Civil Engineering 2024 Paper II 50 marks Solve

Paper II — Q4

(a) The network of a certain project is shown in the figure given below with the estimated time duration (in days) of various…

(a)
(i)

The network of a certain project is shown in the figure given below with the estimated time duration (in days) of various activities. Determine the following : Earliest event time

(ii)

Latest event time

(iii)

Earliest start and finish time of each activity

(iv)

Latest start and finish time of each activity

(v)

Total float of each activity

(vi)

Critical path of the network (Assume the scheduled completion time of the project equal to the earliest event time of the last event) [Network: 1→2: 6; 1→3: 8; 2→5: 7; 3→4: 7; 3→6: 10; 4→5: 6; 4→7: 9; 5→7: 8; 6→7: 9; 7→8: 6 days.] 20 marks

(b)

Prepare a detailed analysis of rate (₹ per sq. m) for a 25 mm thick cement concrete (1 : 2 : 4) floor with cement finishing coat. 15 marks

(c)

Sketch a typical diamond crossing and label all its components. Design a diamond crossing between two B.G. tracks crossing each other at an angle of 1 in 8·5. 15 marks

हिंदी में प्रश्न पढ़ें
(i)

(क) नीचे दिए गए चित्र में एक निश्चित परियोजना का जाल (नेटवर्क) विभिन्न क्रियाओं की आकलित समयावधि (दिनों में) के साथ दर्शाया गया है। निम्नलिखित को ज्ञात कीजिए : यथाशीघ्र घटना काल

(ii)

यथाविलम्बित घटना काल

(iii)

प्रत्येक क्रिया का यथाशीघ्र प्रवर्तन काल एवं यथाशीघ्र समाप्ति काल

(iv)

प्रत्येक क्रिया का यथाविलम्बित प्रवर्तन काल एवं यथाविलम्बित समाप्ति काल

(v)

प्रत्येक क्रिया का कुल प्लव (फ्लोट)

(vi)

जाल का क्रांतिक पथ (परियोजना का निर्धारित समाप्ति काल अंतिम घटना के यथाशीघ्र घटना काल के बराबर मान लीजिए) [नेटवर्क: 1→2: 6; 1→3: 8; 2→5: 7; 3→4: 7; 3→6: 10; 4→5: 6; 4→7: 9; 5→7: 8; 6→7: 9; 7→8: 6 दिन।] (20 अंक)

(ख) सीमेंट के आखिरी लेप के साथ बनाए गए एक 25 मिमी मोटे सीमेंट कंक्रीट (1 : 2 : 4) फर्श के लिए विस्तृत विश्लेषण दर (₹ प्रति वर्ग मीटर) तैयार कीजिए। (15 अंक)

(ग) एक प्रारूपिक समचतुर्भुज (डायमंड) क्रॉसिंग का चित्र बनाइए और उसके सभी घटकों को नामांकित कीजिए। यदि दो बड़ी लाइन (बी.जी.) रेलपथ एक-दूसरे को 8·5 में 1 के कोण पर क्रॉस करते हैं, तो उनके बीच एक समचतुर्भुज (डायमंड) क्रॉसिंग का अभिकल्पन कीजिए। (15 अंक)

Q4 of the 2024 UPSC Mains Civil Engineering Paper II, as printed
The question as printed in the 2024 Civil Engineering paper

The figure this question refers to, in words

The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.

(a) A project network diagram with 8 numbered event nodes (circles 1 through 8) and directed activity arrows labeled with time durations t in days:

  • Arrow from Node 1 to Node 2: t = 6
  • Arrow from Node 1 to Node 3: t = 8
  • Arrow from Node 2 to Node 5: t = 7
  • Arrow from Node 3 to Node 4: t = 7
  • Arrow from Node 3 to Node 6: t = 10
  • Arrow from Node 4 to Node 5: t = 6
  • Arrow from Node 4 to Node 7: t = 9
  • Arrow from Node 5 to Node 7: t = 8
  • Arrow from Node 6 to Node 7: t = 9
  • Arrow from Node 7 to Node 8: t = 6

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) Activities (days): 1–2 = 6, 1–3 = 8, 2–5 = 7, 3–4 = 7, 3–6 = 10, 4–5 = 6, 4–7 = 9, 5–7 = 8, 6–7 = 9, 7–8 = 6.

(i) Earliest event times (forward pass, maximum at merge events):

E₁ = 0; E₂ = 6; E₃ = 8; E₄ = 8 + 7 = 15; E₅ = max(6 + 7, 15 + 6) = 21; E₆ = 8 + 10 = 18; E₇ = max(21 + 8, 15 + 9, 18 + 9) = 29; E₈ = 29 + 6 = 35 days.

(ii) Latest event times (backward pass from L₈ = 35, minimum at burst events):

L₈ = 35; L₇ = 29; L₆ = 29 − 9 = 20; L₅ = 29 − 8 = 21; L₄ = min(21 − 6, 29 − 9) = 15; L₃ = min(15 − 7, 20 − 10) = 8; L₂ = 21 − 7 = 14; L₁ = min(14 − 6, 8 − 8) = 0.

(iii)–(v) Activity times (EST = E_i, EFT = EST + t, LFT = L_j, LST = LFT − t, total float TF = LST − EST):

  • 1–2: EST 0, EFT 6, LST 8, LFT 14, TF 8
  • 1–3: EST 0, EFT 8, LST 0, LFT 8, TF 0
  • 2–5: EST 6, EFT 13, LST 14, LFT 21, TF 8
  • 3–4: EST 8, EFT 15, LST 8, LFT 15, TF 0
  • 3–6: EST 8, EFT 18, LST 10, LFT 20, TF 2
  • 4–5: EST 15, EFT 21, LST 15, LFT 21, TF 0
  • 4–7: EST 15, EFT 24, LST 20, LFT 29, TF 5
  • 5–7: EST 21, EFT 29, LST 21, LFT 29, TF 0
  • 6–7: EST 18, EFT 27, LST 20, LFT 29, TF 2
  • 7–8: EST 29, EFT 35, LST 29, LFT 35, TF 0

(vi) The zero-float activities form the critical path 1–3–4–5–7–8, duration 8 + 7 + 6 + 8 + 6 = 35 days.

(b) Rate analysis for 10 m² (quantities per 10 m², then reduced to 1 m²). Item: 25 mm thick cement concrete 1:2:4 (cement : coarse sand : 10–12.5 mm graded stone aggregate) floor, finished with a floating coat of neat cement. Market rates are assumed.

Materials:

  • Wet volume = 10 × 0.025 = 0.25 m³; dry volume = 1.52 × 0.25 = 0.38 m³ (sum of ratio = 7).
  • Cement = 0.38/7 = 0.0543 m³ = 78 kg ≈ 1.56 bags; neat cement floating coat ≈ 2.2 kg/m² = 22 kg = 0.44 bag; total ≈ 2.0 bags × ₹400 = ₹800.
  • Coarse sand = 2 × 0.38/7 = 0.109 m³ × ₹1,800 = ₹196.
  • Stone aggregate 10–12.5 mm = 4 × 0.38/7 = 0.217 m³ × ₹1,600 = ₹347.
  • Materials total = ₹1,343.

Labour (for 10 m²):

  • Mason 1.0 day × ₹900 = ₹900.
  • Mazdoor (beldar) 1.5 days × ₹650 = ₹975.
  • Bhisti 0.5 day × ₹700 = ₹350.
  • Labour total = ₹2,225.

Rate:

  • Materials + labour = ₹3,568.
  • Water charges @ 1% = ₹36; tools, plant and sundries @ 1% = ₹36; sub-total = ₹3,640.
  • Contractor's profit and overheads @ 15% = ₹546.
  • Cost of 10 m² = ₹4,186, so rate ≈ ₹419 per m² (excluding GST).

Notes: the aggregate size is kept to about one-third of the 25 mm thickness; the floor is laid in panels, tamped, floated with neat cement while green and cured for at least 7 days. The rate changes in proportion to the local material and labour rates.

(c) A diamond crossing is provided where two tracks cross at an angle. Its four rail intersections form a rhombus: two acute crossings at the ends of the long diagonal (each with a nose, wing rails and check rails) and two obtuse crossings at the ends of the short diagonal (each with point rails, knuckle rails and check rails). Sleepers are common to both tracks and laid square to the long diagonal.

Design for two B.G. tracks (G = 1.676 m) crossing at 1 in 8.5:

  • Crossing angle α = cot⁻¹ 8.5 = 6°42′35″; obtuse crossing angle = 180° − α = 173°17′25″; α/2 = 3°21′17″.
  • Side of the diamond (length along each track between its crossings) = G cosec α = 1.676 × 8.559 = 14.34 m.
  • Long diagonal (between the acute crossings) = G cosec(α/2) = 1.676/0.0585 = 28.64 m.
  • Short diagonal (between the obtuse crossings) = G sec(α/2) = 1.676 × 1.0017 = 1.68 m.
  • 1 in 8.5 is the flattest diamond crossing permitted on B.G.: as the angle flattens, the unguarded gap opposite the obtuse crossings becomes longer and a wheel may take the wrong flangeway, so check rails are raised and clearances kept tight at the obtuse crossings.
Diagram: CPM network with event times and critical path, and a diamond crossing layout with its components

How this answer will be evaluated

Framework: CPM (Critical Path Method)

What earns full marks: Complete CPM calculations with correct critical path; detailed rate analysis with all components; accurate diamond crossing sketch with correct angle calculation.

Evaluation rubric

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

  1. (a) Determine CPM parameters (EET, LET, ES, EF, LS, LF, Float) and identify the critical path. 20 marks

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

    Must cover

    • Earliest event times E = 0, 6, 8, 15, 21, 18, 29, 35 for events 1–8 (forward pass, maximum at merges)
    • Latest event times L = 0, 14, 8, 15, 21, 20, 29, 35 (backward pass from 35, minimum at bursts)
    • EST/EFT and LST/LFT of every activity; total float = L_j − E_i − t: 1-2 and 2-5 = 8, 4-7 = 5, 3-6 and 6-7 = 2, others 0
    • Critical path 1-3-4-5-7-8 with project duration 35 days

    Loses marks

    • Arithmetic errors in time calculations
    • Failure to identify the correct critical path
    • Missing units (days) in final results

    Earns more

    • Correct calculation of Earliest Start/Finish times
    • Correct calculation of Latest Start/Finish times
    • Clear tabular presentation of activity parameters
    • Explicit statement of project duration

    Extra mark

    • Neatly drawn network diagram with EET/LET labelled on nodes
    • Highlighting of critical activities on the diagram
  2. (b) Prepare a detailed rate analysis for 25mm cement concrete floor with finishing coat. 15 marks

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

    Must cover

    • Quantities for 10 m² (or 1 m²): wet volume 0.25 m³, dry volume ≈ 1.52 × wet = 0.38 m³
    • 1:2:4 split: cement 0.0543 m³ (≈ 78 kg, 1.56 bags), sand 0.109 m³, 10–12.5 mm aggregate 0.217 m³, plus a neat cement floating (finishing) coat of about 2.2 kg/m²
    • Labour (mason, mazdoor, bhisti) at stated day-rates, with water charges and tools and plant
    • Contractor's profit and overheads (about 15%) and the final rate per m² (≈ ₹420/m² at the assumed rates)

    Loses marks

    • Omission of cement finishing coat cost
    • Incorrect mix ratio calculation (1:2:4)
    • Failure to state units (per sq. m)

    Earns more

    • Inclusion of contractor's profit and overheads
    • Correct application of wastage factors
    • Clear breakdown of material vs labor costs
    • Use of standard IS code mix proportions

    Extra mark

    • Reference to specific IS code for concrete mix
    • Inclusion of tool and plant charges if applicable
  3. (c) Sketch a diamond crossing and design one for B.G. tracks at 1 in 8.5 angle. 15 marks

    describe— define → structure or process in order → labelled diagram → significance

    Must cover

    • Labelled sketch: two acute crossings (nose, wing rails, check rails), two obtuse crossings (point, knuckle and check rails), common sleepers
    • Crossing angle α = cot⁻¹ 8.5 = 6°42′35″; obtuse crossing angle 173°17′25″
    • With G = 1.676 m: side of diamond G cosec α = 14.34 m, long diagonal G cosec(α/2) = 28.64 m, short diagonal G sec(α/2) = 1.68 m
    • Note that 1 in 8.5 is the flattest diamond permitted on B.G., because the unguarded gap at the obtuse crossings grows as the angle flattens

    Loses marks

    • Missing labels on the sketch
    • Incorrect calculation of crossing angle
    • Failure to specify track gauge (B.G.)

    Earns more

    • Correct identification of frog, check rails, and flangeways
    • Accurate crossing angle (6°42′35″ ≈ 6.71°)
    • Proper dimensioning of the diamond crossing
    • Clear distinction between main and crossing tracks

    Extra mark

    • Reference to IS code for diamond crossing design
    • Uses the B.G. gauge 1676 mm
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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