1. (a)

2. (c)

3. (d)

4. (c)

5. (a)

6. (c)

7. (b)

8. (b)

9. (c)

10. (b)

11. (a)

12. (b)

13. (b)

14. (b)

15. (a)

16. (a)

17. (b)

18. (a)

19. (a)

20. (a)

21. (d)

22. (c)

ESE-2018 PRELIMS TEST SERIESDate: 15th October, 2017

23. (a)

24. (b)

25. (b)

26. (c)

27. (c)

28. (a)

29. (a)

30. (b)

31. (a)

32. (c)

33. (c)

34. (d)

35. (c)

36. (c)

37. (c)

38. (d)

39. (b)

40. (d)

41. (c)

42. (a)

43. (d)

44. (a)

45. (b)

46. (b)

47. (b)

48. (b)

49. (a)

50. (b)

51. (d)

52. (c)

53. (b)

54. (c)

55. (b)

56. (d)

57. (c)

58. (b)

59. (a)

60. (d)

61. (d)

62. (b)

63. (b)

64. (b)

65. (d)

66. (a)

ANSWERS

67. (c)

68. (d)

69. (d)

70. (a)

71. (d)

72. (c)

73. (c)

74. (c)

75. (d)

76. (c)

77. (b)

78. (d)

79. (c)

80. (d)

81. (d)

82. (b)

83. (c)

84. (b)

85. (b)

86. (a)

87. (b)

88. (b)

89. (d)

90. (c)

91. (a)

92. (d)

93. (a)

94. (a)

95. (b)

96. (c)

97. (b)

98. (b)

99. (a)

100. (a)

101. (d)

102. (d)

103. (b)

104. (c)

105. (a)

106. (c)

107. (a)

108. (c)

109. (d)

110. (a)

111. (d)

112. (b)

113. (d)

114. (d)

115. (c)

116. (d)

117. (d)

118. (a)

119. (c)

120. (c)

121. (a)

122. (d)

123. (c)

124. (c)

125. (d)

126. (b)

127. (d)

128. (a)

129. (d)

130. (b)

131. (a)

132. (a)

133. (b)

134. (a)

135. (a)

136. (b)

137. (a)

138. (b)

139. (a)

140. (a)

141. (c)

142. (a)

143. (a)

144. (b)

145. (b)

146. (b)

147. (b)

148. (a)

149. (a)

150. (b)

(2) (Test - 04)-15th October 2017

Sol–1: (a)

From first law of thermodynamics

Q = dU W

So Q W = dU = change in internalenergy

Hence (a) is correct.

Sol–2: (c)

According to clausius inequality, for a cycle

dQT 0

Which is dQT = 0 for reversible cycle

and dQT < 0 for irreversible cycle.

Hence (c) is correct.

Sol–3: (d)

Using T-ds relation we know

Tds = dh – vdp

dh = Tds + vdp ...(1)

From T-ds partial differential equation weknow

Tds = CpdT – T

P

V .dPT

Putting this in (1)

dh = CpdT– T

P

VT dP + Vdp

dh = Cpdt + P

VV T dPT

Integrating both side

h2 – h1 =

2 2

1 1

T P

pPT P

VV TC dT dPT

Hence (d) is correct.

Sol–4: (c)

Given

m = 1 kg100

Q = – 1.0 kW

Negative since heat is removed

Applying first law of thermodynamics

Q =

U W

So

U =

Q W = 7.165 kW

Assuming air as ideal gas

U = vdTmCdt

CV = 0.7165 kJ/kgk for air

So dTdt =

v

UmC

dTdt = 1000 K/s

Sol–5: (a)

For perfect gas

PV = mRT

dVdT = mR

P

So slope of V – T graph

dVdT

1P

Since slope m2 > m1

Hence P1 > P2

Hence (a) is correct.

Sol–6: (c)

PVn = C

for n = 1

PV = C

Which represents constant temperatureprocess.

Hence (c) is correct

(3) (Test - 04)-15th October 2017

Sol–7: (b)

p

v

CC =

and = 21f

where f- degree of freedom for triatomicgases

f = 6

= 216

= 1.33

Hence (b) is correct.

Sol–8: (b)

Vander waal’s equation of state of a gas isgiven by

2aP v bv

= RT

Sol–9: (c)

Sol–10: (b)A gas with negative Joule-Thompsoncoefficient will become warmer uponthrottling.

Sol–11: (a)

cycle = net

inp.

WQ

By 1st law of thermodynamics, for a cycle

dQ = dW

In T-s diagram, area gives net heat

W = Area of ABC

= 1

500 300 7 32= 400 unit

Heat is added during BC process only, asentropy is increasing only in BC.So, Qinput = QBC = Area under (BC)

= Area ( ABC) Area

ACN +Area below AN

= 1400 4 300400 300 7 32

= 400 + 200 + 1200= 1800 unit

= 400 0.222 22.2%

1800

Sol–12: (b)

The fan is run by battery Electrical work W = P × t = (100 × 10 × 60)J

= 60 kJ

Since, Q = du W

As electrical energy stored in a battery asinternal energy is high grade energy, it canbe completely converted into work. du = – 60 kJ

Q = – 60 + 60 = 0 kJ

Sol–13: (b)

Area [ADCB] = 50

AB × BC = 50...(i)

AE = BE = BGand BF = FC = BH (given) WEFGH = Area of ellipse

= EB BF

= AB BC2 2

=

AB BC4

= 50 12.5 kJ4

Sol–14: (b)

carnot =

273 271

273 627

= 3001 0.667 66.67%900

actual , Actual efficiency = work done

heat supplied

(4) (Test - 04)-15th October 2017

=

45360000

3600

= 0.9 = 90%

actual > carnot

Not possible

Sol–15: (a)

engine =

273 2 31273 17 58

Qsupplied = 72058 696 W60

engine = s

WQ

W = 3 696 36 W

58

Sol–16: (a)

T1

QsW

HE

WR

T3

T2

Q

HE =

1 22

11

T TT1 TT

(COP)R = 3

2 3

TT T

HE =s

WQ

(COP)R = QW

Combined COP = RHEs

QCOPQ

=

31 2

1 2 3

TT TT T T

Sol–17: (b)

727°K1000 K

Q1

E1

E2

27°C300K

T

W

Q2

Q2

W

Q3

1 = 1

W T1Q 1000 ..(i)

2 = 2

W 3001Q T ...(ii)

From (i) and (ii),

1TQ 1

1000 =

2300Q 1T

1

2

Q T1Q 1000 =

3001T

1 2

1

Q QFor reversible processT T

1T T1T 1000 =

3001T

100 T1T 1000 =

3001T

T = 1000 3002

= 650 K = 377°C

(5) (Test - 04)-15th October 2017

Sol–18: (a)

P

v

b

a1

2

1–a–2 (Isentropic Process)

1 2W = – 6000 J

1 2Q = 1 2 1 2U W

O = 1 2U 6000

1 2U = 6000 J

2 1U = – 6000 J

2–b–1 (Non-adiabatic process)

2 1Q = 2 1 2 1U W

2 1Q = 2000 J

2000 = 2 16000 W

2 1W = 8000 J

Sol–19: (a)

PCritical

point22.1

Triplepoint

D

CPres

sure

0.6A

RB

0 001°C 374°CTemperature

Case-I ice at 0.5 kPa will be somewhereclose to point R and heating isobaricallywill increase temperature with pressureconstant. Hence it will follow AB or par-allel to it.

Case-II water vapour at 400°C will bein superheated state close to point Dand upon compressing isothermallypressure increases with temperatureremain constant hence follow path CD.

Sol–20: (a)

Polytropic

PVn = C

n1 1P V = n

2 2P V

n =

1 2

2 1

12

21

P Plog logP PVV loglog VV

Hence ‘a’ is correct

Sol–21: (d)According to first law of TD

Q = dU W

(dU = 0, for isothermal process)

Q = W, given that W ve(for expansion)

Q = +veHence, heat is supplied to the system

Sol–22: (c)

Sol–23: (a) = o uT ( s) = o genT s

Sol–24: (b)

E W = 1kW

Q1

Q2

T1

T2

=1

WQ 0.2 =

1

1Q

Q1 = 5 kW

Area = 50.5 = 10 m2

(6) (Test - 04)-15th October 2017

Sol–25: (b)

T

S

P = C

V = C

Sol–26: (c)

Sol–27: (c)

Sol–28: (a)

Sol–29: (a)

Tf = Ti

Sol–30: (b)

Sol–31: (a)

Sol–32: (c)

Sol–33: (c)

Work done by pump = p

vdp

=

2000 100959 0.92 = 2153.5 J/Kg

Sol–34: (d)Sol–35: (c)

Work = PdV

where pressure is intensive propertyand volume is extensive property

Sol–36: (c)Heat generated by fricitonal forcesis 1 hour = 5 × 3600 kJ

surrS = 5 3600

293

= 61.4kJ/K

Since all the heat generated byfrictional forces is transferred to thesurroundings,

sysS = 0

univS = sys surrS S

= 0 + 61.4 = 61.4 kJ/K

Sol–37: (c)

(COP)max= 2

1 2

T 270 9= =T T 300 270

(COP)max = 2Qminimum power

Minimum power = 2

max

Q(COP)

= 65 10 kW

3600 9

= 154.3 kW

T1 = 300 K

Q1 R

T = 270 K2

Q2

W

Sol–38: (d)

Advantages of line organization is

Simplicity

Division of authority andresponsibilities

Effective command and control

Rigid discipline

Defined responsibilities at all level.

Unified control

Prompt decision

Flexibility

Sol–39: (b)

Linear programming mathematical tools isused to find best use of limited resources inan optimum manner for maximizing theprofit or minimizing the loss upto twovariables.

Sol–40: (d)

According to Gaussian distribution (Normaldistribution) generally random errors aredistributed.

Hence (d) is correct.

Sol–41: (c)

GivenTs = 3 hourTw = 2 hour

(7) (Test - 04)-15th October 2017

R = Rs. 20/hourAssuming 33.5% incentiveWage of worker is given by

= w s wincentiveRT T T R

100

= 33.520 2 3 2 20100

=

2020 2 3 23 = 46.67 Rs.

Sol–42: (a)

Acceptance sampling is used in batchproduction

Sol–43: (d)

Maxmum Z = 5 A + 8 B

A + B < 6, 5A + 9B < 45, A, B < O

0 1 2 3 4 5 6 7 B(6, 0)1

23456789

10

(0,6)

(0,9)

(5,0)

A

Sol–44: (a)

If the dual of LP problem is minimizationof objective function with 5 variable and 2constraints

Primal of LP problem is maximization ofobjective function with 2 variable and 5constraints. Solution (1 and 3)

Sol–45: (b)

1

2

M 15 32 8 27 11 16M 6 19 13 20 14 7

Job P Q R S T U

According to Johnson’s RuleR T S Q U P

Machine-I Machine-IIIn Out In Out

R 0 8 8 21T 8 19 21 35S 19 46 46 66Q 46 78 78 97U 78 94 97 104P 94 109 109 115

Jobs

Optimal make span time is 115 minutes.

Sol–46: (b)

Expected time, TE = 0 p mT T 4T6

= 8 4 14106

= 8 40 146

= 10.33 min

Sol–47: (b)

Total demand = 400 × 52 = 20800

Ordering cost (C0) = 75/order

Carrying cost (Cc) = 7.5% of product cost= 7.5/100 × 50 = 3.75

EOQ = 2 20800 753.75

= 912 units

Sol–48: (b)

Z = S ET T

E36 T2 = Z

from chart of normal distribution Z = 0 forprobability of 50%.

Therefore TE = TS = 36

Sol–49: (a)

Branch and Bound technique - Integerprogramming (2)

Expected value approach - PERT & CPM(1)

Smoothing and levelling - decision theory

(8) (Test - 04)-15th October 2017

(4) Exponential distribution - Queuingtheory (3)

Sol–50: (b)

Ft = t 1 t 1 t 1F D F

FA = M M MF 0.25 D F

= 20 + 0.25 (20 – 20) = 20FMay = FApril + 0.25 (DA – FA)

= 20 + 0.25 (24 – 20)= 21

FJune = 21 + 0.25 (29 – 21)= 21 + 0.25 × 8 = 23units

Sol–51: (d)

Routing means sequence of operation to beperformed.

Sol–52: (c)

Standard time = Normal time +allowance

0.99 = N.T + 0.20 N.T

N.T = 0.99 0.825 min1.20

N.T = observed time × rating factor

0.825 = 110 O.T100

O.T = 0.825 100 =0.75 min110

Sol–53: (b)

According to question using weighted meanaverage:

0.4 6 0.15 y 7 15 10 101

y = 18.667

Sol–54: (c)

Free float = Ej – (Ei + tij) = 42 – (20 + 19)= 3

Total float = Lj – (Ei + tij) = 56 – (20 + 19)= 17

Independent float = Ej – (Li + tij) = 42 – (22+ 19) = 1

Interfering float = Head event slack = 14

3 1

14 17

= 0.15

Sol–55: (b)

10 15

= 10 215 3

At least 2 customer in the queue = At least3 customer in system

P(at least 3 customer in the system) =P( 3)

P( 3) = 1 – (P0 + P1 + P2)

= 20 0 01 P P P

= 201 P 1

= 21 1 1

= 31 11 11

=3

3 23

= 0.296

Sol–56: (d)

0.8

0.9

0.4

R1= 1–{(1–0.8) (1–0.9)} = 1–(0.2 × 0.1) = 0.98

R =0.42

R1 and R2 in series Reqv. = R1R2 = 0.98 × 0.4

= 0.392

Sol–57: (c)

35 45060 95 10585 70 110 15 40090 100 80 10 20 0

(9) (Test - 04)-15th October 2017

1

2

3

A S Rs. 60B S Rs. 70C S Rs. 80

Rs. 210

Sol–58: (b)

Selling Price = Fixed cost + variable cost +Profit S = F + V + P

sx = F (x) P

(s )x = F + P

x = 10,000 4000F P

s 800 100

= 20 units

Sol–59: (a)

TWC = Total work content =8+6+9+10+12+12=57

No. of stations = 6.

Let total cycle time = (Tsi)max = 12 min =(TC)

L = 57TWC

TC 6 12

= 0.792 79.2%

Sol–60: (d)

No. of units; x = 100

Total cost in each process:

(TC)I = F x 20 3 100 Rs.320

(TC)II = F x 35 2 100 Rs. 235

III

IV

F x 10 1.75 100 Rs.185 BothTCF x 35 1.5 100 Rs.185 PossibleTC

Sol–61: (d)

It is a case of N jobs on 3 machines

To solve this sequencing problem minimumone condition should be satisfiedMin (Ai) Max (Bi)Min (Ci) Max (Bi)

X A B Y B Ca 13 9b 16 15c 8 10d 10 9e 15 9

Johnson’s Rule

c b e a d

Sol–62: (b)D = 15,000C = Rs. 200

Ch = 12.5% of C = 0.125 × 200= Rs. 25

C0 = Rs. 2500

Q* = 0

h

2DCC

= 2 20000 250025

= 2000

N =Q * 2000D 20000

= 0.1 year= 1.2 months

Sol–63: (b)

Annual requirement, N = 3285 units

Let No. of days in year = 365

Daily requirement = N 3285 9

365 365

Reorder point = Lead Time × dailyrequirement

= 8 × 9= 72

Sol–64: (b)

3

2 5

4

1 6A,4 F,9

B,3 D,14

C,7 E,4

(10) (Test - 04)-15th October 2017

A B D F 30 daysA C E F 24 days

Critical path is A – B–D–S = 30 days

Minimum project completion time is 30 daysSol–65: (d)

Sol–66: (a)

P 2 4Q 5 18R 3 8S 4 4T 6 20U 4 24

Job Production Delivery DateSequence (Time (days) (days)

0 + 2 = 2 02 + 5 = 7 0

7 + 3 = 10 10 8 210 + 4 = 14 14 4 1014 + 6 = 20 020 + 4 = 24 0

Job LatenessFlow Time

Average Lateness = 10 2 2

6

Sol–67: (c)

Value = Function or performance or quality

Cost

= 3000 3000 6000 1.5

4000 4000

Sol–68: (d)Total work content (TWC) = 7.2 min

Cycle time (TC) = 7.5 60 1.5 min

300

number of work station ‘n’ = 6Balance delay (BD%) =

C

C

nT TWC 6 1.5 7.2100 100 20nT 6 1.5

Line efficiency = (100 – BD%) = 100 –20 = 80%

Sol–69: (d)Different method to charge depreciationare1. Straight line method (SLM)2. Declining balance method (DBM)3. Double declining balance method

(DDBM)4. Sum of year digits method (SYD)

Sol–70: (a)

Arrival rate ' ' = 20 jobs/hour,Service rate ' ' = 30 jobs/hour

Utilization = ' ' = 20 230 3

Average number of in the queue

‘Lq’ = 2

1 =

2243

2 313

Average waiting time of the jobs in thequeue

Wq= q

4L 13 hour 4 min

20 15

Sol–71: (d)

NT = 80 1000.6 0.5 0.98

100 100

ST = NT + Allowances = NT + 10% of NT= 1.1 NT = 1.1 × 0.98 = 1.078

Sol–72: (c)

Knowledge based system Providesreasoning techniques

Decision support system Uses statisticalrule of inference

Management information system Provides recomendations

Data mining Respond to queries withreports

(11) (Test - 04)-15th October 2017

Sol–73: (c)

Sensitivity analysis of LP models Reduced cost of basic variables are zero at

optimality Constraints are binding when shadow

prices are zero.

Sol–74: (c)

Given,

S = Rs 4,00,000, P = Rs. 50,000, ‘PV’ratio = 25% = 0.25

Margin of safety = Profit

PV ratio

S – SBep = P

PV ratio

SBep = P 50,000S 4,00,000

PV ratio 0.25

= 2,00,000Sol–75: (d)

Winter’s model explicitly accounts forseasonality of demand.

Sol–76: (c)M Markovian (poisson) for arrivalpattern exponential services pattern.

Sol–77: (b)

Average number of defect 'C' = Cn

1 3 7 4 10 6 1 5 4 3 64 2 7 4 2 9 8 5 2 93 4.65

20 20

Upper control limit UCL = C 3 C

= 4.65 3 4.65

= 11.11915

Sol–78: (d)

Chemical reaction taking place

CH4 + 2O2 CO2 + 2H2O

Molecular at 16 64

Ratio =1664 = 1

4

So for 1 kg CH4; O2 needed = 4 × 1 = 4 kg

Hence (d) is correct.

Sol–79: (c)

= 0.5 kg/m3

nozzle dia, d = 50 cmv = 1200 m/s

Thrust mV = AV2

= 0.5 × 4 × 0.52 × 12002

= 141.3 kN 140 kN (closest option)Hence ‘c’ is correct.

Sol–80: (d)

Rocket engines are not air-breathingengines as they carry their own fuel andoxidising agent. Due to this they canoperate in vacuum and at very highaltitude. Hence only rocket engine can bepropelled to space.

So (d) is correctSol–81: (d)

V1

Vr1

u1

Minlet = r1VC

Vr1 = 2 21 1V U

Vr1 = 2 2200 150 250

C = Sonic velocity = 300

M = V 250 0.83C 300

Hence ‘d’ is correct

(12) (Test - 04)-15th October 2017

Sol–82: (b)

rp)opt =2( 1)

max

min

TT

for maximum work output.

Sol–83: (c)

0.1 mPa

1

T

S

2'2x2

0.8 MPa

1–2 is an isentropic process, so S1 = S26.6628 = 1.3026 + x2(7.3594 – 1.3026) x2 = 0.88

Sol–84: (b)

Efficiency increases because meantemperature of heat addition increases

Sol–85: (b)

Sol–86: (a)

Power input factor = Actual power

Theoretical power

Sol–87: (b)

. Efficiency, c = 2s 1

2 1

T – TT T

Polytropic efficiency of compressor,

pc = n 1

n 1

2

1

T

S

2s

2

1

TT =

1

pc

n–1 1n2

p1

P (r )P

2s

1

TT

= –1

2

1

PP

c =

1

1

pc

2s

p11

2

1 p

T 1 (r ) 1TT 1T (r ) 1

Sol–88: (b)

Total to total means stagnation tostagnation state i.e. (kinetic energy offluid is considered)

= 02 03

02 03s

T TT T

Sol–89: (d)Volumetric efficiency of reciprocatingcompressor,

=1/n

2

1

P1 C 1P

C – clearance ratio (P2/P1) – pressureratio and ‘n’ index of compression andexpansion.

Sol–90: (c)

Propulsion efficiency

=

a

2 2

m uThrust power V u= 1Propulsive power m V u2

= 2× flight velocity

Jet velocity + flight velocity

= 2u

V uSol–91: (a)Sol–92: (d)

The correct sequence of flue gases insteam generator is-Superheater/reheater economiser

Dust collector Air preheater

Induced draft fan AtmosphereNote: (i) The dust collectors are many

types one of them iselectrostate precipitator. Oth-ers are fabric filters andbaghouses.

(13) (Test - 04)-15th October 2017

(ii) Induced draft fans areprovided after dust collector.This location requires lesspower and less erosion of fanblades.

Sol–93: (a)In hydrostatic lubrication, an externalpressure supply is used to continuouslyforce fluid lubricant through inlet channelsinto chambers between the bearing surfaceare always separated by a thin lubricantfilm which prevents any friction betweenthe bearing surface.Hence ‘a’ is correct.

Sol–94: (a)Acme threads are trapezoidal in shape.

Sol–95: (b)Given,

dp = 0.25

t =P d d1 1 0.25

P p

= 0.75

Sol–96: (c)P1 = 10 KNL1 = 8000 hrs.P2 = 20 KN

We know

L1 =3

1

CP

1

2

LL =

32

1

PP

L2 =33

1 132

L P 108000P 20

= 1000 hrs.Hence, ‘c’ is correct.

Sol–97: (b)

t

tt sin 45°

Stress = Force FArea wtsin 45°

Hence ‘c’ is correct

Sol–98: (b)

Condition for self-locking screw

tan

– friction co-efficient

– load angle

Hence ‘b’ is correct.

Sol–99: (a)

Fatigue is the weakening of a materialcaused by repeatedly applied loads. It isthe progressive and localized structuraldamage that occurs when material issubjected to cyclic loading. The nominalmaximum stress values that cause suchdamage may be much less than the yieldstrength of the material.

Sol–100: (a)

From the below fig. we can see Soderbergcriteria results the most conservativeestimate of design among the three givencriteria for same factor of safety.

Gerber criterionGoodman criterionSoderberg criterion

Syt Sut m

a )

Se

Sol–101: (d)

Given,

D = 200 mm, t = 1.5 mm, Pmin = 3 MPa,Pmax = 9 MPa

(14) (Test - 04)-15th October 2017

y = 800 MPa,

ut e950 MPa , 600 MPa

In case of this cylindrical pressure vessel

max = maxP D 9 200 600 MPa2t 2 1.5

min = minP D 3 200 100 MPa4 t 4 1.5

mean = max minm 350 MPa

2

amp = max mina 250 MPa

2

According to soderberg criteria

a m

e y

= 1

FOS …… [FOS = Factor of

safety]

or250 350600 800

= 1

FOS

or FOS = 48 1.1741

Sol–102: (d)

50 mm a

b

±P ±P

Stress concentration factor

Kt = max

0

a1 2b

but a = b

So, Kt = 1 + 2 = 3

Sol–103: (b)

With increase in size, chances of occurance

of defects increases, So, with increase insize the endurance limit of the componentdecreases.

Sol–104: (c)

Given, = 150 MPa, h = 15 mm, L = 75mm

Allowable axial load (F) = 2 × 0.707 × h ×L ×

F = 2 × 0.707 × 15 × 75 × 150

or F = 238612.5 N

or F = 238.61 kN

Sol–105: (a)

Given, T 0.5

Tearing effeciency of plate is given by

t = t

t

(P d)ttearing resistance of plateStrength of solid plate P t

t = P d d1P P

or d1 0.5P

or dP = 0.5

Sol–106: (c)

Parallel fillet fail due to shear at the throatarea of the weld at angle 45° to the legdimension.

Sol–107: (a)

Given,

d = 25 mm, t = 15 mm, P = 75 mm, n = 2

t = 400 MPa, = 320 MPa, c 640 MPa

Pmax =

tearing tP (P d)t (75 25)15 400

(15) (Test - 04)-15th October 2017

Pmax = Ptearing = 300000 N = 300 kN

Sol–108: (c)

Life of ball bearing is given by

L90 =3

er

CP

or

LP3 = constant

31 1L P =

31

2PL2

L2 = 8 L1

Sol–109: (d)

Tapered roller bearing Carrying bothradial and thrust

Spherical roller bearing Self-aligning property

Ball bearing light load

Needle roller bearing heavy loadwith oscillatory motion.

Sol–110: (a)

Given, C 1r 100 , = 28 × 10–3 Pa-sec, ns =

40 rps

p = 1.4 MPa = 1.4 × 106 Pa

Sommerfeld number is given by

s =2

sn rp C

= 3

26

28 10 40 1001.4 10

or, s = 8 × 10–3

Sol–111: (d)

Reduction in toque due to friction is givenby

Tf = mWRsin

= mWR cosec

Where Rm = 1 2r r2

...[Considering uniform wear rate]

Rm =3 3

1 22 2

1 2

r r23 r r

...[Considering uniform pressure theory]

Sol–112: (b)

When the clearance ratio changes for thesatisfactory operation of the bearing,sommerfeld number should remainconstant.

1

1

21 s 1

d1

n dcp

= 2

2

22 s 2

d2

n dcp

1

21

1d

dc

=

2

22

2d

dc

...[ As the

speed and the bearing pressure remain thesame]

0.06 × 275

0.2

=2

275

0.15

or 2 = 0.06 × 20.15

0.2

= 0.03375 kg/m-sec

Sol–113: (d)

Sol–114: (d)

Sol–115: (c)

For long shoe in a block brake where angleof contact greater than 60°, the equivalentcoefficient of friction is given by

4 sin2 sin2

Sol–116: (d)

Net horizontal force ‘Fx’ = 250 + 150 = 400N

Net vertical force ‘Fy’ = 250 + 50 = 300 N

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Resultant force ‘Fr’

= 2 2 2 2x yF F (400) (300) 500 N

Resultant moment due to vertical force

= (250 – 50) × 0.25 = 50 N-m (C.C.W)

Resultant moment due to horizontal force

= (250 – 150) × 0.5 = 50 N-m (C.W)

Net moment = 50 – 50 = 0 N-m

Sol–117: (d)

A. Lewis form factor is used to calculatebeam strength of gear and is given by

0.912Y 0.154Z

B. Reynold’s equation used for slidingcontact or journal bearing.

C. Wahl factor 'K' = 4C 1 0.6154C 4 C

used for springs.

D. Unwins formula is given by

'd ' 6 t and is used for rivets.

Sol–118: (a)

Given, m = 4 mm, Zp = 21, P = 12kW, Np = 900 rpm.

V.R = 4 : 1

V.R = g

p

Z4

Z

Zg = 4 × 21 = 84

Ratio factor is given by

Q =g

g p

2 dd d

=g

g p

2 m Zm(z Z )

… [ d = m × z]

Q =g

g p

2 Z 2 84Z Z 84 21

Q = 1.6

Sol–119: (c)

Given, A = 48 MPa

dB = 2dA

DB = 2DA

Shear stress, developed is given by

= 3 38PD D

d d

B

A

=

3B A

A B

D dD d

B =3148 2

2

B = 12 MPa

Sol–120: (c)

Sol–121: (a)

Carnot cycle is not used in vapour powercycles because

1. It is difficult to control condensationprocess precisely

2. It is not possible to design a pump whichcan handle the liquid-vapour mixturephase.

Sol–122: (d)In adiabatic process, the heat transferacross system boundary is zero. But theentropy may increase due to internalirreversibility or friction inside thesystem. So adiabatic process is notconstant entropy or isentropic process.

Sol–123: (c)

T

T0

SSince,

dS = dQT

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On T-S diagram, the area under T-Scurve denotes heat involved in process.The maximum work obtainable dependsupon surrounding temperature T0 andchange in entropy not per degree dropin temperature.

Sol–124: (c)Upon increasing temperature, the rateof increase of cP in small as comparedto rate of increase of cV. So the value ofratio of Cp/Cv decreases.

Sol–125: (d)In order to achieve cooling in throttling,the initial temperature must be belowmaximum inversion temperature.

T

Heatingregion

Coolingregion

P

B

AMaximuminversion

temperature

Sol–126: (b)Basically thermodynamic system is afixed quantity of working substanceupon which heat and work interactionsare studied. According to first law, heatand work are completely mutuallyconvertible. But according to second lawof thermodynamics, work can beconverted completely into heat, but heatcan not be converted into workcompletely but partially.

Sol–127: (d)

The actual shape of Mollier diagram ofwater

TT

TP P

h

s

CP

P

From figure, it is clear that in lowpressure zone, the enthalpy ofsuperheated steam is approximatelysame as saturation enthalpy. But in highpressure zone (near critical point ‘CP’)the enthalpy initially increases atconstant pressure and then remainsconstant.

The reason for constant enthalpy in lowpressure zone is that the vapour behavesas ideal gas and so–

h = u + Pv= u + RT

( Pv = RT)= f(T)

So assertion is wrong at high pressurezone.

Sol–128: (a)In a throttling process, enthalpy (h) = cFor an ideal gas, h = f(T) For an ideal gas in a throttling process,T = C

Sol–129: (d)Sol–130: (b)Sol–131: (a)Sol–132: (a)Sol–133: (b)

The entropy increase of a substance duringa process as a result of irreversibilities maybe offset by a decrease in entropy as a resultheat loss.

Sol–134: (a)

Bending stresses are least for radial blades.So, used in compressors running at highrpm.

(18) (Test - 04)-15th October 2017

Sol–135: (a)

Sol–136: (b)

Sol–137: (a)

Effectiveness of regenerator is directlyproportional to difference of compressoroutlet temperature and turbine exhausttemperature.

Sol–138: (b)

Sol–139: (a)Sol–140: (a)Sol–141: (c)Sol–142: (a)Sol–143: (a)Sol–144: (b)Sol–145: (b)Sol–146: (b)Sol–147: (b)

MixedOR

Semi FluidLubri cation

EC

B

DA

Stable RegionUnstableRegion

fmin

Thin filmlubrication

Thick filmlubrication

Very narrowtransition zone

f

Coefficientof friction

(Bearing modulus)

sn(Bearing characteristic no.)

p

where Bearing modulusSol–148: (a)

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–149: (a)Motions of hands or arms should besymmetrical, simultaneous andopposite. The purpose is to economisethe use of motions and thus performthe work in shortest duration of timeand with minimum of fatigue.

Sol–150: (b)

Both ‘A’ and ‘R’ are true but the ‘R’ isnot the reason for ‘A’.