CE Test 10 Obj Solution

8/18/2019 CE Test 10 Obj Solution

1/9

I E S M

A S T E

R

CE (Test-10), Objective Solutions, 2 nd April 2016 (1)

1. (a)

2. (c)

3. (c)

4. (a)

5. (a)

6. (a)

7. (c)

8. (a)

9. (c)

10. (d)

11. (b)

12. (a)

13. (b)

14. (d)

15. (c)

16. (b)

17. (b)

18. (c)

19. (d)

20. (a)

21. (c)

22. (b)

23. (c)

24. (c)

25. (b)

26. (a)

27. (b)

28. (a)

29. (c)

30. (a)

31. (b)

32. (c)

33. (b)

34. (c)

35. (b)

36. (d)

37. (b)

38. (d)

39. (a)

40. (b)

41. (b)

42. (c)

43. (d)

44. (b)

45. (b)

46. (c)

47. (d)

48. (b)

49. (c)

50. (c)

51. (c)

52. (a)

53. (a)

54. (b)

55. (b)

56. (d)

57. (a)

58. (a)

59. (d)

60. (a)

61. (c)

62. (d)

63. (a)

64. (b)

65. (a)

66. (d)

67. (a)

68. (d)

69. (d)

70. (d)

71. (c)

72. (b)

73. (d)

74. (d)

75. (d)

76. (d)

77. (b)

78. (d)

79. (d)

80. (a)

81. (d)

82. (d)

83. (c)

84. (d)

85. (d)

86. (a)

87. (a)

88. (d)

89. (b)

90. (c)

91. (b)

92. (c)

93. (d)

94. (c)

95. (a)

96. (c)

97. (d)

98. (d)

99. (c)

100. (b)

101. (d)

102. (c)

103. (a)

104. (d)

105. (d)

106. (c)

107. (b)

108. (c)

109. (a)

110. (a)

111. (a)

112: (c)

113. (a)

114. (c)

115. (a)

116. (a)

117. (a)

118. (b)

119. (a)

120. (a)

Conventional Question Practice ProgrameDate: 2 nd April, 2016

ANSWERS


2/9

I E S M

A S T E

R

(2) CE (Test-10), Objective Solutions, 2 nd April 20161. (a)

2. (c)

In case of sway,

CDF 2

6E 2I 12EI 4EIM

9 33

BAF 26E I 6EI 3EI

M 4 22

BACD

F

F

3EIM 92

4M 8EI3

3. (c)

T = 8 years

Risk =

31

1 1T

=

31

1 1 0.3338

4. (a)

Peak flow Q P is achieved at time of concentration t C. The peak value of runoff isgiven by

Q P = CAi; for Ct t

If rainfall continues beyond t C , the runoff willbe constant and at the peak value Q P .

5. (a)

Vertical deflection of A

= Vertical deflection of A keeping B fixed

+ Angular rotation of B × l

= 2 h

2EI EI

ll =

h

EI 2

l l

6. (a)

7. (c)

8. (a)

L/3 L/3 L/3

PL/3

BMDSlope at centre C is 0.

Rotation at end A = Area ofMEI

diagram

between A and C

= 1 PL L PL L2 3EI 3 3EI 6

=2PL

9EI9. (c)

2

2

Pabl

2

2

Pba

l

P

a b

a + b = l

M AB =

2

2

5 3 22.4 kNm

5

MBA =

2

25 2 3

3.6 kNm5

10. (d)

C0 + C 1 + C 2 = 1 C 2 = 1 – 0.05 = 0.55 = 0.4

Given : I 1 = 20 m3/s, I 2 = 40 m

3/s,Q 1 = 11 m

3/s

Hence, Q 2 = C 0I2 + C 1I1 + C 2Q 1= 0.05 × 40 + 0.55 × 20 + 0.4 × 11

= 17.4 m 3/s

11. (b)

Effective Principal stress ratio =

100 103

40 1012. (a)

Treatment System Detention TimeGrit chamber 60 sec

PST 2-2.5 hr

ASP 5-8 days

SST 1.5-2 hr

Sludge Thickner 4 hr

Sludge Digestion 30 days

Septic Tank 24 hr

Imhoff Tank 2 hr

13. (b)

BOD left over = 0.55 × 0.45 = 24.75%

BOD removal efficiency = 100 – 24.75

= 75.25%

14. (d)

Surface loading rate =Flow rate

Surface area

=3

21620 m /hr

15 1.8 m

= 60m 3/hr/m 2

= 1000 l/min/m 2

Detention time =Volume of chamber

Flow rate

= 3

315 1.8 0.8m

1620 m /hr

= 48 seconds


3/9

I E S M

A S T E

R

CE (Test-10), Objective Solutions, 2 nd April 2016 (3)15. (c)

A sl udg e tha t exhi bi ts po or se tt lingcharacteristics is called a bulking sludge.Filamentous micro-organisms have beenfound to be responsible for a bulked sludge.Large surface area to volume ratios of thesemicro-organisms retard their settling velocities.Fungi are the most familiar filamentous micro-organisms. These bacteria are usually foundto develop in activated sludge systems whichare characteristics by the low or variablenutrient concentrations.

16. (b)

The different actions that take place inanaerobic decomposition process are

(i) Acid fermentat ion stage or Acidproduction stage: In this first stage of

sludge digestion, the fresh sewage-sludgebegins to be acted upon by anaerobicand facultative bacteria called acidformers. These organisms solubilize theorganic solids through hydrolysis. Thesoluble products are then fermented tovolatile acids and organic alcohols of lowmolecular weight like propionic acid,acetic acid, etc. Gases like methane,carbon dioxide and hydrogen sulphide arealso evolved. Intensive acid productionmakes the sludge highly acidic and lowersthe pH value to less than 6.

(ii) Acid : regress ion s tage : In thisintermediate stage, the volatile organicacids and nitrogeneous correspounds of the first stage are attacked by the bacteriaso as to form acid carbonates andammonia compounds. Small amount of hydrogen sulphide and carbon-dioxidegases are also given off. The decomposedsludge has a very offensive odour and itspH value rises a little and to be about

6.8.(iii) Alkaline fermentation stage: In this final

stage of sludge digestion, more resistantmaterials like proteins and organic acidsare attacked and broken up by anaerobicbacteria called methane formers intosimple substances like ammonia, organicacids and gases. During this stage, theliquid separates out from the solids andthe digested sludge is formed. Digestedsludge is alkaline in nature. The pH value

during this stage rises to a little above 7in the alkaline range.

17. (b)

MBA = BAF B A

2EI 3M 2

L l

= B4EI

03 {

A 0, 0}Similarly,

MBC = 2

B C100 6 2E 4I 32

12 6 l

= B4EI

3003

{ C B }

Now, M BA + M BC = 0

300 = B8EI

3

B9008EI

MBA = B4EI 4EI 900

3 3 8EI = 150 kN.m

18. (c)

T = 50 years

P =1

50

Required probability =

45

41 49

C50 50

19. (d)

20. (a)

21. (c)

21

0.18 1 1T

10.81 1T

10.9 1T

10.1

TT = 10 year

22. (b)

23. (c)

45°

P

B

L L

A

C

BB2

45°

2 L


4/9

I E S M

A S T E

R

(4) CE (Test-10), Objective Solutions, 2 nd April 2016

MBC =

B

B C3 22EI

2( 2 L) 2 L

=

BB C

32EI2

L L

Displacement er to BC is B2 and it is in

anticlockwise direction.24. (c)25. (b)

For rectangular channel E C = C3

y2

2.7 = C3

y2

yC = 1.8 m

Acutal depth of flow = 1.6 m < 1.8 mFlow is supercritical.

26. (a)27. (b)

Pondage required = 12m 3/s × 6 hr = 12 × 6 × 3600 m 3

= 0.259 mm 3

28. (a)29. (c)

Supercritical flow occurs when depth of flowis below critical depth of flow.

CDL

NDL

Steep slope

S1

S2

S3CDL

NDL

Mild slope

M1

M2

M3

30. (a)

Re lative Total DistributionJoint Member

Stiffness Stiffness Factor 4EI

B BA EI 2EI 1/ 24

3EIBC EI 1/ 2

3Fixed end moment

AB BAM M 0 2

BC CBM6 3

M 4.5 kNm12

Moment Distribution :

0 0

2.251.125

–4.5 4.5

2.25

1.125 2.25 –2.25 4.5 Total

FEM

BalancingC/o

1/2 1/2B

A C

Hence final and moment at A = 1.125 kNm

31. (b)

TotalRelative Distribution

Joint Member RelativeStiffness factor

Stiffness4E(2I)

CB 8 / 118

3E(I) 11EIC CD 3 / 11

8 8CE 0 0

Since end E is free end, member CE haszero relative stiffness

32. (c)

33. (b)

H = m v H0 ( –

)

2 = m v H0 (160 – 80)

v 02

m H

802

H (320 –160) 4cm80

34. (c)

35. (b)

36. (d)

37. (b)

38. (d)

39. (a)

40. (b)

41. (b)

42. (c)

43. (d)

44. (b)

45. (b)

46. (c)

47. (d)

48. (b)

49. (c)

50. (c)

51: (c)

Head (H)

Input Power, P

Speed = Constant

Output Power

Discharge, Q

E f f i c i e n c y ,

P o w e r ,

P

H e a

d ,

H

Efficiency,


5/9

I E S M

A S T E

R

CE (Test-10), Objective Solutions, 2 nd April 2016 (5)52. (a)

h =

2

2 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 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

53. (a)

Pelton is impulse turbine, so has zeroDegree of reaction.

54. (b)

The centrifugal pump will start delivering water if the pressure rise in the impeller is morethan or equal to manometric head (H m).

55. (b)

u = 2gH 0.44 2 9 .81 300 = 33.76 m/s

DN60

= u

D =60u 60 33.76

1.07mN 600

56. (d)

In turbines, only reaction turbines aresubjected to cavitation. In reaction turbinesthe cavitation may occur at the outlet of therunner or at the inlet of the draft-tube wherethe pressure is considerably reduced (i.e.,which may be below the vapour pressure of the liquid flowing through the turbine). Due tocavitation, the metal of the runner vanes anddraft-tube is gradually eaten away, whichresults in lowering the efficiency of the turbine.Hence, the cavitation in a reaction turbinecan be noted by a sudden drop in efficiency.In order to determine whether cavitation willoccur in any portion of a reaction turbine, thecritical value of Thoma’s cavitation factor iscalculated.

57. (a)

Francis turbine is used for medium head.Kaplan is used for low heads and Peltonturbine is used for high heads.

58. (a)

For hydroelectric power uses, a surge tank

is an additional storage space or reservoir fitted between the main storage reservoir andthe power house. It should be as close to thepower house as possible. Surge tanks areprovided in high or medium head plants whenthere is a considerable distance beween thewater source and the power unit,necessitating a long penstock. The mainfunctions of the surge tank are

(i) when the load increases, the water movesbackwards and gets stored in it

(ii) when the load decreases, additionalsupply of water will be provided by surgetank.

59. (d)

In centrifugal pumps the cavitation may occur at inlet of the impeller of the pump, or at thesuction side of the pumps, where thepressure is considerably reduced. Hence if the pressure at the suction side of the pumpdrops below the vapour pressure of the liquidthen the cavitation may occur. The cavitationin a pump can be noted by a sudden drop in

efficiency and head. In order to determinewhether cavitation will occur in any portion of the suction side of the pump, the critical value

of Thoma’s cavitation factor ( ) is calculated.

60. (a)

2 2m

gHN D = 2 2 p

gH

N D

In case of multi-stage pumps, since the sameliquid flows through each impeller, thedischarge of a multi stage pump is same as

the discharge passing through each impeller of the series.

61. (c)

Mass of water m = 50 × 10 –3 × 1000Kg/s = 50 Kg/s,

Head H = 40 m

so power = mgH = 50 × 9.81 × 40 = 19.6kW.

62. (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


6/9


7/9

I E S M

A S T E

R

CE (Test-10), Objective Solutions, 2 nd April 2016 (7)97. (d)

CV =v w

Km ·

K = C V · m v · w

A

B

KK =

v A V A

V B V B

(C ) ·(m )(C ) ·(m )

=4 4

4 43.8 10 3.6 101.9 10 1.8 10

= 4.0

98. (d)

99. (c)

100. (b)

101. (d)

102. (c)

The principal advanates of dual-media andmixed-media filters over sand filters are higher filtration rates and longer filter sums becauseof the increased volume for floc storage withinthe filter.

103. (a)

During the geological history, clays at depthmight experience higher overburden pressure

(so that also higher effective over burdenpressure ) than the present overburden( = z). For example, erosion of the surfacesoil layers can make a clay layer becomeshallower and closer to the surface. Whenthe effective pressure applied is less than themaximum effective over burden pressure inthe field, a small amount of elastic pressurewill occur. When the effective pressure appliedis greater than the maximum effectiveoverburden pressure in the field, the changein void ratio is much larger and make thee-log curve is practically linear and steeper.

elastic expansionof solid particles

e 0

e 1

v o

i d r

a t i

o

Effective pressure, ´ (logscale)

e 2

In the initial phase, relatively great change in

pressure only results in less change in voidratio e. The reason is part of the pressure gotto compensates the expansion of the solid

particles when the soil specimen wassampled. In the following phase e changes ata great rate.

This phenomenon can be verified by the load-unload-reload process in lab consolidationexperiment.

v o

i d r

a t i

o ,

e

effective pressure, (log scale)´

reloadf

unload

a

d

bload

c

In the unload portion, segment cd containselastic deformation and segment da is thepart that can not be recovered so that is theplastic part of the deformation.

Effect of disturbance on void ratio-pressure relationship

virginconsolidationcurve

Laboratoryconsolidationcurve

Consolidationcurve for re mouldedspecimen

f 0.4e

e 0 C 32

g

1d

v o

i d r

a t i

o ,

e

a

0 c =

Normall consolidated clay of low tomedium plasticity

Laboratoryconsolidationcurve

Virginconsolidationcurve

Labora toryreboundcurve

0.4e 0

e 0f

2

c ag

h j i

3 v o

i d r

a t i

o ,

e

d bpressure, (logscale)

Over consolidated clay of low to mediumplasticity


8/9

I E S M

A S T E

R

(8) CE (Test-10), Objective Solutions, 2 nd April 2016104. (d)

Secondary Consolidation

The reduction in volume of a soil mass causedby the application of a sustained load to themass, due principally to the adjustment of the internal structure of the soil mass after

most of the load has been transferred fromthe soil water to the soil solids.

Immediate Compression

The immediate compression occurs as soonas the load from the building is applied onthe foundation soil and is a result of expulsionof air from voids of soil grains.

Primary Consol idation

The force applied on the soil strata alsoincreases the pressure in the pore water. Thisis higher as compared to the pore water pressure in the surrounding soil, thus ahydraulic gradient is developed causing theflow of water to the surrounding areas. Thetime taken for the expulsion of pore water from a soil depends on the permeability of that soil.

105. (d)

Process of consolidation

There are four main stages as follows:

1. Initially: equilibrium (or steady state) porewater pressure, u 0 is constant (and issimply the head of water, or hydrstaticpressure = wz)

2. Load applied to soil surface increasestotal stress on soil sample, whichgenerated a rise in pore water pressure

– Soil part icles try to mov e closentogether– but prevented by incompresiblepore water. Water pressure rises u(excess pore water pressure), to equalthe total stresses increase;

u = w z + applied stress = u 0 + u3. Dissipation of p ore water pressure:

Over a period of time (Months to years),the excess pwp, u, dissipates (drainsslowly out of the clay due to lowpermeability under sustained load – i.esqueezing from voids of pressurised porefluid). The plate like clay paricles take upnew positions, resulting in settlement andincrease in effective stress:

total ( )

constanteffective ( )

increasing

effective ( )

increasingtotal ( )

constant

p.w.p (u)

decreasing – p.w.p (u)

decreasing

+

=

=

4. Full dissipation of excess pwp ( u = 0):In the long term, excess pwp becomeszero and there is a maximum increase ineffective stress, and pwp returns to itsoriginal value, u 0.

106. (c)

As per Terzaghi’s theory, for the giv enpressure increment, the process of consolidation should be over after thedissipation of excess pore water pressure.But some compression is noticed even after the primary consolidation ceases. This isknown as secondary consolidation and maybe due to the highly viscous water betweenthe points of contact of soil particles beingforced out, change in arrangement of particlesand possible fracture of particles because of creep. The magnitude of secondaryconsolidation in inorganic soils is so smallcompared to the primary consolidation, it isusually ignored. But in organic soils, thesecondary consolidation is predominant.

The secondary consolidation is not governedby Terzaghi’s consolidation theory. It is oftenobserved that the laboratory consolidation testresults are in agreement with Terzaghi’stheory upto U = 60%. It means that thesecondary compression takes place evenbefore the primary consolidation is completed.

107. (b)

108. (c)

109. (a)

110. (a)

111. (a)

112: (c)

The specific speed of pelton wheel in low.

113. (a)

114. (c)

The dead load factor which is taken as 0.8 or 0.9 while considering stability against over turning or sliding or while considering reversedof stresses when dead loads are combinedwith wind / earthquake loads. In such cases,underestimating the counteracting effects of dead load results in greater safety.


9/9

I E S M

A S T E

R

CE (Test-10), Objective Solutions, 2 nd April 2016 (9)115. (a)

Code specifies C 1.5 and s 1.15 .

116. (a)

117. (a)

118. (b)

119. (a)

This is so, because regardless of whether the beam is under-reinforced or over-reinforced,

collapse invariably occurs by the crushing of concrete.

120. (a)

Lower values of p t and f st are indicative of lower design loads and lower stains distributedacross the cross-section and hence lower

curvatures and lesser deflections.

CE Test 10 Obj Solution

Documents

Transcript of CE Test 10 Obj Solution