Paper I — 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…
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
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
What are the reasons for carrying inventories in production industries ? 10 marks
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
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
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)
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)
उत्पादन उद्योगों में सामग्री-सूची हस्तगत (कैरी) करने के क्या कारण हैं ? (10 अंक)
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)
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)
Model answer
Drafted with AI 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. How we build rubrics and answers.
(a) TIG welding: power and melting rate
(i) Arc power P = VI = 12 × 200 = 2400 W.
(ii) Heat used for melting = 0.70 × 2400 = 1680 W.
(iii) Volume of weld metal produced per second = 1680 J/s ÷ 20 J/mm³ = 84 mm³/s.
The 5 mm thickness is not needed here; it would matter only for finding welding speed from the weld cross-section.
(b) Tool life by Taylor's equation
VTⁿ = C with n = 0.27. At V₁ = 60 m/min the life is T₁ = 3 h.
(i) At V₂ = 80 m/min: T₂ = T₁(V₁/V₂)^(1/n) = 3 × (0.75)^(1/0.27) = 3 × 0.3445 = 1.034 h, about 62 minutes.
(ii) For T₂ = 2 h: V₂ = V₁(T₁/T₂)ⁿ = 60 × (1.5)^0.27 = 60 × 1.1157 = 66.9 m/min.
These hold for the same tool, work, feed, depth of cut and coolant.
(c) Reasons for carrying inventories
Inventory decouples supply, production and demand. Raw material stock covers supplier lead time and delays; work-in-process stock lets successive operations run independently despite breakdowns or unequal speeds; finished goods stock gives immediate delivery and customer service. Safety stock guards against demand and lead-time uncertainty and avoids stock-outs. Cycle stock arises from buying or making in lots, which cuts ordering and set-up cost and earns quantity discounts. Anticipation stock levels production against seasonal demand or expected price rises. Pipeline stock is unavoidable for goods in transit, and maintenance, repair and operating (MRO) supplies keep machines running. The cost is holding cost, obsolescence and tied-up capital, so levels must balance service against cost.
(d) Sales forecasts
Sales: 5, 17, 29, 41, 39.
(i) Last value method: forecast = 39 units. Averaging method: (5 + 17 + 29 + 41 + 39)/5 = 131/5 = 26.2 units. Three-month moving average: (29 + 41 + 39)/3 = 109/3 = 36.3 units.
(ii) Sales rose by 12 units a month for four months and then dipped slightly, so there is a strong trend. The averaging method is clearly inappropriate: it gives equal weight to the old low values and forecasts 26.2, far below the recent 41 and 39. The moving average also lags the trend, and the last value method reacts to a single observation that may be a random dip. A trend-adjusted method (regression or Holt's exponential smoothing) suits this pattern better.
(e) Plug gauge for a hole of 30 (+0.06, 0) mm
Work tolerance = 0.06 mm, so gauge tolerance = wear allowance = 0.10 × 0.06 = 0.006 mm. Hole limits: 30.000 and 30.060 mm. A plug gauge checks a hole.
GO gauge: basic size = low limit + wear allowance = 30.006 mm, with gauge tolerance on the plus side: 30.006 to 30.012 mm.
NO-GO gauge: basic size = high limit = 30.060 mm, with gauge tolerance on the minus side: 30.054 to 30.060 mm.
No wear allowance is given to the NO-GO gauge, because its wear only rejects slightly more good holes and never accepts an oversize one.
How this answer will be evaluated
What earns full marks: All parts solved with correct formulas, units, and clear justification; no calculation errors.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Find arc power, melting power and volume rate of weld metal produced 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Arc power 2400 W
- Melting power 1680 W
- Volume rate 84 mm³/s
Loses marks
- Applying the 70% factor to the wrong quantity
- Unit slip between W and J/s
Earns more
- 70% heat utilisation applied correctly
- Energy per unit volume used to convert power to volume rate
Extra mark
- Notes that thickness is not needed for this result
- (b) Use Taylor's equation to find tool life at 80 m/min and speed for 2 h life 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Taylor equation VTⁿ = C
- Tool life at 80 m/min ≈ 1.03 h (about 62 min)
- Cutting speed for 2 h life ≈ 66.9 m/min
Loses marks
- Inverting the exponent
- Not converting hours and minutes consistently
Earns more
- Ratio form V₁T₁ⁿ = V₂T₂ⁿ applied correctly
- Exponent 1/n used for T and n for V
Extra mark
- States validity conditions of Taylor's equation
- (c) Give the reasons for carrying inventories in production industries 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- Buffering supply, production and demand (raw material, work-in-process, finished goods)
- Safety stock against uncertainty
- Cycle or lot-size benefits and quantity discounts
Loses marks
- Listing only one or two reasons
- Reasons without explanation
Earns more
- Anticipation or speculative stock
- Pipeline and MRO inventories
Extra mark
- Notes the holding costs and balance needed
- (d) Forecast next month's sales by three methods and comment on their suitability 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Last value 39, averaging 26.2, 3-month moving average 36.3
- Trend in the data identified
- Averaging method judged inappropriate with a reason
Loses marks
- Arithmetic error in a forecast
- Calling all methods suitable without reasoning
Earns more
- Moving average and last value shown to lag or be noisy
- Trend-adjusted method suggested
Extra mark
- Names exponential smoothing with trend or regression
- (e) Design a plug gauge with unilateral tolerance for a 30 (+0.06) mm hole 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Work tolerance 0.06 mm and gauge tolerance and wear allowance 0.006 mm
- GO gauge 30.006–30.012 mm
- NO-GO gauge 30.054–30.060 mm
Loses marks
- Applying wear allowance to NO-GO
- Placing gauge tolerance outside the work limits
Earns more
- Correct limits of the hole
- Wear allowance given to the GO gauge only
Extra mark
- Explains why NO-GO needs no wear allowance
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.
Practice this exact question
Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.
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