IES M

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CE (Test-15), Objective Solutions, 16th April 2016 (1)

1. (b)

2. (b)

3. (d)

4. (c)

5. (a)

6. (c)

7. (d)

8. (a)

9. (d)

10. (c)

11. (d)

12. (b)

13. (b)

14. (b)

15. (b)

16. (b)

17. (b)

18. (b)

19. (c)

20. (a)

21. (a)

22. (c)

23. (b)

24. (d)

25. (a)

26. (b)

27. (d)

28. (c)

29. (c)

30. (a)

31. (a)

32. (d)

33. (b)

34. (b)

35. (c)

36. (c)

37. (c)

38. (a)

39. (b)

40. (d)

41. (d)

42. (b)

43. (d)

44. (b)

45. (a)

46. (a)

47. (b)

48. (c)

49. (c)

50. (b)

51. (d)

52. (d)

53. (c)

54. (a)

55. (b)

56. (b)

57. (d)

58. (a)

59. (b)

60. (a)

61. (b)

62. (d)

63. (b)

64. (b)

65. (d)

66. (c)

67. (b)

68. (d)

69. (a)

70. (a)

71. (b)

72. (a)

73. (a)

74. (d)

75. (b)

76. (c)

77. (d)

78. (b)

79. (b)

80. (b)

81. (a)

82. (d)

83. (a)

84. (d)

85. (b)

86. (c)

87. (d)

88. (d)

89. (a)

90. (b)

91. (b)

92. (a)

93. (d)

94. (c)

95. (d)

96. (c)

97. (c)

98. (d)

99. (b)

100. (b)

101. (a)

102. (a)

103. (a)

104. (c)

105. (a)

106. (b)

107. (a)

108. (b)

109. (b)

110. (c)

111. (a)

112. (a)

113. (b)

114. (a)

115. (a)

116. (d)

117. (a)

118. (a)

119. (d)

120. (d)

Conventional Question Practice ProgrameDate: 16th April, 2016

ANSWERS

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(2) CE (Test-15), Objective Solutions, 16th April 20161. (b)

S.F. = Shrunk length Shrunk scaleOriginal length Original scale

24 Shrunk scale25 1/ 2400

Shrunk scale = 1

2500

2. (b)

sin

Cg = L(1 cos )

Cg = 22 sin / 2

3. (d)4. (c)

If CP is the correction for pull, we have

CP = 0(P P ) LAE

where

P = Pull applied during measurement (N)

P0 = Standard pull (N)

L = Measured length (m)

A = Cross-sectional area of the tape (cm2)

E = Young’s modulus of elasticity(N/cm2)

Here L = 1500 m, P0 = 100 N, P = 150 N

CP = (150 100) 1500 50 1500

AE AE

5. (a)The additive constant, c = f + d

Multiplying constant, K = fi

where,

f = focal length of objective

d = horizontal distance between opticalcentre and vertical axis oftacheometer

Here, i = 5 mm, f = 25 cm, d = 15 cm

c = 0.25 + 0.15 = 0.4 m

K =f 25

0.5

i = 50

6. (c)

f = 20 cm; d = 10 cm

i = 4 mm; s = 2.500 m.

C = f + d = 0.2 + 0.1 = 0.3

K =f 20i 0.4 = 50

Staff interept,

S = 2 × (2.5 – 1) = 3.0 m

Horizontal distance between the staff stationand instrument station,

D = Ks + C

= 50 × 3 + 0.3

= 150.3 m

7. (d)

% of error between 3 under gussian lawof distribution = 99.7%

No. of errors that are expected to exceedthe limit of 3 = (1 – 0.997) × 1000= 3 × 10–3 × 1000 = 3

8. (a)

9. (d)

If a computed quantity is a function of two ormore observed quantities, its probable erroris equal to the square root of summation ofthe squares of the probable errors. of theobserved quantity multipl ied by i tsdifferentiation with respect to that quantity.

Let y = computed quantity

x1, x2, x3 etc. = observed quantities

Such that y = f(x1, x2, x3 etc.)

Then

ey = 1 2 3

2 2 2

x x x1 2 3

dy dy dye e edx dx dx

where ey = Probable error of the computedquantity 1 2 3x x xe , e , e = Probable errors ofobserved quantities. Let area A = a · b

eA = 2 2a be · b e · a

= 2 2(0.05 180) (0.06 120)

= ± 11.53 sqm.

where eA = Probable error in the area.

IES M

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CE (Test-15), Objective Solutions, 16th April 2016 (3)10. (c)

The local attraction at any station isdetected by observing the fore and backbearings of the line.

If local difference between them is 180º,both the end stations are considered tobe free from local attraction, provided thecompass is not having any instrumentalerrors.

If the difference is not 180º, thediscrepancy introduced may be becauseof presence of local attraction at either orboth of the station.

310º 30 – 135º 15 180º

200º 15 – 17º 45 180º

246º 0 – 65º 15 180º

11. (d)

180 19 47 13 160 12 47

12. (b)If the magnetic declination at a place at thetime of observation is known the true bearingof a line can be determined from its magneticbearing and vice-versa. If declination is west,True bearing = Magnetic bearing – declination 89º = magnetic bearing – 1º Magnetic bearing = 90º

Magnetic bearing of BA = 180º – 90º = 90º13. (b)14. (b)

RP

Q129º30

59º

Line FB BBPQ 59º0 239º0QR 129º30 309º30 PQR = –FB of line QR + BB of PQ

= –129º30’ + 239º0’ = + 109º30’(interior included angle)

15. (b)The line of collimation should be parallel tothe axis of the tube when the vertical circlereading is zero.

The axis of the altitude level tube is trulyhorizontal when the bubble is in the centre.

16. (b)17. (c)18. (b)

Ist sub chord = 2220 – 2002.48 = 17.52 m

19. (c)Radius of curvature of the bubble tube

= R = ndL

s

=35 2 10 100 20m

0.05

where

n = 5

d = 2 mm

L = 100 m

s = 0.05 m

20. (a)The importance of the two peg test result isthat even when the levelling instrument is notin correct adjustment, the difference in heightmeasured between two points by a levelequidistant from each gives the true differencein height.

21. (a)Last sub chord = 2303.39 – 2300 = 3.39 m

22. (c)23. (b)

Contour lines are imaginary lines passingthrough points of equal elevations.

24. (d)

Contour lines close together indicate a steepslope.

25. (a)

Width of ground to be covered

= (1 – 0.6) × 0.23 × 20,000 = 1840 m.

26. (b)

Relief displacement of a point,

d =r.hH

where h = height of the object above datum,

H = flying height above the datum,

r = radial distance of the image of thetop of the object from principalpoint.

d = 90 500 9mm

5000

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(4) CE (Test-15), Objective Solutions, 16th April 201627. (d)

28. (c)

29. (c)

Principal Point : It is a point where aperpendicular dropped from the front nodalpoint strikes the photograph.

Isocentre : It is the point in which thebisector of the angle of tilt meets thephotographs.

Crab : It is the term used to designate theangle formed between the flight line andthe edges of the photograph in the directionof flight.

Drift : It is caused by the failure of theaeroplane to stay on the predeterminedflight line.

30. (a)

The Right Ascension (RA) of a celestial bodyis the angular distance measured along thecelestial equator between the first point of Ariesand declination circle of the body.

31. (a)

Lehmann’s method or Trial and error methodin the field is used to find out the positionof the station of a plane table.

32. (d)

R

T1 T2D

V

C

Apex distance, VC = R (sec /2–1)

Length of long chord = T1 DT2 = 2R sin/2

Mid ordinate (M) = Length CD

= R (1–cos/2)

= R versin /2

33. (b)Normal

1

1

2

Soil-1 (K )1

Flowlines 2

Soil-2 (K )2

Flow linesNormal

When two different soils are used in a soilmass, thus making it non-homogeneous. Theflow lines and equipotential lines get deflectedat the interface. The flow net thus get modified.

K1 and K2 are related as

1

2

KK =

1

2

tantan

34. (b)

h = 6 m, Nf = 6; Nd = 18

K = 4 × 10–5 m/min = 4 × 10–5 × 60 × 24

= 0.0576 m/day

Seepage discharge = Khf

d

NN

= 0.0576 × 6 × 6

18= 0.1152 m3/d per m length

35. (c)

e = 0.60

For quick sand condition,

ic = G 11 e

ic = cG 1 1.6 i G 1

1 0.6

G = 1.6 ic + 1

36. (c)

G = 2.60, n = 0.33

ic = (G –1) (1 – n)

= ( 2.60 – 1) (1 – 0.33) = 1.072

37. (c)

Fs = s

tan ' tan ' tan30tan itan i F 1.732

= 0.333

i = tan–1(0.333)

38. (a)

B = 2 m, Df = 2m

C = 30 kN/m2 ; sat = 20 kN/m3

As per skempton's theory,

fDB

= 2 12

NC = 5Df B1 0.2 1 0.2B L

= 5(1 + 0.2 × 1) ( 1 + 0) = 6.0

qnu = CuNc = 30 × 6 = 180 kN/m2

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CE (Test-15), Objective Solutions, 16th April 2016 (5)39. (b)

= 0.70Cu = 40 kN/m2

F.S. = 2.50

B = 0.3 m

D = 10 m

Qu = qbAb + .c. (p D)

= (9 × 40) ×4

× 0.32

+ 0.7 × 40 ( × 0.3 × 10)

= 289.194 kN

Qa = 4Q 289.194 115.68 kNF.S. 2.5

40. (d)Total stress in multi-layered soil :The total stress at depth z is the sum of theweights of soil in each layer thickness above.

Vertical total stress at depth z,

v = 1 1 2 2 3 1 2d d (z d d )

where,

1 2 3, , = unit weights

of soil layers 1, 2, 3, etc.

respectively

z

d1

d2

1

2

3

1

2

3

The increase in vertical stress at a depthof 1 m is 36 kN/m2, this will be the increasein vertical stress every where below the surfacelevel.

41. (d)The permeability

K =3

2w eC D1 e

where C = a constant depending upon shapeof c/s of flow.

= dynamic viscosity of fluid

e = void ratio

D = grain size

42. (b)

= sub h (Assuming 3sub 10kN/m )=

50 = 10.h

h = 5m

43. (d)Q = KAi

QA = 2 hK D4 2L

QB = 2 hK 4D4 L

A

B

QQ =

2

2

hK D 14 2L =h 8K 4D4 L

= 0.125

44. (b)K = 10-3 cm/s

h = 1m

L = 2m

A = 100cm2

Q = KA hi K A=L

dVdt =

hK A.L

V = (10-3cm/s) (100cm2)

×1m2m

× 60 sec = 3ml

45. (a)

Q = hK.A.L

(For constant headpermeameter)

Q hL

QQ =

2h 1 4h LL / 2 (H / L) L h

Q = 4Q

46. (a)As per Allen Hazen’s formula

K = 210C.D

Where K = Coeff. of s permeability (cm/sec)

D10 = effective size (cm)

C = Constant, with a value between100 and 150.

Ratio of permeability = 2.6 4

0.3

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(6) CE (Test-15), Objective Solutions, 16th April 201647. (b)

Kh = 3m/day

KV =1 m/day3

K = h V1K .K 3 1m/day3

48. (c)Failure by piping or underminingWhen the seepage water retains sufficientresidual force at the emerging downstream endof the work, it may lift up the soil particles.This leads to increaed porosity of the soil byprogressive removal of soil from beneath thefoundation. The structure may ultimatelysubside into the hollow so formed, resulting inthe failure of the structure.The exist gradient is said to be critical whenthe upward distributing force on the grain isjust equal to the submerged weight of thegrain at the exit. When a factor of safety equalto 4 or 5 is used, the exit gradient. Can thenbe taken as safe In other words, an exit

gradient equal to 14 to

15 of the critical exit

gardient is ensured, so as to keep the structuresafe against piping.

49. (c)In case of a falling head permeameter,

K = 1

102 1 2

h2.303La logA t t h

where L = 10 cma = 1 cm2

A = 50 cm2

(t2 – t1) = 1 hour = 3600 sec.

1

2

hh =

80 240

Hence K =2.303 10 1 0.3

50 3600

= 3.84×10–5 cm/sec.50. (b)

H = 10 mDf H = 15 m

Df =15 1.5010

Sn =c

CF H

0.164 = c35 F 1.15

Fc 18.5 10

51. (d) The steady seepage condition is critical

for the d/s slope of an earth dam. The critical condition for the stability of the

u/s slope of an earth dam is when there isa sudden drawndown in the reservior u/s.

52. (d)

Cm =C 15 10

1.5 1.5

Sn = 0.046 =10H

H =10 11.5 m

19 0.046

53. (c)

Swedish circle method: The actual shapeof a slipsurface in the case of finite slopes iscurvilinear. For convenience,it is approximatedas circular. The assumption of a circular slipsurface and its application for stability analysisof slopes was developed in sweden. Themethod is known as the swedish circle methodor the method of slices.Stability member (Sn): It is defined as

Sn = m

H c

C CF H

The reciprocal of the stability number is knownas stability factor. The stability number is adimensionless quantity. The stability numbercan be used to determine the factor of safetyof a given slope.Sudden drawdown conditions: When thewater standing on the slope is suddenly andquickly removed, the water pressure (U)disappears. However, if there is no time fordrainage to occur from the soil in the slope,the soil remains submerged as before and thenatural part of the weight is still acting. Thus,the equilibrium of the neutral force is distrubed,although the equilibrium of the inter granularforces remains unaffected.Critical void ratio: It may be observed thatthe void ratio of an initial loose sand decreaseswith as increase in shear strain whereas thatfor the initially dense sand increases with anincrease in strain. The void ratio at which thereis no change in it with an increase in strain isknown as the critical void ratio. If the sandinitially is at critical void ratio, there would bepractically no change in volume with anincrease in shear strain.Where there is soft clay there is a chance ofbase failure

IES M

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CE (Test-15), Objective Solutions, 16th April 2016 (7)54. (a)

The following are the important points ofcomparison of coulomb’s theory with Rankine’stheory: Coulomb considers a retaining wall and

the backfill as a system; he takes intoaccount the friction between the wall andthe backfill, while Rankine does not.

The backfill surface may be plane or curvedin coulomb’s theory, but Rankine’s allowsonly for a plane surface.

In coulomb’s theory, the total earth thrustis first obtained and its position anddirection of the earth pressure are assumedto be known; linear variation of pressurewith depth is tacitly assumed and thedirection is automatically obtained from theconcept of wall friction. In Rankine’s theory,plastic equilibrium inside a semi-infinite soilmass is considered, pressure evaluated, aretaining wall is imagined to be interposedlater, and the location and magnitude ofthe total earth thrust are establishedmathematically.

Coulomb’s theory is more versatile thanRankine’s in that it can take into accountany shape of the backfill surface, break inthe wall face or in the surface of the fill,effect of stratification of the backfill, effectof various kinds of surcharge on earthpressure and the effects of cohesion,adhesion and wall friction. It lends itself toelegant graphical solution and gives morerel iable results, especial ly in thedetermination of the passive earthresistance; this is in spite of the fact thatstatic equilibrium condition does not appearto be satisfied in the analysis.

Rankine’s theory is relatively simple andhence is more commonly used, whilecoulomb’s theory is more rational andversatile although cumbersome at times,therefore, the use of the later is called forin important situation or problems.

55. (b)When a wall moves away from the backfill,some portion of the backfill located immediatelybehind the wall tries to break away from therest of the soil mass. This wedge–shapedportion, known as the failure wedge or thesliding wedge moves downward and outwards.The lateral earth pressure exerted on the wallis a minimum in this case. The soil is at theverge of failure due to a decrease in the lateralstress.

The horizontal strain required to reach the activestate of plastic equilibrium is very small. Ithas been shown that in dense sand, thehorizontal strain required is about 0.5%

It is thus found that very little horizontal strain(about 0.50%) is required to reach one halfthe maximum passive pressure in dense sandbut much more horizontal strain (about 5% indense and and 15% in loose sand) is requiredto reach the full passive pressure.

It may be summarised that the state ofshear failure corresponding to the minimumearth pressure is the active state and thatcorresponding to the maximum earth pressureis the passive state.

56. (b)

Unsupported height = 4C 4 5 1m

20

57. (d)

Ka = 21 sin tan= /21 sin 4

Kp = 21 sin tan= 45 /21 sin

p

a

KK =

2

2tan 45º /2tan 45º – /2

aK =p

1K

p

a

KK =

2

4

2

tan 45 /2 tan= 45 /21

tan 452

58. (a)

Given that, KA = 2 1tan 45º2

=

14

1tan 45º2

=

12

1tan 45º2

= 2

C1 = 20 kN/m2

IES M

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(8) CE (Test-15), Objective Solutions, 16th April 2016

H1 = 1 1

1

4C tan 45º2

H1 = 4 20 2 8m

20

H2 = 2

2

4C 4 40 8m20

H2/H1 = 8 18

59. (b)

For square footing,

Net ultimate bearing capacity,

qnu = (5.7 c) 1.3

= 5.7 × 1.3 × 1 kg/cm2

= 7.41 kg/cm2

= 74.1 t/m2

60. (a)The allowable bearing capacity,

Q1 = 15 t/m2 for allowable settlement,

S1 = 25 mmm

The allowable bearing capacity, Q2 for allowablesettlement, S2 = 40 mm

But Housel’s method,

1

1

QS = 2

2

QS

Q2 = 21

1

S QS

= 240 15 24 t/m25

61. (b)The ultimate bearing capacity of purelycohesive soil is given by qu = CNC+ q

Where q = surcharge due to increase in depthof foundation.

As we can observe from the above equationthat bearing capacity of footing on purelycohesive soil is independent of size of footing.

62. (d)Pile foundations are used in the followingcondition:

When the strata at or just below the groundsurface in highly compressible and veryweak to support the load transmitted bythe structure.

When the plans of the structure is irregularrelative to its outline and load distribution.It would cause non-uniform settlement if ashallow foundation is constructed. A pile

foundation is required to reduce differentialsettlement.

Pile foundation are required for thetransmission of structural loads throughdeep water to a firm stratum.

Pile forces are used to resist horizontalforces in addition to support the verticalloads in earth-retaining structures and tallstructures that are subjected to horizontalforces due to wind and earthquake.

Piles are required when the soil conditionare such that a washout, erosion or scourof soil may occur from underneath ashallow foundation.

In case of expansive soils, such as blackcotton soil, which swell or shrink as thewater content changes, piles are used totransfer the load below active zone.

63. (b)

Raft foundations on sands are quite useful.Their bearing capacity failure is generally outof question, because the bearing capacity insands increases with the width of the footingand this width in rafts is quite large. Thedifferential settlements in rafts are alsogenerally smaller as compared to isolatedfootings even for the same load intensitybecause a raft eliminates the influence of localloose soils.

Soil Type

Permissible total settlement

Permissible differential settlement

For isolated footings

For raft foundation

For isolated footing

For raft foundation

Sandy 4 cm 4 to 6.5 cm

2.5 cm 2.5 cm

Clayey 6.5 cm 6.5 to 10 cm

4 cm 4 cm

64. (b)65. (d)66. (c)

General shear failureGeneral shear failure is seen in dense andstiff soil. The following are characteristics ofgeneral shear failure :

Continuous well defined and distinct failuresurface develops between the edge offooting and ground surface.

Dense or stiff soil that undergoes lowcompressibility experiences this failure.

IES M

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CE (Test-15), Objective Solutions, 16th April 2016 (9) Continuous bulging of shear mass adjacent

to footing is visible.

Failure is accompanied by tilting of footing.

Failure is sudden and catastrophic withpronounced P- curve.

The length of disturbance beyond the edgeof footing is large.

State of plastic equilibrium is reachedinitially at the footing edge and spreadsgradually downwards and outwards.

General shear failure is accompanied bylow strain (<5%) in a soil with considerable ( > 36º) and large N (N > 30) havingrelative density (ID > 70%)

Local shear failure This type of failure is seen in relatively loose

and soft soil. The following are somecharacteristics of general shear failure.

A significant compression of soil below thefooting and partial development of plasticequilibrium is observed.

Failure is not sudden and there is no tiltingof footing

Failure surface does not reach the groundsurface and slight bulging of soil aroundthe footing is observed.

Failure surface is not well defined.

Failure is characterised by considerablesettlement.

Well defined peak is absent inp curve.

Local shear failure is accompanied by largestrain (> 10 to 20%) in a soil withconsiderably low ( <28º) and low N (N<5)having low relative density (ID>20%).

Punching shear failure of foundation soilsThis type of failure is seen in loose and softsoil and at deeper excavations. The followingare some characteristics of general shearfailure :

This type of failure occurs in a soil of veryhigh compressibility.

Failure pattern is not observed

Building of soil around the footing is absent.

Failure is characterised by very largesettlement.

Continuous settlement with no increase inp is observed in p- curve.

67. (b)Observed reading of SPT No. = No

= 6 + 9 = 15

Overburden pressure = = 18×3

= 54 kN/m2 < 280 kN/m2

Hence we apply overburden correction.

N1 = 0350N

70

= 35015 42.354 70

Water table lies at the same depth. Hencewater table correction is also applied.

N2 = 115 42.3 152

= 28.67 2968. (d)

Qu = u base c b u s(C N ) A ( . C ) A

where = adhesion factorCu = undrained cohesion in

embedded length of pile.Cu base = undrained cohesion at the base

of the pileNC = 9

Qu = 2(4 9) 0.5 2 0.5 104

= 2.25 10 12.25

=49 tonnes.

4

69. (a)It has been observed that group efficiency ofdriven piles in loose or medium dense sand is > 1. This is because soil arround andbetween the piles get compacted due to thevibration caused during the driving operations.Whereas in dense sand above phenomenonis not true.

70. (a)The load carrying capacity of a driven pile canbe estimated from the resistance againstpenetration developed during driving operation.The methods give fairly good results only inthe case of free-draining sands and hard claysin which high pore water pressures do notdevelop during the driving of piles. In saturatedfine grained soils, high pore water pressuredevelops during the driving operation and thestrength of the soil is considerably changedand the methods do not give reliable results.

IES M

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(10) CE (Test-15), Objective Solutions, 16th April 201671. (b)

Types of FoundationSpread Footings Used for most buildings where the loads

are light and or there are strong shallowsoils.

Generally suitable for low rise buildings (1–4 stories).

Requires firm soil condition that are capableof supporting the building on the area ofthe spread footings.

These are most widely used because theyare most economical.

Spread footings should be above the watertable.

Drilled piers or Caissons For expansive soils with low to medium

loads or high loads with rock not too fardown, drilled caissons (piers) and gradebeams can be used.

The caissons might be straight or belledout at bottom to spread the load.

Caissous deliver the load to soil of strongercapacity which is located not too far down.

Piles

For expansive soils or soils that arecompressive with heavy loads where deepsoils can not take building load and wheresoil of better capacity is found deep below.

There are two types of piles

1. Friction piles–used where there is noreasonable bearing stratum and they relyon resistance from skin of pile againstthe soil.

2. End bearing– which transfers directly tosoil of good bearing capacity.

Cast in situ piles are composed of holedrilled in earth and then filled with concrete,it is used for light loads on soft groundwhere drilling will not cause collapse.

Mat Foundations

Reinforced concrete raft or mats can beused for small light load buildings on veryweak or expansive soils such as clays.

They are often post tensioned concrete.

They allow the building to float on or in thesoil like a raft.

Can be used for buildings that are 10–20stories tall where it provides resistanceagainst overturning.

Can be used where soil requires such alarge bearing area and the footing might bespread to the extant that it becomes moreeconomical to pour one large slab moreeconomical-less form-work.

It is used in lieu of driving piles because itcan be less expensive and less obstrusive.

Usually used over expansive clays, silts tolet foundation settle without greatdifferences.

72. (a)73. (a)74. (d)75. (b)

NB sscX f

HRT

=

8 24 20

6 = 640 mg/l

76. (c)Bacteria produces highest biomass amongother micorbs.

77. (d)78. (b)79. (b)

h = Sy + Sr

0.40 = Sy + 0.15 Sy = 0.25where Sy = Specific yield Change in groundwater storage of theaquifer= 0.25(23 – 20) × 150 = 112.5 ha.m

80. (b)Q = 2700 lit/min = 0.04533 m3/sFor confined aquifer,

Q = 1 2

2

1

2 T S Srlnr

here r1 = 10 m; S1 = 3m r2 = 100 m; S2 = 0.5 m

0.04533 = 2 T 3 0.5100ln10

T = 6.6487 × 10–3 m2/s = 574.44 m2/day81. (a)

When the flow in normal to the stratification,the equivalent permeability Ke of the aquifer

Ke =

n

i1

ni

i1

L2 4 3 9 12.7 m

2 4 3 1 1 1L6 16 24 3 4 8K

IES M

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CE (Test-15), Objective Solutions, 16th April 2016 (11)82. (d)

When the flow is parallel to the stratification,the equivalent permeability Ke

Ke =

n

i i1

n

i1

K B3 30 2 10 5 20 21m/day

3 2 5B

Transmissibility, T = KB = 21 × 10

= 210 m2/day83. (a)84. (d)85. (b)

Rainfall = 2.7 cmLoss @ 0.3 cm/h for 3 h = 0.9 cm

Effective rainfall, ER = 2.7 – 0.9 = 1.8 cm DRH of the storm has peak = 200 – 20

= 180 m3/s

3h UH has peak = 1801.8 = 100 m3/s

86. (c)Let rainfall excess be ER (in cm) Volume of surface runoff

= 1 60 3600 362

= 6 ER300 10100

ER = 1.08 cm = 10.8 mm

87. (d)ER of 1h UH = 1 cmVolume of runoff = Area under 1h UH

= 1 1 1 1 1 11 3600 3 3 5 5 4 4 2 2 1 12 2 2 2 2 2

= 54,000 m3

Catchment area represented by this UH

= 54,000 100

1

= 5.4 × 106 m2

= 5.4 km2

Time (h)

(1)

Ordinateof 1h UH

(m /s)(2)

3

Ordinate of1h UH lagged

by 1 h(3)

012345678

Ordinate of1h UH lagged

by 2 h(4)

col. 2 +col. 3

+ col. 4(5)

Ordinate of3h UH =(col. 5)/3

(6)0354210

—0354210

——0354210

038

12117310

01

2.674

3.672.33

10.33

0

88. (d)Total rainfall of 4h storm = 7 cm

Loss @ 0.25 cm/h for 4h = 1 cm

Effective rainfall of the storm = 7 – 1 = 6 cm

Peak ordinate of 4h UH = 80 m3/s

Peak ordinate of 4h DRH = 80 × 6

= 480 m3/s

Base flow = 20 m3/s

Peak of the flood discharge due to thestorm = 480 + 20 = 500 m3/s

89. (a)Area of catchment A = 360 km2

Duration of unit hydrograph = 4h

Equilibrium discharge of S-curve

QS = 2.778 A/D m3/s

= 3602.778

4 = 250.02 m3/s

90. (b)

h =

22 V u 1 cos u

V

The efficiency will be maximum for a givenvalue of V when

hd

du = 0

or

22u V u 1 cosd

du V

= 0

or 22

1 cos d 2uV 2uduV

= 0

or 2d 2uV 2udu

= 0 21 cos 0

V

or 2V – 4u = 0 or u = V2

91. (b)In centrifugal pumps, the cavitation may occurat the inlet of the impeller of pump. If thepressure at the suction side of the pump dropsbelow vapour pressure of liquid, then cavitationmay occur.Cavitation is the localised formation andsubsequent collapse of cavities, or bubbles ina liquid. Cavitation is usually caused byinsufficient NPSHA.

In a pump, cavitation will first occur at theinlet of the impeller. Denoting the inlet by i,

IES M

ASTER

(12) CE (Test-15), Objective Solutions, 16th April 2016NPSHA at this point is defined by

NPSHA = 2

vi i

g

pp vg 2g

On the other hand, in a reaction turbine,cavitation will first occur at the outlet of theimpeller, at the entrance of the draft tube.

92. (a)

1.

2.

3.4.

10 to 35

35 to 60

60 to 300300 to 1000

8.5 to 30

30 to 51

51 to 225225 to 860

Pelton wheel with single jetPelton wheel with two or more jetsFrancis turbineKaplan or Propeller turbine

S.No.

Specifiedspeed

Types ofturbine

(M.K.S.) S.I.

93. (d)

Shear due to torque = 1.6T

b

=1.6 9 48 kN

0.3

Equivalent shear = 1.6T V

b

= 48 + 20

= 68 kN

94. (c)

At

dsn

d-nN A

B

d

Take moment about neutral axis

dsB ds n2

= m At × (d–n)

95. (d)

beff = 0w fb 6D

6

l

Where l0 = distance between points ofcontraflexure.

beff = 3600 300 6 100

6 = 1500mm

96. (c)

Z = 22 300 600bD6 6

= 18×106mm3

fcr = ck0.7 f = 0.7×5 = 3.5MPa

Mcr = fcr × Z = 3.5×18 kNm = 63 kNm97. (c)

Kb = 1

r1m

r = st

cbc

1400 2850

Hence, Kb = 0.39Now for a balanced section

cbc b0.5 b K d = st,b stA

Pt(%) = cbcb

st

50 K

= 150 0.39 0.696%

28

98. (d)

Tensile force = 2t d4

Bond stress resistance = bS d l

bond stress resistance = Tensile force

bS d l = 2t d4

l = b

dt4S

99. (b)Diagonal tension is produced in beam due toshear force which is predominent at the endsof beam. So, to counter this diagonal tensionwe require bent up hooked bars at the endsof beam.

100. (b)Hook requires minimum extension of 4d a headof curvature of Hook.

101. (a)The line of collimation of a theodolite must beperpendicular to the horizontal axis at itsintersection with the vertical axis. If thiscondition exists, the lime of sight will generatea vertical plane when the telescope is rotatedabout the horizontal axis.If the line of sight is not perpendicular to thetrunnion axis of the telescope, it will not revolvein a plane when the telescope is raised orlowered but instead, it will trace out the surfaceof a cone.

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CE (Test-15), Objective Solutions, 16th April 2016 (13)102. (a)

The direction of a survey line can either beestablished (i) with relation to each other or(ii) with relation to any meridian. The first willgive the angle between two lines while thesecond will give the bearing of the line.

103. (a)

Feature of contour of terrain:

All points on a contour line are of the sameelevation.

No two contourlines can meet or crosseach other except in the rare case of anoverhanging vertical cliff.

Closely spaced contour lines indicatesteep slope.

Widely spaced contour lines indicate gentleslope

Equally spaced contour lines indicateuniform slope.

Closed contour lines with reducing levelstowards the centre indicate pond or otherdepression.

Contour lines of ridge show higher elevationwithin the loop of the contours. Contourlines cross ridge at right angles.

Contour lines of valley show reducingelevation within the loop of the contours.contour lines cross valley at right angles.

All contour lines must close either withinthe map boundary or outside.

Contour lines cannot merge or cross oneanother on map except in the case of anoverhanging cliff.

Contour lines never run into one anotherexcept in the case of a vertical cliff. In thiscase, several contours coincide and thehorizontal equivalent becomes zero.

104. (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 over variedterrain.

Objects that rise above the surface awayfrom the principal point.

Objects extending below the surface leantowards the principal point.

Displacement increases with the height ofthe object and or distance from the principalpoint.

Relief displacement, d = rhH

wherer = radial distance from principal point

to displaced image pointh = height above surface of the object

pointH = flying height above the surface.

105. (a)Meridian Circle : It is a great circle whichpasses through the zenith and Nadir as wellas through the poles.Vertical Circle : A vertical circle of the celestialsphere is great circle passing through thezenith and Nadir.

106. (b)107. (a)108. (b)109. (b)

The characteristics of f low net can besummarised as under :–

The fundamental condition that is to besatisfied is that every intersection betweena flow line and an equipotential line shouldbe at right angles.

The second condition to be satisfied is thatthe discharge between any two adjacentflow lines is constant and the drop of headbetween the two adjacent equipotential linesis constant.

The ratio of the length and width of eachfield is constant. The ratio is generally takenas unity for convenience. In other words,the flow net consists of approximatesquares.

110. (c)In the Fellenius analysis, the horizontal andvertical forces on the slice boundaries areassumed to be equal and opposite. This istrue if the slice is reduced to the width of aline but as the width of a slice increases theassumption is partially untrue since the twosides will be very different in size. Thus, if thesoil is divided into many slices, as can bedone using modern computers, then areasonably accurate factor of safety can befound. However for manual analysis, thenumber of slices that can realistically be used

IES M

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(14) CE (Test-15), Objective Solutions, 16th April 2016(i.e., upto 10 approximately) limits theaccuracy of the method.Because the forces on the vertical boundariesare assumed to cancel out as shown in figbelow, the total normal force on the base ofthe slice is equal to the component of theweight in the direction of the normal force.N = wcos

N

b

w

The fellenium equation is fairly simple to solveand yields conservative results (lower thanactual factor of safety) especially where theslip surface is deep or where the pore waterpressures are high. In both these cases thefault lies with the neglect of the intersliceforces.

It is also known as swedish circle method. Inthis method, the equilibrium of each slice isdetermined and factor of safety found bysumming the resisting forces and dividing bythe driving forces. The operation is repeatedfor the circles until the lowest safety factor isfound. The method does not consider all theforces acting on a slice, as it omits the shearand normal stresses and pore water pressuresacting on the sides of the slice but usually(although not always) it yields conservativeresults. However, the conservatism may behigh.

111. (a)The Rankine’s theory assumed that the wallsurface is smooth whereas in practice, a lotof friction may develop between the wall surfaceand the soil fill. This friction will depend uponthe wall material. This friction leads to thedevelopment of smaller active pressure andlarger passive pressure than that estimatedby Rankine’s theory.

Thus, the estimation of the active pressureusing Rankine’s theory will be slightly higherthan the actual (reduced due to friction) Passivepressure will be slightly lower.

112. (a)Shear strength is a term used in soilmechanics to describe the magnitude of theshear stress that a soil can sustain. The shearresistance of a soil is a result of friction andinterlocking of particles and possiblycementation or bonding at particle contacts.The shear strength of soil depends on theeffective stress, the drainage conditions, thedensity of particles, the rate of strain and thedirection of the strain.

The drained shear strength is the shearstrength of the soil when pore fluid pressures,generated during the course of shearing thesoil, are able to dissipate during shearing. Italso applies where no pore water exists in thesoil and hence pore fluid pressures arenegligble. It is commonly approximated usingthe Mohr-coulomb equation.

In terms of effective stresses, the shearstrength is often approximated by

= tan c

where ( u) is the effective stress. istotal stress applied normal to the shear planeand u is the pore water pressure acting on thesame plane.

= effective stress friction angle.

c = cohesion

113. (b)Principal factors that influence ultimate bearingcapacities are type and strength of soil,foundation width and depth, soil weight in theshear zone and surcharge.

The depth to the water table influences thesubsurface and surcharge soil weights.

If the water table is below the depth of thefailure surface then the water table has noinfluence on the bearing capacity and effectiveunit weight is equal to the wet unit height ofthe soil.

If the water table is above the failure surfaceand beneath the foundation base, then effectiveunit weight of the soil gets reduced.

114. (a)A relative movement between a pile and soilproduces shear along the interface of the pileand the soil. Such movement can be inducedby a push-load on the pile pressing it downinto the soil or by a pull-load moving it upward.A relative movement can also be induced when

IES M

ASTER

CE (Test-15), Objective Solutions, 16th April 2016 (15)the soil settles in relation to the pile or inswelling soils when the soil moves upward inrelation to the pile. By definiiton, if themovement of the pile is downwrd i.e. the shearstress induced in the pile is upward, thedirection of the shear is positive. If themovement of the pile is upward, the shearstress direction is negative.

In order terminology, the induced shear alonga pile was called skin friction. The negativeand positive skin friction is shear stressinduced by settling or swelling soil respectively.Large ground settlement due to consolidationof soft soils will drag down piles and inducenegative skin friction (NSF) along the pile-soilinterfaces. NSF will induce additional drag loadon the piles which may cause structural failureof piles due to overstress or downdragsettlement which may compromise theserviceability of super structure. It has beenrecognised that pile groups posses beneficialeffects in alleviating NSF on piles.

115. (a)Dynamic formulae for piles :The dynamic formulae are based on theassumption that the kinetic energy deliveredby the hammer during driving operation is equalto the work done on the pile. Thus,

Wh h = R × S

where W = weight of hammer (kN),

h = height of ram drop (cm),

h = efficiency of pile hammer,

R = pile resistance (kN), taken equalto Qu and S = pile penetrationper blow (cm)

The load carrying capacity of a driven pile canbe estimated from the resistance againstpenetration developed during driving operation.

The methods give fairly good results only inthe case of free draining sands and hard claysin which high pore water pressures does notdevelop during driving of piles. In saturatedfine-grained soils, high pore water pressuredevelops during the driving operation and thestrength of the soil is considerably changedand the methods do not give reliable results.The methods can not be used for submerged,uniform fine sands which may be loose enoughto become quick temporarily and show a muchless resistance.

116. (d)

117. (a)

Slug is a higher strength and higher volumewaste discharged in short period of time.

118. (a)

119. (d)

A draft tube is a tube or pipe of graduallyincreasing area which is used for dischargingwater from the exit of the turbine to the tailrace. This is because the pressure at the exitof the summer of a reaction turbine is generallyless than atmosphereic pressure. Thus, thewater at exit cannot be directly discharged tothe tail race.

Also, the draft tube converts a large proportion

of the kinetic energy 22V

2g

rejected at the

outlet of the turbine into useful pressure energy.Without the draft tube, the kinetic energyrejected at the outlet of turbine will go wasteto the tail race.

Hence by using draft tube, net head on turbineincreases. The turbine develops more powerand also efficiency of turbine increases.

120. (d)

Minimum grade of concrete in RCC isRestricted to M20 as per IS 456:2000