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Civil Engineering 2024 Paper II 50 marks Explain

Paper II — Q3

(a) (i) Explain spot speed, running speed, space-mean speed, time-mean speed and average speed. (10 marks) (ii) The consolidated…

(a)
(i)

Explain spot speed, running speed, space-mean speed, time-mean speed and average speed. 10 marks

(ii)

The consolidated data collected from speed and delay studies by floating car method on a stretch of urban road of length 2·5 km, running East-West are given below. Determine the average values of volume, journey speed and running speed of the traffic stream, along either direction. (10 marks) Mean values of speed and delay: - East-West run: mean journey time 4·85 min; mean stopped delay 1·60 min; vehicles overtaking 4·5; vehicles overtaken 6·5; vehicles met in the opposite direction 376 - West-East run: mean journey time 8·36 min; mean stopped delay 1·90 min; vehicles overtaking 4·0; vehicles overtaken 5·5; vehicles met in the opposite direction 280

(b)
(i)

Two points A and B having elevations of 585 m and 250 m respectively, above datum, appear on a vertical photograph obtained with a camera of focal length of 250 mm and flying altitude of 2300 m above datum. Their correlated photographic coordinates are: A: x = +5·65 cm, y = +3·75 cm; B: x = −3·45 cm, y = +8·55 cm. Determine the ground coordinates of points A and B and calculate the horizontal distance between points A and B. 5 marks

(ii)

The lengths, latitudes and departures of a closed traverse ABCD are as follows: - AB: length 350·8 m, latitude +303·03, departure +176·6 - BC: length 408·5 m, latitude −336·7, departure +231·4 - CD: length 285·4 m, latitude −211·3, departure −191·9 Calculate the length and bearing of the line DA. 10 marks

(c)
(i)

For what specific operations are the following equipment used at construction sites? Sheep foot roller

(ii)

Bulldozer

(iii)

Dragline excavator

(iv)

Tower crane

(v)

Hoe 15 marks

हिंदी में प्रश्न पढ़ें
(ii)

(क) (i) स्थानिक (स्पॉट) चाल, धावन (रनिंग) चाल, स्थान-माध्य चाल, समय-माध्य चाल और औसत चाल की व्याख्या कीजिए। (10 अंक) पूर्व-पश्चिम दिशा में 2·5 किमी लंबी शहरी सड़क के एक भाग पर फ्लोटिंग कार विधि द्वारा गति और विलंब अध्ययनों से एकत्रित संगठित आँकड़े नीचे दिए गए हैं। किसी भी दिशा में यातायात प्रवाह की मात्रा, यात्रा चाल और धावन (रनिंग) चाल के औसत मान ज्ञात कीजिए। (10 अंक) गति और विलंब के औसत मान: - पूर्व-पश्चिम: औसत यात्रा समय 4·85 मिनट; औसत विरामी विलंब 1·60 मिनट; ओवरटेक करने वाले 4·5; ओवरटेक किए गए 6·5; विपरीत दिशा में 376 - पश्चिम-पूर्व: औसत यात्रा समय 8·36 मिनट; औसत विरामी विलंब 1·90 मिनट; ओवरटेक करने वाले 4·0; ओवरटेक किए गए 5·5; विपरीत दिशा में 280

(ख) (i) निर्देश (डेटम) के ऊपर, क्रमशः 585 मीटर और 250 मीटर की ऊँचाई वाले दो बिंदु A और B हैं जो 250 मिमी की फोकस दूरी और निर्देश से 2300 मीटर की उड़ान ऊँचाई वाले कैमरे से प्राप्त एक ऊर्ध्वाधर फोटोग्राफ पर दिखाई देते हैं। उनके सहसंबद्ध फोटोग्राफिक निर्देशांक: A: x = +5·65 सेमी, y = +3·75 सेमी; B: x = −3·45 सेमी, y = +8·55 सेमी। बिंदु A और B के ग्राउंड निर्देशांक ज्ञात कीजिए तथा बिंदु A और B के बीच क्षैतिज दूरी की गणना कीजिए। (5 अंक)

(ii)

एक संवृत चक्रम ABCD की लम्बाई, अक्षांश और प्रस्थान निम्नानुसार हैं: - AB: लम्बाई 350·8 मीटर, अक्षांश +303·03, प्रस्थान +176·6 - BC: लम्बाई 408·5 मीटर, अक्षांश −336·7, प्रस्थान +231·4 - CD: लम्बाई 285·4 मीटर, अक्षांश −211·3, प्रस्थान −191·9 रेखा DA की लम्बाई और दिक्मान का परिकलन कीजिए। (10 अंक)

(ग) निर्माण स्थलों पर निम्नलिखित उपकरणों का प्रयोग किन विशिष्ट कार्यों हेतु किया जाता है?

(i)

मेष पाद वेल्लन (शीप फुट रोलर)

(ii)

बुलडोजर

(iii)

संकर्ष-रेखी उत्खनित्र (ड्रैगलाइन एक्सकेवेटर)

(iv)

टावर क्रेन

(v)

कुदाल (हो) (15 अंक)

Q3 of the 2024 UPSC Mains Civil Engineering Paper II, as printed
The question as printed in the 2024 Civil Engineering paper
Evaluate your answer to this question → See all 2024 Civil Engineering questions

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) Table titled 'Mean Values of Speed and Delay'. Columns: Direction, Mean Journey Time (Minutes), Mean Stopped Delay (Minutes), Number of Vehicles (Mean Value) [sub-columns: Overtaking, Overtaken, In Opposite Direction]. Row 1: East-West, 4.85, 1.60, 4.5, 6.5, 376. Row 2: West-East, 8.36, 1.90, 4.0, 5.5, 280.

(b) Table of Traverse Data. Columns: Line, Length (m), Latitude, Departure. Row 1: AB, 350.8, + 303.03, + 176.6. Row 2: BC, 408.5, - 336.7, + 231.4. Row 3: CD, 285.4, - 211.3, - 191.9.

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)(i) Speeds used in traffic studies:

  • Spot speed: the instantaneous speed of a vehicle at a specified point, measured by radar meter, enoscope or timing over a short base. It is used for speed limits, signal design and accident studies.
  • Running speed: the average speed while the vehicle is actually moving, i.e. distance divided by running time (journey time minus stopped delays).
  • Journey (overall) speed / average speed: distance divided by the total journey time including all stops and delays. For a stream, the average speed is taken as the mean over vehicles or trips.
  • Time-mean speed (v_t): the arithmetic mean of the spot speeds of vehicles passing a point during a period, v_t = Σv_i/n.
  • Space-mean speed (v_s): the average speed over a length of road at an instant, equal to the distance divided by the average travel time, i.e. the harmonic mean of spot speeds, v_s = n/Σ(1/v_i). It is the speed used in q = kv_s.

Relation: v_t = v_s + σ_s²/v_s, so the time-mean speed is always slightly higher than the space-mean speed (faster vehicles pass a point more often).

(a)(ii) Floating-car (Wardrop) method, L = 2.5 km. For the flow in a direction:

q = (x + y)/(t_a + t_w), and mean journey time T = t_w − y/q,

where x = vehicles met while the test car runs against that stream, y = vehicles overtaking minus vehicles overtaken while it runs with that stream, t_a and t_w = journey times against and with the stream.

East-West stream (t_w = 4.85, t_a = 8.36 min; x = 280 met on the W-E run; y = 4.5 − 6.5 = −2):

  • q_EW = (280 − 2)/(8.36 + 4.85) = 278/13.21 = 21.0 veh/min = 1263 veh/h.
  • T = 4.85 − (−2/21.04) = 4.945 min; journey speed = 2.5 × 60/4.945 = 30.3 km/h.
  • Running speed = 2.5 × 60/(4.945 − 1.60) = 44.8 km/h.

West-East stream (t_w = 8.36, t_a = 4.85 min; x = 376 met on the E-W run; y = 4.0 − 5.5 = −1.5):

  • q_WE = (376 − 1.5)/13.21 = 374.5/13.21 = 28.35 veh/min = 1701 veh/h.
  • T = 8.36 − (−1.5/28.35) = 8.413 min; journey speed = 2.5 × 60/8.413 = 17.8 km/h.
  • Running speed = 2.5 × 60/(8.413 − 1.90) = 23.0 km/h.

Since the test car was overtaken more often than it overtook, it ran slightly slower than the stream, so the corrected mean times are a little longer than the test-car times.

(b)(i) For a vertical photograph, each point has its own scale f/(H − h), so X = x(H − h)/f and Y = y(H − h)/f, with f = 25 cm.

  • A (h = 585 m): (H − h)/f = 1715/25 = 68.6; X_A = 5.65 × 68.6 = 387.6 m, Y_A = 3.75 × 68.6 = 257.3 m.
  • B (h = 250 m): (H − h)/f = 2050/25 = 82.0; X_B = −3.45 × 82.0 = −282.9 m, Y_B = 8.55 × 82.0 = 701.1 m.
  • Horizontal distance AB = √[(387.6 + 282.9)² + (701.1 − 257.3)²] = √(670.5² + 443.8²) = 804.1 m.

(b)(ii) In a closed traverse ΣL = 0 and ΣD = 0.

  • ΣL (AB, BC, CD) = 303.03 − 336.7 − 211.3 = −244.97, so latitude of DA = +244.97 m (northing).
  • ΣD (AB, BC, CD) = 176.6 + 231.4 − 191.9 = +216.10, so departure of DA = −216.10 m (westing).
  • Length DA = √(244.97² + 216.10²) = 326.7 m.
  • Reduced bearing = tan⁻¹(216.10/244.97) = 41°25′; latitude +, departure − puts DA in the NW quadrant: bearing N 41°25′ W (whole-circle bearing 318°35′).
Diagram: Ground positions of photo points A and B from a vertical photograph, and closing line DA of traverse ABCD

(c) Equipment and their specific operations:

  • Sheepsfoot roller: compaction of cohesive (clayey) soils in layers; the projecting feet penetrate and knead the soil, compacting from the bottom of the layer upwards. Used for earth dams, embankments and canal banks.
  • Bulldozer: a tractor with a front blade, used for clearing jungle and debris, stripping topsoil, pushing earth over short distances (up to about 100 m), spreading fill, backfilling trenches and rough grading; also as a pusher for scrapers.
  • Dragline excavator: a bucket suspended from a long boom and dragged towards the machine; it digs below its own level and in soft, wet or submerged ground, and casts the spoil at a distance. Used for canals, drainage channels, ponds, borrow pits and river-bed excavation.
  • Tower crane: a fixed or rail-mounted crane with a tall mast and jib, used for vertical and horizontal movement of materials (reinforcement, formwork, concrete buckets, precast units) in multi-storey buildings, bridges and congested sites.
  • Hoe (backhoe): a hydraulic excavator whose bucket digs by pulling towards the machine; used for trenches, basements, footings and pipelines below track level, and for loading trucks.

How this answer will be evaluated

What earns full marks: All parts fully answered with correct methods, units, and clear presentation.

Evaluation rubric

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

  1. (a)(i) Definitions of spot, running, space-mean, time-mean, and average speed. 10 marks

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

    Must cover

    • Spot speed: instantaneous speed of a vehicle at a specified point (radar, enoscope, short base)
    • Running speed: distance divided by running time, i.e. journey time minus stopped delays
    • Space-mean speed: distance divided by the average travel time over a length (harmonic mean of spot speeds); time-mean speed: arithmetic mean of spot speeds at a point
    • Relation v_t = v_s + σ²/v_s (so v_t ≥ v_s); average (journey/overall) speed = distance divided by total journey time including stops

    Loses marks

    • Confusing running speed with space-mean speed
    • Omitting the definition of average speed

    Earns more

    • Mention relationship between time-mean and space-mean
    • Provide formula for each speed type

    Extra mark

    • Example calculation of speed types
  2. (a)(ii) Average volume, journey speed, and running speed for both directions. 10 marks

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

    Must cover

    • Wardrop: q = (x + y)/(t_a + t_w), where x = vehicles met while travelling against the stream and y = overtaking minus overtaken while travelling with it
    • q_EW = (280 − 2)/13.21 = 21.0 veh/min ≈ 1263 veh/h; q_WE = (376 − 1.5)/13.21 = 28.4 veh/min ≈ 1701 veh/h
    • Mean journey time T = t_w − y/q: 4.945 min (E-W) and 8.413 min (W-E); journey speeds 30.3 and 17.8 km/h
    • Running speeds = 2.5 × 60/(T − stopped delay): 44.8 km/h (E-W) and 23.0 km/h (W-E)

    Loses marks

    • Not subtracting the stopped delay for running speed, or skipping the y/q correction to the test-car time
    • Incorrect formula for volume calculation

    Earns more

    • Correct unit conversion for time and distance
    • Clear tabulation of results

    Extra mark

    • Comparison of speeds between directions
  3. (b)(i) Ground coordinates of A and B and horizontal distance between them. 5 marks

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

    Must cover

    • Point scale: X = x(H − h)/f and Y = y(H − h)/f with f = 25 cm
    • A: (H − h)/f = 1715/25 = 68.6, so X_A = 387.6 m and Y_A = 257.3 m
    • B: (H − h)/f = 2050/25 = 82.0, so X_B = −282.9 m and Y_B = 701.1 m
    • AB = √(670.5² + 443.8²) ≈ 804.1 m

    Loses marks

    • Incorrect scale factor calculation
    • Using a single average scale for both points

    Earns more

    • Uses a separate scale for each point because their elevations differ
    • Clear step-by-step coordinate calculation

    Extra mark

    • Sketch showing coordinate system
  4. (b)(ii) Length and bearing of line DA in closed traverse ABCD. 10 marks

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

    Must cover

    • ΣLatitudes = 0 and ΣDepartures = 0 for a closed traverse
    • Latitude of DA = +244.97 m and departure of DA = −216.10 m
    • Length DA = √(244.97² + 216.10²) = 326.7 m
    • Bearing DA = N 41°25′ W (whole-circle bearing 318°35′)

    Loses marks

    • Incorrect sign in latitude or departure sum
    • Wrong quadrant for bearing calculation

    Earns more

    • Correct sign convention for latitudes and departures
    • Verification of closure

    Extra mark

    • Tabular presentation of traverse calculations
  5. (c) Specific operations for five construction equipment items. 15 marks

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

    Must cover

    • Sheepsfoot roller: compaction of cohesive (clayey) soils in layers, e.g. earth dams and embankments, by kneading
    • Bulldozer: land clearing, stripping, short-haul (up to about 100 m) pushing, spreading and rough grading
    • Dragline: excavating below track level and in wet ground or under water (canals, ponds, borrow pits), casting spoil at a distance
    • Tower crane: lifting and placing materials in multi-storey construction; hoe (backhoe): trenches, basements and foundations below track level with a pull-in stroke

    Loses marks

    • Confusing dragline with shovel excavator
    • Vague descriptions without specific operations

    Earns more

    • Explains that the hoe is a hydraulic backhoe digging below track level, not a manual tool
    • Specify typical applications for each equipment

    Extra mark

    • Mention capacity or size ranges for equipment
What "Explain" is asking you to do

Make the working of something clear — what sets it off, what follows from what, and what it produces. Explain is the Commission's mechanism word: it dominates the technical papers and the “explain why” stems, where the marks sit in the causal chain and not in the label.

Structure that answers it

State what it is → the initiating condition → the chain of cause, step by step → an instance where it plays out → what the chain produces

Where marks are lost

Describing what something looks like instead of why it works that way. Naming the stages without linking them reads as description too.

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