1. (c)

2. (a)

3. (d)

4. (b)

5. (c)

6. (a)

7. (c)

8. (b)

9. (b)

10. (d)

11. (a)

12. (a)

13. (d)

14. (a)

15. (d)

16. (b)

17. (a)

18. (d)

19. (d)

20. (c)

21. (b)

22. (b)

ESE-2020 PRELIMS TEST SERIESDate: 22nd September, 2019

23. (b)

24. (d)

25. (b)

26. (a)

27. (b)

28. (a)

29. (b)

30. (a)

31. (c)

32. (d)

33. (a)

34. (d)

35. (d)

36. (a)

37. (c)

38. (c)

39. (b)

40. (b)

41. (c)

42. (d)

43. (c)

44. (b)

45. (d)

46. (d)

47. (a)

48. (d)

49. (d)

50. (a)

51. (b)

52. (c)

53. (b)

54. (b)

55. (b)

56. (a)

57. (c)

58. (c)

59. (b)

60. (b)

61. (c)

62. (a)

63. (b)

64. (c)

65. (a)

66. (a)

67. (b)

68. (d)

69. (d)

70. (d)

71. (c)

72. (c)

73. (a)

74. (d)

75. (b)

76. (c)

77. (a)

78. (b)

79. (c)

80. (a)

81. (a)

82. (d)

83. (c)

84. (a)

85. (b)

86. (b)

87. (d)

88. (c)

89. (b)

90. (b)

91. (d)

92. (c)

93. (b)

94. (d)

95. (a)

96. (c)

97. (b)

98. (b)

99. (a)

100. (c)

101. (b)

102. (a)

103. (c)

104. (d)

105. (c)

106. (d)

107. (d)

108. (b)

109. (d)

110. (d)

111. (d)

112. (a)

113. (b)

114. (b)

115. (c)

116. (a)

117. (d)

118. (d)

119. (b)

120. (a)

121. (c)

122. (b)

123. (c)

124. (c)

125. (b)

126. (b)

127. (d)

128. (b)

129. (c)

130. (b)

131. (b)

132. (c)

133. (a)

134. (d)

135. (b)

136. (d)

137. (b)

138. (b)

139. (b)

140. (c)

141. (b)

142. (a)

143. (a)

144. (a)

145. (c)

146. (b)

147. (a)

148. (b)

149. (c)

150. (a)

ANSWERS

IES M

ASTER

(2) SOM+TH+RSE+MD

1. (c)The paddle wheel work on the system, due to100 kg mass dropping 5 m.

W1 = –(F)(d) = –(100)(9.81)(5)

= – 4905 J

The work done by the system on piston is positive.

W2 = 3P V 200 10 V J

The net work done,

Wnet = W1 + W2

– 905 = –4905 + 200 × 103 × VV = 0.02 m3

2. (a)Thermodynamic or absolute temperature scale isindependent of any working substance, The factthat the efficiency of a reversible heat enginecycles depends only on the temperature of thetwo reservoirs makes it possible to establish sucha scale.

This absolute temperature scale is based onsecond law of thermodynamics.

The zeroth law of thermodynamic is basis fortemperature measurement of a body.

3. (d)m = 0.5 kg, u1 = 25.4 kJ/kg

As the vessel is rigid therefore work done shallbe zero.

From first law of thermodynamics:

Q = 2 1(U U ) W

= m(u2–u1)+ 0

5.4 = 0.5(u2 – 25.4)

u2 = 36.2 kJ/kg

4. (b)First law of thermodynamics

U = Q – W .For process 1 – 2

Q = 300 + (– 200)Q or C = 100 kJ

For a closed cycle,

U = 0300 – 100 + U3 – 4 + 50 = 0

U3 – 4 or A = – 250 kJFor process 3 – 4,

W = 300 –(– 250)

W or B = 550 kJ

5. (c)Hyperbolic process is also known as isothermalprocess. (T = constant)

Heat interaction,

Q = dU W

= 21

1

V0 W RT nV

l

Change in internal energy for isothermal processis zero.

6. (a)h1 = 3050 kJ/kg and

1 = 0.025 m3/kg

Enthalpy h = u + P

u1 =5100 10 0.0253050

1000

= 2800 kJ/kg

Also, u2 = uf + ufg

= 152 + 2450 = 2602 kJ/kgSince the cylinder is perfectly insulated then noheat lost to or from the steam during expansion;the process is therefore adiabatic.

W = u1 – u2

(For adiabatic non-flow process)= 2800 – 2602 = 198 kJ/ kg

Work done by steamW = 198 × 2 = 396 kJ

7. (c)

Steady flow energy equation for diffuser

21 1

1h V2

= 22 2

1h V2

21

1 V2

= (3230 – 3150) × 103

2V 0

V1 = 400 m/s

The mass flow rate,

m = AV AV /

A =0.2 0.6

400

= 3 × 10–4m2 or 3 cm2

IES M

ASTER

[ME], ESE-2020 PRELIMS TEST SERIES PAPER-II (TEST-01) (3)

8. (b)Internal energy is defined as the sum of all themicroscopic forms of energy of a system. It isrelated to the molecular structure and the degreeof molecular activity.

The internal energy associated with the phase ofa system is called the latent energy.

The portion of the internal energy of a systemassociated with the kinetic energies of themolecules is called the sensible energy.

9. (b)Both the Kelvin-Planck and the Clausiusstatements of the second law are negativestatements, and a negative statement cannot beproved. Like any other physical law, the secondlaw of thermodynamics is based on experimentalobservations. To date, no experiment has beenconducted that contradicts the second law, andthis should be taken as sufficient proof of itsvalidity.

A process is called totally reversible, or simplyreversible, if it involves no irreversibilities withinthe system and its surroundings.

Energy degrades as it f lows f rom hightemperature to low temperature or under finitetemperature gradient. This degradation is calledas irreversibility.

10. (d)Heat pump has to deliver 2000 W as the heatrejection

Heat pump, COP =Heat rejected

work input

3 =in

2000W

Win = 666.67 W

So the heat pump requires an input of 667.67 W

thus work saved,

Wsaved = 2000 – 666.67 = 1333.33W

11. (a)Carnot engine efficiency,

c = L

H

T1T

TH = L

C

T 280 1400K1 1 0.8

The COP of refrigerator,

COP =L

H L

T 280T T 1400 280

= 0.25

12. (a)Slow heating of water by electric heater isirreversible because heating is done by heateri.e. temperature of heater is more than water, solack of thermal equilibrium.

Mixing of ideal gases at constant pressure isirreversible because there is concentration gradienti.e. lack of chemical equilibrium.

Theoretical isothermal compression is a reversibleprocess.

13. (d)The equation TdS = dH – VdP relates only theproperties of the system. There is no path functionterm in the equation. Hence the equation holdsgood for any process.

In adiabatic mixing process there is alwaysincrease in entropy accompanied by increase inamount of irreversibility. So adiabatic mixing isan irreversible process.

All the spontaneous process in nature occur onlyin one direction from a higher to lower potentialand these are accompanied by an entropyincreases of the universe.

The exergy of an isolated system can neverincrease, but always decreases. For reversibleprocess, exergy remain constant.

14. (a)Transfer of heat through finite temperaturedifference, then

univS =1 2

Q QT T

AS =3000 3000600 400

= 2.5 kJ/k

BS =5000 50001000 700

= 2.14 kJ/K

A BS S , so process A is more irreversiblethan process B.

IES M

ASTER

(4) SOM+TH+RSE+MD

15. (d)The internal irreversibility is caused by the internaldissipative effects like friction, turbulence,electrical resistance, magnetic hysteresis, etc.,within the system. The external irreversibility refersto the irreversibility occurring at the systemboundary like heat interaction with the surroundingsdue to a finite temperature gradient.

Work is energy transfer without associatedentropy transfer and heat is energy transfer withassociated entropy transfer.

16. (b)Heat transfer during the process,

Q = m hfg = 2(2260) = 4520 kJ

steamS = fgfg

hmS m

T

=22602373

= – 12.12 kJ/k

surrS =0

Q 2 2260T 300

= 15.06 kJ/K

univS = system surrS S

= –12.12 +15.06 = 2.94kJ/K

17. (a)Compressor is an open system.

Compressor

1

2

Steady flow equation,

Wc = h1 –h2 + qrev ....(i)

Entropy Equation,

S2 = 1 gendQS ST

S2 =rev

10

qS 0T

qrev = T0(S2 – S1) ...(ii)

(Wc)min = h1 – h2 + T0(S2 – S1)

= 385.4 – 445.8 + 300

(1.80 – 1.75)

= – 45.4 kJ/kg

18. (d)Kelvin scale, Q T

2

1

QQ =

2

1

TT

logarithmic scale, Q exp

2

1

QQ =

2

1

expexp

By comparison,

2

1

TT =

2

1

expexp

cycle =2 2

1 1

T exp1 1T exp

19. (d)Saturated liquid lines Critical

state

L + VTriplepoint line

Saturatedsolid line

p

S'S + V

V

V

Saturated vapour line

VL

S+L

Fig. p-v diagram of a pure substance other than water,whose volume increases on melting

It is clear that the liquid phase does not existbelow triple point line.

During phase change temperature remainsconstant.

The slopes of the sublimation and vaporisationcurves on P-T diagram for all substances arepositive. The slope of the fusion curve for mostsubstances is positive, but for water, it is negative.

20. (c)A general principle that states that for a simplepure substance the state is defined by twoindependent property.

The saturated liquid and saturated vapour statesof a pure substance have the same pressure andthe same temperature, but they are definitely not

IES M

ASTER

[ME], ESE-2020 PRELIMS TEST SERIES PAPER-II (TEST-01) (5)

the same state. In a saturation state, therefore,pressure and temperature are not independentproperties. Two independent properties, such aspressure and specific volume or temperature andspecific volume or pressure and quality, arerequired to specify a saturation state of a puresubstance.

21. (b)Since the vessel is rigid, the specific volume doesnot change. Hence the specific volume at apressure of 800 kPa is also 0.08 m3/kg.

= f fgx

0.08 = 0.0012 + x × 0.4

x = 0.197

Total mass of water,

m =V 10 125 kg

0.08

mass of water vapour,

m v = x m

= 0.197 × 125

= 24.62 kg

22. (b)Compressibility factor is defined for both idealand real gases. z = 1 for ideal gas. For real gasz can be greater than or less than unity

Universal gas constant is independent of molecularweight. Characteristic gas constant decreaseswith increase in molecular weight.

Fig. At very low pressures, all gases approach ideal-gas behaviour

(regardless of their temperature)

Idealgas

Realgas

asP 0

23. (b)The value of is dependent on molecular structureof gases.

= 1.67 for monatomic gas = 1.4 for diatomic gas

24. (d)

w = 0.4( q)

Using first law,

q = u w

vw c T w

0.41.5 w = cv × 20

vcw

= 1.5 100 7.5%

20

25. (b)

P = C

PV = C Pv = C

P

V=C VPVk = C

Differentiate, k k 1dP V Pk V dV 0

dP PkdV V

For isochoric process, k dPdV

For isothermal expansion process, k = 1

dPdV

=

PV

For adiabatic expansion process, k =

dPdV

=

PV

For isobaric process, k = 0

dPdV

= zero

26. (a)

Carnot efficiency, 2c

1

T1T

= 3731 0.5746

Efficiency of engine, 2

1

Q1Q

= 90012100

= 0.571

Since carnot engine is most efficient engineworking between same temperature differences.So, the given engine is not possible.

IES M

ASTER

(6) SOM+TH+RSE+MD

27. (b)

Heat transfer, Q mC T80 mC(37 7)

8mC kW / K3 ...(i)

Entropy generation, f

i

T dQds mC nT T

=

8 310 80n

3 280 320

=

38 100.1 0.253 = 16.67 W/K

28. (a)For isothermal process,

P1V1 = P2V2

V1 = 2

2 21

P V 10VP

V1 = 10 × 0.045 = 0.45 m3

For adiabatic process

P1V1 = P2V2

V2 = 1

11

2

P VP

V2 = 1

1.4 31 0.45 0.086m10

Alternate

P

V

2 2

1

PV = C PV = C

2V > V2

2V > 0.045

Only option (a) matches with this condition.

29. (b)100°C

0°C

–40°C

300°N

50°N

x

300 x300 50

= 100 40

100 0

x = –50°N

30. (a)

Q = mCp T= 1000 × 0.5 × (1200 – 600)

= 300 × 103 kJ

S = mCp ln 1

2

TT

= 1000 × 0.5 × ln 1200600

= 0.35 × 103 kJ/K

Available energy = Q – T0 S

= 300 × 103 – 300 × 0.35 × 103

= 195 MJ

31. (c) W = PdV = P(V2 – V1)

= 2 1nRT nRTPP P

= nR (T2 – T1) = 1 × 1.9 × (227 – 27)

= 380 cal32. (d)

A flat plate with an elliptical hole is subjected totensile force is shown in below fig.

a

b

Where, a = half width of the ellipseperpendicular to thedirection of the load

b = half with of the ellipse inthe direction of the load.

Theoretical stress concentration factor,

Kt =a1 2b

As the width of the elliptical hole in the directionof load approaches zero (b 0) , the stressconcentration factor becomes infinity,

IES M

ASTER

[ME], ESE-2020 PRELIMS TEST SERIES PAPER-II (TEST-01) (7)

33. (a)

Endurancelimit stress

3 4 5 6 7 8log N10

log S10 e

log S10 f

++++++++++++

Failure points

For ferrous materials like steels, the S-N curvebecomes asymptotic at 106 cylces, whichindicates the stress amplitude corresponding toinfinite number of stress cycles. The magnitudeof this stress amplitude at 106 cycles representsthe endurance limit of the material. The S-N curveshown in figure is valid only for ferrous metals

For non-ferrous metals, like aluminium and otherferrous metals such as cast iron this curve neverbecomes horizontal. Hence, these materials donot have any specific endurance limit.

No. of cycles

Fatig

ue s

tress

Other ferrousmetals

Non-ferrousmetals

Fig. Fatigue curve

34. (d)

According to the distortion energy theory, for bi-axial stress

Syt = 2 21 2 1 2 ...(1)

For a component subjected to pure shear stress.

1 = – 2 xy 3and 0

substituting these values in equation (1)

Syt = xy3

Replacing xy syby S ,

Yield strength in shear, Ssy = yt

yt

S0.577S

3

35. (d)According to maximum principal strain theory

1 2 3 ytS / fos

150 – 0.3 (–100 + 50) = 350/fos

165 = 350/fos

fos = 2.12

36. (a)(i) Welded joints have poor resistance to

vibrations and impact load. A riveted joint isideally suitable in such situations.

(ii) Riveting process creates more noise thanwelding due to hammer blows.

(iii) Riveted joints are used for metals with poorweldability like aluminium alloys.

37. (c)Compared with solid shaft, hollow shaft offersfollowing advantages:

(i) The stiffness of the hollow shaft is morethan that of solid shaft with same weight.

(ii) The strength of hollow shaft is more thanthat of solid shaft with same weight.

(iii) The natural frequency of hollow shaft ishigher than that of solid shaft with sameweight.

Compared with solid shaft, hollow shaft has thefollowing disadvantages:

(i) Hollow shaft is costlier than solid shaft.

(ii) The diameter of hollow shaft is more thanthat of solid shaft and requires more space.

38. (c)Deep groove ball bearing takes loads in the radialas well as axial direction.

Deep groove ball bearing is not self-aligning.Accurate alignment between axes of the shaftand the housing bore is required.

Deep groove ball bearing has poor rigiditycompared with roller bearing. This is due to thepoint contact compared with the line contact incase of roller bearing.

39. (b)

SpeedElement time Number of revolut-P N

rpmminute ions in element time4000 0.5 400 2006000 0.5 400 200

IES M

ASTER

(8) SOM+TH+RSE+MD

Equivalent load for complete work cycle in ballbearing,

Pe =

13 3 3

1 1 2 2

1 2

N P N PN N

=

13 3 3200 4000 200 6000

200 200

Pe = 5192.5 N

40. (b)Sut = 600 N/mm2,

Syt = 450 N/mm2

Se = 300 N/mm3, FOS = 5

Permisible mean stress,

Sm = 300 N/mm2

Gerber equation:2

a m

e ut

S SS S

= 1

2aS 300

300 600

= 1

Permissible amplitude

Sa = 225 N/mm2

Mean stress, a = Sa/FOS = 2255

= 45 N/mm2

Mean amplitude, m = Sm/FOS

=3005 = 60N/mm2

a = max min 452

...(i)

m = max min 602

...(ii)

From eq. (i) and (ii) ,

max = 105 N/mm2,

min = 15 N/mm2

Alternate:

Mean stress, m = Sm/FOS = 3505

= 60N/mm2

m = max min 602

max min = 120

Only option (B) is match with above equation.

41. (c)For components made of ductile materials andthose subjected to external fluctuating forces,endurance limit is considered to be the criterionof failure. Such components fail on account offatigue. Fatigue failure depends upon theamplitude of fluctuating stresses and the numberof stress cycles.

42. (d)The pitch (P) of the rivet is defined as the distancebetween the centre of one rivet to the centre ofthe adjacent rivet in the same row.

The margin (m) is the distance between the edgeof the plate to the centreline of rivets in the nearestrow.

Diagonal pitch (Pd) is the distance between thecentre of one rivet to the centre of the adjacentriver located in the adjacent row.

pt

pd

p

m

43. (c)

l/4

l/2 l

PP

Maximum permissible tensile stress,

t =Load

Width thickness of plate

t = P2d t l

t = t

P2d l

44. (b)A bolted joint subjected to tensile force P is shownin figure. The cross-section at the core diameter

IES M

ASTER

[ME], ESE-2020 PRELIMS TEST SERIES PAPER-II (TEST-01) (9)

dc is the weakest section. The maximum tensilestress in the bolt at this cross-section is givenby,

t =2c

P

d4

...(i)

The height of the nut h can be determined byequating the strength of the bolt in tension withthe strength in shear.

Substituting,

t = ytS

fsIn eq. (i), then strength of the bolt in tension isgiven by,

P = yt2c

Sd

4 fs

...(ii)

The threads of the bolt in contact with the nut aresheared at the core diameter dc. The shear areais equal to cd h , where h is the height of thenut. The strength of the bolt in shear is given by,

P = syc

Sd h

fs

= ytc

Sd h

2fs

...(iii)

Equating (ii) and (iii),

h = 0.5 dc

45. (d)Torsional moment,

Mt = 200N-m = 200 × 103N-mm

G = 73 GPa,

= 0.5° per metre length

Shaft diameter on rigidity basis.

d4 = t584MG

l

=5

3584 2 10 1000

73 10 0.5

d = 42.29 mm

46. (d)Pitch Circle Diameter: The pitch circle diameteris the diameter of the pitch circle. The size of thegear is usually specified by the pitch circlediameter. It is also called pitch diameter. Thepitch circle diameter is denoted by d .

Circular Pitch: The circular pitch (Pc) is thedistance measured along the pitch circle betweentwo similar points on adjacent teeth. Therefore,

Pc =dz

...(i)

Where z is the number of teeth.

Diametral Pitch: The diametral pitch (Pd) is theratio of the number of teeth to the pitch circlediameter. Therefore,

Pd =zd

...(ii)

From Eqs (i) and (ii)

Pc × Pd =

47. (a)

Gear ratio, i =Number of teeth on gear

Number of teeth on pinion

i =g

p

zz

The ratio factor Q for internal gears is given by,

Q =g

g p

2zz z =

2ii 1

The ratio factor Q for external gears is given by,

Q =g

g p

2zz z =

2ii 1

48. (d)Compared with hydrostatic bearins, hydrodynamicbearings are simple in construction, easy tomaintain and low in initial as well as maintenancecost. Hydrostatic bearings , although costly, offerthe following advantages:

(i) High load carrying capacity even at lowspeeds;

(ii) No starting friction; and

(iii) No rubbing action at any operating speedor load.

49. (d)

Wr

IES M

ASTER

(10) SOM+TH+RSE+MD

Bearing pressure,

P =W

Projected area of bearing

=W

2r

P =4 1000

2 25 60

= 1.33 N/mm2 or MPa

50. (a)Equivalent load, P = XFr + YFa

= 0.6 × 10 + 1.5 × 4

= 12 kN

Bearing life, L10 =3C

P

=3100

12

= 578.7 million rev.

51. (b)

1 = 240 rad/s, 0

Energy absorbed by brake

E = 2 21 2

1I2

= 21 4 40 0 3200 J2

The average velocity during the braking period is 1 2 1 + /2 or /2 . Therefore,

= 1 40t 2 40 rad2 2

Torque, Mt =E 3200 80 N-m

40

52. (c)

R

Drum

ab

cP

Block

A block brake consists of a simple block, whichis pressed against the rotating drum by means ofa lever as shown in figure. The friction betweenthe block and the brake drum causes theretardation of the drum . This type of brake iseither rigidly attached to the lever or, in someapplications, pivoted to the lever. The angle ofcontact between the block and the brake drum isusually small. When it is less than 45°, theintensity of pressure between the block and brakedrum is uniform.

53. (b)Time required to bring output shaft to the ratedspeed,

t =

1 2 1 2

1 2 t

I II I M

...(1)

Here 1 = angular velocity of drivingshaft

2 = angular velocity of drivenshaft

Torque, Mt =Power

angular speed

=315 10 50 N-m

300

t =

300 0 0.2 0.60.2 0.6 50

= 0.9 sec

54. (b)The rotating rim is subjected to a uniformlydistributed centrifugal force PC, which acts in aradially outward direction. This induces a tensileforce P in the cross-section of the rim acting intangential direction and a bending Moment M.The rim is subjected to direct tensile stress dueto P, and bending stresses due to M. Theresultant stresses in the rim are given by,

t =P MyA I

55. (c)

Mass of disc, m = 2R t

Mass moment of inertia of disc,

I =2

4mR tR2 2

= 47800 0.021 0.52

= 16.08 kg-m2

IES M

ASTER

[ME], ESE-2020 PRELIMS TEST SERIES PAPER-II (TEST-01) (11)

Maximum fluctuation of energy,

E = 2SI C

= 16.08 × 202 × 0.2

= 1286.4 J

56. (a)In hot riveting, when the rivet cools, the reductionin the length of the shank is prevented by theheads resting against the connected members.Therefore, the shank portion of the rivet issubjected to tensile stress while the connectedparts are compressed. This is illustrated in belowfigure

P P

P

P

(a) (b) (c)

Fig. (a) Tendency of Shank to contract (b) Partsin compression (c) Shank in Tension

57. (c)

According to uniform wear theory,

The operating force, F = p d (D d)2

Where, d = inner diameter

D = outer diameter.

F = 22 2 105 205 1057 2

F = 33,000 N = 33 kN

58. (c)

A dry clutch has higher coefficient of friction. Inwet clutches, the coefficient of friction is reduceddue to oil. The coefficient of friction for dryoperation is 0.3 or more, while it is 0.1 or lessfor wet operation.

The torque capacity of dry clutch is high comparedwith the torque capacity of wet clutch of thesame dimensions.

In wet clutches, the clutch facings are groovedto provide for passage of lubricant. This reducesthe net face area for transmitting torque.

59. (b)

r

p2

p4

p1

p3

T

p1.r + p2.r + p3.r + p4.r = T

due to symmetry,p1 = p2 = p3 = p4 = p4.p.r = T

Load on each bolt, p = 385 1000

4 50 = 1925 N

shear stress in a bolt is,

= 2

P PA d

4

= 2

1925 4 722 7

= 50 MPa

60. (b)Primary resources : These include resourcesembodied in nature prior to undergoing any human-made conversions or transformations. This onlyinvolves extraction or capture. Examples of primaryenergy resources are coal, crude oil, sunlight,wind, running rivers, vegetation and radioactivematerial like uranium etc.

Secondary resources : The energy resourcessupplied directly to consumer for utilization afterone or more steps of transformation are knownas secondary or usable energy, e.g. electricalenergy, thermal energy (in the form of steam orhot water), refined fuels or synthetic fuels suchas hydrogen fuels, etc.

61. (c)The performance of a flat plate collector can beimproved by suppressing or reducing the heatlost from the collector by convention andconduction. This is done by having vaccum aroundthe absorber.Absorber plate surfaces which exhibit thecharacteristics of high value of absorptivity forincoming solar radiation and low value of emissivityfor outgoing re-radiation are called selectivesurfaces. Such surfaces are desirable becausethey maximize the net energy collection. Someexamples of selective surface layers are copperoxide, nickel black and black chrome.

IES M

ASTER

(12) SOM+TH+RSE+MD

62. (a)In PV cell, a material with lower band gap willabsorb more number of photons producing highershort circuit current. Thus short curcuit current ofsolar cell decreases with increases in band gap.

Reverse saturation current,

I0 = 1.5 × 105 e–Eg/kT

and Voc=L

T0

IV n 1I

l

Here Voc open circuit voltage

VT is voltage equivalent of temperature

IL is light generated current

The above expressions reveals that with increasein band gap, I0 decreases, which results theincrease in Voc.

63. (b)Usually air is circulated through the bed to add orremove energy.

Their is no corrosion and freezing problem inpacked bed storage

In this system heat addition and removal from thestorage can not be carried out simultaneously.

It can be easily used for thermal storage attemperature higher than 100°C.

64. (c)Reversible voltage E of a cell is given by

E = revW GnF nF

Since 2 electrons are transferred per molecule ofH2, so n = 2.

Now, E =3237.3 10 1.23 volt

2 96500

65. (a)Bio-ethanol is produced from wet biomasscontaining sugars, starches or cellulose.

Vegetable oils is used to produce Bio-diesel.Organic wastes from plants, animals and humansis used to produce Bio-gas

Crop residues such as straw, rice husk etc areuse to produce producer gas.

66. (a)The radiation incident normal to the surface,

n bnI I cos

= 0.9 × cos 60° = 0.45 kW/m2

Power collected, P = In × area of surface

= 0.45 × 50 = 22.5 kW

67. (b)Thermo-chemical storage is a type of chemicalenergy storage system.

68. (d)The conversion efficiency of solar cell

oc sc

i

FF V IP

Here Pi is incident solar power

Pi = 0.8 0.5 0.8 1.07 W

0.3

69. (d)70. (d)

Tip speed ratio, Speed of tip of rotor bladefree stream velocity

= 60 415

Torque coefficient, CT = PC 0.48 0.124

71. (c)Increased energy output can be obtained bypumping water during high tide to increase thebasin level and therefore the generation head alsoincreases.

72. (c)

Sunset hour angle, s = 60°

Solar day length, td = s2

15 (in hours)

=2 60 8 hours

15

73. (a)Concentration ratio,

C =effective aperture area

absorber tube area

=

o

o

(w D )LD L

=2

22.5 8 10

8 10

= 9.63

74. (d)

Ig = 4500 kJ/m2-hr, b = 3000 kJ/m2-hr

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[ME], ESE-2020 PRELIMS TEST SERIES PAPER-II (TEST-01) (13)

Diffuse radiation,

d = g b

= 4500 – 3000

= 1500 kJ/m2-hr

Total radiation flux incident on tilted surface,

T = b b d d b d rr r ( )r 0

= 3000 × 0.9 + 1500 × 0.6

= 3600 kJ/m2-hr

75. (b)

a2

Axis of parabola 2

A DAxis of CPC

Axis

of p

arab

ola

1

Parabola 1Parabola 2

O

B Cx

E

H

y

Acceptance angle a(2θ ) : The angle betweenaxis of CPC and line connecting the focus ofone of the parabolas to the opposite edge ofaperture is called half acceptance angle a( ).The angle made by lines joining focus of eachparabola to their opposite aperture i.e. AED iscalled acceptance angle.

The concentration ratio of a cpc is related toacceptance angle by,

CR =a

1sin

asin =1 1

1.154 2 / 3

a =1 3sin

2

= 60°

Acceptance angle = a2 = 120°

76. (c) The ICS and thermosiphon are passive solarwater heaters where fluid circulation occurs bynatural convection. Active system use electricpumps, valves and controllers to circulate wateror other heat transfer fluids through thecollectors. E.g. drain-back, drain-down systemetc.

In open-loop systems, the water that is pumpedthrough the collectors is same hot water to beused. These active open-loop systems arecalled drain-down system.

Drain-down refers to draining the collector fluidout of the system; drain-back refers to drainingthe collector fluid back into the storage tank.Drain-back systems are called closed-loopactive system.

77. (a) Zenith angle: is the angle between the sun’sray and the normal to the horizontal plane.

Solar azimuth angle: is the angle on ahorizontal plane, between the line due southand the projection of the sun’s ray on thehorizontal plane.

Incident angle: is the angle between the sun’sray incident on the plane surface (collector)and normal to that surface.

78. (b)An efficient way to reduce thermal losses fromthe absorber plate of a solar heating panel isby using selective absorber coatings.

A selective surface is a surface that has a highabsorptance for short wave radiaton (less than2.5 m ) and a low emittance of long wave

radiation (more than 2.5 m ).

79. (c)

Solar furnace are ideal tools to study thechemical, optical, electrical and thermodynamicproperties of the materials at high temperatures.It is basically an optical system in which solarradiations are concentrated over a small area.

Some of the advantages of a solar furnace are:(i) heating without contamination, (ii) easycontrol of temperature, (iii) working is simple,(iv) high heat flux is obtainable, (v) continuousobservation possible and (vi) absence ofelectromagnetic field. In spite of manyadvantages of solar furnaces, these have notbecome popular in industries due to followingreasons:

(i) Its use is limited to sunny days and that toofor 4–5 hours only.

(ii) Its cost is high.

(iii) Very high temperatures are obtained only overa very small area.

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80. (a) The solar air collector has following advantageover liquid flat plate collector:

It is compact, simple in construction andrequires little maintenance.

Corrosion is completely eliminated.

Leakage of air from the duct is less severe.

Possibility of freezing of working fluid is alsoeliminated.

Major disadvantage of air collector are:

Large amount of fluid is to be handled due tolow density. As a result the electrical powerrequired to blow the air through the systemcan be significant if the pressure drop is notkept within prescribed limits.

Heat transfer between absorber plate and air ispoor.

There is less storage of thermal energy due tolow heat capacity.

81. (a) In a module, a number of cel ls areinterconnected, it is very important that thesecells should match as closely as possible.That means Voc, Isc, Vm and Im (or fill factor)for all cells must be exactly same. Anymismatch in the characteristics of these cellsleads to additional mismatch loss. Therefore,peak power of the combination is always lessthan the sum of individual peak power of thecells. Only under ideal case when all cells areexactly identical that the resultant peak powerwould be equal to arithmetic sum of that of itsconstituents.

82. (d)

Rotor radius, r =100

2 = 50 m

uo = 20 m/s and n = 4Tip speed ratio for optimum output,

o

4 4n 4

Tip speed ratio,

o =

o

ru

Rotor speed, = 20

50 = 1.25 rad/s

83. (c) In tidal energy plant, the turbines must operateat low head with large flow rates. For examplebulb and kaplan turbine.

84. (a) The main disadvantage of fuel cells are theirhigh initial cost and low service life.

85. (b) Flat plate solar collectors have low efficiency.

86. (b)

F2

F3

F1

60°

F sin60°2

F cos3

F cos60°2

F sin3

3000 + 1000 = 4000

5000cos

1732.05 + 5000sin

As per the given condition, [Fy = 0]

5000cos = 4000

cos = 0.8

= 36.869°

87. (d) The branch of science, which deals with thestudy of a body when the body is at rest, isknown as Statics while the branch of sciencewhich deals with the study of a body when thebody is in motion, is known as Dynamics.Dynamics is further divided into kinematics andkinetics. The study of a body in motion, whenthe forces which cause the motion are notconsidered, is called kinematics and if theforces are also considered for the body inmotion, that branch of science is calledkinetics.Engineering Mechanics

Statics(Body is at rest)

Dynamics(Body is in motion)

Kinematics(The forces which cause

motion are not considered)

Kinetics(Forces which cause

motion are considered)

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[ME], ESE-2020 PRELIMS TEST SERIES PAPER-II (TEST-01) (15)

88. (c)

v =

r [Vector product]

= ˆ ˆ(10 5t)k 1 i

= ˆ ˆ(10 5t) k i

= ˆ(10 5t) j

At t = 0, ˆv 10 j

i

k

j

Remember

v ra r Vector product

r F

89. (b)

It is given that constant angular velocity

P

V

R

also V = R [because of pure rolling]

Vnet is 2 V

V45°

2 VV

It is given that constant angular velocity

Angular acceleration ( ) = 0

Tangential acceleraton (at) = r = 0

Centrepetal acceleration = 2R

anet is 2R

2R

Angle between Vnet and anet = 45°

90. (b)y

x

W

FBD

W

fr

Wsin Wcos

N

As weight W resting on a rough inclined plane,

then xF = 0

yF = 0

using xF = 0

rW sin f = 0

Friction force, fr = W sin

91. (d)Mass moment of inertia about x-x' axis

MOI (x –x') = 2dm r

rdm

x-axis

x -axis'which depends on mass of the body and itsgeometry or shape, and the axis of rotation ofthe body

92. (c)From equation of motion,

s = 21ut at2

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300 = 214u a(4 )2

...(1)

800 = 218u a(8 )2

...(2)

So, (equations 2) – (equation 1) × 2 gives

800 – 300 × 2 = 2 21 a(8 4 2)2

a = 12.5 m/s2

93. (b)The composite section is divided into two simplefigures, a triangle and a rectangle.

Now, area of triangle, A1 = 1 3 42 = 6 m2

area of rectangle, A2 = 6 × 4 = 24 m2

Total area, A = 30 m2

The coordinates of centroids of these two simplefigures are:

x1 =16 33

= 7 m

y1 =4 m3

x2 = 3m

y2 = 2 m

x =1 1 2 2A x A xA

= 6 7 24 3

30 = 3.8

y =1 1 2 2A y A yA

=

46 24 23

30 = 1.87

94. (d)

Triangle law shows the force polygon closesfor a particle to keep it in equilibrium.

Equilibrium condition is Fnet = 0

If Fnet = 0 then acceleration is also zero.( Fnet = ma)

95. (a)

When a body is just about to move, then staticfriction acts, and at static friction condition friction force is maximum

Graph

m Fks

fr

FoF

At Fo body is just about to move, and we cansee friction (fr) is maximum at this point.

96. (c)

Range = 2u sin2

2g

For maximum range 45

So according to question.2u

2g = 2000 ...(given) ...(1)

Now,2

ou sin22g = 1414.2 ...(2)

Using (1) and (2),

osin2 = 0.707 = 12

o2 = 45°

o = 22.5°

97. (b)FBDCRATE

1000 Nf = = 0.25 × 1000 = 250N

r 1W

aN2

N1

CART

700N

aN2

W2N1

500NW1

N2

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[ME], ESE-2020 PRELIMS TEST SERIES PAPER-II (TEST-01) (17)

(or)

700N

N2

1700N

500N250N

255N

For finding acceleration of carta

500N255N

700N

1000N

500 – 255 = 1700 a

g

{using force equation}

a = 1.41 m/s2

98. (b)Initial velocity, u = 200 m/s

Final velocity, v = 0

(when bullet reached to a depth of 50 cm)

Distance travelled by bullet,

s = 50 cm = 0.5 m

From equation of motion,

v2 = u2 + 2as

0 = (200)2 + 2a × 0.5

a = – 40000 m/s2

99. (a)Weight of train, W = 1500 kN

Train speed, v = 10 m/s

Resistance due to friction,

F = 5 N per kN weight of train

= 5 × 1500 = 7500 N

Friction forceF

P

As engine is moving with uniform speed, theacceleration of engine will be zero.

Net force = ma = 0

P – F = 0

P = F = 7500 N

Power = Force × Velocity

= 7500 × 10

= 75000 W or 75 kW

100. (c)

2 m/s

0.5 kg 1 kg

m1 m2

From conservation of linear momentum,

m1u1 + m2u2 = m1v1 + m2v2

0.5 × 2 + 1 × 0 = v(0.5 + 1)

Here, v1 = v2 = v,

Because both block moves together as a singlebody after collision.

so, v =1

1.5 = 2 m / s3

Initial kinetic energy,

KEi = 2 21 1 2 2

1 1m u m u2 2

= 21 0.5 2 02 = 1 J

Final kinetic energy,

KE f = 21 2

1 (m m ) v2

=21 2(0.5 1)

2 3

=1 3 42 2 9 =

1 J3

Loss in energy due to the collision,

KE = KEi – KEf = 113

= 23

101. (b)Let, mass of the particle is m which is projectedat speed V.

V

60°

V cos60°im

V sin60° f

Highest point

V cos60°

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(18) SOM+TH+RSE+MD

Kinetic energy at the time of projection,

KE = 21mv2

At the highest point, there will be no verticalcomponent of velocity, so, the kinetic energyof the particle will be due to its horizontalcomponent of velocity.

so, kinetic energy at the highest point,

KE f = 21m(v cos60)2

= 21 1mv2 4

=21 1 1mv KE

4 2 4

i.e. KE f =KE4

102.(a)

Moment of inertia, = 22 mR5

R is changing, so will also change

If there is no external torque acting on thebody then, angular momentum does notchange. This is the principle of “conservationof angular momentum.”

extdL 0dt

i.e. L = constant

Angular momentum, L =

L is constant and is chaing, so will alsochange.

Kinetic energy, KE = 212 =

2 2 2L2 2

L constant and changing so KE will alsochange.

103. (c)104. (d)

Volumetric strain, v = x y z

= 0.003 + 0.005 + 0.009

= .017 units

Bulk modulus, K = v

P =

100 MPa0.017

= 5882.35 MPa

105. (c)From the SFD shown above the following loadingdiagram is obtained :

BA4m

18t

8m 4m CV =25tB

3t

DV =14tA

1.5 t/m

x

x

B.M. at a c/s x from B along BD

Mx = 3 (4 + x) + 1.5 (4 + x)

4 x 25 x2

Mx = 0.75 x2 – 16x + 24.To obtain point of contraflexurePut Mx = 00.75x2 – 16x + 24= 0

x = 1.623 m

Hence option C is correct.

106. (d)Point of contraflexure is the point where bendingmoment changes it sign. If bending moment iszero, it can not be says as point of contraflexure(Eg. at the end of cantilever beam).

107.(d)Strain energies stored in beam (i) & (ii) is:

L 2x

10

M dxU

2E

2L 2

10

1 wxU dx2E 2

=

2 5w L40 E ...(i)

U2 = L 2

x

0

Mdx

2 E

U2 =

2L 3

0

1 wx dx2E 6L

U2 =

2 5w L504 E ...(ii)

Divided eq. (i) from (ii),

1

2

UU =

2 5

2 5w L 504 E40 E w L =

635

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[ME], ESE-2020 PRELIMS TEST SERIES PAPER-II (TEST-01) (19)

108. (b)

l2

1

1

2

(1)

(2)

dx

x

l1dx

x

21

1. .4

= Deflection due to self weight of rod (1) +Deflection due to self weight of rod (2) +Deflection of rod (1) due to self weight of rod (2).

= 2 2

2 2 11 2

1

A L LL L2E 2E A E

= 1 1

5 58 10 1000 1000 8 10 1000 1000

2 2 10 2 2 10

1 2

2 5

8 10 70 1000 10004

100 2 104

= 5.96 mm

109. (d)

We know

dMdx = V

Under a transverse laod, slope of BMD

changes and for maximum B.M. dM 0dx

In distributed load region, S.F.D is of 1° ormore, so BMD also of 2nd degree or morehence non-linear shape.

In continuous beam, it is not neccessarythat at every support B.M. is hogging (–ve)value and nature of B.M. will depend uponsupport & loading condition.

110. (d)

P

Q R

S

l/3

l/2

P P

l/3 l/3

T U

Q R

l/3

P P

l/3 l/3Bending Moment at Q & R both zero.

111. (d)

Given, P = 18 kN = 18000 N

A0 = 2 22 9.6 mmd4 4

0 = 400 mm

1 = 401.5 mm

= 20

P 18000A 9.6

4

249 MPa

= i 0

0 0

401.5 400400

0.0037

112. (a)We know,

= Lateral strain d / d

longitudinal strain /

= 0.001 / 100.10 / 100

= 0.1

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(20) SOM+TH+RSE+MD

113. (b)

Bending stress due to bending moment,’M’

max =

4

M D / 2MyI D

64

D diameter of circular shaft4DI

64

332 M

D

Maximum shear stress due to torsion

max =

r4

T D / 2TJ D

32

= 316T

D

max

min=

3

332M / D 2M

T16T / D

max

min=

max

max

2 100 450 T

114. (b)

Eccentricity at A – A will be 'd'2

maximum tensile

stress,

max =P MyA I

=

3P d / 2 · d / 2P

bd b d12

=P 3Pbd bd

max4Pbd

115. (c)

Cross section Shear stress distribution

A.

B.

C.

D.

116. (a)We know,

hoopd pd 1

d 4tE

At 5 MPa

2

55 1000

d 1 0.254 10 2 10

d 0.46875 mm

At 1 MPa

2

51 1000

d 1 0.254 10 2 10

= 0.09375 mm

Net change in diameter due to leakage = 0.46875– 0.09375 = 0.375 mm

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[ME], ESE-2020 PRELIMS TEST SERIES PAPER-II (TEST-01) (21)

117. (d)

By using the property,

1 2 x y

250 = 30 + (–20)

2 = –50 + 10

2 40 MPa

118. (d)

1 2, =x y x y 2

xy2 2

= 2

240 0 40 0 802 2

= 2 220 20 80

= 20 ± 82.5

1 = 102.5 MN/m2

2 = 20 – 82.5

= –62.5 MN/m2

abs. max =1 2 2 1Max , ,

2 2 2

=

102.5 62.5 62.5 102.5Max , ,2 2 2

=165Max , 31.25, 51.25

2

= Max 82.5, 31.25, 51.25

2max, abs 82.5 MN / m

119. (b)

Hoop & longitudinal stresses are equal andvaries from maximum at the inner face tominimum at outer face.

Statement 2 is correct

120. (a)

CB

A

CB

M MEI

B M

EI

A

l l

l

l

Horizontal deflection of C = horizontal deflectionof B

=2M

2EIl

121. (c)

Normal stress = External Load

Cross sec tional Area

For normal stress to be maximum, crosssectional area should be minimum.

1

Mohr Circle for uniaxial loading condition.

From Mohr Circle, shear stress is zero on aplane having maximum normal stress.

122. (b)Expansion prevented = Free expansion – Gapbetween rails

= TL

s

TLf EL

= 6 3

53

11.7 10 30 10 10 2 2 1010 10

= 30.2 MPa

123. (c)

We know that strain energy U = 1 T.2

and

T G TLJ L GJ

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(22) SOM+TH+RSE+MD

21 T LU

2 GJ

2

44

1000 1000 10001U2 8 10 100

32

2000 Nmm

124. (c)125. (b) In case of rectangular section,

max avg32

In case of circular section,

max avg43

% excess shear stress in rectangular sectioncompare with circular section

= avg avg

avg

3 4 1002 3

43

=

3 4 9 82 3 6100 100

4 43 3

=

116 100 100 12.5%

4 83

126. (b)Only 1 and 2 are correct.

Theory of Pure Torsion is valid only for circularcross section. For non circular cross-section,Torsion formula does not hold good.

127. (d)

Flexural stiffness = Flexural Rigidity

LengthThus flexural stiffness is directly proportional toflexural rigidity & inversely proportional to length.

128. (b)Max shear stress theory is also known as Tresca,Guest, Columb theory.

It is not suitable for hydrostatic loading because

max = min 'P '

Thus max = max min 02

Max. strain energy theory is called as Beltramihaigh theory

129. (c)

Meq = 2 21

M M T2

= 2 21 kNm30 30 402

= 40 kNm

130. (b)

=

M P / 23EI 3EI

l ll

= 3P.

2 12EIl l

131. (b)

Shear area, As = dt

= 21 10 = 660 mm2

Average shear stress in plate,

avg =s

PA =

3100 10660

= 51.51 MPa132. (c)

P R30°30°

60°

220 kN

Q

P P2P cos60°

Force balance in Y-direction,

2P cos60° = 220P = 220 kN

Stress in each rod = 2

P

d4

= 24 220

d

= 2

2280 kN / mmd

Equating this stress to permissible stress,

2280 1000d

= 140

d = 44.72 mm

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[ME], ESE-2020 PRELIMS TEST SERIES PAPER-II (TEST-01) (23)

133. (a)Weight of log = 0.5 × 0.5 × 20 × 6 = 30 kNWeight of water displaced by log when justimmersed = 0.5 × 0.5 × 20 × 10 = 50 kN

Total weight required on the log = 50 – 30= 20 kN

O

8 kN P

20 kN x5m

20 mForce balance in vertical direction,

8 + P = 20P = 12 kN

The upthrust acts through O; the beam will behorizontal only if the C.G. of the two weightsis also at O.

8 × 5 = 12 × (10 – x)x = 6.67 m

134. (d) In the theory of simple bending discussedabove, the following assumptions have beenmade:

(1) Normal sections of the beam, which were planebefore bending, remain plane after bending.

(2) The material is homogeneous and isotropic, sothat it has the same elastic properties in alldirections.

(3) The elastic limit is not exceeded so thatHooke’s law is applicable.

(4) Each layer is free to expand or contract in alldirections independently of the layers aboveand below it.

(5) The value of E is the same in tension andcompression.

135. (b)A B

6cm

8cm

4m

Maximum bending moment,

M =w 2 44 4

= 2 kN-m

Section modulus,

Z =2bd

6

=2 2 28 10 (6 10 )

6

= 48 × 10–6 m3

Maximum bending stress,

=MZ

= 3

62 10

48 10

= 41.67 MPa

136. (d)

Mean torque, T =60 P2 N

=360 44 10

2 100

= 4200 Nm or 42×104N-cm

Maximum torque,

Tmax = 1.5 × 42 × 104

= 63 × 104 N-cm

Maximum shear stress,

max = 316T

d = 4

316 63 10

10

= 3208.5 N/cm2

The greatest angle of twist on 1 m length ismax

r=

G

= 63208.5 100

10 5

( = 1 m = 100 cm)

= 0.064 rad.

137. (b)

Pressure, P = gh

= 1000 × 9.81 × 100

= 981 kPa

Hoop stress, h =Pd2t

t =h

Pd2 = 3

981 1.52 20 10

= 0.0367 m or 36.7 mm

138. (b)If process is reversible and no kinetic of potentialenergy changes, then the reversible flow work isgiven by,

W = vdp , Here v is specific volume.

The liquid has a small value of v compared tolarger value of a gas.

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139. (b)(A) For steady flow, mass flow rate is constant.

m V constant

Incompressible flow, m V constant

V constant but in compressible flow,, m V constant V constant as

vary.(R) For steady flow devices, the control volume

is constant, thus there is no boundary workinvolved.

During a steady-flow process, no intensiveor extensive properties within the controlvolume change with time. Thus, the volumeV, the mass m, and the total energy contentE of the control volume remain constant. Asa result, the boundary work is zero for steadyflow systems (since VCV = constant).

140. (c) Entropy is a property and property depends

only on states not on process. dQT

on

irreversible path is not a property.141. (b) Dalton’s law of additive pressures: The pressure

of a gas mixture is equal to the sum of thepressures each gas would exert if it existedalone at the mixture temperaure and volume.

Amagat’s law of additive volumes: The volumeof a gas mixture is equal to the sum of thevolumes each gas would occupy if it existedalone at the mixture temperature and pressure.

The pressure fraction,

iPP

= i i(nRT / V) nn(nRT / V)

= Mole fraction

The volume fraction

iVV

= i i(n RT / P) nn(n RT / P)

= Mole fraction

Therefore,

iPP

= i iV nV n

142. (a)The allowable load Pall per mm length of theparallel fillet weld is given by,

Pall = 0.707 h max ...(1)

The allowable load Pall per mm length of thetransverse fillet weld is given by,

Pall = 0.8284 h max ...(2)

Here h is leg length of weld

max is maximum permissible shear stress

It is observed from Eqs, that the allowable loadfor a transverse fillet weld is more than that of aparallel fillet weld. Or,

all max.

all. max.

P for transverse weld 0.8284h 1.17P for parallel weld 0.707

The strength of transverse fillet weld is 1.17 timesof the stength of parallel fillet weld. Many timestransverse fillet weld is designed by using thesame equations of parallel fillet weld. Such designis on the safer side with the additional advantageof simple calcultaions.

143. (a)Connecting rod bolts: The bolts are subjected toinitial preload plus fluctuating external load. Themean and alternating components of load is givenby,

(Pb)m = i1P P

22

(Pb)a =1 P

22

Here Pb represents resultant load on bolt.

Pi is initial pre-load

P is external load

Connecting rod bolts are tightened with initialtension grater than external load due to thefollowing reasons:

(i) When the connecting rod bolts are tigthenedup with initial pre-load greater than externalload, the term (Pi) is much more than theterm [(1/22)P]. The fluctuating term [(1/22)P]is small and neglected. The total load onthe bolt will be almost static and the boltcan be designed on the basis of staticfailure.

(ii) If the initial pre-load is not high enough, theterm [(1/22)P] is comparalble with the term(Pi). The resultant load on the bolt will beaffected by the external load, which isfluctuating and it is required to desing thebolt on the basis of endurance limit to

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[ME], ESE-2020 PRELIMS TEST SERIES PAPER-II (TEST-01) (25)

design the bolt on the basis of endurancelimit to avoid fatigue failure.

It is desirable to design the bolts on thebasis of static strength rather than on thebasis of endurance limit.

Therefore, in automotive applications, theconnecting rod bolts are tightened with veryhigh pre-load with the stress approachingthe yield point.

144. (a) Animal feed on plants and plants growthrough photosynthesis process using solarenergy. A small portion of the solar radiationis captured and stored in the plant duringphotosynthesis process. These plants arethe source of biomass.

145. (c) Power extracted by the turbine,

PT =

2 31 o

14a(1 a) A2

Here uo is upstream wind velocity.

a is interference factor.

u1 is wind velocity at turbine.

Interference factor,

a =o 1

o

u uu

When the wind speed at the turbine is reducedto zero (i.e. u1 = 0), ‘a’ becomes unity and nopower is extracted. This state is known as(complete) stall state of blades.

146. (b)

147. (a) As conversion of energy of fuel to electricalenergy occurs directly without intermediatethermal stage, the efficiency of conversion isbetter and not limited by carnot efficiency ofthermal stage.

148. (b) Both statements are correct but secondstatement is not the correct explanation of firststatement.

Principle of conservation of momentum holdtrue for all kinds of collisions (either elastic orinelastic collision) but conservation of kineticenergy does not hold for inelastic collisionsdue loss of some kinetic energy during collision.

149. (c) To use moment area method continuity of slopeof elastic curve must be maintained.

150. (a) Due to Torsion, maximum normal stress is at45° to the axis of specimen and cast iron isbrittle in nature and is weak in normal stress.