ESE-2017 PRELIMS TEST SERIESDate: 16th October, 2016
24. (b)
25. (d)
26. (a)
27. (c)
28. (b)
29. (a)
30. (a)
31. (b)
32. (d)
33. (c)
34. (c)
35. (c)
36. (b)
37. (b)
38. (d)
39. (c)
40. (c)
41. (b)
42. (a)
43. (d)
44. (b)
45. (a)
46. (c)
47. (a)
48. (c)
49. (b)
50. (a)
51. (c)
52. (d)
53. (a)
54. (c)
55. (b)
56. (c)
57. (d)
58. (b)
59. (a)
60. (a)
61. (b)
62. (c)
63. (a)
64. (c)
65. (d)
66. (c)
67. (a)
68. (b)
69. (c)
70. (b)
71. (c)
72. (a)
73. (b)
74. (c)
75. (b)
76. (d)
77. (b)
78. (b)
79. (d)
80. (a)
81. (b)
82. (b)
83. (b)
84. (d)
85. (c)
86. (a)
87. (c)
88. (d)
89. (b)
90. (a)
91. (a)
92. (b)
93. (d)
94. (a)
95. (b)
96. (c)
97. (c)
98. (b)
99. (c)
100. (a)
101. (a)
102. (b)
103. (c)
104. (d)
105. (b)
106. (d)
107. (c)
108. (a)
109. (c)
110. (b)
111. (b)
112. (b)
113. (b)
114. (c)
115. (b)
116. (d)
117. (c)
118. (d)
119. (b)
120. (c)
121. (c)
122. (a)
123. (c)
124. (a)
125. (b)
126. (c)
127. (b)
128. (d)
129. (d)
130. (c)
131. (c)
132. (a)
133. (c)
134. (c)
135. (d)
136. (b)
137. (d)
138. (d)
ANSWERS
1. (d)
2. (a)
3. (b)
4. (d)
5. (b)
6. (a)
7. (b)
8. (b)
9. (d)
10. (b)
11. (b)
12. (b)
13. (b)
14. (a)
15. (c)
16. (b)
17. (c)
18. (c)
19. (c)
20. (b)
21. (d)
22. (b)
23. (a)
IES M
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(2) ME (Test-A3), Objective Solutions, 16th October 2016
139. (d)
140. (b)
141. (d)
142. (c)
143. (d)
144. (d)
145. (d)
146. (a)
147. (b)
148. (b)
149. (c)
150. (d)
151. (a)
152. (b)
153. (a)
154. (a)
155. (a)
156. (b)
157. (a)
158. (c)
159. (a)
160. (d)
161. (b)
162. (d)
163. (a)
164. (a)
165. (a)
166. (a)
167. (a)
168. (a)
169. (a)
170. (a)
171. (c)
172. (d)
173. (c)
174. (a)
175. (a)
176. (c)
177. (c)
178. (d)
179. (b)
180. (c)
IES M
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(3) ME (Test-A3), Objective Solutions, 16th October 2016
Sol–1: (d)
T = I
2000 = 3000
= 22000 0.6 rad/s3000
Angular velocity after 10 s = t
= 0.6 × 10 = 6 rad/s
Kinetic energy = 21 I2
= 21 3000 62
= 3000 × 18= 54 kJ
Sol–2: (a)Apply momentum conservation,
600 × V1 = (600 + 150) V2
where V1 = Velocity of hammerV2 = Combined velocity of
hammer and pile
600 V1 = 750 V2
4 V1 = 5 V2
Initial kinetic energy KE1 = 21
1 600 V2
Final kinetic energy KE2 = 22
1 750 V2
1
2
KEKE =
2 21 1
2 22
1 600 V V42 =1 V5750 V2
=24 5 1.25=5 4
Sol–3: (b)In a plastic impact (inelastic collision),the kinetic energy of the bodies is notconsserved, because some amount of thekinetic energy is transformed into heator internal energy of the bodies. However,the total momentum of the two bodiesremains the same.
Coefficient of restitution e =
2 2
1 1
v uv u
where 1 1,v u are initial velocities of the
two bodies, while 2 2and v u are the finalvelocities of two bodies after collision. Ina perfectly plastic collision, e = 0
Sol–4: (d)At the hinged support A, moment will bezero.
A B
60°30°
T
CWl/2 l/2
wT cos 60° × AC = W × AB + w × AC
T2 2 = W w
2
T2 = 2W + w
T =
w2(2W w) 4= W2
Sol–5: (b)Let the distance travelled by the ball beforeit starts falling be xThen, 0 = 302 – 2gx
x = 2 230 30 900 50m= = =2 g 2 9 2 9
Total height = 50 100m=x
Let velocity with which it strikesthe ground = v
Then, v2 = 2g × 100 = 2 × 900
v = 2 30 42= m/s
Sol–6: (a)100 N
m
N
mg
200 N
N = mg + 100
Friction = N (mg 100)
IES M
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(4) ME (Test-A3), Objective Solutions, 16th October 2016
Total force = (mg 100) 200
Work done = (mg 100) 200 10
Sol–7: (b)Since, the rectangular plate is held inequilibrium, hence the moment abouthinge should be zero.
P × 4 + 150 × 2 = 100 × 1
P × 4 = 100 – 300 = – 200
P = 200 50kN=4
Sol–8: (b)Force along the slope opposing the motioncos F = R(mgsin F )
where resistance due to friction, air etc.FR = 30 N/kg × 200 kg
= 6000 N
sin tan = 10200
F =10200 10 6000200
= 100 + 6000 = 6100N Power = F × V
= 6100 × 10 = 61 kWSol–9: (d)Sol–10: (b)
40 NA CB
F F F
10 cm 20 cmTaking moment about point B to be zero,
40 × 0.1 = 2F × 0.2
F = 40 0.1 10 N=0.4
Sol–11: (b)
D Alembert showed that one cantransform an accelerating rigid body intoan equivalent static system by adding theinertia force. The system can then be
analyzed exactly as a static systemsubjected to this inertia force and theexternal forces.
Sol–12: (b)Vertical distance travelled
= 2 2u sin
2g =
2318
22 9.81
= 81 3 12.4m2 9.81
Sol–13: (b)
T = x W
where = weight per unit length ofthe rope
W = weight of bucketT = tension in the rope at a
distance x from bottom.
Tdx
x
W
Work done = L
0Tdx
= L
0dxx W =
L2
0
x Wx2
= 2L WL2
= 21 20 10 20 400 Nm=2
Sol–14: (a)The work done in stretching an elastic
string gets stored as the energy = 21 kx2
where k = constantx = elongation/extension
Sol–15: (c)Work will be done against the frictionalforceHence Power = Friction force × Velocity= 20 × 103 × 10 = 200 kW
IES M
ASTER
(5) ME (Test-A3), Objective Solutions, 16th October 2016
Sol–16: (b)Work done in making the log to standvertical = potential energy of the verticallog
= 21 20 10 500
= 21 200 50 10
Sol–17: (c)Resulting force in horizontal direction =250 + 150 = 400 N.Resulting force in vertical direction =250 + 50 = 300 N
Hence, Resultant force = 2 2400 300
= 500 NThe resulting moment at the centre ofpulley= 250 × 0.5 + 50 × 0.25 – 250 × 0.25 –150 × 0.5 = 100 × 0.5 – 200 × 0.25 = 0
Sol–18: (c)Let the person pull the rope with a forceF. Then, 2F = W
W
F F
F = W 500 250N= =2 2
Sol–19: (c)Rotational kinetic energy
= 2 2
2 2 21 1 mr mrI = =2 2 2 4
Since the circular disc rolls withoutslipping, so v r.=
Rotational KE = 2mv
4
Total energy = 2 21 1I mv2 2
= 2 2 2mv mv 3mv=
4 2 4
Rotational KETotal energy =
2
2
mv14 =3mv 3
4Sol–20: (b)
• Simple harmonic motion is a type ofperiodic motion where the restoringforce is directly proportional to thedisplacement.
• Once the mass is displaced from itsequilibrium position, it experiences a netrestoring force. As a result, it acceleratesand starts going back to the equilibriumposition. When the mass moves closer tothe equilibrium position, the restoring forcedecreases. At the equilibrium position, thenet restoring force vanishes. However atx = 0, the mass has momentum becauseof the impulse that the restoring force hasimparted. Therefore, the mass continuespart the equilibrium position. Thus, inSHM, the motion is not of uniform velocity.
Sol–21: (d)
B
C
W
P
180°–2
90°–
A
The moment about A will be zero.
ABW cos 90 180 22
= P × AC cos
ABW cos 2 902
= P ACcos
ABW sin22
= P ACcos
ABW 2sin cos2
= P ACcos
P = Wsin
IES M
ASTER
(6) ME (Test-A3), Objective Solutions, 16th October 2016
Sol–22: (b)
mg sin
500 N
mg cos
30°
For equilibrium along the plane,mgsin = mgcos 500
mg(sin cos ) = 500
mg = 500sin cos
= 500 2500N=1 3 32 5 2
Sol–23: (a)Considering the forces at point A, onlyvertical force is 8t and there is nohorizontal force at A. Hence, force inmember AC = zero.
B
A C
8tSol–24: (b)Sol–25: (d)
Taking moment about fulcrum B,200 × 0.3 = W × 0.075
W = 200 0.3 800 N=0.075
Sol–26: (a)S = t2(t–1) + 2
Velocity
V = dsdt = t2 + (t – 1) × 2t
= t2 + 2t2 – 2t = 3t2 – 2tAcceleration
a = dvdt = 3 × 2t – 2 = 6t – 2
Velocity will be maximum, when dvdt = 0
6t – 2= 0
t = 2/6 = 13
For coming to rest, V = 0So, 3t2 – 2t = 0 t(3t – 2) = 0 t = 2/3Distance travelled in t = 2/3 will be
S =22 2 1 2
3 3
=4 1 29 3
= 4 502 =27 27
Sol–27: (c)
Given kA = 2 kBForce carried by spring at A
= FA = A Ak B A2k Force carried by spring at B
= FB = B Bk
Deflected shape
A
A
B
Ba 2a
From similar triangles
A3a = B
2a A = B1.5
A
B
FF = A A
B B
kk
= b A
b B
2kk
= B
B
2 1.5 3
Sol–28: (b)
Stress developed in a bar due totemperature
= EL (Deformation prevented)
IES M
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(7) ME (Test-A3), Objective Solutions, 16th October 2016
= E L tL = E t
A
B
=
A
B
E tE t = 3E 3 t
E t
= 9
Sol–29: (a)Square block ABCD is subjected to shearstrain
R
A
C
D
C
Mohr’s circle for shear strainLongitudinal strain
90°
/2
Normal strain/2
/2
Strain at 45° to the side AD (along AC )At 2 = 90°
Diagonal strain = 2
Alternate solution:
A
C
D
C
L
C
45°
L
BB LL
C C =LL cos45
2
AC = L 2
Diagonal strain = L22 L 2
Alternate Solution:
= x y x y xycos2 sin22 2 2
Here,x = y 0
and = 45°
45 = sin 902
= 2
Sol–30: (a)
Rigid plastic
Ductilematerial
LargePlastic zone
Brittle material
No plastic zone
Inelastic material does not have strainrecovered on unloading.
Sol–31: (b)
We know, E = 2G(1 + ) … (i)E = 3K(1 – 2) …(ii)
Equations (i) (ii)
1 =2 G (1 )3 K (1 2 )
Given K = 2G
1 =2 G (1 )3 2G (1 2 )
=27
Sol–32: (d)
Volumetric strain, x y zV
V
= 800 × 10–6 + 400 × 10–6 + (– 1200) × 10–6
= 0
IES M
ASTER
(8) ME (Test-A3), Objective Solutions, 16th October 2016
Sol–33: (c)
Coefficient of thermal expansion of brassis more than steel, so brass will tryto expand more but it’s expansion willbe restrained by steel because both willmove together by same amount. Thereforebrass will be in compression and steel intension.
Sol–34: (c)
In larger diameter column, at the level ofB, area resisting the load will be equal tothe area of shorter diameter column.After load dispersion whole area of columnBC will resist the load.
(9) ME (Test-A3), Objective Solutions, 16th October 2016
C.L
w
AwLR =4
Maximum BM will occur
= AL L L LR w2 2 3 2
maxM = 2
AL w(L / 2)R2 6
= 2wL L wL
4 2 24 =
2wL12
D.w/m
AwLR =2
L
maxM = AL L 1 LR w2 2 2 2
= 2 2wL L wL wL
2 2 8 8
Sol–40: (c)
2P
P
BMD
Sol–41: (b)
20 kN
A CB
10 kN/m
RBRA
2 m 2 m
RA = 10 4 20 30 kN2
RB = RA = 30 kNdVdx = w = –10 kN/m
Slope of SFD will be (–) ve and constant.There will be sudden drop in SFD at point Cequal to 20 kN.
30 kN
30 kNSFDA C B
dMdx = V
Slope of BMD will be (+) ve and decreasing inpart AC and slope will be (–) ve and increasingin part CB.
40 kN-m
Kink
BMDA C B
Mmax = Area of SFD in AC
= 1 (30 10) 22
= 40 kN-m
Sol–42: (a)
Let us assume point of contraflexure liesat a distance x from end A.
M(x) = 8(0.75 – x) – 5(1 – x) At point of contraflexure moment is zero
0 = 6 – 8x – 5 + 5x
x = 13
= 0.333 m
Sol–43: (d)
w/unit length
L/2 L/2LAs loading is udl, so SFD will be linear andBMD will be parabolic in shape.
Due to symmetry RB = RC = 2wL2 = wL
BM at B = 2 2wL L wL
2 4 8 Hogging
Bending moment at mid point of BC
MBC = 2
Bwx LR
2 2
= 2 2w(L) wL
2 2
= 0Hence option (d) is correct.
IES M
ASTER
(10) ME (Test-A3), Objective Solutions, 16th October 2016
Sol–44: (b)
100 kN/m
RA RB
25 kN75 kN
SFD
1 m 1 m
A
B
0.5R 100 25 kN2
R 100 25 75 kN
Sol–45: (a)
Total load=120 kN
3 mTotal load will act through the C.G of loadingdiagram so equivalent structure for calculatingsupport reaction is
120 kN
R
M
1 m 2 m
Support reaction R = 120 kNBending moment, M = 120 × 1
= 120 kNm (hogging)
Sol–46: (c)
dVdx = w
Shear force is uniform throughout the span sodVdx = 0
w = 0hence no vertical loading should be there inthe span. A couple any where in the beam willcause equal and opposite support reactions inthe beam, so SFD will be rectangular oruniform, throughout the beam.
Sol–47. (a) w w
R = wA R = wB
w
wSFD
L/3 L/3 L/3
In portion CD shear force is zero so it is inpure bending.
Sol–48. (c)Radius of Mohr Circle
R =
2x y 2
xy2
=2120 40 30
2
= 50 MPa
Principal Stress
1 = y R2
x = 120 40 50
2
= 130 MPa
2 = y R2
x = 120 40 50
2
= 30 MPa
Sol–49: (b)
Radius of Mohr’s circle,
R = 2
21 22
=
2220 20 0
2
= 0
Radius of Mohr’s circle is zero, Mohr’s circlehas been reduced to a point so at all positionsof orientation of x and y axes shear stress willbe zero.
Sol–50: (a)
Stresses are given on two mutuallyperpendicular planes and associated shearstress on these planes is zero so givenstresses are principal stresses.
Sol–51: (c)
Principal stresses are 1000 MPa, – 1000MPa.
IES M
ASTER
(11) ME (Test-A3), Objective Solutions, 16th October 2016
It may be the case of pure shear withshear stress equal to radius of Mohr’s circlei.e. = 1000 MPa which is representedby option (c).
Sol–52: (d)
x = 800 × 10–6
y = 400 × 10–6
xy = 300 × 10–6
Maximum shear strain
max2
=
2 2x y xy
2 2
= 250 × 10–6
max = 500 × 10–6
Sol–53: (a)
In circular shaftMaximum normal stress due to Bendingmoment
1 = 332 M
dMaximum shear stress due to Torsion T
= 316T
dMaximum normal stress due to combinedaction of Bending moment and Torque
max =2
21 12 2
= 2 23
16M M Td
According to maximum principal stresstheory, for no failure
max Allowable stress
2 23
16M M Td
3d 2 216 [M M T ]
Sol–54: (c)
According to maximum shear stresstheory, for yielding
1 22
= yf
2
200 – 2 = 2502 = – 50 MPa
= 50 MPa (Compressive)Sol–55: (b)
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 8m/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 toreasonably 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.
Sol–56: (c)Steam spray into air: Heating andhumidification
IES M
ASTER
(12) ME (Test-A3), Objective Solutions, 16th October 2016
Air passing over a coil carryingsteam: Sensible heatingAir passing over a coil havingtemperature less than dew point:Cooling and dehumidificationAir passing over a coil havingtemperature above the dew pointbut below the wbt: Sensible cooling
• Dehumidification will take place onlywhen the cooling coil has temperaturelower than the dew point temperatureof air.
Sol–57: (d)During chemical dehumidification, asthe air comes in contact with thesechemicals, the moisture gets condensedout of air and gives up latent heat. Dueto condensation, the specific humiditydecreases and the heat of condensationsupplies sensible heat for heating theair and thus increasing its dry bulbtemperature. Thus, wet bulbtemperature remains constant. Sincewbt line and enthalpy line are parallelto each other in psychrometric charthence enthalpy also remains constant.
Sol–58: (b)(COP)VA <(COP)SE < (COP)VC
Sol–59: (a)For heat engine
= 2
1
T1 0.4T 2
1
TT
= 0.6
For refrigeration,
COP =
2
2 1
21 2
1
TT T= TT T 1
T
= 0.6 0.6 6 1.5= = =
1 0.6 0.4 4Sol–60: (a)
Air, which has extremely low normalboiling and critical points remains as anon-condensable gas, if present inrefrigerant condensers. The presence of
even one percent of air by volume canreduce the condensing coefficient by asmuch as 50%. This is because air tendsto cling to the surface and because of itslow thermal conductivity, it introduces alarge thermal resistance. Anotherdetermintal effect of air is the increase ofhead pressure. Firstly, it is because of adecrease in the condensing coefficient anda consequent operation at a highertemperature differential T and henceresults in a higher pressure pk in thecondenser. Secondly, the head pressure isfurther increased over and above thesaturation pressure pk of the refrigerantby an amount equal to the partial pressureexerted by air or non-condensables present.
Sol–61: (b)From the energy balance,
1 21 w 2 w1 2m m m mh h
= 33 w 3m m h
1 21 w 2 wp 1 p 2m m m mc t c t
= 33 w p 3m m c t
t3 =
1 2
3
1 w 2 w1 2
3 w
m m m mt tm m
Sol–62: (c)To dehumidify, the coil must be at atemperature lower than the dew pointtemperature of the increasing stream.
Sol–63: (a)
1tdp1
dry bulb temp.
spec
ific h
umid
ity
The point at which, a line parallel tothe dbt line or at constant humidityratio, cuts the saturation curve is calleddew point temperature. For saturatedair, dbt, dpt and wbt are same.Dehumidification can be achieved alongwith both heating and cooling. Atconstant enthalpy, wbt remains constanteven with addition of moisture because
IES M
ASTER
(13) ME (Test-A3), Objective Solutions, 16th October 2016
constant enthalpy line and constant wbtlines are parallel in psychrometric chart.
Sol–64: (c)
HE
T 1 = 500 K
T 2
W
= 2
1
T1T
= 0.25
2T1500 = 0.25
2T
500 = 0.75
2T = 375 K
RCOP = 2
1 2
TT T
=
375 3500 375
R
T 2 = 375 K
T 1 = 500 KQ 1
Q 2
W
Sol–65: (d)
T
s
T =T2 33 2
14
s =s4 3 s =s1 2
COP of the refrigeration system workingon reversed Carnot cycle
(COP)R = 1
2 1
TT T
where T1 = lower temperatureand T2 = higher temperature
Thus, COP of the reversed Carnot cycledecreases on(i) increasing the higher temperature
Sol–67: (a)Since the desired condition requirescooling from 45ºC dbt to 25° dbt andreducing relative humidity from 70% to50%, the practical arrangement wouldbe cooling and dehumidification.
Sol–68: (b)• The solvent power of R22 for water is
almost 23 times that of R12. In R22systems, therefore, ice formation generallydoes not occur. Thus, R22 is used in lowtemperature food refrigeration plantsbecause moisture choking would not occureven at freezing temperatures.
• Presence of lubricating oil in evaporatorwill decrease the heat transfer coefficient.
• Refrigerants that are not miscible withoil, such as ammonia or carbon dioxidedon’t cause oil choking. In such a case,an oil seperator is installed a little awayfrom the compressor in the discharge lineand the seperated oil is continuouslyreturned to the crankcase of thecompressor.
• R11 is high boiling and low pressurerefrigerant requiring large suction volume.It is suited for use with centrifugalcompressor in large capacity refrigeratingmachines.
Sol–69: (c)
IES M
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(14) ME (Test-A3), Objective Solutions, 16th October 2016
= v
s
p 0.5p
Since the air is compressed isothermallyso the temperature remains the sameand hence the saturation pressurecorresponding to the temperature willremain the same
=v
s
pp
But sp = ps
Since the total pressure is doubled, sovapour pressure will double, hence
vp = 2pv
= v
s
2p 2 2 0.5p
= 1 or 100%So, the sample of air will becomesaturated.
Sol–70: (b)Wet bulb temperature is the lowesttemperature that can be reached byevaporating water into the air. On a hotday, when the wbt is low, rapidevaporation and hence cooling takes placeat the skin’s surface. As the wbtapproaches the air temperature, lesscooling occurs and the skin temperaturemay begin to rise. When the wet bulbtemperature exceeds the skin’stemperature, no net evaporation occursand the body temperature can rise quiterapidly. Fortunately, the wet bulbtemperature is almost always considerablybelow the temperature of the skin.
Sol–71: (c)Approach (A) is defined as thedifference between the exit temperatureof cooling water and the wet bulbtemperature of the ambient air.
A = 2c wbt t
Range (R) is defined as the differencebetween the temperature of the incomingwarm water
1c(t ) and the exiting cooled
water 2c(t ).
R = 1c wbt t
wet bulb depression = dry bulbtemperature – wet bulb temperature
= db wbt tAccording to the question,
db wbt t = 2 1 2c wb c ct t t t
dbt = 1ct
Sol–72: (a)Automatic expansion valve maintainsconstant evaporator pressure regardlessof the load on the evaporator. It is usedwith dry expansion evaporators wherethe load is relatively constant.
Sol–75 (b)Rotary compressors have two rotatingelements, like gears, between which therefrigerant is compressed. Thesecompressors are very efficient becausethe actions of taking in refrigerant andcompressing refrigerant occursimultaneously. They have very fewmoving parts, low rotational speeds.Thus, they are limited to smallervolumes of the gas and produce lesspressure than other types of compressor.
Sol–76: (d)Vapour absorption system areadvantageous where a large quantity oflow grade thermal energy is availablefreely at the required temperature.However, it will be seen that for therefrigeration and heat rejectiontemperatures, the COP of vapourcompression refrigeration system will bemuch higher than the COP of anabsorption system as a high grade
IES M
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(15) ME (Test-A3), Objective Solutions, 16th October 2016
mechanical energy is used in the former,while a low grade thermal energy isused in the latter.
VCRSCOP = e
c
QW
VARSCOP = e e
g p g
Q QQ Q Q
Sol–77: (b)
T2K
Q2
RW
Q1
T1K
T1 = –43°C = 230K
1
2
QQ = 1
2
TT
Q1 = 900 kJ/min
= 900 kW60
= 15 kWQ2 = Q1 + Winput
= 900 5 20kW60
1520 =
2
230T
T2 = 306.67 KT2 = 33.67°C
Sol–78: (b)Refrigerating effect= h1 – h4 = 190–85 = 105 KJ/Kg1 ton of refrigeration = 210 kJ/min m × RE = 210
m = 210 kJ/min 2 kg/min=105 kJ/kg
Sol–79: (d)
COP = 1
1 2
TT T =
303
303 293
= 30310 = 30.3
Sol–80: (a)Heat removed from refrigerator QR = 1kJ
so, COP = heat removedwork done
5 = RQW
W = 0.2 kJ
Efficiency of heat engine = work doneheat input
0.4 = Wheat input
Heat input = W 0.2 0.5kJ= =0.4 0.4
Sol–81: (b)h3 = 100 kJ/Kg = h4
h2 = 220 kJ/Kgh1 = 200 kJ/Kg
COP =Refrigerating effectcompressor work
=
1 4
2 1
h hh h
=
200 100 100 5= =220 200 20
Sol–82: (b)
p
h
1
23
4
The refrigerant is compressed in thecompressor (1 2) in which itspressure and temperature both increase.Thereafter, the refrigerant rejects heatin the condenser (2 3). Then, therefrigerant expands in the valve
(3 4). The refrigerant extracts heatin the evaporator (4 1).
Sol–83: (b)Latent heat of ammonia is significantlyhigher compared to other refrigerants
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(16) ME (Test-A3), Objective Solutions, 16th October 2016
hence it reduces mass of refrigerantrequired per tonne of refrigeration.
Sol–84: (d)BPF of the cooling coil
= Temperature at supply condition ADP
Temperature of outdoor air ADP
In the given figure, BPF = BCAC
Sol–85: (c)
Cooling andHumidifying
Humidifying
SensibleHeatingR
Cooling andDehumidifying
Sol–86: (a)
Relative humidity = v
s
p 0.01 0.5p 0.02
or 50%Humidity ratio,
= 0.622 v
s v
p 0.01 0.622p p 1.01 0.01
= 0.00622Sol–87: (c)
• The presence of moisture is very criticalin refrigeration systems operating below0°C. If more water is present than whichcan be dissolved by the refrigerant, thenthere is danger of ice formation andconsequent choking in the capillary tubeused for throttling is the system.
• Thermostatic expansion value maintainsa constant degree of superheat at the exitof evaporator.
• Automatic expansion valve maintains aconstant pressure and hence a constanttemperature in the evaporator.
• The float valve is a type of expansion valvewhich maintains the level of liquidrefrigerant constant in a vessel or an
evaporator. Hence, the mass flow rate ofthe refrigerant through the expansionvalve is proportional to the evaporationrate of the refrigerant in the evaporator,which in turn is proportional to the load.
Sol–88: (d)
1wt
2wt
2 1
1–2 Sensible coolingRelative humidity will increase, wetbulb temperature will decrease.Humidity ratio will remain constant.
Sol–89: (b)For dehumidification to take place, thetemperature of the cooling coil must bebelow the dew point temperatrure of air.However, the temperature of the coolingcoil must be above the freezing pointtemperature, otherwise the refrigerantwill not flow, rather it may freeze.
Sol–90: (a)
Adia
batic
Satu
ratio
n2
1
Thus vacancy from the figure i.e. duringthe process 1-2, the enthalpy and wetbulb temperature remains constant,while the dry bulb temperaturedecreases. The humidity ratio andrelative humidity will increase.
Sol–91: (a)Thermostatic expansion valve controlsthe amount of refrigerant flow into theevaporator, thereby controlling thedegree of superheat at the outlet ofevaporator.
Sol–92: (b)The amount of heat rejected in condenseris more than the heat extracted in theevaporator.
QC = W + QE
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(17) ME (Test-A3), Objective Solutions, 16th October 2016
QC = heat rejected in condenserW = work input to the compressorQE = heat extracted in the evaporator
Sol–93: (d)Sol–94: (a)
For liquid vapour regenerative heatexchanger,
1 1h h = 3 3h h = 4 4h h
1 1h h = 4 4h h
1 4h h = 1 4h h
p
h
3 2
14
3
4 1
Thus, for system A, heat extracted = 1 4h h .
For system B, heat extracted = 1 4h h ,Since the vapour is not beingsuperheated in the evaporator and israther being superheated in the liquidvapour regenerative heat exchanger.Hence, since other conditions i.e. powerrequirement for the compressor is same,so
(COP)A = (COP)B
Sol–95: (b)T2
T1
(T2>T )1HP
(CoP)HP = 5
2
2 1
TT T = 5
For heat engine,
= 1 2 1
2 2
T T T1T T
= 1 0.2
5
Sol–96: (c)The presence of graphite flakes in thestructure of grey cast iron gives thismaterial the capacity to dampen vibrationscaused by internal friction andconsequently the ability to dissipateenergy, hence used for making machinetool beds. Grey cast iron has negligibleductility and it is weak in tension,although strong in compression.
Sol–97: (c)An isotropic material have same physicalproperties irrespective of thecrystallagraphic direction in which they aremeasured. Anisotropic materials show di-rection dependent properties.
Sol–98: (b)Cast Iron has excellent vibration dampingproperties due to graphite flakes. Itscompressive strength is higher comparedto that of steel. Cast Iron being a brittlematerial does not undergo large plasticdeformation and fracture at much lowerstrains.
Sol–99: (c)In low carbon steels, presence of smallquantities of sulphur improvesmachinability. Sulphur producesmanganese sulphide nodules which helpin chip breaking thus improvingmachinability.
Sol–100: (a)A composite is a multiphase material thatexhibits a significant proportion ofproperties of both constituents. Theseconstituents phases must be chemicallydis-similar and separated by a distinctinterface. A composite has a surroundingmatrix and a reinforced phase andthermosetting epoxy resin is used as amatrix for some kind of composite.
Sol–101: (a)Ceramics are inorganic non metallicmaterial of a metal and a non-metal. It issolid and inert, brittle, hard, strong incompression, weak in shearing andtension. Ceramics can withstand very high
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(18) ME (Test-A3), Objective Solutions, 16th October 2016
Sol–103: (c)In case of rubber, vulcanization is theprocess of producing a cross linked polymerby addition of sulphur atoms. This crosslinking makes rubber hard, which can beused in various applications such asautomobile tyres etc.
Sol–104: (d)The strength of the fibre reinforced plasticproducts depends upon the fibre and plastic.The strength is higher in the direction ofalignment of fibre and less in thetransverse direction, i.e. it is anisotropic.
Sol–105: (b)Thermosetting plastics are formed bycondensation polymerization. Oncehardened, they cannot be taken back totheir original condition. However, they canbe modulated by heating and cooling.
Sol–106: (d)Polymers are deteriorated by
• Weather• Temperature• Radiation
Sol–107: (c)Both temperature and solute concentrationplay vital role in corrosion process.
Sol–108: (a)Nano robots do not exist presently.
Sol–109: (c)Buckminsterfullerene is C60 molecule.
Sol–110: (b)Sol–111: (b)
The laminar flow over flat plate,
constanth
xx
U
y
x
T >TsPlate
Boundary layer
Local convectivecoefficient
The co-relation for laminar flow overconstant temperature flat plate-
Nu = 0.5 0.330.332Re Prx
hkx x
=
0.50.33V0.332 Prx
xh =
0.5 0.33
0.5V Pr0.332k
x
xh x1/21
Sol–112: (b)In horizontal direction, the characteristicdimension used in Grashoff number andNusselt number is diameter. In verticalorientation, it is length. Hencecharacteristic dimension is decided by flowpattern of natural flow current sets influid due to density variation or buoyancy.
Sol–113: (b)Nusselt number in fully developedturbulent flow in pipe,Nu = 0.023 Re0.8·Prn
wheren = 0.4 when fluid in pipe is being heated.n = 0.3 when fluid in pipe cools down. For heating
Nu = 0.8 0.40.023 Re Pr
This expression is also called as DittusBoelter equation or co-relation
Sol–114: (c)Flow pattern over heated vertical cylinder,
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(19) ME (Test-A3), Objective Solutions, 16th October 2016
T 10°C=
Pipe
d
T =150Cs
L
The heat loss from wall pipe-
Q = h A ΔT
Q = h× d × LΔT ...(i)
Q1 = 1 1h dL T ...(ii)
Q2 = 2 2h dL T ...(iii)
For laminar flow,
Nu = 1/4L rC × Gr P
hLk = 1/43
1C K L
h 1/41
L
Dividing these two equations
1
2
QQ =
3/41
2
LL
L2 =
3/43
2
640 2L
1/4
2
640L = 2
L2 = 640 40 cm16
Sol–115: (b)The relationship between convective heattransfer and distance in laminar flow overflat plate,
hx = 1/2C
x
x
hx
A
hx = 1/2C
xAverage value of heat transfer coefficientfor length x,
ha = x
x0
1 . h .dxx
h0 = x
x0
1 h dxx =
x
10 2
C dxx
x = 1/22C x
x
ha = 2 hx
ahh x
= 2
Sol–116: (d)Area of air heater
A = 60 m2
Overall heat transfer coefficientU = 100 W/m2K
Heat capacity rate of both hot and coldfluid,
Ch = Cc = 2000 W/K Number of Transfer Units i.e. NTU
NTU = min
UAC = 100 60
2000 = 3
Sol–117: (c)We know convective heat transfercoefficient for flat plate is locationdependent, so average vlaue
h = L
0
1 . h x .dxL
The variation of h(x) along flate platelength,
xCh =xh( )x
x
Boundary layer
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(20) ME (Test-A3), Objective Solutions, 16th October 2016
Sol–118: (d)
xNux
U
Local Nusselt number at distance x fromleading edge of plate of length l
Nux = 0.5 0.330.332 Re Pr
Now the average Nusselt number over theplate length l,
Nu = L
x01 Nu dL
x
Nu = 0.5 0.33L0.664 Re Pr
Nu = 2 Nux
Sol–119: (b)The heat capacity ratio–
C = min
max
CC
Since heat lost by hot fluid is equal toheat gained by cold fluid. From thisconcept it is concluded that a fluid hav-ing small specific heat has maximumtemperature change so
hT = 160 – 85 = 75°C
cT = 135 – 35 = 100°C
Heat transferred from hot fluid to coldfluid,
h hC T = c cC T
h
c
CC =
c
h
TT
cT > hT
hC < cC Cc is minimum and Ch maximum
heat capacity ratio
C =
c h
h c
C T 75= =C T 100 = 0.75
Sol–120: (c)
Thi
Th0
TciTc0
Length
Tem
pera
ture
Since hot fluid is cooled and cold fluid isat constant temperature along length ofHE so it is boiler
Sol–121: (c)The capacity ratio in heat exchanger isdefind as–
C = min
max
CC
For phase change phenomenon i.e.evaporation or condensation, heat capac-ity is very very large i.e.Cmax Capacity ratio-
C =minC 0=
Then the effectiveness expression forboth counter and parallel flow configu-rations changes to
In a parallel system if all component thenonly the system fails
R =1 0.9 R = 0.92 R = 0.93
C1 C2 C3
Rsystem = R1 × R2 × R3= 0.93 = 0.729
So, system with series componetns hasreliability lower than its individualcomponent reliability.
Sol–134: (c)Reliability of a system depends upon
• The reliability of each of its components• The type of system, whether series, parallel
or combination,Sol–135: (d)
Several methods are in use for growingC-60 (i) arc process, (ii) laser ablation and(iii) chemical vapour deposition (CVD).
Sol–136: (b)Pitting corrosion occurs when oxide filmon a metal breaks due to abrasion.
Sol–137: (d)Following are some methods used toprevent corrosion:
• Use of noble metals• Use of corrosion and oxidation resistant
materials. (Cu + Zn), (Cu + Sn) etc.• Use of protective coating• Use of inhibitors• Design of components to avoid formation
of galvanic cell• Deaeration of water, and• Cathodic protection
Sol–138: (d)Wet corrosion occurs due to following:
• Phase cell or galvanic cell• Concentration cell• Stress cell
Sol–139: (d)With three components in series eachhaving reliability 0.9, the system reliabilityis
Rs,3 = (0.90)3
= 0.729• With two more components added, totalling
to 5 components in series, Rs5 = (0.90)5 =0.59049
With 6 components in series,Rs–6 = 0.53144
With 7 components in series,Rs–7 = 0.4783
Thus, with series systems, reliabilitydecreases as the number of increases.
Sol–140: (b)Rs = (0.96)4
C1 C2 C3 C4Input
0.96 0.96 0.96 0.96Output
Sol–141: (d)The probability that any component isfunctioning is 0.999. Since the productoperates only if all 50 components areoperational, the probability that the 50components are functioning is
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(23) ME (Test-A3), Objective Solutions, 16th October 2016
Rs = (0.999)50
Sol–142: (c)• The vibration is strongly directional• Rumbles during runup or coastdown as
harmonics pass through the naturalfrequency.
• The vibration is very bad i.e. highamplitude.
Sol–143: (d)• Lack of lubrication may result in
vibrations.Sol–144: (d)
Forced vibration on machines are correctedby mass balancing, aligning and changingthe bad parts.
Sol–145: (d)All these cause excessive vibration in gears.
Sol–146: (a)There are five known fixes for resonance:
• Change speed• Change the natural frequency of the
responding part with added mass atstiffness.
• Add damping• Reduce the source vibration input• Dynamic absorber
Sol–147: (b)Vibration monitoring is used for rotatingmachines to prevent failure and thus helpin increasing their normal life span.
Sol–148: (b)For a rotating machine, the rotationalspeed of the machine is not constant;during start-up and shutdown and evenunder steady state the rotational speedwill vary. So, time frequency analysistechnique use rotational speed to detectfaults.
Sol–149: (c)The oxygen rich region becomes cathodicwith respect to oxygen lean region, thusresulting crevice corrosion.
Sol–151: (a)Reliability applies to a specified period oftime means that a system has a specifiedchance that it will operate without failurebefore time t. So, reliability engineeringensures that components and materialswill meet the requirements during thespecified time.
Sol–152: (b)Both statements are individually correct.In preventive maintenance the regularcheck up is kept to maintain the machinesin good working conditions.
Sol–153: (a)In the late life of the product the failurerate increases, as age and wear lead tohigher chances of failure of the product.This late life represents wear-out failures.
Sol–154: (a)
If the material is homogeneous & isotropic,magnitude of deformation will be same ifE & µ are same in all direction.
x =
yx zE E E ... (i)
x = y zx
E E E... (ii)
Magnitude of x in (i) as well as (ii) is same.Sol–155: (a)
In mild steel tension specimen failure is cup& cone failure where failure system is at45° with the longitudinal axis
45°hence statement A is correct.
From the Mohr circle diagram of directtension member.
f
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(24) ME (Test-A3), Objective Solutions, 16th October 2016
(0, f/2)
Normal to max shear plane
Normal to max principal plane(f, 0)
2
max shear plane is at with the normalto may principal plane when 2 90= 45 .Thus max shear at 45° to the axis.Hence reason is also correct and reason isthe correct justification for assertion.
Sol–156: (b)
Rate of change of bending moment
dM Vdx
Sol–157: (a)
A B
AbR Psin
a b
BaR Psin
a b
P
a b
Sol–158: (c)
Macaulay’s method is based upon themodification in double integration method.This method can be used for any kind ofdiscontinuous loading by using principleof superposition. Hence R is wrong.
Sol–159: (a)
Let 1 2and be two stress
n =
1 2 1 2 cos22 2
for < 45° cos 2 will always be +ve
So, n > 1 2
2
R = 1 2
2
P ( , )PP
21O2
P =
1 2 1 2 cos22 2
Sol–160. (d)
Plane stress Plane strainStress 0, 0, 0 0, 0z xz yz xz yz
, and , , and x y xy x y z xymay be non-zero may be non-zero
Strain 0, 0 0, 0, 0xz yz z xz yz, , and , , x y z xy x y xy
may be non-zero may be non-zero
yx zx E E E
y z xy E E E
yz xz E E E
If z = 0, that doesnot mean z = 0 [except,in case when = 0 i.e. ideal material orwhen x = –y]Similarly,
If z = 0, that doesnot mean z = 0 [exceptin case when = 0 i.e. ideal materialor when x = –y]
Thus note that plain stress and plainstrain components are not same.
In term of stress, z = 0 in plane stressbut z may not be zero in plane strain.
Sol–161: (b)This is a case of inelastic collision, wherethe linear momentum remains conserved.However, the initial and final kineticenergies are different.Also the linear momentum in verticaldirection will not be conserved because ofthe force of gravity.
Sol–162: (d)The particle under equilibrium willremain at rest or move with constantvelocity. Since the net force is zero, hencehere will be no acceleration. Hence, theparticle may have constant velocity.
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(25) ME (Test-A3), Objective Solutions, 16th October 2016
Sol–163: (a)As the ball moves upwards, then it willbe retarded both by the force of gravityas well as the resistance of air. The totalenergy of the ball will decrease due to theresistance of air.But, when the ball starts falling down,the force of gravity will act in its directionof motion, while the resistance to air willbe in opposite direction. Hence, themagnitude of acceleration will be less inthis case when compared with themagnitude of retardation when the ballmoves upwards in the earlier case. Thus,time t1 will be less than the time t2. Also,the average speed while coming down willbe less than the average speed while goingup, because of the loss of energy inresistance to air.
Sol–164: (a)An object may be subjected to severalforces. If the size and direction of forcesare exactly balanced, then there is no netforce acting on the object and the objectis said to be in equilibrium.
Sol–165: (a)Total energy = kinetic energy + potentialenergy. During SHM, velocity becomeszero twice at the end points, hence thekinetic energy also becomes zero at thetwo end points. Thus, total energy getsconverted into potential energy whichthus reaches its maximum value.
Sol–166:(a)3´
3 2´2
14´4ps
pD
pD´
Enthalpy
h = h h = hf3 4 f3´ 4´ h2 h2́h1
Thus we can see from the figure that asthe condenser pressure increases from2 to 2', the work of compression
increases from 1-2 to 1–2' and therefrigerating effect decreases from 4–1to 4'–1.
Sol–167:(a)As the temperature required in the iceplant are lower, hence the suctionpressure decreases. With decrease insuction pressure, refrigerating effectdecreases and work required forcompression increases and hence COPdecreases.
Sol–168:(a)
T2
Q2W R,HP
Q1T1
(COP)HP = (COP)R + 1
(COP)HP = 2 1R
Q Q; COPW W
where W = Q2 – Q1
Sol–169: (a)Although R-12 is non-toxic, nonflammableand non-explosive, it disintegrates intotoxic products after coming into contactwith flame or electrical heatingelements. Thus, whenever there isleakage, it is advised to put all theflames off and keep the door open sothat it can escape to the openatmosphere.It is used in household appliancesbecause it is normally non-toxic, highlystable, suitable for wide range ofoperating conditions and miscible withoil. It does not disintegrate even underextreme operating conditions.
Sol–170: (a)Decrease in pressure in suction pipe linemeans increase in pressure ratio andhence volumetric efficiency of reciprocatingcompressor reduced.
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(26) ME (Test-A3), Objective Solutions, 16th October 2016
Increase in temperature, reduces densityof vapour refrigerant.Both these factorcontributes to reduce in refrigerant massflow rate. Hence refrigerating capacity ofthe system reduces.
Sol–171: (c)First the high pressure refrigerant leaksand then the low-pressure refrigerant.High pressure refrigerants have lowerboiling point.Hence statement (I) is true but statement(II) is false.
Sol–172: (d)The coating on window glass is done onoutside or external surface. The materialof coating is Indium oxide (In2O3). Theexternal coating will reduce convectivecooling load due to reduced inside surfacetemperature.Ordinary window panes are transparentto solar radiations. Coating is done toreduce this transparency.
Sol–173: (c)
We know that,
(COP)HP = 1 + (COP)R
Hence, COP of heat pump is more thanthe COP of its refrigerator. The workrequirement is same in both the caseswhether the device is used as a heat pumpor a refrigerator.
Sol–174: (a)Natural convection current when hotsurface is upward.
Convection Current
Natural
Hot surface upward
Plume
Now hot surface faces down.
Hot surface downward
Plume
In this situation heat loss is mainly byconduction of air which is not effective asconvection.
Sol–175: (a)From figure it is clear that the movementof air particle is highest near (forward)stagnation point (or in vicinity) and thelocal heat transfer coefficient (convectioncoefficient) is highest.
S(forward)
Stagnationpoint.
Separationpoint
Separationpoint
Sol–176: (c)In order to have high effectiveness of fins,the fins are provided on gas side.The resistance offered by convective heattransfer,
= 1hA
For low ‘h’, i.e., gas side, the resistance ishigh. But at the same time, this conditionis desirable from effectiveness of fin pointof view.
Sol–177: (c)In Counter flow HE, the heat transfer fromhot fluid to cold is at more uniformtemperature difference than parallel. Inparallel flow configuration the temperaturedifference at inlet is very high and thereis huge entropy generation i.e. loss ofavailable energy which can never berecovered. So due to less entropygeneration, the counter flow HE are moreefficient thermodynamically.At the same time, the LMTD of counterflow configuration is more than parallelflow.
Sol–178: (d)In step growth polymerization polymer isproduced by bi-functional ormultifunctional monomers react to formdimers than trimers and so on. In additionpolymerization reaction there must exist
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(27) ME (Test-A3), Objective Solutions, 16th October 2016
at least one double bond for reaction.Sol–179: (b)
Ceramic tools have very brittle tool tips,that is why they are prone to impactloads. These tools are used on hard tomachine work material such as cast ironas they are highly wear & abrasionresistant.
Sol–180: (c)
When subjected to compressive load, castiron fails at an oblique plane due to shear.The shear strength of cast iron is morethan tensile strength but considerably lessthan its compressive strength.
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