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Civil Engineering 2024 Paper I 50 marks Compulsory Calculate

Paper I — Q5

(a) A 125 mm diameter vertical cylinder rotates concentrically inside a fixed cylinder of diameter 130 mm. Both cylinders are 325…

(a)

A 125 mm diameter vertical cylinder rotates concentrically inside a fixed cylinder of diameter 130 mm. Both cylinders are 325 mm long. Find the dynamic viscosity of the liquid that fills the space between the cylinders, if a torque of 0·92 Nm is required to maintain a speed of 70 r.p.m. 10 marks

(b)

Calculate the friction drag on a flat plate 15 cm wide and 45 cm long placed longitudinally in a stream of oil of relative density 0·925 and kinematic viscosity 0·9 stoke, flowing with a free stream velocity of 6 m/s. Also find the thickness of the boundary layer and shear stress at the trailing edge. Take density of water 1000 kg/m³. 10 marks

(c)

A large tank as shown in the above figure has a vertical pipe 70 cm long and 2 cm in diameter. The tank contains an oil of density 920 kg/m³ and viscosity 1·5 poise. Find the discharge through the pipe when the height of the oil level of the tank is 0·80 m above the pipe inlet. [Figure: oil surface A is 80 cm above the bottom of the tank, where the 2 cm pipe begins; the pipe runs 70 cm vertically down and discharges freely at B.] 10 marks

(d)

A field density test was conducted by core-cutter method and the following data was obtained: Weight of empty core-cutter = 23 N Weight of soil and core-cutter = 50 N Dimensions of the core-cutter dia = 90 mm and height = 180 mm Weight of wet sample for moisture determination = 55×10⁻² N Weight of oven dry sample = 52×10⁻² N Specific gravity of soil grains = 2·70 Determine its dry density, void ratio and degree of saturation. 10 marks

(e)

Two plate load tests were conducted at a site – one with a 300 mm square plate and other with a 600 mm square test plate. For a settlement of 25 mm the loads were found to be 21·6 kN and 64·8 kN respectively in the two tests. Determine the allowable bearing pressure of the sand and the load which a square footing 1·5 m×1·5 m can carry with the settlement not exceeding 25 mm. 10 marks

हिंदी में प्रश्न पढ़ें

(क) एक 125 mm व्यास का ऊर्ध्वाधर बेलन, एक 130 mm व्यास के आबद्ध बेलन के अन्दर संकेन्द्र: घूमता है । दोनों बेलनों की लम्बाई 325 mm है । यदि 70 r.p.m की गति बनाए रखने के लिए 0·92 Nm बल आघूर्ण की आवश्यकता होती है, तो बेलनों के बीच की जगह में भरे हुए द्रव की गतिक श्यानता ज्ञात कीजिए । (10 अंक)

(ख) आपेक्षिक घनत्व 0·925 और शुद्ध गतिक श्यानता 0·9 स्टोक वाले एक तेल की धारा, जो एक 6 m/s के स्वतंत्र धारा वेग से प्रवाहित है, में 15 cm चौड़ी और 45 cm लम्बी, अनुदैर्ध्यवत रखी गई, एक चपटी पट्टिका पर घर्षण विकर्ष की गणना कीजिए । सीमान्त परत की मोटाई और अनुगामी किनार पर अपरूपण प्रतिबल भी ज्ञात कीजिए । जल का घनत्व 1000 kg/m³ लीजिए । (10 अंक)

(ग) नीचे चित्र में दर्शाई गई एक विशाल टंकी में 70 cm लम्बा और 2 cm व्यास का एक ऊर्ध्वाधर पाइप लगा है । टंकी में, 920 kg/m³ घनत्व एवं 1·5 पॉइज़ श्यानता वाला तेल है । जब टंकी में तेल सतह की ऊँचाई, पाइप के अन्तर्गम से 0·80 m ऊपर हो तो पाइप में से निस्सरण को ज्ञात कीजिए । [चित्र: तेल सतह A टंकी के तल से 80 cm ऊपर है, जहाँ 2 cm पाइप आरम्भ होता है; पाइप 70 cm ऊर्ध्वाधर नीचे जाकर B पर स्वतंत्र रूप से निस्सरित होता है।] (10 अंक)

(घ) क्रोड कर्तन (कोर कटर) विधि द्वारा एक क्षेत्र घनत्व परीक्षण किया गया जिससे निम्नलिखित आंकड़े प्राप्त हुए : खाली कोर कटर का वजन = 23 N मृदा एवं कोर कटर का वजन = 50 N कोर कटर की विमाएं : व्यास = 90 mm एवं ऊँचाई = 180 mm जलांश निर्धारण के लिए नम-नमूने का वजन = 55×10⁻² N भट्टी में सुखाए गए नमूने का वजन = 52×10⁻² N मृदा कणों का विशिष्ट घनत्व = 2·70 इसके लिए शुष्कघनत्व, रिक्ति अनुपात, एवं संतृप्ति की मात्रा ज्ञात कीजिए । (10 अंक)

(ङ) एक स्थल पर दो पट्टिका-भार-परीक्षण किए गए एक 300 mm की वर्गाकार पट्टिका से एवं दूसरा 600 mm की वर्गाकार परीक्षण पट्टिका से । इन दो परीक्षणों में 25 mm निषदन के लिए बलों का मान क्रमशः 21·6 kN एवं 64·8 kN प्राप्त हुआ । रेत के लिए अनुज्ञेय-धारक-दाब; और निषदन के 25 mm से अधिक नहीं होने की स्थिति में एक 1·5 m×1·5 m की वर्गाकार पाद द्वारा वहन किए जाने वाले भार को निर्धारित कीजिए । (10 अंक)

Q5 of the 2024 UPSC Mains Civil Engineering Paper I, as printed
The question as printed in the 2024 Civil Engineering paper
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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.

(c) A large open tank containing oil ('तेल / Oil') with a vertical discharge pipe attached to its bottom. The liquid surface in the tank is marked with a liquid level symbol and a dashed horizontal reference line to label (A). The depth of the oil from the free surface (A) to the bottom of the tank (inlet of the vertical pipe) is 80 cm. A vertical pipe extends downwards from the bottom of the tank with a length of 70 cm to its discharge exit, which is aligned with a dashed horizontal reference line to label (B). The pipe is labeled with 'diameter = 2 cm' ('व्यास = 2 cm'). An arrow pointing downwards from the exit of the pipe is labeled 'Q'.

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) Data: R₁ = 62.5 mm (rotating), R₂ = 65 mm (fixed), L = 0.325 m, N = 70 rpm, T = 0.92 N·m.

  • ω = 2πN/60 = 2π × 70/60 = 7.33 rad/s; surface speed of the inner cylinder u = ωR₁ = 7.33 × 0.0625 = 0.458 m/s.
  • Clearance t = (130 − 125)/2 = 2.5 mm. The gap is small, so the velocity profile is taken as linear: τ = μu/t = μ × 0.458/0.0025 = 183.3μ N/m².
  • Shear force on the inner cylinder = τ × 2πR₁L = 183.3μ × 2π × 0.0625 × 0.325 = 23.39μ N.
  • Torque T = force × R₁ = 23.39μ × 0.0625 = 1.462μ N·m.
  • Setting T = 0.92 gives μ = 0.92/1.462 = 0.63 Pa·s (6.3 poise).

Refinement: the exact solution for flow in an annulus, T = 4πμLωR₁²R₂²/(R₂² − R₁²), gives μ = 0.59 Pa·s. The 6% difference comes from the thin-gap assumption; the thin-gap value is the one usually expected. End effects are neglected.

(b) Data: ρ = 0.925 × 1000 = 925 kg/m³; ν = 0.9 stoke = 0.9 cm²/s = 9 × 10⁻⁵ m²/s; U = 6 m/s; plate 0.15 m wide and L = 0.45 m long.

  • Re_L = UL/ν = 6 × 0.45/(9 × 10⁻⁵) = 3.0 × 10⁴ < 5 × 10⁵, so the boundary layer is laminar over the whole plate (Blasius solution).
  • Average skin-friction coefficient: C_f = 1.328/√Re_L = 1.328/173.2 = 0.00767.
  • Drag on one side: F = C_f × ½ρU² × bL = 0.00767 × (0.5 × 925 × 36) × (0.15 × 0.45) = 0.00767 × 16 650 × 0.0675 = 8.62 N. If both faces are wetted, the total is 17.2 N.
  • Boundary-layer thickness at the trailing edge: δ = 5L/√Re_L = 5 × 0.45/173.2 = 0.0130 m = 13.0 mm.
  • Wall shear stress at the trailing edge: τ₀ = 0.332ρU²/√Re_L = 0.332 × 925 × 36/173.2 = 63.8 N/m².

(c) Data: D = 0.02 m, L = 0.70 m, ρ = 920 kg/m³, μ = 1.5 poise = 0.15 Pa·s. The oil surface A is 0.80 m above the pipe inlet, and the pipe discharges freely at B, 0.70 m lower.

Energy equation from A (surface, velocity ≈ 0) to B (outlet, velocity V), assuming laminar flow:

H = 0.80 + 0.70 = 1.5 m = V²/2g + h_f, where h_f = 32μVL/(ρgD²) (Hagen–Poiseuille).

  • 32μL/(ρgD²) = 32 × 0.15 × 0.70/(920 × 9.81 × 0.02²) = 0.931 s/m, so h_f = 0.931V.
  • 1/(2g) = 0.051 s²/m.
  • 0.051V² + 0.931V − 1.5 = 0, so V = [−0.931 + √(0.931² + 4 × 0.051 × 1.5)]/(2 × 0.051) = 1.49 m/s.
  • Q = V × πD²/4 = 1.49 × 3.142 × 10⁻⁴ = 4.68 × 10⁻⁴ m³/s (0.468 L/s).
  • Check: Re = ρVD/μ = 920 × 1.49 × 0.02/0.15 = 183 < 2000, so the laminar assumption holds.

If the exit velocity head is neglected, h_f = 1.5 m and Q = 5.06 × 10⁻⁴ m³/s, but the jet leaving at B does carry kinetic energy, so 4.68 × 10⁻⁴ m³/s is the correct value. Entry loss is neglected.

(d)

  • Weight of soil in the cutter W = 50 − 23 = 27 N.
  • Volume V = (π/4) × 0.09² × 0.18 = 1.145 × 10⁻³ m³.
  • Bulk unit weight γ = 27/1.145 × 10⁻³ = 23 580 N/m³ = 23.58 kN/m³.
  • Water content w = (0.55 − 0.52)/0.52 = 0.0577 = 5.77%.
  • Dry unit weight γ_d = γ/(1 + w) = 23.58/1.0577 = 22.29 kN/m³, i.e. dry density ρ_d = 22.29/9.81 = 2.27 Mg/m³ (2272 kg/m³).
  • Void ratio e = Gγ_w/γ_d − 1 = 2.70 × 9.81/22.29 − 1 = 0.188.
  • Degree of saturation S = wG/e = 0.0577 × 2.70/0.188 = 0.828 = 82.8%.

Phase check on the 1145 cm³ sample: solids 964 cm³ (25.53 N), water 150 cm³ (1.47 N), air 31 cm³; e = (150 + 31)/964 = 0.188 and S = 150/181 = 0.83.

(e) Two plates tested to the same settlement (25 mm) are analysed by Housel's method: the load is carried partly by bearing under the area and partly by shear at the perimeter, Q = mA + nP, where m is the bearing pressure (kN/m²) and n the perimeter shear (kN/m).

  • 300 mm plate: A = 0.09 m², P = 1.2 m: 21.6 = 0.09m + 1.2n.
  • 600 mm plate: A = 0.36 m², P = 2.4 m: 64.8 = 0.36m + 2.4n.
  • Doubling the first and subtracting: 21.6 = 0.18m, so m = 120 kN/m² and n = (21.6 − 10.8)/1.2 = 9 kN/m.

For the 1.5 m × 1.5 m footing (A = 2.25 m², P = 6 m), at the same 25 mm settlement:

  • Q = 120 × 2.25 + 9 × 6 = 270 + 54 = 324 kN.
  • Allowable bearing pressure q = 324/2.25 = 144 kN/m².

The plate pressures (240 and 180 kN/m²) fall as the plate grows because the perimeter-to-area ratio falls (q = m + n P/A); the large footing therefore carries less pressure than either plate for the same settlement.

Diagram: Rotating-cylinder viscometer, laminar boundary layer, tank-pipe energy balance, soil phase diagram and Housel plate-load method

How this answer will be evaluated

What earns full marks: All parts solved with correct formulas, units, and clear steps.

Evaluation rubric

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

  1. (a) Dynamic viscosity of the liquid filling the annular space. 10 marks

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

    Must cover

    • ω = 2π × 70/60 = 7.33 rad/s; surface speed u = ωR₁ = 0.458 m/s
    • Radial clearance t = (130 − 125)/2 = 2.5 mm; τ = μu/t
    • Torque T = τ × (2πR₁L) × R₁ = 1.462μ
    • μ = 0.92/1.462 ≈ 0.63 Pa·s (6.3 poise); the exact annulus formula gives 0.59 Pa·s, also acceptable

    Loses marks

    • Using diameter instead of radius in torque equation
    • Ignoring unit conversion for length or time

    Earns more

    • Explicit calculation of surface area
    • Unit consistency check (Pa.s or N.s/m²)

    Extra mark

    • Sketch of concentric cylinders with dimensions
  2. (b) Friction drag, boundary layer thickness, and shear stress at trailing edge. 10 marks

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

    Must cover

    • ν = 0.9 stoke = 9 × 10⁻⁵ m²/s, ρ = 925 kg/m³; Re_L = UL/ν = 3.0 × 10⁴ (laminar throughout)
    • Blasius: C_f = 1.328/√Re_L = 0.00767; drag on one side = C_f × ½ρU² × bL = 8.62 N (17.2 N for both faces)
    • δ at the trailing edge = 5L/√Re_L = 13.0 mm
    • τ₀ at the trailing edge = 0.332ρU²/√Re_L = 63.8 Pa

    Loses marks

    • Using wrong formula for Reynolds number
    • Confusing kinematic and dynamic viscosity

    Earns more

    • Explicit calculation of dynamic viscosity (μ)
    • Verification of flow regime (Laminar/Turbulent)

    Extra mark

    • Sketch of boundary layer profile
  3. (c) Discharge through the vertical pipe. 10 marks

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

    Must cover

    • Energy equation from the oil surface A to the pipe outlet B: H = 0.80 + 0.70 = 1.5 m
    • H = V²/2g + h_f with Hagen–Poiseuille h_f = 32μVL/(ρgD²) and μ = 1.5 poise = 0.15 Pa·s
    • Solving 0.051V² + 0.931V − 1.5 = 0 gives V = 1.49 m/s
    • Q = 4.68 × 10⁻⁴ m³/s (0.468 L/s); Re ≈ 183 confirms laminar flow

    Loses marks

    • Ignoring the 70 cm pipe length in head loss
    • Dropping the exit velocity head without comment (gives 5.06 × 10⁻⁴ m³/s instead of 4.68 × 10⁻⁴)

    Earns more

    • Explicit calculation of pipe area
    • Check for laminar flow assumption

    Extra mark

    • Sketch of energy grade line
  4. (d) Dry density, void ratio, and degree of saturation. 10 marks

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

    Must cover

    • Bulk unit weight γ = 27 N / 1.145 × 10⁻³ m³ = 23.58 kN/m³
    • w = 0.03/0.52 = 5.77%
    • γ_d = γ/(1 + w) = 22.29 kN/m³ (ρ_d ≈ 2.27 g/cm³)
    • e = Gγ_w/γ_d − 1 = 0.188; S = wG/e = 82.8%

    Loses marks

    • Confusing wet and dry weights
    • Incorrect volume calculation from dimensions

    Earns more

    • Explicit calculation of wet and dry weights
    • Use of standard soil mechanics formulas

    Extra mark

    • Sketch of soil phase diagram
  5. (e) Allowable bearing pressure and load for 1.5m footing. 10 marks

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

    Must cover

    • Plate pressures: 21.6/0.09 = 240 kN/m² and 64.8/0.36 = 180 kN/m²
    • Housel's method Q = mA + nP: 21.6 = 0.09m + 1.2n and 64.8 = 0.36m + 2.4n give m = 120 kN/m² and n = 9 kN/m
    • Footing 1.5 × 1.5 m: Q = 120 × 2.25 + 9 × 6 = 324 kN
    • Allowable bearing pressure for 25 mm settlement = 324/2.25 = 144 kN/m²

    Loses marks

    • Ignoring the difference in plate sizes
    • Using wrong formula for settlement

    Earns more

    • Explicit calculation of plate areas
    • Recognises that two plates at equal settlement call for Housel's perimeter–area method, not the Terzaghi–Peck size ratio

    Extra mark

    • Sketch of plate load test setup
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