conventional question practice programe ce (test-8), objective...
Post on 29-Mar-2021
1 Views
Preview:
TRANSCRIPT
IES M
ASTER
CE (Test-8), Objective Solutions, 26th March 2016 (1)
1. (a)
2. (a)
3. (a)
4. (b)
5. (d)
6. (b)
7. (b)
8. (b)
9. (d)
10. (c)
11. (b)
12. (a)
13. (c)
14. (d)
15. (d)
16. (b)
17. (b)
18. (c)
19. (a)
20. (c)
21. (b)
22. (d)
23. (c)
24. (c)
25. (c)
26. (a)
27. (d)
28. (b)
29. (d)
30. (b)
31. (d)
32. (b)
33. (d)
34. (a)
35. (b)
36. (c)
37. (b)
38. (d)
39. (b)
40. (b)
41. (a)
42. (c)
43. (d)
44. (a)
45. (c)
46. (d)
47. (c)
48. (d)
49. (c)
50. (b)
51. (b)
52. (d)
53. (d)
54. (b)
55. (b)
56. (c)
57. (b)
58. (c)
59. (a)
60. (a)
61. (c)
62. (c)
63. (c)
64. (b)
65. (c)
66. (b)
67. (d)
68. (c)
69. (c)
70. (d)
71. (a)
72. (a)
73. (b)
74. (a)
75. (b)
76. (a)
77. (b)
78. (a)
79. (c)
80. (c)
81. (b)
82. (c)
83. (a)
84. (c)
85. (b)
86. (b)
87. (c)
88. (c)
89. (d)
90. (a)
91. (d)
92. (d)
93. (b)
94. (a)
95. (d)
96. (a)
97. (a)
98. (c)
99. (a)
100. (b)
101. (d)
102. (d)
103. (a)
104. (a)
105. (a)
106. (d)
107. (c)
108. (a)
109. (a)
110. (b)
111. (a)
112. (c)
113. (d)
114. (a)
115. (a)
116. (b)
117. (a)
118. (a)
119. (d)
120. (a)
Conventional Question Practice ProgrameDate: 26th March, 2016
ANSWERS
IES M
ASTER
(2) CE (Test-8), Objective Solutions, 26th March 2016
1. (a)2. (a)3. (a)
xy =
u vy x = (8 – 12)10–6
= –4 × 10–6 units
4. (b)b = 120 mm
t = 100 mm
R = 10 m = 10000 mm
M EI R
M =EIR
=
35 bt(2 10 )
12R
=5 3(2 10 )(120 10 )
12 10000
= 2 × 105 N-mm5. (d)
Rpermissible
MZ
15 N/mm2 = 3
(12 2.5) kN md
32
d = 400 mm
6. (b)
Stress = MZ
Z = 2bd
6 =
215 306
= 2250
M = 2.25 × 105 kgcm
Stress = 52.25 10
2250 = 100 kg/cm2
7. (b)
b
h
y
= 23
12S (hy y )bh
= 2
312 25000 (100 25 25 )100 100
= 5.625 N/mm2
8. (b)
Modular ratio = 210 2010.5
In equivalent section, width of material 1 will
reduce to 10020 = 5 mm. (When material 1 is
converted to material 2.)
20mm
100mm
5mm
y
N A2
2100mm
Transformed section
Let y be depth of neutral axis, then
(5 × 100 + 20 × 100) y
= 5 × 100 × 50 + 20 × 100 × 110
y = 98 mm (from top)
9. (d)In non-circular solid sections the shearstresses developed are unsymmetrical,therefore warping may occur along with twisting.As a result plane section will not remain planeafter twisting. Hence theory of pure torsion isnot valid.
bA
A
d
In rectnagular sections maximum shear stressdevelops on the middle surface of longer side,i.e. at A or A if b > d.
Shear stress at corners = 0
Shear stress distribution is non-linear.
10. (c)
Let twist angle = .
Let length of shaft = L.
then, 1
1 1
P
T GL
I
IES M
ASTER
CE (Test-8), Objective Solutions, 26th March 2016 (3)
T1 =
4
2
D22G
32 L
=
42
9G D2
L2
Similarly T2 = 4
42
DG D32 2 L
= 4
29
G D15L2
T1 + T2 = T
Required fraction = 1T 2 2T 2 15 17
11. (b)The properties of the fluid at all locations withinthe system are the same and are identical tothe properties of the discharge or effluentstream in ideal completely mixed flow regime.
12. (a)
x y = 1 2
( 2 minor principal stress)
28 – 8 = 40 + 2
2 = – 20 MPa
–20 4010 28
30
Radius of Mohr’s circle = 30 MPa
Centre of Mohr’s circle = 10 MPa
xy = 2 230 18 = 24 MPa
13. (c)Sanitary sewage or domestic wastewater isthe liquid wastewater from sanitary facilitieswithin a building.
14. (d)15. (d)
RB = RC = 0.3 KN
Mmax = MB = MC = 0.3 × 3
= 0.9 KN – m
= 0.9 × 106 N – mm= 9 × 105 N – mm
Between point B and C, Bending moment ishogging and constant. So
max = maxM YI
= 5
26
(9 10 ) 70 21N / mm3 10
16. (b)
Resilience = 80 × 0.005 × 12 × 106
= 20 × 104
Toughness =
480 12020 0.01 10
2= [ 20 + 100 ] × 104
= 120 × 104
17. (b)18. (c)
Chi factor should be greater than 500 for flowto become unstable.
19. (a)20. (c)
Repelled jump occurs when tailwater depth yt< y2.wherey2 is the sequent depth corresponding to afree jump at Vena contracta.
y2 = 2aa
y 1 1 8F2
y2 = 0.4 1 1 8 102
= 1.6 m
Hence for yt < 1.6 m repelled jump occurs.21. (b)
Standard step method is different from directstep method. Standard step method is usedfor natural channel whereas direct step methodis used for prismatic channel Standard stepmethod is similar to the slope area method inhydrology. Here we use only energy equationand continuity equation to solve the problemby trial and error approach.
22. (d)23. (c)24. (c)
C = gy 9.8 9.8 9.8 m/s
IES M
ASTER
(4) CE (Test-8), Objective Solutions, 26th March 2016Velocity of wave moving u/s = C – V= 9.8 – 3 = 6.8 m/s
25. (c)26. (a)
P
P/2
P/2
A B
C
P/2LL
AM = 0 (for potion to the left of the section)
(P – F) × L = 0 F = PForce in ‘a’ = P,Force in ‘b’ = P/2,
Force in ‘c’ = 027. (d)28. (b)
The total organic cabon (TOC) is a measure ofthe organic materials in a wastewater in termsof the amount of carbon in the organic material.
29. (d)30. (b)
Studies have shown that i f the seedmicroorganims are acclimated to the waste,then the removal of organic material is apseudo-first order reaction.
dc kcdt
where,dcdt
= rate of removal of organicmaterial
K = rate constant of the base e
C = concentration of organicmaterial remaining at time ‘t’
31. (d)32. (b)33. (d)34. (a)35. (b)36. (c)
22
4E 2I 4E IK3 4 =
11EI3 = 3.67 EI
I I I I I 33
4E 4EK E 2E 3E4 2
23
2E I EIK 0.5EI4 2
37. (b)Muller Breslau principle suggests two differentmethods by which influence lines for redundantstress and reaction can be obtained (1) byexperiment (2) by computations.ILD for deflection will always be a curved line.Muller Breslau principle is not applicable fordeflecion.Muller Breslau principle can be proved usingvirtual work principle.
38. (d)39. (b)40. (b)
The absolute maximum bending momentoccurs near the mid span and occurs under150 kN load.Hence, for the condition of absolute maximumbeinding moment the load system should beso placed on the span that the resultant of allthe wheel loads and the 150 kN load areequidistant from the middle point of girderTaking moments about 70 kN load,
70 0 +150 ×1 + 60 ×1.5 +120 × 2400x = ×
1.2mx from 70 kN load. Distance betweenresultant load and 150 kN load = 1.20 – 1.00= 0.2 mTherefore, for the condition of absolutemaximum bending moment, the 150 kN loadshould be placed 0.1 m on the right side ofthe centre of girder.
b bV 10 400 4.9 V 196kN,
aV 400 196 204kN
Absolute maximum bending moment for thegriden = BM under 150 kN load
= 196 × 4.9 – 70 × 1= 890.4 kNm.
41. (a)
Reaction coefficient is directly related to 1.33V
H,
where V = flow velocity and H = stream detph.Larger is the rate of reaeration, larger is theassimilation rate. Lesser is the BOD or BODrate constant, greater is the assimilation rate.
42. (c)Synder selected three parameters fordevelopment of SUH. They are :-
IES M
ASTER
CE (Test-8), Objective Solutions, 26th March 2016 (5)(i) Basin time width T
(ii) Peak discharge QP
(iii) Lag time i.e., basin lagtime tP
trunit duration
tPQP
TBase width
Pnet = 1cm
43. (d)
SVI = 200 ml 200 ml ml50
4000 mg 4g g
44. (a)45. (c)
Shear in Panel LoL1
1Lo
L1 L2 L3
R = 1 R = 0
SF in Panel LoL1 = 0
LoL1 L2 L3
1
R =34
SF in Panel LoL1 = 34
Lo
L1 L2 L3
R =12
1
SF in Panel LoL1 = 12
L1 L2 L3
R = 14
1L0
SF in Panel LoL1 = 14
Lo
L1 L2 L3
R = 0
1
SF in Panel LoL1 = 0
R = F cos
F
F Rsec
Moment atL1 = R LoL1
= F cos . LoL1
1L
o 1
MF
Cos L L
ILD for force in member LoU1 isobtained bymultiplying the ordinate of ILD forshear in panel LoL1 by sec also by dividingthe ordinate of ILD for moment at L1 bycos ×LoL1
46. (d)The theoretical oxygen demand (TOD) is theamount of oxygen theoretically required toconvert to elements in the organic material tostable end products.
47. (c)48. (d)49. (c)
Small eddy size Shorter wave size =Smaller losses.
Lower viscosity Smaller laminar sublayer Boundary behaves as hydrodynamicallyrough Larger losses.
Large intensity of turbulence - larger losses.50. (b)51. (b)52. (d)53. (d)54. (b)
Absorption area req = o
2
Q ( / day)204 / m / day
t
l
l.
t = Percolation rate in min
= 210000 m204
t
= 250000 245m204
IES M
ASTER
(6) CE (Test-8), Objective Solutions, 26th March 201655. (b)
Total solid Produced = Solid production in PST+ Solid Production In SSTSolid Production in PST
= 0.6 × 200 mg/l×150×103 l/day= 180 kg/day
Solid production in T.F.= (200 mg/l BOD×1500×103 l/day)
× 0.7 × 0.5= 300 × 0.7 × 0.5 = 105 kg /day
Total solid production = 285 kg/day.56. (c)
1 kg of BOD stabilised produces 0.35 m3 CH4or 0.25 kg CH4 at STP.
57. (b)
c = w u
VxQ X
QwXu = c
Vx
= 1000m3×6
33000 10
10
kg/m3 ×
110days
= 300 kg/day
sWW
= (1 – 0.95)
W = sW0.05
= 20Ws = 20 × 300 = 6000 kg/day
58. (c)
Incomming BOD = 3 6
3 3m 400 10 kg2000day 10 m
800 kg/day
VSS produced = 0.25 × 0.9 × 800= 180 kg/day
Phosphorus removal along with
VSS = 2 180
100 = 3.6 kg/day
59. (a)60. (c)61. (a)
Aerobic digestion plant has lower capital costbut higher operational cost.It is generally adopted for smaller plants.
62. (c)Sequence is1. Hydrolysis2. Acid formation3. Methane formation
63. (c)64. (b)
Nitrogen removal is achieved by denitrification.65. (c)66. (b)
R
C
A
P
BO
The horizontal arm BO being rigid will not storestrain energy. The rigid arm will transmit avertical force P at B and a moment PR at B.
PRBO
R
X
A
C
PAt any section X whose radius vector makesan angle with OB, the bending moment atthe section
= M = PR – PR (1 – cos) = PR cos
Strain energy stored
2 2 2 2
0
M ds P R cos RdU2El 2El
/22 32
0
P R 2 cos d2El
2 3P REl 4
2 3
iP Rw
4El
Vertical deflection of the point B
3 3U 2P R P RP 4El 2El
67. (d)The factors that influence sulphide generationin sewer includes.
(1) Temperature of sewage - Temp below20ºC will not cause appreceable sulphidebuild up.
(2) Velocity of sewage - Higher velocity leadsto lesser H2S generation
(3) Age of sewage - Higher age of sewageleads to greater H2S formation.
IES M
ASTER
CE (Test-8), Objective Solutions, 26th March 2016 (7)(4) PH of sewage - At PH 6 - 8, rate of
generating of H2S is not affected by PH butfor pH < 5.5 & > 9. It is affected.
(5) Sulphate Concentration - More sulphatemore H2S formation.
(6) Ventilation of sewer - More ventilation lessH2SO4 generation. Due to ventilation bothH2S and moisture are ventilated away.
(7) Strength of sewage - BOD less than 80mg/l will cause ceasing of H2S formation.
68. (c)
Iron salts reduces the concentration ofSulphides by converting the dissolved Sulphideto insoluble iron Sulphide.
69. (c)
Flushing manhole is provided on the head end
Drop manhole provided if the gap mentioned inquestion is > 600 mm.
70. (d)
Invert of sewer refers to bottom of sewer.
Most important characteristic of corrosion isthat corrossion occurs above the water line.i.e. It generally occurs is crown portion andhence the phenomenon is known as crowncorrosion.
H2s is produced by reduction of SO42.
71. (a)
Pisciculture refers to fish culture.
Sea water contains 20% less oxygen. Howeverdilution is more sea water.
5
Parameter Into River Into SeaBOD 20 mg/ 100 mg/
SS 30 mg/ 100 mg/l ll l
72. (a)
73. (b)
ILD ordinate y will be equal to 1
1
1
x
y =
2 3
3
1(L x)x 1 x2EI 3EI
1 L3EI
y = 2
3x (3L x)
2L
74. (a)
A Bc
c f11
f211
Unit force at 1
3
11Lf3EI
and
2
21cc´ LfL 3EI
A BC
f12
f22
1C
Unit force at 2
f12 = –cc´´ = 3 2I L L L
3EI 3EI
C
C
1
2
f22 = 1 2CC 1 L
L 3EI
= 2L 1 L 2L
3EI L 3EI 3EI
75. (b)Capacity of each aerator
= 768 20 HP
0.8 24 2
76. (a)
A B2EI
B = 1
L
MB =
B4 2EI
L
77. (b)2
2d 2
d
1 2W dC AV W2 4C V
d = 2d
8WC V
d = 5.95 m
IES M
ASTER
(8) CE (Test-8), Objective Solutions, 26th March 201678. (a)
R
dt
R = Radius of curvatureL = Length of rodd = Centre dipt = Transverse distance of point from N.A.
L
R = 2L
8d =
2208 1
= 50 cm
= tR =
0.5cm50cm = 1 × 10–3
79. (c)80. (c)81. (b)82. (c)
Examples of Force MethodCastigliano’s Theorm (Method of Least Work)Strain energy method.Claperon’s three moment equations (used incontinous beam analysis)Column analogy method (used in rigid frameswith fixed supports)Flexibility matrix methodExamples of displacement MethodSlope deflection methodMoment distribution methodStiffners matrix methodKani’s Method
83. (a)Although not considered to be a major nitrogenremoval process, oxidation ponds (stabilizationponds) can ef fect removal a nitrogencompounds.
84. (c)
D
BE
FC
A 45°2m2m
2t
2m 2mRA = 1t RB = 1t
MA = 0 2 × 2 = 4RB RB = 1 t () RA = 1 t ()
FFD = FDB = 0At joint B,FFB sin 45° = 1
FFB = 1 2
sin45
t
Again, FFB cos 45° = FEB
FEB = 12 1 t2
(Tensile)
85. (b)
9 m
1000 kNC
HAA
VA2000 kN 550 kN
4 m 6 m 6 m4.5 m
1000 kND
BEF
MB = 0 VA × 16 = 1000 × 9 + 1000× 4.5 + 2000 × 12 + 550 × 6 VA = 2550 kNH = 0 HA = 1000 + 1000 = 2000 kN
86. (b)By symmetry, both the support reactions = 10kN each.Consider joint E, forces in member FE can’tbe counterbalanced FFE = 0Consider joint at B,FBF sin 45° = 10
FBF = 10 2 kN (comp.)
45°
2 m
20 kN
2 m 2 m10 kN10 kN
A
C F D
EB
FBF cos45° = FEB
FEB = 110 22
= 10 kN (Tension)
IES M
ASTER
CE (Test-8), Objective Solutions, 26th March 2016 (9)87. (c)
2 kN
2 kN
4 kNE F
CD
3 m
3 m
4 m4.5 kN 4.5 kN
1
1
BA
EM 0
F3 2 2 6 4R
F18R 4.5kN4
Consider joint E,
EF EF4kN F F = 4kN (Tensile)
Take a section (1) – (1)Taking moments about C, and consider rightside (down) part
DF EF4F 4.5 4 F 3 0
or DF4F 4.5 4 3 4 0
DFF 1.5 (compression)
88. (c)3 × 2 + 2 × 6 + 2 × 9 = 4 RH
3m
3m
3m
H
F
D
BA2kN
C2kN
E2kN
G6kN
4m
9kN 9kN
6 kN
RH = 9 kN
tan = 34
cos = 45
Consider the joint, H
EHF cos = GHF
EHF = GHF
cos = 645
= 6 5
4
= 7.5 kN (comp)89. (d)
cos = 45
sin = 0.6 = 36.87º
6m
4m
B
2
= 45°
E
5m
CA
D
4m
Fabrication error case (Lack of fit case)Member lengths are generally made longer orshorter than actual required length in order tointroduce member forces which will compensatefor deflection due to the dead load.Due to this lack of fit, joint deflection fromoriginal level will take place. This joint deflectionin this case is calculated using virtual workprinciple.
1 = i iu dL
Where dLi = fabrication error in ith memberSign convention
dLi = +ve if member is longerthan normally expected
dLi = –ve if member is shorter thannormally expected
ui = Force in member i due to unitload 1 in direction of .
Tension = +veCompression = –ve
= +ve if deflection in the directionof applied unit load
IES M
ASTER
(10) CE (Test-8), Objective Solutions, 26th March 2016
= –ve if deflection opposite to thedirection of applied unit load.
To find out vertical deflection at point c, a unitload at C in vertical direction is applied andmember forces one found out.
At joint C,
FCD sin = CD11 F 1.667
0.6
and FCD cos = CD CBF F 1.33
At joint B,
FAB = FBC = 1.33
At Joint D,
FED = FCD = 1.667
MemberCD –1.667 –10 +16.67CB +1.33 0 0BD 0 0 0AB +1.33 0 0AD 0 0 0DE –1.667 0 0
u dL (mm) udL(mm)
= + 16.67 mm
+ve means deflection is in direction of appliedunit load, i.e. downwards.
90. (a)
M+Pl
P M+Pl
P
PM+Pl
l
l
CB
A
B
C BH H
where, 2
B(M P )H
2EI
l l
91. (d)K11 = force induced in direction 1 due to unitdisplacement indirection 1 while restringing allother displacements
4m2I
M1
M3 5m
1.5I
M2
4mI
31 2 M 5M 4 M 41
4E(2I) 4E(I) 4E(1.5I)
K11 = M1 + M2 + M3
4E(2I) 4E(I) 4E(1.5I)4 4 5
= 4.2EI
Similarly K21 = force induced in direction 2due to unit displacement in direction 1 whilerestraining all other displacements
5mM
M2
= 1
M(5) 14E(1.5I)
M = 1.2EI
21MK 0.6EI2
K22 = Force induced in direction 2 due to unitdisplacement in direction while restraining allother displacements
Displacement restrained in direction 1
M
M(5) 14E(1.5I)
M = 12EI
K22 = M = 1.2EI
92. (d)
93. (b)
(u)shear = 38 MPa
avg = cs
P(A )
P = avg × Acs
= 38 × (80 × 30)
= 91200 N
= 91.2 kN
IES M
ASTER
CE (Test-8), Objective Solutions, 26th March 2016 (11)94. (a)
P
l/3 l
Pl/3 l
P /3l PP
P /3l
ll/3
3P( /3)
3EIl
2P P3 3EI 9EI
l l l
3P81EI
ll
= 3 3 3 3P P P 10P
81EI 9EI 81EI 81EI
l l l ll
95. (d)96. (a)97. (a)
L3
U2
3m
4m
2/64 3sin , cos5 5
2Fsin 06
F = 2 2
46sin 65
F = 1024
4mU2
3m L3
F
1/2
1Fsin 02
1 1 5F42 sin 825
5/8 = 15/24
–10/24
= Tension= Compression
30 1.2524
30 1.2524
98. (c)99. (a)
50t cm 10t 0.5 cm
5 150 10
100. (b)101. (d)
1 = 1 2E
2 = 2 1E
1
2
1
2
102. (d)103. (a)104. (a)105. (a)106. (d)107. (c)108. (a)109. (a)110. (b)111. (a)
112. (c)
1. – external virtual work done
md – internal virtual work done
113. (d)
=
MMP dx
EI
114. (a)
IES M
ASTER
(12) CE (Test-8), Objective Solutions, 26th March 2016115. (a)116. (b)117. (a)
A single-peaked hydrograph, figure consists of(i) a rising limb, (ii) the crest segment, and (iii)the recession or failling limb. The rising limb(or concentration curve) of a hydrographrepresents continuous increase in discharge(or runoff) at the watershed outlet. During theinitial periods of the storm, the increase inrunoff is rather gradual as the fal lingprecipitation has to meet the initial losses inthe form of high infiltration, depression storageand gradual building up of storage in channelsand over the waershed surface.
As the storm continues, losses decrease withtime and more and more rainfall excess fromdistant parts of the watershed reaches thewatershed outlet. The runoff, then, increasesrapidly with time. When the runoff from all partsof the watershed reaches the watershed outlet.simultaneously, the runoff attains the peak (i.e.,maximum) value. This peak flow is representedby the crest segment of the hydrograph. Therecession limb of the hydrograph starts at thepoint of inflection (i.e., the end of the crestsegment) and continues ti l l thecommencement of the natural ground waterflow.
118. (a)
A surge can be positive or negative. The onewhich produces an increase in flow depth is
called positive surge and the one which causesa decrease in flow depth is called negativesurge.
When a gate is suddenly opened at the headof a channel or a gate is suddenly closed atthe tail, wave will be generated which will movedownstream in first case and upstream insecond case. In both the cases surge wavewill increase the depth of water hence abovecases are an example of positive surge.
119. (d)The tapered aeration activated sludge processis only a modification of the conventionalprocess.
120. (a)Waste stabilization ponds is an open flowthrough earthern bars which provides longdetention periods during which water getsstabilized by the action of natural forces.Here, there exists a symbiotic relationshipbetween algae and bacteria where algae in thepresence of sunlight produces oxygen byphotosynthesis which is used by bacteria foroxidising the waste organic matter, the endproducts of which are CO2, NH3, PO4, H2Owhich are required by algae to grow andcontinue producing oxygen.Hence correct option is (a).
top related