ESE-2015Detailed Exam Solutions (Objective Paper - II) Mechanical solutions Explanation of Mechanical Engg. Paper-II (ESE - 2014) SET - D Codes : Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I) (b) Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I) (c) Statement (I) is true but Statement (II) is false (d) Statement (I) is false but Statement (II) is true Statement (I) : The cam in contact with a follower is a case of complete constraint. 3. Sol. 2. S 4. IE Sol. Involute pinion can not have any number of teeth because a minimum number of teeth are decided by interference phenomenon. Both involute and cycloidal teeth satisfy constant velocity ratio condition. Because velocity ratio depends upon ratio of number of teeth or ratio of pitch diameters. Statement (I) : Hooke’s joint connects two nonparallel non-intersecting shafts to transmit motion with a constant velocity ratio. The cam in contact of follower is case of successful constant. Because spring force is required to maintain the contact. This spring force does not guarantee the contact all time because after certain speed, the follower losses contact with can due to inertia force. Statement (I) : Involute pinions can have any number of teeth. Statement (II) : Involute profiles in mesh satisfy the constant velocity ratio condition. Statement (II) : Hooke’s joint connects two shafts the axes of which do not remain in alignment while in motion. Ans. (c) Statement (II) : The pair, cam and follower, by itself does not guarantee continuity of contact all the time. Ans. (c) R Sol. AS (a) M 1. Ans. (c) TE Directions : Each of the following twenty (20) items consists of two statements, one labelled as ‘Statement (I)’ and the other as ‘Statement (II)’. Examine these two statements carefully and select the answers to these items using the codes given below. The Hooke’s joint connects two non-parallel shafts but intersecting. For constant velocity ratio there are two Hooke’s joints in particular torks orientation Statement (I) : Lewis equation for design of involute gear tooth predicts the static load capacity of a cantilever beam of uniform strength. Statement (II) : For a pair of gears in mesh, pressure angle and module must be same to satisfy the condition of interchangeability and correct gearing. Ans. (a) Sol. Both Statement I & II are correct. 5. Statement (I) : Tensile strength of CI is much higher than that of MS. Statement (II) : Percentage of carbon in CI is more than 1.5. Office Address : F-126, Katwaria Sarai, New Delhi - 110 016. Telephone : 011-41013406, Mobile : 8130909220, 9711853908, 7838813406 Web : www.iesmaster.org E-mail :
[email protected] Ans. (d) Statement (I) : Centrifugal clutches are designed to provide automatic and smooth engagement of load to driving member. Sol. Main strength of composite (i.e. febreglan) comes from glass fibres and not the polymer. 10. Statement (I) : Industrial rotors will not have uniform diameter throughout their lengths. TE 6. Ans. (c) R Sol. Tensile strength of cast iron is very less compared to mild steel. However, percentage of carbon in cast iron is more than 2%. Statement (II) : Fiberglass acquires strength from then polymer and flexibility from the glass. Statement (II) : Since the operating centrifugal force is a function of square of angular velocity, the friction torque for accelerating a load is also a function of square of speed of driving member. AS Ans. (a) Ans. (a) Statement (I) : Heating the steel specimen in the furnace up to austenitize temperature f ollowed by f urnance cooling is termed annealing. M 7. Statement (II) : These rotors will have to accommodate transmission elements like pulleys and gears and supports like anti-friction bearings. Sol. Difference diameter on shafts are to support transmission elements. Shoulder are provided to restrict axial movement of bearings. 11. Statement (II) : Annealed steel specimen possesses fine pearlitic structure. Statement (II) : Cored induction furnace, though most efficient, requires a liquid metal charge while starting. S Ans. (c) 8. IE Sol. Annealed steel specimen has a “coarse pearlite” structure and not “fine pearlite” due to slow cooling in furance. Statement (I) : The susceptibility of a ferromagnetic material decreases with an increase in Curie temperature. Statement (II) : A critical temperature at which the alignment of magnetic moments vanishes is called Curie temperature. Statement (I) : Cored induction furnace cannot be used for intermittent operation. Ans. (a) Sol. Cored induction furnace requires a liquid metal charge while starting. Therefore, they cannot be used for intermittent operations. 12. Statement (I) : Low-carbon steel has high weldability and is more easily welded. Statement (II) : Higher carbon contents tend to soften the welded joints resulting in development of cracks. Ans. (a) Sol. Magnetic suspetibility occurs above curie temperature only. Thus, higher the curie temperature lower is the suseptibility. 9. Statement (I) : Fiberglass is a polymer composite made of a plastic matrix containing fine fibers of glass. Ans. (c) Sol. High carbon content tends to make the weld more brittle and thus more prone to cracking. Low carbon steels have excellent weldeability due to low carbon content. Office Address : F-126, Katwaria Sarai, New Delhi - 110 016. Telephone : 011-41013406, Mobile : 8130909220, 9711853908, 7838813406 Web : www.iesmaster.org E-mail :
[email protected] Statement (I) : For cutting multi-start threads, the speed ratio is expressed in terms of the lead of the job thread and lead of the lead screw threads. 16. Statement (II) : The speed of the job is reduced to one-third or one-fourth of the job speed used in the turning operation. Statement (I) : In chain drives, angle of articulation through which link rotates during engagement and disengagement, is greater for a small number of teeth. Statement (II) : The greater angle of articulation will increase the life of the chain. Ans. (c) 14. Sol. For greater life of chain, the angle of articulation should be reduced to minimize wear of chain & fatigue of rollers. TE Ans. (b) Statement (I) : The Bauschinger effect is observed in tension test of mild steel specimen due to loss of mechanical energy during local yielding. 17. AS Statement (II) : The Bauschinger effect is a function of section modulus of specimen under test. Ans. (c) The Baeeschinger effect refers to a property of materials where the material’s stress strain characteristics change as a result of the microscopic stress distribution of material. It is observed in tensile test of mild steel specimen. Statement (II) : Ceramic tools can be used on hard-to-machine work material. Ans. (b) Sol. Ceramic tools have very brittle tool tips, that is why they all prone to impact loads. These tools are used on hard to machine work material such as cast iron as they are highly wear & abrasion resistant. Statement (II) : The DNC is a manufacturing process in which a number of process machines are controlled by a computer through direct connection and real time analysis. Sol. Both Statement I and II are correct, but Statement II does not explain Statement I, 18. This effect is a material property and not a geometric property. So it is not dependent on section modulus. Statement (I) : The ceramic tools used in machining of material have highly brittle tool tips. IE 15. Statement (I) : The CNC is an NC system utilizing a dedicated stored program to perform all numerical control functions in manufacturing. Ans. (b) S M Sol. R 13. Statement (I) : In interference fit, the outer diameter of the shaft is greater than the inner diameter of the hole. Statement (II) : The amount of clearance obtained from the assembly of hole and shaft resulting in interference fit is called positive clearance. Ans. (c) Sol. Clearance obtained in an interference fit is a negative clearance. 19. Statement (I) : One of the most commonly used techniques for testing surface integrity of material is metallography. Statement (II) : Surface integrity of a material does not contribute for the mechanical and metallurgical properties. Office Address : F-126, Katwaria Sarai, New Delhi - 110 016. Telephone : 011-41013406, Mobile : 8130909220, 9711853908, 7838813406 Web : www.iesmaster.org E-mail :
[email protected] 7838813406 Web : www. Surface integrity affects mechanical properties such as fatigue strength of material. Statement (I) : The change in critical path requires rescheduling in a PERT network.110 016. Ans. Under an axial load.iesmaster. the stress in the steel tube is 100 N/mm 2.Ans. (c) Sol. If Es = 2Ec.org E-mail : ies_master@yahoo. (d) AS A copper rod of 2 cm diameter is completely encased in a steel tube of inner diameter 2 cm and outer diameter 4 cm. Katwaria Sarai. (d) 300 N/mm2 = avg 2cm For copper d = 2cm S = For steel 23. x = 900 N/mm2 y = 100 N/mm2 xy = 300 N/mm2 Office Address : F-126. 9711853908. Normal stress 50 N/mm2 N/mm2 (b) M (c) 75 N/cm2 tensile Ans. A system under biaxial loading induces principal stresses of 100 N/cm2 tensile and 50 N/cm2 compressive at a point. New Delhi .co. (c) c = 50 N/mm2 20. The maximum d0 = 4 cm Es = 2Ec 1 2 100 50 = 2 2 = 25 N/cm2 (Tensile) = 20 mm IE Sol.33 N/mm2 On maximum shear stress plane. normal stresses are x 900 N / mm2 .in . 100 Ans. then the stress in the copper rod is (a) (a) TE Ans. (a) 21. 2 = – 50 N/cm2 (b) 33. Statement (II) : Some of the activities cannot be completed in time due to unexpected breakdown of equipment or non-availability of raw materials. Telephone : 011-41013406. (a) 50 N/cm2 compressive (c) 100 N/cm2 tensile (d) 25 N/cm2 tensile 1 = 100 N/cm2 Sol. Mobile : 8130909220. = 20 mm y 100 N / mm2 principal stress is = 40 mm Steel s = 100N/mm2 4cm (a) 800 N / mm2 (b) 900 N / mm2 (c) 1000 N / mm2 (d) 1200 N / mm2 Strain in steel = strain in copper c s = Ec Es c = 100 Ec Es and shear stress 300 N / mm2 . 22. The normal stress at that point on the maximum shear stress plane is R Sol. Copper di = 2 cm In a biaxial state of stress. = 500 1 = 250 MPa Which of the above statements are correct? (a) 1. 2. y 1 1 = x 2 2 AS A constitutional diagram shows relationship among which of the following combinations in a particular alloy system? by x 100 MPa and y 200 MPa . The magnitudes of the other principal stress and 100 = 50 3 MPa and 50 MPa (b) 100 MPa and 50 3 MPa (c) 50 MPa and 50 3 MPa (d) 50 3 MPa and 100 MPa x y 2 42xy 1 100 200 2 42xy 2 40000 = 10000 + 4 2xy 4 2xy = 30000 xy = 50 3 MPa 26.org E-mail : ies_master@yahoo. One of the principal stresses 1 250 MPa . 7838813406 Web : www. 4 2xy x = 100 MPa y = 200 MPa 2 2 900 100 1 900 100 4 300 = 2 2 1 500 = 1000 N/mm2 2 R = 500 1 2 = x y 2 = 100 200 250 TE 1 640000 360000 2 (a) Temperature and composition (b) Temperature and phases present (c) Temperature. the shearing stress xy are respectively (a) = 50 MPa 100 200 1 2 100 200 42xy 250 = 2 2 M 24. Consider the following statements regarding powder metallurgy : 1. (c) S Sol. Refractory materials made of tungsten can be manufactured easily. 4. The state of stress at a point in a body is given IE 25. composition and phases present (d) Temperature and pressure Ans.Maximum principal stress x y 1 1 = 2 2 x y Ans. Telephone : 011-41013406. control of grain size results in relativ ely much unif orm structure. 3. 9711853908. The powder heated in die or mould at high temperature is then pressed and compacted to get desired shape and strength. 2 and 3 only Office Address : F-126.110 016. New Delhi . Katwaria Sarai.co. composition and phases present.iesmaster. the metal powder is gradually heated resulting in coherent bond. Mobile : 8130909220. In metal powder.in . In sintering. (c) 2 Sol. A constitutional diagram gives information regarding temperature. Consider the following sttements related to Mohr’s circle for stresses in case of plane Office Address : F-126.110 016.0 = [25 MPa. The magnitudes of the shearing stress on a plane on which the normal stress is 200 MPa tensile and the normal stress on a plane at right angle to this plane are M 27.iesmaster.000 2 = 125 MPa 29. xy 0 and 150 MPa (d) x 75 MPa. 0] y = (250 – 150) – 200 Radius of Mohr’s circle = 1 2 x y 2 4 2xy y = – 100 MPa 200 100 1 2 200 100 42xy 1 = 2 2 200 100 1 2 200 100 42xy 250 = 2 2 400 = 300 2 42xy = 1 100 50 2 4 1002 2 = 1 22500 40. 3 and 4 2xy = 17500 xy = 50 7 MPa 28. New Delhi . Ans. 9711853908. Mobile : 8130909220. 3 and 4 only (d) 1. (c) (a) x 75 MPa. (d) R (b) The state of stress at a point is given by x 100 MPa . x = 100 MPa y = – 50 MPa 1 = 250 MPa S Sol. The centre of Mohr’s circle and The magnitudes of principal stresses at a point are 250 MPa t ensil e and 150 MPa compressive. (b) Sol. xy 0 and 75 MPa (b) x 25 MPa.co. 7838813406 Web : www. its radius will be AS (a) and xy 100 MPa . xy 0 and 125 MPa (c) x 25 MPa.org E-mail : ies_master@yahoo. Telephone : 011-41013406. Sol. xy 0 and 125 MPa Ans. 2 and 4 only (c) 2. Katwaria Sarai.1. y 50 MPa xy = 100 MPa 2 = – 150 MPa IE x y centre = 2 x = 200 MPa 1 2 = x y .in . All options are correct TE 50 7 MPa and 100 MPa (tensile) (b) 100 MPa and 100 MPa (compressive) (c) 50 7 MPa and 100 MPa (compressive) (d) 100 MPa and 50 7 MPa (tensile) Ans. 2. stress : (c) 120 MPa and 120 MPa 1. Telephone : 011-41013406.56 N When the temperature of this piece increases by 50 C . Katwaria Sarai. New Delhi .2 105 50 4 4 4 202 102 202 = 18849. L A t LB T Lc T R LC RLB RL A – E A E A E A = 0 A B C LA = LB = LC = L 3L T = L R R R E A A A B A C C B 20 10 20 20 20 20 IE 30. E = 2 × 105 MPa The diameter represents the difference between the two principal stresses. The piece is held between two rigid surfaces X and Y. the stresses in sections A and B are (a) 120 MPa and 480 MPa (b) 60 MPa and 240 MPa A = R = 60 MPa 2 20 4 Office Address : F-126. 9711853908. The construction is for variations of stress in a body.org E-mail :
[email protected]. TE 3. The lengths of three sections A.co.2 105 / C and Young’ss modulus E 2 105 MPa for steel : S A C B 20mm 10mm 20mm X 20mm 20mm 20mm B R Since supports are rigid. 7838813406 Web : www. (a) The figure shows a steel piece of diameter 20 mm at A and C.110 016. 2 and 3 (b) 2 and 3 only (c) 1 and 3 only (d) 1 and 2 only T = 50°C A R AS Ans.2 × 10–5/°C Sol. The coefficient of linear M expansion 1. B and C are each equal to 20 mm. Mobile : 8130909220. Ans. The radius of the circle represents the magnitude of the maximum shearing stress.in . and 10 mm at B. (d) 60 MPa and 120 MPa 2. = 1. (b) R Which of the above statements are correct? (a) 1. X 1 1 1 3E T = R 2 2 2 20 10 20 4 4 4 R = Y 3 2 105 1. 0004 Smallest normal stress = 0 and (a) 40 MPa and 40 MPa A cantilever of length 1. 9711853908. 1 = 0. The maximum stress due to bending is not to exceed (b) 0 MPa and 40 MPa 100 N / mm2 .in .6 K = 1.0004 0.iesmaster. Mobile : 8130909220. The value for bulk modulus is (c) 1.2 × 105 MPa AS (a) 1 = 1 2 80 0 = 2 2 = 40 MPa 1 0. 7838813406 Web : www. 2 10 4 = 240 MPa Sol. The beam is of rectangular cross-section with breadth equal to half the depth.2 105 MPa respectively.2 × 105) = 3. 1 2 = 0 MPa K = 2 × 105 MPa 32.32 Maximum shar stress = 105 At a point i n a body.00012 . (b) R For a material following Hooke’s law.2 m carries a concentrated load of 12 kN at the free end. The minimum depth of the beam (c) 80 MPa and 0 MPa should be (d) 0 MPa and 80 MPa (a) 120 mm (b) 60 mm (c) 75 mm (d) 240 mm 2 0.2 105 3K 1.5 105 MPa 5 2. If E 2 105 MPa and 0. Katwaria Sarai.00012 R B = Ans.31. = 2 105 1 0.co.5 10 MPa (b) 2 105 MPa (d) 3 105 MPa Ans.3 2 = 9 K 1. the smallest normal stress and the largest shearing stress are 33. E [(–0. the values E = 2 × 105 of ealstic and shear moduli are 3 105 MPa TE = 0.2 105 S IE 2 1 2 105 1 0.00012 2 = 9KG E = 3K G 3 × 105 = E 1 2 1 2 = 80 MPa M G = 1.0004 2 = – 0.00012) + {0. Office Address : F-126. New Delhi .110 016.org E-mail :
[email protected] .2 × 0.0004}] 3K + (1. (b) E = 3 × 105 MPa Sol. Telephone : 011-41013406.3 and 1.3 0.3 × 0.6 K 0. Mmax = WL 1 = = (12 × 103) = 14.3 At point A.0013 + {0.0013)} = – 200 MPa 6 1 2 Maximum shear stress max = 2 bh2 S 100 = = 14.32 [(– 0.0013) +{0. h3 = 172.3 × (–0.0013 compressive respectively.4 × 104 IE h 2 h = 86. the largest normal and shearing stress at this point are (a) 200 MPa and 200 MPa (b) 400 MPa and 200 MPa 200 200 = 200 MPa 2 A beam ABCD.2m 2 = – 0. 9711853908.8 × 104 h = 120 mm 34.110 016. Two strain gauges fixed along the principal directions on a plane surface of a steel member recorded strain values of 0. Telephone : 011-41013406. W =12 kN A b (c) 260 MPa and 260 MPa (d) 260 MPa and 520 MPa Ans. 6 m long.3 .4 10 E 1 2 1 2 2 105 1 0. New Delhi . 3 m apart with overhangs AB = 2 m and CD = 1 m.0013 Sol. is supported at B and C.4 × 104 2 = 35. 7838813406 Web : www. It carries a uniformly distributed load of 100 kN/m over its entire length : A B 2m 100 kN/m 3m C D 1m The maximum magnitudes of bending moment and shear force are Office Address : F-126. Katwaria Sarai.0013 tensile and 0.32 [0.2) N-m bh3 12 6 14.co.Ans.in . R L=1.iesmaster.0013 E = 2 × 105 MPa = 0.0013)} = 200 MPa 2 = = E 1 2 2 1 2 105 1 0. (a) Sol.4 106 h / 2 M 100 = Mmax y I AS = 14.4 × 106 N-mm max = TE b = h/2 bh2 = 6 × 14. (a) h 3 1 = 0. Mobile : 8130909220. Given that the value of E 2 105 MPa and 0.3×(0.org E-mail :
[email protected] × 103 N × (1. M.C = – 100 + (200) = 100 kN (S.F.M.B = –(100 × 2) = – 200 kN S.200 kN-m and 250 kN (b) 200 kN-m and 200 kN (c) 50 kN-m and 200 kN (d) 50 kN-m and 250 kN 100 kN/m C B A RB 2m D RC 1m 3m R (a) +200 kN 100 kN/m C Sol.)D = 0 Point P.iesmaster.M.co. of diameter 100 mm carries a shear force S S.in .M. (d) Sol. New Delhi .F. d = 100mm V = 10 kN = 104 N max = 4 avg 3 Office Address : F-126. Mobile : 8130909220.F.C = 200 – (100 × 3) = – 100 kN 36.F.F.D B A +ve P C D M=–50 kN-m Mmax=–200 kN-m S.)D = RB x 2 100 x 2 4 = (400 × 2) – 100 4 2 =0 A solid circular cross-section cantilever beam (a) 4 Pa 3 (b) 3 Pa 4 (c) 3 Pa 16 (d) 16 Pa 3 Ans. – (100 x) + RB = 0 100x = 400 x = 4m (B.)D = 0 x * (B.F.)C = – 100 5 + (400 × 3) 2 = – 50 N-m (B.)A = 0 (B. B A RB 2m P 3m TE Ans.F. 9711853908. (b) A D RC 1m x M AS Taking moment about point B. Katwaria Sarai.org E-mail : ies_master@yahoo.)A = 0 +ve B –ve P –ve C D –100kN –200kN S.M. (Rc × 3) – (100 × 6 × 1) = 0 Rc = 200 kN RB + Rc = 600 RB = 400 kN (S.)B = – (100 × 2 × 1) = – 200 N-m of 10 kN at the free end.110 016. The maximum shear stress is IE 5 (B.B = – 200 + 400 = 200 kN S. Telephone : 011-41013406. 7838813406 Web : www. in For detailed and updated information visit us at Web : www. 9711853908 Web : www. 2015[Afternoon] 27th June. 2015[Eve.co. 011-26522064 Mobile : 8130909220.iesmaster.IES MASTER Institute for Engineers (IES/GATE/PSUs) Announces New Batch for IES GATE PSUs Civil Regular Weekend Regular 1st July. 2015 Electrical Weekend 20th June. 2015 Admission Open for Session (2015-16). Apply Now For Excellent Study Material CALL NOW : 09711853908 Ph : 011-41013406.org E-mail :
[email protected] SCROLL DOWN .iesmaster. 2015 24th June.] Mechanical Regular Weekend 22nd June. 2015 20th June. 625 8 max 4 1max = 1.3 W and 0. If the load is substituted by a WL y 8 = I W1 = R P R Q L W1L3 = 48EI TE 37. concentrated load W1 at mid-span such that W1L3 5 WL3 = 48EI 384 EI W1 = 0. keeping the width.25 max L RA = RB = Mmax y I W L y = 0. Let load intensity is . length and loading same. Telephone : 011-41013406.co. Katwaria Sarai.110 016.= 4 104 N 3 1002 mm2 4 max 16 MPa 3 A beam of length L simply supported at its ends carrying a total load W unif ormly distributed over its entire length deflects at the centre by and has a maximum bending stress .iesmaster. b = M y M h/2 = I bh3 /2 b = 6M bh2 Office Address : F-126. 9711853908.3 W and 0. New Delhi .6 (c) 0.org E-mail :
[email protected] .625 W the deflection at the centre remians unchanged. Mobile : 8130909220. (*) = IE S Sol.6 W and 0.625 4 I M (a) 1max = For a rectangular section beam.3 = Ans.6 (d) 0. For rectangular section. if the beam depth is doubled. AS the magnitude of the load W1 and the maximum bending stress will be 0. (b) Sol.3 (b) 0. A RA L = W C 38. 7838813406 Web : www. B RB L 2 deflection at point C = 5 L4 5 WL3 = 384 EI 384 EI L L L L Mmax = 2 2 2 4 = L2 WL = 8 8 W1L y 4 I 0.6 W and 0. the bending stress is decreased by a factor (a) 2 (b) 4 (c) 6 (d) 8 Ans. 110 016. The volume of flash land and flash gutter should be about 20% .M.5 K S Stiffness of each spring (k ) = 2k Mmax and then these springs are arranged in parallel k = k k x 1 = (RA × 1) – 1 1 2 = 3 kN-m = 3 × 106 N-mm IE y= = 2k 2k b = =4 k A beam AB simply supported at its ends A and B. 2m 4 150 = 75mm 2 300 10 4 mm 4 = 75 MPa 41. 3 m long. Katwaria Sarai. the maximum value of bending stress is Mmax y I 3 106 N-mm 75mm Which of the following statements apply to provision of flash gutter and flash land around the parts to be forged? 1.)c Sol. Small cavities are provided which are directly outside the die impression. 9711853908.iesmaster. 2.5 kN Mmax = (B. RB taking moment about point A (a) so 1kN/m 2m x A helical compression spring of stiffness K is cut into two pieces. The equivalent spring stiffness of this new arrangement is equal to Ans.h = 2h b b = 6M 6M = 2 2 bh b 2h (b) 85 MPa (c) 125 MPa (d) 250 MPa h TE A RA b 4 C 1m AS 4 K (b) 2 K (c) K (d) 0. Telephone : 011-41013406. 75 MPa Ans.in . each having equal number of turns and kept side-by-si de under compression. 40. (a) 1 6M = 2 4 bh b = (a) R if Office Address : F-126. (a) x 3kN Sol. When spring is cut into two pieces. at 1 m from A : = 3 kN 1 kN/m B A 1m If ISJB 150 with IXX 300 cm is used for the beam.5 RB = 2. Mobile : 8130909220. M 39.org E-mail : ies_master@yahoo. C B 3 (RB × 3) – (3 × 1) – 3 1 = 0 2 3RB = 3 + 4. carries a uniformly distributed load of 1 kN/m over its entire length and a conventrated load of 3 kN.co. New Delhi .25% of the volume of forging.5 kN RA + RB = 3 + (1 × 3) RA + RB = 6 RA = 3. 7838813406 Web : www. Low limit of shaft = 24. Sol. 2 and 3 (d) 2 and 3 only 2. small positive clearnace or negative clearance (Interference) between the shaft and hole member is employable. high limit of hole = 25.02 mm x + 1. Mobile : 8130909220. Gutter is provided to ensure complete filling of die cavity and not closing of die. Consider the following statements : R In case of assembly of mating parts Ans. 200. high limit of hole = 25. Variable cost for producing the product is Rs. Market survey indicated that the product selling price could be Rs.99 mm and low limit of shaft = 24.02 mm and a minimum clearance of 0. (a) High limit of hole = 25.110 016.01 = 0. 100.co.01 = 0. (a) 1.5 x + 0. (c) M (b) (d) Which of the above statements is/are correct? AS (a) S 42. The break-even quantity is (a) 1000 (b) 2000 (c) 500 (d) 900 Ans. the difference between hole size and shaft size is called allowance in transition fit. upper limit of shaft = 25 mm and low limit of shaft = 24. Fixed cost for producing the product is estimated as Rs.5 times the shaft tolerance. and are to have a clearance fit with a maximum clearance of 0. upper limit of shaft = 24.org E-mail : ies_master@yahoo. Telephone : 011-41013406.000. upper limit of shaft = 24. A hole and a shaft have a basic size of 25 mm. upper limit of shaft = 24. a car manufacturing company predicted the demand for that year as 1040 cars. The limits of both hole and shaft using hole basis system will be low limit of hole = 25 mm.006 mm.006 High limit of shaft = 24. Only statement (2) is correct. 7838813406 Web : www. small positive or negative clearance between the shaft and hole member is employable TE 2. high limit of hole = 25.01 x = 43. If the company’s Office Address : F-126. New Delhi . • In transition fit. (a) 1 and 2 only (b) 1 and 3 only (c) 1.3.006 mm.99 mm 1 only (b) Both 1 and 2 (c) 2 only (d) Neither 1 nor 2 Sol. IE low limit of hole = 25. high limit of hole = 25 mm.986 Also tolerance of shaft + minimum clearance + tolerance of hole = 0. The actual sale was found to be 1140 cars.99 mm and low limit of shaft = 25 mm Ans.01 mm.004mm 25 An organization has decided to produce a new product. 44. • Allowance is a planned deviation between an actual dimension and a nominal or theoretical dimension. The hole tolerance is to be 1.iesmaster.8 mm and low limit of shaft = 24.76 mm (c) low limit of hole = 24 mm.in . For hole basis system Low limit of hole = 25 mm (a) Ans.986 mm low limit of hole = 25 mm.006 mm.99 mm Sol. Gutter is provided to ensure complete closing of the die.5 x = 0.026 mm. 0. BEQ = F 100000 = = 1000 S V 200 100 Usi ng ex ponential smoot hening.00. 9711853908. (a) Sol. 1. Katwaria Sarai.02 45. 5 = 28. 2 per hour.4 46. The number of furnaces required will be (c) coordinator (d) director (a) 3 (b) 2 (c) 1 (d) 4 IE Sol.5 + 3. (c) (b) Rs.4 × 1. 8. a part is made from solid brass rod of 38 mm diameter and length 25 mm.6 (c) 0. coordinated movement of separately driven axes motion is required to achieve the desired path of tool relative to workpiece. Factory overheads are 50% of direct labour cost. 7838813406 Web : www. Ft = dt 1 1 Ft 1 1080 = 1140 1 1040 = 1140 1040 1040 Coarse feed.2 48. Ans. (a) Sol.40 2 Sol. (a) Ans.4 (b) 0.5 Total factory cost = 3 + 1. low cutting speed and insufficient cooling help produce continuous chips in ductile materials (b) discontinuous chips in ductile materials (c) continuous chips with built-up edges in ductile materials (d) discontinuous chips in brittle materials S M (a) Ans.4 N Cost of rod = 2. 4.6 gm per cubic cm and its cost is Rs.in . New Delhi .7 (d) 1. 1. The machining time taken to finish the part is 90 minutes and labour rate is Rs. low cutting speed & i nsuf f icient cooling hel p to produce continuous chips with built-up edges in ductile materials. The factory cost of the part will be R forecast for the next year is 1080. 9711853908. 3 Factory overhead = 0.iesmaster. (b) Sol. low rake angle.80 (d) Rs.org E-mail : ies_master@yahoo. Rs. Mobile : 8130909220. The density of material is 8.89 Cost of labour = 2 × 1.co. Volume of rod = /4d = 0. (a) Office Address : F-126.6 = 243.5 = Rs. 8. The generation of these reference signals is accomplished through a device called 49.110 016.40 (a) approximator (b) interpolator A company wants to expand the solid propellant manufacturing plant by the addition of more than 1 ton capacity curing furnace.80 (c) Rs. Telephone : 011-41013406.36 cm3 Mass of rod = 28. 18.625 Rs. Katwaria Sarai.243kg weight of rod = 2. Ans.(a) 0. 3. Each ton of propellant must undergo 30 minutes of furnace tim e including loading and unloading operations. 47. 1.36 × 8.82 2.40 = /4×3. The required output for the new layout is to be 16 tons per shift (8 hours). Coarse feed.625 per newton. Interpolator coordinate the motion of tool relative to workpiece.84 grms or 0. 14.89 = Rs. what is the value of the smoothening constant? In NC machining. Plant (system) efficiency is estimated at 50 percent of system capacity.5 × 3 = Rs. Furnace is used only 80 percent of the time due to power restrictions. (a) AS 40 = 100 TE Ans. low rate angle. TE Furnace required = Consider the following statements : 53. • Arrivals are markov or memoryless • Serivce is also memory less. (b) Sol. Required output 16 tons in 8 hours or 8 × 60 = 480 minutes. EOQ = 2 80000 4000 40 = 4000 unit (a) 1. New Delhi . Telephone : 011-41013406.in . Maximum level of percentage defectives that wi ll make the lot def init ely unacceptable 4. the arrivals is described as a Poisson distribution 3. In a single-server queueing model Operating time of furnace = 0. (c) Sol. 4. 2 and 3 (d) 2 and 3 only Ans. 16 ton propellent requires = 16 × 30 = 480 minutes of furnace. 480 = 2. uncertainty concering the demand for service exists (a) 1 and 2 only (b) 1 and 3 only (c) 1. (d) Office Address : F-126. (b) S Sol. The consumer is willing to take lots of quality level (LTPD) even though they are unacceptable 10% Ans.org E-mail : ies_master@yahoo. • Arrivals follow poisson distribution To construct an operating characteristic curve. The cost of placing an order is Rs. Katwaria Sarai. 2. 2 and 4 only (d) 2.iesmaster. 3 and 4 (c) 1. Maximum proportion of defectives that will make the lot definitely unacceptable 2. 7838813406 Web : www.5 × 382 = 192 minutes (a) avoid work from rubbing against tool (b) control chip flow (c) strengthen tool edge (d) break chips Ans.5 or 3 192 the arrivals is a memoryless process Which of the above statements are correct? AS = Total furnace time Total plant time 1. Mobile : 8130909220.000 and the inventory cost of each item is Rs.52.8×480 = 384 minutes. 3 and 4 only Ans. What is the economic order quantity? IE 51. 9711853908. an agreement has to be reached between producer and consumer through which of the following points? 1. The producer is willing to aceept that some of satisfying the quality level (AQL) will rejected 5% 3. In a single server queuing model The purpose of providing side rake angle on the cutting tool is to M 50. 2 and 3 only (b) 1.110 016. Plant efficiency (in minutes) = 0. (a) 2000 (b) 4000 (c) 5656 (d) 6666 2. R Sol. The annual demand of a commodity in a supermarket if 80000. 40.co. Purpose of side rake angle of cutting tool is to control chip flow. 9711853908. It is not a binding constraint.Sol.org E-mail : ies_master@yahoo. 7838813406 Web : www. Mobile : 8130909220.iesmaster. To construct an OC curve requirements are x 0 • AQL for which produce risk ( = 0. 15 Constraint 4 x y 350 Office Address : F-126. The number of possible solutions is (a) 15 (b) 225 (c) 6435 (d) 150 Ans.110 016. Constraint 1 Max (p) = 8x + 5y Constraint 1 2x y 1000 Constraint 1 3x 4y 2400 Constraint 3 x y 800 800 x From figure it is clear that constraint -3 does not have effect on solution because.co. (b) Sol. For a 3 × 5 transportation problem. Telephone : 011-41013406. (a) Sol.e.in . New Delhi . Katwaria Sarai.1) or 10% Constraint 3 Constraint 2 Constraint 4 600 : Y 0 Which of the above constraints is a redundant one and does not have any effect on the solution? 500 S Constraint-1 (b) Constraint-2 (c) Constraint-4 (d) Constraint-5 and Constraint-6 IE (a) Ans. the following are the constraints laid out for maximizing the profit : Maximize profit (P) = 8X + 5Y Constraint-6 x y 1 500 1000 • Lot tolerance percent deflective (LTPD) 54. 55. A transportation problem consists of 3 sources and 5 destinations with appropriate rim conditions. The number of possible solutions are (5 × 3) i. R • Consumers risk for LTPD ( = 0.05) is 5% y0 subject to : 2X Y 1000 Constraint-2 : 3X 4Y 2400 Constraint-3 : X Y 800 Constraint-4 : X Y 350 AS Constraint-1 y y 1 800 600 x y 1 800 800 x y 1 350 350 y 100 800 : X 0 M Constraint-5 TE Assuming X and Y are the two control variables. : 011-41013406 | M.org | E-mail : ies_master@yahoo. New Delhi . the decision of the management of IES MASTER shall be final.co. 8130909220 Web : www. 2. The student must be of Classroom program/Online Test Series/ Interview Guidance/ Classroom Test Series/ Conventional Question Practice Program of IES MASTER .110 016 | Ph. Katwaria Sarai. : 9711853908.iesmaster.in SCROLL DOWN .2015 1 st 2 nd 3 rd 4 st th -20 th th 21 -100 will get consolation prizes NOTE : 1. Students who are enrolled in more than one course will be entitled only to a single prize based on their rank.IES MASTER Institute for Engineers (IES/GATE/PSUs) A wards for THE ACHIEVERS IN ESE . F-126. In all cases. S.OUR TOP RESULTS IN ESE-2014 IES MASTER Institute for Engineers (IES/GATE/PSUs) Congratulation to All AIR AIR AIR 16 2 ME 21 AIR 32 23 AIR 49 ME ASHISH CHAUHAN PRAKHAR GUPTA AJAY CHAUDHARY HIMRAJ KUNAL AIR AIR AIR AIR AIR AIR 60 63 82 104 105 115 ME DEEPAK JHALANI ME ME SURYA PRAKASH DWIVEDI KRISHAN LALCHANDANI SANDEEP KUMAR ME ME 54 ME ME VAIBHAV AGRAWAL PARIGYAN SINGH ME ME ARNAV SINHA AIR 51 ME ME ME ABHINAV TRIVEDI AIR AIR and many more. SINGH SHUBHAM KHURANA CE AIR AIR MAYANK SAXENA CE CE CE ANKUR SINGH CE CE 40 CE CE 45 CE ANIL MITTAL AIR 70 KORRAPATI D. 9711853908 Web : www.. : F-126. BHARDWAJ AIR HIMANSHU AGGARWAL DEEPIKA GILL AIR AIR AIR 13 19 CE CE SUBHANKAR ROY ANKIT KUMAR GOYAL AIR AIR 35 30 CE CE AKASHY MISHRA ASHWINI KR.org | E-mail :
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[email protected] . (c) Ans. the link moves with an angular velocity of 20 rad/s. 0) (12.iesmaster. 9) E is given chemical equivalent = (12. (a) M (b) S 59. 5) Feasible region R H x1=12 TE Sol. New Delhi . The magnitude and direction of Coriolis component of acceleration with respect to angular velocity are (a) (b) Which one of the following properties of work materials is responsible for the material removal rate in electrochemical machining? (a) Hardness (b) Atomic weight (c) Thermal conductivity (d) Ductility 30 m / s2 and direction is such as to rortate slider velocity in the opposite sense as the angular velocity (c) 60 m / s2 and direction is such as to rotate slider velocity in the same sense as the angular velocity Sol. Compiler’s function is to translate high-level progress into object code. (12. 9) x1=9 (0. AS Z is maximized at any of the corners of feasible region i. 0) 22 x1 Z is valency of anode material 61. (b) Sol. 5) (d) 60 m / s2 and direction is such as to rotate slider velocity in the opposite sense as the angular velocity Ans. 9711853908. m IE = t F A Z where A is atomic mass of workpiece (10. Which of the following defines the compiler’s function correctly? (a) It translates high-lev el l anguage programs into object code It translates object code into a high-level language (c) It translates object code into assembly language instructions (d) In a crank and slotted lever type quick return mechanism. Mobile : 8130909220. (b) x2 where I is current in amperes. while the slider moves with a linear velocity of 1. 60.110 016. Katwaria Sarai. MRR in ECM is given as Office Address : F-126. Telephone : 011-41013406. F = Faraday’s constant (4. 7838813406 Web : www. 30 m / s2 and direction is such as to rotate slider velocity in the same sense as the angular velocity it translates assembly language instructions into object code IE Ans. The displacement of a follower of a cam in a printing machine is represented by the expression x 10 120 2 1500 3 2000 4 2500 5 1. The basic schematic of Ackerman steering mechanism. the right. 3 and 6 (c) 2. • This steering satisfy fundamental equation w of steering Cot cot = in three = 0 and two positions 25 toward left and right • Since very less sliding surfaces (turning pairs only) So longer life. New Delhi .4. 2. C • All pairs (A. If slider moves away from centre. (a) R Sol. C.org E-mail : ies_master@yahoo. W 63. 2. 4. B. and D) are turning pair in four bar mechanism A. Mechanically correct in all positions (b) 9000 3 Office Address : F-126. 5 and 6 (d) 1. V • Steering control is provided in front wheels only in all steering mechanisms.iesmaster. If the slider moves toward centre of rotation the Coriolis acceleration will be opposite to link rotation as shown below. B. 3 and 5 V C 2V TE Ans. Telephone : 011-41013406. The Coriolis acceleration ac = 2V = 2× 1. Mobile : 8130909220. Has only turning pairs 6. Has both sliding and turning pairs where is the angle of rotation of the cam.co. 3. A Which of the following are associated with Ackerm an steering mechanism used in automobiles? B D positions only namely in straight motion C Since nothing is mentioned about velocity of slider (away or toward centre) so assume it moves away from centre. 7838813406 Web : www. Less friction and hence long life (a) 900 3 48000 3 1500003 2 3. 9711853908.in .110 016.5 × 20 AS = 60 m/sec2 A M Direction: The direction of Coriolis component of acceleration depends upon velocity of slider ‘C’. Controls movement of two front wheels (a) 2. Due to longer life it is used generally despite it is not correct mathematically. 5 and 6 (b) 1. Mathematically not accurate except in three positions 5. C and D. The schematic of quick return mechanism Sol. the Coriolis acceleration will be in the direction of link rotation. The jerk given by the system at any position is 2. V 2V S C IE 2V A 62. Katwaria Sarai. Its acceleration after one seond is TE Ans. (a) Sol. When the crank angle is 45 . 7838813406 Web : www. Katwaria Sarai.org E-mail : ies_master@yahoo. Since the jerk is third derivative of follower displacement Jerk. v = ds = 6t 2 – 6t + 2 dt dv d2 s dt dt IE = 12t – 6 d2 x = (0 240 9000 dt d 240002 500003 ) dt dt3 12 m / s2 = (6 × 2t) – 6 + 0 = 2 (240 9000 240002 500003 ) (c) a = Acceleration.110 016. = 6 m/sec2 The crank shaft of a reciprocating engine having a 20 cm crank and 100 cm connecting rod rotates at 210 r. Telephone : 011-41013406. Displacement of follower in cam x = 10 1202 15003 20004 25005 AS culure = is angle of rotation of cam. dx 2 = (10 240 4500 dt 80003 125004 ) d3 x 6 m / s2 …(i) dt3 80003 125004 ) (a) Ans.(c) 240 2 9000 2 240002 2 3 2 = (9000 48000 150000 ) 64. 9711853908. the velocity of piston is nearly (a) 1.iesmaster. dx (10 240 45002 dt S d dt (d) 3 m / s2 Displacement of body s = 2t 3 – 3t 2 + 2t + 1 Velocity.m.in .co. J = d3 x Sol.8 m/s (b) 1. New Delhi . Mobile : 8130909220. (a) M A body starting from rest moves in a straight line with its equation of motion being R 500003 3 a = 12 × 1 – 6 65.9 m/s (c) 18 m/s (d) 19 m/s 2 = (0 9000 48000 1500002 ) d dt Office Address : F-126.p. (d) 480003 1500003 2 s 2t3 3t2 2t 1 where s is displacement in m and t is time in s. at t = 1 sec Assuming cam rotates with uniform angular velocity Jerk (b) 2 m / s2 Acceleration. (b) M The velocity of piston-P. Mobile : 8130909220. (b) Sol. 67.110 016.iesmaster.Ans.1) = 1. D At the instant shown in figure. At the instant shown. Telephone : 011-41013406. The schematic of mechanism sin2 v r sin 2n C 1=5rad/sec sin90 v = 0. the linear velocity of point B and C will be same VBA = VCD 1.org E-mail : ies_master@yahoo. Katwaria Sarai.7756 m/sec 66. New Delhi . (a) Ans.co.8 m/sec While designing a cam. AB is rotating at 5 rad/s and CD is ratating at 2 rad/s: P R r = 20 cm x B = 45 Angular velocity of crank TE Piston C A 2N 2 210 = 22 rad/sec 60 60 Ratio of length of connecting rod to crank length l 100 5 r 20 The length of AB is AS n D (a) 10 cm (c) 30 cm (b) 20 cm (d) 40 cm Ans. 9711853908. Reciprocating engine mechanism cm 100 = A four-bar mechanism is as shown in the figure bleow.2 sin 45 25 S B A 2 =2rad/sec IE 1 1 = 0. pressure angle is one of the most important parameters which is directly proportional to (a) pitch circle diameter Since the difference between AB and CD (b) prime circle diameter (c) lift of cam (d) base circle diameter 1AB 2CD 5AB = 2CD …(i) CD – AB = 30 From equation (i) Office Address : F-126.4 × (0. AB is shorter than CD by 30 cm.2 22 2 10 = 4.707 + 0. 7838813406 Web : www.in . Sol. 9711853908. Weight of flywheel.57 rad/sec Maximum fluctuation of energy e = Emax – Emin = 1 2 1 2 Imax Imin 2 2 Office Address : F-126. A governor is said to be isochrnous when the equilibrium speed is AS 68. This isochronism character is possible in spring controlled governors only not in dead weight types. the maximum fluctuation of energy is TE 3 AB 30 2 R CD = = 2 (125 120) rad/sec 60 = (13. Mobile : 8130909220. AB = 2 30 = 20 cm 3 70. 7838813406 Web : www. Telephone : 011-41013406. Katwaria Sarai. The flywheel of a machine having weight of 4500 N and radius of gyration of 2 m has cyclic fluctuation of speed from 125 r. (d) Sol.m.iesmaster.co. (a) Sol. to 120 r. w = mg = 4500 N Radius of gyration. the axis of at least one gear rotates (not fixed) about to the fixed frame. The isochronous governor has zero range of speed.org E-mail : ies_master@yahoo. (d) a relative motion of axes and none of the axes of gears has relative motion with the frame 5 AB 2 5 AB AB 30 2 Ans. A governor is said to be isochronous when equilibrium speed is same or constant for all radii of rotation of balls within working range.09 rad/sec min = 12.110 016.p. S M (a) variable for different radii of ratation of governor balls (b) constant for all radii of ratation of the balls within the working range (c) constant for particular radii of ratation of governor balls (d) constant for only one radius of ratation of governor balls Ans.in .p. A planetary gear train is a gear train having (a) 12822 N-m (c) 14822 N-m 69. Assuming g = 10 m/s2.57) rad/sec max = 13.09 – 12.m. New Delhi . k = 2 m Speed fluctuates between = 125 – 120 rpm IE Sol. (a) a relative motion of axes and the axis at least one of the gears also moves relative to the frame (b) no relative motion of axes and no relative motion of axes with respect to the frame (c) no relative motion of axes and the axis of at least one of the gears also moves relative to the frame (b) 24200 N-m (d) 12100 N-m Ans. (b) In planetary gear train or compound epicyclic gear train. The beauty of this train is that it ensures large speed reduction in small space. Sol. 7838813406 Web : www..52 71. 3 and 4 only (d) 1. (b) 1.m. Ans.e. For complete balancing or simply balancing of m echan i cal syst em .in . 2. Alumina doped with magnesia will have reduced thermal conductivity because its structure becomes (a) (b) (c) (d) There should not be any couple in the system System should be statically balanced i. no force should airse due to rotation. Support reactions due to forces are zero but not due to couples 4. (2) AS 73. then there will be dynamic balance also (b) If it is under dynamic balance.110 016.iesmaster. (a) 1.66 × 0. The accelerometer is used as a transducer to measure earthquake in Richter scale. M amorphous free of pores crystalline mixture of crystalline and glass Ans. New Delhi . Centre of masses of the system lies on the axis of rotation.57) = 12009 N. requires static balancing (centre of mass of system at axis of rotation) of system as pre condition. (1) So the nearest answer is 12100 N. (c) For complete dynamic balancing of rotating systems.e. R = = 900 × 25.m. 9711853908. Its design is based on the principle that (a) its natural f requency is v ery low in comparison to the frequency of vibration (b) its natural frequency is very high in comparison to the frequency of vibration (c) its natural frequency is equal to the frequency of vibration Office Address : F-126.e. T he dynam i c balancing i.. (d) 1 I(max min )(max min ) 2 TE = 74.. then there will be static balance also (c) Both static as well as dynamic balance have to be achieved separately Ans. Telephone : 011-41013406. 2. The system is automatically statically balanced Hence in complete dynamic balance.57) 2 10 (13.co. Which of the following statements is correct about the balancing of a mechanical system? (d) None of the above IE 72. centre of mass of the system should be on axis of rotation S W hi ch of the f ol lowing statements are associated with complete dynamic balancing of rotating systems? Sol. Resultant couple due to all inertia forces is zero 3. 2 and 3 only (c) 2. no couple. 3 and 4 (b) 1. Mobile : 8130909220.1 mk2 (max min )(max min ) 2 = 1 4500 2 2 (13. the reaction on the support are due to weight of system and remain constant during rotation of system i.09 12. 3 and 4 only (a) If it is under static balance. Katwaria Sarai.org E-mail :
[email protected] 12. gm M ge = 6 ms M M Xms 3 T = 2 2 k k l T m = 2 g m l = 2 g e 6 77. Katwaria Sarai. Taking Young’s Office Address : F-126. 7838813406 Web : www.iesmaster. (c) S (a) 1 2 (b) 1 3 (c) 1 4 (d) 3 4 Ans. New Delhi . As compared to the time period of a simple pendulum on the earth.co. the effect of the mass of the spring is accounted for by adding X times its value to the mass. = Te 6 gravity in ms 1 th of acceleration due to gravity 6 k at earth. (b) Sol.org E-mail : ies_master@yahoo. 6l = 2 g e Ans.(d) measurement of vibratory motion is without any reference point. Mobile : 8130909220.110 016. where X is The period of oscillation of spring-mass system including the mass of spring. 6 e The natural frequency of accelerometer should be kept around frequency of vibration/excitation in order to read the very weak earthquake. the acceleration due to IE Sol.in . (c) R l = 2 g . its time period on the moon will be (a) 6 times higher (b) 6 times lower 6 times higher (d) 6 times lower M (c) Ans. Telephone : 011-41013406. X = 1 3 A block of mass 10 kg is placed at the free end of a cantilever beam of length 1 m and second moment of area 300 mm 4. …(i) Since on Moon. The time period of simple pendulum on earth l Te = 2 g e While calculating the natural frequency of a spring-mass system. 9711853908. 76. TE In case of resonance the peak is control by proper dampnig. AS 75. iesmaster. The speed rating for turbine rotors is invariably R (a) 30 2 rad s 78. TE The cantilever vibrator. 7838813406 Web : www.co. M Young’s modulus. = 2 (/n ) Hence force transmitted to foundation by turbine is less than it impress upon isolator.0 S mg3 st 3EI 'Tr' Stiffness constant IE Sol. New Delhi . Katwaria Sarai. Ans. Mobile : 8130909220. the transmissibility T r is less than one as shown below The deflection at free end Turbine rotor speed range 1. So t he v i brat i ons are i sol at ed f rom surroundings.110 016. (b) AS Second moment of inertia.modulus of the beam material as 200 GPa. Telephone : 011-41013406.org E-mail : ies_master@yahoo. the natural frequency of the system is (b) 2 3 rad s (c) 3 2 rad s (d) 20 3 rad s more than M =10 kg I = 300 mm 4 E = 200 GPa Sol. mg 3EI k = 3 st k = 1. The magnitude of swaying couple due to partial balance of the primary unbalnacing force in locomotive is (a) inversely proportional to the reciprocating mass Office Address : F-126. = 1m 2 times its natural frequency to (a) increase stability under heavy load and high speed (b) isolate vibration of the system from the surrounding (c) minimize deflection under dynamic loading as well as to reduce transmissbility of force to the surrounding (d) none of the above Ans. (c) The rotating speed of turbine rotor is more than 2 times of natural frequency (n ) of system. 9711853908.in . Because beyond this speed.0 3 200 109 300 1012 3 = 600 × 300 × 10–3 = 180 N/m Frequency of oscillation n = k 180 = 3 2 rad/sec m 10 79. 5.org E-mail : ies_master@yahoo. New Delhi . Copper has FCC structure.9×103 8. AS (1 c) mr2 cos Cylinder–1 a/2 x M x a/2 Cylinder-2 The power of governor is defined as = sleeve displacement × effort TE (d) directly proportional to the distance between the centrelines of the two cylinders (d) each governor ball for given percentage change of speed R (b) directly proportional to the square of the distance between the centrelines of the two cylinders (c) inversely proportional to the distance between the centrelines of the two cylinders 2 S (1 c) mr cos 90 (a) (b) (c) (d) 8.46 × 10–30 m3 a 2 = (1 c)mr cos 2 = a (1 c)mr2 cos( 90) 2 = (1 c) 80.110 016. y and z . (c) Sol. The density of copper will be 82.co.9×103 8.9×103 8. Telephone : 011-41013406. Density (g/m3) = Volume × Avagadro's Number 3 Volume of a unit cell = a3 = (2 2R) = 47. Ans.28 A and atomic mass is 63. 9711853908. 81.Ans.023 1023 = 8. 7838813406 Web : www. mr a (cos sin ) 2 The power of a governor is the work done at (a) the governor balls for change of speed (b) the sleeve for zero change of speed (c) the sleeve for a given rate of change of speed 4 63. (c) IE The swaying couple due to unbalance force about centre line x – x The effort of governor is defined as average force on sleeve for a given change of speed. 8.46 1030 6.iesmaster.9 × 106 g/m3 or 2 kg/mm3 kg/cm3 kg/m3 g/mm3 Ans.9 × 103 kg/m3 A plane intersects the coordinate axes at 2 1 1 .5 47. its atomic radius is 1. What is the Miller index 3 3 2 of this plane? x (a) 932 (c) 423 (b) 432 (d) 364 Office Address : F-126. Katwaria Sarai. So the power can be defined as work done at sleeve for given rate of change of speed. (d) Sol.in . Mobile : 8130909220.9×103 Mass × No. of atoms in a unit cell Sol. Katwaria Sarai. (d) but Sol.org E-mail :
[email protected] R also R = 2 1 1 Intercepts are 3 3 2 3 Take reciprocal intercepts i.in .134 0. M 83.25 (c) 0. Consider the following staements regarding the behaviour of dislocations: 1.0684 a Doctahedral = 2 × Roctahedral = 0. Only edge dislocation and mixed dislocation can have glide motion. 3 a 4 (b) 0.136 a If the atomic radius of aluminium is r.110 016.5 IE Sol. (a) 84. 9711853908.iesmaster.Ans. Aluminium has FCC crystal structure a = 2 2r a volume of unit cell = a3 = (2 2r)3 = 8×2× 2r 3 = 16 2r 3 a 3 4r or 2 a The void formed is an octahedral void.e. 2 2 R R = 1 R 3 3 R 1/3 TE 2 3 = 4 R for BCC will only 3 4r (d) 3 Ans.433 Ans. 3 2 2 Roctahedral = AS Converting to whole number by multiplying by 2 miller indices = (3 6 4) What is the diameter of the largest sphere in terms of lattice parameter . (b) Sol. which will fit the void at the centre of the cube edge of a BCC crystal? (a) 0. Telephone : 011-41013406. Office Address : F-126. what is its unit cell volume? 3 3 2r (a) 2 4r (b) 2 3 S 3 2r (c) 3 (d) 0.158 3 a 4 = 0. The largest radius that can jet in it a Roctahedral = R 2 85.co. 7838813406 Web : www. a = 1/2 Roctahedral = 0. New Delhi . Mobile : 8130909220. New Delhi .2 wsolid (at 20% Ni) = C C = 0. What is the weight ratio of solid phase and liquid pahse? (a) (b) (c) (d) M • Dislocations move perpendicular to direction of shear stress along a slip plane containing Burger’s vector & direction vector. 0% 100% C2 C0 0. then the compositions of the two solid phases are IE S • For screw dislocation there is no specific glide plane defined. 20 30 20 30 wt% wt% wt% wt% B B B B and and and and 90 90 80 80 wt% wt% wt% wt% B B B B Ans.110 016. 1. A screw dislocation cannot have glide motion.4 or 40% w solid wliquid = 0.6 3 = 0. 1. Cu-Ni phase diagram is given as (It is a binary isomorphos system) 87. Mobile : 8130909220.co.16 2 1 • Only edge & mixed dislocation can glide along a specific plane defined. (b) Sol. Dislocat ion mov es in t he direct ion perpendicular to that of shear stress. the eutectic phase contains equal amounts (by wt) of two solid phases.2. 86.26 0. Elements A and B form eutectic type binary phase diagram and the eutectic composition is 60 wt% B. Katwaria Sarai. 1. If just below eutectic temperature. S Cu100% 80%Cu Ni0% 20%Ni %composition AS (a) (b) (c) (d) C1=16% Ni {Given C2=26% Ni data} C1 1453°C L+S 1085°C TE 4. 2. 3. (a) 1 : 1 (b) 3 : 2 (c) 2 : 3 (d) 1 : 2 Ans. the composition of Office Address : F-126. 9711853908. 7838813406 Web : www. (d) Sol. (b) Sol. Equilibrium phase diagram for a eutectic alloy is eutectic point L L L x A 100% B 0% 40% A 60% B y A 0% B 100% Just below eutectic point.org E-mail :
[email protected] . Which of the above statements are correct? L Temp. Dislocation R C0 C2 wliquid = 1 – wsolid = 0. 3 and 4 Ans. Telephone : 011-41013406.4 2 A binary alloy of Cu and Ni containing 20 wt% Ni at a particular temperature coexists with solid phase of 26 wt% Ni and liquid phase of 16 wt% Ni. Motion of disloacation occurs on slip plane that contains Burger’s vector and direction vector.26 0.iesmaster. 2 and 3 2 and 4 only 3 and 4 only 2. 67% B 50% B and 50% of eutectic 40% B and 60% of eutectic Ans. 3 and 4 (b) 10% (d) 14% Ans.6 TE 0.5 yx 0. Pearlite contains how much cementite? R 89.8% C. M There. It is a reversible reaction which can take place both ways.6 x = 0.org E-mail : ies_master@yahoo. At room temperature 88. 9711853908.5x (x + y) = 1.2 Sol. the phase diagram is Ans. at eutectic temperature.110 016. & ) be x and y L respectively.Sol.5 x Amount of ferrite × %carbon in ferrite + Amount of cementite × % carbon in cementite = overall % carbon in pearlite.3% of eutectic and 16. 1.2 Similarly 0.1199 or 12% 6.6 (a) 8% (c) 12% x+y = 1. cementite contains 6. 2 and 3 only 3 and 4 only 2 and 4 only 2.67 Two metals A and B are completely immiscible in sloid and liquid state. The 50% B alloy contains.5y 0. (b) Sol.co. They form eutectic at 200 ºC with 40% B and 60% A.6 = 0.67% C and pearlite contains 0.67 = 0.5y + 0.5(given) wd = yx y – 0. 7838813406 Web : www. liquids of two metals will change into two solids Which of the above statements are correct? (a) (b) (c) (d) 1.5y + 0. (a) (b) (c) (d) 83.in .iesmaster.6 – x = 0.8 = 0.5y – 0. (c) AS w = At room temperature.6 = 0. 0. eutectic reaction is irreversible 4. 30 wt% B and 90 wt% B is and phases.5 x 0. 1. For no miscibility in liquid and solid state.33% B and 16.67% of eutectic 83. as % carbon in ferrite is negligible at toom temperature. Mobile : 8130909220. 1. Melting point of A is 800 ºC and melting point of B is 600 ºC. Katwaria Sarai. Eutectic reaction is two phases (i.5 x = 0. the freezing point of the alloy is minimum IE 2. eutectic mixture solidifies at a constant temperature like pure metal 3. y 0. -iron contains negligible amount of carbon. Telephone : 011-41013406. (c) Office Address : F-126.5(x + y) = 0. Consider the following statements : In a binary phase diagram S 1.e. Amount of cementite × 6. y – 0.e. is only one option satisfying the conditions i.6 = 0. New Delhi .8 cementite = 90. 762 A = 1.e.246A° (given) TE C1 = 4R 8R2 round bars and tubes billets dies rectangular blocks Ans.iesmaster.24 Å and d(111) = 1.5 Å. Inteplaner spacing is given as d = for FCC a 2 h k 2 2 d111 = 3. Katwaria Sarai. M C2 C0 100 50 Quantity of eutectic = C C = 100 40 2 1 92.524A° AS Composition at point A i. Consider the following statements: In shell moulding 1.762 Å and d(111) = 2. Eutectic point a2 600°C R A a = 2 2R 200°C A 100°C B% C0 60%A 50%A 40%B 50%B C2 0% 100% %Composition d200 = d220 = C0 C1 50 40 10 Quantity of B = C C = 100 40 = 60 2 1 = 16.246 A? (a) d(200) = 1. (220). (a) Sol.24 Å.6 Å Ans. minimum rounding radii of 2. a single parting plane should be provided for mould 2.246 = 3.524 2 2 22 02 3. Rotary swaging is used to shape round bars and tubes such as gun barrels 93. d(220) = 1.034 Å (b) d(200) = 1. d(220) = 4.co. 3 and 4 only (b) 1. a R = 1.33% = 60 91. draft angles of not less than 1° should be used Which of the above statements are correct? (a) 1.a2 + a2 = (4R)2 800°C 2a2 = 16R2 Temp. d(220) = 1. Telephone : 011-41013406. (50%B – 50%A) can be found oil by lever rule.org E-mail : ies_master@yahoo. IE S What is the interplanar spacing between (200).034 A Rotary swaging is a process for shaping (a) (b) (c) (d) 50 = 83.24 Å and d(111) = 2.5 mm to 3 mm should be used 4.524 2 1 12 12 = 1.246 A = 2.2 Å and d(111) = 2.67% of B. d(220) = 1.in . (111) planes in an FCC crystal of atomic radius 1. 9711853908. a a = 2 2 × 1. Mobile : 8130909220.110 016. 2 and 3 only Office Address : F-126.034 Å.762 Å (d) d(200) = 2. detachable pattern parts and cores could be included 3. 7838813406 Web : www.524 22 0 2 0 2 3.762 Å.034 Å (c) d(200) = 2. New Delhi . (a) Sol. 3 and 4 only (c) 2. Ans.32 N (c) 1 12 and density core is 0. 2.077 0. 2 and 3 (b) 1 and 2 only (c) 1 and 3 only (d) 2 and 3 only Ans. 3 and 4 only 2. It causes the weld bead and the surrounding area to have a scummy appearance. to be produced by using a core of 10 cm diameter and 200 cm long.62 N (d) 350. to correct minor surface imperfections 3. The density metal is 0. Which of the above statements are correct? (a) 1. Oxidising flame • Forms hard oxides which protect the weld metal Office Address : F-126. • In shell molding • Single parting plane should be provided • Detachable pattern parts & cores should be avoided.4 25.32 N Ans. pitch in mm = Gear ratio = 25. 1.077 N/cm3 M Sol. At high temperature.org E-mail : ies_master@yahoo. A big casting is to have a hole.5 to 3 mm • Draft angle should not be less than 1°. 3 and 4 25. Centre lathe is to be used to cut inch thread of 4 threads per inch. Lapping gives • Geo metrically true surface • Corrects minor surface imperfection • Improves dimensional accuracy Sol.110 016.4 pitch in mm 25.• Very close fit between mating surface (d) 1. Mobile : 8130909220.0165 N/cm 3. to produce geometrically true surface IE 127 60 30 127 (d) 20 80 Sol. (d) Ans.5 N (c) 950. (b) AS (a) 200. to improve dimensional accuracy 4.co. 7838813406 Web : www. 9711853908.4 254 or = 43 12 120 127 60 Consider the following statements in respect of the oxidizing flame due to excess of oxygen in welding: or 97. (a) Sol. 3 and 4 only (d) 1. (a) Sol. 2.iesmaster.in .4 = TPI 4 Also pitch (mm) = Gear ratio × pitch of leads crew. 2 and 3 only (b) 1. • Minimum rounding radii is 2. 2. New Delhi .0165 = 950. Katwaria Sarai. Threads per inch (TPI) = (b) 1100. 3.4 25. What is the upward force acting on the core prints? (a) TE 94. it combines with many metals to form hard and brittle oxides. Buoyant force acting on core. to provide very close fit between the contact surface Which of the above are correct? (a) 1. 3 and 4 96. It has good welding effect in welding of copper-base metal. (c) (b) Ans. (metal core ) × /4×d2 2 = /4× 10 200 0. Then change gear tobe used is R (c) 2. Telephone : 011-41013406. The purpose of lapping process is 1.33 N Consider the following : S 95. Lead screw of lathe has 3 mm pitch. • Has good welding effect on copper base alloys.(ii) Comparing (i) & (ii) AS Ans. (d) . Taylor’s tool life equation is vTn = C v 1 × (T1)0. Y = 120 mm.iesmaster. If n = 0.2s 250 0. The X and Y axes have maximum speed of 10000 mm/min and 5000 mm/min respectively.110 016..75 16 T1 9 Increase in tool life 16 T1 T1 T2 T1 = 100 100 = T1 T1 = For x axis cutter has to travel 150 mm at 10000 = T2 = 250 mm/sec i.75v1 T20.. A cutter tip is initially at X = 10 mm.e. 4.20 s (d) 2.78% or 78% 9 100. New Delhi .75 3 9 (160 .in . It provides better accuracy during machining.5 and C = 300 for the cutting speed and the tool life relation.9 + 1.(i) when speed is reduced by 25% TE 98.. what is the time it will take to reach the destination? (a) 0. y V1 T10. It causes dimensional changes in workpiece and affects accuracy of machining. (a) 1 and 2 (b) 2 and 3 (c) 3 and 4 only (d) 1. = = 0. the tool life will be increased by (a) 100% (c) 78% (b) 95% (d) 50% 1 0.co.1 s 99.5 = 300 Sol. when cutting speed is reduced by 25%.75v 1 × (T2)0. Telephone : 011-41013406.5 T2 T 1 M S x For y axis cutter has to travel 100 mm at 5000 IE 500 150 3 mm/s i. It adversely affects the properties of tool material. 3. 9711853908.. using G00 code.5 = 0. it will take = mm/mm 3 100 3 = 1. It can distort the accuracy of machine tool itself. it moves to X = 160 mm. Ans. 7838813406 Web : www. 2.2 = 2. Operating at maximum speed.org E-mail : ies_master@yahoo. 120) (10. Y = 20 mm. manganese steels and east iron. Mobile : 8130909220.90 s (b) 1.9 s mm/min 3 500 Total time = 0. In a rapid motion. 20) . Which of the following statements are correct for temperature rise in metal cutting operation? 1. Katwaria Sarai.5 2 2 T2 16 1 4 = = = T1 0. (c) R • Gives dirty & scummy appearance to weld bead 700 = 77.5 = 300 0.e. 3 and 4 Office Address : F-126.08 s (c) 1.16 s Sol. (d) 101. (c) Thin weld is designed for shear strength of weldment throat. (a) (c) squre key (d) rectangular key with width equal to onefourth the thickness Ans.iesmaster.110 016. Temperature rise does not give better accuracy during machining. shear strength Which of the above is/are correct? (a) 4 only (b) 3 only (c) 2 and 3 (d) 1 and 4 R Sol. bending strength 4.707h The force are parallel to minimum threat area so it is designed for shear strength only. If the permissible crushing stress for the material of a key is double the permissible shear stress. 102.Ans. it should be a square key Office Address : F-126. It can also distort the tool due to thermal expansion. Consider the following: F F A 0. the thickness of threat.707 h Area of threat. tensile strength 2. HIgh temperature affects hardnen of cutting edge. Given c = 2 S F M The parallel fillet welds along with force F. Telephone : 011-41013406.org E-mail : ies_master@yahoo. compressive strength 3. IE Sol. AS Ans. Sol. 9711853908. then the sunk key will be equally strong in sheaering and crushing if the key is a TE The parallel fillet welded joint is designed for 1. = 1 or = t t Thus. (a) rectangular key with width equal to half the thickness (b) rectangular key with width equal to twice the thickness t = 2h = 2 h k h h 2 = 0.in . 7838813406 Web : www.co. Also crushing force = c Crushing torque = c t 2 t d 2 2 Shearing torque = d 2 for equal strength in shearing & crushing d t d = c 2 2 2 = c t 2 A = t = 0. Mobile : 8130909220. New Delhi .707 h Shear stress. Katwaria Sarai. org E-mail : ies_master@yahoo. Aquaplaning occurs in vehicle tyres when there is continuous film of fluid between the tyre and the wet road. (c) AS Sol.iesmaster. Very small quantity of carbon in iron as in steels forms interstitial solid solution mainly because atomic size(s) of Ans. (b) (b) 64% (d) 44% Ans. 9711853908. the width of the cotter at the centre is 5 cm.in . (a) Sol.15 between screw and nut. 104.2 cm.15 0. is given by R 103.15 tan 45 = 11 0. In this situation. T hi s oper at i on happens due t o ax i al movement of spline hub or shaft ember. For a power screw having square threads with lead angle of 45° and coefficients of friction of 0.110 016. the efficiency of the power screw. The shaft in spline is under pure torsion i. 106. It leads to (a) oscillatory motion of the vehicle (b) jamming the brakes of the vehicle (c) jamming the steering mechanism of the vehicle (d) loss of control of the vehicle (a) torsional misalignment (b) parallel misalignment (c) angular misalignment (d) substantial axial movements between shafts Ans. Mobile : 8130909220. Example clutch. The shear stress developed in the cotter is Ans. shear. 7838813406 Web : www. These couplings can connect Ans. Aquaplaning or hydroplaning is the condition in vehicle or aircraft when there is water layer between tyre and road surf ace. loss of control of Office Address : F-126.15 P 60 103 60000 = = = 2 b t 12 50 600 IE c S Sol. neglecting collar friction. The gear and spline couplings are used in torque transmission when key fails to serve the purpose. In a cotter joint.TE (a) carbon and iron are almost same (b) iron is very much smaller than that of carbon (c) carbon is very much smaller than that of iron (d) None of the above or disconnect the driving and driven shaft according to wish of operator of requirement. New Delhi .15 = 0. (d) Sol. while its thickness is 1. Efficiency of power screw is = 90 = 45 (b) 100 N/mm2 (d) 200 N/mm2 M (a) 50 N/mm2 (c) 120 N/mm2 (a) 75% (c) 54% = tan 1 tan tan tan = tan tan tan = tan 45 1 tan 45 0. Carbon forms interstitial solid solution with iron as size of carbon atom is very small as compared to size of iron atoms.739 or 74% 1.. The use of straight or curved external gear teeth in mesh with internal teeth in ‘gear and spline couplings’ is specifically employed to accommodate. Telephone : 011-41013406. 107..e. Shear stress in cottor = 100 N/mm2 105.e. The load acting on the cotter is 60 kN. there is excessive skidding or slipping of vehicle i. Katwaria Sarai.co. (d) Sol. If the angle of wrap on smaller pulley of diameter 250 mm is 120° and diameter of larger pulley is twice the diameter of smaller pulley. (b) r1 = 2r2 M or TE (b) 750 mm (d) 250 mm V2 100 = v2 = 1 1 0.9% (d) 57. 9711853908.05 tan60 1 0.110 016.04 rpm. Helix angle of worm = 30° Load angle of worm ( ) = 90° – 30° = 60° efficiency of worm gear = 1 tan 1 /tan = 1 0. Two shafts A and B of same material. This situation can be very dangerous when al l t ype h av e equapl ani ng simultaneously. (a) Sol.04 rpm (b) 102.9% Ans. then the centre distance between the pulleys for an open belt drive is Sol. The efficiency of the worm gear set at an interface friction of 0. then considering an elastic creep of 2% the speed of the driven pulley is (a) 104. Velocity of driver R = Ans. If the velocity ratio for an open belt drive is 8 and the speed of driving pulley is 800 rpm. The torque that can be transmitted by A is (a) 2 times that of B (c) 4 times that of B (b) 8 times that of B (d) 6 times that of B Sol.05/tan 60° = 0.8877 1.vehicle. (d) D1 = 2D2 sin = r1 r2 x S wrap angle = 180 – 2 = 120° 2 = 60° = 30° IE or sin 30° = x = 110. Given Sol. 7838813406 Web : www.05 is (a) 87. A worm gear set is designed to have pressure angle of 30° which is equal to the helix angle. Katwaria Sarai.9% (c) 67.04 rpm (c) 100.04 rpm Ans. (b) 250 125 x 125 = 250mm sin30 109. New Delhi .02 AS (a) 1000 mm (c) 500 mm v1 800 = = 100 rpm V R 8 Due to creep 108. Torque transmitted by a shaft T d3 where x is the center distance between the pulleys for smaller pulley = 102.98 Ans. Telephone : 011-41013406.04 rpm TA = 2 3 TB = 8TB 111.9134 = 0.9% (b) 77.co. 0. and A is twice the diameter of B.iesmaster.in . Mobile : 8130909220. v2 = 100 = 102..0289 Office Address : F-126.04 rpm (d) 98.org E-mail : ies_master@yahoo. or 88.in .77% contact ratios are not equal. both radial & tangential stream are compressive. 112. In case the number of teeth on two bevel gears in mesh is 30 and 60 respectively. Sol. 2. Large axial contact ratio will cause larger axial force component. Axial contact ratio & transverse contact ratio. Consider the following statements: In case of helical gears. In an interference fit between a shaft and a hub. They are less noisy. Pitch cone angle of gear (for metregears i. 3. Bevel gears are inherently non-interchangeable.iesmaster. Katwaria Sarai.co. 7838813406 Web : www. shaft is considered externally pressed radial = – Pf (external pressure) M Ans. Mobile : 8130909220.e. 2. (d) Sol.org E-mail : ies_master@yahoo. Helix angle introduces another ratio called axial contact ratio.110 016. For interference fit between hub & shaft. Both Office Address : F-126. Large contact ratio’s lead to engagement of multiple teeth. The most common pressure angle for spiral bevel gear is 20°. 3. 4.5 2 (b) tan1 2 2 (d) tan–1 0. the state of stress in the shaft due to interference fit is R The axes of spiral bevel gear are non-parallel and intersecting Sol. (d) rf2 rL2 rf = outer radius 113. Pf rf2 rL2 fi = inner radius Thus. The most common spiral angle for spiral bevel gear is 35°. 115.e. Spiral bev el gears are generally interchangeable. Which of the above statements are correct? (a) 1 and 4 (b) 1 and 2 (c) 2 and 3 (d) 3 and 4 114. Spirals are noisy and recommended for low speeds of 10 m/s. Large transverse contact ratio does not allow multiple teeth to share the load. Transverse contact ratio is equal to axial contact ratio in helical gears. then the pitch cone angle of the gear will be (a) tan–1 2 (c) tan1 0. (b) Ans. teeth are cut at an angle to the axis of rotation of the gears. tan gential = – S IE 1. Consider the following statements : (a) only compressive radial stress (b) a tensile radial stress and a compressive tangential stress (c) a tensile tangential stress and a compressive radial stress (d) a compressive tangential stress and a compressive radial stress AS TE 1. Telephone : 011-41013406. Which of the above statements are correct? (a) 1 and 2 (b) 2 and 3 (c) 1 and 4 (d) 3 and 4 Ans. where is angle between two shafts. 9711853908. Helical gears have two contact ratios i. (a) Sol. Most common pressure angle is 20°C and most common spiral angle is 35°.5 Ans. 4. New Delhi . org E-mail : ies_master@yahoo. Mobile : 8130909220. Consider that modern machines mostly use short bearings due to the following reasons: IE S 1. Twinning occurs either as mechanical twinning or Annealing twins during annealing heat treatment. Twinning increases strength & reduces ductility as twin planes hinders the movement of dislocations. outer race and retainer are readily separate in thrust bearings. 119. and teeth are curved. Shaft deflection and misalignment do not affect the operation 4. Double-row thrust ball bearing is not possible. The behaviour of metals in which strength of a metal is increased and the ductility is decreased on heating at a relatively low temperature after cold-working is known as (a) clustering (c) twinning (b) strain aging (d) screw dislocation Ans. l/d of the most modern bearings is in the range of 1/4 to 2 2. New Delhi . (c) Sol.in . Twi nni ng i s a pl ane def e ct where arrangement of atoms on either side of a twin plane are identical. (d) Sol.iesmaster. and teeth are straight (c) intersecting. Cylindrical thrust bearings have higher coefficient of friction than ball thrust bearings. If the equivalent load in case of a radial ball bearing is 500 N and the basic dynamic load rating is 62500 N.P2 leakage of oil is not a problem. On the other hand. Consider the following statements in connection with thrust bearings: = tan–1 (2) 1. end leakage is a problem. 4. and teeth are curved and oblique (d) intersecting. and teeth are curved and can be ground AS TE 3. 120. However. Which of the above statements are correct? (a) 1 and 2 (b) 2 and 3 (c) 3 and 4 (d) 1 and 4 Ans. and teeth are curved (b) non-parallel and non-intersecting. TG 60 = tan1 = tan 30 TP 1 118. (c) Sol. Lower race. long bearings have greater load carrying capacity and end Ans. Taper rollers cannot be employed for thrust bearings.co. Telephone : 011-41013406. Katwaria Sarai. 9711853908. In skew bevel gears.110 016. 7838813406 Web : www. (a) Sol. the axes are non-parallel and non-intersecting. 2. • A long bearing is where d ( = length d = diameter) A shaft bearing has advantages such as (i) shaft deflection and misalignment do not affect operation (ii) compact design (iii) Run cooler. Can be applied to both hydro-dynamic and hydrostatic cases Which of the above are correct? (a) 1 and 4 (b) 2 and 3 (c) 1 and 3 (d) 2 and 4 Ans. 116. • Cylindrical thrust bearings have higher coefficeint of friction than ball thrust bearings • Both taper roller & double row thrust ball bearings are used. the axes are R (a) non-parallel and non-intersecting. M 117. In skew bevel gears. No end leakage of oil from the bearing 3. then L10 life of this bearing Office Address : F-126. 756 6.953125 million of revolution Office Address : F-126. New Delhi .is million million million million of of of of revolutions revolutions revolutions revolutions R 1.765 Ans.org E-mail : ies_master@yahoo. (a) Sol. Use of a ball bearing is given as: c L10 = w k AS where L10 = life in million revolutions TE (a) (b) (c) (d) c = Basic dynamic load rating w = equivalent dynamic load k = 3 for ball bearing 3 M 62500 L10 = 500 = (125)3 = 1953125 IE S 1.953 3. Telephone : 011-41013406.in .953 9. 7838813406 Web : www.co. 9711853908. Katwaria Sarai.110 016.iesmaster. Mobile : 8130909220.