Mechanical Engineering 2022 Paper I 50 marks Calculate

Q7

Q7. (a) A firm has identified four operations, which are to be conducted in succession for an order to be processed. The tolerance and mean time of each operation are given in the following table. Tolerance is independent of each other and the time is normally distributed. | Operation | Mean Time (hours) | Tolerance (hours) | |-----------|-------------------|-------------------| | 1 | 7 | 7 ± 0·6 | | 2 | 5 | 5 ± 0·6 | | 3 | 9 | 9 ± 0·8 | | 4 | 6 | 6 ± 0·3 | (i) Find the natural tolerance limits for order completion time. (20 marks) (ii) If the company sets a goal of 27·5 hours, what proportion of the orders will fail to satisfy the goal ? (iii) Find an appropriate capability index and comment. (iv) Using a technique, management has improved the operation 3 to a mean time of 8 hours. What proportion of the orders will now meet the goal ? (Use Standard Normal Distribution table given on the last page) (b) (i) Discuss the factors influencing the facility location selection. (10 marks) (ii) For expansion of a car manufacturing plant, three new locations are to be considered based on three factors : Availability of labour, Proximity to the suppliers, and Proximity to the markets. The weightage of these factors are given as 40%, 35%, and 25% respectively. The rating (on 100-point scale) of the locations against these factors are given in the following table : | Location | Availability of labour | Proximity to the suppliers | Proximity to the markets | |----------|------------------------|---------------------------|-------------------------| | X | 70 | 60 | 45 | | Y | 60 | 45 | 90 | | Z | 55 | 95 | 50 | Find the best and worst location for the new plant. (10 marks) (c) Spot welding of two steel sheets each of 1 mm thickness is performed using 20,000 A welding current supplied for 0·15 seconds. Assume that : (i) interface contact resistance is 200 micro-ohms, (ii) heat required for melting unit volume of steel is 10 J/mm³, and (iii) only 60% of heat generated is used for melting of metal at the interface. Calculate : (10 marks) (I) Heat generated, J (II) Volume of the weld nugget, mm³

हिंदी में प्रश्न पढ़ें

Q7. (a) एक फर्म ने चार प्रक्रियाओं की पहचान कर रखी है जिनको एक ऑर्डर को पूरा करने के लिए एक अनुक्रम में करना है । निम्नलिखित सारणी में प्रत्येक प्रक्रिया के लिए सहिष्णुता व औसत समय दिए गए हैं । सहिष्णुता एक-दूसरे से स्वतंत्र है तथा समय प्रसामान्य बंटित है । | प्रक्रिया | औसत समय (घंटे) | सहिष्णुता (घंटे) | |:---:|:---:|:---:| | 1 | 7 | 7 ± 0·6 | | 2 | 5 | 5 ± 0·6 | | 3 | 9 | 9 ± 0·8 | | 4 | 6 | 6 ± 0·3 | (i) ऑर्डर पूरा होने के समय की प्राकृतिक सहिष्णुता सीमाओं को ज्ञात कीजिए । (ii) यदि कंपनी एक 27·5 घंटे का लक्ष्य निर्धारित करती है, तो ऑर्डरों का कौन-सा भाग, लक्ष्य को संतुष्ट करने में असफल होगा ? (iii) एक उपयुक्त क्षमता गुणक ज्ञात कीजिए तथा टिप्पणी कीजिए । (iv) एक तकनीक का प्रयोग करके प्रबंधन ने प्रक्रिया 3 में सुधार करके औसत समय को 8 घंटे कर दिया है । ऑर्डरों का कौन-सा अनुपात अब लक्ष्य को प्राप्त करेगा ? (मानक प्रसामान्य बंटन सारणी अंतिम पृष्ठ पर दी गई है) (b) (i) सुविधा स्थान निर्धारण के चयन को प्रभावित करने वाले कारकों की विवेचना कीजिए । (ii) एक कार मैन्युफैक्चरिंग संयंत्र के प्रसार के लिए, तीन नए स्थान-निर्धारणों पर विचार तीन कारकों के आधार पर करना है : श्रम की उपस्थिति, आपूर्तिकर्ताओं से निकटता, तथा बाजारों से निकटता । इन कारकों की भारिता क्रमशः: 40%, 35% व 25% है । स्थान-निर्धारणों की इन कारकों के सापेक्ष रेटिंग (एक 100-बिंदु के पैमाने पर) निम्नलिखित सारणी में दी गई है : | स्थान | श्रम की उपलब्धता | आपूर्तिकर्ताओं से निकटता | बाजारों से निकटता | |-------|---------------|------------------------|----------------| | X | 70 | 60 | 45 | | Y | 60 | 45 | 90 | | Z | 55 | 95 | 50 | नए संयंत्र के लिए सबसे अच्छे व सबसे खराब स्थान-निर्धारण को ज्ञात कीजिए । (c) दो इस्पात चादरों जिनमें प्रत्येक 1 mm मोटाई की है का स्थानिक (स्पॉट) वेल्डन 20,000 A वेल्डन विद्युत धारा की 0·15 सेकंड आपूर्ति करके की जाती है । यह मान लीजिए कि : (i) अंतरापृष्ठ पर संपर्क प्रतिरोध 200 micro-ohms है, (ii) इस्पात के इकाई आयतन को पिघलाने के लिए आवश्यक ऊष्मा 10 J/mm³ है, तथा (iii) अंतरापृष्ठ पर धातु को पिघलाने के लिए उत्पादित ऊष्मा का केवल 60% भाग ही प्रयोग में आता है । गणना कीजिए : (I) उत्पादित ऊष्मा, J में (II) वेल्ड नगेट का आयतन, mm³ में

Evaluate your answer to this question → See all 2022 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 the statistical parameters for sequential operations in (a) using variance addition for independent normal distributions, then apply standard normal tables for probability calculations. For (b), apply weighted factor rating method for facility location selection. For (c), use resistance welding heat generation formula Q = I²Rt with efficiency factor. Allocate approximately 50% time to part (a) due to 20 marks and four sub-parts, 25% to part (b), and 25% to part (c).

Key points expected

  • Part (a)(i): Total mean = 27 hours, total variance = 0.6² + 0.6² + 0.8² + 0.3² = 1.45, σ = 1.204 hours; natural tolerance limits = 27 ± 3×1.204 = 23.388 to 30.612 hours
  • Part (a)(ii): Z = (27.5-27)/1.204 = 0.415; P(X > 27.5) = 1 - Φ(0.415) ≈ 0.339 or 33.9% orders fail
  • Part (a)(iii): Cpk = min[(USL-μ)/3σ, (μ-LSL)/3σ] with appropriate specification limits; comment on process capability
  • Part (a)(iv): New mean = 26 hours, same σ; Z = (27.5-26)/1.204 = 1.246; P(X ≤ 27.5) = Φ(1.246) ≈ 0.893 or 89.3% meet goal
  • Part (b)(i): Discuss quantitative factors (transportation cost, labour cost, utilities) and qualitative factors (community attitude, climate, political stability, environmental regulations)
  • Part (b)(ii): Weighted scores - X: 70×0.4 + 60×0.35 + 45×0.25 = 59.25; Y: 60×0.4 + 45×0.35 + 90×0.25 = 62.25; Z: 55×0.4 + 95×0.35 + 50×0.25 = 66.75; Best = Z, Worst = X
  • Part (c)(I): Heat generated Q = I²Rt = (20000)² × 200×10⁻⁶ × 0.15 = 12,000 J
  • Part (c)(II): Effective heat = 0.6 × 12000 = 7200 J; Volume = 7200/10 = 720 mm³

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%8Correctly applies additive property of variances for independent normal variables in (a); recognizes tolerance as ±3σ; applies weighted factor rating method correctly in (b); uses proper resistance welding heat equation with contact resistance in (c); distinguishes between heat generated and heat utilized.Correctly identifies normal distribution addition but confuses tolerance with standard deviation; applies factor rating with minor weighting errors; uses correct heat formula but misses efficiency factor.Adds standard deviations instead of variances; treats tolerances as uniform distributions; applies unweighted scoring in (b); uses wrong formula (e.g., VIt instead of I²Rt) in (c).
Numerical accuracy20%8All calculations precise: total variance = 1.45, σ = 1.204; Z-scores correct to 3 decimal places; probability values correctly interpolated from tables; weighted scores: X=59.25, Y=62.25, Z=66.75; heat = 12000 J, volume = 720 mm³.Correct final answers but minor arithmetic slips in intermediate steps (e.g., variance calculation correct but square root rounded prematurely); Z-score calculation correct but probability lookup approximate.Major calculation errors: adds tolerances directly (0.6+0.6+0.8+0.3=2.3); wrong Z-score formula; heat calculation misses micro-ohm conversion or time unit errors; final answers orders of magnitude wrong.
Diagram quality20%8Sketches normal distribution curves for (a)(ii) and (a)(iv) showing mean, goal line, and shaded rejection regions; draws clear factor rating comparison bar chart for (b); illustrates spot welding nugget geometry in (c) with dimensions implied.Basic normal curve sketched without proper labelling; simple table format for factor comparison; minimal or no welding diagram.No diagrams despite visual nature of probability problems; messy or irrelevant sketches; no comparison chart for location selection.
Step-by-step derivation20%8Shows explicit formula for variance of sum: σ²_total = Σσ²_i with σ_i = tolerance_i/3; writes Z = (X-μ)/σ with substitution; shows weighted score formula S = Σw_i×r_i before substitution; displays Q = I²Rt with unit conversions (μΩ to Ω, A, s).Shows some intermediate steps but jumps from formula to final answer; unit conversions implied but not shown explicitly; table format for factor rating without showing formula.Bare answers with no working; no formula stated; no unit conversion shown; impossible to trace errors.
Practical interpretation20%8Comments on high failure rate (34%) in (a)(ii) suggesting process improvement needed; interprets Cpk < 1.33 as inadequate capability; notes 89.3% success after improvement justifies the effort; relates facility location to Make in India/Automotive hubs (Chennai, Pune, Gurgaon); discusses weld nugget size adequacy for 1mm sheets and automotive body-in-white applications.Brief comment on process capability without specific threshold; mentions improvement is beneficial; generic location factors without industry context; basic statement on weld quality.No interpretation of any results; purely mathematical exercise; no connection to real manufacturing scenarios; ignores practical significance of 33.9% rejection rate.

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 2022 Paper I