solution - ies master

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ME

IES MASTER

General Aptitude

1. Select the word that fits the analogy:White: Whitening : : Light: ____(a) Lightning (b) Lightening

(c) Lighting (d) Enlightening

Sol–1: (c)

White : Whitening

Light : Lighting2. While I agree _____ his proposal this time, I do

not often agree _____ him.

(a) to, with (b) with, with

(c) to, to (d) with, to

Sol–2: (a)

“Agree to” is typically used to show consentto a request.

“Agree with” is most often used when twopeople have the same opinion about some-thing.

3. It was estimated that 52 men can complete astrip in a newly constructed highway connectingcities P and Q in 10 days. Due to an emergency,12 men were sent to another project. How manynumber of days, more than the original estimate,will be required to complete the strip?(a) 5 days (b) 3 days

(c) 13 days (d) 10 days

Sol–3: (b)

52 Men can complete a strip = 10 days

12 Men were sent to another project,

Total men available for work on construc-tion of strip = 52 – 12 = 40

52 × 10 = 40 × x

x = 13 days

Number of days, more than the originalestimate,

= 13 – 10 = 3 days

4. In one of the greatest innings ever seen in 142years of Test history, Ben Stokes upped thetempo in a five-and-a-half hour long stay of 219balls including 11 fours and 8 sixes that sawhim finish on a 135 not out as England squaredthe five-match series.Based on their connotations in the given passage,which one of the following meanings DOES NOTmatch?(a) tempo = enthusiasm(b) squared = lost(c) saw = resulted in(d) upped = increased

Sol–4: (b)Square means level or draw (in cricket).So, here squared does not mean lost, it meanslevel or draw.

5. The two pie-charts given below show the data oftotal students and only girls registered indifferent streams in a university. If the totalnumber of students registered in the universityis 5000, and the total number of the registeredgirls is 1500; then, the ratio of boys enrolled inArts to the girls enrolled in Management is______ .Percentage of students enrolled in different

streams in a University

Arts20%

Engineering30%

Commerce15%

Science20%

Arts30%

Commerce20%

Science25%

Percentage of girls enrolled indifferent streams

(a) 9 : 22 (b) 11 : 9(c) 2 : 1 (d) 22 : 9




ME

IES MASTER

Sol–5: (d)

Ratio of boys enrolled in Arts tothe girls enrolled in management =

Boys (Arts)Girls (Management)

Let, Number of boys in Arts = x

Number of girls in Management =

15 1500 225

100

Number of Students in Arts =

20 5000 1000

100

Number of girls in Arts = 30 1500 450

100

Number of boys in Arts = 1000 – 450 =550

Boys (Arts) 550 110 22

Girls (Management) 225 45 9

6. There are five levels (P, Q, R, S, T), in a linearsupply chain before a product reaches customers,as shown in figure.

P Q R S T Customers

At each of the five levels, the price of the productis increased by 25%. If the product is producedat level P at the cost of Rs. 120 per unit, whatis the price paid (in rupees) by the customers?

(a) 292.96 (b) 366.21

(c) 234.38 (d) 187.50

Sol–6: (b)

Let, product cost at P = x = Rs. 120

Q R S T Customer

x 1.25x (1.25) x2P

(1.25) x3 (1.25) x4 (1.25) x5

Price paid by the customer = (1.25)5x

= (1.25)5 × 120 = Rs. 366.21

7. Find the missing element in the following figure.

5

nt h

x?

9

(a) w (b) e(c) d (d) y

Sol–7: (c)

5ht

?

n

x9

(1) (2) (3) (4) (5) (6) (7) (8) (9)a b c d e f g h i

(10) (11) (12) (13) (14) (15) (16) (17) (18)j k l m n o p q r

(19) (20) (21) (22) (23) (24) (25) (26)s t u v w x y z

5820

?

14

249

14

20

4

9

24

8

5

14–6

14+1014–10

14+6

4 is denoted by d so, the correct option is (c)

8. Climate change and resilience deal with twoaspects - reduction of sources of non-renewableenergy resources and reducing vulnerability ofclimate change aspects. The terms ‘mitigation’and ‘adaptation’ are used to refer to these aspects,respectively.




ME

IES MASTER

Which of the following assertions is bestsupported by the above information?(a) Mitigation deals with actions taken to reduce

the use of fossil fuels.(b) Adaptation deals with causes of climate

change.(c) Adaptation deals with actions taken to

combat green-house gas emissions.(d) Mitigation deals with consequences of climate

change.Sol–8: (a)

• Mitigation is an intervantion to reduce theemissions sources or enhance the sinks ofgreenhouse gases.

• Adaptation is an ‘adjustment in natural orhuman systems in response to actual orexpected climatic stimuli or their effects.i.e., mitigation addresses the causes of cli-mate change (accumulation of greenhousegases in the atmosphere), whereas adapta-tion addresses the impact of climate change.So, the correct option is (a)

9. An engineer measures THREE quantities X, Yand Z in an experiment. She finds that theyfollow a relationship that is represented in thefigure below: (the product of X and Y linearlyvaries with Z)

(X, Y)

O ZThen, which of the following statements isFALSE?(a) For fixed X; Z is proportional to Y(b) XY/Z is constant(c) For fixed Z; X is proportional to Y(d) For fixed Y; X is proportional to Z

Sol–9: (c)(X, Y)

O Z

The product of X and Y linearly varies withZ.

XY = aZ + b, where a and b are constants

For fixed X, Z is proportional to Y.

XYZ is constant,

For fixed Z, X is inversely proportional toY.

For fixed Y, X is proportional to Z.

10. The recent measures to improve the outputwould ____the level of production to oursatisfaction.(a) decrease (b) equalise

(c) increase (d) speed

Sol–10: (c)

The recent measures to improve the outputwould increase the level of production to oursatisfaction.

Mechanical Engineering

1. A bolt head has to be made at the end of a rodof diameter d = 12 mm by localized forging(upsetting) operation. The length of theunsupported portion of the rod is 40 mm. Toavoid buckling of the rod, a closed forgingoperation has to be performed with a maximumdie diameter of ______ mm.

Sol–1: (18)

In upsetting forging operation, if the stock islarger than 3 times the diameter then diecavity should not be wider than 1.5 timesthe stock diameter.

1.5d d=12mm

Hence, diameter of die = 1.5 d= 1.5 × 12= 18mm

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IES MASTER

2. Which one of the following statements about aphase diagram is INCORRECT?(a) It indicates the temperature at which

different phases start to melt

(b) Solid solubility limits are depicted by it

(c) It gives information on transformation rates

(d) Relative amount of different phases can befound under given equilibrium conditions

Sol–2: (c)

• Phase diagram is a plot on the temperaturecomposition space showing the stability ofvarious phases.

• Phase diagram tells us what will be themelting point of alloy.

• Relative amount of different phases can befound under given equilibrium condition bylever rule.

• Phase diagram tells about solid solubilityit does not gives the information ontransformation rate.

3. Which of the following conditions is used todetermine the stable equilibrium of all partiallysubmerged floating bodies?

(a) Metacentre must be at a higher level thanthe centre of gravity

(b) Centre of buoyancy must be above the centreof gravity

(c) Centre of buoyancy must be below the centreof gravity

(d) Metacentre must be at a lower level thanthe centre of gravity

Sol–3: (a)

For partially submerged floating bodies:

• For stable equilibrium metacentre (M) mustbe at higher level than the centre of gravity(g)

• For unstable equilibrium metacentre (M)must be at lower level than the centre ofgravity (g).

• For neutral the metacentre (M) and the centreof gravity (g) coincide to each other.

4. In a furnace, the inner and outer sides of thebrick wall (k1 = 2.5 W/m.K) are maintained at

1100°C and 700°C, respectively as shown infigure.

1100°C 700°C 200°C

Heat flux = 2500 W/m2

Insulation, k2

L = L /42 1L2L1

Brick wallk =2.5W/m.K

1

The brick wall is covered by an insulatingmaterial of thermal conductivity k2.The thickness of the insulation is 1/4th of thethickness of the brick wall. The outer surface ofthe insulation is at 200°C. The heat flux throughthe composite walls is 2500 W/m2.The value of k2 is _____ W/m.K (round off toone decimal place).

Sol–4: (0.5)1100°C 700°C 200°C

Heat flux = 2500 W/m2

Insulation, k2

L = L /42 1L2L1

Brick wallk =2.5W/m.K

1

q = 2500 W/m2

1 R12 R2

3

The heat flux q =

eq. 1

T TR R both resistance

in series soR = R1 + R2

R = 1 2

1 1 2 2

L Lk A k A

Then q =

eq. 1

T TR R

2500 = 1100 7004

2.5




ME

IES MASTER

L1 = 0.4L2 = L1/4 = 0.1

2500 =

2

1100 200q 0.4 0.12.5 k

k2 = 0.5 W/mK5. In the space above the mercury column in a

barometer tube, the gauge pressure of the vapouris(a) negative(b) positive, but more than one atmosphere(c) zero(d) positive, but less than one atmosphere

Sol–5: (a)In the space above the mercury column in abarometer tube the gauge pressure of the vapouris negative.The absolute pressure is positive but less thanone atmosphere.

6. Let I be a 100 dimensional identity matrix andE be the set of its distinct (no value appearsmore than once in E) real eigenvalues. Thenumber of elements in E is _________.

Sol–6: (1)

I =

100 100

1 0 0 00 1 0 00 0 1 0

0011

Since, diagonal elements are eigen values ofidentity matrix.So, E has only 1 element i.e., 1.

7. Two plates, each of 6 mm thickness, are to bebut-welded. Consider the following processes andselect the correct sequence in increasing orderof size of the heat affected zone.1. Arc welding2. MIG welding3. Laser beam4. Submerged arc welding

(a) 3 - 2 - 4 - 1 (b) 4 - 3 - 2 - 1(c) 1 - 4 - 2 - 3 (d) 3 - 4 - 2 - 1

Sol–7: (a)For plates each of 6mm thickness are to bebutt-welded the increasing order of size ofheat affected zone is laser beam welding thenMIG welding then submerged arc weldingthen arc welding.

8. The sum of two normally distributed randomvariables X and Y is(a) normally distributed, only if X and Y have

the same mean(b) normally distributed, only if X and Y have

the same standard deviation(c) always normally distributed(d) normally distributed, only if X and Y are

independentSol–8: (d)

The sum of two normally distributed ran-dom variables X and Y is normally distrib-uted only if X and Y are independent or jointlynormal.

9. A beam of negligible mass is hinged at supportP and has a roller support Q as shown in thefigure.

P5m 5m

R

1200 N

Q

4 m

A point load of 1200 N is applied at point R.The magnitude of the reaction force at supportQ is ______ N.

Sol–9: (1500)

5m 5m

1200 N RQH

4 m

RPH

RPV




ME

IES MASTER

Taking moment about P.RQH × 4 – 1200 × 5 = 0

RQH = 1200 54

RQH = 1500 N

10. Let I =

21 x 2x 0 y 0

xy dydx. Then, I may also be

expressed as

(a)

1 y 2y 0 x 0

yx dxdy (b)1 1 2y 0 x y

xy dxdy

(c)

1 y 2y 0 x 0

xy dxdy (d)

1 1 2y 0 x y

yx dxdy

Sol–10: (b)Given that

I =

21 x 2x 0 y 0

xy dydx

Then by changing the order of integration

x=0 x=1y = x

2

y=0

Limit x = 0 to x = 1y = 0 to y = x2

By changing the order of the integration thestrip change from vertical to horizontal.The new limity = 0 to y = 1

x = y to x = 1

So the integration

I =

1 1 2y 0 x y

xy dxdy

11. Consider the following network of activities, witheach activity named A-L, illustrated in the nodesof the network.

78

5

FI

J

K6

FINISHSTART2A B C E

4104

6 7 9D G H

2L

The number of hours required for each activityis shown alongside the nodes. The slack on theactivity L, is ______ hours.

Sol–11: (2)

A B C E

D G H

F

I

JK

LFinish

Start 2 4 10 4

7

8

5

6

2976

LFT = 42EFT = 40ST = LFT – EFT = 42 – 40 = 2

12. The values of enthalpies at the stator inlet androtor outlet of a hydraulic turbomachine stageare h1 and h3 respectively. The enthalpy at thestator outlet (or, rotor inlet) is h2. The condition(h2 – h1) = (h3 – h2) indicates that the degree ofreaction of this stage is(a) 50% (b) 75%(c) 100% (d) zero

Sol–12: (a)Given that enthalpy at stator inlet h1enthalpy at rotor outlet h3 enthalpy at thestator outlet h2.

(h2 – h1) = (h3 – h2)

Degree of reaction =

Rotor

Rotor stator

hh h

=

3 2

3 2 2 1

h hh h h h

=

3 2

3 2

h h2 h h

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IES MASTER

= 1 1002 = 50%

13. A machine member is subjected to fluctuatingstress 0 cos .8 t The endurance limit ofthe material is 350 MPa. If the factor of safetyused in the design is 3.5 then the maximumallowable value of 0 is ______MPa (round offto 2 decimal places).

Sol–13: (100)

Given = 0 cos 8 t

Endurance limit Se = 350 MPaFactor of safety, F.O.S = 3.5

am

yt eS S

= 1FOS

m = 0

a350

= 13.5

a = o

o = 100 MPa

14. A circular disk of radius r is confined to rollwithout slipping at P and Q as shown in thefigure.

vP

r

Q 2v

If the plates have velocities as shown, themagnitude of the angular velocity of the disk is

(a) vr (b) 3v

2r

(c) v2r (d) 2v

3rSol–14: (b)

vP

r

Q 2v

Radius r and roll without slipping at P and Q

For pure rolling condition P and Q velocity shouldbe zeroAssume roller moves with linear velocity withv0 and rotate with angular velocity

So for P

v0 + v = r

v0 – r = –v ...(i)

For Q, 0v r = 2 v ...(ii)

By solving equation (i) and (ii)

We get = 3v2r

15. An attempt is made to pull a roller of weight Wover a curb (step) by applying a horizontal forceF as shown in the figure.

F

The coefficient of static friction between the rollerand the ground (including the edge of the step)is . Identify the correct free body diagram(FBD) of the roller when the roller is just aboutto climb over the step.

(a)

W

F

N2N1 N2

N1

(b)

F

RW

(c)

F

N2WN2




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IES MASTER

(d)F

W

RSol–15: (b)

F

Roller of weight W over a crub and the force Fhorizontal.Coefficient of static friction between the rollerand ground is r.When the roller is just about to climb over thestep there are no friction force from the belowsurface.

F

RW

16. If a reversed Carnot cycle operates between thetemperature limits of 27°C and –3°C, then theratio of the COP of a refrigerator to that of aheat pump (COP of refrigerator / COP of heatpump) based on the cycle is _______ (round offto 2 decimal places).

Sol–16: (0.9)

Reversed Carnot cycle operate betweentemperature limit 27°C and –3°C

COP of the refrigerator, (COP)R = L

H L

TT T

TL = – 3 + 273 = 270 K

TH = 27 + 273 = 300 K

(COP)R =

270 9300 270

COP of heat pump

(COP)H.P = H

H L

TT T

=

300 10300 270

R

HP

COPCOP =

9 0.910

17. A closed vessel contains pure water, in thermalequilibrium with its vapour at 25°C (Stage # 1),as shown.

Not to scale

Valve AWater vapour

Water

Ambient airat 25°C, 1 atm

Stage #1

25°C

Isothermal oven at 80°C

Valve A

Air at 80°C, 1 atmStage #2

The vessel in this stage is then kept inside anisothermal oven which is having an atmosphereof hot air maintained at 80°C. The vesselexchanges heat with the oven atmosphere andattains a new thermal equilibrium (Stage #2).If the Valve A is now opened inside the oven,what will happen immediately after opening thevalve?(a) All the vapor inside the vessel will

immediately condense(b) Hot air will go inside the vessel through

Valve A(c) Nothing will happen - the vessel will

continue to remain in equilibrium(d) Water vapor inside the vessel will come out

of the Valve ASol–17: (b)

Water

Water vapour

Valve

25°C, 1 atm ambient air




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IES MASTER

Initially when water and water vapour mixtureis at 25°C the maximum vapour pressure thatcan be at 25°C in the saturation pressure ofvapour at 25°C.The saturation pressure at 25°C is very lessthan pressure at 1 atmosphere.Now when this vessel will be kept in oven at80°C the saturation pressure of water is stillless than 1 atm. It is around 47.2 kPa.The vapour pressure will reach 1 atm whentemperature is 100°C. Hence at 80°C also thepressure will be less than 1 atm. At 80°C whenvalve is opened air will enter the valve.

18. The figure below shows a symbolic representationof the surface texture in a perpendicular layorientation with indicative values (I through VI)marking the various specifications whosedefinitions are listed below.P: Maximum Waviness Height (mm);Q: Maximum Roughness Height (mm);R: Minimum Roughness Height (mm);S: Maximum Waviness Width (mm);T: Maximum Roughness Width (mm);U: Roughness Width Cutoff (mm),

II

I

III- IV

V

VI

The correct match between the specificationsand the symbols (I to VI) is(a) I-U, II-S, III-Q, IV-T, V-R, VI-P(b) I-Q, II-U, III-R, IV-T, V-S, VI-P(c) I-R, II-P, III-U, IV-S, V-T, VI-Q(d) I-R, II-Q, III-P, IV-S, V-U, VI-T

Sol–18: (d)R = minimum Roughness heightQ = maximum Roughness heightP = maximum waviness heightS = maximum waviness widthu = Roughness width cutoffT = maximum Roughness width

19. The equation of motion of a spring-mass-dampersystem is given by

2

2d x dx3 9x 10 sin(5t)

dtdtThe damping factor for the system is(a) 0.25 (b) 3(c) 2 (d) 0.5

Sol–19: (d)Given the equation of motion of spring-massdamper system.

2

2d x 3dx 9x

dtdt = 10 sin (5t)

The general equation of forced vibration system

mx Cx Sx = 0F sin t

C Sx x xm m

= 0F sin tm

2

2d x C dx S x

m dt mdt = 0F sin tm

By comparing from the general equation.

C 3m ,

S 9m

C = 3m, S = 9mFrom relation,

C = 2 Sm

3m = 2 9m m

= 1 0.52

20. The number of qualitatively distinct kinematicinversions possible for a Grashof chain with fourrevolute pairs is(a) 4 (b) 2(c) 1 (d) 3

Sol–20: (d)

Total qualitative inversion when Grashof’s lawsatisfied.

(i) Crank-Crank mechanism when shortest link atthe fixed position

(ii) Crank-rocker mechanism when shortest link atadjacent to the fixed position.

(iii) Rocker-Rocker mechanism when shortest linkopposite to the fixed link that is coupler position.

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IES MASTER

21. The solution of

2

2d y y 1,dt

which additionally satisfies

t 0t 0

dyy 0dt

inthe Laplace s-domain is

(a)1

s 1 (b)

1s s 1

(c)

1s s 1 s 1 (d)

1s s 1

Sol–21: (c)

Given 2

2d y y 1dt

t 0

dyy t 0dt = 0

2

2d y ydt

= 1

y (t) y(t) = 1Taking Laplace both side

Ly t Ly(t) = L(1)

2S y(s) sy y y0 0 s = 1s

Given y(0) = 0, y 00

Then s2y(s) – y(s) = 1s

y(s) (s2 – 1) =1s

y(s) =

1s s 1 s 1

22. A matrix P is decomposed into its symmetricpart S and skew symmetric part V.

If

4 4 2S ,4 3 7 / 2

2 7 / 2 2

0 2 3V 2 0 7 / 2

3 7 / 2 0

then matrix P is

(a)

2 9 / 2 11 81 / 4 112 45 / 2 73 / 4

(b)

4 6 12 3 05 7 2

(c)

4 2 56 3 71 0 2

(d)

4 6 12 3 05 7 2

Sol–22: (c)

The matrix P is decomposed its symmetric part.

S =

4 4 24 3 7 / 22 7 / 2 2

Skew symmetric part

V =

0 2 32 0 7 / 23 7 / 2 0

For matrix P its symmetric part is

S = TP P2

For matrix P its skew symmetric part is

V = TP P2

P = S + V

=

4 4 2 0 2 34 3 7 / 2 2 0 7 / 22 7 / 2 2 3 7 / 2 0

4 2 56 3 71 0 2

23. For an air-standard Diesel cycle.(a) heat addition is at constant pressure and

heat rejection is at constant pressure(b) heat addition is at constant volume and heat

rejection is at constant volume(c) heat addition is at constant pressure and

heat rejection is at constant volume(d) heat addition is at constant volume and heat




ME

IES MASTER

rejection is at constant pressureSol–23: (c)

Air standard diesel cycle:

P2 3

Qadd

Qrej

4

1V

1–2 Isentropic compression2–3 Constant pressure heat addition3–4 Reversible adiabatic (Isentropic)expansion4–1 Constant volume heat addition.

24. The process, that uses a tapered horn to amplifyand focus the mechanical energy for machiningof glass, is(a) electrochemical machining(b) abrasive jet machining(c) electrical discharge machining(d) ultrasonic machining

Sol–24: (d)

25. In Materials Requirement Planning, if theinventory holding cost is very high and the setupcost is zero, which one of the following lot sizingapproaches should be used?(a) Lot-for-Lot(b) Economic Order Quantity(c) Base stock Level(d) Fixed Period Quantity, for 2 periods

Sol–25: (a)

In MRP if the inventory holding cost is veryhigh and setup cost is zero then the lot sizingapproaches should be used lot-for-lot.

26. The directional derivative of f(x, y, z) = xyz atpoint (–1, 1, 3) in the direction of vector

ˆˆ î 2j 2k is

(a) 73 (b) ˆˆ ˆ3i 3j k

(c) 7 (d) 73

Sol–26: (a)

Directional Derivative = af

f(x) = xyz

a = ˆˆ î 2j 2k

a =

2 22

ˆˆ â i 2j 2k|a| 1 22

= 1 ˆˆ î 2j 2k3

Gradient of (f) =

ˆˆ î j k ff x y z

= ˆˆ ˆyzi xzj xykDirectional derivative

= 1ˆ ˆˆ ˆ ˆ ˆyzi xzj xyk i 2j 2k3

DD = 1 yz 2xz 2xy3

DD at (–1, 1, 3) is,

DD = 1 71 3 2 3 2 11 13 3

27. The sun (S) and the planet (P) of an epicyclicgear train shown in the figure have identicalnumber of teeth.

RP

S A

B

SR

If the sun (S) and the outer ring (R) gears arerotated in the same direction with angular speed s R and , respectively, then the angular speedof the arm AB is

(a) R S3 14 4 (b) R S

1 12 2

(c) R S3 14 4 (d) R S

1 34 4

Sol–27: (a)




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IES MASTER

R

S

S A

BP R

Number of teeth on sun gear (S) = Number ofteeth on planet gear (P)

TS = TP

Angular speed of sun gear = S

Angular speed of outer ring gear = R

RS + 2RP = RR

Module of all gear is same,

SP

mT mT2

= RmT2

TS + 2TP = TR( TS = TP)

TR = 3 TP

Let, revolution made by (S) = xRevolution made by arm (a) = y

Line

1.

2.

3.

Action

Arm fixed, S+1 rev.

Arm fixed, S+x revAdd y

Rev. of arm

0

x

y

Rev. of S1

x

x+y

Rev. of P Rev. of R

S

P

TT

S

P

T xT

S

P

Ty xT

S P

P R

T TT T

S

R

T xT

S

R

Ty xT

Angular speed of (S) = S x y

Angular speed of (R) = SR

R

T xy x yT 3

x + y = S ...(i)

x y3 = R ...(ii)

By substracting equation (ii) from equation (i)

xx3 = S R

x = S R34

Putting value of x in equation (i)

y = S S R34

S

R3y

4 4 , angular speed of the arm AB.

28. A fair coin is tossed 20 times. The probabilitythat ‘head’ will appear exactly 4 times in thefirst ten tosses, and ‘tail’ will appear exactly 4times in the next ten tosses is ______(round offto 3 decimal places).

Sol–28: (0.042)

Fair coin is tossed 20 times,

Probability (Head exactly appear 4 times in thefirst 10 tosses)

P(X) =

4 610

41 1C 0.2052 2

Probability (Tail exactly appears 4 times in thenext 10 tosses)

P(Y) =

4 610

41 1C 0.2052 2

Total probability, P(T) = P(X) × P(Y)= 0.205 × 0.205 = 0.042

29. Consider a flow through a nozzle, as shown inthe figure below.

streamline

1 2

p = p2 atmv = 50 m/sA = 0.02 m

2

22

p 1v

A = 0.2 m1

12

The air flow is steady, incompressible andinviscid. The density of air is 1.23 kg/m3.The pressure difference, (p1 – patm) is ______kPa(round off to 2 decimal places).

Sol–29: (1.52)

Flow through a nozzle,

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1 2

P VA = 0.2m

1 1

12

P = PatmV = 50 m/sA = 0.02 m

2

2

22

Air flow is steady, incompressible and inviscid.Density of air ( ) = 1.23 kg/m3

Apply Bernoulli equation between (1) and (2)

21 1

1P V zg 2g =

22 2

2P V zg 2g

1 2P Pg =

2 22 1

2 1V Vz z

2g

P1 – Patm = 2 2

2 1 2 1g z z V V2

0

=

22 12

2

VV 12 V

=

22 22

1

AV 12 A

=

221.23 0.0250 1

2 0.2

= 1522 Pa = 1.52 kPa30. Water (density 1000 kg/m3) flows through an

inclined pipe of uniform diameter. The velocity,pressure and elevation at section A are VA = 3.2m/s, pA = 186 kPa and zA = 24.5 m, respectively,and those at section B are VB = 3.2 m/s, pa =260 kPa and zB = 9.1 m, respectively. Ifacceleration due to gravity is 10 m/s2 then thehead lost due to friction is _____ m (round offto one decimal place).

Sol–30: (8)

Density of water ( ) = 1000 kg/m3

g = 10 m/s2

Inclined pipe of uniform diameter,

A B

A B

A B

Section A Section BV 3.2 m / s V 3.2 m / sP 186 kPa P 260 kPaZ 24.5 m Z 9.1m

A

B

Let, Head lost due to friction = hL

Applying Bernoulli principle between section Aand B,Total head at section A,

HA =

2A A

AP V Z

g 2g

= 2186 1000 3.2 24.5

1000 10 2 10

= 43.612 mTotal head at section B,

HB =

2B B

BP V zg 2g

=

2260 1000 3.2 9.11000 10 2 10 = 35.612 m

Head lost due to friction hL = HA – HB

= 43.612 – 35.612 = 8 m31. A steel spur pinion has a module (m) of

1.25 mm, 20 teeth and 20° pressure angle. Thepinion rotates at 1200 rpm and transmits powerto a 60 teeth gear. The face width (F) is 50 mm,Lewis from factor Y = 0.322 and a dynamicfactor Kv = 1.26. The bending stress inducedin a tooth can be calculated by using the Lewisformula given below.If the maximum bending stress experienced bythe pinion is 400 MPa, the power transmittedis ______ kW (round off to one decimal place).

Lewis formula: t

vK W ,FmY

where Wt is the

tangential load acting on the pinion.Sol–31: (10)

Spur pinion, module (m) = 1.25 mm




ME

IES MASTER

Pinion teeth (Zp) = 20

Pressure angle ( ) = 20°

Rotating speed (Np) = 1200 rpm

p =

p2 N125.66 rad / s

60Gear, Teeth (ZG) = 60Face width (F) = 50 mmLewis form factor (Y) = 0.322Dynamic factor (kv) = 1.26Maximum bending stress ( ) = 400 MPaLet, Wt = Tangential load acting on the pinionP = Power transmitted

Bending stress ( ) = t

vk WFmY

400 =

t1.26 W50 1.25 0.322

Wt = 6388.89 NPower transmitted (P) = Torque × Speed

= pinionT

Torque applied on pinion gear

= pt tp

mZW r W

2

T =

1.25 206388.892

Torque, (T) = 79861.125 N.mm= 79.86 N.m

P = T 79.86 125.66= 10035.5 watt = 10 kW

32. One kg of air in a closed system undergoes anirreversible process from an initial state of p1 =1 bar (absolute) and T1 = 27°C, to a final stateof p2 = 3 bar (absolute) and T2 = 127°C. If thegas constant of air is 287 J/kg.K and the ratioof the specific heats 1.4, then the change inthe specific entropy (in J/kg.K) of the air in theprocess is(a) – 26.3(b) 28.4(c) indeterminate, as the process is irreversible(d) 172.0

Sol–32: (a)

Closed system undergoing irreversible process,Air, m = 1kg, R = 0.287 kJ/kgk, = 1.4Initial state Final stateP1 = 1 bar (absolute) P2 = 3 bar (absolute)T1 = 27°C = 300K T2 = 127°C = 400 KChange in the specific entropy of the air in theprocess,

S = S2 – S1

=

2 2

p1 1

T PC n R nT P

=

400 31.005 n 0.287 n300 1

= kJ J0.0262 26.2

kgk kgk33. There are two identical shaping machines S1

and S2. In machine S2, the width of theworkpiece is increased by 10% and the feed isdecreased by 10% with respect to that of S1. Ifall other conditions remain the same then theratio of total time per pass in S1 and S2 will be_______(round off to one decimal place).

Sol–33: (0.8)

Two identical shaping machine, S1 and S2.Let width of workpiece for machine S1 = b1

Feed of workpiece for machine S1 = f1

For machine S2, width of workpiece,b2 = 1.1 b1

Feed of workpiece, f2 = 0.9 f1Shaper total time of machining,

tm

b 1 Rf

1

2

m

m

tt = 1 2

2 1

b fb f =

0.91.1

= 0.818 0.834. In a steam power plant, superheated steam at

10 MPa and 500°C, is expanded isentropicallyin a turbine until it becomes a saturated vapour.It is then reheated at constant pressure to500°C. The steam is next expanded isentropicallyin another turbine until it reaches the condenserpressure of 20 kPa. Relevant properties of steamare given in the following two tables. The workdone by both the turbines together




ME

IES MASTER

is________kJ/kg (round off to the nearestinteger).Superheated Steam Table:

Pressure, p(MPa)

Temperature, T(°C)

Enthalpy, h(kJ/kg)

Entropy, s(kJ/kg.K)

10

1

500

500

3373.6

3478.4

6.5965

7.7621

Saturated Steam Table:

Pressure, p Sat. Temp.T (°C)sat

Enthalpy, h(kJ/kg)

Entropy, s(kJ/kg.K)

1

20 kPa

MPa 179.91

60.06

762.9

251.38

2.1386

0.8319

2778.1

2609.7

6.5965

7.9085

h r hg s f sg

Sol–34: (1513)Process, (1) (2), isentropic expansion(2) (3), constant pressure reheating(3) (4), isentropic expansion

p = 1MPa3

p = 10MPa

6

p = 20kpa5 4

T(°C)

S(kJ/kgK)

1

2

From state (1) to (2), Isentropic expansionAt state (1),h1 = 3373.6 kJ/kg, S1 = 6.5965 kJ/kg-kAt state (2),h2 = hg = 2778.1 kJ/kg, S2 = 6.5965 kJ/kg-kAt state (3),h3 = 3478.4 kJ/kg, S4 = 7.7621 kJ/kg-kAt state (4),h4 = hf + xhfg , S3 = S4 (Isentropic exp)S3 = 7.7621 = 0.8319 + x(7.9085 – 0.8319)x = 0.979h4 = 251.38 + 0.979 (2609.7 – 251.38)= 2560.91 kJ/kgWork done by both the turbine = WT = W1–2 +

W3–4

= (h1 – h2) + (h3 – h4)= (3373.6 – 2778.1) + (3478.4 – 2560.91)= 1513 kJ/kg

35. Keeping all other parameters identical, theCompression Ratio (CR) of an air standard dieselcycle is increased from 15 to 21. Take ratio ofspecific heats = 1.3 and cut-off ratio of the cyclerc = 2.The difference between the new and the oldefficiency values, in percentage,

new CR 21 old CR 15( | ) ( | ) = % (round off toone decimal place).

Sol–35. (4.9)

=

c1

c

r 111r 1r

old CR r 15 =

1.3

1.3 11 2 11

2 1 1.315= 0.5 or 50%

New CR R 21 =

1.3

1.3 11 2 11

2 1 1.321

= 0.5487 or 54.87 %

new oldCR 21 CR 15 54.87 50

= 4.87 4.9%36. The turning moment diagram of a flywheel fitted

to a fictitious engine is shown in the figure.

10001500200025003000

Mom

ent (

Nm

)

2

32

2

Angle (radians)The mean turning moment is 2000 Nm. Theaverage engine speed is 1000 rpm. Forfluctuation in the speed to be within 2% ofthe average speed, the mass moment of inertiaof the flywheel is __________ kg.m2.

Sol–36: (3.58)

https://iesmaster.org/testseries/ssc-je




ME

IES MASTER

10001500200025003000

Mom

ent (

Nm

)

2

32

2

Angle (radians)Mean turning moment, Tmean = 2000 Nm

Average engine speed, N = 1000 rpm

Fluctuation in speed, CS = 0.04Let, I = mass moment of inertia of the fly-wheel (kg-m2)

Maximumg fluctuation of energy, E ,

E = (T – Tmean) × (Time period)

= (3000 – 2000) ×

2 = 1570.8 Joule

E = 2sI C

1570.8 =

22 1000I 0.0460

I = 3.58 kg-m2

37. The forecast for the monthly demand of a productis given in the table below.

Month Forecast Actual Sales1 32.00 30.002 31.80 32.003 31.82 30.00

The forecast is made by using the exponentialsmoothing method. The exponential smoothingcoefficient used in forecasting the demand is(a) 1.00 (b) 0.10(c) 0.40 (d) 0.50

Sol–37: (b)

Ft = t 1 t 1 t 1F D F

F2 = 1 1 1F D F

31.8 = 32 30 32

= 0.138. For a single item inventory system, the demand

is continuous, which is 10000 per year. The

replenishment is instantaneous and backorders(S units) per cycle are allowed as shown in thefigure.

STime

Q

Inventory

As soon as the quantity (Q units) ordered fromthe supplier is received, the backordered quantityis issued to the customers. The ordering cost isRs. 300 per order. The carrying cost is Rs. 4per unit per year. The cost of backordering isRs. 25 per unit per year. Based on the total costminimization criteria, the maximum inventoryreached in the system is_________(round off tonearest integer).

Sol–38: (1137)

STime

Q

Inventory

Demand, D = 10,000Ordering cost, Co = Rs 300Carrying cost, Ch = Rs 4 per unit per yearBack ordering cost, Cb = Rs 25 per unit peryearBased on the total minimization criteria,Economic order quantity

Q* =

o b h

h b

2DC C CC C

Q* =

2 10000 300 25 44 25 = 1319.09

Back order quantity,

*S * =

b h

h

C CC




ME

IES MASTER

S* =

h

b h

Q *C 1319.09 4C C 25 4

= 181.94Maximum inventory reached in the system

*mQ = Q * S *

*mQ = 1319.09 – 181.94

= 1137.15 113739. A thin-walled cylinder of radius r and thickness

t is open at both ends, and fits snugly betweentwo rigid walls under ambient conditions, asshown in the figure.

Rigidwall

Pressure vessel

rRigidwall

The material of the cylinder has Young’smodulus E, Poisson’s ratio , and coefficient ofthermal expansion . What is the minimumrise in temperature T of the cylinder (assumeuniform cylinder temperature with no bucklingof the cylinder) required to prevent gas leakageif the cylinder has to store the gas at an internalpressure of p above the atmosphere?

(a)3 prT2 tE

(b)

1 prT4 t E

(c)prTtE

(d) 1 prT2 t E

Sol–39: (c)Longitudinal strain due to hoop stress gen-erated form inside pressure,

= prtE

...(i)

Rigidwall

Pressure vessel

rRigidwall

Here, –ve sign represents decrease in length.This longitudinal strain in (i) must be equalto longitudinal strain due to rise in tempera-ture (magnitude wise) in order to preventleakage of gas.It is worth noting that at this temperature

the cylinder just touches the wall and doesnot result in generation of longitudinalstresses.Hence, strain due to change in temp = Straindue to hoop stress.

T = prtE

prTtE

40. Moist air at 105 kPa, 30°C and 80% relativehumidity flows over a cooling coil in an insulatedair-conditioning duct. Saturated air exits theduct at 100 kPa and 15°C. The saturationpressures of water at 30°C and 15°C are 4.24kPa and 1.7 kPa respectively. Molecular weightof water is 18 g/mol and that of air is 28.94 g/mol. The mass of water condensing out fromthe duct is ___________ g/kg of dry air (roundoff to the nearest integer).

Sol–40: (10)Moist air,State (1), P1 = 105 kPa, T1 = 30°C, (R.H)1 =80%

1State 2State

State (2), P2 = 100 kPa, T2 = 15°C, (R.H) =100%(Ps)@30°C = 4.24 kPa(Ps)@15°C = 1.7 kPa(M.W.)water = 18 g/mol, (M.W)air

= 28.94 g/molVapour pressure at state (1),PV1 = (R.H.)1 ×(PV)S1 = 0.8×4.24 = 3.392 kPaSpecific humidity,

V11

1 V1

0.622 P 0.622 3.392P P 105 3.392

= 0.02076 kg of vapour/kg.d.aVapour pressure at state (2),




ME

IES MASTER

PV2 = (PVS)2 = 1.7 kPaSpecific humidity

V2

22 V2

0.622P 0.622 1.7P P 100 1.7

= 0.010756 kg of vapour/kg of d.a.Mass of water condensing out from the duct.

1 2M

= 20.76 – 10.756 = 10.1 g/kg 10 g/kg

41. A mould cavity of 1200 cm3 volume has to befilled through a sprue of 10 cm length feedinga horiozntal runner. Cross-sectional area at thebase of the sprue is 2 cm2. Consider accelerationdue to gravity as 9.81 m/s2. Neglecting frictionallosses due to molten metal flow, the time takento fill the mould cavity is __________ seconds(round off to 2 decimal places).

Sol–41: (4.28)Volume = Discharge rate × time

time(t) =Volume

Discharge rate

= Volume

Area Velocity

=

12002 2 981 10

= 4.28 seconds42. A hollow spherical ball of radius 20 cm floats in

still water, with half of its volume submerged.Taking the density of water as 1000 kg/m3, andthe acceleration due to gravity as 10 m/s2, thenatural frequency of small oscillations ofthe ball, normal to the water surfaceis______radians/s (round off to 2 decimal places).

Sol–42: (8.66)

R0Ri

x

(a) (b)

Assuming G = Specific gravity of sphereIn case (a): When hollow sphere is in equilib-rium (when half submerged) weight of sphere= upward buoyant force

3 30 i w

4 R R G g3 = 3

0 wg1 4 R2 3

3 30 iR R G = 3

01 R2 ...(i)

In case (b): When a small downward displace-ment of ‘x’ is givenNet upward force = Additional buoyant forcedue to increase in submergence

Fnet = netm a = 20 wR x g ...(ii)

From (ii)

in.anet = 20 wR x g

3 30 i w net

4 R R G a3 = 2

0 wR x g

From (i), putting 3 30 iR R G =

30R

2, we get

30

w netR4 a

3 2= 2

0 wR x g

o net2 R a3 = x.g ...(ii)

Compairing with SHM equation

anet = 2x

=o

3g2R =

3 102 0.2

= 8.66 rad/sec

43. Uniaxial compression test data for a solid metalbar of length 1 m is shown in the figure.

PR

Q130

100

0.002 0.005 (mm/mm)O

(MPa)

The bar material has a linear elastic responsefrom O to P followed by a nonlinear response.

https://iesmaster.org/testseries/cil




ME

IES MASTER

The point P represents the yield point of thematerial. The rod is pinned at both the ends.The minimum diameter of the bar so that itdoes not buckle under axial loading beforereaching the yield point is __________ mm(round off to one decimal place).

Sol–43: (56.94)

Pcr =2

2e

EL

...(1)

Pcr = y A

Where y is yield strength which is equal to100 MPa from figure and A is the cross-sectionarea of bar,

A = 2d4

I = 4d64

E = y 100 50000 MPa0.002

From (1), we get

2d1004

=

42

2

d5000064

1

2

2 4100 d 64

4 50000 d= 1

2 2100 1650000 d = 1

2d =

32

16 3.2422 10500

d = 0.05694 = 56.94 mm

44. For the integral /2

0(8 4cosx)dx

, the absolute

percentage error in numerical evaluation withthe Trapezoidal rule, using only the end points,is__________ (round off to one decimal place).

Sol–44: (5.18)Conventional Method (I)

I =

/2

0

8 4 cosx .dx

= /208x 4sinx

= 4 4 16.566Numerical Method using trapezoidal rule,

II = 0 nh y y2

where, h = 0

2 2y0 = 8 4 cos 0 12

yn =

8 4 cos 8

2

II = 12 8 15.708

4

% error =

II I 100I

=

II 1 100I

=15.708 1 10016.566

= –5.18= 5.18 (absolute %)

45. The spectral distribution of radiation from ablack body at T1 = 3000 K has a maximum atwavelength max . The body cools down to atemperature T2. If the wavelength correspondingto the maximum of the spectral distribution atT2 is 1.2 times of the original wavelength max ,then the temperature T2 is ____________ K(round off to the nearest integer).

Sol–45: (2500)From Wein’s displacement Law

maxT = b

So, max 1 1T = b ..(1)

and max 2 2T = b ..(2)

From eqn. (1) divided by eqn. (2), we get

max 11

max 22

T.T = 1

Given, T1 = 3000 K

and max 2 = 1.2 × max 1

So,

max 1

max 21

30001.2 T = 1




ME

IES MASTER

T2 = 3000 2500K1.2

46. A point ‘P’ on a CNC controlled XY-stage ismoved to another point ‘Q’ using the coordinatesystem shown in the figure below and rapidpositioning command (G00).

Q (800, 600)

P(200, 300)

100 200 300 400 500 600 700 800 900 1000X

100200300400500600700

Y

A pair of stepping motors with maximum speedof 800 rpm, controlling both the X and Y motionof the stage, are directly coupled to a pair oflead screws, each with a uniform pitch of 0.5mm. The time needed to position the point ‘P’to the point ‘Q’ is__________minutes (round offto 2 decimal places)

Sol–46: (1.5)

100200300400500600700

100 200 300 400 500 600 700 800 9001000

(800,600)y = 300

P(200,300) x = 600

x

y

Maximum speed of N = 800 rpm,Pitch = 0.5 mm

time taken in x direction,

tx = x 600P N 0.5 800

= 1.5 mintime taken in y direction

ty =

y 300

P N 0.5 800

= 0.75 min

For complete movement 1.5 minute time

needed x-axis motor. x-axis motor and y-axismotor both wil work 0.75 min then y-axismotor will stop and x axis motor will run0.75 min more Therefore total time for theoperation will be 1.5 min.

47. Two rollers of diameters D1 (in mm) and D2 (inmm) are used to measure the internal taperangle in the V-groove of a machined component.The heights H1 (in mm) and H2 (in mm) aremeasured by using a height gauge after insertingthe rollers into the same V-groove as shown inthe figure.

D1

H1

D2

H2

D > D21

Which one of the following is the correctrelationship to evaluate the angle as shownin the figure?

(a) 1 2

1 2

(H H )sin(D D )

(b) 1 2 1 2

1 2

(H H ) (D D )cosec2(D D )

(c) 1 2

1 2 1 2

(D D )cos2(H H ) 2(D D )

(d) 1 2

1 2 1 2

(D D )sin2(H H ) (D D )

Sol–47: (d)

Diameter (D )1

y

h1 H1

(I)




ME

IES MASTER

y

h2H2

(II)

From Figure (1)

sin =1

1

D2h

where h1 = 11

DH y2

11

DH y .sin2 = 1D

2Let sin = K

11

DH y2 = 1D

2K

–y = 1 11

D D H2K 2 ...(1)

From Fig. (II)

sin =

2 2

222

D D2 2

Dh H y2

K =

2

1 1 22 1

D2

D D DH H2K 2 2

1 1 22 1

D D DK H H2K 2 2

= 2D

2

2 12 1

D DK H H2 2 = 2 1D D

2 2

K =

2 1

2 12 1

D D2

D DH H2

K =

1 2

1 2 1 2

D D2 H H D D

sin =

1 2

1 2 1 2

D D2 H H D D

48. Water flows through a tube of 3 cm internaldiameter and length 20 m. The outside surfaceof the tube is heated electrically so that it issubjected to uniform heat flux circumferentiallyand axially. The mean inlet and exittemperatures of the water are 10°C and 70°C,respectively. The mass flow rate of the water is720 kg/h. Disregard the thermal resistance ofthe tube wall. The internal heat transfercoefficient is 1697 W/m2K. Take specific heatCp of water as 4.179 kJ/kgK. The inner surfacetemperature at the exit section of the tubeis___________°C (round off to one decimal place).

Sol–48: (85.7)

T =10°Cbi m 720kg / h T =70°Cbe

Te

qo

L = 20 m

d=3 cm=0.03m

Cp = 4.179 kJ/kgK

h = 1697 W/m2K

m = 1720 0.2Kg s

3600 Heat given to the tube = Heat gain by water

in tube.

q d L = P be bim C T T

q =

0.2 4.179 70 10

0.03 20= 26.604 kW/m2

= 26604.34 W/m2

At the end of tube,

q = e beh T T

26604.34 = e1697 T 70 Te = 85.67 85.7°C

49. Air is contained in a frictionless piston-cylinderarrangement as shown in the figure

https://iesmaster.org/results/ese




ME

IES MASTER

Air Q

Stop

8 cm

8 cm

The atmospheric pressure is 100 kPa and theinitial pressure of air in the cylinder is105 kPa. The area of piston is 300 cm2. Heat isnow added and the piston moves slowly from itsinitial position until it reaches the stops. Thespring constant of the linear spring is12.5 N/mm. Considering the air inside thecylinder as the system, the work interaction is_________ J (round off to the nearest integer)

Sol–49: (544)

k

Stop

P = 105 kPa1

x3x2

x1

1

2

3

Air Q x

8cm

8cm

Process (1) – (2) is isobaric (constant pressure) P1 = P2

Process (2) – (3) is a line as shown in P-Vdiagram

P3

P1 = P2

P

VA B C

21

3

Work interaction in process 1-2 is the areabetween 1-2-B-A (W1-2).

= 1 B AP V V

Now, B AV V = (Area of Piston) × (x2 – x1)

= 3 34

300 8 2.4 10 m10010

1 2W = 3 3105 2.4 10 10 252JNow, W2 – 3 is the area between 2-3-C-B

W2-3 = 2 3 C B1 P P V V2

= 2 3 3 21 P P A x x2

where A is the area of piston.

P3 = 2kxPA

=

3 23

412.5 10 8 10105 10

300 10

= 3415 10 Pa3

2 3W =

3 3

4 2

1 415105 10 102 3

300 10 8 10= 292 J

WTotal = 1 2 2 3W W

= 252 292= 544 J

50. A helical spring has spring constant k. If thewire diameter, spring diameter and the numberof coils are all doubled then the spring constantof the new spring becomes(a) k/2 (b) 8 k(c) 16 k (d) k

Sol–50: (d)Since spring constant (k)

=4

3Gd8D n

...(1)

where,G = modulus of rigidity of spring materialD = Diameter of springd = Diameter of wiren = No. of coilsAfter doubling the wire diameter, spring diam-eter and the number of coils,

D' = 2D




ME

IES MASTER

d' = 2d

n' = 2nPutting these values in (1),

k' =

4

3G 2d

8 2D 2n

=

4 4

3 3Gd 2

8 D n 2 2

=4

3Gd k8D n

51. Bars of 250 mm length and 25 mm diameterare to be turned on a lathe with a feed of 0.2mm/rev. Each regrinding of the tool costs Rs.20. The time required for each tool change is 1min. Tool life equation is given as VT0.2 = 24(where cutting speed V is in m/min and tool lifeT is in min). The optimum tool cost per piecefor maximum production rate is Rs. __________(round off to 2 decimal places).

Sol–51: (27)

Length of bar (L) = 250 mmDiameter (D) = 25 mmFeed (f) = 0.2 mm/revTime required for tool change (Tc) = 1 min.Tool life equation, VT0.2 = 24For maximum production rate,

T =

C

1 n 1 0.2T 1n 0.2

= 4 min

V = 0.2

24 18.18m / min4

Total machining time

=

L L 250V 18.18 1000fN f 0.2D 25

= 5.4 minNumber of tool changes per component

= Total machining time 5.4 1.35Tool life of cutting tool 4

Regarding tool cost = Rs 20Optimum tool cost per component = 20 × 1.35= Rs 27

52. A cylindrical bar with 200 mm diameter is being

turned with a tool having geometry 0° - 9° - 7°- 8° - 15° - 30° - 0.05 inch (Coordinate system,ASA) resulting in a cutting force Fc1. If the toolgeometry is changed to 0° - 9° - 7° - 8° - 15° -0° - 0.05 inch (Coordinate system, ASA) and allother parameters remain unchanged, the cuttingforce changes to Fc2. Specific cutting energy (inJ/mm3) is Uc = U0(t1)–0.4, where U0 is thespecific energy coefficient, and t1 is the uncutthickness in mm. The value of percentagechange in cutting force Fc2, i.e.

c2 c1

c1

F F 100F

, is ___________ (round off to

one decimla place).Sol–52: (–5.59)

Given diameter, d = 200 mmInitial geometry (ASA)

0 9 7 8 15 30 0.05 inch

After some time

0 9 7 8 15 0 0.05 inch

We know that specific energy consumption

uc = (e) = cF1000fd

f =

tsin

1 = 90° – 30° = 60°

2 = 90 – 0 = 90°

Specific cutting energy,

uc = 0.4 co 1

Fu t1000fd

Fc = 0.440 f sin 1000 fd

0.4cF sin

c2 c1

c1

F F 100F

=

0.42

1

sin 1 100sin




ME

IES MASTER

=

0.4sin90 1 100sin60

= –5.59 %

53. A rigid block of mass m1 = 10 kg having velocityv0 = 2 m/s strikes a stationary block of massm2 = 30 kg after traveling 1 m along africtionless horizontal surface as shown in thefigure.

m2m1v0

k

Mass-less

1m 0.25mThe two masses stick together and jointly moveby a distance of 0.25 m further along the samefrictionless surface, before they touch the mass-less buffer that is connected to the rigid verticalwall by means of a linear spring having a springconstant k = 105 N/m. The maximum deflectionof the spring is __________ cm (round off to 2decimal places).

Sol–53: (1)

m1 m2vo

0.25m1m

k

Let velocity of combined masses (m1 + m2) justafter collision is v m/s.Since, masses are moving together after colli-sion. So, it is a case of inelastic collision.

1 0 2m v m 0 = 1 2m m v

10v0 = 10 30 v

v = 0 010v v 2 0.5m s40 4 4

m +1 m2k

Frictionless

v

Note: Velocity v0 will be constant beforecollision, i.e. during travelling of 1 m (beforecollision) because the horizontal surface isfrictionless.Similarly, velocity v will also be constant for0.25 m before collision of combined masses with

the spring.Now, According to conservation of mechanicalenergy.Initial K.E. + Initial P.E. = Final K.E. + FinalP.E.when final velocity of the system (combinedmasses and spring) is zero. Final K.E will bezero and there will be maximum deflection inspring i.e. A.

So, 21 2

1 m m v 02

= 210 kA2

21 10 30 0.5 02 = 5 210 10 A

2 240 0.5 = 5 210 A A = 0.01 m = 1 cm

54. The function f(z) of complex variable z = x + iy,

where i 1 , is given as f(z) = (x3 – 3xy2) +i v(x, y). For this function to be analytic,v(x, y) should be(a) (3x2y2 – y3) + constant(b) (x3 – 3x2y) + constant(c) (3x2y – y3) + constant(d) (3xy2 – y3) + constant

Sol–54: (c)

Given: u = 3 2x 3xy

For function to be analytic it should followC-R equations

ux = y y xv and u v ...(2)

Now, ux = vy

xu =

2 2u 3x 3y

x

= yvvy

v = 2 213x 3y dy f x C

v = 3

21

3y3x y f x C3

..(3)

vx = 6xy 0 f x

vx = 6xy f x ...(4)Now,

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ME

IES MASTER

uy =

u 0 6xyy ..(5)

From (2)

uy = xv

6xy = xv [From (5)]

6xy = 6xy f x [From (4)]

6xy = 6xy f x

f x = 0

f(x) = Constant ....(6)So, from (3)

v = 2 313x y y Constant C

[From (6)]

v = 2 33x y y C [Let C =Constant + C1]

55. A cantilever of length l, and flexural rigidityEI, stiffened by a spring of stiffness k, is loadedby a transverse force P, as shown.

k

Pl, EI

The transverse deflection under the load is

(a)3

3P 3E3E 3E 2k

ll (b)

3 3P 6E k3E 6E

l l

(c)3 3P 3E k

3E 3E

l l(d)

3

3P 3E3E 3E k

ll

Sol–55: (d)

P

k

, EI

Since, the deflection in spring and cantileverbeam is same.So, stiffness of both (beam and spring) are inparallel combination.For parallel combination,

Equivalent stiffness (keq) = bk k (Stiffness ofbeam)

kb = 33EIl

[For point load at cantilever beam]

keq = 33EIk l

=3

3k 3EIl

l

Deflection =eq

Pk

= 3

3

Pk 3EI

ll

=3

3P

k 3EI

ll

=3

3P 13EI k 1

3EI

ll

=3

3P 3EI3EI k 3EI

ll

solution - ies master

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