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 MARKRead More

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