Civil Engineering 2023 Paper II 50 marks Solve

Paper II — Q2

(a) For a project consisting of several activities, the allotted time and the dependencies of the activities are presented below…

(a)
(i)

For a project consisting of several activities, the allotted time and the dependencies of the activities are presented below. The activities are: P (5 days, no predecessor); Q (4 days, no predecessor); R (6 days, predecessor Q); S (5 days, predecessor P); T (7 days, predecessor P); U (4 days, predecessors T and R). Prepare a network and mark critical path in it.

(ii)

Calculate Float, Earliest start, Earliest finish, Latest start and Latest finish times. 20 marks

(b)

Trains of different speeds are to be run on a 2° curve on a broad gauge. The average speed of trains to be run on the track is 80 kmph. Calculate the value of equilibrium cant. Also calculate the maximum permissible speed on the track allowing the maximum cant deficiency. 15 marks

(c)
(i)

An aircraft flew at the altitude of 5000 m above the mean sea level. Two consecutive photographs were taken with the camera of focal length 300 mm on the flat ground having elevation of 2000 m above mean sea level. The longitudinal overlap is 65% and photograph print size is 300 mm × 300 mm. Calculate the scale of the photograph and distance between the two consecutive exposure stations. 5 marks

(ii)

What is spectral reflectance curve ? Explain its significance in Remote Sensing. 10 marks

हिंदी में प्रश्न पढ़ें
(i)

(क) विभिन्न क्रियाओं से बनी एक परियोजना के लिए क्रियाओं की आवंटित समय अवधि और पराश्रयता को नीचे दर्शाया गया है । क्रियाएँ: P (5 दिन, कोई पूर्ववर्ती नहीं); Q (4 दिन, कोई पूर्ववर्ती नहीं); R (6 दिन, पूर्ववर्ती Q); S (5 दिन, पूर्ववर्ती P); T (7 दिन, पूर्ववर्ती P); U (4 दिन, पूर्ववर्ती T और R)। एक जाल का निर्माण कीजिए और उसमें क्रांतिक पथ को दर्शाइए ।

(ii)

प्लव (फ्लोट), यथाशीघ्र प्रारंभ, यथाशीघ्र समाप्ति, यथाविलम्बित प्रारंभ और यथाविलम्बित समाप्ति समयों की गणना कीजिए । (20 अंक)

(ख) एक बड़ी लाइन पर 2° वक्र पर विभिन्न गति की रेलगाड़ियों को चलाया जाना है । रेलपथ पर चलाई जाने वाली रेलगाड़ियों की औसत गति 80 kmph है । संतुलन आनति (कैन्ट) के मान की गणना कीजिए । अधिकतम आनति न्यूनता की अनुमति के साथ रेलपथ पर अधिकतम अनुज्ञेय गति की गणना भी कीजिए । (15 अंक)

(ग) (i) एक वायुयान माध्य समुद्र तल से 5000 m की ऊँचाई पर उड़ा । एक समतल मैदान, जो माध्य समुद्र तल से 2000 m की ऊँचाई पर है, से एक कैमरा जिसकी फोकस दूरी 300 mm है, से दो क्रमागत तस्वीरें ली गईं । अनुदैर्ध्य अतिव्यापन 65% है एवं तस्वीर का प्रिन्ट आमाप 300 mm × 300 mm है । तस्वीर के स्केल की गणना कीजिए । दोनों क्रमागत उद्भासन स्टेशनों के बीच की दूरी बताइए । (5 अंक)

(ii)

वर्णक्रमीय परावर्तकता वक्र क्या है ? सुदूर संवेदन में इसकी सार्थकता की व्याख्या कीजिए । (10 अंक)

Q2 of the 2023 UPSC Mains Civil Engineering Paper II, as printed
The question as printed in the 2023 Civil Engineering paper

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) CPM network, critical path and float

(i) Network (activity-on-node): P and Q have no predecessor. P is followed by S and T; Q is followed by R; U follows T and R. So the paths are P–S, P–T–U and Q–R–U.

Forward pass (ES, EF): P 0–5, Q 0–4, R 4–10, S 5–10, T 5–12, U max(12, 10) = 12 to 16. Project duration = 16 days.

Path lengths: P–T–U = 5 + 7 + 4 = 16; Q–R–U = 4 + 6 + 4 = 14; P–S = 10. Critical path: P → T → U, 16 days.

(ii) Backward pass (LF, LS), with S and U ending at 16: U 12–16; T 5–12; R 6–12; S 11–16; P: LF = min(11, 5) = 5, LS = 0; Q: LF = 6, LS = 2.

Total float = LS − ES:

P: ES 0, EF 5, LS 0, LF 5, float 0. Q: ES 0, EF 4, LS 2, LF 6, float 2. R: ES 4, EF 10, LS 6, LF 12, float 2. S: ES 5, EF 10, LS 11, LF 16, float 6. T: ES 5, EF 12, LS 5, LF 12, float 0. U: ES 12, EF 16, LS 12, LF 16, float 0.

The zero-float activities P, T and U confirm the critical path.

Diagram: Activity-on-node CPM network with critical path P-T-U and ES, EF, LS, LF, float

(b) Equilibrium cant and maximum permissible speed

For a 2° curve, R = 1750/D = 1750/2 = 875 m. For broad gauge the dynamic gauge is G = 1750 mm, so

Equilibrium cant e = GV²/(127R) = 1750 × 80²/(127 × 875) = 100.8 mm (about 10 cm).

This is the cant provided on the track. The maximum cant deficiency for broad gauge is 75 mm. At the maximum speed the equilibrium cant needed is e + C_d = 100.8 + 75 = 175.8 mm.

V_max² = 175.8 × 127 × 875/1750 = 11 163, so V_max ≈ 105.7 km/h, about 105 km/h. (The IR form V = 0.27√[(C_a + C_d)R] = 105.9 km/h agrees.)

(c)(i) Photograph scale and exposure interval

Flying height above ground H − h = 5000 − 2000 = 3000 m. Scale = f/(H − h) = 0.300/3000 = 1 : 10 000.

Ground coverage along the flight line = 0.300 m × 10 000 = 3000 m. With 65% overlap, the air base = (1 − 0.65) × 3000 = 1050 m between consecutive exposure stations.

(c)(ii) Spectral reflectance curve

A spectral reflectance curve plots the percentage of incident radiation that a surface reflects against wavelength. It is the spectral signature of the material.

Its significance in remote sensing: different features reflect differently, so the curve allows them to be told apart. Healthy vegetation has low reflectance in the blue and red bands, a small peak in the green, and a sharp rise in the near infrared. Clear water reflects little and almost nothing beyond the near infrared. Dry soil shows a steady rise with wavelength. The curves guide the choice of sensor bands, support classification and indices such as NDVI, and allow monitoring of crop stress, water quality and land-cover change.

Diagram: Typical spectral reflectance curves of green vegetation, dry bare soil and clear water

How this answer will be evaluated

What earns full marks: All parts fully answered with correct formulas, units, and clear diagrams; no calculation errors; code references where applicable.

Evaluation rubric

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

  1. (a) Draw the network, mark the critical path and compute float, ES, EF, LS and LF. 20 marks

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

    Must cover

    • Dependencies: P→S, P→T, Q→R, T and R→U
    • Forward pass gives 16 days
    • Backward pass and float: P0, Q2, R2, S6, T0, U0
    • Critical path P → T → U

    Loses marks

    • Computing float without the backward pass.

    Earns more

    • Marks for the network, forward and backward passes, float table and critical path.

    Extra mark

    • Showing path lengths to confirm the critical path.
  2. (b) Find the equilibrium cant and the maximum permissible speed. 15 marks

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

    Must cover

    • R = 1750/D = 875 m
    • e = GV²/127R = 100.8 mm with G = 1750 mm
    • Maximum cant deficiency 75 mm for BG
    • V_max ≈ 105.7 km/h

    Loses marks

    • Using the 1676 mm track gauge as G or omitting the cant deficiency.

    Earns more

    • Marks for radius, cant, deficiency and speed.

    Extra mark

    • Checking with V = 0.27√[(Ca + Cd)R].
  3. (c)(i) Calculate the photograph scale and the distance between consecutive exposure stations. 5 marks

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

    Must cover

    • H − h = 3000 m
    • Scale = f/(H − h) = 1:10 000
    • Ground coverage 3000 m
    • Air base = 0.35 × 3000 = 1050 m

    Loses marks

    • Using 5000 m as height above ground.

    Earns more

    • Marks for height above ground, scale, coverage and air base.

    Extra mark

    • Stating the answer with units.
  4. (c)(ii) Explain the spectral reflectance curve and its significance in remote sensing. 10 marks

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

    Must cover

    • Reflectance vs wavelength curve is the spectral signature
    • Vegetation, water and soil differ in shape
    • Helps identify features and classify land cover
    • Guides sensor bands and indices such as NDVI

    Loses marks

    • Only naming the curve without any example.

    Earns more

    • Marks for definition and examples of the curves.

    Extra mark

    • Mentioning crop stress or water quality monitoring.
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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