CE Test 15 Objective Solution

8/18/2019 CE Test 15 Objective Solution

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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 Programe

Date: 16th April, 2016

ANSWERS


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

1. (b)

S.F. = Shrunk length Shrunk scale

Originallength Original scale

24 Shrunk scale

25 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 ) L

AE

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 =f

i

where,

f = focal length of objective

d = horizontal distance between optical

centre and vertical axis of

tacheometer

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 20

i 0.4 = 50

Staff interept,

S = 2 × (2.5 – 1) = 3.0 m

Horizontal distance between the staff station

and instrument station,

D = Ks + C

= 50 × 3 + 0.3

= 150.3 m

7. (d)

% of error between 3 under gussian law

of distribution = 99.7% No. of errors that are expected to exceed

the 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 or

more observed quantities, its probable error

is equal to the square root of summation of

the squares of the probable errors. of the

observed quantity multiplied by itsdifferentiation with respect to that quantity.

Let y = computed quantity

x1, x

2, x

3 etc. = observed quantities

Such that y = f(x1, x

2, x

3 etc.)

Then

ey = 1 2 3

2 2 2

x x x1 2 3

dy dy dye e e

dx dx dx

where ey = Probable error of the computedquantity

1 2 3x x xe , e , e = Probable errors of

observed quantities. Let area A = a · b

e A

= 2 2

a be · b e · a

= 2 2(0.05 180) (0.06 120)

= ± 11.53 sqm.

where e A = Probable error in the area.


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10. (c)

The local attraction at any station isdetected by observing the fore and back

bearings of the line.

If local difference between them is 180º,both the end stations are considered to

be free from local attraction, provided the

compass is not having any instrumental

errors.

If the difference is not 180º, thediscrepancy introduced may be because

of presence of local attraction at either or

both 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 the

time of observation is known the true bearing

of a line can be determined from its magnetic

bearing 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 = 90º

13. (b)14. (b)

RP

Q

129º30

59º

Line FB BB

PQ 59º0 239º0

QR 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 to

the axis of the tube when the vertical circle

reading is zero.

The axis of the altitude level tube is truly

horizontal 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 10020m

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 is

that even when the levelling instrument is not

in correct adjustment, the difference in height

measured between two points by a level

equidistant from each gives the true difference

in height.

21. (a)

Last sub chord = 2303.39 – 2300 = 3.39 m

22. (c)

23. (b)

Contour lines are imaginary lines passing

through points of equal elevations.

24. (d)

Contour lines close together indicate a steep

slope.

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.h

H

where h = height of the object above datum,

H = flying height above the datum,

r = radial distance of the image of the

top of the object from principal

point.

d =90 500

9mm5000


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27. (d)

28. (c)

29. (c)

Principal Point : It is a point where aperpendicular dropped from the front nodal

point strikes the photograph.

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

photographs.

Crab : It is the term used to designate theangle formed between the flight line and

the edges of the photograph in the direction

of flight.

Drift : It is caused by the failure of theaeroplane to stay on the predetermined

flight line.

30. (a)

The Right Ascension (RA) of a celestial body

is the angular distance measured along thecelestial equator between the first point of Aries

and declination circle of the body.

31. (a)

Lehmann’s method or Trial and error method

in the field is used to find out the position

of 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)

N o r m a l

1

1

2Soil-1 (K )1

Flowlines 2

Soil-2 (K )2

Flow lines

Normal

When two different soils are used in a soil

mass, thus making it non-homogeneous. The

flow lines and equipotential lines get deflected

at the interface. The flow net thus get modified.

K1 and K2 are related as

1

2

K

K=

1

2

tan

tan

34. (b)

h = 6 m, Nf = 6; N

d = 18

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

= 0.0576 m/day

Seepage discharge = Khf

d

N

N

= 0.0576 × 6 ×6

18

= 0.1152 m3/d per m length

35. (c)

e = 0.60 For quick sand condition,

ic =

G 1

1 e

ic = cG 1

1.6 i G 11 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 i

tan 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,

f D

B =

21

2

NC = 5

Df B1 0.2 1 0.2

B L

= 5(1 + 0.2 × 1) ( 1 + 0) = 6.0

qnu = CuNc = 30 × 6 = 180 kN/m2


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39. (b)

= 0.70

Cu = 40 kN/m2

F.S. = 2.50

B = 0.3 m

D = 10 m

Qu = qb Ab + .c. (p D)

= (9 × 40) ×4

× 0.32

+ 0.7 × 40 ( × 0.3 × 10)

= 289.194 kN

Qa =4Q 289.194 115.68 kN

F.S. 2.5

40. (d)

Total stress in multi-layered soil :

The total stress at depth z is the sum of the

weights 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 depth

of 1 m is 36 kN/m2, this will be the increase

in vertical stress every where below the surface

level.

41. (d)

The permeability

K =

32w eC D

1 e

where C = a constant depending upon shape

of c/s of flow.

= dynamic viscosity of fluid

e = void ratio

D = grain size

42. (b)

= sub h (Assuming3

sub10kN/m )=

50 = 10.h

h = 5m

43. (d)

Q = KAi

Q A

=2 hK D

4 2L

QB

=2 h

K 4D4 L

A

B

Q

Q=

2

2

hK D

14 2L =h 8

K 4D4 L

= 0.125

44. (b)

K = 10-3 cm/s

h = 1mL = 2m

A = 100cm2

Q = KAh

i K A =L

dV

dt=

hK A.

L

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

×1m

2m

× 60 sec = 3ml

45. (a)

Q =h

K.A.L

(For constant head

permeameter)

Q h

L

Q

Q

=

2h 1 4h L

L / 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 between

100 and 150.

Ratio of permeability =2

.64

0.3


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47. (b)

Kh

= 3m/day

KV

=1

m/day3

K = h V1

K .K 3 1m/day3

48. (c)

Failure by piping or undermining

When the seepage water retains sufficient

residual force at the emerging downstream end

of the work, it may lift up the soil particles.

This leads to increaed porosity of the soil by

progressive removal of soil from beneath the

foundation. The structure may ultimately

subside into the hollow so formed, resulting in

the failure of the structure.

The exist gradient is said to be critical when

the upward distributing force on the grain is

just equal to the submerged weight of the

grain at the exit. When a factor of safety equalto 4 or 5 is used, the exit gradient. Can then

be taken as safe In other words, an exit

gradient equal to1

4 to

1

5 of the critical exit

gardient is ensured, so as to keep the structure

safe against piping.

49. (c)

In case of a falling head permeameter,

K =

110

2 1 2

h2.303Lalog

A t t h

where L = 10 cm

a = 1 cm2

A = 50 cm2

(t2 – t

1) = 1 hour = 3600 sec.

1

2

h

h=

802

40

Hence K =2.303 10 1

0.350 3600

= 3.84×10 –5 cm/sec.

50. (b)H = 10 m

Df H = 15 m

Df =15

1.5010

Sn =c

C

F H

0.164 = c35

F 1.15Fc 18.5 10

51. (d)

The steady seepage condition is criticalfor the d/s slope of an earth dam.

The critical condition for the stability of theu/s slope of an earth dam is when there is

a sudden drawndown in the reservior u/s.

52. (d)

Cm

=C 15

10

1.5 1.5

Sn = 0.046 =10

H

H =10

11.5 m19 0.046

53. (c)

Swedish circle method: The actual shape

of a slipsurface in the case of finite slopes is

curvilinear. For convenience,it is approximated

as circular. The assumption of a circular slip

surface and its application for stability analysis

of slopes was developed in sweden. Themethod is known as the swedish circle method

or the method of slices.

Stability member (Sn): It is defined as

Sn =m

H c

C C

F H

The reciprocal of the stability number is known

as stability factor. The stability number is a

dimensionless quantity. The stability number

can be used to determine the factor of safety

of a given slope.

Sudden drawdown conditions: When the

water standing on the slope is suddenly and

quickly removed, the water pressure (U)

disappears. However, if there is no time for

drainage to occur from the soil in the slope,

the soil remains submerged as before and the

natural part of the weight is still acting. Thus,

the equilibrium of the neutral force is distrubed,

although the equilibrium of the inter granular

forces remains unaffected.

Critical void ratio: It may be observed that

the void ratio of an initial loose sand decreaseswith as increase in shear strain whereas that

for the initially dense sand increases with an

increase in strain. The void ratio at which there

is no change in it with an increase in strain is

known as the critical void ratio. If the sand

initially is at critical void ratio, there would be

practically no change in volume with an

increase in shear strain.

Where there is soft clay there is a chance of

base failure


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54. (a)

The following are the important points of

comparison of coulomb’s theory with Rankine’s

theory:

Coulomb considers a retaining wall andthe backfill as a system; he takes into

account the friction between the wall and

the 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 and

direction of the earth pressure are assumed

to be known; linear variation of pressure

with depth is tacitly assumed and thedirection is automatically obtained from the

concept of wall friction. In Rankine’s theory,

plastic equilibrium inside a semi-infinite soil

mass is considered, pressure evaluated, a

retaining wall is imagined to be interposed

later, and the location and magnitude of the total earth thrust are established

mathematically.

Coulomb’s theory is more versatile thanRankine’s in that it can take into account

any shape of the backfill surface, break in

the wall face or in the surface of the fill,

effect of stratification of the backfill, effect

of various kinds of surcharge on earth

pressure and the effects of cohesion,

adhesion and wall friction. It lends itself to

elegant graphical solution and gives more

reliable results, especially in thedetermination of the passive earth

resistance; this is in spite of the fact that

static equilibrium condition does not appear

to be satisfied in the analysis.

Rankine’s theory is relatively simple andhence is more commonly used, while

coulomb’s theory is more rational and

versatile although cumbersome at times,

therefore, the use of the later is called for

in important situation or problems.

55. (b)

When a wall moves away from the backfill,

some portion of the backfill located immediately

behind the wall tries to break away from the

rest of the soil mass. This wedge–shaped

portion, known as the failure wedge or the

sliding wedge moves downward and outwards.

The lateral earth pressure exerted on the wall

is a minimum in this case. The soil is at the

verge of failure due to a decrease in the lateral

stress.

The horizontal strain required to reach the active

state of plastic equilibrium is very small. It

has been shown that in dense sand, the

horizontal strain required is about 0.5%

It is thus found that very little horizontal strain

(about 0.50%) is required to reach one half

the maximum passive pressure in dense sand

but much more horizontal strain (about 5% in

dense and and 15% in loose sand) is required

to reach the full passive pressure.

It may be summarised that the state of

shear failure corresponding to the minimum

earth pressure is the active state and that

corresponding to the maximum earth pressure

is the passive state.

56. (b)

Unsupported height =4C 4 5

1m20

57. (d)

Ka =21 sin

tan= /21 sin 4

Kp =

21 sintan= 45 /2

1 sin

p

a

K

K =

2

2

tan 45º /2

tan 45º – /2

a

K =p

1

K

p

a

K

K =

24

2

tan 45 /2tan= 45 /2

1

tan 452

58. (a)

Given that, K A

= 2 1tan 45º2

= 1

4

1tan 45º2

=

1

2

1tan 45º2

= 2

C1 = 20 kN/m

2


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H1 =

1 1

1

4Ctan 45º

2

H1 =

4 202 8m

20

H2 =

2

2

4C 4 408m

20

H2/H

1 =

81

8

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/m

2 for allowable settlement,

S1 = 25 mmm

The allowable bearing capacity, Q2 for allowable

settlement, S2 = 40 mm

But Housel’s method,

1

1

Q

S =

2

2

Q

S

Q2 =2

11

SQ

S

=240 15 24 t/m

25

61. (b)

The ultimate bearing capacity of purely

cohesive soil is given by qu = CN

C+ q

Where q = surcharge due to increase in depth

of foundation.

As we can observe from the above equation

that bearing capacity of footing on purely

cohesive soil is independent of size of footing.

62. (d)

Pile foundations are used in the following

condition:

When the strata at or just below the groundsurface in highly compressible and very

weak to support the load transmitted by

the structure.

When the plans of the structure is irregular relative to its outline and load distribution.

It would cause non-uniform settlement if a

shallow foundation is constructed. A pile

foundation is required to reduce differential

settlement.

Pile foundation are required for thetransmission of structural loads through

deep water to a firm stratum.

Pile forces are used to resist horizontalforces in addition to support the vertical

loads in earth-retaining structures and tall

structures that are subjected to horizontal

forces due to wind and earthquake.

Piles are required when the soil conditionare such that a washout, erosion or scour

of soil may occur from underneath a

shallow foundation.

In case of expansive soils, such as blackcotton soil, which swell or shrink as the

water content changes, piles are used to

transfer the load below active zone.

63. (b)

Raft foundations on sands are quite useful.

Their bearing capacity failure is generally out

of question, because the bearing capacity in

sands increases with the width of the footing

and this width in rafts is quite large. The

differential settlements in rafts are also

generally smaller as compared to isolated

footings even for the same load intensity

because a raft eliminates the influence of local

loose soils.

SoilType

Permissible totalsettlement

Permissibledifferential

settlement For

isolatedfootings

For raftfoundation

Forisolatedfooting

For raftfoundation

Sandy 4 cm 4 to6.5 cm

2.5 cm 2.5 cm

Clayey 6.5 cm 6.5 to10 cm

4 cm 4 cm

64. (b)

65. (d)

66. (c)

General shear failure

General shear failure is seen in dense and

stiff soil. The following are characteristics of

general shear failure :

Continuous well defined and distinct failure

surface develops between the edge of

footing and ground surface.

Dense or stiff soil that undergoes low

compressibility experiences this failure.


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Continuous bulging of shear mass adjacent

to footing is visible.

Failure is accompanied by tilting of footing.

Failure is sudden and catastrophic with

pronounced P- curve.

The length of disturbance beyond the edge

of footing is large.

State of plastic equilibrium is reached

initially at the footing edge and spreads

gradually downwards and outwards.

General shear failure is accompanied by

low strain ( 36º) and large N (N > 30) having

relative density (ID > 70%)

Local shear failure

This type of failure is seen in relatively loose

and soft soil. The following are some

characteristics of general shear failure.

A significant compression of soil below thefooting and partial development of plastic

equilibrium is observed.

Failure is not sudden and there is no tilting

of footing

Failure surface does not reach the ground

surface and slight bulging of soil around

the footing is observed.

Failure surface is not well defined.

Failure is characterised by considerable

settlement.

Well defined peak is absent inp curve.

Local shear failure is accompanied by large

strain (> 10 to 20%) in a soil with

considerably low ( 1. This is because soil arround andbetween the piles get compacted due to the

vibration caused during the driving operations.

Whereas in dense sand above phenomenon

is not true.

70. (a)

The load carrying capacity of a driven pile canbe estimated from the resistance against

penetration developed during driving operation.

The methods give fairly good results only in

the case of free-draining sands and hard clays

in which high pore water pressures do not

develop during the driving of piles. In saturated

fine grained soils, high pore water pressure

develops during the driving operation and the

strength of the soil is considerably changed

and the methods do not give reliable results.


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71. (b)

Types of Foundation

Spread Footings

Used for most buildings where the loadsare light and or there are strong shallow

soils.

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

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

These are most widely used because theyare most economical.

Spread footings should be above the water table.

Drilled piers or Caissons

For expansive soils with low to mediumloads or high loads with rock not too far down, drilled caissons (piers) and grade

beams can be used.

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

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

Piles

For expansive soils or soils that arecompressive with heavy loads where deep

soils 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 rely

on resistance from skin of pile against

the soil.

2. End bearing– which transfers directly to

soil 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 ground

where drilling will not cause collapse.

Mat Foundations

Reinforced concrete raft or mats can beused for small light load buildings on very

weak 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 resistance

against overturning.

Can be used where soil requires such alarge bearing area and the footing might be

spread to the extant that it becomes more

economical 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 great

differences.

72. (a)

73. (a)

74. (d)

75. (b)

NB ss

cX f

HRT

=

8 2420

6 = 640 mg/l

76. (c)

Bacteria produces highest biomass among

other micorbs.

77. (d)

78. (b)

79. (b)

h = Sy + S

r

0.40 = Sy + 0.15 S

y = 0.25

where 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/s

For confined aquifer,

Q = 1 2

2

1

2 T S S

r ln

r

here r 1 = 10 m; S1 = 3m

r 2 = 100 m; S2 = 0.5 m

0.04533 = 2 T 3 0.5100

ln 10

T = 6.6487 × 10 –3 m2/s = 574.44 m2/day

81. (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 m2 4 3 1 1 1L6 16 24 3 4 8K


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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/day3 2 5

B

Transmissibility, T = KB = 21 × 10

= 210 m2/day

83. (a)

84. (d)

85. (b)

Rainfall = 2.7 cm

Loss @ 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 10

100

ER = 1.08 cm = 10.8 mm

87. (d)

ER of 1h UH = 1 cm

Volume of runoff = Area under 1h UH

= 1 1 1 1 1 1

1 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 of 1h UH lagged

by 1 h(3)

0

1

2

3

4

5

6

7

8

Ordinate of 1h UH lagged

by 2 h(4)

col. 2 +col. 3

+ col. 4(5)

Ordinate of 3h UH =(col. 5)/3

(6)

0

3

5

4

2

1

0

—

0

3

5

4

2

1

0

—

—

0

3

5

4

2

1

0

0

3

8

12

11

7

3

1

0

0

1

2.67

4

3.67

2.33

1

0.33

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

=360

2.7784

= 250.02 m3/s

90. (b)

h =

2

2 V u 1 cos u

V

The efficiency will be maximum for a given

value of V when

hd

du = 0

or

2

2u V u 1 cosd

du V

= 0

or 2

2

1 cos d2uV 2u

du V

= 0

or 2d

2uV 2udu

= 0 21 cos

0 V

or 2V – 4u = 0 or u = V

2

91. (b)

In centrifugal pumps, the cavitation may occur

at the inlet of the impeller of pump. If the

pressure at the suction side of the pump dropsbelow vapour pressure of liquid, then cavitation

may occur.

Cavitation is the localised formation and

subsequent collapse of cavities, or bubbles in

a liquid. Cavitation is usually caused by

insufficient NPSHA.

In a pump, cavitation will first occur at the

inlet of the impeller. Denoting the inlet by i,


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NPSHA at this point is defined by

NPSHA =

2vi i

g

pp v

g 2g

On the other hand, in a reaction turbine,

cavitation will first occur at the outlet of the

impeller, at the entrance of the draft tube.

92. (a)

1.

2.

3.

4.

10 to 35

35 to 60

60 to 300

300 to1000

8.5 to 30

30 to 51

51 to 225

225 to860

Pelton wheel with

single jet

Pelton wheel with

two or more jets

Francis turbine

Kaplan orPropeller turbine

S.No.

Specified

speedTypes of turbine

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

93. (d)

Shear due to torque =1.6T

b

=1.6 9

48 kN0.3

Equivalent shear =1.6T

Vb

= 48 + 20

= 68 kN

94. (c)

At

dsn

d-n

N A

B

d

Take moment about neutral axis

dsB ds n

2

= m At × (d–n)

95. (d)

beff

= 0 w f b 6D6

l

Where l0 = distance between points of

contraflexure.

beff

=3600

300 6 1006

= 1500mm

96. (c)

Z =

22 300 600bD

6 6

= 18×106mm3

f cr = ck0.7 f = 0.7×5 = 3.5MPa

Mcr = f cr × Z = 3.5×18 kNm = 63 kNm

97. (c)

Kb =

1

r 1m

r =st

cbc

140028

50

Hence, Kb = 0.39

Now for a balanced section

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

Pt(%) =cbc

b

st

50 K

= 150 0.39 0.696%28

98. (d)

Tensile force =2

t d4

Bond stress resistance = bS d l

bond stress resistance = Tensile force

bS d l =2t d

4

l =b

dt

4S

99. (b)

Diagonal tension is produced in beam due to

shear force which is predominent at the ends

of beam. So, to counter this diagonal tension

we require bent up hooked bars at the ends

of beam.

100. (b)

Hook requires minimum extension of 4d a head

of curvature of Hook.

101. (a)

The line of collimation of a theodolite must beperpendicular to the horizontal axis at its

intersection with the vertical axis. If this

condition exists, the lime of sight will generate

a vertical plane when the telescope is rotated

about the horizontal axis.

If the line of sight is not perpendicular to the

trunnion axis of the telescope, it will not revolve

in a plane when the telescope is raised or

lowered but instead, it will trace out the surface

of a cone.


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102. (a)

The direction of a survey line can either be

established (i) with relation to each other or

(ii) with relation to any meridian. The first will

give the angle between two lines while the

second 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 an

overhanging vertical cliff.

Closely spaced contour lines indicatesteep slope.

Widely spaced contour lines indicate gentleslope

Equally spaced contour lines indicate

uniform slope.

Closed contour lines with reducing levelstowards the centre indicate pond or other

depression.

Contour lines of ridge show higher elevationwithin the loop of the contours. Contour

lines 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 l ines must close either withinthe map boundary or outside.

Contour lines cannot merge or cross oneanother on map except in the case of an

overhanging cliff.

Contour lines never run into one another except in the case of a vertical cliff. In this

case, several contours coincide and the

horizontal equivalent becomes zero.

104. (c)

Relief displacement : It is caused by changes

in the distance between the ground and the

camera 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 of the object and or distance from the principal

point.

Relief displacement, d =rh

H

where

r = radial distance from principal point

to displaced image point

h = height above surface of the object

point

H = flying height above the surface.

105. (a)

Meridian Circle : It is a great circle which

passes through the zenith and Nadir as well

as through the poles.

Vertical Circle : A vertical circle of the celestial

sphere is great circle passing through the

zenith and Nadir.

106. (b)

107. (a)

108. (b)

109. (b)

The characteristics of flow net can be

summarised as under :–

The fundamental condition that is to besatisfied is that every intersection between

a flow line and an equipotential line should

be at right angles.

The second condition to be satisfied is thatthe discharge between any two adjacent

flow lines is constant and the drop of head

between the two adjacent equipotential lines

is constant.

The ratio of the length and width of eachfield is constant. The ratio is generally taken

as unity for convenience. In other words,

the flow net consists of approximate

squares.

110. (c)

In the Fellenius analysis, the horizontal and

vertical forces on the slice boundaries are

assumed to be equal and opposite. This is

true if the slice is reduced to the width of a

line but as the width of a slice increases the

assumption is partially untrue since the two

sides will be very different in size. Thus, if the

soil is divided into many slices, as can be

done using modern computers, then a

reasonably accurate factor of safety can be

found. However for manual analysis, the

number of slices that can realistically be used


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(i.e., upto 10 approximately) limits the

accuracy of the method.

Because the forces on the vertical boundaries

are assumed to cancel out as shown in fig

below, the total normal force on the base of

the slice is equal to the component of the

weight in the direction of the normal force.

N = wcos

N

b

w

The fellenium equation is fairly simple to solve

and yields conservative results (lower than

actual factor of safety) especially where the

slip surface is deep or where the pore water

pressures are high. In both these cases the

fault lies with the neglect of the interslice

forces.

It is also known as swedish circle method. In

this method, the equilibrium of each slice is

determined and factor of safety found by

summing the resisting forces and dividing by

the driving forces. The operation is repeated

for the circles until the lowest safety factor is

found. The method does not consider all theforces acting on a slice, as it omits the shear

and normal stresses and pore water pressures

acting on the sides of the slice but usually

(although not always) it yields conservative

results. However, the conservatism may be

high.

111. (a)

The Rankine’s theory assumed that the wall

surface is smooth whereas in practice, a lot

of friction may develop between the wall surface

and the soil fill. This friction will depend uponthe wall material. This friction leads to the

development of smaller active pressure and

larger passive pressure than that estimated

by Rankine’s theory.

Thus, the estimation of the active pressure

using Rankine’s theory will be slightly higher

than the actual (reduced due to friction) Passive

pressure will be slightly lower.

112. (a)

Shear strength is a term used in soil

mechanics to describe the magnitude of the

shear stress that a soil can sustain. The shear

resistance of a soil is a result of friction and

interlocking of particles and possibly

cementation or bonding at particle contacts.

The shear strength of soil depends on the

effective stress, the drainage conditions, the

density of particles, the rate of strain and thedirection of the strain.

The drained shear strength is the shear

strength of the soil when pore fluid pressures,

generated during the course of shearing the

soil, are able to dissipate during shearing. It

also applies where no pore water exists in the

soil and hence pore fluid pressures are

negligble. It is commonly approximated using

the Mohr-coulomb equation.

In terms of effective stresses, the shear

strength is often approximated by

= tan c

where ( u) is the effective stress. istotal stress applied normal to the shear plane

and u is the pore water pressure acting on the

same plane.

= effective stress friction angle.

c = cohesion

113. (b)

Principal factors that influence ultimate bearing

capacities are type and strength of soil,

foundation width and depth, soil weight in the

shear zone and surcharge.

The depth to the water table influences the

subsurface and surcharge soil weights.

If the water table is below the depth of the

failure surface then the water table has no

influence on the bearing capacity and effective

unit weight is equal to the wet unit height of

the soil.

If the water table is above the failure surface

and beneath the foundation base, then effective

unit weight of the soil gets reduced.

114. (a)

A relative movement between a pile and soil

produces shear along the interface of the pile

and the soil. Such movement can be induced

by a push-load on the pile pressing it down

into the soil or by a pull-load moving it upward.

A relative movement can also be induced when


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CE (Test-15), Objective Solutions, 16th April 2016 (15)the soil settles in relation to the pile or in

swelling soils when the soil moves upward in

relation to the pile. By definiiton, if the

movement of the pile is downwrd i.e. the shear

stress induced in the pile is upward, the

direction of the shear is positive. If the

movement of the pile is upward, the shear

stress direction is negative.

In order terminology, the induced shear along

a pile was called skin friction. The negativeand positive skin friction is shear stress

induced by settling or swelling soil respectively.

Large ground settlement due to consolidation

of soft soils will drag down piles and induce

negative skin friction (NSF) along the pile-soil

interfaces. NSF will induce additional drag load

on the piles which may cause structural failure

of piles due to overstress or downdrag

settlement which may compromise the

serviceability of super structure. It has been

recognised that pile groups posses beneficial

effects in alleviating NSF on piles.115. (a)

Dynamic formulae for piles :

The dynamic formulae are based on the

assumption that the kinetic energy delivered

by the hammer during driving operation is equal

to 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 equal

to Qu and S = pile penetration

per blow (cm)

The load carrying capacity of a driven pile can

be estimated from the resistance against

penetration developed during driving operation.

The methods give fairly good results only in

the case of free draining sands and hard clays

in which high pore water pressures does not

develop during driving of piles. In saturated

fine-grained soils, high pore water pressure

develops during the driving operation and the

strength of the soil is considerably changed

and the methods do not give reliable results.

The methods can not be used for submerged,

uniform fine sands which may be loose enough

to become quick temporarily and show a much

less resistance.

116. (d)

117. (a)

Slug is a higher strength and higher volume

waste discharged in short period of time.

118. (a)

119. (d)

A draft tube is a tube or pipe of gradually

increasing area which is used for dischargingwater from the exit of the turbine to the tail

race. This is because the pressure at the exit

of the summer of a reaction turbine is generally

less than atmosphereic pressure. Thus, the

water at exit cannot be directly discharged to

the tail race.

Also, the draft tube converts a large proportion

of the kinetic energy22 V

2g

rejected at the

outlet of the turbine into useful pressure energy.

Without the draft tube, the kinetic energy

rejected at the outlet of turbine will go wasteto the tail race.

Hence by using draft tube, net head on turbine

increases. The turbine develops more power

and also efficiency of turbine increases.

120. (d)

Minimum grade of concrete in RCC is

Restricted to M20 as per IS 456:2000

CE Test 15 Objective Solution

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