GATE SOLVED PAPER-ME GATE-2012 GATE Notes | EduRev

GATE : GATE SOLVED PAPER-ME GATE-2012 GATE Notes | EduRev

 Page 1


No part of this publication may be reproduced or distributed in any form or any means, electronic, mechanical, 
photocopying, or otherwise without the prior permission of the author.
GATE SOLVED PAPER
Mechanical Engineering
2012
Copyright © By NODIA & COMPANY
Information contained in this book has been obtained by authors, from sources believes to be reliable.  However, 
neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor its 
authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book 
is published with the understanding that Nodia and its authors are supplying information but are not attempting 
to render engineering  or other professional services.
NODIA AND COMPANY
B-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, Jaipur 302039
Ph :  +91 - 141 - 2101150
www.nodia.co.in
email : enquiry@nodia.co.in
Page 2


No part of this publication may be reproduced or distributed in any form or any means, electronic, mechanical, 
photocopying, or otherwise without the prior permission of the author.
GATE SOLVED PAPER
Mechanical Engineering
2012
Copyright © By NODIA & COMPANY
Information contained in this book has been obtained by authors, from sources believes to be reliable.  However, 
neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor its 
authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book 
is published with the understanding that Nodia and its authors are supplying information but are not attempting 
to render engineering  or other professional services.
NODIA AND COMPANY
B-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, Jaipur 302039
Ph :  +91 - 141 - 2101150
www.nodia.co.in
email : enquiry@nodia.co.in
GATE SOLVED PAPER - ME
2012
© www.nodia.co.in
Q. 1 In abrasive jet machining, as the distance between the nozzle tip and the work 
surface increases, the material removal rate
(A) increases continuously. (B) decreases continuously.
(C) decreases, becomes stable and then increases.
(D) increases, becomes stable and then decreases.
Sol. 1 Option (D) is correct.
Graph for abrasive jet machining for the distance between the nozzle tip and 
work surface () l and abrasive flow rate is given in figure.
It is clear from the graph that the material removal rate is first increases because 
of area of jet increase than becomes stable and then decreases due to decrease in 
jet velocity.
Q. 2 Match the following metal forming processes with their associated stresses in the 
workpiece.
Metal forming process Types of stress
1. Coining P. Tensile
2. Wire Drawing Q. Shear
3. Blanking R. Tensile and compressive
4. Deep Drawing S. Compressive
(A) 1-S, 2-P, 3-Q, 4-R (B) 1-S, 2-P, 3-R, 4-Q
(C) 1-P, 2-Q, 3-S, 4-R (D) 1-P, 2-R, 3-Q, 4-S
Sol. 2 Option (A) is correct.
Metal forming process Types of stress
1. Coining S. Compressive
2. Wire Drawing P. Tensile
3. Blanking Q. Shear
4. Deep Drawing R. Tensile and compressive
Hence, correct match list is, 1-S, 2-P, 3-Q, 4-R
GATE ME 2012
ONE MARK
GATE ME 2012
ONE MARK
Page 3


No part of this publication may be reproduced or distributed in any form or any means, electronic, mechanical, 
photocopying, or otherwise without the prior permission of the author.
GATE SOLVED PAPER
Mechanical Engineering
2012
Copyright © By NODIA & COMPANY
Information contained in this book has been obtained by authors, from sources believes to be reliable.  However, 
neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor its 
authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book 
is published with the understanding that Nodia and its authors are supplying information but are not attempting 
to render engineering  or other professional services.
NODIA AND COMPANY
B-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, Jaipur 302039
Ph :  +91 - 141 - 2101150
www.nodia.co.in
email : enquiry@nodia.co.in
GATE SOLVED PAPER - ME
2012
© www.nodia.co.in
Q. 1 In abrasive jet machining, as the distance between the nozzle tip and the work 
surface increases, the material removal rate
(A) increases continuously. (B) decreases continuously.
(C) decreases, becomes stable and then increases.
(D) increases, becomes stable and then decreases.
Sol. 1 Option (D) is correct.
Graph for abrasive jet machining for the distance between the nozzle tip and 
work surface () l and abrasive flow rate is given in figure.
It is clear from the graph that the material removal rate is first increases because 
of area of jet increase than becomes stable and then decreases due to decrease in 
jet velocity.
Q. 2 Match the following metal forming processes with their associated stresses in the 
workpiece.
Metal forming process Types of stress
1. Coining P. Tensile
2. Wire Drawing Q. Shear
3. Blanking R. Tensile and compressive
4. Deep Drawing S. Compressive
(A) 1-S, 2-P, 3-Q, 4-R (B) 1-S, 2-P, 3-R, 4-Q
(C) 1-P, 2-Q, 3-S, 4-R (D) 1-P, 2-R, 3-Q, 4-S
Sol. 2 Option (A) is correct.
Metal forming process Types of stress
1. Coining S. Compressive
2. Wire Drawing P. Tensile
3. Blanking Q. Shear
4. Deep Drawing R. Tensile and compressive
Hence, correct match list is, 1-S, 2-P, 3-Q, 4-R
GATE ME 2012
ONE MARK
GATE ME 2012
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2012
Q. 3 In an interchangeable assembly, shafts of size 25.000 mm
0.040
. 0 010
+
-
 mate with holes of 
size 25.000 mm
.
.
0 030
0 020
+
+
. The maximum interference (in microns) in the assembly is
(A) 40 (B) 30
(C) 20 (D) 10
Sol. 3 Option (C) is correct.
An interference fit for shaft and hole is as given in figure below.
 Maximum Interference 
lim lim Maximum it of shat Minimum it of hole =-
  (25 0. 40) (25 0.020) 0 =+ - +
  0.02 mm = 2microns 0 =
Q. 4 During normalizing process of steel, the specimen is heated
(A) between the upper and lower critical temperature and cooled in still air.
(B) above the upper critical temperature and cooled in furnace.
(C) above the upper critical temperature and cooled in still air.
(D) between the upper and lower critical temperature and cooled in furnace
Sol. 4 Option (C) is correct
Normalizing involves prolonged heating just above the critical temperature to 
produce globular form of carbine and then cooling in air.
Q. 5 Oil flows through a 200 mm diameter horizontal cast iron pipe (friction factor, 
. f 0 0225 = ) of length 500 m. The volumetric flow rate is 0.2 / ms
3
. The head loss 
(in m) due to friction is (assume 9.81 / ms g
2
= )
(A) 116.18 (B) 0.116
(C) 18.22 (D) 232.36
Sol. 5 Option (A) is correct.
From Darcy Weischback equation head loss
 h f
D
L
g
V
2
2
##
= ...(1)
Given that 500 , m h = 0.2 m D
1000
200
== , . f 0 0225 =
Since volumetric flow rate
 n
o
 () Area velocity of flow V
#
=
 V 
(.)
.
6.37 /
Area
ms
4
02
02
2
#
n
p
== =
o
Hence, h .
..
(. )
0 0225
02
500
2981
637
2
##
#
=
 h 116.33 m = 116.18 m -
GATE ME 2012
ONE MARK
GATE ME 2012
ONE MARK
GATE ME 2012
ONE MARK
Page 4


No part of this publication may be reproduced or distributed in any form or any means, electronic, mechanical, 
photocopying, or otherwise without the prior permission of the author.
GATE SOLVED PAPER
Mechanical Engineering
2012
Copyright © By NODIA & COMPANY
Information contained in this book has been obtained by authors, from sources believes to be reliable.  However, 
neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor its 
authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book 
is published with the understanding that Nodia and its authors are supplying information but are not attempting 
to render engineering  or other professional services.
NODIA AND COMPANY
B-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, Jaipur 302039
Ph :  +91 - 141 - 2101150
www.nodia.co.in
email : enquiry@nodia.co.in
GATE SOLVED PAPER - ME
2012
© www.nodia.co.in
Q. 1 In abrasive jet machining, as the distance between the nozzle tip and the work 
surface increases, the material removal rate
(A) increases continuously. (B) decreases continuously.
(C) decreases, becomes stable and then increases.
(D) increases, becomes stable and then decreases.
Sol. 1 Option (D) is correct.
Graph for abrasive jet machining for the distance between the nozzle tip and 
work surface () l and abrasive flow rate is given in figure.
It is clear from the graph that the material removal rate is first increases because 
of area of jet increase than becomes stable and then decreases due to decrease in 
jet velocity.
Q. 2 Match the following metal forming processes with their associated stresses in the 
workpiece.
Metal forming process Types of stress
1. Coining P. Tensile
2. Wire Drawing Q. Shear
3. Blanking R. Tensile and compressive
4. Deep Drawing S. Compressive
(A) 1-S, 2-P, 3-Q, 4-R (B) 1-S, 2-P, 3-R, 4-Q
(C) 1-P, 2-Q, 3-S, 4-R (D) 1-P, 2-R, 3-Q, 4-S
Sol. 2 Option (A) is correct.
Metal forming process Types of stress
1. Coining S. Compressive
2. Wire Drawing P. Tensile
3. Blanking Q. Shear
4. Deep Drawing R. Tensile and compressive
Hence, correct match list is, 1-S, 2-P, 3-Q, 4-R
GATE ME 2012
ONE MARK
GATE ME 2012
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2012
Q. 3 In an interchangeable assembly, shafts of size 25.000 mm
0.040
. 0 010
+
-
 mate with holes of 
size 25.000 mm
.
.
0 030
0 020
+
+
. The maximum interference (in microns) in the assembly is
(A) 40 (B) 30
(C) 20 (D) 10
Sol. 3 Option (C) is correct.
An interference fit for shaft and hole is as given in figure below.
 Maximum Interference 
lim lim Maximum it of shat Minimum it of hole =-
  (25 0. 40) (25 0.020) 0 =+ - +
  0.02 mm = 2microns 0 =
Q. 4 During normalizing process of steel, the specimen is heated
(A) between the upper and lower critical temperature and cooled in still air.
(B) above the upper critical temperature and cooled in furnace.
(C) above the upper critical temperature and cooled in still air.
(D) between the upper and lower critical temperature and cooled in furnace
Sol. 4 Option (C) is correct
Normalizing involves prolonged heating just above the critical temperature to 
produce globular form of carbine and then cooling in air.
Q. 5 Oil flows through a 200 mm diameter horizontal cast iron pipe (friction factor, 
. f 0 0225 = ) of length 500 m. The volumetric flow rate is 0.2 / ms
3
. The head loss 
(in m) due to friction is (assume 9.81 / ms g
2
= )
(A) 116.18 (B) 0.116
(C) 18.22 (D) 232.36
Sol. 5 Option (A) is correct.
From Darcy Weischback equation head loss
 h f
D
L
g
V
2
2
##
= ...(1)
Given that 500 , m h = 0.2 m D
1000
200
== , . f 0 0225 =
Since volumetric flow rate
 n
o
 () Area velocity of flow V
#
=
 V 
(.)
.
6.37 /
Area
ms
4
02
02
2
#
n
p
== =
o
Hence, h .
..
(. )
0 0225
02
500
2981
637
2
##
#
=
 h 116.33 m = 116.18 m -
GATE ME 2012
ONE MARK
GATE ME 2012
ONE MARK
GATE ME 2012
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2012
Q. 6 For an opaque surface, the absorptivity () a , transmissivity () t and reflectivity () r 
are related by the equation :
(A) ar t += (B) 0 rat ++ =
(C) 1 ar += (D) 0 ar +=
Sol. 6 Option (C) is correct.
The sum of the absorbed, reflected and transmitted radiation be equal to 
 art ++ 1 =
, Absorpivity a = Reflectivity r = , Transmissivity t =
For an opaque surfaces such as solids and liquids
 , 0 t =
Thus, ar + 1 =
Q. 7 Steam enters an adiabatic turbine operating at steady state with an enthalpy of 
3251.0 / kJ kg and leaves as a saturated mixture at 15 kPa with quality (dryness 
fraction) 0.9. The enthalpies of the saturated liquid and vapour at 15 kPa are 
225.94 / kJ kg h
f
= and 2598.3 / kJ kg h
g
= respectively. The mass flow rate of 
steam is 10 / kg s. Kinetic and potential energy changes are negligible. The power 
output of the turbine in MW is
(A) 6.5 (B) 8.9
(C) 9.1 (D) 27.0
Sol. 7 Option (B) is correct.
For adiabatic expansion steam in turbine.
Given 3251.0 / , kJ kg h
1
= 10 / kg s m = , 0.9( ) dryness fraction x =
At 15 kPa
Enthalpy of liquid,  h
f
 225.94 / kJ kg =
Enthalpy of vapour,  h
g
 2598.3 / kJ kg =
Since Power output of turbine.
 P () mh h
12
=-
o
        
(. . ) K E and P E are negligible ...(i)
 h
2
 () hxh h xh h
ffg f g f
=+ =+ -
  225.94 0.9(2598.3 225.94) =+ -
  2361.064 / kJ kg =
From Eq. (i)
 P (. . ) 10 3251 0 2361 064
#
=-
  8899 kW = 8.9 MW =
GATE ME 2012
ONE MARK
GATE ME 2012
ONE MARK
Page 5


No part of this publication may be reproduced or distributed in any form or any means, electronic, mechanical, 
photocopying, or otherwise without the prior permission of the author.
GATE SOLVED PAPER
Mechanical Engineering
2012
Copyright © By NODIA & COMPANY
Information contained in this book has been obtained by authors, from sources believes to be reliable.  However, 
neither Nodia nor its authors guarantee the accuracy or completeness of any information herein, and Nodia nor its 
authors shall be responsible for any error, omissions, or damages arising out of use of this information. This book 
is published with the understanding that Nodia and its authors are supplying information but are not attempting 
to render engineering  or other professional services.
NODIA AND COMPANY
B-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, Jaipur 302039
Ph :  +91 - 141 - 2101150
www.nodia.co.in
email : enquiry@nodia.co.in
GATE SOLVED PAPER - ME
2012
© www.nodia.co.in
Q. 1 In abrasive jet machining, as the distance between the nozzle tip and the work 
surface increases, the material removal rate
(A) increases continuously. (B) decreases continuously.
(C) decreases, becomes stable and then increases.
(D) increases, becomes stable and then decreases.
Sol. 1 Option (D) is correct.
Graph for abrasive jet machining for the distance between the nozzle tip and 
work surface () l and abrasive flow rate is given in figure.
It is clear from the graph that the material removal rate is first increases because 
of area of jet increase than becomes stable and then decreases due to decrease in 
jet velocity.
Q. 2 Match the following metal forming processes with their associated stresses in the 
workpiece.
Metal forming process Types of stress
1. Coining P. Tensile
2. Wire Drawing Q. Shear
3. Blanking R. Tensile and compressive
4. Deep Drawing S. Compressive
(A) 1-S, 2-P, 3-Q, 4-R (B) 1-S, 2-P, 3-R, 4-Q
(C) 1-P, 2-Q, 3-S, 4-R (D) 1-P, 2-R, 3-Q, 4-S
Sol. 2 Option (A) is correct.
Metal forming process Types of stress
1. Coining S. Compressive
2. Wire Drawing P. Tensile
3. Blanking Q. Shear
4. Deep Drawing R. Tensile and compressive
Hence, correct match list is, 1-S, 2-P, 3-Q, 4-R
GATE ME 2012
ONE MARK
GATE ME 2012
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2012
Q. 3 In an interchangeable assembly, shafts of size 25.000 mm
0.040
. 0 010
+
-
 mate with holes of 
size 25.000 mm
.
.
0 030
0 020
+
+
. The maximum interference (in microns) in the assembly is
(A) 40 (B) 30
(C) 20 (D) 10
Sol. 3 Option (C) is correct.
An interference fit for shaft and hole is as given in figure below.
 Maximum Interference 
lim lim Maximum it of shat Minimum it of hole =-
  (25 0. 40) (25 0.020) 0 =+ - +
  0.02 mm = 2microns 0 =
Q. 4 During normalizing process of steel, the specimen is heated
(A) between the upper and lower critical temperature and cooled in still air.
(B) above the upper critical temperature and cooled in furnace.
(C) above the upper critical temperature and cooled in still air.
(D) between the upper and lower critical temperature and cooled in furnace
Sol. 4 Option (C) is correct
Normalizing involves prolonged heating just above the critical temperature to 
produce globular form of carbine and then cooling in air.
Q. 5 Oil flows through a 200 mm diameter horizontal cast iron pipe (friction factor, 
. f 0 0225 = ) of length 500 m. The volumetric flow rate is 0.2 / ms
3
. The head loss 
(in m) due to friction is (assume 9.81 / ms g
2
= )
(A) 116.18 (B) 0.116
(C) 18.22 (D) 232.36
Sol. 5 Option (A) is correct.
From Darcy Weischback equation head loss
 h f
D
L
g
V
2
2
##
= ...(1)
Given that 500 , m h = 0.2 m D
1000
200
== , . f 0 0225 =
Since volumetric flow rate
 n
o
 () Area velocity of flow V
#
=
 V 
(.)
.
6.37 /
Area
ms
4
02
02
2
#
n
p
== =
o
Hence, h .
..
(. )
0 0225
02
500
2981
637
2
##
#
=
 h 116.33 m = 116.18 m -
GATE ME 2012
ONE MARK
GATE ME 2012
ONE MARK
GATE ME 2012
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2012
Q. 6 For an opaque surface, the absorptivity () a , transmissivity () t and reflectivity () r 
are related by the equation :
(A) ar t += (B) 0 rat ++ =
(C) 1 ar += (D) 0 ar +=
Sol. 6 Option (C) is correct.
The sum of the absorbed, reflected and transmitted radiation be equal to 
 art ++ 1 =
, Absorpivity a = Reflectivity r = , Transmissivity t =
For an opaque surfaces such as solids and liquids
 , 0 t =
Thus, ar + 1 =
Q. 7 Steam enters an adiabatic turbine operating at steady state with an enthalpy of 
3251.0 / kJ kg and leaves as a saturated mixture at 15 kPa with quality (dryness 
fraction) 0.9. The enthalpies of the saturated liquid and vapour at 15 kPa are 
225.94 / kJ kg h
f
= and 2598.3 / kJ kg h
g
= respectively. The mass flow rate of 
steam is 10 / kg s. Kinetic and potential energy changes are negligible. The power 
output of the turbine in MW is
(A) 6.5 (B) 8.9
(C) 9.1 (D) 27.0
Sol. 7 Option (B) is correct.
For adiabatic expansion steam in turbine.
Given 3251.0 / , kJ kg h
1
= 10 / kg s m = , 0.9( ) dryness fraction x =
At 15 kPa
Enthalpy of liquid,  h
f
 225.94 / kJ kg =
Enthalpy of vapour,  h
g
 2598.3 / kJ kg =
Since Power output of turbine.
 P () mh h
12
=-
o
        
(. . ) K E and P E are negligible ...(i)
 h
2
 () hxh h xh h
ffg f g f
=+ =+ -
  225.94 0.9(2598.3 225.94) =+ -
  2361.064 / kJ kg =
From Eq. (i)
 P (. . ) 10 3251 0 2361 064
#
=-
  8899 kW = 8.9 MW =
GATE ME 2012
ONE MARK
GATE ME 2012
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2012
Q. 8 The following are the data for two crossed helical gears used for speed reduction :
Gear I : Pitch circle diameter in the plane of rotation 80 mm and helix angle 
30c.
Gear II : Pitch circle diameter in the plane of rotation 120 mm and helix angle 
. 22 5c.
If the input speed is 1440 rpm, the output speed in rpm is
(A) 1200
(B) 900
(C) 875
(D) 720
Sol. 8 Option (B) is correct.
For helical gears, speed ratio is given by
 
N
N
2
1
 
cos
cos
D
D
1
2
1
2
b
b
#
= ...(i)
 1440 , rpm N
1
= 80 , mm D
1
= 12 mm D 0
2
=
 , 30
1
c b = . 22 5
2
c b =
Hence from Eq. (i)
 N
2
 
cos
cos
D
D
N
2
1
2
1
1 ##
b
b
=
  
. cos
cos
120
80
22 5
30
1440
##
c
c
=
 N
2
 899.88 900 rpm - =
Q. 9 A solid disc of radius r rolls without slipping on a horizontal floor with angular 
velocity w and angular acceleration a. The magnitude of the acceleration of the 
point of contact on the disc is
(A) zero (B) r a
(C) () ( ) rr
222
aw + (D) r
2
w
Sol. 9 Option (D) is correct.
For A solid disc of radius () r as given in figure, rolls without slipping on a 
horizontal floor with angular velocity w and angular acceleration a.
The magnitude of the acceleration of the point of contact (A) on the disc is only 
by centripetal acceleration because of no slip condition.
 v r w = ...(i)
By differentiating Eq. (1) w.r.t. () t
 
dt
dv
 r
dt
d
r:
w
a == ,
dt
d
dt
dv
a
w
a ==
bl
GATE ME 2012
ONE MARK
GATE ME 2012
ONE MARK
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