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 2009 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 2009 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 2009 © www.nodia.co.in Q. 1 For a matrix // / M x 35 45 35 = 6 > @ H, the transpose of the matrix is equal to the inverse of the matrix, MM T 1 = - 66 @@ . The value of x is given by (A) 5 4 - (B) 5 3 - (C) 5 3 (D) 5 4 Sol. 1 Option (A) is correct. Given : M x 5 3 5 4 5 3 =>H And [] M T [] M 1 = - We know that when AA 1 = T - 66 @@ then it is called orthogonal matrix. M T 6@ M I = 6@ MM T 66 @@ I = Substitute the values of M & M T , we get x x 5 3 5 4 5 3 5 3 5 4 5 3 . > > H H 1 1 0 0 =>H x x x 5 3 5 3 5 4 5 3 5 3 5 3 5 4 5 3 5 4 5 4 5 3 5 3 2 # # # ## + + + + b b b bb l l l ll R T S S S S V X W W W W 1 1 0 0 =>H x x x 1 25 9 2 25 12 5 3 25 12 5 3 + + + >H 1 1 0 0 =>H Comparing both sides a 12 element, x 25 12 5 3 + 0 = x 25 12 3 5 5 4 # =- =- Q. 2 The divergence of the vector field 32 xz xy yz ij k 2 +- at a point ( ,,) 111 is equal to (A) 7 (B) 4 (C) 3 (D) 0 Sol. 2 Option (C) is correct. Let, V 32 xz xy yz ij k 2 =+ - We know divergence vector field of V is given by () V 4: GATE ME 2009 ONE MARK GATE ME 2009 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 2009 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 2009 © www.nodia.co.in Q. 1 For a matrix // / M x 35 45 35 = 6 > @ H, the transpose of the matrix is equal to the inverse of the matrix, MM T 1 = - 66 @@ . The value of x is given by (A) 5 4 - (B) 5 3 - (C) 5 3 (D) 5 4 Sol. 1 Option (A) is correct. Given : M x 5 3 5 4 5 3 =>H And [] M T [] M 1 = - We know that when AA 1 = T - 66 @@ then it is called orthogonal matrix. M T 6@ M I = 6@ MM T 66 @@ I = Substitute the values of M & M T , we get x x 5 3 5 4 5 3 5 3 5 4 5 3 . > > H H 1 1 0 0 =>H x x x 5 3 5 3 5 4 5 3 5 3 5 3 5 4 5 3 5 4 5 4 5 3 5 3 2 # # # ## + + + + b b b bb l l l ll R T S S S S V X W W W W 1 1 0 0 =>H x x x 1 25 9 2 25 12 5 3 25 12 5 3 + + + >H 1 1 0 0 =>H Comparing both sides a 12 element, x 25 12 5 3 + 0 = x 25 12 3 5 5 4 # =- =- Q. 2 The divergence of the vector field 32 xz xy yz ij k 2 +- at a point ( ,,) 111 is equal to (A) 7 (B) 4 (C) 3 (D) 0 Sol. 2 Option (C) is correct. Let, V 32 xz xy yz ij k 2 =+ - We know divergence vector field of V is given by () V 4: GATE ME 2009 ONE MARK GATE ME 2009 ONE MARK © www.nodia.co.in GATE SOLVED PAPER - ME 2009 So, V 4: x i y j z kxz xy yz ij k 32 2 : 2 2 2 2 2 2 =+ + + - c ^ m h V 4: zxyz 32 2 =+ - At point ( ,,) P 111 () V ( ,,) P 111 4: 31 21 211 #### =+ - 3 = Q. 3 The inverse Laplace transform of /( ) ss 1 2 + is (A) e 1 t + (B) e 1 t - (C) e 1 t - - (D) e 1 t + - Sol. 3 Option (C) is correct. Let () fs ss 1 L 1 2 + = - ;E First, take the function ss 1 2 + & break it by the partial fraction, 1 ss 2 + () () ss s s 1 11 1 1 = + =- + () Solve by s s A s B 1 1 1 + =+ + *4 So, ss 1 L 1 2 + - cm () s s 1 1 1 L 1 =- + - ;E 1 ss 1 1 LL 11 =- + -- :: DD e 1 t =- - Q. 4 If three coins are tossed simultaneously, the probability of getting at least one head is (A) 1/8 (B) 3/8 (C) 1/2 (D) 7/8 Sol. 4 Option (D) is correct. Total number of cases 28 3 == & Possible cases when coins are tossed simultaneously. H H H T H T T T H H T H T H T T H T H H T T H T From these cases we can see that out of total 8 cases 7 cases contain at least one head. So, the probability of come at least one head is 8 7 = Q. 5 If a closed system is undergoing an irreversible process, the entropy of the system (A) must increase (B) always remains constant (C) Must decrease (D) can increase, decrease or remain constant GATE ME 2009 ONE MARK GATE ME 2009 ONE MARK GATE ME 2009 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 2009 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 2009 © www.nodia.co.in Q. 1 For a matrix // / M x 35 45 35 = 6 > @ H, the transpose of the matrix is equal to the inverse of the matrix, MM T 1 = - 66 @@ . The value of x is given by (A) 5 4 - (B) 5 3 - (C) 5 3 (D) 5 4 Sol. 1 Option (A) is correct. Given : M x 5 3 5 4 5 3 =>H And [] M T [] M 1 = - We know that when AA 1 = T - 66 @@ then it is called orthogonal matrix. M T 6@ M I = 6@ MM T 66 @@ I = Substitute the values of M & M T , we get x x 5 3 5 4 5 3 5 3 5 4 5 3 . > > H H 1 1 0 0 =>H x x x 5 3 5 3 5 4 5 3 5 3 5 3 5 4 5 3 5 4 5 4 5 3 5 3 2 # # # ## + + + + b b b bb l l l ll R T S S S S V X W W W W 1 1 0 0 =>H x x x 1 25 9 2 25 12 5 3 25 12 5 3 + + + >H 1 1 0 0 =>H Comparing both sides a 12 element, x 25 12 5 3 + 0 = x 25 12 3 5 5 4 # =- =- Q. 2 The divergence of the vector field 32 xz xy yz ij k 2 +- at a point ( ,,) 111 is equal to (A) 7 (B) 4 (C) 3 (D) 0 Sol. 2 Option (C) is correct. Let, V 32 xz xy yz ij k 2 =+ - We know divergence vector field of V is given by () V 4: GATE ME 2009 ONE MARK GATE ME 2009 ONE MARK © www.nodia.co.in GATE SOLVED PAPER - ME 2009 So, V 4: x i y j z kxz xy yz ij k 32 2 : 2 2 2 2 2 2 =+ + + - c ^ m h V 4: zxyz 32 2 =+ - At point ( ,,) P 111 () V ( ,,) P 111 4: 31 21 211 #### =+ - 3 = Q. 3 The inverse Laplace transform of /( ) ss 1 2 + is (A) e 1 t + (B) e 1 t - (C) e 1 t - - (D) e 1 t + - Sol. 3 Option (C) is correct. Let () fs ss 1 L 1 2 + = - ;E First, take the function ss 1 2 + & break it by the partial fraction, 1 ss 2 + () () ss s s 1 11 1 1 = + =- + () Solve by s s A s B 1 1 1 + =+ + *4 So, ss 1 L 1 2 + - cm () s s 1 1 1 L 1 =- + - ;E 1 ss 1 1 LL 11 =- + -- :: DD e 1 t =- - Q. 4 If three coins are tossed simultaneously, the probability of getting at least one head is (A) 1/8 (B) 3/8 (C) 1/2 (D) 7/8 Sol. 4 Option (D) is correct. Total number of cases 28 3 == & Possible cases when coins are tossed simultaneously. H H H T H T T T H H T H T H T T H T H H T T H T From these cases we can see that out of total 8 cases 7 cases contain at least one head. So, the probability of come at least one head is 8 7 = Q. 5 If a closed system is undergoing an irreversible process, the entropy of the system (A) must increase (B) always remains constant (C) Must decrease (D) can increase, decrease or remain constant GATE ME 2009 ONE MARK GATE ME 2009 ONE MARK GATE ME 2009 ONE MARK © www.nodia.co.in GATE SOLVED PAPER - ME 2009 Sol. 5 Option (A) is correct. We consider the cycle shown in figure, where A and B are reversible processes and C is an irreversible process. For the reversible cycle consisting of A and B. T dQ R # T dQ T dQ 0 AB 1 2 2 1 =+ = ## or T dQ A1 2 # T dQ B2 1 =- # ...(i) For the irreversible cycle consisting of A and C , by the inequality of clausius, T dQ # T dQ T dQ 0 < C A 2 1 1 2 =+ # # ...(ii) From equation (i) and (ii) T dQ T dQ C B 2 1 2 1 -+ # # 0 < T dQ B2 1 # T dQ > C2 1 # ...(iii) Since the path B is reversible, T dQ B2 1 # ds B2 1 = # Since entropy is a property, entropy changes for the paths B and C would be the same. Therefore, ds B2 1 # ds C2 1 = # ...(iv) From equation (iii) and (iv), ds C2 1 # T dQ > C2 1 # Thus, for any irreversible process. ds T dQ > So, entropy must increase. Q. 6 A coolant fluid at 30 C c flows over a heated flat plate maintained at constant temperature of 100 C c . The boundary layer temperature distribution at a given location on the plate may be approximated as 30 70 ( ) exp Ty =+ - where y (in m) is the distance normal to the plate and T is in C c . If thermal conductivity of the fluid is 1.0 / WmK, the local convective heat transfer coefficient (in / Wm K 2 ) at that location will be (A) 0.2 (B) 1 (C) 5 (D) 10 Sol. 6 Option (B) is correct. Given : T 1 30 C c = , T 2 100 C c = , 1.0 / WmK k = T () exp y 30 70 =+ - ...(i) GATE ME 2009 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 2009 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 2009 © www.nodia.co.in Q. 1 For a matrix // / M x 35 45 35 = 6 > @ H, the transpose of the matrix is equal to the inverse of the matrix, MM T 1 = - 66 @@ . The value of x is given by (A) 5 4 - (B) 5 3 - (C) 5 3 (D) 5 4 Sol. 1 Option (A) is correct. Given : M x 5 3 5 4 5 3 =>H And [] M T [] M 1 = - We know that when AA 1 = T - 66 @@ then it is called orthogonal matrix. M T 6@ M I = 6@ MM T 66 @@ I = Substitute the values of M & M T , we get x x 5 3 5 4 5 3 5 3 5 4 5 3 . > > H H 1 1 0 0 =>H x x x 5 3 5 3 5 4 5 3 5 3 5 3 5 4 5 3 5 4 5 4 5 3 5 3 2 # # # ## + + + + b b b bb l l l ll R T S S S S V X W W W W 1 1 0 0 =>H x x x 1 25 9 2 25 12 5 3 25 12 5 3 + + + >H 1 1 0 0 =>H Comparing both sides a 12 element, x 25 12 5 3 + 0 = x 25 12 3 5 5 4 # =- =- Q. 2 The divergence of the vector field 32 xz xy yz ij k 2 +- at a point ( ,,) 111 is equal to (A) 7 (B) 4 (C) 3 (D) 0 Sol. 2 Option (C) is correct. Let, V 32 xz xy yz ij k 2 =+ - We know divergence vector field of V is given by () V 4: GATE ME 2009 ONE MARK GATE ME 2009 ONE MARK © www.nodia.co.in GATE SOLVED PAPER - ME 2009 So, V 4: x i y j z kxz xy yz ij k 32 2 : 2 2 2 2 2 2 =+ + + - c ^ m h V 4: zxyz 32 2 =+ - At point ( ,,) P 111 () V ( ,,) P 111 4: 31 21 211 #### =+ - 3 = Q. 3 The inverse Laplace transform of /( ) ss 1 2 + is (A) e 1 t + (B) e 1 t - (C) e 1 t - - (D) e 1 t + - Sol. 3 Option (C) is correct. Let () fs ss 1 L 1 2 + = - ;E First, take the function ss 1 2 + & break it by the partial fraction, 1 ss 2 + () () ss s s 1 11 1 1 = + =- + () Solve by s s A s B 1 1 1 + =+ + *4 So, ss 1 L 1 2 + - cm () s s 1 1 1 L 1 =- + - ;E 1 ss 1 1 LL 11 =- + -- :: DD e 1 t =- - Q. 4 If three coins are tossed simultaneously, the probability of getting at least one head is (A) 1/8 (B) 3/8 (C) 1/2 (D) 7/8 Sol. 4 Option (D) is correct. Total number of cases 28 3 == & Possible cases when coins are tossed simultaneously. H H H T H T T T H H T H T H T T H T H H T T H T From these cases we can see that out of total 8 cases 7 cases contain at least one head. So, the probability of come at least one head is 8 7 = Q. 5 If a closed system is undergoing an irreversible process, the entropy of the system (A) must increase (B) always remains constant (C) Must decrease (D) can increase, decrease or remain constant GATE ME 2009 ONE MARK GATE ME 2009 ONE MARK GATE ME 2009 ONE MARK © www.nodia.co.in GATE SOLVED PAPER - ME 2009 Sol. 5 Option (A) is correct. We consider the cycle shown in figure, where A and B are reversible processes and C is an irreversible process. For the reversible cycle consisting of A and B. T dQ R # T dQ T dQ 0 AB 1 2 2 1 =+ = ## or T dQ A1 2 # T dQ B2 1 =- # ...(i) For the irreversible cycle consisting of A and C , by the inequality of clausius, T dQ # T dQ T dQ 0 < C A 2 1 1 2 =+ # # ...(ii) From equation (i) and (ii) T dQ T dQ C B 2 1 2 1 -+ # # 0 < T dQ B2 1 # T dQ > C2 1 # ...(iii) Since the path B is reversible, T dQ B2 1 # ds B2 1 = # Since entropy is a property, entropy changes for the paths B and C would be the same. Therefore, ds B2 1 # ds C2 1 = # ...(iv) From equation (iii) and (iv), ds C2 1 # T dQ > C2 1 # Thus, for any irreversible process. ds T dQ > So, entropy must increase. Q. 6 A coolant fluid at 30 C c flows over a heated flat plate maintained at constant temperature of 100 C c . The boundary layer temperature distribution at a given location on the plate may be approximated as 30 70 ( ) exp Ty =+ - where y (in m) is the distance normal to the plate and T is in C c . If thermal conductivity of the fluid is 1.0 / WmK, the local convective heat transfer coefficient (in / Wm K 2 ) at that location will be (A) 0.2 (B) 1 (C) 5 (D) 10 Sol. 6 Option (B) is correct. Given : T 1 30 C c = , T 2 100 C c = , 1.0 / WmK k = T () exp y 30 70 =+ - ...(i) GATE ME 2009 ONE MARK © www.nodia.co.in GATE SOLVED PAPER - ME 2009 Under steady state conditions, Heat transfer by conduction = Heat transfer by convection kA dy dT - hA T D = A" Area of plate (30 70 ) kA dy d e y -+ - hA T D = On solving above equation, we get () kA e 70 y -- - hA T D = At the surface of plate, y 0 = Hence, 70kA hA T D = h 70 AT kA T k 70 DD == () 100 30 70 1 # = - 1/ Wm K 2 = Q. 7 A frictionless piston-cylinder device contains a gas initially at 0.8 MPa and 0.015 m 3 . It expands quasi-statically at constant temperature to a final volume of 0.030 m 3 . The work output (in kJ) during this process will be (A) 8.32 (B) 12.00 (C) 554.67 (D) 8320.00 Sol. 7 Option (A) is correct. Given : 0.8 p 1 = MPa, 0.015 m 1 3 n = , 0.030 m 2 3 n = , tan Cons t T = We know work done in a constant temperature (isothermal) process W ln p 11 1 2 n n n = ak (0.8 10 )(0.015) . . ln 0 015 0 030 6 # = bl (0.012 10 ) 0.6931 6 ## = 8.32 kJ = Q. 8 In an ideal vapour compression refrigeration cycle, the specific enthalpy of refrigerant (in kJ/kg) at the following states is given as: Inlet of condenser :283 Exit of condenser :116 Exit of evaporator :232 The COP of this cycle is (A) 2.27 (B) 2.75 (C) 3.27 (D) 3.75 Sol. 8 Option (A) is correct. First of all we have to make a ph - curve for vapour compression refrigeration cycle GATE ME 2009 ONE MARK GATE ME 2009 ONE MARKRead More

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