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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
ASTER
[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
ASTER
[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
ASTER
[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
ASTER
[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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[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-7) (13)
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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(14) RCC + OCF + Transport
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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[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II (TEST-7) (15)
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.