ESE-2019 PRELIMS TEST SERIESDate: 4th November, 2018

ANSWERS

1. (b)

2. (b)

3. (a)

4. (b)

5. (c)

6. (a)

7. (c)

8. (a)

9. (c)

10. (c)

11. (b)

12. (b)

13. (b)

14. (c)

15. (c)

16. (c)

17. (a)

18. (c)

19. (b)

20. (b)

21. (d)

22. (b)

23. (d)

24. (a)

25. (c)

26. (c)

27. (b)

28. (d)

29. (b)

30. (b)

31. (c)

32. (a)

33. (c)

34. (a)

35. (a)

36. (c)

37. (d)

38. (b)

39. (d)

40. (d)

41. (a)

42. (a)

43. (a)

44. (c)

45. (a)

46. (b)

47. (a)

48. (b)

49. (c)

50. (c)

51. (d)

52. (a)

53. (a)

54. (c)

55. (a)

56. (b)

57. (a)

58. (d)

59. (b)

60. (b)

61. (c)

62. (a)

63. (c)

64. (b)

65. (c)

66. (b)

67. (d)

68. (c)

69. (a)

70. (d)

71. (d)

72. (a)

73. (a)

74. (c)

75. (a)

76. (c)

77. (d)

78. (c)

79. (d)

80. (b)

81. (c)

82. (b)

83. (b)

84. (c)

85. (d)

86. (a)

87. (a)

88. (b)

89. (c)

90. (c)

91. (c)

92. (b)

93. (d)

94. (b)

95. (a)

96. (d)

97. (b)

98. (c)

99. (b)

100. (d)

101. (a)

102. (a)

103. (a)

104. (d)

105. (a)

106. (b)

107. (b)

108. (b)

109. (c)

110. (b)

111. (c)

112. (a)

113. (a)

114. (a)

115. (d)

116. (b)

117. (b)

118. (c)

119. (a)

120. (b)

121. (b)

122. (d)

123. (a)

124. (a)

125. (c)

126. (a)

127. (b)

128. (a)

129. (d)

130. (a)

131. (a)

132. (c)

133. (d)

134. (a)

135. (c)

136. (a)

137. (a)

138. (c)

139. (a)

140. (a)

141. (b)

142. (b)

143. (d)

144. (b)

145. (d)

146. (b)

147. (a)

148. (b)

149. (d)

150. (c)

IES M

ASTER

(2) RCC + OCF + Transport

1. (b)

The maximum value of strain 0.0035 is for flexuralcompression.

2. (b)

In limit state the failure criterion of column andbeam is based on maximum principal straintheory.

3. (a)

Effective width (bf)

bf = 0w f

lb 6D

6

bf 1 2w

l lb

2

l0 = 2 m, bw = 0.3 m = 300 mm

Now, bf = 2000

6 + 300 + 6 × 150 = 1533 mm

and 1 2w

l lb

2

= 300 + 2400 2400

2

= 300 + 2400

f fb 2700 mm B 1533 mm

4. (b)

5. (c)

When shear force is constant, the span is calledshear span. The flexure crack that usually formsfirst in the beam.

6. (a)

Nominal shear force

(Vu) = 380 10

200 300

= 1.33N/mm2

Permissible shear stress = 0.25 N/mm2

Design shear stress = 1.33 – 0.25 = 1.08 N/mm2

The design shear force = 1.08 × 200 × 300

= 64800 N = 64.8 kN

7. (c)

Ld = y

bd

0.87f4

Ld = development length

= diameter of bar = 12 mm

bd = bond strength = 2.1 MPa

For Fe415

Ld = y

bd

0.87f4 1.6

= 12 0.87 4154 2.1 1.6

= 322.36 mm

8. (a)

At the point of inflection anchorage length L0 ismaximum of ‘d’ and 12

d = 600 mm

12 = 12 × 20 = 240 mm

So anchorage length = 600 mm

9. (c)

The lap length in compression shall be equal toLd in compression but not less than 24 .

The lap length shall be calculated on the basisof dia of smaller bar when different dia bar are tobe spliced.

10. (c)

The anchorage value fo bend shall be taken4 times the dia of bar for each 45° bend.

Mimimum turning radius for mid steel is 2(plain), and 4 for cold worked deformedbar where is dia of bars.

Hooks should be provided for plane bars intension. (IS code 456: 200, clause - 26.2.2.1)

11. (b)

Equivalent shear (Ve)

Ve = uu

TV 1.6

b

Vu = factored shear forceTu = factored Torsional moment

So Ve = 50100 1.60.4

eV 300 kN

Width of beam b = 400 mm = 0.4 m

overall depth D = 650 + 50 = 700 mm = 0.7 m

Now equivalent bending moment

Me = Mu + uT D11.7 b

= 50 0.7200 11.7 0.4

= 280.8 kN-m

IES M

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[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-7) (3)

12. (b)

For simply supported beam

spandepth

20 (upto 10 m span)

depth = span20 =

1020 = 0.5 m

13. (b)

Minimum tensile reinforcement

stAbd =

y

0.85f

minstA = 0.85 350 450 50415

minstA = 286.74 mm2

Maximum tensile reinforcement

= 0.04 BD = 0.04 × 350 × 450 = 6300 mm2

ratio of Maximum/minimum reinforcement

= 6300

286.74 = 21.9

14. (c)

The maximum spacing in case of inclined stirrupsat 45° will be equal to effective depth of beami.e. ‘d’.

15. (c)

The horizontal distance between two parallel barsshould not be less than the following :

The diameter of the bar

5 mm more than the nominal size of coarseaggregate.

The diameter of largest bar if different barsare used

Minimum distance = max (16 mm, 25 + 5mm) = 30 mm

16. (c)

n depends upon u

uz

PP

For u

uz

P0.2

P then n 1.0

For u

uz

P0.8

P then n 2.0

17. (a)

18. (c)

Isolated footing bends in saucer-like shapeunder cantilever action.

In strap footings dowels are used such thatstrap and footing acts as a unit.

19. (b)

Combined footing generally used for relativelyheavily loaded column resting on soil with lowsafe bearing capacity.

20. (b)

For Backfill with sloping surface, the active earthpressure coefficient is

ka = 2 2

2 2

cos cos coscos

cos cos cos

= angle of repose of soil = 45°

= angle of inclination of backfill = 30°

ka = 2 2

2 2

cos30 cos 30 cos 45 cos30cos30 cos 30 cos 45

= 0.232

21. (d)

1m

0.5m

Toe

W1

W21m

pa

W1 = 24 × 3 × 0.5 × 1 = 36 kN at 1.25 m fromtoe

W2 = 124 3 1 12

= 36 kN at 2/3 m from toe

Pa = 2a soil

1 k H2

= 21 1 16 32 3 = 24 kN at from toe

Restoring moment RM = 1 22W 1.25 W3

= 236 1.25 363

= 69 kN-m

IES M

ASTER

(4) RCC + OCF + Transport

Overturning moment 0M = 24 × 1 = 24 kN-m

FOS = R

0

M 69M 24

= 2.875

22. (b)

Given

W = 500 kN/m (considering unit width inside)

b = 10 m, h = 5m

FR

x

x = 3m

e = b x 2m2

Maximum pressure intensity at base

= W 6e1B B

=

500 6 2110 10

= 50 × 2.2

= 110 kN/m2

23. (d)

Prestressing is economical for members longspan.

Prestressing wires in electric poles isconcentric.

In pipes & tanks circular prestressing isadopted.

24. (a)

Weinbery clip and Dorland clip are mostcommonly used anchoring devices in thepretensioning system.

25. (c)

For point load, cable profile should betriangular.

Sharp angle in cable induce concentratedloads.

26. (c)

Stress due to prestress :

z = 2 2bd 120 300

6 6

= 18 × 105 mm3

PA

= 5 N/mm2

Pez

= 3

5180 10 50

18 10

= 5 N/mm2

Stress due to loads

M = 2 2wl 4 6

8 8

= 18 kN-m

Mz

= 6

518 1018 10

= 10 N/mm2

fb = 5 + 5 – 10 = 0 N/mm2

fcr = 5 N/mm2

Extra moment required for cracking = 5 × 18 ×105 = 9 kN-m

Cracking moment = 27 kN-m

27. (b)

A

2m

8mWl 2 = 32 kN Wl

2 = 32 kN

Bending moment at A = 8 2 132 2

2

= 48 kN-m

Shift of C line, d = 348 10

480

= 100mm

28. (d)

Refer to code IS : 1343 – 2012 (Clause 18.6.1)

29. (b)

1m 1m2m

pe

Using moment area theorem : deflection at centre

= e ec

p p 1 21 1.5EI EI 2 3

c = e11p6EI

c = 711 1200 1000 50

6 2 10

= 5.5 mm.

30. (b)

Suitable control on deflection is very essentialfor the following reasons:

IES M

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[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-7) (5)

1. Exceesive deflections are likely to causedamage to finishes, partitions & associatedstructures.

2. Large deflections under dynamic effects &under the influence of variable loads maycause discomfort to the uses

3. Excessive, sagging of principal structuralmembers is not only unsighty, but at times,also renders the floor unsuitable for theintended use.

31. (c)

As per IS456 : 2000

Reinforcement in central bandwidth 2Total reinforcement in short direction 1

Where Longer dimension of footingShorter dimension of footing

= 32

= 1.5

Therefore, percentage of reinforcement in central

width = 2 100 80%

1.5 1

32. (a)

Axial capacity of column = 0.4fckAc

= 0.4 × 20 × (300)2 × 10–3 kN

= 720 kN > 600 kN

Hence, column is self sufficient to resist the load.Therefore, area of concrete required to resist direct

stress = 3600 10

0.4 20

= 75000 mm2

As per clause 26.5.3.1(b),

Area of reinforcement required

= 0.875000100

= 600 mm2

33. (c)

Prestress at the level of steel2

cP PefA I

= 2

3(100 600) (100 600) (50) 12100 300 100 (300)

= 2.67 N/mm2

Prestress loss in second cable = mfc

= 2.67 × 6 = 16 N/mm2

34. (a)

35. (a)

36. (c)

Refer clause 8.2.8 of IS456 : 2000.

37. (d)

Critical section forbending moment

t/4

t

38. (b)

39. (d)

= 5 mm, L = 20 m, Es = 200 GPa

Loss = 3

s 35 200 10E 50 MPa

L 20 10

% loss = 50 100 5%

1000

40. (d)

Refer IS456 : 2000, clause 16.1 and table 11.

41. (a)

y1 = 0.3m, y2 = 0.6m Vw = 3 m/s

2w 1 1 2

1

y1V V g (y y )2 y

1w 1V V 0.5 9.81 0.9 2

1V 3 3 V1 = 0 m/sBy continuity equation

1 w 1 2 w 2y (V V ) y (V V )

20.63 (3 V )0.3

2V 1.5 m/sec.

42. (a)

IES M

ASTER

(6) RCC + OCF + Transport

43. (a)

32

1L

21

1

y 1yE

yy 4y

3L

1

E (4) 16y 4 5 5

... (1)

and 2

1 1

1

E F1

y 2

21

1

E 101y 2

1

1

E 51y

... (2)

L

1

E 16 0.0627 6.27%E 5 51

44. (c)

12 2 32 2

2 2 1 21 3 3 3

21 3 1 12

2 2

F y Fq qFy 15gy y ygy

y y

Also 221

1

y 1 1 8F 1y 2

2

1

y 1 1 8 15 1 5y 2

12 32F1

5 15

2F 0.35

45. (a)

221

1

y 1 1 1 8Fy 2

2

1

y 1 1 1 8 10y 2

2

1

y 1 ( 1 9) 4y 2

32

31L

1 2

1

y 1yE 3 27

y 4 4 16y4y

116 10y 5.926 m

27

2y 23.7 m

46. (b)

Hydraulic jump increases weight on apron andthus reduces uplift pressure under masonarystructure by raising the water depth on apron.

47. (a)

Using chezy formula =

2

00 2

c

y1yy S

x y1y

But y0 = yc 0y Sx

The water surface profile are straight lines.

48. (b)

49. (c)

3Q 1m /s , B = 5m, so = 0.0001, y = 1m

v = 1 0.2 m/s5

R = 5 1 5 m5 2 7

sf = 2 2 2 2

54 4

3 3

v (0.02) 0.2 2.67 105R7

5 5

2 2

3 3

dy 0.0001 2.67 10 0.0001 2.67 10dx Q T 1 51 1

gA 10 5

= 5

57.33 10 7.36 100.996

50. (c)

In ventriflume, velocity of flow at throat is smallerthan critical velocity.

IES M

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[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-7) (7)

In standing flume, velocity of flow at throat ingreater than critical flow.

51. (d)

V = QA

V = 10 m/s7

F = V 10 4 20A 7 9.81 7 7 68.67g.T

= 0.34

52. (a)2 2

2 2v QE y y2g 2g B y

2

2102 y

2 10 25 y

2 = y + 21

5y

y = 0.348m, 1.947m

53. (a)

A = 2D ( sin2 )

8P = Wetted perimeter

= r

R =

25 2 38 3 2A

2P 2.53

R = 0.73 m

54. (c)

K =2/3AR

nA = y2

R =2y y

2 2y 2 2

K =2/3

21 yyn 2 2

K =1/3

8/31 1 yn 8

K = 8/31 y2n

C =1

2n

55. (a)

For hydraulically efficient section

Be = 2ye

Be = 2 × 2

Be = 4m

56. (b)

dyT

Area of flow, dA = Tdy

A = y

0Tdy

= y

0k y dy

A = 3/2y 2 2K K y y Ty

3/2 3 3

3AT2y

2 28 y 3A 8 yP T

3A3 T 2y 32y

P = 33A 16y

2y 9A

dpdy =

2

23A 48y 0

9A2y

2

23A 48y

9A2y

A = 24 2 y T 2 2y3

hydraulic radius, R = AP

IES M

ASTER

(8) RCC + OCF + Transport

=

2

2

4 2y3 0.5 y

8y2 2y3.2 2y

57. (a)

By manning’s equation

2/3 1/2n

1V R Sn

... (1)

CV C RS ... (2)

Comparing (1) - (2)

1/61C Rn

58. (d)

2m2

1

Area of flow, 1A 2 22

A = 2m2

2 2P 2 1 2 2 5m

A 1RP 5

2/31Q AR sn

=

12 232 1 10.02 100005

= 0.59 m3/s

59. (b)

60. (b)

0 0.5 1.5 0.5q .10 0.12 2

1 1.5 0.5 10.1 0.12 2

0.5 0 0.5 00.1 0.12 2

1 0.50.12

= 0.25 m3/s/m.

61. (c)

62. (a)

If the change of depth in a varied flow in gradual,so that the curvature of streamline is notsignificant, it is called gradually varied flow.

63. (c)

V =2/3

1/21 A (s)n P

=2/3

1/21 9 (0.0025)0.015 9

=3.33 m/sec.

Q =AV = 9 × 3.33 = 30 m3/sec.

64. (b)

At critical condition, 2

3Q Tg A = 1

2

2 310 (6d) 1

10(3d ) , where d is the depth of flow..

Critical depth = 1/520 d

9

Critical velocity = 2/510 m/s2039

65. (c)

Q = 2/3 1/21A (R) (s)n

Here n and s is fixedQ1 =Q2

2/38

4 28

=

2/32 DD48

D =1/811

34

.

66. (b)

Fr = 3/2 1/2V Q Q

B(y) (g)gy By gy

3/2rF y

Factor =3/2 3/22 3

3 2

IES M

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[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-7) (9)

= 31.5 1.842 .

67. (d)

We know, yc = 1/32q

g

23 ×9.81 = q2

discharge per unit width,

q = 38 9.81 8.86 (m /s)/m

Emin = c3 3y 2 3m2 2

critical velocity, vc = c

q 8.86 4.43 m/sy 2

.

68. (c)

Q = 40 m3/s = AC RS

40 =4 2(4 2)C 0.0025

8

C =100

Using interpolation

60 – 120 60 – 1000.030 – 0.015 0.030 n

n =0.020.

69. (a)

y2

y1

F

If energy loss is neglected

F = 33

1 2

1 2

(y y )1 1 10 3 1.52 y y 2 4.5

F 3.75 kN/m

70. (d)

In defining a froude numbers applicable tochannel flows, the length parameter used ishydraulic mean depth.

After hydraulic jump subcritical flow occurand has another name is tranquil or streamingflow.

The strength of jump is governed by theupstream froude number.

71. (d)

Area of flow, A = 21(2 2) 12

= 5.57 m2

Top width, T = 2 m

Hydraulic depth = A 5.57 2.78 mT 2

72. (a)

73. (a)

Forebays is provided with intake structure,to direct water into the penstocks.

Its function is to store temporarily the waterrejected by the plant when the electrical load isreduced and also to meet the instantaneousincreased demand of water due to suddenincreases in load.

74. (c)

75. (a)

Q1 =

1 cosQ

2

Q2 =

1 cosQ

2

= angle of jet from plate = 60°

1

2

QQ =

111 cos 2 311 cos 12

76. (c)

v

Fn= Force exerted by jet on plate perpendiculardirection to the plate.

IES M

ASTER

(10) RCC + OCF + Transport

Fn=2Q (0 – V sin ) AV sin

100 = 1000 × (0.05) (4) sin

sin = 12

30º .

inclination from vertical = 60°.

77. (d)

Ns = 5/4N P(H)

= 5/4

640 625 640 25 500.3216

(a) Pelton wheel for Ns = 10 to 35 for single jet

(b) Francis tubine Ns = 60 to 300

(c) Propeller turbine Ns = 300 to 600

(d) Kaplan turbine Ns = 600 to 1000.

78. (c)

For minimum starting speed

2 22 1

mu u

H2g

2 22 1

m

D N D N60 60 H

2g

4N2 – N2

2

210 2 10 60

N2 2

2200 60

3

N 60 200 600 2

3 3

79. (d)

Acceleration head (Ha) = 2A r cosag

l .

where

A = area of plunger

l = length of suction/delivery pipe

a = area of suction/delivery pipe

w = rotatioal speed of crank

r = radius of crank = L/2

L = Crank Length

80. (b)

m =

m

2 22

gH 10 100.41 1200V u V

60

2V = 100 12.5 m/s

8 .

81. (c)

Stay ring support the weight of non-rotatingparts of turbine.

Turbine guide vanes remains fixed throughoutthe operation after adjustment has been done.

Guide vane is airfoil shape so that hitting ofwater does not cause energy loss.

82. (b)

Specific speed Ns = 54

N P

H

N P = constant (Ns and H are same)

800 320 N 80

N = 800 4

= 1600 rpm

83. (b)

Pumps in parallel is used to increase the flowrate while head remains constant. Total flow rateis the sum of individual flow rates.

84. (c)

Cavitation factor = a v sH H HH

Ha = 9m, Hv = 1.2 m, Hs = ?, H = 45m

0.15 = s9 1.2 H45

Hs = 1.05 m

85. (d)

Reciprocating pumps are used for high head andlow discharge.

86. (a)

Ns = 3 34 4

N Q 1200 0.36 26.67H (81)

IES M

ASTER

[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-7) (11)

87. (a)

Plant use factor = Maximum demand

Station capacity

88. (b)

% slip = act

th

Q1 100

Q

3 = 3

th

18 101 100Q

18 × 10–3 = 0.97 Qth

18 × 10–3 = 0.97 · ALN60

N = 3

418 10 60 193.3 rpm

0.97 320 10 0.18

89. (c)

90. (c)

SSD = vt + 2v

2gf

2(.278 65).278 65 2.5 92.72m2 9.81 0.35

ISD = 2 × SSD

= 2 × 92.72

= 185.45 m

91. (c)

Too steep slope is undesirable because tendencyof most of the vehicle to travel along the centreline.

92. (b)

Length of transition curve is given by

LS = eNW

mins1L 150 10 75m

20

93. (d)

Extra widening = 2nl V

2R 9.5 R

= 24 6.1 65

2 225 9.5 225

= 0.787 m

94. (b)

Time mean speed = i i

i

q vq

= 1 3 3 7 4 11 7 14

1 3 4 7

= 11.06 m/s = 39.84 km/hr

95. (a)

Condition diagram is prepared to scale showingall the important physical condition of an accidentto be studied. While collision diagram showapproximate path of vehicles and pedestraininvolved in accident.

96. (d)

97. (b)

u = 12 – 0.6 k

Flow, q = ku

= 12 k – 0.6 k2

For maximum flow, dq 0 12 1.2 kdk

k = 10

qmax = 12 × 10 – 0.6 × 102 = 60 VPh

98. (c)

99. (b)

a 60 35 25 40

b 60 15 45 40 40

c = 40 – 40 = 0

d = 10 – 10 = 0

GI = 0.2a + 0.005ac + 0.01bd

= 0.2 × 25 + 0 + 0 = 5

100. (d)

101. (a)

As per IRC, the minimum width of median inrural areas is 5 m.

T-intersection is a type of intersection atgrade while cloverleef interchange is a typeof grade separated intersection.

102. (a) Design speed = 100 kmph

f = .7 × .5 = .35

IES M

ASTER

(12) RCC + OCF + Transport

2

tVSSD .278V

254 f .01n

2100SSD .278 100 2.5

254 .35 .02

SSD = 175.9 m

103. (a)

Flangeway clearence is meant for providing afree passage to wheel flanges.

104. (d)

105. (a) Diameter of standard plate for plate bearingtest = 75 cm.

We have K1 a1 = K2 a2

Where K = modulus of subgrade reaction

a = dia of the plate

6 × 75 = K2 × 30

K2 = 36 75 15 kg/cm30

106. (b)

Assuming the pavement to consist of single layerof base course material only, the pavementthickness is given by

2 1/32 s

bs b

E3p yT a2 E E

= 1/3

23 4100 1.5 0.9 100152 100 0.25 400

= 65.9 cm

107. (b)

The radius of curve

2VR125f

For supersonic transport minimum radius is 180m

Now

60 60R 221.5 m 180 m125 0.13

Hence R = 221.5 m

108. (b)

In the wind rose diagram radial line indicate winddirection and circle represent the duration of windto a certain scale.

The wind blowing direction is measured from TrueNorth

wind rose Type-I show direction and duration ofwind and Type-II shows direction, duration andintensity of wind.

109. (c)

Pavement thickness, t 15

1

C

15

2 1

1 2

t Ct C

t2 = 152520 18.2 cm

40

110. (b)

111. (c)

CSI = S 10H

20

where S = Strength index at 12% moisturecontent.

H = Hardness index at 12% moisture content.

112. (a)

Rail length = 13 m

No. of sleepers in 13 m = 13. + 9 = 19

No. of sleepers in 700 m = 19 70013

= 1023.1 1024

113. (a)

Steel sleepers requires less no. of fastenings.The cost of laying the track is less.

114. (a)

According to ICAO, basic runway length shouldbe increased at a rate of 7% per 300 m rise inelevation from MSL.

Corrected length = 1.07 × 1500 = 1605 m

115. (d)

As per ICAO recommendations,

Maximum longitudinalgradient

1.5% (A,B,C type)2% (D, E type)

116. (b)

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Centrifugal force, P = 2WV

gR

As other conditions P and R remains same

WV2 = Constant

1VW

117. (b)

118. (c)

As per IS : 2386

Flakiness index = 100 100 20%500

Elongation index = 60 100 15%

(500 100)

Total index = 20 + 15 = 35%

119. (a)

Stability of mix increases with increase in bitumencontent upto optimum value after that itdecreases with binder content.

120. (b)

s = 12 m, t = 3 sec, V = 0

From equation of motion

V2 = u2 + 2as

and V = u + at

We get, (at)2 = 2as

a = 22st

also a = fg

f = 22sgt =

2 12 0.2710 9

121. (b)

Elongation for Fe500 is less than Fe415 due tothe ductility, Fe500 is less ductile than the Fe415.

122. (d)

Higher the dimension speed ratio lower will bethe PCU.

123. (a)

As its common practice to design footing withoutshear reinforcement. Therefore, the minimumthickness (depth) of the footing base slab is mostoften dictated by the need to check shear stress.

124. (a)

In yield line method only plastic deformation areconsidered so entire deformation takes placealong yield line.

125. (c)

In case of ribbed bar, the bearing pressurebetween rib and the concrete is inclined to thebar axis. This introduces radial forces in concretei.e. wedging action causing circumferentialstresses in the concrete.

126. (a)

Upto collapse the stress-strain curve is not linearso that margin of safety will not equal to factorof safety.

127. (b)

Due to earthquake, stresses are induced inrandom direction, since helical column has nosharp edges, so stress concentration does notoccur, chances of opening up of ties is not thecase in helically reinforced column. Moreover,they provide better confinement to reinforcement.

128. (a)

There is a significant loss of energy duringhydraulic jump therefore energy equation is notfeasible to be applied. Rather momentum equationis used to solve for parameters.

129. (d)When yycthen Fr 1

o f2r

S sdydx 1 f

Hence at yyc the water surface meets criticaldepth line vertically.

130. (a)

yc

Ec Specific energy (E)

Dep

th o

f flo

w

As the curve is almost vertical near the critical

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depth, a slight change in energy would changedepth to a much greater alternate depth.

131. (a)

The minimum perimeter section will representhydraulically efficient section as it conveys themaximum discharge.

132. (c)

Since n (manning’s coefficient) is proportional to(Es)1/6. A large variation in the absolute roughnessmagnitude of a surface causes correspondinglya small change in the value of n.

133. (d)

d

y

o

y cos

A

2Ap d cos

= ycos ( y d cos )

134. (a)

Jet ratio (m) = Mean diameter of runnerLeast diameter of Jet

Smaller value of ‘m’ will mean smaller dia ofwheel. (The jet dia remaining constant) Thiswill result in close spacing of buckets withwhich the turbine efficiency decreases. If ‘m’becomes large, wheel will become bulk andefficiency will decrease.

For maximum efficiency m should be from11 to 14.

135. (c)

Motion of volute casing with vortex chamber is afree vortex i.e. Vr = constant .

Water moves away from center, velocity ofwhirl decreases thus building up pressure atthe cost of velocity.

136. (a)

P

Q

> 90º(Forward curvedvane)

= 90º

< 90º (Backward curvedvane)

For small increase in discharge more powerrequired in case of forward curved blade centrifugalpump.

137. (a)

Reciprocating pump should be primed in startingso that all air is expelled from the system. If anyair remains, it will create separation and hencehuge pressure wave will be created when this airis compressed. This pressure wave may lead tobursting of pipe or may cause damage to theequipment.

138. (c)

At each stroke, there is acceleration at thebeginning and retardation at the end.

139. (a)

140. (a)

141. (b)

142. (b)

raising outer edge with respect to centre line

s eeNL (W W )2

Pavement is rotated about inner edge

s eL eN (W W )

143. (d)

The speed at which the greatest proportion ofthe vehicles move is called modal speed.

144. (b)

145. (d)

The R value of material is determined when thematerial is in a state of saturation.

146. (b)

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147. (a)

148. (b)

Moorum are non plastic material.

149. (d)

Amount of camber also depends upon type ofpavement. The amount of friction between waterand pavement material is an important factor todecide the amount of camber.

150. (c)

Viscosity of VG-10 at 60°C is 1000 ± 200 poise.