ce test 15 objective solution
TRANSCRIPT
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8/18/2019 CE Test 15 Objective Solution
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1. (b)
2. (b)
3. (d)
4. (c)
5. (a)
6. (c)
7. (d)
8. (a)
9. (d)
10. (c)
11. (d)
12. (b )
13. (b)
14. (b)
15. (b)
16. (b)
17. (b )
18. (b)
19. (c)
20. (a)
21. (a)
22. (c)
23. (b )
24. (d)
25. (a)
26. (b )
27. (d )
28. (c)
29. (c)
30. (a)
31. (a)
32. (d )
33. (b)
34. (b)
35. (c)
36. (c)
37. (c)
38. (a)
39. (b)
40. (d)
41. (d)
42. (b)
43. (d)
44. (b)
45. (a)
46. (a)
47. (b)
48. (c)
49. (c)
50. (b)
51. (d)
52. (d)
53. (c)
54. (a)
55. (b)
56. (b)
57. (d)
58. (a)
59. (b)
60. (a)
61. (b)
62. (d)
63. (b)
64. (b)
65. (d)
66. (c)
67. (b)
68. (d)
69. (a)
70. (a)
71. (b)
72. (a)
73. (a)
74. (d)
75. (b)
76. (c)
77. (d)
78. (b)
79. (b)
80. (b)
81. (a)
82. (d)
83. (a)
84. (d)
85. (b)
86. (c)
87. (d)
88. (d)
89. (a)
90. (b)
91. (b)
92. (a)
93. (d)
94. (c)
95. (d)
96. (c)
97. (c)
98. (d)
99. (b)
100. (b)
101. (a)
102. (a)
103. (a)
104. (c)
105. (a)
106. (b)
107. (a)
108. (b)
109. (b)
110. (c)
111. (a)
112. (a)
113. (b)
114. (a)
115. (a)
116. (d)
117. (a)
118. (a)
119. (d)
120. (d)
Conventional Question Practice Programe
Date: 16th April, 2016
ANSWERS
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1. (b)
S.F. = Shrunk length Shrunk scale
Originallength Original scale
24 Shrunk scale
25 1/ 2400
Shrunk scale =
1
2500
2. (b)
sin
Cg = L(1 cos )
Cg = 22 sin / 2
3. (d)4. (c)
If CP is the correction for pull, we have
CP =0(P P ) L
AE
where
P = Pull applied during measurement (N)
P0
= Standard pull (N)
L = Measured length (m)
A = Cross-sectional area of the tape (cm2)
E = Young’s modulus of elasticity
(N/cm2)
Here L = 1500 m, P0 = 100 N, P = 150 N
CP =
(150 100) 1500 50 1500
AE AE
5. (a)
The additive constant, c = f + d
Multiplying constant, K =f
i
where,
f = focal length of objective
d = horizontal distance between optical
centre and vertical axis of
tacheometer
Here, i = 5 mm, f = 25 cm, d = 15 cm
c = 0.25 + 0.15 = 0.4 m
K =f 25
0.5
i = 50
6. (c)
f = 20 cm; d = 10 cm
i = 4 mm; s = 2.500 m.
C = f + d = 0.2 + 0.1 = 0.3
K =f 20
i 0.4 = 50
Staff interept,
S = 2 × (2.5 – 1) = 3.0 m
Horizontal distance between the staff station
and instrument station,
D = Ks + C
= 50 × 3 + 0.3
= 150.3 m
7. (d)
% of error between 3 under gussian law
of distribution = 99.7% No. of errors that are expected to exceed
the limit of 3 = (1 – 0.997) × 1000
= 3 × 10 –3 × 1000 = 3
8. (a)
9. (d)
If a computed quantity is a function of two or
more observed quantities, its probable error
is equal to the square root of summation of
the squares of the probable errors. of the
observed quantity multiplied by itsdifferentiation with respect to that quantity.
Let y = computed quantity
x1, x
2, x
3 etc. = observed quantities
Such that y = f(x1, x
2, x
3 etc.)
Then
ey = 1 2 3
2 2 2
x x x1 2 3
dy dy dye e e
dx dx dx
where ey = Probable error of the computedquantity
1 2 3x x xe , e , e = Probable errors of
observed quantities. Let area A = a · b
e A
= 2 2
a be · b e · a
= 2 2(0.05 180) (0.06 120)
= ± 11.53 sqm.
where e A = Probable error in the area.
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10. (c)
The local attraction at any station isdetected by observing the fore and back
bearings of the line.
If local difference between them is 180º,both the end stations are considered to
be free from local attraction, provided the
compass is not having any instrumental
errors.
If the difference is not 180º, thediscrepancy introduced may be because
of presence of local attraction at either or
both of the station.
310º 30 – 135º 15 180º
200º 15 – 17º 45 180º
246º 0 – 65º 15 180º
11. (d)
180 19 47 13 160 12 47
12. (b)
If the magnetic declination at a place at the
time of observation is known the true bearing
of a line can be determined from its magnetic
bearing and vice-versa. If declination is west,
True bearing = Magnetic bearing – declination
89º = magnetic bearing – 1º
Magnetic bearing = 90º
Magnetic bearing of BA = 180º – 90º = 90º
13. (b)14. (b)
RP
Q
129º30
59º
Line FB BB
PQ 59º0 239º0
QR 129º30 309º30
PQR = –FB of line QR + BB of PQ= –129º30’ + 239º0’ = + 109º30’
(interior included angle)
15. (b)
The line of collimation should be parallel to
the axis of the tube when the vertical circle
reading is zero.
The axis of the altitude level tube is truly
horizontal when the bubble is in the centre.
16. (b)
17. (c)
18. (b)
Ist sub chord = 2220 – 2002.48 = 17.52 m
19. (c)
Radius of curvature of the bubble tube
= R =ndL
s
=
35 2 10 10020m
0.05
where
n = 5
d = 2 mm
L = 100 m
s = 0.05 m
20. (a)
The importance of the two peg test result is
that even when the levelling instrument is not
in correct adjustment, the difference in height
measured between two points by a level
equidistant from each gives the true difference
in height.
21. (a)
Last sub chord = 2303.39 – 2300 = 3.39 m
22. (c)
23. (b)
Contour lines are imaginary lines passing
through points of equal elevations.
24. (d)
Contour lines close together indicate a steep
slope.
25. (a)
Width of ground to be covered
= (1 – 0.6) × 0.23 × 20,000 = 1840 m.
26. (b)
Relief displacement of a point,
d =r.h
H
where h = height of the object above datum,
H = flying height above the datum,
r = radial distance of the image of the
top of the object from principal
point.
d =90 500
9mm5000
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27. (d)
28. (c)
29. (c)
Principal Point : It is a point where aperpendicular dropped from the front nodal
point strikes the photograph.
Isocentre : It is the point in which thebisector of the angle of tilt meets the
photographs.
Crab : It is the term used to designate theangle formed between the flight line and
the edges of the photograph in the direction
of flight.
Drift : It is caused by the failure of theaeroplane to stay on the predetermined
flight line.
30. (a)
The Right Ascension (RA) of a celestial body
is the angular distance measured along thecelestial equator between the first point of Aries
and declination circle of the body.
31. (a)
Lehmann’s method or Trial and error method
in the field is used to find out the position
of the station of a plane table.
32. (d)
R
T1 T2D
V
C
Apex distance, VC = R (sec /2–1)
Length of long chord = T1 DT2 = 2R sin/2
Mid ordinate (M) = Length CD
= R (1–cos/2)
= R versin /2
33. (b)
N o r m a l
1
1
2Soil-1 (K )1
Flowlines 2
Soil-2 (K )2
Flow lines
Normal
When two different soils are used in a soil
mass, thus making it non-homogeneous. The
flow lines and equipotential lines get deflected
at the interface. The flow net thus get modified.
K1 and K2 are related as
1
2
K
K=
1
2
tan
tan
34. (b)
h = 6 m, Nf = 6; N
d = 18
K = 4 × 10 –5 m/min = 4 × 10 –5 × 60 × 24
= 0.0576 m/day
Seepage discharge = Khf
d
N
N
= 0.0576 × 6 ×6
18
= 0.1152 m3/d per m length
35. (c)
e = 0.60 For quick sand condition,
ic =
G 1
1 e
ic = cG 1
1.6 i G 11 0.6
G = 1.6 ic + 1
36. (c)
G = 2.60, n = 0.33
ic = (G –1) (1 – n)
= ( 2.60 – 1) (1 – 0.33) = 1.072
37. (c)
Fs =s
tan ' tan ' tan30tan i
tan i F 1.732
= 0.333
i = tan –1(0.333)
38. (a)
B = 2 m, Df = 2m
C = 30 kN/m2 ; sat = 20 kN/m3
As per skempton's theory,
f D
B =
21
2
NC = 5
Df B1 0.2 1 0.2
B L
= 5(1 + 0.2 × 1) ( 1 + 0) = 6.0
qnu = CuNc = 30 × 6 = 180 kN/m2
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39. (b)
= 0.70
Cu = 40 kN/m2
F.S. = 2.50
B = 0.3 m
D = 10 m
Qu = qb Ab + .c. (p D)
= (9 × 40) ×4
× 0.32
+ 0.7 × 40 ( × 0.3 × 10)
= 289.194 kN
Qa =4Q 289.194 115.68 kN
F.S. 2.5
40. (d)
Total stress in multi-layered soil :
The total stress at depth z is the sum of the
weights of soil in each layer thickness above.Vertical total stress at depth z,
v = 1 1 2 2 3 1 2d d (z d d )
where,
1 2 3, , = unit weights
of soil layers 1, 2, 3, etc.
respectively
z
d1
d2
1
2
3
1
2
3
The increase in vertical stress at a depth
of 1 m is 36 kN/m2, this will be the increase
in vertical stress every where below the surface
level.
41. (d)
The permeability
K =
32w eC D
1 e
where C = a constant depending upon shape
of c/s of flow.
= dynamic viscosity of fluid
e = void ratio
D = grain size
42. (b)
= sub h (Assuming3
sub10kN/m )=
50 = 10.h
h = 5m
43. (d)
Q = KAi
Q A
=2 hK D
4 2L
QB
=2 h
K 4D4 L
A
B
Q
Q=
2
2
hK D
14 2L =h 8
K 4D4 L
= 0.125
44. (b)
K = 10-3 cm/s
h = 1mL = 2m
A = 100cm2
Q = KAh
i K A =L
dV
dt=
hK A.
L
V = (10-3cm/s) (100cm2)
×1m
2m
× 60 sec = 3ml
45. (a)
Q =h
K.A.L
(For constant head
permeameter)
Q h
L
Q
Q
=
2h 1 4h L
L / 2 (H / L) L h
Q = 4Q
46. (a)
As per Allen Hazen’s formula
K = 210C.D
Where K = Coeff. of s permeability (cm/sec)
D10 = effective size (cm)
C = Constant, with a value between
100 and 150.
Ratio of permeability =2
.64
0.3
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47. (b)
Kh
= 3m/day
KV
=1
m/day3
K = h V1
K .K 3 1m/day3
48. (c)
Failure by piping or undermining
When the seepage water retains sufficient
residual force at the emerging downstream end
of the work, it may lift up the soil particles.
This leads to increaed porosity of the soil by
progressive removal of soil from beneath the
foundation. The structure may ultimately
subside into the hollow so formed, resulting in
the failure of the structure.
The exist gradient is said to be critical when
the upward distributing force on the grain is
just equal to the submerged weight of the
grain at the exit. When a factor of safety equalto 4 or 5 is used, the exit gradient. Can then
be taken as safe In other words, an exit
gradient equal to1
4 to
1
5 of the critical exit
gardient is ensured, so as to keep the structure
safe against piping.
49. (c)
In case of a falling head permeameter,
K =
110
2 1 2
h2.303Lalog
A t t h
where L = 10 cm
a = 1 cm2
A = 50 cm2
(t2 – t
1) = 1 hour = 3600 sec.
1
2
h
h=
802
40
Hence K =2.303 10 1
0.350 3600
= 3.84×10 –5 cm/sec.
50. (b)H = 10 m
Df H = 15 m
Df =15
1.5010
Sn =c
C
F H
0.164 = c35
F 1.15Fc 18.5 10
51. (d)
The steady seepage condition is criticalfor the d/s slope of an earth dam.
The critical condition for the stability of theu/s slope of an earth dam is when there is
a sudden drawndown in the reservior u/s.
52. (d)
Cm
=C 15
10
1.5 1.5
Sn = 0.046 =10
H
H =10
11.5 m19 0.046
53. (c)
Swedish circle method: The actual shape
of a slipsurface in the case of finite slopes is
curvilinear. For convenience,it is approximated
as circular. The assumption of a circular slip
surface and its application for stability analysis
of slopes was developed in sweden. Themethod is known as the swedish circle method
or the method of slices.
Stability member (Sn): It is defined as
Sn =m
H c
C C
F H
The reciprocal of the stability number is known
as stability factor. The stability number is a
dimensionless quantity. The stability number
can be used to determine the factor of safety
of a given slope.
Sudden drawdown conditions: When the
water standing on the slope is suddenly and
quickly removed, the water pressure (U)
disappears. However, if there is no time for
drainage to occur from the soil in the slope,
the soil remains submerged as before and the
natural part of the weight is still acting. Thus,
the equilibrium of the neutral force is distrubed,
although the equilibrium of the inter granular
forces remains unaffected.
Critical void ratio: It may be observed that
the void ratio of an initial loose sand decreaseswith as increase in shear strain whereas that
for the initially dense sand increases with an
increase in strain. The void ratio at which there
is no change in it with an increase in strain is
known as the critical void ratio. If the sand
initially is at critical void ratio, there would be
practically no change in volume with an
increase in shear strain.
Where there is soft clay there is a chance of
base failure
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54. (a)
The following are the important points of
comparison of coulomb’s theory with Rankine’s
theory:
Coulomb considers a retaining wall andthe backfill as a system; he takes into
account the friction between the wall and
the backfill, while Rankine does not.
The backfill surface may be plane or curvedin coulomb’s theory, but Rankine’s allowsonly for a plane surface.
In coulomb’s theory, the total earth thrustis first obtained and its position and
direction of the earth pressure are assumed
to be known; linear variation of pressure
with depth is tacitly assumed and thedirection is automatically obtained from the
concept of wall friction. In Rankine’s theory,
plastic equilibrium inside a semi-infinite soil
mass is considered, pressure evaluated, a
retaining wall is imagined to be interposed
later, and the location and magnitude of the total earth thrust are established
mathematically.
Coulomb’s theory is more versatile thanRankine’s in that it can take into account
any shape of the backfill surface, break in
the wall face or in the surface of the fill,
effect of stratification of the backfill, effect
of various kinds of surcharge on earth
pressure and the effects of cohesion,
adhesion and wall friction. It lends itself to
elegant graphical solution and gives more
reliable results, especially in thedetermination of the passive earth
resistance; this is in spite of the fact that
static equilibrium condition does not appear
to be satisfied in the analysis.
Rankine’s theory is relatively simple andhence is more commonly used, while
coulomb’s theory is more rational and
versatile although cumbersome at times,
therefore, the use of the later is called for
in important situation or problems.
55. (b)
When a wall moves away from the backfill,
some portion of the backfill located immediately
behind the wall tries to break away from the
rest of the soil mass. This wedge–shaped
portion, known as the failure wedge or the
sliding wedge moves downward and outwards.
The lateral earth pressure exerted on the wall
is a minimum in this case. The soil is at the
verge of failure due to a decrease in the lateral
stress.
The horizontal strain required to reach the active
state of plastic equilibrium is very small. It
has been shown that in dense sand, the
horizontal strain required is about 0.5%
It is thus found that very little horizontal strain
(about 0.50%) is required to reach one half
the maximum passive pressure in dense sand
but much more horizontal strain (about 5% in
dense and and 15% in loose sand) is required
to reach the full passive pressure.
It may be summarised that the state of
shear failure corresponding to the minimum
earth pressure is the active state and that
corresponding to the maximum earth pressure
is the passive state.
56. (b)
Unsupported height =4C 4 5
1m20
57. (d)
Ka =21 sin
tan= /21 sin 4
Kp =
21 sintan= 45 /2
1 sin
p
a
K
K =
2
2
tan 45º /2
tan 45º – /2
a
K =p
1
K
p
a
K
K =
24
2
tan 45 /2tan= 45 /2
1
tan 452
58. (a)
Given that, K A
= 2 1tan 45º2
= 1
4
1tan 45º2
=
1
2
1tan 45º2
= 2
C1 = 20 kN/m
2
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H1 =
1 1
1
4Ctan 45º
2
H1 =
4 202 8m
20
H2 =
2
2
4C 4 408m
20
H2/H
1 =
81
8
59. (b)
For square footing,
Net ultimate bearing capacity,
qnu = (5.7 c) 1.3
= 5.7 × 1.3 × 1 kg/cm2
= 7.41 kg/cm2
= 74.1 t/m2
60. (a)
The allowable bearing capacity,Q1 = 15 t/m
2 for allowable settlement,
S1 = 25 mmm
The allowable bearing capacity, Q2 for allowable
settlement, S2 = 40 mm
But Housel’s method,
1
1
Q
S =
2
2
Q
S
Q2 =2
11
SQ
S
=240 15 24 t/m
25
61. (b)
The ultimate bearing capacity of purely
cohesive soil is given by qu = CN
C+ q
Where q = surcharge due to increase in depth
of foundation.
As we can observe from the above equation
that bearing capacity of footing on purely
cohesive soil is independent of size of footing.
62. (d)
Pile foundations are used in the following
condition:
When the strata at or just below the groundsurface in highly compressible and very
weak to support the load transmitted by
the structure.
When the plans of the structure is irregular relative to its outline and load distribution.
It would cause non-uniform settlement if a
shallow foundation is constructed. A pile
foundation is required to reduce differential
settlement.
Pile foundation are required for thetransmission of structural loads through
deep water to a firm stratum.
Pile forces are used to resist horizontalforces in addition to support the vertical
loads in earth-retaining structures and tall
structures that are subjected to horizontal
forces due to wind and earthquake.
Piles are required when the soil conditionare such that a washout, erosion or scour
of soil may occur from underneath a
shallow foundation.
In case of expansive soils, such as blackcotton soil, which swell or shrink as the
water content changes, piles are used to
transfer the load below active zone.
63. (b)
Raft foundations on sands are quite useful.
Their bearing capacity failure is generally out
of question, because the bearing capacity in
sands increases with the width of the footing
and this width in rafts is quite large. The
differential settlements in rafts are also
generally smaller as compared to isolated
footings even for the same load intensity
because a raft eliminates the influence of local
loose soils.
SoilType
Permissible totalsettlement
Permissibledifferential
settlement For
isolatedfootings
For raftfoundation
Forisolatedfooting
For raftfoundation
Sandy 4 cm 4 to6.5 cm
2.5 cm 2.5 cm
Clayey 6.5 cm 6.5 to10 cm
4 cm 4 cm
64. (b)
65. (d)
66. (c)
General shear failure
General shear failure is seen in dense and
stiff soil. The following are characteristics of
general shear failure :
Continuous well defined and distinct failure
surface develops between the edge of
footing and ground surface.
Dense or stiff soil that undergoes low
compressibility experiences this failure.
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Continuous bulging of shear mass adjacent
to footing is visible.
Failure is accompanied by tilting of footing.
Failure is sudden and catastrophic with
pronounced P- curve.
The length of disturbance beyond the edge
of footing is large.
State of plastic equilibrium is reached
initially at the footing edge and spreads
gradually downwards and outwards.
General shear failure is accompanied by
low strain ( 36º) and large N (N > 30) having
relative density (ID > 70%)
Local shear failure
This type of failure is seen in relatively loose
and soft soil. The following are some
characteristics of general shear failure.
A significant compression of soil below thefooting and partial development of plastic
equilibrium is observed.
Failure is not sudden and there is no tilting
of footing
Failure surface does not reach the ground
surface and slight bulging of soil around
the footing is observed.
Failure surface is not well defined.
Failure is characterised by considerable
settlement.
Well defined peak is absent inp curve.
Local shear failure is accompanied by large
strain (> 10 to 20%) in a soil with
considerably low ( 1. This is because soil arround andbetween the piles get compacted due to the
vibration caused during the driving operations.
Whereas in dense sand above phenomenon
is not true.
70. (a)
The load carrying capacity of a driven pile canbe estimated from the resistance against
penetration developed during driving operation.
The methods give fairly good results only in
the case of free-draining sands and hard clays
in which high pore water pressures do not
develop during the driving of piles. In saturated
fine grained soils, high pore water pressure
develops during the driving operation and the
strength of the soil is considerably changed
and the methods do not give reliable results.
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71. (b)
Types of Foundation
Spread Footings
Used for most buildings where the loadsare light and or there are strong shallow
soils.
Generally suitable for low rise buildings (1– 4 stories).
Requires firm soil condition that are capableof supporting the building on the area of the spread footings.
These are most widely used because theyare most economical.
Spread footings should be above the water table.
Drilled piers or Caissons
For expansive soils with low to mediumloads or high loads with rock not too far down, drilled caissons (piers) and grade
beams can be used.
The caissons might be straight or belledout at bottom to spread the load.
Caissous deliver the load to soil of stronger capacity which is located not too far down.
Piles
For expansive soils or soils that arecompressive with heavy loads where deep
soils can not take building load and wheresoil of better capacity is found deep below.
There are two types of piles
1. Friction piles–used where there is noreasonable bearing stratum and they rely
on resistance from skin of pile against
the soil.
2. End bearing– which transfers directly to
soil of good bearing capacity.
Cast in situ piles are composed of holedrilled in earth and then filled with concrete,
it is used for light loads on soft ground
where drilling will not cause collapse.
Mat Foundations
Reinforced concrete raft or mats can beused for small light load buildings on very
weak or expansive soils such as clays.
They are often post tensioned concrete.
They allow the building to float on or in thesoil like a raft.
Can be used for buildings that are 10–20stories tall where it provides resistance
against overturning.
Can be used where soil requires such alarge bearing area and the footing might be
spread to the extant that it becomes more
economical to pour one large slab moreeconomical-less form-work.
It is used in lieu of driving piles because itcan be less expensive and less obstrusive.
Usually used over expansive clays, silts tolet foundation settle without great
differences.
72. (a)
73. (a)
74. (d)
75. (b)
NB ss
cX f
HRT
=
8 2420
6 = 640 mg/l
76. (c)
Bacteria produces highest biomass among
other micorbs.
77. (d)
78. (b)
79. (b)
h = Sy + S
r
0.40 = Sy + 0.15 S
y = 0.25
where Sy = Specific yield
Change in groundwater storage of theaquifer
= 0.25(23 – 20) × 150 = 112.5 ha.m
80. (b)
Q = 2700 lit/min = 0.04533 m3/s
For confined aquifer,
Q = 1 2
2
1
2 T S S
r ln
r
here r 1 = 10 m; S1 = 3m
r 2 = 100 m; S2 = 0.5 m
0.04533 = 2 T 3 0.5100
ln 10
T = 6.6487 × 10 –3 m2/s = 574.44 m2/day
81. (a)
When the flow in normal to the stratification,
the equivalent permeability Ke of the aquifer
Ke =
n
i1
ni
i1
L2 4 3 9
12.7 m2 4 3 1 1 1L6 16 24 3 4 8K
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82. (d)
When the flow is parallel to the stratification,
the equivalent permeability Ke
Ke =
n
i i1
n
i1
K B3 30 2 10 5 20
21m/day3 2 5
B
Transmissibility, T = KB = 21 × 10
= 210 m2/day
83. (a)
84. (d)
85. (b)
Rainfall = 2.7 cm
Loss @ 0.3 cm/h for 3 h = 0.9 cm
Effective rainfall, ER = 2.7 – 0.9 = 1.8 cm
DRH of the storm has peak = 200 – 20
= 180 m3/s
3h UH has peak = 1801.8
= 100 m3/s
86. (c)
Let rainfall excess be ER (in cm)
Volume of surface runoff
=1
60 3600 362
=6 ER300 10
100
ER = 1.08 cm = 10.8 mm
87. (d)
ER of 1h UH = 1 cm
Volume of runoff = Area under 1h UH
= 1 1 1 1 1 1
1 3600 3 3 5 5 4 4 2 2 1 12 2 2 2 2 2
= 54,000 m3
Catchment area represented by this UH
=54,000 100
1
= 5.4 × 106 m2
= 5.4 km2
Time (h)
(1)
Ordinateof 1h UH
(m /s)(2)
3
Ordinate of 1h UH lagged
by 1 h(3)
0
1
2
3
4
5
6
7
8
Ordinate of 1h UH lagged
by 2 h(4)
col. 2 +col. 3
+ col. 4(5)
Ordinate of 3h UH =(col. 5)/3
(6)
0
3
5
4
2
1
0
—
0
3
5
4
2
1
0
—
—
0
3
5
4
2
1
0
0
3
8
12
11
7
3
1
0
0
1
2.67
4
3.67
2.33
1
0.33
0
88. (d)
Total rainfall of 4h storm = 7 cm
Loss @ 0.25 cm/h for 4h = 1 cm
Effective rainfall of the storm = 7 – 1 = 6 cm
Peak ordinate of 4h UH = 80 m3/s
Peak ordinate of 4h DRH = 80 × 6
= 480 m3/s
Base flow = 20 m3/s
Peak of the flood discharge due to thestorm = 480 + 20 = 500 m3/s
89. (a)
Area of catchment A = 360 km2
Duration of unit hydrograph = 4h
Equilibrium discharge of S-curve
QS = 2.778 A/D m3/s
=360
2.7784
= 250.02 m3/s
90. (b)
h =
2
2 V u 1 cos u
V
The efficiency will be maximum for a given
value of V when
hd
du = 0
or
2
2u V u 1 cosd
du V
= 0
or 2
2
1 cos d2uV 2u
du V
= 0
or 2d
2uV 2udu
= 0 21 cos
0 V
or 2V – 4u = 0 or u = V
2
91. (b)
In centrifugal pumps, the cavitation may occur
at the inlet of the impeller of pump. If the
pressure at the suction side of the pump dropsbelow vapour pressure of liquid, then cavitation
may occur.
Cavitation is the localised formation and
subsequent collapse of cavities, or bubbles in
a liquid. Cavitation is usually caused by
insufficient NPSHA.
In a pump, cavitation will first occur at the
inlet of the impeller. Denoting the inlet by i,
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NPSHA at this point is defined by
NPSHA =
2vi i
g
pp v
g 2g
On the other hand, in a reaction turbine,
cavitation will first occur at the outlet of the
impeller, at the entrance of the draft tube.
92. (a)
1.
2.
3.
4.
10 to 35
35 to 60
60 to 300
300 to1000
8.5 to 30
30 to 51
51 to 225
225 to860
Pelton wheel with
single jet
Pelton wheel with
two or more jets
Francis turbine
Kaplan orPropeller turbine
S.No.
Specified
speedTypes of turbine
(M.K.S.) S.I.
93. (d)
Shear due to torque =1.6T
b
=1.6 9
48 kN0.3
Equivalent shear =1.6T
Vb
= 48 + 20
= 68 kN
94. (c)
At
dsn
d-n
N A
B
d
Take moment about neutral axis
dsB ds n
2
= m At × (d–n)
95. (d)
beff
= 0 w f b 6D6
l
Where l0 = distance between points of
contraflexure.
beff
=3600
300 6 1006
= 1500mm
96. (c)
Z =
22 300 600bD
6 6
= 18×106mm3
f cr = ck0.7 f = 0.7×5 = 3.5MPa
Mcr = f cr × Z = 3.5×18 kNm = 63 kNm
97. (c)
Kb =
1
r 1m
r =st
cbc
140028
50
Hence, Kb = 0.39
Now for a balanced section
cbc b0.5 b K d = st,b st A
Pt(%) =cbc
b
st
50 K
= 150 0.39 0.696%28
98. (d)
Tensile force =2
t d4
Bond stress resistance = bS d l
bond stress resistance = Tensile force
bS d l =2t d
4
l =b
dt
4S
99. (b)
Diagonal tension is produced in beam due to
shear force which is predominent at the ends
of beam. So, to counter this diagonal tension
we require bent up hooked bars at the ends
of beam.
100. (b)
Hook requires minimum extension of 4d a head
of curvature of Hook.
101. (a)
The line of collimation of a theodolite must beperpendicular to the horizontal axis at its
intersection with the vertical axis. If this
condition exists, the lime of sight will generate
a vertical plane when the telescope is rotated
about the horizontal axis.
If the line of sight is not perpendicular to the
trunnion axis of the telescope, it will not revolve
in a plane when the telescope is raised or
lowered but instead, it will trace out the surface
of a cone.
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102. (a)
The direction of a survey line can either be
established (i) with relation to each other or
(ii) with relation to any meridian. The first will
give the angle between two lines while the
second will give the bearing of the line.
103. (a)
Feature of contour of terrain:
All points on a contour line are of the sameelevation.
No two contourlines can meet or crosseach other except in the rare case of an
overhanging vertical cliff.
Closely spaced contour lines indicatesteep slope.
Widely spaced contour lines indicate gentleslope
Equally spaced contour lines indicate
uniform slope.
Closed contour lines with reducing levelstowards the centre indicate pond or other
depression.
Contour lines of ridge show higher elevationwithin the loop of the contours. Contour
lines cross ridge at right angles.
Contour lines of valley show reducingelevation within the loop of the contours.
contour lines cross valley at right angles.
All contour l ines must close either withinthe map boundary or outside.
Contour lines cannot merge or cross oneanother on map except in the case of an
overhanging cliff.
Contour lines never run into one another except in the case of a vertical cliff. In this
case, several contours coincide and the
horizontal equivalent becomes zero.
104. (c)
Relief displacement : It is caused by changes
in the distance between the ground and the
camera as the plane flies over the ground.
Characteristics of relief displacement
Characteristics of aerial images over variedterrain.
Objects that rise above the surface awayfrom the principal point.
Objects extending below the surface leantowards the principal point.
Displacement increases with the height of the object and or distance from the principal
point.
Relief displacement, d =rh
H
where
r = radial distance from principal point
to displaced image point
h = height above surface of the object
point
H = flying height above the surface.
105. (a)
Meridian Circle : It is a great circle which
passes through the zenith and Nadir as well
as through the poles.
Vertical Circle : A vertical circle of the celestial
sphere is great circle passing through the
zenith and Nadir.
106. (b)
107. (a)
108. (b)
109. (b)
The characteristics of flow net can be
summarised as under :–
The fundamental condition that is to besatisfied is that every intersection between
a flow line and an equipotential line should
be at right angles.
The second condition to be satisfied is thatthe discharge between any two adjacent
flow lines is constant and the drop of head
between the two adjacent equipotential lines
is constant.
The ratio of the length and width of eachfield is constant. The ratio is generally taken
as unity for convenience. In other words,
the flow net consists of approximate
squares.
110. (c)
In the Fellenius analysis, the horizontal and
vertical forces on the slice boundaries are
assumed to be equal and opposite. This is
true if the slice is reduced to the width of a
line but as the width of a slice increases the
assumption is partially untrue since the two
sides will be very different in size. Thus, if the
soil is divided into many slices, as can be
done using modern computers, then a
reasonably accurate factor of safety can be
found. However for manual analysis, the
number of slices that can realistically be used
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(i.e., upto 10 approximately) limits the
accuracy of the method.
Because the forces on the vertical boundaries
are assumed to cancel out as shown in fig
below, the total normal force on the base of
the slice is equal to the component of the
weight in the direction of the normal force.
N = wcos
N
b
w
The fellenium equation is fairly simple to solve
and yields conservative results (lower than
actual factor of safety) especially where the
slip surface is deep or where the pore water
pressures are high. In both these cases the
fault lies with the neglect of the interslice
forces.
It is also known as swedish circle method. In
this method, the equilibrium of each slice is
determined and factor of safety found by
summing the resisting forces and dividing by
the driving forces. The operation is repeated
for the circles until the lowest safety factor is
found. The method does not consider all theforces acting on a slice, as it omits the shear
and normal stresses and pore water pressures
acting on the sides of the slice but usually
(although not always) it yields conservative
results. However, the conservatism may be
high.
111. (a)
The Rankine’s theory assumed that the wall
surface is smooth whereas in practice, a lot
of friction may develop between the wall surface
and the soil fill. This friction will depend uponthe wall material. This friction leads to the
development of smaller active pressure and
larger passive pressure than that estimated
by Rankine’s theory.
Thus, the estimation of the active pressure
using Rankine’s theory will be slightly higher
than the actual (reduced due to friction) Passive
pressure will be slightly lower.
112. (a)
Shear strength is a term used in soil
mechanics to describe the magnitude of the
shear stress that a soil can sustain. The shear
resistance of a soil is a result of friction and
interlocking of particles and possibly
cementation or bonding at particle contacts.
The shear strength of soil depends on the
effective stress, the drainage conditions, the
density of particles, the rate of strain and thedirection of the strain.
The drained shear strength is the shear
strength of the soil when pore fluid pressures,
generated during the course of shearing the
soil, are able to dissipate during shearing. It
also applies where no pore water exists in the
soil and hence pore fluid pressures are
negligble. It is commonly approximated using
the Mohr-coulomb equation.
In terms of effective stresses, the shear
strength is often approximated by
= tan c
where ( u) is the effective stress. istotal stress applied normal to the shear plane
and u is the pore water pressure acting on the
same plane.
= effective stress friction angle.
c = cohesion
113. (b)
Principal factors that influence ultimate bearing
capacities are type and strength of soil,
foundation width and depth, soil weight in the
shear zone and surcharge.
The depth to the water table influences the
subsurface and surcharge soil weights.
If the water table is below the depth of the
failure surface then the water table has no
influence on the bearing capacity and effective
unit weight is equal to the wet unit height of
the soil.
If the water table is above the failure surface
and beneath the foundation base, then effective
unit weight of the soil gets reduced.
114. (a)
A relative movement between a pile and soil
produces shear along the interface of the pile
and the soil. Such movement can be induced
by a push-load on the pile pressing it down
into the soil or by a pull-load moving it upward.
A relative movement can also be induced when
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swelling soils when the soil moves upward in
relation to the pile. By definiiton, if the
movement of the pile is downwrd i.e. the shear
stress induced in the pile is upward, the
direction of the shear is positive. If the
movement of the pile is upward, the shear
stress direction is negative.
In order terminology, the induced shear along
a pile was called skin friction. The negativeand positive skin friction is shear stress
induced by settling or swelling soil respectively.
Large ground settlement due to consolidation
of soft soils will drag down piles and induce
negative skin friction (NSF) along the pile-soil
interfaces. NSF will induce additional drag load
on the piles which may cause structural failure
of piles due to overstress or downdrag
settlement which may compromise the
serviceability of super structure. It has been
recognised that pile groups posses beneficial
effects in alleviating NSF on piles.115. (a)
Dynamic formulae for piles :
The dynamic formulae are based on the
assumption that the kinetic energy delivered
by the hammer during driving operation is equal
to the work done on the pile. Thus,
Wh h = R × S
where W = weight of hammer (kN),
h = height of ram drop (cm),
h = efficiency of pile hammer,
R = pile resistance (kN), taken equal
to Qu and S = pile penetration
per blow (cm)
The load carrying capacity of a driven pile can
be estimated from the resistance against
penetration developed during driving operation.
The methods give fairly good results only in
the case of free draining sands and hard clays
in which high pore water pressures does not
develop during driving of piles. In saturated
fine-grained soils, high pore water pressure
develops during the driving operation and the
strength of the soil is considerably changed
and the methods do not give reliable results.
The methods can not be used for submerged,
uniform fine sands which may be loose enough
to become quick temporarily and show a much
less resistance.
116. (d)
117. (a)
Slug is a higher strength and higher volume
waste discharged in short period of time.
118. (a)
119. (d)
A draft tube is a tube or pipe of gradually
increasing area which is used for dischargingwater from the exit of the turbine to the tail
race. This is because the pressure at the exit
of the summer of a reaction turbine is generally
less than atmosphereic pressure. Thus, the
water at exit cannot be directly discharged to
the tail race.
Also, the draft tube converts a large proportion
of the kinetic energy22 V
2g
rejected at the
outlet of the turbine into useful pressure energy.
Without the draft tube, the kinetic energy
rejected at the outlet of turbine will go wasteto the tail race.
Hence by using draft tube, net head on turbine
increases. The turbine develops more power
and also efficiency of turbine increases.
120. (d)
Minimum grade of concrete in RCC is
Restricted to M20 as per IS 456:2000