Mathematics 2021 Paper I 50 marks Prove

Q7

(a) Verify Gauss divergence theorem for $\vec{F} = 2x^2y\hat{i} - y^2\hat{j} + 4xz^2\hat{k}$ taken over the region in the first octant bounded by $y^2 + z^2 = 9$ and $x = 2$. (20 marks) (b) Find all possible solutions of the differential equation: $y^2 \log y = xy\dfrac{dy}{dx} + \left(\dfrac{dy}{dx}\right)^2$. (15 marks) (c) A heavy particle hangs by an inextensible string of length $a$ from a fixed point and is then projected horizontally with a velocity $\sqrt{2gh}$. If $\dfrac{5a}{2} > h > a$, then prove that the circular motion ceases when the particle has reached the height $\dfrac{1}{3}(a + 2h)$ from the point of projection. Also, prove that the greatest height ever reached by the particle above the point of projection is $\dfrac{(4a-h)(a+2h)^2}{27a^2}$. (15 marks)

हिंदी में प्रश्न पढ़ें

(a) प्रथम अष्टांश में $y^2 + z^2 = 9$ तथा $x = 2$ द्वारा परिबद्ध क्षेत्र पर $\vec{F} = 2x^2y\hat{i} - y^2\hat{j} + 4xz^2\hat{k}$ के लिए गॉस अपसरण प्रमेय को सत्यापित कीजिए। (20 अंक) (b) अवकल समीकरण: $y^2 \log y = xy\dfrac{dy}{dx} + \left(\dfrac{dy}{dx}\right)^2$ के सभी संभव हल ज्ञात कीजिए। (15 अंक) (c) एक भारी कण $a$ लम्बाई की अवितान्य डोरी से एक स्थिर बिंदु से टंगा है तथा $\sqrt{2gh}$ वेग से क्षैतिज दिशा में प्रक्षेपित किया जाता है। यदि $\dfrac{5a}{2} > h > a$ है, तो सिद्ध कीजिए कि प्रक्षेपण बिंदु से $\dfrac{1}{3}(a + 2h)$ ऊँचाई पहुँचने पर कण की वृत्तीय गति समाप्त हो जाती है। यह भी सिद्ध कीजिए कि उस कण द्वारा प्रक्षेपण बिंदु से ऊपर प्राप्य अधिकतम ऊँचाई $\dfrac{(4a-h)(a+2h)^2}{27a^2}$ है। (15 अंक)

Directive word: Prove

This question asks you to prove. The directive word signals the depth of analysis expected, the structure of your answer, and the weight of evidence you must bring.

See our UPSC directive words guide for a full breakdown of how to respond to each command word.

How this answer will be evaluated

Approach

Begin with a clear statement of intent to verify, solve, and prove across all three parts. For part (a), set up the volume integral of divergence and surface integrals over the cylindrical region in the first octant, allocating approximately 35-40% of effort due to its 20 marks. For part (b), identify the Clairaut form and solve by substitution p = dy/dx, spending about 25-30% of time. For part (c), apply energy conservation and circular motion dynamics to derive the height conditions, using ~30-35% of effort. Conclude each part with boxed final answers and explicit verification statements.

Key points expected

  • For (a): Correct computation of div F = 4xy - 2y + 8xz and proper setup of volume integral over 0 ≤ x ≤ 2, 0 ≤ y ≤ 3, 0 ≤ z ≤ √(9-y²) in first octant
  • For (a): Evaluation of surface integrals over all five faces (x=0, x=2, y=0, z=0, and cylindrical surface y²+z²=9) with correct normal vectors
  • For (b): Recognition of equation as Clairaut's form y = xp + f(p) where p = dy/dx, leading to substitution and factorization
  • For (b): Complete solution including general solution y = cx + c²/log y and singular solution obtained by eliminating c
  • For (c): Application of energy conservation ½mv² = ½m(2gh) - mga(1-cos θ) and tension condition T = 0 for circular motion cessation
  • For (c): Derivation that circular motion ceases at height (a+2h)/3 using the condition 5a/2 > h > a
  • For (c): Calculation of projectile motion after string slackens to reach greatest height (4a-h)(a+2h)²/(27a²)

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Setup correctness20%10For (a): Correct region description with proper limits for cylindrical first-octant volume; for (b): Accurate identification of Clairaut form with correct substitution p=dy/dx; for (c): Proper energy equation setup with correct initial conditions and geometric constraintsRegion description partially correct with minor limit errors; form recognized but substitution unclear; energy equation present but initial conditions slightly wrongMajor errors in region boundaries; fails to identify equation type; incorrect energy setup or missing constraint 5a/2 > h > a
Method choice20%10For (a): Appropriate choice of cylindrical coordinates for surface integrals; for (b): Systematic application of Clairaut's method with complete case analysis; for (c): Correct use of Lagrangian dynamics or Newtonian mechanics with proper tension analysisCorrect general approach but suboptimal coordinate choices; method applied with gaps in case analysis; mechanics approach valid but tension condition poorly handledWrong theorem application for (a); attempts wrong method for (b); uses incorrect dynamical principles for (c)
Computation accuracy20%10For (a): Exact equality of volume and surface integrals demonstrated numerically; for (b): Correct algebraic manipulation leading to explicit solution families; for (c): Precise trigonometric calculations yielding exact height formulas without arithmetic errorsMinor arithmetic slips in integration constants; algebraic errors in solving for p; small calculation errors in final height expressionsMajor computational errors preventing verification in (a); unsolvable quadratic in p; completely wrong numerical results in (c)
Step justification20%10For (a): Clear justification of outward normal directions and orientation; for (b): Explicit reasoning for rejecting extraneous solutions and validating singular solution; for (c): Rigorous proof that T=0 condition applies and physical interpretation of projectile phaseNormals stated without justification; solution verification mentioned but not proved; physical reasoning present but gaps in connecting T=0 to heightNo justification of surface orientations; missing verification steps; no physical interpretation of motion phases
Final answer & units20%10For (a): Explicit statement 'Hence Gauss divergence theorem is verified' with both sides equal; for (b): Complete solution set clearly enumerated; for (c): Both required heights proved with exact algebraic forms and boxed final expressionsVerification statement present but numerical values not explicitly compared; solutions listed but incomplete; heights derived but final forms not simplifiedNo verification conclusion; missing solution branches; incorrect final expressions or no attempt at second height proof

Practice this exact question

Write your answer, then get a detailed evaluation from our AI trained on UPSC's answer-writing standards. Free first evaluation — no signup needed to start.

Evaluate my answer →

More from Mathematics 2021 Paper I