1. (d)

2. (b)

3. (c)4. (c)

5. (c)

6. (d)

7. (b)

8. (c)

9. (a)

10. (a)

11. (d)

12. (a)

13. (c)

14. (c)

15. (a)

16. (b)

17. (a)

18. (b)

19. (c)

20. (d)

21. (a)

22. (c)

23. (b)

24. (b)

25. (c)

26. (d)

27. (c)

28. (b)

29. (d)

30. (a)

31. (c)

32. (d)

33. (a)

34. (c)

35. (d)

36. (c)

37. (b)

38. (c)

39. (d)

40. (a)

41. (c)

42. (c)

43. (b)

44. (d)45. (c)

46. (d)

47. (d)

48. (c)

49. (d)

50. (d)

51. (d)

52. (c)

53. (d)

54. (a)

55. (c)

56. (b)

57. (a)

58. (d)

59. (d)

60. (d)

ESE-2018 PRELIMS TEST SERIESDate: 30th December, 2017

ANSWERS

61. (a)

62. (a)

63. (b)

64: (c)

65. (c)

66. (c)

67. (d)

68. (a)

69. (a)

70. (b)

71. (b)

72. (b)

73. (b)

74. (c)

75. (d)76. (b)77. (c)

78. (b)

79. (b)

80. (c)

81. (c)

82. (b)

83. (c)

84. (d)

85. (c)

86. (b)

87. (c)

88. (c)

89. (d)

90. (a)

91. (d)

92. (b)

93. (c)

94. (d)

95. (d)

96. (a)

97. (b)

98. (a)

99. (d)

100. (c)

101. (c)

102. (b)

103. (c)

104. (c)

105. (b)

106. (c)

107. (d)

108. (d)

109. (d)

110. (a)

111. (b)

112. (d)

113. (d)

114. (d)

115. (c)

116. (b)

117. (a)

118. (d)

119. (c)

120. (d)

121. (d)

122. (d)

123. (c)

124. (d)

125. (a)

126. (d)

127. (d)

128. (c)

129. (c)

130. (b)

131. (d)

132. (d)

133. (b)

134. (c)

135. (c)

136. (b)

137. (c)

138. (a)

139. (b)

140. (b)

141. (b)

142. (a)

143. (a)

144. (a)

145. (b)

146. (a)

147. (b)

148. (b)

149. (b)

150. (c)

(2) (Test-15)-30th Dec. 2017

Sol–1: (d)

=

3K 2G6K 2G

K = 3G

=

9G 2G 7G 7 0.35

18G 2G 20G 20

Sol–2: (b)

Sectional area of cylinder = dt= 100 4= 1256.637 mm2

Circumferential stress

= 1pdf2t

=

2P 100 12.5PN mm

2 4

Longitudinal stress

= 2Pd Pf4t A

=

P 100 500004 4 1256.63

= 2N mm6.25P 39.79

f1 and f2 are the principal stress.

We know qmax = 45N/mm2

1 2f f2 = 45

12.5P 6.25P 39.79

2 = 45

6.25P + 39.79 = 90 P = 8.034 N/mm2

Sol–3: (c)

22x y x y

xy1 102 2

10 =

2210 5 10 5

2 2

2 2510 7.52

2 212.5 7.5

2156.25 56.25

10MPa

Sol–4: (c)

30°60 N/mn2

40 N/mn2

120 N/mn2

x =

x y x y cos22 2 + xy sin2

=

60 120 60 120 cos602 2 +

40 × sin 60

= 30 45 20 3

= 19.64 MPa

x y =

x yxy

( )sin2 cos2

2

=

( 60 120) 32 2

+ 40 × 12

= 340 20

2

= 97.942 MPaSol–5: (c)

Sol–6: (d)

Maximum bending momentM = F × 4 = 24 KN-m

Since the beam is designed mosteconomical, depth of neutral axis

max = MyI

68 =

y100 y

y = 42.85 mm

I = 624 10 42.856

= 17 × 10–5 m4

(3) (Test-15)-30th Dec 2017

Sol–7: (b)

1. D Dt to10 15

3. Critical stress is hoop stress.

4. hoop stress = pd2t

Sol–8: (c)

= E and PA

PA = E

= PAE

Poisson’s ratio

= Stress in lateral direction

Strain in longitudinal direction

elongitudinal = laterale

elateral =

PAE =

3

2 9

(0.42 120 10 )

0.08 3 104

e = 3.3422 × 10–3

elateral = 33.3422 10

D

= 33.33422 10 8 0.026cm

Sol–9: (a)

5 KN/mA

25 25

Shear force V = 25 – 2.5 × 5 = 12.5 KN

NA50

100

Shear Stress

=

3

3VAy 12.5 10 150 100 100Ib 300150 150

12

= 0.3703 mpa

Sol–10: (a)

max = 0.01 radian

D = 300 mmL = 7.5 m

max =

42 N/mm2

TJ =

max G

R L

T = G . JL

=

7 4

310 0.01 300

7.5 10 32= 10602.8 kN-m

T =

2 4maxJ 42N mm 300

300R 322

= 70.875 × 106

T = 70.875 kNm

Sol–11: (d)

P = 2

2eff

EIl

P d4 4I d

PP =

4

4d

0.9 d

PP = 4

10.90

P = 0.94 PP = 0.6561 P

Sol–12: (a)

Dependent variable should not beselected as repeating variable.

Sol–13: (c)

For sharp crested rectangular weir

Q = 3/22Cd 2gLH3

dQ =

1/23 2 Cd 2gLH dH2 3

(4) (Test-15)-30th Dec. 2017

dQQ =

3 dH2 H

dQ 100Q

=

1 2

1

H H3 1002 H

=

3 (0.26 0.25) 1002 0.26 = 6%

Sol–14: (c)

1.6 m

Initial water level

Z = 2 2r1.62g

1.6 2 10 = 22 12 10

16 = 2 2w 10 = 40

2 N60 = 40

N =

60 40 1200

2

Sol–15: (a)Pgas = atmP gh

= 3101.3 10 13534 10 0.1= 87766 Pa

Sol–16: (b)

o

vV =

1/7y

=Moment thickness

Boundary layer thickness

=

mm 1 m 2

m = 7 =

7 7

8 9 72

Sol–17: (a)

In turbulent flow through rough pipes.

maxUU = 1 1.33 f

= 1 1.33 0.04= 1.27

Sol–18: (b)

1

2

PP =

DK1tE

=

3

5600 4.2 101

3 2.1 10= 1 4 5

Sol–19: (c)

Kinematic viscosity of gases decreasesas density is proportional to pressure.

Sol–20: (d)

Sol–21: (a)

Qa = 3ALN60

35000 1060

= 23 0.40 N0.24

60N = 132.6 rpm

Sol–22: (c)Sol–23: (b)

a

cross-sectional area A a

WF ag

Wa h

F = P A1 1

h

yF 0

1WP A a Wg

W Ah

1

AhP A a Ahg

1aP h 1g

=

3510 1 1 1010

(5) (Test-15)-30th Dec 2017

= 15 kPaSol–24: (b)

Power required for pumping

P = fQh =

2

5f QQ

12.1 dl

51 1 2

2 2 1

P f dP f d

51

2

P 0.025 200P 0.02 100 = 40

Sol–25: (c)

zCirculation 2w vorticity

AreaVelocity potential function exists only forirrotational flow.

Sol–26: (d)

Chip equivalent,

(q) =

ss

s

d R(1 sin C ) fR Ccos C 2 2

fd

Where d, f, R, Cs are depth of cut, feed,nose radius, side cutting edge anglerespectively

Sol–27: (c)

As percent of Co increase from 5%compressive strength, hardness wearresistance tends to decrease.

Sol–28: (b)

Sol–29: (d)

Hobbing process cannot be used formachining

Internal gears

Bevel gears

Gears having adjacent shoulder

Sol–30: (a)

Grinding wheel is smaller than regulatingwheel.

Sol–31: (c)

CO2 produces lowest arc temperature and

shallowest penetration

Sol–32: (d)

Sol–33: (a)

Sol–34: (c)

Sol–35: (d)

Draw spike is a tool to remove pattern frommould.

Sol–36: (c)

Sol–37: (b)

Sol–38: (c)

Sol–39: (d)

Sol–40: (a)

Sol–41: (c)

Sol–42: (c)

Sol–43: (b)

Pneumatic actuator have slower responsebecause of compressibility of air

Sol–44: (d)Range of total duration = x 3

= 25 15 3 (3) 5 75 20

55 < Range < 95

Sol–45: (c)

CB

C

ED

New activity can not start with onlydummy activity as its predecessor.

Sol–46: (d)Data given,

S = 6,00,000P = 60,000

pv ratio = 30% = 0.30

Margin of safety = Profit

pv ratio

(6) (Test-15)-30th Dec. 2017

S – SBep =P

pv ratio

SBep = PS

pv ratio

= 6,00,000 – 60,0000.30

= 6,00,000 – 200,000= 4,00,000

Sol–47: (d)Normal time =

70 600.70 0.80

100 100= 0.49 + 0.48= 0.97

Standard time = NT + Allowances= NT + 15% of NT

= 0.97 + 0.97 15100

= 1.115Sol–48: (c)

Data given

LSL = 16.2 kg, USL = 18 kg, 2 = 0.09 kg2

Process capability index Cpk =

USL LSL6

= 18 16.2

6 0.09 = 1

Sol–49: (d)

Value = Function or performance or qualityCost

= 4000 40006000

= 80006000 = 1.33

Sol–50: (d)

Sol–51: (d)

Sol–52: (c)

Sol–53: (d)

Sol–54: (a)Sol–55: (c)

I. Pelton wheel is suited for high head andlow discharge.

II. Since water is incompressible fluid andkinetic energy can be converted into pres-sure energy only in diverging section. So,the straight cylindrical type draft tube isnot serve the purpose of improving hy-draulic efficiency of turbine.

III. The main function of draft tube in reac-tion turbine, is to convert the kinetic en-ergy head in to pressure energy. The de-sign condition of the draft tube is onlyconcerned about the cross section of tubemust increases in flow direction.

Sol–56: (b)The same pump is connected to two different

motors. So model and prototype are same.The power coefficient.

13 51 1

PN D

= 23 52 2

PN D

3 5

21630 D

=

23 5

P3260 D

P2 = 2 ×

332601630

P2 = 2 × 23 = 16 kw.

Sol–57: (a)

I. Employing superheated steam inturbine, the quality of steam at the turbineexit will be high i.e. low water content. Thewater content in steam is responsible forblade erosion. Hence using superheatedsteam decreases the chance of blade erosion.

II. In impulse turbine, the enthalpy droptakes place in only in nozzle. In stator bladeor rotor blades the flow only changes itsdirection and no change in enthalpy andpressure.

III. Since the speed of steam turbine is veryhigh to brings this speed in practicalapplication range compounding is done.Here either the pressure of steam is reducedin steps (pressure compounding) or highvelocity is used in steps (velocity

(7) (Test-15)-30th Dec 2017

compounding)

Sol–58: (d)

The forced draught or induced draught fanis not allowed in locomotive boilers becausethey consume power. Draught is producedin locomotive boilers by creating lowpressure in exit zone of flue gases by flowinghigh pressure steam at high velocity.

Sol–59: (d)

I. Because of intercooling, the heatgenerated is taken away and does not heatthe compressed gas so it reheat factor isless than one. So stage efficiency is morethan adiabatic efficiency.

II. Vaneless space is provided between tipof impeller and diffuser inlet to ensure tosmooth or shockless entry of air in diffuser.

Diffuser

Vanelessspace

III. Increased clearance volume result inreduced efficiency i.e, reduced volume flowrate. Because higher clearance ratio, i.e,higher V3 and V4 volumetric flow rate isproportional to (V1 – V4). So at higherclearance V4 increases and(V1–V4) reduces.

P3 2

14

V

1 4v

1 3

V VhV V

Sol–60: (d)

I. Reheat leads to improvement in specificwork output

II. At high velocity (supersonic) the

efficiency of rocket engine is more thanturbojet

III. The thrust is force and force is rate ofchange of momentum. So rate of change ofmomentum can be due to change in velocityor change in mass or both. The expression

of thrust for rocket is F = p jm v where pmpropeller consumption rate and vj is the jetvelocity from rocket nozzle exit

Sol–61: (a)

Sol–62: (a)

Polytropic efficiency,

P =

( 1) nn 1

=

1.4 1 1.5

1.4 1.5 1

P = 0.8571 = 85.71%

Sol–63: (b)

Propulsive efficiency of rocket engine,

P =

2 2j a

2Vj VaV V

p =

2 2

2 150 120 360 40 97.55%369 41(150 120 )

Sol–64: (c)

Sol–65: (c)

v =

2

1

P1 C CP

Hence, the volumetric efficiency decreasesas the pressure ratio is increased

n 1n2

1 1 41

PnW P (V V ) 1n 1 P

Thus if pressure ratio is increased, shaftpower also increases

Sol–66: (c)

(8) (Test-15)-30th Dec. 2017

Cp = i e

i

P PP

0.55 =i

33P

Pi = 33 60 kW0.55

Sol–67: (d)

Sol–68: (a)Sequential order is given as:Reduction of silicon dioxide to silicaresult in production of MG -Si Purification of MG-Si intomulticrystalline semiconductor gradesilicon conversion of multicrystallinesilicon into single crystal form in formof rod rod is converted into wafersby sawing process doping of waferswith boron.

Sol–69: (a)

Resolver sensors are used for conversion ofangular position of a shaft into cartesianco-ordinates.

Sol–70: (b)

Sol–71: (b)

Range of pressure = 30 – 0 = 30 bar

% error = calibration pressure

range of pressure

= 0.12510030

=0.416%

Sol–72: (b)

Sol–73: (b)Sol–74: (c)

I. Entropy transfer is associated with heattransfer only not with the work transfer.

II. Entropy change of the universe in allreversible processes is zero

III.Increase in entropy always degradesenergy the degradation of energy isgiven

= Tosgenwhere To= surround temperature andSgen = entropy generation.

Sol–75: (d)I. Entropy has tendency to maximise in

spontaneous process. In equilibriumstate, the spontanity of process is zeroand entropy value is maximum.

II. The throtiling process is very fast andthere is no moving part. So the processis adiabatic and no work transfer. Butentropy increasing in throttlingexpansion

III.The entropy of an isolated systemalways increases.

Sol–76: (b)Sol–77: (c)

I. Internal energy of ideal gas is functionof temperature alone.

II. The specific heat depends upon numberof atoms

for diatomic gas = 1.4

Molal specific heat

cp =

R1

For monoatomic gas = 1.67

So, it dpends upon number of atoms inmolecule.

III.The pressure fraction

ipp =

i Tn R vnR T v

= inn = Mole fraction

Volume fraction

ivv =

in R T pnR T p

= inn = mole fraction

Sol–78: (b)

(9) (Test-15)-30th Dec 2017

G = H–TSSol–79: (b)

Clausius inequality is dQ 0,T

thenthe cycle is irreversible and possible.

Sol–80: (c)Conditions for the fins to be effectiveare:

1. Thermal conductivity should be large2. Heat transfer coefficient should be small3. Thickness of the fin should be small

Sol–81: (c)

Sol–82: (b)The peak spectral emissive power shifttowards a smaller wavelength at ahigher temperature.

Sol–83: (c)

45°C

t =30°C1

tC2

Heat last by the refrigerant in condenser= Heat transferred to air

4.2 × 1000 = UA m

4.2 × 1000 = UA × 12

UA = 420012 = 350 W/K

Sol–84: (d)

q =

thickness

kAdT TkAdx L

q = sameL = SameA = Same (assuming)

hence, T will be more for the materialwhich love lowest thermal conductivityMaterial Thermal conductivity (K)Silver 410

Aluminium 225Cast iron 55 – 65Glass (window) 0.75

Sol–85: (c)

A cotter joint is used to connect two co-axial rods, which are subjected to eitheraxial tensile force or axial compressive forceand it is not used for connecting shafts thatrotate and transmit torque.

Sol–86: (b)

Maximum value of fatigue stressconcentration factor, Kf = Kt when materialis fully sensitive to notches (q = 1).

Sol–87: (c)

The teeth of driving and driven gears aresubjected to repeated stress while teeth ofidler gear are subjected to reversed stress.

Sol–88: (c)Sol–89: (d)

The factor ZN/p is termed as bearingcharacteristic number and is adimensionless number. The factor ZN/p helps to predict the performance ofa bearing.

Sol–90: (a)Sol–91: (d)Sol–92: (b)

In closed loop control system a standardvalue of a parameter in maintained bycomparing its value with input orreference value. So1. Flywheel is not a closed loop control

system but is open loop controlsystem because it releases andabsorbs energy without anycomparison.

2. Governor is closed loop controlsystem which controls the speed bycontroling fuel supply based on load.

3. Thermostat of refrigerator controlsthe flow of refrigerant based ontemperature of evaporator. The reliefvalue opens and closes as pressureinside boiler drum changes.

Sol–93: (c)Let energy at point

A = E

(10) (Test-15)-30th Dec. 2017

then energy at pointB = E + 50

then energy at pointC = E + 50 – 20 = E + 30then energy at pointD = E + 30 + 5 = E + 35then energy at pointE = E + 35 –35 = E

Maximum fluctuation of energy

E = max energy–min energy= (E + 50) – E = 50 J

Sol–94: (d)By using modified involute orcomposite systemThe flank portion of pinion tooth lyinginside the base circle and the matingportion of gear tooth face may be madecycloidal in place of involute shape.Remaining portion of the pinion toothmay be of involute profile.By modifying addendum of GeartoothWhen conditions favouring interferenceexist, tip of gear tooth comes in contactwith non-involute portion of flankbetween dedendum circle and basecircle. To avoid interference, an obviousoption can be to chop-off interferingportion of the face of the gear tooth.Increasing centre distanceThe centre distance between twoinvolute gears may be increased withinlimits, without disturbing correctnessof gearing, and this step will preventthe tip of gear tooth from mating withnon-involute flank portion of the pinion.In case of two gears with involute teeth,the centre distance should be increasedwithin limits, means centre distanceshould be increased in such a way thatthe points of contact lie within themaximum allowed path of contact.Thus, the law of gearing is not isolated.Also by increasing the centre distance,the net effect is that the pressure angleis increased.

Sol–95: (d)For two masses to be dynamicallyequivalent

m1 + m2 = m1 1m = 2 2m

2 21 1 2 2m m = 2

amk

Sol–96: (a)II. flame velocity in SI engine combustion

is maximum when mixture strength is10% becuase in this condition, thetemperature is maximum andcombustion is very good.

Sol–97: (b)Sol–98: (a)

Sol–99: (d)

Sol–100: (c)

Sol–101: (c)

Sol–102: (b)

T = –10°C3

T = 35°C1 T = 15°C2

BPF =

2 3

1 3

T TT T

=

15 ( 10)35 ( 10)

= 0.55

Sol–103: (c)

Sol–104: (c)

Sol–105: (b)

Slope of the controlling force diagramshould be more than that of speed linefor stability.

Sol–106: (c)

Sol–107: (d)

Sol–108: (d)

Sol–109: (d)

(11) (Test-15)-30th Dec 2017

In case of a horizontal vibrating system,the gravity has no effect on its motionand thus, the system vibrates about theoriginal equilibrium position.

Sol–110: (a)

C = pI= 2500 × 0.22 × 335 × 0.1= 3350 Nm

Sol–111: (b)Sum of longest + shortest link 12+3 =15 mmSum of other two links = 10 + 8 = 18mmSum of longest + shortest link < sum ofother two links, and link adjacent toshortest is fixed, hence chain will act ascrank-rocker mechanism.

Sol–112: (d)For a cycloidal follower motion,

amax =

2

22h

at = 4

Sol–113: (d)For more power transmission and lesserpressure on the bearings, the pressureangle must be kept small.

Sol–114. (d)Pearlite is a phase mixture of ferriteand cementite and it is said to bemechanical mixture as ferrite and cementiteare separated by a boundary while retainingthe properties of the constitutent elements.Similarly ledeburite is a phase mixture ofausenite and cementite and it is also amechanical mixture.

Sol–115. (c)

1. Ceramic are mostly carbides, nitrides,oxides of metal or non metal.

2. Partially stabilized zirconia, (PSZ) is aceramic consisting of Zr, and oxides of Ca,Y, Mg etc.

3. The ultimate tensile strength (UTS) ofceramic increases with decrease in grainsize.

4. The hexagonal crystal structure of silica

exhibits piezoelectric effect.Sol–116: (b)

I. In reaction turbine, the regulation of flowwithout loss is not possible because regu-latory equipments disturb the flow por-tion i.e. loss.

Sol–117: (a)Sol–118: (d)

I. Priming is a process of filling liquid inimpeller casing and suction pipe to re-move air bubble. If any pocket of air isleft in the pump casing, pump will not beable to lift liquid because of cavitation orbreak in flow.

II. In centrifugal pump, the best perfor-mance and flexibility is with the back-ward blades. Becuae absolute velocity ofwater is less so the friction losses in back-ward curved vanes be less Eddy and tur-bulent lossess are also less in back curvedvaries. Hence centrifugal pumps givesmaximum efficiency when the blades arebent backward.

III. Overloading in centrifugal pump meanshigher discharge and higher head whenspeed of pump increases. This results intrippling of motor due to excess powerdraw. In forward curved vanes, head anddischarge both increases simulteneously,so to avoid this condition backward var-ies are prefered over other because asdischarge increases the head decreasesi.e no overloading.

Sol–119: (c)

x = 28t cos t

y = – 34t sin t

u = 2dx 16t.cos 8t . sint tdt

v = 2 3dy 12t sin 4t . cost tdt

Velocity vactor at t = 0.5 secV = ui + Vj

= 2 i16 0.5cos 8 0.5 8 sin8 0.5 8 0.5

(12) (Test-15)-30th Dec. 2017

2 3 j12 0.5 sin 4 0.5 8 cos8 0.5 8 0.5

as sin n = 0

= 8cos4 i 4 cos j4

cos n = 1 if n = even

= 8i j4

Sol–120: (d)Drawing F.B.D.

5kgF

N =5g11 1N

10kg

2 2N N =(10+5)g2

1 1N

For motion to be start

F 2 2N

F 2 15 g

For both block to move together

F 1 1N

F 1 5g

For both blocks to move together

F 1 1N

F 1 5g

Hence

2 15 g 1F 5 g

or 2 15 g 1 5 g

or

12

3

So minimum value of

12

= 3

Hence ‘d’ is correct.

Sol–121: (d)Sectional dimension is chosen such thatthe ratio of distance from NA to extremefibre in tension and compression isexactly same as the ratio of allowablestresses in tension and compression.Under this provision section will be mosteconomically utilized.

Sol–122: (d)Unsymmetrical section will always twistunder non uniform bending except whenload passes through the shear centre.

Sol–123: (c)In I-section more area is located fartheraway from neutral axis. Generally morethan 80% of BM is resisted by theflanges only.

Sol–124: (d)

Sol–125: (a)

Sol–126: (d)

In langrangian method, a coordinationsystem is attached to a single fluidparticle to describe its velocity andacceleration.

Sol–127: (d)

Apparent viscosity decrease with dudy for

pseudo plastic.

Sol–128: (c)

In swiss type automatic screw machine 1or 2 tool may act simultaneously.

Sol–129: (c)

Sol–130: (b)

Sol–131: (d)

Hot working depends upon recrystallisationtemperature not on red hot condtion. Eg:Molybdenum in red hot condition is coldworked because its recrystallisationtemperature is high.

Sol–132: (d)

Interferometers are optical instrumentsused for measuring flatness anddetermining the length of slip gauges bydirect reference to the wavelength of light.

(13) (Test-15)-30th Dec 2017

Sol–133: (b)

Rotors having low solidity use lift force andturn at higher speed.

Sol–134: (c)

Sol–135: (c)Water has higher heat capacity than soiland rock so the ocean takes much longerto heat and to coal than land. That’swhy coastal are will have moderatetemperatures than inland area.

Sol–136: (b)Velocity of flame propagation ismaximum and ignition lag is minimumwhen the mixture is 10% richer thanstoichiometric. These two arecontradictory, but the ignition lag effectvery much predominates, henceknocking tendency is found to bemaximum.

Sol–137: (c)

Sol–138: (a)

For reversible heat engine runningbetween source and sink temperature Tand T0 respectively

max = 0 maxT d W1T d Q

Exergy Wmax = 0T1 dQT

Sol–139: (b)

Maximum inversiontemperature

Heating zone

T

Coolingzone, j 0

Inversionline

P

h=c j 0

Sol–140: (b)

Sol–141: (b)

Sol–142: (a)

Sol–143: (a)

Back work ratio, BWR = Pump workTurbine work

Sol–144: (a)

Sol–145: (b)

Sol–146: (a)

Sol–147: (b)

Sol–148: (b)Sol–149: (b)

h = 2895N

The sensitiveness of watt governor ispoor at high speeds, hence it is used atlow speeds.

Sol–150: (c)Cam and follower is a higher pairbecause there is point or line contactbetween the cam and the follower.Surface contact takes place between twolinks of a lower pair.