Mechanical Engineering 2025 Paper I 50 marks Calculate

Q7

(a) HSS cutting tool with 9° rake angle, the following data were observed in an orthogonal machining process of medium carbon steel workpiece: Feed rate = 0·25 mm/rev Cutting speed = 250 m/min Depth of cut = 1·5 mm Chip thickness ratio = 0·30 Vertical cutting force = 1100 N Horizontal cutting force = 600 N Calculate the: (i) Shear force along the shear plane (ii) Normal force on the shear plane (iii) Friction force along the rake surface (iv) Normal force along the rake surface (v) Friction angle (vi) Work done in shear (vii) Work done in friction (20 marks) (b) The annual requirement of an item is 2400 units. Each item costs the company ₹ 6. The manufacturer offers a discount of 5% if 500 or more quantities are purchased. If the ordering cost is ₹ 32 per order and inventory cost is 16%, determine whether it is advisable to accept the discount. (20 marks) (c) The following cost-related data has been collected from a company: | Cost Element | Variable Cost | Fixed Cost | | Direct material | 32·8 | – | | Direct labour | 28·4 | – | | Factory overheads | 12·6 | 1,89,900 | | Distribution overheads | 4·1 | 58,400 | | General administrative overheads | 1·1 | 66,700 | | Budgeted sales | – | 18,50,000 | Determine the following: (i) Break even sales volume (ii) Profit at the budgeted sales volume (iii) Profit if the actual sales I. Drop by 10%, and II. Increase by 5% from budgeted sales. (10 marks)

हिंदी में प्रश्न पढ़ें

(a) HSS कतन औजार के साथ 9° नति (रैक) कोण का उपयोग करते हुए, मध्यम कार्बन इस्पात के कार्यखंड की लांबिक मशीन प्रक्रिया में निम्नलिखित आँकड़े पाए गए: प्रभरण दर = 0·25 mm/rev कतन गति = 250 m/min कतन की गहराई = 1·5 mm छीलन की मोटाई का अनुपात = 0·30 उर्ध्वाधर कतन बल = 1100 N क्षैतिज कतन बल = 600 N गणना कीजिए: (i) अपरूपण तल के साथ अपरूपण बल (ii) अपरूपण तल पर प्रसामान्य बल (iii) नति (रैक) पृष्ठ के साथ घर्षण बल (iv) नति (रैक) पृष्ठ के साथ प्रसामान्य बल (v) घर्षण कोण (vi) अपरूपण में किया गया कार्य (vii) घर्षण में किया गया कार्य (20 marks) (b) एक वस्तु की वार्षिक आवश्यकता 2400 इकाई है। कंपनी के लिए प्रत्येक वस्तु की लागत ₹ 6 है। निर्माता 500 अथवा अधिक मात्राओं की खरीद पर 5% छूट की पेशकश करता है। यदि आदेश लागत ₹ 32 प्रति आदेश है तथा सामग्री सूची (इन्वेंट्री) लागत 16% है, तो निर्धारित कीजिए कि क्या छूट स्वीकार करना उचित है। (20 marks) (c) एक कंपनी से लागत-संबंधी निम्नलिखित आंकड़े एकत्र किए गए हैं: | लागत तत्व | परिवर्तनीय लागत | अचल लागत | | प्रत्यक्ष सामग्री | 32·8 | – | | प्रत्यक्ष श्रम | 28·4 | – | | कारखाना उपरिव्य | 12·6 | 1,89,900 | | वितरण उपरिव्य | 4·1 | 58,400 | | सामान्य प्रशासनिक उपरिव्य | 1·1 | 66,700 | | बजट में निर्धारित बिक्री | – | 18,50,000 | निम्नलिखित निर्धारित कीजिए: (i) समतोड़ (ब्रेक-ईवन) बिक्री मात्रा (ii) बजट में निर्धारित बिक्री मात्रा पर लाभ (iii) लाभ यदि वास्तविक बिक्री I. 10% गिरने पर, और II. बजट में निर्धारित बिक्री से 5% की वृद्धि होने पर। (10 marks)

Evaluate your answer to this question → See all 2025 Mechanical Engineering questions

Directive word: Calculate

This question asks you to calculate. The directive word signals the depth of analysis expected, the structure of your answer, and the weight of evidence you must bring.

See our UPSC directive words guide for a full breakdown of how to respond to each command word.

How this answer will be evaluated

Approach

Calculate all numerical sub-parts systematically, allocating approximately 40% time to part (a) given its 7 sub-questions and 20 marks, 35% to part (b) for EOQ with discount analysis, and 25% to part (c) for break-even analysis. Begin each part with stated assumptions and formulae, show complete step-wise calculations with units, and conclude with clear final answers for each sub-part.

Key points expected

  • Part (a): Shear angle φ = arctan(r·cosα/(1-r·sinα)) = arctan(0.3×cos9°/(1-0.3×sin9°)) ≈ 16.7°; then F_s = F_c·cosφ - F_t·sinφ and F_n = F_c·sinφ + F_t·cosφ
  • Part (a): Friction force F = F_c·sinα + F_t·cosα and normal force N = F_c·cosα - F_t·sinα; friction angle β = arctan(F/N)
  • Part (a): Work done in shear = F_s × V_s where V_s = V·cosα/cos(φ-α); work done in friction = F × V_c where V_c = V·sinφ/cos(φ-α)
  • Part (b): EOQ = √(2×D×S/(I×C)) = √(2×2400×32)/(0.16×6) ≈ 400 units; compare total costs at EOQ vs discount lot size 500
  • Part (b): Total cost = DC + (D/Q)S + (Q/2)IC; verify if 5% discount on ₹5.70 with Q=500 beats EOQ total cost at ₹6
  • Part (c): Contribution margin ratio = (Sales - Variable cost)/Sales; BEP sales = Fixed cost/CM ratio
  • Part (c): Profit calculations at budgeted, 90% and 105% sales using P = (Sales × CM ratio) - Fixed costs
  • Part (c): Total variable cost per unit = ₹79.0; total fixed costs = ₹3,15,000; CM ratio = (18,50,000 - (2400×79))/18,50,000

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies Merchant's circle theory for part (a) with proper force component resolution; uses correct EOQ model with discount evaluation for (b); applies marginal costing and CVP analysis correctly for (c); all fundamental relationships (shear angle, chip thickness ratio, friction angle) correctly stated.Uses correct formulae but with minor conceptual gaps (e.g., confuses friction angle with rake angle, or applies EOQ without checking discount threshold); reaches correct final answers despite some theoretical imprecision.Fundamental errors such as treating chip thickness ratio as cutting ratio incorrectly, confusing fixed vs variable costs in break-even, or applying simple average instead of weighted costs for inventory decisions.
Numerical accuracy25%12.5All seven values in (a) correct to 2 decimal places (F_s ≈ 880 N, F_n ≈ 1210 N, F ≈ 763 N, N ≈ 1035 N, β ≈ 36.5°, shear work ≈ 2.42 kW, friction work ≈ 1.91 kW); (b) correctly shows discount acceptance with cost savings ~₹144; (c) BEP ≈ ₹10,50,000, budgeted profit ≈ ₹2,35,000, 90% sales profit ≈ ₹1,46,500, 105% sales profit ≈ ₹2,79,250.Most final answers correct but 1-2 calculation errors (e.g., arithmetic slip in shear velocity or EOQ rounding); correct methodology but wrong decimal placement in cost comparisons.Multiple numerical errors (>3 parts wrong), unit inconsistencies (N vs kN, minutes vs seconds), or order-of-magnitude errors in cost calculations making conclusions invalid.
Diagram quality15%7.5Clean Merchant's circle diagram for (a) showing all force components (F_c, F_t, F_s, F_n, F, N) with angles α, φ, β clearly marked; orthogonal cutting schematic with chip formation; cost comparison graph for (b) showing EOQ vs discount total cost curves; break-even chart for (c) with sales revenue, total cost, fixed cost lines and margin of safety indicated.Merchant's circle drawn but missing some force labels or angle markings; simple tabular comparison for (b) without graphical representation; break-even point indicated arithmetically without diagram.No diagrams despite (a) requiring force visualization; or incorrect diagrams (e.g., wrong angle relationships, forces not resolved properly); diagrams drawn without scale or labels.
Step-by-step derivation25%12.5Explicit derivation of shear angle from chip thickness ratio; complete force balance equations shown before substitution; EOQ formula derived from d(TC)/dQ = 0; total cost comparison table with all three cost components for both scenarios; contribution format income statement shown for (c) with clear PV ratio derivation.Key formulae stated but some intermediate steps skipped (e.g., direct use of Merchant's circle relations without showing vector resolution); final answers correct but working condensed.Final answers stated without formulae or derivation; no working shown for force components; jumps to conclusions without showing cost comparisons; missing units throughout calculations.
Practical interpretation15%7.5For (a): comments on specific energy consumption and tool life implications of high friction work; for (b): discusses inventory holding vs ordering cost trade-off, suggests review of discount structure; for (c): interprets margin of safety (~43%), discusses operating leverage and sensitivity analysis, relates to MSME pricing decisions.Brief comment on each part's practical significance without depth; mentions tool wear for (a), stockout risk for (b), and safety margin for (c) but no quantitative linkage.No interpretation provided; treats as pure mathematics exercise; or irrelevant commentary not tied to manufacturing/industry context (e.g., generic statements about 'efficiency').

Practice this exact question

Write your answer, then get a detailed evaluation from our AI trained on UPSC's answer-writing standards. Free first evaluation — no signup needed to start.

Evaluate my answer →

More from Mechanical Engineering 2025 Paper I