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Ese-2017 Prelims Test Series Answers ESE-2017 PRELIMS TEST SERIES Date: 6th November, 2016 ANSWERS 1. (c) 24. (d) 47. (c) 70. (c) 93. (c) 116. (c) 2. (d) 25. (a) 48. (c) 71. (a) 94. (c) 117. (d) 3. (c) 26. (b) 49. (a) 72. (b) 95. (b) 118. (a) 4. (c) 27. (a) 50. (a) 73. (d) 96. (b) 119. (d) 5. (c) 28. (c) 51. (b) 74. (b) 97. (d) 120. (d) 6. (a) 29. (b) 52. (d) 75. (c) 98. (c) 121. (b) 7. (b) 30. (d) 53. (a) 76. (b) 99. (c) 122. (d) 8. (b) 31. (b) 54. (c) 77. (d) 100. (d) 123. (b) 9. (d) 32. (b) 55. (c) 78. (c) 101. (a) 124. (b) 10. (d) 33. (d) 56. (b) 79. (d) 102. (c) 125. (c) 11. (c) 34. (b) 57. (c) 80. (d) 103. (d) 126. (a) 12. (c) 35. (c) 58. (c) 81. (b) 104. (b) 127. (a) 13. (d) 36. (b) 59. (b) 82. (a) 105. (c) 128. (b) 14. (d) 37. (b) 60. (b) 83. (c) 106. (d) 129. (d) 15. (b) 38. (d) 61. (c) 84. (b) 107. (a) 130. (c) 16. (b) 39. (b) 62. (a) 85. (d) 108. (c) 131. (b) 17. (b) 40. (c) 63. (b) 86. (c) 109. (b) 132. (d) 18. (b) 41. (c) 64. (d) 87. (b) 110. (c) 133. (b) 19. (c) 42. (d) 65. (c) 88. (d) 111. (c) 134. (b) 20. (d) 43. (d) 66. (c) 89. (c) 112. (b) 135. (d) 21. (d) 44. (c) 67. (c) 90. (b) 113. (b) 136. (d) 22. (b) 45. (a) 68. (b) 91. (a) 114. (c) 137. (a) 23. (c) 46. (c) 69. (c) 92. (c) 115. (a) 138. (d) Office : F-126, Katwaria Sarai, New Delhi-110016 (Phone : 011-41013406, 7838813406, 9711853908) Website : www.iesmaster.org E-mail: [email protected] (2) ME (Test-7), Objective Solutions, 6th November 2016 139. (b) 147. (c) 155. (d) 140. (a) 148. (a) 156. (b) 141. (d) 149. (c) 157. (a) 142. (d) 150. (a) 158. (a) 143. (d) 151. (c) 159. (a) 144. (c) 152. (d) 160. (a) 145. (a) 153. (b) 146. (a) 154. (a) IES MASTER (3) ME (Test-7), Objective Solutions, 6th November 2016 Sol–1: (c) resistivity. Here, shortest link OA = 3 cm and the • A governor is said to hunt if the speed of link adjacent to the shortest link is fixed, the engine fluctuates continuously above hence it would result in crank-rocker and below the mean speed. This is caused mechanism. OA will act as a crank. by a too sensitive governor which changes Sol–2: (d) the fuel supply by a large amount when a small change in the speed of rotation Sol–3: (c) takes place. V = 2 2 = 2 2 • A governor is said to be stable when for B/A VVBA 30 40= 50 m/s every speed within the working range, Sol–4: (c) there is only one radius of rotation of the Radial component of acceleration of point governor balls at which the governor is in C. equilibrium. V2 10 2 • For a stable governor, if the equilibrium = = = 5 CD CD speed increases, the radius of governor balls must also increase. In the case of an 102 100 CD = = = 20cm unstable governor, the radius of rotation 5 5 decreases, as the speed increases. Sol–5: (c) The greater the lift of the sleeve Coriolis component of acceleration = corresponding to a given fractional change in speed, the greater is the sensitiveness 2 60 2 v = 2 × 10 = 40 cm / s2 of the governor. 60 Sol–11: (c) Sol–6: (a) Considering the case of a stable governor, Inversions of single slider crank F = ar –b mechanism are: Pendulum pump, c oscillating cylinder engine, crank and 1600 = 400 a – b ...(i) slotted lever, whitworth mechanism, 800 = 240 a – b ...(ii) snome engine. 1–2 800 = 160 a Sol–7: (b) a = 5 and b = 400 Sol–8: (b) Since, both ‘a’ and ‘b’ are positive, hence Sensitiveness of the governor our assumption of this being a stable NN 2NN governor is correct. = 2 1 = 2 1 N NN 1 2 Now, for isochronous governors FC = ar Sol–9: (d) Hence, initial tension must be increased by 400 N. IESV V MASTER B = C = BE CE BC Sol–12: (c) 0.5 V The given arrangement has no. of links = c 0.25 0.1 = 5 and no. of joints j = 5, no. of pairs p = 5. Vc = 0.2m/s Sol–10: (d) Applying the equation, = 2p 4, • A governor is said to be isochronous when LHS = 5, RHS = 2 × 5 – 4 = 6 the equilibrium speed is constant for all Thus, LHS < RHS radii of rotation of the balls. The isochronism is a stage of infinite (4) ME (Test-7), Objective Solutions, 6th November 2016 Sol–14: (d) 3 Further, applying the equation, j = 2 2 Inversion of single slider crank chain are: 3 LHS = 5, RHS = 5 2= 5.5 Pendulum pump; oscillating cylinder 2 engine; Rotory internal combustion Hence, LHS < RHS engine, crank and slotted lever quick Since, LHS < RHS, hence, it is return motion mechanism and whitworth unconstrained kinematic chain, i.e. the quick return motion mechanism. relative motion is not completely Inversion of Four bar mechanism constrained. This type of chain is of little are: practical importance. Beam engine; coupling rod of a locomotive; Sol–13: (d) Watt’s indicator mechanism The product of crank pin effort F and T Inversions of Double slider crank crank pin radius r known as crank effort. mechanism are: Crank effort T = F × r T Oldham’s coupling; Elliptical trammel and scotch yoke mechanism. F sin = s r cos Sol–15: (b) Mass moment of inertia of flywheel F r 2 2 2 = s sin cos tan I = mk = 300 × 1 = 300 kg-m We know that Angular velocity of flywheel sin = rsin 100 2 10 = = rad/s rsin sin 360 3 sin = = n Angular velocity of precession 2 p = 6 rad/s 2 sin cos = 1 sin = 1 n2 Gyroscopic couple = Ip 1 2 2 10 = n sin = 100 6= 6 kNm. n 3 sin sin n Sol–16: (b) tan = = cos n n2 sin 2 Beam engine mechanism is an inversion sin of four-bar mechanism, whereas = whitworth quick return mechanism is an n2 sin 2 inversion of single slider crank cos sin mechanism. Oldham coupling and sin IEST = Fs MASTER r 2 2 elliptical trammel mechanisms are n sin inversions of double slider crank sin 2 mechanism. = F r sin s 2 2 2 n sin Sol–17: (b) Since sin2 is very small compacted to A governor is said to be isochronous when n2, hence neglecting sin2 we have the equilibrium speed is constant for all radii of rotation of the balls within the sin2 sin working range, neglecting friction. T = Fs r = F s OM 2n (5) ME (Test-7), Objective Solutions, 6th November 2016 Sol–18: (b) Sol–20: (d) Toggle mechanism has many applications For a Porter governor, where it is necessary to overcome a large M tan resistance with a small driving force. A M 1 2 tan 895 stone crusher utilizes this mechanism to N2 = m h overcome a large resistance with a small 1 force. It can be used in numerous toggle Thus, h N2 clamping devices, for holding work pieces. dh 1 Hence, . The figures of Hart, watt and beam engine dN N3 mechanism are: Sol–21: (d) Beam Engine Mechanism A body in motion will be subjected to Lever (Link 4) Corioli’s acceleration when that body is restrained to rotate while sliding over E another body. D Sol–22: (b) Piston C rod A round bar in a round hole does not Link 3 form a constraint pair as round bar can Cylinder rotate as well as slide. A square bar in a A square hole forms a sliding pair. A vertical Frame (Link 1) B shaft in a footstep bearing may rotate in Crank the bearing or it may move upwards. This (Link 2) is a case of incompletely constrained Watt Mechanism motion. Since load is placed on the shaft A to prevent axial upward movement of the shaft, hence the motion of the pair is said O1 A to be successfully constrained motion. 1 P P Sol–23: (c) B Applied load convert an incompletely O B constrained motion into successfully constrained motion. The motion of piston Hart Mechanism and cylinder in IC engine is successfully E constrained because it can only reciprocate. Its rotation is restricted by means of gudgeon pin between piston and B C A connecting rod. D Sol–24: (d) O IESP MASTERN O M 1 Q A Slider Bar (Link 3) F (Link 2) Sol–19: (c) P Slotted plate (Link 4) This is Roberts mechanism and point P will trace an approximate straight line. B Slider (Link 1) (6) ME (Test-7), Objective Solutions, 6th November 2016 Y Sol–28: (c) A t Position and u Joint Forward p orientation of t angles Kinematics u end-effector O s P t Q u p y n I Figure: Forward Kinematics X' X Link R O Scheme B x Parameters Y' t Position and Joint angles u Let us take OX and OY as horizontal and Inverse p orientation of for t vertical axes and let the link BA be end-effector Kinematics manipulator u O s t inclined at an angle with the horizontal, u p n as shown in figure. Now the co-ordinates I Figure: Inverse Kinematics of the point P on the link BA will be Link Parameters Scheme x = PQ AP cos ; and y = PR BP sin Sol–29: (b) x y Typical contact type robotic sensors or = cos ; and sin AP BP include: Squaring and adding, • Force sensors x2 y 2 • Torque sensors 2 2 2 2 = cos sin 1 AP BP • Touch sensors This is the equation of an ellipse.
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