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CE (Test-24), Objective Solutions, 14 May 2016 (1)Objective Question Practice ProgrameDate: 14 May, 2016

ANSWERS

1. (c)

2. (d)

3. (c)

4. (d)

5. (c)

6. (d)

7. (d)

8. (a)

9. (d)

10. (b)

11. (d)

12. (a)

13. (d)

14. (d)

15. (c)

16. (a)

17. (c)

18. (d)

19. (b)

20. (c)

21. (d)

22. (d)

23. (c)

24. (b)

25. (d)

26. (a)

27. (d)

28. (d)

29. (c)

30. (c)

31. (b)

32. (b)

33. (d)

34. (b)

35. (b)

36. (a)

37. (a)

38. (a)

39. (c)

40. (c)

41. (d)

42. (c)

43. (d)

44. (a)

45. (c)

46. (d)

47. (c)

48. (d)

49. (a)

50. (b)

51. (a)

52. (b)

53. (b)

54. (b)

55. (b)

56. (c)

57. (c)

58. (b)

59. (d)

60. (d)

61. (b)

62. (b)

63. (d)

64. (c)

65. (b)

66. (d)

67. (c)

68. (a)

69. (d)

70. (d)

71. (d)

72. (b)

73. (b)

74. (b)

75. (c)

76. (a)

77. (d)

78. (d)

79. (b)

80. (d)

81. (a)

82. (b)

83. (b)

84. (a)

85. (c)

86. (a)

87. (d)

88. (b)

89. (d)

90. (d)

91. (a)

92. (b)

93. (b)

94. (c)

95. (b)

96. (d)

97. (b)

98. (c)

99. (a)

100. (c)

101. (d)

102. (a)

103. (c)

104. (b)

105. (b)

106. (c)

107. (a)

108. (c)

109. (a)

110. (d)

111. (d)

112. (c)

113. (a)

114. (b)

115. (a)

116. (a)

117. (b)

118. (a)

119. (c)

120. (c)

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(2) CE (Test-24), Objective Solutions, 14 May 2016Sol. 1. (c)

The rotation component z of any fluid elementin the flow field is

z = 1 v u2 x y

For flow to be irrotational, z 0

uy

x = 0 vx =

uy

Try with option (a)

vx = Clog y

x

x = C yy

x = Cx

uy =

Cy y

x= + 2

Cy

x

vx

uy not

irrotationalTry with option (b)

uy = 2A B y

y

x x = – Bx

vx =

x 212Axy By

2

= – 2Ay

uy

vx not irrotational

Try with option (c)

vx = Cy

x = 0

uy = y

(Cx) = 0

uy =

x it is rotational

Try with option (d)

uy = Cln y

y

x = cy x

x = cy

vx

= xCy

x =

cy

Sol. 2. (d)Bernoulli’s equation states that in a steady,irrotational flow of an incompressible fluid thetotal energy at any point is constant.Accordingly, if Bernoulli’s equation is applied

between any two points in a steady irrotationalflow of an incompressible fluid then we get

21 1

1p v z

2g

=

22 2

2p v z

2g

The above equation has been derived for anideal fluid for which there is no loss of energy.

Sol. 3. (c) u = x – 4yv = – y – 4x

x = v = – y – 4x ... (i)

y = – u = – x + 4y ... (ii)

From equation (i)

= – yx – 2x2 + f(y) ... (iii)

From equation (ii) = – yx + 2y2 + f(x)... (iv)

Comparing equation (iii) & (iv); = – xy – 2x2 + 2y2

Sol. 4 (d) u = 6xy – 2x2

v = ?For a two-dimensional flow of incompressiblefluid, the continuity equation may be expressedas

u vx y

= 0

6y – 4x+vy = 0

or,vy = 4x– 6y

or, v = 4xy – 3y2

Sol. 5. (c)Uniform flow: When the velocity of flow offluid doesnot change, both in magnitude anddirection, from point to point in the flowing fluidfor any given instant of time, the flow is saidto be uniform. In Mathematical form,

VS

= 0

Irrotational flow: A flow is said to be irrota-tional if the fluid particles while moving in thedirection of flow do not rotate about their masscentre. If may however be stated that a trueirrotational flow exists only in the case of flowof an ideal fluid for which no tangential or shear

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CE (Test-24), Objective Solutions, 14 May 2016 (3)stresses occur. But the flow of practical fluids,may also be assumed to be irrotational if theviscosity of the fluid has little significance.

Sol. 6. (d)Vx = 6 + 2xy + t2

Vy = 0 & Vz = 0

ax = u u u uu. V. wt x y z

here u = 6 + 2xy + t2

and v = w = 0

ut

= 2t

ux = 2y

ax = ut

+ u.ux

+ 0 + 0

= 2t + (6 + 2xy + t2). 2yat (3, 1, 2), and t = 2ax = 2 × 2 + (6+2×3×1+22) × 2 × 1

= 4 + 16 × 2 = 36 unitsSol. 7. (d)

Unit of angular deformation, dudy is s–1

unit of velocity gradient, dudy is s–1

unit of speed in rpm is s–1

Sol. 8. (a)The velocity potential, is defined as a scalarfunction of space and time such that itsnegative derivative with respect to any directiongives the fluid velocity in that direction.The stream function is defined as a scalarfunction of space and time such that its partialderivative with respect to any direction givesthe velocity component at right angles to thatdirection.

Sol. 9. (d)

= uy vx

xu = – y

u y = x

ux = u

u x

V tanu

=

and u y = – vThus it is a uniform flow inclined to x-axis.Here ux = velocity component in x-direction

uy = velocity component in y-directionSol. 10. (b)

V = 10m/sy

Vhhmax

V = 20 m/s

30º

02 = Vy2 – 2g hmax

hmax = 210

2g = 5.097 m

Sol. 11. (d)In lagrangian method, single fluid particle istaken and behaviour of this fluid particle isanalysed at different instant of time whereasin Eulerian Method, certain section is taken inspace and behaviour of various fluid particlesis analysed at this section.

Sol. 12. (a)V = 2 2ˆ ˆ ˆyz i xy j xy 2xyz k

here u = yz2

v = xy2

& w = xy – 2xyz

At (1, 2, 3)u = 2 × 32 = 18v = 1 × 22 = 4

and w = 1 × 2 – 2 × 1 × 2 × 3 = – 10

at 1,2,3V = 2 2 2u v w

= 22 218 4 10

= 20.976 21

Sol. 13. (d)u = 2x2 + z2 + 6v = y2 + 2z2 + 7It must satisfy continuity equation

u v

y z

x = 0

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(4) CE (Test-24), Objective Solutions, 14 May 2016

4x + 2y + z

= 0

z

= – (4x + 2y)

= – (4x + 2y)z + f(x,y)

= – 4xz – 2yz + f(x,y)

Sol. 14. (d) Streamlines: A streamline is an imagi-

nary curve drawn through a flowing fluid insuch a way that the tangent to it at anypoint gives tha direction of the velocity offlow at that point. Since a streamline iseverywhere tangent to the velocity vector,there can be no component of the veloc-ity at right angles to the streamline andhence there can be no flow of fluid acrossany streamline. In steady flow, since thereis no change in direction of the velocityvector at any point, the flow pattern is notchanging. But in unsteady flow, since thedirection of the velocity vector at any pointmay change with time, the flow patternalso changes with time. Therefore, for asteady flow the streamline pattern remainsthe same at different times, but for anunsteady flow, the streamline pattern maychanges from time to time.

Pathline: A pathline may be defined asthe line traced by a single fluid particle asit moves over a period of time. thus apathline will show that direction of veloc-ity of the same fluid particle at succes-sive instants of time. In steady flow thepathlines and streamlines are identical. Inunsteady flow, pathlines and streamlinesare different.

Streak line: A streakline may be definedas a line that is traced by a fluid particlepassing through a fixed point in a flowfield. In steady flow, a streak-line, astreamline and a pathline are all identical.In unsteady flow, a streak-line at any in-stant is the locus of end points of particlepaths (pathlines) that started at the sameinstant when particle passed through theinjection point.

Equipotential lines: It represents a curvefor which the velocity potential is same atevery point.

Sol. 15. (c) = 2xy

x

= v and y uy

v = yx

= 2y

v at (3, 4) = 2 × 4 = 8

Again, u = 2xy

u at (3, 4) = – 2 × 3 = –6 Resultant velocity at (3, 4) ,

V= 22 2 2u v 8 –6

= 10 m/sSol. 16 (a)

The force exerted on the plate by the jet =rate of change of momentum of the jet

Sol. 17. (c)Force, F = (av)(v – 0) = av2

v2 = 3F 1 10

a 1000 0.03

v = 5.7735 m/sSol. 18. (d)

It is more convenient to express themomentum flux flowing through any cross-section in terms of the mean velocity of flow.But the actual momentum flux is always greaterthan that computed by using the mean velocityof flow. Hence in order to account for thisdifference in the values of the momentum fluxdue to the non-uniform velocity distribution atany cross-section a factor called momentumcorrection factor is introduced so that themomentum flux computed by using the meanvelocity which would be equal to the actualtotal momentum flux through the entire cross-section.

Sol. 19. (b)

uv

c/s area of jet = a = 0.015 m2

v = 15 m/su = 5 m/s

Force on the plate,F = a (v – u) {(v – u) – 0}

= a (v – u)2

= 1000 × 0.015 (15 – 5)2

= 1500 N

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CE (Test-24), Objective Solutions, 14 May 2016 (5)Sol. 20. (c)Sol. 21. (d)

The discharge through a V notch is given by

Q = 528 Cd 2g tan .H

15 2

Taking log on both sides,

LnQ = 8 5Ln Cd 2g tan LnH15 2 2

On differentiating,

dQQ =

5 dH2 H

By question; dHH

= 0.01

dQQ =

5 0.01 0.025 25%2

Error in discharge measurements= 2.5%

Sol. 22. (d)Proportional weir or sutro weir

12x 2 y1 tanL a

xy

xO

y

a

L

H

In sutro weir, discharge Q varies linearly withhead H over the weir crest.The shape of the proportional weir profile hav-ing Q H as

1/2x y

The modified proportional weir profile as givenby sutro which has its sides diverging down-ward in the form of hyperbolic curves havingthe equation

2xL

= 12 y1 tana

Where a and L are respectively the height andwidth of the small rectangular shaped aperturewhich forms the base of the weir.

The discharge through this weir is given by

Q = aK H3

Where K = Cd L (2ga)1/2

The coefficient of discharge Cd for such weirsvaries from 0.60 to 0.65. Proportional weir isvery useful as a control device, especially, inchemical dosing and sampling.Cipolletti weir: A cipolletti weir is particulartype of trapezoidal weir, the sloping sides ofwhich have an inclination of 1 horizontal to 4

vertical i.e. 142

.

14°14°H

L

The discharge for the cipolletti weir is given by

Q = 32

d2 C 2gLH3

On the basis of experiment, the discharge overa cipplletti weir is Q = 1.86 LH3/2.Rectangular weir (Without and treatment)

Q = 32

d2 C 2gLH3

and Q = 3/2d

2 C 2g (L 0.1nH)3

with ‘n’ end contractionsParabolic weir:

2Q H

Sol. 23.(c)Applying Bernoulli’s equation (without loss ofenergy)

p =

22

w

v 2pv2g s.

v = w

2ps.

v = 2 100 15.81m / s

0.8

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(6) CE (Test-24), Objective Solutions, 14 May 2016Sol. 24. (b)

In a pitot tube, the velocity of the stream isgiven by

v = c 2g h

where c = coefficient of the instrument withususal values between 0.98 to 1.0.

h = 0pps

= 0.012 9810

1.2 9.81

= 10m of air

column.

V = 1 2 9.81 10 = 14.007 m/swhereps = pressure at stagnation pointp0 = pressure of the approaching flow

Sol. 25. (d)The loss of head HL between a section upstream of

the orifice and the vena contracta is

HL = 2vH 1 C

= 2 (1 – 0.982)= 0.0792 m 0.08 m

Sol. 26. (a)The discharge through a 90° V-notch weir isgiven by

Q = 5/2d

8C 2g H15

5/2Q H

5/2

0Q 0.15Q 0.30

5/2

0

Q 0.3Q 0.15

= 5.66

where Q is the new discharge

Q0 = Original discharge.Sol. 27. (d)

Discharge through a circular orfice

Q = Cd. a 2gh

where a = area of orificeand h = head above the orifice

Cd 21d

4

. 2g 0.25 = Cd.22d 2g 1

4

21

2

dd

= 1

0.25

1

2

dd = 1.414 = 2

Sol. 28. (d) Reynold number is used for flow of

incompressible fluid in closed pipes,motion of submarines completely underwater, motion of air planes, flow aroundstructures and other bodies immersedcompletely under moving fluids.

Froude number is used for free surfaceflows such as flow over spillways, sluicesetc in which gravity is a motivating force,flow of jet from an orifice or nozzle,problems in which waves are likely to beformed on the surface and problems inwhich fluids of different mass densitiesflow over one another.

Sol. 29. (c)

Discharge scale ratio, Qr = 2mr r

p

Q L VQ

here Froude model law is valid

Vr = rL

Qr = 2 2.5r r rL . L L

m

p

QQ =

2.5116

Qm = 1024 × 2.51

16 = 1 cumec

Sol. 30. (c)here , Lr = 1000 & Dr = 100

Qm = 0.1 m3/s; Qp = ?

Q = A.V. = (L.D)VFurther, according to Froude Law,

Vr = 1/2rD

P

m

QQ = Lr . Dr . 1/2 1.5

r r rD L .D

Qp = 0.1 × 1000 × (100)1.5

= 105

Sol. 31. (b)Discharge scale ratio

Qr =

3m

32m m rr r3

p rp

p

LQ T L

L .uQ TL

T

Froude model holds good in a hydraulic model

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CE (Test-24), Objective Solutions, 14 May 2016 (7)of a spillway and ur = rL

here by question, QP = 2048 cumec

and Lr = 1

16

mQ

2048 = 1221 1

16 16

Qm = 2.52048 2.016

= cumecs

Sol. 32. (b)Sol. 33. (d)

The model test results can be utilised to obtainin advance the useful information about theperformance of the prototype only if there exista complete similarity between the model andprototype.There are in general three types of similaritiesto be established for complete similarity toexist between the model and its prototype.(i) Geometric Similarity : Geometric

similarity exists between the model andthe prototype if the ratio of correspondinglength dimensions in the model and theprototye are equal.

(ii) Kinematic Similarity : Kinematicsimilarity exists between the model andthe prototype if (a) the paths of thehomologous mov ing particles aregeometrically similar, and (b) if the ratiosof the velocities as well as acceleration ofthe homoglous particles are equal.Homologous point means correspondingpoint in the model and prototype.

(iii) Dynamic Similarity : Dynamic similarityexists between the model and theprototype which are geometically andkinematically similar if the ratio of all theforces acting at homologous points in thetwo systems viz., the model and theprototype are equal. Thus for flows to bedynamically similar, the ratios of thevarious forces acting on the fluid particlesin one flow system should be equal to theratios of similar forces at correspondingpoints in the other flow system.

In compressible flow system, Machnumber should be same in the twosystem.

Sol. 34. (b) Reynolds number

= Re = inertia force

viscous force = vL =

vL

For dynamic similarity where viscous forcesare predominant

m

vL

= meR = peR = p

vL

Froude Number

= Fr =

1/2

1/2inertia force

gravity force = vgL

For dynamic similarity where gravity forcesare predominant

m

vgL

= r m(F ) = r p(F ) = p

vgL

Mach Number,

M =

1/2

1/2inertia force

compressibility force

= v v

cE /

where c = velocity of sound in the mediumwhere compressibility effects predominate, fordynamic similarity,

m

vc

= mM = PM = P

VC

Weber number = W = inertia force

surface tension

= 2v L

when surface tension ef fects

predominate in addition to inertia force(w)m = (w)p

Euler number = E =

1/2

1/2inertia force

pressure force

= vp /

In a fluid system where supplied pressuresare the controlling forces in addition to the

IES M

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(8) CE (Test-24), Objective Solutions, 14 May 2016inertia force and the other forces are eitherentirely absent or are insiginifcant, Euler modellaw is applied.

Sol. 35. (b)In this case the total resistance depends bothon the Reynolds number and on the Froudenumber.Thus, in this case for complete similaritybetween the prototype and its model theReynolds number as well as the Froude numbermust be the same for each i.e.

m m m P P P

m P

V L V L

and m P

m m P P

V Vg L g L

In practice, gm = gp

The combination of the above equation gives3/2

3/2m mr r

P p

Lor L

L

2/3

3/2r r

0.9 1 1L L7.2 8 4

Sol. 36. (a)The specific speed should be same in bothcases.

NS = 5/4N PH

5/4 5/4250 60 250 P

8 40

5/440P 60 57.915

8

P = 3,354.1 kWSol. 37. (a)

For the dynamic similitude conditions,Reynolds number and Mach number must besame for both model and prototype.As per reynolds number,

m m m

m

V L =

A P P

P

V L

m

p

LL =

m P P

P m m

VV

..... (1)

Again, as per mach number,

m

m

m

VK

= P

P

P

VK

P

m

VV =

P

P

m

m

K

K

..... (2)

From (1) and (2)

m

p

LL =

m P P m

P m m P

KK

= m

P

P P

m m

KK

Lr = r

r rK .

Sol. 38. (a)Action of fan is similar to a pump.

Specific speed Ns = 3/4N QH

should remain

constant even in geometrically similar fan.

1 1 2 2N Q N Q for same head

21800 4.5 1200 Q

21800 4.5Q

1200

Q2 = 10.125 m3/sSol. 39. (c)

By question,

Lr =1

4800tP = 12 hr

& tm = 3 minutesFroude law should hold true here

Vr = Dr1/2

where Dr is the horizontal scale

Vr = r

r

Lt

Dr1/2 =

14800

36012

Dr1/2 =

148003 160 12

= 1 60 12

4800 3

= 0.05

Dr = 2.5 × 10–3

= 1

400

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CE (Test-24), Objective Solutions, 14 May 2016 (9)Sol. 40. (c)

If y1 and y2 be conjugate depths or sequentdepths in a hydraulic jump.

2

1

yy =

2r1

1 –1 1 8 F2

or, 16.48 = 2r1

1 –1 1 8 F2

or, 33.96 = 2r11 8 F

or, 2r1F = 144

Fr1=12 >1 supercritical flowSol.41. (d)

The discharge per meter width at the foot ofspillwayq = 10m3/s/mVelocity, v = 20 m/s

v, y1 = 10

y1 = 10 0.5 m20

F1 = 1

V 20 9.03gy 9.81 0.5

y2 = 211

y 1 1 8F2

= 20.5 1 1 8 9.032

= 6.14 mSol. 42. (c)

q = 2/3 1/2y R sn

where R = by yb

q = 3/2 1/2y y sn

or, q = 5/3

1/2y sn

or, q y5/3

or, y q3/5

or, y q0.6

Sol. 43. (d)Here Q = 10 m3/sArea of cross-section of channel = 2.5 × 2 =5m2

Velocity of flow, v = 105 = 2 m/s

Specific energy, E = 2vy

2g

= 222 2.2 m

2g

Sol. 44 : (a)Tree System : In this system, there is onemain supply pipe, from which originates(generally at right angles) a number of submainpipes. The distribution network can be solvedeasily and it is possible to easily and accuratelycalculate the discharges and pressures atdifferent points in the system.Grid-Iron System : In this system, the mains,submains and branches are all inter-connectedwith each other. This system requires morelength of pipe lines and a larger number ofsluice valves (i.e. cut off valves).Circulation System : In this system, a closedring either circular or rectangular, of the mainpipes, is formed around the area to be served.Radial System : In this system water issupplied through radially laid distribution pipes.

Sol. 45. (c)While designing a pipe network, no directmethod is available to compute the pipe sizesof a distribution network and hence a trialprocedure is adopted where in the sizes areassumed and terminal pressures are computedto compare them with the designed minimumand maximum pressures. If the requiredmatching is not obtained, then the sizes arechanged and the procedure is repeated till therequired matching is obtained. Since, thedesign involves the use of various trial valuesof the pipe diameters and that too for variouspipes of the network, it is often convenient touse a Hazen-Williom’s chart instead of usingthe formula. This will avoid cumbersomecalculations. Since the pipes of the distributionsystem are made of cast iron, for which CH =100, the chart is available in the design officeand can be conveniently used. The method ofusing this chart is simple.

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(10) CE (Test-24), Objective Solutions, 14 May 2016Sol. 46. (d)

In this method, the pressures at various pointsalong a suspected pipe line are measured andhydraulic line is plotted. The appearance ofany kink or change in the slope of the hydraulicgradient line will indicate the location of a leakin the pipe line.

Sol. 47. (c)Sol. 48. (d)

Intensity of irrigation i.e.% of area irrigated= 200 – (72 + 66)= 62%

Sol. 49. (a)Sol. 50. (b)

D =

8.64B

For kor period of 25 days

D =8.64 25

0.15

= 1440 ha/(m3/s)

Discharge, Q =28001440 = 2 m3/s

= 2 m3/sSol. 51. (a)Sol. 52. (b)

Normal scour depth

= 1.35

1/32qf

yq2/3

y =

2/3307.410

= 15.39 mSol 53. (b)

w

d

hF.C PWP

d

d

w

dh (FC PWP)

= (0.3 – 0.1) × 1 × 1.5= 0.3 m = 300 mm

Sol. 54. (b)

Sol. 55. (b)Vc = 0.55 mD0.64

= 0.55 × 1.1 × 1.50.64

= 0.784 m/sSol. 56. (c) Floor thickness,

t = h

G 1

h = 9 – (3.4 + 1.8) = 3.8 m

t = 3.8

2.6 1

2.375 m

Sol. 57. (c)Sol. 58. (b)Sol. 59. (d)

2 t/m

AB

C

6 m 6 mEI is constant

MFBA = 2 2w 2 6 6tm

12 12

l

MFBC = 6tmSol. 60. (d)

Al

B

2w30 2w

20

w per unit length

FBA

FAB

MM =

2

2

w3020 3

220w30

Sol. 61. (b)

Sol.D C

B

L/2

w/m

A

L

L

I

2I

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CE (Test-24), Objective Solutions, 14 May 2016 (11)

MFBC = –2 2

FCBWL WL& M12 12

Using slope deflection equation for the memberBC. of the frame,

MBC = – 2

B CWL 2E(2I) 212 L

= – 2

B CWL 4EI 212 L

...(i)

Sol. 62 (b)

10kN 10kNC D

4m 1m

6m 2I

AB

2I

I1m

supports in the frame are not symmetrical,the frame will sway towards right.

Sol. 63. (d)

MBA = MFBA + 2EI4

B

324

(This is for lateral translation towards rightwhich is as per loading conditions)

MFBA = + 60× 4

8 = + 30 kNm

MBA = + 30 + B

2EI 324 4

Δ

Sol. 64. (c)

A C G PZ

D

B

The nature of the forces in members have beendepicted in the above figure.

Sol. 65. (b)

AM 0C

15kN

4m

BA

3m RB

B15 4 3R

BR 20kNSol. 66. (d)

15 kN5 m 5 m

15 kN30 kN

A

B

C D

E

FG

Considering joint G;FGB sin 30 + FGE sin 30 = 30

2FGBsin30° = 30 (by symmetry)

FGB = 30 kN

Again considering joint B,FBC cos 30 = FGB cos 30

FBC = FGB = 30 (compression)

Sol. 67. (c)

tan = 3 453

p = 6 sin45° = 62

Taking moments about point 5,

1 23F 50 6 100kN (Tensile)

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(12) CE (Test-24), Objective Solutions, 14 May 2016

F1-21

p

50 kN

54

6m

3m

Taking moments about point 4,

1 2 1 53 F p.F

1 53 100 300F 2 50 2

p 6

(comp)

Sol. 68. (a)

A B

50kN

50 kNC

D

Consider the joint D,

vF = 0

FCD = 50kN (Tension)

Sol. 69. (d)

Sol. 70. (d)

4tan3

54

3A

B

C60 kN

4 m

3 m

DM 0

VA × 3 = 60 × 4VA = 80 KN

FABsin = VA

FAB = 80 × 54

= 100 KN

Sol. 71. (d)

P

P1 P2

3 3

1 2Ph P h3(3EI) 3(2EI)

1 2P P3 2

2 12P P3 (i)

P1 + P2 = P (ii)From (i) and (ii)

13P P5

3 3

1Ph Ph3(3EI) 15EI

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CE (Test-24), Objective Solutions, 14 May 2016 (13)Sol. 72. (b)

MB =

F CBBC

2EI4EIM10 10

=

4 2 2000 300

10 10

= 120 + 40 = 160 KNMSol. 73. (b)

A

B C

D

50

40

36

36

X

3 m

D

C

50H

40

X

A

2 m

2 m

B

36

36

H

DM = 0

40 + 50 = H × 3

H = 30 kN

FBD of ‘AB’

AM = 0

36 + 30 x 4 = 36 + x × 2

x =120 60

2 kN.

Sol. 74. (b) Gross dia of rivets= 20 + 1.5 = 21.5 mm

For most critical sectionAnct = t × (b – n × d)

Anct = 1.2 × (15 – 2 × 2.15) = 12.84 cm2

Maximum tension in the flat

= 2150 12.84 101000

= 192.6 kN

Sol. 75. (c)Throat thickness = 0.7 × 6 = 4.2 mm

Length of weld = × d = 100 mm

Strength of weld

= 110 × 4.2 × 100= 46200N

Twisting moment =dF2

= 46200 × 50 = 2.31 × 106 N-mm

= 2.31 × 106 N-mm

Sol. 76. (a)Effective thickness of throat of weld

= 5 88 = 5 mm

Safe load = 5 × 120 × 142 = 85.2 kNSol. 77. (d)Sol. 78. (d)

At plastic moment condition neutral axis isequal area axis

B

D1

D

1 1

1 BD 1 B2 D D2 2 D

1DD2

Sol. 79. (b)

P

P

M

M

PM

2 Mp

4MP = WL2

W = P8ML

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(14) CE (Test-24), Objective Solutions, 14 May 2016Sol. 80. (d)

Failure in left or right spanW

0.6 MP

MPMP

LW2

= Mp × + Mp + 0.6 Mp ×

5.2MpWL

Failure in the middle span

W

0.6 MP

0.6 MP0.6 MP

0.6 MP

W L2

= 0.6MP + 0.6MP + 0.6MP +

0.6MP

4.8MpWL

Hence, Wc =P4.8M

LSol. 81. (a)

27400 600

2.4 106

z

2 216 672kNm

8 8WlM

Stress at the top extreme fibre at mid span

ePP MA Z Z

3 3 6

5 7 7960 10 960 10 100 72 102.4 10 2.4 10 2.4 10

4 – 4 + 3 = 3N/mm2

Sol. 82. (b)

Loss due to anchorage slip = sEL

= 5

22 1 10 5 52 520000

. . N / mm

% loss = 52 5 100 4 37 4 41200

. . . %

Sol 83. (b)Sol. 84. (a)

High temperature steam curing would result ina concrete containing less ettringite and moremonosulphate aluminate phase and a greateramount of unhydroted expansive componentsDelayed expansion in the concrete wasattributed to the transformation of metatablemonosulphate aluminate phase and remainingunhydrated expansive components to ettringiteat normal temperature and moist curingconditional. Hence lead to cracking.

Sol. 85. (c)Sol. 86. (a)Sol. 87. (d)Sol. 88. (b)

In concrete aggregate used are in saturatedsurface dry condition because surface ofparticles are dry but inter-particle void aresaturated with water in this condition aggregateswill not affect the free water content of acomposite material.

Sol. 89. (d)Membrane curing is a process of controllingthe curing of concrete by sealing the moisturethat would be lost to evaporation the processis accomplished either by spraying a sealeron the surface or by covering the surface witha sheet film.

Sol. 90. (d)The dynamic modulus is used primarily toevaluate soundness of concrete in durabilitytests, it is more appropriate value to use whenconcrete is to be used in structure subjectedto dynamic loading i.e. impact or earthquake.

Sol. 91. (a)Sol. 92. (b)

Iron pyrites in brick earth causes the brick toget crystallized and disintegrated duringburning, because of the oxidation of the ironpyrites. It also discolorise the bricks.

Sol. 93. (b)Sol. 94. (c)Sol. 95. (b)

Laminations are caused by the entrapped airin the voids of clay. Laminations produce thinlamina on the brick faces which weather outon exposure such bricks are weak in structure

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CE (Test-24), Objective Solutions, 14 May 2016 (15)Sol. 96. (d)

The process of removal of moisture contentfrom wood so as to make it useful forconstruction and other uses is called drying ofwood or seasoning of wood. This reduces thechances of decay, improve load bearingproperties, reduces weight and exhibits morefavourable properties like thermal & electricalinsulation, glue adhesive capacity & easypreservative treatment etc.

Sol. 97. (b)Sol. 98. (c)

Dry rot is a wood decay caused by certainspecies of fungi that digest parts of the woodwhich give the wood strength and stiffnesswhich resulted in a darkly colored deterioratedand cracked condition.

Sol. 99. (a)Creosote are a category of carbonaceouschemicals formed by the distillation of varioustars and by pyrolysis of plant - derived materialsuch as wood or fossil fuel

Sol. 100. (c)Sol. 101. (d)

Hydration of PPC is a slower process thanhydration of ordinary cement, resulting intoslower heat generation and lower internalstresses in concrete. Thus PPC become idealcement for mass concrete like dams retainingwalls, large foundation etc.

Sol. 102. (a)Streamlines are a family of curves that areinstantaneously tangent to the velocity vectorof the flow. These show the direction a fluidelement will travel in at any point in time. Bydefinition, different streamlines at the sameinstant in a flow do not intersect, because afluid particle cannot have two different velocitiesat the same point.

Sol. 103. (c)Ideal fluids are those fluids which have noviscosity and surface tension and they areincompressible. As such for ideal fluids noresistance is encountered as the fluid moves.

A flow is said to be rotational if the fluidparticles while moving in the direction offlow rotate about their mass centres. Theliquid in the rotating tanks illustratesrotational flow where the velocity of eachparticle varies directly as the distance fromthe centre of rotation.

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

Sol. 104. (b)Distorted models are those in which one ormore terms of the model are not identical withtheir counterparts in the prototype. Since thebasic condition of perfect similitude is notsatisfied, the results obtained with the help ofa distorted model are liable to distortion andhave more qualitative value only. A distortedmodel may have either geometrical distortionor material distortion or distortion of hydraulicquantities or a combination of there. Whendifferent scale ratios are adopted for thelongitudinal, transverse and vertical dimensionsthen it is said to be a distortion of dimensions.Distortion of dimensions is frequently adoptedin river models where a different scale ratio fordepth is adopted.

Scale effect in modelsIf complete similitude does exist between amodel and its prototype there will be somediscrepancy between the results obtained fromthe model tests and those which will beindicated by the prototype af ter i tsconstruction. This discrepancy or disturbinginfluence is called scale effect. Some of thefactors which may lead to the departure fromcomplete similarity and hence result in thedevelopment of the scale effect. After it maynot be possible to correctly simulate all theconditions (e.g. roughness) in the model asthat of the prototype. This may also result indeveloping scale effect if any of these conditionshas a pronounced effect on the phenomenon.In order to detect the presence of suchdisturbing influences the proposed work maybe tried in models with different scales andthe resulting scale effects judged from thecomparative results so obtained. Besides thisthe observation collected on modelsconstructed to different scales will also provideare empirical relationship between scale effectand size of model, which may be utilised tocorrect the results of the model tests.

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(16) CE (Test-24), Objective Solutions, 14 May 2016Sol. 105. (b)

depth

yC

sp. energy

At critical flow, the Froude number, Fr = 1. Asthe tangent at the critical point, is vertical inthe curve between depth vs specific energy,even with slight change in specific energy, itwill cause major change in depth.

Sol. 106. (c)Let us consider a case in which the specificenergy is kept constant and the discharge Qis varied.

Es – y = 2

2Q

2gA

or Q = sA 2g (E y)

where Es = specific energyQ = discharge through the channelA = C/s area of the channeland y = depth of flow Q2 = A2 (2g) (Es – y) = 2gA2Es – 2gA2yFor a given specific energy, the discharge will

be maximum if dQ 0,dy

differenting above

equation yields

dQQdy = 2

sdA dA2gE 2A 2g 2yA Ady dy

dAdy = A 2gEs (2AT) – 2g (2yAT) – 2gA2 = 0

or, 4 EsT – 4yT = 0 2T (Es – y) = A

Es = Ay2T

But Es = 2

2Qy

2gA

2

2Qy

2gA =

Ay2T

or 2Q

g = 3A

T

This equation is for the critical depth. Thus fora given specific energy, the discharge in agiven channel is a maximum when the flow isin the critical state. The depth correspondingto the maximum discharge is the critical depth.

ycy

Qmax

QSol. 107. (a)Sol. 108. (c)

The stream function, is defined as a scalarfunction of space and time such that its partialderivative with respect to any direction givesthe velocity component at right angles (in thecounter-clockwise direction) to this direction

u = and vy x

and wz = 2 2

2 212 x y

For irrotational flow,

wz = 0 2 2

2 2 0x y

which is

laplace equation for .equation of continuity for two dimensionalsteady flow of an incompressible fluid.

x y y x

= 0

2

x y

= 2

y x

This will be true if is a continuous functionand its 2nd deriviative exists. Therefore, it maybe stated that any function which iscontinuous is a possible case of fluid flow(which may be rotational or irrotational) sincethe equation of continuity is satisfied.

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CE (Test-24), Objective Solutions, 14 May 2016 (17)However, if the function is such that it satisfiesLaplace equation thus it is a possible case ofan irrotational flow.

Here, stream function,

= 3 2 2 32 3x y x y3 2

= (x, y) is a two dimensional function

y = 3 2 24 9x y x y

3 2

x

= 2 2 32x y 3xy

2

x y

= 2 24x y 9xy

2

y x

= 2 24x y 9xy

2

x y

= 2

y x

here is a continuous function and itsderivative exists. There fore, it be may statedthat it is a possible case of fluid.

Again, 2

2x

= 2 34xy 3y

2

2y

= 3 24 x 9x y3

2 2

2 2x y

= 2 3( 4xy 3y ) 3 24 x 9x y3

0

It is a rotational flow.Sol. 109. (a)

A flow net is a graphical solution to theequations of steady ground water flow. A flownet consists of two sets of lines which mustalways be orthogonal (perpendicular to eachother). Flow lines, which show the direction ofground water flow, and equipotentials (lines ofconstant head), which show the distribution ofpotential energy. Flow nets are usuallyconstructed through trial and error sketching.In an isotropic medium the hydraulic gradienthas to be the steepest possible meaning

thereby that the length of flow shall be theshortest. Thus, flow lines have to crossequipotential lines orthogonally. The spacebetween two adjacent flow lines is called flowpath and the figure formed on the flownetbetween any two adjacent flow lines andadjacent equipotential lines is referred to asfield.

Sol. 110. (d)The discharge through a triangular weir isgiven by

Q = 5/2d

8 C 2g tan H15 2

where Cd is the discharge coefficientH is the head of flow is the vertex angle

H

Area of the triangular notch = 1 (2H tan /2) H2

= H2 tan 2

Avg. velocity =5/2

d8 C 2g H tanQ 15 2

A H tan2

= d8 C 2gH

15

Sol. 111. (d)A positive surge is one which results in anincrease in the depth of flow and a negativesurge causes a decrease in the depth offlow.

y2

y1

Positive surge(Advancing downward)

Vw

y2

y1

Positive surge(Advancing upstream)

Vw

v2v1

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(18) CE (Test-24), Objective Solutions, 14 May 2016

y2

y1

Negative surge(Retreating downstream)

Vw

v2v1

y2

y1

Vw

v2v1

In running irrigation canal when the regulatinggate is partially closed, such a movement givesrise to a positive surge travelling upstream anda negative surge rises in downstream.

Sol. 112. (c) Specific energy,

E =2

2Qy

2gA

For simplicity, consider a rectangular channelof width b, for which A = by. Then E =

2

2y2gy

, expressing E as a function of the

discharge q and the depth, y or

g y 2g (E y) , expressing q as a function

of E and y. The curve of q as a function of y fora fixed E is plotted.

yc

c

q

d

ab

y

Sub-critical qmax

Super-critical

We notice that q is a double valued function ofy and has a maximum possible value m. Fordepth smaller then the critical depth the velocityis greater than the critical velocity. Flow in thisregion is called super critical. The samedischarge is possible with given e in eitherregion.Suppose that there is a lateral constriction inthe channel, reducing its area and increasingdischarge with per unit width (q). Assume thereis no head loss, the specific energy does notchange, so the flows in the constriction arerepresented by points b and d. we note that in

subcritical flow, the depth of flow decreaseswhile in super crictical flow the depth increases.We can also plot y as a function of E for aconstant discharge, q. The point a correspondsto an upper stage or transquil flow.

yc

c

E

d

a

by

E2 E1

Let us suppose there is a hump in the bed ofthe channel that decreases the specific energyE1 to E2. The height of the hump will be thedecrease in the specific energy. If the flow issubcritical, we see that depth will decreaseslightly to point b. If the flow is supercritical thedepth will, on the other hand, increase slightlyto point e. This is exactly the same as theresponse to a lateral constriction. If the humpis high enough, the flow may become critical atit. For any larger hump, the specific energycan not decrease further, and instead theupstream depth must increase to keep the flowcritical over the hump. For this reason, such apoint maybe called a control section, since itcontrols the upstream depth. The vertical tangentat critical depth means that small changes in Ewill cause large changes in y, so the surfacemay appear disturbed.

Sol. 113. (a)Fluid flow is said to be steady if at any pointin the flowing fluid various characteristics suchas velocity, pressure, density, temperature, etcwhich describes the behaviour of the fluidmotion, do not change with time

u 0;t

v 0;t

w 0;t

p 0;t

0;

t

When the velocity of flow of fluid does notchange, both in magnitude and direction, frompoint to point in the flowing fluid, for any giveninstant of time, the flow is said to be uniform.

v 0s

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CE (Test-24), Objective Solutions, 14 May 2016 (19)Sol.114. (b) Kani’s method

Gaspar kani’s method of structural analysis issimilar to cross moment distribution in thatboth these methods use Gauss-seidell iterationprocedure to solve the slope deflection equationwithout explicitly writting them down. However,whereas the moment distribution methodobtains the unknowns (i.e., the end momentsof the structural members) by iterating theirincrements, Kan’s method iterates theseunknowns themselves. This method essentiallyconsists of a single, simple numerical operationperformed repeatedly by the joints of a structurein a chosen sequence. Results of any desiredaccuracy may be obtained by peforming thisoperation a sufficient number of times usingthe required number of significant digits. Kani’smethod is specially useful for the analysis ofmultistorey frames. It has the advantages ofsimplicity, speed, economy of time, labour andspace and of accuracy. However, perhaps, thetwo most attractive features of this method areas follows:

It has a built-in error elimination so thatcomputational errors automatically disappearin subsequent operations. This also makespossible the introduction of any changes inloads or member lengths that may becomenecessary during calculations withoutnecessitating a new analysis. Such changesare inserted in the computational schemewherever required and the analysis simplycontinued.

It requires only one table of calculations evenfor highly irregular frames with multiple sidesway. Compared to most other methods,Kani’s method involves substantially less labourand time in the analysis of such frames.The above advantages make Kani’s methodone of the most powerful techniques applicableto all types of continuous beams and frames.Moment distribution method.This method consists of solving slope deflectionequations by successive approximation thatmay be carried out to any desired degree ofaccuracy. Essentially, the method begins byassuming each joint of a structure is fixed.

Then by unlocking and locking each joint insuccession, the internal moments at the jointsare distributed and balanced until the jointshave rotated to their final or nearly finalpositions. This method of analysis is bothrepetitive and easy to apply. Manual analysisof gable frames mostly uses the momentdistribution or slope deflection methods. Thesemethods are usually lengthy and have no built-in-error elimination capability.

Sol. 115. (a)Sol. 116. (a)Sol. 117. (b)Sol. 118. (a)Sol. 119. (c)Sol. 120. (c)

Velocity potential,

= 4 2 2y 6x y

The continuity equation is given by

ux y

= 0

Any flow must satisfy the continuity equation.

u = –x

and v = –y

u = 2( 12xy )

and v = 3 2(4y 12x y)

ux dy

= 12xy2 – 4y3 + 12x2y 0

it is not a valid potential function.Potential function satisfies the condition ofirrotational flow. The existence of velocitypotential function means that a possible flowmust be irrotational.

for irrotational flow, u

x y

must be equal

to zero

24 xy – 24 xy = 0 it is satisfyingirrotational condition.