Mathematics 2022 Paper II 50 marks Compulsory Solve

Paper II — Q5

(a) It is given that the equation of any cone with vertex at (a, b, c) is f((x-a)/(z-c), (y-b)/(z-c)) = 0. Find the differential…

(a)

It is given that the equation of any cone with vertex at (a, b, c) is f((x-a)/(z-c), (y-b)/(z-c)) = 0. Find the differential equation of the cone. 10 marks

(b)

Solve, by Gauss elimination method, the system of equations 2x + 2y + 4z = 18 x + 3y + 2z = 13 3x + y + 3z = 14 10 marks

(c)
(i)

Convert the number (1093·21875)₁₀ into octal and the number (1693·0628)₁₀ into hexadecimal systems.

(ii)

Express the Boolean function F(x, y, z) = xy + x'z in a product of maxterms form. 10 marks

(d)

A particle at a distance r from the centre of force moves under the influence of the central force F = -k/r², where k is a constant. Obtain the Lagrangian and derive the equations of motion. 10 marks

(e)

The velocity components of an incompressible fluid in spherical polar coordinates (r, θ, ψ) are (2Mr⁻³cosθ, Mr⁻²sinθ, 0), where M is a constant. Show that the velocity is of the potential kind. Find the velocity potential and the equations of the streamlines. 10 marks

हिंदी में प्रश्न पढ़ें
(a)

दिया गया है कि शीर्ष (a, b, c) वाले किसी शंकु का समीकरण f((x-a)/(z-c), (y-b)/(z-c)) = 0 है। शंकु का अवकल समीकरण ज्ञात कीजिए। (10 अंक)

(b)

गॉस विलोपन विधि द्वारा समीकरण निकाय 2x + 2y + 4z = 18 x + 3y + 2z = 13 3x + y + 3z = 14 को हल कीजिए। (10 अंक)

(c)
(i)

संख्या (1093·21875)₁₀ को अष्टाधारी तथा संख्या (1693·0628)₁₀ को षोडश-आधारी पद्धति में बदलिए।

(ii)

बूलिय फलन F(x, y, z) = xy + x'z को योगद (मैक्सटर्म) के गुणन के रूप में अभिव्यक्त कीजिए। (10 अंक)

(d)

एक कण, जो बल-केंद्र से r दूरी पर है, केंद्रीय बल F = -k/r², जहाँ k एक स्थिरांक है, के प्रभाव में गतिमान है। लैग्रांजियन निकालिये तथा गति के समीकरणों को व्युत्पन्न कीजिये। (10 अंक)

(e)

किसी असंपीड्य तरल के गोलीय ध्रुवी निर्देशांकों (r, θ, ψ) में वेग-घटक (2Mr⁻³cosθ, Mr⁻²sinθ, 0) है, जहाँ M एक स्थिरांक है। दर्शाइये कि वेग, विभव प्रकार का है। वेग विभव तथा धारारेखाओं के समीकरण ज्ञात कीजिये। (10 अंक)

Q5 of the 2022 UPSC Mains Mathematics Paper II, as printed
The question as printed in the 2022 Mathematics paper

Model answer

Written 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.

(a) Let u = (x−a)/(z−c), v = (y−b)/(z−c). The given cone is f(u,v)=0. Differentiating totally,

f_u du + f_v dv = 0.

Now du = [(z−c)dx − (x−a)dz]/(z−c)², dv = [(z−c)dy − (y−b)dz]/(z−c)².

Substituting,

f_u[(z−c)dx − (x−a)dz] + f_v[(z−c)dy − (y−b)dz] = 0.

Hence

(z−c)(f_u dx + f_v dy) = [(x−a)f_u + (y−b)f_v]dz.

Let z = z(x,y), so dz = p dx + q dy, where p=∂z/∂x, q=∂z/∂y. Comparing coefficients,

p = (z−c)f_u/D, q = (z−c)f_v/D, where D = (x−a)f_u + (y−b)f_v.

Therefore (x−a)p + (y−b)q = (z−c).

Final answer: (x−a)∂z/∂x + (y−b)∂z/∂y = z−c, equivalently (x−a)p + (y−b)q = z−c.

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(b) The augmented matrix is

[2 2 4 | 18] [1 3 2 | 13] [3 1 3 | 14]

Interchange R1 and R2:

[1 3 2 | 13] [2 2 4 | 18] [3 1 3 | 14]

Perform R2 ← R2 − 2R1:

[1 3 2 | 13] [0 −4 0 | −8] [3 1 3 | 14]

Perform R3 ← R3 − 3R1:

[1 3 2 | 13] [0 −4 0 | −8] [0 −8 −3 | −25]

From R2: −4y = −8, so y = 2.

From R3: −8y − 3z = −25. Putting y = 2, −16 − 3z = −25 ⇒ −3z = −9 ⇒ z = 3.

From R1: x + 3y + 2z = 13. x + 6 + 6 = 13 ⇒ x = 1.

Final answer: (x, y, z) = (1, 2, 3).

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(c)(i) Convert 1093.21875₁₀ to octal.

Integer part: 1093 ÷ 8 = 136 remainder 5 136 ÷ 8 = 17 remainder 0 17 ÷ 8 = 2 remainder 1 2 ÷ 8 = 0 remainder 2

Reading remainders upwards: 2105₈.

Fractional part: 0.21875 × 8 = 1.75 ⇒ digit 1, remainder 0.75 0.75 × 8 = 6.00 ⇒ digit 6, remainder 0

So fractional part is .16₈.

Final answer: (1093.21875)₁₀ = (2105.16)₈

Now convert 1693.0628₁₀ to hexadecimal.

Integer part: 1693 ÷ 16 = 105 remainder 13 = D 105 ÷ 16 = 6 remainder 9 6 ÷ 16 = 0 remainder 6

Reading upwards: 69D₁₆.

Fractional part: 0.0628 × 16 = 1.0048 ⇒ digit 1 0.0048 × 16 = 0.0768 ⇒ digit 0 0.0768 × 16 = 1.2288 ⇒ digit 1 0.2288 × 16 = 3.6608 ⇒ digit 3 0.6608 × 16 = 10.5728 ⇒ digit A 0.5728 × 16 = 9.1648 ⇒ digit 9 0.1648 × 16 = 2.6368 ⇒ digit 2 0.6368 × 16 = 10.1888 ⇒ digit A and so on.

Thus the hexadecimal fraction is non-terminating: 0.0628₁₀ = .1013A92A...₁₆.

Final answer: (1693.0628)₁₀ = (69D.1013A92A...)₁₆, approximately (69D.1013A9)₁₆ to six hexadecimal places.

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(c)(ii) F(x,y,z) = xy + x′z.

We need the product of maxterms, i.e. zeros of F.

Truth table:

x y z | xy | x′z | F 0 0 0 | 0 | 0 | 0 0 0 1 | 0 | 1 | 1 0 1 0 | 0 | 0 | 0 0 1 1 | 0 | 1 | 1 1 0 0 | 0 | 0 | 0 1 0 1 | 0 | 0 | 0 1 1 0 | 1 | 0 | 1 1 1 1 | 1 | 0 | 1

F is zero at minterms 0, 2, 4, 5.

Therefore the maxterms are:

M₀ = x + y + z M₂ = x + y′ + z M₄ = x′ + y + z M₅ = x′ + y + z′

Final answer: F(x,y,z) = Π M(0,2,4,5) = (x + y + z)(x + y′ + z)(x′ + y + z)(x′ + y + z′).

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(d) Use plane polar coordinates (r, θ). Let the mass be m.

Kinetic energy: T = 1/2 m(ṙ² + r²θ̇²).

For central force F = −k/r², the potential V(r) satisfies F = −dV/dr. So −dV/dr = −k/r² ⇒ dV/dr = k/r² ⇒ V = −k/r up to an additive constant.

Thus the Lagrangian is L = T − V = 1/2 m(ṙ² + r²θ̇²) + k/r.

For r:

∂L/∂ṙ = mṙ, d/dt(∂L/∂ṙ) = m r̈.

∂L/∂r = m r θ̇² − k/r².

Euler–Lagrange equation: d/dt(∂L/∂ṙ) − ∂L/∂r = 0.

So m r̈ − m r θ̇² + k/r² = 0, or m(r̈ − r θ̇²) = −k/r².

For θ:

∂L/∂θ̇ = m r² θ̇, d/dt(∂L/∂θ̇) = m(2rṙθ̇ + r²θ̈).

Since ∂L/∂θ = 0, d/dt(m r² θ̇) = 0.

Hence r² θ̇ = h = constant.

Final answer: Lagrangian: L = 1/2 m(ṙ² + r²θ̇²) + k/r.

Equations of motion: m(r̈ − rθ̇²) = −k/r² and r²θ̇ = h = constant.

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(e) As printed, the θ-component is v_θ = M r⁻² sinθ. Check whether this is compatible with an incompressible potential flow.

Divergence in spherical polars:

∇·v = (1/r²)∂(r² v_r)/∂r + (1/(r sinθ))∂(sinθ v_θ)/∂θ + (1/(r sinθ))∂v_ψ/∂ψ.

With v_r = 2M r⁻³ cosθ, v_θ = M r⁻² sinθ, v_ψ = 0,

∇·v = (1/r²)∂(2M r⁻¹ cosθ)/∂r + (1/(r sinθ))∂(M r⁻² sin²θ)/∂θ = (1/r²)(−2M r⁻² cosθ) + (1/(r sinθ))(2M r⁻² sinθ cosθ) = −2M r⁻⁴ cosθ + 2M r⁻³ cosθ ≠ 0.

So the field as printed is not even incompressible.

Also check irrotationality. The ψ-component of curl v is

(∇×v)_ψ = (1/r)[∂(r v_θ)/∂r − ∂v_r/∂θ].

For v_θ = M r⁻² sinθ,

(∇×v)_ψ = (1/r)[∂(M r⁻¹ sinθ)/∂r − ∂(2M r⁻³ cosθ)/∂θ] = (1/r)[−M r⁻² sinθ + 2M r⁻³ sinθ] = M(2−r)r⁻⁴ sinθ ≠ 0.

Thus, as printed, the velocity is not of potential kind.

The standard intended form of this problem uses v_θ = M r⁻³ sinθ. With that intended exponent, let

v = (2M r⁻³ cosθ, M r⁻³ sinθ, 0).

Then (∇×v)_ψ = (1/r)[∂(M r⁻² sinθ)/∂r − ∂(2M r⁻³ cosθ)/∂θ] = (1/r)[−2M r⁻³ sinθ + 2M r⁻³ sinθ] = 0.

Also the other curl components vanish and the divergence is zero, so v is irrotational and incompressible, hence of potential kind. Write v = ∇φ.

Then ∂φ/∂r = 2M r⁻³ cosθ. Integrating, φ = −M r⁻² cosθ + f(θ,ψ).

Also (1/r)∂φ/∂θ = M r⁻³ sinθ ⇒ ∂φ/∂θ = M r⁻² sinθ.

But ∂/∂θ(−M r⁻² cosθ) = M r⁻² sinθ, so f is independent of θ. Since v_ψ = 0, f is constant. Taking it as zero,

φ = −M r⁻² cosθ.

For streamlines, dr/v_r = r dθ/v_θ = r sinθ dψ/v_ψ.

Since v_ψ = 0 and the motion is axisymmetric, ψ = constant. Then

dr/(2M r⁻³ cosθ) = r dθ/(M r⁻³ sinθ).

Thus dr/(2 cosθ) = r dθ/sinθ ⇒ dr/r = 2 cotθ dθ.

Integrating, ln r = 2 ln sinθ + ln C, so r = C sin²θ.

Final answer: As printed with v_θ = M r⁻² sinθ, the velocity is not of potential kind. For the standard intended component v_θ = M r⁻³ sinθ: Velocity potential: φ = −M r⁻² cosθ. Streamlines: r = C sin²θ, ψ = constant.

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.

All UPSC directive words, compared →

How this answer will be evaluated

Approach

(a) derive: given > assumptions > stepwise derivation > result > check | (b) calculate: given > formula > substitution > result with units > interpretation | (c(i)) calculate: given > formula > substitution > result with units > interpretation | (c(ii)) derive: given > assumptions > stepwise derivation > result > check | (d) derive: given > assumptions > stepwise derivation > result > check | (e) derive: given > assumptions > stepwise derivation > result > check Full marks: All parts fully derived with correct notation, verification, and clear step-by-step logic.

Key points expected

  • Define p = ∂z/∂x and q = ∂z/∂y
  • Differentiate f((x-a)/(z-c), (y-b)/(z-c)) = 0
  • Eliminate arbitrary function f
  • State final PDE: (x-a)p + (y-b)q = z-c
  • Form augmented matrix
  • Perform row operations to upper triangular form
  • Solve via back-substitution
  • State final values for x, y, z

Evaluation rubric

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

  1. (a) Differential equation of a cone with vertex (a, b, c). 10 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Define p = ∂z/∂x and q = ∂z/∂y
    • Differentiate f((x-a)/(z-c), (y-b)/(z-c)) = 0
    • Eliminate arbitrary function f
    • State final PDE: (x-a)p + (y-b)q = z-c

    Loses marks

    • Failing to define p and q
    • Skipping the elimination of f
    • Incorrect partial differentiation of arguments

    Earns more

    • Explicitly state chain rule application
    • Verify result by substituting a simple cone equation

    Extra mark

    • Mention general solution form
  2. (b) Solution of the 3x3 linear system via Gauss elimination. 10 marks

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

    Must cover

    • Form augmented matrix
    • Perform row operations to upper triangular form
    • Solve via back-substitution
    • State final values for x, y, z

    Loses marks

    • Arithmetic errors in row operations
    • Skipping back-substitution steps
    • Not stating final solution clearly

    Earns more

    • Show intermediate matrices clearly
    • Verify solution by substitution into original equations

    Extra mark

    • Mention determinant check for uniqueness
  3. (c(i)) Convert (1093.21875)₁₀ to octal and (1693.0628)₁₀ to hexadecimal.

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

    Must cover

    • Convert integer part of 1093 to octal
    • Convert fractional part 0.21875 to octal
    • Convert integer part of 1693 to hex
    • Convert fractional part 0.0628 to hex

    Loses marks

    • Incorrect base conversion for integer part
    • Error in fractional part conversion
    • Not separating integer and fractional parts

    Earns more

    • Show division/multiplication steps for each part
    • Label integer and fractional conversions separately

    Extra mark

    • Verify by converting back to decimal
  4. (c(ii)) Express F(x,y,z) = xy + x'z in product of maxterms.

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Find minterms of F(x,y,z)
    • Identify missing minterms (maxterms)
    • Write product of maxterms form
    • Use standard notation ΠM(...)

    Loses marks

    • Confusing minterms with maxterms
    • Incorrect identification of missing minterms
    • Not using standard ΠM notation

    Earns more

    • Show truth table or K-map for minterm identification
    • Clearly list which minterms are absent

    Extra mark

    • Verify by expanding product of maxterms
  5. (d) Lagrangian and equations of motion for central force F = -k/r². 10 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Write kinetic energy T in polar coordinates
    • Write potential energy V = -k/r
    • Form Lagrangian L = T - V
    • Apply Euler-Lagrange equations for r and θ

    Loses marks

    • Incorrect potential energy sign or form
    • Failing to apply Euler-Lagrange equations
    • Missing either radial or angular equation

    Earns more

    • Show explicit form of T = ½m(ṙ² + r²θ̇²)
    • Derive both radial and angular equations of motion

    Extra mark

    • Mention conservation of angular momentum
  6. (e) Show velocity is potential; find potential and streamlines. 10 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Verify curl of velocity field is zero
    • Integrate to find velocity potential φ
    • Derive streamline equations from dV/φ = 0
    • State final potential and streamline equations

    Loses marks

    • Failing to verify irrotationality
    • Incorrect integration for potential
    • Not deriving streamline equations

    Earns more

    • Show explicit curl calculation in spherical coordinates
    • Integrate each component to find φ

    Extra mark

    • Sketch or describe streamline geometry

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