(a) A low carbon steel stock of thickness 25 mm is to be rolled in two stages. In the first stage, the reduction is to be from 25 mm to 15 mm and in the second stage from 15 mm to 5 mm. Determine the minimum diameter of the rolls for the two stages i

Question

(a) A low carbon steel stock of thickness 25 mm is to be rolled in two stages. In the first stage, the reduction is to be from 25 mm to 15 mm and in the second stage from 15 mm to 5 mm. Determine the minimum diameter of the rolls for the two stages if the maximum angle of bite is 35° for the first stage and 25° for the second stage. Also, calculate the required coefficient of friction in both the stages. 10 marks

(b) In an arc welding process of steel with a potential of 15 volt, current of 150 Amp and travel speed is 5 mm/sec, the cross-sectional area of joint is observed as 15 mm². If heat required to melt the steel is 10 J/mm³ and heat transfer efficiency is 0·75, then calculate the melting efficiency. 10 marks

(c) In an automobile manufacturing industry the demand for a specific part was 250 in April, 100 in May and 200 in June. The forecast for April was 150. Calculate the forecast for the month of July with a smoothing constant of 0·25 and using first order exponential smoothing. 10 marks

(d) Five different products are manufactured in a mixed model production line. The time required for each task is given below:

Each product requires a set of tasks. Calculate the total number of work stations for this mixed-model assembly line, if the cycle time is 15 seconds. 10 marks

(e) For the given dimensions of mated parts, determine the values of allowance, hole tolerance and shaft tolerance using the basic hole system.

Hole : 57·50 mm Shaft : 57·47 mm
57·52 mm 57·45 mm

10 marks

UPSC Answer Check · Accepted Answer

Calculate each sub-part systematically, allocating approximately 20% time to each part given equal 10-mark weighting. Begin with rolling mechanics (a), proceed through welding efficiency (b), exponential smoothing (c), assembly line balancing (d), and tolerance analysis (e). Present derivations first, then substitute values, and conclude with boxed final answers for each sub-part.
- Part (a): Apply bite angle formula α = cos⁻¹(1 - Δh/D) to find D₁ = 68.5 mm and D₂ = 131.4 mm; coefficient of friction μ = tan(α) giving μ₁ = 0.70 and μ₂ = 0.47
- Part (b): Calculate heat supplied H_s = η_t × V × I / v = 337.5 J/mm; heat required H_r = 150 J/mm; melting efficiency η_m = H_r/H_s = 44.4%
- Part (c): Apply exponential smoothing F_t = F_{t-1} + α(A_{t-1} - F_{t-1}) sequentially: May=175, June=156.25, July=167.19 (or 167.2)
- Part (d): Determine total task time per product, compute weighted average or sum of task times, then apply N = Σt_i / C to find minimum workstations (typically 4-5 stations depending on task precedence)
- Part (e): Identify hole limits (57.50, 57.52) and shaft limits (57.45, 57.47); calculate allowance = 0.03 mm, hole tolerance = 0.02 mm, shaft tolerance = 0.02 mm; classify as transition fit
- All parts: Show SI units consistently, state assumptions clearly, and verify answers against physical feasibility (e.g., roll diameter > reduction, efficiency < 100%)

Dimension	Weight	Max marks	Excellent	Average	Poor
Concept correctness	20%	10	Correctly applies rolling geometry (α = cos⁻¹(1-Δh/D)), distinguishes heat transfer vs melting efficiency in welding, uses first-order exponential smoothing formula properly, understands mixed-model line balancing with cycle time constraint, and applies basic hole system correctly for tolerance analysis.	Uses correct formulas but confuses heat transfer efficiency with melting efficiency, or applies simple average instead of exponential smoothing; basic hole system understood but allowance calculation may have sign error.	Confuses bite angle with contact angle, treats welding efficiencies as additive, uses simple moving average for forecasting, or applies basic shaft system instead of basic hole system.
Numerical accuracy	20%	10	All five sub-parts yield correct numerical answers: D₁≈68.5mm, D₂≈131.4mm, μ₁≈0.70, μ₂≈0.47; η_m≈44.4%; July forecast≈167.2; workstations=4 or 5 (with justification); allowance=0.03mm, tolerances=0.02mm each; 3-4 significant figures maintained.	Correct final answers for 3-4 parts with minor arithmetic slips in one part (e.g., rounding errors in exponential smoothing or workstation calculation); correct methodology but calculator error in roll diameter.	Major computational errors in multiple parts (e.g., wrong formula application leading to D < Δh, efficiency >100%, or negative allowance); inconsistent decimal places; missing units.
Diagram quality	15%	7.5	Clear sketch for part (a) showing roll geometry, bite angle α, and reduction Δh with labels; part (e) shows tolerance zone diagram with hole and shaft distributions; part (d) may include precedence diagram for assembly tasks.	Rough sketch for rolling geometry without labels, or tolerance zone diagram missing nominal size reference; diagrams present but lack clarity on critical dimensions.	No diagrams despite geometric nature of parts (a) and (e); or completely incorrect diagrams (e.g., showing extrusion instead of rolling).
Step-by-step derivation	25%	12.5	Explicit derivation for each part: (a) solves D = Δh/(1-cos α) before substituting; (b) shows H_s = η_tVI/v and H_r = A×H_m separately; (c) writes recursive formula and computes month-by-month; (d) shows task time summation and division by C; (e) identifies limits clearly before computing tolerances.	Shows key formulas with substitution but skips some intermediate steps; jumps from given data to final answer in 1-2 parts; working legible but condensed.	Final answers only with no working; or incorrect formula stated but somehow correct answer obtained (suspected copying); illegible or disorganized working across parts.
Practical interpretation	20%	10	Interprets results practically: (a) notes that larger second-stage roll accommodates smaller bite angle for finishing; (b) comments on why melting efficiency <100% (heat losses to HAZ, spatter); (c) explains α=0.25 gives stable but lagging forecast; (d) discusses line balancing efficiency; (e) confirms transition fit suits precision automotive assembly.	Brief comment on 1-2 parts (e.g., 'efficiency is reasonable' or 'fit is tight') without elaboration; mostly computational treatment.	No interpretation whatsoever; or absurd comments (e.g., 'negative allowance means extra clearance') showing fundamental misunderstanding of manufacturing implications.

Q5

Directive word: Calculate

How this answer will be evaluated

Approach

Key points expected

Evaluation rubric

Practice this exact question

More from Mechanical Engineering 2025 Paper I