ANSWERS - IES Master= 5.04 × 10–3 Ns/m2. 6. (c) No-slip condition is a consequence of fluid viscosity not fluid pressure. 7. (d) For vaporisation P P abs v p As, Absolute pressure

1. (b)

2. (c)

3. (c)

4. (d)

5. (c)

6. (c)

7. (d)

8. (b)

9. (b)

10. (c)

11. (d)

12. (c)

13. (c)

14. (c)

15. (c)

16. (a)

17. (b)

18. (c)

19. (c)

20. (c)

21. (a)

22. (b)

23. (c)

24. (d)

25. (d)

ESE-2019 PRELIMS TEST SERIESDate: 28th October, 2018

ANSWERS

26. (d)

27. (d)

28. (c)

29. (c)

30. (c)

31. (b)

32. (c)

33. (b)

34. (a)

35. (b)

36. (c)

37. (c)

38. (b)

39. (d)

40. (d)

41. (b)

42. (c)

43. (a)

44. (c)

45. (b)

46. (d)

47. (d)

48. (c)

49. (b)

50. (c)

51. (b)

52. (b)

53. (a)

54. (b)

55. (b)

56. (c)

57. (d)

58. (b)

59. (d)

60. (b)

61. (b)

62. (c)

63. (d)

64. (a)

65. (b)

66. (b)

67. (c)

68. (c)

69. (c)

70. (d)

71. (d)

72. (c)

73. (c)

74. (c)

75. (c)

76. (b)

77. (a)

78. (c)

79. (a)

80. (a)

81. (d)

82. (b)

83. (d)

84. (c)

85. (a)

86. (d)

87. (c)

88. (b)

89. (b)

90. (b)

91. (d)

92. (d)

93. (c)

94. (c)

95. (b)

96. (b)

97. (c)

98. (c)

99. (c)

100. (a)

101. (c)

102. (c)

103. (d)

104. (d)

105. (d)

106. (d)

107. (b)

108. (b)

109. (d)

110. (a)

111. (a)

112. (d)

113. (b)

114. (b)

115. (d)

116. (c)

117. (a)

118. (b)

119. (c)

120. (a)

121. (d)

122. (a)

123. (c)

124. (a)

125. (d)

126. (c)

127. (a)

128. (d)

129. (c)

130. (b)

131. (a)

132. (a)

133. (c)

134. (a)

135. (c)

136. (b)

137. (a)

138. (a)

139. (b)

140. (b)

141. (a)

142. (b)

143. (d)

144. (d)

145. (a)

146. (a)

147. (c)

148. (c)

149. (a)

150. (d)

IES M

ASTER

(2) Environment + FM + Surveying

1. (b)

Gradual closure

Force = Rate of change of momentum.

Thus, mdvdFdt

P A = dvmdt

P =

AL dvA dt

P

5 0L dV 1000g dt 10 30

50 16.67 m3

2. (c)

Loss at sudden contraction is = 2C 2V V

2g

22C2

2

V V 12g V

222

C

V 10.3 12g C

2 2

C

2 g 10.3 12g C

2

C

4 10.3 12 C

2

C

10.15 1C

CC = 0.72

3. (c)

In V-notch, Q = 5/2d

8 C 2g tan H15 2

5/28 0.5 2 10 tan30 115

= 0.7 m3/s

4. (d)

At no-slip condition, velocity of fluid particle

at solid stationary surface is zero however atdistance away from the surface it has certainvelocity.

In Ideal fluid as there is no viscosity hencethere won’t be any no-slip condition existing.

Wetting phenomenon occurs because ofsurface tension.

5. (c)

Shear stress, dudy

Viscous force = dl

= vdl

t

= 2 2

–3 –25 20 10 12.5 10

2 10 10

= 6250

By Newton's second law of motion,

30 6250 = 300.510

31.56250

= 5.04 × 10–3 Ns/m2.

6. (c)No-slip condition is a consequence of fluidviscosity not fluid pressure.

7. (d)

For vaporisation abs v pP P

As, Absolute pressure = Atmospheric pressure+ Gauge pressure.

100 (10 x) 2.7

x < – 9.73 m head of water..

8. (b)Total force that may act will have a normalcomponent and a tangential component.

Tangential component impart shear stresswhich will cause the fluid to move.

9. (b)A perfect vacuum would be empty space with noatoms or any particles in it. But there is no such

IES M

ASTER

[CE], ESE-2019 PRELIMS TEST SERIES PAPER-II | TEST-6 (3)

thing because virtual particles are always poppingin and out of existence.

10. (c)

P

h

dh w

P+dP

For small element

w = dh × dA ×

PdA + w – (P+ dP) dA = 0

dp dh As, = Constant

P = gh c

At, h = 0

P = Patm

P = Patm + gh

gaugeP gh .

11. (d)

Five mechanisms of deterioration have beenattributed to air pollution i.e. abrasion, deposition andremoval, direct chemical attack, indirect chemical attackand electrochemical corrosion.

12. (c)13. (c) For an activated sludge plant,

VX =

0 0 c

d c

Q S S y1 K

So – S = d c

0 c

VX 1 KQ y

Now, 0

VQ = HRT

[S] = [S0] – d C

C

[X] (1 K )y

=

42900 (1 0.05 10)24150

10 0.5

= 5 mg/L

14. (c) The rotating biological discs serve the followingpurpose.

1. They provide media for the buildup of attachedmicrobial growth.

2. They bring the growth into contact with thewastewater.

3. They aerate the wastewater and the suspendedmicrobial growth in the reservoir in RBCs as.

15. (c)

Maintenance of a good growth of organism on thefilter media is the main target

Organic food is oxidised only by organism whichare present in T.F.

16. (a) Recycling in trickling filter is done by treatedeffluent whereas settled sludge is recycledin ASP.

17. (b)

Peak flow Q (using peaking factor = 3)

Q = 3 × 0.438

= 1.314 m3/sec

Volume = 1.314 4 60

2

= 157.7 m3

18. (c)Primary treatment is often called clarificationsedimentation. Objective of primary treatment ofwastewater is to produce a liquid effluent of suitablyimproved quality for treatment stage i.e. secondarybiological treatment.

19. (c)Manholes provide an access to the sewer forinspection and maintenance operations.

They also serve as ventillation, multiple pipeintersections and pressure relief.

Manholes are commonly located at junctions ofsanitary sewers, at changes in grades or alignmentexcept in curved alignment and at locations thatprovide ready access to the sewer for preventivemaintenance.

IES M

ASTER


20. (c)Land treatment is the controlled application ofwaste water to the land at rates compatible withnatural physical, chemical and biologicalprocesses that occur in soil.

Organic matter in form of nitrogen, phosphorousand micronutrients in waste water are generallyharmful when discharge to lakes and streams, butthese constituents have a positive economic valuewhen applied under properly controlled conditionsto vegetated soils.

21. (a)Set Square : A right-angled triangular plate fordrawing lines, especially at 30°, 60°, 45° or 90°.

Ceylon Ghat Tracer : A reliable instrument forsetting out a grade contour.

Sextant : An instrument used for measuring the

angular distance.

22. (b)

Option (b) is correct.

23. (c)

Option (c) is correct.

24. (d)25. (d)

Option (d) is correct.

26. (d)

All above statement are correct.

27. (d)

Area = 1 2 3 n 1O O O O L

(n 1)

(As per

average ordinate rule)L = nd = 2 × 3 = 6

Area = 6 3.1 3.15 4.5 3.254

= 21 m2.


28. (c)Area of triangle = Base × half of the perpendicularheight

Area of trapezoid = half of sum of parallel sides

× perpendicular height

29. (c)Given, n1 = +1/50, n2 = –1/30

N = 1 1 4

50 30 75

As, Equation of parabola, y = ax2 (in general)

& For summit curve ; y = 2Nx

2L

=

2 24 4x x75 75

2 400 800

Summit point is at a distance of:

1n LN

= 1

1 400n L 50 150 m4N

75

y=

24 (150)75 1.5 m

800

R.L of summit point

= (R.L of starting point + n1L–y)

= 10 + 150 × 1/50 – 1.5

= 11.5 m Ans.

So, option (c) is correct.

30. (c)

6

A

B

D

C

6m

6m

ICG = 3

42 6 3 27 m12

Depth of centre of pressure.

yP = yc + CG

c

I 276 6.250 m1Ay 6 6 62

Centre of pressure below centroid = 6.250 – 6= 0.250 m.

31. (b)

yc = 1 + 3 23

IES M

ASTER


2m

3m

1m

Pressure force on the plate = F = A y

= 10 × 12

× 2 × 3 × 2 = 60 kN.

32. (c)x

y

dy

B C

AF v = Weight of water in portion ABC.

= Specific weight X volume of liquid inportion ABC

= V = 10 × 9

0(xdy) 2

= 20 9 1/20

3 y dy

3/2(9) 40 3020 3 (27)3 32

= 360 3 kN.

33. (b)

2m

5m

4m

G2

G1B

4m

1.6 m

O

From the principle of floatation

Total weight = weight of water displaced.

450 + 350 = (10 × 5 × x) × 10

800 = 500 x

x = 8 1.6m5 .

OB = 1.6 0.8 m2

OG = 1 1 2 2

1 2

w OG w OG 450 1 350 4w w 800

= 2.3125m

Distance b/w B and G.

BG = OG – OB = 2.3125 – 0.8

= 1.5125 m

BM =

310 512I 1.3 m

V 10 5 1.6GM = BM – BG = –0.2125

GM < 0

34. (a)GM = MB – BG

GM = I – BGV

0.15 = 0.5 – BG

BG = 0.35

2 – x = 0.35

x 1.65m .

35. (b)Time period is given by

IT 2W GM

Comfort is more when time period is largefor ships. Hence for greater comfort for a given I,

GM should be less, but for small GM willdecrease the stability of ship. Hence withoutcompromising on stability, if comfort is to beincreased, the size of the ship should be madelarger so that moment of inertia I increases andthus T (time period) increases.

For passenger ships, GM is less—Balance b/wstability and comfort is ensured.

For cargo ship, GM is more— stability is primeconcern.

IES M

ASTER


36. (c)

1.35

0.9

R=0.3 m

0.9 m

0.9 = 2 2(0.3)2 10

= 10 2 = 14.14 rad/sec.

37. (c)The dimensionality of flow depends on the choiceof co-ordinate system and its orientation forexample, a uniform pipe flow is two dimensionalflow in Cartesian coordinate system (i.e.)dependent on y and z coordinate) but onedimensional in polar co-ordinate system(dependent on r- only).

A fluid may also be in rotational motion inthe absence of viscosity in a flow field due tosome rotational motion given to it earlier i.e.before entering the flow field.

38. (b)

Given = 2x + y

u =

2x

v =

1

y

For stream line function.

v 1x

u 2y

2y f x

Now f xx

1 f (x)

f(x) = –x + c

= 2y – x + c

39. (d)

Stream function satisfied laplace equationonly when flow is irrotational flow.

Flow net are drawn in such a way thatdifference between stream function of twosuccessive stream lines are same. Thusdischarge between two successive stream lineis constant in a flownet.

40. (d) Oligotrophic lakes have a low level of productivitydue to a severely limited supply of nutrients tosupport algal growth. In this case, the euphoticzone often extends into the hypolimnion, whichis aerobic.

Eutrophic lakes have a high productivitybecause of an abundant supply of algalnutrients. The algae cause the water to behighly turbid, so the euphotic zone may extendonly partially into the epilimnion. Highlyeutrophic lakes may also have large mats offloating algae that typically impart unpleasanttastes and odours to the water.

Senescent lakes are very old, shallow lakeswhich have thick organic sediments and rootedwater plants in great abundance . These lakeseventually become marshes.

41. (b)

C = 20 mg/ls C = ?m

Q = ?m

Q = 5 m /s

C = 40 mg/lw

w

3

Q = 10 m /ss3

Cw = 40 mg/l

Qw = 5 m3/sec

Cm =

s s w w

s w

C Q C QQ Q =

20 10 40 510 5

= 26.67

mg/l

42. (c) C4H10 + O2 CO2 + H2O

Balancing,

2C4H10 + 13O2 8CO2 + 10H2O

C4H10 = 4 × 12 + 10 × 1 = 58 g/mol

CO2 = 12 + 2 × 16 = 44 g/mol

IES M

ASTER


2 moles of butane produce 8 moles of CO2

2 × 58 grams of butane produce 8 × 44grams of CO2

100 g of butane will produce

8 442 58

× 100

g of CO2

Mass of CO2 produced = 303.45 grams

43. (a)

log

of n

umbe

rs o

f bac

teria lag

phase

Exp

onen

tial

phas

e Death phase

Stationary phase

Time

Bacterial growth curve/kinetic curveThe growth of bacteria in batch culture can bemodeled with four different phase.

(A) lag phase

(B) log phase or exponential phase

(C) Stationary phase

(D) Death phase

1. During lag phase, bacteria adaptthemselves to growth conditions.

2. The log phase (sometimes called thelogarithmic phase or exponential phase)is a period characterised by all doubling.

3. The stationary phase is often due to agrowth l imiting factor such as thedepletion of an essential nutrient and/orthe formation of an inhibitory product suchas organic acid.

4. At death phase (decline phase), bacteriadie. This could be caused by lack ofnutrients, environmental temperature aboveor below the tolerance band for thespecies.

44. (c)KT = K20(1.047)15–20 = 0.183/day

L0 =

0.23 5150 219.5mg

1 el

At 15ºC

BOD5 = 5K0L 1 e

= 0.183 5219.5 1 e = 131.5 mg/ l

45. (b)Bio-chemical oxygen demand is used as ameasure of the quantity of oxygen required foroxidation of bio-degradable organic matter presentin water sample by aerobic biochemical actiononly.

46. (d)COD : It is the amount of oxygen required

to carry out the decomposition of biodegradable andnon biodegradable organic matter.

BOD : It is the amount of oxygen requiredto carry out the decomposition of biodegradable organicmatter present in the system.

47. (d)The compute BOD at ‘t’ number of days,

Lt = L0 (1 – 10–KDt)

DK tt

u 0

LBOD (1 10 )BOD L

0.064 55

u

BOD(1 10 ) 0.52

BOD

48. (c) DOmin = 1.1 2 8.7 8.3

1.1 8.7

= 7.6 mg/l

The saturated value of dissolved oxygen DOsat 20°C is given as 9.2 mg/l

Initial deficit would be

Do = 9.2 – 7.6 = 1.6 mg/l

49. (b) The flow will be split between the two paralleltanks so the tank volumeVeach tank = Q × tR

=

6 3 390 10 10 m 15 sec2 24 60 60 sec = 7.8125 m3

P = G2 V = 7502 × 1.139 × 10–3 × 7.8125= 5 kW

As, = 0.7

Pmixer = 5

0.7 = 7.15 kW

50. (c)Statement (i), (iii) & (iv) are correct and statement

IES M

ASTER


(ii) vertical curve has simplicity in computationwork.


51. (b)In tape and theodolite method angularmeasurement by theodolite and l inearmeasurement by tape.

In tacheometer, both angular and linearmeasurements are done.

Two theodolite methodno linear measurementssare made. only angular measurements.

So option (b) is correct.

52. (b)Point of tangency: At this point alignment ofroute changes from a curve to a straight line.

Point of curvature: At this point alignment ofroute changes from a straight line to a curve.

Point of intersection or vertex: When backtangent and the forward tangent, when extended,intersect at a point called as the vertex (V) orthe point of intersection.

So, option (b) is correct.

53. (a)A = 30°00'20" ... (1)

2A = 60°00'00" ... (2)

In equation (1) the coefficient of A is 1 and theweight of the observation is 2. Therefore, multipleit with 2 × 1 = 2 to get

2A = 60°00'40" ... (3)

Similarly multiply equation (2) by 2 × 1 = 2

4A = 120°00'00" ... (4)

Adding (3) and (4)

2A + 4A = 180°00'40"

A = 30°00'6.66"

54. (b)

As, semi circumference = d

2

22C d

de ed 2

eC = de2

eC = 0.32

= 0.15

so, option (b) is correct.

55. (b) The origin of photochemical smog is thephotolytic composition of nitrogen dioxide. Asmall amount of NO2 can trigger subsequentreactions through its decomposition, formingwhat is called the NO2 photolytic cycle. Thecycle starts as follows:

NO2 + h NO + O

The atomic oxygen liberated then reacts withmolecular O2, producing ozone as follows:O + O2 O3

Ozone subsequently reacts with NO toregenerate NO2 and molecular O2, thuscompleting the cycle. 3 2 2O NO NO O

The above reaction comprises NO2 photolyticcycle. However, presence of hydroxyl radicalsleads to its reaction with carbon monoxide,hydrocarbons and carbonyl compounds, thisproduces peroxy radicals. The peroxy radicalsreact with NO to regenerate NO2, thusterminating the normal NO2 cycle, leaving O2undisturbed and allowing it to accumulate.

56. (c)

The branch of surveying which deals withwater bodies is known as hydrographicsurveying.

Sounding in hydrography is defined as thedepth of water body at the point ofmeasurement.

57. (d)Option (d) is correct.

58. (b)Statement (ii) & (iii) are incorrect, so option (b)is correct.

59. (d)Area covered on one photograph = 250 ×(1–0.6) × 250 × (1–0.3)mm2

Area on ground = 250 × (1 – 0.6) × 250 × (1 –0.3) × 25× 25 × 106 × 10–12 km2

= 10.9375 km2


60. (b)So, option (b) is correct.

IES M

ASTER


61. (b)

z =v u1x y2

= 2 214y x 2x2

= 2y2x – x2

= x (2y2 – x).

62. (c)

Circulation ( ) = vorticity × Area

Given :

x2 + y2 – 2y – 4x + 1 = 0 Represent a circle ofradius = 2 unit.

(x – 2)2 + (y – 1)2 = 22.

Vorticity = v ux y

= (0 – 2) = – 2

Circulation ( ) = (– 2) 2(2)

= 8 .

63. (d)

When all forces are taken into account theequation of motion is called Newton's equation ofmotion.

When compressibility and other minor forcesare neglected, it is called Reynold's equation ofmotion.

64. (a)

For Bernoullis equation :

1. Effect of friction is negligible i.e., when fluidis ideal or viscosity has negligible effect.

2. Flow is essentially one-dimensional i.e.,along a stream line. However, the Bernoulli'sequation can be applied across streamlines itthe How is irrotational.

3. Flow may be uniform or non uniform but flowshould be steady.

65. (b)

P2

Rx

Ry

P1

Vertical component of force to keeps the elbowin position is Ry in Y direction.

As we know,

Not acting Force = change in momentum flux

Ry – P2 A sin 45º = 2Q(V sin 45º 0)

Ry – 22

3130 105 2 104 2

= 1000 22 1030 10 10

24

Ry – 0.1125 × 1000 = 1591 yR 1703.5 N

66. (b)

In case of orificemeter only small length ofpipe is affected hence if there is a spacerestriction, orificemeter can be used in place ofventurimeter.

Head losses are more in orificemeter becauseof flow separation.

67. (c)

=r dP2 dx

500 = 0.25 P

2 20

P 80 kPa .

68. (c)

Q = 31 dPB12 dx

Q = 3

41 20(300)

100012 2 10

= 1 m3/sec.

69. (c)

Head loss (hL) = 2

264 LV

fLV Re2gD 2gD

=232 LV

Re gD .

a1V1 = A2V2

IES M

ASTER


22 × Vend = 202 × 10

1000

Vend = 1 m/sec

hL = 2

3

32 0.08 (1) 0.128 m1000 10 2 10

.

70. (d)Q = Vavg × Area

For circular pipe

Vavg = maxV 2 1m/sec.2 1

Q = 1 × 2 3(0.2) m /s4 100

.

71. (d) Coagulation occurs predominantly by twomechanisms: adsorption of the solublehydrolysis species on the colloid and theredestabilization, the reaction in adsorption-destabilization are extremely fast.

Jar test data may be used to identify whetheradsorption-destabil ization or sweepcoagulation is predominant.

If charge reversal is apparent from the doseturbidity curve, then adsorption-destabil ization is the predominantmechanism. If the dose-turbidity curve doesnot show charge reversal, then sweepcoagulation is predominant.

72. (c)

As, Al2(SO4)3·18 H2O + 3Ca(HCO3)2

3CaSO4 + 2Al(OH)3 + 8CO2 + 18H2O Molecular weight of alum = 2 × 27 + 3 (32 +64) + 18 (2 × 1 + 16) = 666 gm

Molecular weight of alum Ca(HCO3)2 = 40 + 2(1 + 12 + 48) = 162 gm

1 gm eq of alum = 1 gm eq of Ca(HCO3)2

20 y666 / 6 162 / 2

alkalinity = 14.59 mg/l as Ca(HCO3)2

73. (c) The amount of alum which react with alkalinity

1 gm eq/l of alum = 1 gm eq/l of CaCO3

50111

= x

50

x = 22.52 mg/l

In addition to natural alkalinity, lime required =22.52 – 10 = 12.52 mg/l

74. (c) When water does not have sufficient totalalkalinity to react with alum, lime, or soda ashis usually also dosed to provide the requiredalkalinity.

75. (c) As per stoke’s law,

sV' = 2g G 1 d

18 v

=

23

39.81 2.35 1 0.010 10

18 101000

= 7.3575×10–5 m/s

Now, overflow rate (Vs) = QBL

= 0.093800

= 1.1625 × 10–4 m/s

Removal efficiency = s

s

V 100V

'

=

5

47.3575 10 1001.1625 10

= 63.29%

76. (b) For treatment works with a high quality rawwater, it may be possible to omit sedimentationand perhaps flocculation. This modification iscalled direct filtration.

Disinfection is the addition of chemicals (usuallychlorine) to kill or reduce the number ofpathogenic organisms. The color and turbidityconsume disinfectant thus requiring the use ofexcessive amounts of chemical. In addition,the presence of turbidity may shield thepathogens from the action of the disinfectantand thereby prevent efficient destruction.

77. (a)Average count for 500 mL

= 350 340 370 340

4

= 350

Coliforms per 100 mL = 350 100500

= 70

IES M

ASTER


78. (c)

Total alkalinity = 3 3CO HCO OH

23CO

= 360 10

60

= 10–3 mol/l[H+] = 10–10 as pH = 10

3HCO = 3 102

311

10 10CO HK 5 10

= 2 × 10–3 mol/l

= 122 mg/l

3HCO as CaCO3 =

12261

× 50 = 100 mg/l as

CaCO3

23CO

= 60 mg/l

= 60 50 100mg / l30

as CaCO3

OH = 10–4 mol/l 10–4 × 17 = 1.7 mg/l

= 5 mg/l as CaCO3

Total alkalinity =

3 3CO HCO OH

= 100 + 100 + 5

= 205 mg/l as CaCO3

79. (a)Weight of total suspended solids

= 142 × 10–3 × 260 × 10–3

= 0.037 g

Weight of non volatile suspended solids =

= 47.27 – 47.25

= 0.02 g

Weight of volatile suspended solids

= 0.037 – 0.02

= 0.017 g

80. (a)Plane table surveying is used where the greataccuracy is not required.

In geodetic surveying, measurement does notrepresent the projection on the horizontal plane.

81. (d)The scale expressed as RF is given by

fR

H h

201 100

8000 H 1500

H – 1500 = 1600

H 3100 m above MSL

82. (b)From trigonometry,

Height of vane above the line of collimation;

H = D tan = 1000 × tan (9°45') = 172 m

We know that,

R.L. of vane = R.L. of line of collimation + H

R.L. of station B = R.L. of vane – 2

= 180.30 + 172 – 2

= 350.3 m

So, correct option is (b).

83. (d)All of the above statements are correct. So, option

(d) is correct.

84. (c)

B.S. – F.S. = 12.33 – 12.05

= 0.28 m

Last R.L. – first R.L. = 20.57 – 20.38 = 0.19

both results must be same but in this case theyare not.

So, error = 0.19 – 0.28

Closing error = 0.09 m


85. (a)Let ‘d’ is the distance of visible horizon.

d = 3.854 h ( ‘h’ in metres & 'd' in kms)

h = 1.67 + 25 = 26.67 m

d = 3.854 26.67= 19.9 km 20 kms Ans.

So correct option is (a).

IES M

ASTER


86. (d)Difference between staff readings of 1st & 4th

point

= 2.125 m – 0.650 m = 1.475 m

R.L. of 1st point = 82.3 m + 1.475 m

= 83.775 m

So correct option is (d).

87. (c)

We know, nd SR D

n = 5, d = 1.5 mm, D = 150 m and S = 0.05m

or 1.5 × 10–3 m

R = 35 1.5 10 150

0.05

R 22.5 m so correct option is (c).

88. (b)Statement (iii) is correct.

Tertiary triangulation is of lower grade thansecondary triangulation.

From primary triangulation, points are pickupfor control in secondary triangulation.

So, option (b) is correct.

89. (b)

LNorth – LSouth > 0

DWest – DEast > 0 Closing error will be in the N-W quadrant

D

L

EW

S

N

0

A

So, correct option is (b)

90. (b)In this method, a uniform average correction isapplied to all angles.

We know that,

Theoretical sum = (2n – 4) × 90° = (2 × 4 – 4)×90°

= 4 × 90° = 360°00'00''

& Measured sum = 360°1'00''

Angle of misclosure = Measured sum –theoretical sum = 1'

error = – correction

correction = 1'

Correction per angle = 60 154

per angle


91. (d)92. (d)93. (c)

= 2

2avg

u dAAV

O to H velocity variation = 0V yH

H to 2H velocity variation = V0 (constant)

=

2H 2H 2000 H

2

0

V dy B V dy ByH

3B (2H) V4

=

H2 3 2H20

0 H20

20

V y 1V y 13H 399(2H) V816

=

4323

9 278

.

94. (c)Flexible joint is recommended for the placeswhere settlement of the pipe line may occurlike in river with uneven beds.

95. (b)Shear stress at the boundary in case of turbulentflow, is more than that in laminar. This is becauseat boundary, the shear is primarily due toviscosity and velocity gradient is greater forturbulent flow.

Velocity profile in case of turbulent flow ismuch flatter. The profile becomes more flat athigher Reynolds number.

IES M

ASTER


Eddy viscosity is characteristic of flow not afluid property. Hence it may vary from point topoint.

96. (b)

Velocity defect = (Umax – u) = 5.75 u*× log Ry

As u* = w

= 5.75 4000 20log1000 10

= 11.5 log 2 = 3.45 m/sec.

97. (c)

0

*

KKuK K

11.6' 11.6 11.6u

=4

610 250 100 4.31

1000 11.6 211.6 10

As, 0.25 < K 6 transition zone.

'

98. (c)

Given :

1 13 m

0

v y yv

for m = 3

m 3 3

(m 1) (m 2) (3 1) (3 2) 20

203

99. (c)Von karman integral equation

020

ddxV

Assumptions :

(1) 2-D flow

(2) Valid for Laminar boundary layer andapproximately true for turbulent boundary layer

(3) Steady flow

(4) Incompressible flow

(5)dP 0dx

.

100. (a)

V0 = 90 kmph = 90 5 25m/s

18

Re = –5VD VD 25 3

v 1.5 10

= 5 × 106 > 5 × 105

Take Cd = 0.3

Drag force = Cd × 20

1 A V2

= 0.3 × 21.2 (40 3) (25)2

= 13.5 kN.

101. (c)

Q = 2 1

2

1

2 KD H H 2 K 30 2.5 0.1r 1802.3log2.3log

20r

K = 29.12 m/day

102. (c)All are correct. So, option (c) is correct.

103. (d)Design discharge = Maximum of

(Max daily demand+ fire demand) ORMax. hourly demand

Fire demand = 3182 P [As per kuchling formula]

where P is in thousands

= 3182 5000 = 225001.4 l/min

= 0.225 Million Litre/Min.

For 1 hour = 0.225 × 60 = 13.5 MLD

Max. daily demand = 1.8 × 18 = 32.4 MLD

Max. hourly demand = 2.7 × 18 = 48.6 MLD

Design discharge = Maximum

of32.4 13.5 45.9 MLD48.6 MLD

= 48.6 MLD

IES M

ASTER


104. (d)Longidunal temperature stresses createdwhen pipes are laid above the ground.

Longitudinal stresses created due tounbalanced pressure at bends or at points ofchanges of cross-section.

Flexural stresses produced when pipes aresupported over trestles etc.

105. (d)Poliomyelitis is caused by polio virus and notbacteria.

106. (d)All are component of typical manhole provided insewer System Design.

107. (b)Sound pressure level

(dB) = rms10

p20log

20

As, sound pressure is inversely to distance ‘r’.

rms1Pr

80 = rms 45

10p

20log20

4rms 45p 20 10 Pa

rms 45

rms 22.5

P(P ) =

22.5 145 2

(prms)22.5 = 40 × 104 µPa

D22.5 = 4

1040 1020log

20

= 420 log2 log10

= 4.3 × 20 = 86 dB108. (b)

tan = 100 0 10080 20 60

again, tan = 60

100 60L 20

Here, 60

40 100L 20 60

L60 = 44 dBA

109. (d)

Area of circular sewer = QV

2d4 =

3.144

d = 1 m

Dia of egg-shaped sewer ( d ) = 0.84 d

= 0.84 × 1

= 0.84 m

Height of egg-shaped sewer = dd2

= 0.840.84

2

= 1.26 m

110. (a)Possible reasons are :

– Low estimates of the average and maximumflows.

– Large scale infiltration of storm water.

– Unforeseen increase in population or waterconsumption and the consequent increasein sewage production.

111. (a)Statement (i) is correct

Statement (ii) magnetic needle is horizontalat equator & vertical at poles.

Statement (iii)Dip varies between 0° to 90°.

So, option (a) is correct.

112. (d)All of the above are correct

So, option (d) is correct.

113. (b)

Option (b) is correct i.e., true bearing.

114. (b)

IES M

ASTER


W E

N

SN 88°30

1°30WE

N

S

50°30

50°30

50°30

deflection angle

A

C B

Deflection angle = 1°30' + 90° 00' + 50°30'= 142° 00'


115. (d)116. (c)

Area of field = AFB AEC BFD CED

= 1 1250 150 500 1502 2

+ 1 1500 150 250 1502 2

= 112500 m2

= 11.25 Ha Ans.

117. (a)

118. (b)For the first 1000 m,

Mean length of tape, I'

= 30m 30.05 30.025 m

2

Correct length

= actual length of tape measured

distanceNominal lengh of tape

= 30.025 1000 m 1000.833 m

30

For remaining length = 765 m,

Mean length = 30.05 30.15 30.10 m

2

Correct distance = (767.55 + 1000.833)m

= 1768.383 m

Correct option is (b).

119. (c)(i), (ii) & (iv) are correct.

(iii) Distance of object from survey line is called

offsets.

120. (a)

Field Area (f) = 2Map Area

(R.F.)

f = 2

11

(500000)

f = 1 × (5,00,000)2 cm2

= 25 × 1010 cm2

= 25 × 1010 × 10–10 km2

Field area of rectangular plot = 25 km2

So option (a) is correct.

121. (d)Water hammer occurs due to rapid and slowclosure, magnitude of pressure wave is differentfor both conditions.

122. (a)If pipe lengths are large then there will be morefriction losses which are major losses at thattime we can neglect minor losses. As thing perunit length decreases.

123. (c)RCC pipes are not corrosion resistant. Somenuisance bacteria reduces sulphureous materialpresent in sewage into sulphate which formsuphuric acid consequently lead to corrosion ofpipes.

124. (a)

Froude’s low is valid

Thus, r rV L

r rT L

r1T6

PT 6 1.5 9 hours

IES M

ASTER


125. (d)Drag experienced by ship consists of waveresistance and frictional or viscous resistance.The phenomenon is influenced by existence ofinertia, gravity and viscous forces and thusdynamic similarity will be achieved, if the Reynoldand Froude number both are same in the modeland its prototype.

126. (c)Model can be smaller, bigger or equal toprototype. In case of hard disk drives, model areeven bigger than prototype.

Model is a release of a product built to test aconcept or process or to act as a thing to bereplicated or learned from.

127. (a)In triangular weir, head is large even for smalldischarge. Hence flow is not affected much bysurface tension and viscosity (opposite to thishappens for rectangular weir).

128. (d)The viscous forces in a fluid are the outcome ofintermolecular cohesion and molecularmomentum transfer. In fluids like water (liquid)the molecules are comparatively more closelypacked, but not in fluid like air molecular activityis rather small and so the viscosity is primarilydue to molecular cohesion. So, with increase intemperature viscosity change depends on whetherfluid is liquid or gas.

129. (c)Ideal fluids are those fluids which have noviscosity and surface tension and they areincompressible. As such for ideal fluids noresistance is encountered as the fluid moves.

A flow is said to be rotational if the fluidparticles whle moving in the direction offlow rotate about their mass centres.

A flow is said to be irrotational if the fluidparticles while moving in the direction offlow do not rotate about their masscentres. It may however be stated that atrue irrotational flow exists only in thecase of flow of an ideal fluid for which notangential or shear stresses occur. Butthe flow of practical fluids, may also beassumed to be irrotational if the viscosityof the fluid has little significance.

130. (b)131. (a) A high f/m ratio in activated sludge pro-

cess causes dispersed growth such that micro-organism do not settled out of solution by gravity.

Second there is excess unused organic matterin solution that can’t be removed by sedimenta-tion.

132. (a)Both (A) & (R) is correct and (R) is correctexplanation of (A).


133. (c)The loss due to sudden expansion is greaterthan the loss due to a correspondingcontraction. This is because diverging pathsof the flow tend to encourage the formation ofeddies with the flows. Moreover, separation ofthe flow from the wall of the pipe inducespockets of eddying turbulence outside the flowregion. In converging flow there is a dampeningeffect on eddy formation and the conversionfrom pressure energy to kinetic energy is quiteefficient. The loss of energy due to suddenenlargement,

hL =2

1 2(v v )2g

The loss of energy due to sudden contraction,

hL =2v0.5

2g134. (a) Rapid changes in waste water quantity and

quality make operation of a waste watertreatment plant difficult. This is because ofthe necessity to constantly adjust the processparameters in the treatment units tocompensate for the unsteady conditions.Under such circumstances, provision of anequilization tank dampens the rapid changesand hence the treatment plant can operateunder steady conditions.

135. (c)

When chlorine is added to water, it formshypochlorous acid or hypochlorite ions, whichhave an immediate and disastrous effect onmost forms of microscopic organisms.

pH 52 2Cl H O HOCl HCl

The hypochlorous acid is unstable and maybreak into hydrogen ions and hypochlorite ions

IES M

ASTER


pH 8pH 7

HOCl H OCl

The hypochlorous acid is the most destructivebeing about 80 times more effective than thehypochlorite ions.

136.(b) The major deleterious effect of phosphorous isthat it serves as a vital nutrient for the growthof algae. If the phosphorous availability meetsthe growth demands of the algae, there is anexcessive production of algae. When the algaedie, they become an oxygen demanding organicmaterial as bacteria seek to degrade them.This oxygen demand frequently overtakes theDO supply of the water body and as aconsequence causes fish to die.

137. (a) According to Le Chatelier's principle, anyreaction reaching equilibrium will seek newequilibrium when a variable (temperature)influencing that equilibrium is altered. Increasedtemperature causes an increase in kinetic energy.The higher kinetic energy causes more motion inthe gas molecules which break intermolecularbands and help them escape from the solution.

138. (a)Both (A) and (R) are correct & (R) is correctexplanation of (A).


139. (b)Both (A) & (R) are correct but (R) is not correctexplanation of (A).

So, option (B) is correct.

140. (b)

In mixing process of coagulation, baffles areinstalled to reduce short circuiting.

141. (a)Iron bacteria oxidize inorganic ferrous iron Fe+2 toFe+3

Fe+2 (Ferrous) + Oxygen Fe+3 (Ferric) +Energy, Fe(OH)3 is responsible for yellowish orreddish colored slimes.

142. (b)Triangulation stations to which observations arerequired to be taken should be intervisible andstations should form well conditioned triangles.In general no angle should be less than 30° ormore than 120°.

To define exact position of a triangulation stationso that it can be observed from other stations,signals are used. It should be conspicuous andclearly visible over ground.

It should be accurately centered over the stationmark.

143. (d)The solution of a three point problem in planetable surveying is aided by Lehmann’s rule.

144. (d)Air foil body is a well stream line body and henceflow can be separated at the rear end of theairfoil.

145. (a)

For an ideal fluid the flow pattern is symmetricalabout longitudinal axis, and the pressure at thecorresponding points on the front and rear aresame 'No unbalanced pressure force acts on thecylinder and consequently pressure drag is 'zero'.

146. (a)

In combined sewers, the variations indischarge are enormous, because the stormrunoff is generally 20 to 25 times that of thesewage discharge. Hence, combined sewerswill have to run at low discharges of about

120 to

125 times the maximum, during non-

monsoon periods.

Egg shaped sewers maintain hydraulic depthnearly uniform for low discharges.

Smaller circular portion of egg shaped sewerwill be effective during dry weather and fullsection is effective during maximum rain waterflow.

147. (c)More than 1.5 mg/l may cause mottling of teethand upto 1 mg/l may prevent dental cavities.

148. (c)Relief displacement : It is caused by changesin the distance between the ground and thecamera as the plane flies over the ground.Characteristics of relief displacement

Characteristics of aerial images overvaried terrain.

Objects that rise above the surface awayfrom the principal point.

IES M

ASTER


Objects extending below the surface leantowards the principal point.

Displacement increases with the heightof the object and or distance from theprincipal point.

Relief displacement, d = rhH

Wherer = radial distance from principal point todisplaced image pointh = height above surface of the object point

H = flying height above the surface.

149. (a)Both Assertion & Reason are correct & Reasonis correct explanation of Assertion.

150. (d)Tacheometer is a transit theodolite fitted withstadia diaphragm and anallactic lens.

Anallatic lens : Additional convex lens isprovided between the eye piee and the objectglass at a fixed distance from the object glass.

The purpose of providing anallatic lens is tomake additive constant (f+d) exactly zero.

Anallatic lens is provided in external focusingtelescope, not required in internal focusingtelescope.

Notes:Advantage

It simplifies the calculation by makingthe additive constant zero.

Therefore, there is only one constantwhich is multiplying constant.

Distances are directly calcualted bymultiplying the difference of stadia hairreadings by 100.

Disadvantage

Brightness of the image is much reduceddue to absorption of light.

The anallatic lens can not be cleanedeasily

It increases the cost of the instrumentbecause of one extra lens.

Documents

ANSWERS - IES Master= 5.04 × 10–3 Ns/m2. 6. (c) No-slip condition is a consequence of fluid viscosity not fluid pressure. 7. (d) For vaporisation P P abs v p As, Absolute pressure