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Mechanical Engineering 2021 Paper I 50 marks Calculate

Paper I — Q6

(a) Five jobs A, B, C, D and E need to be processed on a machine. Processing time (in days) and due date (from now) are given…

(a)

Five jobs A, B, C, D and E need to be processed on a machine. Processing time (in days) and due date (from now) are given below :

JobsProcessing Time (in days)Due Date (from now)
A65
B410
C515
D820
E730

Determine the total completion time (in days) for all jobs, average flow time, average number of jobs in system per day and average tardiness using Shortest Processing Time (SPT) rule so as to establish appropriate sequence for processing of jobs. 15 marks

(b)

Three cities namely Chennai, Delhi and Kolkata are being considered as potential locations for a new plant. Estimated data of annual fixed cost, variable cost and revenue per unit for each potential location are given below in the table. Determine the most attractive location for the plant if the estimated annual production volume desired is 40000 units. Also determine break-even production volume for each location. 15 marks

HeadCity
ChennaiDelhiKolkata
Fixed cost (₹)800000600000500000
Variable cost per unit (₹)304050
Revenue per unit (₹)606060
(c)

Orthogonal turning of a metal is performed using single point cutting tool having rake angle of 10°. The turning operation produces chip-thickness ratio of 0·2, horizontal cutting force (F_H) of 1400 N and vertical cutting force (F_V) of 2000 N. Determine the shear plane angle, normal force, friction force on rake face and friction angle. 20 marks

हिंदी में प्रश्न पढ़ें
(a)

पाँच कृत्यों A, B, C, D व E का मशीन पर प्रक्रमण होना है । प्रक्रमण समय (दिनों में) व नियत तिथि (अभी से) निम्नवत हैं :

कृत्यकप्रक्रमण समय (दिनों में)नियत तिथि (अभी से)
A65
B410
C515
D820
E730

सभी कृत्यों के लिए कुल समापन समय (दिनों में), औसत प्रवाह समय, प्रति दिन निकाय में कृत्यों की माध्य संख्या तथा न्यूनतम प्रसंस्करण समय (SPT) नियम से माध्य विलंबन ज्ञात कीजिए, जिससे कि कृत्यों के प्रक्रमण के लिए सटीक क्रम निर्धारित किया जा सके । (15 अंक)

(b)

तीन शहरों चेन्नई, दिल्ली व कोलकाता को एक नए संयंत्र के लिए संभावित लोकेशन माना जा रहा है । प्रत्येक संभावित लोकेशन के लिए वार्षिक स्थिर लागत, परिवर्त्य लागत तथा आय प्रति इकाई के अनुमानित आँकड़े नीचे सारणी में दिए गए हैं । संयंत्र के लिए अनुमानित 40000 इकाइयों की वार्षिक उत्पादन मात्रा के लिए सबसे आकर्षक लोकेशन का निर्धारण कीजिए । प्रत्येक लोकेशन के लिए उत्पादन की संतुलन स्तर मात्रा भी ज्ञात कीजिए । (15 अंक)

शीर्षशहर
चेन्नईदिल्लीकोलकाता
स्थिर लागत (₹)800000600000500000
परिवर्त्य लागत प्रति इकाई (₹)304050
आय प्रति इकाई (₹)606060
(c)

एक एकल बिंदु कतन औजार जिसका रेक कोण 10° है, से एक धातु का लंबकोणीय खरादन किया जाता है । खरादन प्रक्रम छीलन मोटाई अनुपात 0·2, क्षैतिज कतन बल (F_H) 1400 N तथा उद्वधार कतन बल (F_V) 2000 N उत्पन्न करता है । अपरूपण तल कोण, अभिलंब बल, रेक फलक पर घर्षण बल तथा घर्षण कोण ज्ञात कीजिए । (20 अंक)

Q6 of the 2021 UPSC Mains Mechanical Engineering Paper I, as printed
The question as printed in the 2021 Mechanical Engineering paper
Evaluate your answer to this question → See all 2021 Mechanical Engineering questions

Model answer

Written by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key.

(a) Using the SPT (Shortest Processing Time) rule, arrange jobs in increasing processing time:

  • B: 4 days
  • C: 5 days
  • A: 6 days
  • E: 7 days
  • D: 8 days

So the sequence is B-C-A-E-D.

Completion times:

  • C_B = 4 days
  • C_C = 4 + 5 = 9 days
  • C_A = 9 + 6 = 15 days
  • C_E = 15 + 7 = 22 days
  • C_D = 22 + 8 = 30 days

Sum of completion times = 4 + 9 + 15 + 22 + 30 = 80 job-days. Makespan, i.e. time to complete all jobs = 30 days.

Average flow time = 80/5 = 16 days.

Using Little’s formula for scheduling, average number of jobs in system = total flow time / makespan = 80/30 = 8/3 ≈ 2.67 jobs.

Tardiness = max(0, completion time − due date):

  • B: max(0, 4 − 10) = 0
  • C: max(0, 9 − 15) = 0
  • A: max(0, 15 − 5) = 10
  • E: max(0, 22 − 30) = 0
  • D: max(0, 30 − 20) = 10

Total tardiness = 20 days. Average tardiness = 20/5 = 4 days.

Hence SPT sequence is B-C-A-E-D; total completion sum = 80 job-days, makespan = 30 days, average flow time = 16 days, average jobs in system = 8/3 ≈ 2.67 jobs, average tardiness = 4 days.

(b) For break-even analysis, break-even volume is Q_BE = Fixed cost / (Revenue per unit − Variable cost per unit). Profit at volume Q = (Revenue − Variable cost)Q − Fixed cost.

Chennai: Contribution per unit = 60 − 30 = ₹30. Total cost at 40000 units = 800000 + 30 × 40000 = ₹20,00,000. Revenue = 60 × 40000 = ₹24,00,000. Profit = 24,00,000 − 20,00,000 = ₹4,00,000. Q_BE = 800000/30 = 80000/3 ≈ 26,666.67 units.

Delhi: Contribution per unit = 60 − 40 = ₹20. Total cost = 600000 + 40 × 40000 = ₹22,00,000. Profit = 24,00,000 − 22,00,000 = ₹2,00,000. Q_BE = 600000/20 = 30,000 units.

Kolkata: Contribution per unit = 60 − 50 = ₹10. Total cost = 500000 + 50 × 40000 = ₹25,00,000. Profit = 24,00,000 − 25,00,000 = −₹1,00,000, i.e. loss of ₹1,00,000. Q_BE = 500000/10 = 50,000 units.

At the desired volume of 40000 units, the most attractive location is Chennai, because it gives the highest profit, ₹4,00,000, and the lowest break-even volume, 26,666.67 units.

(c) For orthogonal cutting, use the chip-thickness ratio relation and Merchant’s force circle.

Given: Rake angle α = 10°, chip-thickness ratio r = t₁/t₂ = 0.2. Horizontal cutting force F_H = 1400 N, vertical cutting force F_V = 2000 N.

Shear plane angle φ is found from tan φ = r cos α / (1 − r sin α).

Substitute values: tan φ = 0.2 cos 10° / (1 − 0.2 sin 10°) = 0.2 × 0.9848078 / (1 − 0.2 × 0.1736482) = 0.1969616 / 0.9652704 = 0.204048.

Therefore, φ = tan⁻¹(0.204048) = 11.53°.

Friction force on rake face: F = F_H sin α + F_V cos α = 1400 sin 10° + 2000 cos 10° = 1400 × 0.1736482 + 2000 × 0.9848078 = 243.107 + 1969.616 = 2212.72 N.

Normal force on rake face: N = F_H cos α − F_V sin α = 1400 cos 10° − 2000 sin 10° = 1400 × 0.9848078 − 2000 × 0.1736482 = 1378.731 − 347.296 = 1031.43 N.

Friction angle β = tan⁻¹(F/N) = tan⁻¹(2212.72/1031.43) = tan⁻¹(2.1453) = 65.01°.

Alternatively, β = α + tan⁻¹(F_V/F_H) = 10° + tan⁻¹(2000/1400) = 10° + 55.01° = 65.01°.

Thus: Shear plane angle φ = 11.53°, Normal force N = 1031.43 N, Friction force F = 2212.72 N, Friction angle β = 65.01°.

These results assume ideal orthogonal cutting, continuous chip formation, and validity of Merchant’s force-circle analysis.

What "Calculate" is asking you to do

Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.

Structure that answers it

Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units

Where marks are lost

Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.

All UPSC directive words, compared →

How this answer will be evaluated

Approach

Framework: Mechanical Engineering, Paper 1. (a) calculate: given > formula > substitution > result with units > interpretation | (b) calculate: given > formula > substitution > result with units > interpretation | (c) calculate: given > formula > substitution > result with units > interpretation Full marks: Correct sequence/logic, all formulas stated, accurate calculations, clear diagrams/tables, and physical interpretation of results.

Key points expected

  • Jobs sorted by ascending processing time (B, C, A, E, D)
  • Flow time and completion time calculated for each job
  • Tardiness calculated as max(0, completion - due date)
  • Average flow time and average tardiness computed
  • Break-even formula: Fixed Cost / (Revenue - Variable Cost)
  • Break-even volume calculated for Chennai, Delhi, and Kolkata
  • Total cost or profit calculated at 40,000 units for each city
  • Most attractive location identified based on lowest cost or highest profit

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) SPT sequence and four performance metrics (total completion, avg flow, avg jobs, avg tardiness). 15 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Jobs sorted by ascending processing time (B, C, A, E, D)
    • Flow time and completion time calculated for each job
    • Tardiness calculated as max(0, completion - due date)
    • Average flow time and average tardiness computed

    Loses marks

    • Incorrect SPT sequence (e.g., sorting by due date)
    • Tardiness calculated as negative values instead of zero
    • Missing units (days) in final metrics

    Earns more

    • Average number of jobs in system derived from total flow time
    • Explicit table showing cumulative completion times

    Extra mark

    • Comparison of SPT result with EDD or FCFS rule
  2. (b) Break-even volume for each city and most attractive location at 40,000 units. 15 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Break-even formula: Fixed Cost / (Revenue - Variable Cost)
    • Break-even volume calculated for Chennai, Delhi, and Kolkata
    • Total cost or profit calculated at 40,000 units for each city
    • Most attractive location identified based on lowest cost or highest profit

    Loses marks

    • Using revenue instead of contribution margin in break-even formula
    • Failing to compare all three cities at the specific volume
    • Arithmetic errors in cost calculation

    Earns more

    • Cost-volume-profit (CVP) graph or table showing comparison
    • Explicit statement of contribution margin per unit

    Extra mark

    • Sensitivity analysis on fixed cost or volume
  3. (c) Shear plane angle, normal force, friction force, and friction angle for orthogonal turning. 20 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Shear plane angle calculated using chip thickness ratio and rake angle
    • Resultant force and friction angle derived from horizontal and vertical forces
    • Normal and friction forces resolved on the rake face
    • Friction angle calculated from force components

    Loses marks

    • Confusing rake angle with friction angle
    • Incorrect force resolution (e.g., using wrong trigonometric functions)
    • Missing units (N, degrees) in final answers

    Earns more

    • Force vector diagram (Merchant's circle) clearly labelled
    • Correct application of trigonometric identities for force resolution

    Extra mark

    • Calculation of shear stress on the shear plane

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