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Mechanical Engineering 2022 Paper I 50 marks Compulsory Solve

Paper I — Q5

(a) An HSS drill during its life can drill 150 through holes in a 10 mm thick brass plate at a drill speed of 400 rpm. Another…

(a)

An HSS drill during its life can drill 150 through holes in a 10 mm thick brass plate at a drill speed of 400 rpm. Another drill of same type can make only 75 holes when the drill speed is increased to 600 rpm. How many holes will be produced by another drill of same type if its speed is raised to 800 rpm ? Take the feed in all the cases as same. 10 marks

(b)

Write the functions of coating on Shielded Metal Arc Welding (SMAW) electrode. 10 marks

(c)

Discuss the Expansionist strategy and Wait-and-See strategy for capacity timing and sizing concerning the capacity planning. 10 marks

(d)

A product is to be processed from its raw form to finished form through a number of workstations. The production lead time is given as 3 days. The daily demand requirement is 500 units. Safety stock is required for one day. Container's capacity is 400 units. Determine the number of Kanbans (containers) required. 10 marks

(e)

Derive the expression for Reorder point when demand is variable and lead time is constant. Suppose the average demand is 18 units per week with a standard deviation of 5 units. The lead time is constant at 2 weeks. Determine the safety stock and reorder point if management wants a 95% customer service level. (Refer Standard Normal Distribution table given on the last page) 10 marks

हिंदी में प्रश्न पढ़ें
(a)

एक HSS ड्रिल अपने जीवन काल में 10 mm मोटी पीतल की प्लेट में 400 rpm ड्रिल चाल से 150 पारगामी छिद्र कर सकता है । दूसरा ड्रिल जो उसी प्रकार का है और जिसकी ड्रिल चाल बढ़ाकर 600 rpm की गई केवल 75 छिद्र कर सकता है । एक अन्य ड्रिल से, जो उसी प्रकार का है, यदि उसकी चाल बढ़ाकर 800 rpm कर दी जाए तो कितने छिद्रों का उत्पादन होगा ? सभी मामलों में प्रवरण समान लीजिए । (10 अंक)

(b)

परिरक्षित धातु आर्क वेल्डन (SMAW) इलेक्ट्रोड पर विलेपन के कार्य लिखिए । (10 अंक)

(c)

क्षमता प्लानिंग के संदर्भ में क्षमता काल समंजन व आमापन के लिए विस्तारवादी युक्ति व प्रतीक्षा व देखना युक्ति की विवेचना कीजिए । (10 अंक)

(d)

एक उत्पाद को उसके अपरिष्कृत रूप से परिष्कृत रूप तक कई क्रियक स्टेशनों से गुजारकर तैयार किया जाता है । उत्पादन अग्रता काल 3 दिन दिया गया है । प्रतिदिन की मांग आवश्यकता 500 इकाइयों की है । रक्षित स्टॉक की एक दिन की आवश्यकता है । कंटेनर की क्षमता 400 इकाइयों की है । आवश्यक कानबन (Kanbans) कंटेनरों की संख्या ज्ञात कीजिए । (10 अंक)

(e)

पुनरादेश बिन्दु के लिए व्यंजक व्युत्पन्न कीजिए जबकि मांग परिवर्ती है तथा अग्रता काल स्थिर है । मान लीजिए कि औसत मांग 18 इकाई प्रति सप्ताह है जबकि मानक विचलन 5 इकाई का है । अग्रता काल दो सप्ताह पर स्थिर है । यदि प्रबंधन 95% ग्राहक सेवा स्तर चाहता है तो सुरक्षा स्टॉक व पुनरादेश बिन्दु ज्ञात कीजिए । (अंतिम पृष्ठ पर दी गई मानक प्रसामान्य बंटन सारणी देखिए) (10 अंक)

Q5 of the 2022 UPSC Mains Mechanical Engineering Paper I, as printed
The question as printed in the 2022 Mechanical Engineering paper
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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) Number of holes at 800 rpm (Taylor tool life)

For a given drill diameter, V ∝ N, so Taylor equation VTⁿ = C becomes NTⁿ = C₁. Time per through hole = L/(fN), so holes per tool life H = TfN/L, i.e. T = HL/(fN). Substituting, N(HL/fN)ⁿ = C₂, and with L and f constant: N^(1−n) Hⁿ = constant.

From the two tests: 400^(1−n) × 150ⁿ = 600^(1−n) × 75ⁿ (1 − n) ln(2/3) + n ln 2 = 0 ⟹ n = ln 1.5/ln 3 = 0.369.

At 800 rpm: H₃ = 150 × (400/800)^((1−n)/n) = 150 × (0.5)^1.7095 = 45.9 ≈ 45 holes (45 complete holes before the drill must be reground).

(b) Functions of the SMAW electrode coating

Diagram: SMAW electrode
  • Arc stabilisation: potassium, sodium, calcium and titanium compounds ionise the arc gap, easing arc striking and steadying the arc.
  • Gas shielding: the coating decomposes to CO₂, CO and water vapour, which keep atmospheric oxygen and nitrogen away from the molten pool.
  • Slag formation: the flux forms a slag that protects the solidifying metal from oxidation, slows cooling and supports the bead in vertical and overhead welding.
  • Deoxidation and refining: Mn, Si, Ti and Al deoxidise the pool and carry sulphur, phosphorus and oxides into the slag.
  • Alloying: ferro-alloys add Cr, Ni, Mo, Mn or V to improve strength, toughness or corrosion resistance of the deposit.
  • Bead control: slag viscosity and surface tension shape the bead, control penetration, spatter and metal transfer.
  • Insulation and arc direction: the coating insulates the core wire and confines the arc to the tip.
  • Higher deposition: iron-powder coatings raise deposition rate and arc efficiency.
  • Binding and protection: silicate binders hold the coating and protect the core wire from corrosion and damage.

(c) Expansionist versus Wait-and-See capacity strategies

Diagram: Capacity expansion strategies

Capacity timing and sizing decide when and how much capacity to add.

Expansionist (lead) strategy: capacity is added ahead of demand in large increments. It captures market share, avoids shortages, deters competitors and gains economies of scale and better service. It suits steadily growing demand, available capital and markets where delay means permanent loss. Risks are excess capacity, idle facilities, high fixed cost and losses if demand does not materialise.

Wait-and-See (lag) strategy: capacity is added only after demand has risen and is confirmed, in small and frequent increments. It lowers the risk of overbuilding, keeps utilisation high, limits capital exposure and keeps flexibility under uncertainty. It suits volatile demand, unreliable forecasts, fast-changing technology or scarce capital. Its costs are lost sales, poorer service, inability to meet surges and market share lost to expansionist rivals, while later expansion may cost more.

Timing and sizing: the expansionist leads demand with large chunks, the wait-and-see firm follows demand with small, cautious steps. In practice firms often use an intermediate (average-capacity) path, with overtime, subcontracting, inventories and flexible plants absorbing the gap.

(d) Number of Kanbans

N = (D × L + S)/C, with D = 500 units/day, L = 3 days, S = 1 day of demand = 500 units, C = 400 units. N = (500 × 3 + 500)/400 = 2000/400 = 5 Kanbans (containers) (a fractional result would be rounded up).

(e) Reorder point with variable demand and constant lead time

Let weekly demand have mean μ and standard deviation σ, independent from week to week, and lead time L constant. Lead-time demand X = d₁ + … + d_L has E(X) = Lμ and Var(X) = Lσ², so σ_X = σ√L. For service level p, choose z with P(Z ≤ z) = p and set P(X ≤ ROP) = p: (ROP − Lμ)/(σ√L) = z ⟹ ROP = Lμ + zσ√L, with safety stock SS = zσ√L.

Given μ = 18, σ = 5, L = 2 weeks, 95% service level ⟹ z = 1.645 (normal table). Lμ = 36 units, σ_X = 5√2 = 7.07 units. SS = 1.645 × 7.07 = 11.6 ≈ 12 units, ROP = 36 + 11.6 = 47.6 ≈ 48 units.

How this answer will be evaluated

What earns full marks: Shows complete method with all governing equations, correct calculations, and physical interpretation

Evaluation rubric

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

  1. (a) Determine the number of holes produced at 800 rpm using the Taylor tool life relationship. 10 marks

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

    Must cover

    • State Taylor's tool life equation (VT^n = C)
    • Establish relationship between tool life (time) and number of holes
    • Calculate exponent n using the two given data points
    • n ≈ 0.369 and about 45.9, i.e. 45 holes at 800 rpm

    Loses marks

    • Plugging numbers without stating Taylor's equation
    • Failing to relate tool life to number of holes
    • Incorrect calculation of exponent n

    Earns more

    • Correctly identifies that tool life is proportional to number of holes
    • Shows dimensional consistency in calculations
    • States assumptions clearly (same feed, same material)
    • Interprets result physically

    Extra mark

    • Draws schematic showing drill and plate
    • States Taylor's equation explicitly with all variables defined
  2. (b) List and explain the functions of coating on SMAW electrode. 10 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Identify flux function (shielding, slag formation)
    • Explain arc stabilizing function
    • Describe deoxidizing and alloying functions
    • Mention current carrying capacity function

    Loses marks

    • Listing functions without explaining their purpose
    • Confusing coating functions with electrode core functions
    • Omitting key functions like arc stabilization

    Earns more

    • Explains how coating prevents atmospheric contamination
    • Describes slag's role in protecting weld pool
    • Mentions specific fluxing agents (e.g., calcium carbonate)
    • Explains how coating affects arc characteristics

    Extra mark

    • Draws labelled diagram of SMAW electrode showing coating
    • Mentions specific coating composition examples
  3. (c) Compare Expansionist and Wait-and-See strategies for capacity planning. 10 marks

    discuss— intro → 3-4 dimensions → example → balanced close

    Must cover

    • Define Expansionist strategy (lead demand)
    • Define Wait-and-See strategy (lag demand)
    • Compare timing and sizing aspects of both
    • Discuss risks and benefits of each approach

    Loses marks

    • Defining strategies without comparing them
    • Omitting timing or sizing aspects
    • One-sided discussion favoring one strategy

    Earns more

    • Provides examples of when each strategy is appropriate
    • Discusses market conditions favoring each strategy
    • Mentions financial implications of each approach
    • Balances discussion with both advantages and disadvantages

    Extra mark

    • Draws capacity-demand timeline diagram
    • References specific industry examples
  4. (d) Determine the number of Kanbans (containers) required. 10 marks

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

    Must cover

    • State Kanban formula: N = (D × L + S) / C
    • Identify all given values with units
    • Substitute values correctly into formula
    • Calculate and round up to nearest whole number

    Loses marks

    • Using incorrect Kanban formula
    • Failing to round up to whole number
    • Incorrect substitution of values

    Earns more

    • Shows dimensional consistency in calculation
    • Explains meaning of each variable
    • States assumptions (e.g., demand is constant)
    • Interprets result in context of production system

    Extra mark

    • Draws Kanban system schematic
    • Mentions safety stock calculation method
  5. (e) Derive reorder point expression and calculate safety stock and reorder point. 10 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Derive reorder point formula for variable demand, constant lead time
    • Calculate standard deviation of demand during lead time
    • Determine z-value for 95% service level
    • SS = 1.645 × 5√2 ≈ 11.6 units, ROP ≈ 47.6 ≈ 48 units

    Loses marks

    • Failing to derive formula before calculating
    • Using incorrect z-value for 95% service level
    • Incorrect calculation of standard deviation during lead time

    Earns more

    • Shows step-by-step derivation with clear assumptions
    • Uses correct standard normal distribution table value
    • States all given values with units
    • Interprets safety stock in context of inventory management

    Extra mark

    • Draws inventory level diagram showing reorder point
    • Mentions assumptions about demand distribution
What "Solve" is asking you to do

Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.

Structure that answers it

Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity

Where marks are lost

Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.

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