ANSWERS - IES Master · According to IRC-94-1986 Binder content = 4.5 – 6% Grades = 30/40 to...

1. (c)

2. (c)

3. (b)

4. (c)

5. (c)

6. (d)

7. (b)

8. (c)

9. (b)

10. (d)

11. (c)

12. (a)

13. (c)

14. (a)

15. (a)

16. (d)

17. (a)

18. (b)

19. (c)

20. (a)

21. (a)

22. (b)

23. (a)

24. (d)

25. (a)

26. (b)

27. (a)

28. (b)

29. (b)

30. (b)

ESE-2018 PRELIMS TEST SERIESDate: 05 November, 2017

ANSWERS

31. (a)

32. (a)

33. (d)

34. (a)

35. (c)

36. (b)

37. (b)

38. (b)

39. (d)

40. (d)

41. (a)

42. (c)

43. (a)

44. (d)

45. (a)

46. (b)

47. (a)

48. (b)

49. (c)

50. (b)

51. (c)

52. (c)

53. (b)

54. (b)

55. (d)

56. (c)

57. (b)

58. (a)

59. (a)

60. (a)

61. (a)

62. (b)

63. (c)

64. (b)

65. (a)

66. (b)

67. (d)

68. (b)

69. (c)

70. (c)

71. (d)

72. (b)

73. (c)

74. (b)

75. (b)

76. (c)

77. (d)

78. (c)

79. (d)

80. (d)

81. (c)

82. (b)

83. (c)

84. (c)

85. (c)

86. (a)

87. (b)

88. (a)

89. (c)

90. (c)

91. (d)

92. (a)

93. (b)

94. (b)

95. (c)

96. (a)

97. (b)

98. (b)

99. (b)

100. (b)

101. (d)

102. (d)

103. (b)

104. (d)

105. (a)

106. (c)

107. (b)

108. (a)

109. (c)

110. (a)

111. (c)

112. (*)

113. (c)

114. (c)

115. (d)

116. (c)

117. (c)

118. (d)

119. (c)

120. (c)

121. (d)

122. (d)

123. (d)

124. (a)

125. (a)

126. (a)

127. (a)

128. (a)

129. (a)

130. (a)

131. (a)

132. (a)

133. (b)

134. (c)

135. (a)

136. (d)

137. (d)

138. (b)

139. (a)

140. (c)

141. (d)

142. (c)

143. (a)

144. (c)

145. (d)

146. (a)

147. (c)

148. (d)

149. (c)

150. (d)

IES M

ASTER

(2) (Test - 07)-05 November 2017

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1. (c)Waterways run from storage to the power houseand include draft tube and tail race which comeafter turbine.

2. (c)

If the average daily traffic count and thetraffic growth rate are ‘A’ and ‘r%’respectively, then the volume of traffic inthe nth year can be calculated as

Traffic(nth year) = 365A(1 + r)n

Therefore, the total traffic flow in ‘n’ years,can be obtained as the summation of theterms of a geometric progression series.Thus

Design traffic = n(1 r) 1365A

r

Load Safety Factor (LSF) is multiplied bythe total equivalent standard axle loadrepetition, to take care of the variations intraffic axle load measurements andunpredicted heavy truck load.

LSF is generally used in design of concretepavements to take into account thesudden nature of failure in concretepavements.

Vehicles don’t travel uniformly along thewidth of road. Majority of vehicles travelthrough middle of road than edges of road.Thus Lateral Distribution Factor (LDF)converts all the vehicle repetitions alongtheir respective traverse line to theequivalent repetition along the critical line.

3. (b)The code (Cl 22.6.2.1) recommends a sectionat distance d from the face of the support ascritical section for shear design.

Design shear force = 5100 0.52

= 200 kN4. (c)

We know,

Load factor of turbine = Average load

Peak load

=

30000 800002

80000

= 0.6875

5. (c)

Larger rebound deflection indicates weakerpavement structure.

Generally deflections are recorded after therainy season. If they are taken in dryseason, they should be multiplied by 2 forclayey soils, 1.2 to 1.3 for sandy soilsand interpolated for soils in between.

The amount of deflection depends upon anumber of factors such as

(i) Wheel load,

(ii) Pavement thickness, composition andcondition

(iii) Soil strength

(iv) Surface temperature

6. (d)Loss in stress due to slip at jacking is givenby

sEL

= 34 210 1025 1000

= 33.6 N/mm2

% loss =33.6 1001200

= 2.8%

7. (b)

1rv u1

11 wv v

v22fv

2wvu2

2rv

2 2w rv v cos u

1r 1v v u

F = 1 2w wQv Qv

P = 1 2w wQ(v v ) u

= 1 1Q(v (v u)cos u) u

= 1Q(v u)(1 cos ) u

P =2

1A(v u) (1 cos ) u

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Efficiency of jet = OuputInput

= 2

3

A uV u 1 cos1 AV2

= 2

32 uV u 1 cos

V

For maxima, ddu

= 0

2 2

3

2 V 4uV 3u 1 cos0

V

2 2V 4uV 3u = 0

v = u or 3u

V u because = 0 V 3u

8. (c)IRC suggest following formula for the design ofoverlay thickness equivalent to grannularmaterial of WBM layer

h0 = c10

a

D550 log mm;

D

h0 = thickness of WBM overlay in‘mm’

Dc = D

Da = 1 if projected design traffic is1500 to 4500.

Dc = 1.5 + 0.5 = 2 mm

Da = 1

h0 = 102550 log1

= 165 mm (WBM)

9. (b)10. (d)

We know,

Angular moment flux at outlet

=2w 2Q V R

= 1000 × 6 × 14 × 1

= 84 × 103 Nm

11. (c)12. (a)

Minimum vertical spacing = Max of

215 mm, Maximum aggregate size, 3

Maximum bar dia

= max 215, 16,123

= max (15, 10.67, 12)

= 15mm

13. (c)Shaft power + Mechanical loss = Runner Power

Overall efficiency = Mechanical efficiency ×Hydraulic efficiency × Volumetric efficiency.

14. (a)

According to IRC-94-1986

Binder content = 4.5 – 6%

Grades = 30/40 to 80/100

Maximum water absorption = 2%

15. (a)

P = cc g sc cc scA ( )A

= 40 × 400 + (300 – 40) × 4 × 20

= 36800 kg

16. (d)

Whirl velocity leads to energy losses at exit ofthe turbine and hence the flow is desired to beradial.

u

vvr

vw

> 90°

v opposite to u become negative

w

vw

17. (a)d1 = 0.278 Vbt = (0.278×60) × 2 = 33.33 m

d2 = VbT + 2aT

2;

S = 0.2Vb + 6 = 0.2 × (60) + 6 = 18 m

T =4S 4 18 10 seca 0.72

d2 = (60 × 0.278)(10) + 0.5(0.72) (10)2

= 202.8 m

d3 = 60 × 0.278 × 10 = 166.8 m

OSD = d1 + d2 = 236.13 m

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18. (b)

Rise (R)

Tread (T)

w

Sloping Length = 2 2R T

Total load on slope (Assuming unit width)

Corresponding load per metre square onhorizontal area

= 2 2w R T = 2 2w R TT =

2 2

2R Tw

T

19. (c)

20. (a)

Track drainage is the interception,collection and disposal of water from, uponor under the track. It is accomplished bysurface and sub-surface drainage system.

Proper drainage of sub-grade is very vitalas excess water reduces the bearingcapacity of the soil and as well as itsresistance to shear.

21. (a)

10kN 20kN

3m

2m 1m5m

yx

As in the case of combined footing, Toeliminate eccentricity,

C-G of load = C.G of footing

x + 2 = y +1

x – y = –1

x + y = 5 – 3 = 2

x = 0.5 y = 1.5

x : y = 1 : 3

22. (b)The turbine would be selected by specificspeed criteria.

Ns = 5/4N PH

= 5 410 940

16Here, Ns = 574.86 (> 300)

for Ns > 300, Kaplan turbine is suitable.

23. (a)24. (d)25. (a)

We know

f =PN60

where, P = number of pair of poles

50 =12 N

60

N = 250 rpm

26. (b)

(i) 23V 2 gS

V3 = 2 gS

= 2 0.35 10 28 14 m/s

(ii) mA×V2+mB×0 = (mA+mB)×V3

V2 =A B

3A

m m Vm

A A3 3

A

m 2m V 3Vm

V2 = 3 × 14 = 42 m/s

27. (a)

50cm

30cm

3m

10cm

l1l2

3m

Effective span of beam, (lo) = 6m

B f = ow f

lb 6d

6 < 1 2

wl lb

2

bw = breadth of web, df = depth of flange

=

6000 300 6 1006

< 3000 30003002

= 1000 + 300 + 600

= 1900 mm < 3300 mm

Hence,B f = 1.9m

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28. (b)We know,

u1 Vr1

V =1 Vw1

Vr2

30° V2

Vw2

u2

= 1V u 1 Kcos u

gH

V1 = 2gH

= 2 10 20

= 20 m/sec.

u = 7 m/sec.

= 150º and K = 1

= 320 7 1 7

210 20

= 0.849

= 84.9%

29. (b)On the basis of rate of introduction of superelevation

2 2

SV 2.7 80L 2.7 57.6 mR 300

30. (b)

The diameter of dowels shall not exceed thediameter of column bars by more than 3 mm.

31. (a)We know

3/2P H 1

2

PP =

3 21

3 22

H

H

P

900=

3/2

3/2169

P =64 900 2133.33kW27

32. (a)

Extra widening = Mechanical widening +Psychlogical widening

Etotal = Emech + Epsy

=2nl V

2R 9.5 R

=22 6 95 0.59 m

2 400 9.5 400

33. (d)34. (a)

Straight cylindrical will not convert kinetic headinto pressure head as inlet and outlet kinetichead will be nearly same.

35. (c)

Excessive filler at the surface causes flair-like cracker. Hair line cracker appear asshort and time cracker at close internalson the surface.

Streaking is characterised by theappearance of alternate lean and heavylines of bitumen either in longitudinal ortraverse direction.

Brittleness of the binder due to ageing ofbinder causes alligator cracks.

Alligator crack appears as interconnectedcracking forming a series of small blockwhich resemble the skin of an alligator.

36. (b)The grade of concrete used in prestressedstructures are high due to the fact that thisconcrete will provide high resistance to tension,shear, bond and bearing.

37. (b)Efficiency of a draft tube is given by

= 2 2

1 22

1

V V1001 K V

Here, K = 0,V1 = 6 m/s & V2 = 3 m/s

= 2 2

21 6 3 100

6

= 75%

38. (b)39. (d)

The minimum FOS against overturning is

2 Static load

1.5 Earthquake load

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40. (d)Due to negative slip, Cd > 1.

Negative slip occurs at high speeds.

Reciprocating pumps are a class of positive-displacement pump.

41. (a)42. (c)

w = 28Pe l

22.2 = 28 1500 e

8

e = 118.4 mm

General equations of parabolic profile for UDL

y = 2

4ex xll

= 4 0.1184 x 8 x64

= 7.4 × 10–3x xl

43. (a)We know,

3 5P

N D = Constant

3

PN

= Constant (in this case)

P =3800 10kW

1000

= 5.12 kW

44. (d)Indian railways are mostly using method (ii) atpresent for measuring rail stresses.

45. (a)

Minimum eccentricity, e = MaxD

500 30 20mm

l,

= Max3000 300500 30

20mm

,

= 20 mm

46. (b)

Break horse power is the power delivered tothe pump shaft.

Break power =mech mech

gHQ mgH

=37.5 10 20 W

0.75

= 10 kW

47. (a)P P

Deflectedshape

Distributionof load

Max. + veBM

(Sagging)Max. – ve

BM(Hogging)

48. (b)

Ec = ck5000 f1

10,000 = 5000 251

= 1.5

49. (c)

Power delivered = QH 10 1 25 = 250 kW

Efficiency =250 100%300

= 83.33%

50. (b)51. (c)

Ve = V + 1.6T

B

= 120 + 1.6 11.5

0.23

= 200 kN

Shear resisted by stirrups = Ve – Vc

= 200 – 0.48 × 230 × 400 × 10–3

= 155.84 kN

52. (c)For no cavitation

> critical

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Where =m

NPSHH =

300 3010

and critical =4/3

sN0.1031000

=4 3640000.103

1000

and c = 0.103 × 256 = 26.368

since, c30 26.368 , pump issuitable for use.

53. (b)The common practice has been to use paraboliccurve in summit curve.

54. (b)

Min. reinforcement = 0.8 BD100

= 20.8 450 600mm100

= 2160 mm2

Max. reinforcement 6% of BD

=6 450 600

100

= 16200 mm2

55. (d)

1

H

2

Between 1 & 2,

atmP =

22 2

fsP V H h

2g

For max. H, P2 PV

H 3610.3 0.3

2 10

= 10 –1.8 m

Hmax = 8.2 m

56. (c)n1 = –0.02, n2 = 0.03

Assume, L > S

L = 2 2NS 0.05 80

2(h S tan ) 2[0.7 80 tan(1 )]

= 76.2m 80m

Let L < S

L =2(h S tan )2S 76 m

N

57. (b)Ties are provided in order to prevent thepossibi l i ty of buckling of longitudinalreinforcement.

58. (a)If the flow properties such as depth of flow anddischarge remain constant along the length ofchannel then the flow is said to be uniform.

Flow

Velocity profiles

Velocity is zero at the boundaries and graduallyincreasses with the distance. Maximumvelocity of flow occurs at a certain depth belowthe flow surface, due to production onsecondary currents. Although the velocity mayvary across the depth as shown in the abovefigure.

59. (a)60. (a)

bd = design bond stress

= 1.50 N/mm2

Development length for deformed bars incompression

= y

bd

0.87f4 1.60 1.25

=bd

45 = 30 = 600 mm

61. (a)Lower limb has low depth (shallow depth) i.e.supercritical (fast).

y

y1

yc2

yc1

y2

EC2

E

q1 q >q2 1

y = 2/3y

1

2

E-y plot

EC2

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Upper limb has deep flow (high depth) andvelocity is slow (subcritical).

Curve shifts right as we increase the q(descharge per unit width)

q =QB

and vice-versa.

62. (b)

U =ba lnK

As we know,

q = KU = bKaln aK lnb lnKK

for max. volume, dq 0dK

1a(lnb lnK) aK 0K

ln b – ln K – 1 = 0

bln ln Ke

K = be (density at max. capacity)

qmax = b b e abalne b e

(max capacity)

63. (c)To enhance bond strength, large number ofsmaller bar diameters are used and terminationof longitudinal reinforcement in tension zone isavoided.

64. (b)With datum, z distance below the bottom ofthe channel.

Ay

B

C

z

Datum

AP

BP

CP

Piezometric head = pz z y

for point

A, B and C will have same piezometrichead = z + y

The irregularity of the section does notaffect the hydrostatic pressure distribution.

We can write energy equation as E = y +2v

2g

If datum is taken at bottom of the channel(z = 0).

Peizometric head = y

65. (a)

66. (b)Simply supported beam of span = 16m

spaneffective depth = 20 (up to span of 10m)

In case span>10m;

spaneffective depth

= 20 × 10 Modification factors

span

So,

spaneffective depth = 20 ×

1016

× 0.8 × 1.2 = 12

Effective depth = span12

= 16 m12

= 1.33 m

67. (d)In an ideal uniform flow channel.

If no obstacle is there then there won’t beany energy loss and any drag force fromthe bottom.

In sluice gate, there is a drag force actingon sluice gate but there is no energy loss.

In simple hydraulic jump, there is energyloss but no drag force because we assumesmooth bottom.

If there is some obstacle in hydraulic jump,there would be both energy loss and adrag force from the obstacle.

68. (b)

In the super-elevation formula, e = 2GV

127R, G

is the dynamic gauge of the track which isequal to the sum of the gauge and the widthof the rail head.

G = 1676 + 74 = 1750 mm for BG track.

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69. (c)Rectangular footing slab 2m × 3m

Central band width is equal to width of thefooting.

So,

Area of central band = 2m × 2m = 4m2

EndBand

CentralBand

EndBand B

L B2 L B

2

B

Reinforcement in central band widthTotal reinforcement in short direction

=

21

Where =Long side of footingShort side of footing =

32

=2 2 4

1.5 1 2.5 5

70. (c)

M = 2 2q y

gy 2

y

y = 0M

E =2Vy

2g

y

y = 0E

E = y

Asymptotes for M–y curve y = 0

Asymptotes for E–y curve E = y andy = 0

71. (d)

V = 4.4 R 70

= 4.4 470 70

= 88 km/h

72. (b)Unsupported length of the given column

=22750.65

= 3500mm

exmin = greater of 3500 450 ,20mm500 30

= 22 m

eymin = greater of 3500 390 ,20mm500 30

= 20 mm

min

min

x

y

ee =

2220

= 1.1

73. (c)

74. (b)

Grade compensation for a 4° curve = 0.04% ×4 = 0.16%

Steepest gradient = 1 in 250 = 0.4%

Ruling gradient = 0.16 + 0.4 = 0.56%

= 1 in 179 1 in 180

75. (b)As per IS 456 – 2000

Where an increased load on account the columnon the strength of helical reinforcement isallowed, the pitch of helical turns shall be

Pitch 75mm,core dia

6

whichever smaller

Pitch 25mm,3

whichever greater

Core Diameter = 440 – 2 × 40 = 360mm

So,

Range of allowed pitch

(25mm, 60mm)

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76. (c)d = CcY = 0.95 × 1 m = 0.95 m

7.0 = d + 2tV

2g

(7 – 0.95) × 20 = 2tv

Vt = 11 m/s

Vactual = CvVt = 10.78 m/s

Q = dVacutal

= 0.95 × 10.78 (m3/s)/m

= 10.241 (m3/s)/m

77. (d)

Q =

e P280W 1 1W 3W1L

Qmin will be corresponding to max P

e = 1 2e e2

e1 entry width

e2 exit width

Q =

7 0.9280 12 1 112 3 2979.2 PCU/hr12148

78. (c)Strength & structural properties of concretemay vary greatly depending upon its ingredients& method of manufacture.

79. (d)

Case (a) E = y + 2v

2g

= 2 +

252 10

= 3.25 m

yc =2E3

= 2 3.253

= 2.166 m

Case (b)

q = vy

= 5 × 2 = 10 (m3/s)/m

yc =

1 32qg = 101/3

= 2.154 m

80. (d)

V = a d0.27 (C C ) R

= 0.27 (135 100) 1000

= 130.8 km/h

81. (c)Concrete members are very rigid. Due togreater mass & stiffness, vibrations are seldoma problem.

Concrete has very less tensile strength.

Concrete has relatively low strength per unitweight or volume compared to steel.

82. (b)

Fr =vgy

=Q

By gy

= 3 2 1 2Q

B y g

3 2rF y

y2 = 1y3

r r2 1

r 1

F F100

F

= 3/2 3/2

2 13/2

1

y y 100y

= 3/21/ 3 1 100

= 419.61%

83. (c)Radius of contact area = a,

Tyre pressure = 2Wheel load

a

a =Wheel load

Tyre pressure

= 662.8 1000 m 0.2m

0.5 10

p = 0.5

Deflection = p a1.5

E

=6

61.5 0.5 10 20

20 10

= 0.75 cm

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84. (c)Final deflections due to all loads including long-term effects of L = span

Temp, creep & shrinkage = L

250

Final deflection including long-term effects oftemperature, creep & shrinkage that occur after

errection of partition L

350 or 20mm which

ever is lesser

85. (c)Maximum discharge will occur when flowconditions are critical

yc = c2E3

= 2 2.4m3

= 1.6 m

Vc = cgy

= 10 1.6

= 16 = 4 m/s

Qm = Maximum discharge

= AcVc = (BycVc)

= (2 × 1.6 × 4)

= 12.8 m3/s

86. (a)a = 20 cm, h = 15 cm

ah

= 2015 = 1.33 < (1.724)

b = 2 21.6a h –0.675 h

19.28 cm 20 cm

87. (b)

Design resistance Design load

nR nY

Rn characteristic resistance / strength

n characteristic loads

For ULM of design uncertainties in loads areconsidered but not for material strength.

Hence, = 1

Y > 1

Y 1

88. (a)A = y2.5

For critical flow,

Fr = 1

i.e.2

3Q TA g = 1

2

2QA g =

AT

...(1)

where, T =dAdy = 2.5 y1.5

Ec =2

c 2Qy

2gA

Ec = cAy2T

=

2.5

1.50.50.5

2 2.5 0.5

= 0.6m

89. (c)n = 3

L = 3×4 = 12 sec.

C =

1.5L 5 1.5 12 51 Y 1 (0.2 0.3 0.25)

= 92 sec.

90. (c)

400mm

200mm

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fcr = ck0.7 f

= 3.5 N/mm2

For cracking

Mcr = crg

fI

y

=

33.5 200 400 18.67kN-m200 12

91. (d)For a sharp crested weir,

Q = 3/2d eff

2 C L 2gH3

H = 0.5 m

Cd = 0.63

Leff = L – 0.1 nH

Leff = 1.6 – 0.1 × 2 × 0.5 = 1.5 m

Q = 3/22 0.63 1.5 2 9.81 0.53

Q = 0.986 m3/s

92. (a)G-IType of subgrade soil0-1 Good

2-4 Fair

5-9 Poor

10-20 Very poor

93. (b)Climiting = 0.36 fck b xu

Limiting depth of neutral axis =xu = kd = 0.48 × 400 = 192 mm

CL = 0.36 × 20 × 200 × 192

= 276.48 kN

94. (b)

RVF GVF RVF GVF RVF GVF

Hydraulic jump

95. (c)

96. (a)Fcompression = FTension

0.36 fck B xu, lim = 0.87 fy Ast

0.36×30×B×0.48×d = 0.87×415×Ast

stABd =

0.36 30 0.480.87 415

stA

%Bd = 1.43%

97. (b)To accomodate a proper GVF profile ahydraullic jump followed by an S1 curve willoccur in the upstream.

98. (b)Coefficient of hardness

=loss in weight in gm20

3

=2120 133

99. (b)(i) Over reinforced section shows brittle failure

while under reinforced section showsexcessive deflection & cracking givingwarning before failure.

(ii) Compression controlled section meansstrain in tension steel is less than 0.002means over reinforced failure by crushingof concrete hence doubly reinforcedsection is to be adopted.

(iii) Flexural members are subjected to bendingcausing tension in concrete in one sidehence tension controlled sections areadopted.

100. (b)The steady-state momentum equation takes asimple form if the channel is assumed to behorizontal and friction less.

Specific force = 1 1 2 2P M P M

Hence, specific force represents the sum ofthe pressure force and momentum flux perunit weight of the fluid at a section.

101. (d)102. (d)

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Stirrups limit the opening of diagonal crackswithin the elastic range.

103. (b)EA = EB

2A

AVy2g

=2B

BVz y2g

As yA = Bz y

(to get water surface elevationunchanged)

VA = VB

AA = AB

3.5 × yB = 5 × 2

yB = 2.857 m

1y z = y2

z = 0.857 m

= z +ve means rise in bedelevation

104. (d)Airport reference temperature

= 125 40 25 30 C3

105. (a)(i) Bent-up bars are widely spaced & few in

number, hence a crack may not beintercepted by more than one bar.

(ii) Some of the bars are bent up, thenremaining bars are subjected to higherstresses.

(iii) Concrete at bends may be subjected tosplitting force resulting in web cracking.

(iv) They do not confine the concrete in shearregion.

106. (c)22q

g = y1 y2 (y1 + y2)

q2 = 10 250.5 2.0 2.52 2

q =5

1.414

Q = 35qB 1.2 4.24m /s1.414

jE =

3 32 1

1 2

y y 1.5 0.84375m4 y y 4 0.5 2.0

107. (b)108. (a)

If v (nominal shear stress) < 0.5 c (shearstrength of concrete)

then, no shear resisforcement provided

If c v c0.5 , min shear reinforcement isprovided.

109. (c)y1 downstream depth, y2 upstreamdepth

1

2

yy = 2

2

11 1 8Fr2

22Fr =

2 2

2

V 607.746 15gy 42

1y0.4

= 11 1 8 15

2

1y 0.4 5 2m

So, downstream depth is 2.0 m

110. (a)Heading and benching method is used whenthe tunnel section is very large and quality ofrock is not very satisfactory i.e., the rock isunstable.

In self supporting rocks, the top headingadvance are round ahead of the bottomheading. In unstable. (broken or non selfsupporting rocks) the bench provides plateformfor timber supports to the heading.

111. (c)112. (*)

3cy =

2

1 2 1 2q 1 1y y 0.5 4 4.5y y 0.5 4g 2 2

q = 34.5 10 6.71 / mm / s

yc =1 32q 1.65m

g

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Lj = 2 16.9 6.9 24.15my y 4 0.5

Ej =

3 32 1

1 2

y y 3.5 5.36m4y y 8

113. (c)114. (c)

Seismic intensity Zone factorLow 0.1

Moderate 0.16

Severe 0.24

Very severe 0.36

115. (d)For most economical trapezoidal channel,

R =y2

,

A = 23 y

Q = 2/3 1/21SA Rn

17.32 =

2/3

1/221 y1.732y 0.0016

0.016 2

0.160.04

=

8/3

2/3

y2

22(2)2/3 = y8/3

y 2m

116. (c)Height of wave as given by Stevenson

h = 1.5 F

where F is in km and h is in m.

h = 1.5 576 36 m

117. (c)C = Celerity = wave speed

= 21 2

1

yg y yy2

Given: y1 = 1.5 m

y = 1.5 m

y2 = 1y y 3.0m

C = 10 3 453 1.52 1.5

= 6.71 m/s

118. (d)Min. superelevation required = cant deficiency– S.e. actually provided

10 = 75 – S.e. actually provided

Superelevation actually provided = 65 mm.

119. (c)yn = 1.20 m

q = 3 39.6 m / s 6.4 / mm / s1.50

yc =

1/3 1/32 2q 6.4 1.60m

g 10

As, yc > yn

So, channel slope is steep.

120. (c)121. (d)

If the bucket is kept too small, the jet is notdeflected smoothly due to turbulence.

122. (d)In reaction turbines, while moving through therunner vane both KE and pressure decreasesand since flow inside the reaction turbine isunder pressure only a part of energy is availableas KE at inlet and remaining as pressure.

123. (d)

Kaplan turbine works at higher specific speedthan modern Francis turbine.

124. (a)125. (a)

The effective length is reduced, which reducesthe inertia head and acceleration head.

126. (a)Open graded bituminous mixes have minimalamounts of fine aggregate (sand) and thereforeare highly permeable to water.

127. (a)

128. (a)

129. (a)

130. (a)

131. (a)

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Modelling driver-driver interactions entailsmodelling how a driver reacts to the actionstaken by another driver. On the other hand,modelling driver-road interaction entailsmodelling how a driver reacts to various featuresas the road such as narrow road width, curvesand so.

132. (a)

133. (b)

134. (c)Design of tie bars is based on the stressesdue to friction.

135. (a)136. (d)

Mean velocityShear veolocity =

vV *

= 8f

137. (d)The choking and backing up is independent ofenergy dissipation and would occur even if thewalls of contraction were streamlined so as toeliminate the energy loss.

138. (b)

A transition may be definied as a changeeither in direction, slope or cross sectionof the channel that produces a change inthe state of the flow.

All controls are transitions, although notall transitions are control.

139. (a)From experimental results it was assured thatdune formation is accompained by a sorting ofthe sediment into finer material on the dunecrests and coarser material in the trough andin fact that on this account dunes are morelikely to be stable in non-uniform than inuniform sand.

140. (c)Specific energy is not constant as there issome energy loss during jump.

141. (d)

Development length Ld = st

bd

0.874

For mild steel For HYSD bar

Ld = bd

250 .874

Ld =

bd

.87 4154 1.6

Ld = 54.375bd

Ld =

bd56.41

142. (c)

Increase in w/c ratio, increases permeability.Efficient compaction can be achieved byvarious procedures such as vibration.

143. (a)144. (c)

Plasticizers reduces the water demand forsame degree of workability.

Strength of concrete increases with reducedwater content.

145. (d)With higher height to width ratio, the effect oflateral restraints decreases, resulting in loosingstrength at middle height which is relativelyweaker section.

146. (a)

Slab

C

TBeam

Monolithic action will resist buckling

147. (c)Providing compression reinforcement willincrease moment resisting capacity incompression hence it may change mode offailure of over reinforced section.

If pt > Ptlim section is over reinforced section& by providing compression reinforcementductile failure may occur hence it may makethe section from over reinforced to underreinforced which may change mode of failure.

148. (d)

max =

D/2

20

3

V y.dAV BD V1.5

BDI.B 8 BDB12

avg. =V

BD max avg.1.5

D

B

NmaxA

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149. (c)Circular columns are better at resisting loadsas compared to rectangular/square columns.Their usage is less because of difficulty inerection.

150. (d)It is possible to reduce the eccentricity bylaterally shifting the footing base relative to thecolumn when eccentricity magnitude is knownand its direction is fixed.

ANSWERS - IES Master · According to IRC-94-1986 Binder content = 4.5 – 6% Grades = 30/40 to...

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Transcript of ANSWERS - IES Master · According to IRC-94-1986 Binder content = 4.5 – 6% Grades = 30/40 to...