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Conventional Question Practice ProgramDate: 9th April, 2016

1. (c)

2. (b)

3. (c)

4. (d)

5. (a)

6. (d)

7. (c)

8. (a)

9. (b)

10. (a)

11. (d)

12. (c)

13. (b)

14. (d)

15. (c)

16. (d)

17. (d)

18. (c)

19. (b)

20. (b)

21. (c)

22. (d)

23. (d)

24. (d)

25. (c)

26. (a)

27. (d)

28. (c)

29. (c)

30. (c)

31. (a)

32. (a)

33. (c)

34. (c)

35. (d)

36. (c)

37. (a)

38. (c)

39. (c)

40. (d)

41. (c)

42. (c)

43. (d)

44. (a)

45. (a)

46. (b)

47. (b)

48. (a)

49. (c)

50. (d)

51. (a)

52. (a)

53. (b)

54. (c)

55. (c)

56. (b)

57. (d)

58. (b)

59. (b)

60. (d)

61. (b)

62. (a)

63. (c)

64. (c)

65. (a)

66. (b)

67. (c)

68. (b)

69. (b)

70. (b)

71. (c)

72. (b)

73. (d)

74. (d)

75. (a)

76. (b)

77. (b)

78. (a)

79. (d)

80. (b)

81. (a)

82. (c)

83. (b)

84. (b)

85. (c)

86. (a)

87. (c)

88. (a)

89. (c)

90. (c)

91. (c)

92. (c)

93. (a)

94. (a)

95. (b)

96. (a)

97. (a)

98. (a)

99. (b)

100. (b)

101. (b)

102. (d)

103. (a)

104. (b)

105. (b)

106. (d)

107. (d)

108. (a)

109. (a)

110. (c)

111. (c)

112. (b)

113. (a)

114. (a)

115. (b)

116. (a)

117. (c)

118. (d)

119. (c)

120. (d)

ANSWERS

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(2) ME (Test-12), Objective Solutions, 9th April 2016

Sol–1: (c)Given,

DBT1 = 40°CAir-washer circulating water with no heataddition or rejection the, WBT remainsconstant.

WBT1 = WBT2 = 20°C

Problem statement says,(DBT2 – WBT2) = 0.25 (DBT1 – WBT1)

DBT2 = 20 + 0.25(40 – 20)= 25°C

DBT

25°C 40°C

21

2DBT 25 C

Sol–2: (b)Given,

1

Q

h = 67 kJ/kg

2

At secton 1,

1m = 3 kg/sh1 = 85 kJ/kg d.a.

1 = 19 gm/kg d.a.At section 2,

2m = 3 kg/sh2 = 43 kJ/kg d.a.

2 = 8 gm/kg d.a.Let ‘Q’ be the required cooling capacity ofthe coil in kW.Moisture removed per kg

= 1 – 2= (19 – 8)

= 11 gm/kg d.a.Total moisture removed

= Mass flow rate × 11= 33 gm/s

From energy balance,Input energy (kW) = Output energy (kW)

11h m = 22Q h m 33 h

85 × 3 = Q + 43 × 3 + 33 × 10–3 × 67

Q 123.8 kW

Sol–3: (c)Given,

RTH = 100 kWRSHF = 0.75

We know that room latent heat is givenby,

RLH = 50 × w × Volume flow ratein m3/min

RSHF= RSHRTH

0.75 = RSH100

RSH = 75 kW

RLH = RTH – RSH = 100 – 75= 25 kW

Now,

RLH = 50 × ( 1 – s ) × 100

25 = 50 × (0.01 – s ) × 100

s = 0.01 – 0.005

s = 0.005 kg/kg of dry airHence specific humidity of the supply airis 0.005 kg/kg of dry air

Sol–4: (d) Azeotropes are mixture of refrigerants

such that constituents of mixtureshould not separate out duringtemperature variation and behave likepure substance.

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Isomers have same chemical formulabut different molecular structures.

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

Sol–5: (a)A good refrigerant should have large latentheat of vapourisation because for a givencapacity of the plant large latent heat willresults in smaller mass flow rate. Lowoperating pressure means light evaporatorand condenser.

Sol–6: (d)1. High pressure angle gives broader tooth

base i.e. stronger tooth. But higherpressure angle results in low contact ratioand rough operation.

2. The axial thrust (Fsin ) increases.

3. Chances of interferences reduces withhigher pressure angle.

4. Backlash has nothing to do with pressureangle.

Sol–7: (c)A. Spur Gear: Connecting two parallel,

coplanar shafts and teeth are parallel toshaft axis. So A–2.

B. Bevel Gear: It connect two non parallelcoplanar and intersecting shafts. Butthese two intersecting shafts needs notbe perpendicular. So B–1.

C. Helical Gear: Two parallel, coplanershafts. Teeth are inclined with shaft axiscalled as helix angle. So C–3.

D. Mitre Gear: Connecting mutuallyperpendicular shafts. The size of two gearsi.e. bevel are equal. So D–4.

Sol–8: (a)

Speed ratio =

1

2

Rotation of input gear NRotation of output gear N

N1 T1 = N2 T2

where T is teeth and N is rpm,

1

2

NN = 2

1

TT

Speed ratio,

= 40 70 50× ×20 35 25

= 2 × 2 × 2 = 8Sol–9: (b)

Follower velocity in various cases,

1. Cycloidal motion, V = 2hω

j

2. SHM, V = πωh ωh= 1.5702j j

3. Uniform velocity motion.

V =hωj

So the cycloidal motion has maximumvelocity, then SHM and the least velocityis in uniform velocity motion.

Sol–10: (a)For SHM cam, the cam displacement :

S = h h π1 - cosβ = 1 - cos .θ2 2 j

Velocity, V = dsdt

Acceleration,

a =2

2d s πhω π π= cos θ. .ω

j j θdt

amax =2 hπ ω

j 2

Sol–11: (d)Sensitiveness of governor is defined in two

cases.1. Governor as single entity i.e. Governor

not fitted in prime mover,

Sensitiveness-

=

1 2

1 2

N N Mean Speed=2 N N Range

2. Governor fitted in prime mover-

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Sensitiveness

=

2 1

2 1

2 N N RangeN N Mean speed

In question, it is not defined whethergovernor is single entity or fitted in primemover. Assuming single entity the correctoption is ‘b’.

Sol–12: (c)

12

Driving shaft

Driven sh

aft

The velocity ratio in Hooke's joint,

2

1

= 2 2

cos1 sin cos

For equal speed; i.e. 2 1

cos = 2 21 sin .cos

2 2sin cos = 1 cos

2 21 cos cos = 1 cos

2cos =

1 cos

1 cos 1 cos

=1

1 cos

2sin = 21 cos

= 1 cos1

1 cos 1 cos

2tan =2

2sin cos= 1 coscos

11 cos

= cos

tan = cos

Sol–13: (b)The schematic of quick return mechanism-Distance between fixed centers-

AB = 150 mm

75m

C90°

A

B

150

mm

Length of driving crank,BC = 75 mm

In triangle ABC,

cos = BC 75 1AB 150 2

= 60°Since cutting of metal happens to be inforward stroke and angle rotated by crankBC during cutting

= 360 2

= 360 – 120 = 240°The angle turned by crank during returnstroke.

B = 2

= 120° Ratio of cutting stroke time to return stroke

time,

=240 2= =120

Sol–14: (d)

The radius of prime circle is equal to thesum of the base circle radius and the radiusof the roller follower.

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The pressure angle varies at all instants ofthe follower motion.

Sol–15: (c)

When the shortest link of a Grashof’s chainis fixed, it gives rise to a double crankmechanism in which both the links connectedto the frame rotate continuously. When thelink opposite to the shortest link is fixed, adouble rocker mechanism results. None ofthe two links i.e. driver and driven connectedto the frame can have complete revolutionbut the coupler link can have full revolution.

When any of the two remaining links(adjacent to the shortest link) is fixed, acrank rocker mechanism results and one ofthe two links (driver or driven) directlyconnected to the frame, is capable of havingfull revolution.

Sol–16: (d)n = 3 2j1

where = no. of links, j = no. of binaryjointsn = no. of degrees of freedomFor a ternary joint, the equivalent numberof binary joints = 2, and for a quarternaryjoint, equivalent number of binary joints= 3Hence, for (a)n = 3 × (6 – 1) – 2 × 8 = 15 – 16 = –1for (b), n = 3 × (6 – 1) – 2 × 10 = 15 – 20= – 5for (c), n = 3 × (6 – 1) – 2 × 9 = 15 – 18= – 3for (d), n = 3 × (6 – 1) 2 × 7 = 15 – 14 =1

Sol–17: (d)When one of the links is fixed in akinematics chain, it is called a mechanism.So we can obtain as many mechanisms asthe number of links in a kinematics chainby fixing, in turn, different links in akinematic chain. This process of obtainingdifferent mechanisms by fixing differentlinks in a kinematic chain is calledinversion of the mechanism. However, therelative motions between various links isnot changed in any manner through the

process of inversion, but their absolutemotions (those measured with respect tothe fixed link) may be changed drastically.This is simply because relative motionbetween different links is a property of theparent kinematic chain.Elliptical trammels are mechanisms withtwo prismatic pairs and two revolute pairs.

Sol–18: (c)

Velocity of sliding s 1 2v QP= . Sincethe velocity of sliding is proportional to thedistance of the contact point from the pitchpoint, so the velocity of sliding at the pitchpoint is zero.

Sol–19: (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, radialcomponent of acceleration of connecting rod,tangential component of acceleration of theconnecting rod and the acceleration of thepiston.From this, the angular acceleration of theconnecting rod can be found out.

Sol–20: (b)Maximum efficiency of spiral gears =

1 cos1 cos

Sol–21: (c)Considering the case of a stable governor,

Fc = ar –b 1600 = 400 a – band 800 = 160 a a = 5 and b = 400Since, both ‘a’ and ‘b’ are positive, henceour assumption of this being a stablegovernor is correct.Now, for isochronous governors FC = arHence, initial tension must be increased

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by 400 N.Sol–22: (d)

In whitworth quick return motionmechanism, the smallest link is fixed andthe connecting rod acts as crank.

Sol–23: (d)T1 + T2 + T3 = 0

1 1 2 2 3 3T T T = 0

3 = 0, 1 1T = 100 kW,

1 100 rad/s=

T1 = 100kW 1kN=100

2 = –10 rad/s 1 × 100 + T2 × (–10) = 0 T2 = 10 kNm T3 = –T1 – T2 = –1 – 10

= – 11 kNmSol–24: (d)

Unstable

F= ar

+ b

c

Isoch

ronou

s F

c=ar

Stable

F c=ar

– b

Cont

rolli

ng fo

rce

(F) C

Radius of rotation (r)

Sol–25: (c)For reversible worm, coefficient of friction tan .A worm drive can be made reversible byincreasing numbers of start. Because num-ber of starts increases lead angle.

Sol–26: (a)In ammonia vapour absorption refrigerationsystem, ammonia is absorbed in water atlow temperature and after being heated bygenerator the ammonia vapour is releasedfrom the aqua ammonia solution. This isbecause the solubility of ammonia in waterdecreases as the temperature increases.Ammonia is used as the refrigerant andwater as absorbent.

Sol–27: (d)Sol–28: (c)

COP = refrigerating effect

work done in compressor

= 1 4

2 4

300 150 150 5= =330 300 30

h hh h

p

3

41

W

2Q

h

1Q

Sol–29: (c)

The name of refrigerant m n p qC H Cl F

would be 1 1R m n q where

2 2n p q m

Inorganic refrigerants are designated byadding 700 to the molecular mass of thecompound.

R 11 3CCl F

R 12 2 2CCl F

R 13 3CClF

R 22 2CHClF

Sol–30: (c)Refrigerant Boiling

TemperatureAmmonia (R717) – 33.3Carbon dioxide (R 744) – 73.6Dichlorodifluoromethane (R 12) – 29Tetrafluoromethane (R 14) – 127.8° Boiling points in descending order isR 12 > R 717 > R 744 > R 14

Sol–31: (a)

Relative humidity = v

s

p 0.01 0.5p 0.02

or

50%

Humidity ratio,

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W = 0.622 v

s v

p 0.01 0.622p p 1.01 0.01

= 0.00622Sol–32: (a)

Test for SO2: An ammonia soaked clothtied at the end of a stick is placed adjacentto the joints or placed where leaks mayoccur. A thick white smoke forms at theplace of leak.Test for NH3: Sulphur candle producesthick white smoke if it comes in contactwith leaking ammonia.Test for Freon: Leaks can be tested withhalide torch. Leaking gas, when enters intothe intake tube of the halide torch, gives agreen hue.Test for Propane: Soap and water test.

Sol–33: (c)

The presence of moisture is very critical inrefrigeration systems operating below 0°C.If more water is present than which can bedissolved by the refrigerant, then there isdanger of ice formation and consequentchoking in the capillary tube used forthrottling is the system.

Thermostatic expansion value maintains aconstant degree of superheat at the exit ofevaporator.

Automatic expansion valve maintains aconstant pressure and hence a constanttemperature in the evaporator. The floatvalve is a type of expansion valve whichmaintains the level of liquid refrigerantconstant in a vessel or an evaporator. Hence,the mass flow rate of the refrigerant throughthe expansion valve is proportional to theevaporation rate of the refrigerant in theevaporator, which in turn is proportional tothe load.

Sol–34: (c)Sol–35: (d)

Since the specific heat of air is one fourthof that of water and density is onethousandth of that of water, volume flowrates required are very large in case of air-

cooled condensers. The thermal conductivityof air is available at dbt while water isavailable at a lower temperature which is2 to 3°C above the wbt. The temperaturerise of air is much larger than that ofwater, therefore the condenser temperaturebecomes large and COP reduces.

Sol–36: (c)

= v

s

p 0.5p

Since the air is compressed isothermally sothe temperature remains the same andhence the saturation pressure correspond-ing to the temperature will remain the same

=v

s

pp

But sp = ps

Since the total pressure is doubled, so vapourpressure will double, hence

vp = 2pv

= v

s

2p 2 2 0.5p

= 1 or 100%So, the sample of air will become saturated.

Sol–37: (a)The schematic of psychrometric chart withdesired information,

Dry bulb temperatureConst. DBT lines H

umid

ity ra

tio (

)

= 10

0%

Con

stan

t () l

ines

Sol–38: (c)Human body losses heat in the form ofradiation and convection which comprisesensible load. But the moisture loss comesin latent heat load catagory.

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Sol–39: (c)Sol–40: (d)

Friction at the tool-chip interface can bereduced by increasing the cutting speed.

Sol–41: (c)Large positive rake angle reducesmagnitude of cutting force.Large positive rake angle reduces heatdissipation.

Sol–42: (c)Using Taylor’s tool life equation.

nVT = CAccording to problem,

n2

1

TT

= 1

2

VV

n18

=12

1n3

Sol–43: (d)High cutting speed, high shear angle andzero negative rake angle.

Sol–44: (a)

Total Cost

Cost

per

pie

ce

Cutting Speed

1

2

3

1. As cutting speed increases, the time takenfor machining reduces, thereforemachining cost will be reduced.

2. Non productive cost never depends on thecutting speed.

3. As cutting speed increases, the tool lifereduced and hence the tool is required tobe changed many a times and toolchanging cost increases.

Sol–45: (a)fm = f Z N

where, fm is feed rate in mm/minfm = 0.1 × 8 × 150

mf 120 mm/min

Sol–46: (b)Sol–47: (b)

Flange portion will not have any stressbecause there is no blank-holding force.

Sol–48: (a)String Milling: In this, small workpieceswhich are to be milled are fed into themilling cutter one after the other. In otherwords, a number of the workpieces will bekept on the machine table in a line andhence called ‘string milling’ or ‘line milling’.The main advantage is that if individualworkpieces are milled, the milling cutterwill have to have the approach distance,which is substantial. By having a numberof workpieces kept in line, the approachdistance will be only at the beginning andend of the line, thus considerably savingthe machine time.Gang Milling: In gang milling, a numberof milling cutters are fastened to the arbour,to suit the profile of the workpiece to bemachined, for example, two side and facemilling cutters with a slab milling cutterat the centre to mill an inverted U-shape.The advantage of gang milling is that sev-eral surfaces are machined at the sametime. It is also possible to combine formcutters along with the general-purpose cut-ters.One of the major problem is the choice ofthe cutting speed, which is determined bythe largest cutter diameter. Hence, it isdesirable that all the cutters should besimilar in size and shape to allow for largerspeeds and feeds. In production millingoperations, gang milling is more generallypreferred.

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Side and face milling cutter

Slab mill

Fig. Gang millingStraddle Milling: Straddle is a specialform of gang milling where only side andface milling cutters are used.A typical sequence of processes used andthe milling cutters required fora componentmachined in a milling machine is shownin fig.

Slab millSlot mill

1 2Finished

part

5Anglemill

Endmill

43

Shell endmill

Side and Face Milling Cutters: Thesehave the cutting edges not only on the facelike the slab milling cutters, but also onboth the sides. As a result, these cuttersbecome more versatile since they can beused for side milling as well as for slotmillingStaggered-tooth side milling cutters are avariation where the teeth are arranged inalternate helix pattern. This type isgenerally used for milling deep slots, sincethe staggering of teeth provides greater chipspace.

Another variation of the side and face cutteris the half-side milling cutter, which hasthe cutting edges only on one side. Thisarrangement will allow providing a positiverake angle and will be useful for machiningon only one side. These have much smoothercutting action and long tool life. The powerconsumed is also less for these cutters.

Sol–49: (c)Ceramic tools have low thermalconductivity.

Sol–50: (d)

Chip Tool

Shear planeC

ttc

A

Workpiece

Chip thickness ratio

= c

t sin=t cos

t = uncut chip thicknesstc = chip thickness = shear angle = rake angle

Sol–51: (a)As the cutting speed increases, wear ratealso increases, so the same wear crition isreached in less time i.e., the tool life de-creases with cutting speed.

VTn = CTool life for a given cutting speed will in-crease with increase in back rake angleupto an optimum value, after which it willdecrease again.

Sol–52: (a)Maximum heat is carried away by chip andtool and w/p shear almost equal heat i.e.,10% each while chip contains 80% of totalheat.

Sol–53: (b)Sol–54: (c)

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Flank wear mainly occurs on the tool noseand front and side relief faces. It occursdue to abrasion between the tool flank andthe workpiece and excessive heat generatedas a result of the same.

Sol–55: (c)Maximum BM will occur at support

A.L

w

2

maxL wLwLM = =3 62

B.L

w/m

maxM = 2L wLwL =2 2

C.L

w

AwLR4

Maximum BM will occur

= AL L L LR w2 2 3 2

maxM = 2

AL w(L / 2)R2 6

= 2 2wL L wL wL

4 2 24 12

D.L

w/m

AwLR2

maxM = AL L 1 LR w2 2 2 2

= 2 2wL L wL wL.

2 2 8 8

Sol–56: (b)

A C

P

B

3a 2a

A B M = P.2a

PIII

At hinged end B, M moment isapplied, for end A is fixed so half ofthe moment will be transferred to A

MA =M Pa2

A

B 2PaPa

Reaction at A = A BM Ml =

Pa 2Pa3a

= P (downward)Sol–57: (d)

Resultant of loading will act at thecentroid of loading diagramResultant R = Area of loading diagram

= 1 WLL W2 2

WL/2

RA RB2L/3 L/3

A B

RA = WL (L / 3) WL2 L 6

RB =WL 2L / 3 WL

2 L 3

Let us assume shear force is zero at a distancex from end A SFA = Area of loading diagram upto x

[V = wdx ]

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WW xL

x

SFA = 1 Wxx2 L

WL6 =

2Wx2L

x =L3

Sol–58: (b)Deflection of free end = 15 mm

4wL

8EI = 15×10–3 ... (i)

Slope of free end = 0.02 radian

3wL

6EI = 0.02 ...(ii)

Dividing equation (i) by (ii), we get

6 L8 =

315 100.02

L = 1 mSol–59: (b)

CA B

2P/3 P/3

BMD

Pl29

2 /3ll/3

x

P

A BC

BM is maximum under the load Deflection will be maximum, where slope

of elastic curve is zero.

Assuming slope is zero at point D at a dis-tance x from B, in the region BC.

Using Moment Area Method

diagram

2P /9EIlPx/3EI

M EIlD x

AB

CBetween point A and B

A = B B l + 1 2P 22 9EI 3

l l 1 2.3 3 3

l l

Sol–60: (d)

a b c d e

Using dF dM and S.F.=dx dx

(D)

a b d e

SF diagram

a b d eSF diagram

(B)

In the BM diagram, between C and e thegraph is symmetrical which is possiblewith (D) i.e. ud , not with (B), i.e. uv .

Sol–61: (b)Bending moment is zero or we can saybending moment changes sign at the pointof contraflexure.

Sol–62: (a)The risk of radioactive hazard is greatestin boiling water reactor because thecoolant water boils in the core of thereactor and no heat exchanger is used.

Sol–63: (c)Percentage of uranium -235 is about 72%in natural uranium.Plutonium is not found naturally insignificant quantities. It is produced in anuclear reactor through the absorption of

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neutrons by uranium 238. The plutoniumemerges from a nuclear reactor as part ofthe mix is spent nuclear fuel, along withunused uranium and other highlyradioactive fission products. To getplutonium into a usable form, a secondkey facility, a reprocessing plant, is neededto chemically seperate out the plutoniumfrom the other materials in spent fuel.

Sol–64: (c)The reactor is called a critical reactor,when the rate of production of neutronsis equal to rate of loss of neutrons. Thus,for a critical reactor

k = rate of production of neutrons 1rate of loss of neutrons

Sol–65: (a)Sodium graphite reactor uses sodium ascoolant and graphite as moderator.

Sol–66: (b)Surging is unsteady, periodic and reversedflow condition in dynamic compressor. Inthis phenomenon, as pressure ratioincreases beyond a limit, the steady flowbreaks and delivery pressure drops auto-matically and this dropped pressure againincreases to limiting value. This cyclehappens again and again till deliverypressure is reduced. This is dangerous ifit happen for long period.

Sol–67: (c)The highest Mach number in centrifugalcompressor occur at inlet to impeller i.e.,at tip of eye of impeller. The Machnumber at this location should be equalto or less than 0.9 to avoid flowseparation. The expression for Machnumber at tip of eye of impeller or inletinducer,

M = w

1

V 0.9RT

Sol–68: (b)

s

T3 2

12´

For maximum efficiency, the interme-diate pressure of intercooling,

P2 = 2 1 3P P P=

Intercooling should be perfect i.e.

T1 = 2T and T2 = T3

Compression work in each stagew1 = cp (T2–T1)

w2 = cp 3 2(T T )

= cp 2 1(T T )

Hence the work of compression in eachstage is equal.

Sol–69: (b)Theoretically, the forward curved bladesare the best option but these blades donot ensure stable operation so from op-eration and performance point of viewand high speed of rotation of compres-sor, radial is most popular from all pointof view, i.e. performance, operation andmanufacturing. There is no undesirablestress in radial blades.

Sol–70: (b)Due to complete axial and open flow infan, the pressure ratio is least. So,For fan = 1.1Blower = 2.5Centrifugal Compressor = 4.0Axial Compressor = 10The highest pressure ratio of axial com-pressor (10) is due to multistages.The difference between pressure ratio offan and blower is due to confined flowin blower.

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Sol–71: (c)

Friction factor f = 02

8v

Coefficient of friction = 02

2ff4 v

Sol–72: (b)

In steady laminar flow through circularpipe shear stress is maximum at the wall

of pipe and it’s magnitude is equal to 8 VD

Shear stress at the centre line is zero Maximum velocity is two times the

average velocity From Hagen Poiseuille equation for

laminar flow

P = 4128 QL

D

Q 1

Slope of hydraulic gradient line

= fhL =

PL

= 4128 Q

D

fhL = 2

32 VD

fhL V

Sol–73: (d)In turbulent flow through rough pipes

maxu 1 1.33 fu

Sol–74: (d)The velocity profile over a solid surfaceexhibits the following characteristicsFor attached flow,

y 0

uy

= +ve

At the verge of separation,

y 0

uy

= 0

Separated flow,

y 0

uy

= –ve

Displacement thickness,

* = 0u1 – dyU

Momentum thickness,

= 0u u1 – dyU U

Sol–75: (a)Given,Distance from leading edge, x = 3 cmlocal reynold’s number, Re = 105 < 5 ×105

Boundary layer is laminarThickness of laminar boundary layer

= 5xRe

= 55 3 0.047 cm 0.47 mm10

Sol–76: (b)Terminal velocity of a sphere settling in aviscous fluid is expressed by stoke’s law

Vt =2

s wgd ( )18

Vt µ d2

Sol–77: (b)The efficiency of screw jack

= tan

tan

For self locking

tantan2

2tan 1 tan

2tan 21 tan2

21 1 tan

2 2

In ideal case i.e. zero friction

= 0

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1 02

max = 0.5 = 50%

Sol–78: (a)In the given options, disc or plate clutch,cone clutch and centrifugal clutch, thepower is transmitted by friction between-1. The plane surfaces of plates.2. Conical surface of truncated cones.3. Surfaces of centrifugal shoe and drum.But jaw clutch is positive drive clutchwhere the jaws of driving and driven shaftengage.

Sol–79: (d)The free body diagram of pin in flexiblecoupling.

A

B

Bush

Portion ‘A’ is in driving flange and ‘B’ indriven flange So the pin is subjected tofollowing stresses.

1. Bearing stress on cylindrical surfacesA & B.

2. Bending stress because the forces fromdriving and driven flanges are not col-linear.

Sol–80: (b)The number of contact surfaces in themultiplate clutch is given by,

= n1 + n2 – 1where n1 is number of disc on driving shaftand n2 on driven shaft.The above formula is explained by follow-ing illustration

1 2 3 4Driving

n =21

Drivenn =32

So driven shaft has 3-plates anddriving shaft 2 plates. Totalnumber of contact plates-= n1 + n2 – 1 = 2 + 3 – 1 = 4

Sol–81: (a)The schematic of isometric square threadwith dimension proportions,

p

d

0.5p

where ‘p’ is pitch and ‘d’ is depth of thread d = 0.5 p

Sol–82: (c)The schematic of V-threads

NNcos

N sinθ

W

The normal reaction

N =W

cos

Hence angle increases the normalreaction. So high friction is obtained in‘V’ thread. As this angle increases Nincreases very fast. So to manage this highfriction, thread angle (2 ) is kept about29°–30°.

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Another force N sin also resultsperpendicular to motion which createsbursting pressure.The difficulty in manufactuirng processis encounterd in square threads.

Sol–83: (b)

The shoe and drum enagementin centrifugal clutch is-

DrumShoe

Spring

O

rd

krg

Fs

Fc

G

In equilibrium of shoe,Centrifugal force (Fc) = Springforce (Fs)

Fc = 2gmr

Fs = k × spring extension(s)

k.s = 2gmr

s =2

gmrk

Because the spring extension isconstant after the mating of shoeand drum. So as stiffness of spring‘k’ increases, the enagement speed' ' increases.

Sol–84: (b)Sol–85: (c)

The Lorenz number is the ratio of thermalconductivity to electrical conductivity asLorenz Number,

L = kT

= Thermal conductivity

Electrical conductivity× temperature

Sol–86: (a)

d = 100 mm

Steam

The Nusselt number,Nu = 25

Nu = hdk

25 = h 0.10.03

h = 25 × 0.030.10

= 7.5 W/m2KSol–87: (c)

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

Nu = 0.14 (Gr.Pr)1/3

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

Sol–88: (a)The Reynold number,

Re = Vd

The velocity is doubled and diameter ishavled, the Reynold number will remainconstant i.e., laminar flow exist Nusselt Number

Nu = 0.664 0.5 0.33Re .PrThe heat transfer coefficient

h = Nu.kd

h = 0.5 0.330.664Re .Prd

Since diameter is halved so h will bedoubled.

Sol–89: (c)The intensity of direction-

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dA1

d

The intensity of radiation in direction from normal is–

I = 1

dqdA cos d

Where 1dA cos is projected area ofradiating surface in direction ' ' .So intensity is defined” radiant energyper unit time (dq) per unit solid angle(d ) per unit projected area in given

direction 1(dA cos ).

Sol–90: (c)

2

1

Sum of view factors, F11 + F12 = 1

F22 + F21 = 1 F22 = 0 F21 = 1Reciprocity theorem,

A1 F12 = A2 F21

F12 = 221

1

A FA

=2

2r 1

2 r

= 12

F11 = 1 – F12

= 1 – 12

= 12

Sol–91: (c)1. Materials such as metal non-metal e.g.

plastic etc are opaque (except glass).The transmissivity of opaque materialis zero: so for opaque material

absorptivity ' ' + reflectivity ' ' = 1

absorptivity ' ' = emissivity ' '

+ = 12. In general emissivity of non-metal

(non-conductor) is more than conduc-tor, so reflectivity ' ' will be low fornon-metal because both are opaque.

3. So metals have low absorptivity andhigh reflectivity and at same timenon-conducting material has highabsorptivity and low reflectivity.

4. So reflectivity and absorptive ishighly dependent on surface texturei.e. polished surface has highreflectivity and low absorption/emis-sivity.Since gaseous radiations are volumet-ric phenomena so reflectivity is verylow. They have high transmissionalso.

Sol–92: (c)

d/2 ld/4

d/4

FShearing area

Torque, T = dF.2

Average stress,

avg = FArea =

2Tdd4

l = 2

8Tdl

avg 28Td

l

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Sol–93: (a)

A B

CD

E

F

Plane ABCD represents shear plane.Plane BCFE represents bearing plane.A key is likely to fail in shear or bearingand is designed for shear and crushingstrength.

Sol–94: (a)Sol–95: (b)

Local Nusselt Number,

Nux = y 0

w

t

Ty

T T

w

w

T TT T

=3

t t

3 y 1 y2 2

Ty

= 2

w3t t

3 1 y T T2

At, y = 0

y 0

Ty

= w

t

T T1.5

Nux =

w

t

w

t

1.5 T T

T T

= 1.5

xNu 1.5

Sol–96: (a)Sol–97: (a)Sol–98: (a)Sol–99: (b)Sol–100: (b)Sol–101: (b)Sol–102: (d)Sol–103: (a)

Sol–104: (b)Sol–105: (b)Sol–106: (d)

Ackermann steering gear is preferred toDavis steering gear because it consists ofturning pairs, hence less friction and so lesswear and tear.

Sol–107: (d)Sol–108: (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 from 2 to2', the work of compression increases from1-2 to 1–2' and the refrigerating effectdecreases from 4–1 to 4'–1.

Sol–109: (a)High carbon tool steels are usually plaincarbon steels containing 0.6 to 1.5% carbon.Carbon steel tools are easy to manufactureand their cutting edge can be easilysharpened. However, their chief draw backis their low wear resistance and low hothardness. They lose their hardness rapidlyat temperatures greater than 200°C. Highcarbon steel can therefore be used for onlythose applications where cuttingtemperatures remain well below 200°C, e.g.in making cutters, twist drills, turning andform tools for machining soft or free cuttingmaterials like wood, magnesium, brass andaluminium.

Sol–110: (c)Carbide tools are weaker in tension thanin compression so shank should have suf-ficient thickness to minimize tensilestresses.Carbide tipped tools usually have negativerake angles.

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1. To increase strength of tool2. To decrease temperature rise of tool due

to flow of chipsSol–111: (c)

Point Angle for Cast Iron 118°Point Angle for Ductile Material(Soft material) 140°Normal rake angle

1 2r / D tantansin

where ,r – radius of the point on the cutting edgeD – Nominal diameter of the drill – Half point angle – Helix angle

Sol–112: (b)

Rate of change of bending moment dM Vdx

Sol–113: (a)Inlet velocity triangle at tip of eye,

V1Vw1

u1

The Mach Number at eye,

M = 1wVRT

By providing inlet guide vanes, 1wV

is reduced to keep M < 0.9.Sol–114: (a)

For laminar flow, in circular pipePower consumption

P = fQh ... (i)

hf = 4128 QL

D

... (ii)

p = 4128 QLQ

D

p =2

4128 Q L

D

... (iii)

From (ii) and (iii), it is evident that both thestatements are correct and R is the correctexplanation of (A).

Sol–115: (b)If Reynold’s number is low, flow islaminar. In laminar flow momentumtransfer between adjacent layer of flow isnegligible and on the other hand ifReynolds number is high flow is turbulentand there is significant momentumtransfer between adjacent layers. Hencewe can say that Reynolds number of afluid flow is indicative of the relativedominance of the effect of momentumtransfer between adjacent layers of theflow over the viscous stresses.If Reynolds number is high, flow isturbulent and velocity profile islogarithmic.

Sol–116: (a)Drag forces are much larger for turbulentthan for laminar flow.

1.

2.

3.

4.

5.

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Here the figure shows the air flow pass acylinder as the air speed increases. Thefirst two pictures show laminar flow at avery low speed. In the third and fourthpicture, the flow has become turbulent.In the fourth picture a turbulent wakehas formed. The drag now is very large.As the airspeed increases further, theturbulent region works itself forward. Wehave what is called a turbulent boundarylayer. The flow lines now separate fromthe cylinder and follow the turbulentboundary layer as shown in the fifthpicture. We have something similar tolaminar flow around an object of adifferent shape. Drag is drasticallyreduced. The transition from the situationshown in picture 4 to the situation shownin picture 5 occurs at lower speeds forlonger surfaces. That is why a golf ball isnot smooth.

Sol–117: (c)Since good absorbers are also good emitter.The nose of aeroplane is painted black toemit the heat generated due toaerodynamic heating not to absorb it.

Sol–118: (d)For the isentropic flow of an ideal gasthrough a convergent-divergent nozzle,mass flow rate per unit area is maximumwhen M = 1. Further, M = 1 occurs onlyat the throat and nowhere else and thishappens only when the discharge is

maximum. The nozzle is said to be chokedbecause it is incapable of allowing moredischarge even with further decrease inexhaust pressure.If the convergent-divergent duct acts asnozzle, in the divergent part also, thepressure will fall continuously to yield acontinuous rise in the velocity. Thus, fora certian mass flow rate, with the decreasein pressure, density decreases at a ratefaster than the rate at which areaincreases, as a result of which velocitycontinues to increase. This is true onlyfor a compressible fluid.Thus when the pressure ratio in a nozzlereaches critical pressure ratio, thedischarge becomes maximum.

Sol–119: (c)The basic function of differential gears inmotor vehicle is to run the back wheels atdifferent speed at turns. Generally it isprovided at back axle.Hence it plays its role on turn. It has noth-ing to do with bump etc.

Sol–120: (d)Helical gears are used for transmittingmotion and power between parallel shaftswhile straight bevel for intersecting shaftat 90°. The helical gear teeth are inclinedwith axis of gear while teeth of bevel gearstapered in height and thickness towardcentre.

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