ESE-2017 PRELIMS TEST SERIES

1. (c)

2. (d)

3. (b)

4. (d)

5. (b)

6. (b)

7. (a)

8. (b)

9. (c)

10. (d)

11. (b)

12. (b)

13. (c)

14. (b)

15. (b)

16. (d)

17. (c)

18. (d)

19. (b)

20. (c)

21. (d)

22. (d)

23. (b)

24. (a)

25. (b)

26. (c)

27. (d)

28. (c)

29. (d)

30. (a)

31. (a)

32. (d)

33. (d)

34. (a)

35. (d)

36. (d)

37. (c)

38. (a)

39. (a)

40. (a)

41. (d)

42. (d)

43. (c)

44. (b)

45. (a)

46. (c)

47. (a)

48. (b)

49. (c)

50. (d)

51. (b)

52. (c)

53. (a)

54. (c)

55. (d)

56. (d)

57. (a)

58. (b)

59. (a)

60. (b)

61. (c)

62. (c)

63. (a)

64. (a)

65. (a)

66. (c)

67. (a)

68. (b)

69. (a)

70. (a)

71. (d)

72. (d)

73. (c)

74. (d)

75. (b)

76. (b)

77. (d)

78. (d)

79. (c)

80. (a)

81. (a)

82. (d)

83. (c)

84. (d)

85. (b)

86. (c)

87. (c)

88. (d)

89. (a)

90. (b)

91. (b)

92. (c)

93. (b)

94. (c)

95. (a)

96. (d)

97. (d)

98. (b)

99. (c)

100. (a)

101. (b)

102. (c)

103. (a)

104. (a)

105 (c)

106. (d)

107. (d)

108. (d)

109. (b)

110. (a)

111. (b)

112. (b)

113. (b)

114. (a)

115. (c)

116. (d)

117. (a)

118. (b)

119. (c)

120. (d)

121. (b)

122. (b)

123. (d)

124. (a)

125. (a)

126. (b)

127. (b)

128. (b)

129. (d)

130. (b)

131. (c)

132. (c)

133. (d)

134. (c)

135. (a)

136. (c)

137. (d)

138. (a)

139. (a)

140. (b)

141. (d)

142. (d)

143. (b)

144. (a)

145. (a)

146. (d)

147. (b)

148. (c)

149. (a)

150. (b)

151. (a)

152. (b)

153. (d)

154. (a)

155. (c)

156. (b)

157. (b)

158. (c)

159. (a)

160. (c)

ESE-2017 PRELIMS TEST SERIESDate: 6th November, 2016

ANSWERS

IES M

ASTER

(2) (Test-7) 6th November 2016

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

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

1. (c) 145.8 haVolume of water supplied by tube well ina year

= 4 3350 10 250 87.5 10 m

Area irrigated = 4

2Vol. 87.5 10 mdepth 0.6

= 145.83 ha

2. (d) 0.6 × 0.8 = FC 15 0.8100 9.81

FC 39.24%

3. (b) Plan growth is a function of soil moisturestress i.e., soil moisture tension andosmotic pressure. When the moisturecotent in the soil is excessive, the poresof soil are completely filled with waterand there is no air available in the rootzone. Because, for satisfactory plangrowth, the presence of air in the rootzone is essential, absence of air retardsthe plan growth.

4. (d) Minimum size = 10.8 RS

= 1.810.8 0.0052 10002

[R= y/2 for hydraullically efficientrectangular channel]

mind 50.54 mm

5. (b) Initial cost is high but maintenance costis low in lined canal.Because of the higher velocity andsmaller exposed area, the evaporationlosses are low>

6. (b) = 0gRS

= 1000 × 9.81 × (0.88) × 0.004 [forwide channle R = y]

234.53 N/m

7. (a) u/s cutoff is provided to reduced the upliftpressure while d/s cutoff is provided toprevent piping.

8. (b) Slope, S = 5/3

1/61 f

3340 Q

= 5/3

1/61 (1.1)

3340 (64)

= 1 1

5699 5700

9. (c) P = 4.75 Q

1.5 1.5

18P = 18 + 1.5 × 2 = 21 m

21 = 4.75 Q

Q = 21

4.75

Q = 19.55 20 m3/s

10. (d) In Lacey’s theory, the procedure followed is

1. f = mm1.76 d , S = 5 3

1 6f

3340Q

2. v =1 62Qf

140

3. Q = A.V

4. P = 4.75 Q

11. (b) The uplift pressure on syphon aqueductis more when canal is dry and drainageis full. This leads to maximum uplift forceand minimum counter acting weight ofcanal.

12. (b) Critical velocity ratio was used bykennedy not Lacey.

13. (c) Percolation and absorption will bemaximum in sandy soil and coarseaggregate soil.

14.(b) Maximum principal stress

H G C 1

For dam to be safe, 22499 kN m

2499H

9.8 2.4 c 1to be on safer side, uplift is neglected C = 0

2499H9.8 3.4

H 75 m

IES M

ASTER

(Test-7) 6th November 2016 (3)



15. (b) Principal stress is given by

= 2 ' 2vP sec P tan

[Where P' is pressure intensity due totail water]

for no tail wate case, 'P 0

= 2vP sec

Also shear stress,

= 'vP P tan

for no tail water 'P 0

= vP tanHence, we can see that both principaland shear stress decrease because ofpresence of tail water.

16. (d) Inertia force due to acceleration

= w 1.962g vf m

acceleration is vertically downwardsit will reduce the effective wt. of the dam Net effective weight of the dam

= ww 1.962g

= w – 0.2 w= 0.8 w

17. (c) Duty of water on the field =

8.64 120.36 0.12 =

432 ha/cumecTherefore duty at the head of watercourse D = 432 × 0.7 = 302.4 ha/cumec Discharge required

Q = 756

302.4 = 2.5 m3/s

18. (d) nd =

y1 100d

Average depth of water stored = d

= 1.5 1.12

= 1.3 m

Numerical deviation from depth ofpenetrationAt head end of the field = 1.5 – 1.3= 0.2 mat tail end of the field = 1.3 – 1.1= 0.2 mAverage numerical dev iation

y = 0.2 0.22 = 0.2

nd =

0.21 1001.3 = 84.62%

19. (b) Fetch is > 32 km

Height of wave hw = 0.032 V FWhere V is in KmphF is in kmand hw is in m

hw = 0.032 90 40= 0.032 × 60= 1.92 m

20. (c) In reservoir empty condition the onlysingle force acting on it is the self weight

(w) of dam at a distance of B3 from the

heel.

Pmax =

V 6e1B B

e = B6

max2wPB

wB 3

B

toeO

heel

2wB

and Pmin =

V 6e1B B

for e = B6

minP 0

IES M

ASTER




21. (d) H1 = 285 – 205 = 80m, H2 = 211 – 205= 6m

U1 U2

2 1 21H H H3

= 300.84 kN

H= 58.86 kN

29.81×80= H= 784.80 kN

1

Total uplift force acting on the dam

= 1 8 784.8 300.842

1 48 300.84 58.862

= 12975 kN

22. (d) CIR = Cu – Re

= 3.5 × 15 – 40 = 12.5 mmNIR = CIR + Water for field preparation+ Water for nursery bed requirement= 12.5 + 15 + 10= 37.5 mm

FIR = a

NIR =

37.5100130

= 48.75 mm

23. (b) Wind pressure, wave pressure, freeze &Thawing

24. (a)25. (b)

Reduction coefficient,

Cr = eff

eff

1.25 , or48b

1.25 , which ever is smaller48d

l

l

Cr = 0.9375Cr 0 9375. = 0.94

26. (c)Maximum permissible tensile stress at timeof initial tensioning is 0.80 × ultimate tensilestress.

27. (d)

All assumptions given holds for pre stressedconcerte.Stress in tendons is assumed to beconstant along its length.Strain variation is linear.Variation of stress in tendons due to changein external loading is not considered.

28. (c)Exposure condition Minimum Crack

Mild M 20Moderate M 25Severe M 30Very severe M 35Extreme M 40

29. (d)

At support, there will be hogging movementand hence tensile reinforcement will be providedon top of beam i.e., in web portion of beam.As the concrete below neutral axis does nottake part in resisting the bending movement,it is best to design it as rectangular beam atsupports.

30. (a)2 2

SCA 6 25 2945mm4

Core dia,

CD 400 2 40 2 8 304 mm

Core area, 2 2

kA 304 72546mm4

Gross area, 2

gA 400 125664mm4

²

Concrete area,2

CA 125664 2945 122719mm

K Cu yC C SP = 1 . 0 5 ( 0 . 4 f A + 0 . 6 7 f A ) 1.05 (0.4 20 122719

0.67 415 2945)

1890640N

Safe load 1890640 N 1260.427KN

1.50

31. (a)

CC SA 400 400 A

4P 1.50 1200 1800kN

K CC c y S0.40 f A 0.67f A

C

3S1800 10 0.40 20 (160000 A )

+ 0.67 × 415 × Asc

270.05 Asc = 520000Asc = 1926 mm²

32. (d) In load balancing design, the externalbending moment is balanced by the bendingmoment due to eccentricity of the axialprestressing force.

IES M

ASTER




33. (d) Cube is always tested on sides i.e., facein touch with mould. Strength of cube isexpressed to nearest 0.5 N/mm2. Individualvariation should not be more than 15% ofaverage.

34. (a) Table 1 (clause 6.4) is 456 : 2000

35. (d) min = ck

y

f0.24

f

= 250.24250

= 0.0048

36. (d)

37. (c) Metakoline is obtained by calcination ofpure or ref ined kalimatic clay attemperature between 650°C and 850°Cfollowed by grinding to achieve finenessof 700 to 900 m3/kg

38. (a)

39. (a) In freyssinet system large number of wirescan be tensioned at a time.

40. (a)

Minimum vertical reinforcementminimum horizontal reinforcement =

0.0015 30.0025 5

0.0012 30.0020 5

41. (d) Fig. 13, IS 456: 2000

42. (d) Clause 20.1 of IS 456 : 2000

43. (c)M20 — 1.2 MPaM25 — 1.3 MPaM 30 — 1.5 MPaM 35 — 1.6 MPa

44. (b) For 1 way continuous slab (both endcontinous) — 26For 1 way continuous slab (one endcontinous) — 23

45. (a)

d = 2

cq 1 sin1 sin

= 2300 1 sin 30

18 1 sin 30

= 1.852 m

46. (c)1. A circular section utilizes minimum

quantity of material and providesmaximum area for a given perimeter andtherefore most efficient.

2. A circular section, being of uniformcurvature all round provides no cornersand hence less chances of deposits ofsludge solids.

47. (a)As per IS 1742 - 1983 following colorindicators are used

1. Sewers - Red2. Waste water and Rain water pipes - Blue3. Casting works - Blacks

48. (b)As per government of India Manual and IS-1742-1983 (Code of Practice far buildingdrainages), the minimum recommended selfcleansing velocity is 0.75 m/s.

49. (c)IS-1742 recommends that it is undesirable toemploy gradients giving a velocity of flowgreater than 2.4 m/s as it will cause scouringof pipes, where it is unavoidable unavailable,the cast iron pipes are to be used.

50. (d)Cast iron pipes should be used in followingsituation.

1. In bad or unstable ground where soilmovement is expected

2. In made up or tipped ground3. To provide far increased strength where

a sewer is laid at insufficient depths,where it is exposed or where it is carriedover the ground.

4. Under buildings and where pipes aresuspended in basement.

5. In reaches where velocity is more then2.4 m/s

6. For crossing of water courses.

IES M

ASTER




51. (b)

Sewer pipes are laid starting from their outfalltowards to their starting end to ensure thefunctioning even during initial stages.The alignment of sewer may be done either bysight rail method or by reference line method.In sight rail method, the sight rails are fixed insuch a way that upper edges coincide withline of sight.The length of the boning rod adjusted accordingto the height as noted in sight rail.

52. (c)

53. (a)

Even when the sewer line runs straight, themanholes, are to be provided at regular intervals.the spacing between the manholes, in such acase, however, depends mainly upon the sizeof sewer line. The larger is diameter, the greaterwill be spacing between man holes.

54. (c)

The length of the branch sewers between thevertical pipe and the plug is known as inspectingarm; and can be used for inspecting andclamping the branch sewer after opening theplug.

Inspectingarm

Branchsevers

Verticalpipes

55. (d)By inspection :BG, CH 3rd member meeting at a co-linearunloaded jointDE, EJ 2 member meeting at a unloadedjoint

At point A V AFF 0 F 0

At joint J

FDJ

FIJ

0

P3

VF 0 , DJ 3 DJ 3F cos P cos F P

HF 0 , DJ IJ 3F sin F P sin 0

FIJ = 0Now at joint I

HF 0 , FHI + FIJ = 0FHI = 0

So, total = 756. (d)

DS = 1 taking HD = as reductant themember forces will be 1 i DP H

Pi = 0 as no external forces actingNow, applying unit load at D in horizontaldirection

AD 1

CD

AD 1 , all other = 0

HD =

i ii i i i

i i2

i

i i

Pl l TA F

lA E

So, HD = 0So, member force FAC = 0

57. (a)

58. (b)FED : FBE : FCD = 15 : 10 : 5

= 3 : 2 : 1

59. (a)

z = IF qN= 0.005 × 10 × 10= 0.5 t/m2

60. (b)

Pa = a zk 2c ka

p0 at surface Z = 0When vertical surcharge of q kN/m2 is appliedthen,

Kaq – 2c ka = 0Pa = Kaq

q = 2c / ka

IES M

ASTER




61. (c)

w = 2 1

3 4

m m G – 1 – 1 100m m G

= 1000

2000 1480

× 2.67 – 1 – 1 100

2.67

= 20.28%

d = / 1+w = 1.70 gm/ml

1.7 = wG(1 e)

e = 0.57 S =

wG 95%e

62. (c)

63. (a) 1 45

5

54

41

25cos41

3 32 1 2q 1 cos q 1 cos

3

23

1 5100 1 100 1 12.258 kN/m412 2

64. (a) a = 75 – 35 = 40 [P is 78 whichis more than 75, so limited to 75]b = 55 – 15 = 40 [P is 78 whichis more than 55 so limited to 55]c = 55 – 40 = 15d = 24 – 10 = 14G.I. = 0.2 × 40 + 0.005 × 40 × 15 +0.01 × 40 × 14

= 16.6 = 17

65. (a) 0F

1

A 15A1 E 1 0.25

= 20 sq. cm.

66. (c)

67. (a)

68. (b) 150 = 50 tan² 45° + 2C tan 45°50 = C

C = 50 kN/m²

69. (a) 1

33

1 150

Deviator stress = 150 – 50 = 100 kN/m²

70. (a) f = 452

70 = 45 502

–1 =

– 23 tan 45 2Ctan 45

2 2

C = 0

–3 300 = – 2

3 tan 70 3= 45.8

71. (d) A =2

4 20.038 11.3 10 m4

,

10.8 0.0176

A =4

0

1

A 11.3 101 1 0.01

= 11.4 × 10–

4 m²

3

21 P 4

P 127 10 111.4 kPaA 11.4 10

P = 111.4 02 = 55.7 kPa

72. (d) Normal precipitation of all station A,B,Care not within 10% of normal precipitation atstation D.

830830 × 1.1 = 913

830 × 0.9 = 747Hence we will use normal ratio method.

DP830

= 1 875 1020 9103 1120 900 760

PD = 860.973

73. (c) Mass curve is plot of accumulated precipitationvs. time.

(a) total precipitation in AB is 20 mm(b) No precipitation during BC & DE.

(c) Average intensity in C – D = 20 1mm/hr20

(d) Average intensity in A – B = 20 1mm/hr20

IES M

ASTER




74. (d)

Total precipitation = 420 6 100 10100

= 120000 m3

Runoff volume = 40000 m3

Precipitation not available to runoff= 12000 – 40000 = 80000 m3

Area = 100 hec.

So = 480000 0.08 m 8 cm

100 10

75. (b)

76. (b)

77. (d)

78. (d)

79. (c)

n = 2

vC

Cv = 27100 100108X

Cv = 25Allowable error = 100 × accuracy= 100 – 90 = 10

n = 225

10

= 6.25

n 7

80. (a) For borrow pit soil

3d

17 16.2 kN / m1 0.05

Volm of borrow pit soil required per

m3 = d

d

( ) compacted soil( ) borrow pit soil

sd s

w and W remains sameV

= 318 1.11m16.2

Swell factor = 1.2Volm required = swell factor × Volm of borrowpit soil required= 1.2 × 1.11 = 1.33 m3

81. (a)

82. (d) Ip = 30.5%Ip from A – line = 0.73 (53.9 – 20) = 24.747

Soil is clay and L.L > 50%

Soil is CH

83.(c) Cu = uniformity coefficient = 60

10

DD

= 0.40.05

= 8

Cc =

20.2 .04 20.4 0.05 .020

Cu > 61 < Cc < 3

Well Graded Sand

84. (d)

85. (b)

86. (c)

87. (c)

3 = 100 kN/m2, d = 350 kN/m2, U = –40

1 = 3 + d = 100 + 350 = 450 kN/m2

1 = 450 –(–40) = 490 kN/m2

3 = 100 –(–40) = 140 kN/m2

Sin = Sin cu = 490 140 350490 140 630

33.74

88. (d)

89. (a)The sandwich plate are provided with 2 wedgeshaped grows on its two faces. The wiresare taken two in each groove and tightened.Then a steel wedge is driven betweentightened wires to anchor them against theplate.

90. (b) 4 2A 300 600 18 10 mm

2P 800 5 12 452389N4

e 300 220 80 mm

P Pe M 0A Z Z

IES M

ASTER




4 6 6

452389 452389 80 M 018 10 18 10 18 10

6M2 51 2 01 0

18 10. .

6M 4 52 18 10 Nmm.

81 36 kNm.

91. (b)

92. (c)The bar dia 1

10 iwall thickness.

1 150 15mm10

bar spacing

1 horizontal length53 wall thickness450 mm

93. (b)Members stronger in shear than in flexure arerequired.

94. (c)shear wall resist moment about the horizontalaxis perpendicular to plane of wall (strongaxis bending)

95. (a)Cl. 32.5 of I 456 : 2000.

96. (d)The unsupported length is taken as cleardistance between floor & lower extermingof capital drop panel or slab, whichever isthe least.

97. (d)where crack control is of special importance,reinforcement in excess of minimumdistribution reinforcement is provided.

98. (b)

tie dia,

4lons,maxt

28/4 7mm

6mm

t 8 mm

99. (c)For bearing walls and shear walls, mint maybe taken smaller of

/25 5500/25 220 mmuHu/25 4000/25 160mml

but 150 mm

mint 160 mm

100. (a)

101. (b)

9

1

VV =

1/791

V9 = 15 × (9)1/7 = 20.53 Km/hr

E = Km (ew – ea) ×

9V116

= 0.5 × (17.54 – 7.02) ×

20.53116

= 0.5 × 10.52 × 2.2831 = 12 mm/dayVolume of water evaporated in 3 days= 250 × 104 × 12 × 10–3 × 3 = 9 ×104 m3

102. (c)

103. (a)A = area of catchmentFor rainfall A

At 15 minutes, 1525 A will be contributing

QA = C × 35 × 1525 A = 21 AC

For rainfall B

QB = C × 45 × 1025 A = 18 AC

For C full area A will be contributing astactual > tcQC = C × 10 × A = 10 ACFor D full area will be contributingQD = C × 15 × A = 15 ACHence (A) gives the highest peak flow

104. (a) Total rainfall (P) = 1.6 + 5.4 + 4.1= 11.1 cmIntial loss (L) = 0.6 cmRunoff (R) = 4.7

-index =

P R 0.267 cm/hr24

But rainfall of 1.6 cm (0.2 cm/hr) will notcontribute to -index as rainfall intensity is

less than -index.

5.4 4.1 4.7 0.3 cm/hr16

105 (c)

IES M

ASTER




106. (d)Rainfall mass curve is a curve betweencumulative rainfall and time.

rai

nfal

l

time

107. (d)

2 = 3/2

21q 1

1 (R / z)

2 at P =

3/2

21100 1

1 (2 / 2.5)

= 252.39KW / m

108. (d)Horizontal stresses can be more thanvertical stresses as in case at passive earthpressure.Zone of influence of strip footing is muchdeeper than the circular footing.

109. (b)

Effective angle of friction for coarse grainedsoils have high value of angle of friction

As angle of friction increases as densenessof packing goes on increasing C and aremeasures of shear strength. Higher thevalues, higher the shear strength.

110.(a)For problems of short term-stability offoundations excavation and earth damsUU(Unconsolidated undrained) tests areappropriate. For problems of long-termstability CD test is appropriate, dependingupon the drainage conditions of soil.

111.(b)

n =

3(1 sin )1 sin

n =

20 (1 sin35 )1 sin30

= 260 t/m

112. (b)For an unconfined compression test

100S C2

= 2 250KN / m .05 / mm

A laterally unconfined compression test isconducted on soils which can stand withoutconfinement. Also, for clay angle of internalfriction 0 .

113. (b)

114. (a)

115. (c)

116. (d)

117. (a)The concrete part above the N.A. will try toexpand due to poission’s effect and theportion below N.A. will try to contract butthese deformation are restrained byadjoining slab portion and in the transversedirection there will be non strain.In beams there will be three dimensionalstrain condition.

118. (b)

leff 20d leff = shorter span

leff =

ll

0

0

d which ever isminimumsupport width

Support width = 300mm

Assuming d < 300mm

3000 d 20d

20d – d > 3000

d > 157.89 mm 158mm < 300 mm

119. (c)

IES M

ASTER




For 8mm spacing =

2

st

1000 d4

A

Ast =

21000 8

480

For 10mm

Spacing =

2

2

1000 10 804

1000 84

= 125mm

120. (d)Annex – D of IS 456-2000.

Clause D 1.4, 1.5, 1.6

121. (b)• Minimum recommended thickness of flat

slab is 125mm• In flat slab longer span is used to check

the deflection criteria• Flat slabs have highest thickness among

all types of slab. Because of the absenceof beams.

• Flat slabs are directly supported on columnswithout beams.

122. (b) Clause 34.2.4 of IS 456123. (d) Clause 34.1.2 of is 456.124. (a)

K

0

C

100qtan 0.9 1

f

3

6100 250 10tan 0.9 1

20 10

= 1.35

125. (a)

Precipation = 4 311150×10 165000 m100

Runoff = 2.5 × 10 × 3600 = 90000 m3

Runoffprecipation =

90000 0.545165000

126. (b) Minimum dia of tie reinforcement,

length 4t

| 40/46mm

= 10 mm

Pitch tt

25mmS 3

Pitch tt

25mmS 3 3 10 30 mm

St = 30 mm

127. (b) Hoyer system is adopted for production ofpretensioned members on a large scaleother are system for post tensioning.

128. (b) Impact factor = 4.5

6 span (in meter)

129. (d)

20kN

10kN

FAC

C E

HGF

20kN

10kN

Taking moment about E10 × 6 – 20 × 3 + FAC × 6 = 0

ACF 0

130. (b)

B

A

C1

1

B

A

C

120160 kN

200

CH = ui pi iAi Ei l

=1.(–160) 4000 2mm

1600 200

(Towards left)

IES M

ASTER




131. (c)As there is no external force on the structure inthe horizontal direction, so horizontal force atsupport A = 0

Force in AE = 0

F

D

A

G

E

B

C

H

W– 2W

– 2WW

W

2WW132. (c)

50B

50 kN

A

0

5050

45º

50º E

50 2

45º

FBA

FAE

50º

BABA

F 50 0 F 50 22

FAE = FBA

2 = 50

45º

50

45º

50 2

FBC

FBE

B

FBE + 50 2 502

= 0

BEF 0

133. (d)

Force in spring = K × l

FBE = 4 × 2 1

B

FBE

FCA

A

D

45º

45ºC

FBA

W

Taking moment about A

Am = 0

W × 2 = BEF 12

W = BEF2 2 =

4 2 42 2

W = 2– 2 = 0.586 kN

W 586 N

134. (c)

T + 80 = 170 kN

T 90 kN

M

TFB = (2) × 10 = 80 kN3

170 kNRA × 9 = 90 × 12

AR 120 kN

RAA

B C4 4 4D

90 kN

90 kN

Resultant reaction

at A = 2 2(90) (120) = 150 kN

135. (a) Truses Generally have very large degree off reedom and less degree of staticindeterminacy So best method is forcemethod

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136. (c) A is correct but R is falseThe base period is the period betweenthe first watering and the last watering.While crop period is the period betweenthe time of sowing and the time ofharvesting the crop.

137. (d) A is false and R is trueIn case of lift irrigation, duty is highbecause the commanded area is closethe the well, transmission losses areless.On the other hand, in case of canalirrigation, the transmission losses arehigh and duty is low.

138. (a) A and R both are true and R is correctexplanation of A.Calcium and magnesium are beneficialfor the soil because they keep it in acoagulated clod form. This occursbecause the silicates and aluminates ofcalcium and magnesium are insolublein water and they possess cementingproperties also. After the exchange ofcalcium and magnesium ions by sodiumions of irrigation water, soluble sodiumsilicates are formed which do not havecementing properties. The clods of soilstherefore cruble and fine sillica grainsare released which clog the pores of thesoil, resulting in reduction of permeabilityand destruction of soil structure.

139. (a)

140. (b) Both A & R are correctBed load: Sediment load moves alongthe bed with occasional jumps into thechannel.Suspended load: Sediment load ismaintained in suspension due toturbulence of the flowing water.Sediment load: is the burden ofsediment carried by the flowing water ina canal. It includes both bed load &suspended load.

141. (d) Maximum limit of sulphates as SO3 is400 mg / l as per table 1 clause (5.4) ofIS 456 : 2000

142. (d) The creep coefficient of concrete decreaseswith the age at loading

Age at loading Creep coefficient7 days 2.228 days 1.61 years 1.1

143. (b)

Pipes laid in ordinary or softer grounds areembeded in concrete to with stand the force oftransport sewage and its turbulence.

For pipes not embed in concrete, half of theportion should atleast be in firm soil having thesame shape as that of the pipe.

144. (a)

The sewer pipe lengths are usually laid fromthe lowest point with their socket end facingupstream. In this way the spigot of each pipecan be easily inserted in the socket of alreadylaid pipes. After fitting the socket and spigotinto each other, the proper joining is done.

145. (a)

146. (d) In convective precipitation when air rises

upward in atmosphere, its gets cooledadiabatically to form a cloud.

147. (b) Precipitation occurs when a local portion

of the atmosphere becomes saturatedwith water vapour, so that the watercondenses and precipitates. Forprecipitation to occur there must besuf f icient nuclei present to aidcondensation.

148. (c) High strength concrete is less susceptibleto shrinkage.

149. (a)

150. (b) Due to poisson effect, portion of slab aboveN. A expands under compression andportion below N.A. contracts” under tension.This induces secondary stress in lateraltransverse direction. The transversereinforcement resists these stress alongwith effects of shrinkage and temperature.

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151. (a) 2° torsion is induced by rotations appliedat one or more points along length ofmember through interconnected members,rather than by directly applied loads on it.These are statically indeterminate andrequire compatibility condition.

152. (b)

153. (d)

154. (a)

155. (c)

156. (b)

157. (b)

158. (c)

W/2 W/2

W1Pratt-truss

1

1

W/2 W/2

W

Howe-trussIn pratt truss diagonal member (which is alonger member) carries tension while inHowe truss it carries compression.Thus if longer member carries compression,there is likely chance of buckling of trussmember.

Pratt truss is better than Howe truss.

159. (a) Water content of inorganic soi ls isdetermined by heating the soil at atemperature of 105°C – 110°C becauseabove 110°C structural water may be lost.

160. (c)

ESE-2017 PRELIMS TEST SERIES

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