ANSWERS - IES Master - Coaching for ESE, GATE and PSUs …€¦ · · 2017-10-01m3 0.0167 60 60 s...

1. (a)

2. (b)

3. (a)

4. (b)

5. (b)

6. (c)

7. (c)

8. (b)

9. (d)

10. (d)

11. (d)

12. (a)

13. (d)

14. (d)

15. (c)

16. (a)

17. (c)

18. (c)

19. (d)

20. (b)

21. (c)

22. (b)

23. (a)

24. (d)

25. (a)

26. (b)

27. (d)

28. (b)

29. (c)

30. (c)

ESE-2018 PRELIMS TEST SERIESDate: 01 October, 2017

ANSWERS

31. (a)

32. (d)

33. (b)

34. (a)

35. (c)

36. (c)

37. (c)

38. (c)

39. (d)

40. (d)

41. (d)

42. (c)

43. (c)

44. (d)

45. (c)

46. (d)

47. (c)

48. (c)

49. (a)

50. (b)

51. (c)

52. (a)

53. (c)

54. (d)

55. (b)

56. (c)

57. (b)

58. (d)

59. (c)

60. (d)

61. (d)

62. (c)

63. (a)

64. (b)

65. (c)

66. (b)

67. (c)

68. (a)

69. (d)

70. (d)

71. (d)

72. (c)

73. (a)

74. (d)

75. (d)

76. (b)

77. (a)

78. (a)

79. (c)

80. (d)

81. (b)

82. (c)

83. (d)

84. (d)

85. (b)

86. (c)

87. (a)

88. (a)

89. (b)

90. (c)

91. (d)

92. (d)

93. (b)

94. (b)

95. (c)

96. (d)

97. (a)

98. (a)

99. (a)

100. (d)

101. (a)

102. (d)

103. (b)

104. (a)

105. (d)

106. (b)

107. (b)

108. (b)

109. (d)

110. (c)

111. (b)

112. (a)

113. (b)

114. (a)

115. (a)

116. (a)

117. (c)

118. (d)

119. (a)

120. (c)

121. (d)

122. (a)

123. (a)

124. (b)

125. (c)

126. (c)

127. (b)

128. (b)

129. (b)

130. (d)

131. (c)

132. (a)

133. (d)

134. (a)

135. (c)

136. (a)

137. (d)

138. (a)

139. (d)

140. (a)

141. (d)

142. (c)

143. (b)

144. (a)

145. (d)

146. (c)

147. (c)

148. (d)

149. (c)

150. (a)

IES M

ASTER

(2) (Test - 02)-01 October 2017

Mobile : E-mail: 8010009955, 9711853908 [email protected], [email protected]. office : Phone : F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 011-26522064

Web : iesmasterpublications.com, iesmaster.org

1. (a)2. (b)

To increase the comfort, time period ofoscillation must be increase.

T =I2

wGM

Action I will increase moment of inertia andaction.

II will increase metacentric height.

Hence option (b) is correct.

3. (a)

Exit gradient=8

d = 5 m

H = 2.5 m

Exit Gradient (GE) = H 1d

= 21 1

2

= b/d

18

= 2.5 15

= 1.621

2b1 1d

2

= 1.621

b = 10.034 m

4. (b)5. (b)

Neglecting the weight of wire,

F = L

Since there is resistance on the inside andoutside of the ring

F = (2 d)

= 2 × 0.075 × × 0.10 N

F = 0.0470 N6. (c)

Normal scour depth = 1/32q1.35

f

q =QL

L = length of wire = 200 m

Q = 35000m / s

q =5000200

= 25

f = mm1.76 d

f = 1.76 0.4 = 1.111

Normal scour depth = 11.137 m.

7. (c)Tracer should not be decomposed in streamwater and also not retained or absorbed bysediments, plants or organisms.

8. (b)Liquid with high vapour pressure evaporatesreadily. Mercury has very low vapour pressure.Hence it can be used in pressure measuringequipments so that it can measure even lowpressure without evaporating.

9. (d)

A = 0.06 hectare = 40.06 10 = 2600m

Q =3m0.0167 60 60

s

= 360.12m / hr = 360m / hr

Time (t) = 50 min = 0.833 hrInfiltration capacity (f) = 5 cm/hr = 0.05 m/hr

t = 2.303 10Y Qlogf Q fA

0.833 = 10Y 602.303 log

0.05 60 0.05 600

0.833 = 10Y 602.303 log

0.05 30

Y = 0.06 mY = 6 cm

10. (d)Sunken pan : Sunken pan has advantagethat certain boundary effects such as directsolar radiation on side walls and the heatexchange through side walls are negligible.Disadvantages of this pan are : (i) it tends tocollect trash and difficult to maintain; (ii) difficultto install this pan.

Floating pan : The use of these pans are notvery common. There are many operationaldifficulties associated with this type of pan.e.g., it is not easily accessible and there maybe splashing of water.

Surface pan : The most common type ofevaporation pan. The advantages of this panare

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(i) Stable pan coefficient

(ii) Easy access for observation

(iii) More stability compared to floating pan

(iv) Relative freedom from direct and trash

11. (d)(H – 0.15) × 9.81 = (0.20) × 13.6 × 9.81

(H – 0.15) = 13.6 × 0.20

(H – 0.15) = 2.72 m

H = 2.87 m

12. (a)For d > 6 m

According to Shield

c

0.056

= sd(S 1) (for d > 6 mm)

39.81KN / m

c = 33

109.81KN / m 0.056 (2.7 1)10

c = 3 29.339 10 KN / m = 29.339N / m

c = 29.34N / m

13. (d)14. (d)

If the fluid is in motion, there must be tangentialforce along with the normal force. So, the totalforce (resultant) will not be normal to thesurface.

F

W

Rm

Area = AArea = a

W =F Aa

15. (c)Bligh assumed that :

(i) The hydraulic slope or gradient is constantthroughout the impervious length of theapron.

(ii) The percolating water to creep along thecontact of the base profile of the apronwith the sub-soil, losing head enroute,proportional to the length of its travel.

16. (a)

S = 5/3

1/6f

3340Q

Q = 3230400m / hr

Q = 364m / sec

S =5/3

1/6(1)

3340 (64)

S =1

6680

S = 41.497 10

17. (c)18. (c)

Let W = width of block and L = length of block

Fb = W

water water(aLW) 13.6 (bLW)

= w8 a b (LW)

a + 13.6b = 8a + 8b

ab

= 0.80

19. (d)Assume capacity of pump = Q m3/hrTotal amount of water required

= 4 2 103 10 m m100

= 33000m

Water given in 15 days = 3(15 20) Qm where Q is in m3/hr

300Q = 3000

Q = 310m / hr

= 80%

Q =100.8

= 312.5m / hr

20. (b)21. (c)

F = Q (V u)(1 K cos )

= deflection angle = 180°

K = 1 (no friction loss)Q = AVr

Vr = 50Q = 0.2 × 50 = 10 m3/secF = 1000 × 10 × (75 – 25)(1 – cos(180))1 = 1 × 103 kN

22. (b)Total Creep length (L) = 5 5 30 8 8 =56

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Hydraulic gradient = LhL

556

So, head loss upto C = Lh 30L

2.678 m

So, head at Point C = 5 – 2.678 = 2.322 m

23. (a)24. (d)

For triangular weir,

Q tan2

dQQ =

d

2 sin cos2 2

dQ d Where d is in radian.Q sin

dQQ =

2º180º 100%

sin 90º

dQQ = 3.49% 3.50%.

25. (a)

Creep length = 30(6 2) (10 2)3

= 42m

Head causing seepage flow = 10 m

Hydraulic gradient = LH 10L 42

So, uplift pressure at point C

= 12 10 15 1010

42 3 42

5.952 m

26. (b)Saturation vapour pressure = 42.43 mbar = es

h = 80%

Actual vapour pressure

e = 0.8es = 0.8 × 42.43 mbar = 33.944 mbar

Saturation deficit = (es – e)

= (42.43 – 33.544)mbar = 8.486 mbar27. (d)

mrF = prF Fr =

VgL

Vm = VPm

P

LL

= 40 × 2.5160

=408 = 5 Kmph

mrF =

5 0.27710 2.5

= 0.277.

28. (b)

P = 4.75 Q

Q = 3(m / s) = 14400 43600

P = 4.75 4 = 9.5m

29. (c)For calculating wind velocity, the power lawprofile is usually expressed as

0

vv =

0

zz

,

for most of the meteorological forecast purpose

=1 0.157

v =

0.15

00

zv

z

Where, v0 is the observed velocity at height z0and v is the velocity at any other height z

30. (c)

FD =2

D1 C AV2

Here, FD and A and CD is same.

So, V .

Hence, V3 =0

3

× V0

=1.20.96

× 19.67

= 22 m/s

31. (a)

B =c

HS c

B =803 1

802

= 40 2

56.56 m

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32. (d)Total volume of flow

= (7 + 27 + 58 + 41 + 31 + 20 + 13) = 197

= 197 × 86400 m3

= 17020800 m3 = 1702.08 ha-m

33. (b)

Power =3

dAV C2

If power, , A remains constant, then

CdV3 = constant

1

3D 1C V =

2

3D 2C V

2

1

VV =

1/3 1/3100 475 3

V2 = 1.10 V1

Increase in speed = 1 1

1

1.10V V100%

V

= 10%

34. (a)

f = mm1.76 d

mmd = 1 mm

f = 1.76

R = 25 V

2 f=

2.5 0.82 1.76 = 0.91

35. (c)

Area of pan = 1224 = 11689.87 cm2

Volume of water removed = 4.75 litres = 4750cm3

Depth of water removed = 4750 cm

11689.87= 0.406 cm = 4.06 mm

Evaporation = rainfall – depth of water removed

= (8.8 – 4.06)mm = 4.74 mm

36. (c)

AN

2 m

3 m

6 m

B

a = 4.905 m/sx2

There will be higher pressure at the rearer sideas compared to front due to acceleration.

tan = xa 1g 2

AN = 3 tan 1.5 m, BN 2 1.5 3.5 m

Height of water level in the tube = 3.50 m

37. (c)

Utilisation factor = Max.powerutilized

Max.power available

=3500044000

= 0.795

i.e. = 79.5%

38. (c)Total rainfall = Intensity of rainfall × duration

= 1.5 × 6 = 9 cm

Volume of runoff = 21.6 × 106m3

Area of basin = 300km2 = 300 × 106m2

Depth of runoff = volume of runoffarea of the basin

=

6

621.6 10300 10 = 0.072 m = 7.2 cm

Total infiltration = (9 – 7.2) cm = 1.8 cm

Average infiltration rate = 1.86

= 0.3 cm/hr = 3 mm/hr

39. (d)

H = 2

5f(L) (Q)12.1(d) for Ist Pipe

For 2nd pipe;

Head loss = 2

5f(2L) (2Q)12.1(d) =

2

5fLQ8

12.1d

fh 8H

40. (d)

Discharge at canal head = 312m / s

Loss of water in transit = 25%

Discharge at field = 12 × 0.75 = 39m / s

Duty at the field = 1250 hectare cumec

Area irrigated = 1250 × 9 = 11250 hectare41. (d)

v = dhkdx

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=(206.25 210.5)12.5m / day

350

= 0.1518 m/day

va = vn

= 0.1518 m / day

0.15 = 1.012 m/day

The actual velocity in the aquifer

= 1.012 m/day

42. (c)If x0 = Length of the pipe having peak pressurein a closure time

T < 2LC

, then

x0 =CTL2

2L3

=CTL2

CT2

=L3

T 12L 3C

43. (c)

Kor-watering is the first watering after plantshave grown a few centimeters high.

44. (d)

Qs = 2.778 A/D = 33602.778 m /s2

A = area of basin in km2

D = Duration of UH

= 500.04 m3/s = 500 m3/s

45. (c)NR = Reynonlds number

= 4VL 1000 V D

10 10

For laminar flow, Assume NR < 2000.

2000 = 3

410 V 0.15

10 10

V = 0.0133 m/s

= 1.33 cm/s46. (d)

Q = 5 m3/s

Area = 20 ha

Time = 5 hrs

Water stored in root zone = 0.4 m

app = Water storedinRoot Zone 100

Water suppliedtofield

= 4

0.4 1005 5 3600

20 10

= 88.89%

47. (c)

Qp

6hr

35hr

Total rainfall = p1 35 3600 Q2

Qp in m3/s

1 cm rainfall excess all over area of 378 km2

So, p1 35 3600 Q2 = 6 1378 10

100Qp = 60 m3/s

For 10 cm of rainfall,

Peak of DRH = 10 × 60 = 600 m3/s

48. (c)

Re = VD VD

250000 = 5V 0.12

1.5 10

V = 31.25 m/s

Velocity of air flow for transition is 31.25 m/s.

49. (a)

W.C. before irrigation = 153 138 100

138

= 10.9% by weight

Depth of water required to be applied to bringthe moisture upto its field capacity.

=18.3 10.91.25 1.2100 100

= 0.111 m

= 111 mm

50. (b)Compactness co-efficient

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= Perimeter of the basin

Circumference of a circle whose areais equal to the area of the basin

Let re be the radius of the equivalent circle.

A = 2er

Circumference = eA2 r 2 2 A

Compactness co-efficient =

P2 A

=

177.242 1936

=

177.242 1.7724 44

=

1002 44

= 5044

= 2522

= 1.136

51. (c)Characteristic length = Hydraulic diameter

= w

4AP =

2 20 i

0 i

4( r r )2 r 2 r

= 0 i2 (r r )

52. (a)

Tractive force c = w S0.056 d(S 1) (Taking

SS = 2.65) = w w1 1d d

10.8 11

c = tractive force = wd11

0 = wRS (force on the bed due to steadyflowing water)

For no movement of sediment

0 = c

wRS < wd11

R <d

11S

R <

38 10111

4000

R < 2.91

R = 2.91

53. (c)

FP

T = Basin time width

tp = Lag time = 6hr

Basin time width = 72 + 3tP= (72 + 3 × 6)hr = 90hr

54. (d)

hm = 2kV

2g , where V is velocity of flow in

smaller pipe

K = 22 2

12

2

A (1) 91 1A 16(2)

Head loss = 29 4

16 2 10

= 0.45 m

55. (b)56. (c)

Transmissivity, T = kD

Equivalent k in x-direction,

keq = kDD

Hence, Teq = kD

keq(D1 + D2 + D3) = kD DD

= kD

= (2 × 30 + 1 × 10 + 3 × 5 + 6 × 10 + 5 × 20)

= 245 m2/day

57. (b)h = distance between plates

L = length between head loss

Vavg =2ph

12 L

P = avg2

12 L V

h

= 212 (0.25) (100) (3)

(0.05)

P 360 kPa pressure drop

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hf = head loss = P 360000 40 mg 900 10

58. (d)(i) Kennedy’s theory considered a trapezoidal

section and Lacey’s theory considered asemi-elliptical section.

(ii) Kennedy’s theory did not give equationfor bed slope whereas Lacey’s did.

(iii) Kennedy’s theory is applicable only foralluvial channel and lacey’s theory isapplicable for alluvial channel and riverboth.

59. (c)Volume of water that can be extracted by theforce of gravity from a unit volume of aquifermaterial is known as specific yield (Sy).

Sy = Volume of water extracted or lost

Area fall in water table

0.25 =

w6

v200 10 0.25

vw = 12.5 × 106 m3 = 12.5 million cubic meter

60. (d)61. (d)62. (c)

True runoff is stream flow in its naturalcondition, i.e. without human intervention. Sucha stream flow unaffected by works of man,such as reservoirs and diversion structures ona stream is called natural flow or virgin flow.

63. (a)Graph 1, Newtowon fluid, viscosity is constantand it doesn’t depend on deformation.

Graph 2, Should be Pseudo plastic because.It is non-newtonian fluid and it is shear thinningfluids so, apparent viscosity decreases withincrease in deformation rate

Graph 3, Should be dilatant since it is shearthickening fluid so, apparent viscosity increaseswith increases in deformation rate.

64. (b)Decreasing order of keeping velocity and bedslope of river in various stages

Rocky stage > Boulder stage > Trough stage> Delta stage

In delta stage, the bed slope and velocity arereduced so much that it is unable to carry itssediment load. It drops down the sediment

and gets divided into channels on either sideof the deposit resulting in the formation of adelta.

65. (c)

8h 12h 20h

Q

30 m /s3

Q =

(30 0)0 (8 0)(12 0)

Q = 20 m3/s

For effective rainfall of 2.5 cm

Qe = 20 × 2.5 = 50 m3/s

66. (b)n = Flow behaviour index

k = consistency index.

67. (c)

Undersluices are opening which are fullycontrolled by gates and these are located onthe same side of the offtaking canal.

Undersluices scour the silt deposited on theriver bed in pocket upstream of the canal headregulator.

68. (a)A front is the interface between two distinct airmasses. Under certain favourable conditionswhen a warm air mass and cold air massmeet, the warmer air mass is lifted over thecolder one with the formation of a front.

In convective precipitation, a packet of air whichis warmer than the surrounding air due tolocalised heating rises because of its lesserdensity. Air from cooler surroundings flows totake up its place thus setting up a convectivecell. The warm air continues to rise, undergoescooling and result in precipitation.

69. (d)

(i) S-1 va (x, y, z, t) ( .V)V

dt

(ii) S – 2 Zero for steady flow

(iii) S–4 2nd term accounts for the effectof the fluid particles moving to a newlocation in the flow field.

70. (d)A groyne which damps the velocity of flow andthus causes depositon of sediment carried by

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the river without repelling or deflecting the flow,is called sedimenting groyne. Permeablegroynes are usually classified as sedimentinggroynes. So from the above dissussion it isclear that the sedimenting groynes are bestsuited for rivers carrying considerable amountof sediment in suspension.

71. (d)For setting a raingauge the fol lowingconsiderations are important:

1. The ground must be level and in the openand the instrument must present ahorizontal catch surface.

2. The gauge must be set as near the groundas possible to reduce wind effects but itmust be sufficiently high to preventsplashing, flooding etc.

3. The instrument must be surrounded byan open fenced area of atleast 5.5 m ×5.5 m. No object should be nearer to theinstrument than 30 m or twice the heightof the obstruction.

72. (c)73. (a)

Bed level(Canal)

1.5 m

103.5

105.0

Bed level(Drainage)

2.5 m

100.5

102.5

Canal Bed Level > Drainage Bed level

So, either aqueduct or syphon aqueduct hasbeen provided since canal is passing over thedrain.

Now, (102.5)

HFLof drainage < (105.0)

FSLof canal

So, aqueduct will be the correct options,hence, option (a) will be correct.

74. (d)In tipping-bucket type, the water from the tippedbucket is collected in a storage can. Thetipping actuates an electrically driven pen totrace a record on clockwork-driven chart. Therecord from the tipping bucket gives data onthe intensity of rainfall. Further, the instrument

is ideally suited for digitizing of the outputsignal.

75. (d)S(i) friction forces are neglected

S(ii) correct

S(iii) Correct

S(iv) 2

stagnationVP P2

76. (b)77. (a)

In the precipitation reaching the surface of acatchment, the major abstraction is frominfiltration process. However, two otherprocesses, though small in magnitude, operateto reduce the water volume available for runoffand thus act as abstractions. These are (i) theinterception process, and (ii) the depressionstorage and together they are called the initialloss.

78. (a)

S-1 actual K.E

average K.E

S-2 It is dimension less

S-3 more significant in case of laminar flow ascompared to turbulent flow.

79. (c)For improving duty of water, rotation of cropsmust be practised.

80. (d)In small streams of shallow depth, the currentmeter is held at the requisite depth below thesurface in a vertical by an observer who standsin the water. The arrangement called wading isquite fast but is applicable only to smallstreams.

81. (b)(i) Dimension is a measure of physical

quantity without physical value

(ii) Seven primary dimensions [Mass, length,time, temp, Electric current, Amount ofintensity, amount of matter]

(iii) correct

(iv) Based on principle of homogeneity

82. (c)Montague type fall was formed to eliminate allthe effects of glacis type fall in which hydraulic

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jump was the basic phenomenon for thedissipation of energy. But with these glacistype of fall there was serious trouble whichwas even after the formation of hydraulic jumpthere was considerable surplus energy in water.Another cause of trouble was due to too rapidexpansion after fluming, eddies were developedwhich caused deep scours. So after furtherresearch the straight glacis of glacis type fallswere replaced by parabolic glacis commonlyknown as montague profile.

83. (d)Direct runoff is that part of runoff which entersthe stream immediately after the rainfall. Itincludes surface runoff, throughflow and rainfallon the surface of the stream. In the case ofsnowmelt, the resulting flow entering the streamis also a direct runoff.

A part of the precipitation that infilterates andmoves laterally through upper crusts of thesoil and returns to the surface at some locationaway from point of entry into the soil. Thiscomponent of runoff is known variously asinterf low, throughflow, storm seepage,subsurface storm flow or quick return flow.

84. (d)(i) Cavitation number

= Local absolute pressure Vapour pressure

Inertial pressure

(ii) Lift coefficient = Lift force

Dynamic force

(iii) Froude number = Inertial force

Gravitational force

(iv) Reynolds number = Inertial forceViscous force

(v) Strouhal number

= Characteristic flow time

Period of oscillation

(vi) Momentum correction factor

= Actual momentum

Average momentum

85. (b)In Border flooding, borders are long, uniformlygraded strips of land seperated by earth bunds.These bunds are to guide the flow of waterdown the field.

Basin flooding is mainly adopted for orchardtrees.

Sprinkler irrigation method is best suited forvery light soils as deep percolation losses areavoided.

In drip irrigation method, water & fertilizer isslowly & directly applied to the root zone ofthe plants in order to minimise the losses dueto evaporation & percolation.

86. (c)87. (a)

S-1 -ve derivative w.r.t to any direction givesvelocity in that direction

S-2 = eixst for ideal and irrotational flow

88. (a)dw = depth of available water/m

d = depth of soil (1 m)

F = Field capacity

yd = dry unit weight of soil

w = unit weight of water

dw\ =d

w

F.d.y

F =w w

d

dd y =

0.35 9.81 14

= 24.5%

89. (b)

-index =P Rt

P = Precipitation

R = run off

P = 0.16 m

R =

8

61.2 10

1200 10 = 0.1

-index = 0.16 0.1

6 = 0.01 m/h = 1 cm/h

90. (c)

S-2 Stream function exist for both rotationaland irrotational flow

S-2 Velocity potential exist only for irrotationalflow

S-3 Correct

S-4 For irrotational incompressible flow

u = x y

... (i)

V = y x

... (ii)

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91. (d)

Depth of water needed = 16

0.8 (1 0.2)

= 25 cm

92. (d)

Risk = n11 1

T

Risk = 10

100 = 0.1, n = 1 [Any year]

0.1 = 111 1

T

T = 10 years

93. (b)

-ve and decreasing slope

+ve and decreasing slope

Temperature

Visc

osity

gasesLiquids

94. (b)

L.R. = ci

cd

E 1.5 100E 20

= 7.5%

95. (c)Gumbel’s distribution has the property whichgives T = 2.33 years for the average of theannual series when N is very large. Thus thevalue of flood with T = 2.33 years is calledmean annual flood.

96. (d)

97. (a)98. (a)

P = 2d

= 0.073 N/cm = 0.073 × 102 N/m

d = 0.1 mm = 0.1 × 10–3 m

P = 2

23

2 0.073 10 146 kN/m0.1 10

So, absolute pressure inside = 146 + 101.3 =247.3 kN/m2

99. (a)100. (d)

The contribution of the runoff would be asfollows:

2nd hour : 2 cm/h rain on A

4th hour : 2 cm/h rain on A and B

5th hour : 2 cm/h rain on A, B and half on C6h

4h

2h

C

B

A

20 ha

50 ha

100 ha

A = 100 ha

B = 50 ha

C = 20 ha

Q = C i A

Q = 0.4 × 22 10

60 60

420 10100 502

Q = 3.55 m3/s

101. (a)

PA

h

5 m

4–h

h

5 m

PB

For same pressure, height of water at bothpoints should be same

h5 =

4 h5

h = 2

tan = x

eff

a a 2g 9.81 5

a = 22 9.81 3.924 3.93 m/s5

102. (d)Raising the water level in the river is the mainfunction of a diversion head works provided atthe off take of a canal from the river.

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103. (b)Volume of runoff due to effective rainfall of 1cm = Area of unit hydrograph.

1 × 10–2 × A = [5×6+2.5×4] × 60 × 60A = 14.40 km2

Volume of runoff = 3.5 × 10–2 ×14.40× 106

= 50.4 × 104 = 50.4 ha104. (a)

= 0.98 × 10–3 Poise = 1390 kg/m3

V = 3 1

8 20.98 10 10 7.05 10 m /s1390

V = 7.05 × 10–4 cm2/s = 7.05 × 10–4 stokes

105. (d)The crest level of a canal diversion head workdepends upon the FSL of the canal, thedischarge parameters & the pond level.

106. (b)Surface Runoff = [(6 – 3) + (9 – 3) + (5 – 3)]× 1 = 11 cm

107. (b)

h

Buoyant weight =Weight of ice + Weight of person(h × 8) × 9.81 × 103 =

320 8 0.9 9.81 10 980100

78480 h = 14126.4 + 980

78480 h = 15106.4

h = 0.1924 m

h = 19.248 cm

h = 19.25 cm

108. (b)The basic factors controlling process ofmeandering are :

(i) Valley slope

(ii) Silt grade and Silt charge

(iii) Discharge

(iv) Bed & side resistance

109. (d)110. (c)

Depth of submergence (y)

y =13c

wH

111. (b)

Piping failures can be prevented by :

(i) Providing sufficient length of the imperviousfloor.

(ii) Providing pile at downstream ends.

112. (a)

Number of events = 4 + 2 + 5 = 11

Years of record = 36 + 25 + 48 = 109

The event is equalled or exceeded 11 times in109 years of record.

T = 109 1 10 years

11

Note: When long records are not available,records at two or more stations are combinedto get one long record for the purposes ofrecurrence interval calculation. This method isknown as station year method.

113. (b)

1 atm = 10.3 m of water

3 atm = 30.9 m of water require

extra height of water required at A = 30.9 – 5.9= 25 m

Z = 25 m = 2 2w r

2g

25 = 2 2w (5)

2g

w = 2g114. (a)

2

3 3

Time 2-h UH S -curve Remarks(h) Ordinate ordinate

(m /s) (m /s)0 0 02 5 54 8 136 5 188 3 21

10 1 22 Equilibrium Q

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115. (a)

A

1 m 3 m

Z2

Z1

Z2 – Z1 = 5;2 2

11

rZ2g

r1 = 1m, r2 = 3m 2 2

22

rZ2g

2 2 2 2(3) (1)2g 2g

= 5

2 292g 2g

= 5, 28 5

2g

2 = 12.2625 rad/sec116. (a)

High water training : also called training fordischarge. The river is trained to providesuf f icient & ef ficient c/s area for theexpenditious passage of max. flood.

Low water training : Also called training fordepth. In this case, the river is trained to providesufficient depth for navigation during low stageof river.

Mean water training : Also called training forsediment. In this case, the river is traimed tocorrect the configuration of river bed for theefficient transport of sediment load in order tokeep the channel in good shape.

117. (c)

Continuity equation

u vx y = 0

(2) + (–1) = 1 0

i.e., flow is not possible.

118. (d)

Discharge passing through is A B| |

A = 73.5

B = 552.9

discharge B A| | = 479.4

119. (a)In case of parallel pipes, loss will be same

hf1 = hf2

Assuming laminar flow hf = 4128 QL

D

1 1 14

1 1

128 Q LD

= 2 2 2

42 2

128 Q LD

1 = 2 1 2, L L

1 = 2

141

QD =

42 14

22

Q Q 0.1 0.4823Q 0.12D

Q1 + Q2 = 12 lt/sec

Q1 = 0.483 Q2

Q2 = 8.096 lt/sec, Q1 = 3.904 lt/sec

120. (c)

KS = 0.4 mm

= *11.6

U

*

avfU U8

U* = 0.032 0.1225

8

= 10–6 m2/sec

= 6

511.6 10 9.469 100.1225

SK

= 3

50.4 10 4.224

9.469 10

0 < SK6 transition state

121. (d)

f = e

64 0.032R

avg

max

UU

=

1 1 0.8081.2371 1.325 f

0.81

122. (a)

0

vV

=

1/my

Shape factor = *

*

= m 2

m

1m

= 0.8 m = 1 10 5, m

0.8 8 4

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*

=

5 24 2.6

54

123. (a)

IDF can be used to get average intensity for arainfall of particular frequency and duration.

124. (b)If after a heavy rainfall, no further rain occurs,stream flow may be supplied entirely by base-flow, till water-table falls below the lowest bed-level of the stream channel. The hydrograph ofthe river from time when water-table is atmaximum height to the time when there is noflow is nothing but the hydrograph of ground-water inflow in the river and is known as groundwater depletion curve

Time

DischargeGround-water

depletion curve

125. (c)When the rain intensity is less, than prevailinginfiltration rate is approximately equal to rainfallrate. Hence, actual prevailing infiltration ratemay be equal to or less than the infiltrationcapacity.

126. (c)

At higher altitudes, the atmospheric pressureis less and hence evaporation should normallybe higher. However, this is not exactly sobecause of the decrease of temperature athigher altitudes, which reduces the evaporation.

127. (b)128. (b)129. (b)

The time of concentration is more in fern leafcatchments since the discharge is distributedover a long-period, as is evident from figure.

Tributaries

Main-stream

130. (d)The peak of outflow hydrograph will occur atthe point of intersection of the inflow and outflowhygrographs.

131. (c)Wind aids in removing the evaporated watervapour from the zone of evaporation andconsequently creates greater scope forevaporation. However, if the wind velocity, islarge enough to remove all the evaporated watervapour, any further increase in wind velocitydoes not influence the evaporation. Thus, therate of evaporation increases with the windspeed up to a critical speed beyond whichany further increase in the wind velocity hasno influence on the evaporation rate.

132. (a)133. (d)

The limiting case of a unit hydrograph of zeroduration is known as instantaneous unithydrograph (IUH). The main advantage of IUHis that it is independent of the duration of ERHand thus has one parameter less than a D-hunit hydrograph.

134. (a) Self-explanatory

135. (c)If Reynold’s no is high, flow is turbulent andvelocity profile is logarithmic.

136. (a)137. (d)

If a ship is safe in rolling then it will be safein pitching also because stability of ship isdirectly propotional to metacentric height andmetacentric height is greater in case of pitchingas compared to rolling because moment ofinertia is max in pitching so, if ship is safe inrolling then it will also be safe in pitching butvice-versa is not always true.

138. (a)139. (d)

Velocity of pressure wave is equal to thevelocity of sound in an infinite expansion ofmedium.

140. (a)141. (d)

Corrugations are in general similar to furrowsbut the main difference between the two beingthat the furrows are the channels of relativelylarger cross section compared to corrugations.

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142. (c)In sprinkler irrigation method, powerrequirement is high since continuous pumpingof water is required.

143. (b)The main advantage of drip irrigation over othermethod of irrigation is the excellent control ofwater application which it provides. Since inthis case water is directly applied to the plantsthrough sub-surface irrigation system due towhich evaporation losses and deep percolationboth can be minimized as a result duty of dripirrigation increases.

Drip irrigation method is particularlyadvantageous when the water is saline. Thisis so because the unwanted salts move to theouter edge of the wetted zone while within theregion of the root zone relatively less salinewater is made available.

Unwanted salts move up to the surface throughcapillary action, since in this case water isapplied to the root of plants.

144. (a)In case of sandy soil, capillary pores are notthere so water holding capacity is not theredue to which better plant growth is not possible.

In case of clayey soil, even due to presenceof more no. of capillary pores, water holdingcapacity is much even higher than loamy soilbut their is a greater resistance againstextraction of water in case of clayey soil. So,it is not much suitable for cultivation purpose.

Thus, an ideal soil for irrigation is that whichhas its pores spaced almost equally dividedbetween capillary and non-capillary pores.Such a soil has enough small pores to provideadequate water holding capacity and alsoenough large pores to permit adequate drainageand easy extraction of water by the roots ofthe plants. The loam soil is ideal for all theseproperties.

145. (d)146. (c)

Crop period : Time between sowing of a crop& its harvesting.

Base period : Time between first & lastwatering.

Consideration of base period is essential fordetermining the total water requirement of acrop.

147. (c)The liberation of plant food is dependent uponthe activity of soil bacteria which requiresadequate amount of oxygen in the air for properfunctioning.

148. (d)A flexible outlet or semi-modular outlet is theone in which the discharge is affected by thefluctuations in the water level of the distributingchannel while the fluctuations in the water levelsof the field channel do not have any effect onits discharge.

Ex. Kennedy’s gauge outlet, pipe outlet, venturiflume.

149. (c)In non gravity type weir, the floor thickness iskept lesser than that in gravity type weir & theuplift pressure is largely resisted by the bendingaction of the reinforced concrete floor.

150. (a)Rupture of floor due to uplift can be preventedby :

(i) Providing impervious floor of sufficientlength

(ii) Providing impervious floor of appropriatethickness.

(iii) Providing pile at the u/s ends so that theuplift pressure to the d/s is reduced.

ANSWERS - IES Master - Coaching for ESE, GATE and PSUs …€¦ · · 2017-10-01m3 0.0167 60 60 s...

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Transcript of ANSWERS - IES Master - Coaching for ESE, GATE and PSUs …€¦ · · 2017-10-01m3 0.0167 60 60 s...