GATE SOLVED PAPER-ME GATE-2010 GATE Notes | EduRev

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GATE : GATE SOLVED PAPER-ME GATE-2010 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
2010
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
2010
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
2010
© www.nodia.co.in
Q. 1 The parabolic arc yx = , x 12 ## is revolved around the x -axis. The volume 
of the solid of revolution is
(A) /4 p (B) /2 p
(C) / 34 p (D) / 32 p
Sol. 1 Option (D) is correct.
We know that the volume of a solid generated by revolution about x -axis bounded 
by the function () fx & limits between a to b is given by the equation.
 V ydx
a
b
2
p =#
Given y x = & a 1 = , b 2 =
Therefore, V () xdx
2
1
2
p =# xdx
1
2
p = #
On integrating above equation, we get
  
x
2
2
1
2
p =
:D
Substitute the limits, we get
 V 
2
4
2
1
2
3
p
p
=- =
:D
Q. 2 The Blasius equation, 
d
df f
d
df
2
0
3
3
2
2
h h
+= , is a
(A) second order nonlinear ordinary differential equation
(B) third order nonlinear ordinary differential equation
(C) third order linear ordinary differential equation
(D) mixed order nonlinear ordinary differential equation
Sol. 2 Option (B) is correct.
Given: 
d
df f
d
df
2
3
3
2
2
h h
+ 0 =
Order  " It is determined by the order of the highest derivation present in it.
So, It is third order equation but it is a nonlinear equation because in linear 
equation, the product of f with / dfd
22
h is not allow.
Therefore, it is a third order non-linear ordinary differential equation.
Q. 3 The value of the integral 
x
dx
1
2
+ 3
3
-
# is
(A) p - (B) /2 p -
(C) /2 p (D) p
Sol. 3 Option (D) is correct.
Let  I 
x
dx
1
2
=
+ 3
3
-
#
 I tan x
1
=
3
3 -
-
6@
GATE ME 2010
ONE MARK
GATE ME 2010
ONE MARK
GATE ME 2010
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
2010
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
2010
© www.nodia.co.in
Q. 1 The parabolic arc yx = , x 12 ## is revolved around the x -axis. The volume 
of the solid of revolution is
(A) /4 p (B) /2 p
(C) / 34 p (D) / 32 p
Sol. 1 Option (D) is correct.
We know that the volume of a solid generated by revolution about x -axis bounded 
by the function () fx & limits between a to b is given by the equation.
 V ydx
a
b
2
p =#
Given y x = & a 1 = , b 2 =
Therefore, V () xdx
2
1
2
p =# xdx
1
2
p = #
On integrating above equation, we get
  
x
2
2
1
2
p =
:D
Substitute the limits, we get
 V 
2
4
2
1
2
3
p
p
=- =
:D
Q. 2 The Blasius equation, 
d
df f
d
df
2
0
3
3
2
2
h h
+= , is a
(A) second order nonlinear ordinary differential equation
(B) third order nonlinear ordinary differential equation
(C) third order linear ordinary differential equation
(D) mixed order nonlinear ordinary differential equation
Sol. 2 Option (B) is correct.
Given: 
d
df f
d
df
2
3
3
2
2
h h
+ 0 =
Order  " It is determined by the order of the highest derivation present in it.
So, It is third order equation but it is a nonlinear equation because in linear 
equation, the product of f with / dfd
22
h is not allow.
Therefore, it is a third order non-linear ordinary differential equation.
Q. 3 The value of the integral 
x
dx
1
2
+ 3
3
-
# is
(A) p - (B) /2 p -
(C) /2 p (D) p
Sol. 3 Option (D) is correct.
Let  I 
x
dx
1
2
=
+ 3
3
-
#
 I tan x
1
=
3
3 -
-
6@
GATE ME 2010
ONE MARK
GATE ME 2010
ONE MARK
GATE ME 2010
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2010
 I [( ) ( )] tan tan
11
33 =+ - -
--
 I 
22
pp
p =-- =
ak
 () () tan tan
11
qq -=-
--
Q. 4 The modulus of the complex number 
i
i
12
34
-
+
bl
 is
(A) 5 (B) 5
(C) / 15 (D) 1/5
Sol. 4 Option (B) is correct.
Let, z 
i
i
12
34
=
-
+
Divide & multiply z by the conjugate of () i 12 - to convert it in the form of abi +
.
So, z 
i
i
i
i
12
34
12
12
#
=
-
+
+
+
() ( )
()()
i
ii
12
34 12
22
=
-
++
  
i
ii
14
310 8
2
2
=
-
++
()
i
14
310 8
=
--
+-
  
i
i
5
510
12 =
-+
=- +
 z () () 12 5
22
=- + = aib a b
22
+= +
Q. 5 The function yx 23 =-
(A) is continuous xR 6 ! and differentiable xR 6 !
(B) is continuous xR 6 ! and differentiable xR 6 ! except at / x 32 =
(C) is continuous xR 6 ! and differentiable xR 6 ! except at / x 23 =
(D) is continuous xR 6 ! except x 3 = and differentiable xR 6 !
Sol. 5 Option (C) is correct.
 () yfx = 
23
0
(2 3 )
if
if
if
xx
x
xx
3
2
3
2
3
2
<
>
=
-
=
--
Z
[
\
]
]
]
]
]
]
Checking the continuity of the function.
at x
3
2
= , () Lf x limfh
3
2
h 0
=-
"
bl
  
2
lim h 23
3 h 0
=- -
"
bl
  lim h 223
h 0
=-+
"
  0 =
and () Rf x limfh
3
2
h 0
=+
"
bl
  lim h 3
3
2
2
h 0
=+-
"
bl
  lim h 23 2 0
h 0
=+ -=
"
Since () lim Lfx
h 0 "
 () lim Rfx
h 0
=
"
So, function is continuous xR 6 !
GATE ME 2010
ONE MARK
GATE ME 2010
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
2010
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
2010
© www.nodia.co.in
Q. 1 The parabolic arc yx = , x 12 ## is revolved around the x -axis. The volume 
of the solid of revolution is
(A) /4 p (B) /2 p
(C) / 34 p (D) / 32 p
Sol. 1 Option (D) is correct.
We know that the volume of a solid generated by revolution about x -axis bounded 
by the function () fx & limits between a to b is given by the equation.
 V ydx
a
b
2
p =#
Given y x = & a 1 = , b 2 =
Therefore, V () xdx
2
1
2
p =# xdx
1
2
p = #
On integrating above equation, we get
  
x
2
2
1
2
p =
:D
Substitute the limits, we get
 V 
2
4
2
1
2
3
p
p
=- =
:D
Q. 2 The Blasius equation, 
d
df f
d
df
2
0
3
3
2
2
h h
+= , is a
(A) second order nonlinear ordinary differential equation
(B) third order nonlinear ordinary differential equation
(C) third order linear ordinary differential equation
(D) mixed order nonlinear ordinary differential equation
Sol. 2 Option (B) is correct.
Given: 
d
df f
d
df
2
3
3
2
2
h h
+ 0 =
Order  " It is determined by the order of the highest derivation present in it.
So, It is third order equation but it is a nonlinear equation because in linear 
equation, the product of f with / dfd
22
h is not allow.
Therefore, it is a third order non-linear ordinary differential equation.
Q. 3 The value of the integral 
x
dx
1
2
+ 3
3
-
# is
(A) p - (B) /2 p -
(C) /2 p (D) p
Sol. 3 Option (D) is correct.
Let  I 
x
dx
1
2
=
+ 3
3
-
#
 I tan x
1
=
3
3 -
-
6@
GATE ME 2010
ONE MARK
GATE ME 2010
ONE MARK
GATE ME 2010
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2010
 I [( ) ( )] tan tan
11
33 =+ - -
--
 I 
22
pp
p =-- =
ak
 () () tan tan
11
qq -=-
--
Q. 4 The modulus of the complex number 
i
i
12
34
-
+
bl
 is
(A) 5 (B) 5
(C) / 15 (D) 1/5
Sol. 4 Option (B) is correct.
Let, z 
i
i
12
34
=
-
+
Divide & multiply z by the conjugate of () i 12 - to convert it in the form of abi +
.
So, z 
i
i
i
i
12
34
12
12
#
=
-
+
+
+
() ( )
()()
i
ii
12
34 12
22
=
-
++
  
i
ii
14
310 8
2
2
=
-
++
()
i
14
310 8
=
--
+-
  
i
i
5
510
12 =
-+
=- +
 z () () 12 5
22
=- + = aib a b
22
+= +
Q. 5 The function yx 23 =-
(A) is continuous xR 6 ! and differentiable xR 6 !
(B) is continuous xR 6 ! and differentiable xR 6 ! except at / x 32 =
(C) is continuous xR 6 ! and differentiable xR 6 ! except at / x 23 =
(D) is continuous xR 6 ! except x 3 = and differentiable xR 6 !
Sol. 5 Option (C) is correct.
 () yfx = 
23
0
(2 3 )
if
if
if
xx
x
xx
3
2
3
2
3
2
<
>
=
-
=
--
Z
[
\
]
]
]
]
]
]
Checking the continuity of the function.
at x
3
2
= , () Lf x limfh
3
2
h 0
=-
"
bl
  
2
lim h 23
3 h 0
=- -
"
bl
  lim h 223
h 0
=-+
"
  0 =
and () Rf x limfh
3
2
h 0
=+
"
bl
  lim h 3
3
2
2
h 0
=+-
"
bl
  lim h 23 2 0
h 0
=+ -=
"
Since () lim Lfx
h 0 "
 () lim Rfx
h 0
=
"
So, function is continuous xR 6 !
GATE ME 2010
ONE MARK
GATE ME 2010
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2010
Now checking the differentiability :
 () Lf x l lim
h
fh f
3
2
3
2
h 0
=
-
--
"
bb ll
  lim
h
h 23
3
2
0
h 0
=
-
-- -
"
bl
  lim lim
h
h
h
h 223 3
3
hh 00
=
-
-+
=
-
=-
""
And () Rf x l lim
h
fh f
3
2
3
2
h 0
=
+-
"
bb ll
  lim lim
h
h
h
h
3
3
2
20
23 2
hh 00
=
+- -
=
+-
""
bl
  3 =
Since Lf
3
2
l
bl
 Rf
3
2
! l
bl
, () fx is not differentiable at x
3
2
= .
Q. 6 Mobility of a statically indeterminate structure is
(A) 1 # - (B) 0
(C) 1 (D) 2 $
Sol. 6 Option (A) is correct.
Given figure shows the six bar mechanism.
We know movability or degree of freedom is 3( 1) 2 nl jh =- - -
The mechanism shown in figure has six links and eight binary joints (because 
there are four ternary joints ,, & ABC D, i.e. , l 6 =  j 8 =  h 0 =
So,  n () 36 1 2 8
#
=- - 1 =-
Therefore, when n 1 =- or less, then there are redundant constraints in the 
chain, and it forms a statically indeterminate structure.
So, From the Given options (A) satisfy the statically indeterminate structure 
n 1 # -
Q. 7 There are two points P and Q on a planar rigid body. The relative velocity 
between the two points
(A) should always be along PQ
(B) can be oriented along any direction
(C) should always be perpendicular to PQ
(D) should be along QP when the body undergoes pure translation
Sol. 7 Option (C) is correct.
GATE ME 2010
ONE MARK
GATE ME 2010
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
2010
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
2010
© www.nodia.co.in
Q. 1 The parabolic arc yx = , x 12 ## is revolved around the x -axis. The volume 
of the solid of revolution is
(A) /4 p (B) /2 p
(C) / 34 p (D) / 32 p
Sol. 1 Option (D) is correct.
We know that the volume of a solid generated by revolution about x -axis bounded 
by the function () fx & limits between a to b is given by the equation.
 V ydx
a
b
2
p =#
Given y x = & a 1 = , b 2 =
Therefore, V () xdx
2
1
2
p =# xdx
1
2
p = #
On integrating above equation, we get
  
x
2
2
1
2
p =
:D
Substitute the limits, we get
 V 
2
4
2
1
2
3
p
p
=- =
:D
Q. 2 The Blasius equation, 
d
df f
d
df
2
0
3
3
2
2
h h
+= , is a
(A) second order nonlinear ordinary differential equation
(B) third order nonlinear ordinary differential equation
(C) third order linear ordinary differential equation
(D) mixed order nonlinear ordinary differential equation
Sol. 2 Option (B) is correct.
Given: 
d
df f
d
df
2
3
3
2
2
h h
+ 0 =
Order  " It is determined by the order of the highest derivation present in it.
So, It is third order equation but it is a nonlinear equation because in linear 
equation, the product of f with / dfd
22
h is not allow.
Therefore, it is a third order non-linear ordinary differential equation.
Q. 3 The value of the integral 
x
dx
1
2
+ 3
3
-
# is
(A) p - (B) /2 p -
(C) /2 p (D) p
Sol. 3 Option (D) is correct.
Let  I 
x
dx
1
2
=
+ 3
3
-
#
 I tan x
1
=
3
3 -
-
6@
GATE ME 2010
ONE MARK
GATE ME 2010
ONE MARK
GATE ME 2010
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2010
 I [( ) ( )] tan tan
11
33 =+ - -
--
 I 
22
pp
p =-- =
ak
 () () tan tan
11
qq -=-
--
Q. 4 The modulus of the complex number 
i
i
12
34
-
+
bl
 is
(A) 5 (B) 5
(C) / 15 (D) 1/5
Sol. 4 Option (B) is correct.
Let, z 
i
i
12
34
=
-
+
Divide & multiply z by the conjugate of () i 12 - to convert it in the form of abi +
.
So, z 
i
i
i
i
12
34
12
12
#
=
-
+
+
+
() ( )
()()
i
ii
12
34 12
22
=
-
++
  
i
ii
14
310 8
2
2
=
-
++
()
i
14
310 8
=
--
+-
  
i
i
5
510
12 =
-+
=- +
 z () () 12 5
22
=- + = aib a b
22
+= +
Q. 5 The function yx 23 =-
(A) is continuous xR 6 ! and differentiable xR 6 !
(B) is continuous xR 6 ! and differentiable xR 6 ! except at / x 32 =
(C) is continuous xR 6 ! and differentiable xR 6 ! except at / x 23 =
(D) is continuous xR 6 ! except x 3 = and differentiable xR 6 !
Sol. 5 Option (C) is correct.
 () yfx = 
23
0
(2 3 )
if
if
if
xx
x
xx
3
2
3
2
3
2
<
>
=
-
=
--
Z
[
\
]
]
]
]
]
]
Checking the continuity of the function.
at x
3
2
= , () Lf x limfh
3
2
h 0
=-
"
bl
  
2
lim h 23
3 h 0
=- -
"
bl
  lim h 223
h 0
=-+
"
  0 =
and () Rf x limfh
3
2
h 0
=+
"
bl
  lim h 3
3
2
2
h 0
=+-
"
bl
  lim h 23 2 0
h 0
=+ -=
"
Since () lim Lfx
h 0 "
 () lim Rfx
h 0
=
"
So, function is continuous xR 6 !
GATE ME 2010
ONE MARK
GATE ME 2010
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2010
Now checking the differentiability :
 () Lf x l lim
h
fh f
3
2
3
2
h 0
=
-
--
"
bb ll
  lim
h
h 23
3
2
0
h 0
=
-
-- -
"
bl
  lim lim
h
h
h
h 223 3
3
hh 00
=
-
-+
=
-
=-
""
And () Rf x l lim
h
fh f
3
2
3
2
h 0
=
+-
"
bb ll
  lim lim
h
h
h
h
3
3
2
20
23 2
hh 00
=
+- -
=
+-
""
bl
  3 =
Since Lf
3
2
l
bl
 Rf
3
2
! l
bl
, () fx is not differentiable at x
3
2
= .
Q. 6 Mobility of a statically indeterminate structure is
(A) 1 # - (B) 0
(C) 1 (D) 2 $
Sol. 6 Option (A) is correct.
Given figure shows the six bar mechanism.
We know movability or degree of freedom is 3( 1) 2 nl jh =- - -
The mechanism shown in figure has six links and eight binary joints (because 
there are four ternary joints ,, & ABC D, i.e. , l 6 =  j 8 =  h 0 =
So,  n () 36 1 2 8
#
=- - 1 =-
Therefore, when n 1 =- or less, then there are redundant constraints in the 
chain, and it forms a statically indeterminate structure.
So, From the Given options (A) satisfy the statically indeterminate structure 
n 1 # -
Q. 7 There are two points P and Q on a planar rigid body. The relative velocity 
between the two points
(A) should always be along PQ
(B) can be oriented along any direction
(C) should always be perpendicular to PQ
(D) should be along QP when the body undergoes pure translation
Sol. 7 Option (C) is correct.
GATE ME 2010
ONE MARK
GATE ME 2010
ONE MARK
© www.nodia.co.in
GATE SOLVED PAPER - ME 2010
Velocity of any point on a link with respect to another point (relative velocity) 
on the same link is always perpendicular to the line joining these points on the 
configuration (or space) diagram.
 v
QP
 = Relative velocity between P & Q
 v
QP
 vv
PQ
=- always perpendicular to PQ.
Q. 8 The state of plane-stress at a point is given by 200 , MPa
x
s =- 100 MPa
y
s = 
100 MPa
xy
t = . The maximum shear stress (in MPa) is
(A) 111.8 (B) 150.1
(C) 180.3 (D) 223.6
Sol. 8 Option (C) is correct.
Given : 
x
s 200 MPa =- , 
y
s 100 MPa = , 
xy
t 100 MPa =
We know that maximum shear stress is given by,
 
max
t ()
2
1
4
xy xy
22
ss t =- +
Substitute the values, we get
 
max
t () ()
2
1
200 100 4 100
22
#
=- - +
  
2
1
90000 40000 =+ . 180 27 = 180.3 MPa -
Q. 9 Which of the following statements is INCORRECT ?
(A) Grashof’s rule states that for a planar crank-rocker four bar mechanism, the 
sum of the shortest and longest link lengths cannot be less than the sum of 
the remaining two link lengths
(B) Inversions of a mechanism are created by fixing different links one at a time
(C) Geneva mechanism is an intermittent motion device
(D) Gruebler’s criterion assumes mobility of a planar mechanism to be one
Sol. 9 Option (A) is correct.
According to Grashof’s law “For a four bar mechanism, the sum of the shortest 
and longest link lengths should not be greater than the sum of remaining two link 
GATE ME 2010
ONE MARK
GATE ME 2010
ONE MARK
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