GATE SOLVED PAPER-ME GATE -2006 GATE Notes | EduRev

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GATE : GATE SOLVED PAPER-ME GATE -2006 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
2006
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
2006
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
2006
© www.nodia.co.in
Q. 1 Match the items in column I and II.
Column I Column II
P. Gauss-Seidel method 1. Interpolation
Q. Forward Newton-Gauss method 2. Non-linear differential equations
R. Runge-Kutta method 3. Numerical integration
S. Trapezoidal Rule 4. Linear algebraic equations
(A) P-1, Q-4, R-3, S-2 (B) P-1, Q-4, R-2, S-3
(C) P-1. Q-3, R-2, S-4 (D) P-4, Q-1, R-2, S-3
Sol. 1 Option (D) is correct.
Column I
P. Gauss-Seidel method 4. Linear algebraic equation
Q. Forward Newton-Gauss method 1. Interpolation
R. Runge-Kutta method 2. Non-linear differential equation
S. Trapezoidal Rule 3. Numerical integration
So, correct pairs are, P-4, Q-1, R-2, S-3
Q. 2 The solution of the differential equation 2
dx
dy
xy e
x
2
+=
-
 with (0) 1 y = is
(A) (1 ) xe
x
2
+
+
 (B) (1 ) xe
x
2
+
-
(C) (1 ) xe
x
2
-
+
 (D) (1 ) xe
x
2
-
-
Sol. 2 Option (B) is correct.
Given : 
dx
dy
xy 2 + e
x
2
=
-
 and () y01 =
It is the first order linear differential equation so its solution is
 (. .) yIF (. .) QIF dx C =+ #
So,  .. IF ee
Pdx xdx 2
==
##
 
()
compare with
dx
dy
Py Q +=
  e
xdx 2
=
#
ee
x
x 2
2
2
2
==
#
The complete solution is,
 ye
x
2
 eedxC
xx
22
#
=+
-
#
 ye
x
2
 dx C =+ # xC =+
 y 
e
xc
x
2
=
+
 ...(i)
Given (0) y 1 =
At x 0 = y 1 & =
Substitute in equation (i), we get
 1 
C
C
1
1 & ==
GATE ME 2006
ONE MARK
GATE ME 2006
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
2006
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
2006
© www.nodia.co.in
Q. 1 Match the items in column I and II.
Column I Column II
P. Gauss-Seidel method 1. Interpolation
Q. Forward Newton-Gauss method 2. Non-linear differential equations
R. Runge-Kutta method 3. Numerical integration
S. Trapezoidal Rule 4. Linear algebraic equations
(A) P-1, Q-4, R-3, S-2 (B) P-1, Q-4, R-2, S-3
(C) P-1. Q-3, R-2, S-4 (D) P-4, Q-1, R-2, S-3
Sol. 1 Option (D) is correct.
Column I
P. Gauss-Seidel method 4. Linear algebraic equation
Q. Forward Newton-Gauss method 1. Interpolation
R. Runge-Kutta method 2. Non-linear differential equation
S. Trapezoidal Rule 3. Numerical integration
So, correct pairs are, P-4, Q-1, R-2, S-3
Q. 2 The solution of the differential equation 2
dx
dy
xy e
x
2
+=
-
 with (0) 1 y = is
(A) (1 ) xe
x
2
+
+
 (B) (1 ) xe
x
2
+
-
(C) (1 ) xe
x
2
-
+
 (D) (1 ) xe
x
2
-
-
Sol. 2 Option (B) is correct.
Given : 
dx
dy
xy 2 + e
x
2
=
-
 and () y01 =
It is the first order linear differential equation so its solution is
 (. .) yIF (. .) QIF dx C =+ #
So,  .. IF ee
Pdx xdx 2
==
##
 
()
compare with
dx
dy
Py Q +=
  e
xdx 2
=
#
ee
x
x 2
2
2
2
==
#
The complete solution is,
 ye
x
2
 eedxC
xx
22
#
=+
-
#
 ye
x
2
 dx C =+ # xC =+
 y 
e
xc
x
2
=
+
 ...(i)
Given (0) y 1 =
At x 0 = y 1 & =
Substitute in equation (i), we get
 1 
C
C
1
1 & ==
GATE ME 2006
ONE MARK
GATE ME 2006
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2006
Then y ()
e
x
xe
1
1
x
x
2
2
=
+
=+
-
Q. 3 Let x denote a real number. Find out the INCORRECT statement.
(A) {: 3} Sxx > = represents the set of all real numbers greater than 3
(B) {: 0} Sxx <
2
= represents the empty set.
(C) {: } and Sxx A x B !! = represents the union of set A and set B.
(D) {: } Sxa x b << = represents the set of all real numbers between a and b, 
where a and b are real numbers.
Sol. 3 Option (C) is correct.
The incorrect statement is, {: } and Sxx A x B !! = represents the union of set
A and set B.
The above symbol () ! denotes intersection of set A and set B. Therefore this 
statement is incorrect.
Q. 4 A box contains 20 defective items and 80 non-defective items. If two items are 
selected at random without replacement, what will be the probability that both 
items are defective ?
(A) 
5
1
 (B) 
25
1
(C) 
99
20
 (D) 
495
19
Sol. 4 Option (D) is correct.
Total number of items = 100
Number of defective items = 20
Number of Non-defective items = 80
Then the probability that both items are defective, when 2 items are selected at 
random is,
 P 
C
CC
100
2
20
2
80
0
= 
!!
100!
!!
!
98 2
18 2
20
=
  
20 19
2
100 99
2
#
#
= 
495
19
=
Common Data For Q.Alternate method
Here two items are selected without replacement.
Probability of first item being defective is
 P
1
 
100
20
5
1
==
After drawing one defective item from box, there are 19 defective items in the 99 
remaining items.
Probability that second item is defective,
 P
2
 
899
19
=
then probability that both are defective
 P PP
12 #
=
 P 
5
1
99
19
495
19
#
==
GATE ME 2006
ONE MARK
GATE ME 2006
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
2006
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
2006
© www.nodia.co.in
Q. 1 Match the items in column I and II.
Column I Column II
P. Gauss-Seidel method 1. Interpolation
Q. Forward Newton-Gauss method 2. Non-linear differential equations
R. Runge-Kutta method 3. Numerical integration
S. Trapezoidal Rule 4. Linear algebraic equations
(A) P-1, Q-4, R-3, S-2 (B) P-1, Q-4, R-2, S-3
(C) P-1. Q-3, R-2, S-4 (D) P-4, Q-1, R-2, S-3
Sol. 1 Option (D) is correct.
Column I
P. Gauss-Seidel method 4. Linear algebraic equation
Q. Forward Newton-Gauss method 1. Interpolation
R. Runge-Kutta method 2. Non-linear differential equation
S. Trapezoidal Rule 3. Numerical integration
So, correct pairs are, P-4, Q-1, R-2, S-3
Q. 2 The solution of the differential equation 2
dx
dy
xy e
x
2
+=
-
 with (0) 1 y = is
(A) (1 ) xe
x
2
+
+
 (B) (1 ) xe
x
2
+
-
(C) (1 ) xe
x
2
-
+
 (D) (1 ) xe
x
2
-
-
Sol. 2 Option (B) is correct.
Given : 
dx
dy
xy 2 + e
x
2
=
-
 and () y01 =
It is the first order linear differential equation so its solution is
 (. .) yIF (. .) QIF dx C =+ #
So,  .. IF ee
Pdx xdx 2
==
##
 
()
compare with
dx
dy
Py Q +=
  e
xdx 2
=
#
ee
x
x 2
2
2
2
==
#
The complete solution is,
 ye
x
2
 eedxC
xx
22
#
=+
-
#
 ye
x
2
 dx C =+ # xC =+
 y 
e
xc
x
2
=
+
 ...(i)
Given (0) y 1 =
At x 0 = y 1 & =
Substitute in equation (i), we get
 1 
C
C
1
1 & ==
GATE ME 2006
ONE MARK
GATE ME 2006
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2006
Then y ()
e
x
xe
1
1
x
x
2
2
=
+
=+
-
Q. 3 Let x denote a real number. Find out the INCORRECT statement.
(A) {: 3} Sxx > = represents the set of all real numbers greater than 3
(B) {: 0} Sxx <
2
= represents the empty set.
(C) {: } and Sxx A x B !! = represents the union of set A and set B.
(D) {: } Sxa x b << = represents the set of all real numbers between a and b, 
where a and b are real numbers.
Sol. 3 Option (C) is correct.
The incorrect statement is, {: } and Sxx A x B !! = represents the union of set
A and set B.
The above symbol () ! denotes intersection of set A and set B. Therefore this 
statement is incorrect.
Q. 4 A box contains 20 defective items and 80 non-defective items. If two items are 
selected at random without replacement, what will be the probability that both 
items are defective ?
(A) 
5
1
 (B) 
25
1
(C) 
99
20
 (D) 
495
19
Sol. 4 Option (D) is correct.
Total number of items = 100
Number of defective items = 20
Number of Non-defective items = 80
Then the probability that both items are defective, when 2 items are selected at 
random is,
 P 
C
CC
100
2
20
2
80
0
= 
!!
100!
!!
!
98 2
18 2
20
=
  
20 19
2
100 99
2
#
#
= 
495
19
=
Common Data For Q.Alternate method
Here two items are selected without replacement.
Probability of first item being defective is
 P
1
 
100
20
5
1
==
After drawing one defective item from box, there are 19 defective items in the 99 
remaining items.
Probability that second item is defective,
 P
2
 
899
19
=
then probability that both are defective
 P PP
12 #
=
 P 
5
1
99
19
495
19
#
==
GATE ME 2006
ONE MARK
GATE ME 2006
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2006
Q. 5 For a circular shaft of diameter d subjected to torque T , the maximum value of 
the shear stress is
(A) 
d
T 64
3
p
 (B) 
d
T 32
3
p
(C) 
d
T 16
3
p
 (D) 
d
T 8
3
p
Sol. 5 Option (C) is correct.
From the Torsional equation
 
J
T
 
r l
G tq
==
Take first two terms,
 
J
T
 
r
t
=
 
d
T
32
4 p
 
d
2
t
= J = Polar moment of inertia
 
max
t 
d
T 16
3
p
=
Q. 6 For a four-bar linkage in toggle position, the value of mechanical advantage is
(A) 0.0 (B) 0.5
(C) 1.0 (D) 3
Sol. 6 Option (D) is correct.
  
 . MA 
T
T
R
R
PA
PD
2
4
4
2
w
w
== =                   from angular 
velocity ratio theorem
Construct BA l and CD l perpendicular to the line PBC. Also, assign lables b 
and g to the acute angles made by the coupler.
 
R
R
PA
PD
 
sin
sin
R
R
R
R
BA
CD
BA
CD
b
g
==
l
l
So, .. MA 
sin
sin
T
T
R
R
BA
CD
2
4
4
2
w
w
b
g
== =
When the mechanism is toggle,then 0c b = and 180c.
So  . MA 3 =
GATE ME 2006
ONE MARK
GATE ME 2006
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
2006
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
2006
© www.nodia.co.in
Q. 1 Match the items in column I and II.
Column I Column II
P. Gauss-Seidel method 1. Interpolation
Q. Forward Newton-Gauss method 2. Non-linear differential equations
R. Runge-Kutta method 3. Numerical integration
S. Trapezoidal Rule 4. Linear algebraic equations
(A) P-1, Q-4, R-3, S-2 (B) P-1, Q-4, R-2, S-3
(C) P-1. Q-3, R-2, S-4 (D) P-4, Q-1, R-2, S-3
Sol. 1 Option (D) is correct.
Column I
P. Gauss-Seidel method 4. Linear algebraic equation
Q. Forward Newton-Gauss method 1. Interpolation
R. Runge-Kutta method 2. Non-linear differential equation
S. Trapezoidal Rule 3. Numerical integration
So, correct pairs are, P-4, Q-1, R-2, S-3
Q. 2 The solution of the differential equation 2
dx
dy
xy e
x
2
+=
-
 with (0) 1 y = is
(A) (1 ) xe
x
2
+
+
 (B) (1 ) xe
x
2
+
-
(C) (1 ) xe
x
2
-
+
 (D) (1 ) xe
x
2
-
-
Sol. 2 Option (B) is correct.
Given : 
dx
dy
xy 2 + e
x
2
=
-
 and () y01 =
It is the first order linear differential equation so its solution is
 (. .) yIF (. .) QIF dx C =+ #
So,  .. IF ee
Pdx xdx 2
==
##
 
()
compare with
dx
dy
Py Q +=
  e
xdx 2
=
#
ee
x
x 2
2
2
2
==
#
The complete solution is,
 ye
x
2
 eedxC
xx
22
#
=+
-
#
 ye
x
2
 dx C =+ # xC =+
 y 
e
xc
x
2
=
+
 ...(i)
Given (0) y 1 =
At x 0 = y 1 & =
Substitute in equation (i), we get
 1 
C
C
1
1 & ==
GATE ME 2006
ONE MARK
GATE ME 2006
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2006
Then y ()
e
x
xe
1
1
x
x
2
2
=
+
=+
-
Q. 3 Let x denote a real number. Find out the INCORRECT statement.
(A) {: 3} Sxx > = represents the set of all real numbers greater than 3
(B) {: 0} Sxx <
2
= represents the empty set.
(C) {: } and Sxx A x B !! = represents the union of set A and set B.
(D) {: } Sxa x b << = represents the set of all real numbers between a and b, 
where a and b are real numbers.
Sol. 3 Option (C) is correct.
The incorrect statement is, {: } and Sxx A x B !! = represents the union of set
A and set B.
The above symbol () ! denotes intersection of set A and set B. Therefore this 
statement is incorrect.
Q. 4 A box contains 20 defective items and 80 non-defective items. If two items are 
selected at random without replacement, what will be the probability that both 
items are defective ?
(A) 
5
1
 (B) 
25
1
(C) 
99
20
 (D) 
495
19
Sol. 4 Option (D) is correct.
Total number of items = 100
Number of defective items = 20
Number of Non-defective items = 80
Then the probability that both items are defective, when 2 items are selected at 
random is,
 P 
C
CC
100
2
20
2
80
0
= 
!!
100!
!!
!
98 2
18 2
20
=
  
20 19
2
100 99
2
#
#
= 
495
19
=
Common Data For Q.Alternate method
Here two items are selected without replacement.
Probability of first item being defective is
 P
1
 
100
20
5
1
==
After drawing one defective item from box, there are 19 defective items in the 99 
remaining items.
Probability that second item is defective,
 P
2
 
899
19
=
then probability that both are defective
 P PP
12 #
=
 P 
5
1
99
19
495
19
#
==
GATE ME 2006
ONE MARK
GATE ME 2006
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2006
Q. 5 For a circular shaft of diameter d subjected to torque T , the maximum value of 
the shear stress is
(A) 
d
T 64
3
p
 (B) 
d
T 32
3
p
(C) 
d
T 16
3
p
 (D) 
d
T 8
3
p
Sol. 5 Option (C) is correct.
From the Torsional equation
 
J
T
 
r l
G tq
==
Take first two terms,
 
J
T
 
r
t
=
 
d
T
32
4 p
 
d
2
t
= J = Polar moment of inertia
 
max
t 
d
T 16
3
p
=
Q. 6 For a four-bar linkage in toggle position, the value of mechanical advantage is
(A) 0.0 (B) 0.5
(C) 1.0 (D) 3
Sol. 6 Option (D) is correct.
  
 . MA 
T
T
R
R
PA
PD
2
4
4
2
w
w
== =                   from angular 
velocity ratio theorem
Construct BA l and CD l perpendicular to the line PBC. Also, assign lables b 
and g to the acute angles made by the coupler.
 
R
R
PA
PD
 
sin
sin
R
R
R
R
BA
CD
BA
CD
b
g
==
l
l
So, .. MA 
sin
sin
T
T
R
R
BA
CD
2
4
4
2
w
w
b
g
== =
When the mechanism is toggle,then 0c b = and 180c.
So  . MA 3 =
GATE ME 2006
ONE MARK
GATE ME 2006
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2006
Q. 7 The differential equation governing the vibrating system is
(A) () 0 mx cx k x y ++ - =
po
(B) () ( ) 0 mx y cx y kx -+ - + =
pp o o
(C) () 0 mx c x y kx +- + =
poo
(D) ( ) () () 0 mx y cx y kx y -+ - + - =
pp o o
Sol. 7 Option (C) is correct.
Assume any arbitrary relationship between the coordinates and their first 
derivatives, say xy > and xy >
oo
. Also assume x 0 > and 0 x >
o
.
A small displacement gives to the system towards the left direction. Mass m is 
fixed, so only damper moves for both the variable x and y.
Note that these forces are acting in the negative direction.
Differential equation governing the above system is,
 F /
 m
dt
dx
c
dt
dx
dt
dy
kx 0
2
2
=- - - - =
bl
 () mx c x y kx +- +
poo
 0 =
Q. 8 A pin-ended column of length L , modulus of elasticity E and second moment of 
the cross-sectional area is I loaded eccentrically by a compressive load P. The 
critical buckling load (P
cr
) is given by
(A) P
L
EI
cr
22
p
= (B) P
L
EI
3
cr
2
2
p
=
(C) P
L
EI
cr
2
p
= (D) P
L
EI
cr
2
2
p
=
Sol. 8 Option (D) is correct.
GATE ME 2006
ONE MARK
GATE ME 2006
ONE MARK
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