Download Me Paper II Obj.sol.(Ese 2015) PDF

TitleMe Paper II Obj.sol.(Ese 2015)
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Page 1

ESE-2015
Detailed Exam Solutions

solutions

(Objective Paper - II)

Mechanical

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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.

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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

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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)

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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

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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

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