NODIA AND COMPANY. GATE SOLVED PAPER Mechanical Engineering Copyright By NODIA & COMPANY

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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. GAE SOLVED PAPER Mechanical Engineering 9 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. his 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 B8, Dhanshree ower Ist, Central Spine, Vidyadhar Nagar, Jaipur 9 Ph : enquiry@nodia.co.in

2 GAE SOLVED PAPER ME 9 / / Q. For a matri 6 M@ > H, the transpose of the matri is equal to the inverse / of ONE MARK the matri, 6M@ 6M@. he value of is given by Sol. (A) (C) (B) Option (A) is correct. Given : M > H And [ M] [ ] M (D) We know that when 6A@ 6A@ then it is called orthogonal matri. 6 M@ I 6 6M@ 6M@ I Substitute the values of M & M, we get > H. > H > H R V S b # l b # l W S S Sb # l b # l b # W W > H lw X 9 > H > H Comparing both sides a element, # Q. he divergence of the vector field zi yj yz k at a point (,, ) is equal to (A) 7 (B) ONE MARK (C) (D) Sol. Option (C) is correct. Let, V zi yjyz k We know divergence vector field of V is given by ( :V)

3 GAE SOLVED PAPER ME 9 So, : V i y j k z y yz z i j c : m ^ k h : V z yz At point P(,, ) ( : V ) P(,, ) # # # # Q. he inverse Laplace transform of /( s s) is t (A) e (B) e ONE MARK t (C) e (D) e t t Sol. Option (C) is correct. Let fs () First, take the function s So, s L ; s se & break it by the partial fraction, s s ss ( ) s ( s ) L c s s m L ; s ( s ) E L L : sd : s D e t Solve by A ( s ) s s B * Q. If three coins are tossed simultaneously, the probability of getting at least one head is ONE MARK (A) /8 (B) /8 (C) / (D) 7/8 Sol. Option (D) is correct. otal number of cases 8 & Possible cases when coins are tossed simultaneously. H H H H H H H H H H H H 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 7 8 Q. If a closed system is undergoing an irreversible process, the entropy of the system (A) must increase ONE MARK (B) always remains constant (C) Must decrease (D) can increase, decrease or remain constant

4 GAE SOLVED PAPER ME 9 Sol. 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. dq dq dq A B or dq dq # A # B...(i) For the irreversible cycle consisting of A and C, by the inequality of clausius, dq dq dq # < # A # C...(ii) From equation (i) and (ii) dq dq # B # < C dq dq # > B # C...(iii) Since the path B is reversible, dq # ds # R# B # # B Since entropy is a property, entropy changes for the paths B and C would be the same. herefore, # ds # ds...(iv) B C From equation (iii) and (iv), dq # ds > # C hus, for any irreversible process. dq ds > C So, entropy must increase. Q. 6 A coolant fluid at c C flows over a heated flat plate maintained at constant temperature of c C. he boundary layer temperature distribution at a given ONE MARK location on the plate may be approimated as 7 ep( y) where y (in m) is the distance normal to the plate and is in c C. If thermal conductivity of the fluid is. W/ mk, the local convective heat transfer coefficient (in W/ m K ) at that location will be (A). (B) (C) (D) Sol. 6 Option (B) is correct. Given : cc, cc, k. W/ mk 7 ep( y)...(i)

5 GAE SOLVED PAPER ME 9 Under steady state conditions, Heat transfer by conduction Heat transfer by convection ka d had A " Area of plate dy ka d ( 7 e y ) had dy On solving above equation, we get ka(7e y ) had At the surface of plate, y Hence, 7kA had h 7kA 7k AD D 7 # ( ) W/ m K Q. 7 A frictionless pistoncylinder device contains a gas initially at.8 MPa and. m. It epands quasistatically at constant temperature to a final volume of ONE MARK. m. he work output (in kj) during this process will be (A) 8. (B). (C).67 (D) 8. Sol. 7 Option (A) is correct. Given : p.8 MPa, n. m, n. m, Constant We know work done in a constant temperature (isothermal) process W n ln n 6 an k (.8 )(.) ln. # b. l p (. # 6 ) # 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: ONE MARK Inlet of condenser :8 Eit of condenser :6 Eit of evaporator : he COP of this cycle is (A).7 (B).7 (C).7 (D).7 Sol. 8 Option (A) is correct. First of all we have to make a p h curve for vapour compression refrigeration cycle

6 GAE SOLVED PAPER ME 9 he given specific enthalpies are Inlet of condenser h 8 kj/ kg Eit of condenser h 6 kj/ kg h From p h curve Eit of evaporator h kj/ kg Refrigerating effect Now, COP h h Work done h h Substitute the values, we get COP Q. 9 A compressor undergoes a reversible, steady flow process. he gas at inlet and outlet of the compressor is designated as state and state respectively. Potential WO MARK and kinetic energy changes are to be ignored. he following notations are used : n Specific volume and p pressure of the gas. he specific work required to be supplied to the compressor for this gas compression process is (A) # pdn (B) # ndp Sol. 9 (C) n ( p p ) (D) ( n n) Option (B) is correct. p Steady flow energy equation for a compressor (Fig a) gives, h dq h dw...(i) Neglecting the changes of potential and kinetic energy. From the property relation ds dh ndp For a reversible process, ds dq So, dq dh ndp...(ii) If consider the process is reversible adiabatic then dq

7 GAE SOLVED PAPER ME 9 From equation (i) and (ii), h h dw & dh h h dw...(iii) And dh ndp...(iv) From equation (iii) and (iv), dw ndp W # ndp Negative sign shows the work is done on the system (compression work) for initial & Final Stage W # n dp Q. A block weighing 98 N is resting on a horizontal surface. he coefficient of friction between the block and the horizontal surface is m.. A vertical cable attached ONE MARK to the block provides partial support as shown. A man can pull horizontally with a force of N. What will be the tension, (in N) in the cable if the man is just able to move the block to the right? Sol. (A) 76. (B) 96. (C) 8. (D) 98. Option (C) is correct. Given : W 98 N, m. First of all we have to make a FBD of the block Here, R N Normal reaction force ension in string Using the balancing of forces, we get In direction S F m N R N R N N m. and S F y or downward forces upward forces W R N WR N 98 8 N

8 GAE SOLVED PAPER ME 9 Q. If the principal stresses in a plane stress problem are ONE MARK the magnitude of the maimum shear stress (in MPa) will be (A) 6 (B) (C) (D) s MPa, s MPa, Sol. Option (C) is correct. Given : s MPa, s MPa We know, the maimum shear stress for the plane comple stress is given by t ma s s t ma 6 MPa Q. A simple quick return mechanism is shown in the figure. he forward to return ratio of the quick return mechanism is :. If the radius of crank O Pis mm, ONE MARK then the distance d (in mm) between the crank centre to lever pivot centre point should be Sol. (A). (B) 6. (C). (D). Option (D) is correct. Given O P r mm Forward to return ratio : We know that, ime of cutting ( forward) stroke b 6 ime of return stroke a a a Substitute the value of Forward to return ratio, we have

9 GAE SOLVED PAPER ME 9 6 a a a 6 a & a c And angle RO O a c 6c Now we are to find the distance d between the crank centre to lever pivot centre point ( OO ). From the DRO O sin 9c a OR a k r OO OO sin( 9c 6c) r OO OO r mm sin c / Q. he rotor shaft of a large electric motor supported between short bearings at both the ends shows a deflection of.8 mm in the middle of the rotor. Assuming ONE MARK the rotor to be perfectly balanced and supported at knife edges at both the ends, the likely critical speed (in rpm) of the shaft is (A) (B) 7 (C) 8 (D) Sol. Option (B) is correct. Given d.8 mm.8 m he critical or whirling speed is given by, g w c d pnc g N 6 d C Critical speed in rpm g N c p d.. 8. # rpm Q. A solid circular shaft of diameter d is subjected to a combined bending moment M and torque,. he material property to be used for designing the shaft using ONE MARK the relation 6 M is pd (A) ultimate tensile strength ( S u ) (B) tensile yield strength ( S y ) (C) torsional yield strength ( S sy ) (D) endurance strength ( S e ) Sol. Option (C) is correct. We know that, for a shaft of diameter d is subjected to combined bending moment M and torque, the equivalent orque is, e M Induced shear stress is, t 6 6 # M pd pd

10 GAE SOLVED PAPER ME 9 Now, for safe design, t should be less than N S sy Where, S sy orsional yield strength and N Factor of safety Q. he effective number of lattice points in the unit cell of simple cubic, body centered cubic, and face centered cubic space lattices, respectively, are ONE MARK (A),, (B),, (C),, (D),, Sol. Option (B) is correct. ^i h Simple cubic ^iihbcc Effective Number of lattice 8 # 8 Effective Number 8 # 8 ^iiihfcc Effective Number of lattice 8 8 # #6 Q. 6 Friction at the toolchip interface can be reduced by (A) decreasing the rake angle (B) increasing the depth of cut ONE MARK (C) decreasing the cutting speed (D) increasing the cutting speed Sol. 6 Option (D) is correct. he cutting forces decrease with an increase in cutting speed, but it is substantially smaller than the increase in speed. With the increase in speed, friction decreases at the tool chip interface. he thickness of chip reduces by increasing the speed. Q. 7 wo streams of liquid metal which are not hot enough to fuse properly result into a casting defect known as ONE MARK (A) cold shut (B) swell (C) sand wash (D) scab

11 GAE SOLVED PAPER ME 9 Sol. 7 Option (A) is correct. wo streams of liquid metal which are not hot enough to fuse properly result into a casting defect known as cold shut. his defect is same as in sand mould casting. he reasons are : (i) Cooling of die or loss of plasticity of the metal. (ii) Shot speed less. (iii) Airvent or overflow is closed. Q. 8 he epected time ( t e ) of a PER activity in terms of optimistic time t, pessimistic time ( t p ) and most likely time ( t l ) is given by ONE MARK (A) t e to tl tp (B) t 6 to tp tl 6 to tl tp (C) te to tp tl (D) te Sol. 8 Option (A) is correct. Under the conditions of uncertainty, the estimated time for each activity for PER network is represented by a probability distribution. his probability distribution of activity time is based upon three different time estimates made for each activity. hese are as follows. t o the optimistic time, is the shortest possible time to complete the activity if all goes well. t p the pessimistic time, is the longest time that an activity could take if every thing goes wrong t l the most likely time, is the estimate of normal time an activity would take. he epected time ( t e ) of the activity duration can be approimated as the arithmetic mean of ( to tp)/ and t l. hus ( t e ) ( to tp) to tl tp tl : D 6 Q. 9 Which of the following is the correct data structure for solid models? (A) solid part " faces " edges " vertices ONE MARK (B) solid part " edges " faces " vertices (C) vertices " edges " faces " solid parts (D) vertices " faces " edges " solid parts e Sol. 9 Option (C) is correct. Correct data structure for solid models is given by, Vertices " edges " faces " solid parts Q. Which of the following forecasting methods takes a fraction of forecast error into account for the net period forecast? ONE MARK (A) simple average method (B) moving average method (C) weighted moving average method (D) eponential smoothening method

12 GAE SOLVED PAPER ME 9 Sol. Option (D) is correct. Eponential smoothing method of forecasting takes a fraction of forecast error into account for the net period forecast. he eponential smoothed average u t, which is the forecast for the net period ( t ) is given by. u t ay a( a) y... a( a) y... t n t t n n ay t ( a)[ ay t a( a) y t... a( a) y t ( n )...] u a( y u ) t t t ut aet where et ( ytut ) is called error and is the difference between the least observation, y t and its forecast a period earlier, u t. he value of a lies between to. Q. An analytic function of a comple variable z iy is epressed as fz () uy (, ) ivy (, ) where i. If u y, the epression for v should be WO MARK Sol. (A) ( y) k (B) y k (C) y ( y) k (D) k Option (C) is correct. Given : z iy is a analytic function fz () uy (,) ivy (,) u y..(i) We know that analytic function is satisfy the CauchyRiemann equation. u v & u v y y So from equation (i), u y & v y y u y & v Let vy (,) be the conjugate function of uy (,) dv v d v dy ( d ) ( ydy ) y Integrating both the sides # dv # d # ydy v y k ( y ) k dy Q. he solution of d WO MARK y with the condition y() 6 is (A) y (B) y (C) y (D) y

13 GAE SOLVED PAPER ME 9 Sol. Option (A) is correct. dy Given y d dy d b l y...(i) dy It is a single order differential equation. Compare this with Py Q & we d get P Q And its solution will be yif (..) QIF (..) d C # # # IF.. e Pd loge e e And complete solution is given by, y # # d C # d C C...(ii) And y() 6 at & y 6 From equation (ii), 6 # C C 6 d hen, from equation (ii), we get y & y Q. A path AB in the form of one quarter of a circle of unit radius is shown in the figure. Integration of ( y) on path AB traversed in a counterclockwise sense WO MARK is (A) p (B) p (C) p (D) Sol. Option (B) is correct. he equation of circle with unit radius & centre at origin is given by, y

14 GAE SOLVED PAPER ME 9 Finding the integration of ( y) on path AB traversed in counterclockwise sense So using the polar form Let: cos q, y sin q r So put the value of & y & limits in first quadrant between to p /. Hence, # p/ ( sin q) dq p/ I ( cos q sin q) dq # # p/ On integrating above equation, we get p/ q cos q : D ( cos q sin q sin qcos q) dq cos cos ; a p p k b le b p lb l p Q. he distance between the origin and the point nearest to it on the surface z y WO MARK (A) (B) Sol. (C) (D) Option (A) is correct. he given equation of surface is z y...(i) Let Pyz (,,) be the nearest point on the surface (i), then distance from the origin is d ( ) ( y ) ( z) d y z z d y...(ii) From equation (i) & (ii), we get d y y d y y Let fy (,) d y y...(iii) he fy (,) be the maimum or minimum according to d maimum or minimum. Differentiating equation (iii) w.r.t & y respectively, we get f f y or y y

15 GAE SOLVED PAPER ME 9 f f Applying maima minima principle & put & equal to zero, f f y y or y y on solving these equations, we get, y So, y is only one stationary point. f Now p f q y f r y or pr q > and r is positive. So, fy (,) d is minimum at (,). Hence minimum value of d at (,). d y y d or fy (,) So, the nearest point is z y z! Q. he area enclosed between the curves y and y is (A) 6 (B) 8 WO MARK Sol. (C) (D) 6 Option (A) is correct. Given : y & y Draw the curves from the given equations, he shaded area show the common area. Now finding the intersection points of the curves. y y 8 y y From second curve Squaring both sides y 8 # 8 # y & y( y 6) y & Similarly put y in curve y # & And Put y # 6

16 GAE SOLVED PAPER ME 9 So,, herefore the intersection points of the curves are (,)& (.,) So the enclosed area is given by A ( y y ) d # Put y & y from the equation of curves y & y A # b d l Integrating the equation, we get A / : D : D Substituting the limits, we get A / # # # d d # b d l # # Q. 6 he standard deviation of a uniformly distributed random variable between and is WO MARK (A) (B) (C) (D) 7 Sol. 6 Option (A) is correct. he cumulative distribution function Z, # a ] f () a [, a < < b b a ], $ b \ and density function, a # # b f () * b a, a >, > b b / a Mean E () f() a b Variance f () f ()6 f ()@ Substitute the value of f () b / a b / a b b d d b a ) b a ( b a) G > ) ( b a) H a b a ( b a ) ( b a) ( b a) ( b a)( b ab a ) ( b a) ( ba) ( b a) ( b a) ( b ab a ) ( a b) a

17 GAE SOLVED PAPER ME 9 a b aba b 6ab/ ( b a) Standard deviation Variance Given : b, a ( b a) So, standard deviation ( b a) Q. 7 Consider steady, incompressible and irrotational flow through a reducer in a horizontal pipe where the diameter is reduced from cm to cm. he pressure WO MARK in the cm pipe just upstream of the reducer is kpa. he fluid has a vapour pressure of kpa and a specific weight of kn/ m. Neglecting frictional effects, the maimum discharge ( in m / s) that can pass through the reducer without causing cavitation is (A). (B).6 (C).7 (D).8 Sol. 7 Option (B) is correct. Given : p V kpa, w kn/ m rg Consider steady, incompressible & irrotational flow & neglecting frictional effect. First of all applying continuity equation at section () & (). AV AV p ( d ) # V p ( d ) # V Substitute the values of d & d, we get p ( ) V # p ( ) V # V V & V V...(i) Cavitation is the phenomenon of formation of vapor bubbles of a flowing liquid in a region where the pressure of liquid falls below the vapor pressure [ pl < pv] So, we can say that maimum pressure in downstream of reducer should be equal or greater than the vapor pressure. For maimum discharge p V p kpa Applying Bernoulli s equation at point () & () p V z r g g p V z rg g

18 GAE SOLVED PAPER ME 9 Here z z for horizontal pipe & w rg kn/ m V g ( V ) From equation (i) V V g 6V V g g V g V. # 9 8. m/sec And V V #..6 m/ sec Maimum discharge, Q ma AV p ( d) V p ( ). 6 # # p. 6 # #.6 m / sec Q. 8 In a parallel flow heat echanger operating under steady state, the heat capacity rates (product of specific heat at constant pressure and mass flow rate) of the WO MARK hot and cold fluid are equal. he hot fluid, flowing at kg/ s with cp kj/kg K, enters the heat echanger at c C while the cold fluid has an inlet temperature of c C. he overall heat transfer coefficient for the heat echanger is estimated to be kw/m K and the corresponding heat transfer surface area is m. Neglect heat transfer between the heat echanger and the ambient. he heat echanger is characterized by the following relations: e ep( NU) he eit temperature (in c C) for the cold fluid is (A) (B) (C) 6 (D) 7 Sol. 8 Option (B) is correct. Given : Co h C o c, mo h kg/sec, c ph kj/ kg K, t h cc, t c cc U kw/ m K, A m he figure shown below is for parallel flow. C o / h mc o h ph kj sk he heat echanger is characterized by the following relation, ep( NU) e..(i) For parallel flow heat echanger effectiveness is given by ep[ NU( C)] e...(ii) C

19 GAE SOLVED PAPER ME 9 On comparing equation (i) and equation (ii), we get capacity ratio C Cc Cmin...(iii) Ch Cma On applying energy balance for a parallel flow Ch( th th) Cc( tctc) Cc th th From equation(iii) Ch tc tc th th tctc Number of transfer units is given by, NU UA #. Cmin ep(. ) Effectiveness, e #. 8.6 Maimum possible heat transfer is, Q ma Cmin( thtc) # 6 ( 7 ) ( 7 8 kw But Actual Heat transfer is, Q a eq ma. 6 # 8 6 kw And Q a Cc( tctc) 6 ( tc ) t c cc Q. 9 In an airstandard Ottocycle, the compression ratio is. he condition at the beginning of the compression process is kpa and 7c C. Heat added WO MARK at constant volume is kj/ kg, while 7 kj/ kg of heat is rejected during the other constant volume process in the cycle. Specific gas constant for air.87 kj/ kgk. he mean effective pressure (in kpa) of the cycle is (A) (B) (C) (D) Sol. 9 Option (D) is correct. Given : r, p kpa, 7 cc (7 7) K K Q s kj/ kg, Q r 7 kj/ kg, R.87 kj/ kg K Mean Effective pressure Net work output p m...(i) Swept Volume Swept volume, n n n ( r ) where n otal volume and n Clearance volume r n n & n v...(ii) Applying gas equation for the beginning process, p n R n R..86 m / kg p 87 # From equation (i) n n m / kg W net Q Q s r ( 7) kj/ kg K 8 kj/ kg K

20 GAE SOLVED PAPER ME 9 p m n 8 8 ( r ). 86( ) 8.9 kpa b kpa. 779 Q. An irreversible heat engine etracts heat from a high temperature source at a rate of kw and rejects heat to a sink at a rate of kw. he entire work output WO MARK of the heat engine is used to drive a reversible heat pump operating between a set of independent isothermal heat reservoirs at 7c C and 7c C. he rate (in kw) at which the heat pump delivers heat to its high temperature sink is (A) (B) (C) (D) 6 Sol. Option (C) is correct. We know that coefficient of performance of a Heat pump for the given system is, Q Q ( COP ) HP.. Q Q W For a reversible process, Q Q ( COP ) Q HP.. W 8 Q 8 9 Q 8 # K 8 Q. You are asked to evaluate assorted fluid flows for their suitability in a given laboratory application. he following three flow choices, epressed in terms of the WO MARK two dimensional velocity fields in the yplane, are made available. P : u y, v Q : u y, v R : u, v y Which flow(s) should be recommended when the application requires the flow to be incompressible and irrotational? (A) P and R (B) Q (C) Q and R (D) R Sol. Option (D) is correct. Given : P : u y, V Q : u y, V

21 GAE SOLVED PAPER ME 9 R : u, V y For incompressible fluid, u v w...(i) y z For irrotational flow z z, z z v u c y m v u c y m v u...(ii) y From equation (i) & (ii), check P, Q & R For P : u y v u v y v u y u v & (Flow is incompressible) y Or, v u y &! (Rotational flow) For Q : u y v u y v y v u y u v & y! (Compressible flow) y Or, v u y &! (Rotational flow) For R : u v y u v y v u y u v y Or, & (Incompressible flow) v u y & (Irrotational flow) So, we can easily see that R is incompressible & irrotational flow. Q. Water at c C is flowing through a. km long. G.I. pipe of mm diameter WO MARK at the rate of.7 m / s. If value of Darcy friction factor for this pipe is. and density of water is kg/ m, the pumping power (in kw) required to maintain the flow is (A).8 (B) 7. (C). (D).

22 GAE SOLVED PAPER ME 9 Sol. Option (A) is correct. Given : L km m, D mm. m, Q.7 M / sec f., r kg/ m Head loss is given by, flv fl Q h f D# g D g # c pd m Q pd V # 6fLQ 8fLQ p D # g p Dg 8. (. 7) # # # m of water (. ) #(. ) #( 98. ). Pumping power required, P rgq # h f # 9. 8 #. 7 # kw..8 kw Q. Consider steadystate conduction across the thickness in a plane composite wall (as shown in the figure) eposed to convection conditions on both sides. WO MARK Sol. Given : hi W/m K, ho W/m K;,,i cc;,o cc, k W/mK; k W/mK; L. m and L. m. Assuming negligible contact resistance between the wall surfaces, the interface temperature, (in cc ), of the two walls will be (A). (B).7 (C).7 (D). Option (C) is correct. he equivalent resistance diagram for the given system is, R eq L L ha ka ka ha i

23 GAE SOLVED PAPER ME 9 R eq # A L L h k k h i m K/ W Q Heat flu, q D Q A AR D eq / R Under steady state condition, q i o hi( i) AReq k( ) k( )...(i) L L q i o ( ) W/ m...(ii) AReq. 88 q i From equation(i) hi ( ). &. 7.cC Again from equation(i), k( ) q L (. 7 ). Common Data For Q.Alternate : &.7cC Under steady state conditions, Heat flow from I to interface wall Heat flow from interface wall to O (, i ) (,o) L L ha i ka ka ha, i, o L L hi k k ho ( ) ().. ( ).. ( ). 86( ) cC 86 Q. he velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in the figure, is given by the epression WO MARK ur () R dp r mbd lc R m dp Where is a constant. he average velocity of fluid in the pipe is d

24 GAE SOLVED PAPER ME 9 Sol. (A) R dp 8mbd l (B) R dp mbd l (C) R dp mbd l (D) R dp m bd l Option (A) is correct. ur () R dp r mbd lc R m herefore, the velocity profile in fully developed laminar flow in a pipe is parabolic with a maimum at the center line and minimum at the pipe wall. he average velocity is determined from its definition, R V avg # urrdr () R R dp r # rdr R d mb lc R m dp r r dr R mbd l # c R m dp r r R dp R R mbd l; R E mbd l; R E dp R R dp mbd l # 8mbd l Common Data For Q.Alternate Method Now we consider a small element (ring) of pipe with thickness dr & radius r. We find the flow rate through this elementry ring. dq ( pr) # dr# u( r) Put the value of ur () dq ( r) dr R dp p r # # c m mb d lc R m Now for total discharge integrate both the rides within limit. Q & toq and R & tor So Q dq # R dp R p r m b d l c 6 Q@ Q Now put the limits r R R R dp p r r m bd l; R E # mdr

25 GAE SOLVED PAPER ME 9 hen Q R dp p R R m bd l; R E Q R dp p R R m bd l: D Q R dp p R c m mbd l: D Q R dp p 8m bd l Now Q Area # Average velocity A# V avg. Q V R dp avg. p A 8m bd l# pr V R dp avg. 8mbd l Q. A solid shaft of diameter d and length L is fied at both the ends. A torque, is applied at a distance L from the left end as shown in the figure given below. WO MARK Sol. he maimum shear stress in the shaft is (A) 6 (B) pd (C) 8 pd Option (B) is correct. (D) pd pd First, the shaft is divided in two parts () & () and gives a twisting moment (in counterclockwise direction) & (in clock wise direction) respectively. By the nature of these twisting moments, we can say that shafts are in parallel combination. So,...(i) From the torsional equation, t Gq & J r l GJq l But, here G G q q For parallel connection J J Diameter is same So, l l L # # L

26 GAE SOLVED PAPER ME 9 Now, From equation (i), And Here > So, maimum shear stress is developed due to, t J ma & t r ma J # r Substitute the values, we get t d ma b l # # p d 8 d p # pd Q. 6 An epicyclic gear train in shown schematically in the given figure. he run gear on the input shaft is a teeth eternal gear. he planet gear is a teeth WO MARK eternal gear. he ring gear is a teeth internal gear. he ring gear is fied and the gear is rotating at 6 rpm CCW (CCWcounterclockwise and CWclockwise). he arm attached to the output shaft will rotate at (A) rpm CCW (B) rpm CW (C) rpm CW (D) rpm CCW Sol. 6 Option (A) is correct. Given Z eeth, Z eeth, Z eeth, N, N 6 rpm( CCW) If gear rotates in the CCW direction, then gear rotates in the clockwise direction. Let, Arm will rotate at N rpm. he table of motions is given below ake CCW ve, CW ve

27 GAE SOLVED PAPER ME 9 S. No. Condition of Motion Revolution of elements Sun Gear Planet Gear Arm Ring Gear N N N N. Arm fied and sun gear rotates rpm (CCW). Give rpm to gear (CCW). Add y revolutions to all elements. otal motion. y Note : Speed of driver Speed ratio Speed of driven Z Z Z Z Z Z Z Z y y y y y Z Z y No. of teeth on dirven No. of teeth on driver Ring gear is fied. So, N y Z From the table Z y Z...(i) Z Given, N 6 rpm( CCW) y 6 From table 6 # rpm And from equation (i), y rpm( CCW) From the table the arm will rotate at N y rpm( CCW) Z # Z y Z Z Q. 7 A forged steel link with uniform diameter of mm at the centre is subjected to an aial force that varies from kn in compression to 6 kn in tension. he WO MARK tensile ( S u ), yield ( S y ) and corrected endurance ( S e ) strengths of the steel material are 6 MPa, MPa and MPa respectively. he factor of safety against fatigue endurance as per Soderberg s criterion is (A).6 (B).7 (C). (D). Sol. 7 Option (A) is correct. Given : S u or s u 6 MPa, S y or s y MPa, S e or 6 kn (ension), F min kn (Compression) F ma Maimum stress, Minimum stress, s ma s min s e MPa, d mm F ma 6 # A p ( ) 6.7 MPa F min # 6.6 MPa A p ( ) #

28 GAE SOLVED PAPER ME 9 Mean stress, 8.9 MPa Variable stress,. MPa s m s v s s ma ma smin smin 6. 7 (6. 6) From the Soderberg s criterion, FS.. So, FS.. s m sy s s v e FS Q. 8 An automotive engine weighing kg is supported on four springs with linear characteristics. Each of the front two springs have a stiffness of 6 MN/ m while WO MARK the stiffness of each rear spring is MN/ m. he engine speed (in rpm), at which resonance is likely to occur, is (A) 6 (B) (C) (D) 9 Sol. 8 Option (A) is correct. Given k k 6 MN/ m, k k MN/ m, m kg Here, k & k are the front two springs or k and k are the rear two springs. hese springs are parallel, So equivalent stiffness k eq k k k k MN/ m We know at resonance w w m k n pn k eq N Engine speed in rpm 6 m N 6 keq p m 6 96 # 6 p 6 p # # 6. 6 rpm

29 GAE SOLVED PAPER ME 9 Q. 9 A vehicle suspension system consists of a spring and a damper. he stiffness of the spring is.6 kn/ m and the damping constant of the damper is Ns/ m. If WO MARK the mass is kg, then the damping factor (d ) and damped natural frequency ( f n ), respectively, are (A).7 and.9 Hz (B).7 and 7.8 Hz (C).666 and. Hz (D).666 and 8. Hz Sol. 9 Option (A) is correct. Given k.6 kn/ m, c Ns/ m, m kg We know that, Natural Frequency w n m k. 6 # 8.8 rad/ sec...(i) And damping factor is given by, d or e c c c c km. 7 #. 6 Damping Natural frequency, w d ewn #. 6 # # d p f d ewn fd w p n f d w p # e 8. 8 (. 7). #.9 Hz # Q. A frame of two arms of equal length L is shown in the adjacent figure. he fleural rigidity of each arm of the frame is EI. he vertical deflection at the WO MARK point of application of load P is (A) PL EI (C) PL EI (B) (D) PL EI PL EI

30 GAE SOLVED PAPER ME 9 Sol. Option (D) is correct. We have to solve this by Castigliano s theorem. We have to take sections XX and YY along the arm BC and AB respectively and find the total strain energy. So, Strain energy in arm BC is, L L U M BC d EI P # # ^ h d M P EI # Integrating the equation and putting the limits, we get L U P PL BC EI : D 6EI Similarly for arm AB, we have L M L U AB # dy PL dy EI # My P L EI # y Integrating the above equation and putting the limit, we get U PL AB EI So, total strain energy stored in both the arms is, U U U AB BC PL PL PL EI 6EI EI From the Castigliano s theorem, vertical deflection at point A is, d A du dp d A d PL PL dp b EI l EI Q. A uniform rigid rod of mass M and length L is hinged at one end as shown in the adjacent figure. A force P is applied at a distance of L/ from the hinge so that WO MARK the rod swings to the right. he reaction at the hinge is

31 GAE SOLVED PAPER ME 9 Sol. (A) P (B) (C) P/ (D) P/ Option (B) is correct. When rod swings to the right, linear acceleration a and angular acceleration a comes in action. Centre of gravity (G ) acting at the midpoint of the rod. Let R be the reaction at the hinge. Linear acceleration a r. a L # a a L a...(i) And about point G, for rotational motion / M G I G # a R L P L b ML a l b6l b L l From equation (i) R P Ma a M R M P...(ii) By equilibrium of forces in normal direction to the rod / F m P R Ma M R P b M M l From equation (ii) P R R P

32 GAE SOLVED PAPER ME 9 & R So, reaction at the hinge is zero. Q. Match the approaches given below to perform stated kinematics/dynamics analysis of machine. WO MARK Analysis Approach P. Continuous relative rotation. D Alembert s principle Q. Velocity and acceleration. Grubler s criterion R. Mobility. Grashoff s law S. Dynamicstatic analysis. Kennedy s theorem (A) P, Q, R, S (B) P, Q, R, S (C) P, Q, R, S (D) P, Q, R, S Sol. Option (B) is correct. Analysis Approach P. Continuous relative rotation. Grashoff law Q. Velocity and Acceleration. Kennedy s heorem R. Mobility. Grubler s Criterion S. Dynamicstatic Analysis. D Alembert s Principle So, correct pairs are P, Q, R, S Q. A company uses units of an item annually. Delivery lead time is 8 days. he reorder point (in number of units) to achieve optimum inventory is WO MARK (A) 7 (B) 8 (C) 6 (D) 6 Sol. Option (C) is correct. In figure, Let the reorder quantity be Now from the similar triangles ROP Reorder point L Lead ime 8 days otal ime 6 days q stock level units

33 GAE SOLVED PAPER ME 9 DABC & DBDE q L & 6 8 Alternate method Given, Demand in a year D 8 6Units 6 # Units Lead time 8 days Now, Number of orders to be placed in a year Number. of days in a year N Lead ime 6 orders 8 Now, quantity ordered each time or reorder point. Demand in a years Q Number of orders 6 Units 6 8 Q. Consider the following Linear Programming Problem (LPP): Maimize Z WO MARK Subject to # # 6 # 8 $, $ (A) he LPP has a unique optimal solution (B) he LPP is infeasible. (C) he LPP is unbounded. (D) he LPP has multiple optimal solutions. Sol. Option (D) is correct. Given Objective function and constraints are Z ma #...(i) # 6...(ii) # 8...(iii) $ $ Plot the graph from the given constraints and find the common area.

34 GAE SOLVED PAPER ME 9 Now, we find the point of intersection E & F. For E, 8 & (iii)) So, 6 8 For F, 8 So, Hence, Now at point E(,) 6 At point F(,) # 8 E(,) 6 or F(,) (E is the intersection point of equation. (ii) Z # # 6 8 Z # # 8 he objective function and the constraint (represent by equation (iii)) are equal. Hence, the objective function will have the multiple solutions as at point E & F, the value of objective function ( Z ) is same. Q. Si jobs arrived in a sequence as given below: WO MARK Jobs Processing ime (days) I II 9 III IV V 6 VI 8 Average flow time (in days) for the above jobs using Shortest Processing time rule is

35 GAE SOLVED PAPER ME 9 (A).8 (B).6 (C). (D) 9. Sol. Option (A) is correct. In shortest processing time rule, we have to arrange the jobs in the increasing order of their processing time and find total flow time. So, job sequencing are I III V VI II IV Jobs Processing ime (days) Flow time (days) I III 9 V VI 8 8 II 9 9 IV Now otal flow time 9 Average flow time otal flow time Number of jobs average 6.8 days Q. 6 Minimum shear strain in orthogonal turning with a cutting tool of zero rake angle is WO MARK (A). (B). (C). (D). Sol. 6 Option (D) is correct. Given : a c We know that, shear strain s cot f tan( f a) a c So, s cot f tan f...(i) For minimum value of shear strain differentiate equation (i) w.r.t. f ds d ( cot f tan f) cosec f sec f...(ii) df df Again differentiate w.r.t. to f, d s cosec f# ( cosec fcot f) sec f# ( sec ftan f) df cosec fcot f sec ftan f...(iii) Using the principle of minima maima and put ds in equation(ii) d f cosec sec f sin f cos f cos f sin f sin f# cos f

36 GAE SOLVED PAPER ME 9 From equation (iii), at f cos f sin f cos f f cos () p f p p e d s d o cosec p cot p sec tan p p # f f p e d s d o # # # # 8 f f p d s e d o > f f p herefore it is minimum at f p, so from equation (i), () s min cot p tan p Q. 7 Electrochemical machining is performed to remove material from an iron surface of mm # mm under the following conditions : WO MARK Inter electrode gap. mm Supply voltage (DC) V Specific resistance of electrolyte W cm Atomic weight of Iron.8 Valency of Iron Faraday s constant 96 Coulombs he material removal rate (in g/s) is (A).7 (B).7 (C).7 (D) 7. Sol. 7 Option (A) is correct. Given : L. mm, A mm # mm mm, V Volt r Wcm # W mm, Z.8, v, F 96 Coulombs We know that Resistance is given by the relation rl R # #.. W A # I Rate of mass removal mo So, mo V A R. F I # Z v #.7 g/ sec

37 GAE SOLVED PAPER ME 9 Q. 8 Match the following: WO MARK NC code Definition P. M. Absolute coordinate system Q. G. Dwell R. G. Spindle stop S. G9. Linear interpolation (A) P, Q, R, S (B) P, Q, R, S (C) P, Q, R, S (D) P, Q, R, S Sol. 8 Option (C) is correct. NC code Definition P. M. Spindle stop Q. G. Linear interpolation R. G. Dwell S. G9. Absolute coordinate system So, correct pairs are, P, Q, R, S Q. 9 What are the upper and lower limits of the shaft represented by 6 f 8? Use the following data : WO MARK Diameter 6 lies in the diameter step of 8 mm. / Fundamental tolerance unit, i in m m.d.d Where D is the representative size in mm; olerance value for I8 i, Fundamental deviation for f shaft.d. (A) Lower limit 9.9 mm, Upper limit 9.97 mm (B) Lower limit 9.9 mm, Upper limit 6. mm (C) Lower limit 9.97 mm, Upper limit 6.6 mm (D) Lower limit 6. mm, Upper limit 6.6 mm Sol. 9 Option (A) is correct. Since diameter 6 lies in the diameter step of 8 mm, therefore the geometric mean diameter. D # mm Fundamental tolerance unit. / i. D. D /. ( 6. 6). # 6. 6

38 GAE SOLVED PAPER ME 9.86 mm.86 mm Standard tolerance for the hole of grades 8 (I8) i # mm Fundamental deviation for f shaft e f.d.. ( 6. 6 ).. mm. mm Upper limit of shaft Basic size Fundamental deviation mm Lower limit of shaft Upper limit olerance Q. Match the items in Column I and Column II. WO MARK Sol. Column I Column II P. Metallic Chills. Support for the core Q. Metallic Chaplets. Reservoir of the molten metal R. Riser. Control cooling of critical sections S. Eothermic Padding. Progressive solidification (A) P, Q, R, S (B) P, Q, R, S (C) P, Q, R, S (D) P, Q, R, S Option (D) is correct. Column I Column II P. Metallic Chills. Progressive solidification Q. Metallic Chaplets. Support for the core R. Riser. Reservoir of the molten metal S. Eothermic Padding. Control cooling of critical sections So, correct pairs are P, Q, R, S Common Data for Question and : he inlet and the outlet conditions of steam for an adiabatic steam turbine are as indicated in the figure. he notations are as usually followed.

39 GAE SOLVED PAPER ME 9 Q. If mass rate of steam through the turbine is kg/ s, the power output of the WO MARK turbine (in MW) is (A).7 (B).9 (C) 68. (D) Sol. Option (A) is correct. Given : h kj/ kg, V 6 m/ sec, z m p mpa, mo dm kg/ sec dt It is a adiabatic process, So dq Apply steady flow energy equation [ SFEE....] at the inlet and outlet section of steam turbine, h V dq zg h V zg dw dm dm dq dq So dm And h V zg h V zg dw dm dw ( h h ) V V ( z z ) g dm b l ( 6) ( ) ( 6) # ; ( 6)9.8 E dw Jkg / 67.8 kj/ kg dm Power output of turbine P Mass flow rate dw # dm # 67.8 # m o kg/ sec P.7 MJ/ sec.7 MW Q. Assume the above turbine to be part of a simple Rankine cycle. he density of water at the inlet to the pump is kg/ m. Ignoring kinetic and potential WO MARK energy effects, the specific work (in kj/ kg) supplied to the pump is (A).9 (B). (C).9 (D). Sol. Option (C) is correct. Given : r kg/ m Here given that ignoring kinetic & potential energy effects, So in the steady flow energy equation the terms V /, Z g are equal to zero and dq is also zero for adiabatic process. S.F.E.E. is reduces to, dw h h p Here, W dm p represents the pump work where h Enthalpy at the inlet of pump and h Enthalpy at the outlet of the pump. dw p h h dh...(i) dm

40 GAE SOLVED PAPER ME 9 For reversible adiabatic compression, dq dh ndp ( dq ) dh ndp...(ii) From equation (i) and (ii), we get dw p ndp dm dw p ( p p ) v dm r r dw ( 7)kPa p dm 9 kpa.9 kpa Common Data for Questions and : Radiative heat transfer is intended between the inner surfaces of two very large isothermal parallel metal plates. While the upper plate (designated as plate ) is a black surface and is the warmer one being maintained at 77c C, the lower plate (plate ) is a diffuse and gray surface with an emissivity of.7 and is kept at 7c C. Assume that the surfaces are sufficiently large to form a twosurface enclosure and steadystate conditions to eits. StefanBoltzmann constant is given as.67 # 8 W/m K Q. he irradiation (in kw/ m ) for the plate (plate ) is (A). WO MARK Sol. (B).6 (C) 7. (D) 9. Option (D) is correct. 8 Given : s b.67 # W/ m K, (7 7) K K (77 7) K K Let, a " he absorptivity of the gray surface E " he radiant energy of black surface E " he radiant energy of gray surface Now, Plate emits radiant energy E which strikes the plate. From it a part

41 GAE SOLVED PAPER ME 9 a E absorbed by the plate & the remainder ( E ae) is reflected back to the plate. On reaching plate, all the part of this energy is absorbed by the plate, because the absorptivity of plate is equal to one (it is a black surface). Irradiation denotes the total radiant energy incident upon a surface per unit time per unit area. Energy leaving from the plate is, E E ( a) E...(i) Hence, E is the energy emitted by plate. 8 E es b.7 #.67 # # () E es b #. 67 # # 6 # 8.6 W/ m And fraction of energy reflected from surface is, ( a)e ( as ). 67 # 8 (. 7) #( ) 7 W/ m Now, otal energy incident upon plate is, E E ( a) E W/ m 9.9 kw/ m, 9. kw/ m Q. If plate is also diffuse and gray surface with an emissivity value of.8, the net radiation heat echange (in kw/m ) between plate and plate is WO MARK (A) 7. (B) 9. (C). (D).7 Sol. Option (D) is correct. Given : e 8., e 7. As both the plates are gray, the net heat flow from plate to plate per unit time is given by, Q ee sb ( ) s e e b( ) ee e e. [( ) ( ) ] # 67 # # # 6. 6 W/ m.7 kw/ m Common Data for Questions and 6: Consider the following PER network: he optimistic time, most likely time and pessimistic time of all the activities are given in the table below:

42 GAE SOLVED PAPER ME 9 Activity Optimistic time (days) Most likely time (days) Pessimistic time (days) Q. he critical path duration of the network (in days) is (A) WO MARK (B) (C) 7 (D) 8 Sol. Option (D) is correct. Make the table and calculate the ecepted time and variance for each activity Activity Optimistic time (days) t o Most likely time (days) t m Pessimistic time (days) t p Epected ime (days) to tm tp te 6 Variance tp to s b 6 l 8 6 b 6 l b 6 l b 6 l b 6 l b 6 l b 6 l b 6 l b l 9 Now, the paths of the network & their durations are given below in tables.

43 GAE SOLVED PAPER ME 9 Paths Epected ime duration (in days) i Path ii Path iii Path 7 6 Since path 67 has the longest duration, it is the critical path of the network and shown by dotted line. Hence, he epected duration of the critical path is 8 days. Q. 6 he standard deviation of the critical path is (A). (B). WO MARK (C).77 (D).66 Sol. 6 Option (C) is correct. he critical path is 67 Variance along this critical path is, We know, s s s s6 s Standard deviation Variance ( s ) he most appropriate answer is Statement for Linked Answer Questions 7 and 8 : In a machining eperiment, tool life was found to vary with the cutting speed in the following manner : Cutting speed (m/min) ool life (minutes) Q. 7 he eponent ( n ) and constant ( K ) of the aylor s tool life equation are (A) n. and K (B) n and K 86 WO MARK (C) n and K.7 (D) n. and K. Sol. 7 Option (A) is correct. Given : V 6 m/ min, 8 min, V 9 m/ min, 6 min. From the aylor s tool life Equation For case (I), For case (II), V n Constant ( K) V n K 6 # ( 8) n K...(i) V n K

44 GAE SOLVED PAPER ME 9 9 # ( 6) n K...(ii) By dividing equation (i) by equation (ii), n 6 # ( 8) n K 9 # ( 6) K b6 8 l n 9 n b l 9 6 b l aking (log) both the sides, n log 9 b l log b l n #.. 76 n. Substitute n. in equation (i), we get K 6 # ( 8 ). So, n. and K Q. 8 What is the percentage increase in tool life when the cutting speed is halved? (A) % (B) % WO MARK (C) % (D) % Sol. 8 Option (C) is correct. ake, n. {from previous part} From aylor s tool life equation V n C C V. V...(i) Given that cutting speed is halved V V & V V Now, from equation (i), V V & Now, percentage increase in tool life is given by # # # %

45 GAE SOLVED PAPER ME 9 Statement for Linked Answer Questions 9 and 6 A c full depth involute spur pinion of mm module and teeth is to transmit kw at 96 rpm. Its face width is mm. Q. 9 he tangential force transmitted (in N) is (A) (B) 6 WO MARK (C) 776 (D) Sol. 9 Option (A) is correct. Given : m mm, Z, P kw # W N 96 rpm, b mm, f c Pitch circle diameter, D mz # 8mm angential Force is given by, F...(i) r Power transmitted, P pn & 6 6 pn P hen F 6 pn P # r r Pitch circle radius 6 # #. 96 # # # #.6 N N Q. 6 Given that the tooth geometry factor is. and the combined effect of dynamic load and allied factors intensifying the stress is.; the minimum allowable stress WO MARK (in MPa) for the gear material is (A). (B) 66. (C). (D) 7. Sol. 6 Option (B) is correct. From Lewis equation Fp d s F b p by b# y# m p m m d p c p p s b # #. # # s b MPa Minimum allowable (working stress) s W s b# Cv #. 66. MPa s W

46 GAE SOLVED PAPER ME 9 Answer Sheet. (A). (B). (A) 7. (A) 9. (A). (C). (C) 6. (A) 8. (A). (D). (C). (B) 7. (B) 9. (A). (A). (D) 6. (D) 8. (B). (D). (C). (A) 7. (A) 9. (D). (B). (D) 6. (B) 8. (A). (C). (B). (D) 7. (A) 9. (C). (D). (C). (D) 8. (A). (D). (A). (D) 6. (C) 9. (B). (C). (C). (A) 7. (A). (C). (A). (A) 6. (D) 8. (C). (C). (B). (B) 7. (A) 9. (A). (D). (A) 6. (A) 8. (C) 6. (B)