ESE-2017 PRELIMS TEST SERIES - IES Masteriesmaster.org/public/archive/2016/IM-1483268055.pdf · 2017-05-26 · 21. (d) Under-reamed pile is a special type of bored pile which is provided

1. (b)2. (b)3. (c)4. (b)5. (a)6. (a)7. (b)8. (d)9. (d)10. (d)11. (a)12. (c)13. (a)14. (a)15. (b)16. (c)17. (a)18. (b)19. (c)20. (a)21. (d)22. (a)23. (b)24. (c)25. (d)26. (c)27. (c)28. (b)29. (b)30. (a)31. (b)32. (d)33. (b)

34. (c)35. (a)36. (b)37. (b)38. (b)39. (b)40. (b)41. (b)42. (d)43. (c)44. (c)45. (a)46. (d)47. (a)48. (b)49. (b)50. (d)51. (c)52. (c)53. (c)54. (b)55. (d)56. (d)57. (a)58. (b)59. (a)60. (b)61. (d)62. (c)63. (b)64. (b)65. (a)66. (a)

67. (d)68. (c)69. (a)70. (c)71. (a)72. (a)73. (d)74. (b)75. (b)76. (a)77. (c)78. (a)79. (a)80. (d)81. (b)82. (b)83. (b)84. (a)85. (c)86. (a)87. (d)88. (a)89. (b)90. (b)91. (d)92. (d)93. (c)94. (b)95. (b)96. (a)97. (a)98. (d)99. (c)

100. (c)101. (b)102. (a)103. (a)104. (b)105. (b)106. (c)107. (a)108. (b)109. (c)110. (b)111. (b)112. (b)113. (b)114. (d)115. (c)116. (b)117. (c)118. (d)119. (d)120. (c)121. (c)122. (c)123. (a)124. (d)125. (a)126. (d)127. (d)128. (b)129. (b)130. (c)131. (b)132. (a)

133. (b)134. (c)135. (b)136. (a)137. (d)138. (d)139. (c)140. (a)141. (b)142. (c)143. (a)144. (a)145. (c)146. (a)147. (c)148. (a)149. (b)150. (a)151. (a)152. (d)153. (c)154. (c)155. (a)156. (a)157. (a)158. (d)159. (d)160. (b)

ESE-2017 PRELIMS TEST SERIESDate: 1th January, 2017

ANSWERS

IES M

ASTER

(2) (Test-20 FL-3) 1st January 2017

8010009955, 9711853908 [email protected], [email protected]. : E-mail:

Regd. office : Phone : F-126, (Upper Basement), Katwaria Sarai, New Delhi-110016 011-41013406

1. (b) Burls or excrescence are formed when treereceives injury in its young age.

2. (b) Ammonium chloride is hydrating agentwhile others are dehydrating agents.

3. (c) In belt conveyer, tendency to segregateconcrete is more.

4. (b) Water cement ratio = Weight of free waterWeight of cement

5. (a)

6. (a) Blown bitumen is obtained by passing airunder pressure at higher temperature.Such bitumen can be used as roofing anddamp-proofing felts, in manufacture of pipeasphalts and joints fillers, and as heatinsulating material.

7. (b)

fwd = wp p wk

mw mw

f k fy y

= wkwk

0.85 f 0.567f1.5

8. (d) Terrazzo is artificial stone.

Terrazzo is used for bathrooms, residentialbuildings, temples, etc.

9. (d) Aluminium is an excellent reflector ofelectro magnetic and sound waves. Dueto high corrosion resistance of aluminum,the maintenace cost is negligible.

10. (d)

11. (a)

12. (c) PI Consistency

O Non-plastic

< 7 Low-plastic

7-17 Medium plastic

> 17 High plastic

13. (a)1K

decreases with increase in temperture

K increases with increase in temperature.

14. (a) Under the point load, the vertical stress isgiven by

= 20.4775Q

Z(from Bossinesq’s equation

2

0.4775Q 0.254z

15. (b)

i = Q 0.06 0.6A 0.002 50

at middle = (19.13 – 9.81) × 0.06 – 0.06× 0.6 × 9.81

0.206 0.21 kN/m2

16. (c)

Sn =C

CH F

0.116 =20

2.5 20 H

H =1

2.5 0.116 = 3.448 m

17. (a)

18. (b) sin = 1 3

1 3

( ) / 2( ) / 2

sin = 1

1

100100

1 1100 2 200

21 300KN / m

2d 300 100 200KN / m

19. (c) The width of test pit is kept 5 times thewidth of the plate.

20. (a) Flow duration curve of a stream is a plotof discharge against percentage of timethe flow was equalled or exceeded.A flow duration curve based on daily flowdata will be steeper than a curve basedon monthly flow data because the largerinterval data will smoothen out thevariations in shorter interval data.

21. (d) Under-reamed pile is a special type ofbored pile which is provided with a bulb/pedestal at the end. The usual size of suchpiles are 150 to 200 mm shaft diameter,3 to 4m long. Its bulb diameter is 2 to 3times the shaft diameter.

IES M

ASTER

(Test-20 FL-3) 1st January 2017 (3)



22. (a) The rails are to be connected at their ends.This is achieved by means of a pair offish plates per tail. The holes are drilledthrough the plates and the web of railsand then the fish bolts and nuts areprovided in these holes. when the boltsand nuts are tightened up, the rails arejoined together and a continuous track isformed.

23. (b)

24. (c)

25. (d) Excavation of small sized heading (drift)is done centrally at top or bottom of the facein the drift method. In heading and benchingmethod the top portion is driven in advance ofthe bottom portion.

In full face method the whole section isexcavated once for all. It is suitable for tunnelsof small cross-sections area say upto 3mdiameter.

26. (c)

27. (c)

28. (b) Astigmation is optical lens defect whileaplanation, definition and acromatism isdesirable optical characterstics.

29. (b)

30. (a)

31. (b)

32. (d)

33. (b) The beam is designed as a balancesection.

So, N.A 0.48 d

= 0.48 × 450 = 216 mm

Df = 100 mm

f7 D3 =

7 1003

= 233.33 mm > 216 mm

So the flange depth will be stressed withboth rectangular and parabolic portion ofstress block.

So modified Df = yf

= 0.15 x4 + 0.65 DF

= 0.15×216 + 0.65×100

= 97.4 mm

34. (c) The modification factor to ld ratio

decrease with increase in tensilereinforcement and grade of steel.

The permissible ld ratio is decrease and

it is a indicator of increased deflection.

Higher grade and higher tensilereinforcement are indicator of higherstresses and higher loads which giveshigher deflection.

35. (a) Minimum bar dia for Fe415 in slab = 8mm

Maximum dia = D8

= 150

8 = 18.75

So, Minimum is 8 mm

and Maximum is 18 mm

36. (b) In two way slab.

Middle strip = l0.75

Edge strip = l0.125

0.125 lx

0.75 lx

0.75 ly0.125 ly

ly

lx

Middle strip in shorter direction= 0.75 ly= 0.75 × 6= 4.5 m

Edge strip in longer direction= 0.125 lx= 0.125 × 4= 0.5 m

37. (b) upward force by tendon atcentre 2Psin 2P tan

upward force due to prestress

downward load=

0 152 P3.

IES M

ASTER




For balancing, 0 152 P w 90KN

3.

3 1P 90

0 15 2.

P 900KN

38. (b) In WSM failure occurs when materialreaches it’s permissible stress. So failureis stress controlled. In LSM failure occurswhen material reaches its limiting valueof strain. So failure is strain controlled.

39. (b) A given x section (balance) with givengrade of concrete and steel will give largerMOR in LSM due to larger area of steel inbalance section in LSM.

40. (b) to check the servicebility criteria

Combination (i)

= 1 × (DL + LL)

= 1 × (5 + 10) = 15 kN/m

Combination (ii)

= 1 D.L 0.8 L.L. E.L

= 5 0.8 10 10

= 21 kN/m

So design load will be 21 kN/m

41. (b)

42. (d) D = i sC C 1100 100 2005 5

B2 = 1100 – 2 × 200 = 700 Rs

B3 = 1100 – 3 × 200 = 500 Rs

B2 + B3 = 1200 Rs.

43. (c) Hindered settling occurs in secondarysedimentation tank of the activated sludgeprocess.

44. (c) The steps involved in vermi compostingare:

1. Dig a small pit.2. Line the pit with straw or dried leaves

and grass.3. Organise the disposal of organic

domestic waste into the pit as and whengenerated.

4. Introduce a culture of worm that is nowproduced commercially.

5. Cover the pit contents daily, bysprinkling of dried leaves and soil everyday.

6. Water the pit once or twice a week tokeep it moisture.

7. Turn over the content of pit every 15days.

45. (a) Some time seeded water is used fordilution in BOD test. The seeded waterwith seeding of mixed bacterial culture isused, but it not used to decomposedifferent types of organic matter. Seededwater is used when there is difficiency ofbacteria in the sample, then we introducebacteria from out sides.

46. (d)

(1) The rate is more in shallow depth(2) Rate is more in a running stream than in

a quiscent pond(3) More is the saturation deficit, more is the

rate of oxygen supplied(4) More is temperature, more is the rate

47. (a)

(1) the flow of waste water in sewer issteady and uniform as per assumptionand the unsteadiness and non-uniformityof flow characterstics are accounted forin the design by proper sizing ofmanholes.

(2) Design of sewers is based on peak flowdischarge i.e. maximum hourlydischarge.

48. (b) Particles with velocity above or equal toSOR will be completely removed andthose with settling velocity below SOR are

removed in proportional, VSOR

where V is

the settling velocity.Thus overall removal =

(0.5 100 0.1 100 0.05 100)1001

= 165 mg/L

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49. (b) Volume of sample pollutant = 0.5 ml

Volume of diluted pollutant = 100 ml

Dilution ratio = 1000.5 = 200

Consumption of oxygen = 2 mg/l BOD5 = Consumption of oxygen × Dilution

factor= 2 × 200 = 400 mg/l

50. (d) Recirculation factor

F =

21 R

(1 0.1R)

F =

21 1.5 1.89

(1 0.15)

=

1

100w1 0.0044

V FW1 =2000 × 1000 × 1.50 × 10–6 kg/dayW1 = 300 kg/dayV1 = 2000 m3 = 0.2 hactare-m

=

1003001 0.0044

0.2 1.89 = 88.97 89%

51. (c) Suspended particulate matter causesproblem in respiration. Carbon monoxidecombine with haemoglobin of blood to formcarboxyhaemoglobin which reduces theoxygen carrying capacity of the blood.SO2 causes discolouration, leaching ofbuilding materials. NO, a principlecomponent of acid rain affects vegetation.

52. (c)

53. (c) Equating head to efficients

1 1

1

N DH =

2 2

2

N DH

2D = 16 40025

= 320 mm

54. (b) 2 = 2D N60

=3.14 0.1 3000

60

= 15.7 m/s

For radial vane tips w2V = 2u

= 15.7 m/s

H =w2 2

gHV u

and

w2 2

gHV u = 0.8

H =220.8 u

9.81

= 20.8 15.79.81

= 20.10 m

55. (d)

Area = myc2

Top width = 2 myc

2

3Q TgA = 1 for critical condition

1m

yc

2myc

QC =2 3c

c

g (my )2 my

Qc yc 52

triangular

for rectangular

yc

C B

yc

12 3q

g

and for shallow parabolic 2cQ y

IES M

ASTER




56. (d)

57. (a) For given specific energy the discharge ismaximum when flow is in critical state

Thus depth of flow Y = Yc

Y = Yc

= 2 2E 1.8 1.2 m3 3

For rectangular channel –

Yc = 1

2 3qg

1.2 =1

2 3qg

q =1

3 2(1.2) 9.81 q = 4.12 m3/s per m

Total discharge,

3mQ 4.12 5 20.6 s

58. (b)Vertical component weight of liquidover the curved surface

= 2

w w w12 4 4 (8 )

4

Horizontal component force on verticalprojection of the curved surface

wh A

= w w w2 0.5 1 4 2.5 4 10

So, ratio = 810

59. (a)By equating the pressure in inclinedtube and container at the level ofinterface of oil and mercury.

P + h × 0.8 w = wsin 13.6 l

P = w w10.2 13.6 0.05 0.82

P = (1.36 – 0.04) w = 1.32 w

= 1.32 m of water

oil

P =

1.320.8 = 1.65m

So pressure = 1.65m of liquid

60. (b)

2mm2cm

A B

By blancing the forces at point B.

upward force at A = 0.02 equilent

pressure = 2

0.02 40.020.02

4

this pressure

will be transferred to B and equivalent down

word force at B = 24 0.0020.02 4

So, 2 2w

4 0.002 0.002 H0.02 4 4

= 0.002

wH = 4 4

0.002 0.02

H =

4 0.075 1 1 100cm10000 0.002 0.02

= 1.35 cm

61. (d)u = 2x2 + z2 + 6v = y2 + 2z2 + 7It must satisfy continuity equation

u v

y z

x = 0

4x + 2y + z

= 0

z

= – (4x + 2y)

= – (4x + 2y)z + f(x,y)

= – 4xz – 2yz + f(x,y)

62. (c)

Rectangular notch = 3/2d

2 C 2g LH3

2.95 CdLH3/2

Broad crested weir = 1.705 CdLH3/2

% increase

2.95 1.705 100

1.705

73.19%

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63. (b)Assuming the flow in pipe is laminar.

RU

r

0 =

R P2 x

=

3 350 10 50 102 10

=250

2 N/m2

= 0.125 kPa

64. (b)

Oil

water

Since the interface is shown clearly i.e.no intermixing. The stress at interface mustbe same from both sides i.e. oil and waterside.

oil = water

00

dudy

= ww

dudy

0 w

du dudy dy

= w

0

The shear stresses from both sides aresame so it should be continuous andvelocity is discontinuous as shown abovei.e. velocity gradient are in reverse ratioof their viscosity i.e. different.

65. (a) Weber Model is used for Flow over weirs involving very low heads

Very thin sheet of liquid flowing over asurface

Capillary waves in channels, etc.Mach Model is used in Aerodynamic testing

In phenomena involving velocitiesexceeding the speed of sound.

Hydraulic model testing for the cases ofunsteady flow, especially water hammerproblems.

Froude Model is used in Free surface flows such as flow over

spillways, sluices, etc Flow of jet from an orifice or nozzle Problems in which waves are likely to

be formed on the surface. Problems in which fluids of different

mass densities flow over one another.66. (a) Porosity of soil can’t exceed 100% as

VV V and VV / V

Water content of fine grain soil is generallymore than coarse grained soils. Specificgravity of coarse grained soil is generallyless than that of fine grained soil. Sandbath method is f ield method fordetermination of water content.

67. (d) Momentum equation is not used in Hardy-cross method.

68. (c)

LReservoir

A B

Pressure distribution at B.

time

2LC

2LC

2LC

2LC

O

+ph

–ph

T0 =2LC

=2 2000

1000

= 4 sec.Thus, negative water hammer pressure at

the valve will last for 4 – 8 sec.

69. (a) Similar to splices in flanges, webs alsomay need splicing or cuts outs maybecome necessary in webs for routing ofservices. Such web splices and cut outsare best provided at locations away fromhigh shear and large concentrated loads.

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Splices in web are designed to withstandthe shear and moments at the splicedsection.

70. (c) The length of weld connecting each edgeof the batten plate to the member shall, inaggregate, be not less than half the depthof the batten plate. At least one-third ofthe weld shall be placed at each end ofthe edge. In addition, the welding shall bereturned along the the other two edges ofthe plates transeversely to the axis of mainmember for a length not less than theminimum lap specified:

71. (a) Gross diameter of rivet = 20 + 1.5= 21.5 mm

Strength of rivet in bearing = b × d × t= 300 × 21.5 × 16 = 103.20 kNStrength of rivet in double shear

=

22 d4

=

22 100 21.54

= 72.61 kN

Strength of plate per pitch

= at (P – d) t

= 150 × (90 – 21.5) × 16 = 164.40 kN Strength of joint per pitch = Minimum of

above three value = 72.610 kNStrength of plate per pitch = 150 × 90 ×16 = 216 kN

Efficiency = 72.61 100216

= 33.61%

72.(a)

73. (d) In most cases of the connections subjectedto tension forces, the flexibility of theconnected parts can lead to deformationsthat increases the tension applied to thebolts. The additional tension is calledprying force.

74. (b)

C

B

A 125 mm leg

75 mm leg

55

79

50 50 50dhole = 22

Path AC

Net area

= 1164 – 2 × 22 × 6 = 900 mm2

Path ABC

=

2 250 501164 3 22 6 64 55 4 79

= 883.65 mm2

75. (b)

76. (a)Assume te = 24 hActual rainfall = (1.6 + 5.4 + 4.1)

= 11.1 cmRunoff = 4.7 cm

Loss = 6.4 cm

trial = 6.43 8 = 0.267 cm/h

Loss in first 8h = 8 × 0.267= 2.133 cm > actual rainfall of 1.6 cm/hHence te < 24 h2nd trial : Let te = 16 hInfiltration loss in remaining 16 h

= 6.4 – 1.6 = 4.8 cm

=4.8 0.3cm / h16

Losses in 2nd and 3rd time block = 2.4 cmwhich is less than actual rainfall values.

Hence = 0.3 cm/h

77. (c) The plastic hinges are formed first at thesections subjected to the greatestdeformation (curvature).

78. (a)(1) The contractor whose tender has been

accepted is to deposite 10% of the tenderestamount as security money with thedepartment inclusive of the earnest moneyalready deposited.

(2) This amount is kept as a check so that thecontractor fulfils all the terms and conditionand caries out due work satisfactory with ina time limit. The security money also servesas a security against the materials or duetools and machineries provided by thedepartment.

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(3) The security money is refunded to thecontractor after satisfactorily working of theproject upto 6 months or one rainy season.

79. (a) Multiplying constant

K = fi

K = 24012

K = 20

80. (d) Item rate contract is also known as unitprice contract or schedule contract. Acontractor takes work on the basis of itemrate basis. He is required to quote rate forindividual item of work on the basis of billof quantities furnished by the department.The amount received by contractorsdepends upon the quantities of workactually performed. The payment to thecontractor is made on the basis of thedetailed measurement of different itemsof work actually executed.

81. (b)

82. (b)Free float FF = Total float – Sj

= (48 – 12) – (48 – 36)= 22 – 12= 10

83. (b)

84. (a) Power shovel is used in excavation ofearth is confined area or pit.Hoe is used in excavation of trenchesbecause trenches need precise control ofdimensions and hoe is suitable for thispurpose.

Clamshell is used in rehauling of loose orexcavated material.

Bull dozers are used in clearing andscrubbing worksite.

85. (c) When the size of the bucket is to bechanged the length of the boom also needsto be changed as the maximum liftingcapacity is limited by the force that willcause the tilting of the machine.

The weight of the bucket + the load it cancarry is called the capacity of the dragline.

A large boom will be required toincrease the reach of the draglien butin order to keep the tilting torqueconstant the size of bucket must bereduced.

86. (a)

87. (d)

P

F FF F

At the plastic load, all four rods are stressedto the yield stress.

F= YA

Sum of forces in the vertical direction andsolve for the load:

PP = 2F + 2F sin

PP = Y2 A(1 sin )

88. (a) The fire resistance of RCC buildings ismeasured as time (in hours) of safeexposure to fire

89. (b) Thin walls are susceptible to instability(bucking) at regions of high compressivestress. (Minimum t = 150 mm)

90. (b)b = 300 mm and d = 500 mm

2u uu ck

x xM 0.36 f bd= 1 0.42d d

= 0.36 × 20 × 0.48 (1 – 0.42 × 0.48) × 300 × 5002

= 206.945 kNm

uM 137.96 138 kNm=1.5

91. (d) Bond strength for M25 grade concrete =1.4 N/mm2

(This is corresponding to mild steel bar intension)

If HYSD bars are used, the bond strengthis increased by 60% and due tocompressive force it is further increasedby 25%.

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Modified bond strength bd = 1.6 × 1.25 ×1.40 N/m2

bd = 2.8 N/mm2

Minimum development length required

Ld = y

bd

0.87f4

Ld = 0.87 415 12

4 2.8

Ld 386.8

dL 390 mm

92. (d) Shear resistance of shear reinforcement,

Vus = 0 87 y sv

v

. f A dS

= 20 87 415 2 8 350

4 200.

= 63.52 kN

Shear resistance of concrete,Vuc = cbd

= 0.5 × 200 × 350 = 35 kN Total shear capacity = Vus + Vuc

= 63.52 + 35= 98.52 kN

93. (c) If no tension occur in the column section,maximum compressive strain at highlycompressed extreme fiber in the concretesubjected to axial compression andbending will be

= 0.0035 – 0.75 × Strain in leastcompression fiber

CH = 0.0035 – 0.75 × CL

A B

DC

CH

CH = 0.0035 – 0.75 × 0

CH 0.0035

94. (b)H = 200 m Head of water

P = gH

= 3 6 210 9.81 200 1.962 10 N m

= 21.962N mm 1.962MPa

h =Pd2t

20 = 1.962 1000

2t t = 49.05 mm

95. (b)b = 60 cm

A

B

C

D

b = 60 cm0

Square will follow 1/3rd middle rule

AC = b 60BD 20cm3 3

AO = 20OC BO OD 10cm2

BC = 2 2BO OC= 2 210 10

= 14.14 cm96. (a)

3m

7m

CBA

–6 kN

14 kN

+

–

MB = –9 kN/m

MC – MB = C

BV dx

= 1 14 72

MC – MB = 49

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MC + 9 = 49MC = 40 kN-mOption (a) is correct.

97. (a)98. (d)

Given,

1 = – 100 MPa

2 = – 10 MPaApplying maximum shear stress theory(MSST).

Abs max < Sys

1

2

< yf2

fy > –100 MPa

yf 100 MPa

99. (c)

d

d/3

d/3

d/3

y

y

y

y

Moment of resistance capacity ofrectangular beam

M = 2

ybd6

Stress at the extreme fiber of the middlethird of section.

yy

dd/2 6

yy 3

Moment of resistance capacity of middle thirdsection

M = y z

M =

2

y

db3

3 6

M =2

y1 bd9 3 6

= 2

y1 bd27 6

1M M27

100.(c) If two shaft are connected in series. Theywill carry same torque

T = 50 kN-m

Power transmitted by shaft =2 NT

60

where N = number of rotations per min.P = 2NT

N = No. of rotations per second.P = 2 × 5 × 50P = 1570 kW

101. (b)

Critical slenderness ratio = 2

cc

E

where E = Young’s modulus of mild steel

= 2.1 × 105 N/mm2

c = Crushing strength

= 3300 kg/cm2

= 323.73 N/mm2

c =2 52.1 10 80

323.73

Closest option is 78.

102. (a)

Jam density kj = 1000 153.84 1546.5

Maximum flow =

fV kj 80 1544 4

= 3080 veh/hr.

103.(a)

Y1 =200 1/5

1000

Y2 =250 1/4

1000

Y = Y1 + Y2 + Y3

= 39 Y20

L = 4 × 3 = 12 sec.

C = 60 sec.

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C =

3

1.5L 5 1.5 12 51– Y 1 9/20 Y

Y3 = 1/6

If green time for 3rd phase

=

1/ 660 – 12 12.97 13sec.1 1 15 4 6

104.(b)105. (b)106. (c)

2S 80 80R 80 m8M 8 10

107. (a)108. (b)

Volume of sewage treated = 4.5 M-l/day

Since suspended solids amount to 275 mg/l,we have the mass of suspended solidspresent in sewage.

= 66

275 4.5 1010

= 1237.5 Kg/day

Since 50% of sol ids are removed insedimentation tank we have, the mass ofsolids removed in sedimentation tank

= 501237.5

100 = 618.75 Kg/day

When moisture content of sludge is 98% then2 kg of solids (dry sludge) will make = 100kg of wet sludge

618.65 kg of solids (dry sludge) will make

= 100 618.75

2

= 30940 Kg= 30.94 tones

109. (c)

De chlorinating agentSodium thiosulphateActivated carbonSulphur dioxide

Fluoridation agentSodium silica fluoride

Hydro silicic acidSodium fluoride

110.(b)

111. (b) Only if sodium alkalinity is absent thenonly we can say that the entire alkalinityis due to presence carbonates andbycarbonates of calcium and Magnesium.In this case only we can say that carbonatehardness is equal to the alkalinity.

112.(b)113. (b) Re = 0.05

laminar flow

So Cd = 24Re

Cd = 24

0.05Cd = 480

114. (d)

Vw = C – V1

Vw = Absolute velocity of surge = ?y1 = 1.6 m, V1 = 2.4 m/s

C = gy

= 10 1.6

C = 4Vw = 4 – 2.4 = 1.6 m/s.

115. (c) 525R 0.90 0.91580

116. (b)

x = Depth of cutting

D–x

GL3m

A

Level of cutting

D = 30m

Thin Impervious layer

Previous Soil

W.T

w satH (D x)

w satH (D x)

27 × 9.81 = 18.84 (30 – x)

x = 30 – 14.06 = 15.94m

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117. (c) w = 2 1

3 4 s

m m G 1 1m m G

= 250 2.65 1 1130 2.65

=

1.1973 – 1

= 0.1973 = 19.73%

118. (d) tfield =

2 2

lab field2 2

lab

t (H ) 15 10(H ) 0.01

= 15 × 106 min

= 28.54 years

119. (d)

C D

VD

2m

As the joint C is a hinge so B.M at C = 0.So from right sideBM = VD × 2 = 0VD = 0

120. (c)

A B

MABR

20

Stiffness of joint B = KBA + KBD + KBC

= EI 3EZ 0l

l

So, DFBA =

EI1l

4EZ 4l

So, mBA = 180 20kN m4

Now by considering equilibrium of memberAB

Bm 0

20 – mAB = 0mAB = 20

121. (c)By virtual work theorem

B B C C E E D DP P P P

20 × 5 + 30 × 10 + 40 × 10 = PD × 10

PD = 80 kN

122. (c)

40

4m

HA

MBC + MBA = 0

MBA = –MBC = +40

So, HA = 40 10kN( )4

So, HD = 10 kN( ) [HA + HD = 0]

So,

10

3m

MCD

MCD = +30

123. (a)

DFBC = FBA

3EI13 D3EI 4EI 2

3 4

So, MBC = 120 102

So,

103mR

R R =

10 (tension)3

124.(d) Weight of =1

1 13 4

= 127 .

125. (a) Arches generally transfer loads mainly byaxial copression. The axial compressionis highest near supports. So an arch shouldbe made thickest at support and thinnestat crown.

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126. (d)At joint 5

F25

40 kN

cos = 7.5 1

7.5 2 2

F = 0

2540cos F = 0

F25 = 402

= 20 2

127.(d) Cable will take only axial tension so ifloading is in upward direction then cablewill not carry any load and structure willbe determinate.

But due to gravity loads cable will alwayscarry tension and structure wil l beindeterminate to 1 degree.

So for gravity load structure can not bedetermiante.

128.(b) If the end point of traverse does notconcide exactly with the starting pointowing to the errors in the f ieldmeasurements of angles and distances,such error is known as closing error.This error is distributed to all stationsproportionally by shifting the station parallelto the direction of closing error.

129.(b) l/2 l/2R

A

(1) When load is at,

RA

CRR2

AR = 0RC = 0

(2) When load is at a distance x fromA,

RA

CRR2

mA = 0

x R2

= 0

R = 2x

RC =xl

When load at the joint of springand beam AB

x = 2l

Rc = 12

(3) When load at the end of AB,

mA = 0

R2

l = 1 l

R = 2Rc = l

So, ILD

1

CD

130. (c) When TWC coincides with JRC no extraconstruction is required to dissipate energyor protect the basin.

131.(b) Angular value of 1 div. (sensitivity)

= S 206265

nD

= 0.08 206265 41.25sec.=

5 80

132. (a)

dw = (28 12) 15 0.8

100 9.81

Depth of water applied = 15 0.80.16

9.81 0.8

= 244.65 mm

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133. (b)

Cutting

"Berm"

Bank

134. (c) To counter act the uplift pressure,The buoyant weight of concrete = Theseepage force(G – 1) × t × 1 = 4.8 × 1

t =4.8 3 m1.6

135. (b)

2

6 43

mD 10 8.64 10 Bm /s

days

D = 8.64 B m100

D = 8.64

cm

136. (a) Due to less requirement of mortor, thinmortor in brickwork renders it more durablethan stonework.

137. (d) Excess lime will make cement unsound.

Higher Ca(OH)2 is not desirable as it is solublein water & gets leached out making concreteporous.

138. (d) Limiting values of slenderness ratio forcolumn (=12) is less than that of walls(=27). Reason is correct.

139. (c) There is vey little tendency to allow waterand swell due to strong hydrogen bond.

140. (a)

141.(b)

142. (c)

143. (a) The given dimension of beam & slabs areconvinient in construction of form work.

144. (a) With increase in depth of NA, depth ofuncracked concrete increases, crackformed are smaller which improvesaggregate interlocking.

145. (c) Inertial forces usually follow parabolicdistribution in regular multi-story buildingswith maximum values at top floor levels.

146. (a)

147. (c) Once formed, lake stratification is verystable and can be broken by voilent stormsonly. As the summer progresses, thisstability increases as the epilimnioncontinues to worm while the hypolimnionremains at a fairly constant temperature.

148. (a) If the sewage is disposed on land, someof it evaporates but some of it percolatesinto the soil. If the soil is parous, thesewage may easily percolate and theconditions inside the soil still remainsaerobic while in the case of sticky andfine grained soil, the sewage percolatingmay clog the pores, thereby cutting of theair and hence anaerobic conditions maydevelope which may lead to evolution offoul gases and smell.

149. (b) The main idea of providing screens is toprotect the pump and other equipmentsfrom possible damage due to floatingmatter of sewege. These floating matter ifnot removed, will choke pumps.Screens should preferably be placedbefore grit chambers. However if thequality of ‘grit’ is not very important screensmay even be placed after grit chambers.They may sometimes be accomodated inthe body of grit chamber itself.

150. (a)151. (a)

Streamlines are a family of curves that areinstantaneously tangent to the velocityvector of the flow. These show the directiona fluid element will travel at any point oftime. By definition, different streamlines atthe same instant in a flow do not interesect,because a fluid particle cannot have twodifferent velocities at the same point.

152. (d)

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153. (c)Total strength of the web is made of webstrength before the buckling and postbuckling strength developed due to tensionfield action. Thus the provision of suitablyspaced sti ffeners called intermediatestiffness can be used to develope tensionfield action and improve shear capacities ofthe web. The main purpose of these stiffnessis to provide stiffners to the web which resistthe applied loads.

154. (c)

155. (a)

156. (a) A single-peaked hydrograph, consists of(i) a rising limb, (ii) the crest segment,and (iii) the recession or failling limb. Therising limb (or concentration curve) of ahydrograph represents continuous increasein discharge (or runoff) at the watershedoutlet. During the initial periods of thestorm, the increase in runoff is rathergradual as the falling precipitation has tomeet the initial losses in the form of highinfiltration, depression storage and gradualbuilding up of storage in channels and overthe waershed surface.

As the storm continues, losses decreasewith time and more and more rainfallexcess from distant parts of the watershedreaches the watershed outlet. The runoff,then, increases rapidly with time. Whenthe runoff from all parts of the watershedreaches the watershed outlet.simultaneously, the runoff attains the peak(i.e., maximum) value. This peak flow isrepresented by the crest segment of thehydrograph. The recession limb of thehydrograph starts at the point of inflection(i.e., the end of the crest segment) andcontinues till the commencement of thenatural ground water flow.

157. (a)

158. (d) Elements may be rotations or deflectionsso all elements would not have same units

159. (d) The ductility calls for lower grade of steelas it has clearly defined and longer yieldplateau and hence, plastic higher formedwill have larger rotation capacities leadingto greater energy dissipation.

160. (b) Ductile materials are weak in shear so wecan apply max distortion energy theory.

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ESE-2017 PRELIMS TEST SERIES - IES Masteriesmaster.org/public/archive/2016/IM-1483268055.pdf · 2017-05-26 · 21. (d) Under-reamed pile is a special type of bored pile which is provided