mechanical engineering...aux = ? overall = g m boiler ism aux aux = 0.33 0.97 0.95 0.92 0.42 92.68%...

MECHANICAL ENGINEERING

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Detailed SolutionSET - A

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1. In various solar energy storage systems,pumped hydro-electric storage system fallsunder which one of the following categories?

(a) Thermal energy storage

(b) Electrical energy storage

(c) Mechanical energy storage

(d) Electromagnetic energy storage

Ans. (c)Sol. Mechanical energy storage

Energy can be stored in various forms andthe storage methods are classified on thebasis of the form in which it is stored.1. Mechanical storage

(a) Pumped storage(b) Compressed air storage(c) Flywheel storage

2. Chemical energy storage(a) Batteries storage(b) Hydrogen storage(c) Reversible chemical reaction storage

3. Electromagnetic energy storage4. Electrostatic energy storage5. Thermal energy storage6. Biological storage

2. What is the standard value of solar constantadopted by World Radiation Centre?

(a) 1192 W/m2 (b) 1084 W/m2

(c) 1927 W/m2 (d) 1367 W/m2

Ans. (d)Sol. Solar constant value adopted by world

Radiation centre is 1367 W/m2.The solar constant is defined as the energyreceived from the sun per unit time, on a unitarea of surface perpendicular to the direction ofpropagation of the radiation at the top of theatmosphere and at the earth’s mean distancefrom the sun.

3. What is the tip speed ratio of savonius windturbine rotor?

(a) 1 (b) 3

(c) 5 (d) 7

Ans. (a)

Sol. Tip-speed ratio is defined as

=0

speed of the rotor blade R=speed of incoming air u

Rotor Type Tip speed ratio1. Propeller 6 – 20

(1-3 Blades)2. Sailwing 43. Chalk multiblade 3 - 44. American multiblade 15. Dutch type 2 - 36. Savoanious 17. Darrieus 5 - 68. Musgrove and Evan 3 - 4

4. What is the solidity of American multibladewind turbine rotor?

(a) 0.4 (b) 0.7

(c) 0.9 (d) 1

Ans. (b)Sol.

Rotor Type SoliditySavonious rotor 1American multiblade 0.7Solidity is defined as the ratio of the projectedarea of the rotor blades on the rotor plane to theswept areas of the rotor.

5. The energy density of Bio-ethanol is

(a) 8.3 MJ/kg (b) 14.6 MJ/kg

(c) 26.9 MJ/kg (d) 34.7 MJ/kg

Ans. (c)

6. The percentage of hydrogen in producer gasis

(a) 34% (b) 27%

(c) 18% (d) 8%

Ans. (c)Typical gas composition of producer gas is19 percent CO, 18 percent H2, 1 PercentCH4, 11 percent CO2 and the rest N2.





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7. In single basin, double effect scheme, poweris generated

(a) during filling. (b) during emptying.

(c) on ebb only. (d) on both flood and ebb.

Ans. (d)Sol. In single basin, double effect scheme, power

is generated on both flood and ebb. Two- way(reversible) hydraulic turbines are used.In single effect scheme, power is generated eitherduring filling or emptying the basin.

8. The operating temperature of alkaline fuel cellis

(a) 39°C (b) 90°C

(c) 127°C (d) 192°C

Ans. (b)Sol. The operating temperature of alkaline fuel cell

is 90°CCell type Operating temperaturePhosphoric Acid 150 – 200°Cfuel cellAlkaline fuel cell 90°CSolid polymer 40 – 60 °Cfuel cellDirect methanol 50 – 120 °Cfuel cellMolten Carbonate 600 – 700 °Cfuel cellSolid oxide fuel cell 600 –1000°C

9. The ideal emf produced by polymer electrolytemembrane fuel cell at 25°C is

(a) 3.57 V (b) 2.94 V

(c) 1.23 V (d) 0.73 V

Ans. (c)Sol. Polymer electrolyte membrane fuel cell (PEMFC)

or solid polymer fuel cell (SPFC) operates at40 – 60°C. The ideal emf produced is 1.23 V at25°C.

10. Which one of the following fuel cells hashighest efficiency?

(a) PAFC (b) MCFC

(c) PEMFC (d) AFC

Ans. (d)Sol. Characteristics of various fuel cells

Fuel Cell FuelOperatingtemp. Efficiency

1. PEMFC2. AFC3. PAFC4. MCFC5. SOFC

40-60°C90°C

150-200°C600-700°C

600-1000°C

H2

HH

H & CoH & Co

2

2

2

2

48-58%64%42%50%

60-65%

11. How many kilograms of steam per day isproduced by 15 m diameter community solarcooker developed by Centre for ScientificResearch, Auroville (Puducherry)?

(a) 100 kg (b) 300 kg

(c) 600 kg (d) 1000 kg

Ans. (c)Sol. A 15 m diameter community solar cooker has

been developed at the Centre for ScientificResearch (CSR), Auroville. Around 600 kg ofsteam per day could be generated from this bowl.

12. In a solar passive space heating system, thesouth-facing thick wall is called

(a) Vent wall (b) Trombe wall

(c) Damper wall (d) Ventilation wall

Ans. (b)Sol. The south facing thick wall in a solar passive

heating system, called Trombe wall is made ofconcrete, adobe, stone or composites of brickblocks and sand, designed for thermal storage.

13. All power plants use superheated steam dueto which of the following advantages?

1. Superheating is mostly done from waste heatof boiler without additional cost of fuel.

2. The plant efficiency increases due to highertemperature of steam.

3. There is less corrosion and erosion of equip-ment due to absence of moisture in the steam.

Select the correct answer using the code givenbelow:

(a) 1 and 2 only (b) 2 and 3 only

(c) 1 and 3 only (d) 1, 2 and 3





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Ans. (d)

Sol.

T 23

1 5 5

44

Superheating

s• 1 – 2 – 3 – 4 – 5 is normal Rankine cycle

• 1– 2 – 3 – 4 – 5 is superheated Rankinecycle

Because of superheating mean temperature (Tm)of heat addition .

As = f(Tm)

So, for superheated cycle

• at point 5 , dryness fraction is more than at5, so moisture content is less after the turbineexit, so less erosion and corrosion is there insuperheated Rankine cycle than Rankinecycle.

• Superheaters (CSH) are placed after thereheater and before the economizer whichuses waste heat releasing to the stack. Furthersuperheated steam for convectivesuperheater (CSH) goes to desuperheater andthen to radiant superheater (again usesradiating energy waste) and then pendentsuperheater (PSHt) and final superheatedsteam supplied to the turbine.

14. What are the effects of regenerative feedwaterheating for the same turbine output?

1. It significantly increases the cycle efficiencyand reduces the heat rate.

2. It increases the steam flow rate.

3. It increases the steam flow to the condenser.

4. If there is no change of boiler output, theturbine output drops.


(a) 1, 2 and 3 only (b) 1, 2 and 4 only

(c) 1, 3 and 4 only (d) 2, 3 and 4 only

Ans. (b)Sol. Effect of regeneration:

(i) & Heat rate Due to in meantemperature of heat addition.

(ii) For the same turbine output steam flow (bigger boiler is required) due to bleeding ofsteam to heat the feed water.

(iii) Steam flow rate in Condenser is reduces aseach stage of regeneration, steam is bleedingoff because of this less amount of steamcondense in condenser.

(iv)If boiler size is fixed and same amount ofsteam is produced then because of bleeding,less amount of steam available at turbine ateach stage leads to in turbine output.

T

23

5

s

1

46

71-y1-y kg

y kg

• Due same boiler output

= TW reducedQ more reduced

• Since the amount of heat input in the boilerreduced significantly compared to work lost,the efficiency .

15. Which of the following are the advantages ofpulverized coal firing?

1. Higher boiler efficiency.

2. Fast response for no load changes.

3. Ability to use low preheated air reducinginternal losses.

4. Ability to release large amounts of heatenabling it to generate about 2000 t/h of steamin one boiler.



(c) 1 and 4 only (d) 2, 3 and 4 only





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Ans. (c)Sol. Pulverized coal firing system has following

advantages:• It has better and far response over firing with

load variation.• Ability to use high preheated air reducing

exhaust losses.• Ability to release large amount of heat

enabling it to generate about 2000 t/h ofsteam or more in one boiler.

16. A fuel consists of 92% carbon, 7% hydrogenand remaining residual matter by mass.Working f rom first principles, the highercalorific value of the fuel is

(a) 40176 kJ/kg (b) 41176 kJ/kg

(c) 40876 kJ/kg (d) 41678 kJ/kg

Ans. (b)

Sol. Heat released by carbon is combustion

=Ckg 407000

12

=0.92 407000

12

= 31203.64 kJ/g

Heat released by hydrogen in combustion

= 143000 × 0.07

= 10010 kJ/kg

So, HHV= 31203.64 × 10010

= 41213.64 kJ/kg 41176

17. In order to burn a fuel completely, which ofthe following basic conditions must be fulfilled?

1. Supply enough air for complete combustionof fuel.

2. Secure low turbulence for thorough mixing offuel and air.

3. Maintain a furnace temperature high enoughto ignite the incoming fuel air mixture.

4. Provide a furnace volume large enough toallow time for combustion to be completed.




Ans. (c)

Sol. More turbulence is required for the propermixing of fuel and air due to better surfacecontact.

18. The efficiency of any cycle increases with

(a) the decrease of maximum pressure and theconstant of exhaust pressure.

(b) the decrease of maximum pressure and thedecrease of exhaust pressure.

(c) the increase of maximum pressure and thedecrease of exhaust pressure.

(d) the increase of maximum pressure and theconstant of exhaust pressure.

Ans. (c)

Sol.

T

SP

P

P T S m 0

, for maximum& P T (Heat rejection)decreases,

m

19. In a power plant, the efficiencies of the electricgenerator, turbine (mechanical), boiler, cycleand the overall plant are 0.97, 0.95, 0.92,0.42 and 0.33, respectively. What is theefficiency of auxiliaries?

(a) 98.14% (b) 92.68%

(c) 83.41% (d) 75.14%

Ans. (b)

Sol.

g = 0.97

m = 0.95

boiler = 0.92





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isen = 0.42

overall = 0.33

aux = ?

overall = g m boiler ism aux

aux =0.33

0.97 0.95 0.92 0.42

92.68%

20. Consider the following statements for analysisof steam cycles:

(1) A steam power plant continuously convertsthe energy stored in fossil fuels or fissile fu-els into shaft work.

(2) Steam power plants work on Brayton cycle.

(3) In supercritical steam cycle, steam is gener-ated in a ‘once-through’ boiler at a pressureabove the critical point of 27.5 bar.

(4) Deaerator is used for the purpose of deaerat-ing the feedwater.

Which of the above statements are correct?


(c) 1 and 4 only (d) 2 and 4 only

Ans. (c)

Sol. Steam power plant works on Rankine cycle.

A steam power plant continuously convertsthe energy stored in fossil fuels on fissile fu-els into shaft work.

Deaeratar is used for the purpose of removalof air (deaerating) the feed water entering tothe boiler.

Critical point of water is 22.12 MPa or 220bar and 373.14°C.

21. Consider the following statements for solidfuels:

(1) Peat is the first stage in the formation of coalfrom wood.

(2) The average calorific value of bituminous coalis 1524 kJ/kg.

(3) Anthracite is very hard coal and has a shin-

ing black lustre.

(4) Wood charcoal is obtained by destructive dis-tillation of wood.




Ans. (b)

Sol. Anthracite is the highest grade of coal, it isa shiny black dense, hard, brittle coal thatborders on graphite.

The calorific value of coals range from 25600to 32600 kJ/kg

Peat is considered as the first geological stepin coal’s formation.

Wood charcoal is obtained by destructive dis-tillation of wood.

22. Consider the following statements for fluidizedbed boilers:

(1) Fluidized bed boilers produce steam from fos-sil and waste fuels by using a technique calledfluidized bed combustion.

(2) Cyclone separators are gas cleaning devicesthat utilize the centrifugal force created by aspinning gas stream to separate particles froma gas.

(3) In a pressurized fluidized bed boiler, the com-bustion process takes place in a pressurizedenvironment resulting in a compact furnaceand improved combustion efficiency.



(c) 1 and 3 only (d) 1, 2 and 3

Ans. (d)

Sol. All statements are correct as per their definition.

23. Consider the following statements for steamturbines:

(1) The ratio of actual enthalpy drop to isentropicenthalpy drop is known as mechanical effi-ciency.

(2) The ratio of enthalpy drop in moving bladesto enthalpy drop in the stage is known as de-





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gree of reaction.

(3) Rateau turbine is the example of reaction tur-bine.

(4) Curtis turbine is the example of impulse tur-bine.



(c) 2 and 3 only (d) 1, 2, 3 and 4

Ans. (a)

Sol. Rateau turbine is a pressure compounding

multistage impulse turbine.

Curtis turbine is velocity compounding multi-stage impulse turbine.

moving blade

moving blade fixed blade

( h)R =

( h) + ( h)

=

moving blade

stage

hh

act

isenisen

( h)Actual enthalpy drop= =isentropic enthalpy drop ( h)

24. Consider the following statements for coolingtowers:

(1) Cooling tower is an artificial device used tocool the hot cooling water coming out of con-denser more effectively.

(2) The amount of water usually lost with induceddraft cooling tower ranges from 5% to 6% byevaporation.

(3) The amount of water usually lost with induceddraft cooling tower ranges from 7% to 8% bydrift losses.

(4) The rate of evaporation of water and its cool-ing effect on the remaining water dependsupon the relative humidity of air passingthrough the tower.



(c) 2 and 3 only (d) 1, 2, 3 and 4

Ans. (a)

Sol. The amount of water usually lost with induceddraft cooling towers ranges from 1 to 2% byevaporation and 0.5 to 2% by drift losses.

25. A single-acting reciprocating pump, runningat 50 rpm delivers 0.00736 m3/s of water. Thediameter of the piston is 200 mm and strokelength is 300 mm. What is the percentageslip of the pump?

(a) 5.29% (b) 6.29%

(c) 7.29% (d) 8.29%

Ans. (b)Sol. Single acting reciprocating pump

N = 50 rpmQ = 0.00736 m3/sD = 200 mmL = 300 mm

% slip = ?

% slip =th act

th

Q – Q ×100Q

Qth = 2ALN 50= ×0.30×0.2060 4 60

= 0.00785 m2/s

% slip =0.00785 – 0.00736 ×100

0.00785

= 6.25%

26. A pump discharges a liquid into a tank at therate of 0.032 m3/s. The tank, 1.5 m in diameterand 4.20 m in height, can hold 3500 kg ofliquid. The density of the liquid and mass flowrate of the liquid handled by the pump arerespectively,

(a) 471.57 kg/m3 and 16 kg/s

(b) 471.57 kg/m3 and 15 kg/s

(c) 481.57 kg/m3 and 16 kg/s

(d) 481.57 kg/m3 and 15 kg/s

Ans. (b)Sol. Assuming the tank to be cylindrical,

Volume of the tank = 2d ×h4

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2= × × 4.201.54

= 7.418 m3

Density of liquid handled by pump ( )

= Mass 3500= = 471.80 kg/s

volume 7.418

Mass flow rate of liquid handled by the pump = volume flow rate=0.032 × 471.80 = 15.097 kg/s

27. In Francis turbine, as the water discharge isradial at the outlet, the velocity whirl at theoutlet becomes

(a) 1 (b) 0

(c) (d) 0.5

Ans. (b)Sol. In Francis turbine,

Discharge is radial at the outlet and hence whirlvelocity (Tangential component of absolutevelocity) at the outlet is zero.

V2

u2

2rV

1rV

28. A pump impeller is 375 mm in diameter andit discharges water with velocity componentsof 2 m/s and 12 m/s in the radial and tangentialdirections respectively. The impeller issurrounded by a concentrical cyl indricalchamber with parallel sides, the outer diameterbeing 450 mm. If the flow in the chamber isa free spiral vortex, what are the tangentialvelocity and radial velocity at the outlet of thechamber respectively?

(a) 12 m/s and 1.67 m/s

(b) 10 m/s and 1.67 m/s

(c) 12 m/s and 1.76 m/s

(d) 10 m/s and 1.76 m/s

Ans. (b)Sol. Pump

D2 = 375 mm

2wV = 12 m/s

2fV = 2 m/s

Free vortex motion;

wV r = constant

2w 2V r = 3w 3V r

37512×

2=

3w450V ×2

3wV = 10 m/sFurther;

2

2

w

f

VV =

3

3

w

f

VV

3fV =

32

2

wf

w

VV ×

V

=102× =1.67m12

29. Which one of the following types of impellersis used to handle highly solid laden liquidslike concrete and slurry?

(a) Fully open impeller

(b) Semi-Enclosed Impeller

(c) Fully-Enclosed Impeller

(d) Quarter Open Impeller

Ans. (a)Sol. Fully open impeller is used to handle high solid

laden liquids like concrete and slurry.

30. In a single reciprocating pump without airvessel, the ratio of the average frictional headto the maximum frictional head in the deliverypipe is

(a) 1/2 (b) 1/3

(c) 2/3 (d) 3/4

Ans. (c)Sol. In single reciprocating pump without air vessel;

the ratio of average frictional head to maximumfrictional head in delivery pipe is 2/3.





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31. A turbine develops 800 kW when running at1000 rpm. The head on the turbine is 30 m.If the head is reduced to 18 m, what is thespeed developed by the turbine?

(a) 67.46 rpm (b) 95.24 rpm

(c) 54.67 rpm (d) 77.46 rpm

31.

Ans. (d)Sol. P1 = 8000 kW

N1 = 1000 rpmH1 = 30 m

if H2 = 18 m; N2 = ?

Nu =NH

2

1

NN =

2

1

HH

N2 =2

11

H 18N =1000×H 30

= 774.60 rpmNone of the given option is correct[There seems a typographical error in the optionsand if 77.46 rpm is read as 774.6 rpm, option(d) may be marked as correct.

32. The steam turbine can be governed by thefollowing methods except

(a) Reaction governing

(b) Throttle governing only

(c) Nozzle control governing only

(d) Combination of throttle and nozzle controlgoverning.

Ans. (a)

Sol. Steam turbine is governed by

(i) Throttle governing

(ii) Nozzle control governing

(iii) Combination of throttle and nozzle govern-ing.

Reaction governing is done to increase theefficiency of turbine by the use of reactionblade.

33. In a gas turbine plant, heat supplied is 667.2kJ/kg, and heat rejected is 391.43 kJ/kg. Whatis the thermal efficiency of the plant?

(a) 57.29% (b) 72.51%

(c) 41.33% (d) 32.83%

Ans. (c)

Sol.667.2 – 391.43=

667.2 = 41.33%

34. The constant pressure gas turbine works on

(a) Stirling Cycle (b) Atkinson Cycle

(c) Rankine Cycle (d) Brayton Cycle

Ans. (d)

Sol. Brayton a cycle is constant pressure gas cycleon which gas turbine works.

35. In hydraulic turbines, if the energy availableat inlet is only kinetic energy, energy, thenthat type of turbine is

(a) Reaction turbine

(b) Impulse turbine

(c) Francis turbine

(d) Kaplan turbine

Ans. (b)

Sol. In impulse turbines, the energy at the en-trance to the runner is in the form of kineticenergy only. The pressure remains atmo-spheric throughout the runner.

36. A centrifugal pump has an impeller of 30 cmouter diameter. The vane tips are radial atthe outlet. For a rotative speed of 1450 rpm,what is the manometric head developed?(Assume a manometric efficiency of 82% andtake g = 9.81 m/s2)

(a) 37.24 m

(b) 43.38 m

(c) 29.46 m

(d) 32.88 m

Ans. (b)

Sol.





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V1

u1

Vr1

Vr2

u = V 2 w2

V2

D2 = 30 cmVane tips are radial at outletN = 1450 rpm

mano = 82%

Hm = ?

u2 = 2D N × 0.30 ×1450=60 60

= 22.765 m/s

mano = 2

m m2

w 2 2

gH gH=

V u u

mano = 0.82 = m2

9.81×H(22.765)

mH = 43.32 m

37. Lenoir cycle is used for

(a) Gas turbines (b) Pulse jet engines

(c) S.I engines (d) C.I engines

Ans. (b)

Sol.

Gas turbine Brayton cycle

S.I. engine Otto cycle

C.I engine Diesel cycle

Pulse jet engine Lenoir cycle.

38. A diesel engines has a compression ratio of20 and cut-off takes place at 5% of the stroke.What is the cut-off ratio

(a) 1.21 (b) 1.47

(c) 1.73 (d) 1.95

Ans. (d)

Sol. P2 3

1V

1

2

VV = 20 = r

V3 – V2 = 1 25 V – V

100

3

2

V – 1V

=1

2

V5 – 1V100

or3

2

VV =

1 19+1= +120 – 120 20

= 1.95

39. The cubic capacity of a four-stroke over-square spark-ignition engine is 275 cc. Theclearance volume is 25 cc. What is thecompression ratio of the engine?

(a) 8 (b) 10

(c) 12 (d) 14

Ans. (c)

Sol. P

VC

3

2 4

1

1V

VS = 275 = V1 – V2 = V1 – VC ...(i)VC = 25 = V2 ...(ii)

So from (i) & (ii) V1 = VS + VC = 275 + 25 = 300

r =1

2

V 300= =12V 25

40. The mechanical efficiency of a single-cylinderfour-stroke engine is 60%. The frictional poweris estimated to be 30 kW. What is theindicated power?

(a) 120 kW (b) 75 kW





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(c) 150 kW (d) 130 kW

Ans. (b)Sol.

m =bp = 0.6ip

or bp = 0.6 ipor, ip – fp = 0.6 ip

ip =fp 30= = 75kW

0.4 0.4

41. A four-stroke petrol engine at full load delivers100 kW. It requires 10 kW to rotate it withoutload at same speed. What is the mechanicalefficiency at half load?

(a) 67.82% (b) 70.24%

(c) 77.32% (d) 83.33%

Ans. (d)Sol. bp = 100 kW at full load

bp =100 = 50kW

2 at half load

fp = constant = 10 kW

So, m =bp 50= = 83.33%ip 50 + 10

42. Freon-12 is used in a simple saturation cycle,with suction saturation temperature of –10°Cand condensing saturation temperature of30°C. If the clearance volume is 6% of thestroke v olume, what is the v olumetr iceff iciency? (Consider specif ic volume atsuction and discharge to be 0.07815 m3/kgand 0.025 m3/kg respectively)

(a) 87.24% (b) 71.31%

(c) 64.85% (d) 55.43%

Ans. (a)

Sol. Volumetric efficiency v( )

= suction

discharge1+ C C

C = Clearance factor = Clearance volume

Swept volume

Stroke volume = Clearance Volume + SweptVolume= 0.06 × stroke volume + swept volume

Swept volume = (0.94 × stroke volume)

C = Clearance volume

Swept volume = 0.06 ×stroke volume0.94 ×stroke volume

= 347

suction = 0.07815 m3/kg

discharge = 0.025 m3/kg

v = 3 3 0.078151+ ×47 47 0.025

= 0.86429 = 86.43% 87.24%AlternatelyConsidering stroke volume = swept volumeC = 0.06

V = suction

discharge1+ C C

= 0.078151+ 0.06 0.06×

0.025

= 0.8724 = 87.24%

43. Relative ozone destruction efficiency of R-12is

(a) 0.29 (b) 0.86

(c) 0.05 (d) 0.57

Ans. (b)

Sol. The relative ozone destruction efficiency ofR-12 ranges in 0.82 – 0.86.

44. An air cooled condenser has 6 m2 of surfacewith a removal of 50 kJ hr–1. m–2.°C–1. Whatis the refrigerant temperature to dissipate 5235kJ/hr, if the room temperature is 25°C?

(a) 24.31°C (b) 35.82°C

(c) 42.45°C (d) 56.94°C

Ans. (c)





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Sol. The removal rate of condenser = 2kJ50

hr m c

Area = 6 m2

Total removal rate = kJ(50× 6)

hr C

= kJ300

hr C

Heat to be removed = 5235 kJ/hr

t = 5235 C300

= 17.45°C

(t – troom) = 17.45°Ct = (17.45 + 25) = 42.45°C

45. The actual and theoretical COP of rollingpiston compressor are 3.6 and 4.7respectively. What is the relative COP?

(a) 8.3 (b) 16.92

(c) 1.3 (d) 0.76

Ans. (d)

Sol. Relative COP = Actual COP

Theoretical COP

= 3.64.7

= 0.76

46. In an absorption type refrigeration system,heat ing in generator, ref r igerat ion inevaporator and cooling by cooling water incondenser, take place at 95°C, –5°C and 30°Crespectively. What is the maximum COP ofthe system?

(a) 1.17 (b) 1.35

(c) 1.52 (d) 1.78

Ans. (b)

Sol.

Temperature of evaporator (TE) = 5 C = 268 k

Temperature of generator (TG) = 95°C = 368 kTemperature of surrounding (To) = 30°C = 303k

Maximum COP = E G D

G O E

T (T T )T (T T )

= 268×(368 303)368×(303 268)

= 1.352

47. Consider the following statements for sensibleheat factor:

1. Sensible heat factor will be negative if sen-sible heat and latent heat are both negative.

2. Sensible heat factor will be negative if sensibleheat is negative and latent heat is positive.

3. Sensible heat factor will be negative if sensibleheat is positive and latent heat is negative.

4. Sensible heat factor will be negative if sensibleheat and latent heat are both positive.




Ans. (a)

Sol.Sensible heat factor (SHF)

= Sensible Heat (SH)

Sensible Heat (SH) +Latent Heat (LH)

(1)when SH is –ve & LH is –ve SHF +ve(4)when SH +ve & LH is +ve SHF +ve

Hence, statement 1 & 4 are incorrect andstatement 2 & 3 are correct.

48. If the air initially at dry bulb temperature 35°Cand wet bulb temperature 26.1°C as it entersan air washer which has a humidify ingefficiency of 85%, then what is the final drybulb temperature of ai r washed wi threcirculated spray water?

(a) 26.81°C (b) 27.43°C

(c) 32.83°C (d) 30.49°C

Ans. (b)

Sol. Humidifying efficiency = outDBT TDBT WBT

0.85 = out35 T(35 26.1 )





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Tout = 35° – 0.85 (35° – 26.1°)

= 27.435°

49. Consider the following statements for Nucleateboiling:

1. For water, the critical heat flux does not ex-ceed 1 MW/m2.

2. Nucleate boiling is the most desirable boilingregime in practice because of high heat trans-fer rates.

3. Heat flux increases at a higher rate with in-crease in temperature

Which of the above statements is/are correct?

(a) 1 only (b) 2 only


Ans. (b)

Sol. For water, critical heat flux > 1 Mw/m2

From heat transfer point of view, nuclearboiling regime is the most desirable range tooperate.

Heat transfer rate increase at a lower rate as

eT increases.

50. In drop-wise condensation, the heat transferrate is

(a) 5 times less than that in f i lm-wisecondensation.

(b) 15 times less than that in f i lm-wisecondensation

(c) 25 times more than that in f i lm-wisecondensation.

(d) 10 times more than that in f i lm-wisecondensation.

Ans. (d)Sol. In drop-wise condensations, the vapours

condense on the surface on drops, which dripsdown the surface. A continuous films of liquid isnot formed on the surface. Thus, more of thebase area at temperature Ts is always exposedto the vapours. Therefore, heat transfer rate ishigher (up to 10 times) in drop-wisecondensation as compared to the value in film-wise condensation.

51. 1 kg of water falls from an altitude of 1000 mabove ground level. What is the change inthe temperature of water at the foot of thefall, if there are no losses during the fall ?(Take specific heat of water as 1 kcal. kg–1.K–1 and g = 9.81 m/s2)

(a) 3.35°C (b) 2.35°C

(c) 3.32°C (d) 4.12°C

Ans. (b)

Sol. Since total energy is conserved

Loss of potential energy = Gain in thermal energy

mgh = pmc T

1 × 9.81 × 1000 = 1 1000 4.18 J T

kg K

T = 2.346889952 °C

T = 2.35 C

52. A stationary mass of gas is compressedwithout friction from an initial state of 0.3 m3

and 0.105 MPa, the pressure remainingconstant during the process. There is a transferof 40 kJ of heat from the gas during theprocess. How much does the internal energyof the gas change ?

(a) –24.25 kJ (b) –19.62 kJ

(c) –15.91 kJ (d) –12.72 kJ

Ans. (a)

Sol.

0.15 m30.3 m3

V

P

0.105 MPa

Work done = P V

= 3–(0.105 0.15) 10 kJ

= –15.75 kJ

Q = –40 kJ





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First law of thermodynamics :

Q = w + u

–40 = –15.75 + u

u = (–40 + 15.75) kJ = – 24.25 kJ

53. The state of a simple compressible puresubstance can be fixed by specifying

(a) one independent property.

(b) two independent properties.

(c) three independent properties.

(d) four independent properties.

Ans. (b)Sol.

The state of pure substance of a given masscan be fixed by specifying two independentintensive properties provided the system is inequilibrium. This is known as 2-properties rule.Temperature and specific volume, for example,are always independent properties, and togetherthey can fix the state of a simple compressiblesystem.

54. In a thermoelectric thermometer for t°Ctemperature, the emf is given as: E = 0.003t– 5 × 10–7 t2 + 0.5 × 10–3 volts. Thermometeris having reference junction at ice point andis calibrated at ice point and steam point.What is the temperature shown by thethermometer for a substance at 30°C?

(a) 33.23°C (b) 36.28°C

(c) 41.23°C (d) 46.28°C

Ans. (a)

Sol. E = 0.003t – 5 × 10–7 t2 + 0.5×10–3

At ice point, when t = 0°C,E = 0.5 × 10–3

At steam point t = 100°C

E = 0.003 × 100 –5 × 10–7 × (100)2

+ 0.5 × 10–3

= 0.2955 V

At 30°C

E = 0.003 × 30 –5 × 10–7 × (30)2

+ 0.5 × 10–3

= 0.09005 V

When the gas thermometer reads 30°C thethermocouple will read

=100 0.09005

0.2955

= 30.44°CThe nearest answer is 33.23°C.

55. Consider the fol lowing statements forcomparison of heat and work:

1. Both heat and work are transient phenomena.

2. Both heat and work are boundary phenom-ena.

3. Both heat and work are path functions andinexact differentials.

Which of the above statements are correct ?

(a) 1 and 2 only (b) 1, 2 and 3


Ans. (b)

Sol. Both heat and work are interactions betweena system and its surroundings. They havefollowing similarities:

Both are recognized only as they cross theboundary of a system, i.e. both are boundaryphenomena.

Systems posses energy, but not heat or work.That is, heat and work are transient phenom-ena.

Both are path functions.

56. A tank containing a fluid is stirred by a paddlewheel. The work input to the paddle wheel is5090 kJ. The heat transfer from the tank is1500 kJ. What is the change in internal energyof this control mass ? (Consider the tank andthe fluid inside a control surface)

(a) – 3590 kJ (b) + 3590 kJ

(c) + 4590 kJ (d) – 4590 kJ

Ans. (b)

Sol. Q = U + W

U = Q – W = –1500 – (–5090)





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= 3590

57. During the charging of a storage battery, thecurrent is 20 A and the voltage is 12.8 V. Therate of heat transfer from the battery is 10W.At what rate is the internal energy increasing?

(a) – 256 J/s (b) + 246 J/s

(c) + 256 J/s (d) – 246 J/s

Ans. (b)

Sol.

Q = U+ W

–10 = U – IR

–10 = U – 20 12.8

U = 246 J/sec

58. A refrigerator operates on Reversed Carnotcycle. What is the power required to drive therefrigerator between temperatures of 42°C and4°C, if heat at the rate of 2 kJ/s is extractedfrom the low temperature region ?

(a) 0.174 kW (b) 0.374 kW

(c) 0.274 kW (d) 0.474 kW

Ans. (c)

Sol.

2 kJ/s

315 k

277 k

T

s

COP = inQPower

=

L

H L

TT T

=277

(315 277)

inQ = 2 kW

2

Power = 27738

Power 2× 38= kW277

= 0.274 kW

59. Entropy generated (Sgen) can be taken as acriterion to indicate feasibility of process.Which of the following conditions are correct?

1. If Sgen = 0, then the process is a reversibleprocess.

2. If Sgen > 0, then the process is an irreversibleprocess.

3. If Sgen < 0, then the process is impossible.



(c) 1 and 3 only (d) 1, 2 and 3

Ans. (d)

Sol. All the statements are correct.

when genS 0 Irreversible process

genS = 0 Reversible process

genS 0 Impossible process

60. What is the critical radius of insulation forasbestos (thermal conductivity = 0.17 W.m–1.°C–1) surrounding a circular pipe and exposedto room air at 20°C with heat transfercoefficient 3 W.m–2. °C–1?

(a) 7.21 cm (b) 6.37 cm

(c) 5.67 cm (d) 6.93 cm

Ans. (c)

Sol.

h = 3 wm–2°C–1

k = 0.1700 wm–1°C–1

Critical radius of insulation = kh

=0.17 1000 mm

3

= 5.667 mm

61. Water is flowing through a pipe of diameter





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200 mm with a velocity of 3 m/s. What is thehead loss due to friction for a length of 5 mif the coefficient of friction is given by f =

0.02 + 0.30.09 ,Re

where Re is Reynolds number?

(Take the kinematic viscosity of water as 0.01stoke, g = 9.81 m/s2 and (6 × 105)0.3 = 54.13)

(a) 0.993 m of water (b) 0.783 m of water

(c) 0.685 m of water (d) 0.552 m of water

Ans. (a)

Sol.

Re = 5–4

VD 3× 0.2= = 6×100.01×10

f = 5 0.30.09 0.090.02 + = 0.02 +

54.13(6 10 )

f = 0.02166

Friction factor = 4f = 4 × 0.02166

Head loss due to friction = hf = 2(4f ) V

2gD

hf =24 0.02166 5 (3) = 0.993 m

2 9.81 0.2

62. Water is flowing through a horizontal pipe ofdiameter 200 mm at a velocity of 3 m/s. Acircular solid plate of diameter 150 mm isplaced in the pipe to obstruct the flow. Whatis the loss of head due to obstruction in thepipe if Cc = 0.62 ? (Take g = 9.81 m/s2)

(a) 3.311 m (b) 4.211 m

(c) 5.211 m (d) 6.211 m

Ans. (a)Sol. Head loss due to obstruction in pipe

=

22

Lc

AV – 1h =C2g A – a

V = Velocity in pipeA = Area of pipea = max. area of obstruction

Cc = Coefficient of contraction

hL =

222

2 20.23 – 1

0.622× 9.81 –0.2 0.15

Lh = 3.311 m

63. Three pipes of length 800 m, 500 m and 400m and of diameters 500 mm, 400 mm and300 mm respectively are connected in series.These pipes are to be replaced by a singlepipe of length 1700 m. What is the diameterof the single pipe ?

(a) (0.007118)0.2 m (b) (0.003609)0.3 m

(c) (0.003609)0.2 m (d) (0.007118)0.3 m

Ans. (a)

Sol.

Q

l1 = 800 m l l2 3

1 2 3

= 500 m = 400 md = 500 m d = 400 m d = 300 m

Q

l1 = 1700 md = ?

Since the two pipes are two be equivalent,hence the discharge and head loss must besame in the two. Assuming same frictionfactor in the two sets of pipes

2

5f Q

12.1d

=22 2

31 25 5 51 2 3

f Qf Q f Q+ +12.1d 12.1d 12.1d

51700

d = 5 5 5

800 500 400+ +(0.5) (0.4) (0.3)

= 239037.1785

d5 = 0.0071118

0.2d = (0.0071118) m

64. The head of water at the inlet of a pipe 2000m long and 500 mm diameter is 60 m. Anozzle of diameter 100 mm at its outlet isfitted to the pipe. What is the velocity of waterat the outlet of the nozzle if f = 0.01 for the





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pipe ? (Take g = 9.81 m/s2)

(a) 30.61 m/s (b) 34.81 m/s

(c) 36.52 m/s (d) 38.36 m/s

Ans. (a)Sol. For

= 2000m

head = 60m

V1d=0.5m, f = 0.01

friction factor = 4× 0.01=0.04

100mm

2

1

Head loss in 2000 m length of pipe = 2

5f Q

12.1d

Applying energy equation between (1) & (2)

60 =2

2 22 f

P V+ Z + +h2g

0 0

[Neglecting loss in nozzle]

60 =2 22 1V f V+

2g 2gd

21V × 0.54

= 22V × 0.14

[Continuity eqn]

V1 = 2V25

60 =22V f1+

2×9.81 d×625

=22V 0.04 ×20001+

2×9.81 0.5 ×625

2V = 30.61 m/s

65. Water is flowing with a velocity of 1.5 m/s ina pipe of length 2500 m and of diameter 500mm. At the end of the pipe, a valve isprovided. What is the rise in pressure if thevalve is closed in 25 seconds ? (Take thevalue of C as 1460 m/s)

(a) 12 N/cm2 (b) 15 N/cm2

(c) 16 N/cm2 (d) 18 N/cm2

Ans. (b)

Sol.

l = 2500 m

V = 1.5 m/s, d = 0.5 m

Valve

Time of closure of valve = t0 = 25 sec

Critical time (tc) = 2 2 2500= sec = 3.42 secC 1460

t0 > tc

it is a slow closure case

P = 20

VL 1000 1.5 2500 N=t 25 m

P = 42 2

N N15 10 = 15m cm

66. If a submerged body is in unstable equilibrium,then

(a) the centre of buoyancy is below the centre ofgravity.

(b) the centre of buoyancy is above the centre ofgravity.

(c) meta-centre is below the centre of buoyancy.

(d) meta-centre is above the centre of buoyancy.

Ans. (a)Sol. For unstable equilibrium of submerged body

centre of gravity should be above centre ofBuoyancy.

67. How much of concrete with = 25 kN/m3

must be attached to a beam having a volumeof 0.1 m3 and specific gravity 0.6 to causeboth to sink in water ? (Take g = 9.81 m/s2)

(a) 0.825 kN (b) 0.745 kN

(c) 0.525 kN (d) 0.645 kN

Ans. (d)Sol. For Sinking of both beam & the attached

concrete, the weight should be greater than orequal to the Buoyant force.Weight = [Vconc × 25 + 0.1 × (0.6 × 9.81)] kNBuoyant force = (Vconc + 0.1) × 9.81 kNHence





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(Vconc × 25) + (0.1 × 0.6 × 9.81) Vconc × 9.81+ 0.981

Vconc 0.02583 m3

Wt of concrete 0.02583 × 25 = 0.6458 kN

(Wt of concrete to be attached)min = 0.6458kN

68. A liquid has a specific gravity of 1.9 and akinematic viscosity of 6 stokes. What is thedynamic viscosity ?

(a) 1.14 Ns/m2 (b) 2.44 Ns/m2

(c) 3.40 Ns/m2 (d) 11.40 Ns/m2

Ans. (a)

Sol. Dynamic viscosity (µ) =

µ = (6 × 10–4) m2/s × (1.9 × 1000) kg/m3

µ = 1.14 Ns/m2

69. Oil of specific gravity 0.8 flows through a 0.2m diameter pipe under a pressure of 100 kN/m2. If the datum is 5 m below the centerlineof the pipe and the total energy with respectto the datum is 35 Nm/N, the discharge is(Take g = 9.81 m/s2)

(a) 0.58 m3/sec (b) 0.47 m3/sec

(c) 0.31 m3/sec (d) 0.22 m3/sec

Ans. (a)

Sol.

B

ZB

Datum

P = 100 kN/mB2

Z = 5 mB

2B B

BP V+ Z +

2g= 35 m

2B B

Bw

P V+ z +0.8 2g = 35 m

2BV100 + 5 + = 35

0.8 9.81 2 9.81 VB = 18.401 m/s

Q =2 2

Bd (0.2)V =18.4014 4

=3m0.578

s

70. Bernoulli’s equation is obtained by

(a) integrating the Euler’s equation of motion.

(b) differentiating the Euler’s equation of motion.

(c) differentiating the Navier-Stokes equations.

(d) integrating energy equation.

Ans. (a)

71. W hich one of the fol lowing is not themethodology of control separation of flow fromboundary in the application of aerofoils ?

(a) Streamlining of blunt body shapes

(b) Fluid ejection from the boundary layer

(c) Suction of fluid from the boundary layer

(d) Creating a motion of the boundary wall

Ans. (b)Sol. The various methods adopted for controlling the

boundary layer separation are1. streamlining the object2. Fluid injection into Boundary layer3. Suction of retarded fluid form Boundary layer4. Creating a motion of boundary wall thereby

giving energy to the retarded fluid.Hence answer should be (b) in which in place ofinjection it is given as ejection.

72. What is the value of mass of the air in aroom of size 4 m × 5 m × 6 m at 100 kPa and25°C? (Take R = 0.287 kPa.m3.kg–1.K–1).

(a) 150 kg (b) 180 kg

(c) 140 kg (d) 130 kg

Ans. (c)Sol. P = RT

=

3P 100kPa=

RT m0.287kPa × x273 + 25kg.K = 1.169235 kg/ms

Mass of air = Volume

= 1.169235 × (4 × 5 × 6) kg= 140.308 kg





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73. A body of weight 100 N is placed on a roughhorizontal plane. what is the coefficient offriction if a horizontal force of 60 N just causesthe body to slide over the horizontal plane ?

(a) 0.4 (b) 0.5

(c) 0.6 (d) 0.9

Ans. (c)

Sol.

60 N

R

µR

100 N

R = 100 N

60 = µ × 100

µ = 0.6

74. A body is moving with a velocity of 2 m/s.After 4 seconds, the velocity of the bodybecomes 5 m/s. The acceleration of the bodyis

(a) 0.55 m/s2 (b) 0.65 m/s2

(c) 0.75 m/s2 (d) 0.45 m/s2

Ans. (c)

Sol. Vi = 2 m/s

Vf = 5 m/s

t = 4 sec

Acceleration = 2f iV – V 3= = 0.75 m/st 4

75. The principal stresses at a point in an elasticmaterial are 60 N/mm2 tensile, 20 N/mm2

tensile and 50 N/mm2 compressive. What isthe volumetric strain by considering Young’sModulus as 100 × 103 N/mm2 and = 0.3?

(a) 1.20 × 10–4 (b) 1.06 × 10–5

(c) 1.30 × 10–3 (d) 1.12 × 10–2

Ans. (a)

Sol.

v = x y zx y z

( + + )+ + = (1– 2 )

E

v = 3(60 + 20 – 50) (1– 2 0.3)

100 10

v = 1.2 × 10–4

76. A man whose weight is 650 N, standing onthe ground, raises a load of 3000 N by meansof single string system of pulleys. There aresix light pulleys in each block. The thrust ofthe man on the ground is

(a) 120 N (b) 135 N

(c) 150 N (d) 175 N

Ans. (c)

Sol.

In single string system of pulleys:

Velocity ratio = n

where, n = number of strings between fixedpulley block and movable pulley block.

Considering 100% efficiency:

Mechanical advantage = Velocity ratio

= n = 6

Mechanical advantage = Weight = 6Tension

Tension = Weight 3000= = 500 N

6 6

So, thrust of man on ground = w - T

= (650 – 500)

= 150 N

wFBD of man

T

77. A particle starts with an initial velocity of 200cm/s and moves with a uniform retardation of10 cm/s2. If it describes 1500 cm in time t,what is/are the possible value(s) of t ?

https://iesmaster.org/testseries/gate-2022





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(a) 10 sec only (b) 10 sec and 30 sec

(c) 30 sec only (d) 5 sec and 10 sec

Ans. (b)

Sol. u = 200 cm/s

g = –10 cm/s2

S = 1500 cm

S = 21ut + gt2

1500 = 200 t – 5t2

2t – 40 t + 300 = 0

t1 = 10 sec and t2 = 30 sec.

78. Consider the following statements for systemof forces:

1. Two or more forces that act at the same pointare called coplanar forces.

2. Two or more forces whose directed arrows liein same plane are called concurrent forces.

3. Varignon’s theorem states that the momentof several concurrent coplanar forces aboutany point O in their plane equals the momentof their resultant about the point O’.

4. Lami’s theorem states that if a body is in equi-librium under the action of three forces, eachforce is proportional to the sine of angle be-tween the other forces.


(a) 1, 2 and 4 only (b) 1 and 4 only


Ans. (c)

79. Consider the following statements related tostress and strain:

1. Shear stress is always tangential to the areaover which it acts.

2. Shear stresses on the transverse pair of facesare known as complimentary shear stresses.

3. Shear strain is defined as the change in theright angle of the element measured in radi-ans.

4. Modulus of rigidity is the ratio of shear strainto shear stress.




Ans. (d)

Sol. Statements 1, 2 and 3 are correct.

Modulus of rigidity = Shear stressShear strain

80. Consider the following statements for stressand strain analysis:

1. The stress components on any inclined planecan easily be found with the help of a geo-metrical construction known as Mohr’s stresscircle.

2. The ratio of longitudinal strain to lateral strainis known as Poisson’s ratio.

3. When a body is acted upon by three mutuallyperpendicular forces, there is change in thevolume of the body which is referred to asdilation of the material.

4. The ratio of original volume to increase involume is known as volumetric strain.



(c) 3 and 4 only (d) 1, 2, 3 and 4

Ans. (a)

Sol. Statement 1 and 3 are correct

Poisson’s ratio = lateral strain

longitudinal strain

Volumetric strain = Change in volumeOriginal volume

81. The stresses on two perpendicular planesthrough a point in a body are 160 MPa and100 MPa, both compressive, along with ashear stress of 80 MPa. What is the normalstress on a plane inclined at 30° to the planeof 160 MPa stress ?

(a) –42.4 MPa (b) –75.7 MPa

(c) –59.1 MPa (d) –86.3 MPa

Ans. (b)





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Sol.

x

y

x

xy

y

x = y xy–160 MPa, = – 100 MPa, = 80 MPa

2 Normal stress on a plane inclined at30° to the plane of 160 MPa stress.

Case-I when the plane is in anticlockwise di-rection

x = x y x yxy

+ –+ cos2 + sin2

2 2

2 = –160 – 100 –160 + 100+ cos (2 30)

2 2

+ 80 sin (2 60)

x = –75.71 MPa

Case-II When the plane is in clockwisedirection i.e., is negative

2 = –160 – 100 –160 +100+

2 2

cos (–60) + 80 sin (–60)

x = – 214.28 MPa

82. Consider the following statements regardingtypes of supports and beams:

1. When both supports of beams are rollersupports, the beam is known as simplysupported beam.

2. A beam with one end fixed and the other endfree is known as fixed beam.

3. A beam with both ends fixed is known ascantilever beam.

4. A beam with one end fixed and the othersimply supported is known as proppedcantilever.

Which of the above statements is/are correct ?

(a) 1 only (b) 1 and 4 only


Ans. (b)

83. Consider the following statements regardingstress in beam:

1. If a member is subjected to equal and oppo-site couples acting in the same longitudinalplanes, the member is said to be in pure bend-ing.

2. The internal stresses developed in the beamare known as flexural stresses.

3. There is an intermediate surface known asneutral surface, at which the stress is zero.

4. An axis obtained by intersection of the neu-tral surface and a cross-section is known asneutral axis.



(c) 3 and 4 only (d) 1, 2, 3 and 4

Ans. (d)

84. Consider the following statements for thesymmetric beam under pure bending:

1. In the elastic range, the normal stress varieslinearly with the distance from the neutralsurface.

2. As long as the stresses remain in the elasticrange, the neutral axis passes through thecentroid of the section.

3. If stresses are in the plastic range, the neu-tral axis passes through the centroid of thesection.




Ans. (c)

Sol. As long as the stresses remain in the linearelastic range, neutral axis passes through thecentroid and normal stress varies linearly withthe distance from the neutral axis.





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If stresses are in the plastic range, it is notnecessary to pass neutral axis through cen-troid.

85. The volume of FCC unit cell in terms of theatomic radius R is

(a) 3cV = 16R 3 (b) 3

cV = 8R 2

(c) 3cV = 16R 2 (d) 3

cV = 8R 3

Ans. (c)

Sol.

For FCC crystal, a= 2 2RVolume of unit cell is

= 33a = 2 2R

= 316R 2

86. Which one of the following alloying ingredientsincreases the hardenability and forms carbidesfor wear resistance ?

(a) Chromium (b) Molybdenum

(c) Nickel (d) Manganese

Ans. (b)

Sol.Carbide forming tendency in Chromium is greaterthan Manganese (Mn) but lesser than tungsten(W). Hardenability increases moderately.Manganese has consider formation tendencygreater than iron (Fe) in steel but lesser thanchromium. Manganese increases hardenabilitymoderately.Nickel leads to formation graphite that it hasnegative effect on carbide formation but itstabilized the Austenite.Molybdenum leads to formation of carbide morethan chromium and increases hardenabilitystrongly.

87. Which one of the following related to the moststable arrangement of atoms in a crystal isnot correct ?

(a) Preserves electrical neutrality

(b) Maximizes strong ion-ion repulsion

(c) Satisfies discreteness of all covalent bonds

(d) Packs the atoms as closely as possible

Ans. (b)

Sol. Stability of atoms in crystal depends on :

• Electrical neutrality

• Atoms should have greater atomic packingfactor.

• The ease of formation of the positive andnegative ions from respective neutral atoms.

• Satisfies discreteness of all the covalentbonds.

• As many anions as possible to surround thecation for maximum reduction in electrostaticenergy.

88. The dielectric constant of rubber variesbetween

(a) 0.5 and 1.0 (b) 1.0 and 1.5

(c) 1.5 and 2.0 (d) 2.5 and 5.0

Ans. (d)

Sol.Dielectric Constant :Rubber - 2 to 4Nylon - 3.4 to 22.4Air - 1 to 1Backelite - 5 to 22Glass - 3.8 to 14.5

89. Consider the following statements for ductilefracture:

1. The material undergoes substantial plasticdeformation with high energy absorption be-fore fracture.

2. Presence of cracks on the surface of mate-rial initiates this type of failure.

3. Fracture occurs due to necking.




Ans. (d)

Sol.Ductile fracture occurs often prolonged plasticdeformation and is accompanied by necking. It





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require higher strain energy.• Ductile fracture begin at the central region

then propagates to the periphery of the neckedregion.

• Crack at surface influence fatigue failure.

90. The TTT diagram shows the times requiredfor isothermal transition from

(a) austenite to pearlite

(b) austenite to ferrite

(c) ferrite to pearlite

(d) martensite to pearlite

Ans. (a)Sol. Time Temperature Transformation (TTT) diagram

is used to indicate time and temperaturetransformation of austenitic stainless steel topearl i te at eutectoid temperature. Thistransformation of austenite is time dependent.

91. Iron at 20°C is BCC with atoms of atomicradius 0.124 nm. What is the lattice constant‘a’ for the cube edge of the iron unit cell ?

(a) 0.2864 nm (b) 0.1496 nm

(c) 0.2173 nm (d) 0.1756 nm

Ans. (a)

Sol. r = 0.124 nm

4r = 3 a

4 × 0.124 = 3 ×a

or a = 0.286365 0.2864 nm

92. Copper has the FCC crystal structure and aunit cell with a lattice constant of 0.361 nm.What is the interplanar spacing ‘d220’ ?

(a) 0.085 nm (b) 0.174 nm

(c) 0.206 nm (d) 0.128 nm

Ans. (d)

Sol. dhkl = 2 2 2

a

h + k +

or, 220d = 2 2

0.361

2 + 2 + 0

= 0.12763 0.128 nm

93. Which one of the following is not a step oflever rule of determination of phase amounts?

(a) A tie line is constructed across the two-phaseregion at the temperature of the alloy.

(b) The overall alloy composition is located onthe tie line.

(c) Perpendiculars are dropped from these inter-sections to the horizontal composition axis,from which the composition of each of therespective phases is read.

(d) The fraction of one-phase is computed by tak-ing the length of tie line from the overall alloycomposition to the phase boundary for theother phase and dividing by the total tie linelength.

Ans. (c)

Sol. The Lever Rule is used to find the fractionsof liquid phase in the binary alloy in the two-phase state. The rule can be obtained byusing the law of conservation of mass.The tie line must be used along with aprocedure that is generally termed as leverrule (or the inverse lever rule) which is asfollows (same is marked in figure).

(1)The tie line is constructed across the twophase region at the temperature of the alloy.

(2)The overall alloy composition is located onthe tie line.

(3)The fraction of one phase is computed bytaking the length of tie line from the overallalloy composition to the phase boundary forthe other phase and dividing by the total tieline length.

(4)The fraction of the other phase is determinedin the same manner.

94. Consider the following statements regardingpolymeric materials :

1. A plastic material that requires heat to makeit formable (plastic) and upon cooling, retainsits shape is known as thermosetting plastic.

2. The chemical reaction in which high molecu-lar mass molecules are formed from two or





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more monomers is called chain polymeriza-tion.

3. A polymer chain consisting of two or moretypes of monomeric units is called copolymer.



(c) 2 only (d) 3 only

Ans. (b)

Sol. Thermoplastic materials are those which

soften on the application of heat with orwithout pressure, but then require cooling toset them to shape.

Thermosetting materials are those plasticwhich require heat and pressure to mold theminto shape. They can not be resoftened asonce they have set and hardened.

95. As per mechanical properties, which one ofthe following microconstituents is soft andductile ?

(a) Bainite (b) Martensite

(c) Spheroidite (d) Tempered martensite

Ans. (c)

Sol. Ductility of spheroidite > Coarse pearlite >fine pearlite > bainite > tempered martensite> martensite.Hardness: Tempered martensite > bainite >fine pearlite > coarse pearlite > spheroidite.

96. The simultaneous compaction and shaping ofa ceramic powder (and binder) by pressureapplied uniformly in all directions is known as

(a) Glaze pressing (b) Porcelain pressing

(c) Slip pressing (d) Isostatic pressing

Ans. (d)Sol. In isostatic pressing, pressure is applied from all

the directions against the powder that arecontained in a flexible mould. Hydraulic pressureis used to achieve compaction. Compaction andshaping are completed simultaneously.

DeformableMould

PressurisedFluid (Gas)

Powder Container

97. An electric motor drives a punching press. Aflywheel fitted to the press has a radius ofgyration of 0.5 m and runs at 250 rpm. Thepress is capable of punching 800 holes perhour with each punching operation taking 1.5seconds and requiring 12,000 N-m of work .The energy delivered by the motor duringpunching operation is

(a) 2000 N-m (b) 3000 N-m

(c) 4000 N-m (d) 5000 N-m

Ans. (c)Sol. Actual punching time, ta = 1.5 sec

Cycle time or time required to complete onerotation, tc = 4.5 sec

800 hole / hr1 hole/4.5 sec

Energy required per operation,Ereq = 12000 N-m or Joule

Pmotor =req

C

E 12000= Joule/secT 4.5

Energy delivered by motor,,E = Pmotor × ta

=12000 ×1.5

4.5

= 4000 Joule

E = 4000 N – m





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98. A linkage has 11 links and 4 loops. What isthe degree of freedom if it has only singleturning pairs ?

(a) 0 (b) 1

(c) 2 (d) 3

Ans. (c)Sol. No. of linkages (N) = 11

No. of loop (L) = 4Degree of freedom (F) = N – (2 L + 1) = 11 –(2 × 4 + 1) = 2

99. Which one of the following is the applicationof first inversion of single-slider-crank chain ?

(a) Hand -pump

(b) Reciprocating engine

(c) Elliptical trammel

(d) Whitworth quick-return mechanism.

Ans. (b)

Sol. First inversion Reciprocating Engine

Cylinder is fixed

100. Consider the following statements regardingcams:

1. A cam in which the follower moves radiallyfrom the centre of rotation of the cam is knownas a disc cam.

2. A globoidal cam is a double-disc cam, the twodiscs being keyed together and are in constanttouch with the two rollers of a follower.

3. A conjugate cam can have two types ofsurfaces, convex or concave.

4. In a spherical cam, the follower oscillatesabout an axis perpendicular to the axis ofrotation of the cam.



(c) 2 and 3 only (d) 1, 2, 3 and 4

Ans. (a)

Sol. 1 (Correct) A cam is which the followermoves radial ly f rom thecentre of rotation of the camis known as radial or disc cam.

2 (Incorrect) The given description is forconjugate cam.

3 (Incorrect) A globoidal can have twotypes of surfaces, convex andconcave.

4 (Correct) In a spherical cam, thefollower oscillates about anaxis perpendicular to the axisof rotation of the cam. (1 and4) (a)

101. Under which of the following conditions isCoriolis component encountered in the relativeacceleration of two points ?

1. The two points are coincident, but on differentlinks.

2. The point on one link traces a path on theother link.

3. The link that contains the path rotates.



(c) 2 and 3 only (d) 1, 2 and 3

Ans. (d)Sol. All the statements are correct coriolis acceleration

is encountered when the point may have it’smotion relative to a moving body system. Foreg: motion of a slider on rotating link.

102. A mass ‘m’ attached to a light spring oscillateswith a period of 2 seconds. If the mass isincreased by 2 kg, the period increases by 1second. What is the value of the mass?

(a) 0.9 kg (b) 1.6 kg

(c) 2.1 kg (d) 2.7 kg





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Ans. (b)Sol. frequency of oscillation:

k

m

n =km

Time period = n

2 m= 2k

Now, Initially,

2 =m2k

...(i)

Later

3 =m + 22

k ...(ii)

Dividing (ii) by (i)

32

=m + 2

m

or,94

=m + 2

m

or, 9m = 4m + 8or, 5m = 8

or, m = 1.6kg

103. What is the critical speed of the shaft if itsnatural frequency of transverse vibration is2.85 Hz?

(a) 171 rpm (b) 285 rpm

(c) 570 rpm (d) 142.5 rpm

Ans. (a)Sol. The critical speed of the shaft is revolution per

second is equal to the natural frequency oftransverse vibration is Hz.Now, fn = 2.85 Hz

Nc = 2.85 rpsN = 60 × Nc = (60 × 2.85) = 171 RPM

104. Consider the following statements regardinggears:

1. The ratio of number of teeth on the gear tothat on the pinion is known as gear ratio.

2. The circle passing through the tips of teeth iscalled dedendum circle.

3. The circle passing through the roots of teethis called addendum circle.

4. Backlash is the difference between the spacewidth and the tooth thickness along the pitchcircle.



(c) 2 and 3 only (d) 1, 2, 3 and 4

Ans. (b)Sol. 1 (Correct)

G = T Teeth on the gear= (correct)t Teeth on the pinion

2 (Incorrect)The circle passing through the tip of teeth isaddendum circle3 (Incorrect)The circle passing through the roots of teethis dedendum circle4 (Correct statement) (1 and 4) (b)

105. The number of teeth of a spur gear is 30 andit rotates at 200 rpm. What is the pitch linevelocity if it has a module of 2 mm?

(a) 341.7 mm/s (b) 497.2 mm/s

(c) 628.3 mm/s (d) 758.5 mm/s

Ans. (c)Sol. Given:

T = 30m = 2mmN = 200 rpm

Pitch line velocity = 2 N × r60

r =mT2

Pitch line velocity = 2 ×200 2×30 mm/s.×

60 2

= 628.32 mm/s

106. The compound gear train shown in the figurebelow consists of compound gear B-C and D-E. All gears are mounted on parallel shafts.The motor shaft rotating at 800 rpm is





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connected to the gear A and the output shaftto the gear F. The number of teeth on gearsA, B, C, D, E and F are 24, 56, 30, 80, 32and 72 respectively. What is the speed of thegear F?

24 30 32 7256

ABC

D

E

80

F

(a) 57.14 rpm (b) 32.51 rpm

(c) 74.63 rpm (d) 69.72 rpm

Ans. (a)Sol.

A

24

BC

3056

DE

3280

F

72

NA = 800 RPM

B

A

NN =

A

B

TT

NB =24 RPM800×56

NB = C24N = RPM800×56

D

C

NN =

C

D

TT

ND =C

CD

T 24 30N = RPM800× ×T 56 80

ND = E24 30N = RPM800× ×56 80

NF = EE

F

T 24 30 32N = 800 × × ×T 56 80 72

=400RPM7

= 57.14 RPM

107. A quick-return mechanism is to be designed,where the outward stroke must consume 1.2sand the return stroke 0.8s. If the cycle time is2.0 s/rev, what is the speed at which the

mechanism should be driven?

(a) 10 rev/s (b) 30 rev/s

(c) 10 rev/min (d) 30 rev/min

Ans. (d)Sol. Cycle time = 2s

So, basically, the mechanism will rotate onefull relocation is 2s.

speed =60s = 30 RPM2s

or, 30 rev/s

108. The following data relate to a single-cylinderreciprocating engine : mass of reciprocatingparts = 40 kg, mass of revolving parts = 30kg at crank radius, speed = 150 rpm, stroke= 350 mm. If 60% of the reciprocating partsand all the revolving parts are to be balanced,what is the balance mass required at a radiusof 320 mm?

(a) 15.27 kg (b) 21.43 kg

(c) 24.96 kg (d) 29.53 kg

Ans. (d)Sol.

mrev

msec

Stroke = 2 × crank radius (r) = 350 mm Massto be balanced (M) = Revolving mass + 60%of reciprocating mass.

= (30 + 0.6 × 40) kg= (30 + 24) kg= 54 kg

Now,2 2

Mass to be balanced Balancedpart mass

Mw r = mw ×R

0.3554×

2

= (m × 0.32)

or, m =54×0.35 kg0.32× 2

= 29.53 kg

https://iesmaster.org/staff-selection-commission





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109. A leaf spring consists of seven steel plates,each 60 mm wide and 6 mm thick. What isthe length of the spring if it is to carry a centralload of 3 kN, without the stress exceeding150 MPa?

(a) 547 mm (b) 498 mm

(c) 494 mm (d) 504 mm

Ans. (d)

Sol. In case of leaf spring, maximum bendingstress is given by:

23 wL=2 nbt

Now, = 150 MPaw = 3 kNn = 7b = 60 mmt = 6 mm

L =2(2 nbt )×

3w

=2 9 6

32×7×60×6 ×10 ×150×10

3×3×10

= 0.504 m= 504 mm

110. According to maximum shear stress theory,with comparison to yield strength in tension,the yield strength in shear is

(a) half the yield strength in tension

(b) same as that of yield strength in tension

(c) double the yield strength in tension

(d) 1.33 times that of the yield strength in tension

Ans. (a)Sol. According to maximum shear stress theory,

1 2– Syt

But in pure shear state of stress,

1 = and 2 = –

– – = Syt

2 = Syt

When yielding occurs in shear under pure shear

state of stress, ys= S

ys

yt

SS =

12

111. In curved beams, normally the nature of stressdistribution is

(a) linear (b) circular

(c) parabolic (d) hyperbolic

Ans. (d)

Sol. In curved beams, normally the nature of stressdistribution is hyperbolic.

112. Consider the following statements regardingcrack:

1. Crack is more likely to occur in the regions ofdiscontinuity such as oil holes.

2. Crack is more likely to occur in the regions ofirregularities in machining operations such asstamp mark.

3. Crack is more likely to occur in the internalcracks due to defects in materials like blowholes.



(c) 1 and 3 only (d) 1, 2 and 3

Ans. (d)Sol. The crank is more likely to occur in the following

regions:(i) Regions of discontinuity, such as oil holes,

keyways, screw threads etc.(ii) Region of irregularities in machining operation

such as scratches on the surface stamp mark,inspection mark etc.

(iii) Internal crack due to defects in materialslike blow holes.

113. The yield strength of bolt material is 300 MPaand factor of safety is 2.5. What is themaximum principal stress using maximumprincipal stress theory?

(a) 750 MPa (b) 120 MPa





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(c) 27.38 MPa (d) 10.95 MPa

Ans. (b)

Sol. Given: Yield strength, yt = 300 MPa

Factor of safety, FoS = 2.5Using maximum principal stress theory,

Maximum principal stress, 1 = yt

N

=3002.5 = 120 MPa

114. Which one of the following theories givessatisfactory results for brittle materials?

(a) Maximum principal stress theory

(b) Maximum shear stress theory

(c) Distortion energy theory

(d) Shear stress energy theory

Ans. (a)

115. A cast steel bar having an ultimate strengthof 120 MPa is subjected to a reversed,repeated, bending load. The bar will bemachined to a rectangular cross-section, 150mm wide × 200 mm high. What is theequivalent diameter?

(a) 14 mm (b) 30 mm

(c) 140 mm (d) 300 mm

Ans. (c)Sol. The ‘effective’ diameter of any non-circular cross-

section is the given by,

de = 95A0.0766

...(i)

where,A95 = portion of cross-sectional area of the non-cylindrical part that is stressed between 95%and 100% of the maximum stressde = effective diameter of the non-cylindricalpartFor a rectangular cross-section having width band depth h,

A95 = 0.05 bh ...(ii)

from equation (i) and (ii),

de =0.05bh = 0.808 bh0.0766

= 0.0808 150×200 = 140 mm

116. Consider the following statements regardingtypical analysis of bolt failure:

1. 15% failure of bolt occur at the fillet underthe head.

2. 50% failure of bolt occur at the end of threadson the shank.

3. 80% failure of bolt occur in the threads thatare in contact with the nut.



(c) 2 and 3 only (d) 1, 2 and 3

Ans. (a)

Sol.

A typical analysis of bolt failures indicate that

15% failures of bolt occur at the fillet underthe head

20% failures of bolt occur at the end ofthreads on the shank.

65% failures of bolt occur in the threads thatare in contact with the nut.

117. Which of the following are the functions oflubrication in a bearing unit?

1. To protect the bearing components from cor-rosion.

2. To absorb heat from the bearing unit.

3. To carry heat away from the bearing unit.



(c) 2 and 3 only (d) 1, 2 and 3

Ans. (d)

Sol. The lubricants are used in bearings to reducefriction between the rubbing surfaces and tocarry away the heat generated by friction. It





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also protects the bearing against corrosion.

118. Consider the following statements regardingclutches:

1. Dry clutch has higher coefficient of frictioncompared to wet clutch.

2. The torque capacity of wet clutch is high com-pared to dry clutch.

3. The engagement in a dry clutch is smootherthan in case of wet clutch.



(c) 2 only (d) 1, 2 and 3

Ans. (a)

Sol. The difference between dry and wet clutchesis as follows:

(i) A dry clutch has higher coefficient of friction.In wet clutches, the coefficient of friction isreduced due to oil. The coefficient of frictionfor dry operation is 0.3 or more, while it is0.1 or less for wet operation.

(ii) The torque capacity of dry clutch is highcompared with the torque capacity of wetclutch of the same dimensions.

(iii) For dry clutch, it is necessary to preventcontamination due to moisture or nearbylubricated machinery, by providing seals Sucha problem is not serious in wet clutches.

(iv) Heat dissipation is more difficult in dryclutches. In wet clutches, the lubricating oilcarries away the frictional heat.

(v)Rate of wear is far less in wet clutchescompared to dry clutches. The wear rate inwet clutches is about 1% of the rate expectedin dry clutches.

(vi) The engagement in wet clutch is smootherthan in the case of dry clutch.

(vii) In wet clutches, the clutch facings aregrooved to provide for passage of lubricant.This reduces the net face area for transmittingtorque.

119. Which of the following factors can causemisalignment of the teeth on the pinion relativeto those on the gear?

1. Inaccurate gear teeth

2. Misalignment of the axes of shafts carryinggears

3. Thermal distortions during operation



(c) 2 and 3 only (d) 1, 2 and 3

119.

Ans. (d)

120. Consider the following statements regardingwelded and riveted joints:

1. Welded assemblies are tight and leak proofas compared with riveted assemblies.

2. Single-welded V-joint is less reliable thansquare butt joint.

3. Welding results in a thermal distortion of theparts, thereby inducing residual stresses.



(c) 2 only (d) 1, 2 and 3

Ans. (b)

Sol. Welded joints are subjected to residual

stresses due to non-uniform heating of theparts being joined. There is always apossibility that localised thermal stresses mayresult from uneven heating and cooling duringfusion and subsequent cooling. This alsoresults in distortion.

Single welded V-joint is more reliable thansquare butt joint. Single welded V-joint withbacking strip is more reliable than singlewelded V-joint without backing strip. Doublewelded V-joint is more reliable than singlewelded V-joint with backing strip. The costalso increases with the reliability of the joint.

121. Which one of the following lubricants is usedin forward hot extrusion of steel?

(a) Molten glass (b) Soap solution

(c) Copper sulphate (d) Vegetable oil





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Ans. (a)Sol. Hot extrusion involves prior heating of the billet

to a temperature above its recrystallizationtemperature, lubrication is critical hot extrusionfor certain metal’s (eg. steels) and speciallubricants have been developed that are effectiveunder harsh conditions in hot extrusion. For hotextrusion glass is an excellent lubricants withsteels, stainless steels and high temperaturesmetals and alloys.

122. Which one of the following statements is notthe correct statement regarding operatingcharacteristics (OC) curve for sampling plan?

(a) It shows ability to distinguish between goodand bad lots.

(b) No sampling plans can give complete protec-tion against acceptance of defectives.

(c) Larger the sample size, steeper is the slopeof the curve

(d) Acceptance number is zero for ideal samplingplan.

Ans. (d)

Sol. Operating curve shows ability to distinguish

between good and bad lots.

No sampling plans can give completeprotection against acceptance of defectives.

Larger the sample size, sleeper is the slopeof the curve.

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

1

2 4 6 80

Percent defective

N = 50

N = 100

N = 250

N = 1000

(Pro

babi

lity

of a

ccep

tanc

e)

Acceptance number need not be zero for idealsampling plan.

123. Which one of the following statements is not

correct for forward or direct extrusion process?

(a) High friction forces must be overcome.

(b) High extrusion forces are required butmechanically simple and uncomplicated.

(c) Low scrap or material waste only 5 – 6% ofbillet weight.

(d) Simple, but the material must slide along thechamber wall.

Ans. (c)Sol.

• A ram drives the entire billet to and througha stationary die and must provide additionalpower to overcome the frictional resistancebetween the surface of the moving billet andthe confining chamber, Hence material mustslide along the chamber wall.

• Huge amount of frictional energy aredeveloped due to which more amount of forceare required for extruction.

Billet

Extruded shape

Die

Ramfrictional force

124. Consider the following statements regardingdefects in forgings:

1. Flakes are internal breaks or ruptures occur-ring in some grades of alloy steel.

2. Die shift is caused by misalignment betweenthe top and bottom forging dies.

3. Fins and rags are small projections or loosemetal driven into the surface of the forging.



(c) 2 and 3 only (d) 1, 2 and 3

Ans. (d)Sol.

• Flakes are basically internal ruptures causedby the improper cooling of the large forging.





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• Die shift is caused by the misalignment ofthe two die halves, making the two halves ofthe forging to be of improper shape.

• Fins and rags are small projection of loosemetal into the forging surface. The possiblecause are improper workmanship, poor diedesign etc.

125. Consider the following statements regardingdesirable properties of cutting fluid:

1. It should get oxidised when left in air.

2. It should react with the materials of machinetool parts.

3. It should wet the surface of cutting tool andworkpiece.



(c) 2 and 3 only (d) 1, 2 and 3

Ans. (b)Sol.

Cutting fluids are used for decreasing powerrequirement and increasing heat dissipation. Thedesirable properties of cutting fluid are• It should not get oxidised when left in air• It should not react with the material of

machine tools parts• Easily available• It should not foam• Low viscosity• It should wet the surface of cutting tool and

workpiece for better heat dissipation.

126. Consider the following statements regardinglimits and fits:

1. Actual size is the standard size of the partand is the same both for the hole and its shaft.

2. Basic size is the dimension as measured onthe manufactured part.

3. Deviation is the algebraic difference betweena size and the corresponding basic size.



(c) 2 only (d) 1, 2 and 3

Ans. (a)Sol.

• Actual size is the actual measured dimensionof the part not the standard size of the partand if is not the same fg both hole and shaft.Hence wrong.

• Basic size is the size of a part to which alllimits of variation are applied. Basic dimensionis theoretical dimension. Hence wrong.

• Deviation is the algebraic difference betweena size (actual max etc) and the correspondingbasic size. Hence correct.

127. Which of the following statements is notcorrect about PERT?

(a) Network is constructed based on the events.

(b) It does not take uncertainties involved in theestimation of times.

(c) Network deals with uncertainities and hencethree time estimations are considered.

(d) As there is no certainty of time, activityduration cannot be reduced.

Ans. (b)

Sol. In case of PERT, three time estimates, optimis-tic time (to), pessimistic time (tp) and mostlikely time (tm) are considered to calculatethe average or expected time.

o m PE

t + 4 t + tt =

6

128. Parallel misalignment is present when

(a) two shafts are parallel to each other but arenot in the same plane.

(b) two shafts are parallel to each other and arein the same plane.

(c) the shafts are not parallel to each other.

(d) the shafts are aligned with each other.

Ans. (b)Sol.

Parallel misalignment is present when the centrelines of both shafts are parallel but they areoffset in same plane.





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129. The major limitation with displacement orproximity probes is

(a) Size (b) Time

(c) Accuracy (d) Cost

Ans. (a)Sol. The sensing range of a proximity probe depends

on the type of metal being detected, its shape,its size used in design. Due to above reasonproximity probes has distance (size) limitations.

130. Which one of the following contains designdata on all products, e.g., their constituentcomponent and parts?

(a) Engineering data master file

(b) Process data master file

(c) Inventory master file

(d) Sales master file

Ans. (a)

131. Which one of the following is the cutter witha curved tooth outline of the same shape asthe profile of the workpiece?

(a) Plain milling cutter

(b) Face milling cutter

(c) End milling cutter

(d) Profile milling cutter

Ans. (d)Sol.

• Plain milling, Face milling and End millingcutter are used where the geometry of themachined surface are not related with thetool shape

• Form or profile milling cutter are used wherethe job profile becomes the replica of thetool form.

132. Which one of the following is a joining processthat may employ acetylene, natural gas,

butane in combination with oxygen to supplythe heat required to melt the filler rod anddiffuse it into the surface of the base metal?

(a) Furnace brazing (b) Torch brazing

(c) Induction brazing (d) Dip brazing

Ans. (b)Sol. Heat source that are used for brazing are a

molten bath of brazing filler metal, oxy-acetylenetorch, controlled atmosphere furnace, electricalresistance heating and inductor heating.• In torch brazing a reducing flame (oxy-

acetylene) is used.• In dip brazing a molten salt bath is used to

supply heat to the base metal.• In induction heating heat is obtained from the

high-frequency current induced in the work.• In furnace brazing a furnace with a controlled

atmosphere is used.

133. Consider the following statements regardingmodulation:

1. The modulation is essential in communica-tion systems, where a weak signal is trans-mitted by the use of a carrier signal.

2. When the frequency of the high frequencysignal is varied in accordance with the inten-sity of the low-frequency weak signal, themodulation is said to be frequency modula-tion.

3. The process of recovering original basebandsignal from the modulated signal is calledphase modulation.



(c) 2 and 3 only (d) 1, 2 and 3

Ans. (b)Sol. Modulation is a process of changing the

parameters of a high frequency signal called acarrier signal, with respect to the intensity of agiven weak signal called an original basebandsignal or modulating signal. The high frequencysignal is usually a sinusoidal signal. Theparameters are simply the amplitude, frequency,and phase. Modulation is essential incommunication systems, where a weak signal is





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transmitted by the use of a carrier signal. Thecarrier signals are usually sinusoidal in nature.There are many forms of modulation such asAmplitude Modulation (AM), FrequencyModulation (FM), and Phase Modulation (PM).When a high frequency signal has amplitudevaried in response to the intensity of a low-frequency weak signal, the modulation is calledAM. When the frequency is varied in accordancewith the intensity of the weak modulating signal,the modulation is referred to as FM. A similardefinition can also be given for PM. The processof recovery of the original baseband signal fromthe modulated signal is called de-modulation.Phase modulation is a type of modulationintended for transmitting communication signals.It changes message signal in accordance withthe carrier signal due to differences in theimmediate phase.

134. Consider the situation where a microprocessorgives an output of an 8-bit word. This is fedthrough an 8-bit digital-to-analog converter toa control valve. The control valve requires6.0 V to be fully open. If the fully open stateis indicated by 11111111, then what is theoutput to the valve for a change of 1 bit?

(a) 0.033 V (b) 0.053 V

(c) 0.043 V (d) 0.023 V

Ans. (d)Sol. The full scale output voltage of 6V will be divided

into 28 intervals. A change of 1 bit is thus achange in the output voltage of

= 86 = 0.023 V2

135. W hich one of the fol lowing is not anapplication of Hall effect sensor?

(a) Magnetic switch for electric transducer

(b) Measurement of current

(c) Measurement of acceleration

(d) Measurement of power

Ans. (c)Sol. Measurement of acceleration

Application of Hall effect sensor

1. Hall sensors are used for proximity switching,positioning, speed detection and currentsensing application.

2. They are used in brushless DC motors todetect the position of the permanent magnet

3. It is used as a magnetic switch for electrictransducer.

4. It is used for the measurement of the position,displacement and proximity.

5. It is used for measurement of current.6. It is used for measurement of power.

136. Consider the following statements regardingmechatronics systems:

1. The anti-lock brakes on a car are a simpleexample of a real time computing system.

2. The completion of an operation after its dead-line is considered useless in soft real timesystem.

3. The hard real time system tolerates latenessand may respond with decreased servicequality.



(c) 3 only (d) 1, 2 and 3

Ans. (a)Sol. Real time control and computing is the study of

hardware and software systems which are subjectto a “real time constraint”.A system is said to be a real time system ifthere is total correctness of an operation. Thisdepends upon not only the operation’s logicalcorrectness, but also the time in which it isperformed and used. The antilock brakes on acar are a simple example of a real-timecomputing system and the real time constraintin the system is the short time in which thebrakes must be released to prevent the wheelfrom locking.Two types of real time system:1. Hard real time system: In this completion of

an operation after its deadline is considereduseless.

2. Soft real time system: It tolerates suchlateness and may respond with decreased





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service quality.

137. Which one of the following materials has leastpiezoelectric charge sensitivity?

(a) Quartz (b) Barium Titanate

(c) PZT (d) PVDF

Ans. (a)Sol. Piezoelectric charge sensitivity of following

materials:

Quartz - 2.2 pC/NBarium Titanate - 130 pC/NLead zirconate Titanate (PZT) - 265 pC/NPVDF - 3.10 pC/N

138. The ideal hydraulic rotary actuator providesshaft torque, T, which is

(a) equal to displaced volume measured.

(b) inversely proportional to the displaced volumemeasured.

(c) proportional to the differential pressure.

(d) inversely proportional to the differential pres-sure.

Ans. (c)

Sol. Proportional to the differential pressure ro-tary actuators are the hydraulic or pneumaticequivalent of electric motors which are usedwhen a twisting or turning motion is required.A rotary actuator based on the principle of agear motor. Fluid enters at the top with highpressure, applying force on the gear facesresulting in rotation.The design and construction of a vane motoris similar to a vane pump. At the entry, avane has high pressure on one side, whereason the other side the pressure will be verylow due to high pressure of f luid. Thisdifference in pressure exerted on the vanewill produce a torque that would result inrotation of vanes.

139. An ammeter requires a change of 3 A in itscoil to produce a change in deflection of thepointer by 12 mm. W hat is the stat icsensitivity?

(a) 36 mm/A (b) 9 mm/A

(c) 4 mm/A (d) 15 mm/A

Ans. (c)Sol. The sensitivity is the relationship indicating how

much output you get per unit input, i.e. output/input

Sensitivity = Change in outputChange in input

=12 = 4mm / A3

140. What is the force needed to apply to a pistonof 2 cm radius in order to result a force of6000 N at the working piston of radius 6 cm?

(a) 1334 N (b) 333 N

(c) 1050 N (d) 667 N

Ans. (d)Sol. Pressure will be same

1

1

FA =

2

2

FA

1

2

F

24 =

26000

64

F1 = 666.67 N

141. The fol lowing symbol in the ladder logicrepresents :

(a) Normally open contacts

(b) Normally closed contacts

(c) Outputs loads

(d) Special instruction

Ans. (c)

142. The settling time for a unit step response ofa second-order system is

(a) proportional to the natural frequency

(b) inversely proportional to the natural frequency

(c) equal to the damping ratio

(d) proportional to the damping ratio

Ans. (b)





IES MASTE

R

Sol. In second order system, the setting time is usedas a measure of the time taken for the oscillationsto die away.

ts n

1

Here n is natural frequency

is damping ratio

143. A typical wrist mechanism with three rotationaljoints would be indicated by

(a) TRL (b) TRT

(c) LLL (d) TRR

Ans. (d)

144. The Analog-to-Digital conversion processinvolves:

1. Quantizing

2. Sampling

3. Encoding

What is the correct sequence?

(a) 2-1-3 (b) 2-3-1

(c) 3-1-2 (d) 3-2-1

Ans. (a)

Directions: Each of the next six (06) items consistsof two statements, one labelled as ‘Statement (I)’and the other labelled as ‘Statement (II)’. You areto examine these two statements carefully and selectthe answers to these items using the codes givenbelow:

Codes :

(a) Both statement (I) and Statement (II) areindividually true and Statement (II) is thecorrect explanation of Statement (I).

(b) Both statement (I) and Statement (II) areindividually true and Statement (II) is not thecorrect explanation of Statement (I).

(c) Statement (I) is true, but Statement (II) is false.

(d) Statement (I) is false, but Statement (II) istrue.

145. Statement (I): Gravity is the driving forcebehind flows through open channels.

Statement (II): Gravity stands to reason thatthe ratio of inertial to gravitational forces willplay a major role in open channel flowanalysis.

Ans. (b)

146. Statement (I ): The v iscosity of l iquidsdecreases with the increase of temperaturewhile the viscosity of gases increases withthe increase of temperature.

Statement (II): The viscous forces in a fluidare due to cohesive forces and molecularmomentum transfer.

Ans. (a)

Sol.

147. Statement (I): If two systems are in thermalequilibrium with a third system, then they arenot in thermal equilibrium with each other.

Statement (II): Zeroth law of thermodynamicsis the basis for temperature.

Ans. (d)

Sol. When a body A is in the thermal equilibriumwith a body B, and separately with body C,then bodies B and C will be in thermal equi-librium with each other. This is known aszeroth law of thermodynamics. It is the basisof the temperature measurement.

148. Statement (I): A thermal energy reservoir is asystem that always remains at constanttemperature even though the heat is added toor removed from it.

Statement (II): A thermal reservoir thatsupplies heat energy is called sink and onethat absorbs the heat energy is called source.

Ans. (c)Sol. A thermal reservoir that supplies energy is termed

as source and one that absorbs the heat energyis called sink.





IES MASTE

R

149. Statement (I): Wear is an issue whenever twocomponents operate with relative motionbetween them or when l iquids or solidsimpinge on a surface at high velocity.

Statement (II): Wear is often cumulative andcan ev entual ly render the componentsincapable of del iv er ing thei r expectedperformance.

Ans. (b)Sol. Wear is the undesirable removal of solids from

a sliding or rolling component. Wear is generallyproportional to the applied load and the amountof sliding erosive wear is the cutting of furrowson a surface by hard particles contained in afluid travelling at high velocity.It is often cumulative and leads to poorperformance of machining parts.

150. Statement (I): Increased productivity, reducedcost of labour and improved quality can beachieved by automation.

Statement (II): Due to automation in processinventory, dependence on operator skills maybe increased.

Ans. (b)

Sol. Increased productivity, reduced cost of labour,

improved qualities can be achieved byautomation.

Due to automation is process inventory,dependence on operator skil ls may beincreased.

mechanical engineering...aux = ? overall = g m boiler ism aux aux = 0.33 0.97 0.95 0.92 0.42 92.68%...

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