##### Document Text Contents

Page 1

ESE-2015

Detailed Exam Solutions

solutions

(Objective Paper - II)

Mechanical

Page 2

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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.

Codes :

(a) 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

1. Statement (I) : The cam in contact with a

follower is a case of complete constraint.

Statement (II) : The pair, cam and follower, by

itself does not guarantee continuity of contact

all the time.

Ans. (c)

Sol. 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.

2. Statement (I) : Involute pinions can have any

number of teeth.

Statement (II) : Involute profiles in mesh satisfy

the constant velocity ratio condition.

SET - D

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

Ans. (c)

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.

3. Statement (I) : Hooke’s joint connects two non-

parallel non-intersecting shafts to transmit

motion with a constant velocity ratio.

Statement (II) : Hooke’s joint connects two

shafts the axes of which do not remain in

alignment while in motion.

Ans. (c)

Sol. 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

4. 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.

Page 21

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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

3

x

2x

+x

=6

1

2

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) 22.5 minutes, 30 minutes and 2.75

technicians

(b) 30 minutes, 22.5 minutes and 2.25

technicians

(c) 22.5 minutes, 22.5 minutes and 2.75

technicians

(d) 30 minutes, 30 minutes and 2.25

technicians

Ans. (b)

Sol. Service rate = 8/hr =

Arrival rate = 6/hr =

=

6 3

8 4

=

Time spent in the system =

1 60

30 minutes

2

= =

Time spent in the queue =

6 60

8 2

=

= 22.5 minutes

Expected technicians in the queue =

2 9/16 9

1 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

Page 22

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Ans. (b)

Sol. x2 x1=12

x1=9

(12, 5)

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

(0, 9)

Feasible

region

H (4, 9)

Z is maximized at any of the corners of feasible

region i.e. (12, 5)

59. Which of the following defines the compiler’s

function 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. Compiler’s function is to translate high-level

progress into object code.

60. 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

Ans. (b)

Sol. MRR in ECM is given as

m

t

=

E

F

I

where I is current in amperes.

F = Faraday’s constant

E is given chemical equivalent =

A

Z

where A is atomic mass of workpiece

Z is valency of anode material

61. 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) 230 m / s and direction is such as to

rotate slider velocity in the same sense

as the angular velocity

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

rortate slider velocity in the opposite

sense as the angular velocity

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

rotate slider velocity in the same sense

as the angular velocity

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

rotate slider velocity in the opposite sense

as the angular velocity

Ans. (c)

Page 41

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P2 =

1 1G

P

T 60

tan tan

T 30

=

= tan–1 (2)

116. In skew bevel gears, the axes are

(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

Ans. (a)

Sol. In skew bevel gears, the axes are non-parallel

and non-intersecting, and teeth are curved.

117. Consider that modern machines mostly use

short bearings due to the following reasons:

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

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 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 (b) strain aging

(c) twinning (d) screw dislocation

Ans. (c)

Sol. Twinning is a plane defect 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

Page 42

Office Address : F-126, Katwaria Sarai, New Delhi - 110 016.

Telephone : 011-41013406, Mobile : 8130909220, 9711853908, 7838813406

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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 revolutions

c = Basic dynamic load rating

w = equivalent dynamic load

k = 3 for ball bearing

L10 =

362500

500

= (125)3 = 1953125

1.953125 million of revolution

ESE-2015

Detailed Exam Solutions

solutions

(Objective Paper - II)

Mechanical

Page 2

Office Address : F-126, Katwaria Sarai, New Delhi - 110 016.

Telephone : 011-41013406, Mobile : 8130909220, 9711853908, 7838813406

Web : www.iesmaster.org E-mail : [email protected]

IE

S

M

AS

TE

R

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.

Codes :

(a) 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

1. Statement (I) : The cam in contact with a

follower is a case of complete constraint.

Statement (II) : The pair, cam and follower, by

itself does not guarantee continuity of contact

all the time.

Ans. (c)

Sol. 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.

2. Statement (I) : Involute pinions can have any

number of teeth.

Statement (II) : Involute profiles in mesh satisfy

the constant velocity ratio condition.

SET - D

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

Ans. (c)

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.

3. Statement (I) : Hooke’s joint connects two non-

parallel non-intersecting shafts to transmit

motion with a constant velocity ratio.

Statement (II) : Hooke’s joint connects two

shafts the axes of which do not remain in

alignment while in motion.

Ans. (c)

Sol. 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

4. 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.

Page 21

Office Address : F-126, Katwaria Sarai, New Delhi - 110 016.

Telephone : 011-41013406, Mobile : 8130909220, 9711853908, 7838813406

Web : www.iesmaster.org E-mail : [email protected]

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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

3

x

2x

+x

=6

1

2

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) 22.5 minutes, 30 minutes and 2.75

technicians

(b) 30 minutes, 22.5 minutes and 2.25

technicians

(c) 22.5 minutes, 22.5 minutes and 2.75

technicians

(d) 30 minutes, 30 minutes and 2.25

technicians

Ans. (b)

Sol. Service rate = 8/hr =

Arrival rate = 6/hr =

=

6 3

8 4

=

Time spent in the system =

1 60

30 minutes

2

= =

Time spent in the queue =

6 60

8 2

=

= 22.5 minutes

Expected technicians in the queue =

2 9/16 9

1 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

Page 22

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Ans. (b)

Sol. x2 x1=12

x1=9

(12, 5)

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

(0, 9)

Feasible

region

H (4, 9)

Z is maximized at any of the corners of feasible

region i.e. (12, 5)

59. Which of the following defines the compiler’s

function 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. Compiler’s function is to translate high-level

progress into object code.

60. 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

Ans. (b)

Sol. MRR in ECM is given as

m

t

=

E

F

I

where I is current in amperes.

F = Faraday’s constant

E is given chemical equivalent =

A

Z

where A is atomic mass of workpiece

Z is valency of anode material

61. 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) 230 m / s and direction is such as to

rotate slider velocity in the same sense

as the angular velocity

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

rortate slider velocity in the opposite

sense as the angular velocity

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

rotate slider velocity in the same sense

as the angular velocity

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

rotate slider velocity in the opposite sense

as the angular velocity

Ans. (c)

Page 41

Office Address : F-126, Katwaria Sarai, New Delhi - 110 016.

Telephone : 011-41013406, Mobile : 8130909220, 9711853908, 7838813406

Web : www.iesmaster.org E-mail : [email protected]

IE

S

M

AS

TE

R

P2 =

1 1G

P

T 60

tan tan

T 30

=

= tan–1 (2)

116. In skew bevel gears, the axes are

(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

Ans. (a)

Sol. In skew bevel gears, the axes are non-parallel

and non-intersecting, and teeth are curved.

117. Consider that modern machines mostly use

short bearings due to the following reasons:

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

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 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 (b) strain aging

(c) twinning (d) screw dislocation

Ans. (c)

Sol. Twinning is a plane defect 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

Page 42

Office Address : F-126, Katwaria Sarai, New Delhi - 110 016.

Telephone : 011-41013406, Mobile : 8130909220, 9711853908, 7838813406

Web : www.iesmaster.org E-mail : [email protected]

IE

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TE

R

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 revolutions

c = Basic dynamic load rating

w = equivalent dynamic load

k = 3 for ball bearing

L10 =

362500

500

= (125)3 = 1953125

1.953125 million of revolution