IES M

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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

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(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

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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

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(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

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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

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(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

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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

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(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

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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

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(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

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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

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(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).