Cast iron pipes are mostly used as trunk mainsand distribution sub-mains or laterals, due totheir strength, durability, corrosion resistance,ease of laying and maintenance. Interior andexterior coatings may be required dependingon water quality. Steel mains are lighter inweight but higher in strength; so they are usedfor larger diameters ranging from 900 to 3000mm. Steel pipes are prone to corrosion andthe life of steel pipes is about half that of castiron pipes.
2. (a)
S-2 for steady flow, temporal acceleration iszero
i.e., Total acceleration = advective acceleration(convective acceleration)
S-3 for uniform flow, convective acceleration iszero
Total acceleration = Temporal acceleration
3. (c)
Let the length be 4x and width be 3x.
No. of photograph per strip
=4x 1
(1 0.6) 4000 250
= 4x 1
10 10
No. of strips =3x 1
(1 0.4) 4000 250
= 4x 1
20 10
From the given options, taking length = 16km
x = 4 km = 4 × 106 mm
No. of photograph per strip
= 40 + 1 = 41
No. of strips = 20 + 1 = 21
Total no. of photographs = 861
4. (d)
In intermittent system of supply, water issupplied only during fixed shifts of the day,This system is adopted when sufficient quantityof water is not available and when there is
inadequate pressure. Water is utilized verycarefully, wastage of water is less and repairscan be done in non-supply hours. Water maynot be available for fire fighting, negativepressure may develop during non-supply hoursand this may result in pollution andcontamination of water through leaking joints.Generally taps remain open and when watersupply starts there is a large wastage of water.
5. (a)S-1
Direction of forcesame in plane surface
S-2
Direction of forcechanging atevery point
S-3 Pressure prism concept is applicablefor plane surface only
The operation of the waste stabilization basinis largely controlled by the weather, designsare normally based on loading factors. Anotherdrawback is the high concentration ofsuspended solids in the effluent of stabilizationbasins. The effluent suspended solids of 35 to100 mg/L will surely not meet the secondarystandard requirement. Some means must beapplied to remove the algae in the effluent.Algae are responsible for the high suspendedsolids. When discharged into a receivingstream, these algae decompose, resulting ina high dissolved oxygen demand.
11. (c)
G1 = 0.8, G2 = 1.2
for 1st liquid,
1 wW V G = 35 N ...(i)
for 2nd liquid,
2 wW VG = 20 N ...(ii)
(i) – (ii)
2 w 1 wV G V G = 15 N
3 3V 1.2 9.81 10 0.8 9.81 10 = 15
V =
33
15 15m L3.9243.924 10 3.82 L
12. (a)Tangent distance = 200 m
60Rtan2
= 200 m
60Rtan2
= 200 m
Tang
ent d
istan
ce
Rtan (
/2)R[sec( /2) –1]
Apex Distance60°
60°
R = 200 3 m
Apex distance = R sec 12
= 200 3 sec30 1
= 2 1200 33
= 400 – 200 3= 53.59 m
13. (b)
In high rate TF, due to high rate of filterationsewage remains fresh and there is little or nofly or odour nuisance but nitrogenous mattermay not get sufficient time for nitrification.
4. Disruption of streamline due to fans andblades inside it
5. Applicable for Mach number less than 0.3,not applicable for Mach number 0.3
15. (d)Let L be line on the plan
Actual length = 4000 L
Measured length = 5000 L
Error in length = 1000 L
Similarly error in area = (5000L)2 – (4000L)2
Percentage error
= 2 2
2(5000 L) (4000 L) 100%
(4000 L)
= 25 16 100%
16
= 9 100%
16
= 56.25%
16. (c)
FM
= 0QSVX
= 600.5 24 60 60 0.92
800 3500
17. (a)
dP =
p pdx .dzx z
pdx
= – xa
pdz
= – za g
ax = g
az = 0
dp = gdx gdz
Integrating both sides
P = – gx gz c
At x = 0, z = 0, P = Patm
C = Patm
P = – atmgx gz P
P – Patm = – gx gz
At free surface
P = Patm
0 = – gx gz
x + z = 0
18. (b)The change in length due to temperaturemust be equal to sag correction and assag correction is –ve, temp should beraised. i.e. the temperature correction mustbe +ve.
|Temperature correction| = |Sag correction|
× L × (T1 – T2) = 2 3
2w L24P
30 × 1 × 10–6 (T – 10°)
=
2 33
23015 10 10
24 100
T – 10 = 84.375
T = 94.38°C
19. (b)
SVI = SV 1000 mL / g
MLSS
= 350 1000 125mL/g
2800
Where SVI = Sludge volume index, mL/g
SV = Volume of settled solids in one-litregraduated cylinder after30 min settling (mL/L)
Aerobic pond is a shallow pond in which lightpenetrates to the bottom, thereby maintainingactive algal photosynthesis throughout theentire system. During the day light hours, largeamounts of oxygen are supplied by thephotosynthesis process ; during the hours ofdarkness, wind mixing of the shallow watermass generally provides a high degree ofsurface reaeration. Anaerobic ponds areparticularly suited for the treatment of hightemperature and high strength wastewaters.An aerobic pond can be converted into ananaerobic.
23. (d)
= 1000 m
V = 1.5 m/s
dia = 0.1 m
h f = 2f V2gd
f =ex
f 64 164 4R 1000
h f =
216 1000 1.51000 2 9.81 0.1
h f = 18.348 m
24. (d)In a clockwise traverse, Included angle =fore bearing of forward line – back bearingof previous line
CDE = Fore bearing of DE – back bearingof CD
DE
CB
A
15°30
116°51220°00
30°45
CDE = 220° – (30°45 + 180°)
= 220 210 45
CDE = 9 15
25. (b)For the concentration of the waste (Lw)reduced by 25%, oxygen deficit shall bereduced by 25%.
1CD = DOs – DO = 10.83 – 6
= 4.83 mg/l
2CD = 254.83
100= 120 mg/l
2CD = DOs – DO
DO = 10.83 – 1.20
= 9.63 mg/l
26. (b)
For laminar flow between parallel plates
hL 3
1B
distance between points where local velocityis equal to mean velocity
=B3
For first system, 1B 0.3m3
B1 = 0.3 3 m
For second system, 2B 0.6m3
B2 = 0.6 3 m
1
2
L
L
hh =
332
1
B 0.6 3B 0.3 3
= 827. (c)
Set squares are used to set outperpendicular lines. Only one optionmatches.
Foaming in sludge digestion tank is due toinsufficient amount of well buffered sludgein the digester, excessive addition of rawsludge (with high volatile content); or poormixing of digester contents; or temperaturetoo low for prolonged periods followed byrise in temperature of digester contents; orwithdrawal of too much digested sludge; orexcessive scum or grit accumulations.
32. (b)
ax =u u u uu v wt x y z
Since,ut
=u uu w 0y z
ax = 2/3
1/3u xu 9 xx 3
ax =1/3
2/3 x3 x3
At, x = 8
ax = 1/32/3 83 8 3
= 2
3 1 11 L4 6 12 T
33. (d)In a clockwise traverse
Included angle = fore bearing of next line –back bearing of previous line
BAC = (F.B)AB – (B.B.)CA
270° = (F.B)AB – 65 35
(F.B)AB = 335 35
3x
A
B C
x
60°30°
335°35’
63°35’
ABC = (FB)BC – (BB)AB
300° = (FB)BC – –180 335 35
(F.B)BC = 95 35
34. (a)Sewerage systems may be classified assanitary sewers designed to receivedomestic sewage and industrial wastesexcluding storm water, storm sewersdesigned to carry off storm water and groundwater but excluding domestic sewage andindustrial wastes, and combined sewersdesigned to receive domestic sewage,industrial wastes and storm water.
35. (a)
h f =2f L V
2gd
40 =20.025 2000 V
2 10 0.25
V 2 m/s
Critical time (T0) = 2LC = 4 seconds
Water hammer pressure developed when actualtime of closure (T) is much greater than T0 isgiven by
LVpT
p =1000 2000 2Pa
10
p 0.4 MPa
36. (d)Lines joining the loci of places having samevalue of dip are known as isoclinic lines,whereas those joining the loci of placeswith no dip is called as aclinic line such asmagnetic equator.
37. (b)When a sewer line dips below the hydraulicgrade line, it is called an inverted siphon.As the siphons are depressed below thehydraulic grade line, maintaining selfcleansing velocity at all flows is veryimportant. Inverted siphons may needcleaning more often than gravity sewers andhence should not have any sharp bendseither horizontal or vertical.An overloaded existing sewer may requirerelief sewer, with the relief sewer constructedparallel to the existing line. Relief sewersare also called supplementary sewers.Sewage may have to be carried to highelevations through force mains.
38. (c)
h f =2kV
2g ,
where V is flow velocity in the smaller pipe
2
1
AA =
1 0.254
From table (by interpolation);
0.16 0.360.38 0.28
= 0.16 0.25
0.38 K
K 0.335
hm =
0.335 16 0.268m2 10
39. (d)
45°
E
S
N
W
S45°W = 225° (whole circle bearing)Lattitude DA = – [lat. AB + lat. BC + lat.CD] = –685.0Departure of DA = –685
40. (a)
The resistance of vitrified clay pipes tocorrosion from most acids and to erosiondue to grit and high velocities gives it anadvantage over other pipe materials inhandling those wastes which contain highacid concentrations.
41. (a)
K = Bulk modulus of elasticity
=p
V / V
2.2 = – 2P 00.006
P2 = 0.0132 GPa or 13.2 MPa
310 MPa 1 GPa
42. (a)
PQ length = 2 240 – 20 –20 – 30
= 53.85 m43. (a)
Flap gates or backwater gates are installedat or near sewer outlets to prevent backflowof water during high tide or at high stagesin the receiving stream.
Sewer scooters arrangement is an improvedversion of the scrapes and consists of twojacks, a controlling rope and the scooterwith a tight fitting shield, the scootercompletely stops any flow of sewage.
44. (b)
In soap bubble (in air), there are two interfacebetween soap film and air. One inside thebubble and other outside and are of nearly thesame radius.
22 r = 2P r
P =4 8r d
150 = 28
0.4 10
0.075 N/m
45. (c)The height of the tunnel along the centreline
= 3.465 + 1.155 = 4.620 m.
46. (a)
Since the resin removes virtually 100 percentof hardness, it is necessary to bypass a portionof the water and then blend in order to obtainthe desired final hardness.
Modified aeration is similar to the conventionalplug-flow process except that shorter aerationtimes and higher F/M ratios are used. BODremoval efficiency is lower than other activatedsludge processes.
Extended aeration process requires low organicloading and long aeration time.
The oxidation ditch consists of a ring or ovalshaped channel and is equipped withmechanical aeration device.
50. (d)
The U-tube manometer mentioned aboveusually requires the reading of fluid levels attwo or more points since a change in pressurecause a rise of the liquid in one limb of themanometer and a drop in other. The difficulty
is however overcome by using single columnmanometer (micro-manometer) which is amodified form of a U-tube manometer in whicha shallow reservoir having a large cross-sectional area (100 times approx.) ascompared to the area of the tube is connectedto one limb of the manometer. Its sensitivityalso increases due to this modification.
51. (d)WCB of line PQ is given by
3°30
W E
MN
TS
TN
Sun location at noon
55°45
Q
P
= 180 55 45
= 235 45
Magnetic declination = 3 30 W
True bearing of PQ line = magnetic bearing– west declination
= 235 45 3 30
= 232 15
True Quadrantal bearing of line PQ is
= S(232 15 180 ) W
= S52 15 W
52. (b)
6 12 6 2 2 2C H O 6O 6CO 6H O
Weight of glucose = 180 g
Weight of oxygen used = 32 × 6g = 192g
Thus, it takes 192 g of oxygen to oxidize 180g of glucose to CO2 and H2O
as the force diagram of water pressure willbe of triangular shape.
54. (c)Plunging also known as transisting orreversing is the process of rotating thetelescope about the horizontal or trunnionaxis.
Swinging is the process of rotating thetelescope about the vertical axis for thepurpose of pointing the telescope in differentdirections. The right-swing is a rotation inthe clockwise direction and the left swingis a rotation in the counterclockwisedirection. Changing of face is done bytransiting the telescope first then turningthrough 180° (swinging)
55. (c)
The amount of oxygen required to oxidizea substance to carbondioxide and watermay be calculated by stoichiometry if thechemical composition of the substance isknown. This amount of oxygen is known asthe theoretical oxygen demand (ThOD). Thechemical oxygen demand (COD), is ameasured quantity that does not dependon knowledge of the chemical compositionof the substances in the water. In the CODtest, a strong chemical oxidizing agent(chromic acid) is mixed with water, sampleand then boiled.
If the oxidation of an organic compound iscarried out by micro organisms using theorganic matter as a food source, the oxygenconsumed is known as biochemical oxygendemand (BOD). The test is bioassay (tomeasure by biological means), that utilizesmicroorganisms in conditions similar tothose in natural water.
56. (d)
Buoyant force (Fb) = Weight of waterdisplaced by stone
= (105 – 67) kN= 38.0 kN
= w × volume of stone
Volume of stone = 38000
1000 10 = 3.8 m3
Specific gravity
= Weight of stone in air
Weight of equal volume of water
=
105 2.7610 3.8
57. (a)Pixel size and spatial resolution are notinterchangeable.If a sensor has a spatial revolution of 20 mand an image from the sensor is displayedat full resolution, each pixel represents anarea of 20 m × 20 m on the ground. In thiscase, pixel size and resolution are thesame. However, it is possible to display animage with a pixel size different than theresolution. Many posters of satellite imagesof the earth have their pixels averaged torepresent larger areas although the originalspatial resolution of the sensor thatcollected the imagery remains the same.
58. (c)
There are literally thousands of naturallyoccurring organic compounds, not all of whichcan be degraded with equal ease. Simplesugars and starches are rapidly degraded andwill therefore have a very large BOD rateconstant. Cellulose (for example, toilet paper)degrades much more slowly. Hair and fingernailsare almost undegradable in the BOD test or innormal wastewater treatment.
59. (c)
As surface tension is directly dependent uponintermolecular cohesive forces, its magnitudefor all liquids decrease as the temperaturerises.
Sometime air is injected into the flow in orderto prevent cavitation damage, because airbubbles prevent the formation of the vapourbubbles of flowing liquid.
60. (b)
If L = true chain length L = measuredchain length.
In a natural environment receiving a continuousdischarge of organic waste, that population oforganisms which can most efficiently utilizethis waste will predominate. However, theculture used to inoculate the BOD test maycontain only a very small number of organismsthat can degrade the particular organiccompounds in the waste. The result is thatthe BOD rate constant is lower in the laboratorytest than in the natural water.
Most biological processes speed up as thetemperature increases and slow down as thetemperature drops. Because oxygen utilizationis caused by the metabolism of micro-organisms.
62. (c)
It is observed that for a free vortex motionalthough the circulation around the variousstreamlines is not zero yet the motion isirrotational. For an irrotional vortex, thecirculation is zero along any closed contourthat does not enclose the vortex axis, and hasa fixed value of circulation for any contour thatdoes enclose the axis once.
63. (c)
In retrograde vernier, n divisions of vernierscale are equal to (n + 1) divisions of mainscale.
64. (d)
The overall reaction for ammonia oxidation is :
microorganisms3 2 3 2NH 2O NO H O HNBOD/gm N
= grams of oxygen used
grams of nitrogen oxidized
=
2 34 16 4.57g O / g(NH N)
14NBOD = 30 × 4.57 = 137 mg O2/L
65. (b)
V =
2
Q 0.01 1A 0.040.04
4
=1
0.125 = 8
Velocity of flow of oil = 8 m/s
Kinetic energy flux = 2QV
2
= 20.80 1000 0.01 8
2
= 256 W66. (c)
A tie line usually enables the surveyor tolocate the interior details which are far awayfrom the main chain lines. Check line isused to check the accuracy of survey andframework.
67. (b)
The rate at which NBOD is exerted dependsheavily on the number of nitrifying organismspresent. In untreated sewage, there are fewof these organisms, while in a well-treatedeffluent, the concentration is high. Whensamples of untreated and treated sewageare subjected to the BOD test, the NBODis exerted after much of CBOD has beenexerted, in the case of untreated sewage.The lag is due to the time it takes for thenitrifying bacteria to reach a sufficientpopulation for the amount of NBOD exertionto be significant compared with that of theCBOD. In the case of the treated sewage,a higher population of nitrifying organismsin the sample reduces the lag time.
68. (b)
Thrust experienced, F = exit exit i iW V W Vg g
= 4500 0200 10 200
10 10
= 94500 N= 94.50 kN
69. (b)Geoid is the surface of earth which iseverywhere normal to the force of gravity.
70. (b)Fumigating plume is observed when aninversion layer occurs at a short distanceabove the top of stack and super adiabaticcondition prevail below the stack then plumeis said to be fumigating. In such casepollutants can not escape above the top ofstack because of inversion layer.
71. (a)The difference in elevations of water in twolimbs in caused due to velocity head offlowing gas i.e. air.
2u2g =
water
air
1 x
where x = manometer reading
210
20=
1000 x11.2
5 = (833 – 1) x
x =5 0.006m
832
Manometer reading 6mm
72. (b)73. (c)
Long term exposure of PM are likely toaffect our respiratory system. It reduceslung function in children.
Large particles are generally filtered in noseand throat via mucus but particular mattersmaller than 10 µm can settle in the bronchiand lungs and can cause health problem.
74. (a)Using Froude’s law
Vr = rL
Using Reynold’s law,
reR =r r r
r
V L
So,
r r
r
L L = 1
3/2
r r
r
L = 1
3/2r rL
m =3/2
mp
p
LL
=
3/26 140 10
4= 6 25 10 m /s
75. (d)
L veD ve
bearing in south-east
quadrant.
But bearing of closing line will be justopposite to it i.e. in north-west quadrant.
76. (c)
Environment lapse rate = 1.09 5.11
202 10
= –0.0209 °C/m= –20.937 °C/km
This is greater than adiabatic lapse rate.
So, the atmosphere is unstable.
77. (d)
1.
2 30 0 0
53 5 3 3
P ML T M L TD LML T
2. 22 3 1
0 0 01 1
D LML T M L TML T
3.2
1 1D T L
4.
3 10 0 0
33 1
Q L T M L TD LT
78. (b)Two theodolite method is most useful whenground is undulating, rough and it is notsuitable for linear measurements. Themethod is based on the property of thecircle that the angle between the tangentat a point and chord is equal to the anglewhich the chord subtends in the oppositesegments.
79. (d)
Complete removal of hardness can not bedone by lime process, 40 mg/l of CaCO3and 10 mg/l of mg (OH)2 as CaCO3 usuallyremains and after some time will slowlyprecipitate and accumulated inside the pipesso recarbonation is necessary.
Since length of sight to B is 1400 m, thecorrection for curvature and refraction mustbe applied to the observed reading on B.
Combined correction, C = –0.0673 × 1.42
= –0.132 m
The corrected reading on B = 3.693 – 0.132= 3.561 m
Hence, the true difference = 3.561 – 0.584= 2.977 m
A O B
50m 1400m
82. (d)
Total hardness (in mg/l) =
[Milliequivalent of Mg++ + Milliequivalent ofCa++] × equivalent weight of Ca(HCO3)2
= 20 60 16224 40
22 2
= 378
83. (a)
K / = relative roughness
K = av. height of roughness projection
As increases, K / decreases.
Thus, plate will first behave as rough,followed by transiiton and downstreamportion behaves as hydrodynamicallysmooth.
If
u k* 5 smooth
u k* 70 Rough
u k*5 70 transition
84. (b)
Instrument at A : Apparent difference of level= 2.005 – 1.275 = 0.730
Instrument at B : Apparent difference of level= 1.660 – 1.040 = 0.620
True difference of level (true rise fromB to A)
= 0.730 + 0.620
2 = 0.675
Instrument at B : True reading on A
= 1.660 – 0.675 = 0.985
But observed reading on A = 1.040
Line of collimation is inclined upwards
e = 1.040 – 0.985 = +0.055 m
85. (c)
Efficiency of trickling filter =100
1 0.44 u
= 100
101 0.441000
= 95.78%
86. (b)
87. (d)
The temporary adjustments of the theodoliteare
1. Setting up the instrument
2. Focussing the eye piece and object glass(elimination of parallax)
3. Levelling up the instrument
88. (a)
The retention of sludge depends on thesettling rate of sludge in SST. If sludgesettles well in SST proper recirculation ofsludge in aeration tank is possible. Thiswill help in maintaining desired SRT in thesystem otherwise. If the sludge has poorsettling properties, it will not settle in theSST and recirculation of sludge will bedifficult and this may reduce SRT in system.
89. (a)
A
B
= L
F(L/2)
F
C 2L 2C L
A = B2 i.e., A B
90. (a)
We have, es where s = scale of map =0.25 mm (given)
NS, AD is running EW and its bearing =270° and BDA = 270° – 240° = 30°).
AB = 25 m
25 1tan BDAAD 3
AD = 25 3
AD = 25 3BDC = 330° – 250° = 90°
and ADC = 90 – 30° = 60°From triangle ADC
CA = AD tan ADC
= 25 3 tan60 75m
100. (b)
101. (a)
dudy = 1
3
v 1.50 500sh 3 10
=2du 0.2 500 100N/m
dy
102. (c)
True error : The true error is the differencebetween observed value and the true valueof a quantity.
Most probable error : The most probableerror is the quantity which, when added toor subtracted from the most probable value,defines the limits within which there isgreater probability that true value may lie.
Residual error : Residual error is thedifference between the observed value andthe most probable value of a quantity.
103. (c)
Settling velocity does not depend on lengthof channel.
104. (d)
1. The coefficient of discharge Cd, is definedas
Cd =th
Actual discharge QTheoretical discharge Q
2. The coefficient of resistance, Cr, is definedas the ratio of loss of kinetic energy as theliquid flow through an orifice and the actualkinetic energy possessed by the flowingfluid.
105. (d)
For sum and differences, the weight is thereciprocal of the sum of the reciprocal
weight =1 12
1 1 74 3
weight of A + B = 127
Similarly, weight of A – B = 127
106. (b)
((BOD)5)mix =
ww5 wastewater
sw5 seeded water
ww sw
VBOD
VBOD
V V = 7
((BOD)5)seeded water = 2 mg/l
7 = 5 wastewater
30 2 270BOD300
(BOD5)wastewater = 52 mg/l
107. (a)
Y
Xx
b
y p
zd c
c
bd
yy
pp xx
px y z x.dy. zy x
= 0
y =
px
i.e., pressure gradient in the direction offlow is equal to the shear gradient in thenormal direction of flow.
The negative sign for the pressure gradient
px
indicates a decrease of fluid pressure
in the direction of flow.
108. (d)
Co-declination : (The complement ofdeclination) is the angular distance of aheavenly body from the nearer pole.
Co-latitude : (The complement of latitude)is the angular distance from the zenith tothe pole.
a heavenly body from the zenith. It is thecomplement of altitude.
Right ascension : The right ascension ofa heavenly body is the angular distancemeasured along, the equator measuredtowards the east from the vernal equinox.
109. (b)
K =
2 1
1 2
Qln(r / r )2 D(S S )
K =
10152.303log 2000
0.152 10 0.6 0.1
= 293.23 m/day
110. (c)
I. In case of laminar flow, the shear stress is entirely due to viscous action andtherefore it may be evaluated by usingNewton’s law of viscosity, according towhich,
=Vµy
II. In order to overcome the shear resistanceto flow, the pressure drops from section tosection in the direction of flow.
111. (d)
Two equations can be set up for the twomeasurements
D =KS Cm
S = stadia interceptm = sum of reading on micrometer
60 =K 1.5 C22.5
...(i)
120 =K 1.5 C11.28
...(ii)
On solving equation (i) and (ii)
K = 904.8 and C = 0.32
112. (b)
Vs = 2g dG 1
18
= 23
2 4
9.8 2.6 1 0.15 1018 10 10
= 0.0196 m/sec
113. (a)
The boundary layer thickness increases asthe distance from leading edge increases.
114. (d)
Electronic tacheometers work on theprinciple of electromagnetic distancemeasurement. The phase differencebetween the transmitted signal and thereceived signal is measured and convertedinto distance measurements.
The error in the calculated horizontaldistances should not exceed 1 in 500 for asingle observation under average conditions.For vertical distances, the error should notexceed 0.1 m.
The accuracy that can be obtained with atacheometer depends on the instrumentsfeatures and also on the natural conditionsof temperature, wind etc.
I. A very wide wake is developed on thedownstream side of the disc to be exposedto a zone having pressure considerablybelow that on the upstream side.
II. With the increase in Reynolds number, theregion in which the influence of viscosity ispredominant is considerably reduced andis restricted only to a very small zone ofboundary layer formed close to the sphere.
121. (c)
The reaction of NH3 with water may berepresented by
3 2 4NH H O NH OH
From this reaction, increasing the pH willdrive the reaction to the left, increasing theconcentration of NH3. This makes ammoniamore easily removed by stripping. Inpractice, the pH is increased to about 10to 11 by using lime. Stripping is done byintroducing the wastewater at the top of thecolumn and allowing it to flow down andcountercurrent of air introduced at thebottom. The stripping medium inside thecolumn may be composed of packings orfillings trays. The liquid flows in thin sheetsaround the medium, thereby allowing moreintimate contact between liquid and thestripping air.
122. (a)
The process of eutrophication is directlyrelated to the aquatic food chain. Algae usecarbondioxide, inorganic nitrogen,orthophosphate and trace nutrients forgrowth and reproduction. These plants serveas food for microscopic animals. Smallfishes feed on microscopic animals andlarge fishes consume small ones. Abundantnutrients unbalance the normal successionand promote blooms of blue green algaethat are not easily utilized as food bymicroscopic animal. Thus, the waterbecomes turbid. Floating masses of algaeare windblown to the shore where theydecomposes. Decaying algae also settleto the bottom, reducing dissolved oxygen.
123. (a)
Duckweeds are very small floating aquaticplant, only a few millimeters in size, it canbe easily swept off from the water surfaceto one side by wind or waves. Therefore, afloating grid system of plastic, bamboo orany other suitable material has to beinstalled to ensure quiescent conditions.
124. (d)
Aerobic decomposition is the method ofchoice for large quantit ies of dilutewastewater (BOD5 less than 500 mg/L)because decomposition is rapid, efficientand has a lower odour potential. For high-strength wastewater, (BOD5 is greater than1,000 mg/L), aerobic decomposition is notsuitable because of the difficulty in supplyingenough oxygen and because of the largeamount of biological sludge produced.
125. (d)
The BOD rate constant for raw sewage is0.15 – 0.30 and well-treated sewage is 0.05– 0.10.The lower rate constants for treatedsewage compared to raw sewage result fromthe fact that easily degradable organiccompounds are removed while less readilydegradable organic compound remains anddecrease the rate constant.
126. (b)
If CO2 concentration is high it will consumemore lime, so the amount of lime availablefor removing carbonate hardness is reduced.Hence CO2 is removed by aeration prior toremoval by lime.
E-coli are present in intestine of humanbeing, the life of E-coli bacteria is greaterthan fecal bacteria, so if E-coli is absent,it can be assumed that fecal bacteria mustbe absent
128. (b)
Proper supply of air and proper design tomaintain endogenous growth phase ofmetabolism will help in avoiding bulking ofsludge.
129. (d)
The pH in aeration basin should be kept at6.5 – 8.5
Lower pH can induce the growth offilamentuous organisms and cause bulkingof sludge
The pH should be adjusted using caustic,lime or magnesium hydroxide
130. (a)
Sludge age is the amount of time, in daysthat solids or bacteria are under aeration.Sludge age is used to maintain the properamount of activated sludge in aeration tanks.
131. (d)
Not all of the liquids completely wet thesurface eg. mercury, but most of them do,because attraction of liquid molecules withsolid surface is greater than that of betweenthe liquid molecules.
132. (c)
Since a streamline is every where tangentto the velocity vector, there can be nocomponent of the velocity at right angles tothe streamline and hence there can be noflow of fluid across any streamline.
133. (c)
For dp 0,dx
the thickness of boundary layer
increase because in case of divergent flowthe fluid is decelerated which assists inthickening of the boundary layer.
134. (d)
At the boundary, the velocity gradient dvdy
is steeper in turbulent than laminar boundarylayer, because in turbulent boundary layernear the boundary layer velocity changeoccurs in a relatively small vertical distance.
135. (a)
Dynamic viscosity varies widely withtemperature. For gases, it increases withincrease in temperature while for liquids itdecreases with increases in temperature,and this is due to fundamental differencebetween inter molecular characteristics.
The cohesive forces decreases when thereis increase of temperature in case of liquidsleading to decrease in viscosity whilerandomness in case of gases increasesleading to increase of dynamic viscosity.
136. (c)
When meta centric height coincides withthe centre of gravity, it results into neutralequilibrium because there will neither beany restoring couple nor an overturningcouple developed when the body isdisplaced slightly, but will simply move toa new position when displaced slightly.
137. (a)
In case of unsteady flow, the velocity vectorchanges from time to time and hence thereis a change in streamline patterns. Whilein case of steady flow there is no changeof velocity vector and hence no change instreamline flow pattern takes place, thereforein case of unsteady flow we generally takeinstantaneous flow pattern.
138. (a)
The actual kinetic energy possessed bythe flowing fluid is greater than thatcomputed by using the mean velocity.
If velocity distribution is non-uniform thenkinetic energy correction factor is introducedto account for the difference between kineticenergy calculated using mean velocity andactual kinetic energy.
139. (b)
The introduction of the equivalent sand grainroughness (Ks) instead of average height ofroughness projection (K) is due to the factthat as compared to K, Ks is morerepresentative parameter of actualroughness. This is so because usually the
actual roughness projection on the surfaceof the plate are of various height and theydo not have any uniform pattern and hencesuch roughness cannot be truelyrepresentative.
140. (b)
141. (a)
In “offsets from chord produced” method ofsetting out curve
Length of any offset = (Length of previouschord + length of chord) divided by ‘2R’
Thus,
O1 = 1O C2R
O2 = 2 1 2C (C C )2R
O3 = 3 2 3C (C C )2R
. .
. .
. .
i.e. On = n n 1 nC (C C )2R
Thus it is noted that offsets of subsequentchord depends on the previous chord aswell, i.e. each successive point dependsupon the accuracy of previous point. Thuserror gets accumulated.
142. (a)
To enable stereoscopic vision, it is essentialto have photographs in main coveringcommon areas. In aerial photography,photographs are taken from two cameraposition with sufficient overlap in thephotographs.
Only the areas seen in both the photographsare amenable to stereoscopic vision
Stereoscopic vision enables one to view thedepth or height of points on the photograph,when the photographs are suitably placedand the left eye views one and the righteye views the other, the common areacomes in relief or in three dimensions dueto stereoscopic fusion.
143. (c)
A traverse is said to be closed when acomplete circuit is made, i.e., when itreturns to the starting point forming a closed
polygon (Fig. A) or when it begins and endsat point whose positions on planes areknown (Fig. B)
A
B
C
E D
Fig. A
A
B
C D
E
Fig. B
144. (d)
Incorrect holding of the chain leads tocompensating error.
145. (d)
We shall always follow whole to part forlocalizing and preventing accumulation oferror.
146. (a)
147. (c)
The line of collimation of a theodolite mustbe perpendicular to the horizontal axis atits intersection with the vertical axis. If thiscondition exists, the line of sight willgenerate a vertical plane when the telescopeis rotated about the horizontal axis.
If the line of sight is not perpendicular tothe trunnion axis of the telescope, it willnot revolve in a plane when the telescopeis raised or lowered but instead, it will traceout the surface of a cone.
148. (d)
Height of instrument method is generallyused where more number of reading can betaken with less number of change points.
149. (c)
Relief displacement : It is caused bychanges in the distance between the groundand the camera as the plane flies over theground.
Characteristics of relief displacement
Characteristics of aerial images over variedterrain.
Objects that rise above the surface awayfrom the principal point.
Objects extending below the surface leantowards the principal point.
