C
( + ) ?
h3
h2
h1
Water level
Closed valve
A
B
C
Saturated soil
Water
(a) 
 
2 3
(2h h ) (b) 
  
1 2 3
(h h h )
(c)    
2 3 sat w
(h h )( ) (d)   
1 2 3 sat
(h h h )
Sol: (a)
h3
h2
h1
Water level
Closed valve
A
B
C
Saturated soil
Water

A = 0

B = 

2
h

C =   

2 3
h +h
  
  
A B c
+ + =   

2 3
2h +h
42. A 100 mg of HNO3 (strong acid) is added to
water, bringing the final volume to 1.0 liter.
Consider the atomic weights of H, N, and O, as
1 g/mol, 14 g/mol, and 16 g/mol, respectively.
The final pH of this water is (Ignore the dissociation
of water.)
(a) 2.8 (b) 6.5
(c) 3.8 (d) 8.5
Sol: (a)

+ –
3 3
HNO H +NO
1 mole of HNO3 = 1 mole of H+
No. of moles of HNO3 =  
100
m mol
1+14 +16×3
= 1.587 m mol
[H+] in 1 lit. of water= 1.587 m mol Lit
= –3
1.587×10 mol lit
pH =  
+
–log H
=  
–3
–log 1.587×10
= 2.8
43. In a city, the chemical formula of biodegradable
fraction of municipal solid waste (MSW) is
C100H250O80N. The waste has to be treated by
forced-aeration composting process for which air
requirement has to be estimated.

Solution
Detailed
Ph:
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Assume oxygen in air (by weight) = 23%, and
density of air = 1.3 kg/m3.
Atomic mass: C = 12, H = 1, O = 16, N = 14.
C and H are oxidized completely whereas N is
converted only into NH3 during oxidation.
For oxidative degradation of 1 tonne of the waste,
the required theoretical volume of air (in m3/tonne)
will be (round off to the nearest integer)
(a) 4749 (b) 8025
(c) 1418 (d) 1092
Sol: (a)

100 250 80 2 2 2 3
247
C H O N+xO 100CO + H O+NH
2
80 + x × 2 =
247
200 +
2
x = 121.75
1 mole of MSW required 121.75 mole of oxygen
No. of moles of MSW
=  
1 ton
100×12 + 250 + 80×16 +14 g
No. of moles of O2 required =
1 tonne
121.75×
2744g
wt. of O2 =
121.74
×32tonne
2744
Mass of air =
3
1.42
×10 Kg
0.23
= 3
6.174×10 kg
Density of air = 1.3 Kg/m3
Volume of air =
3 3
6.174
×10 m
1.3
= 4749.2 m3 per tonne of MSW
44. A single-lane highway has a traffic density of 40
vehicles/km. If the time-mean speed and space-
mean speed are 40 kmph and 30 kmph,
respectively, the average headway (in seconds)
between the vehicles is
(a) 3.00 (b) 2.25
(c) 8.33 × 10–4 (d) 6.25 × 10–4
Sol: (a)
Given that,
Traffic density = 40 Veh/Km.
Time mean speed = 40 Kmph
Space mean speed = 30 kmph
Time Headway (sec) = ?
We know that traffic volume
= Density × Space mean speed
= 40×30 =1200 Veh hr.
and we also know that
q =
t
3600
H
where Ht - average headway (sc)
1200 =
t
3600
H
Ht =
3600
= 3 secs.
1200
45. Let y be a non-zero vector of size 2022 × 1.
Which of the following statement(s) is/are TRUE?
(a) yyT is a symmetric matrix
(b) yTy is an eigenvalue of yyT
(c) yyT has a rank of 2022
(d) yyT is invertible
Sol: (a,b)
Let vector y =

 
 
 
 3 1
4
4
4
yT =   
1 3
4 4 4
yyT =  
 
 
 
 
4
4 4 4 4
4
=

 
 
 
 
 3 3
16 16 16
16 16 16
16 16 16
yTy = [42 + 42 + 42]1 × 1
= [48]1 × 1
(y) =      
T T T
(y ) (yy ) (y y) 1
From above information
 yyT is asymmetric.
 yTy is an eigen value of yyT.
yyT has rank 1
det (yyT) = 0 so, yyT is not invertible

Solution
Detailed
Ph:
CE
I
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M
A
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E
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46. Which of the following statement(s) is/are correct?
(a) If a linearly elastic structure is subjected to
a set of loads, the partial derivative of the
total strain energy with respect to the
deflection at any point is equal to the load
applied at that point.
(b) If a linearly elastic structure is subjected to
a set of loads, the partial derivative of the
total strain energy with respect to the load at
any point is equal to the deflection at that
point.
(c) If a structure is acted upon by two force
system Pa and Pb , in equilibrium separately,
the external virtual work done by a system of
forces Pb during the deformations caused by
another system of forces Pa is equal to the
external virtual work done by the Pa system
during the deformation caused by the Pb
system.
(d) The shear force in a conjugate beam loaded
by the M/EI diagram of the real beam is equal
to the corresponding deflection of the real
beam.
Sol: (a,b,c)
47. Water is flowing in a horizontal, frictionless,
rectangular channel. A smooth hump is built on
the channel floor at a section and its height is
gradually increased to reach choked condition in
the channel. The depth of water at this section is
y2 and that at its upstream section is y1. The
correct statement(s) for the choked and unchoked
conditions in the channel is/are
(a) In choked condition, y1 decreases if the flow
is supercritical and increases if the flow is
subcritical.
(b) In choked condition, y2 is equal to the critical
depth if the flow is supercritical or subcritical.
(c) In unchoked condition, y1 remains unaffected
when the flow is supercritical or subcritical.
(d) In choked condition, y1 increases if the flow
is supercritical and decreases if the flow is
subcritical.
Sol: (a,b,c)
E1 y1
1
1
y2
E2
Z
Hump
Energy line
2
2
Horizontal, frictionless, rectangular channel.
Depth
y
y1
y2
yc
C
R P
Z
E2
E1
Zm
Specific
energy
R
P
Q = const.
1
2
Case-I: Subcritical flow at section 1-1:
 For a subcritical flow, a decrease in specific
energy is associated with a decrease in depth
of flow and increase in velocity.
 When     max
0 Z Z , upstream water level
remains stationary at y1 while the depth of
flow at section 2 decreases with y2 reaching
a minimum value of yc at   max
Z = Z (choking
condition).
 With further increase in the value of Z (i.e.,
   max
Z Z ), y2 will continue to remain at yc
and y1 will increase to 
1
y , to have higher
specific energy 
1
E .
Zmax
Z
y and y
1 2
Depth y1
Depth y2
yc
Case-II: Supercritical flow at section 1-1:
 For a supercritical flow, a decrease in specific
energy is associated with an increase in the
depth of flow and decrease in velocity head.

Solution
Detailed
Ph:
CE
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 When     max
0 Z Z depth of flow y1 at
section 1, is constant while the depth of flow
y2 at section 2 increase upto critical depth yc
at   max
Z = Z (choking condition).
 For    max
Z Z , the depth of flow over hump
y2 = yc will remain constant and the upstream
depth y1 decreases to 
1
y , to have higher
specific energy 
1
E .
Z
yc
Depth y2
Depth y1
y and y
1 2
Zmax
48. The concentration s(x, t) of pollutants in a one-
dimensional reservoir at position x and time t
satisfies the diffusion equation
 

 
2
2
s(x, t) s(x, t)
D
t x
on the domain  
0 x L, , where D is the diffusion
coefficient of the pollutants. The initial condition
s(x, 0) is defined by the step-function shown in
the figure.
x
L
0.4L
0
0
s0
s(x, t = 0)
The boundary conditions of the problem are given
by



s(x, t)
0
x
at the boundary points x = 0 and
x = L at all times. Consider D = 0.1 m2/s, s0 =
5  mol/m, and L = 10 m.
The steady-state concentration
   
 
   
   
 L L
s s ,
2 2
at the center x =
L
2
of the reservoir (in  mol/m)
is _______ (in integer)
Sol: (*)
49. A pair of six-faced dice is rolled thrice. The
probability that the sum of the outcomes in each
roll equals 4 in exactly two of the three attempts
is ______. (round off to three decimal places)
Sol: (0.02)
Event, E = {(1, 3)(3, 1)(2, 2)}
n(E) = 3
n(S) = 36
p = P(E) = 
3 1
36 12
q =   
1 11
P(E) 1
12 12
P(x) = 3C2(p2)(q1)
=
   
   
   
2
1 11
3
12 12
= 0.01909 = 0.02
50. Consider two linearly elastic rods HI and IJ, each
of length b, as shown in the figure. The rods are
co-linear, and confined between two fixed supports
at H and J. Both the rods are initially stress free.
The coefficient of linear thermal expansion is 
for both the rods. The temperature of the rod IJ is
raised by T , whereas the temperature of rod HI
remains unchanged. An external horizontal force
P is now applied at node I. It is given that
 
     
6 1
10 C , T 50 C, b = 2m, AE = 106 N.
The axial rigidities of the rods HI and IJ are 2AE
and AE, respectively.
H
b
P
b
AE
J
2AE I
To make the axial force in rod HI equal to zero,
the value of the external force P (in N) is
____________ (round off to the nearest integer)
Sol: (50)
b
b
AE
J
H 2AE
P
I
For axial force to be zero in rod HI, axial
deformation will be zero i.e., force P balances
the expansion due to temperature in rod IJ.

Solution
Detailed
Ph:
CE
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M
A
S
T
E
R
For rod IJ
P
P
net =  
axial load temp
+ = 0
 
P
– T
AE

 = 0
 P =  

AE T
P = 106(10–6 × 50)
P = 50N
51. The linearly elastic planar structure shown in the
figure is acted upon by two vertical concentrated
forces. The horizontal beams UV and WX are
connected with the help of the vertical linear spring
with spring constant k = 20 kN/m. The fixed
supports are provided at U and X. It is given that
flexural rigidity EI = 105 kN-m2, P = 100 kN, and
a = 5 m. Force Q is applied at the center of
beam WX such that the force in the spring VW
becomes zero.
W 2EI X
a
a
Q
k
V
U 2a
EI
P
The magnitude of force Q (in kN) is _______
(round off to the nearest integer)
Sol: (640)
U
2a
EI
P
V
W
a a
2 EI
X
Q
If force in spring is zero, there will be no
deformation in spring i.e., deflection of point V
will be equal to deflection of point W
2a
EI
P
V
2 EI
W
a a
Q
v = w
 3
P 2a
3EI
=
 
   
3 2
Q a Qa
+ + a
3 2EI 2 2EI

8
P
3
=
 
 
 
1 1
+ Q
6 4
Q =
32
P = 640kN
5
52. A uniform rod KJ of weight w shown in the figure
rests against a frictionless vertical wall at the
point K and a rough horizontal surface at point J.
It is given that w = 10 kN, a = 4 m and b = 3 m.
b
J
K
a
The minimum coefficient of static friction that is
required at the point J to hold the rod in equilibrium
is _______ (round off to three decimal places)
Sol: (0.375)
K
J
3m
4m
FBD
4 m
J
RJ
s J
R
10 kN
Rk
K 1.5 m 1.5 m
x
y
 v
F = 0
 J
R = 10kN
 K
M = 0

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
 s J J
10×1.5 + R × 4 – R ×3 = 0
 s =
10×3 – 10×1.5
= 0.375
10× 4
53. The activities of a project are given in the following
table along with their durations and dependency.
Activities Duration (days) Depends on
A 10 –
B 12 –
C 5 A
D 14 B
E 10 B, C
The total float of the activity E (in days) is
_________ (in integer).
Sol: (1)
1
2 4
3 5
A C
E
D
B
10 5
10
14
(0, 0)
(12, 12)
(10, 11) (15, 16)
(26, 26)
For activity E,
Total float (FT) =
5 4 E
L L
T – T – t
= 26 – 15 – 10 =1day
54. A group of total 16 piles are arranged in a square
grid format. The center-to-center spacing (s)
between adjacent piles is 3 m. The diameter (d)
and length of embedment of each pile are 1 m
and 20 m, respectively. The design capacity of
each pile is 1000 kN in the vertical downward
direction. The pile group efficiency g
( ) is given
by
   
 
   
 
g
(n 1) m (m 1) n
1–
90 mn
where m and n are number of rows and columns
in the plan grid of pile arrangement, and
  
   
 
1 d
tan .
s
The design value of the pile group capacity (in
kN) in the vertical downward direction is
__________ (round off to the nearest integer)
Sol: (11088)
S=3m
n = 4, m = 4
S = 3m, d = 1 m
L = 20 m
 =
   

   
   
–1 –1
d 1
tan = tan =18.435
s 3
ng =
   
 

 
  
n – 1 m+ m – 1 n
1–
90 mn
=
   
 
 
 
18.435 4 –1 ×4 + 4 –1 ×4
1–
90 4×4
= 0.693
Design value of pile group capacity
= 16×1000×0.693
= 11088 KN
55. A saturated compressible clay layer of thickness
h is sandwiched between two sand layers, as
shown in the figure. Initially, the total vertical stress
and pore water pressure at point P, which is
located at the mid-depth of the clay layer, were
150 kPa and 25 kPa, respectively. Construction
of a building caused an additional total vertical
stress of 100 kPa at P. When the vertical effective
stress at P is 175 kPa, the percentage of
consolidation in the clay layer at P is __________.
(in integer)
Sand
Sand
h/2
h/2
P clay
Sol: (50)

Solution
Detailed
Ph:
CE
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Initial effective stress at P

 1
P = 150 – 25
= 125 kPa
After construction,

 2
P = 125 + 100 = 225 kPa
100% consolidation will occur when effective stress
at P reaches 225 kPa.
Percentage of consolidation when vertical effective
stress is at P is 175 kPa.
=
175 –125
×100
225 – 125
=
50
×100
100
= 50
56. A hydraulic jump takes place in a 6 m wide
rectangular channel at a point where the upstream
depth is 0.5 m (just before the jump). If the
discharge in the channel is 30 m3/s and the
energy loss in the jump is 1.6 m, then the Froude
number computed at the end of the jump is
___________. (round off to two decimal places)
(Consider the acceleration due to gravity as 10
m/s2).
Sol: (0.3 to 0.4)
Given, in a rectangular channel
y
B
y1
y2
Q = 30 m3/sec,
B = 6 m
y1 = 0.5 m
EL = 1.6 m
q =
   
   
   
3
Q 30
= = 5 m / sec/ m
B 6
Method-I
We know that,
EL =
 3
2 1
1 2
y – y
= 1.6m
4 y y
...(i)
 3
2
2
y – 0.5
4× y ×0.5
= 1.6
 3 2
2 2 2
y – 1.5y + 0.75y – 0.125 = 3.2y2
 y2 = 2.5 m, – 0.0527 m, – 0.947 m
Hence, y2 = 2.5 m
Post jump Froude’s No. (F2) =
2
2
V
gy
=
 
 
 
30
6× 2.5
= 0.4
10×2.5
This problem can also be solved by using other
given data:
Method-II:
For rectangular horizontal channel,
2
2q
g
=  
1 2 1 2
y + y × y y

 2
2× 5
10
=  
2 2
0.5 + y ×0.5× y
 2
2 2
0.5y + 0.25y – 5 = 0
 y2 = 2.922 m, – 3.422 m
Adopt, y2 = 2.922 m
Post jump Froude’s No. (F2) =
2
2
V
gy
=
 
 
 
30
2.922× 6
10× 2.922
= 0.316
Note: From Method-II result,
EL =
   
3 3
2 1
1 2
y – y 2.922 – 0.5
=
4y y 4×2.922×0.5
= 2.431 m
Thus, it is clearly a case of discrepancy in
given data in the question.

Solution
Detailed
Ph:
CE
I
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S
M
A
S
T
E
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57. A pump with an efficiency of 80% is used to draw
groundwater from a well for irrigating a flat field of
area 108 hectares. The base period and delta for
paddy crop on this field are 120 days and 144
cm, respectively. Water application efficiency in
the field is 80%. The lowest level of water in the
well is 10 m below the ground. The minimum
required horse power (h.p.) of the pump is
________. (round off to two decimal places)
(Consider 1 h.p. = 746 W; unit weight of water =
9810 N/m3)
Sol: (30.82)
Base period (B) = 120 days
Delta  
 = 144 cm = 1.44 m
Duty (D) =

8.64B 8.64×120
=
1.44
D = 720
hec
cumec
Q =
3
Area 108 3
= = m /s
D 720 20
Qapplied =
a
Q 3 3
= =
20×0.8 16
Power required =
mgh
t
=
 
  
 
w
volume
× g×h
t
=  
wQgh = Qh
=
3 73575
9810× ×10 = watt
16 4
Horse power of pump =
Power required
746×efficiency
=
73575
= 30.82 hp
4×746×0.8
58. Two discrete spherical particles (P and Q) of equal
mass density are independently released in water.
Particle P and particle Q have diameters of 0.5
mm and 1.0 mm, respectively. Assume Stokes’
law is valid.
The drag force on particle Q will be________ times
the drag force on particle P. (round off to the
nearest integer)
Sol: (8 to 8)
In case of discrete particle settling and Stoke’s
law valid, at terminal velocity, since there is no
change in velocity, the net force on the body
is zero. Hence,
 
 
 
weight of sphere
–buoyant force
= Drag force (FD)
Vterminal
drag
force (F )
D
s
FB
Buoyant force
Weight
fluid f
= ( )

 FD =
 
 
 
 
 
3 3
s f
D g – D g
6 6
 

 
3
D s f
F = gD –
6
For density of medium  
f and mass density
of sphere  
s constant,
Drag force (FD)  D3
For particle P, diameter (DP) = 0.5 mm
For particle Q, diameter (DQ) = 1 mm

 
 
D Q
D P
F
F =
 
 
 
 
 
3 3
Q
3
P
D 1
= = 8
0.5
D
 (FD)Q = 8 × (FD)P
59. At a municipal waste handling facility, 30 metric
ton mixture of food waste, yard waste, and paper
waste was available. The moisture content of this
mixture was found to be 10%. The ideal moisture
content for composting this mixture is 50%. The
amount of water to be added to this mixture to
bring its moisture content to the ideal condition is
_______metric ton. (in integer)
Sol: (12)
MSW = 30 metric ton
Moisture content of MSW = 10%
Amount of water present

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
=
10
30× = 3metric ton
100
Ideal moisture content of MSW = 50%
Amount of water should be present for composting
=
50
30× =15 metric ton
100
Amount of water to be added to mixture to bring
its M/C to ideal condition
= 15 – 3 =12 metric ton
60. A sewage treatment plant receives sewage at a
flow rate of 5000 m3/day. The total suspended
solids (TSS) concentration in the sewage at the
inlet of primary clarifier is 200 mg/L. After the
primary treatment, the TSS concentration in
sewage is reduced by 60%. The sludge from the
primary clarifier contains 2% solids concentration.
Subsequently, the sludge is subjected to gravity
thickening process to achieve a solids
concentration of 6%. Assume that the density of
sludge, before and after thickening, is 1000 kg/
m3.
The daily volume of the thickened sludge (in m3/
day) will be_________. (round off to the nearest
integer)
Sol: (10)
PST
Q=5000 m /d
3
TSS = 200 mg/L
40% TSS
Gravity
Thickening
60% TSS
2% solids
wt. of TSS at inlet of PST
=
3 –6
L Kg
5000×10 × 200×10
d L
= 1000 Kg/d
wt. of solids in sludge from PST = 0.6×1000
= 600 Kg/d
wt. of sludge before thickening =
600
Kg d
2
100
= 30000 Kg/d
wt. of sludge after thickening =
600
Kg d
6
100
= 10000 Kg/d
Daily volume of thickened sludge
=
2
10000
m
1000
= 10 m3
61. A sample of air analyzed at 25°C and 1 atm
pressure is reported to contain 0.04 ppm of SO2.
Atomic mass of S = 32, O = 16.
The equivalent SO2 concentration (in  g/m3) will
be__________. (round off to the nearest integer)
Sol: (157)
Concentration of SO2 in ppm = 0.04
Let’s equivalent concentration in 
3
g m is x.
 
x g of SO2 present in 1 m3 of air.
.
We know, 1 mole of SO2 (at 0°C, 1 atm) has
volume of 22.4 L.
at 25° C and 1 atm, volume of SO2
=  
22.4
× 273 + 25
273 + 0
= 24.45 lit
–6
x×10
96
mole of SO2 has volume
=
–6
24.45× x×10
lit.
96
in 1m3 of air
–3 3
0.255x×10 m of SO2 present in 6 3
10 m of
air.
 –3
0.255x ×10 = 0.04
x = 
156.86 157
0.04 ppm of SO2 = 
3
2
157 g m of SO
62. A parabolic vertical crest curve connects two road
segments with grades +1.0% and –2.0%. If a 200
m stopping sight distance is needed for a driver
at a height of 1.2 m to avoid an obstacle of height
0.15 m, then the minimum curve length should be
______ m. (round off to the nearest integer)
Sol: (271 to 274)
Given that, 1
n = +1% and 2
n = –2%
n = 1 2
n – n

Solution
Detailed
Ph:
CE
I
E
S
M
A
S
T
E
R
=  
1– –2 = 3%
SSD = 200 m
and h1 = 1.2 m and h2 = 0.15 m
as given N1 up gradient, and n2 - down gradient.
So curve is summit curve.
Assume L > SSD
L =
 
2
2
1 2
NS
2 h + h
=
 
 
2
2
3 200
×
100 2× 1.2 + 0.15
L = 272.91 m > 200 m
L = 272.91 m
63. Assuming that traffic on a highway obeys the
Greenshields model, the speed of a shockwave
between two traffic streams (P) and (Q) as shown
in the schematic is _______ kmph. (in integer)
Direction of traffic
(P)Flow = 1200
vehicles/hour
Speed = 60
kmph
(Q)Flow = 1800
vehicles/hour
Speed = 30 kmph
Sol: (14.5 to 15.5)
Speed of shock wave =
Change in flow
Change in density
Flow = Speed × Density
Density =
Flow
Speed
=
Q P
Q P
Q P
q – q
q q
–
V V
=
1800 –1200
1800 1200
–
30 60
=
600 600
= =15 Kmph
60 – 20 40
64. It is given that an aggregate mix has 260 grams
of coarse aggregates and 240 grams of fine
aggregates. The specific gravities of the coarse
and fine aggregates are 2.6 and 2.4, respectively.
The bulk specific gravity of the mix is 2.3.
The percentage air voids in the mix is
____________. (round off to the nearest integer)
Sol: (7.5 to 8.5)
Given that,
Coarse aggregate = 260 gms
Fine aggregate = 240 gms
GCA = 2.6
GFA = 2.4
Bulk specific gravity (Gm) = 2.3
Percentage air voids in the mix = ?
Gt (Theoretical specific gravity)
=


W
W
G
=
260 + 240
= 2.5
260 240
+
2.6 2.4
% air voids (VV) =
t m
t
G – G
×100
G
=
2.5 – 2.3
×100 = 8%
2.5
VV = 8%
65. The lane configuration with lane volumes in
vehicles per hour of a four-arm signalized
intersection is shown in the figure. There are only
two phases: the first phase is for the East-West
and the West-East through movements, and the
second phase is for the North-South and the
South-North through movements. There are no
turning movements. Assume that the saturation
flow is 1800 vehicles per hour per lane for each
lane and the total lost time for the first and the
second phases together is 9 seconds.
440
460
N
504
504
360
396
The optimum cycle length (in seconds), as per
the Webster’s method, is ____________. (round
off to the nearest integer)
Sol: (36 to 38)
From figure,

Solution
Detailed
Ph:
CE
I
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M
A
S
T
E
R
N–S
y =
 
 
 max
360 396
,
1800 1800
N–S
y =
396
= 0.22
1800
E–W
y =
 
 
 max
504 + 504 440 460
, ,
2×18.0 1800 1800
= 0.28
Optimum Cycle length
=
1.5L + 5
1– y
=  
N–S E–W
1.5×9 + 5
1– y + y
=  
1.5×9 + 5
1– 0.22 + 0.28
C0 = 37 secs.

Afternoon Aptitude and Technical Questions

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