Mechanical Engineering 2022 Paper I 50 marks Calculate

Q4

(a) An automobile weighing 15 kN is moving at a speed of 100 km per hour. The dynamic coefficient of friction between the rubber tyre and concrete road is 0·60. When the driver jams on his brakes, the car goes into skid in the direction of motion. What is the distance the car will move before it comes to rest when (i) the road is flat, (ii) the road is 7° upward inclination, (iii) the road is 7° downward inclination? (15 marks) (b) Explain the principle of annealing with the help of Continuous Cooling Transformation (CCT) diagram. Write the purpose and applications of annealing heat treatment. (15 marks) (c) A shaft carries four masses A, B, C and D which are in complete balance. Masses C and D make angles of 90° and 210° respectively with that of mass B in the counterclockwise direction. The rotating masses A, B, C and D can be assumed to be concentrated at radii of 360 mm, 480 mm, 240 mm and 300 mm respectively. The masses B, C and D are 15 kg, 25 kg and 20 kg respectively and the planes containing B and C are 300 mm apart. Determine the following: (i) Mass A and its angular position (ii) Positions of planes A and D (20 marks)

हिंदी में प्रश्न पढ़ें

(a) एक ऑटोमोबाइल जिसका भार 15 kN है, 100 km प्रति घंटे की चाल से चल रही है । रबर टायर व कंक्रीट की सड़क के बीच गतिक घर्षण गुणांक 0·60 है । जब चालक ब्रेक लगाता है, तो कार गति की दिशा में फिसलती है । विराम अवस्था में आने से पूर्व कार कितनी दूरी चलेगी, जब (i) सड़क चपटी है, (ii) सड़क ऊपर की तरफ 7° नति रखती है, (iii) सड़क नीचे की ओर 7° नति रखती है ? (15 अंक) (b) सतत शीतलन रूपांतरण (CCT) आरेख की सहायता से अनीलन के सिद्धांत की व्याख्या कीजिए । अनीलन उष्मा उपचार का उद्देश्य व उसके अनुप्रयोग लिखिए । (15 अंक) (c) एक शाफ्ट चार द्रव्यमानों A, B, C व D को वहन करता है जो कि पूर्ण रूप से संतुलित हैं । द्रव्यमान C व D, द्रव्यमान B से क्रमशः 90° व 210° के कोण वामावर्त दिशा में बनाते हैं । घूर्णन कर रहे द्रव्यमानों A, B, C व D को माना जा सकता है कि वे क्रमशः 360 mm, 480 mm, 240 mm व 300 mm पर केंद्रित हैं । द्रव्यमान B, C व D क्रमशः 15 kg, 25 kg व 20 kg हैं तथा वे तल, जिनमें B व C हैं, 300 mm की आपसी दूरी पर हैं । निम्नलिखित को ज्ञात कीजिए : (i) द्रव्यमान A व उसकी कोणीय स्थिति (ii) तल A व D की स्थितियाँ (20 अंक)

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Directive word: Calculate

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How this answer will be evaluated

Approach

Calculate the braking distances for all three inclinations in part (a) using work-energy principle with friction and gravity components. For part (b), explain annealing using CCT diagram with proper time-temperature axes and transformation curves, then enumerate purposes and applications. For part (c), solve the balancing problem using vector polygon method and tabular method for planes, allocating approximately 35% time to (a), 25% to (b), and 40% to (c) given the computational complexity of the four-mass balancing problem.

Key points expected

  • Part (a)(i): Flat road - s = v²/(2μg) = (27.78 m/s)²/(2×0.6×9.81) = 65.6 m using work-energy principle
  • Part (a)(ii)-(iii): Inclined roads - include mg sinθ term; uphill s = v²/[2g(μcosθ + sinθ)] = 47.2 m, downhill s = v²/[2g(μcosθ - sinθ)] = 109.5 m
  • Part (b): CCT diagram with time (log scale) vs temperature axes, showing ferrite, pearlite, bainite start/finish curves; slower cooling than TTT, transformation during continuous cooling
  • Part (b): Annealing purposes - relieve internal stresses, soften for machining, improve ductility, refine grain structure; applications - cold-worked steel sheets, castings, welded components
  • Part (c): Vector polygon closure for mass moments: Σmr cosθ = 0, Σmr sinθ = 0; solve for m_A = 10 kg at 143° from B (or 217° counterclockwise)
  • Part (c): Couple polygon for planes: ΣmrL cosθ = 0, ΣmrL sinθ = 0 with reference plane; find L_A = 510 mm from reference, L_D = 255 mm from reference on opposite side of B-C

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies work-energy principle with proper sign convention for inclinations in (a); accurately describes CCT vs TTT distinction with transformation kinetics in (b); sets up complete static and dynamic balance conditions with correct reference plane selection in (c).Uses correct basic formulas but minor errors in inclination sign convention or CCT/TTT distinction; balancing equations set up but reference plane choice suboptimal.Confuses energy methods with kinematic equations; treats CCT as identical to TTT; fails to recognize need for both force and couple balance in rotating masses.
Numerical accuracy20%10All six numerical answers correct: (a) 65.6 m, 47.2 m, 109.5 m; (c) m_A = 10 kg, θ_A = 143° or 217°, plane positions L_A = 510 mm, L_D = 255 mm with consistent significant figures and units.Most answers correct but one or two calculation errors (e.g., wrong angle quadrant, arithmetic slip in couple balance); correct methodology evident.Multiple numerical errors; wrong velocity conversion (100 km/h = 27.78 m/s); incorrect trigonometric evaluation; final answers without units.
Diagram quality20%10CCT diagram properly drawn with log(time) horizontal axis, temperature vertical axis, Ms line, ferrite/pearlite/bainite C-curves, and cooling curves; vector polygons for mass and couple balance drawn to scale with angles measured and values labelled.CCT axes correct but curves approximate; vector polygons sketched not to scale but angles roughly correct; missing some labels.No CCT diagram or confused with iron-carbon diagram; free-body diagrams for braking forces missing; no vector polygons for balancing, only algebraic manipulation.
Step-by-step derivation20%10Full derivation shown: (a) KE = ½mv² = work against friction ± gravity; (b) explains why CCT curves are displaced right/down from TTT due to undercooling; (c) tabular method with columns for m, r, mr, θ, mr cosθ, mr sinθ, L, mrL, mrL cosθ, mrL sinθ.Key steps shown but skips some algebraic manipulation; table present but incomplete; CCT explanation descriptive rather than derived from transformation kinetics.Final answers only with no working; jumps to vector polygon without establishing equilibrium equations; no tabular organization for balancing problem.
Practical interpretation20%10Discusses safety implications: downhill braking distance doubles, requiring lower speeds on descents like Himalayan ghat roads; annealing applications to IS 2062 steel fabrication; balancing significance for automotive crankshafts, turbine rotors, and ISI-certified machine tools.Brief mention of safety for braking; generic annealing applications without specific materials; states balancing is important without naming applications.No physical interpretation; treats all parts as pure mathematics; no connection to engineering practice or Indian industrial standards.

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