answers - ies master · begining of test (t = 0) u = excess pore pressure after the time t. so, u i...

1. (c)

2. (a)

3. (a)

4. (a)

5. (a)

6. (a)

7. (a)

8. (d)

9. (c)

10. (d)

11. (c)

12. (b)

13. (d)

14. (c)

15. (d)

16. (b)

17. (d)

18. (b)

19. (b)

20. (a)

21. (d)

22. (d)

ESE-2019 PRELIMS TEST SERIESDate: 09th December, 2018

23. (a)

24. (d)

25. (b)

26. (a)

27. (b)

28. (b)

29. (b)

30. (d)

31. (d)

32. (b)

33. (c)

34. (a)

35. (c)

36. (c)

37. (c)

38. (d)

39. (c)

40. (c)

41. (b)

42. (c)

43. (b)

44. (c)

45. (a)

46. (c)

47. (d)

48. (d)

49. (c)

50. (a)

51. (a)

52. (c)

53. (a)

54. (c)

55. (a)

56. (b)

57. (d)

58. (c)

59. (c)

60. (b)

61. (b)

62. (a)

63. (b)

64. (d)

65. (c)

66. (a)

ANSWERS

67. (a)

68. (d)

69. (d)

70. (b)

71. (d)

72. (a)

73. (c)

74. (c)

75. (c)

76. (b)

77. (d)

78. (d)

79. (a)

80. (b)

81. (d)

82. (c)

83. (c)

84. (c)

85. (d)

86. (b)

87. (b)

88. (d)

89. (c)

90. (c)

91. (a)

92. (c)

93. (c)

94. (c)

95. (c)

96. (c)

97. (c)

98. (b)

99. (b)

100. (b)

101. (b)

102. (d)

103. (b)

104. (d)

105. (c)

106. (c)

107. (c)

108. (b)

109. (b)

110. (b)

111. (a)

112. (c)

113. (d)

114. (a)

115. (c)

116. (b)

117. (c)

118. (d)

119. (a)

120. (a)

121. (a)

122. (d)

123. (b)

124. (d)

125. (a)

126. (c)

127. (a)

128. (c)

129. (b)

130. (a)

131. (a)

132. (c)

133. (c)

134. (b)

135. (a)

136. (b)

137. (b)

138. (b)

139. (a)

140. (a)

141. (d)

142. (a)

143. (d)

144. (d)

145. (a)

146. (b)

147. (a)

148. (c)

149. (a)

150. (d)

IES M

ASTER

[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-02) (2)

1. (c)

2. (a)

xdu 3dx

yzdw dv 0 0 0dy dz

3. (a)

Modulus of elasticity = Yield stress

strain

= 2240 N / mm

0.0012 = 20 × 104 N/mm2

= 2 × 105 MPa

Modulus of elasticity in compression =Modulus of elasticity of tension.

4. (a)

5. (a)

P MA Z

Comp,max 4

P eP P PedA Ad Ad 2864

P e1A d 8

ten, maxP e1A d 8

comp, max

ten,max

1 e d 8 131 e d 8 3

de5

6. (a)

Cr E

1 1 1 1 11P P P 600 400 240

P = 240 kN

7. (a)

8. (d)

9. (c)

Due to Poisson’s effect, longitudinal strainleads to lateral strain which will lead toanticlastic curvature for an originallyrectangular cross-section.

For a doubly symmetric section, shear

centre always coincide with the centroid ofthe section.

10. (d)

From Mohr circle :

(120, 0)

( , 50)n

(20, 0) (70, 0)

Radius of Mohr Circle = 50 MPa

20MPa

Maximum shear stress = 120 20

2

= 50 MPa

max 50 20 30 MPa

11. (c)

1 2 1 2max, abs max , ,

2 2 2

For thin spherical cells, 1 2and are same equal

to pd4t .

max, abspd8t

12. (b)

Maximum permissible stress Circumferentialstress

pd802t

2t 80p

d

2 50 80p 3.2 MPa

2500

Therefore max internal pressure allowed = 3.2MPa.

13. (d)

Analyse only half beam,100 kNm

B

100 = B4EI5

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B 4100 5

4 2.5 10

= 0.005 radian

14. (c)

M = MOA + MOB + MOC

oOA

4EIM

L

oOB

3EIM

L

oOC

3EIM

L

o10EIM

L

oOC

3EI 3M ML 10

OCc

M L 3M L ML6EI 10 6EI 20EI

15. (d)

Influence line diagram for reaction at ‘C’

100

38

5 1318 8

3 m 8 m 5 m

1 m

120

DCBA

Maximum reaction at support C

= 13 100 120 1 282.5 kN8

When 100 kN load is at A

Reaction at C = 3 100 37.5 kN8

Case II : Reaction at C when 120 kN load is atA

1.5120

100

2/8

3/8

= 3 2120 100 20 kN8 8

16. (b)

No. of members, m = 8

No. of reactions, r = 7

No. of joints, j = 8

Additional DOF = 4 in internal hinge + (3 – 1) injoint hinge = 6

Dk = (3j – r) + additional DOF – No. of inextensiblemembers

= (3 × 8 – 7) + 6 – 9

= 14

17. (d)

Members, AK, BJ, DI, DE, EF, HG and GF arezero force members.

18. (b)

Flemish Bond gives better appearance thanEnglish Bond.

• Construction of Flemish bond requiresgreater skill in comparison to English Bond.

19. (b)

The inner annual ring surrounding the pithconstitute the heart wood. It indicates deadportion of tree and doesn’t actively take part ingrowth of tree. It imparts rigidity to the tree andhence it provides strong and durable timber forvarious engineering purposes.

20. (a)

Initial setting time time limit for handlingthe cement mortar.

Final setting Time beginning of developmentof mechanical strength.

21. (d)

22. (d)

23. (a)

Absorption of FA = 1 846

100 = 8.46 litres

Absorption of CA = 0.5 1152 5.76 litres100

Total amount of water to be used = 150 + 8.46+ 5.76 = 164.2 Kg/m3

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FA = 846 – 8.46 = 837.5 Kg/m3

CA = 1152 – 5.76 = 1146.2 Kg/m3

24. (d)

All of above parameters effects swelling pressureof soil.

25. (b)

26. (a)

Max. dry unit weight obtained by compaction

maxw

dG 2.7 10y

wG .30 2.71 1s .80

= 313.42 kN/m

27. (b)

In electrical analogy method an analogy existsbetween the darcy law and ohm’sn law:

Analogous quantities

Flow of water Flow of current

1.hLaw k AL

E.I k AL

2. Discharge ( ) current (I)

3. Head (h) voltage (E)

4. Length (L) Length (L)

5. Area (A) Area (A)

6. Permeability (k) Conductivity (k)

28. (b)

degree of consolidation given as

i

i

U U%U 100U

where, Ui = initial excess pore pressure at thebegining of test (t = 0)

U = excess pore pressure after thetime t.

So, Ui = 200 kN/m2

U = 60 kN/m2

200 60%U 100 70%200

29. (b)

In U.U test, no drainage is permitted duringthe consolidation stage. The drainage isalso not permited in the shear stage.

30. (d)

It approximates the curved failure envelopeby a straight line, which may not give correctresults.

For some clayey soils, there is no fixedrelationship between the normal and shearstresses on the plane of failure. The theorycannot be used for such soils.

31. (d)

The following assumptions were taken in Tezaghi’sbearing capacity theory

(i) The base of footing is rough.

(ii) The footing is laid at a shallow depth. i.e.

fD B .

(iii) The shear strength of soil above the baseof footing is neglected. The soil above thebase is replaced by a uniform suchargerDf.

(iv) The load on the footing is vertical and isuniformly distributed.

(v) Fooling is long ie. L/B is infinite. where Lis length of footing and B is the width.

32. (b)

= wG 2.7 10 19.31 e 1 0.4

kN/m3

Coefficient of earth pressure = 1 sin at rest(K0) = 0.5

v at 2.7 m = 19.3 × 2.7 kN/m2

= 19.3 = 52.1 kN/m2

H = 0.5v = 26 kN/m2

q = v H1 13.052

kN/m2

P = v H1 39.052

kN/m2.

33. (c)

Side of cube a = 10 mm = 0.01 m

Total surface area of cube A = 6a2 = 6 × 0.012

= 6 × 10–4 m2

IES M

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Volume of cube V = (0.01)3 = 10–6 m3

Specific surface area of cube

= 4

16

6 10 600 m10

34. (a)

Available shearing resistance,

2f subc zcos tan

= 15 + (20 – 10) × 4 cos245° tan30°

= 26.54 kN/m2

Mobilised shearing resistance,

= satzcos sin

= 20 × 4 × cos45° × sin45° = 40

F.O.S. = 26.54 0.66

40

35. (c)

36. (c)

S = 37t100(1 (0.63) )

= 100 [1 – (0.63)1.5] = 50%

37. (c)

For trickling filter

Surface area = 0 eQ QHydraulic Loading Rate(including recirculation)

= 375 2 10

20

= 7500 m2

Surface area excluding recirculation

=

3

275 10 6250m12

Surface area = 750 m2

38. (d)

All the three are possible reasons for increase inacidity of sludge.

39. (c)

Li = 14 (Qs)1/3

= 14 × (27)1/3 = 42 m

40. (c)

– Water pipes are not designed for self

cleansing velocity.

– Sewer pipes are laid on continuous gradientto permit gravity flow.

41. (b)

Air required = air 2

w% of O

=

3

1200 kg / day1.15 kg / m 0.21

= 4968.94 m3/day

O2 transfer efficiency of aeration equipment =10%

Air needed per minute =

4968.940.1 60 24

= 34.50 m3/min

42. (c)

G = PV

= 3

410 10

10.03 10 10.8

= 960.81 sec–1

43. (b)

Vs = d

dt

td = s

dV = 4

34.2 10

td = 1.98 hr

44. (c)

Assume total solids in undigested sludge

= 100 kg

Volatile solids = 60 kg

Fixed solids = 40 kg

Fixed solid remain unchanged after digestionassume volatile solid after digestion = x kg

Given, x = 0.3 (x + 40)

x = 17.15 kg

% volatile reduced to gases

=

60 17.15 100 71.4

60

IES M

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45. (a)

50

50w

5

5w

w w w15 50 53

Seepage :Pressure at Drainage gallery

= w w w1H' [ h H']3

= w w w15 [50 5 ]3

= w20 .

46. (c)

Silt excluder up stream of head regulator

Silt ejector down stream of head regulator..

47. (d)

48. (d)

T.R. =

Total mass transpired by the plant during its full growth

Mass of dry matter produced

Mass of dry matter produced is generally takenas entire mass of plants including its roots.

49. (c)

50. (a)

The integration of the hydrograph gives the flowmass curve and mathematically it is expressedby

V(t) = t

0

Q t dt

51. (a)

AI = PET AET 100

PET

= 150 120 100 20%

150

52. (c)

Probability of exceedence of flood = 1 0.02

50

Probability of occurring at least once in 100 yrs

= 1 – (1 – 0.02)100

= 1 – (0.98)100

= 0.87

53. (a)

54. (c)

Relative height (jump)

= 2 1j

1

y yh (height of jump)

E

Relative initial depth = 1

1

yE

Relative sequent depth = 2

1

yE

55. (a)

1

1

v 81 9.03 9gy 9.81 8.2

2

1

y 1 1 1 8 81y 2

2

1

y0.5 1 649

y

2

1

y 12.23y

y2 = t8.2 12.23 100.35 m y

Free repelled jump.

56. (b)

Using direct step method

0 f

ExS S

2 1

1 2

E Ex

10 (sf sf )2

5 2 3x 30 m1 0.1(0.15 0.5)2

IES M

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57. (d)

q = 35 5m /s/m1

v = 5 2.5 m/s2

2 2v 2.5 0.3125 m2g 2 10

v 2.5 2.5Fgy 10 2 20

= 0.56 < subcritical

E1 = 0.312 + 2 = 2.31 m

E2 = 2.31 – 0.10 = 2.21 m

yc =

1 12 23 3 1

3q 5 (2.5) 1.35g 10

Ec = 1.5 × 1.35 = 2.025 m

Since E2 > EC

Upstream depth will remain unchanged.

58. (c)

59. (c)

60. (b)

Correction = 30.1 30 100 3.33 m

30

Corrected length = 1000 + 3.33 = 1003.33

61. (b)

62. (a)

Quartzite is a non-foliated metamorphic rock.

Parent rock can be any of three types ofrock igneous, sedimentary or metamorphic.

63. (b)

Hypocentre: it is a location of earthquake in acrustal plate. It is also known as focus.

Epicentre: The projection on the surface of earthdirectly above the focus.

64. (d)Limitation of bar chart:

1. It doesn’t indicate the critical activities ofthe project.

2. The financial aspect involved in the projectis unknown from the bar chart. Becausecritical path can’t be identified, there is noquestion of crashing activites.

65. (c)Calculation of expected time,

tE = 0 m pt 4t t

6

EActivity t1 2 52 3 83 4 104 5 53 5 7

1 2 3

4 5

5 8

10

5

7

TE = 28 days

As, Z =

s ET T

=30 282

= 1

P = 85%

66. (a)

Free float, Ff = j i ijEET T t

67. (a)

As intefering float is equal to slack of head eventfor this we have to plot TE & TL in network diagram.

0 0 2

2

77

1010 12 12

9 955

6 8FA

2

2

B3

D5

E5

H 1

I2

G3

C4

ST for F = 10 – 10 = 0

IES M

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68. (d)

T(time) C(cost)

Q(quality)

AB

D

E

at T, Q and C, time quality and cost respectivelyare the only individual dominant factors.

at pt B, time and quality are controlling factorsonly.

at pt. D, time and cost are controlling factorsonly.

at pt. E, quality and cost are controlling factorsonly.

at pt A, a balance is strunk b/w the 3 constraintsof construction project management i.e quality,time and cost.

69. (d)70. (b)71. (d)

For the crawler tractor—

Coefficient of traction is high as 0.9

Speed is low

Most efficient in muddy area and in flatarea it is not efficient.

High pulling affort.

Capacity is high.

72. (a)

Years purchase is defined as the capital sumrequired to be interested in order to receive a netannual income of 1` at a certain rate of interest.

Y.P. = c

100i i (when sinking fund is also recovered)

=

100 12.55 3

73. (c)

Assuming t = 2.5 secs, f = 0.35 for V = 80 kmph

As there is ascending gradient on one side ofsummit and descending gradient on the otherside, the effect of gradients on the SSD isassumed to get compensated and hence ignoredin the calculations.

SSD = 0.278 Vt + 2V

254f

= 0.278 × 80 × 2.5 + 280

254 0.35

= 127.6 m 128 m

N = 0.03 – (– 0.05) = 0.08

Assuming L > SSD

2 2NS .08 128L4.4 4.4

= 297.9 > SSD ok

L = 298 m

74. (c)

75. (c)

Lc = concrete

2f

= 2 16 0.88 m

1.5 24

76. (b)

77. (d)

H.C. = 8 25 0.2 20 tonnes

2

78. (d)

79. (a)

The maximum harbour depth below lowest lowwater is

= Loaded draft + 1.2 m for soft bottom= Loaded draft + 1.8 m for rock bottom

80. (b)

In general, lining of 2.8 cm per 30 cm of borediameter is provided.

Therefore, thickness of lining = 8002.8300

= 74.67 cm 75 cm

81. (d)

Size of shaft is governed by following factors:

(i) Amount of muck to be hoisted.

(ii) Hoisting system to be adopted.

(iii) Size of muck car.

(iv) No of workmen.

(v) Type of equipments to be used.

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82. (c)

The Buoyancy force due to mercury.

1BF = 1 1gV = 13.6 × 1000 × 9.81 × V1

The Buoyancy force due to water

2BF = 2 2gV = 1000 × 9.81 × V2

Given : 1 2V V V total volume of metallic

body.

From the Archimedes principle of floatation.

W = 1 2B BF F

gV = 13.6 × 1000 × 9.81 V1 + 1000 × 9.81× V2.

4.5 × 1000 × 9.81 × V = 13.6 × 1000 × 9.81 ×V1 + 1000 × 9.81 × (V – V1).

4.5 V = 13.6 V1 + V – V1

3.5 V = 12.6 V1

V1 = 27.77% of V

V2 = 72.22% of V.

83. (c)

Electrical analogy method : This method is basedon the fact that the flow of fluids and flow ofelectricity through a conductor are analogous.These two systems are similar in the respectthat

(i) Electric potential is analogous to the velocitypotential

(ii) The electric current is analogous to thevelocity of flow, and

(iii) The homogeneous conductor is analogousto the homogeneous fluid.

84. (c)

Applying Bernoulli's equation when valve is closed.

V1 = V2 = 02 2

1 1 2 11 2

P V P VZ Z2g 2g

rh

o

2 11 2

P PZ Z

3

1 270 10z z 7m

10 1000

Now when valve is open2 2

1 1 2 21 2 L

P V P VZ Z h2g 2g

1 21 2 L

P PZ Z h

1 2V V

23 + 7 = hL Lh 30m .

85. (d)

Dimensionless parameter to specify instability2 du

dy

this is zero at near the wall (y = 0)

and at the centre of pipe du 0dy

. This

instability is maximum somewhere between thesetwo.

86. (b)

Thickness of Hydrodynamically smooth boundary

kinematic wiscosity 1Flow velocity

.

87. (b)

Oil (G = 0.6)

water (G = 1)o

A

B

B

C

D

Downward force on AB

=

20.8 0.8 0.8 1 (10 0.6)

4= 0.82 kN

Upward force on BCD

= 21.6 0.8 1 0.6 10 (1.6) 10 1

8` = 17.73 kN

Net force (vertical) = 17.73–0.82 = 16.9 kN

88. (d)

u = 2y, v = 8x + 2

Equation of streamline, dy vdx u

dydx =

8x 22y

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2y dy = (8x + 2) dx

Integrating both sides, we get

y2 = 4x2 + 2x + c

y2 = 2x (2x + 1) + c

89. (c)

Displacement thickness, * =

00

u1 dyu

= 0

y1 dy

=

2

0

yy2

=2

Momentum thickness, =

0

0 0

u u1 dyu u

= 0

y y1 dy

=

2 3

20

y y2 3

= 6

shape factor =

* / 2 3

/ 6

Alternatively; for

m

o

u y * m 2u m

Hence shape factor =

1 2 3

1

90. (c)

For maximum efficiency, H = 3hf

H = 2

53f l Q12.1d

100 =

2

53 0.03 50 Q

12.1 (0.5)Q = 2.9 cumecs

= 2900 l/sec.

91. (a)

Head loss in pipes connected in parallelwill be same irrespective of their sizes andlengths.

Water hammer pressure is more in rigidpipes than flexible pipes.

92. (c)

Rankine efficiency of hydraulic ram = q(H h)

Qh

=2.5(30 3) 100

40 3

=2.5 27 100

120

= 56.25 %.

93. (c)

Slip = Qth – Qactual

% Slip = th act

th

Q Q100

Q

When the slip becomes 'negative' it is callednegative slip, Qactual > Qtheoritical.

Negative slip occurs when suction pipeis long, delivery pipe is short and speed of rotationis high. In this case the delivery valve opensbefore the suction stroke is completed due toinertial pressure in suction pipe leading to Qactual> Qtheoritical.

94. (c)

The instantaneous velocity Vd in the delivery pipemay be obtained as :

Vd =d

A r sina

(Vd)mean = d2

00

A r sin da

d

= d

A ra

(Vd)max = d d

A A rr sin 90ºa a

d mean

d max

(V ) 1 .(V )

95. (c)

To reduce the acceleration head we needto reduce the length of suction pipe (ls) andlength of delivery pipe (ld) in which fluctuationof velocity occurs. This is done by the fitting

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air vessels closed to cylinder as possible.Thus between air vessel and cylinder onlyfluctuating velocity will occur. Below airvessel on suction side and above air vesselon delivery side velocity will be constant.

Due to movement of piston, when there isexcess velocity above average b/w suctionside air vessel and cylinder, water issupplied from air vessel.

96. (c)

97. (c)

98. (b)

1000critical s/c

200

1 m

The critical section for maximum moment islocated halfway between the centreline andedge of the wall in this case

Given qu = 200 kN/m2

= 1000 200

2 4 = 450 mm

For 1 m width inside, the design bending moment

is given as, Mu = 2

u0.45W 1

2

= 0.2025200 1

2 = 20.25 kNm/m

99. (b)

As maximum hoop tension will develop at bottomand is given by

= wH D

2=

10 3 22

= 30 kN/m

100. (b)101. (b)102. (d)

Loss of stress due to elastic deformationdepends upon modular ratio and it isindependent of relexation of steel.

Loss of stress due to shrinkage is timedependent.

In pretensioned there is only one type ofimmediate loss i.e. elastic shortering.

103. (b)

Initial stress in wires

=

3

2

300 10 1590 MPa2 60

4Initial stress in concrete

=

3300 10 4.8 MPa250 250

Losses:

6 3

210Elastic deformation= 4.8 33.6 MPa30

Shrinkage loss 200 10 210 10 42 MPa8other loss= 1590 127.2 MPa

100202.8 MPa

Effective prestress

= 1590 – 202.8 = 1387.2 N/mm2

104. (d)

Moment due to prestress, M = Pe

Pe Pe

Upward deflection at mid span = 2ML

8EI

= 2PeL

8EI

= 2

3PeL

bd8E12

= 2

33PeL2bd E

105. (c)

MOR = 0.36 fck bx [d – .416 x]

For balance section x = xu = kd = .48 × 500 =240 mm.

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MOR = .36 × 20 × 300 × 240 × [500 – .416 × 240]

= 207.44 kN-m 210 kN-m.

106. (c)

y SV

V

0.87f AS = v C b

u

vV

b dVu = 1.5 × V

v =

300 1000 2MPa500 300

VS =

0.87 415 2 1004 378.1mm

3002 1.5

300 mm and 0.75 × d = 0.75 × 500 = 375 mm

107. (c)

Li = 2

i iB 2

i i

WhVWh …(A)

VB = Ah × Total load

Total load = 1700 kN

Load at each floor level = 1700 425 kN

4

2i iWh = 425 × 82 + 425 × 62 + 425 × 42 + 425

× 22 = 51000 kN

at storey level 2, storey force =

2425 4850

51000= 113.33 kN

108. (b)

Shear wall are very efficient in terms ofconstruction cost and effectiveness in minimizingearthquake damage.

109. (b)

aS 1.36, 0.55 T 4g T

= 1.36 2.2670.6

110. (b)

Damping ratio, 1

j 1

u1 ln2 j u

= 1 50ln

2 5 25

= 0.022

= 2.2%

111. (a)

For springs connected in parallel,

ek k 2k 3k

Equivalent stiffness of entire system,

e

1 1 1 1k k 3k 4k

e

1 (12 4 3)k 12k

e12kk19

112. (c)

Elasticity Idealised Curve

PlasticityStress fy

Strain(A) It will increase the margin of safety

(B) In case of tension members, fractures mayexists.

(C) We Generally neglect the strain hardeningeffect in Plastic analysis.

113. (d)

(i) Members supporting lifting or rolling loads.

(ii) Members subjected to repeated cycles ofstress from vibrating machine.

(iii) Members subjected to wind inducedoscillation for a large numbers of cycles inlife.

(iv) Members subjected to crowd inducedoscillation of a large numbers of cycles inlife.

114. (a)

Bolted connection allow greater slip betweencomponents than rivetted connection.

115. (c)

Structural members subjected to bendingaccompanied by large axial comprensive loadsat the same time are known as beam column.

IES M

ASTER


A Beam column differs from column only bypresence of eccentricity of load application, endmoment, transverse load.

116. (b)

Length of Purlin, L = Spacing of truss = 5m

Maximum moment along principal axis of purlin

= PL (5 6) 5 15 kN-m10 10

117. (c)

Maximum bending moment,

M = PL 80 5 100 kN-m4 4

Since the beam is laterally supported throughout.Hence, required plastic section modulus

= 0m 6

y

1.1M 100 10f 250

= 4.4 × 105 mm3

= 440 cm3

118. (d)

119. (a)

Purlins may be designed as simplysupported, cantilever or continuous beam.

IS: 800 recommends to design purlin ascontinuous beams. Sag rods are designedas tension members.

Shear centre of purlin should coincide withnode points of the truss.

120. (a)

121. (a)

Power Tw 150 2150060

7.5 kW [ Assertion is also correct]

122. (d)1000 mm

10 mm

200 y

100

NA NA Tansformed sec tionI I

combined Transformedy y

Trans

200 100 100 1000 10 205y

200 100 1000 10

= 135 mm

depth of NA from top most fibre

= (200 + 10) – 135 = 75 mm

123. (b)

124. (d)

Support slip causes decrease in the span of thecable resulting in marginal increase in the sag,resulting in corresponding decrease in the cabletension.

125. (a)

A well seasoned timber is a good timber.Seasoning of timber has a variety of advantages.

1. Reduction of shrinkage and warping afterplacement in structure.

2. Increment of strengths, durability andworkability

3. Suitable for painting

4. Reduction in weight

5. Reduction in tendency to split and decay.

126. (c)

Modulus of elasticity of matrix must be lowerthan that of fibre for effective stress transfer.

Use of fibres of low elastic modulus impartgreater degree of toughness and resistanceto impact while fibres of higher elasticmodulus impart strength and stiffness.

127. (a)

128. (c)

Double tangenet method is used in case of plateload test.

129. (b)

Westergaard’s solution is suitable for sedimentarydeposits as it does not consider soil to beisotropic.

130. (a)

131. (a)

IES M

ASTER


Since no direct relationship exist between theintensity of light scattered at 90° and jacksoncandle turbidity meter, calibration of nephelometerin terms of candle or silica unit is not valid.

132. (c)

Sea water normally contain 20% less oxygenthan that contained in fresh water and containslarge amount of dissolved matter as such thecapacity of sea water to absorb sewage solids isnot as high as that of fresh water of stream.

133. (c)

pH 8pH 7

HOCl H OCl

HOCl is 80% more effective than OCl–. Thus, pHof water should be maintained slightly below 7.

134. (b)

Majority of biological treatment unit hastemperature range of 20° to 40° which is suitablefor mesophilic bacteria.

135. (a)

136. (b)

137. (b)

138. (b)T/2 T/2

yy

= 60°

B

m

139. (a)Assertion & reason both are correct and reasonis correct as and engineering equipment loomtheir value at constant rate.

140. (a)Reason is statement of central limit theorem thatcorrectly explains the assertion.

141. (d)

Traffic capacity is the ability of a roadway toaccommodate traffic volume.

Basic capacity is the maximum number ofpassenger cars that can pass a given point on

a lane or roadway during one hour under themost ideal roadway and traffic conditions.

142. (a)

If the water table is very near to the subgrade ofthe road, it will ultimately cause cracking of theroad surface, because the consistency of thesoil will change from plastic to liquid state leadingto its volumetric decrease.

143. (d)

Flow always taken place in the direction fromhigh TEL to low TEL.

144. (d)

Pressure gradient dpdx

is positive due to

decrease in velocity in the region of boundarylayer separation.

145. (a)Large number of smaller diameter bars, welldistributed in tension zone, reduces thecrackwidth.

So smaller dia bar is used in slab.

146. (b)147. (a)

In PSC, a change in external moments in elasticrange results in a shift of the pressure line ratherthan in an increase in the resultant force in thebeam.

148. (c)

In moment resisting frame. Members and jointsare capable of resisting force primarily by flexure.

149. (a)Beta distribution model for PERT

Gives

x = tE = 0 m pt 4t t

6Weightage of to & tp is equal.

150. (d)

Crop rotation means changing the crops to begrown every year in the same field or differencegroups in rotations in the same field.

answers - ies master · begining of test (t = 0) u = excess pore pressure after the time t. so, u i...

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