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MECHANICAL ENGINEERING

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Detailed SolutionSET - D

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1. The time taken to face a workpiece of 80 mmdiameter for the spindle speed of 90 rpm andcross feed of 0.3 mm/rev will be

(a) 4.12 min (b) 3.24 min

(c) 2.36 min (d) 1.48 min

Ans. (d)Sol. Given diameter of workpiece, d = 80 mm

Spindle speed, N = 90 rpm

Cross feed, f = 0.3 mm/rev.

Time taken to face the workpiece,

Tm = LfN

80L 40 mm2

L = —d2

Tool

Then, Tm = 40

90 0.3

Tm = 1.48 min

2. A feed f for the lathe operation is

(a)m

N mm / revL T (b)

m

L mm / revN T

(c) mTmm / rev

N L(d) mT L

mm / revN

where: Tm = Machining time in min.

N = Speed in rpm

L = Length of cut in mm

Ans. (b)

Sol. We know, machining time, Tm = LfN

where, L = length of cut

f = feed

N = rpm

m

LfN T

3. The main advantage of the radial drillingmachine is that

(a) It is very compatible and handy formachining

(b) It is accurate, economical, portable andleast time consuming while machining

(c) Heavy workpieces can be machined in anyposition without moving them

(d) Small workpieces can be machined and itcan be used for mass production as well

Ans. (c)Sol. Advantages of radial drilling:

1. They are generally powerful and can turn alarge diameter drill.

2. They allow a fixed positioning of theworkpiece, and the drill can be easily movedto the required position.

So, the correct answer is option (c).

4. For the purpose of sampling inspection, themaximum percent defective that can beconsidered satisfactory as a process average is

(a) Rejectable Quality Level (RQL)

(b) Acceptable Quality Level (AQL)

(c) Average Outgoing Quality Limit (AOQL)

(d) Lot Tolerance Percent Defective (LTPD)

Ans. (d)Sol. Lot tolerance percent defective (LTPD):

Lot tolerance percent defective (LTPD),expressed in percent defective, is the poorestquality in an individual lot that should beaccepted. The LTPD has a low probability ofacceptance.

5. Hard automation is also called

(a) Selective automation

(b) Total automation

(c) Group technology

(d) Fixed position automation

Ans. (d)Sol. Fixed position Automation is also known as Hard

Automation.

It refers to the use of special purpose equipmentto automate a fixed sequence of processing orassemble operations.

6. The method of CNC programming whichenables the programmer to describe partgeometry using variables is





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(a) Computer assisted part programming

(b) Computer aided drafting programming

(c) Conversational programming

(d) Parametric programming

Ans. (d)Sol. Parametric programming:

It is an enhancement to the methods like manualpart programming, computer assisted partprogramming & conversational programming.

It enables the programmer to describe partgeometry using variables.

Once described, entering specific values for thevariables that unique identify the part generatesan actual tool path CNC program.

7. Revolving joint of the robot is referred to as

(a) L joint (b) O joint

(c) T joint (d) V joint

Ans. (d)Sol. Revolving joint: Revolving joint is generally

known as V-joint. Here the output link axis isperpendicular to the rotational axis and theinput link is parallel to the rotational axis.

8. Repairing of a machine consists of 5 steps thatmust be performed sequentially. Time taken toperform each of the 5 steps is found to have anexponential distribution with a mean of 5minutes and is independent of other steps. Ifthese machines break down in Poisson fashionat an average rate of 2/hour and if there is onlyone repairman, the average idle time for eachmachine that has broken down will be

(a) 120 minutes (b) 110 minutes

(c) 100 minutes (d) 90 minutes

Ans. (c)Sol. Here, number of phases k = 5,

service time per phase = 5 minutes.

Service time per unit = 5 × 5 = 25 minutes.

=1 units / minute

25

=12 units / hour5

= 2 units/hour

Average idle time of the machine = Averagetime spent by the machine in the system.

Ws =k 1 1 12k

=5 1 2 5 1 5

122 5 12 1225

=1 5 5 20 5 hours2 2 12 12 3

= 100 minutes

9. A portion of the total float within which an activitycan be delayed for start without affecting thefloats of preceding activities is called

(a) Safety float (b) Free float

(c) Independent float (d) Interfering float

Ans. (c)Sol. Independent float: It is that portion of the total

float within which an activity can be delayed forstart without affecting the float of the precedingactivities.

10. An oil engine manufacturer purchases lubricantcans at the rate of Rs 42 per piece from avendor. The requirement of these lubricant cansis 1800 per year. If the cost per placement ofan order is Rs. 16 and inventory carryingcharges per rupee per year is 20 paise, theorder quantity per order will be

(a) 91 cans (b) 83 cans

(c) 75 cans (d) 67 cans

Ans. (b)Sol. Given: Demand D = 1800 per year

Ordering cost, Co = 16

Carrying cost Ch = 210 paise/rupee/year

oh

h

2DCEOQ C 0.20 42C

2 1800 16EOQ0.20 42

EOQ 6857.142857

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= 82.8078

EOQ = 83

11. Consider the following data regarding theacceptance sampling process:

N = 10,000, n = 89, c = 2, P = 0.01 andPa = 0.9397

The Average Total Inspection (ATI) will be

(a) 795 (b) 687

(c) 595 (d) 487

Ans. (b)Sol. ATI = n + (1 – Pa) (N – n)

= 89 + (1 – 0.9397) × (10000 – 89)

= 686.633

687

12. The Non-Destructive Inspection (NDI) techniqueemployed during inspection for castings of tubesand pipes to check the overall strength of acasting in resistance to bursting under hydraulicpressure is

(a) Radiographic inspection

(b) Magnetic particle inspection

(c) Fluorescent penetrant

(d) Pressure testing

Ans. (d)Sol. Proof test is a kind of pressure test employed

for pipes in resistance to bursting underhydraulic pressure.

In it hydrostatic pressure shall be applied to theassembly. The actual test pressure prior torupture must be at least equal to the computedproof test pressure defined below.

P =2stD

P = Proof test pressure

S = Actual tensile strength of run pipe

t = Nominal run pipe wall thickness

D = specified outside diameter of the run pipe

Test is considered successful if the pipewithstand rupture. and test pressure 150% isapplied.

13. Consider the situation where a microprocessorgives an output of an 8-bit word. This is fedthrough an 8-bit digital-to-analogue converterto a control valve. The control valve requires6.0 V being fully open. If the fully open state isindicated by 11111111, the output to the valvefor a change of 1-bit will be

(a) 0.061 V (b) 0.042 V

(c) 0.023 V (d) 0.014 V

Ans. (c)Sol. The full-scale output voltage of 6.0 V will be

divided into 28 intervals. A change of 1 bit isthus a change in the output voltage of

86.02

= 0.023 V..

14. Which of the following factors are to beconsidered while selecting a microcontroller?

1. Memory requirements

2. Processing speed required

3. Number of input/output pins

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

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

Ans. (d)Sol. In selecting a microcontroller the following

factors need to be considered:

(1) Number of input/output pins:How many input-output pins are going tobe needed for the task concerned?

(2) Interfaces required:What interfaces are going to be required?For example, is PWM required? Manymicrocontrollers have PWM outputs, e.g.the PIC17C42 has two.

(3) Memory requirements:What size memory is required for the task?

(4) The number of interrupts required:How many events will need interrupts?

(5) Processing speed required:The microprocessor takes time to executeinstructions, this time being determined bythe processor clock.

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15. Which of the following statements regardinginterface circuit are correct?

1. Electrical buffering is needed when theperipheral operates at a different voltage orcurrent to that on the microprocessor bussystem or there are different groundreferences.

2. Timing control is needed when the datatransfer rates of the peripheral and themicroprocessor are different.

3. Changing the number of lines is neededwhen the codes used by the peripheralsdiffer from those used by the micro-processor.


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

Ans. (a)Sol. (i) Electrical buffering is needed when the

peripheral operates at a different voltage orcurrent to that on the microprocessor bussystem or there are different groundreferences.

(ii) Timing control is needed when the datatransfer rates of the peripheral and themicroprocessor are different.

(iii) Code conversion is needed when the codesused by the peripherals differ from thoseused by microprocessor.

16. Alternative paths provided by vertical paths fromthe main rung of a ladder diagram, that is, pathsin parallel, represent

(a) Logical AND operations

(b) Logical OR operations

(c) Logical NOT operations

(d) Logical NOR operations

Ans. (b)Sol. Alternative paths provided by vertical paths from

the main rung of a ladder diagram, that is, pathsis parallel, represents logical OR operations.

An example of an OR gate control system is aconveyer Belt transporting bottled products ofpackaging where a deflector plate is activatedto deflect bottles into a reject bin if either theweight is not within certain tolerances or thereis no cap on the bottle.

17. The resolution of an encoder with 10 tracks willbe nearly

(a) 0.15° (b) 0.25°

(c) 0.35° (d) 0.45°

Ans. (c)Sol. The resolution of an encoder with 10 track for

n track

Resolution = n3602

= 10360 0.352

18. Which of the following features is/are relevantto variable reluctance stepper motors?

1. Smaller rotor mass; more responsive

2. Step size is small

3. More sluggish

(a) 1 only (b) 2 only

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

Ans. (a)Sol. A variable reluctance stepper motor has a

ferromagnetic rotor than a permanent magnetmotor. Motion and holding result from theattraction of stator and poles to positions withminimum magnetic reluctance that allow formaximum magnetic flux. A variable reluctancemotor has the advantage of a lower rotor inertiaand therefore a faster dynamic response.

19. Which of the following statements regardinghydraulic pumps are correct?

1. The gear pump consists of two close-meshing gear wheels which rotate inopposite directions.

2. In vane pump, as the rotor rotates, thevanes follow the contours of the casing.

3. This leakage is more in vane pumpcompared to gear pump.

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

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

Ans. (b)Sol. 1. The gear pump consists of two close-

meshing gear wheels which rotate inopposite directions.

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2. In vane pump as the rotor rotates, the vanefollow the contours of the casing.

The results in fluid becoming trappedbetween successive vanes and the casingand transported round from the inlet port tooutlet port.

3. The leakage is less in vane pump than withthe gear pump.

20. The selection of the right controller for theapplication depends on

1. The degree of control required by theapplication.

2. The individual characteristics of the plants.

3. The desirable performance level includingrequired response, steady-state deviationand stability.

Which of the above statements are correct?


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

Ans. (d)Sol. 1) The degree of control required by the

application.

2) The individual characteristics of the plant.

3) The desirable performance level includingresponse, steady-state deviation andstability.

21. Consider a system described by

x Ax Buy Cx Du

The system is completely output controllable ifand only if

(a) The matrix

2 n 1CB CBA CB A CB A D

is of rank n

(b) The matrix

2 n 1CB CAB CA B CA B D

is of rank m

(c) The matrix

2 n 1BC BAC BA C BA C D

is of rank m

(d) The matrix

2 n 1BC ABC CA B CB A D

is of rank n

where:

x = State vector (n-vector)

u = Control vector (r-vector)

y = Output vector (m-vector)

A = n × n matrix

B = n × r matrix

C = m × n matrix

D = m × r matrix

Ans. (b)Sol. Consider the system described by

x Ax Bu …(i)

y Cx Du …(ii)

where x = state vector (n-vector)

u = control vector (r-vector)

y = output vector (m-vector)

A = n × n matrix

B = n × r matrix

C = m × n matrix

D = m × r matrix

The system described by equation (i) and (ii) issaid to be completely output controllable if it ispossible to construct an unconstrained controlvector u(t) that will transfer any given initialoutput y (t0) to any final output y (t1) in a finitetime interval 0 1t t t .

It can be proved that the condition for completeoutput controllability is as follows: The systemdescribed by equations (i) and (ii) is completelyoutput controllable if and only if the m × (n +1)rmatrix

2 n 1CB CAB CA B CA B D

is of rank m.

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22. Which one of the following symbols is used asthe notation for designating arm and body of arobot with jointed arm configuration?

(a) TRL (b) TLL, LTL, LVL

(c) LLL (d) TRR, VVR

Ans. (d)Sol.

Robot configuration Symbol(Arm & Body)

Polar configuration TRLCylindrical configuration TLL, LTL, LVLCartesian configuration LLLJoint ed arm configuration TRR, VVR

23. A compliant motion control of robots can beunderstood by the problem of controlling of

(a) Position and velocity of joints

(b) Position and acceleration of the end-effector

(c) Manipulator motion and its force interactionswith the environment.

(d) Joint velocities of given end-effector velocity.

Ans. (c)Sol. The role of a compliant motion scheme is to

control a robot manipulation in contract with itsenvironment. By accommodating with theinteraction force.

24. For the vector v = 25 i + 10 j + 20 k, performa translation by a distance of 8 in the x-direction,5 in the y-direction and 0 in the z-direction. Thetranslated vector Hv will be

(a)

1203315

(b)

3315201

(c)

15331

20

(d)

1152033

Ans. (b)

Sol. Given vector, V = ˆ ˆ ˆ25i 10j 20k

Translation in x direction = 8

Translation in y-direction = 5

Translation in z-direction = 0

Translation vector =

1 0 0 8 250 1 0 5 100 0 1 0 200 0 0 1 1

=

3315201

Directions: Each of the next six (06) items consists oftwo statements, one labelled as ‘Statement (I)’ and theother labelled as ‘Statement (II)’. You are to examinethese two statements carefully and select the answersto these items using the codes given below:

Codes:(a) Both Statement (I) and Statement (II) are

individually true and Statement (II) is thecorrect explanation of Statement (I).

(b) Both Statement (I) and Statement (II) areindividually true, but Statement (II) is notthe correct explanation of Statement (I).

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

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

25. Statement (I) : The greater the chemical affinityof two metals, the more restricted is their solidsolubility and greater is the tendency offormation of compound.

Statement (II) : Wider the separation ofelements in the periodic table, greater is theirchemical affinity.

Ans. (a)Sol. The greater the chemical affinity of two metals

the more restricted is their solid solubility andthe greater is the tendency towards compoundformation.

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26. Statement (I) : The size of a memory unit isspecified in terms of the number of storagelocations available; 1 K is 210 = 1024 locationsand thus a 4 K memory has 4096 locations.

Statement (II) : Erasable and programmableROM (EPROM) is a form of memory unit usedfor ROMs that can be programmed and theircontents altered.

Ans. (b)Sol. The size of the memory is determined by the

number of wires in the address bus. Thememory elements in a unit consistessentially of large numbers of storage cellswith each cell capable of storing either a 0or a 1 bit. The storage cells are grouped inlocations with each location capable ofstoring one word. In order to access thestored word, each location is identified bya unique address. Thus with a 4-bit addressbus we can have 16 different addresseswith each, perhaps, capable of storing 1byte, i.e. a group of 8 bits.

The size of a memory unit is specified interms of the number of storage locationsavailable; 1 K is 210 = 1024 locations andthus a 4 K memory has 4096 locations.

The term erasable and programmable ROM(EPROM) is used for ROMs that can beprogrammed and their contents altered.

27. Statement (I) : Microprocessors which havememory and various input/output arrangements,all on the same chip, are called microcontrollers.

Statement (II) : The microcontroller is theintegration of a microprocessor with RAM, ROM,EPROM, EEPROM and I/O interfaces, andother peripherals such as timers, on a singlechip.

Ans. (a)Sol. Systems using microprocessors basically have

three parts: a central processing unit (CPU) torecognize and carry out program instructions(this is the part which uses the microprocessor),input and output interfaces to handlecommunications between the microprocessorand the outside world (the term port is used forthe interface), and memory to hold the programinstruction and data. Microprocessors whichhave memory and various input/output

arrangements all on the same chip are calledmicrocontrollers.

The microcontroller is the integration of amicroprocessor with memory and input/outputinterfaces, and other peripherals such as timers,on a single chip.

28. Statement (I) : Capacitive proximity sensor canonly be used for the detection of metal objectsand is best with ferrous metals.

Statement (II) : One form of capacitive proximitysensor consists of a single capacitor plate probewith the other plate being formed by the object,which has to be metallic and earthed.

Ans. (d)Sol. One form of capacitive proximity sensor consists

of a single capacitor metallic and earthed figure.As the object approaches so the ‘plateseparation’ of the capacitor changes, becomingsignificant and detectable when the object isclose to the probe.

Magnetic field proximity sensors are relativelysimple and can be made using a permanentmagnet. The magnet can be made a part ofobject being detected or can be part of thesensor device. It can only be used for thedetection of metal object and is best with ferrousmetal.

29. Statement (I) : SCARA configuration providessubstantial rigidity for the robot in the verticaldirection, but compliance in the horizontal plane.

Statement (II) : A special version of theCartesian coordinate robot is the SCARA, whichhas a very high lift capacity as it is designed forhigh rigidity.

Ans. (c)Sol. A special version of the jointed arm robot is the

SCARA stands for selective complianceassembly robot arm, and this configurationprovides substantial rigidity of the robot in thevertical direction, but compliance in thehorizontal plane. This makes it ideal for manyassembly tasks.

30. Statement (I) : The stepper motor is a devicethat produces rotation through equal angles,the so-called steps, for each digital pulsesupplied to its input.





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Statement (II) : Stepper motors can be used togive controlled rotational steps but cannot givecontinuous rotation, as a result their applicationsare limited to step angles only.

Ans. (c)Sol. The stepper motor is a device that produces

rotation through equal angles, the so-calledsteps, for each digital pulse supplied to its input.

Stepper motors can be used to give controlledrotational steps but also can give continuousrotation with their rotational speed controlledby controlling the rate at which pulses areapplied to it to cause stepping. This gives avery useful controlled variable speed motorwhich finds many applications.

31. A stone weighs 400 N in air and when immersedin water it weighs 225 N. If the specific weightof water is 9810 N/m3, the relative density ofthe stone will be nearly

(a) 5.9 (b) 4.7

(c) 3.5 (d) 2.3

Ans. (d)

Sol. Let, Density of stone is and volume V..

Case 1: When stone is weighted in air.

Weight, W = agV gV

agV is buoyancy force due to air..

So, 400 = agV gV

[ Density of air, 3a 1.18 kg/m ]

a =400gV ...(i)

Case 2: When stone is weighted in water

225 = wgV gV

wgV is buoyancy force due to water..

So, w =225gV ...(ii)

Divide both equations,

a

w

=

400225

1.181000

=400225

3w

9810 1000 kg/m9.81

= 2284

Relative density,

=2284 2.31000

32. A flat plate 0.1 m2 area is pulled at 30 cm/srelative to another plate located at a distanceof 0.01 cm from it, the fluid separating thembeing water with viscosity of 0.001 Ns/m2. Thepower required to maintain velocity will be

(a) 0.05 W (b) 0.07 W

(c) 0.09 W (d) 0.11 W

Ans. (c)

Sol. Shear stress, dv vdy y

Power required to maintain flow,

P = vA

=v A vy

=2vA

y

Given: = 0.001 N-s/m2

y = 0.01 cm

v = 30 cm/s

A = 0.1 m2

So, the putting the respective values,

P = 22 10300.001 0.10.01

P 0.09 W

The correct answer is (c).

33. When the pressure of liquid is increased from3 MN/m2 to 6 MN/m2, its volume is decreasedby 0.1%. The bulk modulus of elasticity of theliquid will be

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(a) 3 × 1012 N/m2 (b) 3 × 109 N/m2

(c) 3 × 108 N/m2 (d) 3 × 104 N/m2

Ans. (b)Sol.

Bulk modulus of elasticity,

k =P VV

Given: P = 6 – 3 = 3 MN/m2

= 3 × 106 N/m2

VV

= –0.1% = 0.1100

= –10–3

so, k =6

33 1010

= 3 × 109 N/m2

The correct answer is (b).

34. A curve that is everywhere tangent to theinstantaneous local velocity vector, is

(a) Streak line (b) Path line

(c) Normal line (d) Streamline

Ans. (d)Sol.

Streamlines are the lines drawn through theflow field in such a manner that the velocityvector of the field at each and every point onthe streamline is tangent to the streamline atthat instant.

so, A curve that is everywhere tangent to theinstantaneous local velocity vector is‘streamline’.

35. A 120 mm diameter jet of water is dischargingfrom a nozzle into the air at a velocity of 40 m/s. The power in the jet with respect to a datumat the jet will be

(a) 380 kW (b) 360 kW

(c) 340 kW (d) 320 kW

Ans. (b)Sol. Given:

Diameter, d = 120 mm = 0.12 m

Velocity, V = 40 m/s

The power in the jet with respect to a datum atthe jet will be,

P = 21mV2 m aV

= 21 VaV2

= 31 aV2

= 2

31 1000 0.12 402 1000 4

So, P = 361.91 kW

The closest option is (b) i.e. 360 kW

36. Which of the following applications regardingNavier-Stokes equations are correct?

1. Laminar unidirectional f low betweenstationary parallel plates.

2. Laminar unidirectional flow between parallelplates having no relative motion.

3. Laminar flow in circular pipes.

4. Laminar flow between concentric rotatingcylinders

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

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

Ans. (a)Sol.

Navier stokes theorem based on no sliphypothesis.

It can be applied for both steady andunsteady flow.

For flow between two concentric rotatingcylinder if Navier stokes theorem is appliedthen flow must be steady. For it if onescylinders stand still, the solution in the limit

2 1

1

R R0

R

reduces to the Couette flow..

So, in view of the above option 4 is wrong.

37. A crude oi l having a specific gravity of0.9 flows through a pipe of diameter 0.15 m atthe rate of 8 lps. If the value of is0.3 Ns/m2, the Reynolds number will be nearly





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(a) 295 (b) 235

(c) 205 (d) 165

Ans. (c)

Sol. Density of oil, = 0.9 × 1000 = 900 kg/m3

Diameter, d = 0.15 m

Discharge, Q = 8 ps = 8 × 10–3 m3/s

Viscosity, = 0.3 Ns/m2

Reynolds number, Re = V d

V = 24Qd

Re =4 Qd

=34 900 8 10

0.15 0.3

Re 205

According to the given options, closest optionis (c).

38. Two pipes of lengths 2500 m each anddiameters 80 cm and 60 cm respectively, areconnected in parallel. The coefficient of frictionfor each pipe is 0.006 and the total flow is 250litres/s. The rates of flow in the pipes are nearly

(a) 0.17 m3/s and 0.1 m3/s

(b) 0.23 m3/s and 0.1 m3/s

(c) 0.17 m3/s and 0.4 m3/s

(d) 0.23 m3/s and 0.4 m3/s

Ans. (a)

Sol.

Q1 d1=0.8m

Q

Q2 d2=0.6 m

L=2500 m

f1 = f2 = 0.006

Q = 3250 / s 0.25 m /s

Pipes are in parallel so head loss is equal

1fh = 2f

h

21 1 1

51

f L Q12d =

22 2 2

52

f L Q12d

1

2

QQ =

5/21

2

dd

= 2.5

2.50.81.33

0.6

...(i)

Q1 + Q2 = 0.25 ...(ii)

From equation (i) and (ii),

Q1 = 0.17 m3/sec

and Q2 = 0.08 m3/sec 0.1 m3/sec

Method (ii) Shortcut: Pipes are in parallel soflow rate through both pipe i.e. Q = Q1 + Q2check options.

39. A fluid of mass density 1790 kg/m3 and viscosity2.1 Ns/m2 f lows at a velocity of3 m/s in a 6 cm diameter pipe. The head lossover a length of 12 m pipe will be nearly

(a) 62.0 m (b) 54.0 m

(c) 46.5 m (d) 38.5 m

Ans. (d)Sol.

L = 12 m

= 1790 kg/m3

= 2.1 Ns/m2

v = 3 m/s

d = 0.06 m

Head loss = ?

Major head loss is frictional head loss = 2fLV

2gd

hf = 232 VL

gd

[circular pipe]

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= 232 2.1 3 12

1790 9.81 (0.06)

=32 12 21 3 1000

1790 9.81 36

=32 21 10

180

=224 38

6

The closest option is (d).

40. Which of the following characteristics regardingfluid kinematics is/are correct?

1. Streamline represents an imaginary curvein the flow field so that the tangent to thecurve at any point represents the directionof instantaneous velocity at that point.

2. Path lines, streamlines and streak lines areidentical in steady flow.


(c) Both 1 and 2 (d) Neither 1 nor 2

Ans. (c)Sol. 1. Streamline is a line everywhere tangent to

the velocity vector at a given instant

2. Pathline is the actual path traversed by agiven fluid particle.

3. A streakline is the locus of particles thathave earlier passed through a prescribedpoint.

So, the statement 1 is correct.

For steady flow, streamlines, pathlines andstreaklines are identical because

(1) For a steady flow, the velocity vector at anypoint is invariant with time.

(2) The pathline of the particles with differentidentities passing through a point will notdiffer.

(3) The pathline could coincide with oneanother in a single curve which will indicatethe streak line too.

So, both the statements are correct.

41. To maintain 0.08 m3/s flow of petrol with aspecific gravity of 0.7, through a steel pipe of0.3 m diameter and 800 m length, withcoefficient of friction of 0.0025 in the Darcyrelation, the power required will be nearly

(a) 0.6 kW (b) 1.0 kW

(c) 2.6 kW (d) 3.0 kW

Ans. (b)Sol. Given:

Coefficient of friction, f = 0.0025

Discharge, Q = 0.08 m3/s

length, l = 800 m

Diameter, d = 0.3 m

Density, = 0.7 × 1000 = 700 kg/m3

Head loss to friction in pipe, hf = 2(4f ) V

2dgl

Here, remember that friction factor is four timesto that of coefficient of friction.

v = 24Qd

(for pipe, A = 2d

4

)

So, power required, P = fgQh

= 2

2 5gQ (4f ) 16Q

2 d g

l

= 3

2 54f 8 Q

d

l

Putting the respective values,

P = 3

2 5700 4 0.0025 800 8 (0.08)

(0.3) 1000

P 0.956 1kW

42. The diameter of a nozzle d for maximumtransmission of power through it, is

(a)

15 4D

8 fL

(b)

15 2D

8 fL

(c)

15 48D

fL

(d)

15 28D

fL

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where :

D = Diameter of pipe

f = Coefficient of friction

L = Length of pipe

Ans. (a)Sol. Maximum transmission of power, the condition

is given by equation as, fHh3

But hf =24f L V

D 2g

24fLV

D 2g =2H 4fLV or H 3

3 D 2g

But H is also = total head at outlet of nozzle +losses

Equating the two values of H, we get

24fLV3

D 2g

= 2 2 2 2 24fLV 12fLV 4fLV or

2g D 2g D 2g D 2g 2g

or28fLV

D 2g =

2

2g

But from continuity, AV = aa or VA

Substituting this value of V in equation (i), weget

2 2

28fL a

D 2g A

=

2

2g

or 2

28fL a 1D A

2Divide by

2g

...(ii)

or

22

22

d8fL 4 1D

D4

or 4 5

44

8fL d D1 or dD 8fLD

d =1/45D

8fL

43. A piston-cylinder device with air at an initialtemperature of 30°C undergoes an expansion

process for which pressure and volume arerelated as given below:

3

p(kPa) 100 37.9 14.4

V (m ) 0.1 0.2 0.4

The work done by the system for n = 1.4 willbe

(a) 4.8 kJ (b) 6.8 kJ

(c) 8.4 kJ (d) 10.6 kJ

Ans. (d)Sol. Work done in polytropic process is given by,

W = 1 1 2 2P V P Vn 1

Here, n = 1.4 (given)

Using, P1 = 100 kPa, V1 = 0.1 m3

and P3 = 14.4 kPa, V3 = 0.4 m3

So, work done, W = 1 1 3 3P V P Vn 1

=100 0.1 14.4 0.4

1.4 1

W 10.6 kJ

44. A domestic food f reezer maintains atemperature of –15°C. The ambient airtemperature is 30°C. If heat leaks into thefreezer at the continuous rate of 1.75 kJ/s, theleast power necessary to pump this heat outcontinuously will be nearly

(a) 0.1 kW (b) 0.2 kW

(c) 0.3 kW (d) 0.4 kW

Ans. (c)Sol.

T = 303 K2

T = 258 K1

H.RW

W + QR

QR = 1.75 kJ/s

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Least power wil l be required when thisrefrigeration cycle operates on reversed Carnotcycle.

so, (COP)ref. =1

2 1

TT T = RQ

W

258

303 258 =

1.75W

W =1.75 45

258

Power inputs, W 0.30 kW

45. An ideal gas is flowing through an insulatedpipe at the rate of 3 kg/s. There is a 10%pressure drop from an inlet to exit of the pipe.The values of R = 0.287 kJ/kgK and To = 300K. The rate of energy loss for the pressuredrop due to friction, will be nearly

(a) 34 kW (b) 30 kW

(c) 26 kW (d) 22 kW

Ans. (c)Sol. Rate of entropy generation

gen1

PS mRP

= 1

1

0.10P3 0.28 0.0861kw/KP

Rate of exergy loss (Rate of energy loss for thepressure drop due to friction)

0 genT S I

= 300 0.0861 25.83 26 kW

46. A cyclic heat engine operates between a sourcetemperature of 800°C and a sink temperatureof 30°C. The least rate of heat rejection per kWnet output of engine will be nearly

(a) 0.2 kW (b) 0.4 kW

(c) 0.6 kW (d) 0.8 kW

Ans. (b)

Sol. 800°C

HeatEngine

30°C

QH

W

QL

Source

Sink

Reversible heat engine cycle will reject leastrate of heat (QL) per kW net output (W) ofengine.

For reversible condition,

Q T

So,

LQW

=L

H L

T 303T T 770

= 0.393 0.4 kW

47. A fictitious pressure that, if it acted on the pistonduring the entire power stroke, would producethe same amount of net work as that producedduring the actual cycle is called

(a) Quasi equivalent pressure

(b) Mean equivalent pressure

(c) Mean effective pressure

(d) Quasi static pressure

Ans. (c)Sol. Mean effective pressure is defined as a

hypothetical pressure which is thought to beacting on the piston throughout the powerstroke, would produce the same amount ofnet work as that produce during the actualcycle.

48. An ideal cycle based on the concept ofcombination of two heat transfer processes,one at constant volume and the other atconstant pressure, is called

(a) Otto cycle (b) Dual cycle

(c) Diesel cycle (d) Carnot cycle

Ans. (b)





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Sol. In a dual cycle a part of the heat is first suppliedto the system at constant volume and then theremaining part at constant pressure.

49. The ideal thermodynamic cycle for thedevelopment of gas-turbine engine is

(a) Otto (b) Stirling

(c) Ericsson (d) Brayton

Ans. (d)Sol. The Joule or Brayton cycle is the most idealized

cycle for the simple gas turbine engine.

50. If the pressure at exhaust from the turbine isthe saturation pressure corresponding to thetemperature desired in the process heater, sucha turbine is called

(a) Condensing turbine

(b) Extraction turbine

(c) Pass out turbine

(d) Back pressure turbine

Ans. (d)Sol. The pressure at exhaust from the turbine is the

saturation pressure corresponding to thetemperature desired in the process heater, sucha turbine is called back pressure turbine.

51. The purpose of providing fins on heat transfersurface is to increase

(a) Temperature gradient so as to enhance heattransfer by convection

(b) Effective surface area to promote rate ofheat transfer by convection

(c) Turbulence in flow for enhancing heattransfer by convection

(d) Pressure drop of the fluid

Ans. (b)Sol. The purpose of providing fins on heat transfer

surface is to increase effective surface area topromote rate of heat transfer by convection.

52. For fully developed laminar pipe flow, theaverage velocity is

(a) One-half of the maximum velocity

(b) One-third of the maximum velocity

(c) One-fourth of the maximum velocity

(d) Two-third of the maximum velocity

Ans. (a)Sol. For fully developed laminar pipe flow,

ru

umax

u

Velocity distribution:

u = 2 21 PR r4 x

Maximum velocity, umax (r = 0) = 21 P R

4 x

Average velocity, u = 2 maxu1 P R8 2x

The average velocity is one-half of the maximumvelocity.

53. The overall heat transfer coefficient due toconvection and radiation for a steam maintainedat 200°C running in a large room at 30°C is17.95 W/m2K. If the emissivity of the pipesurface is 0.8; the value of = 5.67 × 10–8 W/m2K4; the heat transfer coefficient due toradiation will be nearly

(a) 17 W/m2K (b) 14 W/m2K

(c) 11 W/m2K (d) 8 W/m2K

Ans. (c)Sol. Overall heat transfer coefficient,

h0 = hc + hr

Here, hc is heat transfer coefficient due toradiation.

hr is heat transfer coefficient due to convection.

Now, Q = hr (To – Ti) = 4 4o iT T

hr = 8 4 45.67 10 0.8 (473 303 )(473 303)

= 11 W/m2K

54. Large heat transfer coefficients for vapourcondensation can be achieved by promoting

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(a) Film condensation

(b) Dropwise condensation

(c) Cloud condensation

(d) Dew condensation

Ans. (b)Sol. Due to exposure of large surface area to the

vapours in drop wise condensation heat transfercoefficient is higher (by as much as ten times)than that in film condensation.

55. Which one of the following valves is providedfor starting the engine manually, during coldweather conditions?

(a) Starting jet valve

(b) Compensating jet valve

(c) Choke valve

(d) Auxiliary air valve

Ans. (d)Sol. Auxiliary air valve is provided for cold starting

the engine manually, during cold weatherconditions.

56. A 4-cylinder, 4-stroke single acting petrol engineconsumes 6 kg of fuel per minute at 800 rpmwhen the air-fuel ratio of the mixture suppliedis 9 : 1. The temperature is 650 K and pressureis 12.5 bar at the end of compression stroke.Take R = 300 Nm/kg.K, diameter of cylinder as8 cm, stroke of cylinder as 10 cm. Thecompression ratio will be nearly

(a) 6.2 (b) 5.7

(c) 5.2 (d) 4.6

Ans. (a)Sol. Given:

Mass flow rate of fuel f 6 kg/minm

Note: Practically not possible due to smallsize of cylinder.So it should be in 6 kg/hr

AFR = 9 : 1

Mass flow rate of air am 9 6 54 kJ/hr

Total rate of injection of air and fuel (m) = ma+ mf = 60 kg/hr

P

V

3

2 4

1

T =650 K2P =12.5 bar = 1250 kPaR = 300 Nm/kgK = 0.3 kJ/kgK

2

P2V2 = 2mRT

1250 × V2 = 60 0.3 650

3600

V2 = 2.6 × 10–3 m3/sec

Swept volume, Vs = 1 2ALNKV V60 2

= 2800 40.10.084 60 2

= 0.0134 m3/sec

So V1 = 0.0134 + V2

= 0.016 m3/sec

Compression ratio (r) = 1

2

VV

= 30.016 6.153 6.2

2.6 10

57. Ice is formed at 0°C from water at 20°Ctemperature of the brine is –8°C. Therefrigeration cycle used is perfect reversedCarnot cycle. Latent heat of ice = 335 kJ/kg,and cpw = 4.18. The ice formed per kWh will benearly

(a) 81.4 kg (b) 76.4 kg

(c) 71.8 kg (d) 68.8 kg

Ans. (a)Sol.

Ice at 0°CLatentHeat

Water at 0°C

Water at 20°C





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Qref = 335 × m + 4.18 × (20 – 0)× m

The cycle is working between –8°C and20°C.

So, (COP)ideal = 8 273 265

20 ( 8) 28

COP =refQ

Power Input

26528 =

335 m 4.18 20 m3600

m =265 3600

28 (335 4.18 20)

= 81.39 81.4 kg

58. A Freon 12 simple saturation cycle operates attemperature of 35°C and –15°C for thecondenser and evaporator. If the refrigerationeffect produced by the cycle is 111.5 kJ/kg andthe work required by the compressor is 27.2kJ/kg, the value of COP will be nearly

(a) 4.1 (b) 3.6

(c) 3.1 (d) 2.6

Ans. (a)

Sol. 35°C

Ref

–15°C

QH

W = 27.2 kJ/kg

QL=111.5 kJ/kg

COP = Refrigeration effect produced

Work required

=111.5 4.127.2

59. A cold storage is to be maintained at –5°C whilethe surroundings are at 35°C. The heat leakagefrom the surroundings into the cold storage isestimated to be 29 kW. The actual COP of the

refrigeration plant used is one-third of an idealplant working between the same temperatures.The power required to drive the plant will be

(a) 10 kW (b) 11 kW

(c) 12 kW (d) 13 kW

Ans. (d)Sol.

T = 308 K2

T = 268 K1

H.RPin

QR

(COP)actual = ideal1(COP)3

(COP)ideal for refrigeration = 1

2 1

TT T

R

in

QP =

1 2683 308 268

Pin = 12.98 kW

So, The power required to drive the plant, P =12.98 13 kW .

The closest option is (d).

60. A single acting two-stage air compressor dealswith 4m3/min of air at 1.013 bar and 15°C witha speed of 250 rpm. The delivery pressure is80 bar. If the inter cooling is complete, theintermittent pressure after first stage will be

(a) 9 bar (b) 8 bar

(c) 7 bar (d) 6 bar

Ans. (a)Sol. For minimum total work, the intermediate

pressures are given by the stage pressure ratio,

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r =

1nk

o

PP

where, n is the number of stages.

In a two stage system, this gives a geometricmean value for the intermediate pressure

Pi = k oP P

= 80 1.013

= 81.04 = 9 bar

61. Which of the following are the advantages ofNano-composite materials?

1. Decreased thermal expansion coefficients

2. Higher residual stress

3. Reduced gas permeability

4. Increased solvent resistance



Ans. (b)Sol.

Nano composites are used at hightemperature where thermal stability isrequired. It is used with matrix of butylrubber within which embedded thin plateletsof veumiculite is used.

Nano composites materials size is of nanorange which is comparable to gas moleculessizes. So, gas permeability is less.

Nano composites are resistant to solventso, that they cannot dissolve in it.

They do not have residual stress as thesematerials give excellent mechanicalstrength, toughness etc.

62. A rod of copper originally 305 mm long is pulledin tension with a stress of 276 MPa. if themodulus of elasticity is 110 GPa and thedeformation is entirely elastic, the resultantelongation will be nearly

(a) 1.0 mm (b) 0.8 mm

(c) 0.6 mm (d) 0.4 mm

Ans. (b)

Sol. L = 305 mm

= 376 MPa

E = 110 GPa

Elongation

3L 276 305L

E 110 10

L 0.765 mm

L 0.8 mm

63. A 1.25 cm diameter steel bar is subjected to aload of 2500 kg. The stress induced in the barwill be

(a) 200 MPa (b) 210 MPa

(c) 220 MPa (d) 230 MPa

Ans. (a)Sol. Load M = 2500 kg

diameter d = 1.25 cm

stress 2 2

mg 2500 9.81

d (12.5)4 4

199.85 MPa

200 MPa

64. The maximum energy which can be stored in abody up to the elastic limit is called

(a) Proof resilience

(b) Modulus of resilience

(c) Impact toughness

(d) Endurance strength

Ans. (a)Sol. Proof resilience is the maximum energy stored

in a body up to the elastic limit.

65. A cast iron bed plate for a pump has a cracklength of 100 µm. If the Young’s modulus ofcast iron is 210 GN/m2 and the specific surfaceenergy is 10 J/m2, the fracture strength requiredwill be nearly

(a) 1.0 × 108 N/m2 (b) 1.2 × 108 N/m2

(c) 1.4 × 108 N/m2 (d) 1.6 × 108 N/m2





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Ans. (b)Sol.

Fracture strength for brittle material,

=2E

c

where, E = Young’s modulus

= specific surface energy

c = crack length

Given: E = 210 × 109 N/m2

= 10 J/m2

c = 100 m = 100 × 10–6 m

=9

62 210 10 10

100 10

= 8 21.15 10 N / m

1.2 × 108 N/m2

66. A 13 mm diameter tensile specimen has 50 mmgauge length. If the load corresponding to the0.2% offset of 6800 kg, the yield stress will benearly

(a) 31 kg/mm2 (b) 43 kg/mm2

(c) 51 kg/mm2 (d) 63 kg/mm2

Ans. (c)Sol. 0.2% offset yield strength,

y = y y

2o

P PA d

4

= 26800 4

13

= 51.23 kg/mm2

67. The magnitude of the velocity of any point onthe kinematic link relative to the other point onthe same kinematic link is the product of

(a) A square root of an angular velocity of thelink and the distance between the two pointsunder consideration

(b) An angular velocity of the link and thesquare of distance between the two pointsunder consideration

(c) A square of an angular velocity of the linkand the distance between the two pointsunder consideration

(d) An angular velocity of the link and thedistance between the two points underconsideration

Ans. (d)Sol.

VAO

o

B

A

VBO

BAV AB

68. In a mechanism, the number of Instantaneouscentres (I-centres) N is

(a) n(n 1)2 (b)

n(2n 1)2

(c)2n(n 1)

3

(d)n(2n 1)

3

where :

n = Number of links

Ans. (a)Sol. No. of I-centres in a mechanism containing n

links

n(n 1)N2

69. In cycloidal motion of cam follower, themaximum acceleration of follower motion fmax

at 4

(a)2

2h2

(b)

2

23h

2

(c)2

22h

(d)2

23h

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where:h = Maximum follower displacement = Angular velocity of cam = Angle for the maximum fol lower

displacement for cam rotationAns. (c)Sol.

Acceleration,

f =2

22h 2sin

at =4

f = fmax = 2

22h 2sin

4

2

max 22hf

70. A shaft of span 1 m and diameter 25 mm issimply supported at the end. It carriers a 1.5 kNconcentrated load at mid-span. If E is 200 GPa,its fundamental frequency will be nearly

(a) 3.5 Hz (b) 4.2 Hz

(c) 4.8 Hz (d) 5.5 Hz

Ans. (d)Sol.

Fundamental frequency f = n

2

f = 1 g

2

= 31 g

2 PL48E

3PL for SSB

48E

= 3 4

3 31 9810 48 200 10 25

2 1.5 10 1000 64

f 5.5 Hz

71. A vibrating system consists of mass of 50 kg, aspring with a stiffness of 30 kN/m and a damper.If damping is 20% of the critical value, thenatural frequency of damped vibrations will be

(a) 16 rad/s (b) 20 rad/s

(c) 24 rad/s (d) 28 rad/s

Ans. (c)Sol. m = 50 kg, k = 30 kN/m

= 0.20

n =km

=330 10

50

= 24.49 rad/s

d = 2n1

= 21 0.20 24.49

= 24 rad/s

72. A refrigerator unit having a mass of 35 kg is tobe supported on three springs, each havingspring stiffness s. The unit operates at 480 rpm.If only 10% of the shaking force is allowed totransmit to the supporting structure, the valueof stiffness will be nearly

(a) 2.7 N/mm (b) 3.2 N/mm

(c) 3.7 N/mm (d) 4.2 N/mm

Ans. (a)Sol. m = 35 kg, N = 480 rpm

Stiffness = s

=2 N60

=2 480 50.27 rad/s

60

Transmissibility, T = 0.10

= 21

(1 r )

{when 0, Assume 0 }





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Consider –ive sign

10 = –1 +r2

r2 = 11

r = 11

n

= 3.317

n =50.273.317

eqsm

= 15.16

seq = 15.162 × 35 = 8043.89 N/m

s =eqs3

= 2.68 × 103 N/m

s 2.7 N / mm

73. In which one of the following tooth profiles, doesthe pressure angle remain constant throughoutthe engagement of teeth?

(a) Cycloidal (b) Involute

(c) Conjugate (d) Epicycloid

Ans. (b)Sol. In case of involute tooth profile pressure angle

remains constant.

74. If the axes of the first and last wheels of acompound gear coincide, it is called

(a) Simple gear train

(b) Compound gear train

(c) Epicyclic gear train

(d) Reverted gear train

Ans. (d)Sol.

Reverted gear train

75. In a reciprocating engine, the force along theconnecting rod FQ is

(a)P

2 2

F

n sin (b)

P2 2

F

2 n sin

(c)P

2 2

nF

2 n sin (d)

P2 2

nF

n sin

where

FP = Force on piston

n = Lr

= Angle of crank from IDC

Ans. (d)Sol.

r

FQFP

QF cos = FP

FQ = PFcos

but sin =sin

n where n

r

cos =2

2sin1

n

= 2 2n sin

n

so, FQ =P

2 2

nF

n sin

76. A mass m1 attached to a shaft at radius r1rotating with angular velocity rad/s, can bebalanced by another single mass m2 which isattached to the opposite side of the shaft atradius r2 in the same plane, if

(a) m1r2 = m2r1 (b) m1r1 = m2r2

(c) m1r1 1 = m2r2 2 (d) m1r2 1 = m2r1 2

Ans. (b)

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Sol.

m1

m2

r1

r2

F = m r2 2 22

F = m r1 1 12

For complete balancing F1 = F2

1 1 2 2m r m r

77. For a single cylinder reciprocating engine speedis 500 rpm, stroke is 150 mm, mass ofreciprocating parts is 21 kg; mass of revolvingparts is 15 kg at crank radius. If two-thirds ofreciprocating masses and all the revolvingmasses are balanced, the mass at a radius of150 mm will be

(a) 7.5 kg (b) 10.5 kg

(c) 12.5 kg (d) 14.5 kg

Ans. (d)Sol. N = 500 rpm, stroke = 150 mm

r = 75 mm

mrec. = 21 kg

mrev. = 15 kg

C = 23

rb = 150 mm

mbrb = rev. rec.2m m r3

mb × 150 = 215 21 753

bm 14.5 kg

78. If the axes of the rolling of the ship and of thestabilizing rotor are parallel, it will result in

(a) A higher bow and lower stern

(b) A lower bow and higher stern

(c) Turning towards left

(d) No gyroscopic effect

Ans. (d)Sol.

rotoraxis rolling

axis

or

rolling axis

rotor axis

Since both the axis are parallel.

So, there is no gyroscopic effect

79. Coaxing is a process of

(a) improving the fatigue properties, attainedby under-stressing and then raising thestress in small increments

(b) Decreasing the hardness by full annealing

(c) Increasing the uniaxial tensile strength byheating above recrystallization temperatureand quenching in oil media

(d) Maintaining the ductility of the material bychemical treatment

Ans. (a)Sol. The fatigue resistance of some metals may be

improved by understressing followed by aprocess of gradually increasing the amplitudeof the alternating stress in small increments aprocedure ordinarily called ‘coaxing’.

80. According to the distortion-energy theory, theyield strength in shear is

(a) 0.277 times the yield stress

(b) 0.377 times the maximum shear stress

(c) 0.477 times the yield strength in tension

(d) 0.577 times the yield strength in tension

Ans. (d)Sol. According to the distortion - energy theory, the

yield strength in shear is 0.577 times the yieldstrength in tension.

By

2(Syt)2 = 2 22

2 3 3 11 2

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1 =

2 = –

3 = 0

2(Syt)2 = [( + )2 + (– – 0)2 + (0 – )2]

2(Syt)2 = [(4 2 + 2 + 2)

(Syt)2 = 3 2

=ytS3

yt0.577 S

81. For the prediction of ductile yielding, the theoryof failure utilized is

(a) Maximum strain energy theory

(b) Distortion energy theory

(c) Maximum normal strain theory

(d) Mohr theory

Ans. (b)Sol. For the prediction of ductile yielding distortion

energy theory is used.

82. A steel specimen is subjected to the followingprincipal stresses: 120 MPa tensile, 60 MPatensile and 30 MPa compressive. If theproportionality limit for the steel specimen is250 MPa; the factor of safety as per maximumshear stress theory will be nearly

(a) 1.3 (b) 1.7

(c) 2.3 (d) 2.7

Ans. (b)

Sol. 1 = 120 MPa

2 = 60 MPa

3 = – 30 MPa

Syt = 250 MPa

By maximum shear stress theory

ytSFOS

= 1 3

FOS =yt

1 3

S

FOS =250

120 30

FOS = 1.7

83. For which one of the following loading conditionsis the standard endurance strength multipliedby a load factor, Ke = 0.9?

(a) Reversed beam bending loads

(b) Reversed axial loads with no bending

(c) Reversed axial loads with intermediatebending

(d) Reversed torsion loads

Ans. (b)Sol. The load factor, ke = 0.9 for reversed axial loads

with no bending.

For, reversed torsional/shear loads (brittlematerial), ke = 0.8

and for compressive reversed or bending loads,ke = 1.

84. A 120 mm wide uniform plate is to be subjectedto a tensile load that has a maximum value of250 kN and a minimum value of 100 kN. Theproperties of the plate material are: endurancelimit stress is 225 MPa, yield point stress is300 MPa. If the factor of safety based on yieldpoint is 1.5, the thickness of the plate will benearly

(a) 12 mm (b) 14 mm

(c) 16 mm (d) 18 mm

Ans. (a)Sol.

w = 120 mm

P1 = 250 kN

P2 = 100 kN

Syt = 300 MPa

Se = 225 MPa

FOS = 1.5

Let t is the thickness of plate

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max = 1PA

=3250 10 MPa

120 t

min = 2PA

= 3100 10 MPa

120t

Mean stress,

m = max min

2

=3 3250 10 100 10

2 120 t

=3175 10 MPa

120t

Amplitude stress

A = max min

2

=3 3250 10 100 10

2 120t

=375 10

120 t

By Goodman formula,

m A

yt eS S

=1

FOS

3 3175 10 75 10120 t 300 120t 225

= 1

1.5

t = 11.45 mm

12 mm

85. A steel connecting rod having Sut = 1000 MN/m2, Syt = 900 MN/m2 is subjected to acompletely reversed axial load of 50 kN. Byneglecting any column action, if the values ofke = 0.85, kb = 0.9, ka = 0.82, kt = 1.5, q = 0.6and N = 2, the diameter of the rod will be nearly

(a) 20 mm (b) 23 mm

(c) 26 mm (d) 29 mm

Ans. (c)Sol. Se = KaKbKcKd (0.5 Sut)

Kf = q(kt – 1) + 1

Kf = 0.6 (1.5 – 1) + 1

Kf = 1.3

Kd =f

1 1 0.77K 1.3

Se = 0.82 × 0.9 × 0.85 × 0.77 ×0.5 × 1000

Se = 241.51 MPa

Axial endurance stress

(Se)a = 0.8 × 241.51

(Se)a = 193.2 MPa

e aSFOS

=2

P

d4

d2 = e a

4P FOSS

d =34 50 10 2

193.2

d = 25.67 mm

d 26mm

86. During crushing or bearing failure of rivetedjoints

(a) The holes in the plates become oval shapedand joints become loose

(b) There is tearing of the plate at an edge

(c) The plates will crack in radial directions andjoints fail

(d) The rivet heads will shear out by appliedstress

Ans. (a)Sol. During crushing failure of riveted joints the holes

in the plates becomes oval shaped and jointsbecome loose.





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87. The double riveted joint with two cover platesfor boiler shell is 1.5 m in diameter subjected tosteam pressure of 1 MPa. If the joint efficiencyis 75%, allowable tensile stress in the plate is83 MN/m2, compressive stress is 138 MN/m2

and shear stress in the rivet is 55 MN/m2, thediameter of rivet hole will be nearly

(a) 8 mm (b) 22 mm

(c) 36 mm (d) 52 mm

Ans. (b)Sol. Given:

Pressure, P = 1 MPa

Diameter, d = 1.5 m

Joint efficiency, 0.75

t = 83 MN/m2

c = 138 MN/m2

c = 55 MN/m2

For boiler shell,

thickness of cylindrical shell,

t =t

PD CA2

Here, CA (corrosion allowance)

Putting the respective values,

t =1 1.5 1000 22 83 0.75

= 14.048 mm

t > B

So, using Unwin’s formula,

Diameter of rivet, d = 6 t

= 6 14.048

d = 22.48 mm

So, the closest option is (b).

88. A bearing supports a radial load of 7000 N anda thrust load of 2100 N. The desired life of theball bearing is 160 × 106 revolution at 300 rpm.If the load is uniform and steady, service factoris 1, radial factor is 0.65, thrust factor is 3.5, k

= 3 and rotational factor is 1, the basic dynamicload rating of a bearing will be nearly

(a) 65 kN (b) 75 kN

(c) 85 kN (d) 95 kN

Ans. (a)

Sol. Pe = r aXVF yF

Pe = 0.65 × 1 × 7000+3.5×2100 N

= 11900 N

Lq0 =k

e

CP

C = Pe × (Lq0)1/k

C = 11900 × (160)1/3

C = 64603.14 N

C = 64.6 kN

C 65 kN

89. A solid cast iron disk, 1 m in diameter and 0.2m thick, is used as a flywheel. It is rotating at350 rpm. It is brought to rest in 1.5 s by meansof a brake. If the mass density of cast iron is7200 kg/m3, the torque exerted by the brakewill be nearly

(a) 3.5 kN m (b) 4.5 kN m

(c) 5.3 kN m (d) 6.3 kN m

Ans. (a)Sol. Mass of flywheel

m = 2d t4

= 27200 0.214

= 1130.97 kg

Moment of inertia (I)

I = 2 2MR 1130.97 0.52 2

= 141.37 kg-m2

= 0 t

0 =2 350 1.5

60

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= 22 350 24.43 rad/s60 1.5

Torque (T) = I

= 141.37 × 24.43

= 3453.67 Nm

= 3.45 kNm

T 3.5 kNm

90. The torque transmitting capacity of frictionclutches can be increased by

(a) Use of friction material with a lowercoefficeint of friction

(b) Decreasing the mean radius of the frictiondisk

(c) Increasing the mean radius of the frictiondisk

(d) Decreasing the plate pressure

Ans. (c)

Sol. T = mWR

mT R

So the torque transmitting capacity of frictionclutches can be increased by increasing themean radius of the friction disk.

91. The ideal gas-refrigeration cycle is the sameas the

(a) Brayton cycle

(b) Reversed Brayton cycle

(c) Vapour compression refrigeration cycle

(d) Vapour absorption refrigeration cycle

Ans. (b)Sol. Reversed Brayton Cycle is used in gas-

refrigeration cycle.

3

4

1

2

T

s

92. If the atmospheric conditions are 20°C, 1.013bar and specific humidity of 0.0095 kg/kg of dryair, the partial pressure of vapour will be nearly

(a) 0.076 bar (b) 0.056 bar

(c) 0.036 bar (d) 0.016 bar

Ans. (d)Sol.

Specific humidity,

=v

v

0.622PP P

Given that, P = 1.013 bar

= 0.0095 kg/kg of dry air

0.0095 =v

v

0.622 P1.013 P

Pv = 0.016 bar

93. In air-conditioning systems, air may be cooledand dehumidified by

1. Spraying chilled water to air in the form offine mist

2. Circulating chilled water or brine in a tubeplaced across the air flow.

3. Placing the evaporator coil across the airflow.

Which of the above statements are correct ?


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

Ans. (c)In case of chilled water, it is very difficult tomaintain temperature of chilled water less thanDPT of entry air. Even if it is maintained, aircarries water vapor along with it in the processof heat exchange by which humidity willincrease. Hence this process reducetemperature but humidity goes up.

94. A duct of rectangular cross-section 600 mm ×400 mm carries 90 m3/min of air having densityof 1.2 kg/m3. When the quantity of air in bothcases is same, the equivalent diameter of acircular duct will be nearly

(a) 0.86 m (b) 0.76 m

(c) 0.64 m (d) 0.54 m





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Ans. (d)Sol. There are two methods for finding out the

equivalent diameter for a given cross-section.

(A) The rectangular duct carrying the samequantity of air as circular duct and thepressure drop per unit length of duct in bothcases is same.

D =

13 5(ab)1.265

(a b)

where a and b are the two sides of theduct and D is the diameter of the circularduct.

(B) The air velocity through rectangular duct issame as circular and pressure drop perunit length of duct in both cases is same.

D =2aba b

Now, Using the following equation,

D =

13 5(ab)1.265

(a b)

=

13 5(60 40)1.265

(60 40)

= 54.5 cm or 0.54 m

95. A room having dimensions of 5 m × 5 m × 3 mcontains air at 25°C and 100 kPa at a relativehumidity of 75%. The corresponding value ofps is 3.169 kPa. The partial pressure of dry airwill be nearly

(a) 106 kPa (b) 98 kPa

(c) 86 kPa (d) 78 kPa

Ans. (b)Sol.

Relative humidity,

= 75% or 0.75

Total pressure,

Pt = 100 kPa

Now, =v

s

PP

0.75 = vP3.169

Partial pressure of water vapor,

Pv = 2.37 kPa

Now, Pt = Pa + Pv

100 = Pa + 2.37

Partial pressure of dry air,

Pa = 97.63 kPa

96. A measure of feeling warmth or coolness bythe human body in response to the airtemperature, moisture content and air motionis called

(a) Dry bulb temperature

(b) Effective temperature

(c) Wet bulb temperature

(d) Dew point temperature

Ans. (b)Sol. There is no proper method to measure the

feeling of comfort as it is controlled by numberof variables. Assuring that purity of the air ismaintained as required and proper air motionis provided which will be just sufficient to break-away the film of air formed and vapour adjacentto the body, the comfort feeling mostly dependsupon the atmospheric conditions as dry bulbtemperature and relative humidity. The AmericanSociety of Heating and Ventilating Engineershas made exhaustive tests on different kinds ofpeople subjected to wide variations ofcombinations of temperature, relative humidityand air motion.

Effective Temperature: The effectivetemperature is a measure of feeling warmth orcold to the human body in response to the airtemperature, moistures content and air motion.

97. While designing a Pelton wheel, the velocity ofwheel ‘u’ is

(a) uK gH (b) u2K gH

(c) uK 2gH (d) u2K 2gH

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where :

Ku = Speed ratio

H = Net head on turbine

g = Gravity

Ans. (c)Sol.

Speed ratio,

ku =u

2gH

u = uk 2gH

Here, H is net head on turbine and u is wheelvelocity.

98. The turbines of the same shape will have thesame

(a) Thomas number

(b) Reynolds number

(c) Specific speed

(d) Rotational speed

Ans. (c)Sol. The specific speed of any turbine is the speed

in r.p.m. of a turbine geometrically similar tothe actual turbine.

99. A centrifugal pump is required to lift 0.0125m3/s of water from a well with depth 30 m. Ifrating of the pump motor is 5 kW, and thedensity of water is 1000 kg/m3, the efficiency ofthe pump will be nearly

(a) 82% (b) 74%

(c) 66% (d) 58%

Ans. (b)Sol. Q = 0.0125 m3/s

H = 30 m

Actual power,

Pa = QgH

= 1000 × 0.0125 × 9.81 × 30

= 3.678 kW

Pump motor power,

Pm = 5 kW

Efficiency of pump,

= a

m

P 100%P

=3.678 100%

5

= 73.57%

100. An inward flow reaction turbine has an externaldiameter of 1 m and its breadth at inlet is 250mm. If the velocity of flow at inlet is 2 m/s and10% of the area of flow is blocked by bladethickness, the weight of water passing throughthe turbine will be nearly

(a) 10 kN/s (b) 14 kN/s

(c) 18 kN/s (d) 22 kN/s

Ans. (b)Sol. D1 = 1m: B1 = 0.25 m;

Vf1 = 2 m/s

Blade thickness coefficient (k) = 0.9

Q = 1 1 f1k D B V

= 1.413 m3/s

Weight of water = gQ

= 13.86 kN/s

101. The process of abstracting steam at a certainsection of the turbine and subsequently usingit for heating feed water supplied to the boileris called

(a) Reheating

(b) Regeneration

(c) Bleeding

(d) Binary vapour cycle

Ans. (b)Sol. Bleeding is a process by which we extract the

steam from turbine which has variousapplication in passout turbine regeneration etc.In regeneration extracted steam is used to heatfeed water to boiler.

102. When blade speed ratio is zero, no work isdone because the distance travelled by theblade is zero even if the torque on the blade





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(a) is minimum

(b) is zero

(c) is maximum

(d) remains the same

Ans. (c)Sol. In impulse steam turbine, the maximum

efficiency is 2cos and maximum work done

per kg of steam is 2b2V when cos / 2.

When 0, the work done becomes zero asthe distance travelled by the blade (Vb) is zero,even though the torque on the blade ismaximum.

When 1, the work done is zero as the torqueacting on the blades becomes zero even thoughthe distance travelled by the blade is maximum.

103. In an axial flow turbine, the utilization factorhas an absolute maximum value of unity, forany degree of reaction if the value of nozzleangle is

(a) 270° (b) 180°

(c) 90° (d) 0°

Ans. (c)Sol. In axial flow gas turbine now a days blade

angles is measured from axial, or flow direction.

i.e. Hence angle is measured with respect axis.While in steam turbine it is with respect to tan-gential and direction.

so, 1 = Angle of axial turbine

= Angle of steam turbine

then, 1 = 90

Maximum utilization factor for steam turbine

=2

22cos

1 cos

So, for axial flow turbine it is

=

21

21

2 cos 90

1 cos 90

A/Q.

21

21

2 cos 901

1 cos 90

1 = 90°

104. Which of the following are essential for a goodcombustion chamber of turbojet engine?

1. It should allow complete combustion of fuel.

2. It should maintain suff iciently hightemperature in the zone of combustion inaddition to proper atomization of fuel thusleading to continuous combustion.

3. It should not have high rate of combustion.

4. The pressure drop should be as small aspossible



Ans. (d)Sol. Because of turbine material limitations, only a

limited amount of fuel can be burnt in thecombustion chamber. The exhaust productsdownstream of the turbine still contain aconsiderable amount of excess oxygen.Additional thrust augmentation can be achievedfrom the turbojet engine by providing an afterburner in which additional fuel can be burnt.

105. If mf is the mass of fuel supplied per kg of airin one second, then the mass of gases leavingthe nozzle of turbojet will be

(a) f kg / s1 m (b) f

1 kg / s1 m

(c) f kg / s1 m (d) f

1 kg / s1 m

Ans. (c)Sol.

(m + m )a f

mf

ma

[Leaving mass from Nozzle]

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As per question,

ma is unity ma = 1.

So, ma + mf = (1 + mf) kg/s.

106. Which one of the following may be consideredas a single cylinder two-stroke reciprocatingengine running at 2400 rpm to 2700 rpm forrapid chain of impulses?

(a) Turbo jet (b) Pulse jet

(c) Ram jet (d) Athodyd jet

Ans. (a)Sol.

Athodyel is nothing but an aero-thermodynamic duct.

It is a straight duct type of engine withouta compressor and turbine wheels.

The entire compression is obtained by aram, eliminating the need of turbine.

Ramjet and pulse jet are the type of it.

Turbojet uses turbine and compressor.

107. In jet propulsion of ships, when the inlet orificesare at right angles to the direction of motion ofthe ships, the efficiency of propulsion is

(a)22u

V u(b) 2

2VuV u

(c) 22u

V u(d)

2VuV 2u

where :V = Absolute velocity of the issuing jetu = Velocity of the moving ship

Ans. (b)

Sol.

directionof ship

VForce

Orifice

Vr = V + u

Mass of water / fluid issuing from orifice, backof the ship

= area of orifice relative velocity

= ra V a V u

Propulsive force exerted on the ship

= ra V uV u

= a vV u

So, Thrust power = a V u TPV u

prop =TP

TP K.E. loss

= 2 2

a(V u)V u 2Vua(V u)(V u) V ua V uV u 2

where = density of fluid / water

a = area of orifice

108. 0.8 kg of air flows through a compressor understeady state conditions. The properties of air atentry are : pressure 1 bar, velocity 10 m/s,specific volume 0.95 m3/kg and internal energy30 kJ/kg. The corresponding values at exit are8 bar, 6 m/s, 0.2 m3/kg and 124 kJ/kg.Neglecting change in potential energy, thepower input will be

(a) 117 kW (b) 127 kW

(c) 137 kW (d) 147 kW

Ans. (b)Sol. Ma = 0.8 kg

By steady flow energy equation

2 21 2

1 cvc g

h q2000 1000

= 2 22 2

2 qc g

h w2000 1000

2 21 2

q 1 2c c

W (h h )2000

2 210 6(125 284)2000

= –159 + 0.032

W input = 159 kJ/kg





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Total power input = 159 kJ/kg × 0.8 Kg/s

= 127 kW

2

1

2

1

1 2

1 23 3

1 2

1 2

1 1 1 1 2 2 2 2

P 1bar P 8 barC 10 m/s C 6 m/sv 0.95 m /kg v 0.2 m /msu 30 ms/kg u 124 ms / mgh u P v h u P V

30 (100 0.95) 124 800(0.2)125 ms/mg 284 ms/mg

109. In a power plant, the efficiency of the electricgenerator, turbine, boiler, cycle and the overallplant are 0.97, 0.95, 0.92, 0.42 and 0.33respectively. In the generated electricity, theauxiliaries will consume nearly

(a) 7.3% (b) 6.5%

(c) 5.7% (d) 4.9%

Ans. (a)

Sol. plant Boiler Turbine(mech) Generator

Cycle Auxiliaries

Auxiliaries0.33

0.97 0.95 0.92 0.42

= 0.9268

1 – 0.9268 = 0.0732

or 7.32% of total electricity generated isconsumed by the auxiliaries.

110. The higher power requirements for compressionin a steam power plant working on Carnotvapour cycle

(a) Increases the plant efficiency as well aswork ratio

(b) Reduces the plant efficiency as well as workratio

(c) Does not affect the plant efficiency as wellas work ratio

(d) Increases the plant efficiency and reduceswork ratio

Ans. (b)

Sol. net T c

s s

W W Wx

Q Q

Increase in power input (wc) will reduceefficiency of cycle work ratio

= net T c

T T

W W WW W

As power input increases, work ratio decreases.

111. For the same compression ratio, the Braytoncycle efficiency is

(a) Same as the Diesel cycle efficiency

(b) Equal to the Otto cycle efficiency

(c) More than the Diesel cycle efficiency

(d) Less than the Otto cycle efficiency

Ans. (b)

Sol. Efficiency of Brayton cycle = 1/

P

11r

= 1

11r

Where r = compression ratio

Efficiency of otto cycle = 1

11r

Hence for same compression ratio, braytoncycle efficiency is equal to otto cycle efficiency.

112. An economizer in a steam generator performsthe function of preheating the

(a) Combustion air

(b) Feed water

(c) Input fuel

(d) Combustion air as well as input fuel

Ans. (b)Sol. Economizer heats the feed water supplied to

the boiler drum by utilizing heat of exhaustgases.

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113. Air enters the compressor of a gas turbine plantoperating on Brayton cycle at 1 bar and 27°C.The pressure ratio in the cycle is 6. If the relationbetween the turbine work WT and compressorwork WC is WT = 3WC and 1.4, the cycleefficiency will be nearly

(a) 40% (b) 50%

(c) 60% (d) 70%

Ans. (a)Sol. P1 = 1 bar; T1 = 27°C

rP = 6

WT = 3 WC

1

4

3

2

T

S

Efficiency of Brayton cycle = 1/p

11r

=

1.4 11.4

11 0.40 or 40%

6

114. A fluidized bed combustion system having anoutput of 35 MW at 80% efficiency when usinga coal of heating value 26 MJ/kg with a sulphurcontent of 3.6% requires a particular limestoneto be fed to it at a calcium-sulphur molar ratioof 3.0 so as to l imit emissions of SO2adequately. The limestone used contains 85%CaCO3. The required flow rate of limestone willbe

(a) 2405 kg/h (b) 2805 kg/h

(c) 3205 kg/h (d) 3605 kg/h

Ans. (a)

Sol. Coal burning rate = 35 1.683kg/s

0.80 26

= 6057.7 kg/h

Flow rate of sulphur = 6057.7 0.036

32

= 6.81 kmol/h

For a Ca : S molar ratio of 3.0, the flow rate of

calcium is 6.81 × 3.0 = 20.44 kmol/h

Mass flow rate of CaCO3 required = 20.44× 100

= 2044 kg/h (since molecular weight of CaCO3is 100)

Mass flow of limestone = 2044/0.85 = 2404.7kg/h which is about 40% of the coal burningrate.

115. In Orsat apparatus, when the percentage ofcarbon dioxide, oxygen and carbon monoxideare known, the remaining gas is assumed tobe

(a) Hydrogen (b) Sulphur dioxide

(c) Nitrogen (d) Air

Ans. (c)Sol. An orsat gas analyzer is a laboratory equipment

used to analyse a gas sample for O2, CO2 andCO directly. Nitrogen gas content is calculatedindirectly.

Potassium hydroxide absorbs CO2

Pyrogallic acid absorbs O2

Copper chloride absorbs CO

Any left over gas is assumed to be nitrogen.

116. The partial vacuum created by the fan in thefurnace and flues, draws the products of thecombustion from the main flue and allows themto pass up to the chimney. Such a draught iscalled

(a) Balanced draught (b) Forced draught

(c) Induced draught (d) Artificial draught

Ans. (c)Sol.

Air

Furnance

Induced draught





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Induced draught fans discharge gasesessentially at atmospheric pressure and placethe system under negative gauge pressure inupstream side.

117. Which of the following are applied (used) waysof compounding steam turbines?

1. Pressure compounding

2. Temperature compounding

3. Velocity compounding

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

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

Ans. (d)Sol. There are three types of compoundings used

in steam turbines which are:

(1) Pressure compounding

(2) Velocity compounding

(3) Pressure-Velocity compounding

118. A steam ejector which removes air and othernon-condensable gases from the condenser isknown as

(a) Wet air pump (b) Dry air pump

(c) Centrifugal pump (d) Circulating pump

118.Ans. (b)Sol. Air pumps are provided to maintain a desired

vacuum in the condenser by extracting the airand other non-condensable gases. Dry air pumpwhich removes air and non-condensable gases.Wet air pump which removes condensate andgases.

119. In a heat exchanger, 50 kg of water is heatedper minute from 50°C to 110°C by hot gaseswhich enter the heat exchanger at 250°C. Thevalue of Cp for water is 4.186 kJ/kg.K and forair is 1 kJ/kg.K. If the flow rate of gases is 100kg/min, the net change of enthalpy of air will benearly

(a) 17.6 MJ/min (b) 15.0 MJ/min

(c) 12.6 MJ/min (d) 10.0 MJ/min

Ans. (c)

Sol. 250°C

Hot gasesehT

ecT

icT

2

50 4.18660

H O

100 160

By heat balancing in heat exchanger

Qexchanged by Air (Hot) = Qexchanged by H2O

Qexchanged by Hot air = w w ce cim c T T

= 50 4.186 110 5060

= 209.3 kJ/s = 12.558 MJ/min

12.6 MJ/min

120. The phenomenon that enables cooling towersto cool water to a temperature below the drybulb temperature of air is termed as

(a) Chemical dehumidification

(b) Adiabatic evaporative cooling

(c) Cooling and dehumidification

(d) Sensible cooling

Ans. (b)Sol. The cooling tower utilizes the phenomenon of

evaporative cooling to cool the warm waterbelow the DBT of air. However, water neverreaches the minimum temperature i.e. WBTsince an excessively large cooling tower wouldthen be required.

121. The angle through which the Earth must turn tobring the meridian of a point directly in line withthe Sun’s rays is called

(a) Altitude angle

(b) Hour angle

(c) Solar azimuth angle

(d) Zenith angle

Ans. (b)Sol. Hour angle: The hour angle at any moment is

the angle through which the earth must turn tobring the meridian of the observer directly in

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line with sun’s rays.

122. In which type of collector is solar radiationfocused into the absorber from the top, ratherthan from the bottom ?

(a) Fresnel lens

(b) Paraboloidal

(c) Concentrating

(d) Compound parabolic

Ans. (a)Sol. In linear Fresnel lens collector and circular

Fresnel lens concentrator solar radiationfocused into the absorber from the top, ratherthan from the bottom.

123. A flat plate collector is 150 cm wide and 180cm high and is oriented such that it isperpendicular to the sun rays. Its active area is90% of the panel size. If it is in a location thatreceives solar insolation of 1000 W/m2 peak,the peak power delivered to the area of thecollector will be

(a) 1.23 kW (b) 2.43 kW

(c) 4.46 kW (d) 6.26 kW

Ans. (b)Sol. Area of the flat plate = 150 × 180 = 27000 cm2

= 27000 cm2

= 2.7 m2

Solar insolation = 1000 W/m2

= 1000 × 2.7 × 0.90

= 2700 × 0.90

= 2.43 kW

124. A surface having high absorptance for short-wave radiation (less than 2.5 m ) and a lowemittance of long-wave radiation (more than 2.5

m ), is called

(a) Absorber (b) Emitter

(c) Selective (d) Black

Ans. (c)Sol. A surface having high absorption for short-wave

radiation and a low emittance of long-waveradiation is called selective surface.

125. In a solar tower power system, each mirror ismounted on a system called

(a) Regenerator (b) Linear Fresnel

(c) Dish (d) Heliostat

Ans. (d)Sol. A heliostat is composed of a reflective surface

mirror, mirror support structure, pedestal,foundation and control and drive mechanisms.

126. The ratio of PV cell’s actual maximum poweroutput to its theoretical power output is called

(a) Quantum factor

(b) Fill factor

(c) Quantum efficiency

(d) PV factor

Ans. (b)

Sol. Fill factor = Maximum power

Theoretical power

These power output of both the open circuitvoltage and short circuit current together.

Fill factor measure the quality of the solarcell.

127. With respect to the wind turbine blades, TSRmeans

(a) Tip Swift Ratio (b) Tip Sharp Ratio

(c) Tip Speed Ratio (d) Tip Swing Ratio

Ans. (c)Sol.

TSR = Tip speed ratio

=blade tip speed

wind speed

128. For a wind turbine 10 m long running at 20 rpmin 12.9 kmph wind, the TSR will be nearly

(a) 3.6 (b) 5.8

(c) 7.6 (d) 9.8

Ans. (b)Sol. Given:

Wind speed = 12.9 kmph = 512.9 m/s

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Blade tip speed = DN60

=(10 2) 20 m / s

60

TSR, =blade tip speed

wind speed

=20 20 18

60 12.9 5

5.8

129. Which one of the following is an enclosure orhousing for the generator, gear box and anyother parts of the wind turbine that are on thetop of the tower?

(a) Turbine blade (b) Nacelle

(c) Turbine head (d) Gear box

Ans. (b)Sol. The nacelle houses the generator, the gear box,

the hydraulic system and yawing of wind turbinethat are on the top of tower.

130. The force required for producing tides in theocean is

(a) 70% due to Moon and 30% due to Sun

(b) 30% due to Moon and 70% due to Sun

(c) 45% due to Moon and 55% due to Sun

(d) 55% due to Moon and 45% due to Sun

Ans. (a)Sol. The tides are caused by the combined attraction

of the sun and the moon on the water of therevolving globe. The effect of moon is 2.6 timesmore than that of sun, influencing the tides ofthe ocean.

i.e. 70 2.3 2.630

131. Which of the following are related to the ProtonExchange Membrane Fuel Cell (PEMFC)?

1. Polymer electrolyte

2. Hydrogen fuel and oxygen

3. Pure water and small amount of electricity

4. Nitrogen gas

(a) 1 and 3 only

(b) 2 and 4 only

(c) 1 and 2 only

(d) 3 and 4 only

Ans. (c)Sol.

Hydrogen fuel

Oxygenwith air

Water

Load

AnodeIon exchange membrane

Cathode

22H 4H 4e

2 2O 4H 4e 2H O

4e–

Polymer electrolyte fuel cell

132. Which of the following are the essentialfunctions of fuel cells?

1. The charging (or electrolyser) function inwhich the chemical AB is decomposed to Aand B

2. The storage function in which A and B areheld apart.

3. The charge function in which A and B arecharged with the simultaneous generationof electricity.


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

Ans. (c)Sol. (i) The charging (or electolyser) function in which

the chemical AB is decomposed to A and B.

(ii) The discharge (or fuel cell) function, in whichA and B are re-united, with the simultaneousgeneration of electricity.

133. The position of centroid can be determined byinspection, if an area has

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(a) Single axis of symmetry

(b) Two axes of symmetry

(c) An irregular shape

(d) Centre axes of symmetry

Ans. (b)Sol. If an area has two axes of symmetry, the

position of the centroid can be determined byinspection.

In case of center of symmetry, area is symmetricabout a point. It has no axes of symmetry butthere is a point such that every line drawnthrough that point contacts the area inasymmetrical manner.

134. Which of the fol lowing statements ofD’Alembert’s principle are correct?

1. The net external force F actually acting onthe body and the inertia force Ft togetherkeep the body in a state of fictitiousequilibrium

2. The equation of motion may be written asF + (ma) = 0 and the fictitious force (– ma)is called an inertia force

3. It tends to give solution of a static probleman appearance akin to that of a dynamicproblem.


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

Ans. (b)Sol. D’Alembert principle involves statements:

1. The net external force F actually acting onthe body and the inertia force Ft togetherkeep the body in a state of fictitiousequilibrium.

2. It tends to given the solution procedure ofa dynamic problem an appearance akin tothat of a static problem.

Hence, statement (3) given is incorrect.

So, correct statement are 1 and 2 only.

135. The linear relationship between stress and strainfor a bar in simple tension or compression isexpressed with standard notations by theequation

(a) E (b) E

(c) G (d) G

Ans. (a)Sol. According to Hooke’s law,

Stress strain

E

where E = Young’s modulus of elasticity

A

0

where A = proportional limit

136. A punch is used for making holes in steel plateswith thickness 8 mm. If the punch diameter is20 mm and force required for creating a hole is110 kN, the average shear stress in the platewill be nearly

(a) 139 MPa (b) 219 MPa

(c) 336 MPa (d) 416 MPa

Ans. (b)Sol. In punching operation,

d = 0.02 m

force required = F = t L

where L = length of shearing region

t = thickness = 0.008 m

F = force = 110 kN

F = t d

110 × 103 = 3 3(8 10 ) 20 10

= 3 6110 10 10 Pa

8 20





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= 110 100 MPa 218.83 MPa16 3.14

137. A rod of length 2 m and diameter 50 mm iselongated by 5 mm when an axial force of 400kN is applied. The modulus of elasticity of thematerial of the rod will be nearly

(a) 66 GPa (b) 72 GPa

(c) 82 GPa (d) 96 GPa

Ans. (c)

Sol.

d=50

mm

P =

400

kN

5mm 2m

As we know, under axial load,

Change in axial length = PLAE

L = 2P L 4

d E

E = 24 P L

dL

=3

3 2 64 400 10 2

5 10 3.14 50 10

E =4 400 1000 2 GPa5 2500 3.14

E 81.48 GPa 82 GPa

138. A beam of span 3 m and width 90 mm is loadedas shown in the figure. If the allowable bendingstress is 12 MPa, the minimum depth requiredfor the beam will be

12 kN 5 kN 12 kN

0.6m0.9m0.9m0.6m

(a) 218 mm (b) 246 mm

(c) 318 mm (d) 346 mm

Ans. (b)

Sol.

5 kN 12 kN12 kN

0.6m 0.9m 0.9m 0.6m

RA RB

BAC

maxMZ

To find reactions, AM 0

12 × 0.6 + 5 × 1.5 + 12 × 2.4 – RB × 3 = 0

RB = 14.5 kN

RA + RB = 29 kN RA = RB = 14.5 kN

Bending moment is maximum at point Cbecause shear force changes its sign at pointC.

Mc = 14.5 × 1.5 – 12 × 0.9 = 10.95 kNm3 3

6max 22

10.95 10 10.95 10 612 100.09 dbd

6

d = 246.6 mm

139. A vertical hollow aluminium tube 2.5 m highfixed at the lower end, must support a lateralload of 12 kN at its upper end. If the wall

thickness is 18

th of the outer diameter and the

allowable bending stress is 50 MPa, the innerdiameter will be nearly

(a) 186 mm (b) 176 mm

(c) 166 mm (d) 156 mm

Ans. (d)Sol.

P

d

D

L=2.5m

https://iesmaster.org/results/gate

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t =D8

Allow = 50 Mpa

d = D – 2t

d =2 D 3DD84 4

k =d 3D 4

Mmax = P × L

max = 3 4

M 32 P LZ D 1 k

max =

3

3 4

32 12 10 2.51D 0.75

= 50 × 106

D = 0.207 m

d = Di = 0.207 × 0.75

= 0.15525 156 mm

140. A wooden beam AB supporting twoconcentrated loads P has a rectangular cross-section of width = 100 mm and height = 150mm. The distance from each end of the beamto the nearest load is 0.5 m. If the allowablestress in bending is 11 MPa and the beamweight is negligible, the maximum permissibleload will be nearly

(a) 5.8 kN (b) 6.6 kN

(c) 7.4 kN (d) 8.2 kN

Ans. (d)Sol.

A

P P

C D EB

RA 0.5 0.5m

RB

B=100 mm

H=150 mm

max = 11 MPa

By symmetry,

RA = RB = P

max = maxMZ

Mmax = P × 0.5 = MC = MD = ME

Z =2BH

6

max =

62

P 0.5 611 100.1 0.15

P 8.25 kN

141. Which of the following statements regarding thinand thick cylinders, subjected to internalpressure only, is/are correct?

1. A cylinder is considered thin when the ratioof its inner diameter to the wall thickness isless than 15.

2. In thick cylinders, tangential stress hashighest magnitude at the inner surface ofthe cylinder and gradually decreasestowards the outer surface


(c) Both 1 and 2 (d) Neither 1 nor

Ans. (b)Sol. A cylinder is considered as thin when ratio of

diameter and thickness is greater than 20.

d 20t

Hence, statement (1) is wrong.

Hoop stress is tangential

h = 2b ar

ih max . at r R = 2

i

b aR

h





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As shown in the stress profile, tangential stressis maximum at inner radius and decreasestowards outer radius.

Hence, statement 2 is only correct.

142. A cylindrical storage tank has an inner diameterof 600 mm and a wall thickness of 18 mm. Thetransverse and longitudinal strains induced are255 × 10-6 mm/mm and 60 × 10-6 mm/mm, andif G is 77 GPa, the gauge pressure inside thetank will be

(a) 2.4 MPa (b) 2.8 MPa

(c) 3.2 MPa (d) 3.6 MPa

Ans. (d)Sol.

d D

Given: d = 600 mm, t = 18 mm

= 60 × 10–6 mm/mm

hoop = transverse

= 255 × 10–6 mm/mm

We know that,

=Pd (1 2 )4tE

...(i)

hoop =Pd (2 )4 tE

...(ii)

Dividing equation (i) by equation (ii),

hoop

=

(1 2 ) 60(2 ) 255

= 0.3

E = 2G(1 )

= 2 × 77 × (1.3) = 200.2 GPa

Putting the respective values in equation (i),

60 × 10–6 = 3P 600 (1 2 0.3)

4 18 200.2 10

P = 3.603 MPa 3.6 MPa

143. A compressed air spherical tank having an innerdiameter of 450 mm and a wall thickness of 7mm is formed by welding. If the allowable shearstress is 40 MPa, the maximum permissible airpressure in the tank will be nearly

(a) 3 MPa (b) 5 MPa

(c) 7 MPa (d) 9 MPa

Ans. (b)Sol. Given: di = 450 mm, t = 7 mm

Spherical tank

di

tPi

i ihoop longitudinal

Pd( for spherical tank)

4t

Maximum shear stress hoop i imax

p d2 8t

iP 450408 7

iP 4.977 MPa 5MPa

144. A solid bar of circular cross-section having adiameter of 40 mm and length of 1.3 m issubjected to torque of 340 N.m. If the shearmodulus of elasticity is 80 GPa, the angle oftwist between the ends will be

(a) 1.26° (b) 1.32°

(c) 1.38° (d) 1.44°

Ans. (a)Sol. Given: d = 40 mm

l = 1.3 m = 1300 mm

G = 80 GPa = 80 × 103 N/mm2

d

l T = Torque = 340Nm = 340 × 10 N-mm3

T

Torsion equation is given by

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o

T GR J L

T G TLJ L GJ

where J = 4d32

3

3 4

340 10 1300 0.21983 rad80 10 (40)

32

1.2595 1.26

145. Which of the following statements regardingscrew dislocation is correct?

(a) It lies parallel to its Burgers vector

(b) It moves in the direction parallel to itsBurgers vector.

(c) It initially requires very less force to move.

(d) It moves very fast as compared to edgedislocation

Ans. (a)Sol.

Dislocation Position Direction ofmovement

Edge Perpendicular ParallelScrew Parallel Perpendicular

Relation with burger vector

146. The percentage of pearlite in a slowly cooledmelt of 0.5% of carbon steel is

(a) 48.5% (b) 52.5%

(c) 58.5% (d) 62.5%

Ans. (c)Sol. Fe – c diagram

A B C

Pearlite

Temp.

910°

C

723°C

-ferrite

0.08 0.5 0.8%C

Applying lever rule at temp. 723°C i.e. Alongline ACB

Mass-fraction of pearlite = AC 100AB

= C A

B A

W W 0.5 0.08100 100W W 0.8 0.08

Mass-friction of pearlite = 42 100 58.37%72

147. In the study of phase diagrams, the rule whichhelps to calculate the relative proportions ofliquid and solid material present in the mixtureat any given temperature is known as

(a) Hume-Rothery rule (b) Lever rule

(c) Gibb’s phase rule (d) Empirical rule

Ans. (b)Sol. Liquid

Solid

A mliq mc msolid B

wt % of B

Temp.

By Lever Rule

% liquid = s c

s

m m100

m m

Similarly

% solid = c liquid

s

m m100

m m

148. The phenomenon that artificially increases thedielectric constant of plastics containing fillersis known as

(a) Gamma polarization

(b) Interfacial polarization

(c) Post-forming drawing

(d) Reinforcement drawing

Ans. (b)





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Sol. Interfacial polarization is phenomenon thatartificially increases the dielectric strength ofplastics containing fillers.

In it under the application of applied field someof defects may migrate through the materialtowards two electrode that has opposite polarityto their charge. If they reach the electrode andarea able to discharge.

149. The addition of alloying element nickel to castiron will primarily improve

(a) Wear resistance

(b) Toughness

(c) Carbide formation

(d) Machinability

Ans. (b)Sol. The purpose of Nickel in the engineering

cast irons is to control the structure byretarding austenite transformation,stabil izing pearl i te and maintainingcombined carbon at the eutectoid quantity.

Nickel lowers the critical temperature sopearlite forms at lower temperature sopearlite is finer and tougher in this case.Nickel also reduces the carbon content.These factor increases toughness, plasticityand fatigue resistance

150. A unidirectional fibre-epoxy composite contains65% by volume fibre and 35% epoxy resin. Ifthe relative density of the fibre is 1.48 and ofthe resin is 1.2, the percentage weight of fibrewill be nearly

(a) 70% (b) 75%

(c) 80% (d) 85%

Ans. (a)

Sol. We know that, Mass

volume

Volume =Mass

Density

% volume of fibre = f

epoxy fibre

VV V

...(i)

% volume of epoxy = e

epoxy fibre

VV V

From equation (i) and (ii)

% volume of fibre% volume of epoxy =

f f f

e e e

V m /V m /

0.650.35

=f

e

m / 1.48m / 1.2

f

e

mm = 2.2904 ...(iii)

and mf + me = 100 ...(iv)

From equation (iii) and (iv)

ff

mm2.2904

= 100

mf = 69.609 70

mf = 70%

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