1. (a)

2. (d)

3. (a)4. (a)

5. (a)

6. (b)

7. (d)

8. (c)

9. (a)

10. (b)

11. (b)

12. (b)

13. (b)

14. (d)

15. (d)

16. (d)

17. (c)

18. (c)

19. (b)

20. (d)

21. (c)

22. (d)

23. (c)

24. (b)

25. (c)

26. (c)

27. (b)

28. (c)

29. (c)

30. (d)

31. (d)

32. (d)

33. (d)

34. (d)

35. (b)

36. (a)

37. (a)

38. (a)

39. (c)

40. (c)

41. (a)

42. (a)

43. (c)

44. (c)45. (d)

46. (d)

47. (d)

48. (b)

49. (a)

50. (b)

51. (c)

52. (b)

53. (c)

54. (b)

55. (c)

56. (b)

57. (b)

58. (a)

59. (d)

60. (d)

ESE-2018 PRELIMS TEST SERIESDate: 17th December, 2017

ANSWERS

61. (c)

62. (d)

63. (c)

64: (d)

65. (b)

66. (b)

67. (d)

68. (b)

69. (c)

70. (c)

71. (b)

72. (c)

73. (d)

74. (c)

75. (b)76. (c)77. (b)

78. (b)

79. (d)

80. (c)

81. (a)

82. (d)

83. (d)

84. (b)

85. (d)

86. (c)

87. (d)

88. (c)

89. (c)

90. (a)

91. (d)

92. (d)

93. (b)

94. (c)

95. (b)

96. (c)

97. (b)

98. (a)

99. (d)

100. (a)

101. (a)

102. (d)

103. (b)

104. (c)

105. (b)

106. (d)

107. (d)

108. (d)

109. (d)

110. (a)

111. (c)

112. (a)

113. (c)

114. (d)

115. (c)

116. (b)

117. (b)

118. (d)

119. (d)

120. (d)

121. (b)

122. (a)

123. (a)

124. (b)

125. (a)

126. (a)

127. (b)

128. (b)

129. (b)

130. (a)

131. (c)

132. (b)

133. (d)

134. (b)

135. (b)

136. (a)

137. (b)

138. (a)

139. (a)

140. (d)

141. (b)

142. (a)

143. (a)

144. (c)

145. (a)

146. (a)

147. (b)

148. (d)

149. (b)

150. (b)

(2) (Test - 13)-17th Dec. 2017

Sol–1: (a)

dM vdx

differentiation should be w.r.t 'x' to get SF @ that point

Sol–2: (d)Sol–3: (a)

Max. principal stress theory is applicableto brittle material because brittle materialfail under tension leading to fracture.

Sol–4: (a)

w

x

Mx =2wx

2

2bdz6

=2

2x

wxb d2

6

2xbd

3 = wx2

dx = 3w xb

Hence xd x

Sol–5: (a)Sol–6: (b)

Taking moment about hinge (A)T × 0.5 = 300 × 0.3 + 300 × 0.9

T 720 N

Sol–7: (d)

Sol–8: (c)

= Pl .lAE E

, l, E are sameSo E is also same

Energy stored, E = 1 V2 is same in

both.

Sol–9: (a)

Slope of B.M. diagram = Shear force at thatpoint

dMdx = V

In portion BC, shear force = 0Hence B.M. is non-zero constant.Hence ‘a’ is correct.

Sol–10: (b)

On heating, material will expand in alldirection reducing the size of the hole.Hence ‘b’ is correct.

Sol–11: (b)For simply supported beam

centre 1I [other parameters

constant]

2

1

=

31

32

12I bdI 12 db

2 =2

1db

Hence, ‘b’ is correct.

Sol–12: (b)T

a

mg

When lift moving upwards.T1 = mg + ma

When lift moving downwards.T2 = mg – ma

Given, T1 = 2T2mg + ma = 2mg – 2ma

a =g3

Hence, ‘b’ is correct.Sol–13: (b)

Let velocity of car = VRelative velocity of car coming from

opposite direction = V + 45Using relative velocity concept,

(3) (Test - 13)-17th Dec 2017

10 6V 45 60

V = 55 Km/hrHence, ‘b’ is correct.

Sol–14: (d)

Sol–15: (d)For both ends fixed column

Pcr =2 2 2

2 2 2eff

EI EI 4 EIL LL

2

Hence ‘d’ is correct.Sol–16: (d)

For notch Q = 5/2D

8 c H 2g tan15 2

dQQ = 5 dH

2 H

dQ 100Q

=35 1.5 10 100 0.75%

2 0.5

Sol–17: (c)

For maximum acceleration, oil will reachjust upto the maximum height of the tankwithout being spill out and volume of oilwill remain the same.

3 m

x

10 m

1.5 m

1 x 3 52 = 10 × 1.5 × 5

x = 10 m

therefore tan = xa 3g 10

ax = 3 m/s2

Sol–18: (c)Area of gate = 5 × 3 = 15 m2

5 m

98.1 kN/m2

The equivalent height of water which givesa pressure intensity of 98.1 kN/m2 atbottom

h = w

P 10 m

Centre of pressure h = aIxAx

x = 10 – 2.5 = 7.5 m

I = 3bd

12

h = 33 57.5

12 15 7.5

= 7.77 mSol–19: (b)

Boundary layer will form on both sides ofplate

FD = 2

wD

V2 C A2

5.31 = 3 2

D10 22 C 2 1

2

CD = 6.64 × 10–4

Sol–20: (d)Since the pipes are connected in parallelhence head loss will be same in both thepipes

2

5fLQ

12.1 d = C

2 5Q d

1

2

QQ =

52 5/21

2

d (3) 15.6d

Sol–21: (c)Sol–22: (d)Sol–23: (c)

The continuity for unsteady flow is

Q Ax t

= 0

(4) (Test - 13)-17th Dec. 2017

QAt x

A 0.10t

Sol–24: (b)For homologous pumps the specific speedis same .

NSA = NSB

NS = 3/4N QH

3/4600 0.4

50 = 3/4

B

600 0.3H

HB = 41.28 m

Sol–25: (c)

If 2 = 0

Then flow is considered to be steadyincompressible irrotational as continuityequation in 3-D is

U V Wx y z

= 0

Hence ‘c’ is correct.

Sol–26: (c)

A B

P1, V1 P , V2 2

1 2

1 2

From continuity equation

1 1A V = 2 2A V

V2 = 11

2

A VA

=4

26

1 10 10 1020 10

V2 = 0.5 m/sUsing Bernoulli’s equation between (1) and(2).

21 1

1P V Zg 2g

=

222

2VP Z

g 2g

1 2P Pg =

2 22 1V V

2g

[Given Z1 = Z2]

or 1 2P P = 2 22 1V V2

2 21200 144Pa0.5 0.12

Hence ‘c’ is correct.

Sol–27: (b)

W

100 m/s

D = 60 mm

Weight of ball = Drag force by air on ball

W = 2D D

1F C AV2

= 221 0.8 1.2 1000.062 4

W = 13.5648 NHence ’b’ is correct.

Sol–28: (c)Velocity profile for steady incompressiblelaminar viscous flow between twostationary parallel plate is parabolic asshown.

y Upper plate

Parabolic

x

Bottom Plate

Hence ‘c’ is correct.

Sol–29: (c)

Sol–30: (d)

Sol–31: (d)

Sol–32: (d)

(5) (Test - 13)-17th Dec 2017

Sol–33: (d)

Sol–34: (d)

An articulated robot has all the revolutejoints (RRR). These robots has more degreeof freedom, then any other robot, so are themost versatile.

Sol–35: (b)

Fp = Rx(90°)P

Fp =

1 0 0 0 30 C S 0 40 S C 0 20 0 0 1 1

=

31 0 0 00 0 1 0 40 1 0 0 20 0 0 1 1

Fp =

3241

= [3 –2 4 1]T

Sol–36: (a)

Stepper motors have high vibration levelsdue to stepwise motion

Sol–37: (a)

1 scale division = 80 0.666 mA

120

Resolution = 1 scale division20

= 1 0.66620

=0.033 mA

Sol–38: (a)

Sol–39: (c)

Sol–40: (c)

Shaping is independent of diameter range,hence speed range = cutting range.

Sol–41: (a)

Sol–42: (a)

s btanI tan cos tan sin

btanI tan when 90

i.e. sC 90 Cs = 0

Sol–43: (c)

fn

n

m1

m2

T1

T2T2>Tm m

1

2 1>

Sol–44: (c)

Rolling force, position of neutral plane,length of contact, maximum possible draft,and angle of bite all depends upon rollerdiameter.

Sol–45: (d)

Sol–46: (d)

As fineness increases, permeabilitydecreases.

Sol–47: (d)

Sol–48: (b)

Calendering - shower curtain

Impact extrusion - collapsible tubes

Thermoforming - Bath tub

CO2 moulding - Machine beds

Sol–49: (a)

Forward welding technique allowspreheating of the plate ahead of moltenmetal.

Backward welding techniques is used forjoining thicker plates over 5mm.

(6) (Test - 13)-17th Dec. 2017

Sol–50: (b)

Sol–51: (c)

Short circuit transfer takes place when CO2is used as shielding gas with low current.

Sol–52: (b)

Sol–53: (c)

Sol–54: (b)

G04 - Dwell

G41 - Tool radius compensation left

G42 - Tool radius compensation right

G71 - Programming in mm.

Sol–55: (c)

38100 = %C

171 = %Cr

31 = %Ni

Sol–56: (b)

I. In gas turbine the efficiency of the cycleget decreased for supersonic velocity flowdue to drag force and eddy formation.

II. The thrust developed in turbojet

F = a cm V u

At high altitude density is low so themass flow rate is low. Hence the thrustdecreases with increase in speed and therelative velocity (vi - u) decreases when Uincreases.

III. Before combustion in rocket engine,there is no motion of oxident and fuelmolecules, so stagnation condition exist incombustion chamber of rocket engine.

IV. Due to variation of velocity from rootto tip and centrifugal motion turbine bladesare subjected centrifugal stress.

Sol–57: (b)

III. Surging in centrifugal compressor is dueto either increasing delivery pressure veryhigh or creating some obstacles in air flwo

path. Generally diffuser vanes are 1/3 ofimpeller vane. Mathematically equal vanesin impeller and diffuser does not createsurging but more than the impaller bladesin diffuser stops the free flow of air andcauses surging.

Sol–58: (a)

II. In impulse turbine, the enthalpy droptakes place only in nozzle. In stator bladeor rotor blades the flow only changes itsdirection and no change in enthalpy andpressure.

Sol–59: (d)

In four row velocity compound steamturbines, the ratio of work output from firstto fourth stage 7:5:3:1

So, the power developed in 3rd stage

Stage 3 = 3 8000kW

7 5 3 1

= 1500 kW

and for the last stage

Stage 4 = 1 800016

= 500 kW

Sol–60: (d)

The forced draught or induced draught fanis not allowed in locomotive boilers becausethey consume power. Draught is producedin locomotive boilers by creating lowpressure in exit zone of flue gases by flowinghigh pressure steam at high velocity.

Sol–61: (c)

Sol–62: (d)

Enthalpy drop in moving blade

mh = 12 kJ/kg

Enthalpy drop in fixed blade

fh = 24 kJ/kg

Degree of reaction, of turbine

=

m

m f

hh h

(7) (Test - 13)-17th Dec 2017

= 12 1212 24 36

= 33%

Sol–63: (c)

Subcritical boiler consists of three distinctsection as Economiser, evaporator andsuperheater.

Sol–64: (d)

Lancashire and Babcock and wilcox boilersare natural circulation boilers.

Sol–65: (b)

II. At a given pressure ratio when speedincreases in a centrifugal compressor thenflow increases and efficiency decreases.

Sol–66: (b)The steam drum is provided in allmodern steam generator except oncethrough type, where feed water from theeconomizer is fed.

Sol–67: (d)

Sol–68: (b)1. Blower is placed near the base of the

boiler2. Flow of air is uniform and power

requirement in this type of draught isless as the fan handles cold air.

Sol–69: (c)Effect of reheating on rankine cycle

1. Turbine work increases, hence net workincreases but pump work remainsconstant.

2. Correct statement3. Along with increase in net work heat

supply also increases hence thermalefficiency increase, decrease or remainsconstant.

Sol–70: (c)Sol–71: (b)

Cast iron is strong in compression andweak in tension.

Sol–72: (c)

Given,

QL = 45 kW, TL = –3 + 273 = 270 K

TH = 37 + 273 = 310 K

(COP)R = ideal2 (COP)3

LQW = L

H L

T23 T T

or45W =

2 2703 310 270

W =3 4045 10 kW2 270

Sol–73: (d)

Sol–74: (c)

Sol–75: (b)

In this system compressor and motor areplaced in a welded steel shell, hencemaintainance is very complex.

Sol–76: (c)

Sol–77: (b)Heat conduction is one dimensional.

Sol–78: (b)There is no change of phase either ofthe fluid during the heat transfer.

Sol–79: (d)

Sol–80: (c)

q = kAdTdx

K will be maximum if dTdx (slab) is

minimum because q and A is same forslabs.Hence, for slabs which have lowest slope,thermal conductivity of that slab willbe highest

K2 > K1 > K3

(8) (Test - 13)-17th Dec. 2017

Sol–81: (a)There is no internal heat generation.The temperature profile is linear notparabolic1 and 2 statements are correct

Sol–82: (d)Over heat transfer coefficient (U)

1 0

1 1 L 1U h K h

1 1 0.15 1U 25 0.15 25

1 27 25UU 25 27

Hence, it is closer to the heat transfercoefficient based on the bricks alone.Hence, (d) is the correct answer.

Sol–83: (d)

Sol–84: (b)

Sol–85: (d)

Effectiveness, =

1 2

1 1

h h h

min h c

C T TC T T

Ch = mhCph = 1 × 1 = 1kJ/K = Cmin

=

(420 120) 3001(420 20) 400

= 0.75 = 75%

Sol–86: (c)Parallel flow heat exchanger is moreirreversible than counter flow heatexchanger.

universe universein parallel flow in counter flow

S ( S)

Sol–87: (d)The shape factor is purely a function ofgeometric parameters only.If the radiation surface is subdivided,the shape factor for that surface w.r.tthe receiving surface is not equal to the

sum of the individual shape factors.

Sol–88: (c)

(2, 0)(3, 0) x 01

x 02

(0, 3)

(0, 3)Z = 9

(2, 0)Z = 2

Optimum value is (0, 3)

Sol–89: (c)

Sol–90: (a)

Purchase orders and all kinds of releasesare principal outputs, whereas records andschedule are principal inputs.

Sol–91: (d)

At EOQ, holding cost = ordering cost

Total annual ordering cost,

= 800 × 4 = 3200

Sol–92: (d)

Sol–93: (b)

The effect of radial component is neglected.Sol–94: (c)Sol–95: (b)

A load chart is a type of Gantt chart.It depicts the loading and idle timesof a group of machines or a list ofdepartments. It also shows whencertain jobs are scheduled to start andfinish,and when idle time is expected.SIMO chart and man-machine chartsare used in method study.

Sol–96: (c)

Sol–97: (b)

(9) (Test - 13)-17th Dec 2017

2 2x y xy x y3 6 18 9 365 8 40 25 644 13 52 16 1692 4 8 4 166 15 90 36 22520 46 208 90 510Total

b =

22n xy x yn x x

2

5 208 20 462.4

5 90 20

a = y b xn

= 46 2.4 205

= – 0.4y = a + bxy = – 0.4 + 2.4x

Sol–98: (a)

Lq =2

1

= 5 patients/hr = 6 patients/hr

=56

Lq =

256

516

= 25 patients.6

Sol–99: (d)All assumptions are true.

Sol–100: (a)

S* = * h

b h

CQC C

=1.27850

1.27 0.89

= 499.76Sol–101: (a)

Critical activities with lower possiblecost slope are subjected to crashing.

Sol–102: (d)

Sol–103: (b)

F = 3 1 2j h

= 5 j = 5 h = 1F = 3(5 –1) – 2 × 5 –1 = 1

Sol–104: (c)The velocity of the common I-centrerelative to a fixed third link is the samewhether the I-centre is considered on thefirst or the second link.

V24 = 2 424 12 24 14

4

2

= 24 12

24 14

Sol–105: (b)

cam angle

cam angle

Acce

lera

tion

velo

city

Given velocity variation is ofconstant acceleration and decelerationfollower motion,

(acceleration)max = 2

24h

Sol–106: (d)

Sol–107: (d)

Sol–108: (d)

Sol–109: (d)

S.I. engines are quantative governingengines as air-fuel mixture suppliedthrough carburator so we can not havequalitative governing.

(10) (Test - 13)-17th Dec. 2017

Sol–110: (a)Sol–111: (c)

Sol–112: (a)In rating of SI engine fuel, ISO-octaneis assigned 100 octane number andnormal heptane as zero octane number.

Sol–113: (c)

critical speed= g

where = static deflection of shaft will depend on length and diameter ofshaft.

Sol–114: (d)

Primary force = 2mr cos

Primary couple = 2mr cos

Secondary force =

2mr cos2n

Secondary couple =

2mr cos2n

An in-line four-cylinder four-strokeengine has two outer as well as twoinner cranks in line. The inner cranksare at 180° to the outer throws. Thus,the angular positions for the cranks are: for first, (180 ) for the second,

(180 ) for the third and for thefourth. Primary force

= 2mr [cos cos(180 )

cos cos ] 0=180

Primary couple

= 2 3 1mr cos cos 1802 2

+

1 cos2

3 cos 0=180 2Secondary force

=

2mr cos2 cos 360 2n cos 360 2 cos2

=

24mr cos2n

Secondary couple

= 2 3 1mr cos2 cos 360 22 2

1 3cos cos2360 22 2

= 0

Sol–115: (c)

Sol–116: (b)In case of hunting, the governor tendsto intensify the speed variation insteadof controlling it. Higher thesensitiveness of the governor, moreacute becomes the problem of hunting.

Sol–117: (b)Sol–118: (d)

This type of shape of vapour dome isobtained in pressure enthalpy diagram.This diagram is very useful inrefrigeration cycle i.e. vapourcompression cycle etc.

Sol–119: (d)Sol–120: (d)

Absolute gas temperature scale is not athermodynamic temperature scale, sinceit cannot be used at very lowtemperature (due to condensation) andat very high temperatures (due todissociation)

Sol–121: (b)

Rotors having low solidity use lift force andturn at higher speed.

Sol–122: (a)The idealisation of concept of continuumto a system leads that property is a pointfunction i.e. at every point, property issame P, T, etc.

Sol–123: (a)Sol–124: (b)

In SI engine, the spark is advanced tohave peak pressure around 10° after topdead center during expansion. Becausefirst phase of ignition is time dependentnot speed. So at higher speed, moreangle is rotated in given time. Thisensures good brake power. Theturbulances in chamber are also

(11) (Test - 13)-17th Dec 2017

increased by high speed, so the flamevelocity.

Sol–125: (a)In infrared gas analyzer, two compart-ments containing one reference gasand other sample gases are used. Bycomparing spectra of both compart-ments the composition of gases insample can be analysed.

Sol–126: (a)Sol–127: (b)Sol–128: (b)Sol–129: (b)Sol–130: (a)Sol–131: (c)Sol–132: (b)

Sol–133: (d)

Sol–134: (b)Sol–135: (b)

F = wB

GB B

Stable equilibriumRestoring couple= W.GM sin

Stable equilibrium

B

4

Sol–136: (a)Due to introduction of Air vessel the flowcondition will become more uniform so thefriction losses will reduce and hence thepower required will also reduce.

Sol–137: (b)

Sol–138: (a)

Sol–139: (a)

When DCRP is used, it tends to overheatthe tungsten electrode, hence it can melt.

Sol–140: (d)

LVDT is an electrical comparator whichconsists of 3 symmetrically spaced coilswound around an insulated foam.

Sol–141: (b)

Machine origin or zero is a point set by themanufacturer of machine and normally itcannot be altered.

Program origin is also called home positionof tools and it is the point from where toolstarts for executing a program of instructionand return back to it.

On small CNC machines, machine originmay be made as program origin.

Sol–142. (a)

Sol–143: (a)

Sol–144: (c)The condensation occurs when asaturation vapour comes in contact withsurface whose temperature is lower thanthe saturation temperaturecorresponding to the vapour pressure.

Sol–145: (a)

Sol–146: (a)

The strength of transverse fillet weld is 1.17times of the strength of parallel fillet weld.In order to simplify the design of filletwelds, many times, transverse fillet weldis designed by using the same equations ofparallel fillet weld.

Sol–147: (b)Sol–148: (d)Sol–149: (b)

General purpose machine tools arethose designed to perform on a widerange of components. By the verynature of generalisation, the generalpurpose machine tool, though capableof carrying out a variety of tasks, wouldnot be suitable for large production,since the setting time for any givenoperation is large. Thus, the idle timefor a general purpose machine tool ismore and machine utilisation is poor.Hence, their utility is in job shopscatering to small-batch, large varietyjob production. Examples –lathe, shaperand milling machine.

Sol–150:(b)Fixed position layout is used where it

(12) (Test - 13)-17th Dec. 2017

is economical to move machines, toolsand men on to the job rather thanmoving the job. For example -manufacturing ships, huge aircraftsetc. Capital investment is minimumin the case of fixed position layout