GATE Solved Paper ME 2003 PDF

Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 2 Q. 1 lim sin x is equal to x"0 x GATE ME 2003 ONE MARK (A) 0...

Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 2 Q. 1 lim sin x is equal to x"0 x GATE ME 2003 ONE MARK (A) 0 (B) 3 (C) 1 (D) - 1 Sol. 1 Option (A) is correct 2 2 Let, f (x) = lim sin x = lim sin x # x x"0 x x"0 x x = lim b sin x 2 lim sin x = 1 x l# x x"0 x"0 x Alternate: om = (1) 2 # 0 = 0 c 2 Let f (x) = lim sin x 0 : 0 formD s. x"0 x f (x) = lim 2 sin x cos x Apply L-Hospital rule ta x"0 1 = lim sin 2x = sin 0 = 0 a x"0 1 1 ld Q. 2 The accuracy of Simpson’s rule quadrature for a step size h is GATE ME 2003 (A) O (h2) i (B) O (h3) iv ONE MARK (C) O (h 4) (D) O (h5).C Sol. 2 Option (D) is correct. Accuracy of Simpson’s rule quadrature is O (h5) Q. 3 GATE ME 2003 © ww For the matrix > 4 1 1 4H the eigen values are ONE MARK w (A) 3 and - 3 (C) 3 and 5 (B) - 3 and - 5 (D) 5 and 0 Sol. 3 Option (C) is correct. 4 1 A => 1 4H Let, The characteristic equation for the eigen value is given by, 1 0 I = Identity matrix > 0 1H A - lI = 0 4 1 1 0 >1 4H - l >0 1H = 0 4-l 1 =0 1 4-l (4 - l) (4 - l) - 1 = 0 (4 - l) 2 - 1 = 0 Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 l2 - 8l + 15 = 0 On solving above equation, we get l = 5, 3 Q. 4 The second moment of a circular area about the diameter is given by (D is the GATE ME 2003 diameter). 4 4 (A) pD (B) pD ONE MARK 4 16 4 4 (C) pD (D) pD 32 64 Sol. 4 Option (D) is correct. We know that, moment of inertia is defined as the second moment of a plane area about an axis perpendicular to the area. Polar moment of inertia perpendicular to the plane of paper, 4 J or IP = pD 32 By the “perpendicular axis” theorem, IXX + IYY = IP For circular section IXX = IYY 2IXX = IP 4 IXX = IP = pD = IYY 2 64 Q. 5 A concentrated load of P acts on a simply supported beam of span L at a distance GATE ME 2003 L/3 from the left support. The bending moment at the point of application of the ONE MARK load is given by (A) PL (B) 2PL 3 3 (C) PL (D) 2PL 9 9 Sol. 5 Option (D) is correct. We know that, the simplest form of the simply supported beams is the beam supported on rollers at ends. The simply supported beam and the FBD shown in the Figure. Where, RA & RB are the reactions acting at the ends of the beam. In equilibrium condition of forces, P = RA + RB...(i) Taking the moment about point A, RB # L = P # L 3 RB = P 3 From equation (i), R A = P - R B = P - P = 2P 3 3 Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Now bending moment at the point of application of the load M = RA # L = 2P # L = 2PL 3 3 3 9 Or, M = RB # 2L = 2PL 3 9 Q. 6 Two identical circular rods of same diameter and same length are subjected to GATE ME 2003 same magnitude of axial tensile force. One of the rod is made out of mild steel ONE MARK having the modulus of elasticity of 206 GPa. The other rod is made out of cast iron having the modulus of elasticity of 100 GPa. Assume both the materials to be homogeneous and isotropic and the axial force causes the same amount of uniform stress in both the rods. The stresses developed are within the proportional limit of the respective materials. Which of the following observations is correct ? (A) Both rods elongate by the same amount (B) Mild steel rod elongates more than the cast iron rod (C) Cast iron rod elongates more than the mild steel rods (D) As the stresses are equal strains are also equal in both the rods Sol. 6 Option (C) is correct. Given : Ls = Li , Es = 206 GPa , Ei = 100 GPa , Ps = Pi , Ds = Di , & As = Ai om Where subscript s is for steel and i is for iron rod. We know that elongation is given by,.c DL = PL as AE Now, for steel or iron rod at DLs = Ps Ls DL i Ai Ei = Ei As Es # Pi Li Es ld Substitute the values, we get i DLs = 100 = 0.485 < 1 iv DL i 206 Or, D L s < D L i & D L i > DL s.C So, cast iron rod elongates more than the mild steel rod. © ww Q. 7 The beams, one having square cross section and another circular cross-section, GATE ME 2003 are subjected to the same amount of bending moment. If the cross sectional area ONE MARK as well as the material of both the beams are same then w (A) maximum bending stress developed in both the beams is same (B) the circular beam experience more bending stress than the square one (C) the square beam experience more bending stress than the circular one (D) as the material is same, both the beams will experience same deformation. Sol. 7 Option (B) is correct. Let, a = Side of square cross-section d = diameter of circular cross-section Using subscripts for the square and c for the circular cross section. Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Given : Ms = Mc Ac = As So, p d 2 = a2...(i) 4 From the bending equation, M =s=E & s= M # y I y R I Where, y = Distance from the neutral axis to the external fibre. s =Bending stress For square cross-section bending stress, ss = M4s # a = 6M3 s...(ii) a 2 a 12 And for circular cross-section, sc = Mc # d = 32M c...(iii) p d4 2 d3 om 64 On dividing equation (iii) by equation (ii), we get.csc = 32Mc ss # a3 = 16 a3 6Ms 3 d3 Mc = Ms...(iv) s 3 d a From equation (i), t p 2 3/2 a 4 d k = (a ) = a 2 3/2 3 da a3 = p 3/2 = 0.695 d3 a 4 k il iv Substitute this value in equation (iv), we get sc = 16 0.695 = 3.706.C ss 3 # sc > 1 & s > s © ww ss c s So, Circular beam experience more bending stress than the square section. w Q. 8 The mechanism used in a shaping machine is GATE ME 2003 (A) a closed 4-bar chain having 4 revolute pairs ONE MARK (B) a closed 6-bar chain having 6 revolute pairs (C) a closed 4-bar chain having 2 revolute and 2 sliding pairs (D) an inversion of the single slider-crank chain Sol. 8 Option (D) is correct. A single slider crank chain is a modification of the basic four bar chain. It is find, that four inversions of a single slider crank chain are possible. From these four inversions, crank and slotted lever quick return motion mechanism is used in shaping machines, slotting machines and in rotary internal combustion engines. Q. 9 The lengths of the links of a 4-bar linkage with revolute pairs are p, q, r, and s GATE ME 2003 units. given that p < q < r < s. Which of these links should be the fixed one, for ONE MARK obtaining a “double crank” mechanism ? (A) link of length p (B) link of length q (C) link of length r (D) link of length s Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Sol. 9 Option (A) is correct. Given p < q < r < s “Double crank” mechanism occurs, when the shortest link is fixed. From the given pairs p is the shortest link. So, link of length p should be fixed. Q. 10 Consider the arrangement shown in the figure below where J is the combined GATE ME 2003 polar mass moment of inertia of the disc and the shafts. k1, k2, k 3 are the torsional ONE MARK stiffness of the respective shafts. The natural frequency of torsional oscillation of the disc is given by k1 + k 2 + k 3 k1 k 2 + k 2 k 3 + k 3 k1 m (A) (B) J J (k1 + k2) (C) co k1 + k 2 + k 3 (D) k1 k 2 + k 2 k 3 + k 3 k1 Sol. 10 s. J (k1 k2 + k2 k 3 + k 3 k1) Option (B) is correct. J (k2 + k 3) ta Here k1 & k2 are in series combination & k 3 is in parallel combination with this a series combination. ld So, keq = k1 # k2 + k 3 = k1 k2 + k2 k 3 + k1 k 3 k1 + k 2 k1 + k 2 i iv keq Natural frequency of the torsional oscillation of the disc, wn = J Substitute the value of keq , we get.C wn = k1 k2 + k2 k 3 + k1 k 3 J (k1 + k2) Q. 11 © ww Maximum shear stress developed on the surface of a solid circular shaft under pure torsion is 240 MPa. If the shaft diameter is doubled then the maximum w GATE ME 2003 ONE MARK shear stress developed corresponding to the same torque will be (A) 120 MPa (B) 60 MPa (C) 30 MPa (D) 15 MPa Sol. 11 Option (C) is correct. Given : t1 = tmax = 240 MPa Let, diameter of solid shaft d1 = d , And Final diameter d2 = 2d (Given) From the Torsional Formula, T = t &T= t J J r r # Where, J = polar moment of inertia Given that torque is same, So, t1 J = t2 J r1 # 1 r2 # 2 2t1 J = 2t2 J J = p d4 d1 # 1 d2 # 2 32 Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 t1 p d 4 = t2 p d4 d1 # 32 1 d2 # 32 2 3 t1 # d 13 = t2 # d 23 & t2 = t1 # d 13 d2 Substitute the values, we get t2 = 240 # b d l = 240 # 1 = 30 MPa 3 2d 8 Common Data For Q.Alternate method From the Torsional Formula, t =T r r = d & J = p d4 J 2 32 So, maximum shear stress, tmax = 16T3 =240 MPa pd m Given Torque is same & Shaft diameter is doubled then, tlmax = 16T 3 = 16T3 = tmax = 240 = 30 MPa co p (2d) 8p d 8 8 Q. 12 GATE ME 2003 represent s. A wire rope is designated as 6 # 19 standard hoisting. The numbers 6 # 19 a ONE MARK (A) diameter in millimeter # length in meter at (B) diameter in centimeter # length in meter (C) number of strands # numbers of wires in each strand ild (D) number of wires in each strand # number of strands iv Option (C) is correct. Sol. 12 The wire ropes are designated by the number of strands multiplied by the number.C of wires in each strand. Therefore, © ww 6 # 19 = Number of strands # Number of wires in each strand. Q. 13 A cylindrical body of cross-sectional area A, height H and density rs , is immersed to a depth h in a liquid of density r, and tied to the bottom with a string. The w GATE ME 2003 ONE MARK tension in the string is (A) rghA (B) (rs - r) ghA (C) (r - rs) ghA (D) (rh - rs H) gA Sol. 13 Option (D) is correct. Given : Cross section area of body = A Height of body = H Density of body = rs Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Density of liquid = r Tension in the string = T We have to make the FBD of the block. B = Buoyancy force From the principal of buoyancy, Downward force = Buoyancy force m = rn T + mg = rhAg T + rs ng = rhAg & n = A#H T + rs AHg = rhAg Q. 14 om T = rhAg - rs AHg = Ag (rh - rs H) A 2 kW, 40 liters water heater is switched on for 20 minutes. The heat capacity c p GATE ME 2003.c for water is 4.2 kJ/kgK. Assuming all the electrical energy has gone into heating s ONE MARK the water, increase of the water temperature in degree centigrade is (A) 2.7 (C) 14.3 ta (B) 4.0 (D) 25.25 a ld Sol. 14 Option (C) is correct. i Given : p = 2 kW = 2 # 103 W , t = 20 minutes = 20 # 60 sec , c p = 4.2 kJ/kgK iv Heat supplied, Q = Power # Time = 2 # 103 # 20 # 60 = 24 # 105 Joule.C And Specific heat at constant pressure, Q = mc p DT © ww DT = 24 # 105 = 24 # 100 = 14.3c C 40 # 4.2 # 1000 40 # 4.2 Q. 15 An industrial heat pump operates between the temperatures of 27c C and - 13c C. GATE ME 2003 ONE MARK w The rates of heat addition and heat rejection are 750 W and 1000 W, respectively. The COP for the heat pump is (A) 7.5 (B) 6.5 (C) 4.0 (D) 3.0 Sol. 15 Option (C) is correct. Given : T1 = 27c C = (27 + 273) K = 300 K , T2 =- 13c C = (- 13 + 273) K = 260 K Q1 = 1000 W , Q2 = 750 W Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Q1 1000 So, (COP) H.P. = = =4 Q1 - Q2 1000 - 750 Common Data For Q.Alternate Method From energy balance om Win + Q2 = Q1.c Win = Q1 - Q2 = 1000 - 750 = 250 W And as Q (COP) H.P. = Desired effect = 1 = 1000 = 4 t Win Win 250 a A plate having 10 cm area each side is hanging in the middle of a room of 100 m2 2 Q. 16 d GATE ME 2003 total surface area. The plate temperature and emissivity are respectively 800 K ONE MARK il and 0.6. The temperature and emissivity values for the surfaces of the room are iv 300 K and 0.3 respectively. Boltzmann’s constant s = 5.67 # 10-8 W/m2 K 4. The total heat loss from the two surfaces of the plate is.C (A) 13.66 W (B) 27.32 W (C) 27.87 W (D) 13.66 MW Sol. 16 © ww Option (B) is correct. w Given, for plate : A1 = 10 cm2 = 10 # (10-2) 2 m2 = 10-3 m2 , T1 = 800 K , e1 = 0.6 For Room : A2 = 100 m2 , T2 = 300 K , e2 = 0.3 And s = 5.67 # 10-8 W/m2 K 4 Total heat loss from one surface of the plate is given by, (Q12) = Eb1 - Eb2 (1 - e1) (1 - e2) + 1 + A1 e1 A1 F12 A2 e2 Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 If small body is enclosed by a large enclosure, then F12 = 1 and from Stefan’s Boltzman law Eb = sT 4. So we get s (T 14 - T 24) (Q12) = 1 - e1 + 1 + 1 - e2 A1 e1 A1 A2 e2 5.67 # 10-8 [(800) 4 - (300) 4] = 1 - 0.6 + 1 + 1 - 0.3 10-3 # 0.6 10-3 100 # 0.3 = 22.765 # 103 = 13.66 W 666.66 + 1000 + 0.0233 Q12 is the heat loss by one surface of the plate. So, heat loss from the two surfaces is given by, Qnet = 2 # Q12 = 2 # 13.66 = 27.32 W Q. 17 For air with a relative humidity of 80% GATE ME 2003 (A) the dry bulb temperature is less than the wet bulb temperature ONE MARK (B) the dew point temperature is less than wet bulb temperature (C) the dew point and wet bulb temperature are equal m (D) the dry bulb and dew point temperature are equal o Sol. 17.c Option (B) is correct. We know that for saturated air, the relative humidity is 100% and the dry bulb s temperature, wet bulb temperature and dew point temperature is same. But a temperature. at when air is not saturated, dew point temperature is always less than the wet bulb ld DPT < WBT Q. 18 i For a spark ignition engine, the equivalence ratio (f) of mixture entering the iv GATE ME 2003 combustion chamber has values ONE MARK (A) f < 1 for idling and f > 1 for peak power conditions.C (B) f > 1 for both idling and peak power conditions © ww (C) f > 1 for idling and f < 1 for peak power conditions (D) f < 1 for both idling and peak power conditions w Sol. 18 Option (B) is correct. Equivalence Ratio or Fuel Air Ratio b F l A f = Actual Fuel - Air ratio stoichiometric Fuel air Ratio F bAl = actual F bAl stoichiometric If f = 1, & stoichiometric (Chemically correct) Mixture. If f > 1, & rich mixture. If f < 1, & lean mixture. Now, we can see from these three conditions that f > 1, for both idling & peak power conditions, so rich mixture is necessary. Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Q. 19 A diesel engine is usually more efficient than a spark ignition engine because GATE ME 2003 (A) diesel being a heavier hydrocarbon, releases more heat per kg than gasoline ONE MARK (B) the air standard efficiency of diesel cycle is higher than the Otto cycle, at a fixed compression ratio (C) the compression ratio of a diesel engine is higher than that of an SI engine (D) self ignition temperature of diesel is higher than that of gasoline Sol. 19 Option (C) is correct. The compression ratio of diesel engine ranges between 14 to 25 where as for S.I, engine between 6 to 12. Diesel Engine gives more power but efficiency of diesel engine is less than compare to the S.I. engine for same compression ratio. Q. 20 In Ranking cycle, regeneration results in higher efficiency because GATE ME 2003 (A) pressure inside the boiler increases ONE MARK (B) heat is added before steam enters the low pressure turbine om (C) average temperature of heat addition in the boiler increases (D) total work delivered by the turbine increases Sol. 20.c Option (C) is correct. as at ild iv.C © ww Fig : T - s curve of simple Rankine cycle From the observation of the T - s diagram of the rankine cycle, it reveals that heat is transferred to the working fluid during process 2 - 2' at a relatively low w temperature. This lowers the average heat addition temperature and thus the cycle efficiency. To remove this remedy, we look for the ways to raise the temperature of the liquid leaving the pump (called the feed water ) before it enters the boiler. One possibility is to transfer heat to the feed water from the expanding steam in a counter flow heat exchanger built into the turbine, that is, to use regeneration. A practical regeneration process in steam power plant is accomplished by extracting steam from the turbine at various points. This steam is used to heat the feed water and the device where the feed water is heated by regeneration is called feed water heater. So, regeneration improves cycle efficiency by increasing the average temperature of heat addition in the boiler. Q. 21 Considering the variation of static pressure and absolute velocity in an impulse GATE ME 2003 steam turbine, across one row of moving blades ONE MARK (A) both pressure and velocity decreases (B) pressure decreases but velocity increases (C) pressure remains constant, while velocity increases (D) pressure remains constant, while velocity decreases Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Sol. 21 Option (D) is correct. om.c Easily shows that the diagram that static pressure remains constant, while s velocity decreases. a Q. 22 GATE ME 2003 order is at During heat treatment of steel, the hardness of various structures in increasing ld ONE MARK (A) martensite, fine pearlite, coarse pearlite, spherodite i (B) fine pearlite, Martensite, spherodite, coarse pearlite iv (C) martensite, coarse pearlite, fine pearlite, spherodite.C (D) spherodite, coarse pearlite, fine pearlite, martensite Option (D) is correct. © ww Sol. 22 Steel can be cooled from the high temperature region at a rate so high that the austenite does not have sufficient time to decompose into sorbite or troostite. In w this case the austenite is transformed into martensite. Martensite is ferromagnetic, very hard & brittle. So hardness is increasing in the order, Spherodite " Coarse Pearlite " Fine Pearlite " Martensite Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Q. 23 Hardness of green sand mould increases with GATE ME 2003 (A) increase in moisture content beyond 6 percent ONE MARK (B) increase in permeability (C) decrease in permeability (D) increase in both moisture content and permeability Sol. 23 Option (C) is correct. Permeability or porosity of the moulding sand is the measure of its ability to permit air to flow through it. So, hardness of green sand mould increases by restricted the air permitted in the sand i.e. decrease its permeability. Q. 24 In Oxyacetylene gas welding, temperature at the inner cone of the flame is around GATE ME 2003 (A) 3500c C (B) 3200c C ONE MARK (C) 2900c C (D) 2550c C Sol. 24 om Option (B) is correct. In OAW, Acetylene (C 2 H 2 ) produces higher temperature (in the range of 3200c C c )than other gases, (which produce a flame temperature in the range of 2500c C ). s because it contains more available carbon and releases heat when its components a (C & H) dissociate to combine with O 2 and burn. Q. 25 at Cold working of steel is defined as working (A) at its recrystallisation temperature d GATE ME 2003 l ONE MARK (B) above its recrystallisation temperature i iv (C) below its recrystallisation temperature (D) at two thirds of the melting temperature of the metal Sol. 25.C Option (C) is correct. © ww Cold forming or cold working can be defined as the plastic deforming of metals and alloys under conditions of temperature and strain rate. Theoretically, the working temperature for cold working is below the w recrystallization temperature of the metal/alloy (which is about one-half the absolute melting temperature.) Q. 26 Quality screw threads are produced by GATE ME 2003 (A) thread milling ONE MARK (B) thread chasing (C) thread cutting with single point tool (D) thread casting Sol. 26 Option (D) is correct. Quality screw threads are produced by only thread casting. Quality screw threads are made by die-casting and permanent mould casting are very accurate and of high finish, if properly made. Q. 27 As tool and work are not in contact in EDM process GATE ME 2003 (A) no relative motion occurs between them ONE MARK (B) no wear of tool occurs (C) no power is consumed during metal cutting (D) no force between tool and work occurs Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Sol. 27 Option (D) is correct. In EDM, the thermal energy is employed to melt and vaporize tiny particles of work-material by concentrating the heat energy on a small area of the work-piece. A powerful spark, such as at the terminals of an automobile battery, will cause pitting or erosion of the metal at both anode & cathode. No force occurs between tool & work. Q. 28 The dimensional limits on a shaft of 25h7 are GATE ME 2003 (A) 25.000, 25.021 mm (B) 25.000, 24.979 mm ONE MARK (C) 25.000, 25.007 mm (D) 25.000, 24.993 mm Sol. 28 Option (B) is correct. Since 25 mm lies in the diameter step 18 & 30 mm, therefore the geometric mean diameter, D = 18 # 30 = 23.24 mm We know that standard tolerance unit, i (microns) = 0.45 3 D + 0.001D i = 0.45 3 23.24 + 0.001 # 23.24 = 1.31 microns om Standard tolerance for hole ‘h ’ of grade 7 (IT 7), IT 7 = 16i = 16 # 1.31 = 20.96 microns.c Hence, lower limit for shaft = Upper limit of shaft – Tolerance = 25 - 20.96 # 10-3 mm = 24.979 mm Q. 29 as When a cylinder is located in a Vee-block, the number of degrees of freedom GATE ME 2003 ONE MARK (A) 2 at which are arrested is (B) 4 (C) 7 ild (D) 8 iv Sol. 29 Option (B) is correct..C © ww w We clearly see from the figure that cylinder can either revolve about x -axis or slide along x -axis & all the motions are restricted. Hence, Number of degrees of freedom = 2 & movability includes the six degrees of freedom of the device as a whole, as the ground link were not fixed. So, 4 degrees of freedom are constrained or arrested. Q. 30 The symbol used for Transport in work study is GATE ME 2003 (A) & (B) T ONE MARK (C) > (D) 4 Sol. 30 Option (A) is correct. The symbol used for transport in work study is given by, & Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Q. 31 Consider the system of simultaneous equations GATE ME 2003 x + 2y + z = 6 TWO MARK 2x + y + 2z = 6 x+y+z = 5 This system has (A) unique solution (B) infinite number of solutions (C) no solution (D) exactly two solutions Sol. 31 Option (C) is correct. Given : x + 2y + z = 6 2x + y + 2z = 6 x+y+z = 5 Comparing to Ax = B ,we get R1 2 1V R6V S W S W om A = S2 1 2W, B = S6W S1 1 1W S5W c T X T X. Write the system of simultaneous equations in the form of Augmented matrix, s R1 2 1 : 6V S W ta 6A: B@ = S2 1 2 : 6W S1 1 1 : 5W a T Applying R2 " R2 - 2R1 and R 3 " 2R 3 - R2 d X il R1 2 1 : 6V S W iv = S0 - 3 0 : - 6W S0 1 0 : 4W.C Applying R 3 " 3R 3 + R2 T X R1 2 1 : 6V © ww S W = S0 - 3 0 : - 6W S0 0 0 : 6W It is a echelon form of matrix. T X w Since r 6A@ = 2 and r 5A: B? = 3 r [A] ! r [A: B ] So, the system has no solution and system is inconsistent. Q. 32 The area enclosed between the parabola y = x2 and the straight line y = x is GATE ME 2003 (A) 1/8 TWO MARK (B) 1/6 (C) 1/3 (D) 1/2 Sol. 32 Option (B) is correct. Given : y = x2 & y = x. The shaded area is show the area, which is bounded by the both curves (common area) Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 On solving given equation, we get the intersection points as, y = x2 put y = x x = x2 2 x -x = 0 x (x - 1) = 0 x = 0, 1 Then from y = x For x=0&y=0 & x=1&y=1 om We can see that curve y = x2 and y = x intersects at point (0, 0) and (1, 1).c So, the area bounded by both the curves is x=1 y = x2 # # asA= dydx #at x=0 y=x x=1 y = x2 x=1 # # = dx dy = dx 6y @xx 2 ld x=0 y=x x=0 i After substituting the limit, we have # iv = x=1 (x2 - x).C x=0 Integrating the equation, we get © ww 3 2 1 = :x - x D = 1 - 1 =- 1 3 2 0 3 2 6 = 1 unit2 Area is never negative 6 Q. 33 GATE ME 2003 w The solution of the differential equation dy dx + y2 = 0 is 3 TWO MARK (A) y = 1 (B) y = - x + c x+c 3 (C) cex (D) unsolvable as equation is non- linear Sol. 33 Option (A) is correct. dy + y2 = 0 dx dy =- y2 dx dy - 2 = dx y Integrating both the sides, we have Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 dy -# = # dx y2 y-1 = x + c 1 = x+c &y= 1 y x+c Q. 34 The vector field is F = xi - yj (where i and j are unit vector) is GATE ME 2003 (A) divergence free, but not irrotational TWO MARK (B) irrotational, but not divergence free (C) divergence free and irrotational (D) neither divergence free nor irrational Sol. 34 Option (C) is correct. Given : F = xi - yj First Check divergency, for divergence, om Grade F = 4:F = ; 2 i + 2 j + 2 k E:6xi - yj @ = 1 - 1 = 0.c 2x 2y 2z as So we can say that F is divergence free. Now we checking the irrationality. For irritation the curl F = 0 at Curl F = 4# F = ; 2 i + 2 j + 2 k E # [xi - yj] ild R 2x S i j 2y V k W 2z iv = S 2 2 S2x 2y 2z W 2 W = i [0 - 0] - j [0 - 0] + k [0 - 0] = 0.C S x -y 0 W T X © ww So, vector field is irrotational. We can say that the vector field is divergence free and irrotational. Q. 35 Laplace transform of the function sin wt is GATE ME 2003 TWO MARK w (A) 2 s 2 s +w (B) 2 w 2 s +w (C) s (D) 2 w 2 s2 - w2 s -w Sol. 35 Option (B) is correct. Let f (t) = sin wt From the definition of Laplace transformation L [F (t)] = #0 3e-st f (t) dt = #0 3e-st sin wtdt iwt - e-iwt dt = #0 3e-st b e 2i l iwt -iwt sin wt = e - e 2i = 1 #0 3 (e-st eiwt - e-st e-iwt) dt 2i 2i #0 6 @dt = 1 3 (- s + iw) t e -e - (s + iw) t On integrating above equation, we get Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 (- s + iw) t - (s + iw) t3 = 1 =e - e 2i - s + iw - (s + iw)G (- s + iw) t - (s + iw) t3 0 = 1 =e +e 2i - s + iw (s + iw)G 0 Substitute the limits, we get -0 = 1 =0 + 0 - e e0 + e 2i (- s + iw) s + iw oG =- 1 ; s + iw + iw - s E 2i (- s + iw) (s + iw) =- 1 # 2iw = -w = 2w 2 2i (iw) 2 - s 2 - w2 - s 2 w +s Alternate : From the definition of Laplace transformation L [F (t)] = # e-st sin wtdt 3 0 eat a sin bt - b cos bt a =- s and a + b2 6 @ We know # eat sin btdt = 2 e o b=w -st L [sin wt] = ; 2e 2 ^- s sin wt - w cos wt hE 3 Then, om s +w 0 s +w -3.c -0 = ; 2e 2 (- s sin 3 - w cos 3)E - ; 2 e 2 (- s sin 0 - w cos 0)E s +w as = 0- 2 1 [0 - w] =- 2 1 2 (- w) at L [sin wt] = 2 w s + w2 s + w2 s +w Q. 36 ild A box contains 5 black and 5 red balls. Two balls are randomly picked one after GATE ME 2003 TWO MARK (A) 1/90 iv another form the box, without replacement. The probability for balls being red is (B) 1/2.C (C) 19/90 (D) 2/9 © ww Sol. 36 Option (D) is correct. Given : black balls = 5, Red balls = 5, Total balls=10 Here, two balls are picked from the box randomly one after the other without w replacement. So the probability of both the balls are red is 5 P = C 010# C2 5 n Cr = n C2 r n-r 5! 5! # = 0! # 5! 3!2! = 1 # 10 = 10 = 2 10! 45 45 9 3!2! Common Data For Q.Alternate method Given : Black balls = 5 , Red balls = 5 Total balls = 10 The probability of drawing a red bell, P1 = 5 = 1 10 2 Ball is not replaced, then box contains 9 balls. Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 So, probability of drawing the next red ball from the box. P2 = 4 9 Hence, probability for both the balls being red is, P = P1 # P2 P =1#4=2 2 9 9 Q. 37 A truss consists of horizontal members (AC,CD, DB and EF) and vertical GATE ME 2003 members (CE and DF) having length l each. The members AE, DE and BF are ONE MARK inclined at 45c to the horizontal. For the uniformly distributed load “p” per unit length on the member EF of the truss shown in figure given below, the force in the member CD is om.c (A) pl 2 as (B) pl (C) 0 at (D) 2pl 3 d Sol. 37 Option (A) is correct. il Given : AC = CD = DB = EF = CE = DF = l iv At the member EF uniform distributed load is acting, the U.D.L. is given as “p ” per unit length..C So, the total load acting on the element EF of length l = Lord per unit length # Total length of element © ww = p # l = pl w This force acting at the mid point of EF. We made the FBD of the object. At A & B reactions are acting because of the roller supports, in the upward direction. In equilibrium condition, Upward force = Downward forces Ra + Rb = pl...(i) And take the moment about point A, pl # bl + l l = Rb (l + l + l) 2 pl pl # 3 l = Rb # 3l & Rb = 2 2 Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Substitute the value of Rb in equation (i), we get pl Ra + = pl 2 pl pl Ra = pl - = 2 2 pl So, Ra = Rb = 2 At point A we use the principal of resolution of forces in the y -direction, / Fy = 0 pl FAE sin 45c = Ra = 2 pl pl pl FAE = # 1 = # 2 = 2 sin 45c 2 2 pl 1 = pl And FAC = FAE cos 45c = # 2 2 2 At C , No external force is acting. So, pl FAC = = FCD 2 Q. 38 A bullet of mass “m ” travels at a very high velocity v (as shown in the figure) GATE ME 2003 ONE MARK om and gets embedded inside the block of mass “M ” initially at rest on a rough horizontal floor. The block with the bullet is seen to move a distance “s ” along c the floor. Assuming m to be the coefficient of kinetic friction between the block the bullet ? s. and the floor and “g ” the acceleration due to gravity what is the velocity v of ta a m ild (A) M + m 2mgs (B) M - m 2mgs m (C) iv m (M + m) 2mgs (D) M 2mgs.C m m Sol. 38 Option (A) is correct. © ww Given : Mass of bullet = m Mass of block = M Velocity of bullet = v w Coefficient of Kinematic friction = m Let, Velocity of system (Block + bullet) after striking the bullet = u We have to make the FBD of the box after the bullet strikes, Friction Force (Retardation) = Fr By Applying principal of conservation of linear momentum, dP = 0 or P = mV dt = cons tan t. Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 So, mv = (M + m) u u = mv...(i) M+m And, from the FBD the vertical force (reaction force), RN = (M + m) g Fr = mRN = m (M + m) g - m (M + m) g Frictional retardation a = - Fr = =- mg...(ii) (m + M) M+m Negative sign show the retardation of the system (acceleration in opposite direction). From the Newton’s third law of motion, V f2 = u2 + 2as Vf = Final velocity of system (block + bullet) m =0 co u2 + 2as = 0 u2 =- 2as s. u2 =- 2 # (- mg) # s = 2mgs From equation (ii) ta Substitute the value of u from equation (i), we get mv 2 a M + m k = 2mgs da il m2 v2 = 2mgs iv (M + m) 2 2mgs (M + m) 2.C v2 = m2 v = 2mgs # b M + m l = M + m 2mgs © ww m m Q. 39 A simply supported laterally loaded beam was found to deflect more than a specified value. Which of the following measures will reduce the deflection ? w GATE ME 2003 TWO MARK (A) Increase the area moment of inertia (B) Increase the span of the beam (C) Select a different material having lesser modulus of elasticity (D) Magnitude of the load to be increased Sol. 39 Option (A) is correct. We know, differential equation of flexure for the beam is, d 2y d 2y EI 2 = M & 2 = M dx dx EI Integrating both sides, dy = 1 # Mdx = 1 Mx + c1 dx EI EI Again integrating, 2 y = 1 b Mx l + c1 x + c2...(i) EI 2 Where, y gives the deflection at the given point. It is easily shown from the equation (i), If we increase the value of E & I , then deflection reduces. Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Q. 40 A shaft subjected to torsion experiences a pure shear stress t on the surface. The GATE ME 2003 maximum principal stress on the surface which is at 45c to the axis will have a TWO MARK value (A) t cos 45c (B) 2t cos 45c 2 (C) t cos 45c (D) 2t sin 45c cos 45c Sol. 40 Option (D) is correct. Given figure shows stresses on an element subjected to pure shear. Let consider a element to which shear stress have been applied to the sides AB and DC. om Complementary stress of equal value but of opposite effect are then setup on.c sides AD and BC in order to prevent rotation of the element. So, applied and s complementary shears are represented by symbol txy. a Consider the equilibrium of portion PBC. Resolving normal to PC assuming unit depth. at sq # PC = txy # BC sin q + txy # PB cos q ld = txy # PC cos q + txy # PC sin q cos q i = txy (2 sin q cos q) # PC iv sq = 2txy sin q cos q The maximum value of sq is txy when q = 45c..C sq = 2t sin 45c cos 45c Given (txy = t) Q. 41 GATE ME 2003 © ww For a certain engine having an average speed of 1200 rpm, a flywheel approximated as a solid disc, is required for keeping the fluctuation of speed within 2% about w TWO MARK the average speed. The fluctuation of kinetic energy per cycle is found to be 2 kJ. What is the least possible mass of the flywheel if its diameter is not to exceed 1 m? (A) 40 kg (B) 51 kg (C) 62 kg (D) 73 kg Sol. 41 Option (B) is correct. Given N = 1200 rpm , DE = 2 kJ = 2000 J , D = 1 m , Cs = 0.02 Mean angular speed of engine, w = 2pN 60 = 2 # 3.14 # 1200 60 = 125.66 rad/ sec Fluctuation of energy of the flywheel is given by, 2 DE = Iw2 Cs = 1 mR2 w2 Cs For solid disc I = mR 2 2 Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 m = 22D2E...(i) R w Cs Substitute the values in equation (i), = 2 # 2000 1 2 (125.66) 2 0.02 b2l # # = 4#2# 2000 = 50.66 kg - 51 kg (125.66) 2 # 0.02 Q. 42 A flexible rotor-shaft system comprises of a 10 kg rotor disc placed in the middle GATE ME 2003 of a mass-less shaft of diameter 30 mm and length 500 mm between bearings TWO MARK (shaft is being taken mass-less as the equivalent mass of the shaft is included in the rotor mass) mounted at the ends. The bearings are assumed to simulate simply supported boundary conditions. The shaft is made of steel for which the (A) 60 Hz om value of E 2.1 # 1011 Pa. What is the critical speed of rotation of the shaft ? (B) 90 Hz (C) 135 Hz.c (D) 180 Hz Sol. 42 as Option (B) is correct. at Given m = 10 kg , d = 30 mm = 0.03 m , l = 500 mm = 0.5 m , Eshaft ild = 2.1 # 1011 Pa iv.C © ww We know that, static deflection due to 10 kg of Mass at the centre is given by, d = Wl = 3 mgl 3...(i) w 48EI 48EI The moment of inertia of the shaft, I = p d 4 = p (0.03) 4 = 3.974 # 10-8 m 4...(ii) 64 64 Substitute values in equation (i), we get 10 # 9.81 # (0.5) 3 d = 48 # 2.1 # 1011 # 3.974 # 10-8 = 12.2625 3 = 3.06 # 10-5 m 400.58 # 10 If wc is the critical or whirling speed in r.p.s. then, g g wc = & 2pfc = d d 1 g 1 9.81 fc = = 2p d 2 # 3.14 3.06 # 10-5 = 1 9.81 = 90.16 Hz - 90 Hz 6.28 30.6 # 10-6 Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Q. 43 Square key of side “d/4 ” each and length ‘l ’ is used to transmit torque “T ” from GATE ME 2003 the shaft of diameter “d ” to the hub of a pulley. Assuming the length of the key TWO MARK to be equal to the thickness of pulley, the average shear stress developed in the key is given by (A) 4T (B) 16T ld ld 2 (C) 8T2 (D) 16T3 ld pd Sol. 43 Option (C) is correct. Given : Diameter of shaft = d Torque transmitted = T Length of the key = l We know that, width and thickness of a square key are equal. i.e. w =t=d 4 Force acting on circumference of shaft om F = T = 2T r d (r = d/2) Shearing Area,.cA = width # length = d # l = dl 4 4 as Average shear stress, t = Force = 2T/d = 8T2 Q. 44 at shearing Area dl/4 ld In a band brake the ratio of tight side band tension to the tension on the slack GATE ME 2003 TWO MARK i d side is 3. If the angle of overlap of band on the drum is 180c, the coefficient of l friction required between drum and the band is iv (A) 0.20 (B) 0.25 (C) 0.30 (D) 0.35 Sol. 44.C Option (D) is correct. Let, w T1 " Tension in the tight side of the band, w T2 " Tension in the slack side of the band q "Angle of lap of the band on the drum w Given : T1 = 3 , q = 180c = p # 180 = p radian T2 180 For band brake, the limiting ratio of the tension is given by the relation, T1 = emq or 2.3 log T1 = mq T2 bT2 l 2.3 # log (3) = m # p 2.3 # 0.4771 = m # 3.14 m = 1.09733 = 0.349 - 0.35 3.14 Q. 45 A water container is kept on a weighing balance. Water from a tap is falling GATE ME 2003 vertically into the container with a volume flow rate of Q ; the velocity of the TWO MARK water when it hits the water surface is U. At a particular instant of time the total mass of the container and water is m. The force registered by the weighing balance at this instant of time is (A) mg + rQU (B) mg + 2rQU 2 (C) mg + rQU /2 (D) rQU 2 /2 Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Sol. 45 Option (A) is correct. Given : om Flow rate = Q Velocity of water when it strikes the water surface = U.c Total Mass (container + water) = m Force on weighing balance due to water strike = Change in momentum as DP =Initial Momentum - Final momentum at = rQU - rQ (0) = rQU Final velocity is zero Weighing balance also experience the weight of the container & water. ld So, Weight of container & water = mg i Therefore, total force on weighing Balance = rQU + mg Q. 46 iv In a counter flow heat exchanger, for the hot fluid the heat capacity = 2 kJ/kgK ,.C GATE ME 2003 mass flow rate = 5 kg/s , inlet temperature = 150cC , outlet temperature = 100cC TWO MARK. For the cold fluid, heat capacity = 4 kJ/kgK , mass flow rate = 10 kg/s , inlet w temperature = 20cC. Neglecting heat transfer to the surroundings, the outlet temperature of the cold fluid in cC is w (A) 7.5 (B) 32.5 Sol. 46 w (C) 45.5 Option (B) is correct. (D) 70.0 In counter flow, hot fluid enters at the point 1 & exits at the point 2 or cold fluid enter at the point 2 & exit at the point 1. Given : for hot fluid, o h = 5 kg/ sec , th1 = 150c C , th2 = 100c C ch = 2 kJ/kg K , m and for cold fluid, oc = 10 kg/ sec , tc2 = 20c C , tc1 = ? cc = 4 kJ/kg K , m From the energy balance, Heat transferred by the hot fluid = Heat gain by the cold fluid o h ch (th1 - th2) = m m oc cc (tc1 - tc2) Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 5 # 2 # 103 (150 - 100) = 10 # 4 # 103 (tc1 - 20) 10 4 # 50 = 4 # 10 4 (tc1 - 20) tc1 = 130 = 32.5c C 4 Hence, outlet temperature of the cold fluid, tc1 = 32.5c C Q. 47 Air flows through a venturi and into atmosphere. Air density is r ; atmospheric GATE ME 2003 pressure is pa ; throat diameter is Dt ; exit diameter is D and exit velocity is U. TWO MARK The throat is connected to a cylinder containing a frictionless piston attached to a spring. The spring constant is k. The bottom surface of the piston is exposed to atmosphere. Due to the flow, the piston moves by distance x. Assuming incompressible frictionless flow, x is om.c as at (A) (rU 2 /2k) pD s2 2 (B) (rU 2 /8k) c D 2 - 1m pD s2 ild 2 (C) (rU 2 /2k) c D 2 - 1m pD s2 Dt 4 (D) (rU 2 /8k) c D 4 - 1m pD s2 iv Dt Dt.C Sol. 47 Option (D) is correct. w w w First of all we have to take two section (1) & (2) By applying Bernoulli’s equation at section (1) & (2). p1 V 12 p 2 + + z1 = 2 + V 2 + z 2 r g 2g r g 2g p1 V 12 p 2 + = 2 + V2 z1 = z 2 r 2 r 2 r p1 - p 2 = (V 2 - V 12)...(i) 2 2 Apply continuity equation, we get A1 V1 = A2 V2 Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 p D 2 V = p D2 U V = U. Let at point (1) velocity = 4 t 1 4 2 V1 V1 = b D l # U 2...(ii) Dt Substitute the value of V1 from equation (ii) into the equation (i), r r p1 - p2 = ;U 2 - b D l U 2E = U 2 ;1 - b D l E 4 4...(iii) 2 Dt 2 Dt From the figure, we have Spring force = Pressure force due to air - kx = As (p1 - p2)= p D s2 # (p1 - p2) 4 m r = p D s2 # U 2 ;1 - b D l E From equation (iii) 4 4 2 Dt co kx = p D s2 rU 2 ;b D l - 1E 4 s. 8 rU 2 D 4 Dt ta x = 8k ;b Dt l - 1E pD s2 a Q. 48 Consider a laminar boundary layer over a heated flat plate. The free stream d GATE ME 2003 velocity is U3. At some distance x from the leading edge the velocity boundary TWO MARK il layer thickness is dv and the thermal boundary layer thickness is dT. If the Prandtl iv number is greater than 1, then (A) dv > dT (B) dT > dv.C -1/2 (C) dv. dT + (U3 x) (D) dv. dT + x-1/2 w Sol. 48 Option (A) is correct. The non-dimensional Prandtl Number for thermal boundary layer is given by, w dv = (Pr) 1/3 w dT (i) When Pr = 1 dv = dT (ii) When Pr > 1 dv > dT (iii) When Pr < 1 dv < dT So for Pr > 1, dv > dT Q. 49 Considering the relationship Tds = dU + pdn between the entropy (s), internal GATE ME 2003 energy (U ), pressure (p), temperature (T) and volume (n), which of the following TWO MARK statements is correct ? (A) It is applicable only for a reversible process (B) For an irreversible process, Tds > dU + pdn (C) It is valid only for an ideal gas (D) It is equivalent to Ist law, for a reversible process Sol. 49 Option (D) is correct. The Tds equation considering a pure, compressible system undergoing an internally reversible process. From the first law of thermodynamics (dQ) rev. = dU + (dW ) rev...(i)] By definition of simple compressible system, the work is Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 (dW ) rev = pdn And entropy changes in the form of dQ ds = b T lrev (dQ) rev = Tds From equation (i), we get Tds = dU + pdn This equation is equivalent to the Ist law, for a reversible process. Q. 50 In a gas turbine, hot combustion products with the specific heats c p = 0.98 kJ/kgK, GATE ME 2003 and cv = 0.7538 kJ/kgK enters the turbine at 20 bar, 1500 K exit at 1 bar. The TWO MARK isentropic efficiency of the turbine is 0.94. The work developed by the turbine per kg of gas flow is (A) 689.64 kJ/kg (B) 794.66 kJ/kg (C) 1009.72 kJ/kg (D) 1312.00 kJ/kg Sol. 50 Option (A) is correct. om.c as at ild iv Given : c p = 0.98 kJ/kgK , hisen = 0.94 , cv = 0.7538 kJ/kgK , T3 = 1500 K p 3 = 20 bar = 20 # 105 N/m2 , p 4 = 1 bar = 1 # 105 N/m2.C c g = p = 0.98 = 1.3 cv 0.7538 w Apply general Equation for the reversible adiabatic process between point 3 and w 4 in T - s diagram, g-1 w T3 = p 3 g bT4 l b p4 l 1.3 - 1 1500 = 20 # 105 1.3 = (20) 10..33 T4 c 1 105 m # T4 = 1500 0.3 = 751.37 K (20) 1.3 Actual output T3 - T4l And hisentropic = = Ideal output T3 - T4 0.94 = 1500 - T4l 1500 - 751.37 0.94 # 748.63 = 1500 - T4l T4l = 1500 - 703.71 = 796.3 K Turbine work, Wt = c p (T3 - T4l) = 0.98 (1500 - 796.3) = 698.64 kJ/kg Q. 51 An automobile engine operates at a fuel air ratio of 0.05, volumetric efficiency of GATE ME 2003 90% and indicated thermal efficiency of 30%. Given that the calorific value of the TWO MARK fuel is 45 MJ/kg and the density of air at intake is 1 kg/m3 , the indicated mean Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 effective pressure for the engine is (A) 6.075 bar (B) 6.75 bar (C) 67.5 bar (D) 243 bar Sol. 51 Option (A) is correct. m Given : f = F = f = 0.05 , hv = 90% = 0.90 , hith = 30% = 0.3 A ma CVfuel = 45 MJ/kg , rair = 1 kg/m3 We know that, volumetric efficiency is given by, hv = Actual Volume = nac Swept Volume ns nac = hv ns = 0.90Vs...(i) Mass of air, ma = rair # nac = 1 # 0.9ns = 0.9ns m f = 0.05 # ma = 0.045ns om hith = I.P. m f # CV = im p LAN m f # CV I.P. = pim LAN.c h m pim = ith # f # CV LAN = ns as LAN 0.30 # 0.045 # ns # 45 # 106 = 0.6075 106 t ns # da = 6.075 # 105 Pa = 6.075 bar 1 bar = 105 Pa Q. 52 il For an engine operating on air standard Otto cycle, the clearance volume is 10% iv GATE ME 2003 of the swept volume. The specific heat ratio of air is 1.4. The air standard cycle TWO MARK efficiency is.C (A) 38.3% (B) 39.8% (C) 60.2% (D) 61.7% Sol. 52 w Option (D) is correct. w Given: nc = 10% of ns = 0.1ns w ns = 1 = 10 nc 0.1 And specific heat ratio c p /cv = g = 1.4 We know compression ratio, r = nT = nc + ns = 1 + ns nc nc nc = 1 + 10 = 11 Efficiency of Otto cycle, hOtto = 1 - 1g - 1 = 1 - 1 (r) (11) 1.4 - 1 = 1 - 1 0.4 = 1 - 0.3832 = 0.6168 - 61.7% (11) Q. 53 A centrifugal pump running at 500 rpm and at its maximum efficiency is delivering GATE ME 2003 a head of 30 m at a flow rate of 60 litres per minute. If the rpm is changed to TWO MARK 1000, then the head H in metres and flow rate Q in litres per minute at maximum efficiency are estimated to be (A) H = 60, Q = 120 (B) H = 120, Q = 120 (C) H = 60, Q = 480 (D) H = 120, Q = 30 Sol. 53 Option (B) is correct. Given : N1 = 500 rpm , H1 = 30 meter , N2 = 1000 rpm , Q1 = 60 litres per minute From the general relation, Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 U = pDN = 2gH 60 DN \ H & N\ H D Centrifugal pump is used for both the cases. So D1 = D2 N\ H H1 = N 12 H2 N 22 2 (1000) 2 H2 = N 22 # H1 = # 30 = 120 m N1 (500) 2 The specific speed will be constant for centrifugal pump & relation is given by, N Q Ns = = Constant H 3/4 N1 Q 1 N Q So, 3/4 = 2 3/4 2 For both cases H1 H2 m Q2 = N1 # b H2 l # Q1 = 500 # b 120 l # 60 3/4 3/4 N2 H1 1000 30 o = 1 # (2) 3/2 # 60 c s Squaring both sides. 2 taQ2 = 1 # 8 # 60 = 120 litre/ min 4 Alternate : da l From unit quantities Unit speed i iv N u = N1 = N 2 H1 H2.C N1 = N 2 w H1 H2 H 2 = N 2 H1 w N1 or w 2 H2 = N 2 #2 H1 = N1 2 (1000) # 30 (500) 2 H2 = 120 m Unit discharge Q1 Q2 Qu = = H1 H2 Q1 Q2 = H1 H2 Q1 H 2 or Q2 = = 60 # 120 H1 30 Q2 = 120 litre/ min Q. 54 Hardness of steel greatly improves with GATE ME 2003 (A) annealing (B) cyaniding TWO MARK (C) normalizing (D) tempering Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Sol. 54 Option (B) is correct. Hardness is greatly depend on the carbon content present in the steel. Cyaniding is case-hardening with powered potassium cyanide or potassium ferrocyanide mixed with potassium bichromate, substituted for carbon. Cyaniding produces a thin but very hard case in a very short time. Q. 55 With a solidification factor of 0.97 # 106 s/m2 , the solidification time (in seconds) GATE ME 2003 for a spherical casting of 200 mm diameter is TWO MARK (A) 539 (B) 1078 (C) 4311 (D) 3233 Sol. 55 Option (B) is correct. Given : q = 0.97 # 106 s/m2 , D = 200 mm = 0.2 m From the caine’s relation solidification time, T = q bV l 2 Volume om V = 4 pR 3 3 A Surface Area.c A = 4pR2 So, as 4 pR 3 2 T = 0.97 # 106 f 3 2 p = 0.97 # 106 b R l 2 at = 0.97 # 10 b 4p R 6 0.2 l 2 = 1078 sec 3 d 9 2 il iv Q. 56 A shell of 100 mm diameter and 100 mm height with the corner radius of 0.4 mm GATE ME 2003 is to be produced by cup drawing. The required blank diameter is.C TWO MARK (A) 118 mm (B) 161 mm (C) 224 mm (D) 312 mm Sol. 56 w Option (C) is correct. w Given : d = 100 mm , h = 100 mm , R = 0.4 mm w Here we see that d > 20r If d $ 20r , blank diameter in cup drawing is given by, D = d 2 + 4dh Where, D = diameter of flat blank d = diameter of finished shell h = height of finished shell Substitute the values, we get D = (100) 2 + 4 # 100 # 100 = 50000 = 223.61 mm - 224 mm Visit : www.Civildatas.com  Visit : www.Civildatas.com GATE SOLVED PAPER - ME 2003 Q. 57 A brass billet is to be extruded from its initial diameter of 100 mm to a final GATE ME 2003 diameter of 50 mm. The working temperature of 700cC and the extrusion constant TWO MARK is 250 MPa. The force required for extrusion is (A) 5.44 MN (B) 2.72 MN (C) 1.36 MN (D) 0.36 MN Sol. 57 Option (B) is correct. Given : di = 100 mm , d f = 50 mm , T = 700c C , k = 250 MPa Extrusion force is given by, Fe = kAi ln c Ai m Af