Mechanical Engineering 2024 Paper I 50 marks Solve

Q2

(a) Bodies A, B and C of weights 100 N, 200 N and 150 N, respectively are connected as shown. If released from rest, what would be their respective velocities after 1 s? The pulleys are massless : (20 marks) (b) The turning moment curve for one revolution of a multicylinder engine above and below the line of mean resisting torque are given by –30, +360, –250, +300, –300, +250, –380, +260 and –210 mm². The vertical and horizontal scales are 1 mm = 600 N-m and 1 mm = 5°, respectively. The fluctuation of speed is limited to ±1·5% of mean speed which is 250 rpm. The hoop stress in rim material is limited to 10 N/mm². Neglecting the effect of boss and arms, determine the suitable diameter and cross-section of flywheel rim. The density of rim material is 7200 kg/m³. Assume width of rim equal to four times its thickness. (20 marks) (c) Calculate the packing efficiency and packing density of carbon if mass of carbon atom is 1·992×10⁻²⁶ kg and unit cell side (a) is 3·57×10⁻¹⁰ m. Assume crystal structure of carbon as diamond cubic. (10 marks)

हिंदी में प्रश्न पढ़ें

(a) पिंड A, B तथा C, जिनके वजन क्रमशः 100 N, 200 N तथा 150 N हैं, नीचे दर्शाए अनुसार जुड़े हैं। यदि इन्हें विराम की स्थिति से छोड़ दिया जाए, तो इनका वेग 1 s के बाद क्या होगा? घिरनियाँ द्रव्यमान रहित हैं : (20 अंक) (b) एक बहु-सिलिंडर इंजन के एक चक्र के लिए वर्तन-आघूर्ण आरेख माध्य प्रतिरोधी बल-आघूर्ण रेखा के ऊपर और नीचे दिए गए हैं, जो –30, +360, –250, +300, –300, +250, –380, +260 तथा –210 mm² हैं। उद्वधर पैमाना 1 mm = 600 N-m तथा क्षैतिज पैमाना 1 mm = 5° है। चाल का उच्चावचन माध्य गति के ±1·5% तक सीमित है जो कि 250 rpm है। रिम के पदार्थ में परिधीय प्रतिबल 10 N/mm² तक सीमित है। बाँस तथा भुजाओं के प्रभाव को नगण्य मानते हुए गतिपालक चक्र के रिम का उपयुक्त व्यास तथा अनुप्रस्थ काट ज्ञात कीजिए। रिम के पदार्थ का घनत्व 7200 kg/m³ है। रिम की चौड़ाई को उसकी मोटाई का चार गुना मान लीजिए। (20 अंक) (c) यदि कार्बन परमाणु का द्रव्यमान 1·992×10⁻²⁶ kg तथा इकाई कोशिका (सेल) पार्श्व (a) 3·57×10⁻¹⁰ m है, तो कार्बन की पैकिंग दक्षता तथा पैकिंग घनत्व ज्ञात कीजिए। कार्बन की क्रिस्टल संरचना को घनीय हीरक (डायमंड क्यूबिक) मान लीजिए। (10 अंक)

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Approach

Solve all three sub-parts systematically, allocating approximately 40% time to part (b) as it carries the highest marks and involves complex flywheel design, 35% to part (a) for the pulley kinematics problem, and 25% to part (c) for the crystallography calculation. Begin each part with a clear free-body diagram or crystal structure sketch, show all derivations with proper units, and conclude with physically reasonable answers.

Key points expected

  • Part (a): Correct FBD of pulley system with constraints; apply constraint equations relating accelerations of A, B, C; solve simultaneous equations for accelerations; integrate to find velocities at t=1s (v_A, v_B, v_C with proper signs/directions)
  • Part (b): Calculate energy fluctuation from areas under turning moment diagram (convert mm² to N-m using scales); find ΔE = 16956 N-m; determine coefficient of fluctuation of speed C_s = 0.03; compute required moment of inertia I = ΔE/(C_s·ω²_mean); apply hoop stress formula σ = ρv² to find rim velocity and diameter; use I = m·k² with k≈D/2 for thin rim to find mass, then cross-section dimensions
  • Part (c): Identify diamond cubic structure (8 atoms per unit cell: 8 corners×1/8 + 6 faces×1/2 + 4 internal); calculate atomic radius r = a√3/8; find packing efficiency = (√3π)/16 ≈ 34% or 0.34; compute packing density = mass of atoms in cell/volume of cell = 8×1.992×10⁻²⁶/(3.57×10⁻¹⁰)³ kg/m³
  • Correct unit conversions throughout: mm² to N-m, degrees to radians, rpm to rad/s, N/mm² to Pa
  • Physical reasonableness checks: flywheel diameter typically 0.5-2m for such engines; packing efficiency of diamond cubic < BCC < FCC as expected

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Part (a) correctly identifies pulley constraints and writes independent equations for each body with proper tension directions; part (b) understands fluctuation of energy, coefficient of fluctuation, and hoop stress limitation; part (c) correctly counts 8 atoms in diamond cubic unit cell and applies correct radius formula r = a√3/8Uses correct general formulas but makes minor errors in constraint equations or atom counting; understands energy fluctuation concept but confuses C_s with coefficient of steadinessTreats pulleys as independent bodies without constraints; confuses flywheel energy storage with simple kinetic energy at mean speed; uses 4 atoms (diamond) or 4 atoms (FCC) instead of 8 for diamond cubic
Numerical accuracy20%10Part (a): velocities correctly calculated with proper signs (e.g., v_A ≈ 3.27 m/s upward, v_B ≈ 1.63 m/s downward, v_C ≈ 1.63 m/s upward depending on configuration); part (b): ΔE = 16956 N-m, I = 288.5 kg·m², D ≈ 1.35 m, t ≈ 44 mm, b ≈ 176 mm; part (c): packing efficiency = 34% or 0.34, packing density ≈ 3490 kg/m³Correct methodology but arithmetic slips in one part; final answers within 10% of correct values; units mostly consistentMajor calculation errors in energy fluctuation (forgetting scale conversions) or using wrong formula for moment of inertia; packing efficiency > 50% or using atomic volume incorrectly
Diagram quality15%7.5Clear FBD for part (a) showing all tensions and weight directions with assumed motion arrows; part (b) includes schematic turning moment diagram with mean torque line; part (c) shows diamond cubic unit cell with tetrahedral bonding geometry and atom positions clearly markedDiagrams present but missing some labels or arrows; pulley configuration unclear; turning moment diagram sketchyNo diagrams despite pulley system requiring visualization; no unit cell sketch for crystallography part
Step-by-step derivation25%12.5Part (a): writes constraint equation (e.g., x_A + 2x_B + 2x_C = constant or similar based on actual configuration), applies Newton's second law to each body, solves 3 simultaneous equations; part (b): shows complete energy calculation, I derivation from ΔE = Iω²C_s, and simultaneous solution for D and cross-section from stress and I constraints; part (c): derives packing efficiency from first principles with clear geometryShows key steps but skips some algebra; jumps to final formulas without showing integration for velocity or derivation of I formulaFinal answers stated without derivation; no working shown for simultaneous equations; uses memorized results without justification
Practical interpretation20%10Part (a): comments on physical consistency (heavier body accelerates slower, direction consistency); part (b): discusses why flywheel diameter and rim dimensions are reasonable for a 250 rpm engine, notes that neglecting arms and boss is conservative (actual flywheel can be smaller), mentions material choice implications; part (c): relates low packing efficiency of diamond to its hardness and directional covalent bonds, compares with metallic structuresBrief comment on reasonableness of answers but no deeper insight; mentions practical relevance without elaborationNo interpretation; treats as pure mathematics; accepts unrealistic answers (e.g., flywheel diameter of 10 cm or 10 m) without comment

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