Me Paper II Obj.sol.(Ese 2015)

ESE-2015 Detailed Exam Solutions (Objective Paper - II) Mechanical

solutions

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

Codes :

Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I)

(b)

Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I)

(c)

Statement (I) is true but Statement (II) is false

(d)

Statement (I) is false but Statement (II) is true

Statement (I) : The cam in contact with a follower is a case of complete constraint.

3.

Sol.

2.

S

4.

IE

Sol.

Involute pinion can not have any number of teeth because a minimum number of teeth are decided by interference phenomenon. Both involute and cycloidal teeth satisfy constant velocity ratio condition. Because velocity ratio depends upon ratio of number of teeth or ratio of pitch diameters. Statement (I) : Hooke’s joint connects two nonparallel non-intersecting shafts to transmit motion with a constant velocity ratio.

The cam in contact of follower is case of successful constant. Because spring force is required to maintain the contact. This spring force does not guarantee the contact all time because after certain speed, the follower losses contact with can due to inertia force. Statement (I) : Involute pinions can have any number of teeth. Statement (II) : Involute profiles in mesh satisfy the constant velocity ratio condition.

Statement (II) : Hooke’s joint connects two shafts the axes of which do not remain in alignment while in motion.

Ans. (c)

Statement (II) : The pair, cam and follower, by itself does not guarantee continuity of contact all the time. Ans. (c)

R

Sol.

AS

(a)

M

1.

Ans. (c)

TE

Directions : Each of the following twenty (20) items consists of two statements, one labelled as ‘Statement (I)’ and the other as ‘Statement (II)’. Examine these two statements carefully and select the answers to these items using the codes given below.

The Hooke’s joint connects two non-parallel shafts but intersecting. For constant velocity ratio there are two Hooke’s joints in particular torks orientation Statement (I) : Lewis equation for design of involute gear tooth predicts the static load capacity of a cantilever beam of uniform strength. Statement (II) : For a pair of gears in mesh, pressure angle and module must be same to satisfy the condition of interchangeability and correct 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 is more than 1.5.

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Ans. (d)

Statement (I) : Centrifugal clutches are designed to provide automatic and smooth engagement of load to driving member.

Sol. Main strength of composite (i.e. febreglan) comes from glass fibres and not the polymer. 10.

Statement (I) : Industrial rotors will not have uniform diameter throughout their lengths.

TE

6.

Ans. (c)

R

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

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

Statement (II) : Since the operating centrifugal force is a function of square of angular velocity, the friction torque for accelerating a load is also a function of square of speed of driving member.

AS

Ans. (a)

Ans. (a)

Statement (I) : Heating the steel specimen in the furnace up to austenitize temperature f ollowed by f urnance cooling is termed annealing.

M

7.

Statement (II) : These rotors will have to accommodate transmission elements like pulleys and gears and supports like anti-friction bearings.

Sol. Difference diameter on shafts are to support transmission elements. Shoulder are provided to restrict axial movement of bearings. 11.

Statement (II) : Annealed steel specimen possesses fine pearlitic structure.

Statement (II) : Cored induction furnace, though most efficient, requires a liquid metal charge while starting.

S

Ans. (c)

8.

IE

Sol. Annealed steel specimen has a “coarse pearlite” structure and not “fine pearlite” due to slow cooling in furance. Statement (I) : The susceptibility of a ferromagnetic material decreases with an increase in Curie temperature. Statement (II) : A critical temperature at which the alignment of magnetic moments vanishes is called Curie temperature.

Statement (I) : Cored induction furnace cannot be used for intermittent operation.

Ans. (a) Sol. Cored induction furnace requires a liquid metal charge while starting. Therefore, they cannot be used for intermittent operations. 12.

Statement (I) : Low-carbon steel has high weldability and is more easily welded. Statement (II) : Higher carbon contents tend to soften the welded joints resulting in development of cracks.

Ans. (a) Sol. Magnetic suspetibility occurs above curie temperature only. Thus, higher the curie temperature lower is the suseptibility. 9.

Statement (I) : Fiberglass is a polymer composite made of a plastic matrix containing fine fibers of glass.

Ans. (c) Sol. High carbon content tends to make the weld more brittle and thus more prone to cracking. Low carbon steels have excellent weldeability due to low carbon content.

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Statement (I) : For cutting multi-start threads, the speed ratio is expressed in terms of the lead of the job thread and lead of the lead screw threads.

16.

Statement (II) : The speed of the job is reduced to one-third or one-fourth of the job speed used in the turning operation.

Statement (I) : In chain drives, angle of articulation through which link rotates during engagement and disengagement, is greater for a small number of teeth. Statement (II) : The greater angle of articulation will increase the life of the chain.

Ans. (c) 14.

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

TE

Ans. (b) Statement (I) : The Bauschinger effect is observed in tension test of mild steel specimen due to loss of mechanical energy during local yielding.

17.

AS

Statement (II) : The Bauschinger effect is a function of section modulus of specimen under test. Ans. (c)

The Baeeschinger effect refers to a property of materials where the material’s stress strain characteristics change as a result of the microscopic stress distribution of material. It is observed in tensile test of mild steel specimen.

Statement (II) : Ceramic tools can be used on hard-to-machine work material. Ans. (b) Sol. Ceramic tools have very brittle tool tips, that is why they all prone to impact loads. These tools are used on hard to machine work material such as cast iron as they are highly wear & abrasion resistant.

Statement (II) : The DNC is a manufacturing process in which a number of process machines are controlled by a computer through direct connection and real time analysis.

Sol. Both Statement I and II are correct, but Statement II does not explain Statement I, 18.

This effect is a material property and not a geometric property. So it is not dependent on section modulus. Statement (I) : The ceramic tools used in machining of material have highly brittle tool tips.

IE

15.

Statement (I) : The CNC is an NC system utilizing a dedicated stored program to perform all numerical control functions in manufacturing.

Ans. (b)

S

M

Sol.

R

13.

Statement (I) : In interference fit, the outer diameter of the shaft is greater than the inner diameter of the hole. Statement (II) : The amount of clearance obtained from the assembly of hole and shaft resulting in interference fit is called positive clearance.

Ans. (c) Sol. Clearance obtained in an interference fit is a negative clearance. 19.

Statement (I) : One of the most commonly used techniques for testing surface integrity of material is metallography. Statement (II) : Surface integrity of a material does not contribute for the mechanical and metallurgical properties.

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Ans. (c)

c = 50 N/mm2

20.

22.

Statement (I) : The change in critical path requires rescheduling in a PERT network. Statement (II) : Some of the activities cannot be completed in time due to unexpected breakdown of equipment or non-availability of raw materials.

100

Ans. (a)

50 N/cm2 compressive

(c)

100 N/cm2 tensile

(d)

25 N/cm2 tensile

1 = 100 N/cm2

Sol.

2 = – 50 N/cm2

(b) 33.33 N/mm2

On maximum shear stress plane, Normal stress

50 N/mm2 N/mm2

(b)

M

(c)

75 N/cm2 tensile

Ans. (d)

AS

A copper rod of 2 cm diameter is completely encased in a steel tube of inner diameter 2 cm and outer diameter 4 cm. Under an axial load, the stress in the steel tube is 100 N/mm 2. If Es = 2Ec, then the stress in the copper rod is (a)

(a)

TE

Ans. (a) 21.

A system under biaxial loading induces principal stresses of 100 N/cm2 tensile and 50 N/cm2 compressive at a point. The normal stress at that point on the maximum shear stress plane is

R

Sol. Surface integrity affects mechanical properties such as fatigue strength of material.

(d) 300

N/mm2

 = avg

2cm

For copper d = 2cm

S

=

For steel

23. Copper

di = 2 cm

In a biaxial state of stress, normal stresses are

x  900 N / mm2 ,

= 20 mm

 y  100 N / mm2

principal stress is

= 40 mm

Steel

s = 100N/mm2 4cm

(a)

800 N / mm2

(b) 900 N / mm2

(c)

1000 N / mm2

(d) 1200 N / mm2

Strain in steel = strain in copper c s = Ec Es

c =

100  Ec Es

and

shear stress   300 N / mm2 . The maximum

d0 = 4 cm

Es = 2Ec

1  2 100  50 = 2 2

= 25 N/cm2 (Tensile)

= 20 mm

IE

Sol.

Ans. (c) Sol.

 x = 900 N/mm2

 y = 100 N/mm2 xy = 300 N/mm2

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Maximum principal stress   x  y  1 1 =    2  2

x  y 

Ans. (c) 2

Sol.

 4 2xy

 x = 100 MPa

 y = 200 MPa

2 2  900  100  1  900  100   4   300   =   2  2

1  500  = 1000 N/mm2 2

R

= 500 

1  2 =  x  y 2 = 100  200   250

TE

1 640000  360000 2

(a)

Temperature and composition

(b)

Temperature and phases present

(c)

Temperature, composition and phases present

(d)

Temperature and pressure

Ans. (c)

S

Sol. A constitutional diagram gives information regarding temperature, composition and phases present. The state of stress at a point in a body is given

IE

25.

   y  1 1 =  x   2  2

AS

A constitutional diagram shows relationship among which of the following combinations in a particular alloy system?

by x  100 MPa and y  200 MPa . One of the principal stresses 1  250 MPa . The magnitudes of the other principal stress and

100 =

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

x  y 

2

 42xy

1 100  200 2  42xy 2

40000 = 10000 + 4 2xy 4 2xy = 30000

xy = 50 3 MPa 26.

Consider the following statements regarding powder metallurgy : 1.

Refractory materials made of tungsten can be manufactured easily.

2.

In metal powder, control of grain size results in relativ ely much unif orm structure.

3.

The powder heated in die or mould at high temperature is then pressed and compacted to get desired shape and strength.

4.

In sintering, the metal powder is gradually heated resulting in coherent bond.

the shearing stress xy are respectively (a)

= 50 MPa

 100  200  1  2 100  200   42xy 250 =    2  2

M

24.

= 500 

1 = 250 MPa

Which of the above statements are correct? (a)

1, 2 and 3 only

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1, 2 and 4 only

(c)

2, 3 and 4 only

(d)

1, 2, 3 and 4

2xy = 17500

xy = 50 7 MPa 28.

Ans. (d)

R

(b)

The state of stress at a point is given by x  100 MPa ,

Sol. All options are correct

TE

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)

(a)

 x  75 MPa,

xy  0  and 75 MPa

(b)

 x  25 MPa,

xy  0  and 125 MPa

(c)

 x  25 MPa,

xy  0  and 150 MPa

(d)

 x  75 MPa,

xy  0  and 125 MPa

Ans. (b) Sol.

 x = 100 MPa

 y = – 50 MPa

1 = 250 MPa

S

Sol.

its radius will be

AS

(a)

and

xy  100 MPa . The centre of Mohr’s circle and

The magnitudes of principal stresses at a point are 250 MPa t ensil e and 150 MPa compressive. The magnitudes of the shearing stress on a plane on which the normal stress is 200 MPa tensile and the normal stress on a plane at right angle to this plane are

M

27.

y   50 MPa

xy = 100 MPa

2 = – 150 MPa

IE

   x  y centre =  2 

 x = 200 MPa

1  2 =  x  y

   ,0   

= [25 MPa, 0]

 y = (250 – 150) – 200

Radius of Mohr’s circle =

1 2

x  y 

2

 4 2xy

 y = – 100 MPa  200  100  1  2 200  100   42xy 1 =    2  2  200  100  1  2 200  100  42xy 250 =    2  2

400 =

 300 2  42xy

=

1 100  50 2  4 1002  2

=

1 22500  40,000 2

= 125 MPa 29.

Consider the following sttements related to Mohr’s circle for stresses in case of plane

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stress :

(c)

120 MPa and 120 MPa

1.

The construction is for variations of stress in a body.

(d)

60 MPa and 120 MPa

2.

The radius of the circle represents the magnitude of the maximum shearing stress.

Ans. (b)

R

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

T = 50°C A

R

AS

Ans. (a)

The figure shows a steel piece of diameter 20 mm at A and C, and 10 mm at B. The lengths of three sections A, B and C are each equal to 20 mm. The piece is held between two rigid surfaces X and Y. The coefficient of linear

M

expansion   1.2 105 / C and Young’ss modulus E  2  105 MPa for steel :

S

A

C

B

20mm

10mm

20mm

X 20mm

20mm

20mm B

R

Since supports are rigid.   L A  t   LB T   Lc  T  

R  LC  RLB  RL A – E  A  E  A  E  A  = 0 A B C 



LA = LB = LC = L

3L   T =

L R R R     E  A A A B A C 

C

B

20

10

20

20

20

20

IE

30.

E = 2 × 105 MPa

The diameter represents the difference between the two principal stresses.

TE

3.

 = 1.2 × 10–5/°C

Sol.

X

1 1 1     3E  T = R      2 2 2    20    10    20    4  4   4

R =

Y

 3  2  105  1.2  105  50  4 4   4    202    102    202 

= 18849.56 N When the temperature of this piece increases by 50 C , the stresses in sections A and B are (a)

120 MPa and 480 MPa

(b)

60 MPa and 240 MPa

A =

R = 60 MPa  2   20  4 

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31.

 2   10  4 

= 240 MPa

Sol.

1 = 0.0004 2 = – 0.00012

R

B =

Ans. (b)

R

For a material following Hooke’s law, the values

E = 2 × 105

of ealstic and shear moduli are 3 105 MPa

TE

 = 0.3

and 1.2 105 MPa respectively. The value for bulk modulus is

(c)

1.5 105 MPa 5

2.5 10 MPa

(b) 2 105 MPa (d) 3 105 MPa

Ans. (b) E = 3 × 105 MPa

Sol.

=

 2  105 

1  0.3 2

=

9  K  1.2  105  3K  1.2  105  

S

IE

2   1

 2  105  1  0.32 

Maximum shar stress =

105

At a point i n a body,

1  2 

= 0 MPa

K = 2 × 105 MPa

32.

E

[(–0.00012) + {0.3 × 0.0004}]

3K + (1.2 × 105) = 3.6 K 0.6 K = 1.2 ×

0.0004  0.3   0.00012 

2 =

9KG E = 3K  G

3 × 105 =

E 1   2  1  2

= 80 MPa

M

G = 1.2 × 105 MPa

AS

(a)

1 =

1  2 80  0 = 2 2

= 40 MPa 1  0.0004

Smallest normal stress = 0

and

(a)

40 MPa and 40 MPa

A cantilever of length 1.2 m carries a concentrated load of 12 kN at the free end. The beam is of rectangular cross-section with breadth equal to half the depth. The maximum stress due to bending is not to exceed

(b)

0 MPa and 40 MPa

100 N / mm2 . The minimum depth of the beam

(c)

80 MPa and 0 MPa

should be

(d)

0 MPa and 80 MPa

(a)

120 mm

(b) 60 mm

(c)

75 mm

(d) 240 mm

2   0.00012 . If E  2 105 MPa and   0.3 , the smallest normal stress and the largest shearing stress are

33.

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Ans. (a) Sol.

W =12 kN A

b

(c)

260 MPa and 260 MPa

(d)

260 MPa and 520 MPa

Ans. (a) h

3

1 = 0.0013

Sol.

R

L=1.2m

2 = – 0.0013 E = 2 × 105 MPa  = 0.3

At point A, Mmax = WL

1 =

= (12 ×

103)

= 14.4 ×

103

N × (1.2) N-m

 bh3     12 

6  14.4  10

E

1 2 

 1  2 

 2  105  1  0.32 

[0.0013 +

{0.3 × (–0.0013)} = 200 MPa 2 =

=

E

1  2 

2  1

 2  105  1  0.32 

[(– 0.0013)

+{0.3×(0.0013)} = – 200 MPa

6

 1  2  Maximum shear stress  max  =    2 

bh2 

S

100 =

=

14.4  106   h / 2 

M

100 =

Mmax  y I

AS

= 14.4 × 106 N-mm max =

TE

b = h/2

bh2 = 6 × 14.4 × 104

IE

h 2    h = 86.4 × 104 2

= 35.

h3 = 172.8 × 104 h = 120 mm

34.

Two strain gauges fixed along the principal directions on a plane surface of a steel member recorded strain values of 0.0013 tensile and 0.0013 compressive respectively. Given that the value of E  2 105 MPa and   0.3 , the largest normal and shearing stress at this point are (a)

200 MPa and 200 MPa

(b)

400 MPa and 200 MPa

200  200 = 200 MPa 2

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

A

B 2m

100 kN/m

3m

C

D 1m

The maximum magnitudes of bending moment and shear force are

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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

100 kN/m C

B

A

RB

2m

D

RC 1m

3m

R

(a)

+200 kN

100 kN/m C

Sol. B

A

RB

2m

P 3m

TE

Ans. (b) A

D RC 1m

x

M

AS

Taking moment about point B, (Rc × 3) – (100 × 6 × 1) = 0 Rc = 200 kN RB + Rc = 600 RB = 400 kN (S.F.)A = 0

+ve

B

–ve

P –ve C

D

–100kN

–200kN S.F.D

B

A

+ve

P

C

D

M=–50 kN-m Mmax=–200 kN-m

 S.F.B = –(100 × 2) = – 200 kN  S.F.B = – 200 + 400 = 200 kN  S.F.C = 200 – (100 × 3) = – 100 kN

36.

of diameter  100 mm carries a shear force

S

 S.F.C = – 100 + (200) = 100 kN (S.F.)D = 0 Point P, – (100 x) + RB = 0 100x = 400 x = 4m (B.M.)A = 0 (B.M.)B = – (100 × 2 × 1) = – 200 N-m

of 10 kN at the free end. The maximum shear stress is

IE

5  (B.M.)C = –  100  5   + (400 × 3)  2 = – 50 N-m (B.M.)D = 0 x  * (B.M.)D = RB  x  2    100  x    2 4  = (400 × 2) –  100  4    2 =0

A solid circular cross-section cantilever beam

(a)

4 Pa 3

(b)

3 Pa 4

(c)

3 Pa 16

(d)

16 Pa 3

Ans. (d) Sol.

d = 100mm V = 10 kN = 104 N max =

4   avg  3

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=

4  104  N   3    1002  mm2   4 

max

16 MPa 3 A beam of length L simply supported at its ends carrying a total load W unif ormly distributed over its entire length deflects at the centre by  and has a maximum bending stress  . If the load is substituted by a

 WL   y 8  =  I W1

=

R

P

R

Q

L

W1L3 =  48EI

TE

37.

concentrated load W1 at mid-span such that

W1L3 5 WL3 = 48EI 384 EI W1 = 0.625 W

the deflection at the centre remians unchanged,

AS

the magnitude of the load W1 and the maximum bending stress will be 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. (*)

=

IE

S

Sol. Let load intensity is .

A

RA

L = W

C

38. B RB

L 2

deflection at point C

  =

5 L4 5 WL3 = 384 EI 384 EI

 L L   L L      Mmax =   2 2  2 4

=

L2 WL = 8 8

W1L y 4 I

0.625  8 max 4

1max = 1.25 max

L

RA = RB =

Mmax  y I

W L y =  0.625  4 I

M

(a)

1max =

For a rectangular section beam, if the beam depth is doubled, keeping the width, length and loading same, the bending stress is decreased by a factor (a)

2

(b) 4

(c)

6

(d) 8

Ans. (b) Sol. For rectangular section, b =

M  y M   h/2  = I bh3 /2 

b =

6M bh2

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h = 2h b

b =

6M 6M = 2 2  bh b  2h 

(b) 85 MPa

(c)

125 MPa

(d) 250 MPa

h

TE

A

RA

b 4

C

1m

AS

4 K

(b) 2 K

(c)

K

(d) 0.5 K

S

Stiffness of each spring (k ) = 2k Mmax

and then these springs are arranged in parallel k = k  k

x

1  = (RA × 1) –  1 1   2 = 3 kN-m = 3 × 106 N-mm

IE

y=

= 2k  2k b =

=4 k

A beam AB simply supported at its ends A and B, 3 m long, carries a uniformly distributed load of 1 kN/m over its entire length and a conventrated load of 3 kN, at 1 m from A :

=

3 kN 1 kN/m B

A 1m

If ISJB 150 with IXX  300 cm is used for the beam, the maximum value of bending stress is

Mmax  y I

 3  106 N-mm    75mm 

Which of the following statements apply to provision of flash gutter and flash land around the 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 volume of forging.

2m 4

150 = 75mm 2

300  10 4 mm 4 = 75 MPa

41. C

B

3  (RB × 3) – (3 × 1) –  3  1  = 0  2 3RB = 3 + 4.5 RB = 2.5 kN RA + RB = 3 + (1 × 3) RA + RB = 6 RA = 3.5 kN Mmax = (B.M.)c

Sol. When spring is cut into two pieces.

40.

RB

taking moment about point A

(a)

so

1kN/m

2m

x

A helical compression spring of stiffness K is cut into two pieces, each having equal number of turns and kept side-by-si de under compression. The equivalent spring stiffness of this new arrangement is equal to

Ans. (a)

x

3kN

Sol.

M

39.

75 MPa

Ans. (a)

1  6M  =  2  4  bh  b =

(a)

R

if

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3.

Gutter is provided to ensure complete closing of the die.

(a)

1 and 2 only

(b) 1 and 3 only

(c)

1, 2 and 3

(d) 2 and 3 only

2.5 x = 0.01 x = 43.

Consider the following statements :

R

In case of assembly of mating parts

Ans. (a)

1.

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

the difference between hole size and shaft size is called allowance in transition fit, small positive or negative clearance between the shaft and hole member is employable

TE

2.

A hole and a shaft have a basic size of 25 mm, and are to have a clearance fit with a maximum clearance of 0.02 mm and a minimum clearance of 0.01 mm. The hole tolerance is to be 1.5 times the shaft tolerance. The limits of both hole and shaft using hole basis system will be

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 = 24.986 mm

low limit of hole = 25 mm, high limit of hole = 25.026 mm, upper limit of shaft = 24.8 mm and low limit of shaft = 24.76 mm

(c)

low limit of hole = 24 mm, high limit of hole = 25.006 mm, upper limit of shaft = 25 mm and low limit of shaft = 24.99 mm

1 only

(b) Both 1 and 2

(c)

2 only

(d) Neither 1 nor 2

Sol. • Allowance is a planned deviation between an actual dimension and a nominal or theoretical dimension. • In transition fit, small positive clearnace or negative clearance (Interference) between the shaft and hole member is employable. Only statement (2) is correct. 44.

IE

low limit of hole = 25.006 mm, high limit of hole = 25 mm, upper limit of shaft = 24.99 mm and low limit of shaft = 25 mm

Ans. (a)

Sol. For hole basis system Low limit of hole = 25 mm

(a)

Ans. (c)

M

(b)

(d)

Which of the above statements is/are correct?

AS

(a)

S

42.

0.01 = 0.004mm 25

An organization has decided to produce a new product. Fixed cost for producing the product is estimated as Rs. 1,00,000. Variable cost for producing the product is Rs. 100. Market survey indicated that the product selling price could be Rs. 200. The break-even quantity is (a)

1000

(b) 2000

(c)

500

(d) 900

Ans. (a)

High limit of hole = 25.006 High limit of shaft = 24.99 mm

Sol.

Low limit of shaft = 24.986 Also tolerance of shaft + minimum clearance + tolerance of hole = 0.02 mm x + 1.5 x + 0.01 = 0.02

45.

BEQ =

F 100000 = = 1000 S  V 200  100

Usi ng ex ponential smoot hening, a car manufacturing company predicted the demand for that year as 1040 cars. The actual sale was found to be 1140 cars. If the company’s

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(a)

0.4

(b) 0.6

(c)

0.7

(d) 1.2

48.

Ft = dt 1  1    Ft 1 1080 =   1140  1    1040 = 1140  1040  1040 

Coarse feed, low rake angle, low cutting speed and insufficient cooling help produce 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

S

M

(a)

Ans. (c)

(b) Rs. 4.80

(c)

Rs. 14.80

(d) Rs. 18.40

=  /4×3.82  2.5 = 28.36 cm3 Mass of rod = 28.36 × 8.6 = 243.84 grms or 0.243kg weight of rod = 2.4 N Cost of rod = 2.4 × 1.625 Rs. 3.89 Cost of labour = 2 × 1.5 = Rs. 3 Factory overhead = 0.5 × 3 = Rs. 1.5 Total factory cost = 3 + 1.5 + 3.89 = Rs. 8.40

(a)

approximator

(b) interpolator

A company wants to expand the solid propellant manufacturing plant by the addition of more than 1 ton capacity curing furnace. Each ton of propellant must undergo 30 minutes of furnace tim e including loading and unloading operations. Furnace is used only 80 percent of the time due to power restrictions. The required output for the new layout is to be 16 tons per shift (8 hours). Plant (system) efficiency is estimated at 50 percent of system capacity. The number of furnaces required will be

(c)

coordinator

(d) director

(a)

3

(b) 2

(c)

1

(d) 4

IE

Sol. Coarse feed, low rate angle, low cutting speed & i nsuf f icient cooling hel p to produce continuous chips with built-up edges in ductile materials. 47.

Rs. 8.40

2 Sol. Volume of rod = /4d  

 = 0.4

46.

(a)

Ans. (a)

AS

40 = 100 

TE

Ans. (a) Sol.

a part is made from solid brass rod of 38 mm diameter and length 25 mm. The machining time taken to finish the part is 90 minutes and labour rate is Rs. 2 per hour. Factory overheads are 50% of direct labour cost. The density of material is 8.6 gm per cubic cm and its cost is Rs. 1.625 per newton. The factory cost of the part will be

R

forecast for the next year is 1080, what is the value of the smoothening constant?

In NC machining, coordinated movement of separately driven axes motion is required to achieve the desired path of tool relative to workpiece. The generation of these reference signals is accomplished through a device called

49.

Ans. (b) Sol. Interpolator coordinate the motion of tool relative to workpiece.

Ans. (a)

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52.

In a single-server queueing model

Operating time of furnace = 0.8×480 = 384 minutes. 16 ton propellent requires = 16 × 30 = 480 minutes of furnace. Plant efficiency (in minutes) = 0.5 × 382 = 192 minutes

(a)

avoid work from rubbing against tool

(b)

control chip flow

(c)

strengthen tool edge

(d)

break chips

Ans. (b)

S

Sol. Purpose of side rake angle of cutting tool is to control chip flow. The annual demand of a commodity in a supermarket if 80000. The cost of placing an order is Rs. 4,000 and the inventory cost of each item is Rs. 40. What is the economic order quantity?

IE

51.

(a)

2000

(b) 4000

(c)

5656

(d) 6666

2.

the arrivals is described as a Poisson distribution

3.

uncertainty concering the demand for service exists

(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

The purpose of providing side rake angle on the cutting tool is to

M

50.

480 = 2.5 or 3 192

the arrivals is a memoryless process

Which of the above statements are correct?

AS

=

Total furnace time Total plant time

1.

TE

Furnace required =

Consider the following statements :

53.

• Arrivals are markov or memoryless • Serivce is also memory less. • Arrivals follow poisson distribution To construct an operating characteristic curve, an agreement has to be reached between producer and consumer through which of the following points? 1.

Maximum proportion of defectives that will make the lot definitely unacceptable

2.

The producer is willing to aceept that some of satisfying the quality level (AQL) will rejected    5% 

3.

Maximum level of percentage defectives that wi ll make the lot def init ely unacceptable

4.

The consumer is willing to take lots of quality level (LTPD) even though they are unacceptable    10% 

Ans. (b)

Sol.

R

Sol. Required output 16 tons in 8 hours or 8 × 60 = 480 minutes.

EOQ =

2  80000  4000 40

= 4000 unit

(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

x 0

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

y0

subject to : 2X  Y  1000

Constraint-2

: 3X  4Y  2400

Constraint-3

: X  Y  800

Constraint-4

: X  Y  350

AS

Constraint-1

y y   1 800 600



x y   1 800 800



x y   1 350 350 y

100 800

: X 0

M

Constraint-5



TE

Assuming X and Y are the two control variables, the following are the constraints laid out for maximizing the profit : Maximize profit (P) = 8X + 5Y

Constraint-6

x y   1 500 1000



• Lot tolerance percent deflective (LTPD) 54.

R

• Consumers risk for LTPD ( = 0.1) or 10%

Constraint 3 Constraint 2 Constraint 4

600

: Y 0

Which of the above constraints is a redundant one and does not have any effect on the solution?

500

S

Constraint-1

(b)

Constraint-2

(c)

Constraint-4

(d)

Constraint-5 and Constraint-6

IE

(a)

Ans. (b) Sol.

Constraint 1

Max (p) = 8x + 5y Constraint 1  2x  y  1000 Constraint 1  3x  4y  2400 Constraint 3  x  y  800

800

x

From figure it is clear that constraint -3 does not have effect on solution because. It is not a binding constraint. 55.

A transportation problem consists of 3 sources and 5 destinations with appropriate rim conditions. The number of possible solutions is (a)

15

(b) 225

(c)

6435

(d) 150

Ans. (a) Sol. For a 3 × 5 transportation problem. The number of possible solutions are (5 × 3) i.e. 15

Constraint 4  x  y  350

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56.

Maximize Z  2X1  3X2 subject to

(b)

30 minutes, 22.5 minutes and 2.25 technicians

(c)

22.5 minutes, 22.5 minutes and 2.75 technicians

(d)

30 m inutes, 30 m inutes and 2.25 technicians

X1  X2  3 X1, X2  0

Ans. (b)

(a)

optimal

(b) infeasible

(c)

unbounded

(d) degenerate

Ans. (b) Sol.

Sol.

Arrival rate = 6/hr = 

AS

y

M

6

x

S

–3

=6 +x 2 2x 1

3

IE

Hence, there is no feasible region. Thus solution to LPP is infeasible 57.

A company has a store which is manned by 1 attendant who can attend to 8 technicians in an hour. The technicians wait in the queue and they are attended on first-come-first-served basis. The technicians arrive at the store on an average 6 per hour. Assuming the arrivals to follow Poisson and servicing to follow exponential distribution, what is the expected time spent by a technician in the system, what is teh expected time spent by a technician in the queue and what is the expected number of technicians in the queue? (a)

Service rate = 8/hr = 

TE

The solution to the above LPP is

R

2X1  X2  6

22.5 minutes, 30 minutes and 2.75 technicians

Tim e

 =

6 3 = 8 4

spent

in

the

system

=

1 60 = = 30 minutes  2  6  60 Time spent in the queue =       = 8  2

= 22.5 minutes Expected technicians i n the queue = 2 9/16 9 = = = 2.25 technicians. 1  1/4 4

58.

Objective function Z  5X1  4X2 (Maximize) subject to 0  X1  12 0  X2  9 3X1  6X2  66 X1 , X2  0 What is the optimum value? (a)

6, 9

(b) 12, 5

(c)

4, 10

(d) 0, 9

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Ans. (b) x2

where I is current in amperes. F = Faraday’s constant

(4, 9)

x1=9

(0, 9)

E is given chemical equivalent =

(12, 5)

Feasible region

R

H

x1=12

TE

Sol.

m IE = t F

A Z

where A is atomic mass of workpiece

(10, 0)

(12, 0)

22

x1

Z is valency of anode material

61.

Which of the following defines the compiler’s function correctly? (a)

It translates high-lev el l anguage programs into object code It translates object code into a high-level language

(c)

It translates object code into assembly language instructions

(d)

In a crank and slotted lever type quick return mechanism, the link moves with an angular velocity of 20 rad/s, while the slider moves with a linear velocity of 1.5 m/s. The magnitude and direction of Coriolis component of acceleration with respect to angular velocity are (a)

(b)

Which one of the following properties of work materials is responsible for the material removal rate in electrochemical machining? (a)

Hardness

(b)

Atomic weight

(c)

Thermal conductivity

(d)

Ductility

30 m / s2 and direction is such as to rortate slider velocity in the opposite sense as the angular velocity

(c)

60 m / s2 and direction is such as to rotate slider velocity in the same sense as the angular velocity

Sol. Compiler’s function is to translate high-level progress into object code. 60.

30 m / s2 and direction is such as to rotate slider velocity in the same sense as the angular velocity

it translates assembly language instructions into object code

IE

Ans. (a)

M

(b)

S

59.

AS

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

(d)

60 m / s2 and direction is such as to rotate slider velocity in the opposite sense as the angular velocity

Ans. (c)

Ans. (b) Sol. MRR in ECM is given as

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4.

Mathematically not accurate except in three positions

5.

Has only turning pairs

6.

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

V

C

2V  

TE

Ans. (a)

R

Sol. The schematic of quick return mechanism

Sol. The basic schematic of Ackerman steering mechanism.

The Coriolis acceleration ac = 2V = 2× 1.5 × 20

AS

= 60 m/sec2

A

M

Direction: The direction of Coriolis component of acceleration depends upon velocity of slider ‘C’. If slider moves away from centre, the Coriolis acceleration will be in the direction of link rotation. If the slider moves toward centre of rotation the Coriolis acceleration will be opposite to link rotation as shown below. V

2V

S

C

IE

2V

A

62.

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 w  of steering  Cot   cot  =  in three  

  = 0 and two positions    25  toward left and right • Since very less sliding surfaces (turning pairs only) So longer life. Due to longer life it is used generally despite it is not correct mathematically.

V

• Steering control is provided in front wheels only in all steering mechanisms.

A

Which of the following are associated with Ackerm an steering mechanism used in automobiles?

B

D

positions only namely in straight motion

C

Since nothing is mentioned about velocity of slider (away or toward centre) so assume it moves away from centre, the right.

W

63.

The displacement of a follower of a cam in a printing machine is represented by the expression

x 10  120 2  1500 3  2000 4  2500 5

1.

Has both sliding and turning pairs

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

2.

Less friction and hence long life

(a)

900 3  48000 3   1500003 2

3.

Mechanically correct in all positions (b)

9000 3

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(c)

240 2  9000 2   240002 2

3 2 =  (9000  48000  150000 )

64.

(d)

480003   1500003 2

s  2t3  3t2  2t  1

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

TE

Ans. (a) Sol.

Displacement of follower in cam x = 10  1202  15003  20004  25005

AS

culure  = is angle of rotation of cam.

Since the jerk is third derivative of follower displacement



Jerk, J =

d3 x

Sol.

dx  (10  240  45002  dt

S

d dt

(d) 3 m / s2

Displacement of body s = 2t 3 – 3t 2 + 2t + 1 Velocity, v =

ds = 6t 2 – 6t + 2 dt

dv d2 s  dt dt

IE

= 12t – 6

d2 x = (0  240  9000 dt d 240002  500003 ) dt

dt3

12 m / s2

= (6 × 2t) – 6 + 0

= 2 (240  9000  240002  500003 )



(c)

a =

Acceleration, at t = 1 sec

Assuming cam rotates with uniform angular velocity 

Jerk

(b) 2 m / s2

Acceleration,

dx 2 = (10  240  4500 dt 80003  125004 )

d3 x

6 m / s2

…(i)

dt3

80003  125004 )



(a)

Ans. (a)

M



A body starting from rest moves in a straight line with its equation of motion being

R

 500003 3

a = 12 × 1 – 6 65.

= 6 m/sec2 The crank shaft of a reciprocating engine having a 20 cm crank and 100 cm connecting rod rotates at 210 r.p.m. When the crank angle is 45 , the velocity of piston is nearly (a)

1.8 m/s

(b) 1.9 m/s

(c)

18 m/s

(d) 19 m/s

2

=  (0  9000 48000  1500002 )

d dt

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Ans. (a)

Ans. (b)

Sol.

67.

Reciprocating engine mechanism cm 100 = 

A four-bar mechanism is as shown in the figure bleow. At the instant shown, AB is shorter than CD by 30 cm. AB is rotating at 5 rad/s and CD is ratating at 2 rad/s:



P

R

r = 20 cm

x

B

 = 45

Angular velocity of crank

TE

Piston

C

A

2N 2  210   = 22 rad/sec 60 60

Ratio of length of connecting rod to crank length l 100  5 r 20

The length of AB is

AS

n

D

(a) 10 cm (c) 30 cm

(b) 20 cm (d) 40 cm

Ans. (b)

M

The velocity of piston-P,

Sol.

The schematic of mechanism

sin2   v  r  sin     2n 

C 1=5rad/sec

sin90   v = 0.2  sin 45   25  

S



B

A 2 =2rad/sec

IE

1  1  = 0.2  22    2 10 

= 4.4 × (0.707 + 0.1) = 1.7756 m/sec

66.

D

At the instant shown in figure, the linear velocity of point B and C will be same VBA = VCD

 1.8 m/sec While designing a cam, pressure angle is one of the most important parameters which is directly proportional to



(a)

pitch circle diameter

Since the difference between AB and CD

(b)

prime circle diameter

(c)

lift of cam

(d)

base circle diameter

1AB  2CD 5AB = 2CD

…(i)

CD – AB = 30 From equation (i)

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

(d) a relative motion of axes and none of the axes of gears has relative motion with the frame

5 AB 2

5 AB  AB  30 2

Ans. (a) Sol.

AB =

2  30 = 20 cm 3

70.

A governor is said to be isochrnous when the equilibrium speed is

AS

68.

A governor is said to be isochronous when equilibrium speed is same or constant for all radii of rotation of balls within working range. The isochronous governor has zero range of speed. This isochronism character is possible in spring controlled governors only not in dead weight types. A planetary gear train is a gear train having

(a) 12822 N-m (c) 14822 N-m

69.

(a) a relative motion of axes and the axis at least one of the gears also moves relative to the frame (b) no relative motion of axes and no relative motion of axes with respect to the frame (c) no relative motion of axes and the axis of at least one of the gears also moves relative to the frame

(b) 24200 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

IE

Sol.

S

M

(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)

In planetary gear train or compound epicyclic gear train, the axis of at least one gear rotates (not fixed) about to the fixed frame. The beauty of this train is that it ensures large speed reduction in small space. The flywheel of a machine having weight of 4500 N and radius of gyration of 2 m has cyclic fluctuation of speed from 125 r.p.m. to 120 r.p.m. Assuming g = 10 m/s2, the maximum fluctuation of energy is

TE

3 AB  30 2

R

CD =

=

2 (125  120) rad/sec 60

= (13.09 – 12.57) rad/sec 

max = 13.09 rad/sec min = 12.57 rad/sec



Maximum fluctuation of energy e = Emax – Emin =

1 2 1 2 Imax  Imin 2 2

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1 mk2 (max  min )(max  min ) 2

=

1 4500 2  2 (13.09  12.57) 2 10 (13.09  12.57)

= 12009 N.m.

(2)

AS 73.

M

amorphous free of pores crystalline mixture of crystalline and glass

Ans. (c)

For complete dynamic balancing of rotating systems, (1)

So the nearest answer is 12100 N.m. Alumina doped with magnesia will have reduced thermal conductivity because its structure becomes (a) (b) (c) (d)

There should not be any couple in the system System should be statically balanced i.e., centre of mass of the system should be on axis of rotation

S

W hi ch of the f ol lowing statements are associated with complete dynamic balancing of rotating systems?

Sol.

2. Support reactions due to forces are zero but not due to couples

4. Centre of masses of the system lies on the axis 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

(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

Ans. (b)

1. Resultant couple due to all inertia forces is zero

3. The system is automatically statically balanced

Hence in complete dynamic balance, the reaction on the support are due to weight of system and remain constant during rotation of system i.e., no force should airse due to rotation. Which of the following statements is correct about the balancing of a mechanical system?

(d) None of the above

IE

72.

Sol.

R

=

= 900 × 25.66 × 0.52

71.

Ans. (d)

1 I(max  min )(max  min ) 2

TE

=

74.

For complete balancing or simply balancing of m echan i cal syst em . T he dynam i c balancing i.e., no couple, requires static balancing (centre of mass of system at axis of rotation) of system as pre condition. The accelerometer is used as a transducer to measure earthquake in Richter scale. Its design is based on the principle that (a) its natural f requency is v ery 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 without any reference point.

6l = 2 g e

Ans. (c)

R

l = 2 g . 6 e

The natural frequency of accelerometer should be kept around frequency of vibration/excitation in order to read the very weak earthquake.

TE

In case of resonance the peak is control by proper dampnig.

As compared to the time period of a simple pendulum on the earth, its time period on the moon will be (a) 6 times higher (b) 6 times lower 6 times higher

(d)

6 times lower

M

(c)

Ans. (c)

S

(a)

1 2

(b)

1 3

(c)

1 4

(d)

3 4

Ans. (b) Sol.

The time period of simple pendulum on earth l Te = 2  g e

While calculating the natural frequency of a spring-mass system, the effect of the mass of the spring is accounted for by adding X times its value to the mass, where X is

The period of oscillation of spring-mass system including the mass of spring.

…(i)

Since on Moon, the acceleration due to

IE

Sol.

76.

AS

75.

= Te 6

gravity in

ms

1 th of acceleration due to gravity 6

k

at earth.





gm

M

ge = 6

ms   M   M  Xms 3  T = 2   2 k k

l T m = 2 g m l = 2  g  e    6 



77.

X =

1 3

A block of mass 10 kg is placed at the free end of a cantilever beam of length 1 m and second moment of area 300 mm 4. Taking Young’s

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modulus of the beam material as 200 GPa, the natural frequency of the system is (b) 2 3 rad s

(c) 3 2 rad s

(d) 20 3 rad s

more than

M =10 kg

I = 300 mm 4

E = 200 GPa

Sol.

M

Young’s modulus,

Ans. (b)

AS

Second moment of inertia,

TE

The cantilever vibrator,

 = 1m

2 times its natural frequency to

(a) increase stability under heavy load and high speed (b) isolate vibration of the system from the surrounding (c) minimize deflection under dynamic loading as well as to reduce transmissbility of force to the surrounding (d) none of the above

Ans. (c)

The rotating speed of turbine rotor is more than

2 times of natural frequency (n ) of system. Because beyond this speed, the transmissibility T r is less than one as shown below

The deflection at free end

Turbine rotor speed range

1.0

S

mg3 st  3EI

'Tr'

Stiffness constant



IE

Sol.

The speed rating for turbine rotors is invariably

R

(a) 30 2 rad s

78.

mg 3EI k =   3  st

k =

1.0

3  200  109  300  1012 3

= 600 × 300 × 10–3 = 180 N/m Frequency of oscillation

n =

k 180  = 3 2 rad/sec m 10

79.

= 2

(/n )

Hence force transmitted to foundation by turbine is less than it impress upon isolator. So t he v i brat i ons are i sol at ed f rom surroundings. The magnitude of swaying couple due to partial balance of the primary unbalnacing force in locomotive is (a) inversely proportional to the reciprocating mass

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Ans. (c) Sol.

Ans. (d) Sol.

81.

AS

(1  c) mr2 cos  Cylinder–1

a/2

x

M

x

a/2

Cylinder-2

The power of governor is defined as = sleeve displacement × effort

TE

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

(d) each governor ball for given percentage change of speed

R

(b) directly proportional to the square of the distance between the centrelines of the two cylinders (c) inversely proportional to the distance between the centrelines of the two cylinders

2

S

(1  c) mr cos    90 

(a) (b) (c) (d)

8.9×103 8.9×103 8.9×103 8.9×103

Mass × No. of atoms in a unit cell Sol. Density (g/m3) = Volume × Avagadro's Number 3 Volume of a unit cell = a3 = (2 2R)

= 47.46 × 10–30 m3

a 2 = (1  c)mr cos  2

=

 a (1  c)mr2 cos(  90)     2

= (1  c) 80.

mr a (cos   sin ) 2

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

4  63.5 47.46  1030  6.023  1023

= 8.9 × 106 g/m3 or

2

kg/mm3 kg/cm3 kg/m3 g/mm3

Ans. (c)

IE

The swaying couple due to unbalance force about centre line x – x

The effort of governor is defined as average force on sleeve for a given change of speed. So the power can be defined as work done at sleeve for given rate of change of speed. Copper has FCC structure; its atomic radius is 1.28 A and atomic mass is 63.5. The density of copper will be

82.

8.9 × 103 kg/m3

A plane intersects the coordinate axes at 2 1 1 , y  and z  . What is the Miller index 3 3 2 of this plane? x

(a) 932 (c) 423

(b) 432 (d) 364

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Ans. (d) but

Sol.

a =

1/2

Roctahedral = 0.158 R also R =

 2 1 1 Intercepts are   3 3 2 3 Take reciprocal intercepts i.e.  3 2

 2 

Roctahedral =

AS

Converting to whole number by multiplying by 2 miller indices = (3 6 4)

What is the diameter of the largest sphere in terms of lattice parameter  , which will fit the void at the centre of the cube edge of a BCC crystal? (a) 0.134 

0.158  3 a 4

= 0.0684 a

Doctahedral = 2 × Roctahedral = 0.136 a If the atomic radius of aluminium is r, what is its unit cell volume? 3

3

 2r  (a)    2

 4r  (b)    2

3

S

3

 2r  (c)    3

(d) 0.5 

IE

Sol.

3 a 4

(b) 0.25 

(c) 0.433  Ans. (a)

84.

M

83.

2  2  R R =   1 R 3  3 

R

1/3

TE

2 3

=

4 R for BCC will only 3

 4r  (d)    3

Ans. (b) Sol. Aluminium has FCC crystal structure a = 2 2r

a

volume of unit cell = a3 = (2 2r)3 = 8×2× 2r 3 = 16 2r 3 a 3

 4r  or    2

a

The void formed is an octahedral void. The largest radius that can jet in it a  Roctahedral =   R  2 

85.

Consider the following staements regarding the behaviour of dislocations: 1. Only edge dislocation and mixed dislocation can have glide motion.

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2. A screw dislocation cannot have glide motion. 3. Dislocat ion mov es in t he direct ion perpendicular to that of shear stress.

1, 1, 2, 1,

2 and 3 2 and 4 only 3 and 4 only 2, 3 and 4

Ans. (d) Sol. Dislocation

R C0 C2

wliquid = 1 – wsolid = 0.4 or 40% w solid wliquid

=

0.6 3 = 0.4 2

A binary alloy of Cu and Ni containing 20 wt% Ni at a particular temperature coexists with solid phase of 26 wt% Ni and liquid phase of 16 wt% Ni. What is the weight ratio of solid phase and liquid pahse?

(a) (b) (c) (d)

M

• Dislocations move perpendicular to direction of shear stress along a slip plane containing Burger’s vector & direction vector.

Elements A and B form eutectic type binary phase diagram and the eutectic composition is 60 wt% B. If just below eutectic temperature, the eutectic phase contains equal amounts (by wt) of two solid phases, then the compositions of the two solid phases are

IE

S

• For screw dislocation there is no specific glide plane defined.

(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 binary isomorphos system)

87.

0% 100%

C2  C0 0.26  0.2 wsolid (at 20% Ni) = C  C = 0.26  0.16 2 1

• Only edge & mixed dislocation can glide along a specific plane defined.

86.

S

Cu100% 80%Cu Ni0% 20%Ni %composition

AS

(a) (b) (c) (d)

C1=16% Ni {Given C2=26% Ni data} C1

1453°C

L+S

1085°C

TE

4. Motion of disloacation occurs on slip plane that contains Burger’s vector and direction vector. Which of the above statements are correct?

L

Temp.

20 30 20 30

wt% wt% wt% wt%

B B B B

and and and and

90 90 80 80

wt% wt% wt% wt%

B B B B

Ans. (b) Sol. Equilibrium phase diagram for a eutectic alloy is eutectic point L L L



    x A 100% B 0%

40% A 60% B

y A 0% B 100%

Just below eutectic point, the composition of

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Sol. Eutectic reaction is

two phases (i.e.  & ) be x and y

L  

respectively.

It is a reversible reaction which can take place both ways.

y  0.6 = 0.5(given) wd = yx

y – 0.5y + 0.5 x = 0.6

TE

0.5(x + y) = 0.6

(a) 8% (c) 12%

x+y = 1.2 Similarly

0.6  x = 0.5 yx

0.6 – x = 0.5y – 0.5 x 0.6 = 0.5y + 0.5x (x + y) = 1.2

Sol. At room temperature

88.

M

There, is only one option satisfying the conditions i.e. 30 wt% B and 90 wt% B is  and  phases.

Consider the following statements : In a binary phase diagram

S

1. the freezing point of the alloy is minimum

IE

2. eutectic mixture solidifies at a constant temperature like pure metal 3. eutectic reaction is irreversible 4. at eutectic temperature, liquids of two metals will change into two solids Which of the above statements are correct? (a) (b) (c) (d)

1, 1, 1, 1,

2 and 3 only 3 and 4 only 2 and 4 only 2, 3 and 4

(b) 10% (d) 14%

Ans. (c)

AS

w =

At room temperature,  -iron contains negligible amount of carbon, cementite contains 6.67% C and pearlite contains 0.8% C. Pearlite contains how much cementite?

R

89. y – 0.6 = 0.5y  0.5 x

Amount of ferrite × %carbon in ferrite + Amount of cementite × % carbon in cementite = overall % carbon in pearlite. as % carbon in ferrite is negligible at toom temperature. Amount of cementite × 6.67 = 0.8 cementite =

90.

0.8 = 0.1199 or 12% 6.67

Two metals A and B are completely immiscible in sloid and liquid state. Melting point of A is 800 ºC and melting point of B is 600 ºC. They form eutectic at 200 ºC with 40% B and 60% A. The 50% B alloy contains. (a) (b) (c) (d)

83.33% B and 16.67% of eutectic 83.3% of eutectic and 16.67% B 50% B and 50% of eutectic 40% B and 60% of eutectic

Ans. (b) Sol. For no miscibility in liquid and solid state, the phase diagram is

Ans. (c)

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a2 + a2 = (4R)2

800°C

2a2 = 16R2

Temp.

Eutectic point

a2

600°C

R

A

a = 2 2R

200°C A 100°C B%

C0

60%A 50%A 40%B 50%B

C2 0% 100%

%Composition

d200 =

d220 =

C0  C1 50  40 10 Quantity of B = C  C = 100  40 = 60 2 1

= 16.67% of B.

M

C2  C0 100  50 Quantity of eutectic = C  C = 100  40 2 1

92.

IE

S

What is the interplanar spacing between (200), (220), (111) planes in an FCC crystal of atomic radius 1.246 A? (a) d(200) = 1.762 Å, d(220) = 1.24 Å and d(111) = 2.034 Å (b) d(200) = 1.24 Å, d(220) = 1.762 Å and d(111) = 2.034 Å (c) d(200) = 2.034 Å, d(220) = 1.24 Å and d(111) = 1.762 Å (d) d(200) = 2.5 Å, d(220) = 4.2 Å and d(111) = 2.6 Å

Ans. (a) Sol. Inteplaner spacing is given as d = for FCC

a 2

h  k 2  2

d111 =

3.524 22  0 2  0 2

3.524 2

2  22  02 3.524 2

1  12  12

= 1.762 A 

= 1.246 A 

= 2.034 A 

Rotary swaging is a process for shaping (a) (b) (c) (d)

50 = 83.33% = 60

91.

a

a = 2 2 × 1.246 = 3.524A°

AS

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

a

R = 1.246A° (given)

TE

C1

=

4R

8R2

round bars and tubes billets dies rectangular blocks

Ans. (a) Sol. Rotary swaging is used to shape round bars and tubes such as gun barrels 93.

Consider the following statements: In shell moulding 1. a single parting plane should be provided for mould 2. detachable pattern parts and cores could be included 3. minimum rounding radii of 2.5 mm to 3 mm should be used 4. draft angles of not less than 1° should be used Which of the above statements are correct? (a) 1, 3 and 4 only (b) 1, 2 and 3 only

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• Very close fit between mating surface

(d) 1, 2, 3 and 4 96.

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°. A big casting is to have a hole, to be produced by using a core of 10 cm diameter and 200 cm long. The density metal is 0.077

N/cm3

M

Sol. Buoyant force acting on core. (metal  core ) × /4×d2   2

= /4× 10   200   0.077  0.0165  = 950.33 N Consider the following :

S

95.

The purpose of lapping process is 1. to produce geometrically true surface

IE

127 60

30 127

(d)

20 80

Sol. Threads per inch (TPI) =

(b) 1100.62 N (d) 350.32 N

Ans. (c)

(b)

Ans. (b)

AS

(a) 200.5 N (c) 950.32 N

(c)

1 12

and

density core is 0.0165 N/cm 3. What is the upward force acting on the core prints?

(a)

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94.

Centre lathe is to be used to cut inch thread of 4 threads per inch. Lead screw of lathe has 3 mm pitch. Then change gear tobe used is

R

(c) 2, 3 and 4 only

2. to correct minor surface imperfections 3. to improve dimensional accuracy 4. to provide very close fit between the contact surface Which 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

25.4 pitch in mm

25.4 25.4 = TPI 4 Also pitch (mm) = Gear ratio × pitch of leads crew.

pitch in mm =

Gear ratio =

25.4 25.4 254 or = 43 12 120

127 60 Consider the following statements in respect of the oxidizing flame due to excess of oxygen in welding:

or 97.

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. (d)

Ans. (a)

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

Sol. Oxidising flame • Forms hard oxides which protect the weld metal

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• Has good welding effect on copper base alloys, manganese steels and east iron. A cutter tip is initially at X = 10 mm, Y = 20 mm. In a rapid motion, using G00 code, it moves to X = 160 mm, Y = 120 mm. The X and Y axes have maximum speed of 10000 mm/min and 5000 mm/min respectively. Operating at maximum speed, what is the time it will take to reach the destination? (a) 0.90 s (b) 1.08 s (c) 1.20 s (d) 2.16 s

Sol. Taylor’s tool life equation is vTn = C v 1 × (T1)0.5 = 300

0.75v 1 × (T2)0.5 = 300

Sol. y

V1  T10.5 = 0.75v1  T20.5  T2  T   1

M S

x

For y axis cutter has to travel 100 mm at 5000

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 500  150  3 mm/s  i.e. = = 0.9 s mm/min   3  500

Total time = 0.9 + 1.2 = 2.1 s 99.

If n = 0.5 and C = 300 for the cutting speed and the tool life relation, when cutting speed is reduced by 25%, the tool life will be increased by (a) 100% (c) 78%

(b) 95% (d) 50%

1 0.75

16 T1 9

Increase in tool life  16    T1  T1   T2  T1  =    100   100 =  T1  T1   

=

For x axis cutter has to travel 150 mm at 10000

=

T2 =

 250  mm/sec  i.e. it will take = mm/mm   3 

100  3 = 1.2s 250

0.5

2 2 T2 16  1  4 = =     = T1  0.75  3 9

(160 , 120)

(10, 20)

...(ii)

Comparing (i) & (ii)

AS

Ans. (d)

...(i)

when speed is reduced by 25%

TE

98.

Ans. (c)

R

• Gives dirty & scummy appearance to weld bead

700 = 77.78% or 78% 9

100. Which of the following statements are correct for 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)

101. Consider the following:

F F  A 0.707h

The force are parallel to minimum threat area so it is designed for shear strength only. 102. If the permissible crushing stress for the material of a key is double the permissible shear stress, then the sunk key will be equally strong in sheaering and crushing if the key is a

TE

The parallel fillet welded joint is designed for 1. tensile strength 2. compressive strength 3. bending strength 4. shear strength Which of the above is/are correct? (a) 4 only (b) 3 only (c) 2 and 3 (d) 1 and 4



R

Sol. Temperature rise does not give better accuracy during machining. HIgh temperature affects hardnen of cutting edge. It can also distort the tool due to thermal expansion.

AS

Ans. (a)

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

Ans. (c)

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

Sol. Given c = 2

S

F

M

The parallel fillet welds along with force F.

IE

Sol.

(a) rectangular key with width equal to half the thickness (b) rectangular key with width equal to twice the thickness

t =

2h = 2

h

k h

h 2

= 0.707 h

Area of threat,

Also crushing force = c 

Crushing torque = c 

t  2

t d  2 2

Shearing torque =     

d 2

for equal strength in shearing & crushing     

d t d = c     2 2 2

  = c t 2

A =   t = 0.707 h   

Shear stress,

 = 1 or  = t t

Thus, it should be a square key

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TE

(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

or disconnect the driving and driven shaft according to wish of operator of requirement. T hi s oper at i on happens due t o ax i al movement of spline hub or shaft ember. Example clutch. The shaft in spline is under pure torsion i.e., shear. 106. For a power screw having square threads with lead angle of 45° and coefficients of friction of 0.15 between screw and nut, the efficiency of the power screw, neglecting collar friction, is given by

R

103. Very small quantity of carbon in iron as in steels forms interstitial solid solution mainly because atomic size(s) of

Ans. (c)

AS

Sol. Carbon forms interstitial solid solution with iron as size of carbon atom is very small as compared to size of iron atoms. 104. In a cotter joint, the width of the cotter at the centre is 5 cm, while its thickness is 1.2 cm. The load acting on the cotter is 60 kN. The shear stress developed in the cotter is

Ans. (b)

(b) 64% (d) 44%

Ans. (a) Sol. Efficiency of power screw is   = 90   = 45 

(b) 100 N/mm2 (d) 200 N/mm2

M

(a) 50 N/mm2 (c) 120 N/mm2

(a) 75% (c) 54%

 =

tan  1  tan  tan   tan  = tan      tan   tan 

=

tan 45 1  tan 45  0.15  0.15  tan 45

=

11  0.15  = 0.739 or 74% 1.15

P 60  103 60000 = = = 2  b  t 12  50 600

IE

c

S

Sol. Shear stress in cottor

= 100 N/mm2

105. The use of straight or curved external gear teeth in mesh with internal teeth in ‘gear and spline couplings’ is specifically employed to accommodate.

107. Aquaplaning occurs in vehicle tyres when there is continuous film of fluid between the tyre and the 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

(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 in torque transmission when key fails to serve the purpose. These couplings can connect

Ans. (d) Sol.

Aquaplaning or hydroplaning is the condition in vehicle or aircraft when there is water layer between tyre and road surf ace. In this situation, there is excessive skidding or slipping of vehicle i.e., loss of control of

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vehicle. This situation can be very dangerous when al l t ype h av e equapl ani ng simultaneously.

v2 =

100 = 102.04 rpm. 0.98

Ans. (d) D1 = 2D2

sin =

r1  r2 x

S

wrap angle = 180 – 2 = 120° 2 = 60°  = 30°

IE

or

sin 30° =

x =

110. Two shafts A and B of same material, and A is twice the diameter of B. The torque that can be transmitted by A is (a) 2 times that of B (c) 4 times that of B

(b) 8 times that of B (d) 6 times that of B

Sol. Torque transmitted by a shaft T d3

where x is the center distance between the pulleys for smaller pulley

= 102.04 rpm

Ans. (b)

r1 = 2r2

M

or

TE

(b) 750 mm (d) 250 mm

V2 100 = v2 =   1  1  0.02

AS

(a) 1000 mm (c) 500 mm

v1 800 = = 100 rpm V R 8

Due to creep

108. If the angle of wrap on smaller pulley of diameter 250 mm is 120° and diameter of larger pulley is twice the diameter of smaller pulley, then the centre distance between the pulleys for an open belt drive is

Sol. Given

Sol. Velocity of driver

R

=

Ans. (b)

250  125 x

125 = 250mm sin30

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

(b) 102.04 rpm

(c) 100.04 rpm

(d) 98.04 rpm

TA =  2 3 TB = 8TB 111. A worm gear set is designed to have pressure angle of 30° which is equal to the helix angle. The efficiency of the worm gear set at an interface friction of 0.05 is (a) 87.9% (c) 67.9%

(b) 77.9% (d) 57.9%

Ans. (a) Sol. Helix angle of worm = 30° Load angle of worm (  ) = 90° – 30° = 60° efficiency of worm gear =

1   tan  1   /tan 

=

1  0.05  tan60 1  0.05/tan 60°

=

0.9134 = 0.8877 1.0289

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or 88.77%

contact ratios are not equal. Large contact ratio’s lead to engagement of multiple teeth.

112. Consider the following statements :

(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

AS

TE

1. 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 bev el 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

114. In an interference fit between a shaft and a hub, the state of stress in the shaft due to interference fit is

R

The axes of spiral bevel gear are non-parallel and intersecting

Sol. For interference fit between hub & shaft, shaft is considered externally pressed radial = – Pf (external pressure)

M

Ans. (b)

Ans. (d)

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

tan gential = –

S

IE

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)

rf2  rL2 

rf = outer radius

113. Consider the following statements: In case of helical gears, teeth are cut at an angle to the axis of rotation of the gears.

Pf  rf2  rL2 

fi = inner radius Thus, both radial & tangential stream are compressive. 115. In case the number of teeth on two bevel gears in mesh is 30 and 60 respectively, then the pitch cone angle of the gear will be (a) tan–1 2

(c)

  tan1 0.5 2

(b)

  tan1 2 2

(d) tan–1 0.5

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

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

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P2

leakage of oil is not a problem.

TG   60  = tan1   = tan    30   TP  1 

118. Consider the following statements in connection with thrust bearings:

= tan–1 (2)

1. Cylindrical thrust bearings have higher coefficient of friction than ball thrust bearings. 2. Taper rollers cannot be employed for thrust bearings.

116. In skew bevel gears, the axes are

R

(a) non-parallel and non-intersecting, and teeth are curved (b) non-parallel and non-intersecting, and teeth are straight (c) intersecting, and teeth are curved and oblique (d) intersecting, and teeth are curved and can be ground

AS

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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. (a) Sol.

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

M

117. Consider that modern machines mostly use short bearings due to the following reasons:

IE

S

1. l/d of the most modern bearings is in the range of 1/4 to 2 2. No end leakage of oil from the bearing 3. Shaft deflection and misalignment do not affect the operation 4. Can be applied to both hydro-dynamic and hydrostatic cases Which 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 not affect operation (ii) compact design (iii) Run cooler. However, end leakage is a problem. On the other hand, long bearings have greater load carrying capacity and end

Ans. (d) Sol. • Cylindrical thrust bearings have higher coefficeint of friction than ball thrust bearings • Both taper roller & double row thrust ball bearings are used.

119. The behaviour of metals in which strength of a metal is increased and the ductility is decreased on heating at a relatively low temperature after cold-working is known as (a) clustering (c) twinning

(b) strain aging (d) screw dislocation

Ans. (c) Sol.

Twi nni ng i s a pl ane def e ct where arrangement of atoms on either side of a twin plane are identical. Twinning occurs either as mechanical twinning or Annealing twins during annealing heat treatment. Twinning increases strength & reduces ductility as twin planes hinders the movement of dislocations. 120. If the equivalent load in case of a radial ball bearing is 500 N and the basic dynamic load rating is 62500 N, then L10 life of this bearing

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is million million million million

of of of of

revolutions revolutions revolutions revolutions

R

1.953 3.756 6.953 9.765

Ans. (a) Sol. Use of a ball bearing is given as: c L10 =   w

k

AS

where L10 = life in million revolutions

TE

(a) (b) (c) (d)

c = Basic dynamic load rating w = equivalent dynamic load

k = 3 for ball bearing 3

M

 62500  L10 =    500 

= (125)3 = 1953125

IE

S

1.953125 million of revolution

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