ANSWERS - IES Master · 2017-12-17 · According to IS: 800 – 1984 Cl 6.6 the effective length of...

1. (b)

2. (a)

3. (c)

4. (b)

5. (a)

6. (c)

7. (a)

8. (a)

9. (b)

10. (c)

11. (b)

12. (c)

13. (b)

14. (b)

15. (d)

16. (c)

17. (a)

18. (b)

19. (a)

20. (b)

21. (a)

22. (c)

23. (d)

24. (b)

25. (b)

26. (d)

27. (b)

28. (a)

29. (a)

30. (b)

ESE-2018 PRELIMS TEST SERIESDate: 17 December, 2017

ANSWERS

31. (a)

32. (b)

33. (c)

34. (c)

35. (a)

36. (d)

37. (a)

38. (b)

39. (a)

40. (c)

41. (b)

42. (c)

43. (a)

44. (c)

45. (d)

46. (c)

47. (b)

48. (d)

49. (b)

50. (c)

51. (c)

52. (c)

53. (d)

54. (b)

55. (c)

56. (a)

57. (a)

58. (d)

59. (d)

60. (b)

61. (c)

62. (c)

63. (c)

64. (a)

65. (a)

66. (c)

67. (c)

68. (c)

69. (a)

70. (b)

71. (a)

72. (b)

73. (c)

74. (c)

75. (d)

76. (c)

77. (b)

78. (c)

79. (b)

80. (b)

81. (b)

82. (a)

83. (d)

84. (d)

85. (b)

86. (d)

87. (a)

88. (b)

89. (b)

90. (b)

91. (a)

92. (c)

93. (c)

94. (c)

95. (a)

96. (c)

97. (d)

98. (a)

99. (a)

100. (b)

101. (b)

102. (d)

103. (c)

104. (b)

105. (d)

106. (c)

107. (b)

108. (a)

109. (d)

110. (a)

111. (b)

112. (c)

113. (a)

114. (c)

115. (c)

116. (c)

117. (c)

118. (c)

119. (b)

120. (c)

121. (b)

122. (a)

123. (a)

124. (d)

125. (d)

126. (d)

127. (b)

128. (c)

129. (d)

130. (b)

131. (d)

132. (d)

133. (a)

134. (a)

135. (a)

136. (b)

137. (a)

138. (b)

139. (b)

140. (d)

141. (c)

142. (c)

143. (b)

144. (d)

145. (a)

146. (d)

147. (a)

148. (d)

149. (b)

150. (a)

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1. (b)

Self explanatory.

3. (c)350 mm

800 mm

3000 mm

d = 800 – 40 = 760 mmVu = 135 kN

v =u

uMV tandbd

=

63 202.5 10 450135 10

760 3000300 760

= 0.42 < 0.438 N/mm2

Hence, nominal shear reinforcement

Sumax = y sv0.87 f A0.4 b

= 302 mm > 300 mm

S = 300 mm

4. (b)

Ld =bd4 1.6 1.25

=360

4 1.2 1.6 1.25

= 37.5

5. (a)

bd = design bond stress

= 1.50 N/mm2

Development length for deformed bars incompression

=y

bd

0.87 f4 (1.60 1.25 )

=bd

45 30 600 mm

6. (c)

As per Rankine Formula

d =2

xq 1 sin1 sin

=2981 1 sin30

18 1 sin 30

=981 118 9

=10918

= 6.05 m 6 m

8. (a)Xu max = 0.531 × 550

= 292.1 mmAssumingfsc = fst = 0.87 fy

Cuc + Cus = Tu

0.362 xu × 20 × 300 + (0.87 × 250 – 0.447 × 20) × 982

= 0.87 × 250 × 3054xu = 211.5 mm

sc =500.0035 1

211.5

= y0.00267

= 50.87 250

2 10

= 0.00109

Hence fsc = 0.87 fy

9. (b)

Shift of P line at centre = MP

=6

30.25 68 3 10

540 10

= 95 mm

P line above NA = 95 mm – 60 mm= 35 mm.

Shift of P line at quarter span

=6

30.125 3 68 10

540 10

= 48 mm

P line below NA = 60 – 48= 12 mm

10. (c)

The concept of load balancing is applied fordeterminate structure.

11. (b)(i) Shear walls make building laterally stable(ii) Failure of beams is a localised effect while

column failure may effect the stability ofwhole building, hence it is better to makecolumn stronger than beam.

(iii) Flexible rubber pads are used between thegrooves and the foundation such thatbuilding is isolated from the ground. Due tothis increased flexibility, time period is

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increased this concept is called baseisolation.

12. (c)

Independent float: It gives us an idea aboutthe excess time that exists if the preceedingactivity ends as late as possible and thesucceeding activity starts as early aspossible. The independent float is therefore,defined as the excess of minimum availabletime over the required activity duration.

13. (b)

It is issued by the banks and involves anagreement between the client, contractorand the bank.

With this bond, the bank compensatesclient for their financial loss if the contractordoes not fulfill his obiligation.

14. (b)

Independent float affects the concernedactivity only.

15. (d)Range of total duration = x 3

= 25 15 3 (3) 5

75 20

55 < Range < 95

16. (c)

CB

C

ED

New activity can not start with only dummyactivity as its predecessor.

17. (a)

0

0 0

PH T

= C

For standard conditionPressure (P) = 760 mm of Hg

Temperature = 273 + 25 298 KFor given condition at 32°C

P0 = 740 mm of Hg

Temperature = 273 + 32 = 305 KH0 Power = 76 HP

740

76 305 = P

760H 298

HP = 78.96 HP

18. (b)Plates are jointed using lap joint

Bolts will be in single shearNow length of joint = 3 × 75 = 225 mm

> 15 × 14 = 210 mm

Blj =

2251.075200 14

Blj = 0.9946Grip length lg = 2 × 12 = 24 mm

< 5 × d= 5 × 14 = 70 mm

ig = 1

Here, no packing is used

pkg = 1

Shear capacity of bolt

= 2 4000.78 14 0.9946 1 1

4 3 1.25

= 22.06 kN

19. (a)Net effective area = Total area – deductionof holes for case (b) and (d)

Net effective area = 1 2A KA

K =

1

1 2

3A 13A A

For case (c)

Net effective area = 1 2A kA

K =

1

1 2

5A 15A A

20. (b)There are two possible collapse mechanism

(1)

L/2 L/2

MP

MP

2

P

MP

LP2

= P P PM 2M 2M

P = P10 ML

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(2)

3L/4 L/4

MP

MP

4

3

P

MP

LP2

= P P PM M 4 6M

P = P22 ML

From (1) and (2), collapse load, P = P10 ML

21. (a)

Gross area of angles (Ag)

= (90 + 60 – 6) × 6 × 2= 1728 mm2

Strength governed by yielding

Tug = g y

m0

A f

= 31722 250 1 392.73 kN

1.1 10

.

22. (c)

Slenderness ratio = 3000 154.619.4

Corresponding to this, al lowablecompressive stress = 43 MPaArea of angle = (150 – 5) × 10 × 2

= 2900 mm2

Capacity of section =

43 2900 124.7 kN

100023. (d)

According to IS 800 – 1984

As far as practicable, the lacing systemshall not be varied throughout the length ofthe strut.

Single laced systems on opposite sides ofthe components shall preferably be in thesame direction so that one be the shadowof other.

Rolled sections or tubes of equivalentstrength may be used as lacing bars insteadof flats.

24. (b)

According to IS: 800 – 1984 Cl 6.6 theeffective length of compression flanges withend unrestrained against lateral bendingshall be L. In case, each end is notrestrained against torsion the above valuesof the effective length shall be increased by20 percent.

25. (b)

This stiffner is provided to prevent bucklingof web due to bending compressive stress.

27. (b)

6 7

5

8

4

2 3

1 9

2 3 4 5 6 7

2 3 4 5 6 7 9

2 3 4 54 65 7

56 67

x x x x x xy y y y y y

67 = 56

DK = 20Consider inextensibility,

2 3 4 7y , y , y , y = 0

5y = 6y

2x = 3x

4x = 5x

6x = 7xDk reduces by 8Dk = 20 – 8 = 12

28. (a)

1 m 2 m

1 m 2 m

P2P1

P2P1

3

12 EI(1)

3

12 EI(2)

2

6EI(2)

2

6EI(1)

2

6EI(2)

P

P

yF = 0

P1 = 3

12EI 12 EI1(1)

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P2 = 3

12EI 12 EI8(2)

P = P1 + P2

=

12 EI 12 EI1 8

= (96 12) EI

8 = 13.51 EI

Force required for unit displacement

=

P 13.5 EI

= 13.5 EI

31. (a)

K11 =

2 2AB AC

AB AL

AE AEcos cosL L

2

ADAD

AE cosL

K11 = 6×104 + 3.9 × 104 + 2.12 × 104

K11 12.02 × 104

33. (c)50 kN

E D

AB C

4 m

3 m

4 m

18.75 kN 18.75 kN

50 kN

X

X

Cutting a section through x – x

vF 0

BEF 18.75 = 0

BEF = 18.75 (compressive)For joint A

AEF sin = 18.75

AEF =

18.75sin

=

18.75 5 31.25 kN

3 HF = 0

50 = AE ABF cos F

FAB = 450 31.25 25 kN5

FAB : FBE : FAE = 25 : 18.75 : 31.25

50 50 3 5 50: :

2 8 8

1 3 5: :2 8 8

4 : 3 : 5

34. (c)

No horizontal force so, HA = HB

5 m

10 m

8 mHA

RA

HB

RB

HA = HB

Taking moment about A

B18R H 1 8 1 4 = 0H = 18 RB – 32 ...(i)

Taking moment about C = 010 RB – 5H = 0

RB = 0.5 H ...(ii)From eq. (i) and eq (ii)

H = 18 × 0.5 H – 32H = 4 kN

RB = 2 kN

35. (a)

Aquifuge : Neither porous nor permeableand hence it neither retains nor yieldsground water. Eg. Granite rock

Aquitard : Does not yield water freely towells due to its lesser permeability, althoughseepage is possible through it. Eg. - SandyClay

Aquiclude : Highly porous, containing largequantit ies of water, but essential lyimpervious, so it doesn’t yield water. Eg. -Clay layer

36. (d)

Any one of the five analytical methods notedbelow are accepted for the determination oftotal coliform.

1. Multiple tube fermentation technique (MTF)

2. Membrane filter technique

3. Minimal media ONPG-MUG test (chilliestsystem)

4. Presence-absence colifom test (P-A)

5. Colisure technique.

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37. (a)

At a pH of 7.2, all alkalinity can beassumed to be bicarbonate.

Equivalent weight of Alum (Al2(SO4)3.18H2O)

=666 111 g

6

Alum dose in terms of CaCO3 equivalent

=5040111

= 18.02 mg/L as CaCO3.

This is equal to the amount of alkalinitythat will be destroyed. Thus, the alkalinityremaining is

3HCO remaining = 29.5 – 18.02

= 11.48 mg/L as CaCO3

Note : The valency of alum has been takenequal to 6 because each mole of alum addeduses six moles of alkalinity.

2 2 34 32

2 2 2 43

Al 14H O 6HCOSO2Al .3H IO 6CO 8H O 3SOOH s

38. (b)

The examples of free flow condition are theflows in canals, flumes, grade aqueductsand grade tunnels. Examples of pressureflow conditions are the flows in pressureaqueducts, pressure tunnels, pipeline andforce mains, and depressed pipes “aninverted siphons.”

39. (a)yt = y0(1 – e–Kt)

213 = 318.4(1 – e–K×6)

K =ln0.33

6 = 0.184 day–1

40. (c)

Catabolic reactions are those in which foodis broken down to release energy.

The end products of aerobic catabolism arelow-energy, stable compounds with most ofthe energy being stored in cellular material.

41. (b)

Theoretical carbonaceous oxygen demand= no. of moles × 6 × 32 × 103 mg/l.

No. of moles of 6 15 6C H O N

=3

310 10 0.051 10197

Mass of 6 15 6C H O N

= 6 × 12 + 15 × 1 + 16 × 6 + 14 =197

TCOD = 3 30.051 10 6 32 10

= 9.792mg l

No. of moles of NH3 = 0.051 × 10–3

Nitrogenous oxygen demand

= 3 30.051 10 2 32 10

= 3.264mg l

Total oxygen demand = 13.056mg l

42. (c)

i = CC

75 , 5 T 20T 10

CC

100 , 20 T 100T 20

; TC in minutes

i =100 2cm hr

30 20

43. (a)

A “4.0 percent sludge” contains 4.0 percentby mass of solids and 96.0 percent bymass of water. Assuming that the specificgravity is not appreciably different from thatof water, we can approximate therelationship between volume and percentsolids as follows :

1

2

VV

= 1

2 2

P 100 0.080P V 0.040

V2 = 50 m3/d

44. (c)

Sludge thickening is accomplished in oneof two ways; the solids are floated to thetop of the liquid (floatation) or one allowedto settle to the bottom (gravity thickening).

The goal is to remove as much water aspossible before final dewatering or digestionof the sludge.

45. (d)

The NO2 cycle, whereby, NO2 getdecomposed in the presence of sunlight,resulting in formation of NO and O. Theoxygen reacts with O to form ozone, whichsubsequently reacts with NO to form NO3.The reactions combined constitutes NO2photolytic cycle.

The NO2 cycle does not produce a netincrease in NO2 or O3. Hence if not disturbedand provided that the starting concentrationsof the species are low, there will be no

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undesirable effects. However, humanactivities emit not only NOx to the atmos-phere but also carbon monoxide, thecarbonyl compounds and the hydrocarbons.Carbon monoxide and hydrocarbons by theirreactions with the hydroxyl radical, disturbthe normal NO2 photolytic cycle throughthe formation of peroxy radicals. Theseperoxyl prevents the reaction of O3 withNO, thus terminating the cycle and causingthe O3 to accumulate.

46. (c)

The power of sound(W) is defined as therate of doing work by a travelling soundwave in the direction of propagation of thewave.

The sound intensity is defined as soundpower averaged over the time, per unit areanormal to direction of propagation of thesound wave.

47. (b)

Modulus of toughness

= Strain energy stored upto fracture

Volume of Material

48. (d)

=

3K 2G6K 2G

K = 3G

=

9G 2G 7G 7 0.3518G 2G 20G 20

49. (b)

Sectional area of cylinder = dt

= 100 4

= 1256.637 mm2

Circumferential stress

= 1pdf2t

=

2P 100 12.5PN mm2 4

Longitudinal stress

= 2Pd Pf4t A

=

P 100 500004 4 1256.63

= 2N mm6.25P 39.79f1 and f2 are the principal stress.

We know qmax = 45N/mm2

1 2f f2

= 45

12.5P 6.25P 39.79

2= 45

6.25P + 39.79 = 90

P = 8.034 N/mm2

50. (c)

2

2x y x yxy1 10

2 2

10 = 2

210 5 10 52 2

2 2510 7.52

2 212.5 7.5

2156.25 56.25

10MPa

51. (c)

30°60 N/mn2

40 N/mn2

120 N/mn2

x = x y x y cos22 2

+ xy sin2

=60 120 60 120 cos60

2 2

+

40 × sin 60

= 30 45 20 3

= 19.64 MPa

x y = x yxy

( )sin2 cos2

2

=( 60 120) 3

2 2

+ 40 × 12

=340 20

2

= 97.942 MPa

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53. (d)

Maximum bending moment

M = F × 4 = 24 KN-m

Since the beam is designed mosteconomical, depth of neutral axis

max =MyI

68 =

y100 y

y = 42.85 mm

I =624 10 42.856

= 17 × 10–5 m4

54. (b)

1.D Dt to10 15

3. Critical stress is hoop stress.

4. hoop stress = pd2t

55. (c)

= E and PA

PA

= E

=P

AEPoisson’s ratio

= Stress in lateral directionStrain in longitudinal direction

elongitunal= laterale

elateral =

P

A =

3

2 9

(0.42 120 10 )

0.08 3 104

e = 3.3422 × 10–3

elateral = 33.3422 10D

= 33.33422 10 8 0.026cm

56. (a)

5 KN/mA

25 25

Shear force V = 25 – 2.5 × 5 = 12.5 KN

NA50

100

Shear Stress

= 3

3VAy 12.5 10 150 100 100

300Ib 150 15012

= 0.3703 mpa

57. (a)

max = 0.01 radian

D = 300 mmL = 7.5 m

max = 42 N/mm2

TJ =

max G

R L

T =G . JL

=7 4

310 0.01 300

7.5 10 32

= 10602.8 kN-m

T =2 4

maxJ 42N mm 300300R 32

2

= 70.875 × 106

T = 70.875 kNm

58. (d)

P =2

2eff

EIl

P d4 4I d

PP

=

4

4d

0.85d

PP

= 41

0.85

P = 40.85 P

P = 0.522P59. (d)

L.R = d

i

DD

=

Depth of water drainedout per unit area

Depth of irrigation waterapplied per unit area

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=i

Di CuD

Di = water depth appliedCu = consumptive useDd = Drained out water

=40 8

40

=32 0.840

60. (b)

Bmin =C

HS C

[C = 1 {full reservoir}]

=28 23.66m

2.4 1

min slidingB = C

HS C

= 28 25m

0.8 2.4 1

Bmin = max. 23.66m, 25m

= 25 m

61. (c)

Q = 2/3 1/2A R Sn

A = y2 and 2A y yR

P 2 2y 2 2

50 =

2 2/3

2/3y y 1

0.015 902 2y8/3 = 135

y = 6.3 m

62. (c)

Hygroscopic water is that water which isadsorbed by the particles of dry soil fromthe atmosphere and is held as a very thinfilm on the surface of the soil particles dueto adhesion or attraction between surfaceof particles and water molecules.

63. (c)

A vertical drop fall is not suitable as a meterdue to the formation of partial vacuum underthe nappe.

64. (a)

Dependent variable should not be selectedas repeating variable.

65. (a)

If there is end contraction, then for eachend contraction, ‘L’ should be reduced by0.1 H.

66. (c)

The speed of plane is equivalent to thespeed of air with respect to the plane. Pitottube fixed in the plane measures the speedof air with respect to plane.

The speed, V =

air

2 Pg 2 8000.45

= 59.6 m/s

67. (c)

For sharp crested rectangular weir

Q = 3/22 Cd 2gLH3

dQ =

1/23 2 Cd 2gLH dH2 3

dQQ =

3 dH2 H

dQ 100Q =

1 2

1

H H3 1002 H

=

3 (0.26 0.25) 1002 0.26 = 6%

68. (c)

1.6 m

Initial water level

Z =

2 2r1.62g

1.6 2 10 = 22 12 10

16 = 2 2w 10 = 40

2 N60 = 40

N =60 40 1200

2

69. (a)

Pgas = atmP gh

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= 3101.3 10 13534 10 0.1

= 87766 Pa

70. (b)

o

vV

=

1/7y

=Moment thickness

Boundary layer thickness

=

mm 1 m 2

m = 7

= 7 7

8 9 7271. (a)

In turbulent flow through rough pipes.

maxUU = 1 1.33 f

= 1 1.33 0.04= 1.27

72. (b)

1

2

PP

=DK1tE

=3

5600 4.2 1013 2.1 10

= 1 4 5

73. (c)

Kinematic viscosity of gases decreases asdensity is proportional to pressure.

75. (d)

Relative compaction

= d

d

in field100

Max from proctor test

Relative density =

max nat

max min

e e100

e e

Permeability on wet side (dispersedstructure) has lower value than dry side(floculated structure)

76. (c)

Sn =C

F H

F.O.S. =n

CS H

=30

0.025 18 15

=30 1000270 25

F.O.S. = 4.44

77. (b)

Degree of expansiveness DFS, %

Low 20%

Moderate 20 35

High 35 50

VeryHigh 50

78. (c)

q/meter width of dam

= F

D

NK H 60 60 24N

= 7 436 864005 1012

= 30.5184m day / m

Total Seepage Loss

= 0.5184 300

= 155.52 m3/day

79. (b)

Pa = a aK h 2c K

a1 sin 1HereK1 sin 3

=11 21 h 2 17.32

3 3

Pa @ 3m = 1 17.3221 3 23 1.732

= 1 kPa

80. (b)

Assumptions made in Terzaghi’s 1-Dconsolidated theory.

1. Compression and flow are directional(vertical).

2. Darcy’s law is valid.3. The soil is homogenous.4. The soil is completely saturated.5. Soil grains and water are both incompres-

sible.6. Strains are small, i.e, the applied load

increment produces virtually no changes inthickness, and K and av remain constant.

7. There is a unique relationship, independentof time, between void ratio and effectivestress.

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81. (b)

Friction pile is one that transfers almost allthe structural load to the soil by skin frictionalong a substantial length of pile.

A pile that transfers almost all the structuralload to the soil at the bottom end of thepile is named an end bearing or pointbearing pile.

82. (a)

• Use Mohr’s circle to solve the problem, asshown below :

0 40 80 120 160

40

(kPa)

(kPa)

From this plot, it can be seen that theangle of internal friction is zero degree.

83. (d)

Hydraulic gradient, i = H 100 2.86L 35

Coefficient of permeability; K = QiAt

= 200 0.031cm s

2.86 25 90

= 0.31 mm/s

84. (d)

Groundwater coming up into the casing orhollow stem augers can cause the test zoneto become “quick”. This lossening will reducethe strength of the soil.

85. (b)

1 is incorrect because the primary functionof geo grids is reinforcement.

Geo grids have parallel sets of tensile ribsand hence are used as tension member.

86. (d)

Influence factor for the chart is

= 1

Number of sub. area

= 1 0.005

10 20

87. (a)

Swelling potential of soil increases asplasticity increases.

89. (b)

The error is half of the difference betweenface left and face right observations.

Hence, the difference is 3

Hence, error = 3 1 302

90. (b)

Map is an orthographic projection where asaerial photographs is perspective i.e. centralprojection.

91. (a)

Pitch of screw, p = 1 cm 0.01cm

100

f 22.5 2250P 0.01 = K

m = 3.425 + 3.93 = 7.355

Horizontal distance, D = Ks Cm

= 2250 3 0.425 918.17m7.355

92. (c)

True bearing of line = 6 32 1 7 32

As the true bearing of the line never change,the present true bearing will also be 7°32´.Present true bearing = magnetic bearing +9°42´

7 32 = magnetic bearing + 9 42

Magnetic bearing = 2 10

= 360 2 10

= 357 50

93. (c)

Normal equation for A

3 A 31 0 454 2A 82 43 203 3A 94 32 1520A 208 16 20

A = 208 16 20 10 24 49

20

94. (c)

Slope area method is approx and indirectmethod used when other methods couldn’tbe used. Discharge is estimated byobserving the water surface slope and cross-section area.

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96. (c)

Given catchment area = 1.5 km2

= 6 21.5 10 m

runoff coefficient (C) = 0.42

i(Rainfall intensity) = 48 mm min28

Peak flow (Q) = CiA

= 3

6 348 100.42 1.5 10 m s28 60

= 18 m3/s

97. (d)

For a reservoir, minimum storage capacityrequired is the maximum value of thecumulative excess demand volume.

98. (a)

Mean =

i iPAAi

0.25 8 0.35 7.3 0.1 7.70.13 8.9 0.17 8.5

1

7.93 cm

99. (a)

The probability of collision = Probability ofarriving a vehicle before 5 seconds.

p = n tetn!

= 90 0.025second

3600

p t 5 =

0.025 5

0 e0.025 5 0!

= 0.8825where t = time interval between two vehicles= time head way.

p(t < 5) = 1 p t 5

= probability of collisonp(t < 5) = 1 – 0.8825

= 0.1175 or 1–e–0.125

100. (b)

Bulk specific gravity of mix (Gbm)

= total

solid air w

mV V

mtotal = mass of mix = 1120 gm

volume of mix including air voids = V

As bouyant mass (mb) is given by

= total wm V

wV = w total bV m m

Gbm = Bulk specific gravity of mix

=1120

1120 630= 2.286

101. (b)

=1.5P a

E = 0.5 cmP = 6 kg/cm2

a = radius of contact area.

Contact area = 2aa = 14.75 cm

E = 21.5 6 14.75 265.5kg cm0.5

102. (d)

Lime treatment is generally effective forsolids which contains a relatively highpercentage of clay and silty clay.

Lime may be mixed with the preparedmaterial either in slurry or in dry state.When the lime is used for improving thesubgrade, the soil-mix shall be consideredfor CBR value.

103. (c)

Self explanatory

104. (b)

V =2L

8R

R =211.8 696.2m

8 0.025

106. (c)

We know that,

For most efficient Trapezoidal channel

R =y2

y = 2R

y 0.5 m

And the Radius of circle which can beinscribed inside this Trapezoid is

Radius of Circle = y

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Rcircle = 0.5Dia. of circle = 2×Rcircle

= 2 × 0.5= 1 m

107. (b)

For critical flow2

3Q TgA = 1

Q2 =3gA

T

Q = gAAT

Q =9.81 33 1

3

Q = 39.4 m s

110. (a)

Qa =3ALN

60

35000 1060

=

23 0.40 N0.2460

N = 132.6 rpm

111. (b)

The surface of freshly cut good timber ishard and shining and plane surface is brightand smooth. Dull surface is a sign of poorquality of timber.

112. (c)

The lime containing more than 30%impurities like clay is called poor lime.

113. (a)

The value of fineness modulus is higher forcoarser aggregate.

114. (c)

For very small jobs, premixed concrete canbe an economical alternative to buying theseparate ingredient for mixing.

116. (c)

Ceramic product has a low absorption rate.

118. (c)

Y-alloy is a good all round material whichis good conductor of heat. It is an alloy ofaluminium.

120. (c)

Modified duration in ladder diagram

= 1Max. day + sum of rest three4

= 124 16 12 204

= 36 days

Alternate:

1 2 3 4 5

6 7

14

8

15

9

16

10

17

11

18

12

19

22 23 24 25 26 27 28 29

20 21

13

D4

D3D2D1 6

5555

666

C1 C2 C3 C4

B1

3 3 3 3

B2 B3 B4

A1

4 44 4

A2 A3 A4

Critical path

29 28

1 2 6 7 14 15 22 23 24 25 26 27

No. of days = 36

121. (b)

Unlike creep, shrinkage strains areindependent of the stress condition in theconcrete. Also, shrinkage is reversible to agreat extent, i.e., alternating dry and wetconditions will cause alternating volumechanges in concrete.

122. (a)

In relatively deep beams, tied arch actionmay develop following inclined cracking,thereby transferring a part of the load to thesupport and so reducing the effective shearforce at the section.

123. (a)

At the flexural crack location, the tensionis carried by the reinforcement alone,whereas in between the cracks, concretecarries some tension and thereby partiallyrelieves the tension in the steel bars.

124. (d)

Rolled section like channel, angle T andsolid section are in buckling class C. RolledI section will not always be in buckling classC.

125. (d)

Slope deflection method is a stiffnessmethod.

126. (d)

Moment distribution method uses 2 tablesfor iteration whereas Kani’s method needs

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only one even in case of sway problemshence kani’s method is more efficient.

127. (b)

H+, OCl– and HOCl relate to each otheraccording to the following ionizationreaction, which depends primarily on thepH of the solution.

HOCl H OCl From this reaction, adding H+ on the rightpushes the reaction on the left. This meansthat at lower pH values, more HOCl exists.It has been found that HOCl is a strongerdisinfectant than OCI–. Hence to be moreeffective, the pH of the solution should belowered. At pH 5.5 and below and alltemperatures of 0° to 20°C, the concen-tration of HOCl is practically 10%. At pH <5 chlorine is practically insoluble in water.

128. (c)

In aerobic digestion, because the fractionof volatile matter is reduced, the specificgravity of the digested sludge solids will behigher than it was before digestion. Thus,the sludge settles to a more compact mass,and the clarifier underflow concentration canbe expected to reach 3 percent. Beyondthis, its dewatering properties are terrible.

129. (d)

In activated sludge process, themicroorganisms grow and are mixed bythe agitation of the air, the individualorganisms clump together (Flocculate) toform an active mass of microbes (biologicalfloc). Because the microorganisms aresuspended in the liquid wastewater, thisprocess is also known as a suspendedgrowth process.

130. (b)

The formation of OH Chydroxyl radical)

3 2O hv O OD

2O H O 2OHD

This is the beginning of a complex seriesof reactions in the atmosphere. Thesereactions disturb the normal NO2 photolyticcycle, leading to the formation of thecomponents of the more complexphotochemical smog. Hydroxyl radical, onceformed, can react with the threecompounds : CO, hydrocarbons andcarbonyl compounds. Thus if only these

three compounds are not present, thephotolytic cycle will not be disturbed andthere will be an absence of photochemicalsmog. The peroxy radicals, the end productof hydroxyl radicals with carbon monoxide,carbonyl or hydrocarbon compounds, arethe ones that react with NO to regenerateNO2, thus terminating the normal NO2cycle, leaving O3 undisturbed and allowingit to accumulate.

131. (d)

Sectional dimension is chosen such thatthe ratio of distance from NA to extremefibre in tension and compression is exactlysame as the ratio of allowable stresses intension and compression. Under thisprovision section will be most economicallyutilized.

132. (d)

Unsymmetrical section will always twistunder non uniform bending except whenload passes through the shear centre.

133. (a)

Natural flood-flow takes place across thecontours.

134. (a)

In delta region, velocity of river-flow isdecreased significantly leading to moresilting and shoal formation which results inspilling of water and formation of newchannels.

139. (b)

• F (Modulus of Elasticity) of soil varies acrossthe width of the loaded area, being greaternear the centre than near the edges.

• As a consequence, the displacemenetdistribution below a flexible footing will beof dish pattern and contact pressure will beuniformly distributed.

140. (d)

• Newmark’s chart can be used for thedetermination of vertical stress under anyshape of loaded area not any slope. Henceassertion is wrong.

• Newmark’s charts are based onBoussinesq’s equation.

141. (c)

In theodolite, graduations increases inclockwise direction.

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142. (c) If weighing factor (x) is zero then storage

(S) = m mK xI Q1 x

For x = 0

S = KQm

[For natural channels m = 1]S = KQ

Storage Outflow discharge

143. (b)

Map cracking appears as interconnectedcracking, forming a series of small blockwhich resembles the skin of an alligator.This patters is also called map cracking.

Permanent deformation in any of apavement’s layer or subgrade usuallycaused by consolidation or lateral movementof the material due to traffic loading.

144. (d)

Tidal bore a positive surge travel upstreamin a tidal river with the incoming tide.

146. (d)

Intelligent glass blocks the heat but not thelight.

148. (d)

In langrangian method, a coordinationsystem is attached to a single fluid particleto describe its velocity and acceleration.

ANSWERS - IES Master · 2017-12-17 · According to IS: 800 – 1984 Cl 6.6 the effective length of...

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Transcript of ANSWERS - IES Master · 2017-12-17 · According to IS: 800 – 1984 Cl 6.6 the effective length of...