ACE Engineering Academy

Transcription

Hyderabad |Delhi |Bhopal |Pune |Bhubaneswar |Bengaluru |Lucknow |Patna |Chennai |Vijayawada |Visakhapatnam |Tirupati
:2:
GATE_2016_Forenoon Session
GENERAL APTITUDE
1-5 carry one mark each
01.
If I were you, I ____ that laptop, It’s much too expensive.
(A) won’t buy
(B) shan’t buy
(C) wouldn’t buy
(D) would buy
Ans: (C)
Sol:
In if clause (type2) ‘were’ is in the past tense so the so main clause should be in the conditional
clause (past tense). Therefore ‘C’ is the best answer
02.
He turned a deaf ear to my request.
What does the underlined phrasal verb mean?
(A) ignored
(B) appreciated
(C) twisted
(D) returned
Ans: (A)
Sol:
‘turned a deaf ear’ means ignored
03.
Choose the most appropriate set of words from the options given below to complete the
following sentence.
_________ is a will, ______ is a way.
(A) Wear, there, their
(B) Were, their, there
(C) Where, there, there
(D) Where, their, their
Ans: (C)
Sol:
Where there is a will there is a way. It is a quotation
04.
(x% of y) + (y% of x) is equivalent to _____.
(A) 2% of xy
(B) 2% of (xy/100)
(C) xy % of 100
(D) 100 % of xy
Ans: (A)
Sol:
(x% of y) + (y% of x) =
:3:
x
y
y
x
100
100

xy  xy
100

2xy
100
= 2% of xy
05.
The sum of the digits of a two digit number is 12. If the new number formed by reversing the
digits is greater than the original number by 54, find the original number.
(A) 39
(B) 57
(C) 66
(D) 93
Ans: (A)
Sol:
The new number formed by reversing the digits is greater than the original number is possible in
options A and B only.
Options ‘A’:
Sum of two digits in the number = 3 + 9 = 12
After reversing the two digits number = 93
The difference between the new number formed and original number = 93 – 39 = 54
 option ‘A’ is correct.
Option ‘B’:
original number = 57
After reversing, the number is formed = 75
The difference between these two numbers = 75 – 57 = 18
 It is not.
:4:
:5:
06-10 Carry two marks each
06.
Two finance companies, P and Q, declared fixed annual rates of interest on the amounts invested
with them. The rates of interest offered by these companies may differ from year to year. Yearwise annual rates of interest offered by these companies are shown by the line graph provided
below.
9
8
7
6.5
10
9.5
8
8
9
7.5
8
6.5
6
P
Q
4
2000
2001
2002
2003
2005
2004
2006
If the amounts invested in the companies, P and Q, in 2006 are in the ratio 8:9, then the amounts
received after one year as interests from companies P and Q would be in the ratio:
(A) 2:3
(B) 3:4
(C) 6:7
(D) 4:3
Ans: (D)
Sol:
The amounts invested in the companies of, P and Q in 2006 = 8 : 9
The rate of interest of company ‘P’ in 2006 = 6%
The rate of interest of company ‘Q’ in 2006 = 4%
The amounts received after one year by P and Q companies in 2006 year
P
07.
Q
6% of 8
:
4% of 9
6
8
100
:
4
9
100
4
:
3
Today, we consider Ashoka as a great ruler because of the copious evidence he left behind in the
form of stone carved edicts. Historians tend to correlate greatness of a king at his time with the
availability of evidence today.
Which of the following can be logically inferred from the above sentences?
:6:
(A) Emperors who do not leave significant sculpted evidence are completely forgotten.
(B) Ashoka produced stone carved edicts to ensure that later historians will respect him.
(C) Statues of kings are a reminder of their greatness.
(D) A king’s greatness, as we know him today, is interpreted by historians.
Ans: (D)
Sol:
‘Today, historians correlate greatness of a king at his time with the availability of evidence.’ This
statement leads to the best inference i.e. option ‘D’
08
Fact 1: Humans are mammals.
Fact 2: Some humans are engineers.
Fact 3: Engineers build houses.
If the above statements are facts, which of the following can be logically inferred?
I. All mammals build houses.
II. Engineers are mammals.
III. Some humans are not engineers.
(A) II only.
(B) III only.
(C) I, II and III.
(D) I only.
Ans: (B)
Sol:
From given facts, the following venn diagram is possible.
BH
M
H = Humans
H
M = Mammals
E
E = Engineers
BH = Build houses
 From above diagram, statement III is true.
09.
A square pyramid has a base perimeter x, and the slant height is half of the perimeter. What is
the lateral surface area of the pyramid?
(A) x2
(B) 0.75 x2
(C) 0.50 x2
(D) 0.25 x2
:7:
Ans: (D)
Sol:
Base perimeter of square pyramid = x = p
Slant height 
x
l
2
Lateral surface area of pyramid

1
 Base perimeter (p)  slant height (l)
2
x/2
x 1
  p l
2 2

1
x
x
2
2

x2
4
x/4
x/4
= 0.25x2
10.
Ananth takes 6 hours and Bharath takes 4 hours to read a book. Both started reading copies of
the book at the same time. After how many hours is the number of pages to be read by Ananth,
twice that to be read by Bharath? Assume Ananth and Bharath read all the pages with constant
pace.
(A) 1
(B) 2
(C) 3
(D) 4
Ans: (C)
Sol:
Ananth takes 6 hours to read a book
Bharath takes 4 hours to read a book
L.C.M = 12
The number of pages read by Ananth and Bharath must be 12 (or) multiple of 12 only.
If Ananth read 12 number of pages in 6 hrs
 1 hr =
12
= 2 pages
6
If Bharath read 12 number of pages in 4 hrs
1 hr =
:8:
12
= 3 pages
4
1 hrs
Ananth
Option ‘A’
Read
2 pages
Bharath
Not read
10 pages
Read
3 pages
Not read
9 pages
10 : 9
 it is not
2 hrs
Option ‘B’
Ananth
Read
4 pages
Bharath
Not read
8 pages
Read
6 pages
Not read
6 pages
8 : 6
4: 3
 it is not
Options ‘C’
3 hrs
Ananth
Read
6 pages
Bharath
Not read
6 pages
Read
9 pages
Not read
3 pages
2 :1
 After 3 hours is the number of pages to be read by Ananth, twice that to be read by Bharath.
:9:
:10:
01 – Q. 25 carry one mark each.
01.
 0 1
The eigenvalues of the matrix are 

  1 0
(A) i and –i
(B) 1 and –1
(C) 0 and 1
(D) 0 and –1
Ans: (A)
Sol:
 +  = 0;   = + 1
 = i  = – i
02. The number of solutions of the simultaneous algebraic equations y = 3x + 3 and y = 3x+5 is
(A) zero
(B) 1
(C) 2
(D) infinite
Ans: (A)
Sol:
y = 3x+5 & y = 3x+3 are parallel to each other
No solution is zero solution
03. At x = 0, the function is f ( x )  sin
2x
(– < x < , L >0)
L
(A) continuous and differentiable.
(B) not continuous and not differentiable.
(C) not continuous but differentiable.
(D) continuous but not differentiable.
Ans: (D)
Sol:
 2x 
f  x   sin 
 at x = 0 is having graph
 L 
-2L
-L
0
L
2L
It is continuous at x = 0 but not differential at x = 0
04.
:11:
f x , y   x 3  3xy 2 and g  x, y   3x 2 y  y3
For the two functions
which one of the following options is correct?
(A)
f g

x x
(B)
f
g

x
y
(C)
f
g

y
x
(D)
f g

y x
Ans: (C)
Sol:
f
g
 6xy;
 6xy
y
x

05. The function f Z 
f g

y x
z2  1
is singular at
z2  4
(A) z =  2
(B) z =  1
(C) z = i
(D) z =  2i
Ans: (D)
Sol:
To get singular points let z2+4 = 0
z2 = – 4
z=2i
06. A fair coin is tossed N times. The probability that head does not turn up in any of the tosses is
1
(A)  
2
N 1
1
(B) 1   
2
N 1
1
(C)  
2
N
1
(D) 1   
2
N
Ans: (C)
07. A normal random variable X has the following probability density function
1
f X x  
e
8
(A) 0
  x 12 


 8 

,  x   Then  f X x dx 
1
(B)
1
2
(C) 1 
1
e
(D) 1
:12:
Ans: (B)
Sol:
1
2 2


1
e
1  x 1 
 

2 2 
1 

2
x 1
dx
t 
 dt  dx = 2dt
2
2
1 2
t  V  t2 = 2V
2
2
  t
2
2 
e
dt

2 2 0
1

  dV
1
V
2
=
e
V

2 0
2
=
08.
1
2

e

V
V
1
 
2
0

2t dt = 2 dV  dt 
dV
2V
1
2
The elastic modulus of a rigid perfectly plastic solid is
(A) 0
(B) 1
(C) 100
(D) infinity
Ans: (D)
Sol:
Stress
Strain
09.
Consider the following statements:
(P) Hardness is the resistance of a material to indentation.
(Q) Elastic modulus is a measure of ductility.
(R) Deflection depends on stiffness.
(S) The total area under the stress-strain curve is a measure of resilience.
Among the above statements, the correct ones are
(A) P and Q only.
(B) Q and S only.
(C) P and R only.
(D) R and S only.
:13:
Ans: (C)
Sol:
Hardness of a material will be measured by indentation technique.
If a material undergo more strain (deflection) in elastic region means stiffness is low.
10.
A beam is subjected to an inclined concentrated load as shown in the figure below. Neglect the
weight of the beam.
The correct Free Body Diagram of the beam is
(A)
(C)
(B)
(D)
Ans: (B)
11.
:14:
Consider a circular cam with a flat face follower as shown in the figure below. The cam is
rotated in the plane of the paper about point P lying 5 mm away from its center. The radius of the
cam is 20 mm. The distance (in mm) between the highest and the lowest positions of the flat face
follower is
5 mm
P
(A) 5
(B) 10
(C) 40
(D) 45
Ans: (B)
Sol:
For the highest position the distance between the cam center and follower = (r+5) mm
For the lowest position it is (r –5) mm
So the distance between the two positions
= (r+5) – (r–5) = 10 mm
12.
A vertical cylindrical tank of 1 m diameter is filled with water up to a height of 5 m from its
bottom. Top surface of water is exposed to atmosphere. A hole of 5 mm2 area forms at the
bottom of the tank. Considering the coefficient of discharge of the hole to be unity and the
acceleration due to gravity to be 10 m/s2, the rate of leakage of water (in litre/min) through the
hole from the tank to the atmosphere, under the given conditions, is _______
Ans:
(3)
Sol:
V  2  g  h  2 10  5  10m / s
Q = AV = 510–610
= 510–5 m3/sec = 0.05 liter/sec = 0.05  60 = 3 lit/min
:15:
:16:
13.
The figure below shows an air standard Diesel cycle in p-V diagram. The cut-off ratio is given
by:
2
3
P
4
1
V
(A)
v3
v1
(B)
v2
v1
(C)
v3
v2
(D)
v1
v3
Ans: (C)
14.
:17:
The ratio of press force required to punch a square hole of 30 mm side in a 1 mm thick
aluminium sheet to that needed to punch a square hole of 60 mm side in a 2 mm thick aluminium
sheet is________
Ans: (0.25)
Ratio of force 
Sol:

15.
F1max
F2 max

4x1 t1  u
4x 2 t 2 u
x1t1 30  1 1

  0.25
x 2 t 2 60  2 4
Which one of the following is a natural polymer?
(A) Cellulose
(B) Nylon
(C) Polyester
Ans:
(A)
Sol:
cellulose is made by C, H,O atoms can be extracted from trees.
16.
In powder metallurgy, sintering of the component
(D) Polyvinyl chloride
(A) increases density and reduces ductility (B) increases porosity and reduces density
(C) increases density and reduces porosity (D) increases porosity and reduces brittleness
Ans:
(B)
Sol:
During sintering the solvent added at mixing stage will evaporate and produce porosity which
comes the reduction in density of the component.
17.
A single point right handed turning tool is used for straight turning. The feed is 0.25 mm/rev and
the uncut chip thickness is found to be 0.25 mm. The inclination angle of the main cutting edge
is 10o. The back rake angle (in degrees) is _______
Ans: (100)
Sol:
Turning operation,
i = inclination angle = 10,
:18:
f = 0.25 mm/rev, t1 = 0.25 = f cos Cs
cos Cs 
0.25 0.25

 1  Cs = 0
f
0.25
 = 90 – Cs = 90 – 0 = 90
tan b = sin  tan i + cos  tan 
= sin90 tan10 + cos 90 tan 
= 1 0.176 + 0 = 0.176
b = tan–1(0.176) = 10
18.
Consider the following statements.
(P) Electrolyte is used in Electro-chemical machining.
(Q) Electrolyte is used in Electrical discharge machining.
(R) Abrasive-slurry is used in Ultrasonic machining.
(S) Abrasive-slurry is used in Abrasive jet machining.
(A) P and R only
(B) Q and S only
(C) Q, R and S only (D) P and Q only
Ans: (A)
Sol: In EDM, dielectric fluid is used. Hence ‘Q’ is wrong.
19.
Consider the following statements.
(P) Computer aided process planning (CAPP) takes input from material requirement plan (MRP).
(Q) Production flow analysis helps in work cell formation.
(R) Group technology takes input from choice of machining or cutting parameters.
Among the above statements, the correct one(s) is (are)
(A) P only
(B) Q and R only
(C) P and R only
(D) Q only
Ans: (B)
Sol: CAPP will takes inputs from design but not from MRP. Hence “P” is wrong.
20.
:19:
The limits of a shaft designated as 100h5 are 100.000 mm and 100.014 mm. Similarly, the limits
of a shaft designated as 100h8 are 100.000 mm and 100.055 mm. If a shaft is designated as
100h6, the fundamental deviation (in m) for the same is
(A) –22
(B) zero
(C) 22
(D) 24
Ans: (B)
Fundamental deviation for shaft “h” is always “zero” irrespective of grade of tolerance.
21.
The roughness profile of a surface is depicted below.
Profile height
Sol:
2 m
centre-line of
the profile
X
2 m
5 mm
The surface roughness parameter Ra (in m) is _______.
Ans: (0)
Sol:
Ra 

22.
h
n
2  16  2  16
0
32
The facility layout technique that uses relationship (REL) chart is
(A) CRAFT.
(B) Travel chart.
(C) Partial Set Covering.
(D) ALDEP.
:20:
Ans: (D)
Sol: ALDEP – BASED ON REL REQUIREMENTS
23.
For a random variable, X, let X be the sample average. The sample size is n. The mean and the
standard deviation of X are µ and σ, respectively. The standard deviation of X is
(B) 
(A) n
(C )
Ans: (D)

n
(D)

n

n
Sol:
Standard deviation, s 
24.
ST and NT denote the standard time and the normal time, respectively, to complete a job.
Allowance = LL ST, where 0 < LL<1. Which one of the following relationships is correct?
(A) ST 
NT
1  LL
(C ) ST 
(B) ST=NT(1+LL)
NT
(D) ST=NT(1–LL)
1  LL
Ans: (A)
Sol:
RA = Relaxation allowance = LL  ST
ST = NT + RA
ST = NT + LL  ST
ST (1 – LL) = NT
ST 
25.
NT
1  LL
The throughput rate of a production system is 20 units per hour. The average flow time is 30
minutes and the cycle time is 3 minutes. The average inventory (in units) in the system is
(A) 1.5
(B) 9
(C) 10
(D) 11.33
Ans: (C)
Sol:
Average 

max .production  min .production
2
20  0
 10 units
2
:21:
Q. 26 – Q. 55 carry two marks each.
26.


The range of values of k for which the function f ( x )  k 2  4 x 2  6x 3  8x 4
has a local maxima at point x = 0 is
(B) k  –2 or k 2
(A) k < –2 or k >2
(C) –2 < k < 2
(D) –2  k  2
Ans: (A)
Sol:
f(x) = 32x3+18x2+2(k2–4)x
f(x) = 96 x2 + 36x + 2 (k2–4)
f(0) = 2(k2–4) < 0
k2 – 4 < 0
 k < –2 (or) k > 2
27.
 e5 x  1 

lim x  0 
 x 
2
is equal to ______.
Ans: (25)
Sol:
 e5x  1   e5x  1 
= Lt 
5
5
x 0
 5x
  5x

 e5x  1   e5x  1 
= Lt 
 5  Lt 
5
x 0
 5x
 x 0  5x

= 5 1 51 = 25
28.
To solve the equation 2sinx = x by Newton-Raphson method, the initial guess was chosen to
be x = 2.0. Consider x in radian only. The value of x (in radian) obtained after one iteration will
be closest to
(A) – 8.101
(B) 1.901
(C) 2.099
(D) 12.101
:22:
Ans: (B)
Sol:
f(x) = 2sin x – x
h
f (x) (2sin x  x)

f (x)
2 cos x  1
= – 0.099
x1= 2+h = 2 – 0.099 = 1.901
29.
In linear gas tungsten arc welding of two plates of the same material, the peak temperature T (in
K) is given as T 
c1q
, where q is the heat input per unit length (in J/m) of weld,  is the thermal

diffusivity (in m2/s) of the plate materials and C1 is a constant independent of process parameters
and material types. Two welding cases are given below.
Case I: V = 15 V, I = 200 A, v = 5 mm/s, k = 150 W/mK,  = 3000 kg/m3, C = 900 J/kg-K
Case II: V = 15 V, I = 300 A, v = 10 mm/s, k = 50 W/mK,  = 8000 kg/m3, C = 450 J/kg-K
where, V is welding voltage, I is welding current, v is welding speed, and k, 
and C refer to the thermal conductivity, the density and the specific heat of the plate materials,
respectively. Consider that electrical energy is completely converted to thermal energy. All other
conditions remain same.
The ratio of the peak temperature in Case I to that in Case II is
(A)
1
3
(B)
1
2
(C) 1
(D) 2
Ans: (A)

Sol:
1 
k
C
150
 55.5  10 6
3000  900
2 
50
 13.88 10 6
8000  450
q1 
VI 15  200

 6  105
3
r
5  10
q2 
15  300
 4.5  105
3
10  10
:23:
T1 q1  2


T2 q 2 1
30.

6  105 13.88  10 6

4.5  105 55.5  10 6

6  13.88
1
 0.333 
4.5  55.55
3
A bar of rectangular cross-sectional area of 50 mm2 is pulled from both the sides by equal forces
of 100 N as shown in the figure below. The shear stress (in MPa) along the plane making an
angle 450 with the axis, shown as a dashed line in the figure, is ________________.
100 N
100 N
45
o
Ans: (1)
 100 / 50 

 1MPa
2
2
Sol:
max 
31.
A 1 m × 10 mm × 10 mm cantilever beam is subjected to a uniformly distributed load per unit
length of 100 N/m as shown in the figure below. The normal stress (in MPa) due to bending at
point P is ____________.
:24:
Point P
100 N/m
1m
Ans:
(300)
Sol:
1
M  100 1    50N  m
2
M
50 103


 300 MPa (Tension)
Z  10  102 


 6 
32.
A thin-walled cylindrical pressure vessel of internal diameter 2 m is designed to withstand an
internal pressure of 500 kPa (gauge). If the allowable normal stress at any point within the
cylindrical portion of the vessel is 100 MPa, the minimum thickness of the plate of the vessel (in
mm) is________.
Ans:
Sol:
(5)
P = 500 kPa
= 0.5 MPa
h = 100 MPa, D = 2m
h 
PD
2t
100 
0.5  2000
2 t
thickness, t = 5mm
33.
An engine, connected with a flywheel, is designed to operate at an average angular speed of 800
rpm. During operation of the engine, the maximum change in kinetic energy in a cycle is found
to be 6240 J. In order to keep the fluctuation of the angular speed within ±1% of its average
value, the moment of inertia (in kg-m2) of the flywheel should be ____________.
:25:
Ans: (44.45)
Sol:
Given mean speed N = 800 rpm

2N
rad / sec
60
Fluctuation of energy E = 6240 J
Fluctuation of speed =  1%
Moment of inertia I = ?
We know E = I 2Cs
I
34.
E
6240

= 44.45 kg-m2
2
2
 Cs  2N 

  0.02
 60 
A 2 m × 2 m square opening in a vertical wall is covered with a metallic plate of the same
dimensions as shown in the figure below. Consider the acceleration due to gravity to be 10.0
m/s2. The force (in kN) exerted by water on the plate is _____
Patm
Patm
2m
water 2m
vertical wall
2m  2m
square plate
vertical wall
Ans: (120)
Sol:
F  gAh  1000 10  2  2  (2  1)
= 120 kN
35.
:26:
An ideal gas of mass m is contained in a rigid tank of volume V at a pressure P. During a
reversible process its pressure reduces to P1 . Following statements are made regarding the
process.
(P) Heat is transferred from the gas.
(Q) Work done by the gas is zero.
(R) Entropy of the gas remains constant.
(S) Entropy of the gas decreases.
(A) P and R only
(B) P, Q and R only (C) Q and R only
(D) P, Q and S only
Ans: (D)
Sol:
At constant volume to decrease pressure, temperature has to be decreased. So heat has to be
transferred from the gas.
As the change in volume is zero, work done by the gas is zero. Entropy of the gas decrease as
system is losing heat.
36.
A long slender metallic rod of length L is used as a fin. As shown in the figure below, the left
end of the fin is kept at a constant temperature tb. The fin loses heat by convection to the
atmosphere which is at a temperature ta (ta< tb). Four options of temperature profiles are shown.
Identify the correct option.
convection
tb
x=L
x
atmosphere at ta
(B)
tb
ta
0
temperature
temperature
(A)
x
L
tb
ta
0
x
L
:27:
(C)
tb
ta
0
L
x
temperature
temperature
(D)
tb
ta
0
x
L
Ans: (B)
Sol:
 cosh m (Lc  x)

cosh m Lc
0
d 2
 0 , so angle of inclination of tangents for curve should increase with ‘x’
dx 2
37.
In a fully developed turbulent flow through a circular pipe, a head loss of h1 is observed. The
diameter of the pipe is increased by 10% for the same flow rate and a head loss of h2 is noted.
Assume friction factor for both the cases of pipe flow is the same. The ratio of
(A) 0.34
(B) 0.62
(C) 0.87
h2
is closest to
h1
(D) 1.00
Ans: (B)
Sol:
For turbulent flow h 
h
38.
fLQ 2
12.1d 5
h
d1 5
1
 2 
 0.62
5
d
h1 1.1d1 5
Two cast iron blocks P and Q, each of 500 mm length, have the same cross-sectional area. Block
P has rectangular cross-section of 100 mm × 200 mm. Block Q is of square cross-section. Both P
and Q were cast under the same conditions with all their surfaces enclosed within the mould. The
solidification time of a casting is proportional to the square of the ratio of its volume to its
surface area. The ratio of solidification time of the block P to that of the block Q is ______.
:28:
Ans: (0.9016)
Sol:
LP = LQ= 500,
Block “P” is rectangular = 100200
Block “Q” is square
(Ac)P = (Ac)Q
100  200 = x2
x  100  200  141.42mm
MP 

100  200  500
 29.411
2100  200  200  500  500  100
MQ 

V
As
V
As
141.422  500
2
4  500  141.42  2  141.4
 P  M P 

 Q  M Q 
 30.974
2
2
 29.411 

  0.9016
 30.974 
39.
A 300 mm long copper wire of uniform cross-section is pulled in tension so that a maximum
tensile stress of 270 MPa is developed within the wire. The entire deformation of the wire
remains linearly elastic. The elastic modulus of copper is 100 GPa. The resultant elongation (in
mm) is ________.
Ans: (0.81 mm)
Sol:
Given l = 300 mm
:29:
max = 270 MPa
E = 100 GPa
 max  E


270  100  103 

300
l =81010–3m = 0.81mm
40.
In a single-pass rolling operation, a 200 mm wide metallic strip is rolled from a thickness 10
mm to a thickness 6 mm. The roll radius is 100 mm and it rotates at 200 rpm. The roll-strip
contact length is a function of roll radius and, initial and final thickness of the strip. If the
average flow stress in plane strain of the strip material in the roll gap is 500 MPa, the roll
separating force (in kN) is _______
Ans: (2598)
Sol:
b = 200 mm, H0 = 10mm, H1 = 6mm,
R = 100 mm, N = 200 rpm
y = 500 MPa,
Roll separating force 
2
 L 
 y b L1 

3
 4H 
L  RH  100  4  20
by assuming maximum reduction

H
H
 0 .2
R
10  6
8
2
R.S.F 
2
 0.2  20 
 500  200  201 

48 
3

= 2598076 N 2598.076kN
41.
:30:
Two solid cylinders of equal diameter have different heights. They are compressed plastically by
a pair of rigid dies to create the same percentage reduction in their respective heights. Consider
that the die-workpiece interface friction is negligible. The ratio of the final diameter of the
shorter cylinder to that of the longer cylinder is __________
Sol:
Let d1 = d2 = d
h1 = height of first cylinder
h2 = height of second cylinder
assume h1 < h2
let % reduction in height = 10%
Ist cylinder
h0  hf
 0.1
h0
h0 – hf = 0.1 h0
hf = h0 – 0.1 h0 = 0.9 h0
A0h0 = Af hf
d02h0 = df2 hf
df  d0
h0
h0
 d0
hf
0 .9 h 0
= 1.054 d0 = 1.054 (d0)1
IInd cylinder
A0h0 = Afhf
d02h0 = df2 hf
df  d0
 d0
h0
hf
h0
= 1.054 (d0)2
0 .9 h 0
Ratio 

42.
:31:
d 0 1
d 0 2
1.054d 0 1
1
1.054d 0 2
Two flat steel sheets, each of 2.5 mm thickness, are being resistance spot welded using a current
of 6000 A and weld time of 0.2 s. The contact resistance at the interface between the two sheets
is 200  and the specific energy to melt steel is 10×109 J/m3. A spherical melt pool of diameter
4 mm is formed at the interface due to the current flow. Consider that electrical energy is
completely converted to thermal energy. The ratio of the heat used for melting to the total
resistive heat generated is ______
Ans: (0.2327)
Sol: t = 2.5 mm , I = 6000A,  = 0.2sec
R = 200
Heat required, (H.R/m3)= 10109 J/m3
nugget diameter = d = 4mm (sphere )
Heat generated = I2 R
= 60002  20010–6  0.2 = 1440 J
Heat required = volume  heat required/m3

4
   R 3  H.R / m 3
3

3
4
   2  10 3  10  109
3


= 335.1032 J
H .I 

Heat required
Heat generated
335.1032
 0.2327
1440
43.
:32:
A cylindrical bar of 100 mm diameter is orthogonally straight turned with cutting velocity, feed
and depth of cut of 120 m/min, 0.25 mm/rev and 4 mm, respectively. The specific cutting energy
of the work material is 1×109 J/m3. Neglect the contribution of feed force towards cutting power.
The main or tangential cutting force (in N) is ________
Ans:
(1000)
Sol:
Orthogonal cutting
D = 100 mm, Vc = 120 m/min, f = 0.25 mm/rev
d = 4mm,
Specific cutting energy (Sp.C.E) = 109 J/m3
Specific cutting energy 

power or  Work done
MRR
FC  VC
J / m3
f  d  VC 106
FC = Sp.C.E  f  d 10–6
 109  0.25  4 106  1000
44.
A 60 mm wide block of low carbon steel is face milled at a cutting speed of 120 m/min, feed of
0.1 mm/tooth and axial depth of cut of 4 mm. A schematic representation of the face milling
process is shown below. The diameter of the cutter is 120 mm and it has 12 cutting edges. The
material removal rate (in mm3/s) is _____
 120 mm
60 mm
feed
work piece
+
cutter
schematic representation (top view)
:33:
Ans: (1527.8872)
Sol:
b = 60 mm,
V = 120 m/min
t = 0.1 mmtooth, d = 4 mm
End mill cutter, D = 120 mm
z = 12
fm = ft × z× N
= 0.1×12×
1000 120  1000
 120
= 381.9718 rpm
MRR = bdfm = 604381.9718
= 9176.232 mm3/min = 1527.8872 mm3/sec
45.
In abrasive water jet machining, the velocity of water at the exit of the orifice, before mixing
with abrasives, is 800 m/s. The mass flow rate of water is 3.4 kg/min. The abrasives are added to
the water jet at a rate of 0.6 kg/min with negligible velocity. Assume that at the end of the
focusing tube, abrasive particles and water come out with equal velocity. Consider that there is
no air in the abrasive water jet. Assuming conservation of momentum, the velocity (in m/s) of
the abrasive water jet at the end of the focusing tube is _________
Ans: (680)
Sol:
Vw = 800 m/s
mw = 3.4 kg/min
ma = 0.6 kg/min
 a va  m
 w v w  m
a m
 w V
m
0.6
3.4
 0.6  3.4 
0
 800  
 V
60
60
 60 
45.33 =
1
V
15
V = 45.33×15 = 679.95
= 680 m/sec
46.
:34:
A single axis CNC table is driven by a DC servo motor that is directly coupled to a lead screw of
5 mm pitch. The circular encoder attached to the lead screw generates 1000 voltage pulses per
revolution of the lead screw. The table moves at a constant speed of 6 m/min. The corresponding
frequency (in kHz) of the voltage pulses generated by the circular encoder is________
Sol:
p = 5 mm
1000 pulses  1 rev of motor
 1 rev of lead screw
Velocity of table = 6 m/min
= 6000 mm/min
= 100 mm/sec
1000 pulses  1 rev of lead screw  5 mm
1 pulse 
5
 0.005 mm
1000
BLU = 0.005 mm
Table speed = BLU × Rate of Pulses
Rate of pulses =
100
0.005
= 20000 pulses/sec
= 20000 Hz
= 20 kHz
47.
A helical gear with involute tooth profile has been machined with a disc-type form gear milling
cutter. The helical gear has 30 teeth and a helix angle of 30o. The module of the gear milling
cutter is 2. The pitch circle diameter (in mm) of the helical gear is_________
Ans: (69.28)
Sol: mn = normal module = 2mm
Pitch circle diameter, d 
Z m n 30  2

= 69.28mm
cos  cos 30
48.
:35:
A quality control engineer has collected 5 samples, each of size 30. The numbers of defective
items in the samples are given in the table below.
Sample Number
I
II
III
IV
V
Number of defective items
3
2
4
1
5
The upper three-sigma (3σ) control limit for the proportion of defective items in any sample is
_____
Ans: UCL = 0.2643
Sol:
p
sum of defectives
sample size  number of samples

3  2  4 1 5
5  30

15
1
  0.1
5  30 10
Standard error of proportion
p 1  p 
n
p 

0.1 1  0.1
30

0 .1  0 .9
 0.054772
30
Central line = p = 0.1
Upper control limit = p  3 p
= 0.1 + 30.054772 = 0.2643
Lower control limit = p  3 p
= 0.1 – 30.054772 = 0
49.
:36:
A job consists of two work elements, P and Q. Completion time (in minutes) of each work
element was measured. A pilot study involved collecting a sample of 40 observations. The
results of this pilot study are summarized in the table below.
Work element Mean completion time (in minutes) Standard deviation (in minutes)
P
1
0.50
Q
1
0.05
For the main study, the minimum sample size for the sample mean time of any work element to
be within 0.1 minutes of its true mean time with 95% confidence (corresponding standard normal
value, Z0.025 = 1.96 is ______________
Ans:
(96)
Sol:
Standard deviation = s1 = 0.5 minutes
Error of study = h = 0.1 minutes
Z0.025 = 1.96
 Z  s1 
h1  

 h 
2
2
 1.96  0.5 

 = 96.04 observations
 0.1 
Standard deviation = s2 = 0.05 minutes
Error of study = h = 0.1 minutes
Z0.025 = 1.96
 Z  s2 
h2  

 h 
2
2
 1.96  0.05 

 = 1 observations
0.1


Larger of the two values is 96.
Hence study should be continued upto 96 observations.
50.
:37:
Consider a system with 10 identical components connected in series. The time to failure of each
component is exponentially distributed with a failure rate of 0.10 per 500 days. The reliability of
the system after 400 days of operation is _____
Ans: 0.449
Sol:
Rs = e-nt
Failure rate,  
0.1
500
t = 400 days
n = 10 in series
RS  e
51.
10
0.1
400
500
 0.449
For a process, the quality loss coefficient is 5. The target value on the dimension to be attained
through the process is 50 mm. If the maximum loss permissible (in monetary terms) is INR 80,
the maximum allowable deviation (in mm) from the target is
(A)
1
4
(B)
1
10
(C) 4
(D) 10
Ans: (C)
Sol:
R
A0
 0 2
R = Quality loss coefficient
A0 = Economic consequence
0 = Functional limit or customer tolerance
R = 5 , A0 = 80,
0 
A0
80

 16  4
5
R
52.
:38:
Consider a network with nodes 1, 2, 3, 4, 5 and 6. The nodes are connected with directed arcs as
shown in the table below. The respective costs (in INR) incurred while traversing the directed
arcs are also mentioned.
Directed arcs
1→2 1→3
Cost (in INR) 3
2→4
9
2→5
3
3→2
2
3→4
2
3→5
4
4→5
8
4→6
7
5→6
2
The second shortest path from node 1 to node 6 (i.e. the path that has the second least total cost
and does not use any part of the shortest path) has a total cost (in INR) of
(A) 7
(B) 8
(C) 15
(D) 19
Ans: (C)
Sol:
9
3
2
1
3
8
4
7
3
2
2
4
2
Path
Duration
1-3-4-6
9+4+2 = 15
1-3-2-4-6
9+2+3+2 = 16
1-3-4-5-6
9+4+7+2 = 22
1-3-2-5-6
9+2+2+2 =15
1-3-2-4-5-6
9+2+3+7+2 = 23
1-2-4-6
3+3+2 = 8
1-2-5-6
3+2+2 = 7
1-2-4-5-6
3+3+7+2 = 15
5
6
2
2
53.
:39:
Five jobs need to be processed on a single machine. All the jobs are available for processing at
time t = 0. Their respective processing times are given below.
Jobs
I
II III IV V
Processing times (in minutes) 13 4
7
14 11
The average completion time (in minutes) of jobs as per the Shortest Processing Time rule is
(A) 9.8
Ans: (B)
Sol:
(B) 24.2
(C) 49.0
(D) 121.0
SPT RULE
Job
II
III
V
I
IV
Time
4
7
11
13
14
Completion time (ci)
4
11
22
35
49
Ci = 121
Mean completion Time 
54.
 Ci 121

 24.2 min
n
5
Transportation costs (in INR/unit) from factories to respective markets are given in the table
below. The market requirements and factory capacities are also given. Using the North-West
Corner method, the quantity (in units) to be transported from factory R to market II is
Requirements
Factory
Market
(in units)
P
Q
R
S
I
3
3
2
1
50
II
4
2
5
9
20
III
1
2
1
4
30
Factory capacity(in units) 20 40 30 10
(A) 30
(B) 20
(C) 10
(D) 0
:40:
Ans: (C)
Sol:
Supply = 20+40+30+10 = 100 units
Demand = 50+20+30 = 100 units
FACTORY
P
3
Q
R
3
30
X
X
2
10
5
10
X
20
30
REQ
S
2
1
I
20
II
X
III
X
2
X
1
20
4
10
20
40
30
10
4
1
9
50
Factory R to market II is 10 units.
55.
In a given year, a restaurant earned INR 38,500 in revenues. In that year, total expenses incurred
were INR 30,000 and the depreciation amount was INR 3,200. At 40% tax rate, the net cash flow
(in INR) for that year was _____
Ans: 6380
Sol:
Yearly revenue
(–) Total expenses
(–) Depreciation
Earning before taxes (EBT)
(–) Tax rate @ 40%
Earnings after taxes (EAT)
(+) Depreciation
Net Cash flow = (EAT + Depreciation)
Amount (INR)
38500
30000
3200
5300
2120
3180
3200
6380

ACE Engineering Academy

Transcription

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