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. Hence • Position sensors the path traced by point P is an ellipse Typical non-contact robotic sensors whose semi-major axis is AP and semi- include: minor axis is BP. • Proximity sensors Sol–25: (a) • Electro-optical sensors The initial and installation costs of • Range imaging sensors equipments of robots are quite high. Sol–30: (d) Sol–26: (b) The wide range of gripping methods Yaw, also called, wrist yaw, facilitates include rightward or leftward swivelling movementIES of the wrist. MASTER• Mechanical gripping Sol–27: (a) • Magnetic gripping The cartesian or rectilinear robot also • Vacuum gripping termed as gantry robot, has three Sol–31: (b) mutually perpendicular axes which define The four basic components of a robot a rectangular work volume. system, are: 1. Manipulator 2. Sensory devices 3. Controller 4. Power conversion unit (7) ME (Test-7), Objective Solutions, 6th November 2016
TV camera Manipulator (visual sensory device) 900 Joint C.R. = 10 (contains actuator, 2 transmission, and sensor) = 0.8789mm Link Sol–35: (c) Given: Range of rotation = 3.5 × 360° = 1260° Controller Total control resolution = 0.31 = 0.62°
Power conversion Range of rotation Power to unit Control resolution = actuators on n joints 2
• The manipulator is composed of the 1260 0.62° = following three divisions: (i) The major 2n linkages; (ii) The minor linkages (wrist components); (iii) the end-effector (gripper 1260 2n = 2032.26 or tool). 0.62 • Sensory devices may monitor position, Taking log on both sides, we get speed, acceleration or torque. nlog 2 = log 2032.26 • Controller provides the “intelligence” to e e cause the manipulator to perform in the n = 10.988 11 manner described by its trainer/user. Sol–36: (b) • Power conversion unit contains the Control resolution for translation components necessary to take a singal from the sequencer (either digital or low- Range of movement (H H ) = max min level analog) and convert it into a 2n meaningful power level so that the actuators can move. 1500 750 = Sol–32: (b) 212 Sol–33: (d) = 0.183 mm The inverse kinematics problem can be Sol–37: (b) solved by several techniques, the most commonly used are: Sol–38: (d) • Work envelope depends upon the number (i) Matrix algebraic of types of joints, physical size of the joints (ii) IterativeIES approach MASTERand links and the ranges of various joints. (iii) Geometric approach • Since the arm movements in different robot Sol–34: (b) configuration are different, the work volumes or work envelopes of different Control resolution coordinate systems are also different. Range of movement Sol–39: (b) = n 2 • Usually, a full 360° rotation in is not where n = Number of bits devoted to the permitted, due to restrictions imposed by joint and 2n = Number of addressable points hydraulic, electrical or pneumatic connections or lines. (8) ME (Test-7), Objective Solutions, 6th November 2016
• Also, there is minimum as well as • Load capacity required maximum extension due to mechanical • Adequate number of DoF requirements. Consequently, the overall volume or work envelope is a portion of a • Speed of movement cylinder. • Reach (range of travel) Sol–40: (c) Sol–44: (c) Robots can be categoriesed as follows: Control Resolution Accuracy = 3 2 1. Tele-robots: These are guided by human operators through remote Range of travel control Control resolution = 2n 2. Telepresence robots: This type of robot is similar to a tele-robot with a 800 difference that it is provided with a = 0.78125mm 210 feedback of video, sound and other data. 3. Mobile robots: These robots are so 0.78125 Accuracy = 3 0.06 designed as to navigate and carry out 2 the task with the intervention of = 0.571 mm human beings. Sol–41: (c) 4. Autonomous robots: These robots perform their tasks independently and Number of slots, S = 15 receive their power from their Number of emitter detector pairs, n = 2 environment
Angular control resolution, Ac = 0.06° 5. Androids: These robots are built to mimic humans. Range of rotation = 2 360 6. Static + industrial robots: These are 360 the widely used arms employed in Gear reduction ratio, G = n S Ac 2 factories and laboratories all over the world. 360 Sol–45: (a) = 2 15 0.06 2 • A robot is a lighter and more portable equipment comparatively. = 100 • A robot’s programming is different from Sol–42: (d) the part programming used in NC machine • All these factors govern the speed of tools. movement. Sol–46: (c) • HeavierIES parts and higher placementMASTER • Spherical configuration has a telesocpic accuracy demand slower movement while arm which pivots about a horizontal axis the lighter parts can be moved at faster and also rotates about a vertical axis. speeds. • Owing to mechanical and/or actuator • The speed ‘varies’ from one point to connection limitations, the work envelope another. of such a robot is a portion of sphere. Sol–43: (d) Sol–47: (c) Selection of a robot depends upon the Most popular drive technologies are particular application, for which following factors should be considered. – Electric drives – Hydraulic drives (9) ME (Test-7), Objective Solutions, 6th November 2016
– Pneumatic drives finished quickly and the cylinder is • Many robotic manipulators today use cooled before further extrusion. electric drives in the form of either D.C. Backward Hot Extrusion servomotors or D.C. stepper motors. In order to completely overcome the However, for high-speed manipulation of friction, the backward hot extrusion, is substantial loads, hydraulic drives are used. In this, the metal is confined fully preferred, but their major draw back is by the cylinder. The ram which houses their lack of cleanliness. the die, also compresses the metal against the container, forcing it to flow • Electrically powered robots have the backwards through the die in the hollow advantage of accurate movements and plunger or ram. It is termed backward quiet operation. because of the opposite direction of the • They also offer a wide range of payload flow of metal to that of ram movement. capacity, accompanied by an equally wide Thus, the billet in the container remains range of costs. stationary and hence no friction. Also, Sol–48: (c) the extrusion pressure is not affected by the length of the billet in the The advantages of pneumatic drives are: extrusion press since friction is not Simple construction, relatively inexpensive, involved. The surface quality achieved fast and reliable. The disadvantages of is generally good since there is no hot pneumatic system are smaller payloads, the cracking due to the friction between the mass inertia and delayed response of the billet and the extrusion cylinder robot, arm due to the sponginess and interface. The disadvantage of backward reduced repeatability. extrusion is that the surface defects of Sol–49: (a) the billet would end up in the final By extrusion process, it is possible to product unlike direct or forward make components which have a extrusion where these are discarded in constant cross-section over any length the extrusion container. Though as can be had by the rolling process. advantageous, this process is not extensively used because of the problem Forward Hot Extrusion of handling extruding metal coming out In the forward hot extrusion, the flow of metal is in the forward direction, i.e, through the moving ram. the same as that of the ram. In forward Sol–50: (a) extrusion, the problem of friction is prevalent because of the relative motion Heat input rate P = VI between the heated metal billet and the Heat required for melting will be cylinder walls. This is particularly proportional to volume of metal to be severe in the case of steels because of melted, which in turn would depend on their higher extrusion temperatures. To thickness of the metal. Since the thickness reduceIES this friction, lubricants MASTER are to reduces by half, so the power required be used. At lower temperatures, a would also be half. Thus, keeping the mixture of oil and graphite is generally same voltage, we may decrease the current used. The problem of lubrication gets by half. compounded at the higher operating temperatures. Molten glass is generally Sol–51: (b) used for extruding steels, this stays in The various radii in the die cavity can liquid from at the operating significantly affect the formation of temperatures and provides necessary defects. The material follows a large heat insulation to the hot metal billet corner radius better than a small in addition to lubrication. To reduce the radius. With small radii, the material damage to equipment, extrusion is can fold over itself and produce a lap. (10) ME (Test-7), Objective Solutions, 6th November 2016
Sol–52: (d) (D) Permanent mold casting-kitchenware (E) Die casting-Hand tools Thickness of skin time Sol–60: (b) 1 In true centrifugal casting hollow t = 2 2 cylindrical parts such as pipes, gunbarrels and street lamp posts are produced. In this the molten metal is poured in a t2 = 0.5 0.709 rotating mold. t –t = 0.709 1 0.291 2 1 Sol–61: (c) = – 29% A small draft or taper is provided in sand Hence, the thickness decreases by 29%. mold patterns to enable removal of the Sol–53: (a) pattern without damaging the mold. Good collapsibility in the molding sand Sol–62: (a) allows the casting to shrink while cooling Process planning: Selecting sequence of and the main cause of hot tearing is operations improper solidification during cooling. Routing sheet: Standard time for Sol–54: (c) operations. Inclusions are mainly caused due to CAPP: Generative system reaction of molten metal with the MRP: Inventory control environment or with crucible or mold material. Sol–63: (b) CAPP: Computer aided process planning Sol–55: (c) decreases the lead time hence leading to The microstructure of a particular material higher productivity. contributes to the detrimental factors, such Sol–64: (d) as microporosity, microhardness, compositional variations etc. MRP-II- Manufacturing resource planning, through feedback controls all aspects of Sol–56: (b) manufacturing planning. Sol–57: (c) Sol–65: (c) Both deoxidation and purging are used to Sol–66: (c) remove dissolved gases from the molten metal. A. Just in time production - zero inventory • purging means flushing the metal with an inert gas B. FMS - AGV C. CIM - MRP • in case the dissolved gas is O2, the molten metal is deoxidised. D. Group technology -opitz system Sol–58: (c) IES MASTERSol–67: (c) Undercut is a welding defect caused due CAD/CAM: This combination allows the to reduction of cross sectional thickness of transfer of information from the design base metal. stage into the stage of planning for the Sol–59: (b) manufacture of a product. (A) Investment casting- cams and valves Sol–68: (b) (B) Slush casting- Ornamental and Sol–69: (c) decorative objects. Sol–70: (c) (C) Expendable pattern casting - lost foam In both arc welding and oxyfuel welding, process coalescing of materials takes place by means of heat. (11) ME (Test-7), Objective Solutions, 6th November 2016
Sol–71: (a) 2. Inter axial angles Sol–72: (b) = 90 Sol–73: (d) Sol–77: (d) Twin boundary separates two orientations A. Fluorspar: Simple cubic that are mirror image of one another. Twin boundaries can be annealing twins B. Alpha Iron: Body centred cubic or mechanical twins. C. Silver: Face centred cubic Sol–74: (b) D. Zinc: Hexagonal close packed Number of atoms Sol–78: (c) centered on the plane Planar density = The intercept of the direction vector on x, Area of plane y & z axis are 1/2, 1 and 0 respectively. 2 Taking reciprocal of these intercepts we for (100) FCC crystal plane PD = a2 have (210) Z Sol–79: (d)
Monotectic = Liquid1 Liquid2 + Solid A B Eutectic = Liquid Solid1+Solid2
Eutectoid = Solid Solid1 + Solid2 4 Peritectic = Liquid + Solid Solid X 1 2 C D Sol–80: (d) Hatched plane is (100) plane in a FCC A. Vanadium: Increases endurance crystal structure. strength Number of atoms centered on (100) plane B. Molybdenum: Improves creep resis- = 1/4th of each corner atom + one full tance 1 1 1 1 C. Silicon: Increases resistance to high atom on the face = 1 2 4 4 4 4 temperature oxidation atoms. D. Chromium: Increases hardness. Area of plane (100) = a × a = a2 Sol–81: (b) Chemicals attack atoms within grain 2 Therefore, planer density is = boundaries more because grain bound- a2 aries are in higher energy state than Sol–75: (c) those in the grains because atoms are not as closely packed on grain bound- Both FCC and HCP crystal structure aries. have same coordination number i.e. 12.IES MASTERSol–82: (a) Both have same packing fraction i.e. A. Nickel: Hardenability 0.74. B. Chromium: Corrosion resistance Both are closely packed crystal C. Tungsten: Heat resistance structure where only the stacking D. Silicon: Magnetic permeability sequence is different. Sol–83: (c) Sol–76: (b) Manganese steel or mangalloy or Hadfield For a tetragonal crystal system. steel having manganese content from 11% 1. Axes lengths are to 14% is extremely wear resistant steel a = b c used in earth moving equipments and (12) ME (Test-7), Objective Solutions, 6th November 2016
mining industry. m Sol–84: (b) T = N 1 l g l Addition of silicon to cast iron promotes graphite flake formation; it also increases = 1 mg ...(ii) fluidity of molten metal while casting silicon forces carbon out of solid solution, Comparing eqns. (i) and (ii), this carbon forms graphite flakes. 1 mg = mg Sol–85: (d) In full annealing, hypoeutectoid steel is = heated above upper critical temperature. 1 At this temperature dislocations move out of the body reducing imperfections, Work done when the chain slides down thereby increasing ductility of steel. by small distance dx at length x down Cooling is done inside the furnaces. Final the table is given by microstructure is uniform coarse pearlite. Sol–86: (c) mgx dW = dx l Air has very low cooling rate followed by, fused salt, oil and water in increasing 1 l order. 1l 1 l 2 mgx mg x Fdx dx Sol–87: (b) W = 0 0 l l 2 0 • Normalizing refines the grain
structure by producing fine pearlite. 2 2 mg 1 l2 1 m l g = • Full annealing gives uniform grain l 2 2 structure. Sol–90: (b) • Martempering improves toughness of steel without compromising on its When block C collides elastically with hardness, while ductility decreases. block A, they simply exchange their velocities, i.e., block C comes to rest and • Spherodizing heat treatment gives block A gets velocity v just after the maximum softness of structure. collision. Sol–88: (d) Now the block A moving with velocity v Time, temperature, Transformation (TTT) compresses the spring and the combined diagrams indicates transformation of system moves with speed v’ (say) at the austenitic phase. maximum compression of the spring. Sol–89: (c) IES MASTERFor block A and B: Let be the mass per unit length of the mv + 0 = (m m)v string m/l . v = v/2
1 1 1 m 2 2 2 Weight of hanging part = lg mg Also, mv 0 = m m v kxm l 2 2 2
T = mg ...(i) 2 1 2 1 v 1 2 mv = 2m kxm For the chain on the table, 2 2 4 2 (13) ME (Test-7), Objective Solutions, 6th November 2016
Sol–93: (c) m v xm = 1 1 2k kx2 I 2 = C 2 2 K.E. at maximum compression is k I 2x x 2 2 = 0 12 v 1 2 2 2 K.E. = 2m = mv 2 2 4 Ia kx 2 = 0 Thus, options (2) and (4) are correct. r Sol–91: (a) Ia kx = r2 Let v be the final velocity of the system x I at maximum compression of the spring, = a kr2 mv0 = M m v I 2 T = 2 m kr v = v0 M m mr2 3mr2 I = mr2 = 2 2 1 2 12 1 2 Also, mv0 = M m v kxm 2 2 2 3mr2 3m T = 2 2 2kr2 2k 1m2 v 2 1 1 2 M m 0 kx2 Sol–94: (c) mv0 = 2 m 2 2M m 2 150 103 Mass m = kg 1 1 1m2 v 2 1 mMv 2 9.81 kx2 = mv2 0 0 2 m 20 2 M m 2 M m Spring constant 10 103 = 8 105 N / m mM k = 2 x = v ...(ii) 1.25 10 m 0 M m k 1 1 mv2 = kx2 1 2 2 Final K.E. of the system is M m v2 2 mV2 150 10 3 2 2 x2 = = k 9.81 8 105 1 m2 v 2 0 but its value will be zero in the x = 27.66cm 2 M m Sol–95: (b) centre of mass frame. Momentum = mv, since for the given Sol–92: (c) IES MASTERparticle mass ‘m’ will be fixed, hence, for dm the momentum to double, its velocity has u 2.20 75 165N F = 1 dt to double. Since KE = mv2 , hence KE 2 P = Fv = 165 × 2.2 = 363 W will become four times. K.E. acquired time
1 dm 2 1 2 = v 75 2.2 181.5W 2 dt 2
Rate at which energy is lost = 363 – 181.5W = 181.5W (14) ME (Test-7), Objective Solutions, 6th November 2016
Sol–96: (b) Then Tension T in the rope at a distance 8 kN 10 kN ‘x’ from the bottom = x 100 Work done due to tension in the rope for B A D Tdx C raising by a distance ‘dx’ = 6 R 6 2 A RB x x 100dx = 100x = 0 6 m 4 m 5 m 2 0 R + R = 8 + 10 = 18 A B = 36 100 6= 18 600 Moment of forces about A should be zero. 2 Hence, R × 10 – 8 × 6 – 10 × 15 = 0 20 B = N/m 6 R = 19.8 kN () B 20 Work done = 18 600= 660Nm 6 RA = 1.8 kN () Sol–100: (d) Sol–97: (d) Average acceleration: v / t 10 kN Instantaneous acceleration: lim v / t t 0 Uniform motion: a = 0 Uniformly accelerated motion: a = P 10 kN constant A Sol–101: (a) 6 kN Acceleration due to gravity will be Balancing the forces in vertical direction independent of the mass hence acceleration at point A, we have, the force in member for the two bodies will be same. However, P of the truss = 6 kN tensile. momentum, potential energy, kinetic Sol–98: (c) energy, will be different for different masses. Balancing the force at point P, Sol–102: (c) 1 Let the weight of the box be w. Then since the box moves with constant velocity, P so net force on the box will be zero. w 8 kN F = the force in number (1) will be 8 kN F = 0.4 w tensile. Sol–99: (c) IES MASTER F T H dx w x A w For the box to overturn, moment about A 100 N should be zero. Let the weight per unit length of the rope Here, w × 30 = F × H = N/m. w × 30 = 0.4 w×H (15) ME (Test-7), Objective Solutions, 6th November 2016
w 30 30 40 1 1 1 H = = = 75 mm = mm 0.4 w 0.4 E 16 25 36 Sol–103: (d) Sol–107: (a)
Angular acceleration 1 m 2 m C B Final angular velocity A x2 Deflected shape initial angular velocity x1 = time of rigid beam 20 0 = = 5 rad/s2 From similar triangle 4 x x Moment of inertia of disc = 1 = 2 … (i) mr2 3 2 = 0.5kg m2 Force in spring at C K1 x 1 1 3 3 2 = = = Force in spring at B K x Torque required = I 2 2 2 2 4 1 Sol–108: (c) B = 5= 2.5Nm 2 A + 25 kN C Sol–104: (b) a b V2 = u2 – 2 as b 3L Initial velocity u = 20m/s, final velocity v RA = P 25 15 kN (Tension) a b 2L 3L = 0 a 2L 2 R = P 25 10kN(Comp) Deceleration a = g= 0.8 10 = 8m/s C a b 2L 3L 0 = 202 – 2 × 8 × s Sol–109: (b) SFD is zero along the span it shows that 202 s = = 25m only couple of opposite nature and equal 2 8 magnitude are acting on the member. No Sol–105: (c) point loading or udl is acting any where Stress induced in the bar due to rise in in the span. temperature Sol–110: (c) E = (Deflection prevented) dM L V E dx = (L T 0.02) L Slope of BMD = Shear force 106 Moving from C to A at point B slope of = [25 20 106 100 0.02] 25 BMD is reduced this shows shear force is IES2 2 MASTERreduced means reaction or load at B and = 1200 = 12×10 kg/cm Sol–106: (d) C are in opposite direction. Which is the case only with option (c) hence (c) is the Elongation in bar due to load P correct answer. PL = AE Sol–111: (c) 10 100 4 = A B 100 kN/m E d2 1000 4 1 1 1 a = 2 m a = 2 m = 2 2 2 E (40) (50) (60) (16) ME (Test-7), Objective Solutions, 6th November 2016
Maximum hogging moment will occur at A C B 4 M support A and B. diagram EI EI Maximum sagging moment (if sagging
moment is present) will be at mid span of At central point C, deflection is C and due to the span AB symmetry, deflection at C is maximum and 2 wa slope of elastic curve is zero i.e. C = 0 Maximum hogging moment = 2 Applying moment area equation between point 100 22 C & B = 200 kNm 2 B = CC L Ax wl2 wa 2 4 Maximum sagging moment = 0 = 0 2 1 8 2 C EI 100 82 100 2 2 = = 800–200 = 600 8 8 2 = kNm C EI Hence design moment = Max [Hogging (–) ve sign shows downward deflection. moment, Sagging moment] Alternative : = 600 kNm
A B Sol–112: (b)
RA = RB = 0 2 m 2 m 10 kN W 10 kN At C slope is zero so part AC can be thought A C B as cantilever beam carrying moment M = 4 kNm at end A. As reaction at A is zero so it 2 m 4 m 2 m can be treated as free end.
At support, MA = 10 × 2 = 20 kNm M A C Given : MA = MC W L 20 kNm 20 kNm ML2 4 2 2 8 w/2 w/2 = 4 m A 2 EI 2EI EI B.M. at mid span which is equal to in original structure. WL W 4 C M = 20 = 20 C 4 4 = W – 20 Sol–114: (c) 10 kN/m 20 = W – 20 C IESW = 40 kN MASTERA B Sol–113: (b) 4 m 4 m 4 kN.m 4 kN.m R A B A C C B R 4 m 4 m 4 m 4 4 R = R = = 0 Downward deflection of point C in beam AC = A B 4 Downward deflection of point C in beam CB Using moment area method : wl4 RL 3 RL3 = 8EI 3EI 3EI (17) ME (Test-7), Objective Solutions, 6th November 2016
Sol–117: (d) 10 44 R 4 3 R 43 = y-axis 8 3 3 R = 7.5 kN y -axis (y , xy ) Sol–115: (a) y B y x y , 0 Principal strains 1/2 (y , x y ) D 2 2 2 x y x y xy 0 x = C 2 2 2 2 (x , – x y ) A x 800 106 100 10 6 = x -axis 2 (x , xy ) 2 2 800106 10010 6 80010 6 2 2 x-axis –6 –6 1 = 450 × 10 + 531.507 × 10 Sol–118: (a) –6 A 1 = 981.5 × 10 –6 –6 –6 2 = 450 × 10 – 531.507 × 10 = –81.5 ×10 x y Sol–116: (c) xy 0 Major principal stress, x y B 2 2 x y x y 2 = sin2 cos2 1 2 2 xy xy 2 = sin2 0 = 0 40 0 40 2 = 80 2 2 2 Option (a) is correct. = 20 + 82.462 Sol–119: (d) 2 1 = 102.462 102.5 MN/m
Minor principal stress, 1 2 1 2 x = cos(2 45 ) 2 45 2 2 x y x y 2 = 3 2 2 xy 60 40 2 x 0 = 10 N/mm = 20 – 82.462 45 2 2 –62.5 MN/m Sol–120: (d) IES2 MASTER 1 xy 2 80 tan tan1 According to max. shear stress theory 2 = = ()x y 40 fy 2 = 75.96 FOS = 2 = 37.98° Absolute max. shear stress 38° 280 2 = p p p p Max.1 2 , 1 , 2 2 2 2 (18) ME (Test-7), Objective Solutions, 6th November 2016
140 T 300 = = 100 40 100 40 1000 1000 max , , 2 2 2 2T 140 1 + 0.3 = = 1.3 = 1000 100 40 100 40 max , , 1300 2 2 2 T = = 650K 2 140 = = 2 Sol–124: (b) 70 Sol–121: (b) According to first law of thermodynamics, for a cyclic process, Sol–122: (d) Q = W According to 1st law of thermodynamics, Q + Q + Q + Q for a cyclic process, QW= 12 23 34 41 = W + W + W + W 2 12 23 34 41 220 + (–25) + (–180) + 50
= 15 + (–10) + 60 + W41 1 WA1 = (220 – 25 – 180 + 50 –15 + 10 – 60) = 0 Q12 + Q21 = W12 + W21 Sol–125: (c) 100 + 120 = 50 + W21 Work transfer in free expansion is zero, W = 170 kJ 21 because in free expansion, the pressure Sol–123: (b) 1000 K against which the gas expands is zero. Q Work output of reversible heat 1 Hence, work = P V 0 . engine (HE1) Sol–126: (a) HE1 W T For steady flow process, W = 1 Q1 Q 1000 2 h1 + Q = h2 + W Similarly, for heat engine T Q = (h2 – h1) + W (HE2) Q 2 Q = 30 + (–90) = –60kJ/kg
300 HE W Total heat lost = 60 kJ/kg W = 1 Q2 2 T Heat lost to circulating cooling Q Both the engine produce same 3 water = 40 kJ/kg amount of work, 300 K Heat lost to atmosphere = 60 – 40 = hence, 20 kJ/kg T 300 Sol–127: (a) 1 Q1 = 1 Q2 IES1000 MASTERT 3 1 P P T Q S = T 300 2 = C 1 = 1 C 2 1000 T Q S T 1 2 = 4 4 = C C 1 3 Q2 T Since, = V Q 1000 V 1 Otto cycle Stirling cycle
T 1 300 hence, 1 = 1000 T 1000 (19) ME (Test-7), Objective Solutions, 6th November 2016
2 3 = 106 0.06 0.03 4 1 T P P T = pV =c = = 30 kJ C C pV =c Heat absorbed Q = 84 kJ 3 2 4 1 According to first law of thermodynamics, V V Ericsson cycle Brayton cycle Q = UW Sol–128: (b) 84 = U 30 For adiabatic process, Q = 0 U = 84 – 30 = 54 KJ So, according to 1st law of thermodynamics Change in internal energy
Q = U W= 0 U = 54 KJ Sol–132: (d) U = W= 5000 J According to first law of thermodynamics, When the system returns to original state Q = UW then U = U 5000 J 1 2 3 1 2 3 1 2 3
Q = UW 100 = U1 2 3 60 W= Q U 1000 5000 = 4000 J U1 2 3 = 40 Kg Sol–129: (d) U1 2 3 = U1 4 3 40kJ Since the chamber is rigid, so volume is Q1 4 3 = UW1 4 3 1 4 3 constant, = 40 + 20 = 60 kJ hence work done W = 0 Sol–133: (b) The chamber is insulated, T Maximum efficiency = 1 2 T So, Q = 0 1 T = 873 K From first law of thermodynamics, 1
Q1 Q = dU + W W dU = 0 HE Sol–130: (c) Q2 Heat added Q = 30 + 5 = 35
Net work done W = 3 + 10 – 8 + 0 = 5 T2 = 293 K W 5 EfficiencyIES = = = 0.1433MASTER293 W 1 Q 35 1 = 873 QQ1 1 Sol–131: (b) 873 293 3 Initial volume V1 = 0.03m = 873 Q1 3 Final volume V2 = 0.06m 873 Q = 1.505kW 1 580 Constant pressure P= 1MPa 106 Pa Heat rejection QQW2= 1 Work done = 1.505 – 1 = 0.505 kW
W = P(V2 – V1) (20) ME (Test-7), Objective Solutions, 6th November 2016
Sol–134: (b) cold and hot water mixed up. But below Work done = area under the graph 4°C, water and then ice is lighter than 1 normal water, so it floats at surface. = × (10 + 5) × 1 + 5 × 2 = 17.5J 2 Sol–138: (d) = 2.5 J U The normal drinking water is not a pure Q = U + W = 2.5 + 17.5 = 20J substance because it contains solid Sol–135: (d) contaminants. But at the same time if Pure substance, number of components, water is distilled one, it will be pure. C = 1 Sol–139:(b) Degree of freedom from Gibb’s phase rule, If a machine would continuously pro- duce net work without expenditure of DOF = C – + 2 some other form of energy, then such a In saturated vapour, number of phases, heat engine is called perpetual motion = 1 machine of the second kind (PMM2). A PMM2 is impossible. It violates second DOF = 1 – 1 + 2 = 2 law of thermodynamics. The efficiency For superheated vapour, number of phases, of a heat engine can never be 100%. = 1 There has to be a heat rejection always. DOF = 1 – 1 + 2 = 2 Sol–140: (a) Sol–136: (d) Sol–141: (d) P Sol–142: (d) Saturated Sol–143: (d) Vapour Curve Most of the materials exist in polycrystalline form and not as single 2 1 Isothermal crystal. In poly crystalline material curve individual crystals have different orientations. There are some materials,
V1 /2 V1 which exist in form of single crystals such V as rock salt, sugar etc. Since at time of compression, the working Sol–144: (c) medium is saturated vapour (point 1). It is known that slope of saturation curve is Cast Iron is hard, brittle and wear re- more than the isothermal compression. So sistant. Cast Iron contains more than as soon as compression starts the 2% carbon and percentage of cementite compressed fluid comes in two phase in it is higher. region and condensation at constant Sol–145: (a) pressureIES and temperature starts MASTER and some vapour condenses and pressure and Sol–146. (a) temperture do not change. Shear is maximum at 45° to the direction of the applied tensile force and mild steel Sol–137: (a) is weak in shear so failure takes place in At 4°C, the density of water starts the direction of maximum shear stress, reducing due to hydrogen bond formation. in cup and cone shape fracture. Before 4°C, the water inside of lake is hot and light while at surface it is cooled and heavy. So natural circulation starts and (21) ME (Test-7), Objective Solutions, 6th November 2016
Sol–147: (c) = 9 – 8 – 0 = 1
Assertion is correct but reason is wrong Hence, the mechanism has single degree dM of freedom and the overall movability of because = F = 0 is applicable for load the mechanism is unity. dx intensity, not concentrated load because Sol–151: (c) dM due to concentrated load is not Inversion is obtained by fixing, in turn, dx depended mathematically. different links in a kinematic chain and we can obtain as many mechanisms as the links in a kinematic chain. For example, four inversions of a single slider crank chain are possible.
Connecting Cutting stroke rod Return stroke Line of Ram R1 Tool stroke
R R2 Q 2 P1 PL/4 WL /8 P Slider (Link 1) 2 Sol–148: (a) Crank (Link 2) B q q q A B A B B A q q C q q q q q q B q q 1 B2 C D C D D 90 Fixed q q q C 2 (Link 3) (a) (b) (c) Slotted bar (Link 4) When an element is in a state of pure shear, A maximum direct stresses are induced on Fig. Crank and slotted level quick return mutually perpendicular planes which are 45º motion mechanism. to planes of pure shear. One of the maximum In crank and slotted lever quick return direct stresses is tensile. So block will generate motion mechanism, the link AC i.e. link cracks along this if these stress increases and 3 forming the turning pair is fixed. The other maximum direct stress is compressive so link 3 corresponds to the connecting rod failure may occur due to crushing in this case. of a reciprocating steam engine. Sol–149: (c) Sol–152: (d) Cam and follower is a higher pair because Ackermann steering gear is preferred to there is point or line contact between the Davis steering gear because it consists of cam andIES the follower. Surface MASTER contact turning pairs, hence less friction and so takes place between two links of a lower less wear and tear. pair. Sol–153: (b) Sol–150: (a) In hot working, the material is above Number of degrees of freedom the recrystallisation temperature, hence it is possible to continuously reform the n = 3 1 2j h grains in metal working and if the temperature and the conditions of Here, = 4, j = 4, h = 0 working are properly controlled, a very n = 3(4 – 1) – 2 × 4 – 0 favourable grain size could be achieved giving rise to better mechanical (22) ME (Test-7), Objective Solutions, 6th November 2016
properties. castings are in one mould. Runner/gate Sol–154:(a) control is more suitable and produces a better yield with heavy castings with Sol–155: (d) only a few in a mould. It is important to make sure that the Sol–156:(b) sprue does not act as the choke. This is Hot chamber die casting uses mainly because a condition can be generated in low melting point metals. which iron can leave the gating system faster than it can be poured into the Sol–157: (a) pouring basin. So avoid this, at least as Electric drives are widely used as they much liquid must be capable of passing are clean and quiet with a high degree of through the sprue as through all the accuracy and reliability. gates. This can be expressed mathematically Sol–158: (a) using the equation for the rate of filling derived using the Toricelli’s equation. The In automation, the machine produces a velocity of liquid flow through the top of job following a set operational sequence, the sprue is given by the equation, while a robot can be made to do different jobs at different times and in different
Vs = C 2gh sequences through programming. An where C is a friction factor automated machine has neither a knowledge base nor intelligence, like The rate of flow Q = As C 2gh , where robot. So, robot is more than an
As is the cross-sectional area of the automated machine/equipment. sprue. Sol–159: (a) The sharp change in direction from the sprue to runner causes the formation A robotic manipulator arm consists of of a stationary liquid pool soon after several separate links making a chain pouring has started. This diverts the (simple open chain formed by links liquid flow into the runner. The reduced connected in series). The arm is located pressure in the region can lead to aspi- relative to the ground on either a fixed ration, thus drawing air into the base or a movable base. It has a free-end stream. This air can increase turbu- where an end-deflector or gripper or lence further down the stream causing sometimes a specialised tool holder or any slag to mix with liquid iron rather than powered device is attached. sticking to the top of the runner. A cor- Sol–160: (a) rectly designed runner and gating sys- tem should prevent slag and dross from The load carrying capacity of a robot entering the mould cavity. In this con- depends on its physical size and nection, gating ratio is the relationship construction, so that is why, very light betweenIES the cross-sectional areaMASTER of the models have capacity as low as 1.5 kg, sprue, runner and gate. For example, a while the heavier class of robots have their ratio of 1:2:4 describes a non-pressurised capacities as high as 1000 kg. or sprue/runner - controlled system, whereas a ratio of 1:1:0.7 describes a pressurised or runner/gate-controlled system. Runner/gate control is used for slag forming irons such as malleable and flake irons. Sprue/runner control is used for spheroidal irons. Further, a sprue/runner control produces a higher yield when a large number of small