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ESE-2017 PRELIMS TEST SERIESDate: 25th December, 2016

28. (b)

29. (c)

30. (d)

31. (b)

32. (d)

33. (d)

34. (a)

35. (c)

36. (c)

37. (c)

38. (d)

39. (b)

40. (c)

41. (a)

42. (c)

43. (b)

44. (b)

45. (d)

46. (c)

47. (d)

48. (d)

49. (d)

50. (d)

51. (b)

52. (c)

53. (a)

54. (b)

55. (d)

56. (b)

57. (c)

58. (a)

59. (a)

60. (c)

61. (b)

62. (c)

63. (b)

64. (a)

65. (c)

66. (c)

67. (d)

68. (c)

69. (a)

70. (d)

71. (a)

72. (d)

73. (b)

74. (c)

75. (c)

76. (d)

77. (d)

78. (a)

79. (c)

80. (a)

81. (b)

82. (d)

83. (c)

84. (c)

85. (b)

86. (c)

87. (a)

88. (b)

89. (b)

90. (c)

91. (a)

92. (c)

93. (c)

94. (d)

95. (d)

96. (c)

97. (b)

98. (d)

99. (a)

100. (b)

101. (d)

102. (a)

103. (d)

104. (b)

105. (c)

106. (c)

107. (c)

108. (d)

109. (a)

110. (c)

111. (b)

112. (a)

113. (a)

114. (c)

115. (c)

116. (c)

117. (d)

118. (c)

119. (d)

120. (c)

121. (b)

122. (d)

123. (c)

124. (d)

125. (d)

126. (c)

127. (b)

128. (a)

129. (c)

130. (b)

131. (a)

132. (d)

133. (c)

134. (d)

135. (d)

136. (b)

137. (b)

138. (d)

139. (b)

140. (d)

141. (d)

142. (a)

143. (a)

144. (b)

145. (d)

146. (c)

147. (d)

148. (d)

149. (b)

150. (a)

151. (a)

152. (a)

153. (b)

154. (b)

155. (b)

156. (b)

157. (b)

158. (a)

159. (c)

160. (a)

ANSWERS

1. (b)

2. (c)

3. (d)

4. (d)

5. (a)

6. (c)

7. (b)

8. (c)

9. (d)

10. (d)

11. (c)

12. (a)

13. (c)

14. (d)

15. (b)

16. (c)

17. (d)

18. (a)

19. (a)

20. (d)

21. (c)

22. (c)

23. (a)

24. (b)

25. (d)

26. (c)

27. (c)

IES M

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(2) ME (Test-18), Objective Solutions, 25th December 2016

Sol–1: (b)1. Free expansion is always irreversible

process due to absence of resistanceagainst the expansion.

2. Slow process is a quasi-static processwhich is reversible in nature so slowheating of oil from a constanttemperature source is reversibleprocess.

3. Evaporation process is reversible.4. In isentropic compression, the increase

in entropy due to a irreversibility canbe compensated by heat loss and dueto this entropy decreases and processbecomes isentropic but not reversible.

Sol–2: (c)

T1

4

S

23

The polytropic process–pVn = constant

(A) n = 0 P = constant, i.e. Isobaric A-2 or3

(B) n = 1, i.e.pV = constant, isothermal

B – 4

(C) n = 1.4 .=

pV = constant adiabatic orisentropic

C – 1(D) n = undefined From definition of specific heats

v

TS

= pp p

T T Tand =C S C

Cp > Cv so the slope of constant pressure

curve is less than constant volumecurve.

A is 3 i.e. A–3 and D is 2 i.e. D–2 Final combination

A –3, B–4, C–1, D–2Note: The general process T-S tracesdiagram

T

n=1

s

n=0 (

P=C)n =

n V C=( )=

Sol–3: (d)The series combination of heat engines.Since,

W1 = W2

HE1

HE2W2

T2

W1

T = 180 °C1

T = 20 °C3

293 KQ3

Q2

Q1453 K

Q1 – Q2 = Q2 – Q3

2Q2 = Q1 + Q3

In Carnot engines

1

1

QT = 2

2

QT

= 3

3

QT

= Constant-C 2 CT2 = CT1 + CT3

T2 = 1 3T T2

= 453 313

2

= 383 K= 110°C

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Sol–4: (d)The reversible engine–

T = 800K1

Q =6kJ1

T = 600K2

Q =8kJ2

HE W

Q0

T =100 K0

For reversible heat engine–(Q/T) = 0

01 2

1 2 0

QQ QT T T

= 0

0Q6 8800 600 100

= 0

T0 = 100 K

Q0 = 6 88 6

= 50 25 kJ=24 12

Work outputW = Q1 + Q2 – Q0

= 6 + 8 – 2512

= 11.9167 kJ Thermal efficiency–

th = 1 2

W 11.9167=Q Q 14

= 0.8512 85.12%Sol–5: (a)

1h1

2h2

Q

Compr

h1 = 100 kJ/kgh2 = 200 kJ/kg

heat loss,Q = 40 kJ/kg

Since compressor is a steady flow machine.So from steady flow energy equation

h1 + q= h2 + ww = (h1 – h2) + q

= (100 – 200) – 40= – 140 kJ/kg

for 0.25 kg/sec

w = 0.25 × 140 kW = 35 kWSol–6: (c)

Ideal gas in cylinder piston device executesreversible adiabatic expansion. In suchprocesses, the change in entropy of systemand surrounding will be zero. In additionto this, the entropy change of system andsurroundings will also be zeroindividually.

Sol–7: (b)T/P

S/V

AP

B

CP

Because in two phase region temperatureand pressure are not independent so asone increase other increase or one decreaseother decrease.The enthalpy of evaporation is proportionalto length of line AB. As point P goes fromleft to right, dryness fraction increases.So in conclusion,A. Saturation pressure increase results in

increase of saturation temperature anddecrease in evaporation enthalpy i.e.A-3/4

B. Saturation temperature increasesresults in reduction in enthalpy ofevaporation i.e. B-3.

C. Saturation pressure reduction resultsin increase in specific volume i.e.C-2.

D. Higher dryness fraction results inhigher entropy due to higher heat

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involved.Sol–8: (c)

The Mollier diagram showing throttlingprocess,

h

s

x1x2x3x4

T2

T1

P1P2

T > T1 2

x > x > x > x1 2 3 4

cP

12

During throttling process 1–2, Pressure and temperature reduces Enthalpy remains constant Dryness fraction ‘x’ improves Since dryness fraction ‘x’ improves,

the specific volume increases due toreduction in pressure.

Throttle is entropy generation pro-cess so entropy increases.

Sol–9: (d)Since in adiabatic mixing process totalenergy is conserved i.e. first law ofthermodynamics. The properties in finalstate can be obtained by perfect gasrelationship for mixture.

Sol–10: (d)The difference between Cp and Cv,

Cp – Cv=2

P T

V PTT V

• As absolute temperature T approacheszero, the above expression,

Cp – Cv = 2

P T

V P0T V

= 0 Cp = Cv

• At critical point-

cT T=

PV

= 0

Cp – Cv= 0Cp = Cv

Hence at critical state, the differenceof liquid and gases diminishes.

Sol–11: (c)1

h 500 kJ/kg1 =s 1.1 kJ/kgK1 =

T 300 K0 =

2

s 0.7 kJ/kgK2 = h 100 kJ/kg2 =

Change in availability of flow process= (h1 – h2) – T0 (S1 – S2)= (500 – 100) – 300 × (1.1–0.7)= 400 – 300 × 0.4= 280 kJ/kg

Sol–12: (a) Throttling is constant enthalpy

process. In isentropic process, the change in

entropy of system is zero. Free expansion, the work done by

system in zero. Isothermal process, the temperature

chagne is zero so the internal energy.Sol–13: (c)

The KOH solution absorbs only CO2 fromdry flue gases. In the given questionvolume of gases after KOH solutionabsorption

= 89

CO2 by volume = 100 – 89 = 11% Further the volume of O2 present

= 89 – 84 = 5 cc = 5%Volume of CO = 84 – 82 = 2 cc = 2% Volume of N2

= 100 – (CO2+O2+ CO)= 100 – (11+5+2)=82cc= 82%

Percentage ratioN2 : CO2 : O2 : CO = 82:11:5:2)

Sol–14: (d)The correct sequence of flue gases insteam generator is-

IES M

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Superheater/reheater economiser

Dust collector Air preheater

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

types one of them iselectrostatic precipitator.Others are fabric filters andbaghouses.

(ii)Induced draft fans are providedafter duct collector. This loca-tion requires less power and lesserosion of fan blades.

Sol–15: (b)Velocity -pressure variation in impulseturbine,

Pressure

velocityMBN

In simple impulse turbine (i.e. De-Laval)there is only one stage composed of oneset of nozzle and one set of movingblades. Here entire pressure/enthalpychest drops in nozzle and it is convertedto velocity i.e. kinetic energy. The ve-locity of steam reduces in moving bladesbecause of momentum transfer to tur-bine blades.

Sol–16: (c)The blade speed ratio is–

=blade speed

steam velocity

Vr

Ub

V

= bUV

In impulse turbine for maximum bladeefficiency–

= bU cos=V 2

Sol–17: (d)Reheating of steam results1. More output from same steam circu-

lation i.e. higher specific output.2. It can reduce efficiency but generally

improve if the average temperatureof heat addition is higher.

3. Blade erosion is reduced due lowwater drops i.e. high quality of steamin low pressure stages.

But the possible combination in optionsis (d) – 1 and 4

Sol–18: (a)

v

P

1

2

3

4

For given conditions in question,P1V1 = P2V2 = P3V3 = P4V4 = RT

1–2–3–4–Isothermal.Since in ideal condition work done

W =

n 1n2

1 11

Pn P V 1n 1 P

+

n 1n3

2 22

Pn P V 1n 1 P

+

n 1n4

3 33

Pn P V 1n 1 P

+

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Since,

1

1

PP = 3

2

PP = 4

3

PP

W =

n 1n2

1 11

P3n P V 1n 1 P

=

n 13n4

1 11

Pn P V 1n 1 P

Sol–19: (a)Because of intercooling, the heat gener-ated is taken away and does not heatthe compressed gas. So its reheat factoris less than one and adiabatic efficiencyis less than stage efficiency.But in multistage turbine, adiabaticefficiency.

ad = stageR.F.

RF > 1 in turbine.

So ad > stage

Sol–20: (d)T

Wc

s

Wt

3

42

1

The T-s diagram of simple turbine plantis-The work ratio:

WR = net t c c

t t t

W W W W1= =W W W

For increase in work ratio, compressorwork Wc should decrease and turbinework should increase. For reduction in compressorwork ‘Wc’ : The one step compression(1–2) should be done in steps and thereshould be cooling of gas between stepsof compression. This process is called

intercooling. This intercooling brings theadiabatic compression close to isother-mal i.e. process for minimum compres-sion work.For maximum work in turbine ex-pansion ‘Wt’ : Total adiabatic expan-sion (3–4) is done in steps and there isheating between two successive expan-sion. This process is called reheatingand gives maximum turbine work Wt.Regeneration improves cycle efficiencyreducing heat rejected and heat added.The physics behind this is reducing lossof available energy or reducingirreversibilities involved in cycle.

Sol–21: (c)Assuming equal pressure drop in eachstage, total pressure will be equal toproduct of pressure drop in one stageand number of stages.Steam turbine cycle is more efficientthan Gas turbine.Efficiency of steam turbine, = 35%Efficiency of gas turbine, = 25%.The mean temperature of heat additionin gas turbine is lesser than steamturbine so, efficiency is more for steamturbine. Reason is use of compressor ingas turbine which consumes largeamount of power and steam turbine usespumps for increasing pressure of satu-rated liquid and consumes negligiblepower as compared to compressor.The degree of reaction is the ratio ofenergy due to reaction to total energytransfer in stage.

Sol–22: (c)

1

2

3

4

5

6

P2T

s

24 6

P0

P1

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From figure it is clear that the expan-sion is in two stages i.e. process 3–4and process 5–6. In between these twostage expansion, process 4–5 is reheat.But the adiabatic exit from compressionis state 2 and there is no other pointbetween 2 and 3. Another indication istemperature at 6 is lower than tempera-ture at 2. So no regeneration.

Sol–23: (a)

T

sx

4

31

2

Q1

P = constQ

The reheating in gas turbine power plantcan be represented on T-s diagram byReheating process 2–3Power output without reheatingW = cp(T1 – Tx)Power output with reheatWr = cp(T1–T2) + cp(T3–T4)But Wr is always more than W. So re-heat is provided to increase turbineoutput.

Sol–24: (b)

T5 bar

1 bar 2

s

T =1000k1

Specific heat of gas,

pc 1.0425kJ / kgK

vc 0.7662 kJ / kgK

Heat capacity ratio, p

v

c1.36

c

Temperature after expansion insideturbine,

1 21 1

1 2

T TP P

1 361 362

1 36 1 1 36 1T1000

5 1

0 361 36

21T 10005

= 653 K

Power developed by turbine,

Work done 1 2h h p 1 2c T T

1 0425 1000 653 361 64 kJ / kg

Sol–25: (d)The variation of shear stress in varioustypes of fluids.

Bingham pl

astic

Shea

r str

ess

Non-Newtonian

Newton

ian

Ideal fluid

Velocity gradientDrilling mud and sewage sludge areBingham plastic.

Sol–26: (c)

50cm

56cm45cm

atmBenzene

sp. g.=0.88

Mercurysp. g = 13.6

Water

Taking reference level of pressure of leftlimb of mercury,

P + 0.56g = 0.5 × 0.88g + 0.45 × 13.6 g

Pρg = (0.5×0.88 + 0.45 × 13.6 – 0.56)

= 6 m of waterSol–27: (c)

weight of Cube = weight of volume ofmercury displaced

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(1 × 1 × 1) × 2.26 × g= (1 × 1 × h) × 13.6 × g

h = 2 26 0 166=13 6

Sol–28: (b)

W

BW

WS

Weight of body in submerged condition,WS = W – B

WS = W V g

1. For 30 N

30 = W – V × 0.8 g ...(i)2. For 15 N

15 = W – V × 1.2 g ...(ii) Subtracting these equations

15 = V g 1.20 0.8

V = 315 m1000 0.40 9.81

= 3.823 × 10–3 m3 = 3.823 litSol–29: (c)

Since it is given that flow A and B exist.It irrotational must exist i.e.

x

2 2=

y x y

( u) = ( v)

y x–1 = ±3/2

Hence flow A is rotational.For flow B

2 2(xy ) = (x y)y x

2xy = 2xyIrrotational.

Sol–30: (d)P = 380 Pa

P = 21 V2

V = 2 P

= 2 3801.2 = 25.17 m/sec

Sol–31: (b)When pipes are in series, the equivalentcondition,

hfe = 1 2f fh h

Same discharge in both pipes in thiscondition

2

12 5

32fL Qg d

= 2 2

2 5 2 532fL Q 32fL Q

g d g d

L1 = 2L ...(i)Now pipes are arranged in parallel. Inthis condition, the discharge in both pipeswill be equal to equivalent pipe. Sincediameter is same, so each pipe carry halfof total discharge

L,d

L,d

Q

Q

feh = 1 2f fh h=

222 5

2Q32fLg d

= 2

2 532fL Q

g d

L 2 = L4 ...(ii)

1

2

LL = 2L 8 :1=

L / 4

Sol–32: (d)

Uyu U

=

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u yU

Momentum thickness of Boundary layer

= 0u u1 dyU U

= 0y y1 - dy

2 3 6

Displacement thickness of Boundary layer

* =0

u1 dyU

=0

y1 dy

= 2

16

* 32

Sol–33: (d)Shear stress distribution in pipe,

=r dP2 dx

r = Distance of layer from center(at center, r = 0).

= 0as shown below

U

Sol–34: (a)Because of boundary layer formationroughess is neglected friction factor inlaminar flow

f = 64Re

In fully developed turbulent flow in roughpipe the friction factor 'f' is independent ofReynold number.

Sol–35: (c)Various aspects of Kaplan turbine are-(i) It is a low head, high discharge

reaction turbine

(ii) It is an axial flow turbine not mixednot radial.

(iii)It's blades are adjustable i.e. pitch ofblades, blade inlet and outlet angle areadjusted according to varied dischargeand head. This adjustment is done formaximum efficiency over wideoperating conditions.

Sol–36: (c)Negative slip, i.e. theoretical discharge isless than actual discharge, occurs whenlength of suction pipe is large and deliverypipe small. This happens specially whenspeed of pump is high. The reason for thisis the inertia of water in suction pipe islarge as compared to pressure at deliveryvalve which open it.

Sol–37: (c)The schematic of wall,

200°C

300°C

t = 0.5 m

Wall

The conductivity varies linearly withtemperature with temperature in wall. k = a + bTAt T = 300°C

k = 25 W/mK25 = a + 300b

At T = 200°Ck = 15 W/mK

15 = a + 200bUpon solving these two equations,

a = – 5b = 0.1

Thermal conductivity,k = (–5 + 0.1 T)

We know that steady state heat flux isconstant. So, from Fourier’s law,

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q = dT dTk ( 5 0.1T)dx dx

(5 – 0.1T) dT = q·dxIntegrating it.

200 0.5

300 0(5 0.1T) dT q dx

2 20.15(200 300) (200 300 )2

= q × 0.5

– 500 + 2500 = q2

q = 2000 × 2= 4000 W/m2

= 4 kW/m2

Sol–38: (d)1. Since the fins are not so effective on

hot fluid side because heat has to flowtoward base and against temperaturegradient.

2. Statement ‘2’ is clear correct and doesnot require any explanation.

3. Higher thermal conductivity ensuresuniform temperature throughoutlength of fin and more heat loss. Sohigher thermal conductivity of finmaterial than wall material is correct.

4. Since fins are provided on lowerconvection coefficient side i.e. naturalconvection. So right angle direction offlow is not correct. But this does notmean that fins are not provided onforce convection side.

Sol–39: (b)Since convection problem have bothReynolds number and Prandtl number,so the problem is forced convection andReynolds number.

Re = V

x

For very higher value of Re, shouldbe very low and should be high.Prandtl number is very less–

Pr = pck

Hence ‘k’ is very high. All these factorsindicates that the fluid is liquid metal.So the thermal and hydrodynamicboundary layer relation.

t = h0.33Pr

Pr < < 1 t > h

Hence thermal boundary layer thicknessis more than hydrodynamic boundarylayer.

Sol–40: (c)For free covection over vertical the plateNusselt Number,Nu = 0.54 (Gr.Pr)1/4 for laminar condition.For turbulent condition

Nu = 0.14 (Gr.Pr)1/3

Nusselt number varies as Gr1/4 forlaminar free convection and Gr1/3 forturbulent free convection.

Sol–41: (a)

d = 100 mmSteam

The Nusselt number,Nu = 25

Nu = hdk

25 = h 0.10.03

h = 25 × 0.030.10

= 7.5 W/m2KSol–42: (c)

The equality condition of counter flowconfiguration of HE,

Ch = CcEffectiveness in counter flow configuration.

= NTU1 NTU

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NTU = 0.5 (given)

= 0.5 1 0.331.5 3

Sol–43: (b)Gas to gas counter flow heat exchangertemperature profile can be depicted below.

Cold

Hot

Length

T

Some its feature are–1. Hot gases temperature reduces down

the flow.2. Cold gas temperature increases up the

flow.3. Both temperature profile are very

similar.Sol–44: (b)

R

1

2

In this configuration, view factor of planesurface with respect to itself,

F22 = 0 (Flat surface)F21 = 1

Theorem of reciprocity,A1F12 = A2F21

F12 = 2 21

1

A AA

=2

2r 1

2 r

= 12

Sol–45: (d)Total emissive power of diffused body,

E =0

E .d

=3 m 12 m

0 3 m0.d 150d

25 m

12 m 25 m300d 0.d

= 0 + 150 × (12 – 3)+ 300 (25 – 12) + 0

= 150 × 9 + 300 × 13= 1350 + 3900 = 5250 W/m2

Sol–46: (c)This is the case of heat exchanger invapour compression cycle.

s

T

Com

pres

sor

33

2

14Evaporator

1

Given,

1h = 182 kJ/kgh3 = 78 kJ/kg

h3 = 68 kJ/kgh2 = 230 kJ/kg

Heat Loss = Heat gain by HE

3 3h h = 1 1h h

h1 = 78 – 68 + 182 = 192 kJ/kg

COP = 1 31 4

2 1 2 1

h hh hh h h h

=182 68230 192

= 3

Sol–47: (d)

2

Compre

ssor

3

h

pCondenser

Evaporator4

1

Given,h1 = 195 kJ/kgh2 = 210 kJ/kgh3 = 90 kJ/kgh4 = 90 kJ/kg

m = 0.5 kg/sHeating Capacity of heat pump

Q1 = m (h2 – h3) = 0.5 (210 – 90)

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= 0.5 × 120 = 60 kWSol–48: (d)

Azeotropes are mixture of refrigerantssuch that constituents of mixtureshould not separate out duringtemperature variation and behave likepure substance.

Isomers have same chemical formulabut different molecular structures.

The formula n + p + q = 2m is usedfor unsaturated Hydrocarbons

Sol–49: (d)All the statements are wrong becausevertical lines denotes DBT, horizontal linedenotes spcific humidity not-relativehumidity. Sensible heating or cooling isrepresented by horizontal line.

Sol–50: (d)Sensible heating,

DBT

HumidityRatio

( )

1 2

100%

RH

80%

RH

40%

RH

The following conclussion can be madefrom psychrometric charts. Relative humidity decreases Humidity ratio remains constant Specific enthalpy increases Wet bulb temperature increases

Sol–51: (b)The cooling coil and various temperature,

CoolingCoil

T =5°Cs T2

T =27°C1

Bypass factor,

BPF = 2 s

1 s

T TT T

0.4 =

2 2T 5 T 527 5 32

T2 + 5 = 0.4×32 = 12.8

T2 = 12.8-5 = 7.8°CSol–52: (c)

Given,

am = 10 kg-da/s

apc = 1.02 kJ/kg da Khfg = 2500 kJ/kg v

2

DBT

Humidity Ratioin kgv/kg deg.

2= 0.008

1 = 0.017

t = 16ºC2

1

LH

SH

t = 30ºC1

Sensible heat,SH = am apc t

= am apc (t1 – t2)= 10 × 1.02 × 14 = 142.8 kJ/s

Mass of water vapour,

vm = a 1 2m ( ) = 20 (0.017 – 0.008)= 0.09 kg v/s

Latent heat in kW,

LH = 0.09 kgv kg2500

s kgv

= 225 kJ/sSol–53.: (a)

Q1

WInput

Q = 2.4 kW 2

T = 324KH

R

T = 243KL

Given,

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act.

Re v

(COP)(COP)

= 0.8

(COP)act = 0.8 × (COP)Rev

2

Input

QW = 0.8 × L

H L

TT T

input

2.4W = 0.8 ×

243 2.4324 243

Winput = 1 kWSol–54: (b)

The Inside temperature of the room will bemaintained at 20ºC only when the heatpumped by pump is equal to the heattransfer through the wall (i.e. Q1 = 7.9 kJ/s)

W

Q = 7.9 kJ/s2

T = 20ºCH

HP

T = 5ºCL

Q2

(COP)max = H

H L

TT T

1

min

QW =

293293 278

Wmin = 15 7.9893

= 0.4044 kJ/s

= 404.4 WSol–55: (d)

1. Because of a regular flame front insidechamber of SI engine duringcombustion, the temperature of endgases is very high so these gases areprone to detonation. But there is nosuch front in CI engine and detonationoccurs as combustion start.

2. It is CI engines which do not face theproblem of pre-ignition not SI engine.

3. Low inlet pressure and temperature

reduces over all temperature ofcombustion so the detonation in SIengine. But at the same time lowtemperature reduces rate ofevaporation in CI engine fuel and largequantity of fuel is accumulated in thecylinder before combustion start. Theburning of this large quantity of fuelis detonation in CI engine.

Sol–56: (b)Physical delay means physical prepara-tion of mixture e.g. mixing, heating, fuelvapourizing, atomising and how muchfast this occure. From given factors,compression ratio and turbulence (orswirl) are the only factors which en-hances this physical preparation fast i.e.reduced physical delay.Cetane number is fuel characteristicwhich influences chemical delay.Increased injection advance meanslonger physical delay because it startsfrom start of injection and increasedknock.

Sol–57: (c)In the rating of SI engine fuel, iso-oc-tane is assigned 100 octane number andnormal-heptane as zero octane number.Both of these are reference fuels. Forexample if a fuel has 85% iso-octane and15% normal heptane the octane numberof fuel will be 85.

Sol–58: (a) Supercharging is mandatory for air

craft engine because density of airvaries with altitutde.

Morse test find out friction losses inmulticylinder engine.

Heterogeneous combustion is indiesel (CI) engine because air andfuel injection are separate processes.

The ignition quality of petrol isdirectly related to octane number i.e.high octane number (high selfignition temperature) means longerignition delay.

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Sol–59: (a) Use of high compression ratio and

high degree of supercharging resultsin high temperature during combus-tion so high NOx emission.

It is reduction catalyst (rhodium)which control NOx not oxidation cata-lyst.

So the only way to control NOx iscatalytic conveter in the given op-tions.

Sol–60: (c)

A. Bell Coleman Cycle: It is reverse ofBrayton cycle. So it consists of twoadiabatics and two isobar i.e. A-2.

B. Stirling cycle: It consists of twoconstant volumes and two isothermals.So B-4.

C. Ericsson cycle: It consists of twoisobars and two isothermals. So C -3.

D. Diesel cycle: It is consisting of twoadiabatics/and isentropics, one isobarand one constant volumes. So D-1.

Sol–61: (b)Flat plate collector does not requiredtracking of sun and mounted in stationaryposition.

Sol–62: (c)

Transmittance ( )r for system of Nidentical covers is given as:

r =1 r

1 (2N 1)r

= 1 0.05

1 2 2 1 0.05

r = 1923

Sol–63: (b)

a = zkL/cose

From above expression, k should bedecreased in order to increase value of a.

Sol–64: (a)All above statements are correct.

Sol–65: (c)

P =24f L V

2d

P V2

Sol–66: (c)FR = F F FR = 0.8 × 0.9FR = 0.72

Sol–67: (d)Temperature range over which systemoperates should be minimum.

Sol–68: (c)Sensible heat storage system cannot deliverenergy at constant temperature.

Sol–69: (a)A = Volumetric strain

VV

= y z 1 2E

x

= p 1 2E

B = Strain energy per unit volume

B = 1 stress strain2

= 2p

2E C : E = 3K 1 2

EK = 3 1 2

D : E = 2G (1 )

Sol–70. (d)A DB C

365 450 13045

FBD

A B B C C D45

45

365

320 320

450

13045

Thus BC has 320 (Compressive) force.

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Sol–71. (a)

2 kN

0L

RA = 2 × = 0.8 kN25 RB = = 1.2 kN

25 × 2

2 m 1 m 2 m

M N

B.M. at L = RA × 0 = 0B.M. at left of M = AR 2 0.8 2 1.6kNm = =B.M. at right of M = 1.6 + 2 = 3.6 kNmB.M. at 0 = BR 0 0 =

Sol–72. (d) Beam AP

L/2 L/2

max =3PL

48EI

max A

max B

( )( ) =

3

3

PL48EI5 PL

384 EI

= 85

Beam Bw per unit length

L/2 L/2Given : P = WL

W =PL

max =45 WL

384 EI = 35 PL

384 EI

Sol–73. (b)

We know that1 + 2 = x + y40 + 2 = 32 – 10

2 = –18 MPa

Sol–74. (c)

A/c to maximum normal strain theory.Max. normal strain < fy/ECase (i) Tensile (fy) and 1 > 2

1 2E E

< yfE

180 + 0.25 2 < 2402 < 240 N/mm2

Case (ii) Compressive (fy) and 1 < 2

2 1E E

< yfE

2 < 240 – 0.25 × 1802 < 195 N/mm2

We will take lesser value of case (i) & (ii)So ans. (c)

Sol–75. (c)

Hole should be made at the centre of beam(at neutral axis) otherwise strength of sec-tion will decrease more.

Sol–76. (d)

Flexural stress = MyI

For to be zero either M is zero or y iszero or both are zero.In simply supported beam carry in udl,bending moment is zero at supports.

= 0 at support and at neutral axisy = 0

=M 0I

= 0

Shear stress =VAyIb

If V = 0 (SF is zero at mid span) = 0

If Ay = 0 ( Ay is zero at top and bottom fibre)

= 0Hence both flexure stress and shear stress arezero.(i) At mid span section at neutral axis.(ii) At support section at bottom fibre.

Sol–77. (d)Shafts are joined in series so applied torquewill be same on both the shafts (T1 = T2).From torsion formula

r =

T GJ L

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=TLGJ

1

2

= 1 1 1 1

2 2 2 2

(T L /G J )(T L /G J )

Given L1 = L2 and for same material G1 = G2

1

2

=

42

41

J d /32 1J 16(2d) /32

1

2

= 1:16

Sol–78. (a)

According to maximum shear stress theoryFactor of safety =

Yield stress

2Absolute Max.shear stress

=

y

1 2 1 2

f2

p p p pmax , ,2 2 2

=

2402

80 20 80 20max , ,2 2 2

= 12040 = 3

Sol–79: (c)

ap =m g2 2 p =m g1 1

a

Acceleration of each mass is given by

a = 1 2 1 2

1 2 1 2

(p p ) p p g(m m ) p p

The acceleration of centre of mass is givenby

CMa = 1 1 2 2

1 2

(m a m a )(m m )

= 1 2 1 2

1 2 1 2

p a p a p p a(p p ) p p

= 2

1 2

1 2

p p gp p

Sol–80: (a)Force of friction between B and surface

is :

B 2 1 2f (m m )g =

0.5 (2 8) 10 50 N

m =8kg2 B25 N

m =2kg11=0.2

1=0.5

A

As applied force 25 N is less than thisforce of friction, the system has notendencyto move.Hence, friction between A and B is zero.

Sol–81: (b)

60°30 cm

36 cm

60 cm

36 cmA

B

CDVelocity of point A,

VA = DA ×

= 30 × 2 100

60

= 314.6 cm/secSol–82: (d)

The schematic of mechanism having slideron rotating link is

V

0

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The coriolis component of accelerationacting on slider,

= 2VSol–83: (c)

1. The trace point is reference point onfollower which describes pitch curveand movement of follower. In case ofroller follower, it is center of roller.

2. The curve generated by locus of tracepoint is pitch curve not cam profile.

3. The actual working curve is cam profilenot pitch curve.

Sol–84: (c)1. Sensitiveness of governor is defined as

how much movement is noticed insleeve/balls as speed changes.

2. Isochronous governor as infinitelysensitive because,

Nmax = Nmin

Sensitiveness-

=

max min

max min

N N2 N N

3. Hunting is obtained because of veryhigh sensitivity.

Sol–85: (b)

BO

Ac=nr

Angular velocity of connecting rod,

c = cos4

Max. value =2 3000 cos0

60 5

=2 3000 cos0

60 5

= 78.5 rad/secSol–86: (c)

1. Flywheel stores kinetic i.e. rotationalkinetic energy.

2. Governor controls the speed variationswith load variation.

3. Lead screw in lathe is used to controlthe movement of tool carriage e.g. inthread cutting etc. not to feed.

4. Fixture only locate the work piece andit is Jig which locates the work pieceand guide tools.

Sol–87: (a)1. The pressure angle is constant

throughout the engagement of matinggears in involute profile gears. It isvariable in cycloidal profile teeth.

2. The profile of teeth is single curve i.e.involute curve. Any normal to involuteprofile is tangent to base circle ofinvolute profile. So both face and flankare made of single curve i.e. involute.

3. The centre distance of mating gearscan be varied within the limit to avoidinterference and velocity ratio does notchange.

Sol–88: (b)

A B

Casing

C

The speed of casing ‘C’ is arithmaticmean of speed of wheel A and B at turn

Speed of C = speed of A + speed of B2

Sol–89: (b)Acceleration of reciprocating mass

a = 2 cos2r cosn

Inertia force = acceleration × massF = ma

= 2 cos2mr cosn

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=2

2 mrmr cos cos2n

= FP + FS

Primary disturbing force-

FP = 2mr cos Secondary disturbing force-

FS =2mr cos2

n

Sol–90: (c)110 cm

N

x1 x2

I1 = 100 kg.m2I2 = 10 kg.m2

x1 + x2 = 10At node N, no rotation i.e. it can beassumed fixed. So natural frequency of bothpendulums around node is same.

1 = 2

1 1

GJx I =

2 2

GJx I

where ‘J’ in moment of inertia of rodand I moment of inertia of discs. x1I1 = x2I2

x1×100 = x2×10x2 = 10x1

x1 + x2 = 110x1 + 10x1 = 110

x1 = 110 = 10cm11

x2 = 100 cmSol–91: (a)

62.5100 = P 3

P

0.625 = 31P

P = 3 8cm1 0.625

Sol–92: (c)

Qint = 2i 8i 1 3

6i = 8i – 88 = 2ii = 4

Qext = 2i 2 4i 1 4 1

= 8/5Sol–93: (c)

250 = 4 t k

2

t + k = 5004

= 125 ...(i)

2ii 1 = 1.6

2i = 1 – 6 i + 1.6

i = 4 or t 4 or t 4kk

5k = 125 or k = 25and t = 100

Sol–94: (d)Sol–95: (d)

Flexible shafts are restricted to low power.They have low drive rigidity in bendingmaking them flexible.They have high rigidity in torsion makingthem capable to transmit torque.

Sol–96: (c)During the braking action the mass movesthrough a distance of 0.5 m, the anglethrough which drum rotates during thebraking period is

= Distance movedradius

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= 0.5 50 2 rad0.75 75 3

Energy = Torque

10,000 = T 2 , T 15000 N.m3

Sol–97: (b)

CB = 1/310P L

CR = 3

1010P L

B

R

CC =

313 1010L

CB = 1

30R 10C L

= 26 × 103/30

= 26 × 101/10kNSol–98: (d)

Option d is the result of stribeck’s equationand not the assumption made to derivethe result.

Sol–99: (a)

P

w d = e

s

Sf

Hence t =

3s

e

P f 30 2 10w d S 50 10 50.9

= 29.47 mmSol–100: (b)

Simple cubic: Manganese

Body centered cubic: Iron

Face-centered cubic: CopperHexagonal closed packed: Zinc

Sol–101: (d)A. Isomorphous system: Two met-

als are completely soluble in solidand liquid state.

B. Eutectic system: Complete solubil-ity in liquid state and only partialsolubility in solid state.

C. Perutectic system: Liquid + Solid1 Solid2

D. Monotectic:Liquid1 Liquid2+Solid

Sol–102: (a)A. Chromel: Thermocouple wireB. Babbitt Material: Journal bearingC. Nimonic Alloy: Gas turbine bladeD. High Speed Steel: Milling Cutter

Sol–103: (d)Normalizing is a heat treatment of steelswhere specimen is heated above uppercritical temperature and cooled in air.Normalizing relieves stress, increasesstrength and results in better surfacefinish while machining. Normalizingproduces fine pearlite and it reduces thegrain size.

Sol–104: (b)A. Water produces Martensite structureB. Oil produces very fine pearliteC. Air produces fine pearliteD. Furnace cool produces coarse pearlite.

Sol–105: (c)A. Blade of bulldozer: High wear and

abrasion resistance.B. Gas turbine blades: High creep

strengh and good corroson resistance.C. Drill bit: High wear resistance.D. Spring of automobiles: Low

young’s modulus and high fatiguestrength.

Sol–106: (c)Gray cast Iron gives best performance invibration damping when used in basestructures.

Sol–107: (c)Natural resins, polystyrene and polyvinylchloride are thermoplastics phenolformaldehyde is a thermosetting polymer.

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Sol–108: (d)Squeeze casting involves solidification ofthe molten metal under high pressure.The steering knuckle is made up of squeezecast aluminium.

Sol–109: (a)Filler metals-the metals added to the weldarea during welding. These may or maynot be used.

Sol–110: (c)In brazing, the filler metal placedb/w the two surfaces to be joined,temperature is raised such that the fillermetal metlts but not the workpiece so thatthe molten metal fills the closely fittedspaces by capillary action.(A) Arc welding: fusion welding(B) Resistance welding: solid state welding(C) Brazing: capillary action(D) Adhesive bonding: joining plastics

Sol–111: (b)The filler metal is called braze metal incase of brazing and solder in case ofsoldering.

Sol–112: (a)The cutting ratio is given by:

r = 0

c

tt

t0 depth of cuttc thickness of chip

1

= compression ratio

So, 1

= c

0

tt

Since chip thickness is always greater thanthe depth of cut, hence compression ratiois always greater than unity.

Sol–113: (a)As the friction at the tool chip interfaceincreases, the shear angle decreases andchip becomes thicker.

Sol–114: (c)

t = lf N = 70

0.5 400

= 0.35 min. = 0.35 × 60= 21 sec.

Sol–115: (c)(A) Tracer lathes-contousing lathes(B) Automatic lathes-Chuckers(C) Automatic bar machines-Similar

threaded parts(D) Turret Lathe-Capstan wheel

Sol–116: (c)The grinding wheel comprises of bondedabrusives which perform the finishingoperation. An abrasive is a small, non-metallic hard particle having sharp edgesand an irregular shape.

Sol–117: (d)Dividers are without any graduated scales.They are used to transfer the measuredsize to a direct reading instrument suchas rule.

Sol–118: (c)(A) Boring-Internal surfaces or profiles(B) Drilling-Round holes of various sizes(C) Broaching-Slots and contours(D) Sawing-Straight and contours

Sol–119: (d)A microprocessor is ALU (Arithmetic logicunit) with register unit and control unit on asingle chip.

input output

Memory

Microprocessor

RegisterUnit

Sol–120: (c)Program Counter (PC) is a 16-bit special-purpose register used to hold the memoryaddress of the next instruction to be executed.During the execution of an instruction, themicroprocessor increments the content of theprogram counter so that it points to the addressof the next-instruction to be executed.

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Sol–121: (b)In 8085 lower order address bus and data busare multiplexed to reduce the no. of pins anddemultiplexing is done using ALE pin. WhenALE = 1 then lower order byte work as addressand when 0, it works as data.

Sol–122: (d)Force

analogy

Displacement(x)

Damper (B)

Spring Constant (K)

Mass (M)

Force (F)

Voltageanalogy

Currentanalogy

Voltage (V) Current (I)

Inductor (L) Capacitor (C)

1/C 1/L

R 1/R

Charge (Q) Flux ( )

Sol–123: (c)Since interrupt priority order is given asTRAP > RST 7.5 > RST 6.5 > RST 5.5The highest priority device should be relativelyconnected to higher priority interrupt. HenceP – RST 7.5Q – RST 6.5R – RST 5.5

Sol–124: (d)When the pop instruction is executed, two databytes from stack, transferred back to HLregister pair and stack pointer is incrementedby two.

Sol–125: (d)– Hydraulic system has high force to weight

ratio. It means, a small size actuator candevelop a very large force or, torque.

– In hydraulic system, power is transmittedthrough the action of fluid flow pressure. So,no source with supply and return line isrequired.

– Hydraulic systems are sensitive totemperature changes. It has tendency tobecome sluggish at low temperaturesbecause of increase viscosity of the fluid.

Sol–126: (c)• Sensitivity of closed loop negative feedback

control system is less than sensitivity of openloop control system.

• The overall gain of closed loop system withnegative feedback reduces by a factor of

1

1 G Hs s

.

Sol–127: (b)

R CG1 G2

H1

H2

+ +

G1eq

G1+

H2

G1eq C–

G1eq = 2

2 1

G1 G H

CR =

1 1eq

2 1 1eq

G G1 H G G

=

1 2

2 1

1 2 2

2 1

G G1 G H

G G H11 G H

= 1 2

2 1 1 2 2

G G1 G H G G H

CR = 1 2

2 1 1 2 2

G G1 G H G G H

G1 = 2 1 210 10,G ,H s 3,H 1s s 1

=

10 10.s s 1

10 10 101 .(s 3) . .1s 1 s s 1

=

100s(s 1)

s(s 1) 10s(s 3) 100s(s 1)

= 2 2100

s 1s 10s 30s 100

= 2100

11s 31s 100

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Sol–128: (a)

R(s) C(s)s + 2s + 1

Figure - I

Figure - II

R(s)1

s + 1X

+C(s)

+

from Fig – I

C(s) =s 2R(s).s 1

...(1)

From Fig. – II

C(s) = XR(s) R(s)s 1

C(s) = XR(s) 1s 1

... (2)

From equation (1) and (2),

s 2R(s).s 1 =

s 1 xR(s)s 1

X = 1Sol–129: (c)

Fn+1 = n nD 1 F

= 0.2 × 450 + 0.8 × 500= 490 units

Sol–130: (b)Total cost for first method = fixed cost +variable costLet x be the quantity produced

(TC)1 = 2000 + 20xSimilarly, total cost for second method =1500 + 30xBreak even quantity is that quantitywhere total cost of both methods becomesequal.

2000 + 20x = 1500 + 30x10x = 500

x = 50 unitsSol–131: (a)

Indirect cost are fixed costs that cannotbe directly linked to the product such asinsurance cost etc. = Rs. 4,00,000Direct cost of product is variable cost =Rs. 20/unit.Selling price/unit = Rs. 60

BEP = 40000060 20

= 10,000 products

Sol–132: (d)To solve a LPP having three variable, werequire three set of constraints. Thus, athree variable and two constraint LPPcannot be solved by graphical method.

Sol–133: (c)

1 21 2

E 8 E 32L 26 L 37

Earliest Latest ActivityEvent time time duration1 8 weeks 26 weeks 11 weeks

2 32 weeks 37 weeks (1) (2) 11

for activity 1 – 2 total float is= (L2 – E1) – D12 = (37 –8) – 11 = 18weeks

Sol–134: (d)The system capacity would be equal tothe capacity of the slowest machine i.e.the machine which takes maximum timeto process the Job. In this case it is 3rdmachine processing 160 products per day.

Sol–135: (d)Nondestructive testing is done when

• Testpiece too precious to be destroyed• Testpiece to be reuse after inspection• Testpiece is in service• For quality control purpose• Testpiece cannot be harmed

Sol–136: (b)Rsystem = R1e × R2e × R3e

R1e = 1 – (1 – R1) (1 – R2)= 1 – (1 – 0.7) (1 – 0.7)= 1 – 0.09 = 0.91

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R2e = 1 – (1 – R3) (1 – R4)(1 – R5)= 1 – (1 – 0.8)(1 –0.8) (1–0.8)= 1 – (0.2 × 0.2 × 0.2)= 1 – 0.008 = 0.992

R3e = R6 = 0.90 Rsystem = 0.91 × 0.992 × 0.90

= 0.8125Sol–137: (b)

Basically thermodynamic system is a fixedquantity of working substance upon whichheat and work interactions are studied.According to first law, heat and work arecompletely mutually convertible. Butaccording to second law ofthermodynamics, work can be convertedcompletely into heat, but heat can not beconverted into work completely butpartially. So reason (R) is also correct.

Sol–138: (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 of superheatedsteam is approximately same as saturationenthalpy. But in high pressure zone (nearcritical point ‘CP’) the enthalpy initiallyincreases at constant pressure and thenconstant.

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–139: (b)The index of compression and expansionin reciprocating compressor are differ-ent but in numerical problems both aretaken same for calculation purposesonly.It is well known facts that reciprocat-ing compressors are used for high pres-sure ratio and low discharge. The rea-son being use of non reversal valves.

Sol–140: (d)

Reheating is preferred for higher poweroutput not for efficiency.

T

y4

QA

x2

1s

3

wf

wc

Thermal efficiency

th = t c

A

W WQ

QA can be reduced if some heat fromprocess 4-y-1 is shifted to process 2–x–3 i.e. regeneration.

Sol–141: (d)To have neutral equilibrium, the centerof gravity and center of buoyancy mustcoincide. This happens only whenspherical body is submerged andhomogeneous. So spherical floating bodycan not be always in neutral equilibrium.The possibility of center of gravity abovecenter of buoyancy is in homogeneousspherical floating body.

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Sol–142: (a)

h

1 2

V P

In pitot tube, the velocity head 21 V

2 gchanges to potential head h. Bernoulli’s theorem between 1 and 2.

2P V P hg 2g g

V2 = 2ghSol–143: (a)

For gases (fluids), the Prandtl Number,

Pr = 1Now from Reynold analogy, StantonNumber

St = fC2

Thermal diffusivity, = Kinemetic viscosity

Pr = 1

t = hh0.33Pr

Hydrodynamic and thermal boundarylayers are same. In Reynolds analogy theunity condition of Prandtl number isassumption.

Sol–144: (b)

1hT

c2T1

c1T2

2hT

Because of uniform temperature differencealong length of heat exchanger in case ofcounter flow, LMTD is more for same inlet

and outlet temperature. So the counterflow heat exchanger for same heat transferare short in length.The basic definition of LMTD–

= 1 2

1

2ln

where,

1 - Inlet temperature difference of flu-ids.

1 = 1 2h cT T

2 = Outlet temperature difference offluids

2 = 2h c1T T

Sol–145: (d)

Specific humidity or humidity ratio is theratio of mass of water vapour to the massof dry air not mass of moist air. Hencehumidity ratio,

= v

a

mm

or = v

0 v

P0.622P P

The above expresion can be expressed as,

and = v 0f(P ) P is constant

= f(Ts)

So reason is correct and assertion iswrong.

Sol–146: (c)The definite amount or slight super heatis always desirable because wetcompression represent incompletevapourization of refrigerent, hence liquidrefrigerant may damage compressor valvesand lubricant can washout. Work inputto the compressor increases with theSuperheat because the temperature atthe inlet to the compressor increases.

W =

1

21

1

PmRT 11 P

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Sol–147: (d)Knocking in petrol engine is reduced byrich mixture because flame temperaturecan be kept low thus eliminatingconsiderably tendency of knock. So,assertion is wrong.Knocking increase by compression ratiobecause by high compression ratiodensity of end charge increase hence thetemperature.

Sol–148: (d)No, the working cycle in an IC engine isnot air standard cycle. The air standardcycle is very fundamental cycle using idealgas (air) as working medium with someassumptions. In IC engines, it is actualcycle which is made after suitablemodifications.Since the chemical reaction of nitrogenwith oxygen is negligible and assumed tozero. So nitrogen does not undergo anychemical reaction during combustion.

Sol–149. (b) From torsion formula

r =

T GJ L

=T r

J

d2ds

d1

They have same weight Area of x-section will be same for the two

shaft

2Sd

4 =

2 21 2(d d )4

2Sd = 2 2

1 2d d ... (i)From (i) we can say that d1 > dS

Assuming same max. shear stress in both case

h

s

TT =

hmax

1

Smax

S

J(d / 2)

J(d / 2)

h

s

TT = sh

s 1

dJJ d

h

s

TT =

4 4s1 2

s 1

dd dd 4 d

h

s

TT =

3 41 2 1

3s

d d / dd

...(ii)

From (i) & (ii)We can say that

Th > Tsalso power transmitted by shaft

P = TP T

So Ph > Ps

Sol–150. (a) For bending moment to be maximum,

dMdx = 0

We know, V =dMdx

dMdx = V = 0

For maximum bending moment,shear force is either zero or changes sign(means its value will be zero at once atleast)

dMdx = V

M = Vdx If shear force diagram is horizontal – BM

diagram will be inclined. If shear forcediagram inclined – BM diagram is aparabola of second degree.Reason is true and satisfy the condition thatdM V=dx , hence reason is correct

explanation of Assertion.Sol–151. (a)

Cast iron is brittle material, which failsdue to normal stress. In tension testmaximum normal tensile stress is on across-section perpendicular to the axis ofspecimen, so failure will take place on thisplane.

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Sol–152: (a)The flat faced follower experiences low sidethrust in bearing as compared to knifeedge or roller follower. This side thrust isfurther reduced by proper offsetting thefollower. If this offsetting is in direction ofperpendicular to plane of rotation therewill be some rotation in follower duringtranslation which further reduces slidingfriction in bearing.Rotating follower along with translatingmotion in cam mechanism is shown below:

Sol–153: (b)Both the statements are self explanatoryand correct. Assertion (A) defineshammer blow and Reason (R) describeseffect of hammer blow. But reason (R)is not the explanation of assertionbecause both are self sufficient.

Sol–154: (b)During starting and stopping of journalbearing the speed of bearing is low andhydrodynamic pressure generated byrotation will not be sufficient to supportthe load. So there is metal to metal contactand bearing may damage so to avoid thisdamage the hydrostatic bearing is providedfor these operating condition of Journalbearing.

Sol–155: (b)More is length of arc having oil film, lessis chances of air entrapment in oil. So to

avoid air in lubrication, the oil inhydrodynamic bearing is provided atpressure more than atmospheric pressure.In the converging part of Journal bearing,the oil is pressurised due to Journalrotation to form pressurised oil wedge.

Sol–156: (b)A poly crystal is not a single crystal butgrains of individual single crystallites,which are very small in size. Defects aremore likely in larger grains rather thansmaller ones. Therefore, it is much hardto break a polycrystal due to small grains.

Sol–157: (b)

Rigid setups are necessary for interruptedcutting and difficult to machine materialsas it induces vibrations.

Sol–158: (a)Semi centrifugal casting uses a centralsprue and a core along with centrifugalforce action to produce castings.

Sol–159: (c)Straight polarity is used when less heatis to be generated at electrode (1/3rd oftotal heat), thus controlling rate ofmelting of electrode.Polarity does not affect stability of arc.

Sol–160: (a)PERT is prefered over CPM for the purposeof project evaluation because CPM usesdeterministic time estimates, which arenot possible under uncertainty conditionson the other hand PERT uses three timeestimates viz optimistic, pessimistic andmean time to compute project duration.

answers - ies masteriesmaster.org/public/archive/2016/im-1482668855.pdf · 2017-05-26 · r (2) me...

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