d:mohd. kalim 2018all branch · during reversible non-flow process, work done w = p v p v 1 1 2 2 n...

1. (c)

2. (a)

3. (a)

4. (a)

5. (a)

6. (a)

7. (c)

8. (d)

9. (c)

10. (a)

11. (c)

12. (d)

13. (c)

14. (c)

15. (a)

16. (c)

17. (c)

18. (b)

19. (d)

20. (b)

21. (c)

22. (b)

23. (b)

24. (c)

25. (d)

ESE-2019 PRELIMS TEST SERIESDate: 16th December, 2018

26. (d)

27. (b)

28. (b)

29. (a)

30. (d)

31. (b)

32. (d)

33. (c)

34. (c)

35. (d)

36. (a)

37. (c)

38. (a)

39. (c)

40. (b)

41. (b)

42. (c)

43. (c)

44. (c)

45. (a)

46. (a)

47. (d)

48. (c)

49. (b)

50. (b)

51. (b)

52. (a)

53. (d)

54. (b)

55. (b)

56. (b)

57. (a)

58. (d)

59. (d)

60. (b)

61. (d)

62. (c)

63. (b)

64. (d)

65. (c)

66. (d)

67. (a)

68. (d)

69. (b)

70. (b)

71. (b)

72. (a)

73. (b)

74. (a)

75. (b)

ANSWERS

76. (d)

77. (b)

78. (a)

79. (b)

80. (c)

81. (b)

82. (d)

83. (c)

84. (b)

85. (b)

86. (a)

87. (c)

88. (d)

89. (d)

90. (c)

91. (b)

92. (b)

93. (c)

94. (d)

95. (b)

96. (c)

97. (c)

98. (b)

99. (d)

100. (d)

101. (d)

102. (a)

103. (c)

104. (a)

105. (b)

106. (c)

107. (d)

108. (a)

109. (b)

110. (d)

111. (b)

112. (b)

113. (c)

114. (a)

115. (d)

116. (a)

117. (d)

118. (b)

119. (c)

120. (d)

121. (d)

122. (c)

123. (b)

124. (d)

125. (d)

126. (a)

127. (b)

128. (a)

129. (b)

130. (b)

131. (a)

132. (b)

133. (a)

134. (b)

135. (c)

136. (a)

137. (d)

138. (c)

139. (c)

140. (b)

141. (b)

142. (b)

143. (b)

144. (c)

145. (c)

146. (b)

147. (c)

148. (c)

149. (b)

150. (d)

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[ME], ESE-2019 PRELIMS TEST SERIES PAPER-II (FLT-03) (2)

1. (c)2. (a)

xdu 3dx

yzdw dv 0 0 0dy dz

3. (a)4. (a)

P MA Z

Comp,max 4

P eP P PedA Ad Ad 2864

P e1A d 8

ten, maxP e1A d 8

comp, max

ten,max

8e1 13d8e 31d

de5

5. (a)

Cr E

1 1 1 1 11P P P 600 400 240

P = 240 kN

6. (a)7. (c)

Due to Poisson’s effect, longitudinal strainleads to lateral strain which will lead toanticlastic curvature for an originallyrectangular cross-section.

For a doubly symmetric section, shearcentre always coincide with the centroid ofthe section.

8. (d)From Mohr circle :

(120, 0)

( , 50)n

( , 0) (70, 0)

Radius of Mohr Circle = 50 MPa = 120

2

20MPa

Maximum shear stress = 120 20

2

= 50 MPa

max 50 20 30 MPa

9. (c)

1 2 1 2max, abs max , ,

2 2 2

For thin spherical cells, 1 2and are same equal

to pd4t .

max, abspd8t

10. (a)

5 KN/mA

25 25

Shear force V at point A = 25 – 2.5 × 5

= 12.5 KN

NA50

100

Shear Stress

=

3

3VAy 12.5 10 150 100 100Ib 300150 150

12 = 0.3703 mpa

11. (c)Applying Bernoulli's equation when valve is closed.

V1 = V2 = 02 2

1 1 2 11 2

P V P VZ Z2g 2g

h

o

2 11 2

P PZ Z

3

1 270 10z z 7m

10 1000

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Now when valve is open2 2

1 1 2 21 2 L

P V P VZ Z h2g 2g

1 21 2 L

P PZ Z h

1 2V V

23 + 7 = hL Lh 30m .

12. (d)u = 2y, v = 8x + 2

Equation of streamline, dy vdx u

dydx =

8x 22y

2y dy = (8x + 2) dx

Integrating both sides, we get

y2 = 4x2 + 2x + c

y2 = 2x (2x + 1) + c

13. (c)

Displacement thickness, * =

00

u1 dyu

= 0

y1 dy

=

2

0

yy2

=2

Momentum thickness, =

0

0 0

u u1 dyu u

= 0

y y1 dy

=

2 3

20

y y2 3

= 6

shape factor =

* / 2 3

/ 6

Alternatively; for

m

o

u y * m 2u m

Hence shape factor =

1 2 3

114. (c)

For maximum efficiency, H = 3hf

H = 2

53f l Q12.1d

100 =

2

53 0.03 50 Q

12.1 (0.5)

Q = 2.9 cumecs = 2900 l/sec.

15. (a)

Head loss in pipes connected in parallelwill be same irrespective of their sizes andlengths.

Water hammer pressure is more in rigidpipes than flexible pipes.

16. (c)The instantaneous velocity Vd in the delivery pipemay be obtained as :

Vd =d

A r sina

(Vd)mean = d2

00

A r sin da

d

= d

A ra

(Vd)max = d d

A A rr sin 90ºa a

d mean

d max

(V ) 1 .(V )

17. (c)

To reduce the acceleration head we needto reduce the length of suction pipe (ls) andlength of delivery pipe (ld) in which fluctuationof velocity occurs. This is done by the fittingair vessels closed to cylinder as possible.Thus between air vessel and cylinder onlyfluctuating velocity will occur. Below airvessel on suction side and above air vesselon delivery side velocity will be constant.

Due to movement of piston, when there isexcess velocity above average b/w suctionside air vessel and cylinder, water issupplied from air vessel.

18. (b)When local pressure falls below the vapourpressure, boiling starts taking place which leadsto cavitation.

So (b) is correct

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19. (d)

w wh = 1 1h

h1 =

w

1

w

h 10

20. (b)A. Pitot tube – used to find velocity by measuring

stagnation pressure head (4)

B. Micro-meter – Measures differential pressure(3)

C. Pipe bend meter – measure rate of flow (2)

D. Wall pressure tap – measures flow staticpressure (1)

21. (c)In characteristics curve of a pump, the curve isdrawn between (head/efficiency/input power) onordinate vs discharge (Q) on abscissa.

Hence ‘c’ is correct.

22. (b)Force along the slope opposing the motion isF = R(mgsin F )

where resistance due to friction, air etc. is

FR = 30 N/kg × 200 kg

= 6000 N

sin = 1

200

F =

1200 10 6000200

= 10 + 6000 = 6010N

Power = F × V

= 6010 × 10 = 60.1 kW

23. (b)24. (c)

Maximum efficiency

max = =1000 300 0.7 or 70%1000

Heat input Q = 500 kJ

Maximum work output

Wmax = Q max = 500 × 0.7 = 350 kJ

Minimum heat rejected

= Q – Wmax = 500 – 350 = 150 kJ

25. (d)Total heat supplied

Q1 = =57 1080 1026W60

T1 = 285 K

Q1 HE

Q2 W

T = 275 K2

For reversible heat engine

1 2

1 2

Q QT T

= = = 1

2 21

Q 75Q T 1026 990 WT 285

Work output = Q1 – Q2

= 1026 – 990 = 36W

26. (d)If the gas obeys Van der Waal’s equation at thecritical point, then,

c c c2

c

ap b RT

At critical point, =

cT T

p = 0

and=

c

2

2T T

p = 0

hence, b = c and 3

a = 2c c3p

2c c c

c cc2c

3pp RT3

cc

24p

3 = RTTc

c c8p3

= RTTc

c

c c

RTp = =8 2.67

3

27. (b)I. The flow through insulated constant diameter

pipe

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1 2

Applying steady flow energy equation

h1 + Q = h2 + w

Q = 0 = w

So, h1 = h2 = constant

Its enthalpy which remains constant along withflow.

For an isolated system, isoS 0 .

28. (b)First law for a stationary system in a processgives

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

= 0.15 (0.2 – 0.4)MJ= –30 kJ

and givenQ1–2 = –45 kJ

from (i)–45 kJ = U2 – U1 – 30kJ

U2 – U1 = –45 + 30

= – 15 kJ29. (a)

During reversible non-flow process, work done

W =

1 1 2 2p v p vn 1

As n increases, W decreases.

Further adiabatic mixing is irreversible process.

30. (d)Over heat transfer coefficient (U)

i 0

1 1 L 1U h K h

1 1 0.15 1U 25 0.15 25

1 27 25UU 25 27Hence, it is closer to the heat transfercoefficient based on the bricks alone.

Hence, (d) is the correct answer.

31. (b)32. (d)

L = 10 cm, F12 = 0.1716

F1-1 + F1-2 + F1-3 = 1F1-1 = 0

F1-3 = 1 – F1-2 = 1 – 0.1716 = 0.8284

AA1F1-3 = A3F3-1

F3-1 = 1

1 33

A FA

=

2(10)4 0.8284

10 10= .2071

33. (c)

x x1/3 1/31 Kh h

x x

x

average 1/30

1 Kh dxL x

= 4

1/30

1 K dx4 x =

42/3

0

1 xK243

= 42/30

3 K x8

= 2/33 K(4)8

at x 4 1/3Kh

(4)

1

3

2/3avg 1/3

at x 4

3 K(4)h 38 (4)h 8K 4

34. (c)

n n

x 2

x 2lim 80x 2

n 1

x 2lim nx 80

n 1 5 1

x 2lim nx 5.2

n = 5

n shields

without shields

Q 1 1 1Q n 1 5 1 6

35. (d)

LC = Volume (V)

Surface area (As)

=

3

2

4 R R 0.33 0.053 3 24 R

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Bi =

chL 200 0.05 0.025K 400

Bi < 0.1

Bi = Conduction resis tance (internal)Surface convective resis tance

Bi < 0.1

Hence, (d) will be correct answer.

36. (a)At the insulated surface, heat conducted = 0

so, dTkAdx

= 0 (for insulated face)

dTdx

= 0 (condition for maxima/minima)

T is max at insulated face.

37. (c)Density of air decreases with increase intemperature and hence specific gravity alsodecreases.

38. (a)Thin film orboundary lubrication(unstable)

Thick film lubrication(stable) PS

QR

A

K

p

Coef

f. of

frict

ion

()

(min)

Partial lubrication

N

39. (c)40. (b)

Given, h = 12 mm, = 150 N/mm2

Shearing load is given by

Ps = 0.707 × h × L ×

or sPL

= 0.707 × 12 × 150 = 1272.6 N/mm

= 12.726 kN

cm41. (b)

Compression and transfer molding are used forshaping thermosetting plastics.

42. (c)Flow-line layout – Automobiles

Process layout – Job shop production

Fixed position layout – Aeroplanes

Hybrid layout – Flammable, explosiveproducts

43. (c)

Sensitivity of LVDT = Output currentDisplacement

=

333 10 4 10 A / mm

0.75Sensitivity of instrument = Amplification factor× sensitivity of LVDT

= 4 × 10–3 × 200 = 0.8 A/mm

44. (c)45. (a)

Solidity = Projected area of blades

Swept area

=

NC 20 0.15

R 1.5 = 0.6366

46. (a)Pneumatic sensors may be contact less devicesor contact sensors. They can be used as proximitysensors, touch sensors, or pressure or forcesensors.

47. (d)48. (c)

SO2 can be removed in the combustion processby adding limestone to the fluidized bed,el iminating the need for an externaldesulfurization process.

Because of the reduced combustiontemperature, NOx emission are inherently low.

49. (b)High specific volume of vapour leads to large sizeof compressor and work requirements alsoincreases.

50. (b)51. (b)

Bypass factor =

21 12 0.2530 12 0.7 0.25

= 0.475 kg/kg of dry air

52. (a)

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53. (d)54. (b)

The cooling of air between two stages ofcompression is known as intercooling

Due to intercooling:

(i) Work of compression is reduced

(ii) Specific out of the plant is increased

(iii) Thermal efficiency is reduced, however this losscan be remedied by employing exhaust heatexchanger as in regeneration.

55. (b)In De-laval turbine,

Work done = change in kinetic energy

600 = 2 22

1 1 (40 V )2

V2 = 20 m/s

56. (b)

Specific steam consumption = net

3600W

= T P

3600 3600W W 939.54 3.54

= 3600936

= 3.84

57. (a)

In creep feed grinding MRR is high, feed rateis very slow, grinding wheels are mostly of softergrades, and reduced temperature at the worksurface.

58. (d)

59. (d)

Peak pressure generated by explosive dependsupon

Types of explosive

Weight of explosive

Standoff

Compressibility of energy transmitting medium

Acoustic impedance of energy transmittingmedium, which is equal to the product of itsdensity and sound velocity in the medium

60. (b)

61. (d)

No-Go

Go

Tolerance

Some of the good parts are rejected

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

63. (b)64. (d)

The flue gas passes through the stages ofsuperheater, and then economizer. In the airpreheaters, the flue gases transfer their residualheat to the combustion air, during which theyare cooled to the exit flue gas temperature ofthe steam generator. For further cleaning, theflue gas is conducted through an electrostaticprecipitator to remove dust.

65. (c)Goodman line is safe w.r.t. fatigue failure but thereis small portion inside the Goodman line wherecomponent could fail due to yielding, hencemodified Goodman diagram combines fatigue failurewith failure by yielding.

66. (d)67. (a)

For the transverse vibration of a shaft carrying severalloads, there are two methods of finding naturalfrequency of the system: Dunkerley’s method andenergy method. Dunkerley’s method givesapproximate results but is simple. This is usedwhen the diameter of the shaft is uniform. Theenergy method gives accurate results but involvesheavy calculations in case there are many loads.This is also known as Rayleigh’s method.

68. (d)When a point on one link is sliding along anotherrotating link, such as in quick return motionmechanism, then the coriolis component ofacceleration is taken into account. These is nosuch occurence in the case of slider crank orscotch yoke mechanism.

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69. (b)Mean time between failure (MTBF) = 1200 hr

Mean time to repair (MTTR) = 36 hrs

So, Availability =

MTBF 1200MTBF MTTR 1200 36

= 12001236

= 0.9708

70. (b)Annual demand = 94600 units

Available time = 46 weeks/year × 6 shift/week

× 8hr/shift × line

= 46 × 6 × 8 × 0.98

= 2163.84 hrs

The cycle time TC = Available time

Annual demand

=2163.8494600

= 0.0228 hrs/unit

= 1.372 minutes/unit

71. (b)

Pr ocess Critical ratio time Due Due dateOrder = remaining date

Pr ocessing time(day)P 5 25 5Q 17 35 2.05R 7 22 3.142S 9 27 3

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

72. (a)

Arrival rate = 45 customer/hr = 4560

customer/min

Service ratio, = 60 customer/min

Utilization =

45 360 4

Average number of customer in the queue

Lq =

2

1

=

23 4314

= 2.25

Average waiting time of the customer in the queue

Wq =

qL 2.25

3 4 = 3 min

73. (b)D = 60 units/days, Co = 120/order, Ch = Rs 0.05/unit day, Lt = 5 days

Optimal order quantity = Q = 0

h

2DCC

= 2 60 120

0.05 = 537 units

Safety stock= Lt × d

= 5 × 60

= 300 units

74. (a)

0

/2/2

/2

OO

C B

A

Given,

OA = 300 mm

OB = OC = 150 mm

OB 150cos 0.52 OA 300

60 120

2

Hence Time of cutting strokeTime of return stroke

=

360 360 120 240 2120 120

75. (b)76. (d)77. (b)78. (a)

Part programming generation of cutter locationdata.

Hence ‘a’ is correct.

IES M

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79. (b)

The rotation of the index crank = 40 31 turns28 7

One full rotation + 9 holes in 21 hole circle in platenumber 2.

Hence, ‘b’ is correct.

80. (c)

Tetragonal crystal structure :

a b c, 90

81. (b)

Thermoplastic material have no specific glasstransition temperature.

Thermosetting material have specific glasstransition temperature.

Thermoplastic material has mostly linearstructure Eg : polyethylene.

82. (d)

When polymers are subjected to bending orfolding then polymers may become white incolour and this phenomena is known as stresswhitening.

83. (c)

Piezoelectric transducers are more effectivethan magnetorstrict iv e t ransducers butmagnetorstrictive transducers are more ruggedcan be used at higher powers and have morelife than piezoelectric transducers. Hencemagnetorstrictive transducers are mostly usedin ultrasonic machining.

84. (b)

85. (b)

1) Complete breakdown of the steady throughflow is called Surging

2) In high pressure ratio multistage compressor,the axial velocity is relatively small in theHigher pressure stages.

3) Separation of flow from the blade surfaces isStalling.

86. (a)

87. (c)

88. (d)

The cyclone collection efficiency increases withincreasing (a) particle size, (b) particle density,(c) inlet gas velocity, (d) cyclone body length,(e) number of gas rev olutions, and (f )smoothness of cyclone walls. The collectionefficiency decreases with increasing (a) cyclonediameter, (b) gas outlet duct diameter, and (c)gas inlet area.

89. (d)

90. (c)

91. (b)

For the same output, the weight per horsepowerof a gas turbine is about one third that ofreciprocating piston type internal combustionengine.

92. (b)

The Turgo turbine is an impulse water turbinedesigned for medium head applications.Operational Turgo turbines achieve efficienciesof about 87%.

93. (c)

When n

1

, transmittivity factor is negative.

This means that there is a phase difference of180° between the transmitted force and thedisturbing force o(F cos t) .

94. (d)Critical damping reponse is much more fasterthan Overdamping response.

Damping ratio for critically damped system isequal to one.

Door closing mechanism is a overdampedsystem.

95. (b)When the load increases, the configuration ofthe governor changes and a valve is moved toincrease the supply of the working fluid.

When the load decreases, the engine speedincreases and the governor decreases the supplyof working fluid.

The function of a flywheel in an engine is entirely

IES M

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different from that of a governor. It controls thespeed variation caused by the fluctuations of theengine turning moment during each cycle ofoperation. It does not control the speed variationscaused by a varying load

96. (c)97. (c)

Since the rolling motion takes place between thecontacting surfaces (i.e. the roller and the cam),therefore the rate of wear is greatly reduced. Inroller followers also the side thrust exists betweenthe follower and the guide. The roller followersare extensively used where more space isavailable such as in stationary gas and oil enginesand aircraft engines.

In radial cams or disc cams, the followerreciprocates or oscil lates in a directionperpendicular to the cam axis.

98. (b)99. (d)100. (d)

Primary force = 2mr cos

Primary couple = 2mr cos

Secondary force =

2mr cos2n

Secondary couple =

2mr cos2n

An in-line four-cylinder four-stroke engine hastwo outer as well as two inner cranks in line.The inner cranks are at 180° to the outer throws.Thus, the angular positions for the cranks are:

for f irst, (180 ) for the second,

(180 ) for the third and for the fourth.

Primary force

= 2mr [cos cos(180 )

= cos cos ] 0180

Primary couple

= 2 3 1mr cos cos 1802 2

= 1 3cos cos 0180

2 2

= 3 cos 0180

2

Secondary force

= 2mr cos2 cos 360 2n

cos cos2360 2

= 24mr cos2

n

Secondary couple

= 2 3 1mr cos2 cos 360 22 2

1 3cos cos2360 22 2

= 0

101. (d)S.I. engines are quantative governing engines asair-fuel mixture supplied through carburator so wecan not have qualitative governing.

102. (a)103. (c)104. (a)

In rating of SI engine fuel, ISO-octane is assigned100 octane number and normal heptane as zerooctane number.

105. (b)106. (c)

Demand in December, Ddec =30 units

Fore cast for December, Fdec=35 units

= 0.4

Fore cast for Jan 2017

Fjan = dec dec decF (D F )

= 35 + 0.4(30 – 35)

= 35 + 0.4 x – 5 = 33 units

107. (d)

(TC)I = FI + xVI = 100000 + 75x

(TC)II = FII + xVII = 190000 + 60x

For break even point for both

alternatives

(TC)I = (TC)II100000 + 75x = 190000 + 60x

15x = 90000 x = 6000

IES M

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108. (a)Given, LP

x y 23x 5y 15x,y 0

x y2 2 = 1

x y15 153 5

= 1or x y 15 3

3x + 5y = 15x + y = 2

(2, 0) (5, 0)

(0, 2)

(0, 3)

109. (b)

I

III III

RI = 0.8, RII = 0.85, RIII = 0.9

Reliability of two parallel units

= [1 – (1 – R1)(1 – R1)] = [1 – (1 – 0.8)(1 – 0.8)] = 1 – (1 – 0.8)2

Reliability of series system

= [1 – (1 – 0.8)2] × RII × RIII = [1 – (1 – 0.8)2]× 0.85 × 0.9 = 0.7344

110. (d)When a beam of charged particles passesthrough a magnetic field, forces act on theparticles and the beam is deflected from itsstraight line path. A current flowing in a conductoris like a beam of moving charges and thus canbe deflected by a magnetic field. This effect iscalled the Hall effect.

Magnetic field

Current

Potential differenceproduced by deflection

of electronsFig. : Hall effect

111. (b)A ROM is a non-volatile memory. It storesinformation permanently.

It is not accessible to user, and hence he cannot write anything into it.

112. (b)Maximum wind power

Pmax = 3i

16 1 AV27 2

60 × 1000 = 316 1 1.2 3 A27 2

A = 6250 m2

113. (c)

Revolving joint is represented by letter ‘V’. Sincethe arm movements in dif ferent robotconfigurations are different, the work volumes orwork envelopes of different coordinate systemsare also different.

114. (a)

C30 S30 0 0S30 C30 0 0T 0 0 1 0

0 0 0 1

3 1 0 02 21 3 0 02 20 0 1 00 0 0 1

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115. (d)

116. (a)

Stepper motors have high vibration levels due tostepwise motion

117. (d)

118. (b)Let velocity of car = VRelative velocity of car coming from oppositedirection = V + 45

Using relative velocity concept, 10 6

V 45 60

V = 55 Km/hr

Hence, ‘b’ is correct.

119. (c)The continuity for unsteady flow is

Q Ax t

= 0

QAt x

A 0.10t

120. (d)121. (d)

Sectional dimension is chosen such that the ratioof distance from NA to extreme fibre in tensionand compression is exactly same as the ratio ofallowable stresses in tension and compression.Under this provision section will be mosteconomically utilized.

122. (c)In I-section more area is located farther awayfrom neutral axis. Generally more than 80% ofBM is resisted by the flanges only.

123. (b)124. (d)

Flow always taken place in the direction fromhigh total energy to low total energy.

125. (d)

Pressure gradient dpdx

is positive due to

decrease in velocity in the region of boundarylayer separation.

126. (a)Beta distribution model for PERT

Gives

0 m pE

t 4t tx t

6

Weightage of to & tp is equal.

127. (b)

128. (a)

129. (b)

130. (b)

131. (a)

132. (b)

133. (a)

When Bi < 0.1, the variation of temperature withlocation within the body is slight and canreasonably be approximated as being uniform.

134. (b)

135. (c)136. (a)

Heat cannot be extracted from the flue gassesfor economiser, superheater etc. since theeffective draught will be reduced if the temperatureof the flue gasses is decreased.

137. (d)A is false, R is true

As rake angle ( ) increases, shear angle ( )increases hence shear area decreases and dueto lower shear area, cutting force decreases.

138. (c)In resistance projection welding, multipleprojection can be applied during embossing toplate having higher thickness and higherconductivity material hence multiple spot weldingcan be done simultaneously.

139. (c)Assertion is correct but the moving ion is cationas it is smaller in size and hence more mobilethan anion.

140. (b)A certain water level in the drum should bemaintained to give adequate steam space andreduce Priming.

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141. (b)

Volumetric efficiency =

1nH

L

P1 C 1P

142. (b)In the fixed blade, the velocity obtained in themoving blade is transformed into pressure energy.

143. (b)Rateau/Pressure compounding of impulse turbineis done to decrease high rotational speed ofimpulse turbine and bring it to a more practicallimit.

144. (c)Governor is said to be isochronous when theequilibrium speed is constant (i.e. range of speedis zero) for all radii of rotation of the balls withinthe working range, neglecting friction.

The isochronism is the stage of infinite sensitivity

145. (c)Limiting addendum for pinion is larger than thatof gear for equal addenda, interference takesplace between the tips of the wheel teeth andthe flank of the pinion teeth.

146. (b)

In SI engine, the spark is advanced to have peakpressure around 10° after top dead center duringexpansion. Because first phase of ignition is timedependent not speed. So at higher speed, moreangle is rotated in given time. This ensures goodbrake power. The turbulances in chamber arealso increased by high speed, so the flamevelocity.

147. (c)Free float = Total float - Head event slack.

148. (c)

The tape reader must be omitted in DNC, thusrelieving the system of its least reliable component

149. (b)

150. (d)Wind energy power plants is among indirectmethods of harnassing solar energy.The direct means include only thermal andphotovoltaic.

d:mohd. kalim 2018all branch · during reversible non-flow process, work done w = p v p v 1 1 2 2 n...

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