answers - ies master of the member. secondary ... indicating the relative position of the each...

1. (d)

2. (c)

3. (b)

4. (d)

5. (a)

6. (a)

7. (d)

8. (c)

9. (a)

10. (a)

11. (c)

12. (d)

13. (b)

14. (d)

15. (c)

16. (a)

17. (d)

18. (d)

19. (b)

20. (b)

21. (c)

22. (d)

23. (b)

24. (d)

25. (c)

26. (b)

27. (b)

28. (b)

29. (a)

30. (a)

ESE-2018 PRELIMS TEST SERIESDate: 03 December, 2017

ANSWERS

31. (a)

32. (c)

33. (d)

34. (a)

35. (a)

36. (a)

37. (a)

38. (d)

39. (a)

40. (a)

41. (a)

42. (b)

43. (a)

44. (b)

45. (b)

46. (c)

47. (d)

48. (d)

49. (a)

50. (b)

51. (d)

52. (c)

53. (a)

54. (b)

55. (d)

56. (c)

57. (d)

58. (b)

59. (d)

60. (c)

61. (b)

62. (b)

63. (d)

64. (b)

65. (c)

66. (a)

67. (c)

68. (c)

69. (d)

70. (c)

71. (c)

72. (b)

73. (b)

74. (c)

75. (c)

76. (d)

77. (a)

78. (c)

79. (c)

80. (c)

81. (a)

82. (d)

83. (a)

84. (c)

85. (a)

86. (d)

87. (c)

88. (d)

89. (b)

90. (a)

91. (a)

92. (b)

93. (c)

94. (b)

95. (d)

96. (c)

97. (b)

98. (a)

99. (a)

100. (d)

101. (b)

102. (d)

103. (c)

104. (b)

105. (d)

106. (d)

107. (a)

108. (c)

109. (d)

110. (c)

111. (b)

112. (b)

113. (d)

114. (d)

115. (c)

116. (c)

117. (b)

118. (d)

119. (a)

120. (c)

121. (b)

122. (d)

123. (d)

124. (a)

125. (c)

126. (c)

127. (a)

128. (b)

129. (a)

130. (d)

131. (d)

132. (b)

133. (b)

134. (d)

135. (a)

136. (b)

137. (a)

138. (c)

139. (a)

140. (a)

141. (a)

142. (a)

143. (b)

144. (b)

145. (b)

146. (a)

147. (d)

148. (a)

149. (b)

150. (a)

IES M

ASTER

(2) (Test-11)-03 December 2017

Mobile : E-mail: 8010009955, 9711853908 [email protected], [email protected]. office : Phone : F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 011-26522064

Web : iesmasterpublications.com, iesmaster.org

1. (d)1. Creep increases when

(a) Cement content is high

(b) w/c ratio is high

(c) Aggregate content is low

(d) Air entrainment is high

(e) Relative humidity is low

(f) Temperature is high

(g) Size is small

(h) loading occurs at early stage.

2. Unlike creep, shrinkage strains areindependent of the stress conditions of theconcrete.

3. Chemical shrinkage induces internal voidsand autogenous shrinkage results inelement shortening.

4. In limit state method of design, two factorof safety are used, one to account foruncertainity in load and other to accountfor uncertainity in material.

2. (c)1. It has been seen that the compressive (as

well as tensile) strength of concrete isreduced by the presence of shear stress.Also shear strength of concrete isenchanced by the application of directcompression (except extreme case of veryhigh compression), whereas it is(expectedly) reduced by the application ofdirect tension.

2. The code limits the maximum value of fyto 415, as higher strength reinforcementsmay be rendered brittle at the sharp bendsof the web reinforcement, also a shearcompression failure could precede theyielding of the high strength steel.

3. (b)Beam is subjected to torque = 15 kN-m,

and shear force (factored) = 140 kN.

Calculating equivalent shear force

Ve = uu

1.6TVB

=1.6 151400.230

= 244.348 kN.

Additional shear for which the stirrups haveto be provided

= e cV ( )bd

= 3230 400(244.348) (0.48)

10

= 200.188

= 200.19 kN

200 kN

4. (d)1. Chemical Adhesion: Due to property in

the products of hydration (formed duringthe making of concrete)

2. Frictional resistance: Due to the surfaceroughness of the reinforcement and thegrip exerted by the shrinkage of concrete.

3. Mechanical interlocking : Due to surfaceprotrusions or ‘ribs’ oriented transverselyto the bar axis provided in deformed bar.

5. (a)

Ld =s

bd4 =

20 0.87 4154 1.6 1.25 1.5

=7221 mm12

Ld = 601.75 mm

Ld 602 mm

6. (a)Primary (or equilibrium) torsion is associatedwith twisting moment that are developed in astructural member to maintain static equilibriumwith the external load directly applied onmember, and are independent of the torsionalstiffness of the member.

Secondary (or compatibility) torsion is inducedin a member as a secondary effect, by rotation(twist) applied at one or more points along thelength of the member through interconnectedmembers, rather than by directly applied loadon it.

7. (d)For simply supported beam

Spandepth

modification factor for1020 tension & compression

spanreinforcement

Spandepth

1020 0.9 1.115

depth span

1020 0.9 1.115

depth 1.136 m

mailto:[email protected],

mailto:[email protected]

IES M

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(Test-11)-03 December 2017 (3)



8. (c)9. (a)

Hoyer system is generally used for massproduction (like Railways sleeper, poles, etc).The end abutment are kept sufficient distanceapart, and several members are casted in asingle line. The shuttering is provided at thesides and between the member.

When the level of prestressing is such thatthe tensile stress under service load is withinthe cracking stress of concrete, it is calledlimited prestressing.

10. (a)Area of cross-section = 200 × 250

= 5 × 104 mm2

Modular Ratio = 13

Prestressing force,

P = 1200 × 600 = 720 kN.

Stress in concrete at the level of steel

=P M.yA I

=3

4720 10 05 10

= 14.4 N/mm2

Loss of prestress due to elastic shortening ofconcrete

= 2m 13 14.4 187.2 N/mm

Stress in wire after loss

= 1200 – 187.2 N/mm2

= 1012.8 N/mm2

Stress in concrete corresponding to 1200 N/mm2 Prestress = 14.4 N/mm2

Stress in concrete corresponding to 1012.8 N/mm2 Prestress

=14.4 1012.81200

= 12.15 N/mm2

12 N/mm2

11. (c)1.According to clause 4.7 of IS 4326 – 1993

2.According to clause 4.9 of IS-4326 – 1993

12. (d)In fully rigid design of steel structures,connections are capable of transmitting

moments and the angle between members atthe joint does not change.

13. (b)In class 4.6 bolt,

(i) 4 denotes th1

100 of its ultimate strength

in MPa

(ii) 0.6 denotes ratio of yield strength toultimate strength.

14. (d)

15. (c)

Pressure at the bottom of tank

= w1.75 h

= 1.75 × 9.81× 27

= 463.5 kN/m2

Force per pitch length = d × p

= 2.5 463.5 × 50 × 10–3

= 182 kN

16. (a)Critical section for rupture can be

160

1

1

2

3

2

a

a

1

I. a – 1 – 1 – a

Anet = (160 – 3 × (16 + 2)) × 8 = 848 mm2

II. a – 1 – 2 – 1 – 1 – a

Anet =

22 40160 4 18 84 25

= 960 mm2

III. a – 1 – 2 – 3 – 2 – 1 – a

Anet = 24 40160 5 18 8

4 25

= 1072 mm2

Hence, the critical most section is I

Anet = 848 mm2

17. (d)Lug angles are used to reduce the length ofjoint.



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18. (d)Battens of a built up column are designed forcarrying bending moment and shear arisingfrom a transverse shear V

Where V = 2.5 Axial load100

19. (b)For compression member maximum allowableslenderness ratio is 180

ratio = 180

minrl

= 180

l = min180 r

=180 24 4.32 m

1000

20. (b)

Given maximum shear force = 56 kN

Shear area = height × thickness of beam of web

= 450 × 6 = 2700 mm2

Shear stress = 256 1000 20.74 N/mm2700

21. (c)22. (d)

Schedule contracts are usually used formaintenance or minor job. They are also usedfor conventional building construction projects.

23. (b)

1 2 5

3

4

6

t = 6 t = 8

t = 1

t = 9 t = 5

t = 2t = 4

T = 7

ET = 7 L

T = 0

ET = 0 L

T = 1

ET = 1 L

T = 16

ET = 16 L

T = 18ET = 18 L

T = 11

ET = 11 L

(a) 1 – 2 – 3 – 5 – 6 17 days

(b) 1 – 2 – 3 – 4 – 5 – 6 18 days

(c) 1 – 2 – 4 – 5 – 6 17 days

(d) 1 – 2 – 4 – 3 – 5 – 6 Not valid

Hence critical path is 1 – 2 – 3 – 4 – 5 – 6

24. (d)

1 2 3 5 6

4

6

A B

D E

F10 146

8 5

C

Critical path is A – B – C – F with projectduration of 36 days. Range of project durationis ET 3 .

Standard deviation of project is calculated alongcritical path

2 2 2 2project A B c F

2 2 2 2project 1 2 2 0

project 3

Range of project duration= [36 – 3 × 3] to [36 + 3 × 3]

= (27, 45)

25. (c)The following are important stages in developinga bar chart.

1. Breakdown: The project into its variousactivities or jobs or operation, eachrepresenting manageable unit for planningand control.

2. Decide: The method to be employed inexecution of the project as well as for eachactivity or operation or task; also decideabove the sequence in which the activitiesare to be completed.

3. Assign: Duration of time for the completionof each activity. Once the activities areseperated and choice of method is made,it is possible to estimate the time requiredfor the completion of each activity.

4. Represent: The above information is thebar chart, indicating the relative position ofthe each activity.

26. (b)Scheduling is the deciding of the phasing rateof activities. The starting and completion datesand the sequential relationship among thevarious activities in a project such that workcan be carried out in an orderly and effectivemanner.

27. (b)For spreading of soil

Area to be covered = Volume 80000

Thickness 0.2



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= 4,00,000 m2

Area covered per hour per pass

= Speed × width × OF

=504 1000 360

= 10,000 m2

Total area covered by one grader with 5 passes= 50,000 m2

No. of grader required = 4,00,000 850,000

28. (b)

Kani’s method Displacement method

29. (a)

400 = (100 × 10) + (15 × 300)

= 13.75 mm30. (a)

Stiffness of BC = kBC = 3EI EI3

Stiffness of AB = AB4EIk EI4

D.F for AB = 12

D.F for BC = 12

So, moment in MBA = 20 kN-m

So, At A moment will be = 20 10 kN-m2

31. (a)32. (c)

Maximum B.M will occur when centre of thebeam lies midway between the load and C.Gof the series of load

2.5 m

3 kN3 kN

C.G of load 3 5 2.5

6

For absolute maximum B.M

1.25 m

3 kN3 kN

12 m 12 m

3.75 m

RBRA

RA + RB = 6kN

AM = 0

B24 R = 3 × 8.25 + 3 × 13.25

RB = 2.6875 kN

RA = 3.3125 kN

Mmax = 2.6875 × 10.75 m 28.89 kN-m

= Mz

= 6

3 328.89 10 N-mm16.2 10 mm

= 1783.33 N/mm2

= 1.78 GPa

33. (d)34. (a)

No. of rotation possible = 3 (m – 1) = 3 (5 –1) = 12

35. (a)In plastic region, deformation is caused byslippage of material along oblique surface Thusvolume remain constant in this region.

36. (a)

dM vdx

differentiation should be w.r.t 'x' to get SF @ that point

37. (a)Max. principal stress theory is applicable tobrittle material because brittle material failunder tension leading to fracture.

38. (d)39. (a)

w

x

Mx =2wx

2

2bdz6

=

2

2x

wxb d2

6

2bd x

3

= wx2

dx =

3w xb

Hence xd x



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40. (a)41. (a)42. (b)

Taking moment about hinge (A)

T × 0.5 = 300 × 0.3 + 300 × 0.9

T 720 N

43. (a)

Work done = l 2eq

1 K ( )2

and Keq =1 2

1 2

K KK K

WD =21 2

1 2

K K ( )2 (K K )

l

WD = 2 2100 200 100( ) ( )600 3

l l

44. (b)For a spring, (K × L) is a constant

K L = 2LK3

2L3

= Length of longer piece

3K K2

45. (b)

allwable = 2

2103000 kN/m 41200 kN/m2.5

Allowable shear force

V =

2all allow allowA d

4

=

241200 0.0125 5.056 kN4

Force supported by nuts

F = 2 × V = 2 × 5.056 = 10.11 kN

46. (c)

at C

c =

6

4

Tr 40 10 75J (150)

32

c =

6

4

7530 2 10My 2I 150

64

22c2

max c2 2

120.72 MPa.

47. (d)

Maximum slope will occur at the end wheremoment is applied. Angle is clockwise so itwill be –ve

max = 3

6 6ML 1.13 10 0.63EI 3 1370 10 2.08 10

–0.0793 rad = – 4.54 degree

48. (d)

49. (a)

max 3

My 32 MI D

max 3

r 16TJ D

max

max

2MT

50. (b)

8N = m m w wV V

= 6w w w(13.6) ( ) (28 10 ) ( ) (V )

Vw =

63

8 13.6 28 109.81 10

= 4.34 × 10–4 m3

= 434.69 cm3 3435 cm

51. (d)

For notch Q = 5/2D

8 c H 2g tan15 2

dQQ =

5 dH2 H

dQ 100Q

=35 1.5 10 100 0.75%

2 0.5

52. (c)

For maximum accleration oil will reach justupto the maximum height of the tank withoutbeing spill out and volume of oil will remain thesame.



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3 m

x

10 m

1.5 m

1 x 3 52 = 10 × 1.5 × 5

x = 10 m

therefore tan = xa 3g 10

ax = 3 m/s2

53. (a)Area of gate = 5 × 3 = 15 m2

5 m

195 kN/m2

The equivalent height of water which gives apressure intensity of 195 kN/m2 at bottom

h = w

P 19.87 m

Centre of pressure h = aIxAx

x = 19.87 – 2.5 = 17.37 m

I = 3bd

12

h = 33 517.37

12 14 17.37

= 17.49 m

54. (b)Boundary layer will form on both sides of plate

FD = 2

wD

V2 C A2

5.31 = 3 2

D10 22 C 2 1

2

CD = 6.64 × 10–4

55. (d)Since the pipes are connected in parallel hencehead loss will be same in both the pipes

2

5fLQ

12.1d = C

2 5Q d

1

2

QQ =

52

5/21

2

d(3) 15.6

d

56. (c)57. (d)58. (b)

Entrance length is (0.05 Re)D for laminar flow

f = e

64R

59. (d)Volume of water precipitation

= 500 × 104 × 10 × 10–2

= 0.5 × 106 m3

Volume of water from inflow

= 5 × 30 × 24 × 3600

= 12.96 × 106 m3

Volume of water from outflow

= 2 × 30 × 24 × 3600

= 5.184 × 106 m3

Volume lost due to evaporation

500 × 104 × 8 × 10–2 = 0.4 × 106 m3

Increase in depth

= 6 6 6

40.5 10 12.96 0.4 10 5.184 10

500 10

= 1.5752 m

60. (c)Muskingum equation is used for channelrouting.

Both Saint Venant equation and hydraulicmethod uses continuity and moment equation.While Hydrological method uses continuity andenergy equations.

61. (b)62. (b)

Q =

1

2

SAS lnT S

=

2(11 2.5) 2.54 ln70 1.360

= 1.1 m3/hour



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63. (d)64. (b)65. (c)

From the concept of visible horizon

h

d1 d2

h

d = 1 2d d

d = 2 3.855 h

20.09 = 2 3.855 h

h = 6.78 m

66. (a)

x

10

1

Grade 1 in 10

= 21 10 10.05

Fall in 30 m = 30 1 2.985

10.05

Slope correction Cg =2 2h 2.985 0.15m

2L 2 30

67. (c)

68. (c)69. (d)

x = 0 F

ES S

E1 = 2

1Vy2g

E1 = 2

2101 6 m

1 2 10

2E = 2

2100.85 7.77 m

0.85 2 10

x = 7.77 6

0.015 0.020.0032

x = 122 m

70. (c)Find the continuity for unsteady flow

Q Ax t

= 0

QAt x

A 0.10t

71. (c)

72. (b)For homologous pumps the specific speed issame .

NSA = NSB

NS = 3/4N QH

3/4600 0.4

50 = 3/4

B

600 0.3H

HB = 41.28 m

73. (b)74. (c)

Flow Q = w a

x yt t

x = 200

y = 10 – 20 = –10

tw = ta = 5 1 hrs

60 12

Q = 200 10 1140 Veh hr1 112 12

Flow of vehicles travelling at 80 kmph

= 0.4 1140

= 456 Veh/hr

Time taken for 5 Km by vehicles of speed 80

Kmph = 5 hrs80

So vehicles met by the car when travelling

against the flow = 5 545680 60

= 66.5

vehicles.

75. (c)76. (d)

CBR of subgrade = 5% and thickness = 50cm



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50 cm = tsubbase + tbase + twearing course

CBR of subbase = 20% and thickness = 20cm

20 cm = tbase + twearing course

twearing course = 5 cm

So tbase = 20 - 5 = 15 cm

tsubbase = 50 – 15 – 5 = 30 cm

So ratio = 30 215

77. (a)78. (c)

Ruling gradient = 1 in 150 = 0.67%

Allowable ruling gradient = 1 in 183 = 0.547%

Grade compensation = 0.67 – 0.547 = 0.123%

= 0.04% ×N°

N° 3°

79. (c)In sand,

ds

wd d F.C.

= 30 151.5 20 4.5cm

100

Similarly Clay

dc = 201.2 20 4.8cm

100

Total moisture storage capacity

= 4.8 + 4.5 = 9.3 cm

80. (c)

Grain dia = 6 cm = 0.06 m [> 6 mm]

d 11RS

Rmax d 6 111S 100 11 0.01

Rmax = 0.544 m or 54.4 cm.

81. (a)

Ta = sWater stored in root- zone W

100Water delivered to thefield(Ws)

=

3

3110 10 8 60 60 485 100

110 10 8 60 60

= 84.69%

Closest answer is (a)

82. (d)When water from higher level is supplied tolower level, by the action of gravity then it iscalled Flow Irrigation.

If the water is lifted up by some mechanical ormanual methods and then supplied for irrigation,it is called lift irrigation.

Natural sub irrigation is caused by leakage ofwater flowing in channels.

Artificial sub irrigation is very costly. Hence isemployed only for cash crops of high value.

83. (a)

The length and width of meander varies as thesquare root of the discharge i.e. Q

84. (c)Ground Profile

Soil water zone

Intermediate water zone

Capillary fringe zone

Zone of aeration

Saturation Zone

Internal water

Zone ofSaturation

Zone ofRock

Fracture

Zone ofRock

Flowage

85. (a)Nitrogen in the form of ammonia exerts anoxygen demand and can be toxic to fish.Removal of nitrogen can be accomplished eitherbiologically or chemically. The biologicalprocess is called nitrification / denitrification.The chemical process is called ammoniastripping.

86. (d)Sodium hypochlorite, ozone, chlorine dioxideand UV are used as disinfectant. Sodiumhypochlorite is handled in liquid form thatcontains from 5% to 15% available chlorine.

87. (c)Molecular weight of H2SO4

= 2 1 32 4 16 = 98 gNo. of moles of H2SO4 in 1 Litre of solution =

33100 10 1.02 10 mole L

98 1

The reaction is 22 4 4H SO 2H SO

Therefore, 32 1.02 10 mole/L H+ ion areproduced. The pH is

3pH log 2.04 10 = 2.69 2.7



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(10) (Test-11)-03 December 2017



88. (d)The philosophy for controlling water hammerin pipes is

(i) to minimize the length of the returningwater column causing water hammer.

(ii) to dissipate energy of the water columnlength by air cushion valve.

(iii) to provide a quick opening pressure reliefvalue to relieve any rise in pressure incritical zones.

These objective are achieved by the followingthree values

1. Zero velocity valve

2. Air cushion valve

3. Opposed poppet valve.

89. (b)When bacteria are introduced into a syntheticliquid medium, reproduction takes place bybinary fission, each cell divides producing twonew cells, the increase in population followsgeometric progression,

1 2 4 8 16 32, and so forth.

90. (a)Reaction of chlorine with natural organics suchas fulvic and humic acids produce undesirabledisinfection by-products (DBPs) such astrihalomethanes (THM), the most predominantof which are chloroform andbromachloromethane. These THMs aresuspected carcinogens. Minute quantities ofphenolic compounds react with chlorine tocreate DBPs with severe taste and odour. Toprevent the formation of DBPs, the organics inthe raw water must be removed beforedisinfection or if formed, through the use ofactivated carbon adsorption. The other way isto prevent the formation by avoiding the use ofchlorine and substituting chloramines as thedisinfectant.

91. (a)Molecular weight of glutamic acid

= 5 12 9 1 4 16 14 = 147

Total oxygen used in the reaction

= 208g4.5 2 2 16

147g of glutamic acid requires 208g of oxygen

208ThOD 63 89.14mg L147

ThOD 89 mg/L

92. (b)The DO sag equation has been developed usingoxygen deficit rather than dissolved oxygenconcentration, to make it easier to solve theintegral equation that results from themathematical description of the mass balance.The saturation value of dissolved oxygen isheavily dependent on water temperature - itdecreases as the temperature increases.

93. (c)The activated sludge process is controlled bywasting a portion of the microorganisms eachday in order to maintain the proper amount ofmicroorganism to efficiently degrade the BODs.Wasting means that a portion of themicroorganisms is discarded from the process.

94. (b)Total weight of filterable residue

= gm325.57 325.46 = 0.11 gm

Quantity of filterable residues

= 3

30.11 10200 10

= 550 mg/L.

95. (d)96. (c)97. (b)

Compaction decreases compressibility of soil.

98. (a)

KZ = 4 5 5

124 4 4

1 10 4 10 2 10

= 53.53 10 mm sec

Head Causing Flow = Head of water due toartesian pressure – Head of Water due toground water

[Take top of gravel as datum]

Head Causing Flow = 150 13 2m10

q = KiA = 5 23.53 10 112

= 5 3 20.588 10 mm / sec/ mm .

99. (a)The radius of fictitious circular footing of areaequal to the rectangular footing.



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(Test-11)-03 December 2017 (11)



2eqR = 30 × 12

Req 10.7 m

Z for circular loaded area

Z =

3/2

21

q 1 R1Z

=

3/2

21

1000 1 10.7120

= 1000 × 0.3145

Z = 314.5 N/m2

100. (d)Tetrahedral Central ion is silicon

Octahedral Central ion is Aluminium.

In octahedral iron or magnesium ions mayreplace aluminium ions in some units.

When Aluminium is at centre Gibbsite sheet

When Magnesium is at centre Brucite sheet

101. (b)The unconfined compression strength of thesoil is given by

1 =

0

P 1A

, where P = 360 N

A0 = 2 23.14 12.56cm44

= 0.8 0.18

1 = 360 1 0.1

12.56

2 225.8N cm 258 KN m

102. (d)The effective stress at point A can be thencalculated as 1 1 2 2S H H

S 16 3 1 516 9.81 18 9.81

= S + 95.14

According to the problem, the maximumeffective pressure at point A is equal to 150KPa.

= S + 95.14 = 150

S = 54.86 kN/m2

103. (c)

From this problem, we know the thickness ofthis soil layer is 3 m. We need to determinethe void ratio before and after ultimateconsolidation settlement. They can becalculated based on

e0 = 0

0

n 0.5 11 n 1 0.5

e1 = 1

1

n 0.45 0.821 n 1 0.45

Based on the equation

Ultimate settlement,

Sf = 0 1

0

e e 1 0.82H 3 0.27 m1 e 1 1

104. (b)

Ka =

2 2 1tan tan45 45 1532

The lateral stress at the base for soil.

H = a1K h 100 33.3 KPa3

The lateral force due to soil.

Pa = H1 1H 33.3 52 2

= 83.33 kN/m

105. (d)

wm

When the slope angle of the sand conereaches its maximum, the sand on the slopesurface is in equilibrium

TMOB = TFF

The mobilized shear force along the slopesurface is TMOB = MW sin .

The shearing resistance along the slope surfaceis “ TFF = M sW cos tan

M mW sin W cos tan

tan tan

35

106. (d)Capillarity permeability test is used todetermine the coefficient of permeability ofunsaturated soil.



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(12) (Test-11)-03 December 2017



Capillary flow analog is a method used forconstructing flow Net.

Consolidation test is sometimes used todetermine the coefficient of permeability ofclayey soil.

107. (a)

108. (c)

Vibrofloatation is used for in-situ densificationof loose granular soil.

In graded filter, the filter are provided such thateach layer is coarser than the one below.

109. (d)

The timber, which is the sawn or milled wood,has inherent structural characteristics.

110. (c)

The pith nourishes the plant in its young age.

111. (b)

112. (b)

113. (d)

Aggregates are not just inert filler but theirproperties influence-workability, strength,stiffness, creep, etc. of mortar and concrete.

114. (d)

A 2:1:9 cement lime mortar contains two partof cement, one part of lime and nine parts ofsand by volume.

115. (c)

The compressive strength of concrete isgoverned by its water-cement ratio.

116. (c)

Batching plant operators will have to computethe amount of water to be added bysubstracting the amount of free-water inaggregate from the amount of design free-water.

117. (b)

118. (d)

Angular Unconformity : Unconformity bydifferent inclination and structural features.

Parallel Unconformity : Unconformity such thatstratas remain parallel.

Non-conformity : Unformity based on geologicalorigin of rock formations.

119. (a)

120. (c)

1 2 3 4

5 6

11

7

12

8

13

9

14

10

15 16222

C1 C2 C3

B1

4 4 4

B2 B3

A1

3 33

A2 A3

17

13131153

15111177

3

9630

Alternate:Modified duration:

= 1Max. of given days + sum of other3

= 112 9 63

= 17 days

121. (b)Nominal distribution reinforcement should beprovided mainly to account for secondarymoments due to poisson’s effect and possibledifferential settlement, and also to take care ofshrinkage and temperature effects.

122. (d)Losses in case of pre-tensioned beam is morethan losses is post-tensioned beam.

123. (d)Earthquake load and wind load are notconsidered together while designing tallbuilding.

124. (a)125. (c)

Impact factor is used due to dynamic action ofmoving load. It is multiplied to static load evenwhen gantry girder is effectively restrainedlaterally.

126. (c)No stresses are developed in 3-hinge arch dueto temperature rise.

127. (a)

128. (b)129. (a)

We know, flexural stiffness

= Flexural rigidity

length

and Flexural rigidity of a beam = EI

where, I = MOI (Moment of Inertia)



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(Test-11)-03 December 2017 (13)



130. (d)131. (d)132. (b)

F = wB

GB B

Stable equilibriumRestoring couple= W.GM sin

Stable equilibrium

B

4

133. (b)134. (d)

Two contour do not intersect each other exceptin the cases of an overhanging cliff or a cavepenetrating a hill side.

135. (a)Due to introduction of Air vessel the flowcondition will become more uniform so thefriction losses will reduce and hence the powerrequired will also reduce.

136. (b)Rutting is a longitudinal depression or groovealong the wheel lines.

If rutting is accompanied by adjacent building,it may be a sign of subgrade movement orweak pavement.

137. (a)Construction R.C.C. through (ducts) involvevarious joints like expansion joints, contractionjoints, sliding joints, construction joints.

To prevent leakage through these joints waterbars or water stops made of rubber, PVCgalvanized iron etc. are used.

138. (c)

A water logged soil is unsuitable for cultivation,reclamation of saline and alkaline lands is done.

139. (a)Circular setting basins are circular in plan.Unlike the rectangular basin, circular basinsare easily upset by wind, cross current.Because of its rectangular shape, more energyis required to cause circulation in rectangularbasin; in contrast the content of the circularbasin is conducive to circular stream lining.This condition may cause short circuiting of

the flow. For this reason, circular basins arenormally designed for diameter not exceeding30 m.

140. (a)

All the sewer pipes are generally laid startingfrom their outfall ends, towards their startingends. The advantage gained in starting fromthe tail end is the utilization of the tail lengtheven during the initial period of its construction,thus ensuring that the functioning of thesewerage scheme has not to wait till thecompletion of the entire scheme.

141. (a)It is preferred to carry out thickening beforedigestion, since digestion takes longerdetention time than thickening. The volume ofthe digester would be much, much smallerwhen put after the thickener than when putbefore thickner.

142. (a)An important method to control the productionof leachate is to eliminate the infiltration ofsurface water from the landfill which is themajor contributor to the total volume of theleachate. For this we use an impervious claylayer over the top of the fill at a descent slope,provided with adequate drainage and surfaceinfiltration.

143. (b)

Area ratio = 2 22 1

21

D D 100D

D1 = inside diameter of cutting edge

D2 = outside diameter of cutting edge

< 20 for stiff clay

< 10 for sensitive clay

Rotary samples are double walled tubesamplers with an inner removable liner. Theyare useful for sampling in firm to hard cohesivesoils and rocks.

144. (b)Negative skin friction acts in the same directionas the applied load hence it reduces theallowable load on piles.

As ground water table lowers, effective stressincreases, which can cause consolidation ofsoil which causes downward drag on piles.

145. (b)Effect of capillarity in a sand bed is to increaseit’s effective vertical stress or it’s stiffness.



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(14) (Test-11)-03 December 2017



A plate load test is of short duration. Hencereason is also correct but is not the correctexplanation of statement 1.

146. (a)

147. (d)Glass by definition is not a ceramic materialbecause it is an amorphous solid.

148. (a)149. (b)

The modulus of elasticity of aluminium is68.9Gpa as compared to 206.7Gpa for steel.

150. (a)



answers - ies master of the member. secondary ... indicating the relative position of the each...

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