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Mathematics 2021 Paper II 50 marks Solve

Paper II — Q6

(a) Solve the wave equation a²∂²u/∂x² = ∂²u/∂t², 0<x<L, t>0 subject to the conditions u(0,t)=0, u(L,t)=0 u(x,0)=(1/4)x(L-x)…

(a)

Solve the wave equation

a²∂²u/∂x² = ∂²u/∂t², 0<x<L, t>0

subject to the conditions

u(0,t)=0, u(L,t)=0 u(x,0)=(1/4)x(L-x), ∂u/∂t|ₜ₌₀=0 20 marks

(b)

Obtain the Boolean function F(x, y, z) based on the table given below. Then simplify F(x, y, z) and draw the corresponding GATE network:

xyzF(x, y, z)
1111
1101
1011
1000
0111
0100
0010
0000(15 marks)
(c)

Obtain the Lagrangian equation for the motion of a system of two particles of unequal masses connected by an inextensible string passing over a small smooth pulley. 15 marks

हिंदी में प्रश्न पढ़ें
(a)

तरंग समीकरण

a²∂²u/∂x² = ∂²u/∂t², 0<x<L, t>0

का शर्तों

u(0,t)=0, u(L,t)=0 u(x,0)=(1/4)x(L-x), ∂u/∂t|ₜ₌₀=0

से प्रतिबंधित हल प्राप्त कीजिए। (20 अंक)

(b)

नीचे दी गई सारणी पर आधारित बूलियन फलन F(x,y,z) को निकालिए और तब F(x,y,z) को सरल कीजिए तथा उसके अनुरूप GATE परिपथ खींचिए :

xyzF(x,y,z)
1111
1101
1011
1000
0111
0100
0010
0000(15 अंक)
(c)

एक छोटी चिकनी घिरनी के ऊपर से गुजरने वाली एक अवितान्य डोरी के सिरों से बंधे असमान संहति वाले दो कणों के निकाय की गति के लिए लग्रांजी समीकरण प्राप्त कीजिए। (15 अंक)

Q6 of the 2021 UPSC Mains Mathematics Paper II, as printed
The question as printed in the 2021 Mathematics 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.

(b) Table with four columns: x, y, z, F(x, y, z) Row 1: 1, 1, 1, 1 Row 2: 1, 1, 0, 1 Row 3: 1, 0, 1, 1 Row 4: 1, 0, 0, 0 Row 5: 0, 1, 1, 1 Row 6: 0, 1, 0, 0 Row 7: 0, 0, 1, 0 Row 8: 0, 0, 0, 0

(b) Table with 4 columns and 8 data rows: x, y, z, F(x, y, z) 1, 1, 1, 1 1, 1, 0, 1 1, 0, 1, 1 1, 0, 0, 0 0, 1, 1, 1 0, 1, 0, 0 0, 0, 1, 0 0, 0, 0, 0

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,t)=X(x)T(t). Substituting in a²∂²u/∂x²=∂²u/∂t² gives

a²X''T=XT'', hence X''/X=T''/(a²T)=−λ², say.

Thus X''+λ²X=0 and T''+a²λ²T=0.

The boundary conditions u(0,t)=0, u(L,t)=0 give X(0)=X(L)=0. For X''+λ²X=0,

X(x)=A cos(λx)+B sin(λx).

X(0)=0 gives A=0. X(L)=0 gives B sin(λL)=0. For a non-trivial solution,

sin(λL)=0, so λL=nπ, n=1,2,3,...

Therefore λₙ=nπ/L and Xₙ(x)=sin(nπx/L).

For Tₙ, Tₙ''+a²λₙ²Tₙ=0, so

Tₙ(t)=Cₙ cos(aλₙt)+Dₙ sin(aλₙt).

Since ∂u/∂t at t=0 is 0, Tₙ'(0)=0, hence Dₙ=0. Therefore

u(x,t)=Σ over n=1,2,... Aₙ sin(nπx/L) cos(a nπ t/L).

Using u(x,0)=x(L−x)/4,

Σ Aₙ sin(nπx/L)=x(L−x)/4.

By the Fourier sine coefficient formula,

Aₙ=(2/L)∫ from 0 to L (x(L−x)/4) sin(nπx/L) dx.

Let k=nπ/L. Then

Aₙ=(1/(2L))[L∫ from 0 to L x sin(kx) dx−∫ from 0 to L x² sin(kx) dx].

Now

∫ from 0 to L x sin(kx) dx = L²(−1)^(n+1)/(nπ),

and

∫ from 0 to L x² sin(kx) dx = L³(−1)^(n+1)/(nπ)+2L³((−1)^n−1)/(n³π³).

Hence

L∫x sin(kx) dx−∫x² sin(kx) dx =2L³(1−(−1)^n)/(n³π³).

Thus

Aₙ=L²(1−(−1)^n)/(n³π³).

So Aₙ=0 for even n, and Aₙ=2L²/(n³π³) for odd n. Therefore

u(x,t)=Σ over n=1,3,5,... [2L²/(n³π³)] sin(nπx/L) cos(a nπ t/L).

Equivalently,

u(x,t)=Σ over m=0 to ∞ [2L²/((2m+1)³π³)] sin((2m+1)πx/L) cos(a(2m+1)πt/L).

(b) From the table, F=1 for the minterms

111, 110, 101, 011.

Thus

F=xyz+xyz'+xy'z+x'yz.

Using a K-map, adjacent pairs give:

xyz+xyz'=xy,

xyz+xy'z=xz,

xyz+x'yz=yz.

Therefore the simplified Boolean function is

F(x,y,z)=xy+xz+yz.

This is the majority function: it is 1 when at least two of x,y,z are 1.

Gate network: use three 2-input AND gates and OR them.

  • x and y enter AND₁, giving xy.
  • x and z enter AND₂, giving xz.
  • y and z enter AND₃, giving yz.
  • The outputs xy, xz, yz enter an OR gate, giving F=xy+xz+yz.

If only 2-input OR gates are available, combine as F=(xy+xz)+yz.

(c) Let the masses be m₁ and m₂, with m₁≠m₂. Let x be the downward displacement of m₁ from a fixed reference. Since the string is inextensible and the pulley is small and smooth, m₂ moves upward by the same distance x. Hence both masses have speed magnitude dx/dt.

The kinetic energy is

T=(1/2)m₁(dx/dt)²+(1/2)m₂(dx/dt)²=(1/2)(m₁+m₂)(dx/dt)².

Taking the pulley level as zero and ignoring constants, if m₁ moves down by x, its height decreases by x, while m₂ moves up by x. Thus

V=m₁g(h₀−x)+m₂g(h₀+x)=constant+(m₂−m₁)gx.

So, ignoring the constant,

L=T−V=(1/2)(m₁+m₂)(dx/dt)²+(m₁−m₂)gx.

Lagrange’s equation is

d/dt(∂L/∂(dx/dt))−∂L/∂x=0.

Here

∂L/∂(dx/dt)=(m₁+m₂)(dx/dt),

so

d/dt(∂L/∂(dx/dt))=(m₁+m₂)(d²x/dt²).

Also

∂L/∂x=(m₁−m₂)g.

Therefore

(m₁+m₂)(d²x/dt²)−(m₁−m₂)g=0,

or

(m₁+m₂)(d²x/dt²)=(m₁−m₂)g .

Thus the acceleration is

d²x/dt²=((m₁−m₂)/(m₁+m₂))g,

directed downward for m₁. If m₁>m₂, m₁ accelerates downward; if m₂>m₁, the acceleration is upward for m₁. If m₁=m₂, the system remains in equilibrium.

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.

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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) derive: given > assumptions > stepwise derivation > result > check Full marks: Complete derivations with all steps shown, correct final results, and clear diagrams where required.

Key points expected

  • Apply separation of variables to find eigenfunctions sin(nπx/L)
  • Solve for time-dependent part T_n(t) using initial velocity condition
  • Compute Fourier sine coefficients for initial displacement (1/4)x(L-x)
  • State final series solution u(x,t) with specific coefficients
  • Identify minterms from the truth table (m1, m2, m3, m5, m7)
  • Use Karnaugh map or Boolean algebra to simplify the function
  • Derive the simplified expression (e.g., x + yz)
  • Draw the logic gate network for the simplified expression

Evaluation rubric

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

  1. (a) Solution to the 1D wave equation with given boundary and initial conditions. 20 marks

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

    Must cover

    • Apply separation of variables to find eigenfunctions sin(nπx/L)
    • Solve for time-dependent part T_n(t) using initial velocity condition
    • Compute Fourier sine coefficients for initial displacement (1/4)x(L-x)
    • State final series solution u(x,t) with specific coefficients

    Loses marks

    • Skipping the calculation of Fourier coefficients
    • Failing to apply the zero initial velocity condition
    • Incorrect eigenvalues for the spatial part

    Earns more

    • Explicitly state orthogonality of sine functions
    • Show integration steps for coefficient calculation
    • Verify boundary conditions are satisfied by the solution

    Extra mark

    • Mention physical interpretation of the standing wave modes
  2. (b) Simplified Boolean function F(x,y,z) and its corresponding gate network. 15 marks

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

    Must cover

    • Identify minterms from the truth table (m1, m2, m3, m5, m7)
    • Use Karnaugh map or Boolean algebra to simplify the function
    • Derive the simplified expression (e.g., x + yz)
    • Draw the logic gate network for the simplified expression

    Loses marks

    • Incorrect minterm identification from the table
    • Failing to simplify the Boolean expression
    • Gate diagram not matching the derived expression

    Earns more

    • Show the K-map grouping clearly
    • Label inputs and output on the gate diagram
    • Verify the simplified function against the truth table

    Extra mark

    • Mention alternative gate implementations (e.g., NAND only)
  3. (c) Lagrangian equation of motion for two unequal masses on a pulley. 15 marks

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

    Must cover

    • Define generalized coordinate (e.g., displacement x of one mass)
    • Express Kinetic Energy (T) and Potential Energy (V) in terms of x
    • Formulate the Lagrangian L = T - V
    • Apply Euler-Lagrange equation to find the equation of motion

    Loses marks

    • Incorrect expression for potential energy change
    • Failing to account for the constraint in the Lagrangian
    • Algebraic errors in the Euler-Lagrange derivation

    Earns more

    • Clearly define the coordinate system and reference level
    • Show the constraint relation between the two masses
    • State the final equation in standard form (e.g., (m1+m2)x'' = (m1-m2)g)

    Extra mark

    • Mention the physical meaning of the effective mass

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