me paper ii obj.sol.(ese 2015)

ESE-2015Detailed Exam Solutions

solutions

(Objective Paper - II)Mechanical

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Directions : Each of the following twenty (20) itemsconsists of two statements, one labelled as Statement(I) and the other as Statement (II). Examine thesetwo statements carefully and select the answers tothese 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

1. Statement (I) : The cam in contact with afollower is a case of complete constraint.Statement (II) : The pair, cam and follower, byitself does not guarantee continuity of contactall the time.

Ans. (c)

Sol. The cam in contact of follower is case ofsuccessful constant. Because spring forceis required to maintain the contact. Thisspring force does not guarantee the contactall time because after certain speed, thefollower losses contact with can due to inertiaforce.

2. Statement (I) : Involute pinions can have anynumber of teeth.Statement (II) : Involute profiles in mesh satisfythe constant velocity ratio condition.

SET - DExplanation of Mechanical Engg. Paper-II (ESE - 2014)

Ans. (c)

Sol. Involute pinion can not have any number ofteeth because a minimum number of teethare decided by interference phenomenon.Both involute and cycloidal teeth satisfyconstant velocity ratio condition. Becausevelocity ratio depends upon ratio of numberof teeth or ratio of pitch diameters.

3. Statement (I) : Hookes joint connects two non-parallel non-intersecting shafts to transmitmotion with a constant velocity ratio.Statement (II) : Hookes joint connects twoshafts the axes of which do not remain inalignment while in motion.

Ans. (c)

Sol. The Hookes joint connects two non-parallelshafts but intersecting. For constant velocityratio there are two Hookes joints in particulartorks orientation

4. Statement (I) : Lewis equation for design ofinvolute gear tooth predicts the static loadcapacity of a cantilever beam of uniformstrength.Statement (II) : For a pair of gears in mesh,pressure angle and module must be same tosatisfy the condition of interchangeability andcorrect gearing.

Ans. (a)

Sol. Both Statement I & II are correct.5. Statement (I) : Tensile strength of CI is much

higher than that of MS.Statement (II) : Percentage of carbon in CI ismore than 1.5.



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Ans. (d)

Sol. Tensile strength of cast iron is very lesscompared to mild steel. However, percentageof carbon in cast iron is more than 2%.

6. Statement (I) : Centrifugal clutches aredesigned to provide automatic and smoothengagement of load to driving member.Statement (II) : Since the operating centrifugalforce is a function of square of angular velocity,the friction torque for accelerating a load isalso a function of square of speed of drivingmember.

Ans. (a)

7. Statement (I) : Heating the steel specimen inthe furnace up to austenitize temperaturefol lowed by furnance cooling is termedannealing.Statement (II) : Annealed steel specimenpossesses fine pearlitic structure.

Ans. (c)

Sol. Annealed steel specimen has a coarsepearlite structure and not fine pearlite due toslow cooling in furance.

8. Statement (I) : The susceptibi l i ty of aferromagnetic material decreases with anincrease in Curie temperature.Statement (II) : A critical temperature at whichthe alignment of magnetic moments vanishesis called Curie temperature.

Ans. (a)

Sol. Magnetic suspetibility occurs above curietemperature only. Thus, higher the curietemperature lower is the suseptibility.

9. Statement (I) : Fiberglass is a polymercomposite made of a plastic matrix containingfine fibers of glass.

Statement (II) : Fiberglass acquires strengthfrom then polymer and flexibility from the glass.

Ans. (c)

Sol. Main strength of composite (i.e. febreglan)comes from glass fibres and not the polymer.

10. Statement (I) : Industrial rotors will not haveuniform diameter throughout their lengths.Statement (II) : These rotors will have toaccommodate transmission elements likepulleys and gears and supports like anti-frictionbearings.

Ans. (a)

Sol. Difference diameter on shafts are to supporttransmission elements. Shoulder are providedto restrict axial movement of bearings.

11. Statement (I) : Cored induction furnace cannotbe used for intermittent operation.Statement (II) : Cored induction furnace,though most efficient, requires a liquid metalcharge while starting.

Ans. (a)

Sol. Cored induction furnace requires a liquid metalcharge while starting. Therefore, they cannotbe used for intermittent operations.

12. Statement (I) : Low-carbon steel has highweldability and is more easily welded.Statement (II) : Higher carbon contents tendto soften the welded joints resulting indevelopment of cracks.

Ans. (c)

Sol. High carbon content tends to make the weldmore brittle and thus more prone to cracking.Low carbon steels have excellent weldeabilitydue to low carbon content.



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13. Statement (I) : For cutting multi-start threads,the speed ratio is expressed in terms of thelead of the job thread and lead of the leadscrew threads.Statement (II) : The speed of the job is reducedto one-third or one-fourth of the job speed usedin the turning operation.

Ans. (b)

14. Statement (I) : The Bauschinger effect isobserved in tension test of mild steel specimendue to loss of mechanical energy during localyielding.Statement (II) : The Bauschinger effect is afunction of section modulus of specimen undertest.

Ans. (c)

Sol. The Baeeschinger effect refers to a propertyof materials where the materials stress straincharacteristics change as a result of themicroscopic stress distribution of material. Itis observed in tensile test of mild steelspecimen.

This effect is a material property and not ageometric property. So it is not dependenton section modulus.

15. Statement (I) : The ceramic tools used inmachining of material have highly brittle tooltips.Statement (II) : Ceramic tools can be used onhard-to-machine work material.

Ans. (b)

Sol. Ceramic tools have very brittle tool tips, that iswhy they all prone to impact loads. These toolsare used on hard to machine work materialsuch as cast iron as they are highly wear &abrasion resistant.

16. Statement (I) : In chain drives, angle ofarticulation through which link rotates duringengagement and disengagement, is greater fora small number of teeth.Statement (II) : The greater angle of articulationwill increase the life of the chain.

Ans. (c)

Sol. For greater life of chain, the angle of articulationshould be reduced to minimize wear of chain& fatigue of rollers.

17. Statement (I) : The CNC is an NC systemutilizing a dedicated stored program to performall numerical control functions in manufacturing.Statement (II) : The DNC is a manufacturingprocess in which a number of processmachines are controlled by a computer throughdirect connection and real time analysis.

Ans. (b)

Sol. Both Statement I and II are correct, butStatement II does not explain Statement I,

18. Statement (I) : In interference fit, the outerdiameter of the shaft is greater than the innerdiameter of the hole.Statement (II) : The amount of clearanceobtained from the assembly of hole and shaftresulting in interference fit is called positiveclearance.

Ans. (c)

Sol. Clearance obtained in an interference fit is anegative clearance.

19. Statement (I) : One of the most commonlyused techniques for testing surface integrity ofmaterial is metallography.Statement (II) : Surface integrity of a materialdoes not contribute for the mechanical andmetallurgical properties.



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Ans. (c)

Sol. Surface integrity affects mechanical propertiessuch as fatigue strength of material.

20. Statement (I) : The change in critical pathrequires rescheduling in a PERT network.Statement (II) : Some of the activities cannotbe completed in time due to unexpectedbreakdown of equipment or non-availability ofraw materials.

Ans. (a)

21. A copper rod of 2 cm diameter is completelyencased in a steel tube of inner diameter 2 cmand outer diameter 4 cm. Under an axial load,the stress in the steel tube is 100 N/mm2. If Es= 2Ec, then the stress in the copper rod is(a) 50 N/mm2 (b) 33.33 N/mm2

(c) 100 N/mm2 (d) 300 N/mm2

Ans. (a)

Sol. For copperd = 2cm

= 20 mmFor steel

di = 2 cm= 20 mm

d0 = 4 cm= 40 mm

s = 100N/mm2

Es = 2Ec

Copper

Steel

4cm

2cm

Strain in steel = strain in copper

s

sE

= ccE

c =c

s

100 EE

c = 50 N/mm2

22. A system under biaxial loading induces principalstresses of 100 N/cm2 tensile and 50 N/cm2compressive at a point. The normal stress atthat point on the maximum shear stress planeis(a) 75 N/cm2 tensile(b) 50 N/cm2 compressive(c) 100 N/cm2 tensile(d) 25 N/cm2 tensile

Ans. (d)

Sol. 1 = 100 N/cm2

2 = 50 N/cm2

On maximum shear stress plane,Normal stress

= avg

= 1 2100 50

2 2 =

= 25 N/cm2 (Tensile)23. In a biaxial state of stress, normal stresses are

2x 900 N / mm ,

2y 100 N / mm and

shear stress 2300 N / mm . The maximum

principal stress is

(a) 2800 N / mm (b) 2900 N / mm

(c) 21000 N / mm (d) 21200 N / mm

Ans. (c)

Sol. x = 900 N/mm2

y = 100 N/mm2

xy = 300 N/mm2



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Maximum principal stress

1 = 2x y 2

x y xy1 4

2 2

= 2 2900 100 1 900 100 4 300

2 2

=1500 640000 3600002

= 21500 500 1000 N/mm2

=

24. A constitutional diagram shows relationshipamong which of the following combinations ina particular alloy system?(a) Temperature and composition(b) Temperature and phases present(c) Temperature, composition and phases

present(d) Temperature and pressure

Ans. (c)

Sol. A constitutional diagram gives informationregarding temperature, composition and phasespresent.

25. The state of stress at a point in a body is given

by x 100 MPa and y 200 MPa . One of

the principal stresses 1 250 MPa . Themagnitudes of the other principal stress and

the shearing stress xy are respectively

(a) 50 3 MPa and 50 MPa

(b) 100 MPa and 50 3 MPa

(c) 50 MPa and 50 3 MPa

(d) 50 3 MPa and 100 MPa

Ans. (c)

Sol. x = 100 MPa

y = 200 MPa

1 = 250 MPa

1 2 = x y

2 = 100 200 250

= 50 MPa

1 = 2x y 2x y xy1 4

2 2

250 = 2 2

xy100 200 1 100 200 4

2 2

100 = 2 2

xy1 100 200 42

40000 = 10000 + 2xy4

2xy4 = 30000

xy = 50 3MPa

26. Consider the following statements regardingpowder metallurgy :1. Refractory materials made of tungsten

can be manufactured easily.2. In metal powder, control of grain size

results in relatively much uniformstructure.

3. The powder heated in die or mould athigh temperature is then pressed andcompacted to get desired shape andstrength.

4. In sintering, the metal powder is graduallyheated resulting in coherent bond.

Which of the above statements are correct?(a) 1, 2 and 3 only



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

Ans. (d)

Sol. All options are correct27. The magnitudes of principal stresses at a point

are 250 MPa tensile and 150 MPacompressive. The magnitudes of the shearingstress on a plane on which the normal stressis 200 MPa tensile and the normal stress on aplane at right angle to this plane are

(a) 50 7 MPa and 100 MPa (tensile)

(b) 100 MPa and 100 MPa (compressive)

(c) 50 7 MPa and 100 MPa (compressive)

(d) 100 MPa and 50 7 MPa (tensile)

Ans. (c)

Sol. 1 = 250 MPa

2 = 150 MPa

x = 200 MPa

1 2 = x y

y = (250 150) 200

y = 100 MPa

1 = 2 2

xy200 100 1 200 100 4

2 2

250 = 2 2

xy200 100 1 200 100 4

2 2

400 = 2 2xy300 4

2xy = 17500

xy = 50 7 MPa

28. The state of stress at a point is given by

x 100 MPa , y 50 MPa and

xy 100 MPa . The centre of Mohrs circle and

its radius will be

(a) x xy75 MPa, 0 and 75 MPa

(b) x xy25 MPa, 0 and 125 MPa

(c) x xy25 MPa, 0 and 150 MPa

(d) x xy75 MPa, 0 and 125 MPa

Ans. (b)

Sol. x = 100 MPa

y = 50 MPa

xy = 100 MPa

centre =x y ,0

2

= [25 MPa, 0]

Radius of Mohrs circle = 2 2x y xy1 42

= 2 21 100 50 4 1002

=1 22500 40,0002

= 125 MPa29. Consider the following sttements related to

Mohrs circle for stresses in case of plane



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stress :1. The construction is for variations of stress

in a body.2. The radius of the circle represents the

magnitude of the maximum shearingstress.

3. The diameter represents the differencebetween the two principal stresses.

Which of the above statements are correct?(a) 1, 2 and 3 (b) 2 and 3 only(c) 1 and 3 only (d) 1 and 2 only

Ans. (a)

30. The figure shows a steel piece of diameter 20mm at A and C, and 10 mm at B. The lengthsof three sections A, B and C are each equal to20 mm. The piece is held between two rigidsurfaces X and Y. The coefficient of linear

expansion 51.2 10 / C and Youngss

modulus 5E 2 10 MPa for steel :

20

A

20

C

10

B

20 20 20

X Y

When the temperature of this piece increasesby 50 C , the stresses in sections A and B are

(a) 120 MPa and 480 MPa(b) 60 MPa and 240 MPa

(c) 120 MPa and 120 MPa(d) 60 MPa and 120 MPa

Ans. (b)

Sol. = 1.2 105/CE = 2 105 MPa

T = 50C

20mm 20mm 20mm

R R

BX

AB

C

20mm 10mm 20mm

Since supports are rigid.

A B cL t L T L T

CA B

A B C

R LRL RL 0E A E A E A

=

LA = LB = LC = L

A B C

L R R R3L TE A A A

=

3E T = 2 2 2

1 1 1R20 10 20

4 4 4

R = 5 5

2 2 2

3 2 10 1.2 10 504 4 420 10 20

= 18849.56 N

A =2

R 60 MPa20

4

=



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B =2

R 240 MPa10

4

=

31. For a material following Hookes law, the values

of ealstic and shear moduli are 53 10 MPa

and 51.2 10 MPa respectively. The value for

bulk modulus is

(a) 51.5 10 MPa (b) 52 10 MPa

(c) 52.5 10 MPa (d) 53 10 MPa

Ans. (b)

Sol. E = 3 105 MPaG = 1.2 105 MPa

E =9KG

3K G

3 105 =

5

5

9 K 1.2 103K 1.2 10

3K + (1.2 105) = 3.6 K0.6 K = 1.2 105

K = 2 105 MPa

32. At a point in a body, 1 0.0004 and

2 0.00012 . If 5E 2 10 MPa and 0.3 ,

the smallest normal stress and the largestshearing stress are(a) 40 MPa and 40 MPa(b) 0 MPa and 40 MPa(c) 80 MPa and 0 MPa(d) 0 MPa and 80 MPa

Ans. (b)

Sol. 1 = 0.0004

2 = 0.00012

E = 2 105

= 0.3

1 = 1 22E

1

=

5

22 10 0.0004 0.3 0.000121 0.3

= 80 MPa

2 = 2 12E

1

=

5

22 101 0.3

[(0.00012) + {0.3 0.0004}]= 0 MPa

Maximum shar stress = 1 280 0

2 2 =

= 40 MPaSmallest normal stress = 0

33. A canti lever of length 1.2 m carries aconcentrated load of 12 kN at the free end.The beam is of rectangular cross-section withbreadth equal to half the depth. The maximumstress due to bending is not to exceed

2100 N / mm . The minimum depth of the beamshould be(a) 120 mm (b) 60 mm(c) 75 mm (d) 240 mm



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Ans. (a)

Sol.

L=1.2m3A

b

h

W =12 kN

b = h/2At point A,

Mmax = WL= (12 103) N (1.2)= 14.4 103 N-m= 14.4 106 N-mm

max =maxM y

I

100 = 6

314.4 10 h / 2

bh12

100 = 6

26 14.4 10

bh

bh2 = 6 14.4 104

2h h2

= 86.4 104

h3 = 172.8 104

h = 120 mm34. Two strain gauges fixed along the principal

directions on a plane surface of a steel memberrecorded strain values of 0.0013 tensile and0.0013 compressive respectively. Given that the

value of 5E 2 10 MPa and 0.3 , the

largest normal and shearing stress at this pointare(a) 200 MPa and 200 MPa(b) 400 MPa and 200 MPa

(c) 260 MPa and 260 MPa(d) 260 MPa and 520 MPa

Ans. (a)

Sol. 1 = 0.0013

2 = 0.0013E = 2 105 MPa = 0.3

1 = 1 22E

1

=

5

22 101 0.3

[0.0013 +

{0.3 (0.0013)} = 200 MPa

2 = 2 12E

1

=

5

22 101 0.3

[( 0.0013)

+{0.3(0.0013)} = 200 MPa

Maximum shear stress max = 1 22

=200 200

2

= 200 MPa

35. A beam ABCD, 6 m long, is supported at Band C, 3 m apart with overhangs AB = 2 mand CD = 1 m. It carries a uniformly distributedload of 100 kN/m over its entire length :

2 m 3 m 1 m

100 kN/mA B C D

The maximum magnitudes of bending momentand shear force are



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(a) 200 kN-m and 250 kN(b) 200 kN-m and 200 kN(c) 50 kN-m and 200 kN(d) 50 kN-m and 250 kN

Ans. (b)

Sol.

AB C D

2m 3m 1m

x

RB RC

100 kN/m

P

Taking moment about point B,(Rc 3) (100 6 1) = 0

Rc = 200 kNRB + Rc = 600

RB = 400 kN(S.F.)A = 0

BS.F. = (100 2) = 200 kN

BS.F. = 200 + 400 = 200 kN

CS.F. = 200 (100 3) = 100 kN

CS.F. = 100 + (200) = 100 kN(S.F.)D = 0

Point P, (100 x) + RB = 0

100x = 400x = 4m

(B.M.)A = 0(B.M.)B = (100 2 1) = 200 N-m

(B.M.)C = 5100 52

+ (400 3)

= 50 N-m(B.M.)D = 0

(B.M.)D = *B

xR x 2 100 x2

= (400 2) 4100 42

= 0

AB C D

2m 3m 1mRB RC

100 kN/m

+200 kN

+ve +ve

ve veB P CD

100kN200kN

A

S.F.D

A B C D

M =200 kN-mmaxM=50 kN-m

P

36. A solid circular cross-section cantilever beamof diameter 100 mm carries a shear forceof 10 kN at the free end. The maximum shearstress is

(a)4 Pa

3 (b)3 Pa4

(c)3 Pa16

(d)16 Pa3

Ans. (d)

Sol. d = 100mmV = 10 kN

= 104 N

max = avg43

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=4

22

4 10 N3 mm100

4

=16 MPa3

37. A beam of length L simply supported at itsends carrying a total load W uniformlydistributed over its entire length deflects at thecentre by and has a maximum bendingstress . If the load is substituted by aconcentrated load 1W at mid-span such thatthe deflection at the centre remians unchanged,the magnitude of the load 1W and themaximum bending stress will be

(a) 0.3 W and 0.3

(b) 0.6 W and 0.6

(c) 0.3 W and 0.6

(d) 0.6 W and 0.3

Ans. (*)

Sol. Let load intensity is .

A BC

RBRAL

L W =

RA = BLR

2=

deflection at point C

=4 35 L 5 WL

384 E 384 E =I I

Mmax =L L L L

2 2 2 4

=2L WL

8 8 =

max =

WL y8

I

P QR

L

W1

=3

1W L48EI

31W L

48EI =35 WL

384 EIW1 = 0.625 W

1max =

maxM yI

= 1W L y4 I

= W L y0.625

4

I

= max0.625 8

4

1max = 1.25 max

38. For a rectangular section beam, if the beamdepth is doubled, keeping the width, length andloading same, the bending stress is decreasedby a factor(a) 2 (b) 4(c) 6 (d) 8

Ans. (b)

Sol. For rectangular section,

b =

3M y M h/2

bh /2 =I

b = 26Mbh



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if h = 2h

b

hb = 2 2

6M 6Mbh b 2h

=

= 21 6M4 bh

b =b

4

39. A helical compression spring of stiffness K iscut into two pieces, each having equal numberof turns and kept side-by-side undercompression. The equivalent spring stiffnessof this new arrangement is equal to(a) 4 K (b) 2 K(c) K (d) 0.5 K

Ans. (a)

Sol. When spring is cut into two pieces.

Stiffness of each spring (k ) = 2k

and then these springs are arranged in parallelso k = k k

= 2k 2k

= 4 k40. A beam AB simply supported at its ends A and

B, 3 m long, carries a uniformly distributed loadof 1 kN/m over i ts entire length and aconventrated load of 3 kN, at 1 m from A :

1 m 2 m

1 kN/mA

CB

3 kN

If ISJB 150 with 4XXI 300 cm is used for the

beam, the maximum value of bending stress is

(a) 75 MPa (b) 85 MPa(c) 125 MPa (d) 250 MPa

Ans. (a)

Sol.

A

3kN

B1m 2m

x

RBCRA

x

x1kN/m

taking moment about point A

(RB 3) (3 1) 33 12

= 0

3RB = 3 + 4.5RB = 2.5 kN

RA + RB = 3 + (1 3)RA + RB = 6

RA = 3.5 kNMmax = (B.M.)c

= (RA 1) 11 12

= 3 kN-mMmax = 3 10

6 N-mm

y =150 75mm

2=

b =maxM y

I

= 6

4 43 10 N-mm 75mm

300 10 mm

= 75 MPa

41. Which of the following statements apply toprovision of flash gutter and flash land aroundthe parts to be forged?1. Small cavities are provided which are

directly outside the die impression.2. The volume of flash land and flash gutter

should be about 20% - 25% of the volumeof forging.



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3. Gutter is provided to ensure completeclosing of the die.

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

Ans. (a)

Sol. Gutter is provided to ensure complete filling ofdie cavity and not closing of die.

42. A hole and a shaft have a basic size of 25 mm,and are to have a clearance fit with a maximumclearance of 0.02 mm and a minimumclearance of 0.01 mm. The hole tolerance is tobe 1.5 times the shaft tolerance. The limits ofboth hole and shaft using hole basis systemwill be(a) low limit of hole = 25 mm, high limit of

hole = 25.006 mm, upper limit of shaft =24.99 mm and low limit of shaft = 24.986mm

(b) low limit of hole = 25 mm, high limit ofhole = 25.026 mm, upper limit of shaft =24.8 mm and low limit of shaft = 24.76mm

(c) low limit of hole = 24 mm, high limit ofhole = 25.006 mm, upper limit of shaft =25 mm and low limit of shaft = 24.99 mm

(d) low limit of hole = 25.006 mm, high limitof hole = 25 mm, upper limit of shaft =24.99 mm and low limit of shaft = 25 mm

Ans. (a)

Sol. For hole basis systemLow limit of hole = 25 mmHigh limit of hole = 25.006High limit of shaft = 24.99 mmLow limit of shaft = 24.986Also tolerance of shaft + minimum clearance +tolerance of hole = 0.02 mm

x + 1.5 x + 0.01 = 0.02

2.5 x = 0.01

x =0.01 0.004mm25

=

43. Consider the following statements :In case of assembly of mating parts1. the difference between hole size and

shaft size is called allowance2. in transition fit, small positive or negative

clearance between the shaft and holemember is employable

Which of the above statements is/are correct?(a) 1 only (b) Both 1 and 2(c) 2 only (d) Neither 1 nor 2

Ans. (c)

Sol. Allowance is a planned deviation betweenan actual dimension and a nominal ortheoretical dimension.

In transition fit, small positive clearnace ornegative clearance (Interference) between theshaft and hole member is employable.Only statement (2) is correct.

44. An organization has decided to produce a newproduct. Fixed cost for producing the productis estimated as Rs. 1,00,000. Variable cost forproducing the product is Rs. 100. Market surveyindicated that the product selling price couldbe Rs. 200. The break-even quantity is(a) 1000 (b) 2000(c) 500 (d) 900

Ans. (a)

Sol. BEQ =F 100000 1000

S V 200 100 = =

45. Using exponential smoothening, a carmanufacturing company predicted the demandfor that year as 1040 cars. The actual salewas found to be 1140 cars. If the companys



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forecast for the next year is 1080, what is thevalue of the smoothening constant?(a) 0.4 (b) 0.6(c) 0.7 (d) 1.2

Ans. (a)

Sol. Ft = t 1 t 1d 1 F

1080 = 1140 1 1040

= 1140 1040 1040

40 = 100

= 0.446. Coarse feed, low rake angle, low cutting speed

and insufficient cooling help produce(a) continuous chips in ductile materials(b) discontinuous chips in ductile materials(c) continuous chips with built-up edges in

ductile materials(d) discontinuous chips in brittle materials

Ans. (c)

Sol. Coarse feed, low rate angle, low cutting speed& insuf f icient cooling help to producecontinuous chips with built-up edges in ductilematerials.

47. In NC machining, coordinated movement ofseparately driven axes motion is required toachieve the desired path of tool relative toworkpiece. The generation of these referencesignals is accomplished through a device called(a) approximator (b) interpolator(c) coordinator (d) director

Ans. (b)

Sol. Interpolator coordinate the motion of toolrelative to workpiece.

48. a part is made from solid brass rod of 38 mmdiameter and length 25 mm. The machiningtime taken to finish the part is 90 minutes andlabour rate is Rs. 2 per hour. Factory overheadsare 50% of direct labour cost. The density ofmaterial is 8.6 gm per cubic cm and its cost isRs. 1.625 per newton. The factory cost of thepart will be(a) Rs. 8.40 (b) Rs. 4.80(c) Rs. 14.80 (d) Rs. 18.40

Ans. (a)

Sol. Volume of rod = 2/4d

= 2 3/43.8 2.5 28.36 cm =Mass of rod = 28.36 8.6 = 243.84 grms or0.243kg

weight of rod = 2.4 NCost of rod = 2.4 1.625 Rs. 3.89Cost of labour = 2 1.5 = Rs. 3Factory overhead = 0.5 3 = Rs. 1.5Total factory cost = 3 + 1.5 + 3.89 = Rs. 8.40

49. A company wants to expand the solid propellantmanufacturing plant by the addition of morethan 1 ton capacity curing furnace. Each ton ofpropellant must undergo 30 minutes of furnacetime including loading and unloadingoperations. Furnace is used only 80 percent ofthe time due to power restrictions. The requiredoutput for the new layout is to be 16 tons pershift (8 hours). Plant (system) efficiency isestimated at 50 percent of system capacity.The number of furnaces required will be(a) 3 (b) 2(c) 1 (d) 4

Ans. (a)



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Sol. Required output 16 tons in 8 hours or 8 60= 480 minutes.Operating time of furnace = 0.8480 = 384minutes.16 ton propellent requires = 16 30 = 480minutes of furnace.Plant efficiency (in minutes) = 0.5 382 = 192minutes

Furnace required = Total furnace time

Total plant time

=480 2.5 or 3192

=

50. The purpose of providing side rake angle onthe cutting tool is to(a) avoid work from rubbing against tool(b) control chip flow(c) strengthen tool edge(d) break chips

Ans. (b)

Sol. Purpose of side rake angle of cutting tool is tocontrol chip flow.

51. The annual demand of a commodity in asupermarket if 80000. The cost of placing anorder is Rs. 4,000 and the inventory cost ofeach item is Rs. 40. What is the economicorder quantity?(a) 2000 (b) 4000(c) 5656 (d) 6666

Ans. (b)

Sol. EOQ =2 80000 4000

40

= 4000 unit

52. Consider the following statements :In a single-server queueing model1. the arrivals is a memoryless process2. the arrivals is described as a Poisson

distribution3. uncertainty concering the demand for

service existsWhich of the above statements are correct?(a) 1 and 2 only (b) 1 and 3 only(c) 1, 2 and 3 (d) 2 and 3 only

Ans. (c)

Sol. In a single server queuing model Arrivals are markov or memoryless Serivce is also memory less. Arrivals follow poisson distribution

53. To construct an operating characteristic curve,an agreement has to be reached betweenproducer and consumer through which of thefollowing points?1. Maximum proportion of defectives that will

make the lot definitely unacceptable2. The producer is willing to aceept that

some of satisfying the quality level (AQL)

will rejected 5%

3. Maximum level of percentage defectivesthat wi l l make the lot def initelyunacceptable

4. The consumer is willing to take lots ofquality level (LTPD) even though they are

unacceptable 10%

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

Ans. (d)



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Sol. To construct an OC curve requirements are

AQL for which produce risk ( 0.05) = is 5%

Consumers risk for LTPD ( 0.1) = or 10% Lot tolerance percent deflective (LTPD)

54. Assuming X and Y are the two control variables,the following are the constraints laid out formaximizing the profit :Maximize profit (P) = 8X + 5Ysubject to

Constraint-1 : 2X Y 1000

Constraint-2 : 3X 4Y 2400

Constraint-3 : X Y 800

Constraint-4 : X Y 350

Constraint-5 : X 0

Constraint-6 : Y 0

Which of the above constraints is a redundantone and does not have any effect on thesolution?(a) Constraint-1(b) Constraint-2(c) Constraint-4(d) Constraint-5 and Constraint-6

Ans. (b)

Sol. Max (p) = 8x + 5y

Constraint 1 2x y 1000

Constraint 1 3x 4y 2400

Constraint 3 x y 800

Constraint 4 x y 350

x 0

y 0

x y500 1000

1

y y

800 600 1

x y

800 800 1

x y

350 350 1

y

100

800

600

Constraint 1

Constraint 3Constraint 2

Constraint 4

x800500

From figure it is clear that constraint -3 doesnot have effect on solution because. It is not abinding constraint.

55. A transportation problem consists of 3 sourcesand 5 destinations with appropriate rimconditions. The number of possible solutionsis(a) 15 (b) 225(c) 6435 (d) 150

Ans. (a)

Sol. For a 3 5 transportation problem. The numberof possible solutions are (5 3) i.e. 15

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56. Maximize 1 2Z 2X 3X

subject to

1 22X X 6

1 2X X 3

1 2X , X 0

The solution to the above LPP is(a) optimal (b) infeasible(c) unbounded (d) degenerate

Ans. (b)

Sol. y

6

3

3x

2x+x =6

12

Hence, there is no feasible region. Thussolution to LPP is infeasible

57. A company has a store which is manned by 1attendant who can attend to 8 technicians inan hour. The technicians wait in the queue andthey are attended on first-come-first-servedbasis. The technicians arrive at the store onan average 6 per hour. Assuming the arrivalsto follow Poisson and servicing to followexponential distribution, what is the expectedtime spent by a technician in the system, whatis teh expected time spent by a technician inthe queue and what is the expected number oftechnicians in the queue?(a) 22.5 minutes, 30 minutes and 2.75

technicians

(b) 30 minutes, 22.5 minutes and 2.25technicians

(c) 22.5 minutes, 22.5 minutes and 2.75technicians

(d) 30 minutes, 30 minutes and 2.25technicians

Ans. (b)

Sol. Service rate = 8/hr =

Arrival rate = 6/hr =

=6 38 4

=

Time spent in the system =

1 60 30 minutes2

= =

Time spent in the queue = 6 608 2

=

= 22.5 minutesExpected technicians in the queue =

2 9/16 91 1/4 4

= = = 2.25 technicians.

58. Objective function

1 2Z 5X 4X (Maximize)

subject to

10 X 12

20 X 9

1 23X 6X 66

1 2X , X 0

What is the optimum value?(a) 6, 9 (b) 12, 5(c) 4, 10 (d) 0, 9



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Ans. (b)

Sol. x2 x1=12

x1=9

(12, 5)

(12, 0) 22 x1(10, 0)

(0, 9)

Feasibleregion

H (4, 9)

Z is maximized at any of the corners of feasibleregion i.e. (12, 5)

59. Which of the following defines the compilersfunction correctly?(a) It translates high-level language

programs into object code(b) It translates object code into a high-level

language(c) It translates object code into assembly

language instructions(d) it translates assembly language

instructions into object code

Ans. (a)

Sol. Compilers function is to translate high-levelprogress into object code.

60. Which one of the following properties of workmaterials is responsible for the material removalrate in electrochemical machining?(a) Hardness(b) Atomic weight(c) Thermal conductivity(d) Ductility

Ans. (b)

Sol. MRR in ECM is given as

mt

=EFI

where I is current in amperes.F = Faradays constant

E is given chemical equivalent = AZ

where A is atomic mass of workpieceZ is valency of anode material

61. In a crank and slotted lever type quick returnmechanism, the link moves with an angularvelocity of 20 rad/s, while the slider moves witha linear velocity of 1.5 m/s. The magnitude anddirection of Coriolis component of accelerationwith respect to angular velocity are

(a) 230 m / s and direction is such as to

rotate slider velocity in the same senseas the angular velocity

(b) 230 m / s and direction is such as to

rortate slider velocity in the oppositesense as the angular velocity

(c) 260 m / s and direction is such as to

rotate slider velocity in the same senseas the angular velocity

(d) 260 m / s and direction is such as torotate slider velocity in the opposite senseas the angular velocity

Ans. (c)



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Sol. The schematic of quick return mechanism

V

C 2V

The Coriolis acceleration

ac = 2V = 2 1.5 20

= 60 m/sec2

Direction: The direction of Coriolis componentof acceleration depends upon velocity of sliderC. If slider moves away from centre, theCoriolis acceleration will be in the direction oflink rotation. If the slider moves toward centreof rotation the Coriolis acceleration will beopposite to link rotation as shown below.

V

C

A

C

A

2V2V

V

Since nothing is mentioned about velocity ofslider (away or toward centre) so assume itmoves away from centre, the right.

62. Which of the following are associated withAckerman steering mechanism used inautomobiles?1. Has both sliding and turning pairs2. Less friction and hence long life3. Mechanically correct in all positions

4. Mathematically not accurate except inthree positions

5. Has only turning pairs6. Controls movement of two front wheels(a) 2, 4, 5 and 6 (b) 1, 2, 3 and 6(c) 2, 3, 5 and 6 (d) 1, 2, 3 and 5

Ans. (a)

Sol. The basic schematic of Ackerman steeringmechanism.

A W D

B C All pairs (A, B, C, and D) are turning pair

in four bar mechanism A, B, C and D. This steering satisfy fundamental equation

of steering wCot cot

= in three

positions only namely in straight motion 0 = and two positions 25 towardleft and right

Since very less sliding surfaces (turning pairsonly) So longer life. Due to longer life it isused generally despite it is not correctmathematically.

Steering control is provided in front wheelsonly in all steering mechanisms.

63. The displacement of a follower of a cam in aprinting machine is represented by theexpression

2 3 4 5x 10 120 1500 2000 2500

where is the angle of rotation of the cam.The jerk given by the system at any position is

(a) 3 3 3 2900 48000 150000

(b) 39000



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(c) 2 2 2 2240 9000 24000

3 350000

(d) 3 3 248000 150000

Ans. (a)

Sol. Displacement of follower in cam

x = 2 3 4 510 120 1500 2000 2500

culure = is angle of rotation of cam.

Since the jerk is third derivative of followerdisplacement

Jerk, J = 3

3d xdt

(i)

2

3 4

dx (10 240 4500dt

d8000 12500 )dt

dxdt =

2

3 4

(10 240 4500

8000 12500 )

Assuming cam rotates with uniform angularvelocity

2d xdt

= 2 3

(0 240 9000d24000 50000 )dt

= 2 2 3(240 9000 24000 50000 )

Jerk 3

3d xdt

= 2

2

(0 9000d48000 150000 )dt

= 3 2(9000 48000 150000 )

64. A body starting from rest moves in a straightline with its equation of motion being

3 2s 2t 3t 2t 1

where s is displacement in m and t is time ins. Its acceleration after one seond is

(a) 26 m / s (b) 22 m / s

(c) 212 m / s (d) 23 m / s

Ans. (a)

Sol. Displacement of body

s = 2t3 3t2 + 2t + 1

Velocity, v = dsdt = 6t

2 6t + 2

Acceleration,

a = 2dv d s

dt dt

= (6 2t) 6 + 0

= 12t 6

Acceleration, at t = 1 sec

a = 12 1 6

= 6 m/sec2

65. The crank shaft of a reciprocating engine havinga 20 cm crank and 100 cm connecting rodrotates at 210 r.p.m. When the crank angle is45 , the velocity of piston is nearly

(a) 1.8 m/s (b) 1.9 m/s(c) 18 m/s (d) 19 m/s



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Ans. (a)

Sol. Reciprocating engine mechanism

xP

Piston

100cm

=

45 =r = 20 cm

Angular velocity of crank

2 N 2 21060 60

= 22 rad/sec

Ratio of length of connecting rod to cranklength

l 100n 5r 20

The velocity of piston-P,

sin2v r sin2n

v = sin900.2 sin452 5

= 1 10.2 22

102

= 4.4 (0.707 + 0.1)

= 1.7756 m/sec

1.8 m/sec66. While designing a cam, pressure angle is one

of the most important parameters which isdirectly proportional to(a) pitch circle diameter(b) prime circle diameter(c) lift of cam(d) base circle diameter

Ans. (b)

67. A four-bar mechanism is as shown in the figurebleow. At the instant shown, AB is shorter thanCD by 30 cm. AB is rotating at 5 rad/s and CDis ratating at 2 rad/s:

B C

A

D

The length of AB is(a) 10 cm (b) 20 cm(c) 30 cm (d) 40 cm

Ans. (b)

Sol. The schematic of mechanism

B C

A

D

1 5rad/sec=

2 2rad/sec=

At the instant shown in figure, the linearvelocity of point B and C will be same

VBA = VCD

1 2AB CD

5AB = 2CD (i)

Since the difference between AB and CD

CD AB = 30

From equation (i)



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CD = 5 AB2

5 AB AB 302

3 AB 302

AB = 2 30

3

= 20 cm

68. A governor is said to be isochrnous when theequilibrium speed is(a) variable for different radii of ratation of

governor balls(b) constant for all radii of ratation of the balls

within the working range(c) constant for particular radii of ratation of

governor balls(d) constant for only one radius of ratation of

governor balls

Ans. (b)

Sol. A governor is said to be isochronous whenequilibrium speed is same or constant for allradii of rotation of balls within working range.The isochronous governor has zero rangeof speed. This isochronism character ispossible in spring controlled governors onlynot in dead weight types.

69. A planetary gear train is a gear train having(a) a relative motion of axes and the axis at

least one of the gears also moves relativeto the frame

(b) no relative motion of axes and no relativemotion of axes with respect to the frame

(c) no relative motion of axes and the axis of atleast one of the gears also moves relativeto the frame

(d) a relative motion of axes and none of theaxes of gears has relative motion with theframe

Ans. (a)

Sol. In planetary gear train or compound epicyclicgear train, the axis of at least one gearrotates (not fixed) about to the fixed frame.The beauty of this train is that it ensureslarge speed reduction in small space.

70. The flywheel of a machine having weight of4500 N and radius of gyration of 2 m has cyclicfluctuation of speed from 125 r.p.m. to 120r.p.m. Assuming g = 10 m/s2, the maximumfluctuation of energy is(a) 12822 N-m (b) 24200 N-m(c) 14822 N-m (d) 12100 N-m

Ans. (d)

Sol. Weight of flywheel, w = mg = 4500 N

Radius of gyration, k = 2 m

Speed fluctuates between

= 125 120 rpm

= 2 (125 120) rad/sec60

= (13.09 12.57) rad/sec

max = 13.09 rad/sec

min = 12.57 rad/sec

Maximum fluctuation of energy

e = Emax Emin

= 2 2max min1 1I I2 2



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= max min max min1I( )( )2

= 2 max min max min1mk ( )( )2

= 21 4500 2 (13.09 12.57)

2 10(13.09 12.57)

= 900 25.66 0.52

= 12009 N.m.

So the nearest answer is 12100 N.m.71. Alumina doped with magnesia wil l have

reduced thermal conductiv ity because itsstructure becomes(a) amorphous(b) free of pores(c) crystalline(d) mixture of crystalline and glass

Ans. (c)

72. Which of the fol lowing statements areassociated with complete dynamic balancingof rotating systems?

1. Resultant couple due to all inertia forces iszero

2. Support reactions due to forces are zerobut not due to couples

3. The system is automatically staticallybalanced

4. Centre of masses of the system lies on theaxis of rotation.

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

Ans. (d)

Sol. For complete dynamic balancing of rotatingsystems,

(1) There should not be any couple in thesystem

(2) System should be statically balancedi.e., centre of mass of the systemshould be on axis of rotation

Hence in complete dynamic balance, thereaction on the support are due to weight ofsystem and remain constant during rotationof system i.e., no force should airse due torotation.

73. Which of the following statements is correctabout the balancing of a mechanical system?(a) If it is under static balance, then there will

be dynamic balance also(b) If it is under dynamic balance, then there

will be static balance also(c) Both static as well as dynamic balance have

to be achieved separately(d) None of the above

Ans. (b)

Sol. For complete balancing or simply balancingof mechanical system. The dynamicbalancing i.e., no couple, requires staticbalancing (centre of mass of system at axisof rotation) of system as pre condition.

74. The accelerometer is used as a transducer tomeasure earthquake in Richter scale. Its designis based on the principle that(a) its natural f requency is very low in

comparison to the frequency of vibration(b) its natural frequency is very high in

comparison to the frequency of vibration(c) its natural frequency is equal to the

frequency of vibration



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(d) measurement of vibratory motion is withoutany reference point.

Ans. (c)

The natural frequency of accelerometer shouldbe kept around frequency of vibration/excitationin order to read the very weak earthquake.In case of resonance the peak is control byproper dampnig.

75. As compared to the time period of a simplependulum on the earth, its time period on themoon will be(a) 6 times higher(b) 6 times lower

(c) 6 times higher

(d) 6 times lower

Ans. (c)

Sol. The time period of simple pendulum on earth

Te = el2

g (i)

Since on Moon, the acceleration due to

gravity in 1 th6 of acceleration due to gravity

at earth.

gm = eg

6

Tm = m

l2g

= e

l2g6

= e

6l2g

= e

l2 . 6g

= eT 676. While calculating the natural frequency of a

spring-mass system, the effect of the mass ofthe spring is accounted for by adding X timesits value to the mass, where X is

(a)12

(b)13

(c)14

(d)34

Ans. (b)

Sol. The period of oscillation of spring-masssystem including the mass of spring.

k

M

ms

T = s

s

mM M Xm32 2k k

X = 13

77. A block of mass 10 kg is placed at the free endof a cantilever beam of length 1 m and secondmoment of area 300 mm4. Taking Youngs



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modulus of the beam material as 200 GPa, thenatural frequency of the system is

(a) 30 2 rad s (b) 2 3 rad s

(c) 3 2 rad s (d) 20 3 rad s

Ans. (c)

Sol. The cantilever vibrator,

M =10 kg

1m =

Second moment of inertia,

I = 300 mm4

Youngs modulus,

E = 200 GPa

The deflection at free end

3

stmg3E

I

Stiffness constant

k = 3st

mg 3E

I

k = 9 12

33 200 10 300 10

= 600 300 103

= 180 N/m

Frequency of oscillation

n = k 180m 10

= 3 2 rad/sec

78. The speed rating for turbine rotors is invariablymore than 2 times its natural frequency to

(a) increase stability under heavy load and highspeed

(b) isolate vibration of the system from thesurrounding

(c) minimize deflection under dynamic loadingas well as to reduce transmissbility of forceto the surrounding

(d) none of the above

Ans. (b)

Sol. The rotating speed of turbine rotor is morethan 2 times of natural frequency n( ) ofsystem. Because beyond this speed, thetransmissibility Tr is less than one as shownbelow

'Tr'

1.0

1.0 2=n( / )

Turbine rotorspeed range

Hence force transmitted to foundation byturbine is less than it impress upon isolator.So the v i brat ions are isolated f romsurroundings.

79. The magnitude of swaying couple due to partialbalance of the primary unbalnacing force inlocomotive is(a) inversely proportional to the reciprocating

mass



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(b) directly proportional to the square of thedistance between the centrelines of the twocylinders

(c) inversely proportional to the distancebetween the centrelines of the two cylinders

(d) directly proportional to the distance betweenthe centrelines of the two cylinders

Ans. (d)

Sol.

2(1 c) mr cos

2(1 c) mr cos 90

x x

a/2

a/2

Cylinder1

Cylinder-2

The swaying couple due to unbalance forceabout centre line x x

= 2

2

a(1 c)mr cos2

a(1 c)mr cos( 90)2

= 2mr a(1 c) (cos sin )

2

80. The power of a governor is the work done at(a) the governor balls for change of speed(b) the sleeve for zero change of speed(c) the sleeve for a given rate of change of

speed

(d) each governor ball for given percentagechange of speed

Ans. (c)

Sol. The power of governor is defined as = sleevedisplacement effort

The effort of governor is defined as averageforce on sleeve for a given change of speed.So the power can be defined as work doneat sleeve for given rate of change of speed.

81. Copper has FCC structure; its atomic radius is1.28 A and atomic mass is 63.5. The density ofcopper will be(a) 8.9103 kg/mm3

(b) 8.9103 kg/cm3

(c) 8.9103 kg/m3

(d) 8.9103 g/mm3

Ans. (c)

Sol. Density (g/m3) = Mass No. of atoms in a unit

cellVolume Avagadro's

Number

Volume of a unit cell = a3 = 3(2 2R)

= 47.46 1030 m3

= 30 234 63.5

47.46 10 6.023 10

= 8.9 106 g/m3

or 8.9 103 kg/m3

82. A plane intersects the coordinate axes at

2 1 1x ,y and z3 3 2

. What is the Miller index

of this plane?(a) 932 (b) 432(c) 423 (d) 364



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Ans. (d)

Sol.

1/2

1/323

Intercepts are 2 1 13 3 2

Take reciprocal intercepts i.e. 3 3 22

Converting to whole number by multiplying by2miller indices = (3 6 4)

83. What is the diameter of the largest sphere interms of lattice parameter , which will fit thevoid at the centre of the cube edge of a BCCcrystal?

(a) 0.134 (b) 0.25

(c) 0.433 (d) 0.5

Ans. (a)

Sol.

a

a

a

The void formed is an octahedral void. Thelargest radius that can jet in it

Roctahedral = a R2

but a =4 R3 for BCC will only

=2 2R R 1 R3 3

=

Roctahedral = 0.158 R

also R = 3 a4

Roctahedral = 0.158 3 a

4

= 0.0684 aDoctahedral= 2 Roctahedral = 0.136 a

84. If the atomic radius of aluminium is r, what isits unit cell volume?

(a)32r

2

(b)34r

2

(c)32r

3

(d)34r

3

Ans. (b)

Sol. Aluminium has FCC crystal structure

a = 2 2r

volume of unit cell = a3

= 3(2 2r) = 3 3 82 2r = 16 2r

or 34r

2

85. Consider the following staements regarding thebehaviour of dislocations:

1. Only edge dislocation and mixed dislocationcan have glide motion.



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2. A screw dislocation cannot have glidemotion.

3. Dislocat ion moves in the direct ionperpendicular to that of shear stress.

4. Motion of disloacation occurs on slip planethat contains Burgers vector and directionvector.

Which of the above statements are correct?(a) 1, 2 and 3(b) 1, 2 and 4 only(c) 2, 3 and 4 only(d) 1, 2, 3 and 4

Ans. (d)

Sol. Dislocation Only edge & mixed dislocation can glide

along a specific plane defined. For screw dislocation there is no specific glide

plane defined. Dislocations move perpendicular to direction

of shear stress along a slip plane containingBurgers vector & direction vector.

86. A binary alloy of Cu and Ni containing 20 wt%Ni at a particular temperature coexists with solidphase of 26 wt% Ni and liquid phase of 16wt% Ni. What is the weight ratio of solid phaseand liquid pahse?(a) 1 : 1 (b) 3 : 2(c) 2 : 3 (d) 1 : 2

Ans. (b)

Sol. Cu-Ni phase diagram is given as (It is a binaryisomorphos system)

L

L+S1453C

S

C2C0C1

1085C

0%100%

80%Cu20%Ni

%composition

Cu100%Ni0%

C =16% Ni1C =26% Ni2

{Given data}

Temp.

wsolid (at 20% Ni) = 2 0

2 1

C C 0.26 0.2C C 0.26 0.16

=

wliquid = 1 wsolid = 0.4 or 40%

solid

liquid

ww =

0.6 30.4 2

=

87. Elements A and B form eutectic type binaryphase diagram and the eutectic composition is60 wt% B. If just below eutectic temperature,the eutectic phase contains equal amounts (bywt) of two solid phases, then the compositionsof the two solid phases are(a) 20 wt% B and 90 wt% B(b) 30 wt% B and 90 wt% B(c) 20 wt% B and 80 wt% B(d) 30 wt% B and 80 wt% B

Ans. (b)

Sol. Equilibrium phase diagram for a eutectic alloyis

eutectic point

L L L

xA 100%B 0%

40% A60% B

y A 0%B 100%

Just below eutectic point, the composition of



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two phases (i.e. & ) be x and yrespectively.

wd =y 0.6 0.5(given)y x

=

y 0.6 = 0.5y 0.5 x

y 0.5y + 0.5 x = 0.60.5(x + y) = 0.6

x+y = 1.2Similarly

w =0.6 x 0.5y x

=

0.6 x = 0.5y 0.5 x0.6 = 0.5y + 0.5x

(x + y) = 1.2There, is only one option satisfying theconditions i.e. 30 wt% B and 90 wt% B is

and phases.

88. Consider the following statements :In a binary phase diagram

1. the freezing point of the alloy is minimum2. eutectic mixture solidifies at a constant

temperature like pure metal3. eutectic reaction is irreversible4. at eutectic temperature, liquids of two

metals will change into two solidsWhich of the above statements are correct?(a) 1, 2 and 3 only(b) 1, 3 and 4 only(c) 1, 2 and 4 only(d) 1, 2, 3 and 4

Ans. (c)

Sol. Eutectic reaction is

L It is a reversible reaction which can take placeboth ways.

89. At room temperature, -iron contains negligibleamount of carbon, cementite contains 6.67%C and pearlite contains 0.8% C. Pearlitecontains how much cementite?(a) 8% (b) 10%(c) 12% (d) 14%

Ans. (c)

Sol. At room temperatureAmount of ferrite %carbon in ferrite + Amountof cementite % carbon in cementite = overall% carbon in pearlite. as % carbon in ferrite isnegligible at toom temperature.Amount of cementite 6.67 = 0.8

cementite =0.8 0.1199 or 12%

6.67=

90. Two metals A and B are completely immisciblein sloid and liquid state. Melting point of A is800 C and melting point of B is 600 C. Theyform eutectic at 200 C with 40% B and 60%A. The 50% B alloy contains.(a) 83.33% B and 16.67% of eutectic(b) 83.3% of eutectic and 16.67% B(c) 50% B and 50% of eutectic(d) 40% B and 60% of eutectic

Ans. (b)

Sol. For no miscibility in liquid and solid state, thephase diagram is



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Eutecticpoint

A600C

800C

200C

A 100CB%

0%100%

C2

%Composition

60%A 50%A40%B 50%B

Temp.

C1 C0

Composition at point A i.e. (50%B 50%A)can be found oil by lever rule.

Quantity of B = 0 1

2 1

C C 50 40 10C C 100 40 60

= =

= 16.67% of B.

Quantity of eutectic = 2 0

2 1

C C 100 50C C 100 40

=

=5060

83.33%=

91. What is the interplanar spacing between (200),(220), (111) planes in an FCC crystal of atomicradius 1.246 A?(a) d(200) = 1.762 , d(220) = 1.24 and d(111) =

2.034 (b) d(200) = 1.24 , d(220) = 1.762 and d(111) =

2.034 (c) d(200) = 2.034 , d(220) = 1.24 and d(111) =

1.762 (d) d(200) = 2.5 , d(220) = 4.2 and d(111) =

2.6

Ans. (a)

Sol. Inteplaner spacing is given as

d =2 2 2

a

h k

for FCC

a2 + a2 = (4R)2

2a2 = 16R2

a2 = 8R2

a = 2 2R

4R

a

a

R = 1.246A (given)

a = 2 2 1.246 = 3.524A

d200 = 2 2 23.524 1.762 A

2 0 0

=

d220 = 2 2 23.524 1.246 A

2 2 0

=

d111 = 2 2 23.524 2.034 A

1 1 1

=

92. Rotary swaging is a process for shaping(a) round bars and tubes(b) billets(c) dies(d) rectangular blocks

Ans. (a)

Sol. Rotary swaging is used to shape round barsand tubes such as gun barrels

93. Consider the following statements:In shell moulding1. a single parting plane should be provided

for mould2. detachable pattern parts and cores could

be included3. minimum rounding radii of 2.5 mm to 3 mm

should be used4. draft angles of not less than 1 should be

usedWhich of the above statements are correct?(a) 1, 3 and 4 only (b) 1, 2 and 3 only



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

Ans. (a)

Sol. In shell molding Single parting plane should be provided Detachable pattern parts & cores should be

avoided. Minimum rounding radii is 2.5 to 3 mm Draft angle should not be less than 1.

94. A big casting is to have a hole, to be producedby using a core of 10 cm diameter and 200 cmlong. The density metal is 0.077 N/cm3 and

density core is 0.0165 N/cm3. What is theupward force acting on the core prints?(a) 200.5 N (b) 1100.62 N(c) 950.32 N (d) 350.32 N

Ans. (c)

Sol. Buoyant force acting on core.

metal core( ) 2/4d

= 2/4 10 200 0.077 0.0165

= 950.33 N95. Consider the following :

The purpose of lapping process is1. to produce geometrically true surface2. to correct minor surface imperfections3. to improve dimensional accuracy4. to provide very close fit between the contact

surfaceWhich of the above are correct?(a) 1, 2 and 3 only (b) 1, 3 and 4 only(c) 2, 3 and 4 only (d) 1, 2, 3 and 4

Ans. (d)

Sol. Lapping gives Geo metrically true surface Corrects minor surface imperfection Improves dimensional accuracy

Very close fit between mating surface96. Centre lathe is to be used to cut inch thread of

4 threads per inch. Lead screw of lathe has 3mm pitch. Then change gear tobe used is

(a)1

12(b)

12760

(c)30

127(d)

2080

Ans. (b)

Sol. Threads per inch (TPI) = 25.4

pitch in mm

pitch in mm = 25.4 25.4TPI 4

=

Also pitch (mm) = Gear ratio pitch of leadscrew.

Gear ratio = 25.4 25.4 254 or 4 3 12 120

=

or 12760

97. Consider the following statements in respect ofthe oxidizing flame due to excess of oxygen inwelding:1. At high temperature, it combines with many

metals to form hard and brittle oxides.2. It causes the weld bead and the surrounding

area to have a scummy appearance.3. It has good welding effect in welding of

copper-base metal.Which of the above statements are correct?(a) 1, 2 and 3 (b) 1 and 2 only(c) 1 and 3 only (d) 2 and 3 only

Ans. (a)

Sol. Oxidising flame Forms hard oxides which protect the weld

metal



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Gives dirty & scummy appearance to weldbead

Has good welding effect on copper basealloys, manganese steels and east iron.

98. A cutter tip is initially at X = 10 mm, Y = 20mm. In a rapid motion, using G00 code, itmoves to X = 160 mm, Y = 120 mm. The Xand Y axes have maximum speed of 10000mm/min and 5000 mm/min respectively.Operating at maximum speed, what is the timeit will take to reach the destination?(a) 0.90 s (b) 1.08 s(c) 1.20 s (d) 2.16 s

Ans. (d)

Sol. (160 , 120)

(10, 20)

y

x

For y axis cutter has to travel 100 mm at 5000

mm/mm 250 mm/sec3

i.e. it will take =

100 3 1.2s250

=

For x axis cutter has to travel 150 mm at 10000

mm/min 500 mm/s3

i.e. = 150 3 0.9s

500 =

Total time = 0.9 + 1.2 = 2.1 s99. If n = 0.5 and C = 300 for the cutting speed

and the tool life relation, when cutting speed isreduced by 25%, the tool life will be increasedby(a) 100% (b) 95%(c) 78% (d) 50%

Ans. (c)

Sol. Taylors tool life equation isvTn = C

v1 (T1)0.5 = 300 ...(i)

when speed is reduced by 25%0.75v1 (T2)

0.5 = 300 ...(ii)Comparing (i) & (ii)

0.51 1V T =

0.51 20.75v T

0.52

1

TT

=1

0.75

2

1

TT =

2 21 4 160.75 3 9

= =

T2 = 116 T9

Increase in tool life

= 1 1

2 1

1 1

16 T TT T 100 100T T

=

= 700 77.78% or 78%9

=

100. Which of the following statements are correctfor temperature rise in metal cutting operation?1. It adversely affects the properties of tool

material.2. It provides better accuracy during machining.3. It causes dimensional changes in workpiece

and affects accuracy of machining.4. It can distort the accuracy of machine tool

itself.(a) 1 and 2 (b) 2 and 3(c) 3 and 4 only (d) 1, 3 and 4



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Ans. (d)

Sol. Temperature rise does not give better accuracyduring machining. HIgh temperature affectshardnen of cutting edge. It can also distort thetool due to thermal expansion.

101. Consider the following:The parallel fillet welded joint is designed for1. tensile strength2. compressive strength3. bending strength4. shear strengthWhich of the above is/are correct?(a) 4 only (b) 3 only(c) 2 and 3 (d) 1 and 4

Ans. (a)

Sol. The parallel fillet welds along with force F.

Thin weld is designed for shear strength ofweldment throat, the thickness of threat.

hh

k

F

t = 2h2

= h2

= 0.707 h

Area of threat,

A = t = 0.707 h

Shear stress,

F FA 0.707h

The force are parallel to minimum threat areaso it is designed for shear strength only.

102. If the permissible crushing stress for thematerial of a key is double the permissibleshear stress, then the sunk key will be equallystrong in sheaering and crushing if the key isa(a) rectangular key with width equal to half the

thickness(b) rectangular key with width equal to twice

the thickness(c) squre key(d) rectangular key with width equal to one-

fourth the thickness

Ans. (c)

Sol. Given

c = 2

Also crushing force = ct2

Crushing torque = ct d2 2

Shearing torque = d2

for equal strength in shearing & crushing

d2

= ct d2 2

t

= c2

t

= 1 or t =

Thus, it should be a square key



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103. Very small quantity of carbon in iron as in steelsforms interstitial solid solution mainly becauseatomic size(s) of(a) carbon and iron are almost same(b) iron is very much smaller than that of carbon(c) carbon is very much smaller than that of

iron(d) None of the above

Ans. (c)

Sol. Carbon forms interstitial solid solution with ironas size of carbon atom is very small ascompared to size of iron atoms.

104. In a cotter joint, the width of the cotter at thecentre is 5 cm, while its thickness is 1.2 cm.The load acting on the cotter is 60 kN. Theshear stress developed in the cotter is(a) 50 N/mm2 (b) 100 N/mm2

(c) 120 N/mm2 (d) 200 N/mm2

Ans. (b)

Sol. Shear stress in cottor

c =3P 60 10 60000

2 b t 12 50 600

= =

= 100 N/mm2

105. The use of straight or curved external gearteeth in mesh with internal teeth in gear andspline couplings is specifically employed toaccommodate.(a) torsional misalignment(b) parallel misalignment(c) angular misalignment(d) substantial axial movements between shafts

Ans. (d)

Sol. The gear and spline couplings are used intorque transmission when key fails to servethe purpose. These couplings can connect

or disconnect the driving and driven shaftaccording to wish of operator of requirement.This operat ion happens due to ax ialmovement of spline hub or shaft ember.Example clutch. The shaft in spline is underpure torsion i.e., shear.

106. For a power screw having square threads withlead angle of 45 and coefficients of friction of0.15 between screw and nut, the efficiency ofthe power screw, neglecting collar friction, isgiven by(a) 75% (b) 64%(c) 54% (d) 44%

Ans. (a)

Sol. Efficiency of power screw is 90 45 = =

= tan 1 tan tantan

tan tan tan

=

= tan45 1 tan45 0.15

0.15 tan45

= 1 1 0.15 0.739 or 74%

1.15 =

107. Aquaplaning occurs in vehicle tyres when thereis continuous film of fluid between the tyre andthe wet road. It leads to(a) oscillatory motion of the vehicle(b) jamming the brakes of the vehicle(c) jamming the steering mechanism of the

vehicle(d) loss of control of the vehicle

Ans. (d)

Sol. Aquaplaning or hydroplaning is the conditionin vehicle or aircraft when there is water layerbetween tyre and road surface. In thissituation, there is excessive skidding orslipping of vehicle i.e., loss of control of



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vehicle. This situation can be very dangerouswhen al l type hav e equaplaningsimultaneously.

= 1000.98 = 102.04 rpm.

108. If the angle of wrap on smaller pulley ofdiameter 250 mm is 120 and diameter of largerpulley is twice the diameter of smaller pulley,then the centre distance between the pulleysfor an open belt drive is(a) 1000 mm (b) 750 mm(c) 500 mm (d) 250 mm

Ans. (d)

Sol. Given D1 = 2D2or r1 = 2r2

sin = 1 2r r

x

where x is the center distance between thepulleysfor smaller pulley

wrap angle = 180 2 = 120

2 = 60

or = 30

sin 30 =250 125

x

x =125 250mm

sin30=

109. If the velocity ratio for an open belt drive is 8and the speed of driving pulley is 800 rpm.,then considering an elastic creep of 2% thespeed of the driven pulley is(a) 104.04 rpm (b) 102.04 rpm(c) 100.04 rpm (d) 98.04 rpm

Ans. (b)

Sol. Velocity of driver

v2 =1v 800 100 rpm

V R 8= =

Due to creep

2v = 2V 100

1 1 0.02 =

= 102.04 rpm110. Two shafts A and B of same material, and A is

twice the diameter of B. The torque that canbe transmitted by A is(a) 2 times that of B (b) 8 times that of B(c) 4 times that of B (d) 6 times that of B

Ans. (b)

Sol. Torque transmitted by a shaft 3T d

TA = 3

B B2 T 8T=

111. A worm gear set is designed to have pressureangle of 30 which is equal to the helix angle.The efficiency of the worm gear set at aninterface friction of 0.05 is(a) 87.9% (b) 77.9%(c) 67.9% (d) 57.9%

Ans. (a)

Sol. Helix angle of worm = 30Load angle of worm ( ) = 90 30 = 60

efficiency of worm gear = 1 tan1 /tan

=1 0.05 tan601 0.05/tan 60

=0.9134 0.88771.0289

=



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or 88.77%112. Consider the following statements :

The axes of spiral bevel gear are non-paralleland intersecting1. The most common pressure angle for spiral

bevel gear is 20.2. The most common spiral angle for spiral

bevel gear is 35.3. Spiral bevel gears are generally

interchangeable.4. Spirals are noisy and recommended for low

speeds of 10 m/s.Which of the above statements are correct?(a) 1 and 4 (b) 1 and 2(c) 2 and 3 (d) 3 and 4

Ans. (b)

Sol. Bevel gears are inherently non-interchangeable.They are less noisy. Most common pressureangle is 20C and most common spiral angleis 35.

113. Consider the following statements:In case of helical gears, teeth are cut at anangle to the axis of rotation of the gears.1. Helix angle introduces another ratio called

axial contact ratio.2. Transverse contact ratio is equal to axial

contact ratio in helical gears.3. Large transverse contact ratio does not allow

multiple teeth to share the load.4. Large axial contact ratio will cause larger

axial force component.Which of the above statements are correct?(a) 1 and 2 (b) 2 and 3(c) 1 and 4 (d) 3 and 4

Ans. (d)

Sol. Helical gears have two contact ratios i.e. Axialcontact ratio & transverse contact ratio. Both

contact ratios are not equal. Large contactratios lead to engagement of multiple teeth.

114. In an interference fit between a shaft and ahub, the state of stress in the shaft due tointerference fit is(a) only compressive radial stress(b) a tensile radial stress and a compressive

tangential stress(c) a tensile tangential stress and a compressive

radial stress(d) a compressive tangential stress and a

compressive radial stress

Ans. (d)

Sol. For interference fit between hub & shaft, shaftis considered externally pressed

radial = Pf (external pressure)

tangential =

2 2f f L

2 2f L

P r r

r r

rf = outer radiusfi = inner radius

Thus, both radial & tangential stream arecompressive.

115. In case the number of teeth on two bevel gearsin mesh is 30 and 60 respectively, then thepitch cone angle of the gear will be

(a) tan1 2 (b) 1tan 22

(c) 1tan 0.52

(d) tan1 0.5

Ans. (a)

Sol. Pitch cone angle of gear (for metregears i.e.where is angle between two shafts.



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P2 =1 1G

P

T 60tan tanT 30

=

= tan1 (2)116. In skew bevel gears, the axes are

(a) non-parallel and non-intersecting, and teethare curved

(b) non-parallel and non-intersecting, and teethare straight

(c) intersecting, and teeth are curved andoblique

(d) intersecting, and teeth are curved and canbe ground

Ans. (a)

Sol. In skew bevel gears, the axes are non-paralleland non-intersecting, and teeth are curved.

117. Consider that modern machines mostly useshort bearings due to the following reasons:1. l/d of the most modern bearings is in the

range of 1/4 to 22. No end leakage of oil from the bearing3. Shaft deflection and misalignment do not

affect the operation4. Can be applied to both hydro-dynamic and

hydrostatic casesWhich of the above are correct?(a) 1 and 4 (b) 2 and 3(c) 1 and 3 (d) 2 and 4

Ans. (c)

Sol. A long bearing is where d ( = length d= diameter)A shaft bearing has advantages such as (i)shaft deflection and misalignment do notaffect operation (ii) compact design (iii) Runcooler. However, end leakage is a problem.On the other hand, long bearings havegreater load carrying capacity and end

leakage of oil is not a problem.118. Consider the following statements in connection

with thrust bearings:1. Cylindrical thrust bearings have higher

coefficient of friction than ball thrust bearings.2. Taper rollers cannot be employed for thrust

bearings.3. Double-row thrust ball bearing is not

possible.4. Lower race, outer race and retainer are

readily separate in thrust bearings.Which of the above statements are correct?(a) 1 and 2 (b) 2 and 3(c) 3 and 4 (d) 1 and 4

Ans. (d)

Sol. Cylindrical thrust bearings have highercoefficeint of friction than ball thrust bearings

Both taper roller & double row thrust ballbearings are used.

119. The behaviour of metals in which strength of ametal is increased and the ductility is decreasedon heating at a relatively low temperature aftercold-working is known as(a) clustering (b) strain aging(c) twinning (d) screw dislocation

Ans. (c)

Sol. Twinning is a plane defect wherearrangement of atoms on either side of atwin plane are identical. Twinning occurseither as mechanical twinning or Annealingtwins during annealing heat treatment.Twinning increases strength & reducesductility as twin planes hinders the movementof dislocations.

120. If the equivalent load in case of a radial ballbearing is 500 N and the basic dynamic loadrating is 62500 N, then L10 life of this bearing



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is(a) 1.953 million of revolutions(b) 3.756 million of revolutions(c) 6.953 million of revolutions(d) 9.765 million of revolutions

Ans. (a)

Sol. Use of a ball bearing is given as:

L10 =kc

w

where L10 = life in million revolutionsc = Basic dynamic load ratingw = equivalent dynamic load

k = 3 for ball bearing

L10 =362500

500

= (125)3 = 19531251.953125 million of revolution

me paper ii obj.sol.(ese 2015)

Documents

isfalsed statement

explanation of statement

becausevelocity ratio

ratio of numberof teeth

contact of follower

ratio of pitch diameters

angular velocity

involute profiles