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 MARKRead More

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