Mechanical Engineering 2021 Paper I 50 marks Compulsory Calculate

Q5

(a) TIG welding of two sheets of 5 mm thickness is performed using welding current (I) of 200 A and arc voltage (V) of 12 V. Assume 70% of generated arc heat is utilised for melting of base metals. If the steel being welded needs 20 J/mm³ heat for melting, then determine the following : (i) Power of welding arc (W) (ii) Rate at which energy is delivered for melting (W) (iii) Volume rate (mm³/sec) at which weld metal is produced (10 marks) (b) A high speed steel (HSS) cutting tool during turning of aluminium offers tool life of 3 hours at cutting speed of 60 m/minute. Determine the following using above data and assuming value of n is 0·27 : (i) Life of tool if turning is performed at 80 m/minute cutting speed. (ii) Cutting speed at which cutting tool will have tool life of 2 hours. (10 marks) (c) What are the reasons for carrying inventories in production industries ? (10 marks) (d) An XYZ company launched a new product which had sales of 5, 17, 29, 41 and 39 units respectively in its first five months of launch. The Sales Manager now wants a forecast of sales in the next month. (i) Find out sales forecast by the last value method, the averaging method and the moving average method with the 3 most recent months. (ii) Given the sales pattern so far, do any of the above methods seem inappropriate for obtaining the forecast ? Why ? (10 marks) (e) A mechanic needs a gauge for checking the diameter of holes to be machined to a diameter of 30⁺⁰·⁰⁶ mm. What should be the dimensions of the gauge, if unilateral system of tolerances are incorporated ? Assume gauge tolerance and wear allowance each as 10% of work tolerance. (10 marks)

हिंदी में प्रश्न पढ़ें

(a) 5 mm मोटाई की दो चादरों का TIG वेल्डन 200 A विद्युत धारा (I) तथा 12 V आर्क वोल्टता (V) पर किया जाता है । मान लीजिए कि आर्क द्वारा जनित 70% ऊष्मा का उपयोग मूल धातु के पिघलाने में होता है । यदि इस्पात जिसका वेल्डन होना है उसको पिघलाने के लिए 20 J/mm³ ऊष्मा की आवश्यकता होती है, तो निम्न को ज्ञात कीजिए : (i) वेल्डन आर्क की शक्ति (W) (ii) वह दर जिस पर ऊर्जा गलाने के लिए प्रदान की जाती है (W) (iii) वह आयतन दर (mm³/sec) जिस पर वेल्ड धातु का उत्पादन होता है (10 अंक) (b) एक उच्च चाल इस्पात (HSS) के कतन औजार का ऐलुमिनियम के 60 मी./मिनट की चाल पर खरादन के दौरान औजार जीवन काल 3 घंटे है । n का मान 0·27 मानते हुए तथा उपर्युक्त आँकड़ों का प्रयोग करते हुए निम्न को ज्ञात कीजिए : (i) 80 मी./मिनट कतन चाल पर, खरादन करने पर औजार आयु । (ii) वह कतन चाल जिस पर कतन औजार की आयु 2 घंटे होगी । (10 अंक) (c) उत्पादन उद्योगों में सामग्री-सूची हस्तगत (कैरी) करने के क्या कारण हैं ? (10 अंक) (d) एक XYZ कंपनी ने एक नया उत्पाद अवतरित किया जिसकी अवतरण के प्रथम पाँच महीनों में बिक्री क्रमशः: 5, 17, 29, 41 व 39 इकाइयों की थी । बिक्रय मैनेजर अब अगले महीने में बिक्रय पूर्वानुमान चाहते हैं । (i) निम्नलिखित विधियों से बिक्रय पूर्वानुमान ज्ञात कीजिए : आखिरी मान विधि (Last Value Method), औसत विधि और तीन अति नवीनतम महीनों में चल औसत विधि द्वारा । (ii) अभी तक दिए गए बिक्रय प्रतिरूप के अनुसार क्या उपर दी गई विधियों में कोई भी विधि पूर्वानुमान प्राप्त करने के लिए अनुपयुक्त है ? क्यों ? (10 अंक) (e) एक मैकेनिक को 30⁺⁰·⁰⁶ mm व्यास तक के छिद्रों जिनका मशीनिंग होना है, के व्यास की जाँच के लिए एक गेज की आवश्यकता है । यदि एक पार्श्विक सहिष्णुता प्रणाली का पालन होना हो, तो गेज की विमाएँ क्या होंगी ? मान लीजिए कि गेज सहिष्णुता व निघर्षण छूट प्रत्येक, कार्य सहिष्णुता का 10% है । (10 अंक)

Evaluate your answer to this question → See all 2021 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 numerical solutions for all five parts systematically, spending approximately 20% time each on (a) welding energy, (b) Taylor's tool life, (c) inventory reasons, (d) forecasting methods, and (e) gauge design. For numerical parts, show all formulas, substitutions, and unit conversions; for part (c), enumerate 6-8 well-explained reasons; for part (d), compute all three forecasts and critically evaluate method appropriateness based on trend visibility.

Key points expected

  • Part (a): Arc power = IV = 2400 W; melting power = 0.70 × 2400 = 1680 W; volume rate = 1680/20 = 84 mm³/s
  • Part (b): Using Taylor's VT^n = C, find C = 60×(180)^0.27 ≈ 228.6; then T at 80 m/min = (228.6/80)^(1/0.27) ≈ 0.89 hr; V for T=2 hr = 228.6/(120)^0.27 ≈ 69.4 m/min
  • Part (c): Reasons for inventory: demand variability, supply lead time, quantity discounts, production smoothing, safety stock, work-in-progress, seasonal demand, and speculative buying
  • Part (d)(i): Last value = 39; Averaging = (5+17+29+41+39)/5 = 26.2; 3-month MA = (29+41+39)/3 = 36.33
  • Part (d)(ii): Last value and simple averaging inappropriate due to clear upward trend; 3-month MA better but still lags; trend projection most suitable
  • Part (e): Work tolerance = 0.06 mm; gauge tolerance = wear allowance = 0.006 mm; GO gauge: 30.000 to 30.006 mm; NO-GO gauge: 30.054 to 30.060 mm (unilateral system)
  • Part (e) alternative: GO gauge = 30.006⁺⁰·⁰⁰⁶ mm; NO-GO gauge = 30.060⁻⁰·⁰⁰⁶ mm with wear allowance on GO only

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly applies heat transfer efficiency in welding (part a), Taylor's tool life equation with proper exponent handling (part b), distinguishes inventory types and their strategic purposes (part c), identifies appropriate forecasting methods based on data pattern recognition (part d), and applies unilateral tolerance system with proper gauge tolerance and wear allowance allocation (part e).Uses correct basic formulas but makes minor errors in exponent interpretation for Taylor's equation or confuses bilateral/unilateral gauge systems; inventory reasons listed but not well-categorized.Fundamental errors: treats arc heat as 100% efficient, uses Taylor's equation with n instead of 1/n, or applies bilateral tolerance to gauge design; inventory reasons confused with production planning.
Numerical accuracy20%10All calculations precise: (a) 2400 W, 1680 W, 84 mm³/s; (b) C≈228.6, T≈0.89 hr, V≈69.4 m/min; (d) forecasts 39, 26.2, 36.33; (e) GO 30.006⁺⁰·⁰⁰⁶, NO-GO 30.060⁻⁰·⁰⁰⁶; unit conversions correct throughout.Final answers approximately correct but arithmetic slips in intermediate steps (e.g., C value off by 5-10%, or gauge dimensions with tolerance arithmetic errors); most units correct.Major calculation errors: wrong power formula (uses I²R or V²/R), incorrect time unit conversion in Taylor's equation (mixes minutes and hours), or gauge tolerance applied incorrectly as percentage of nominal size rather than work tolerance.
Diagram quality10%5Clear sketch for part (e) showing GO and NO-GO plug gauges with tolerance zones, wear allowance direction, and work tolerance zone; optionally includes trend plot for part (d) sales data showing why moving average lags; diagrams labelled with all dimensions.Basic gauge sketch present but missing tolerance zone representation or wear allowance indication; no diagram for part (d) trend visualization.No diagrams despite gauge design question explicitly requiring dimensional visualization; or incorrect schematic showing ring gauges instead of plug gauges for hole checking.
Step-by-step derivation30%15Every part shows complete derivation: (a) states P=IV, applies efficiency factor, divides by energy density; (b) derives C from given data, explicitly solves for T and V with logarithmic steps; (c) enumerates 6-8 reasons with brief justification each; (d) shows all three forecast calculations with formulas, then evaluates; (e) calculates work tolerance, allocates 10% each for gauge tolerance and wear allowance, constructs gauge dimensions systematically.Derivations present but some steps skipped (e.g., jumps to final Taylor's result without showing C calculation, or lists inventory reasons without brief explanations); one part may have minimal working.Minimal or no working shown—final answers stated without formulas or substitution steps; part (c) as brief bullet points without elaboration; part (e) states gauge dimensions without showing tolerance allocation calculation.
Practical interpretation20%10Interprets welding volume rate in terms of travel speed and bead geometry; discusses practical implications of reduced tool life at higher cutting speeds for HSS tools; relates inventory reasons to Indian manufacturing context (e.g., supply chain uncertainties, MSME working capital constraints); critically evaluates forecasting methods based on product life cycle stage; explains why unilateral system preferred for gauge design in mass production.Some interpretation present but generic—states that higher speed reduces tool life without discussing HSS limitations, or mentions inventory costs without Indian context; limited critical evaluation of forecasting methods.Purely computational answer with no interpretation; treats all methods as equally valid in part (d) without recognizing trend; no mention of why 10% gauge tolerance is standard or implications of wear allowance placement.

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 2021 Paper I