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ESE-2017 PRELIMS TEST SERIESDate: 13th November, 2016

24. (b)

25. (c)

26. (a)

27. (c)

28. (d)

29. (c)

30. (a)

31. (b)

32. (a)

33. (a)

34. (b)

35. (b)

36. (c)

37. (b)

38. (c)

39. (d)

40. (b)

41. (a)

42. (a)

43. (a)

44. (c)

45. (b)

46. (c)

47. (c)

48. (c)

49. (c)

50. (a)

51. (b)

52. (c)

53. (d)

54. (c)

55. (a)

56. (c)

57. (c)

58. (d)

59. (d)

60. (c)

61. (b)

62. (c)

63. (c)

64. (a)

65. (a)

66. (c)

67. (d)

68. (a)

69. (d)

70. (b)

71. (a)

72. (c)

73. (c)

74. (c)

75. (c)

76. (b)

77. (b)

78. (b)

79. (a)

80. (b)

81. (c)

82. (c)

83. (b)

84. (a)

85. (c)

86. (d)

87. (c)

88. (d)

89. (a)

90. (a)

91. (c)

92. (d)

93. (b)

94. (b)

95. (d)

96. (d)

97. (c)

98. (d)

99. (d)

100. (c)

101. (d)

102. (a)

103. (b)

104. (c)

105. (c)

106. (b)

107. (b)

108. (a)

109. (b)

110. (d)

111. (b)

112. (d)

113. (d)

114. (a)

115. (d)

116. (b)

117. (c)

118. (a)

119. (c)

120. (a)

121. (b)

122. (a)

123. (a)

124. (a)

125. (a)

126. (d)

127. (c)

128. (c)

129. (d)

130. (d)

131. (a)

132. (a)

133. (d)

134. (d)

135. (a)

136. (b)

137. (a)

138. (a)

ANSWERS

1. (d)

2. (a)

3. (a)

4. (c)

5. (b)

6. (a)

7. (b)

8. (a)

9. (c)

10. (c)

11. (b)

12. (c)

13. (d)

14. (d)

15. (d)

16. (c)

17. (c)

18. (c)

19. (d)

20. (c)

21. (b)

22. (a)

23. (b)

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(2) ME (Test-8), Objective Solutions, 13th November 2016

139. (b)

140. (a)

141. (a)

142. (a)

143. (b)

144. (b)

145. (d)

146. (b)

147. (c)

148. (c)

149. (d)

150. (d)

151. (c)

152. (a)

153. (a)

154. (a)

155. (b)

156. (b)

157. (d)

158. (d)

159. (d)

160. (cc)

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Sol–1: (d)At EOQ,annual ordering cost = annual carryingcostAnd annual carrying cost = 4 × 500 =Rs. 2000.

Sol–2: (a)Dispatching is the act of authorizingthe work-man actually to peform thework according to the method outlinedusing prescribed tools at apredetermined standard and schedule.

Sol–3: (a)Objectives of scheduling: Reducing average waiting time of a

batch. To meet the deadline of order

fullfilment. To increase facility utilization Meet customer due dates Minimize job lateness Minimize response time Minimize completion time Minimize time in the system Minimize overtime Maximize machine or labour utilization Minimize idle time Minimize work-in-process inventory

Sol–4: (c)At breakeven point, the profit is zero.Hence, the cost of production = totalsales.

Sol–5: (b)Payback period: It refers to the periodof time required to recoup the fundsexpended in an investment, or to reachthe break-even point. The time value ofmoney is not taken into account.Payback period intuitively measureshow long something takes to “pay foritself”.Net present value: NPV of a timeseries of cash flows, both incoming andoutgoing, is defined as the sum of

present values of the individual cashflows of the same entity. It measuresthe excess or shortfall of cash flow inpresent value terms, above the cost offunds. Each inflow/outlfow is discountedback to its present value. Therefore,NPV is the sum of all terms,

NPV =

tt

R1 i

t = time of cash flowi = discount rateRt = net cash flow at a timeInternal rate of return: It is a rateof return to measure and compare theprofitability of investments. It is alsocalled discounted cash flow rate ofreturn. It can also be defined as thediscount rate at which the present valueof all future cash flow is equal to theinitial investment.

Sol–6: (a)Total factory cost= (Direct cost) + (Factory on cost)= (material cost + labour cost +

direct expense) + (Labour cost × 1.5)= (375 + 245 + 80) + (245 × 1.5)= 1067.50

Sol–7: (b)Flow-line layout is known for itsinflexibility because any breakdown ofequipment along the production line candisrupt the whole system.

Sol–8: (a)Profit-volume chart is based on theassumption that either a single productis being sold, or if there are multipleproducts, these are sold in a constantmix.P-V chart is similar to break-even chartand records the profit or loss at eachlevel of sales, at a given sales price

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Margin ofsafety

SalesrevenueBreakeven

Contribution line

Profit

Loss

Sol–9: (c)Discount: Material cost.Preparation of the machine for aproduct : Set-up cost.Negotiations with vendors:Ordering cost.Rent of the warehouse: Carryingcost.

Sol–10: (c)Inventory Turnover Ratio (ITR)

=

cos t of goods consumed/soldduring the period

Average inventory hold during the period

Thus, ITR indicates the number of timesthe average inventory is consumed andreplenished. A high value of ITR indicatesthat the material in question is a fastmoving one. A low ITR indicates over-investment and looking up of workingcapital on undesirable items.Raw material inventory turnover ratio

=Material consumed

Average raw material inventorySol–11: (b)Sol–12: (c)

A capital expenditure is an amountspent to acquire or improve a long-termasset such as equipment or buildingsusually, capital spending is spent onacquiring property, plant or equipment.

Sol–13: (d)Since, the annual cost of materials isRs. 200000 and the unit cost of materialis Rs. 10, hence the number of products

per year = 200000 20000=10 .

Annual overhead cost = Rs. 120000hence per product, overhead cost =120000 Rs. 6=20000 Total cost of a product= direct material cost + overhead cost+ labour cost= 10 + 6 + 5 = Rs. 21

Sol–14: (d)Sol–15: (d)

Inventory carrying cost includes: Money tied up in inventory, such as

the cost of capital or theopportunity cost of the money

Physical space occupied by theinventory inc luding rent,depreciation, utility costs, insurance,taxes etc.

Cost of handling the items Cost of deterioration and obsolescence

Sol–16: (c)Dispatching function in productionplanning and control refers toauthorising a production work order tobe launched.

Sol–17: (c)Product standardisation consists ofmaking all given products alike; thusparts are interchangeable. It makespossible shorter setup times, quickerprocessing and longer production runswith smoother production flow. Inaddition, product standardisation makesfor simplified inspection requirementswith the application of the principles ofstatistical quality control.Standardisation of product alsosimplifies the functions of storage,marketing and distribution.With a standardised interchangeableproduct, mass production techniques canbe applied and labour can be specialisedand more effectively utilised. Further,the management can decide to investin fewer specialised machines to producethe products.

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Sol–18: (c)FA + nVA = FB + nVB

n = A B

B A

F F 800 1200=V V 0.08 0.10

= 400 20,000=0.02

Sol–19: (d)Let the breakeven volume = xThen, fixed cost + variable cost per unit× x = selling price per unit × x

8,00,000 + 40 x = 200 x

160x = 800000

x = 5000Sol–20: (c)

Flow-line layout – AutomobilesProcess layout – Job shop productionFixed position layout – AeroplanesHybrid layout – Flammable, explosive

productsSol–21: (b)

The major disadvantage of productlayout is its inflexibility and any oneworkstation may become a bottleneckand hold up the flow of work.In product layout, the raw materialenters the production line andoperations are performed on it in asequence until it goes out as a finishedproduct. Thus, there will be lower inprocess inventory and lesser materialhandling. However, in the sequence ofoperations if any one machine breaksdown, then the whole production processmay get affected.

Sol–22: (a)1. Analysis of the product and break-

ing it down into components.2. Determination of the lot sizes.3. Determination of operations and

processing time requirement.4. Taking make or buy decisions.

Sol–23: (b)For break even point,

F + n × V = n × SWhere F = fixed price, V and S arevariable and selling prices per unit, n= no. of units.

n = F 80000 5000= =S V 20 4

Sol–24: (b)Annual Demand D = 6000Number of orders per year = 12

Order size, Q = 600 500=12

Assuming, orders of equal size areplaced at periodical intervals, and theitems against an order are replenishedinstantaneously and the items areconsumed at a constant rate, so Average

Inventory = Q 250=2 .

Sol–25: (c)The Lenoir cycle consists of the followingprocesses: constant volume heat addition(1–2); reversible adiabatic expansion (2–3); and constant pressure heat rejection(3–1). The Lenoir cycle is applicable topulse jet engines.

p

v

Q1

Q23

2

1

pV =C

Sol–26: (a)Sol–27: (c)

Air liquefication Plant: ReversedStirling cycleGas turbine with multistageexpansion: Ericsson cycleFree piston engine/compressor witha gas turbine: Atkinson cyclePulse jet: Lenoir cycle

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Sol–28: (d)Volumetric efficiency is the ratio of actualvolume of gas drawn into the cylinder tothe piston displacement. Volumetricefficiency is less than 100% mainly becauseof the following reasons:Firstly, the gas is heated during admissionto the cylinder. Secondly there is leakagepast valves and piston rings. Thirdly,there is re-expansion of the gas trappedin the clearance volume from the previousstroke.

Sol–29: (c)The work done in compression inreciprocating compressor is minimumwhen the process of compression isisothermal. Hence, the isothermalefficiency of reciprocating compressor isdefined as the ratio of the isothermal workand the actual work done. Thus,isothermal efficiency is used forquantifying the degree of deviation ofactual compression process from idealcompression process.

Sol–30: (a)In proximate analysis of coal, the weightpercentage of carbon, volatile matter,moisture and ash content is determined.While in ultimate analysis, individualpercentage of H, C, N, S, O, moistureand ash content are acertained.

Sol–31: (b)

Pulverised coal is in powder form andlarge fuel particles surface area. Soits requires 5–10% more air

Air contains oxygen 21% by volumeand 23% by mass.

Due to incomplete combustion i.e. CO,CO2 and NOX are formed.

The volumetric composition of fluegases from coal fired furnace containsabout 70% nitrogen because solidcarbon takes gaseous oxygen andproduces gaseous carbondioxide inequal volume.

The molecules of liquids are heavieri.e. bigger in size as compared togases. So in liquid combustion, more

molecules are formed.Sol–32: (a)

Excess air does not control air tempera-ture but results in loss of heat in fur-nace. But (10–15)% excess air is neces-sary for efficient burning. Sulphur re-sults in SO2.

Sol–33: (a)The higher calorific value and lowercalorific value of fuel is related as

LCV = HCV – x fgh

where x - fraction of water vapour inproducts of combustion.When this fraction ‘x’ of water vapourcondensed to liquid water, it liberateslatent heat of vaporization ‘hfg’ and

LCV = HCVThis condensation takes place at nor-mal or initial temperature of fuel.

Sol–34: (b) For maximum compression efficiency

in 2 stage compressorP P3 3

2

P1 1

V Intercooling must be perfect. The intermediate pressure P2 should

be geometrical mean of initial pres-sure P1 and delivery pressure, P3,

P2 = 1 3P .P

Sol–35: (b)As compared to reciprocating compres-sor, the centrifugal compressor flow islarge and pressure ratio is less. So theyare used for small pressure ratio (4:1)but large quantity. As far as rpm isconcerned, the rpm of centrifugal com-pressor is very large as compared toreciprocating compressor i.e. around 8–10 times. But its rpm are less than axialcompressor.

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Sol–36: (c)The axial flow compressor and axial flowturbines are reversible machines. So theaxial flow compressors are called asreverse reaction turbine.In 50% degree of reaction, the velocitytriangles are similar. The reason beingequal inlet and exit velocity triangle i.e.

1 = 2 and

1 = 2

Sol–37: (b)Multistaging means compression of airin small steps so turning moment fluc-tuations of compressor shaft reduces.Multistaging ensures good intercoolingso low compression work. Different cyl-inders are used for different pressure socylinder size is reduced. Due tomultistages, the pressure ratio reducesand volumetric efficiency increases.

Sol–38: (c)Isentropic efficiency

a = TT

...(i)

T

1

2 2´

sEnthalpy change due to Euler work-

(h2 – h1) = 2w s 2u

pc T = 2w s 2u

T = 2

w s 2

p

uc

...(ii)

2P 2u = pc T

From equation (i)

a =2

P 2 P

p w s2

w s 2

p

u =cu

c

P = s w a

Sol–39: (d)The pressure inside axial compressorincreases (not drop) successively throughcontracting passage. The blades areformed from aerofoil shapes, notcircular arc.The compression in vane compressor isdue to back flow pressure of receiverand vane motion due to eccentric rota-tion of rotor.

Sol–40: (b)The density of air increases in laterstages of axial compressor and velocityof flow in constant. So to have constantmass flow rate through all the stages,the area must decreases as

m = AV

Constant = density increase × Area ×constant velocity.To satisfy above equation area mustdecrease.

Sol–41: (a)Consider a centrifugal compressor hav-ing no inlet guide vane and backwardcurved exit.The pressure rise in rotor–

rotPg

= 1 2

2 2 2 22 1 r r

1 u u V V2g

...(i)Assuming constant flow velocity.

1fV = 2f 1V V=

At inlet to impeller,

u1

V1=Vf1

1A B

C

2r1V = 2 2 2 2

f1 1 1 1V u V u …(i)

At outlet of impeller,

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V2

Vf2

u2

Vr2

D E

F

2V

2r2V =

22 2f 2 2 wV (u V )

= 2

2 21 2 wV (u V ) …(ii)

Subtracting equation (ii) from (i),

1 22 2r rV V = 2 2

1 1(V u )

2 22 2 21 2 w 2 w(V u V 2u V )

= 2 2

2 2 22 w w 1 22u V V u u

1 22 2 2 2r r 2 1V V u u =

2 22

2 w w2u V V

...(iii)Total pressure rise in compressor

stagePg

= 2 1w 2 w 1V u V u

= 2 1w 2 wV u ( V 0)

Degree of reaction,

R = rot

stage

PP

= 2 2

2

22 w w

2 w

2u V V2u V

R = 2w

2

V1

2uSol–42: (a)

For all same parameters the forwardcurved vanes gives maximum pressurerise. But they are not preferred becauseof lower efficiency and unstableoperation. Forward curved blades givemaximum blade exit velocity (V2) i.e.more exit kinetic energy to convert inpressure inside diffuser.

Sol–43: (a)In turbo prop engine, a propeller isprovided which sucks large amount of

air from atmosphere. This increasedmass flow rate of air increases thrustdeveloped by engine. But in turbojet,there is no propeller and entire mass ofair (less than turboprop) passes throughcompressor and turbine and wholethrust is developed by nozzle jet. Soturboprop has propeller an additionalcomponent and rest components are sameas turbojet.

Sol–44: (c)Jet engine carry only fuel not oxidantbecause it takes oxidant (air) from at-mosphere while rocket carry both fueland oxidant inside its body. Rocket en-gine has no moving part i.e. compressoror propeller since there is no need of airfrom atmosphere for combustion. Theycan operate in vacuum also. Rocketengine fuel and oxidant can be singlephase-mixture of fuel and oxidant, doublephase fuel solid and oxidant liquid orgas.

Sol–45: (b)The thrust, developed in turbojet,

F = a jm V U

At high altitude, density is low so massflow rate am is low.

Efficiencythrust

U/Vj

a jF/ m V

Where U is flight speed and Vj is jetspeed. Hence the thrust decrease withincrease in speed. The relative velocitywith medium i.e. (Vj–U) decreases as Uincrease.

Sol–46: (c)Jet velocity from nozzle,

Vj = 2700m/secFlight velocity,

u = 1350 m/sec The propulsive efficiency,

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p =2 flight velocity

flight velocity jet velocity

= 2 13501350 2700

= 66.67 %Sol–47: (c)

In wind tunnels, a large reservoir athigh pressure contains air. When tun-nel is fed, the pressure variations dueto flow are negligible. So the stagnationconditons exist in reservoir. Similarconditions are also in rocket combus-tion chamber. There is no limitation onexit velocity of gases i.e. very high exitvelocity.But the point to be noted is that therocket engines are least efficient amongall jet engines.

Sol–48: (c)

Propulsion efficiency

=

a

2 2

m uThrust power V u= 1Propulsive power m V u2

= 2× flight velocity

Jet velocity + flight velocity

= 2u

V u

Sol–49: (c)Sol–50: (a)

2 2m

gHN D

= 2 2p

gHN D

In case of multi-stage pumps, since thesame liquid flows through each impeller,the discharge of a multi stage pump issame as the discharge passing througheach impeller of the series.

Sol–51: (b)An axial flow pump is one in which thefluid enters parallel to the axis ofrotation and leaves in axially tangentialplane. It is used for the cases of high

capacity and low head.Sol–52: (c)

Mass of water m = 50 × 10–3 × 1000kg/s = 50 kg/s,Head H = 40 mSo Power = mgH = 50 × 9.81 × 40

= 19.6 kWSol–53: (d)

A draft tube is a tube or pipe of graduallyincreasing area which is used fordischarging water from the exit of theturbine to the tail race. This is becausethe pressure at the exit of the runner ofa reaction turbine is generally less thanatmosphereic pressure. Thus, the waterat exit cannot be directly discharged tothe tail race.Also, the draft tube converts a large

proportion of the kinetic energy 22V

2g

rejected at the outlet of the turbine intouseful pressure energy. Without the drafttube, the kinetic energy rejected at theoutlet of turbine will go waste to the tailrace.Hence by using draft tube, net head onturbine increases. The turbine developsmore power and also efficiency of turbineincreases.

Sol–54: (c)Sol–55: (a)

Q

Forward

RadialBackward

H

Sol–56: (c)

h max =

1 cos

2Sol–57: (c)

For hydroelectric power uses, a surge tankis an additional storage space or reservoirfitted between the main storage reservoirand the power house. It should be as close

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to the power house as possible. Surgetanks are provided in high or mediumhead plants when there is a considerabledistance beween the water source and thepower unit, necessitating a long penstock.The main functions of the surge tank are(i) when the load increases, the water

moves backwards and gets stored init

(ii) when the load decreases, additionalsupply of water will be provided bysurge tank.

Sol–58: (d)Kaplan and propeller, both are axial flowreaction turbines. When the vanes arefixed to the hub and they are notadjustable, the turbine is known aspropeller turbine, but if the vanes on thehub are adjustable, the turbine is calledKaplan turbine.

Sol–59: (d)Degree of reaction is one in case of lawnsprinkler.Pelton turbine is impulse turbine. Francisis inward flow reaction turbine havingradial discharge at outlet.

Sol–60: (c)Velocity ratio for francis turbine

=u

2gH

where u = peripheral velocity of blades

Sol–61: (b)Pressure due to ram = pressure due toplunger

Force on ramArea of ram = Force on plunger

Area of plunger

2

1000 9.81

12.54

= 2

F

1.254

F =2

21000 9.81 1.25 98.1N=

12.5

Sol–62: (c)

3Q

D N = constant

3

340 10

0.1 1000 =

30.8

0.4 N N = 312.5 rpm

Sol–63: (c)The ratio of the manometric head to thehead imparted by the impeller to thewater is known as manometric efficiency.Mathematically, it is written as

man = Manometric headHead imparted by impeller to

water

=

2 2

m m

w 2 w 2

H gHV u V u

gThe power at the impeller of the pump ismore than the power given to the waterat outlet of the pump. The ratio of thepower given to water at outlet of the pumpto the power available at the impeller, isknown as manometric efficiency.The power given to water at outlet of the

pump = mWH kW1000

The power at the impeller

= Work done by impeller per second kW

1000

=

2w 2V uW kWg 1000

man =

2 2

mm

w 2 w 2

W Hg H1000

V u V uWg 1000

Sol–64: (a)Applying the Bernoulli’s theorem betweenthe jet at the nozzle exit and the jet at anelevation y.

2V2g =

2u y2g y = 2 2V u

2g

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Sol–65: (a)

Head (H)

Input Power, PSpeed = Constant

Output Power

Discharge, Q

Effic

ienc

y,

Pow

er, P

Hea

d, H

Efficiency,

Sol–66: (c)Propeller turbine: Kaplan turbineTangential turbine: Pelton turbineReaction is zero: Impulse turbineReaction turbine: Gas turbinePelton turbine is a tangential flow impulseturbine. In case of pure impulse turbine,degree of reaction will be zero. Propellerand kaplan, both are axial flow reactionturbines.

Sol–67: (d)

Force = 2a(v u)

= 1000 × 0.02 × (20–10)2 = 2000NSol–68: (a)

1.

2.

3.4.

10 to 35

35 to 60

60 to 300300 to 1000

8.5 to 30

30 to 51

51 to 225225 to 860

Pelton wheel with single jetPelton wheel with two or more jetsFrancis turbineKaplan or Propeller turbine

S.No.

Specifiedspeed

Types ofturbine

(M.K.S.) S.I.

Sol–69: (d)Hydraulic coupling is a device used fortransmitting power from driving shaft todriven shaft with the help of fluid. Thereis no mechanical connection between thetwo shafts.

Sol–70: (b)Draft tube: Reaction turbineSurging: Centrifugal pump

Air Vessel: Reciprocating pumpNozzle: Impulse turbine

Sol–71: (a)NPSH indicates the minimum suctionconditions in the case of centrifugalpumps.NPSH = Absolute pressure head at inletof the pump–vapour pressure head +velocity head.If the pressure at the suction side of thepump drops below the vapour pressure ofthe liquid then cavitation may occur.Thus, to avoid cavitation, available NPSHshould be greater than the minimumNPSH.

Sol–72: (c)Sol–73: (c)

High air fuel ratio in gas turbines reducesthe exhaust gas temperature.

Sol–74: (c)

DOR = m

m f

hh h

Sol–75: (c)

1

T

s5 pi

p0

2

3

4

6

Hence, we can see that the heatrejected through the condenserincreases thus reheating leads toincreased condenser require-ments.

Sol–76: (b)Sol–77: (b)

Single stage impulse: De LavalPressure compounding: RateauVelocity compounding: CurtisReaction turbine: Parsons

Sol–78: (b)In the given figure,

T3 = 1000 K

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T5 = 800 KT4 = 600 KT6 = 570 KT2 = 450 KT1 = 300 K

Heat addition Q = (h3 – h2) + (h5 – h4)= Cp [(T3 – T2) + (T5 – T4)]

Turbine work WT = (h3 – h4) + (h5 – h6)Compressor work WC = (h2 – h1) Work output W = WT – WC

= [(h3 – h4) + (h5 – h6) – (h2 – h1)]= Cp[(T3 – T4) + (T5 – T6) – (T2 – T1)]

Efficiency

= p 3 4 5 6 2 1

p 3 2 5 4

C T T T T T TW =Q C T T T T

= 1 2 3 4 5 6

3 2 5 4

T T T T T TT T T T

= 300 450 1000 600 800 5701000 450 800 600

= 480 0.64 or 64%=750

Sol–79: (a)

Regeneration will increase efficiency, buthas no effect on the power output.

Sol–80: (b)Sol–81: (c)

The overall efficiency of steam powerplant will increase by increasing the steamtemperature and providing air preheaters.APH is meant to heat air before com-bustion for increasing thermal effi-ciency. Increasing steam temperaturewill increase mean temperature of heataddition.Increasing the condenser pressure, how-ever, will reduce the work output,thereby reducing thermal efficiency.

Sol–82: (c)Optimum pressure ratio for maximum

specific work output rp = 2 13

1

TT

Sol–83: (b)

VelocityPressure

Nozzle(B )1

MovingBlade(B )2

Nozzle(B )3

MovingBlade(B )4

The solid line shown in the questionrepresents velocity variation, B2 and B4are rotor passages.

Sol–84: (a)Elastomers undergo large deformationsthat are fully recoverable and timedependent. These materials do not obeyhooke’s law. Rubber is an elastomer. Viscoelastic materials on the other hand obeyhooke’s law.

Sol–85: (c)Shore scleroscope test is used to measurehardness of lathe bed. In this test, acylindrical shaped, diamond tippedhammer rebounds from the specimen,when dropped from a certain height. Theheight of rebound is the measure ofhardness.

Sol–86: (d)Understressing is a phenomenon whereendurance limit of a material improveswhen material is subjected to stressesbelow its fatigue strength.Understressing improves the fatiguestrength of metallic material.Overstressing, scratching increases stressconcentration thereby reducing fatiguestrength.

Sol–87: (c)Toughness: Energy absorbed beforefracture in a tension test.Endurance strength: Fatigue loading.Resistance to abrasion: Hardness.Deflection in beam: Moment areamethod.

Sol–88: (d)During tensile-testing of a specimen usingUTM, load cells and extensometersmeasures load and elongation.

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Sol–89: (a)Teflon or PTFE (poly tetra flouroethene)is a thermo plastic polymer having verylow coefficient of friction. It has highdielectric strength and therefore used asan insulating material.

Sol–90: (a)Styrene-butadiene Rubber (SBR) is astandard material for car tyres. SBR hasgood abrasion resistance and good agingstability. About 50% of car tyres madeworldwide are of SBR.

Sol–91: (c)A. Neoprene: Oil seals.B. Bakelite: Electric switches.C. Foamed poly-Urethane: Thermal

insulator.D. Araldite: Adhesive.

Sol–92: (d)In water line corrosion, the corrosion occursin a water storage tank just below the waterline. The concentration of oxygen is higherjust above the water line as compared to thesubmerged portion near the water line. Thelatter zone acts as anode and thereforecorrodes. Water line corrosion is a type ofcrevice corrosion.

Sol–93: (b)Electrodeposition has liquid as its startingphase, while rest three have solid as startingphase.

Sol–94: (b)Sol–95: (d)Sol–96: (d)

Q = TR

R = 1 2

1 1 1 1000 1 1224 k r r 60 1504 357

R = 1000 90440 60 150

= 1 kW440

T = Q.R

= 144000

440= 100°C

Sol–97: (c)rc – r1 = 2

rc = 2 + 1.3 = 3.3 mm

2 Kh = 3.3 × 10–3

3

2 0.113.3 10 = h

h = 66.67 W/m2KSol–98: (d)

34 3 14

3 8 =

2

w3h 4 T T

200

34 3 14

3 8=

w

4h 4 9 T 25

4 10

44 14 10 25800

= wT

wT = 725°C

Sol–99: (d)

2

2T

x= 0

Tx = b

T = bx + aT = a + bx

40 = a + bx128 = a + b x2

– 12 = b (x2 – x1)

120015 = b

40 = a + b × 0

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a = 40

T = 120040 x

15T = 40 – 80 x

= – 80 x + 40Sol–100: (c)

In the analysis of fin we consider constantand uniform heat transfer coefficient over theentire fin surface.

Sol–101: (d)A black body is a diffuse emitter the termdiffuse means independent of direction,whereas confined means in a particulardirection.

Sol–102: (a)Energy emitted in the range

= 0.8

2

0.53 4 d

= 3 3 2 23 1 40.3 0.8 0.5 0.8 0.53 2

= 1.551 W/m2

Sol–103: (b)T1 T2

A Aq

1m=

Cross sectional area cylinder,A = 20 cm2

Heat transfer rate

q = 1 2T TkA

600010 60

= 100 × 10–3 1 2T T1

(T1 – T2) = 16000 6000=

60600 10 = 100° C

Sol–104: (c)The general expression for equivalentemissivity in enclosure and equivalentsystem.

e =1

1 2 2

1A1 1 1A

A. Infinite parallel planes i.e. A1 = A2

e =

1 2

11 1 1

B. Body 1 completely enclosed in body 2but 1 is very small i.e. A1 << A2

e = 1

1 2

1 =1 10 1

C. Two small Gray body exchanging

radiation is given separately as,e = 1 2

D. Two concentric cylinders of largelength,

e =1

1 2 2

1A1 1 1A

Sol–105:(c)

The degree of drawing is expressed interms of the reduction factor in thecross-sectional area.

Thus, D =

2 2

i 0 i 02

i i

A A D DA D

Sol–106: (b)Die Casting: Die casting utilises twoblocks of heat resistant metal machinedto meet along the plane of the partingline and having cavities machinedaccurately and smoothly into each toform opposite halves of the shape to becast around the edges of the mould. Finevents are cut to allow air to escape asthe metal enters. If air is not allowedto escape it would be trapped andproduce blow holes in the casting.In die casting, molten or semi-moltenmetal is either poured under (gravitydie casting) or is forced under highpressure is maintained till solidificationstage (pressure die casting). As a result,a very smooth and accurate part isobtained. Die casting is used for massproduction and is most suitable for non-

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ferrous metals and alloys of low fusiontemperature. By this method, castingsof nearly all sizes can be producedeconomically. The process is rapid andthe surface is smooth consequently littleor no finishing operation is required.The material is dense and homogeneousand has no possibility of sand inclusionsor other impurities. Thin wall can beproduced and uniform thickness can bemaintained.Centrifugal Casting: In this procesthe molten metal is poured in a mouldand allowed to solidify while the mouldis revolving, i.e. the metal solidifiesunder the pressure of centrifugal force.The centrifugal forces cause the metalto take up the impression of the mouldcavity. The pressure is selective, thatis, the greater force is exerted on thedenser components. This is ofconsiderable benefit in eliminating non-metallics and gases during casting. Thisprocess is best suited for massproduction. Cylindrical parts and pipesare most adaptable to this process. Incentrifugal casting, the molten metalis subjected to centrifugal force due towhich if flows in mould cavities easilyand results in the production of highdensity castings. On casting surfacesbetter details of numerals and numberscan be obtained and thin parts of highstrength can be easily produced. Thecastings are produced with promoteddirectional solidifcation as the coldermetal is thrown to outer rotation whichfurther acts as a feeder duringsolidification of metal. No core is neededto form the hole in the middle.Centrifugal casting finds its best usein mass production operation. The useof the machinery and equipment forcentrifugal casting can be justified onlywhen a large quantity of identicalcastings are required.Precision Casting: This process iscalled the lost-wax process or precisioncasting. This process uses wax patternwhich is subsequently melted from the

mould, leaving a cavity having all thedetails of the original pattern. Castingsobtained by this process have very closetolerances or the order of 0.005mm.Generally this process is used forproducing light and intricate parts. Thisprocess does not need a parting line orany form of split mould.Electroslag casting: It is a processtechnique used to produce castingsusing electroslag melting. Theelectroslag remelting process is used toremelt and refine steels and varioussuper-alloys resulting in high-qualityingots. This process uses the as-castalloy as a consumable electrode.

Sol–107:(d)Sol–108: (a)Sol–109: (b)

Flattening: Pressing a workpiecebetween two flat diesDrawing: Thickness is reducedcontinuously at different sections alonglengthFullering: Metal is displaced away fromcentre, reducing thickness in middleand in creasing lengthWire drawing: Rod is pulled througha die

Sol–110: (d)Sol–111: (b)

Rolling of very thin strips presentsdifficulties because roll flattening becomescommensurate with strip thickness.Interface pressures can be reduced withsmall diameter rolls and application offront and back tensions to the strip; evenso, good lubrication is indispensable. Someminimum friction is still needed,otherwise the strip may skid in the rollsand the lateral strip movement becomesdifficult to control.

Sol–112: (d)Flux coating: Shielded metal arcweldingFlux granules: Submerged arcwelding

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CO2: Gas metal arc weldingVacuum: Electron beam welding

Sol–113: (d)The power source characteristic can bewritten as

V = 100 volts100400 I

PowerP = 100I I100400

For maximum power,

dPdI = 0

100 – 100 2I400

= 0

I = 200 A

V = 100100 200 50 V=400

Sol–114: (a)The pipes and tubes can be seamless orwith seam. Tubes with seam aremanufactured by various weldingtechniques.Seamless tubes can be made by extrusionand piercing methods. From piercingmethod, seamless tubes are made in twostages – manufacture from round blankor billet of a thick walled shell in apiercing mill and Rolling of shell into apipe of given diameter and wall thickness.A method for pipe rolling is themanufacture of thick walled shells in apiercing mill and then subsequent rollingin several continuous mills.Hot drawing and cupping is anothertechnique of producing seamless cylindersand tubes. The manufacture of seamlesstubes may consist of four stages: 1.Reduction of billets and blooms 2.Production of hollow 3. Hot finishing 4.Cold finishing.

Sol–115: (d)Defects may occur during deep drawing(a) if the clearance betwen punch and dieis too small (b) punch and die are not

properly aligned (c) if too much force isapplied to the blank, the punch load willincrease and this will lead to fracture ofthe shell wall. (d) If the corner radii ofpunch and die are too small, the cornermay fracture because of the increasedforce required to draw the cup. (e)Scratches, dirt or any surface defect ofthe punch or die increase the requireddrawing force and may cause a shell totear (f) If the blank holder exerts too littlepressure, or if the die radius is too large,wrinkles will appear at the top flange ofthe part.

Sol–116: (b)The radius of prime circle is equal to thesum of the base circle radius and theradius of the roller follower.The pressure angle varies at all instantsof the follower motion.

Sol–117: (c)According to Grashof’s law, for a four barmechanism, the sum of the shortest andlongest link lengths should not be greaterthan the sum of the remaining two linklengths if there is to be a continuousrelative motion between the two links.

Sol–118: (a)

Work done per cycle = mT

= mT 4

= p1 T2

mT 4 = p1 T2

Tp = 8 Tm = 8×10 = 80 NmSol–119: (c)

Since there are four links (i.e. n = 4),therefore the number of instantaneouscentres,

N = n(n 1) 4(4 1) 6= =2 2

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14I atI13

14I at

I24

VBBI23

2 D3

I12

1

I34

AVA4

Sol–120: (a)Module: Ratio of pitch circle diameter inmm to the number of teethAddendum: Radial distance of a toothfrom the pitch circle to the top of thetoothCircular pitch: Distance on thecircumference of the pitch circle from pointof one tooth to the corresponding point onthe next tooth.

Sol–121: (b)Klein’s construction is a graphical methodfor the kinematic analysis of an IC enginemechanism. The velocity diagramrepresents velocity of crank, velocity ofpiston and velocity of piston relative tocrank. From this, the velocity of any pointon the connecting rod can also be foundout. The acceleration diagram is aquadrilateral and the four sides representcentrifugal acceleration of the crank,radial component of acceleration ofconnecting rod, tangential component ofacceleration of the connecting rod and theacceleration of the piston.From this, the angular acceleration of theconnecting rod can be found out.

Sol–122: (a)

N2 = m M 895

m h

h = 2m M 895

m N

Sol–123: (a)A flywheel used in machines serves as areservoir, which stores energy during theperiod when the supply of energy is morethan the requirement, and releases itduring the period when the requirementof energy is more than the supply.In case of steam engines, internalcombustion engines, reciprocatingcompressors and pumps, the energy isdeveloped during one stroke and theengine is to run for the whole cycle onthe energy produced during this onestroke. For example, in internalcombustion engines, the energy isdeveloped only during expansion or powerstroke which is much more than theengine load and no energy is beingdeveloped during suction, compression andexhaust strokes in case of four strokeengines and during compression in caseof two stroke engines. The excess energydeveloped during power stroke is absorbedby the flywheel and releases it to thecrankshaft during other strokes in whichno energy is developed, thus rotating thecrankshaft at a uniform speed. A littleconsideration will show that when theflywheel absorbs energy, its speedincreases and when it releases energy,the speed decreases. Hence a flywheel doesnot maintain a constant speed, it simplyreduces the fluctuation of speed. In otherwords, a flywheel controls the speedvariations caused by the fluctuation ofthe engine turning moment during eachcycle of operation.In machines where the operation isintermittent like punching machine,shearing machines, rivetting machines,crushers, etc., the flywheel stores energyfrom the power source during the greaterportion of the operating cycle and gives itup during a small period of the cycle.Thus, the energy from the power sourceto the machines is supplied practically ata constant rate throughout the operation.The function of a governor in an engineis entirely different from that of a

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flywheel. It regulates the mean speed ofan engine when there are variations inthe load, e.g., when the load on the engineincreases, it becomes necessary to increasethe supply of working fluid. On the otherhand, when the load decreases, lessworking fluid is required. The governorautomatically controls the supply ofworking fluid to the engine with thevarying load condition and keeps themean speed of the engine within certainlimits.As discussed above, the flywheel does notmaintain a constant speed, it simplyreduces the fluctuation of speed. It doesnot control the speed variations causedby the varying load.

Sol–124: (a)

unstable

isochron

ous

stable

F =ar+bC

F =arC

F =ar–bC

FC

rSol–125: (a)

Belt and pulley form a higher pair becausepoint or line contact takes place whereasturning pairs, sliding pairs and screwpairs are lower pairs because surfacecontact takes place.

Sol–126: (d)

x = APcos

y = PR BPsin=

2 2x y

AP BP

= 1

X B RQ

QP(x,y)

yA

2 2

2 2x y

AP BP = 1

This is the equation of an ellipse. Hence,the path traced by point P is an ellipsewhose semi-major axis is AP and semi-minor axis is BP.

Sol–127: (c)Mean effective pressure

=Net work output

Swept volume

= 3384

(2.2 0.20) 10= 192 kPa

Sol–128: (c)P 3

24

1

VThe air standard Otto cycle-compression ratio,

1

2

VV = 4

3

V 6V

=

P1 = 0.1 MPaT1 = 27°C = 300 KT3 = 1569°C = 1842 K

Specific heat capacity ratio, = 1.4

Temperature after isentropic expansion.

T4 =

31 1.4 1

4 3

T 1842=6V /V

= 0.41842 899.56K=6

900 KSol–129: (d)

SI engine burns mixture close tostoichiometric while CI engine alwaysburns lean. Hence SI engine uses richmixture than CI engine.

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The power output from SI engine ishigher than CI engine because it burnsmore fuel in given capacity of cylinder.

Specific output (output power per unitpiston displacement) is about two timesof CI engine.

Because of rich mixture, exhaustpollution is high.

Sol–130: (d)1. Since all the processes in Otto cycle are

assumed to be reversible, so the cycle isreversible.

2. The efficiency of Otto cycle is

= 1

c

11r

If rc is high, the efficiency will be high.Now if the ratio of specific heat is high,the efficiency improves.

3. In old engines i.e. carburetted engines,the maximum limit of compressionratio is 11.

Sol–131: (a)The angular velocity of crank in degree/sec.

= 2 N rad/sec60

= 2 N 180 deg /sec60

= 6 N deg/secThe rotation of crank in 1.5m/sec

= w.t. degree= 6 N × 1.5 × 10–3

= 6 × 2000 × 1.5 × 10–3

= 18° degree.Sol–132: (a)

Stoichiometric Air: Fuel = 14.5:1Note: CI engine always run on lean/weak mixture.A. Air : Fuel = 10:1

So it is rich mixture - This range isused in SI engine Idling. DuringIdling SI engine uses rich mixturebecause of residual gas dilution.

B. Air : Fuel = 16:1 - Lean mixture-This is used in economy range i.e.cruising in SI engine

C. Air : Fuel = 35 : 1 - Very lean andout of ignition limits of SI engine, soit will be for CI engine. We know CIengine always run on lean mixtureand leaness decreases as load in-creases. So it will be CI engine partload.

D. Air : Fuel ratio (12.5:1) - It is richmixture and used for SI engine fullload.

Sol–133: (d)1. Due to presence of throttle valve,

carburettor venturi etc in the pathof air, SI engine volumetric efficiencyis low.

2. SI engine throttled means, air flowis restricted i.e. situation is likeidling. In this condition the BHP isnegligible so the mechanical effi-ciency is low.

3. Fuel consumption per unit kW out-put i.e. specific fuel consumptiondecreases as capacity of engine in-creases.

4. Due to presence of excess oxygen inCI engines, its exhaust temperatureare lower than SI engine.

Sol–134: (d)The order of detonation of fuel is,paraffins straight chain hydrocarbonhave highest detonation tendency thencomes olefins (double bond), then satu-rated cyclic (naphthalenes) and leastdetonation tendency is aromatic.

Sol–135: (a) n-cetane or Hexadecane is assigned

100 cetane number. The word ‘n’ (normal) is not used in

chemical (IUPAC) name. Chemical formula of n-cetane,

CH3(CH2)14CH3

It is a straight chain compound,saturated paraffin series.

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Sol–136: (b)A. Highest useful compression

Ratio (HUCR): It is the maximumcompression ratio used in SI enginewithout knock. So it is limited bydetonation.

B. Dopes: They are chemicals mixedwith fuel which increases knockrating of fuel.

C. Air fuel mixture: is ignited onlyin certain limiting strength range.

D. Delay Period: During this period,certain chemical reactions occures athigh temperature and high pressure.

Sol–137: (a)Manufacturing cell refers to the system oforganising manufacturing process in acluster. Thus, it reduces movement ofmaterials and work-in-progress, andprovides visual control. This reduces costof material handling and supervision andimproves quality of production.

Sol–138: (a)General purpose machine tools are thosedesigned to perform on a wide range ofcomponents. By the very nature ofgeneralisation, the general purposemachine tool, though capable of carryingout a variety of tasks, would not be suitablefor large production, since the setting timefor any given operation is large. Thus,the idle time for a general purposemachine tool is more and machineutilisation is poor. Hence, their utility isin job shops catering to small-batch, largevariety job production. Examples –lathe,shaper and milling machine.

Sol–139: (b)The economic lot size depends on the setupcost. If the setup time can be reduced, thelot size can be reduced. The economic lotsize is not more than a direct relationshipbetween the inventory cost and setup cost.The effect of setup cost decreasesexponentially according to the increase inbatch size. Therefore, in order to calculatethe economic lot size, it is supposed thatthe setup cost is constant i.e. the setup

time is constant. This transitionalhypothesis is based on a constant setuptime i.e. it is not possible to reduce thesetup time. More often than not, though,setup time can be reduced. A setup costdecreases, the economic lot size would alsodecrease until, the unit product lot size isreached.Customer satisfaction depends on variousfactors such as the quality of product,meeting due dates, price, volume andflexibility.

Sol–140:(a)Production planning and control incontinuous production is usually farsimpler than that in job or batchproduction. Extensive effort is requiredfor detailed planning before productionstarts, but both scheduling and controlneed not be elaborate usually. Theoutput is either limited by availablecapacity or regulated within even limitsto conform to production targets basedon periodic sales forecast. Detailedplanning is required for setting the rateof production from line and arrangingthe various production centres on theline accordingly.

Sol–141: (a)Inlet velocity triangle at tip of eye,

V1Vw1

u1

The Mach Number at eye,

M = 1wVRT

By providing inlet guide vanes, 1wV isreduced to keep M < 0.9.

Sol–142: (a)

The mixture of air and fuel used in gasturbine is very lean of the order of50 to 100:1. Fuel more than this cannot be burnt in gas turbine because itis a steady flow machine and peak tem-

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perature also remains constant. So thismaximum value of temperature is lim-ited by material availability.

Sol–143: (b)The turboprop engines fills the gapbetween aircraft driven by reciprocat-ing engine (used for low speed and lowaltitude) and turbojet engine (used forhigh speed and high altitude). So theyare used for medium speed and mediumaltitude with good performance.The reasons for reduction in efficiencyof turboprop engine of higher speed are- profile drag of blades, separation offlow from blade surface, swirl and slipof stream over blades of propeller i.e.shock formation.

Sol–144: (b)At sea level the density is higher due tohigh pressure. So same compressor willdeliver more air mass as compared tocompressor at mountain where pressureand density are low. But capacity rat-ing (m3/min) are given as free air inlocal atmospheric condition.

Sol–145: (d)A draft tube is a tube or pipe of graduallyincreasing area which is used fordischarging water from the exit of theturbine to the tail race. This is becausethe pressure at the exit of the summer ofa reaction turbine is generally less thanatmosphereic pressure. Thus, the waterat exit cannot be directly discharged tothe tail race.Also, the draft tube converts a large

proportion of the kinetic energy 22V

2g

rejected at the outlet of the turbine intouseful pressure energy. Without the drafttube, the kinetic energy rejected at theoutlet of turbine will go waste to the tailrace.Hence by using draft tube, net head onturbine increases. The turbine developsmore power and also efficiency of turbineincreases.

Sol–146: (b)Sol–147: (c)

The specific speed of pelton wheel is low.Sol–148: (c)

Because of higher relative leakages i.e.percentage of leakage as compared to totaldischarge rate, model volumetric efficiencyis lower than prototype.

m(vol) < p(vol)

Again in small machine, the relativecontact area is more than the big machine,so friction losses are more in small turbinethan big one. The mechanical efficiency ofsmall machine is lower

m(mech) < p(mech)

So, overall efficiency of small turbine(model) is less than bigger one (prototype).

Sol–149: (d)In turbines using nozzle governing, thesteam supply from the main valve is splitinto two or three lines, each line feedinga set of nozzles, when the load diminishes,the steam supply to one or more of thesenozzle sets is cutoff by the valve operation.The working nozzles receive steam at fullpressure so that a high efficiency ismaintained at all loads. The loss ofavailable enthalpy drop is less than for athrottle governed turbine. Thus the actualefficiency of and steam rate for a nozzlegoverned turbine will be better.

Sol–150: (d)The maximum efficiency of a singlestage impulse turbine = 2cos

where is the nozzle angle.The lower is the nozzle angle, higher isthe blading efficiency.However, too low a nozzle angle maycause energy loss at blade inlet. There-fore, the nozzle angle has to be main-tained within a certain range, whichvaries from 16° to 22°.

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Sol–151: (c)The internal structure of a material givesinformation about the arrangement ofmaterial atoms. However, the propertiesof bulk matter of all kinds depend stronglyon the nature and distribution ofimperfections especially physical andmechanical properties (strength, hardnessetc.).

Sol–152: (a)In furnance, the wall are reradiating andtemperature is same as incident radiationtemperature. So they emits samewavelength radiation as incident.Sometimes they are also called as adiabaticwall.

Sol–153: (a)The lumped capacity method, thetemperature surface of cooling body isassumed equal to temperature of interiorof body as shown in figure below:

HotBody

T TCoolingmedium

B < 0.1i

This is possible at Biot number less thanone as,

Bi = h 0.1k

= hA

kA = h A

kA 1

l

=conduction resistanceconvective resistance

Since higher convection resistance forcesuniform temperature throughout theinterior of hot body.

Sol–154:(a)Complex shapes are possible in invest-ment casting because pattern is with-drawn by melting it.

Sol–155:(b)In the indirect extrusion, a hollow ramcompresses metal through a die in adirection opposite to ram motion. Either

the ram is moved against a stationarybillet or the billet (hence container) ismade to move against stationary ram.The scrap is less (5 to 6% of billet weight)in indirect extrusion in comparison to thedirect extrusion (18 to 20% of billetweight).

Sol–156:(b)The rolls pull the material into roll gapthrough a net frictional force on thematerial. Hence, the frictional force tothe left side of the neutral point mustbe higher than the friction force to theright. Although, friction is necessaryfor rolling material, energy is dissipatedin overcoming friction. Also, the increas-ing friction increases the rolling forceand power requirements. Furthermore,high friction may damage the surfaceof the rolled products or cause sticking.In practice, a compromise is made byusing effective lubricants to lower thecoefficient of friction.

Sol–157:(d)Hunting is due to high senstivity ofgovernor. A governor is said to be huntif the speed of the engine fluctuatescontinuously above and below the meanspeed. This is caused by a too sensitivegovernor which changes the fuel sup-ply by a large amount when a smallchange in the speed of rotation takesplace.

Sol–158: (d)The two kinematic chains are different infigure 1 and figure 2, although both haveequal number of links and joints.

Sol–159: (d)Power generated by 4-stroke engine isalways less than 2-stroke engine of sameswept volume, speed, temperature andpressure condition. The reason is wellknown that 4-stroke engine gives onepower stroke in two revolution and 2-stroke engine gives one power stroke ineach revolution.

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Sol–160: (c)Due to high compression ratio used inCI engine, the efficiency of CI engine ismore than the SI engine. But for samecompression ratio, the efficiency of Ottoengine (SI) is more than diesel engine.The expression for efficiency-

otto = 1c

11r

diesel = 1

c

1111r

where cut off ratio in dieselcycle.rc – compression ratio.

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