answers - ies master · limestone quick lime 2 2 slaked lime caco cao co h o (slaking) ca(oh) white...

1. (c)

2. (b)

3. (b)

4. (a)

5. (a)

6. (b)

7. (b)

8. (a)

9. (a)

10. (b)

11. (c)

12. (a)

13. (d)

14. (c)

15. (b)

16. (c)

17. (d)

18. (b)

19. (a)

20. (d)

21. (a)

22. (c)

23. (b)

24. (c)

25. (d)

26. (a)

27. (d)

28. (c)

29. (d)

30. (b)

ESE-2018 PRELIMS TEST SERIESDate: 12 November, 2017

ANSWERS

31. (a)

32. (d)

33. (c)

34. (a)

35. (c)

36. (d)

37. (c)

38. (c)

39. (c)

40. (b)

41. (a)

42. (d)

43. (c)

44. (a)

45. (c)

46. (d)

47. (b)

48. (d)

49. (d)

50. (c)

51. (c)

52. (b)

53. (b)

54. (b)

55. (d)

56. (c)

57. (b)

58. (d)

59. (b)

60. (c)

61. (a)

62. (c)

63. (a)

64. (b)

65. (c)

66. (a)

67. (d)

68. (b)

69. (b)

70. (c)

71. (c)

72. (d)

73. (b)

74. (c)

75. (a)

76. (d)

77. (b)

78. (d)

79. (b)

80. (b)

81. (d)

82. (b)

83. (b)

84. (c)

85. (a)

86. (b)

87. (a)

88. (c)

89. (a)

90. (a)

91. (a)

92. (b)

93. (b)

94. (b)

95. (a)

96. (c)

97. (c)

98. (c)

99. (c)

100. (b)

101. (b)

102. (d)

103. (b)

104. (c)

105. (c)

106. (c)

107. (b)

108. (d)

109. (b)

110. (b)

111. (c)

112. (d)

113. (c)

114. (c)

115. (a)

116. (b)

117. (c)

118. (d)

119. (b)

120. (d)

121. (b)

122. (d)

123. (b)

124. (c)

125. (b)

126. (a)

127. (c)

128. (a)

129. (b)

130. (d)

131. (a)

132. (c)

133. (b)

134. (b)

135. (d)

136. (a)

137. (a)

138. (c)

139. (d)

140. (c)

141. (c)

142. (a)

143. (d)

144. (d)

145. (d)

146. (b)

147. (d)

148. (c)

149. (c)

150. (a)

IES M

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1. (c)

Sea shore and river gravels are roundedand the aggregate produced from crushedrock is angular in nature.

2. (b)

Bernoull i’s equation is obtained byintegrating Euler’s equation and the threeheads are constant over a streamline only.If the flow is irrotational then it is constantover the whole field.

3. (b)

4. (a)Wshaft = v

=2(0.02) 4008000 10

4 1000

= 3.2 N

=dudy

= TV0.50 20.20 202 1000

= 5000 VT

At equilibrium;Wshaft = A

3.2 = T20 4005000V

1000 1000

VT = 0.08 m/s Terminalvelocity of shaft

5. (a)

3 2Calcination Quick LimeLimestone

2

2Slaked Lime

CaCO CaO CO

H O (Slaking) Ca(OH)

White Chalk Pure LimestoneKankar Impure Limeston

e

6. (b)

Head loss till Vena contracta

= HL = H 2v1 C

Where H = head available at orifice

CV = coefficient of velocity

HL = a 2v

ph 1 C

= 2183 1 0.980.9 10

= 5 (1– 0.982)= 0.198 m

7. (b)

The inflection point of the falling limb orrecession limb of the hydrograph iscommonly assumed to be the point or timeat which direct runoff ends.

8. (a)

Depending on the size of the kiln, dailyoutputs can be between 10,000 and 28,000bricks, 70 percent of which being of highquality.

9. (a)

hcp = cgcg

cg

Ih

h A

hcg – hcg =cg

cg

Ih A

=

3(1.2)(3)36

2 13 (3) 1.2 33 2

= 0.10 m

10. (b)

Using Darcy’s Law,Q = KiA

Q = (40 34)25 4000 302000

Q = 9000 m3/day

11. (c)Transmissivity = KeqB

Keq = 1 1 2 2 3 31 (K H K H K H )H

=1 (30 5 20 3.5 3.5 45)

12

Keq = 31.46 m/day = 31.5 m/dayT = 31.5 × 12

= 378 m2/day

12. (a)

V =42

6×60×60 = 1.94 × 10–3

VS = nV = 0.2 × 1.94 × 10–3

= 3.88 × 10–4

i =0.8542

= 0.020

K =Vi

= 43.88 10

0.020

= 1.92 cm/s

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13. (d)

Not all the clays and flyashes are suitablefor production of bricks.

14. (c)

If GM is increased, stability will be increaseand time of oscillation will decrease thus itwill decrease the comfort for passengers.

T = I2WGM

15. (b)

The hydrated lime is slow setting cementingmaterial. However, the speed of set can beincreased by using IMPURE limestone inkiln, to form a hydraulic lime. Hydraulic limeswill gain greater strength and at a fasterrate.

Same can be achieved by adding pozzolanicmaterial to the mortar mix.

16. (c)

Q = 5/2d

8C 2g tan (H)15 2

200

3 = 0.50 × 5/28 2 5 1 H

15

H5/2 = 25 5

H = 2/5 5m25 5

Q 5 HQ 2 H

For Triangular weir/notch

QQ

=52

× 0.1 100%5

= 5%.

17. (d)

hf =2fLV

2gd

V =QA

= 2

Q

d4

hf =

2

2

52

fL Q fLQ2gd 12.1dd

4

K = 5fL

12.1d = 5

0.022 30012.1 (0.25)

= 558 s2/m5

18. (b)

19. (a)

Let cross section area of solid be A cm2.

For solid in water

A × 25 × Ss × rw = A × (25 – 16) × Sw × w

Ss =9

25For solid in oil.

A × 25 × Ss × rw = A × (25 – 10) × Soil × w

Ss = oil15 S25

9 25

25 15 = Soil Soil = 0.60

Specific gravity of oil is 0.60.

20. (d)

* Water held in cell pores or cavities = Freewater.

* At FSP, there is no free water, only cell orbound water exists i.e., 25–30%.

* At FSP, wood is classed as unseasonedas its moisture content is 25–30%.

21. (a)

Velocity distribution is more uniform inturbulent flow and flow in porous media isprimarily laminar.

22. (c)

The actual volume of water that can beextracted by the force of gravity from a unitvolume of acquifer material is known as thespecific yield, Sy. The fraction of water heldback in the acquifer is known as specificretention, Sr. Thus, porosity

n = Sy + Sr

23. (b)

24. (c)

Stability of a floating body can be improvedby making width larger which will increaseMoment of Inertia (I) and will thus increasethe metacentric height also by keeping thecentre of mass low and making the hullsmall.

25. (d)

Aggregates occupy 70-80% of the volumeof the concrete.

Aggregates are not just inert filler, but itsproperties influence workability, strength,

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stiffness and shrinkage etc. of mortar andconcrete.

The strength of aggregate should be at leastequal to that of concrete.

26. (a)

Since gravity is the dominant force; Froude'smodel law is applicable here.

Qr = discharge ratio = 5/2 2.5r rL L

3

2.5rr

r

L LT

r rT L

P

m

TT =

P

m

LL

TP = 4 × –11

64

= 4 × 8 = 32 min

27. (d)

Head losses occur when pipe experiencessudden contraction

hf =2CV

2g

For sudden contraction, C = 0.5

=20.5 4

2 9.81

= 0.41 m

28. (c)

29. (d)

0.5 m

0.5 m

0.4 m

2 2r2g

In this case

2 2r2g = 1 m

2 =

21 2g

r

=

2 9.8 10.2 0.2

= 7 10 7 rev s

N =60 30 7 210 rpm2 2

30. (b)

Period = t = 1 year

Input volume = 140 × 106 × 120100 = 168 Mm3

Output volume = 2 × (3 × 30 × 24 × 60 × 60) +3 × (6 × 30 × 24 × 60 × 60) +5 × (3 × 30 × 24 × 60 × 60)

= 101.088 × 106 Mm3

Runoff coefficient =Output volumeInput volume

=

6

6101.088 10

168 10 = 0.60.

31. (a)

Fan-shaped, i.e. nearly semi-circular shapedcatchments give high peak and narrowhydrographs while elongated catchmentsgive broad and low peaked hydrographs.

Small basins behave different from the largeones in terms of the relative importance ofvarious phases of the runoff phenomenon.In small catchments the overland flow phaseis predominant over the channel flow. Hencethe land use and intensity of rainfall haveimportant role on the peak flood. On largebasins these effects are suppressed as thechannel flow phase is more predominant.

The slope of the main stream controls thevelocity of flow in the channel. As therecession limb of the hydrograph representsthe depletion of storage, the stream channelslope will have a pronounced effect on thispart of the hydrograph.

32. (d)

Equilibrium Discharge (Q) = 180 m3/s

Area of catchment (A)21 10A

4 3600

= 180

A = 259.2 km2

For unit hydrograph D = 4 hours

Volume of runoff= 259.2 × 106 × 1 × 10–2

=1 T 3600 302

T = 48 hours

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33. (c)

34. (a)u = –y2 and v = –6x

Equation of streamline

dxu =

uv

2dxy

=dy

6x

6xdx = y2dy3

2 y3x3

= C Putting x = 1, y = 1

C = 133

= 223

32 y3x

3 =

83

9x2 – y3 = 8

35. (c)

Specific heat of an aggregate is the measureof its heat capacity. If the coefficient ofexpansion of coarse aggregate and ofcement paste varies much, a temperaturechange may introduce differential movementwhich may break the bond between theaggregate and the cement paste. Thermalconductivity of aggregate is a measure ofits ability to conduct heat.

36. (d)

Force ratio = Fr =3

r rL

P

m

FF

=3

P P

m m

LL

PF150 =

31250 641000 8

FP = 150 × 1.25 × (8)3

=96000 kN1000

PF 96 kN

37. (c)

Total Runoff Volume = 9.6 + 5.4 + 2.3 +3.5 + 2.3 + 2.2 + 1.4 + 6.4 + 12.4 + 10.9

= 56.4 Mm3

Average flow = 56.440 = 1.41 Mm3/day

Average flow in 4 days = 5.64 Mm3

38. (c)

Limecrete uses hydraulic lime instead ofcement in the concrete mixture along withsand and aggregate.

39. (c)

Time ratio = rH

r

LV

= rH

rv

LL

(As gravity forces are dominant, Froude’smodel is used)

Tr = m

p

TT =

3 112 60 240

Lrv =

22

rH

r

1L 4800

1T240

= 1

400

40. (b)

Creosote is an oil type breservative whichis generally suitable only for outside usesdue to its unpleasant smell.

* AsCu is copper chromate arseniccomposition.

* AsCu (water soluble preservative) treatedtimbers are adopted for inside locations onlyas if applied over outside surfaces, the saltscan be leached by rain water.

* Preservative treatment is only supposed tominimise fungal decay and insects attack.

41. (a)

= 2y du

dy

> 500 Laminar flow

becomes unstable.

42. (d)

Volume of direct runoff= 756 × 106 × 1 × 10–2

= 756 × 104 m3

756 × 104 = P1 70 3600 Q2

QP = 60 m3/s

Peak discharge due to 5 cm rainfall excess= 60 × 5 = 300 m3/s

43. (c)

The coal consumption in the dry process isapproximately one-fourth to that in the wetprocess.

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44. (a)

u = y

= –2x

v = 2yx

uv = –4v = 6

v = 2 2( 4) (6) 52

45. (c)

Fine aggregates proportion = x

Coarse aggregates proportion = y

x + y = 100% = 1

As per question

5.4 =2.8x 6.8y

x y

2.6x = 1.4y

xy =

713

x

1 x=

713 13x = 7 – 7x

x =7

20 = 0.35 = 35%

46. (d)p = f ( .r. )

a b cr. = (constant) p

a

cb –31L MLT

= –1 –2ML T

=a 2, b 3c 1c 1, b 2,

p = (Constant) 2 2r

47. (b)

(10,30)

Time(h) 66

Run

off m

/s

3

Runoff Volume

=1 66 60 60 302

= 3.56 × 106

Runoff Depth =6

63.56 10189 10

= 18.83 × 10–3

= 18.8 mm

48. (d)

49. (d)

The magnitude of the pressure force actingon a plane surface of completely submergedplate is equal to product of pressure atcentroid of surface and area of the surface.

P = Ah

= 1000 × 9.81 × 4

× 22 × 3

= 92.41 kN

50. (c)

Constant rate of withdrawal be 3Qm sec ,

A = 1375 × 104 m2

Input volume = 18.5 × 10–2 × AOutput volume = Q × (30 × 24 × 60 × 60) +

2.5 × 10–2 A + 9.5 × 10–2A

Input volume – Output volume = change instorage

18.5 × 10–2 A – [Q × (30 × 24 × 60 × 60)+ 2.5 × 10–2A + 9.5 × 10–2A] = –0.75A

18.5 × 10–2 –

2 –2

4Q 30 24 60 60 2.5 10 9.5 10

2592 10

= –0.75

Q = 8.15 m3/s.

51. (c)

52. (b)Effective Rainfall = 7.5 – 6 × 0.25

= 6 cmDirect runoff = 120 – 30

= 90 m3/s

Peak Ordinate

(QP) =906

= 15 m3/s

53. (b)

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54. (b)

StemFlow Throughfall

Interception

Evaporation

Net Rainfall

55. (d)

Pycnometer is used to determine specificgravity of aggregates smaller than 10 mm.Wire basket is used for coarse aggregatelarger than 10 mm to determine sp. gr. Theten per cent fines test is a test for measuringcrushing strength. Its value gives a measureof the resistance of an aggregate tocrushing, that is, applicable to al laggregates.

56. (c)

02 5

M

D=

2 –2

2–3 51

ML T

ML (L)T= Mº Lº Tº

3Q

D=

3 –1

51

L T

(L)T = Mº Lº Tº

Others have some dimensions.

57. (b)

Effective Rainfall = 2.7 – 0.3 × 3= 1.8 cm

Direct Runoff = 200 – 20= 180 m3/s

Peak flow in unit hydrograph

=1801.8 = 100 m3/s

58. (d)

Depending upon the particle sizedistribution, the fine aggregate has beendivided into four grading zones. The gradingzones become progressively finer fromgrading zone-I to grading zone-IV.

59. (b)

1. Cloud seeding is the dropping of crystalsinto clouds to cause artificial rain.

2. Rotation of earth causes coriolis effect on

winds crossing equator and changes theirdirection.

3. Easterlies are the air flowing towardsequator.

4. The energy received by Earth’s surfaceinform of short waves is termed as in comingsolar radiation or insolation.

60. (c)

* Exogenous trees grow outward, increasingin bulk with the formation of a ring everyyear.

* Conifers are evergreen trees which yieldsoftwood.

* Endogenous trees grow inwards and fibrousmass is seen in their longitudnal section.

* Decidious trees are non-resinous, dark incolour and heavy weight.

61. (a)

The essential boundary conditions orAt y = 0,

= 0 ...(1)u = 0 ...(2)

At y = ,

u = V0 ...(3)

dudy

= 0 ...(4)

Desirable boundary conduction2

2d udy

= 0 ...(5)

62. (c)

Risk =n11 1

T

n = 20

& Risk =5

100 = 0.05

0.05 =2011 1

T

T = 400 years

63. (a)

The main ingredients of lime concrete areslaked lime (the binding material), sand andcoarse aggregate.

64. (b)

In order to satisfy continuity;

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u vx y

= 0 ux

= –vy

(a)ux = –1,

vy

= 1 satisfy

(b)ux = 0,

vy

= 0 satisfy

(c)ux = 4,

vy = –4 satisfy

(d)ux = 3t,

vy = 3t doesn’t

satisfy

(e)ux

= 8xy3, vy = –8xy3 satisfy

(f)ux

= y, vy

= x doesn’t satisfy

65. (c)

• Plastics combine elastic and viscous orplastic reaction to stress.

• Traditional plastics are not biodegradablei.e. do not decompose by biological action.

• All plastic materials are polymers i.e. longchains of molecules loosely tangledtogether.

• Thermoplastic soften on heating withoutundergoing any chemical change and hardenon cooling. Thus, it can be recycled.

Thermosets harden permanently andundergo an irreversible chemical changewhen heated much. Thus, it cannot berecycled.

66. (a)

force on one parachute

FD =2

DC V A2

=2 21.2 1.2 10 (30)

2 4

= 18 × 302 N

= 55.55 × 302 = 50000 N= 50 kN.

No. of parachutes required

=10050 = (2).

67. (d)

Using mass balance equationQ1 = 4 l/sQ = ??C0 = 0 (Assumed)C1 = 500 g/l

= 5 × 105 mg/lC2 = 4 mg/l

Q1 × C1 + QC0 = (Q + Q1)C2

Q =

1 1 2

2 0

Q (C C )(C C )

= 54(5 10 4)

4Q = (5 × 105 – 4) 5 × 105 l/sQ = 500 m3/s

68. (b)

The most of shrinkage occurs during thefirst 28 days, after this time it is negligible.The foamed concrete is fire resistant andthere is suited to fire rated applications.

69. (b)

max avgu uu *

=154

uavg = max15u u *4

=154 0.44

= 2.5 m/s

Note : max avgu u 15

u * 4

For both smooth and rough pipe.

70. (c)

Coefficient of runoff =Runoff

Rainf allRunoff = 0.2 × 300 × 104 × 0.1

= 60,000 m3

Stream flow rate =60,000

10 60 60 = 1.66 m3/s = 100 m3/min

71. (c)

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72. (d)

From figure,

L

H

d

d

tan =2dL

xag

=2dL

3

9.81 =2dL

d 0.153L

73. (b)

74. (c)

V = 2Q 4QA d

hL =2fLV

2gd =

2

24QfLd

2gd

hL =2

2 58fLQ

gd

Here Q and f is assumed to be constant,then

5Lh d

75. (a)

Mix by weight is not affected by bulking.

76. (d)

RCL =VD 1250 6 0.5

1.5

= 2500 < 5×105

i.e. laminar boundary layer.

CD =CL

1.328 1.328 0.02656R 2500

TotalDF = 2 × 2D

1 C AV2

= 0.02656×1250 0.5 × 1 × 36

= 597.6 N 600 N

77. (b)

Q = Discharge = V.A

V = 0.2 0.8V V2

A =1 2 0.92 = 0.9 m2

V0.2 = 0.6 m/s, V0.8 = 0.4 m/s

V =0.6 0.4

2

= 0.5 m/s

Q = 0.5 × 0.9= 0.45 m3/s

78. (d)

79. (b)

For steady uniform flow,

w =2fV

8 dP r

dx 2

=21000 0.0125 (2)

8

= 6.25 N/m2

80. (b)

Duralumin is the most important alloy ofaluminium. Copper is used so as to havebetter properties of strength, durability ascompared to aluminium. Due to its lightweight and strength, it is used in aerospacestructures. Its strength is comparable withmild steel but relative density is just 2.8comparable with pure aluminium.

81. (d)In an OCF, Froude’s model law will beconsidered. Acceleration is given by ar = gr= 1 (given)

82. (b)

w =2 N 2 (191)60 60

= 20 rad/sec

25 cm

20 cm 20 cm

A

w

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PA = P0 + 2 2r w2

= 1.20×10×0.25+2 21.20 0.2 20

2

= 12.6 KPa.

83. (b)

84. (c)

= 00

rr

where r = distance from centre of pipe

=2fV 10 7

8 10

=1000 0.025 16 3

8 10

= 15 N/m2

85. (a)

Total weight of ingredients = (2400)(15) =36000 kg

Weight of cement = X

X + 1.8X + 3.6X + 0.5X = 36000

X = 5217.39 kg

Water = 0.5X = 2608.69 kg = 2.6 m3

86. (b)

Lift = VLT where T = Circulation,

V = Velocity of wind

T = c2 R V , Vc = velocity due tocirculation. (rotational speed of cylinder)

VC =

2 RN2 R

60

=2 RN2 R

60

=2 2 24 R N 4 10 1 9060 60

= 60 m2/sec.Lift = 1.2 × 50 × 10 × 60 N

= 36000 N= 36 kN.

87. (a)

Volume of Runoff (ER = 1 cm)

= 6 1260 10100

= P1 48 60 60 Q2

Qp = 30 m3/s

9 16

Qt

30

Qt = (30 0)0 9(16 0)

Qt = 16.88 m3/s

For effective rainfall = 4 cmQR = 16.88 × 4QR = 67.52 m3/s

88. (c)

1. For high class brick masonary, modularbricks are used.

2. English type of bond is provided in heavyload masonary.

3. Carbon bricks are made from crushed cokebonded with tar.

89. (a)

2A A

Ap V

Z2g

= 2

B BB L

p VZ h

2g

2AV575 0

10 2g =

2B

LV375 7.2 h

10 2g

A BV V

hL = 12.8 m

90. (a)Abstraction = Precipitation – Runoff

= 20×10–3×6×100×104–30,000= 90,000 m3

Depth of abstraction

= 490,000

100 10 = 0.09 m = 9 cm.

91. (a)For turbulent, 4/5x

0 1/51

x

For laminar x

01x

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92. (b)

ReL = 4–4

VoL 2 0.80 1.6 1010

= 1.6 × 104 < 5 × 105

Laminar Boundary Layer Throughout

Rex = 4–4

2 0.20 0.4 1010

4x

5 50.20 mx Re 0.4 10

15.81mm

avgf 4

L

1.328 1.328C 0.0105Re 1.6 10

.

93. (b)

94. (b)

For water,

weR = w w w

w w

V D VD

For given fluid,

feR =w f

w

0.9 (V )(D)0.3

So, w

w

VD

=w f

w

0.9 V D0.3

fVV3

95. (a)

Maturity = 9 0t(T T )

= 2[(20 + 11) + (21 + 11) + (19+ 11) + (23 + 11)]

= 254°C days

96. (c)F = A

F = At

Therefore,

1

2

FF =

1

L

7881400 = L

2

µi = 3.55 m/sec

97. (c)

The dilution method of flow measurement,also known as the chemical methoddepends upon the continuity principle

applied to a tracer which is allowed to mixcompletely with the flow.

The dilution method has the major advantagethat the discharge is estimated directly inan absolute way. It is particularly attractivemethod for small turbulent streams, suchas those in mountainous areas. However, itis necessary to emphasise here that thedilution method of gauging is based on theassumption of steady flow.

98. (c)

Kinematic viscosity has units of length andtime and it is not influenced by causingforces and kinematics doesn’t involve theirstudy so it is thus named.

Benzene is highly volatile and is notpreferred. In ordinary ranges, pressuredoesn’t influence viscosity.

99. (c)

All the detention ponds provide someamount of dead storage, which is storagebelow the level of the outlet structure. All ofthe above are advantages of dead storageexcept (C). The dead storage should notbe so large or deep so that pollutants fromthe bottom material of the pond are releasedback into the overlying water.

100. (b)

Carbon content should be low becauseunburnt carbon leads to more demand ofwater for workability. High silicon contentprovides better hydrated gel i.e. bonding forconstituents with fly ash. Fineness of flyash should be as high as possible to havebetter workability and initial strength. Thesilica contained in fly ash should be in finelydivided state since it combines slowly overa long period with the lime liberated duringthe hydration of cement.

101. (b)

c = 8gf

102. (d)

Boundary layer concept was first given byL. Prandt.

As the depth of laminar boundary layerincreases, it cannot dissipate the effect ofinstability in flow and hence transition toturbulent boundary layer occurs.

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103. (b)

Water will be initially displaced equal tothe weight of ice cube and weight of nail.But when melting occurs, ice cube willbecome water and will occupy the spacecorresponding to the weight of waterdisplaced due to its weight but density ofnail is greater than that of water hencevolume of water displaced due to it will begreater initially than the volume of nail.Hence level will fall.

104. (c)Q1 = Q2 + Q3

Q1 = V1A1; Q2 = V2A2

V1A1 = V2A2 + Q3

Q3 = V1A1 – V2A2

Q3 = 2 22 0.254 1.5 0.24 4

= 0.054 m3/s

105. (c)

fw = w b mK f f

= 0.8 10 3 4.4 MPa

106. (c)

When angle of contact is less than 90°then adhesion is greater than cohesion andthen liquid is known as wetting liquid i.e.liquid wets solid surface. Adhesion of waterto the walls of vessels can cause an upwardforce on liquid i.e. capillary rise and resultsin meniscus turning upward i.e. concavetop surface.

107. (b)

Total Precipitation

=

2

(140 130) (130 120)5 82 2 10

(120 110) 112

= [135 × 5 + 125 × 8 + 115 × 11] × 10–2

= 29.40 m3

Runoff coefficient = Runoff

Precipitation = 6.5

29.40

= 0.22

108. (d)

All statements are correct.

109. (b)

Bernoulli equation between point 1 (pond)and point 2 (discharge from pump)

21 1

1P V

Z Hr 2g

= 2

2 22 f

P VZ h

r 2g

For pond, p1 = 0 and V1 = 0. Neglect headloss so hf = 0. Since the pump dischargesinto the atmosphere p2 = 0.

Solving for pump head H = 22

2 1V(Z Z )2g

= 24.51.5

2 10

= 2.513 m

Rate of energy being added = QH

= 210 1000 0.2 4.5 2.5324

= 3.55 kW

110. (b)

n 1 22 P 110cm

Cv = n 1 22100 100p 110

= 20%

N =22 20Cv 16

5

.

According to WMO recommendations, atleast 10% of the total raingauges shouldbe of self-recording type. Therefore, 1.6 or2 raingauges shall be of self recording type.

Additional Non recording raingauges= 16 – 7 – 2= 7.

111. (c)Vy2

90º

Vx1

Ry

R

Rx

By symmetry Rx = Ry.

Rx = Ry = QV

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= 1000 × A × V2

= 41000 50 256 1280 N

10

R = 2 1280 N 1810 N

112. (d)

The positive pressure gradient helps in theBoundary layer separation.

If y 0

du0,

dy

the flow will not separate.

For external flow, B.L separation increasespressure drag.

For internal flow, separation leads toincrease in flow losses.

113. (c)

Free surface of a fluid has zero pressure.As the liquid is under pressure, the freesurface would lie above the container level.

114. (c)

The unit hydrograph is a characteristic of aparticular watershed. It is the hydrographresulting from 1 unit of excess rainfall.

115. (a)

Heavy creosote oil is widely used as ananti-termite coating for protection ofwoodens surfaces against termites.

116. (b)

p = (2 – 1) MN/m2

= 1 × 106 N/m2

v = (1000 – 995) cm3

= 5 cm3 (Reduction)

K =p

v / V

= Bulk modulus

=6(10 ) Pa

( 5) / 1000 = 200 MPa

117. (c)

Initial Volume = 8 litres

Weight of water drained by gravity= 110 – 95 = 15 N

Volume of water drained

= 315

9.8×10 = 1.52 litres

Specific yield =1.52 100

8 = 19%

118. (d)

119. (b)

For both pipes in parallel, the head loss ineach pipe is equal

21 1 1

1

f L V2gD =

22 2 2

2

f L V2gD

f1 = f2, D1 = D2

2122

VV

= 2

1

LL

= 0.5 1

2

VV = 0.71

Since diameter and pipe areas are the same

1

2

QQ = 0.71

1

1 2

QQ Q =

0.711 0.71

= 0.41 = 41%

120. (d)

Due to replacement of cement partly bypozzolana in making concrete, there is anincrease in setting time of concrete. It alsomakes concrete impermeable at later agesand more resistant towards sulphate attack.It also leads to reduction in bleeding andsegregation in concrete.

121. (b)

H2S evolved dissolves in the water films ofconcrete if it is not running full.

122. (d)

The desirable content of manganese in mildsteel is between 0.30 to 1.00 percent.

123. (b)

Maximum slenderness ratio for tallerbuilding is lesser because imperfections inworkmanship in regard to verticality arelikely to be more pronounced in case oftaller buildings.

124. (c)

Electrical seasoning is the most rapidmethod of seasoning. This method is basedon the principle that if a high frequencyalternating current is passed through a badconductor, it gets heated up. Due to uniformrise in temperature and consequentlyuniform evaporation of moisture results inuniform quality of timber. Resistance tocurrent increases as the timber dries. Greentimber offers less resistance due to presenceof moisture.

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125. (b)

Both A and R are correct. But R is notcorrect explanation of A.

The modulus of elasticity of aluminium is70 GPa as compared to 210 GPa for steeli.e. one third of steel. Thus the deflection inan aluminium structure due to load will bethree times that in steel with the samecross section.

126. (a)

The total amount of deleterious materials inaggregates should not exceed 5% as perIS code. The sand obtained from seashoreor a river estuary contains salts and some-times as 6% of mass of sand. It can beremoved by washing with fresh water thus,it prevents efflorescence and corrosioncaused by absorption of moisture from air.

127. (c)

Plasticisers are organic materials added tomodify plastic to impart desirablecombination of strength, flexibility andsoftness. Plasticisers is supposed toneutralize a part of the intermolecular forceof attraction between macromolecules ofresins. Consequently the polymeric part ofresin move with greater freedom, therebyincreasing plasticity and flexibility of thecompound material. However, tensilestrength and chemical resistance isreduced.

128. (a)

Annealing : Glass articles are allowedto cool under room temperature by pass-ing through different chambers withdescending temperature. If cooled rapidly,the glass being bad conductor of heat,the superficial layer cools down first andstrain develops in the interior portions,which causes unequal expansion andthe articles are likely to crack due toweakening.

129. (b)

Thermal conductivity increases with density.

130. (d)

TEL for idealised Bernoulli’s flow is alwayshorizontal, because losses are zero in idealflow. In real fluid TEL always falls down.

131. (a)

In triangular weir Cd is fairly constant withdepth and head is large even for smalldischarge. Hence, flow is not affected muchby surface tension and viscosity (as ithappen in rectangular weir) which results inaccurate measurement of discharge.

132. (c)

The turbulent shear stress is due to largescale momentum transfer between differentlayer which gives rise to additional shearover and above the viscous shear. Near wall,viscosity is primarily responsible for shearand away from wall, turbulence is primarilyresponsible for shear.

133. (b)

134. (b)

135. (d)

Hardy cross method is not applicable toopen channel flow to find out the discharge.

136. (a)

137. (a)

138. (c)

If the manometric liquid is mercury (S2= 13.6) and the liquid at A or B is water(S1 = 1), then the difference in pressureheads at the points A & B is 12.6 time thedeflection x of the manometric liquid in thetwo limbs of manometer. As such the useof mercury as manometric liquid in U-tubemanometer is suitable for measuring largepressure difference, but for small pressuredifference it will not give precise result. Forthat purpose liquid which is slightly heavierthan water can be used.

139. (d)

In piezometer the glass tube having aninternal diameter less than 12 mm shouldnot be used. Reason is correct.

140. (c)

In mercury cohesive property is moredominant than its adhesive porperty. Dueto which its show convex meniscus.Whereas in water adhesion is morepredominant and show concave meniscus.

141. (c)

Due to very low vapour pressure mercury isused in a barometer to measure pressure.

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Various volatile liquid like benzene etc. havehigh vapour pressure.

In low vapour pressure liquid, sensitivitytoward changes in the atmosphere is moreand hence rises more quickly to record thechanges in atmospheric pressure.

142. (a)

Hydrological cycle is the movement of waterfrom the earth surface to atmosphere &back to earth surface & oceans by themeans of phase-transformation. Water iscirculated from hydrosphere to atmosphere& being utilised by living organisms ofbiosphere.

143. (d)

Warmer air has more water vapour holdingcapacity.

As temperature decreases this capacityalso decreases & at a point of temperatureair becomes saturated & condensationstarts, that point is called dew point.

144. (d)

Due to dissolution of any substance in aliquid, motion of molecules is restricted &vapour pressure of liquid is decreased.

Due to salinity, saturated vapour pressureover water body is lower than that over purewater which causes relatively lessevaporation.

145. (d)

In Rayleigh’s method of dimensionalanalysis, the dependent variables is writtenas the function of exponential terms ofindependent variables.

146. (b)

The term rainfall is used to describeprecipitations in the form of water drops of

sizes larger than 0.5 mm. The maximumsize of a raindrop is about 6 mm. Any droplarger in size than this tends to break upinto drops of smaller sizes during its fallfrom the clouds.

147. (d)

The mass curve of rainfall is a plot of theaccumulated precipitation against time,plotted in chronological order. Mass curvesof rainfall are very useful in extracting theinformation on the duration and magnitudeof a storm. Also, intensity at various timeintervals in a storm can be obtained by theslope of the curve. The hectograph is derivedfrom the mass curve.

148. (c)

A forest soil rich in organic matter will havea much higher value of infiltration rate underidentical conditions than the same soil inan urban area where it is subjected tocompaction.

The soil temperature will affect thetemperature of water that will affect theviscosity of the water by which in turnaffects the infiltration rate.

149. (c)

An ephemeral stream is one which doesnot have any base-flow contribution. Theannual hydrograph of such a river showsseries of short duration spikes marking flashflows in response to storms. The streambecomes dry soon after the end of the stormflow.

An intermittent stream has l imitedcontribution from the groundwater.

150. (a)

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