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 2006 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 2006 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 2006 © www.nodia.co.in Q. 1 Match the items in column I and II. Column I Column II P. Gauss-Seidel method 1. Interpolation Q. Forward Newton-Gauss method 2. Non-linear differential equations R. Runge-Kutta method 3. Numerical integration S. Trapezoidal Rule 4. Linear algebraic equations (A) P-1, Q-4, R-3, S-2 (B) P-1, Q-4, R-2, S-3 (C) P-1. Q-3, R-2, S-4 (D) P-4, Q-1, R-2, S-3 Sol. 1 Option (D) is correct. Column I P. Gauss-Seidel method 4. Linear algebraic equation Q. Forward Newton-Gauss method 1. Interpolation R. Runge-Kutta method 2. Non-linear differential equation S. Trapezoidal Rule 3. Numerical integration So, correct pairs are, P-4, Q-1, R-2, S-3 Q. 2 The solution of the differential equation 2 dx dy xy e x 2 += - with (0) 1 y = is (A) (1 ) xe x 2 + + (B) (1 ) xe x 2 + - (C) (1 ) xe x 2 - + (D) (1 ) xe x 2 - - Sol. 2 Option (B) is correct. Given : dx dy xy 2 + e x 2 = - and () y01 = It is the first order linear differential equation so its solution is (. .) yIF (. .) QIF dx C =+ # So, .. IF ee Pdx xdx 2 == ## () compare with dx dy Py Q += e xdx 2 = # ee x x 2 2 2 2 == # The complete solution is, ye x 2 eedxC xx 22 # =+ - # ye x 2 dx C =+ # xC =+ y e xc x 2 = + ...(i) Given (0) y 1 = At x 0 = y 1 & = Substitute in equation (i), we get 1 C C 1 1 & == GATE ME 2006 ONE MARK GATE ME 2006 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 2006 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 2006 © www.nodia.co.in Q. 1 Match the items in column I and II. Column I Column II P. Gauss-Seidel method 1. Interpolation Q. Forward Newton-Gauss method 2. Non-linear differential equations R. Runge-Kutta method 3. Numerical integration S. Trapezoidal Rule 4. Linear algebraic equations (A) P-1, Q-4, R-3, S-2 (B) P-1, Q-4, R-2, S-3 (C) P-1. Q-3, R-2, S-4 (D) P-4, Q-1, R-2, S-3 Sol. 1 Option (D) is correct. Column I P. Gauss-Seidel method 4. Linear algebraic equation Q. Forward Newton-Gauss method 1. Interpolation R. Runge-Kutta method 2. Non-linear differential equation S. Trapezoidal Rule 3. Numerical integration So, correct pairs are, P-4, Q-1, R-2, S-3 Q. 2 The solution of the differential equation 2 dx dy xy e x 2 += - with (0) 1 y = is (A) (1 ) xe x 2 + + (B) (1 ) xe x 2 + - (C) (1 ) xe x 2 - + (D) (1 ) xe x 2 - - Sol. 2 Option (B) is correct. Given : dx dy xy 2 + e x 2 = - and () y01 = It is the first order linear differential equation so its solution is (. .) yIF (. .) QIF dx C =+ # So, .. IF ee Pdx xdx 2 == ## () compare with dx dy Py Q += e xdx 2 = # ee x x 2 2 2 2 == # The complete solution is, ye x 2 eedxC xx 22 # =+ - # ye x 2 dx C =+ # xC =+ y e xc x 2 = + ...(i) Given (0) y 1 = At x 0 = y 1 & = Substitute in equation (i), we get 1 C C 1 1 & == GATE ME 2006 ONE MARK GATE ME 2006 ONE MARK © www.nodia.co.in GATE SOLVED PAPER - ME 2006 Then y () e x xe 1 1 x x 2 2 = + =+ - Q. 3 Let x denote a real number. Find out the INCORRECT statement. (A) {: 3} Sxx > = represents the set of all real numbers greater than 3 (B) {: 0} Sxx < 2 = represents the empty set. (C) {: } and Sxx A x B !! = represents the union of set A and set B. (D) {: } Sxa x b << = represents the set of all real numbers between a and b, where a and b are real numbers. Sol. 3 Option (C) is correct. The incorrect statement is, {: } and Sxx A x B !! = represents the union of set A and set B. The above symbol () ! denotes intersection of set A and set B. Therefore this statement is incorrect. Q. 4 A box contains 20 defective items and 80 non-defective items. If two items are selected at random without replacement, what will be the probability that both items are defective ? (A) 5 1 (B) 25 1 (C) 99 20 (D) 495 19 Sol. 4 Option (D) is correct. Total number of items = 100 Number of defective items = 20 Number of Non-defective items = 80 Then the probability that both items are defective, when 2 items are selected at random is, P C CC 100 2 20 2 80 0 = !! 100! !! ! 98 2 18 2 20 = 20 19 2 100 99 2 # # = 495 19 = Common Data For Q.Alternate method Here two items are selected without replacement. Probability of first item being defective is P 1 100 20 5 1 == After drawing one defective item from box, there are 19 defective items in the 99 remaining items. Probability that second item is defective, P 2 899 19 = then probability that both are defective P PP 12 # = P 5 1 99 19 495 19 # == GATE ME 2006 ONE MARK GATE ME 2006 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 2006 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 2006 © www.nodia.co.in Q. 1 Match the items in column I and II. Column I Column II P. Gauss-Seidel method 1. Interpolation Q. Forward Newton-Gauss method 2. Non-linear differential equations R. Runge-Kutta method 3. Numerical integration S. Trapezoidal Rule 4. Linear algebraic equations (A) P-1, Q-4, R-3, S-2 (B) P-1, Q-4, R-2, S-3 (C) P-1. Q-3, R-2, S-4 (D) P-4, Q-1, R-2, S-3 Sol. 1 Option (D) is correct. Column I P. Gauss-Seidel method 4. Linear algebraic equation Q. Forward Newton-Gauss method 1. Interpolation R. Runge-Kutta method 2. Non-linear differential equation S. Trapezoidal Rule 3. Numerical integration So, correct pairs are, P-4, Q-1, R-2, S-3 Q. 2 The solution of the differential equation 2 dx dy xy e x 2 += - with (0) 1 y = is (A) (1 ) xe x 2 + + (B) (1 ) xe x 2 + - (C) (1 ) xe x 2 - + (D) (1 ) xe x 2 - - Sol. 2 Option (B) is correct. Given : dx dy xy 2 + e x 2 = - and () y01 = It is the first order linear differential equation so its solution is (. .) yIF (. .) QIF dx C =+ # So, .. IF ee Pdx xdx 2 == ## () compare with dx dy Py Q += e xdx 2 = # ee x x 2 2 2 2 == # The complete solution is, ye x 2 eedxC xx 22 # =+ - # ye x 2 dx C =+ # xC =+ y e xc x 2 = + ...(i) Given (0) y 1 = At x 0 = y 1 & = Substitute in equation (i), we get 1 C C 1 1 & == GATE ME 2006 ONE MARK GATE ME 2006 ONE MARK © www.nodia.co.in GATE SOLVED PAPER - ME 2006 Then y () e x xe 1 1 x x 2 2 = + =+ - Q. 3 Let x denote a real number. Find out the INCORRECT statement. (A) {: 3} Sxx > = represents the set of all real numbers greater than 3 (B) {: 0} Sxx < 2 = represents the empty set. (C) {: } and Sxx A x B !! = represents the union of set A and set B. (D) {: } Sxa x b << = represents the set of all real numbers between a and b, where a and b are real numbers. Sol. 3 Option (C) is correct. The incorrect statement is, {: } and Sxx A x B !! = represents the union of set A and set B. The above symbol () ! denotes intersection of set A and set B. Therefore this statement is incorrect. Q. 4 A box contains 20 defective items and 80 non-defective items. If two items are selected at random without replacement, what will be the probability that both items are defective ? (A) 5 1 (B) 25 1 (C) 99 20 (D) 495 19 Sol. 4 Option (D) is correct. Total number of items = 100 Number of defective items = 20 Number of Non-defective items = 80 Then the probability that both items are defective, when 2 items are selected at random is, P C CC 100 2 20 2 80 0 = !! 100! !! ! 98 2 18 2 20 = 20 19 2 100 99 2 # # = 495 19 = Common Data For Q.Alternate method Here two items are selected without replacement. Probability of first item being defective is P 1 100 20 5 1 == After drawing one defective item from box, there are 19 defective items in the 99 remaining items. Probability that second item is defective, P 2 899 19 = then probability that both are defective P PP 12 # = P 5 1 99 19 495 19 # == GATE ME 2006 ONE MARK GATE ME 2006 ONE MARK © www.nodia.co.in GATE SOLVED PAPER - ME 2006 Q. 5 For a circular shaft of diameter d subjected to torque T , the maximum value of the shear stress is (A) d T 64 3 p (B) d T 32 3 p (C) d T 16 3 p (D) d T 8 3 p Sol. 5 Option (C) is correct. From the Torsional equation J T r l G tq == Take first two terms, J T r t = d T 32 4 p d 2 t = J = Polar moment of inertia max t d T 16 3 p = Q. 6 For a four-bar linkage in toggle position, the value of mechanical advantage is (A) 0.0 (B) 0.5 (C) 1.0 (D) 3 Sol. 6 Option (D) is correct. . MA T T R R PA PD 2 4 4 2 w w == = from angular velocity ratio theorem Construct BA l and CD l perpendicular to the line PBC. Also, assign lables b and g to the acute angles made by the coupler. R R PA PD sin sin R R R R BA CD BA CD b g == l l So, .. MA sin sin T T R R BA CD 2 4 4 2 w w b g == = When the mechanism is toggle,then 0c b = and 180c. So . MA 3 = GATE ME 2006 ONE MARK GATE ME 2006 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 2006 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 2006 © www.nodia.co.in Q. 1 Match the items in column I and II. Column I Column II P. Gauss-Seidel method 1. Interpolation Q. Forward Newton-Gauss method 2. Non-linear differential equations R. Runge-Kutta method 3. Numerical integration S. Trapezoidal Rule 4. Linear algebraic equations (A) P-1, Q-4, R-3, S-2 (B) P-1, Q-4, R-2, S-3 (C) P-1. Q-3, R-2, S-4 (D) P-4, Q-1, R-2, S-3 Sol. 1 Option (D) is correct. Column I P. Gauss-Seidel method 4. Linear algebraic equation Q. Forward Newton-Gauss method 1. Interpolation R. Runge-Kutta method 2. Non-linear differential equation S. Trapezoidal Rule 3. Numerical integration So, correct pairs are, P-4, Q-1, R-2, S-3 Q. 2 The solution of the differential equation 2 dx dy xy e x 2 += - with (0) 1 y = is (A) (1 ) xe x 2 + + (B) (1 ) xe x 2 + - (C) (1 ) xe x 2 - + (D) (1 ) xe x 2 - - Sol. 2 Option (B) is correct. Given : dx dy xy 2 + e x 2 = - and () y01 = It is the first order linear differential equation so its solution is (. .) yIF (. .) QIF dx C =+ # So, .. IF ee Pdx xdx 2 == ## () compare with dx dy Py Q += e xdx 2 = # ee x x 2 2 2 2 == # The complete solution is, ye x 2 eedxC xx 22 # =+ - # ye x 2 dx C =+ # xC =+ y e xc x 2 = + ...(i) Given (0) y 1 = At x 0 = y 1 & = Substitute in equation (i), we get 1 C C 1 1 & == GATE ME 2006 ONE MARK GATE ME 2006 ONE MARK © www.nodia.co.in GATE SOLVED PAPER - ME 2006 Then y () e x xe 1 1 x x 2 2 = + =+ - Q. 3 Let x denote a real number. Find out the INCORRECT statement. (A) {: 3} Sxx > = represents the set of all real numbers greater than 3 (B) {: 0} Sxx < 2 = represents the empty set. (C) {: } and Sxx A x B !! = represents the union of set A and set B. (D) {: } Sxa x b << = represents the set of all real numbers between a and b, where a and b are real numbers. Sol. 3 Option (C) is correct. The incorrect statement is, {: } and Sxx A x B !! = represents the union of set A and set B. The above symbol () ! denotes intersection of set A and set B. Therefore this statement is incorrect. Q. 4 A box contains 20 defective items and 80 non-defective items. If two items are selected at random without replacement, what will be the probability that both items are defective ? (A) 5 1 (B) 25 1 (C) 99 20 (D) 495 19 Sol. 4 Option (D) is correct. Total number of items = 100 Number of defective items = 20 Number of Non-defective items = 80 Then the probability that both items are defective, when 2 items are selected at random is, P C CC 100 2 20 2 80 0 = !! 100! !! ! 98 2 18 2 20 = 20 19 2 100 99 2 # # = 495 19 = Common Data For Q.Alternate method Here two items are selected without replacement. Probability of first item being defective is P 1 100 20 5 1 == After drawing one defective item from box, there are 19 defective items in the 99 remaining items. Probability that second item is defective, P 2 899 19 = then probability that both are defective P PP 12 # = P 5 1 99 19 495 19 # == GATE ME 2006 ONE MARK GATE ME 2006 ONE MARK © www.nodia.co.in GATE SOLVED PAPER - ME 2006 Q. 5 For a circular shaft of diameter d subjected to torque T , the maximum value of the shear stress is (A) d T 64 3 p (B) d T 32 3 p (C) d T 16 3 p (D) d T 8 3 p Sol. 5 Option (C) is correct. From the Torsional equation J T r l G tq == Take first two terms, J T r t = d T 32 4 p d 2 t = J = Polar moment of inertia max t d T 16 3 p = Q. 6 For a four-bar linkage in toggle position, the value of mechanical advantage is (A) 0.0 (B) 0.5 (C) 1.0 (D) 3 Sol. 6 Option (D) is correct. . MA T T R R PA PD 2 4 4 2 w w == = from angular velocity ratio theorem Construct BA l and CD l perpendicular to the line PBC. Also, assign lables b and g to the acute angles made by the coupler. R R PA PD sin sin R R R R BA CD BA CD b g == l l So, .. MA sin sin T T R R BA CD 2 4 4 2 w w b g == = When the mechanism is toggle,then 0c b = and 180c. So . MA 3 = GATE ME 2006 ONE MARK GATE ME 2006 ONE MARK © www.nodia.co.in GATE SOLVED PAPER - ME 2006 Q. 7 The differential equation governing the vibrating system is (A) () 0 mx cx k x y ++ - = po (B) () ( ) 0 mx y cx y kx -+ - + = pp o o (C) () 0 mx c x y kx +- + = poo (D) ( ) () () 0 mx y cx y kx y -+ - + - = pp o o Sol. 7 Option (C) is correct. Assume any arbitrary relationship between the coordinates and their first derivatives, say xy > and xy > oo . Also assume x 0 > and 0 x > o . A small displacement gives to the system towards the left direction. Mass m is fixed, so only damper moves for both the variable x and y. Note that these forces are acting in the negative direction. Differential equation governing the above system is, F / m dt dx c dt dx dt dy kx 0 2 2 =- - - - = bl () mx c x y kx +- + poo 0 = Q. 8 A pin-ended column of length L , modulus of elasticity E and second moment of the cross-sectional area is I loaded eccentrically by a compressive load P. The critical buckling load (P cr ) is given by (A) P L EI cr 22 p = (B) P L EI 3 cr 2 2 p = (C) P L EI cr 2 p = (D) P L EI cr 2 2 p = Sol. 8 Option (D) is correct. GATE ME 2006 ONE MARK GATE ME 2006 ONE MARKRead More

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