GATE SOLVED PAPER- ME GATE-2005 GATE Notes | EduRev

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 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
2005
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
2005
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
2005
© www.nodia.co.in
Q. 1 Stokes theorem connects
(A) a line integral and a surface integral
(B) a surface integral and a volume integral
(C) a line integral and a volume integral
(D) gradient of a function and its surface integral
Sol. 1 Option (A) is correct.
We know that the Stokes theorem is,
 dr F
C
: # () dS Fn
S
4
#
: =## () Curl dS F
S
: =##
Here we can see that the line integral dr F
C
: # & surface integral () Curl ds F
S
: ## 
is related to the stokes theorem.
Q. 2 A lot has 10% defective items. Ten items are chosen randomly from this lot. The 
probability that exactly 2 of the chosen items are defective is
(A) 0.0036 (B) 0.1937
(C) 0.2234 (D) 0.3874
Sol. 2 Option (B) is correct.
Let,  P = defective items
 Q = non-defective items
10% items are defective, then probability of defective items
 P . 01 =
Probability of non-defective item
 Q .. 101 09 =- =
The Probability that exactly 2 of the chosen items are defective is
  ()( ) CP Q
10
2
28
=
!!
!
(0.1) (0.9)
82
10 28
=
  (.) ( .) 45 0 1 0 9
28
##
= . 0 1937 =
Q. 3 () sin sin xxdx
a
a
67
+
-
# is equal to
(A) 2sinxdx
a
6
0
# (B) 2sinxdx
a
7
0
#
(C) 2( ) sin sin xxdx
a
67
0
+ # (D) zero
Sol. 3 Option (A) is correct.
Let () fx () sin sin xxdx
a
a
67
=+
-
#
 () fx sin sin xdx xdx
a
a
a
a
67
=+
--
##
We know that
 () fxdx
a
a
-
# 
0()();
2() ( ) ();
when odd function
when even function
fx fx
fx f x fx
a
0
=
-=-
-=
*
#
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
GATE ME 2005
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
2005
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
2005
© www.nodia.co.in
Q. 1 Stokes theorem connects
(A) a line integral and a surface integral
(B) a surface integral and a volume integral
(C) a line integral and a volume integral
(D) gradient of a function and its surface integral
Sol. 1 Option (A) is correct.
We know that the Stokes theorem is,
 dr F
C
: # () dS Fn
S
4
#
: =## () Curl dS F
S
: =##
Here we can see that the line integral dr F
C
: # & surface integral () Curl ds F
S
: ## 
is related to the stokes theorem.
Q. 2 A lot has 10% defective items. Ten items are chosen randomly from this lot. The 
probability that exactly 2 of the chosen items are defective is
(A) 0.0036 (B) 0.1937
(C) 0.2234 (D) 0.3874
Sol. 2 Option (B) is correct.
Let,  P = defective items
 Q = non-defective items
10% items are defective, then probability of defective items
 P . 01 =
Probability of non-defective item
 Q .. 101 09 =- =
The Probability that exactly 2 of the chosen items are defective is
  ()( ) CP Q
10
2
28
=
!!
!
(0.1) (0.9)
82
10 28
=
  (.) ( .) 45 0 1 0 9
28
##
= . 0 1937 =
Q. 3 () sin sin xxdx
a
a
67
+
-
# is equal to
(A) 2sinxdx
a
6
0
# (B) 2sinxdx
a
7
0
#
(C) 2( ) sin sin xxdx
a
67
0
+ # (D) zero
Sol. 3 Option (A) is correct.
Let () fx () sin sin xxdx
a
a
67
=+
-
#
 () fx sin sin xdx xdx
a
a
a
a
67
=+
--
##
We know that
 () fxdx
a
a
-
# 
0()();
2() ( ) ();
when odd function
when even function
fx fx
fx f x fx
a
0
=
-=-
-=
*
#
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2005
Now, here sin x
6
 is an even function & sin x
7
 is an odd function. Then,
 () fx sin xdx 20
a
6
0
=+ # sin xdx 2
a
6
0
= #
Q. 4 A is a 34
#
 real matrix and Ax b = is an inconsistent system of equations. The 
highest possible rank of A is
(A) 1
B) 2
(C) 3
(D) 4
Sol. 4 Option (C) is correct.
We know, from the Echelon form the rank of any matrix is equal to the Number 
of non zero rows.
Here order of matrix is 34
#
, then, we can say that the Highest possible rank of 
this matrix is 3.
Q. 5 Changing the order of the integration in the double integral (, ) I f x y dydx
x
4
2
0
8
= # #
leads to (, ) I f x y dxdy
p
q
r
s
= # # What is q ?
(A) 4y
(B) 16 y
2
(C) x
(D) 8
Sol. 5 Option (A) is correct.
Given  I (, ) fxydydx
/4
2
0
8
=
p
# #
Here we can draw the graph from the limits of the integration, the limit of y is 
from y
x
4
= to y 2 =
For x the limit is  x 0 = to x 8 =
Here we use the changing the order of the integration. The limit of x is 0 to 8 
but we have to find the limits in the form of y then x 0 = to xy 4 = & limit of y 
is 0 to 2
So (, ) f x y dydx
/ x 4
2
0
8
# # (, ) fxydxdy
y
0
4
0
2
= # # (, ) fxydxdy
p
q
r
s
= # #
Comparing the limits and get
r 0 = , s 2 = , p 0 = , qy 4 =
GATE ME 2005
ONE MARK
GATE ME 2005
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
2005
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
2005
© www.nodia.co.in
Q. 1 Stokes theorem connects
(A) a line integral and a surface integral
(B) a surface integral and a volume integral
(C) a line integral and a volume integral
(D) gradient of a function and its surface integral
Sol. 1 Option (A) is correct.
We know that the Stokes theorem is,
 dr F
C
: # () dS Fn
S
4
#
: =## () Curl dS F
S
: =##
Here we can see that the line integral dr F
C
: # & surface integral () Curl ds F
S
: ## 
is related to the stokes theorem.
Q. 2 A lot has 10% defective items. Ten items are chosen randomly from this lot. The 
probability that exactly 2 of the chosen items are defective is
(A) 0.0036 (B) 0.1937
(C) 0.2234 (D) 0.3874
Sol. 2 Option (B) is correct.
Let,  P = defective items
 Q = non-defective items
10% items are defective, then probability of defective items
 P . 01 =
Probability of non-defective item
 Q .. 101 09 =- =
The Probability that exactly 2 of the chosen items are defective is
  ()( ) CP Q
10
2
28
=
!!
!
(0.1) (0.9)
82
10 28
=
  (.) ( .) 45 0 1 0 9
28
##
= . 0 1937 =
Q. 3 () sin sin xxdx
a
a
67
+
-
# is equal to
(A) 2sinxdx
a
6
0
# (B) 2sinxdx
a
7
0
#
(C) 2( ) sin sin xxdx
a
67
0
+ # (D) zero
Sol. 3 Option (A) is correct.
Let () fx () sin sin xxdx
a
a
67
=+
-
#
 () fx sin sin xdx xdx
a
a
a
a
67
=+
--
##
We know that
 () fxdx
a
a
-
# 
0()();
2() ( ) ();
when odd function
when even function
fx fx
fx f x fx
a
0
=
-=-
-=
*
#
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2005
Now, here sin x
6
 is an even function & sin x
7
 is an odd function. Then,
 () fx sin xdx 20
a
6
0
=+ # sin xdx 2
a
6
0
= #
Q. 4 A is a 34
#
 real matrix and Ax b = is an inconsistent system of equations. The 
highest possible rank of A is
(A) 1
B) 2
(C) 3
(D) 4
Sol. 4 Option (C) is correct.
We know, from the Echelon form the rank of any matrix is equal to the Number 
of non zero rows.
Here order of matrix is 34
#
, then, we can say that the Highest possible rank of 
this matrix is 3.
Q. 5 Changing the order of the integration in the double integral (, ) I f x y dydx
x
4
2
0
8
= # #
leads to (, ) I f x y dxdy
p
q
r
s
= # # What is q ?
(A) 4y
(B) 16 y
2
(C) x
(D) 8
Sol. 5 Option (A) is correct.
Given  I (, ) fxydydx
/4
2
0
8
=
p
# #
Here we can draw the graph from the limits of the integration, the limit of y is 
from y
x
4
= to y 2 =
For x the limit is  x 0 = to x 8 =
Here we use the changing the order of the integration. The limit of x is 0 to 8 
but we have to find the limits in the form of y then x 0 = to xy 4 = & limit of y 
is 0 to 2
So (, ) f x y dydx
/ x 4
2
0
8
# # (, ) fxydxdy
y
0
4
0
2
= # # (, ) fxydxdy
p
q
r
s
= # #
Comparing the limits and get
r 0 = , s 2 = , p 0 = , qy 4 =
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2005
Q. 6 The time variation of the position of a particle in rectilinear motion is given by 
22 xt t t
32
=+ + . If v is the velocity and a is the acceleration of the particle in 
consistent units, the motion started with
(A) 0, 0 va == (B) 0, 2 va ==
(C) 2, 0 va == (D) 2, 2 va ==
Sol. 6 Option (D) is correct.
Given ;  x tt t 22
32
=+ +
We know that,
 v
dt
dx
= ()
dt
d
tt t 22
32
=++ tt 62 2
2
=+ + ...(i)
We have to find the velocity & acceleration of particle, when motion stared, So
At t 0 = ,  v 2 =
Again differentiate equation (i) w.r.t.  t
 a 12 2
dt
dv
dt
dx
t
2
2
== = +
At t 0 = ,  a 2 =
Q. 7 A simple pendulum of length of 5 m, with a bob of mass 1 kg, is in simple 
harmonic motion. As it passes through its mean position, the bob has a speed of 
5 m/s. The net force on the bob at the mean position is
(A) zero (B) 2.5 N
(C) 5 N (D) 25 N
Sol. 7 Option (A) is correct.
We have to make the diagram of simple pendulum
Here, We can see easily from the figure that tension in the string is balanced by 
the weight of the bob and net force at the mean position is always zero.
Q. 8 A uniform, slender cylindrical rod is made of a homogeneous and isotropic material. 
The rod rests on a frictionless surface. The rod is heated uniformly. If the radial 
and longitudinal thermal stresses are represented by 
r
s and 
z
s , respectively, then
(A) 0, 0
rz
ss == (B) 0, 0
rz
ss == Y
(C) 0, 0
rz
ss == Y (D) 0, 0
rz
ss == YY
Sol. 8 Option (A) is correct.
We know that due to temperature changes, dimensions of the material change. 
If these changes in the dimensions are prevented partially or fully, stresses 
are generated in the material and if the changes in the dimensions are not 
prevented, there will be no stress set up. (Zero stresses).
Hence cylindrical rod is allowed to expand or contract freely.
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
GATE ME 2005
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
2005
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
2005
© www.nodia.co.in
Q. 1 Stokes theorem connects
(A) a line integral and a surface integral
(B) a surface integral and a volume integral
(C) a line integral and a volume integral
(D) gradient of a function and its surface integral
Sol. 1 Option (A) is correct.
We know that the Stokes theorem is,
 dr F
C
: # () dS Fn
S
4
#
: =## () Curl dS F
S
: =##
Here we can see that the line integral dr F
C
: # & surface integral () Curl ds F
S
: ## 
is related to the stokes theorem.
Q. 2 A lot has 10% defective items. Ten items are chosen randomly from this lot. The 
probability that exactly 2 of the chosen items are defective is
(A) 0.0036 (B) 0.1937
(C) 0.2234 (D) 0.3874
Sol. 2 Option (B) is correct.
Let,  P = defective items
 Q = non-defective items
10% items are defective, then probability of defective items
 P . 01 =
Probability of non-defective item
 Q .. 101 09 =- =
The Probability that exactly 2 of the chosen items are defective is
  ()( ) CP Q
10
2
28
=
!!
!
(0.1) (0.9)
82
10 28
=
  (.) ( .) 45 0 1 0 9
28
##
= . 0 1937 =
Q. 3 () sin sin xxdx
a
a
67
+
-
# is equal to
(A) 2sinxdx
a
6
0
# (B) 2sinxdx
a
7
0
#
(C) 2( ) sin sin xxdx
a
67
0
+ # (D) zero
Sol. 3 Option (A) is correct.
Let () fx () sin sin xxdx
a
a
67
=+
-
#
 () fx sin sin xdx xdx
a
a
a
a
67
=+
--
##
We know that
 () fxdx
a
a
-
# 
0()();
2() ( ) ();
when odd function
when even function
fx fx
fx f x fx
a
0
=
-=-
-=
*
#
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2005
Now, here sin x
6
 is an even function & sin x
7
 is an odd function. Then,
 () fx sin xdx 20
a
6
0
=+ # sin xdx 2
a
6
0
= #
Q. 4 A is a 34
#
 real matrix and Ax b = is an inconsistent system of equations. The 
highest possible rank of A is
(A) 1
B) 2
(C) 3
(D) 4
Sol. 4 Option (C) is correct.
We know, from the Echelon form the rank of any matrix is equal to the Number 
of non zero rows.
Here order of matrix is 34
#
, then, we can say that the Highest possible rank of 
this matrix is 3.
Q. 5 Changing the order of the integration in the double integral (, ) I f x y dydx
x
4
2
0
8
= # #
leads to (, ) I f x y dxdy
p
q
r
s
= # # What is q ?
(A) 4y
(B) 16 y
2
(C) x
(D) 8
Sol. 5 Option (A) is correct.
Given  I (, ) fxydydx
/4
2
0
8
=
p
# #
Here we can draw the graph from the limits of the integration, the limit of y is 
from y
x
4
= to y 2 =
For x the limit is  x 0 = to x 8 =
Here we use the changing the order of the integration. The limit of x is 0 to 8 
but we have to find the limits in the form of y then x 0 = to xy 4 = & limit of y 
is 0 to 2
So (, ) f x y dydx
/ x 4
2
0
8
# # (, ) fxydxdy
y
0
4
0
2
= # # (, ) fxydxdy
p
q
r
s
= # #
Comparing the limits and get
r 0 = , s 2 = , p 0 = , qy 4 =
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2005
Q. 6 The time variation of the position of a particle in rectilinear motion is given by 
22 xt t t
32
=+ + . If v is the velocity and a is the acceleration of the particle in 
consistent units, the motion started with
(A) 0, 0 va == (B) 0, 2 va ==
(C) 2, 0 va == (D) 2, 2 va ==
Sol. 6 Option (D) is correct.
Given ;  x tt t 22
32
=+ +
We know that,
 v
dt
dx
= ()
dt
d
tt t 22
32
=++ tt 62 2
2
=+ + ...(i)
We have to find the velocity & acceleration of particle, when motion stared, So
At t 0 = ,  v 2 =
Again differentiate equation (i) w.r.t.  t
 a 12 2
dt
dv
dt
dx
t
2
2
== = +
At t 0 = ,  a 2 =
Q. 7 A simple pendulum of length of 5 m, with a bob of mass 1 kg, is in simple 
harmonic motion. As it passes through its mean position, the bob has a speed of 
5 m/s. The net force on the bob at the mean position is
(A) zero (B) 2.5 N
(C) 5 N (D) 25 N
Sol. 7 Option (A) is correct.
We have to make the diagram of simple pendulum
Here, We can see easily from the figure that tension in the string is balanced by 
the weight of the bob and net force at the mean position is always zero.
Q. 8 A uniform, slender cylindrical rod is made of a homogeneous and isotropic material. 
The rod rests on a frictionless surface. The rod is heated uniformly. If the radial 
and longitudinal thermal stresses are represented by 
r
s and 
z
s , respectively, then
(A) 0, 0
rz
ss == (B) 0, 0
rz
ss == Y
(C) 0, 0
rz
ss == Y (D) 0, 0
rz
ss == YY
Sol. 8 Option (A) is correct.
We know that due to temperature changes, dimensions of the material change. 
If these changes in the dimensions are prevented partially or fully, stresses 
are generated in the material and if the changes in the dimensions are not 
prevented, there will be no stress set up. (Zero stresses).
Hence cylindrical rod is allowed to expand or contract freely.
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2005
So, 0
r
s = and 0
z
s =
Q. 9 Two identical cantilever beams are supported as shown , with their free ends in 
contact through a rigid roller. After the load P is applied, the free ends will have
(A) equal deflections but not equal slopes
(B) equal slopes but not equal deflections
(C) equal slopes as well as equal deflections
(D) neither equal slopes nor equal deflections
Sol. 9 Option (A) is correct.
From the figure, we can say that load P applies a force on upper cantilever and 
the reaction force also applied on upper cantilever by the rigid roller. Due to 
this, deflections are occur in both the cantilever, which are equal in amount. 
But because of different forces applied by the P and rigid roller, the slopes are 
unequal.
Q. 10 The number of degrees of freedom of a planar linkage with 8 links and 9 simple 
revolute joints is
(A) 1 (B) 2
(C) 3 (D) 4
Sol. 10 Option (C) is correct.
Given l 8 = , j 9 =
We know that, Degree of freedom, 
 n () lj 31 2 =- - () 38 1 2 9
#
=- - 3 =
Q. 11 There are four samples P, Q, R and S, with natural frequencies 64, 96, 128 
and 256 Hz, respectively. They are mounted on test setups for conducting 
vibration experiments. If a loud pure note of frequency 144 Hz is produced by 
some instrument, which of the samples will show the most perceptible induced 
vibration?
(A) P (B) Q
(C) R (D) S
Sol. 11 Option (C) is correct.
The speed of sound in air 332 / ms =
For frequency of instrument of 144 Hz, length of sound wave
 L
I
 2.30 m
144
332
==
For sample P of 64 Hz,
 L
P
 5.1875 m
64
332
==
Q of 96 Hz L
Q
 3.458 m
96
332
==
R of 128 Hz L
R
 2.593 m
128
332
==
S of 250 Hz L
S
 1.2968 m
256
332
==
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
GATE ME 2005
ONE MARK
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GATE SOLVED PAPER- ME GATE-2014-2

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pdf

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GATE SOLVED PAPER- ME GATE-2013

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GATE SOLVED PAPER- ME GATE-2005 GATE Notes | EduRev

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