GATE SOLVED PAPER- ME GATE-2004 GATE Notes | EduRev

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GATE : GATE SOLVED PAPER- ME GATE-2004 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
2004
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
2004
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
2004
© www.nodia.co.in
Q. 1 If () sin xaqq =+ and (1 ) cos ya q =- , then 
dx
dy
 will be equal to
(A) sin
2
q
bl
 (B) cos
2
q
bl
(C) tan
2
q
bl
 (D) cot
2
q
bl
Sol. 1 Option (C) is correct.
Given : x () sin aqq =+ , y () cos a 1 q =-
First differentiate x w.r.t. q,
 
d
dx
q
 [] cos a 1 q =+
And differentiate y w.r.t. q
 
d
dy
q
 [( )] sin sin aa 0 qq =-- =
We know,
 
dx
dy
 
/
/
d
dy
dx
d
dx d
dy d
#
q
q
q
q
==
Substitute the values of 
d
dy
q
 & 
d
dx
q
 
dx
dy
 
[]
sin
cos cos
sin
a
a 1
1
1
#
q
q q
q
=
+
=
+
 
cos
sin cos
2
2
2
22
2 q
qq
=
  
cos
sin
2
2
q
q
= tan
2
q
= 12 cos cos
2
2
q
q
+=
Q. 2 The angle between two unit-magnitude coplanar vectors (0.866, 0.500, 0) P and 
(0.259, 0.966, 0) Q will be
(A) 0c (B) 30c
(C) 45c (D) 60c
Sol. 2 Option (C) is correct.
Given : (. ,. ,) P 0 866 0 500 0 , so we can write
 P 0.866 0.5 0 ij k =+ +
(. , . , ) Q 0 259 0 966 0 = , so we can write
 Q 0.259 0.966 0 ijk =+ +
We know that for the coplanar vectors
 PQ : cos PQ q =
 cos q 
PQ
PQ :
=
 PQ : (0.866 0.5 0 ) (0.259 0.966 0 ) ij k i j k : =+ + + +
  .. . . 0 866 0 259 0 5 0 966
##
=+
GATE ME 2004
ONE MARK
GATE ME 2004
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
2004
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
2004
© www.nodia.co.in
Q. 1 If () sin xaqq =+ and (1 ) cos ya q =- , then 
dx
dy
 will be equal to
(A) sin
2
q
bl
 (B) cos
2
q
bl
(C) tan
2
q
bl
 (D) cot
2
q
bl
Sol. 1 Option (C) is correct.
Given : x () sin aqq =+ , y () cos a 1 q =-
First differentiate x w.r.t. q,
 
d
dx
q
 [] cos a 1 q =+
And differentiate y w.r.t. q
 
d
dy
q
 [( )] sin sin aa 0 qq =-- =
We know,
 
dx
dy
 
/
/
d
dy
dx
d
dx d
dy d
#
q
q
q
q
==
Substitute the values of 
d
dy
q
 & 
d
dx
q
 
dx
dy
 
[]
sin
cos cos
sin
a
a 1
1
1
#
q
q q
q
=
+
=
+
 
cos
sin cos
2
2
2
22
2 q
qq
=
  
cos
sin
2
2
q
q
= tan
2
q
= 12 cos cos
2
2
q
q
+=
Q. 2 The angle between two unit-magnitude coplanar vectors (0.866, 0.500, 0) P and 
(0.259, 0.966, 0) Q will be
(A) 0c (B) 30c
(C) 45c (D) 60c
Sol. 2 Option (C) is correct.
Given : (. ,. ,) P 0 866 0 500 0 , so we can write
 P 0.866 0.5 0 ij k =+ +
(. , . , ) Q 0 259 0 966 0 = , so we can write
 Q 0.259 0.966 0 ijk =+ +
We know that for the coplanar vectors
 PQ : cos PQ q =
 cos q 
PQ
PQ :
=
 PQ : (0.866 0.5 0 ) (0.259 0.966 0 ) ij k i j k : =+ + + +
  .. . . 0 866 0 259 0 5 0 966
##
=+
GATE ME 2004
ONE MARK
GATE ME 2004
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2004
So,  cos q 
(. ) ( .) (. ) ( . )
.. . .
0 866 0 5 0 259 0 966
0 866 0 259 0 5 0 966
22 2 2
##
=
++ +
+
  
..
..
..
.
0 99 1 001
0 22429 0 483
0 99 1 001
0 70729
##
=
+
= . 0 707 =
 q (. ) cos 0 707 45
1
c ==
-
Q. 3 The sum of the eigen values of the matrix given below is
 
3 1
1
3
2
5
1
1
1
R
T
S
S
S
S
V
X
W
W
W
W
(A) 5 (B) 7
(C) 9 (D) 18
Sol. 3 Option (B) is correct.
Let A 
3 1
1
3
2
5
1
1
1
=
R
T
S
S
S
S
V
X
W
W
W
W
We know that the sum of the eigen value of a matrix is equal to the sum of the 
diagonal elements of the matrix
So, the sum of eigen values is,
 1+5+1 7 =
Q. 4 The figure shows a pin-jointed plane truss loaded at the point M by hanging a 
mass of 100 kg. The member LN of the truss is subjected to a load of
(A) 0 Newton (B) 490 Newtons in compression
(C) 981 Newtons in compression (D) 981 Newtons in tension
Sol. 4 Option (A) is correct.
First of all we consider all the forces, which are acting at point L.
Now sum all the forces which are acting along x direction,
 F
LK
 F
LM
= Both are acting in opposite direction
Also summation of all the forces, which are acting along y-direction.
 F
LN
 0 = Only one forces acting in y-direction
So the member LN is subjected to zero load.
GATE ME 2004
ONE MARK
GATE ME 2004
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
2004
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
2004
© www.nodia.co.in
Q. 1 If () sin xaqq =+ and (1 ) cos ya q =- , then 
dx
dy
 will be equal to
(A) sin
2
q
bl
 (B) cos
2
q
bl
(C) tan
2
q
bl
 (D) cot
2
q
bl
Sol. 1 Option (C) is correct.
Given : x () sin aqq =+ , y () cos a 1 q =-
First differentiate x w.r.t. q,
 
d
dx
q
 [] cos a 1 q =+
And differentiate y w.r.t. q
 
d
dy
q
 [( )] sin sin aa 0 qq =-- =
We know,
 
dx
dy
 
/
/
d
dy
dx
d
dx d
dy d
#
q
q
q
q
==
Substitute the values of 
d
dy
q
 & 
d
dx
q
 
dx
dy
 
[]
sin
cos cos
sin
a
a 1
1
1
#
q
q q
q
=
+
=
+
 
cos
sin cos
2
2
2
22
2 q
qq
=
  
cos
sin
2
2
q
q
= tan
2
q
= 12 cos cos
2
2
q
q
+=
Q. 2 The angle between two unit-magnitude coplanar vectors (0.866, 0.500, 0) P and 
(0.259, 0.966, 0) Q will be
(A) 0c (B) 30c
(C) 45c (D) 60c
Sol. 2 Option (C) is correct.
Given : (. ,. ,) P 0 866 0 500 0 , so we can write
 P 0.866 0.5 0 ij k =+ +
(. , . , ) Q 0 259 0 966 0 = , so we can write
 Q 0.259 0.966 0 ijk =+ +
We know that for the coplanar vectors
 PQ : cos PQ q =
 cos q 
PQ
PQ :
=
 PQ : (0.866 0.5 0 ) (0.259 0.966 0 ) ij k i j k : =+ + + +
  .. . . 0 866 0 259 0 5 0 966
##
=+
GATE ME 2004
ONE MARK
GATE ME 2004
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2004
So,  cos q 
(. ) ( .) (. ) ( . )
.. . .
0 866 0 5 0 259 0 966
0 866 0 259 0 5 0 966
22 2 2
##
=
++ +
+
  
..
..
..
.
0 99 1 001
0 22429 0 483
0 99 1 001
0 70729
##
=
+
= . 0 707 =
 q (. ) cos 0 707 45
1
c ==
-
Q. 3 The sum of the eigen values of the matrix given below is
 
3 1
1
3
2
5
1
1
1
R
T
S
S
S
S
V
X
W
W
W
W
(A) 5 (B) 7
(C) 9 (D) 18
Sol. 3 Option (B) is correct.
Let A 
3 1
1
3
2
5
1
1
1
=
R
T
S
S
S
S
V
X
W
W
W
W
We know that the sum of the eigen value of a matrix is equal to the sum of the 
diagonal elements of the matrix
So, the sum of eigen values is,
 1+5+1 7 =
Q. 4 The figure shows a pin-jointed plane truss loaded at the point M by hanging a 
mass of 100 kg. The member LN of the truss is subjected to a load of
(A) 0 Newton (B) 490 Newtons in compression
(C) 981 Newtons in compression (D) 981 Newtons in tension
Sol. 4 Option (A) is correct.
First of all we consider all the forces, which are acting at point L.
Now sum all the forces which are acting along x direction,
 F
LK
 F
LM
= Both are acting in opposite direction
Also summation of all the forces, which are acting along y-direction.
 F
LN
 0 = Only one forces acting in y-direction
So the member LN is subjected to zero load.
GATE ME 2004
ONE MARK
GATE ME 2004
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2004
Q. 5 In terms of Poisson’s ratio ( u) the ratio of Young’s Modulus () E to Shear Modulus 
() G of elastic materials is
(A) 2(1 ) u + (B) 2(1 ) u -
(C) (1 )
2
1
u + (D) (1 )
2
1
u -
Sol. 5 Option (A) is correct.
We know that, relation between , E G & u is given by,
 E 2(1 ) G u =+
Where E = young’s modulus
 G = Shear Modulus
 u = Poisson’s ratio
Now,  
G
E
 2(1 ) u =+
Q. 6 Two mating spur gears have 40 and 120 teeth respectively. The pinion rotates at 
1200 rpm and transmits a torque of 20 Nm. The torque transmitted by the gear is
(A) 6.6 Nm (B) 20 Nm
(C) 40 Nm (D) 60 Nm
Sol. 6 Option (D) is correct.
Given : Z
P
40 teeth = , Z
G
120 teeth = , N
P
1200 rpm = , T
P
20 N m
-
=
Velocity Ratio,  
Z
Z
G
P
 
N
N
P
G
=
 N
G
 
Z
Z
N
G
P
P #
=
 N
G
 1200
120
40
#
= 400 rpm =
Power transmitted is same for both pinion & Gear.
 P 
NT NT
60
2
60
2
PP GG
p p
==
 NT
PP
 NT
GG
=
 T
G
 
N
NT
G
PP
= 
400
1200
20
#
= 60 N m
-
=
So, the torque transmitted by the Gear is 60 N m
-
Q. 7 The figure shows the state of stress at a certain point in a stressed body. The 
magnitudes of normal stresses in x and y directions are 100 MPa and 20 MPa 
respectively. The radius of Mohr’s stress circle representing this state of stress is
GATE ME 2004
ONE MARK
GATE ME 2004
ONE MARK
GATE ME 2004
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
2004
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
2004
© www.nodia.co.in
Q. 1 If () sin xaqq =+ and (1 ) cos ya q =- , then 
dx
dy
 will be equal to
(A) sin
2
q
bl
 (B) cos
2
q
bl
(C) tan
2
q
bl
 (D) cot
2
q
bl
Sol. 1 Option (C) is correct.
Given : x () sin aqq =+ , y () cos a 1 q =-
First differentiate x w.r.t. q,
 
d
dx
q
 [] cos a 1 q =+
And differentiate y w.r.t. q
 
d
dy
q
 [( )] sin sin aa 0 qq =-- =
We know,
 
dx
dy
 
/
/
d
dy
dx
d
dx d
dy d
#
q
q
q
q
==
Substitute the values of 
d
dy
q
 & 
d
dx
q
 
dx
dy
 
[]
sin
cos cos
sin
a
a 1
1
1
#
q
q q
q
=
+
=
+
 
cos
sin cos
2
2
2
22
2 q
qq
=
  
cos
sin
2
2
q
q
= tan
2
q
= 12 cos cos
2
2
q
q
+=
Q. 2 The angle between two unit-magnitude coplanar vectors (0.866, 0.500, 0) P and 
(0.259, 0.966, 0) Q will be
(A) 0c (B) 30c
(C) 45c (D) 60c
Sol. 2 Option (C) is correct.
Given : (. ,. ,) P 0 866 0 500 0 , so we can write
 P 0.866 0.5 0 ij k =+ +
(. , . , ) Q 0 259 0 966 0 = , so we can write
 Q 0.259 0.966 0 ijk =+ +
We know that for the coplanar vectors
 PQ : cos PQ q =
 cos q 
PQ
PQ :
=
 PQ : (0.866 0.5 0 ) (0.259 0.966 0 ) ij k i j k : =+ + + +
  .. . . 0 866 0 259 0 5 0 966
##
=+
GATE ME 2004
ONE MARK
GATE ME 2004
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2004
So,  cos q 
(. ) ( .) (. ) ( . )
.. . .
0 866 0 5 0 259 0 966
0 866 0 259 0 5 0 966
22 2 2
##
=
++ +
+
  
..
..
..
.
0 99 1 001
0 22429 0 483
0 99 1 001
0 70729
##
=
+
= . 0 707 =
 q (. ) cos 0 707 45
1
c ==
-
Q. 3 The sum of the eigen values of the matrix given below is
 
3 1
1
3
2
5
1
1
1
R
T
S
S
S
S
V
X
W
W
W
W
(A) 5 (B) 7
(C) 9 (D) 18
Sol. 3 Option (B) is correct.
Let A 
3 1
1
3
2
5
1
1
1
=
R
T
S
S
S
S
V
X
W
W
W
W
We know that the sum of the eigen value of a matrix is equal to the sum of the 
diagonal elements of the matrix
So, the sum of eigen values is,
 1+5+1 7 =
Q. 4 The figure shows a pin-jointed plane truss loaded at the point M by hanging a 
mass of 100 kg. The member LN of the truss is subjected to a load of
(A) 0 Newton (B) 490 Newtons in compression
(C) 981 Newtons in compression (D) 981 Newtons in tension
Sol. 4 Option (A) is correct.
First of all we consider all the forces, which are acting at point L.
Now sum all the forces which are acting along x direction,
 F
LK
 F
LM
= Both are acting in opposite direction
Also summation of all the forces, which are acting along y-direction.
 F
LN
 0 = Only one forces acting in y-direction
So the member LN is subjected to zero load.
GATE ME 2004
ONE MARK
GATE ME 2004
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2004
Q. 5 In terms of Poisson’s ratio ( u) the ratio of Young’s Modulus () E to Shear Modulus 
() G of elastic materials is
(A) 2(1 ) u + (B) 2(1 ) u -
(C) (1 )
2
1
u + (D) (1 )
2
1
u -
Sol. 5 Option (A) is correct.
We know that, relation between , E G & u is given by,
 E 2(1 ) G u =+
Where E = young’s modulus
 G = Shear Modulus
 u = Poisson’s ratio
Now,  
G
E
 2(1 ) u =+
Q. 6 Two mating spur gears have 40 and 120 teeth respectively. The pinion rotates at 
1200 rpm and transmits a torque of 20 Nm. The torque transmitted by the gear is
(A) 6.6 Nm (B) 20 Nm
(C) 40 Nm (D) 60 Nm
Sol. 6 Option (D) is correct.
Given : Z
P
40 teeth = , Z
G
120 teeth = , N
P
1200 rpm = , T
P
20 N m
-
=
Velocity Ratio,  
Z
Z
G
P
 
N
N
P
G
=
 N
G
 
Z
Z
N
G
P
P #
=
 N
G
 1200
120
40
#
= 400 rpm =
Power transmitted is same for both pinion & Gear.
 P 
NT NT
60
2
60
2
PP GG
p p
==
 NT
PP
 NT
GG
=
 T
G
 
N
NT
G
PP
= 
400
1200
20
#
= 60 N m
-
=
So, the torque transmitted by the Gear is 60 N m
-
Q. 7 The figure shows the state of stress at a certain point in a stressed body. The 
magnitudes of normal stresses in x and y directions are 100 MPa and 20 MPa 
respectively. The radius of Mohr’s stress circle representing this state of stress is
GATE ME 2004
ONE MARK
GATE ME 2004
ONE MARK
GATE ME 2004
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2004
(A) 120 (B) 80
(C) 60 (D) 40
Sol. 7 Option (C) is correct.
  
 100 MPa
x
s = (Tensile), 20 MPa
y
s =- (Compressive)
We know that,  
1
s 
22
xy xy
xy
2
2
ss ss
t =
+
+
-
+
ak
 
2
s 
22
xy xy
xy
2
2
ss ss
t =
+
-
-
+
ak
From the figure, Radius of Mohr’s circle,
 R 
2
12
ss
=
-
  2
2
1
2
xy
xy
2
2
ss
t
#
=
-
+
ak
Substitute the values, we get
 R 
()
2
100 20
60
2
=
--
=
:D
Q. 8 F or a mechanism shown below, the mechanical advantage for the given configuration 
is
(A) 0 (B) 0.5
(C) 1.0 (D) 3
Sol. 8 Option (D) is correct.
Mechanical advantage in the form of torque is given by,
 .. MA 
T
T
input
output
output
input
w
w
==
Here output link is a slider, So, 
output
w 0 =
Therefore, .. MA 3 =
Q. 9 A vibrating machine is isolated from the floor using springs. If the ratio of 
excitation frequency of vibration of machine to the natural frequency of the 
isolation system is equal to 0.5, then transmissibility ratio of isolation is
(A) 
2
1
 (B) 
4
3
(C) 
3
4
 (D) 2
GATE ME 2004
ONE MARK
GATE ME 2004
ONE MARK
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