Paper I — Q6
A heavy string, which is not of uniform density, is hung up from two points. Let T₁, T₂, T₃ be the tensions at the intermediate…
A heavy string, which is not of uniform density, is hung up from two points. Let T₁, T₂, T₃ be the tensions at the intermediate points A, B, C of the catenary respectively where its inclinations to the horizontal are in arithmetic progression with common difference β. Let ω₁ and ω₂ be the weights of the parts AB and BC of the string respectively. Prove that (i) Harmonic mean of T₁, T₂ and T₃ = 3T₂/(1 + 2cos β) (ii) T₁/T₃ = ω₁/ω₂ 20 marks
Solve the equation: d²y/dx² + (tan x - 3cos x)dy/dx + 2y cos²x = cos⁴x completely by demonstrating all the steps involved. 15 marks
Evaluate ∫_C F⃗ · dr⃗, where C is an arbitrary closed curve in the xy-plane and F⃗ = (-yî + xĵ)/(x² + y²). 15 marks
हिंदी में प्रश्न पढ़ें
एक भारी डोरी, जिसका घनत्व एक समान नहीं है, दो बिन्दुओं से टंगी हुई है। माना कि T₁, T₂, T₃ क्रमशः कैटिनरी के बीच के बिन्दुओं A, B, C पर तनाव हैं, जिन पर इसके क्षैतिज के साथ आनति कोण, सार्व अंतर β के साथ समांतर श्रेढ़ी में हैं। माना कि डोरी के AB तथा BC भागों के भार क्रमशः ω₁ तथा ω₂ हैं। सिद्ध कीजिए (i) T₁, T₂ तथा T₃ का हरात्मक माध्य = 3T₂/(1 + 2cos β) (ii) T₁/T₃ = ω₁/ω₂ 20 अंक
सभी अंतरस्थ (शामिल) चरणों को दर्शाते हुए समीकरण: d²y/dx² + (tan x - 3cos x)dy/dx + 2y cos²x = cos⁴x को पूर्ण रूप से हल कीजिए। 15 अंक
∫_C F⃗ · dr⃗ का मान निकालिए, जहाँ C, xy-समतल में एक सैच्छिक संयुक्त वक्र है तथा F⃗ = (-yî + xĵ)/(x² + y²) है। 15 अंक
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) Heavy string of non-uniform density
Let the inclinations at A, B, C be θ − β, θ, θ + β. The horizontal tension is constant, H, so T₁cos(θ − β) = T₂cos θ = T₃cos(θ + β) = H, that is, T₁ = H sec(θ − β), T₂ = H sec θ, T₃ = H sec(θ + β).
(i) The harmonic mean of T₁, T₂, T₃ is 3/(1/T₁ + 1/T₂ + 1/T₃) = 3H/[cos(θ − β) + cos θ + cos(θ + β)].
Since cos(θ − β) + cos(θ + β) = 2cos θ cos β, the denominator is cos θ(1 + 2cos β). So the harmonic mean = 3H/[cos θ(1 + 2cos β)] = 3T₂/(1 + 2cos β). ∎
(ii) The weight of an arc equals the change in the vertical component of tension, and T sin ψ = H tan ψ. So ω₁ = H[tan θ − tan(θ − β)] = H sin β/[cos θ cos(θ − β)], and ω₂ = H[tan(θ + β) − tan θ] = H sin β/[cos θ cos(θ + β)].
ω₁/ω₂ = cos(θ + β)/cos(θ − β) = [H sec(θ − β)]/[H sec(θ + β)] = T₁/T₃. ∎
(b) y″ + (tan x − 3cos x)y′ + 2y cos²x = cos⁴x
Put t = sin x, so dt/dx = cos x, y′ = cos x·Y′ and y″ = cos²x·Y″ − sin x·Y′.
Substituting, the terms in sin x·Y′ cancel (tan x · cos x = sin x), leaving cos²x[Y″ − 3Y′ + 2Y] = cos⁴x. Dividing by cos²x:
Y″ − 3Y′ + 2Y = cos²x = 1 − t².
Complementary function: roots 1, 2, so Y_c = C₁eᵗ + C₂e²ᵗ.
Particular integral: Y_p = At² + Bt + C. Matching 2At² + (2B − 6A)t + (2A − 3B + 2C) with 1 − t² gives A = −1/2, B = −3/2, C = −5/4.
Hence y = C₁e^(sin x) + C₂e^(2 sin x) − (1/2)sin²x − (3/2)sin x − 5/4, on intervals where cos x ≠ 0.
(c) ∮ F·dr for F = (−y î + x ĵ)/(x² + y²)
With x = r cos θ, y = r sin θ: F·dr = (−y dx + x dy)/(x² + y²) = r²dθ/r² = dθ.
So ∮_C F·dr is the total change of the polar angle θ along C. For a closed curve not passing through the origin, it equals 2π n, where n is the winding number of C about the origin.
If C does not enclose the origin, the integral is 0 (F is irrotational there). If C encloses the origin once anticlockwise, it is 2π; clockwise, −2π. If C passes through the origin, F is undefined and the integral does not exist in the ordinary sense.
How this answer will be evaluated
What earns full marks: Complete, rigorous derivations with all steps shown and verified.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Prove the harmonic-mean relation for the tensions and T₁/T₃ = ω₁/ω₂. 20 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Constant horizontal tension gives T = H sec ψ
- Prove the harmonic mean equals 3T₂/(1 + 2cos β) using cos(θ−β) + cos(θ+β) = 2cos θ cos β
- Weights as differences of vertical tension components giving ω₁/ω₂ = cos(θ+β)/cos(θ−β) = T₁/T₃
Loses marks
- Taking vertical tension as constant
- Missing proof of the second part
Earns more
- Inclinations θ−β, θ, θ+β set up correctly
- Identity tan u − tan v = sin(u−v)/(cos u cos v)
- (b) Solve the second-order variable-coefficient equation completely. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Substitution t = sin x with correct transformation of y′ and y″
- Reduce to Y″ − 3Y′ + 2Y = 1 − t²
- Complementary function and particular integral giving the general solution
Loses marks
- Wrong transformation of y″
- Omitting a constant of integration
Earns more
- Cancellation of the sin x·Y′ terms shown
- Coefficients A = −1/2, B = −3/2, C = −5/4
Extra mark
- Domain cos x ≠ 0 noted
- (c) Evaluate the line integral over an arbitrary closed curve. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Show F·dr = dθ
- State the value 2π times the winding number about the origin
- Treat the cases: origin outside (0), inside once anticlockwise (2π), through the origin (undefined)
Loses marks
- Applying Green's theorem across the singularity
- Answering 0 or 2π only
Earns more
- Polar substitution
- Reason about irrotational field away from the origin
Extra mark
- Mention the clockwise case −2π
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.
Practice this exact question
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