1. (d)

2. (a)

3. (b)4. (d)

5. (c)

6. (b)

7. (d)

8. (c)

9. (c)

10. (b)

11. (b)

12. (b)

13. (d)

14. (a)

15. (d)

16. (b)

17. (b)

18. (c)

19. (d)

20. (c)

21. (c)

22. (b)

23. (c)

24. (c)

25. (d)

26. (d)

27. (b)

28. (b)

29. (b)

30. (b)

31. (a)

32. (b)

33. (d)

34. (a)

35. (d)

36. (a)

37. (d)

38. (b)

39. (b)

40. (b)

41. (b)

42. (b)

43. (d)

44. (d)45. (c)

46. (b)

47. (a)

48. (b)

49. (d)

50. (c)

51. (c)

52. (d)

53. (a)

54. (d)

55. (c)

56. (d)

57. (a)

58. (d)

59. (b)

60. (d)

ESE-2018 PRELIMS TEST SERIESDate: 24th December, 2017

ANSWERS

61. (c)

62. (b)

63. (c)

64: (b)

65. (b)

66. (d)

67. (a)

68. (b)

69. (c)

70. (d)

71. (c)

72. (a)

73. (b)

74. (a)

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

78. (d)

79. (d)

80. (c)

81. (d)

82. (b)

83. (c)

84. (a)

85. (b)

86. (d)

87. (b)

88. (a)

89. (c)

90. (b)

91. (a)

92. (a)

93. (d)

94. (a)

95. (c)

96. (d)

97. (c)

98. (c)

99. (a)

100. (b)

101. (b)

102. (c)

103. (b)

104. (b)

105. (a)

106. (b)

107. (b)

108. (a)

109. (d)

110. (b)

111. (c)

112. (d)

113. (a)

114. (b)

115. (b)

116. (d)

117. (d)

118. (d)

119. (b)

120. (d)

121. (c)

122. (d)

123. (a)

124. (a)

125. (a)

126. (c)

127. (d)

128. (c)

129. (a)

130. (a)

131. (a)

132. (b)

133. (c)

134. (a)

135. (a)

136. (b)

137. (a)

138. (c)

139. (b)

140. (a)

141. (d)

142. (a)

143. (a)

144. (b)

145. (b)

146. (b)

147. (b)

148. (b)

149. (c)

150. (a)

(2) (Test - 14)-24th Dec. 2017

Sol–1: (d)

=distance from NA

radius of curvature

= y 0.03 RR 0.003

R 10 m

Sol–2: (a)

If the cross-section of the beam is suchthat at every section, moment ofResistance = applied bending moment,then the beam is called fully stressed orbeam is of constant strength. In this casemax-stress (at every cross-section) willbe equal to permissible stress

Sol–3: (b)

Maximum distortion energy of uniaxialloading

=2

1

6G

=2250

6G = 62500

6G

Sol–4: (d) AM = 0

B1 2L 2L(W)(L) (R )2 3 3 = 0

i.e. RB = WL2

Reaction at support A 0 becauseresultant of the above shown uniformly

varying load acts at support B .

AV 0

Maximum moment occur at support B

=

1 2L 2 1 2Lw2 3 3 3 3

= 24 w81

l

Sol–5: (c)

Strain in lateral direction when only 1

acting

lateral =

1

E

Lateral strain when all stresses areacting.

lateral =

2 1 2

E E E

As per question

1

E =

2 1 22

E E E 1 = 2 1 22 2 2 x( ) = 2 (2)( 1 ) 1 = 22 (1 )

1

2 2 (1 )

Sol–6: (b)

Let the distance of the centroidal neutralaxis from the bottom edge by y . Let fcand ft be the max compression and max.tensile stress.

c

t

ff =

h y 3y

h y = 3y

hy4

t

h

t

x

fc

x

y

ft

But from geometry the centroid of thesection

h t(th) (x t) t h2 2(th) (x t) t 4

2hx th 2t

(3) (Test-14)-24th Dec 2017

Putting value of t = 25 mm and h = 100mm

x =

210025 mm100 2 25

= 225 mm

Sol–7: (d)

Axial stress =

32

2100 10 40 N/mm

(100)4

Maximum shear stress due to torque

=

6

3 316T 16 5 10

d 100

80 N/mm2

Maximum shear stress due to combinedeffect

max =

22

2

2 2 2(20) 80 20 17 N/mm

Sol–8: (c)

L

Pcr d

crP = F LCrP = K L

Pcr = KL

Sol–9: (c)Outer dia. = 100 mmRadius (r) = 50 mm

Wall thickness = 25 mm = 125 MPa

TJ =

r

T =

4 4J 125 (100 50 )r 32 50

23.0097 × 106 Nmm

23 kNm

Sol–10: (b)

MEI diagram

+MEI

L/2 CA B

Deflection of C, with respect to tangentat A

tAC =

MArea under diagramEI

× C.G of area about point C

tAC = 2M L L ML

EI 2 4 8EISol–11: (b)

Sol–12: (b)Force along the slope opposing themotion is F = R(mgsin F )

where resistance due to friction, air etc.is

FR = 30 N/kg × 200 kg= 6000 N

sin tan = 1200

F =

1200 10 6000200

= 10 + 6000 = 6010N Power = F × V

= 6010 × 10 = 60.1 kWSol–13: (d)

Coplanar and concurrent forces are theones which lie in one plane or meet atone point.

Sol–14: (a)

(4) (Test - 14)-24th Dec. 2017

Sol–15: (d)

B

D1

Q = 1 1 fD B V

Q = 1 1 fD B V 4 0.8 3= 30.16 m3/s

Sol–16: (b)

Maximum accleration head in suctionpipe

Hasm =

2s

s

L A rg a

=

2 2

2

(20)5 2 30 0.4049.81 60 2(10)

4Hasm = 4.024 m

At limiting conditionHasm + HV + HS = Hatm0

4.024 + 2.5 + Hs = 10HS = 3.475 m

Sol–17: (b)

Applying energy equation at (1) and (2)

H

1

2

12 2

1 2 21 2 2

VP P VZ Z h2g 2g

but V2 = Vc

H + 0 + 0 = 0 + 0 + 2C

LV h2g

hL = 2c

C VVH V C 2gH2g

hL = 2vC (2gH)H

2ghL = 2 2

VH (1 C ) 4 (1 0.94 )= 0.47 m

Sol–18: (c)

FD

FB

V

W

FB + FD = WFD = W – FB

=

3 3sD g D g

6 6 l

= 3sD g ( )

6

= 3 3(3 10 ) 9.81 (1840 840)

6= 13.86 × 10–5 N

Sol–19: (d)

d( )

2dP4

Surface tension

As there is only one interface betweenliquid and air

d =

2d( P)4

P =4

d

=

24 0.0156 1.25 N/m0.05

Sol–20: (c)

y = 4

y = 1

x = 2

A B

x = 4

x

yCD

Vorticity = 2wz

=

V ux y

(5) (Test-14)-24th Dec 2017

= (–5 – 8) = – 13Circulation

Area = VorticityCirculation = – 13 × 6 = –78

Sol–21: (c)

We know for laminar flow through acircular pipe

u =

2

max 2rU 1R

=

2

21000.5 1200

=

1 1.50.5 1 m/s4 4

= 37.5 cm/s

Sol–22: (b)

Qr = 3r

r rr

LA VT

Tr for froude’s law = rL

Qr = 2.5rL

m

P

QQ =

2.51100

Qm =

2.54 3110 0.1 m /s

100

Sol–23: (c)

total =

Viscous shear stress turbulent shear stress

laminar turbulent

1f = 10 e2 log (R f ) 0.8

for 5 × 104 < Re < 4 × 107

(for K 0D ) prandtl equation

Sol–24: (c)

Velocity profile in laminar sub layer isactually parabolic, but as the thicknessis very small, it may be taken as linear.

Sol–25: (d)

Sol–26: (d)

Sol–27. (b)

High specific volume of vapour leads tolarge size of compressor and workrequirements also increases.

Sol–28: (b)

Sol–29: (b)

Bypass factor =

21 12 0.2530 12 0.7 0.25

= 0.475 kg/kg of dry air

Sol–30: (b)

(COP)H.P = H

H L

T 303 5.509T T 303 248

(COP)H.P = Q 5.509W

W = Q 30 5.45 kW

5.509 5.509

Sol–31: (a)

(COP) =

E G o

G o E

T (T T )T (T T )

=

268(473 300) 3.06473(300 268)

Sol–32: (b)1. Low inlet pressure and temperature

reduces overall temperature ofcombustion, hence knocking tendency isreduced.

2. By rich mixture flame temperature canbe kept low hence knocking tendencycan be reduced.

Sol–33: (d)1. Heavier fly wheel is required in 4 stroke

CI engine as compared to 2 strokeengine.

2. Specific fuel consumption

= fuel used in kg / h

power in kw

Power output is higher in CI enginehence specific fuel consumption is lessHence it is a correct statement

(6) (Test - 14)-24th Dec. 2017

3. In CI engine mixing of fuel air occurs inengine cylinder.

Sol–34: (a)Metering rod and economiser devices areused when power requirement is highi.e for power enrichment.

Sol–35: (d)Multistage compression and multistageexpansion in gas turbine areapproximated as isothermal

P = C

P = CT

S

So this turbine cycle becomes close toericsson cycle.

Sol–36: (a)

Sol–37: (d)

Sol–38: (b)

Sol–39: (b)The cooling of air between two stages ofcompression is known as intercoolingDue to intercooling:

(1) Work of compression is reduced(ii) Specific out of the plant is increased

(iii) Thermal efficiency is reduced, howeverthis loss can be remedied by employingexhaust heat exchanger as in reheatcycle.

Sol–40: (b)

in outm (Q Q ) = WT – WP

1.5 (3400 – 2600)= 1218 – WP

wP = 1218 – 1200wP = 18 kW

Sol–41: (b)In De-laval turbine work done

= change in kinetic energy

600 = 2 22

1 1 (40 V )2

V2 = 20 m/s

Sol–42: (b)

Specific steam consumption = net

3600W

=T P

3600W W

=

3600939.54 3.54

= 3600936

= 3.84Sol–43 (d)

From steady flow energy equation

1h Q 2h W (adiabatic Q = 0)

W = h2 – h1Sol–44: (d)

Velocity relationship, c fV V

cos sin

When Vf = Vc

cos sin cos 90

= 90

2 = 90°Sol–45: (c)

Mandrel is used to hold and locate aworkpiece with central hole such asgears and pulleys.

Sol–46: (b)Universal milling machine is suitablefor producing helical grooves and helicalgears as table of machine can beswivelled.

Sol–47: (a)In creep feed grinding MRR is high feedrate is very slow, grinding wheels aremostly of softer grades, and reducedtemperature at the work surface.

Sol–48: (b)SiC is mostly used as abrassive ingrinding wheel for grinding WC tools

Sol–49: (d)

(7) (Test-14)-24th Dec 2017

Sol–50: (c)

2 2 2 2 2CaC 2H O C H Ca OH

The amount of C2H2 produced afterreaction is proportional to the contentof moisture in the sand.

Sol–51: (c)According to Caine’s method of riserdesign, if the casting solidifies infinitelyrapidly, the riser volume should be equalto the solidification shrinkage of thecasting.

Sol–52: (d)

Powerof arc(Pa)

(P )a max

loptlength of arc ( )la

Sol–53: (a)Sol–54: (d)Sol–55: (c)

When f1 > f2 then there exist netfrictional force to the right making therolling operation possible by pulling thestrip into roll gap.

Sol–56: (d)Ideal force required for extrusion doesnot depends upon die angle.

Sol–57: (a)Anisotropy or directionality may bepresent in the plane of sheet (planaranisotropy) and in the thicknessdirection (normal or plastic anisotropy).These behaviours is particularlyimportant in deep drawing as both planeand thickness of sheet gets deformed.

Sol–58: (d)Peak pressure generated by explosivedepends upon

• Types of explosive• Weight of explosive• Standoff• Compressibility of energy transmitting

medium• Acoustic impedance of energy

transmitting medium, which is equal to

the product of its density and soundvelocity in the medium

Sol–59: (b)Sol–60: (d)

No-Go

Go

Tolerance

Some of the good parts are rejected

Sol–61: (c)CIMS offers following benefits• Increased machine utilization• Scheduling flexibility• Lower in-process inventory• Reduced manufacturing lead time• Reduced direct and indirect labour.

Sol–62: (b)Sol–63: (c)

Fusible Plug : sefety precautionagainst low water level in steam boilers.Safety valve : To prevent excessiveinternal pressure.Blow off cock : To remove sludge andother impurites from inside the boiler.

Sol–64: (b)The reheat is done to remove themoisture carried by the steam in thefinal stages of expansion.

Sol–65: (b)The critical pressure ratio is defined asthe ratio of the pressure at the throatto the inlet pressure for choked flow.When M = 1, occurs at the throat,

0

p *p =

nn 12

n 1

Sol–66: (d)

The flue gas passes through the stagesof superheater, and then economizer.In the air preheaters, the flue gasestransfer their residual heat to thecombustion air, during which they arecooled to the exit flue gas temperatureof the steam generator. For furthercleaning, the flue gas is conductedthrough an electrostatic precipitator to

(8) (Test - 14)-24th Dec. 2017

remove dust.Sol–67: (a)

Blow off cock is a boiler mounting whichremoves water from the shell at regularintervals to remove the variousimpurities that may be settled at thebottom of the shell.

Sol–68: (b)

DOR = mb

fb mb

hh h

= 9 9 0.45= =

11 9 20

Sol–69: (c)

Goodman line is safe w.r.t. fatigue failurebut there is small portion inside theGoodman line where component could faildue to yielding, hence modified Goodmandiagram combines fatigue failure withfailure by yielding.

Sol–70: (d)

Sol–71: (c)

(i) Fine threads have greater strengthwhen subjected to fluctuating loads andgreater resistance to unscrewing.

(ii) The static load carrying capacity ofcoarse threads is higher.

(iii) Coarse threads have more even stressdistribution.

(iv) Coarse thread is designated by theletter ‘M’ followed by the value of thenominal diameter in mm. For example M12.

(v) A screw thread of fine series isspecified by the letter ‘M’ followed by thevalues of the nominal diameter and thepitch in mm and seperated by the symbol‘x’. e.g. M12 ×1.25.

Sol–72: (a)

L10 = 10h6

60nL10

=

660 9000 1350

10= 729 milliion revolution

Dynamic Load CapacityC = P(L10n)1/3

= 8000 (729)1/3

= 72000 N

Sol–73: (b)

A fitted bearing is a bearing in which theradius of the journal and the bearing areequal.

Sol–74: (a)For the transverse vibration of a shaftcarrying several loads, there are twomethods of finding natural frequencyof the system: Dunkerley’s method andenergy method. Dunkerley’s methodgives approximate results but is simple.This is used when the diameter of theshaft is uniform. The energy methodgives accurate results but involvesheavy calculations in case there aremany loads. This is also known asRayleigh’s method.

Sol–75: (d)When a point on one link is slidingalong another rotating link, such as inquick return motion mechanism, thenthe coriolis component of accelerationis taken into account. These is no suchoccurence in the case of slider crank orscotch yoke mechanism.

Sol–76: (c)Hammer blow is the maximummagnitude of the unbalanced forcealong the perpendicular to the line ofstroke.

Hammer Blow = 2B b .Sol–77: (b)Sol–78: (d)Sol–79: (d)

Number of instantaneous centres

N = n n 1

2where n = no. of links

N =

9 8 362

Sol–80: (c)Sol–81: (d)

The schematic of flat wall havingvarious parameters is shown in figure.

(9) (Test-14)-24th Dec 2017

q

T °C1

tT °C2

k=0.3kW/mK

Givenk = 0.3 w/mKT2 – T1 = – 20°C

The heat flux through wall

q =

dTkdx

q =

2 1T Tkt

150 =

2000.3

t

t = 200 0.3

150 = 0.4m

Sol–82: (b)Consider a plane element of furnace wallas shown in figure initially.

t T2

k=0.8 W/mK

T1

Thickness of wall, t = 35cmThermal conductivity, k = 0.75W/m-KThe heat loss from furnace per unit area.

q = dTkdx =

2 1T Tk

t

q = 1 2k T T

t

q = 1 20.75 T T35 ...(i)

Now thermal conductivity is changedto 0.15 W/m-K, the heat loss is same-

q = 1 20.15 T T

t ...(ii)

For same heat loss,

1 20.75 T T35 =

1 20.15 T Tt

t =

35 0.15 7cm

0.75

Sol–83: (c)Data, givenConvection coefficeint h = 80 W/m2Karea of slab = 3m2

Hot temperature T 120 C and Ts = 40°C

The heat gain rate by slab

Q = hA T

= shA T T

= 80 × 3 (120 – 40)= 240 × 80= 19200 W= 19.2 kW

Sol–84: (a)Data, given hx = 15W/m2KFluid flowing over slab-

x

Nu = 75= 0.7 mx

P

Nusselt Number-

Nu =hkx x

75 = 15 0.7K

Thermal conductivity of fluid

K =

15 0.7 0.14 W/mK

75

Sol–85: (b)I. The flow through insulated constant

diameter pipe

1 2

Applying steady flow energy equation

(10) (Test - 14)-24th Dec. 2017

h1 + Q = h2 + w Q = 0 = wSo, h1 = h2 = constantIts enthalpy which remains constant alongwith flow.

Sol–86: (d)

closesystem

1 21Q2

system boundary

1Q2 = 45 kJ

U2 – U1 = U= – 15kJ

from first law of thermodynamics1Q2 = 1w2 + (U2 – U1)45 = 1w2 – 15

1w2 = 45 + 15 = 60 kJSol–87: (b)

First law for a stationary system in aprocess gives

Q = U WQ1–2 = U2 – U1 + w1–2 ...(i)w1–2 = P(v2 – v1)

= 0.15 (0.2 – 0.4)MJ= –30 kJ

and givenQ1–2 = –45 kJ

from (i)–45 kJ = U2 – U1 – 30kJ

U2 – U1 = –45 + 30= – 15 kJ

Sol–88: (a)

HE

SinkT = 87°C2

Q2

Q1

Source T1

w

Given T2 = 87 °C= 360 k

Q2 = 0.3Q1For Carnot cycle

1

1

QT = 2

2

QT

T1 = 21

2

T QQ

= 360 × 1

1

Q0.3Q

=3600.3 = 1200k

So, temperature at sourceT1 = 1200 – 273

= 927 °CSol–89: (c)

I. Entropy transfer is associated with heattransfer only not with the work transfer.

II. Entropy change of the universe in allreversible processes is zero

III.Increase in entropy always degradesenergy the degradation of energy isgiven

= Tosgenwhere To= surround temperature andSgen = entropy generation.

Sol–90: (b)It is a throttling process and henceenthalpy is constant.

Sol–91: (a)

PdV is the expression for quasistaticprocess in a closed, non-flow system.

Sol–92: (a)During reversible non-flow process,work done

W = 1 1 2 2p v p vn 1

As n increases, W decreases.Further adiabatic mixing is irreversibleprocess.

Sol–93: (d)As per the third law of thermodynamics,the entropy of a pure substance in

(11) (Test-14)-24th Dec 2017

complete thermodynamic equilibriumbecomes zero at the absolute zerotemperature.

Sol–94: (a)

Sol–95: (c)

Tilt factor for diffused radiation

=

1 cos 1 cos302 2

= 0.93

Tilt factor for reflected radiation

=

1 cos2

=

1 cos 300.302

= 0.020

Sol–96: (d)

Power law coefficient ( )

=

1 2

1 2

n H /Hn V /V

=

n 12 / 48n 4 / 32

= 0.66

Sol–97: (c)

Sol–98: (c)

Sol–99: (a)

Output voltage

EH = KH . IB/t

= –1 × 10–6 × 3 × 0.5/(4 × 10–3)

= –0.375 mV.

Sol–100: (b)

Instruction HLT means Halt and enter waitstate.

Sol–101: (b)

Thermocouples are active transducer

Sol–102: (c)

Jobs processing time Jobs flow time or completionC 4 0 4 4B 5 4 5 9F 7 9 7 16A 12 16 12 28E 13 28 13 41D 15 41 15 56

So, average completion time =

Total completion timeNumber of jobs

= 4 9 16 28 41 56

6 = 25.6 minute

Sol–103: (b)Mean time between failure (MTBF) =

1200 hrMean time to repari (MTTR) = 36 hrs

So, Availability = MTBF

MTBF MTTR =

1200

1200 36 = 12001236 = 0.9708

Sol–104: (b)Given

F = Rs 16000V = Rs 40/ unit

XBep = 400 units

BepS V X = F

S 40 400 = 16000

S = 16000 40

400

= Rs. 80/unitSol–105: (a)

Data givenS = Rs. 160F = Rs 64,000V = 80P = 16,000

Let the minimum quantity to be producedto get the profit of Rs 16,000 is x

X(S – V) = F + P

(12) (Test - 14)-24th Dec. 2017

X = F PS V

=

64,000 16,000160 80

=80,000

80 = 1,000 units

Sol–106: (b)Annual demand = 94600 unitsAvailable time = 46 weeks/year × 6 shift/week × 8hr/shift × line

= 46 × 6 × 8 × 0.98= 2163.84 hrs

The cycle time TC =Available time

Annual demand

=2163.8494600

= 0.0228 hrs/unit= 1.372 minutes/unit

Sol–107: (b)Due dateOrder Process time remaining (day) Due date Critical ratio =

Processing timeP 5 25 5Q 17 35 2.05R 7 22 3.142S 9 27 3

jobs are sequenced on increasing order of theircritical ratio i.e Q – S – R – P

Sol–108: (a)

Arrival rate = 45 customer/hr = 4560 cus-

tomer/min

service rate = 1

= 1 min

= 1 customer/min

Utilization = = =

4560 = 3

4

Average number of customer in the queue

Lq =

2

1

=

23 4314

= 2.25

Average waiting time of the customer in thequeue

Wq =

qL =

2.253 4 = 3 min

Sol–109: (d)

4

5

6

2

3

1

Pr ocessing Works flowWorks Due date tardinesstime time

w 29 30 0 30 30 0w 7 32 30 7 37 37 32 5w 9 34 37 9 46 46 34 12w 13 48 46 13 59 59 48 11w 27 52 59 27 86 86 52 34w 11 55 86 11 97 97 55 42

Hence total number of tardy works are 5Sol–110: (b)

D = 60 units/days, Co = 120/order, Ch = Rs 0.05/unit day Lt = 5 days

Optimal order quantity = Q = 0

h

2DCC

= 2 60 1200.05 = 537 units

Safety stock= Lt × d= 5 × 60= 300 units

Sol–111: (c)

o p m e 0 pActivity t t t Expected time t t 4m t / 6i j1 2 2 4 3 31 3 4 6 5 51 4 2 2 2 22 5 15 3 6 73 5 4 16 7 84 6 9 3 6 65 6 12 4 2 4

1 2 5 6473

4

62 385

1 – 2 – 5 – 6 = 3 + 7 + 4 = 141 – 3 – 5 – 6 = 5 + 8 + 4 = 171 – 4 – 6 = 2 + 6 = 8

Hence the critical path is 1 – 3 – 5 – 6Hence expected project completion timeis 17 days

(13) (Test-14)-24th Dec 2017

Sol–112: (d)

(i) In order to find the instantaneouscentre of a moving body, assume the bodyis purely rotating about a point. If this pointsatisfy all the motion condition, this iscalled instantaneous centre. In this case thepoint of contact of disc with plane isinstantaneous centre.

(iii) The geneva mechanism producesintermittent motion from continuousrotation.

Sol–113: (a)

0

/2/2

/2

OO

C B

A

Given,

OA = 300 mm

OB = OC = 150 mm

cos2 =

OB 150 0.5OA 300

2 = 60° 120

Hence

Time of cutting strokeTime of return stroke

=

360 360 120 240 2120 120

Sol–114: (b)

(i) The curve generated by locus of tracepoint is pitch curve not cam profile.

(ii) For a stable governor, the controllingforce must increase more with increased inradius of rotation.

Sol–115: (b)

(i) When crank is perpendicular to the lineof stroke of the piston, then velocity ofpiston is maximum.

(ii) The acceleration in slider crankmechanism

a =

2 cos2r cos

n

= 0 at inner dead centre

a =

2 1r 1n

At outer dead centre, 180

2 21 1a r 1 r 1n n

Hence acceleration is maximum at innerdead centre i.e 0

Sol–116: (d)

Workdone per second = 48000 J = 48000N-m

For a double acting engine, the number ofworking strokes per minute = 2 × 240 =480

Workdone / stroke

= Work done per sec ond

Number of working stroke / sec ond

=

48000 48000 60 6000 N m

480 / 60 480

Fluctuation of energy E = 6000 × 0.2 = 1200N-m

Coefficient of fluctuation of speed,

(14) (Test - 14)-24th Dec. 2017

K =

1 2 1.02 0.98 0.04

Also, E = 2I K

K =

2 2 2E E

I mK

2 2

12000.04m 0.4 25

2 N60

m = 300 kg

Sol–117: (d)

(i) Rayleigh’s method gives 2 – 2.5% highervalue of frequency than natural frequencyobtained by exact method.

(ii) At node, rotation is zero. so the nodesare assumed fixed. For more clarity considertwo rotor system.

N

I2

I1

Let two rotors are rotating in oppositedirection then there will be point N on shaftwhich has zero rotation. Here point N isnode.

(iii) In the viscous damping, the dampingforce is always proportional to velocity onlyand acts against motion.

Viscous force = dxCdt

Sol–118: (d)

(i) In skew bevel gears, the axes are nonparallel and non-intersecting and teeth arestraight.

(ii) The purpose of differential gear on rearaxle is to alow to back wheels on at differentspeed during turn and at same speed in

straight motion. Because at turn the outerwheels moves faster than inner to covermore distance and inner wheel moves slowto cover less distance.

(iii) The sliding velocity between conjugateprofile teeth gears

= 1 2( ) distance between pitch pointand point of contact

So, it is zero at certain points i.e. pitchpoint.

Sol–119: (b)

(ii) In involute tooth profiles, the line ofaction is fixed, the line of action passesthrough a fixed point (i.e. pitch point) andinclined with prependicular to line joiningcentre of gears at a fixed angle. Hence,pressure angle is constant throughout theteeth engagement.

(iii) Stub tooth are shorter than standardtooth. The addendum of the gear is reducedwhile keeping other parameters same. Sochances of interference are less in stubtooth.

Sol–120: (d)

Given

mn = 16 mm, T1 = 18 1 = 30°

2 = 55° – 30° = 25°

VR = 2 1

1 2

N T13 N T

T2 = 1

R

TV

= 18 54

1 / 3So,

C =

n 1 2

1 2

m T T2 cos cos

=

16 18 542 0.866 0.906

C = 8(20.78 + 59.58)

= 642.8 643 mm

(15) (Test-14)-24th Dec 2017

Sol–121: (c)The magnitude of deflection due to shearforce, axial force and torsion isinsiginificant is comparsion to thedeflection caused due to bending andhence deflection due to bending is onlyconsidered.

Sol–122: (d)Even if there is axis of symmetry butload or bending couple is not in thatplane, then the bending will beunsymmetrical bending. In this caseplane of couple and plane of bending isnot same.

Sol–123: (a)

D Alembert showed that one cantransform an accelerating rigid bodyinto an equivalent static system byadding the inertia force. The system canthen be analyzed exactly as a staticsystem subjected to this inertia forceand the external forces.

Sol–124: (a)

Sol–125: (a)

Sol–126: (c)

Sol–127: (d)

Fluid particle continue to deform anddoes not maintain their identify.

Sol–128: (c)

Sol–129: (a)

As per classius clapeyron equation

fg

e c

h 1 1R T Tc

e

P eP

As hfg increases it leads to higher pressureratio so lower volumetric efficiency.

Sol–130: (a)Heat cannot be extracted from the fluegasses for economiser, superheater etc.since the effective draught will bereduced if the temperature of the fluegasses is decreased.

Sol–131: (a)

Sol–132: (b)Sol–133: (c)

The drill cross section is designed formaximum possible strength within theconstraints of drilling force and torque,as stronger geometry and bigger webmay result in higher drilling force andtorque.

Sol–134: (a)A blind riser lies entirely in the copeand is entirely surrounded by mouldingsand, hence it looses heat slowly.Thus, metal is available in molten statefor longer duration in case of a blindriser. Hence it is usually smaller thanopen riser.

Sol–135: (a)Sol–136: (b)

Lubricants are not used in this operationbecause they can be entrapped in diecavities and thus prevent reproductionof fine die-surface details

Sol–137: (a)Sol–138: (c)

The tape reader must be omitted inDNC, thus relieving the system of itsleast reliable component

Sol–139: (b)

Sol–140: (a)Sol–141: (d)

Entropy is not conserved. It can begenerated but never destroyed.

Sol–142: (a)Sol–143: (a)Sol–144: (b)Sol–145: (b)Sol–146: (b)Sol–147: (b)Sol–148: (b)Sol–149: (c)

Limiting addendum for pinion is largerthan that of gear for equal addenda,interference takes place between the tipsof the wheel teeth and the flank of thepinion teeth.

Sol–150: (a)