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24. (d)

25. (a)

26. (a)

27. (a)

28. (a)

29. (c)

30. (d)

31. (c)

32. (b)

33. (c)

34. (b)

35. (a)

36. (c)

37. (c)

38. (a)

39. (d)

40. (b)

41. (b)

42. (c)

43. (c)

44. (c)

45. (a)

46. (a)

47. (b)

48. (b)

49. (a)

50. (c)

51. (c)

52. (d)

53. (d)

54. (b)

55. (d)

56. (b)

57. (b)

58. (b)

59. (b)

60. (b)

61. (a)

62. (d)

63. (a)

64. (d)

65. (b)

66. (b)

67. (a)

68. (b)

69. (b)

70. (b)

71. (d)

72. (d)

73. (d)

74. (d)

75. (b)

76. (a)

77. (d)

78. (c)

79. (b)

80. (b)

81. (b)

82. (b)

83. (c)

84. (a)

85. (c)

86. (c)

87. (b)

88. (b)

89. (b)

90. (b)

91. (d)

92. (d)

93. (b)

94. (c)

95. (c)

96. (c)

97. (b)

98. (b)

99. (b)

100. (a)

101. (d)

102. (c)

103. (b)

104. (a)

105. (b)

106. (b)

107. (c)

108. (a)

109. (a)

110. (a)

111. (a)

112. (d)

113. (b)

114. (b)

115. (c)

116. (a)

117. (c)

118. (d)

119. (b)

120. (c)

121. (a)

122. (c)

123. (d)

124. (c)

125. (b)

126. (c)

127. (d)

128. (a)

129. (a)

130. (d)

131. (b)

132. (c)

133. (d)

134. (a)

135. (b)

136. (c)

137. (c)

138. (b)

ANSWERS

1. (b)

2. (b)

3. (d)

4. (a)

5. (c)

6. (d)

7. (c)

8. (c)

9. (d)

10. (c)

11. (a)

12. (a)

13. (a)

14. (a)

15. (a)

16. (a)

17. (a)

18. (a)

19. (a)

20. (a)

21. (a)

22. (a)

23. (a)

ESE-2017 PRELIMS TEST SERIESDate: 9th October, 2016

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(2) ME (Test-A2), Objective Solutions, 2nd October 2016

139. (a)

140. (c)

141. (b)

142. (c)

143. (c)

144. (b)

145. (a)

146. (b)

147. (a)

148. (b)

149. (a)

150. (d)

151. (d)

152. (d)

153. (c)

154. (a)

155. (a)

156. (b)

157. (c)

158. (c)

159. (d)

160. (d)

161. (b)

162. (d)

163. (a)

164. (a)

165. (a)

166. (a)

167. (d)

168. (c)

169. (c)

170. (d)

171. (a)

172. (a)

173. (a)

174. (a)

175. (c)

176. (b)

177. (c)

178. (d)

179. (d)

180. (d)

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Sol–1: (b)

= dudy

Sol–2: (b)

Ideal SolidIdeal Plastic F

luid

Non-Newtonian

Fluid

Newtonian Fluid

Ideal Fluid

Velocity Gradient (du/dy)

Shea

r Str

ess

A fluid which is incompressible and ishaving no viscosity is known as idealfluid. Ideal fluid is only an imaginary fluidas all the fluids, which exist, have someviscosity.

Sol–3: (d)The stability of a floating body isdetermined from the position of Meta-centre (M).In case of floating body, theweight of the body is equal to the weightof liquid displaced.Stable Equilibrium. If the point M isabove G, the floating body will be instable equilibrium. If a slight angulardisplacement is given to the floating bodyin the clockwise direction, the centre ofbuoyancy shifts from B to B1 such thatthe vertical line through B1 cuts at M.Then the buoyant force FB through B1and weight W through G constitute acouple acting in the anti-clockwisedirection and thus bringing the floatingbody in the original position.

W M

G

BFB

B1

DisturbingCouple

Unstable Equilibrium. If the Point Mis below G, the floating body will be inunstable equilibrium as shown in figure.The disturbing couple is acting in theclockwise direction. The couple due tobuoyant force FB and W is also acting inthe clockwise direction and thusoverturning the floating body.

W

GM

BB1

FB

Neutral Equilibrium: If the point M isat the centre of gravity of the body, thefloating body will be in neutralequilibrium.

Sol–4: (a)

F = gAh , for a submerged body,where A = area of surface

h = depth of centroidg = specific weight of liquid

Sol–5: (c)

The density of the solid piece = 8.25

Let the volume of solid piece = VHence, the weight of solid piece W =8.25 Vg .

Let the volume of solid piece submergedin mercury be V

Then, buoyance force FB = 13.6 V g

Since, W = FB.

Hence, 8.25 V g = 13.6 V g

V

V = 8.25 0.607=13.6

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Fraction of volume of solid piece abovethe surface of mercury = 1–0.607 = 0.393.

Sol–6: (d)For an incompressible fluid, if the flow issteady then continuity equation is givenby,

u vx y = 0 [for 2-dimensional flow]

u = c lnxy

ux =

c yxy

= cx

vy = c

x

v = cyx

Sol–7: (c)A ship or a boat may have two types ofoscillatory motions viz., rolling andpitching. The oscillatory motion of a shipor a boat about its longitudinal axis isdesignated as rolling. On the other handthe pitching movement or simply pitchingmay be defined as the oscillatory motionof a ship or a boat about its transverseaxis. It may, however, be noted that sincethe moment of inertia of the cross-sectionalarea of the ship or boat at the liquidsurface about its transverse axis is muchmore than the same about its longitudinalaxis, the metacentric height for thepitching motion is invariably greater thanthat for the rolling motion. As such if aship or a boat has a safe metacentricheight for rolling motion then it will besafe in pitching motion also. As regardsthe time period of oscillation it has beenfound that for rolling motion of ships orboats the the theoretical values of T givenby Eq. agree reasonably with theexperimental values, but the agreementis less good for pitching movements.In the case of ships the shifting of cargomay cause the ships to roll. As such alongwith the considerations of the stability ofa ship, its period of roll is also requiredto be determined. This is so because, asindicated earlier, increasing in the

metacentric height reduces the time periodof rolling of the body. A smaller value oftime period of rolling of a passenger shipis quite uncomfortable for the passengers.Further a ship with a smaller time periodof rolling is subjected to undue strainswhich may damage its structure. In thecase of cargo ships the metacentric heightvaries with the loading and hence somecontrol on the value of the metacentricheight as well as the time period of rollingis possible by adjusting the position ofthe cargo. However, in the case ofwarships and racing yachts, the stabilityis more important than comfort, andhence such vessels have largermetacentric height.

Sol–8: (c)

u =

212x 2

x x y

v = 2 3

1 22xy y y y

v = 32ˆ ˆ2i j

y

Sol–9: (d)Steady and Uniform flow have zero mate-rial acceleration .

Sol–10: (c)Cross-section area of block = A

Weight of block = Bouyancy force on it

A 5 x 5000 g = (5 + x 10)

× A × g × 1000 (5 + x) × 5 = (5 + 10x) 25 + 5x = 5 + 10x 20 = 5x x = 4cm

Sol–11: (a)

h =

4gd

d =

34 4 0.05gh 1000 10 1 10

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h = 0.02 mSol–12: (a)

d = 0.01 mm = 0.01 × 10–3mPressure outside = 10 × 104 N/m2

Surface tension = 0.05 N/mPressure inside droplet, in excess of out-side pressure

P =

34 4 0.05d 0.01 10

= 2 N/cm2

Total pressure = P + 10 = 2 + 10= 12 N/cm2

Sol–13: (a)

m mV g = w m0.90 V g

m = 1000 × 0.90

= 900 kg/m3

Sol–14: (a)Sol–15: (a)

= Vorticity × Areav = ˆ ˆ2xi 6xy j

Given 2 2x y 2ay = 0

22x 0 y a = a2

Radius = a

Area of circle = 2a

Vorticity = v u 6yx 2x 6yx y x y

= 26 a y

Sol–16: (a)

Area = 21 6 6base height 18m

2 2

C.G. from surface = 1 6h m 2m3 3

Total pressure = gAh 1000 10 18 2

= 360 kN

Sol–17: (a)Horizontal force exerted by water

Fx = force of vertical area BOC

= g Ah

Area A = 2BOC 6 8 48m ,

h = 1 6 3m2

Fx = 1000 × 10 × 48 × 3= 1440 kN

Sol–18: (a)Stream function is a scalar function ofspace and time.

Sol–19: (a)

= 5(x2 – y2)

x = 10x

y = – 10y

u = 10x and v = 10y, @ (4, 3), u = 40and v = 30

Resultant velocity v = 2 2u v 50 unitsSol–20: (a)

h =

4 cosgd

=

34 0.5 cos 1201360 10 5 10

=

3

4 0.5 0.51360 10 5 10

= – 14.7 mmSol–21: (a)

Ratio =

1 / 2 gH DH 9H H1 / 2 g D3 3

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Sol–22: (a)Dynamic viscosity =

= (0.6 × 1000) × (0.5 × 10–4)= 0.03 Ns/m2

Sol–23: (a)

T =

2GK2

g GM

25 =2GK2

10 0.5

25 5

2= KG

KG = 28mSol–24: (d)Sol–25: (a)

Volume of box = VVolume of box displacing sea water =0.75 VFor equilibriumWeight of body = Buoyant force

metalVg = waterg

metalW V = waterW 10 0.85 V

Wmetal = 8.5 kN/m3

Sol–26: (a)Sol–27: (a)Sol–28: (a)

For dehumidification, the cooling coiltemperature must be below dew pointtemperature of incoming air.

Sol–29: (c)Since the hot water spray is at a highertemperature and humidity than theunsaturated air, so the air will undergoheating and humidification.

Sol–30: (d)

Saturation

curve

Outdoorcondition

Mixtureconditionenteringthe coolingcoil

ADP5

42

6

31

GSHF line

Supply airfrom coolingcoil

RSHFline

Dry bulbtemperature

Fig. Grand sensible heat factorThe intersection of RSHF and GSHFlines gives room supply air condition.In the diagram, point 3 represents themixture condition of air entering thecooling coil. When the mixture conditionenters the cooling coil, it is cooled anddehumidified. The point 4 shows thesupply air from the cooling coil. Whenthe point 3 is joined with point 4, itgives grand sensible heat factor line(GSHF line). This line when producedupto the saturation curve, it gives theapparatus dew point (ADP).

Sol–31: (c)

SHF =

SH 0.8SH LH

SH = 0.8 × 5 = 4 KJ/min

LH =5 – 4 = 1 kJ/minSol–32: (b)

This process is mainly used inindustrial air conditioning and can alsobe used for some comfort airconditioning installation requiring eithera low relative humidity or low dew pointtemperature in the room.

Adiabatic dehumidification

Dry bulb temperature

td1 td3 td2

1

32 Sp

. hum

idity

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In this process, the air is passed overchemicals which have an affinity formoisture. As the air comes in contactwith these chemicals, the moisture getscondensed out of the air and gives upits latent heat. Due to the condensation,the specific humidity decreases and theheat of condensation supplies sensibleheat for heating the air and thusincreasing its dry bulb temperture.The process, which is the reverse ofadiabatic saturation process, is shownby the line 1-2 on the psychrometricchart as shown in Fig. The path followedduring the process is along the constantwet bulb temperature line or constantenthalpy line.

Sol–33: (c)Sol–34: (b)

When saturated warm air is cooled, itfollows process ( A B ) along saturationcurve. Hence specific humidity decreasesand relative humidity remains constant.

SC

DBT

tAtB

1

A

B

WA

WB

Sol–35: (a)Sol–36: (c)Sol–37: (c)

The specific humidity or humidity ratioremains constant during sensibleheating or cooling of moist air. Therelative humidity will decrease, whilewet bulb temperature will increaseduring the sensible heating.

Sol–38: (a)In cooling tower, evaporative coolingtakes place. Water evaporates by takinglatent heat from the water therebycooling it.

Sol–39: (d)

DBT

2 1

Relative humidity will increase whilespecific humidity remains constant.

Sol–40: (b)When air is fully saturated i.e. whenthe relative humidity reaches 100%,DPT, DBT and WBT become same.

Sol–41: (b)

Saturation

curve

Outdoorcondition

Mixtureconditionenteringthe coolingcoil

ADP5

42

6

31

GSHF line

Supply airfrom coolingcoil

RSHFline

Dry bulbtemperature

When the point 3 is joined with point 4,it gives a grand sensible heat factor line(GSHF line). This line when producedupto the saturation curve, it givesapparatus dew point (ADP).

Sol–42: (c)Flash chamber increases therefrigerating effect. The improvement inCOP depends on the choice ofintermediate pressure, the flashchamber pressure for given condenserand evaporator pressure. There existsan intermediate pressure pi at whichthe CoP is maximum.

Sol–43: (c)Psychrometric chart is drawn for fixedatmospheric pressure.

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Sol–44: (c)Temperature and humidity ratio bothdecrease in process AB.

Sol–45: (a)Effective temperature is uniformtemperature of an imaginary enclosurewith the same heat transfer by radiationand convection as in actualenvironment.The degree of warmth or cold felt by ahuman body depends mainly on thefollowing three factors:1. Dry bulb temperature2. Relative humidity, and3. Air velocityIn order to evaluate the combined effectof these factors, the term effectivetemperature is employed. It is definedas that index which correlates thecombined effects of air temperature,relative humidity and air velocity on thehuman body. The numerical value ofeffective temperature is made equal tothe temperature of still (i.e. 5 to 8 m/min air velocity) saturated air, whichproduces the same sensation of warmthor coolness as produced under the givenconditions.The practical application of the conceptof effective temperature is presented bythe comfort chart. This chart is theresult of research made on differentkinds of people subjected to wide rangeof environmental temperature, relativehumidity and air movement by theAmerican Society of Heating,Refrigeration and Air conditioningEngineers (ASHRAE). It is applicable toresonably still air (5 to 8m/min airvelocity) to situations where theoccupants are seated at rest or doinglight work and to spaces whose enclosingsurfaces are at a mean temperatureequal to the air dry bulb temperature.All points located on a given effectivetemperature line in the comfort chartdon’t indicate conditions of equal comfort

or discomfort. The extremely high or lowrelative humidities may produceconditions of discomfort regardless of theexistent effective temperature. The mostdesirable relative humidity range liesbetween 30 and 70 percent. Further, itdoes not take into account the variationsin comfort conditions when there arewide variations in the mean radianttemperature. The effect of mean radianttemperature on comfort is lesspronounced at high temperature thanat low temperatures.Effective temperature takes intoconsideration the air velocity and isapplicable to reasonably still air with airvelocity 5 to 8m/min.

Sol–46: (a)ADP = 12°C t 2 = 15°C

BPF = 2

10 15=

t ADPt ADP

0.15 =1

15 1212t

ADP=12°C

1 2

t1 = 312 12 20 32= =0 15

C

Sol–47: (b)Heating

coil t = 41°CH

t =

28°C

2t =

15°C

1

Coil Efficiency = 2 1

H 1

t tt t

= 28 15 13 0.5 or 50%= =41 15 26

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Sol–48: (b)RSH = 100 kW,RLH = 25 kWOASH = 10 kWOALH = 10 kWBPF = 0.15Hence, Room sensible heat factor

RSHF = RSHRSH RLH

= 100 0.8=100 25

Effective Room sensible Heat (ERSH) =RSH + BPF × OASH = 100 + 0.15 × 10= 101.5 kWEffective Room Latent Heat (ERLH) =RLH + BPF × OALH = 25 + 0.15 × 10= 26.5 kW

Effective Room Sensible Heat factor,

ERSHF = ERSH 101.5=ERSH ERLH 101.5 26.5

= 0.79Sol–49: (a)

DBT

2 1

1 2 : Sensible cooling

2 1 : Sensible heating

During sensible cooling and sensibleheating, dew point temperature remainsconstant.

Sol–50: (c)For dehumidification, the coiltemperature must be below the dew pointtemperature of incoming air.

Sol–51: (c)Critical path has zero float. If a criticalactivity is delayed or extended, it willdelay or extend the completion of project.

Sol–52: (d)

Variance = 2

p 0t t6

= 222 10 4=

6

Sol–53: (d)Sol–54: (b)

It depends on the type of material, loadand path (whether fixed or changing).

Sol–55: (d)•Critical path analysis is used forscheduling and management. It is ameans to create a logical sequence ofproject tasks, and to track project progressand identify delays.• The line of balance can be used forplanning and controlling batch productionwhere activities or processes arerepetitive.It plots the cumulativeproduction against time for each processor activity and compares the actualprogress with the planned figures. LOBhelps the management to identify theactivities which are out of balance and totake the corrective action. LOB techniqueis a method of production control whichcombines the features from a Gantt chartand CPM time chart with graphs ofmaterial requirement. It is mostappropriate for assembly operationsinvolving a number of distinctcomponents.• Dispatching is the routine of settingproduction activities in motion throughthe release of order and instructions inaccordance with previously planned timesand sequence embodied in route sheetsand schedule charts.• Gantt charts are used in scheduling.

Sol–56: (b)Delphi technique is an interactiveforecasting method that relies on a panelof experts.

Sol–57: (b)The number of cycles that must be timedis a function of three things:(i) the variability of observed times(ii) the desired accuracy

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(iii) the desired level of confidenceSol–58: (b)

Type of job Complete sequence of manufacture:

Outline process chart, flow processchart, flow diagram.

Factory layout: movement ofmaterials: flow diagram.

Factory layout: movement ofworkers: String diagram.

Handling of materials: Flow processchart-material type flow diagram.

Workplace layout: Two handedprocess chart.

Gang work: Multiple Activity chartflow process chart-equipment type.

Movement of operations at work:Film analysis, Simo chart,memomotion photographs, micromotionanalysis.

Sol–59: (b)Production cost per unit can be reducedby producing more with the sameinputs.

Sol–60: (b)MTM or Methods Time Measurement isa predetermined motion time system thatis used primarily in industrial settings toanalyze the methods used to perform anytask and as a product of that analysis,set the standard time in which a workershould complete the task. It is a tool ofwork measurement.Standard time = Observed time × RatingFactor (+ allowances)Work sampling is a random samplingmethod designed originally as amethodology for estimating delayallowances. The technique involvesobserving the worker at many randomlyselect moments and observing theproportion of time the worker is activelyworking or idle. The fundamentalprinciple that allows the conversion of

these kinds of observations to percentagedelay allowances and normal time is thefact that the number of observations isproportional to the amount of time spentin the working or idle state.

Sol–61: (a)Hungarian method is used for solving as-signment problems.

Sol–62: (d)All of the mentioned are the advantage ofline balancing other methods.

Sol–63: (a)Utilization factor,

= Mean arrival rate

Service rate

Sol–64: (d)Sol–65: (b)

P-chart is used to monitor the proportionof defective items in a process. It is basedon binomial distribution.

Sol–66: (b)Sol–67: (a)

Standard time = O.T. [1 + 0.1 + 0.01 +0.01]

0.5

1.12 = Observed time = 0.44 min

Sol–68: (b)Sol–69: (b)Sol–70: (b)

Bottleneck of line is 4.1 minLine efficiency

=

2.2 3.4 4.1 1.7 2.7 3.3 2.64.1 7

= 69.7%Balance delay = 1 – line efficiency

= 1 – 0.697 = 30.3%Sol–71: (d)Sol–72: (d)

At critical point of pure substance,

CT T

Pv

= 0

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C

2

2T T

Pv = 0

C

3

3T T

Pv < 0

Sol–73: (d)Let the total mass of the mixture be m1.Then, sum of number of moles of eachconstituent

= 31 2

1 2 3

m Mm m m m ...M M M

If the molecular weight of the mixture= M.Then, number of moles

= 1 2 3m m m m m mM

Hence, 31 2

1 2 3

m mm m m m ...M M M

= 1 2 3m m m m m mM

M = 1 2 3

31 2

1 2 3

m m m m m m ...m mm m m m ...

M M M

= 1 2 3

31 2

1 2 3

m m m ....mm m ...

M M M

= 31 2

1 2 3

1mm m ...

M M M

Sol–74: (d)If gas obeys Vander Waal’s equation atthe critical point, then

c 2 c

c

ap b = RTc

But, at the critical point,

cT T=

p = 0 and c

2

2T T=

p = 0

which gives b =c

3 and a = 2c c3p

Hence,

2c c c

c 2 cc

3pp3 = RTc

cc c

2p 3p3 = RTc

c c8p

3 = RTc

c

c c

RTp = 8 2.67=

3Sol–75: (b)

Among the thermodynamic variables p,V and T, the following relation holds good:

pT V

p V TV T p

= –1

Sol–76: (a)Actual dryness fraction x = (x1) (x2)where x1 = dryness fraction of steam inseperating calorimeter = 0.9and, x2 = dryness fraction of steamentering throttling calorimeter = 0.95 x = 0.9 × 0.95 = 0.855

Sol–77: (d)

pv = 2 3A 1 B p C p D p ...

p 0lim pv

= A = RT

pvRT = 2 31 B p C p D p

= 2 3B RT RT RT1 C D

p v p

pvRT = 2 3

B C D1v v v

Sol–78: (c)

P

STT

= CP

andv

STT

= CV

Cp – CV = R

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Sol–79: (b)Vander Waal’s equation is

2

2anP V nbv

= nRT

Where ‘b’ is the constant for volumecorrection and ‘a’ is the constant formolecular attraction. A higher value of‘a’ reflects increased attraction betweengas molecules. Hence, if a real gas exertsa pressure P, then an ideal gas wouldexert a pressure equal to P + p (p is thepressure lost by the gas molecules due toattraction). The gas having higher valueof ‘a’ can be liquefied easily.

Sol–80: (b)CP – CV = RFor monoatomic gas,

CP = 5R2

VU / T = CV

Joule Thomson coefficient = h

TP

Sol–81: (b)It is a throttling process and henceenthalpy is constant.

Sol–82: (b)Since area under isothermal curve inP–V diagram is minimum, so workrequirement will be minimum

P

V

isothermal

adiabatic

pv = C1.1

pv =C

Sol–83: (c)

Cp = P

STT

Sol–84: (a)

CP – Cv = R

P

V

CC =

= P

P P

C 1=C R 1 R / C

Sol–85: (c)Since, the pure substance is contained ina rigid vessel, hence it will be a constantvolume process.Thus, the process will be along a verticallyupwards line. To pass through criticalpoint, the initial condition should be atwo phase or wet steam.

Critical point

n

n

n

ncc

b

cb

mama

m

mp

aa

b

vx = 0 x = 0.25 x = 0.50

x=0.75 x = 1.0

Sol–86: (c)Adiabatic process

PV = C

lnP + ln V = ln C

dP dVP V

= 0

dPdV = P

V

Isothermal processPV = C

ln P + ln V = ln C

dP dVP V

= 0

dPdV = P

V

Constant Volume ProcessTds = du + Pdv

Tds = du = CV dT

dTdS =

V

TC

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Constant pressure processTds = dh = vdP

Tds = dh = CP dT

dTdS =

P

TC

Sol–87: (b)b will have same dimensions as ‘v’.

Sol–88: (b)Q1 + Q2 + Q3–W = 0

Q1 + Q2 = W – Q3 = 50 –300 = –250Sol–89: (b)

The available energy of a system decreasesas its temperature decreases andapproaches that of the surroundings. Whenheat is transferred from a system, itstemperature decreases and hence thequality of its energy deteriorates. In caseof the isothermal expansion, thetemperature remains constant and hence,there is no degradation of energy

Sol–90: (b)The given process A-B, represented bydotted line appears to be an irreversibleand constant enthalpy process. Freeexpansion is an irreversible processcharacterized by zero work transfer andzero heat transfer.

Sol–91: (d)Mechanical work is a high grade energy,while heat energy is a low grade energy.According to Clausius inequality,

dQT 0

If dQ 0T

, then cycle is reversible

dQ 0T

, then cycle is irreversible

and possible

dQ 0T

, then cycle is impossible,

since it violates the second law ofthermodynamics.

Zeroth law defines temperature.Gibbs function G = H – TSA – 3, B – 1, C – 4, D – 2

Sol–92: (d)Entropy represents the degree ofrandomness. Increase in entropy meansdegradation of energy or decrease inavailable energy

Sol–93: (b)A unit cell is a basic building block of acrystal structure. This unit cell whenrepeated in all 3-dimension viz x, y & zentire crystal structure is formed.

Sol–94: (c)Grain boundary is a surface imperfectionwhich separates crystals of differentorientations in a poly-crystallineaggregate. Fine grained materials havemore grain boundaries compared to coarsematerials.

Sol–95: (c)For FCC crystal structure

R

2R

R

a

a

2RR

R

a2 + a2 = (4 R)2

2a2 = 16R2

a2 = 8R2

a = 2 2 RSol–96: (c)

Relation between atomic radius and edgeelement a is1. Simple cubic : a2 = 4r2

2. BCC : 3a2 = 16r2

3. FCC : a2 = 8r2

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Sol–97: (b)A screw dislocation moves perpendicularto its Burger’s vector and lies parallel toits Burger’s vector. Also, the Burger’svector is parallel to the dislocation line.

Sol–98: (b)The hatched plane has no intercept on yaxis and z axis. The intercept on x-axis is–1. Thus miller indices of the plane are(100) .

Sol–99: (b)Equilibrium phase diagram for a eutecticalloy is

eutectic point

L L L

xA 100%B 0%

40% A60% B

y A 0%B 100%

Just below eutectic point, the compositionof two phases (i.e. & ) be x and yrespectively.

w = y 0.6 0.5(given)=y x

y – 0.6 = 0.5y 0.5x

y – 0.5y + 0.5 x = 0.60.5(x + y) = 0.6x+y = 1.2Similarly

w = 0.6 x 0.5=y x

0.6 – x = 0.5y – 0.5 x0.6 = 0.5y + 0.5x(x + y) = 1.2There, is only one option satisfying theconditions i.e. 30 wt% B and 90 wt% B is

and phases.

Sol–100: (a) Chromium & molybdenum are ferrite

stablizers therefore they shift lowercritical temperature (727°C) to higherside.

Nickel, copper, Manganese areaustenite stablizers and these shiftlower critical temperature to lowerside.

Sol–101: (d)System: Consists of solids, liquids orgases or their combination.Phase: Homogeneous portion of a systemthat has uniform physical characteristics.Phase equilibrium: Free energy is aminimum.Grain boundary: Interface between twograins.

Sol–102: (c) Characteristic of any alloys can be

found by phase diagram. Phasediagram gives amount andcomposition of each phase present inthe phase diagram.

Liquid & gases can be phase withordered crystal structure.

Phase diagrams are equilibriumdiagrams already hence they do notgive any information on how rapidlyequilibrium is reached.

Sol–103: (b)Season cracking occurs in copper basedalloys under residual tensile stresses andcorrosive environments. Season crackingis also known as stress corrosion crackingand environmental cracking.

Sol–104: (a)During peritectic reaction one liquidcombines with a solid to form anothersolid.Liquid +

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Sol–105: (b)The eutectoid of carbon in Iron (0.77%carbon), when cooled below 727°C resultsin ferrite and cementite in a lamellarformation called pearlite.

30.77% C0.77% C 6.67% C

Iron Iron Fe C

Sol–106: (b)Austempering is employed to obtain 100%bainitic structure. It is an isothermaltransformation process.

Sol–107: (c)Cold working is carried out below therecrystallization temperature of thematerial. It results in increases instrength and hardness of material due tostrain hardening.

Sol–108: (a)A. Annealing: Improves ductility.B. Nitriding: Increases surface

hardness.C. Martempering: Improves the

hardness of the whole mass.D. Normalizing: Refined grain

structure.Sol–109: (a)

q = dTkAdx

dT1 = dT2q1 = q2

1dT0.5W/mK×A×1m = 2

2dTk A0.5

k2 = 0.5 × 0.5 = 0.25 W/mKSol–110: (a)

h × A × dT = Qr × A 50 × s(T T ) = 500

40 – T = 10

T = 30°CSol–111: (a)

The heat transfer process from sun toinside is convection on left side/outside,

conduction through wall andconvection on right side/inside. So theequivalent thermal circuit.

1 2

Rco Rwall Rci

Inside

Wall Inside

1 2Outside

Sol–112: (d)Sol–113: (b)

4E T

Sol–114: (b)Only 1 dimension heat conduction existsand no heat generation is considered.

Sol–115: (c)Equivalent emissivity for radiationexchange between completely enclosed bodysmall compared to enclosing body(subscript 1 for enclosed body) = 1

1 = 0.5

Sol–116: (a)

rc = 2Kh

K = cr h 0.375W/mK2

Sol–117: (c)It depends only on the orientation orgeometry and independent of surfaceproperties and temperatures.

Sol–118: (d)Sol–119: (b)Sol–120: (c)Sol–121: (a)

Q = kA T Tb b/kA

Here in our question b = t, Thermal resistance = t/kA

Sol–122: (c)

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Sol–123: (d)We assume interfaces to be perfect i.e.there is no temperature drop at interfaces.

Sol–124: (c)• Conduction in solid occurs due to free

electron transfers.• Thermal conductivity is property of

material• Mercury has the highest conductivity

among liquid,KHg = 8.2 W/mK

• At zero kelvin, body cannot emit thermalradiation

Sol–125: (b)

factor =

3 31 2

1 1 3 3 3 3 2 2

11 11 11 1

F F

F1–3 = F3–2 = 1

Factor = 31 2

1 3 2

111 12 2

= 1 2 3

2 3 1 2 1 3 1 2 32 2

Sol–126: (c)Sol–127: (d)

Irradiation is the total radiation fallingupon a surface per unit per unit area.Radiosity is total radiation leaving per unittime per unit area.

Sol–128: (a)

J = E G

i.e. Radiosity = Emission + Irradiation.Sol–129: (a)

1

2

EE =

41 142 2

T AT A

=4 24000 2 64

1000 4

Sol–130: (d)The fill factor is defined as the ratio ofpeak power to the product of VOC (opencircuit voltage) and ISC (short circuitcurrent). It indicates the quality of solarcell. Fill factor improves by increasingthe photo current.

Sol–131: (b)

= max max

in

V IP

Pin =20.225mW / 0.86cm

0.15= 1.74 mW/cm2

Sol–132: (c)

m =z

1 1cos cos30

= 2 / 3 1.155

Sol–133: (d)At the stagnation state of the absorberplate the useful heat gain is zero

Qu = P LA S Q

O = P LA S Q

QL = P 2

700 430A S100

= 30.1 WSol–134: (a)

= max max

P

V IA S

= OC SC

P

FF V IA S

=3

30.03 0.474 9 10

0.79 0.9 10

= 0.18 or 18%Sol–135: (b)

E = 1.24

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= 1.24 1.24E 2.42

= 0.512 mSol–136: (c)

=OutputInput

= 12 4.18 20820 5.5

= 0.222 or 22.2Sol–137: (c)

u=4m/s

=

= 70°V = 4/tan 70 = 1.45 m/s

Sol–138: (b)

F = L DF cos F sin

= 300 cos [10 + 20] + 100sin [10 + 20]

= 309.8 NSol–139: (a)

Solidity =Projected area of blades

Swept area

= 2 2Nb 50 0.5r 8

= 0.124Sol–140: (c)

1.6 = RU

U = 2 10 6 160 2 1.6

= 1.96 m/s

Sol–141: (b)

PT = w16 P27

60 × 1000 = 316 1 1.2 3 A27 2

A = 6250 m2

Sol–142: (c)

Cp = i e

i

P PP

0.55 =i

33P

Pi = 33 60 kW0.55

Sol–143: (c)PT = PT = FT × VT

and VT = i p max2 V at C3

60 = 300 × VT

VT = 2 m/s Vi = 3m/s

Sol–144: (b)

F = 2 2t i e

1 A V V2

t

FA = 2 2

i e1 V V2

= 1 1.2 81 42

= 46.2 N/m2.Sol–145: (a)

P =2

1

h

hgA hdhE

t T

1000 × 106 =6 2 2

11000 9.81 50 10 10 h2 22350

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h1 = 2.97 m ~ 3m.Sol–146: (b)

= G H T SH H

= T S1H

T S = (dQ)reversible

Sol–147: (a)Amount of dry matter =

4 22.22 kg/dry0.18

To make slurry equal amount of water isadded

Mass of slurry = 22.22×2=44.44 kg/drySol–148: (b)

The digestor of a biogas plant is normallyand not always, below the ground level.

Sol–149: (a)

CR = max

1sin

max =angle of acceptance

2

maxsin = angle of acceptancesin2

= sin [7.5]= 0.131

CR = 0.131–1 = 7.63Sol–150: (d)Sol–151: (a)

Both statements are correct and reasonexplains the assertion.

Sol–152: (d)Elements are classified into metals andnon metals on the basis of ability to gainor lose electrons and not on the basis oftheir atomic weights. Valence electronsdecide whether the bond is ionic orcovalent.

Sol–153: (c)Free machining steels have goodmachinability due to presence of sulphur.Sulphur forms manganese sulphidenodules which helps in breaking chips,thereby making discontinuous chips andimproving machinability.

Sol–154: (a)Carbon forms interstitial solid solutionwhen added to Iron because size of carbonatom is much smaller than the size ofIron atom.

Sol–155: (a)Magnetic suspetibility occurs above curietemperature only. Thus, higher the curietemperature lower is the suseptibility.

Sol–156: (b)This is the case of floating body. Ballastloading of bottom means addition of heavyweight in lower portion of ship. The ballastincreases weight and brings the center ofgravity lower. The stability is related tometacentric height,

GM = I BG 0V

As metacentric height GM increases, thestability increases.

G

B

M

Ballast

Sol–157: (c)

hz

G

P

The depth of center of pressure,

GIh = + ZAZ

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Hence independent of density. Center ofpressure always lies below geometriccenter G.As geometric center of surface goes downand down, 'G' and 'P' approaches closureand closure.

Sol–158. (c)The U-tube manometer connected to aventurimeter in a pipe line can measurevelocity as well as discharge. This alldepends upon calibrations i.e. manometercalibrated to velocity or discharge. Sincethese reading are difference in heads anddynamic or static head can not be readoutdirectly. To measure dynamic or static orboth it pitot tube not U-tube manometer.

Sol–159: (d)

h1

BC

D

F

H

E

G

Ah2

Two parallel surfaces ABCD and EFGHare parallel to free liquid surface. Thehydrostatic head at surface ABCD is h1and EFGH is h2. These two heads are notequal so pressure ( gh ) will not be equalbecause Pascal's law says that pressureis equal in all directions at a particularhead.

Sol–160: (d)The viscosity of liquid decreases while ofgases, it increases with increase intemperature.The reason in question is correctexplanation of above mentioned statementin answer.

Sol–161: (b)The rise and depression of water andmercury in glass tube is due to contactangle. For water it is zero i.e. positivecosine and for mercury it is obtuse angle

i.e. negative cosine.Sol–162: (d)

The general expression for non-Newtonianfluid.

=n

0dudy

So ' ' is not linear function of velocitygradient not distance.This is due to large size of molecules inpolymers etc. and do not behave likemolecules of Newtonian fluid.

Sol–163: (a)The comfort depends on many factors suchas type of clothing, age and sex, durationof stay, kind of activity, density ofoccupants, climatic and seasonaldifferences. Hence the actual inside designtemperatures selected in the airconditioner may not be suitable foroptimum comfort.

Sol–164:(a)Sol–165:(a)

Specific heat of water is higher andhence water has more cooling capacitythan air.

Sol–166: (a)

Heat pump used for heating is a definiteadvancement over the simple electricheater because a heat pump rejects moreheat (Q1 = Q2 + W) as compare to simpleelectric heater (W).Heat pump is more economical because itis giving more output (Q1) with same input(W) to both the devices.

Sol–167: (d)In adiabatic process, the heat transferacross system boundary is zero. But theentropy may increase due to internalirreversibility or friction inside the system.So adiabatic process is not constantentropy or isentropic process.

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Sol–168: (c)

T

T0

SSince,

dS = dQT

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

Sol–169: (c)Upon increasing temperature, the rate ofincrease of cP in small as compared torate of increase of cV. So the value of ratioof Cp/Cv decreases.In reason, the value of CP/Cv decrease ismentioned which is wrong.

Sol–170 (d)In order to achieve cooling in throttling,the initial temperature must be belowmaximum inversion temperature. It theinitial temperature in above of this, thegases are precooled before throttling asshown below

TA

B

Pre cooling

Heatingregion

Coolingregion

PIf the gas at a directly throttled, it willheat up. So to get cooling the gas is firstprecooled (A B), then throttled.

Sol–171: (a)Radiation heat transfer

Q = 4 4T T

Convection heat transfer–

Q = h A T T

At high temperature, radiation transferwill be high due to 4th power of tem-perature. Another requirement of con-vection is medium which strongly effectheat transfer by convection

Sol–172: (a)In lumped capacity method, thetemperature variation within hot body isnegligible i.e. negligible temperaturegradient.This negligible temperature gradient isachieved by having high thermalconductivity k of hot solid body materialand low convective heat transfer coefficient‘h’. The relative magnitude of thermalconductivity ‘k’ and convective coefficientis grouped as Biot number.

Bi = hk

For lumped capacity method to beapplicable, Bi should be less than 0.1i.e.

Bi < 0.1Sol–173: (a)

By providing refractory brick lining insideof pipe carrying hot gases, the heat loss isreduced. The brick lining inside surfacepromotes the formation of stagnant layerof gas and this layer is good insulator inaddition to brick. Hence the brick liningon inside surface provides more thermalresistance and less heat loss as comparedto outside brick lining.

Sol–174: (a)

The convective resistance = 1hA

When insulation is added inside, surfacearea A reduces (because now insulatingsurface come in picture for convection

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which lower the bore of pipe) so resistanceincreases. Along with this, conductionresistance also increases. So insideinsulation increases over all resistance ofpipe and no conflicting terms like outsideinsulation.

Sol–175: (c)For meeting due date requirements,Earliest due date (EDD) rule ispreferred.

Sol–176: (b)Fixed portion layout is used where it iseconomical to move machines, tools andmen on to the job rather than movingthe job. For example - manufacturingships, huge aircrafts etc. Capitalinvestment is minimum in the case offixed position lay

Sol–177: (c)FIFO (First-in-First-out) is simple andfair. But it is not appropriate for manyscheduling situations. In this, thewaiting time depends on the arrivalorder. Short processes are struck

waiting for long processes. FIFO is anon pre-emptive scheduling algorithm.

Sol–178: (d)Assertion is wrong

• Savonius windmill is a vertical axiswindmill which requires large surfacearea.

• Darrieus type wind mill requires lesssurface area. Hence not all vertical axiswindmill requires less surface area.

Sol–179: (d)Wind energy power plants is amongindirect methods of harnassing solarenergy.The direct means include only thermaland photovoltaic.

Sol–180: (d)The drag and lift both are due to pressureand shear forces combined, their onecomponent along the body is drag andother perpendicular to body is lift.Although pressure drag for aerofoil shapewhich is a streamlined, is small not zero.

* * * * *

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