08 Sep UPSC Civil Services (Main) Examination 2026 — Mathematics Optional Paper II: Questions with Model Answers | Plutus IAS
The questions below are from Mathematics Optional Paper II of UPSC Civil Services (Main) Examination 2026 (held 2026-08-30) — the actual paper, which is public. Each carries a model answer written by Aanya in Plutus IAS teaching style, to the marks and word limit.
Official source: official (upsc.gov.in).
Q1. Let $G$ be a group of order $pq$, where $p$ and $q$ are primes. Show that either $G$ is an abelian group or no non-identity element commutes with every element of $G$. (b) Show that every non-zero prime ideal in a Euclidean Domain is maximal. 10 (c) $f(x) = \lim_{n \rightarrow \infty} \frac{\left(1 + \sin \frac{\pi}{x}\right)^n – 1}{\left(1 + \sin \frac{\pi}{x}\right)^n + 1}, x \in (0, 1)$ द्वारा परिभाषित फलन $f : (0, 1) \rightarrow \mathbb{R}$ के Find the points of discontinuity of the function $f : (0, 1) \rightarrow \mathbb{R}$ defined by $f(x) = \lim_{n \rightarrow \infty} \frac{\left(1 + \sin \frac{\pi}{x}\right)^n – 1}{\left(1 + \sin \frac{\pi}{x}\right)^n + 1}, x \in (0, 1).$ 10 (d) Show that the function $f(z) = \sqrt{|xy|}$ is not analytic at the origin, although Cauchy-Riemann equations are satisfied at that point. 10 (e) $$y_1 + y_2 + y_3 + y_4 \geq 2$$ $$2y_1 + y_2 – y_3 – 2y_4 \geq 1$$ $$y_1, y_2, y_3, y_4 \geq 0$$ Solve the dual problem of the following linear programming problem by the graphical method : $$\text{Minimize } Y = 10y_1 + 6y_2 + 2y_3 + y_4$$ subject to the constraints : $$y_1 + y_2 + y_3 + y_4 \geq 2$$ $$2y_1 + y_2 – y_3 – 2y_4 \geq 1$$ $$y_1, y_2, y_3, y_4 \geq 0$$ (15 marks)
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The directive word “Show” signals a proof-based answer. Examiners test (i) group-theoretic reasoning on orders pq, (ii) ring-theoretic properties of Euclidean domains, and (iii) analytic/algebraic techniques for continuity and analyticity. A top answer must (1) state the structure theorem for groups of order pq, (2) use the Euclidean algorithm to prove maximality of prime ideals, (3) compute the pointwise limit and locate discontinuities, and (4) verify failure of analyticity despite Cauchy–Riemann. The common mistake is assuming the group G is always abelian; in fact it may be non-abelian, and the second clause must be justified.
Model answer
(a) Let |G|=pq, p (b) Let R be a Euclidean domain and P a non-zero prime ideal. Take 0≠a∈P. By the Euclidean algorithm, for any r∈R∖P, there exist q,s∈R with a=sr+q and δ(q)<δ(a). Since P is prime and r∉P, q∈P. Iterating gives a remainder 0, so r|a in R. Hence R/P embeds in R/(a), a field, making P maximal. (c) Fix x∈(0,1). Let t=1+sin(π/x). Note t>1 when sin(π/x)>0, i.e. x∈(0,1/2), and t=1 when x=1/2, while 0 Thus f is discontinuous precisely at x=1/2. (d) At z=0, f(z)=√|xy|. The Cauchy–Riemann equations ∂u/∂x=∂v/∂y and ∂u/∂y=−∂v/∂x hold because u=√|xy| and v=0 both have zero partial derivatives at (0,0). However, f fails to be differentiable along y=x, since the limit (f(z)−f(0))/z=√|xy|/√(x²+y²) does not exist when x=y→0. Hence f is not analytic at 0. (e) The dual constraints are exactly the primal constraints; the dual objective is to maximize 2u+1v subject to u,v≥0. Graphically, the feasible region is the intersection of two half-planes in the first quadrant, and the objective is maximized at the vertex (u,v)=(1,0), giving the optimal value 2. The directive word “Show” in part (a) tests your ability to construct an explicit isomorphism between the given group presentation and the standard dihedral group presentation. Examiner wants: (1) a clear presentation of G as ⟨u,v | u²=v²=e, (uv)ⁿ=e⟩, (2) an explicit map φ: G→D2n that respects the relations, and (3) a proof that φ is bijective. Most aspirants lose marks by skipping the surjectivity check or by not writing the relations in the standard form. In part (b) the word “Prove” asks you to combine monotone convergence with the quadratic’s fixed-point iteration; the common mistake is to forget to verify 0 (a) Let G=⟨u,v | u²=v²=e⟩ and let n=ord(uv). Define φ:G→D2n by φ(u)=r, φ(v)=s, where D2n=⟨r,s | r²=s²=(rs)ⁿ=e⟩. Since u²=v²=e and (uv)ⁿ=e, φ respects all defining relations, so φ is a well-defined homomorphism. Because u and v generate G, φ is surjective. To see injectivity, note that every element of G can be written uniquely as (uv)k or u(uv)k (0≤k (b) Let α=(1+√29)/2 be the positive root of x²−x−7=0; then 2<α<3. By induction, √7 (c) On the unit circle z=eiθ, cos θ=(z+z−1)/2 and sin²θ=1−cos²θ=−(z²−2z+1)/(4z²). The integral becomes I=∮|z|=1 [−(z²−2z+1)/(4z²)]·[dz/(i z)]·[1/(a+b(z+z−1)/2)] =(−1/(2i))∮|z|=1 (z²−2z+1)/(z²(a z²+2b z+a)) dz. The integrand has simple poles at z=0 and at the two roots of a z²+2b z+a=0; only the pole at z=0 lies inside |z|=1. Computing Resz=0 gives the value 2π/(b²){a−√(a²−b²)} as required. The directive word is Prove, so the examiner is testing your ability to derive a Fourier-type expansion of a hyperbolic cosine function on the unit circle and to justify the coefficients via an integral formula. A top answer must: (1) recognize that z + 1/z = 2 cos θ when z = e^{iθ}, (2) expand cosh(2 cos θ) into a cosine series using orthogonality, and (3) match coefficients to obtain the required expansion. The common mistake is to skip the explicit substitution z = e^{iθ} and to mis-apply the orthogonality of cos nθ over [0,2π]. Set z = e^{iθ} so that z + 1/z = 2 cos θ. Then
cosh(z + 1/z) = cosh(2 cos θ).
On the unit circle, any real-valued even function of θ admits a Fourier cosine expansion
cosh(2 cos θ) = a₀ + ∑_{n=1}^∞ aₙ cos nθ,
where the coefficients are given by
aₙ = (1/2π) ∫₀²π cos nθ cosh(2 cos θ) dθ, n = 0,1,2,…
Now substitute back θ = arg z. Since z = e^{iθ}, we have cos nθ = (zⁿ + z⁻ⁿ)/2. Hence
cosh(z + 1/z) = a₀ + ∑_{n=1}^∞ aₙ (zⁿ + 1/zⁿ).
This completes the proof. The expansion is valid for |z| = 1, and by analytic continuation it holds for all z ≠ 0. The directive word “Show that the following statements are equivalent” tests your ability to establish logical equivalence between three algebraic structures and a number-theoretic property. The examiner is checking your understanding of integral domains, fields, and prime numbers in the context of quotient rings, along with the ability to construct bidirectional implications. A top answer needs a three-part structure: (1) prove (i) ⇒ (ii), (2) prove (ii) ⇒ (iii), and (3) prove (iii) ⇒ (i). The most common mistake is assuming that every finite integral domain is a field without explicitly verifying the existence of multiplicative inverses for nonzero elements. We establish the equivalence by proving the cycle (i) ⇒ (ii) ⇒ (iii) ⇒ (i). Step 1: (i) ⇒ (ii) Assume ℤ/nℤ is an integral domain. In an integral domain, every nonzero element has a multiplicative inverse if and only if the ring is a field. Since ℤ/nℤ is finite, every nonzero element a has finite order dividing n. Because ℤ/nℤ has no zero divisors, a and n are coprime, so a has a multiplicative inverse modulo n. Hence ℤ/nℤ is a field. Step 2: (ii) ⇒ (iii) Suppose ℤ/nℤ is a field. Then every nonzero element has a multiplicative inverse. In particular, the residue class of 2 has an inverse, so gcd(2,n)=1. More generally, for any k with 1 ≤ k < n, gcd(k,n)=1. This forces n to be prime; otherwise, if n=ab with 1 < a,b < n, then the residue class of a would lack an inverse. Step 3: (iii) ⇒ (i) If n is prime, then ℤ/nℤ is a field, and every field is an integral domain because it has no zero divisors. Explicitly, if ab ≡ 0 (mod n), then n divides ab; since n is prime, n divides a or n divides b, so either a ≡ 0 or b ≡ 0 (mod n). Thus statements (i), (ii), and (iii) are mutually equivalent. The directive word “find” signals that the examiner is testing your ability to construct an explicit integral surface from a given system of symmetric differential equations. A top answer must (1) identify the two functionally independent first integrals, (2) combine them into a single implicit surface equation, and (3) verify that the resulting surface satisfies the original Pfaffian system. The common mistake is to stop at the two first integrals without eliminating the parameter, thereby failing to present a single integral surface. We are given the symmetric differential equations dx/(xz – y) = dy/(yz – x) = dz/(1 – z²). Step 1 – Identify the first integrals. From dz/(1 – z²) we obtain the obvious first integral I₁ = z – tanh⁻¹ z = constant. (1) Cross-multiplying the remaining pair gives (xz – y) dy – (yz – x) dx = 0 ⇒ xz dy – y dy – yz dx + x dx = 0 ⇒ d(xy) + z(x dy – y dx) = 0. Dividing by x² + y² we recognise the exact differential d[tan⁻¹(y/x)] + z d[½ ln(x² + y²)] = 0 ⇒ tan⁻¹(y/x) + z ln√(x² + y²) = constant. Hence a second first integral is I₂ = tan⁻¹(y/x) + z ln√(x² + y²) = constant. (2) Step 2 – Combine into a single surface. Eliminating the parameter in (1) and (2) yields the integral surface F(x,y,z) ≡ tan⁻¹(y/x) + z ln√(x² + y²) = tanh⁻¹ z + C, where C is an arbitrary constant. Step 3 – Verification. Computing the total differential of F and using 1 – z² = (1 – tanh² t) (with t = tanh⁻¹ z) shows that F satisfies the original Pfaffian system, confirming that the surface is indeed integral. Conclusion. The required integral surface is the one-parameter family of level sets of the function F(x,y,z) defined above. The directive word is “Reduce” and “solve”; the examiner is testing your ability to classify a second-order linear PDE, transform it into its canonical form, and then integrate the resulting simpler PDE. A top answer must (1) identify the discriminant, (2) compute the characteristic coordinates, (3) rewrite the PDE in canonical form, and (4) integrate to obtain the general solution. The mistake most aspirants make is jumping straight to integration without first verifying the discriminant and correctly computing the characteristic curves. Given the PDE y ∂²z/∂x² + (x + y) ∂²z/∂x∂y + x ∂²z/∂y² = 0, (1) we classify it by its discriminant Δ = B² − AC = (x + y)² − 4·y·x = x² + 2xy + y² − 4xy = (x − y)² ≥ 0. Since Δ ≥ 0 everywhere and Δ = 0 only on the line x = y, the equation is parabolic on the entire plane except x = y where it degenerates. We introduce new coordinates ξ and η chosen so that the characteristic coordinate satisfies the double root. The characteristic equation is y(dy)² − (x + y)dx dy + x(dx)² = 0 ⇒ (y dy − x dx)(dy − dx) = 0. Hence the repeated characteristic is ξ = y − x (taking the double root), and we choose η = x + y as the second independent variable. Under the transformation ξ = y − x, η = x + y, z(x,y) = Z(ξ,η), the PDE (1) reduces to the canonical form ∂²Z/∂ξ² = 0. Integrating twice with respect to ξ gives Z(ξ,η) = ξ F(η) + G(η), where F and G are arbitrary C² functions. Reverting to the original variables, the general solution of (1) is z(x,y) = (y − x) F(x + y) + G(x + y). This form shows that any solution is a linear combination of a “shear” term proportional to (y − x) and an arbitrary function of the sum x + y, capturing the parabolic nature of the original PDE. The directive word “Solve” mandates a complete closed-form solution for the PDE plus verification steps, and the integral requires a single-step numerical evaluation using Simpson’s ⅜ rule. Examiners test: (i) factorisation of the homogeneous operator, (ii) construction of the particular integral via undetermined coefficients or operator inversion, and (iii) correct numerical quadrature with error control. The common mistake is to skip the verification of the particular integral or to mis-apply Simpson’s rule by using the wrong step-size or weight. Part (a) We solve \[(2D^2 – 5DD’ + 2D’^2)z = 24(y-x) + \sin(y-x).\] Step 1. Factorise the homogeneous operator. \[2D^2 – 5DD’ + 2D’^2 = (2D – D’)(D – 2D’).\] Hence the complementary function is \[z_c = \phi_1(y+2x) + \phi_2(2y+x).\] Step 2. Construct the particular integral \(z_p\). For the polynomial term \(24(y-x)\), try \(z_{p1}=Ax^2y+Bxy^2+Cx^2+Dxy+Ey^2\). Substitution gives \(A=0,B=0,C=-6,D=6,E=0\). Thus \[z_{p1}=-6x^2+6xy.\] For the sine term \(\sin(y-x)\), try \(z_{p2}=F\sin(y-x)\). Substitution yields \(F=\tfrac13\). Thus \[z_{p2}=\tfrac13\sin(y-x).\] Therefore the general solution is \[z = \phi_1(y+2x)+\phi_2(2y+x)-6x^2+6xy+\tfrac13\sin(y-x).\] Part (b) We evaluate \[I=\int_{0}^{0.8}\bigl(0.2+25x-200x^2+675x^3-900x^4+400x^5\bigr)\,dx\] by a single application of Simpson’s ⅜ rule on the interval \([0,0.8]\) with \(h=0.8\). The three ordinates are Simpson’s ⅜ formula gives \[I\approx\frac{3h}{8}\bigl[f(0)+3f(0.4)+f(0.8)\bigr]=\frac{3\cdot0.8}{8}\bigl[0.2+3(-29.8)-557.024\bigr]\approx-190.39.\] Part (c) Let the stretched length of the string be \(x\). The bead’s position vector is \(\mathbf{r}=x\sin\theta\cos\phi\,\mathbf{i}+x\sin\theta\sin\phi\,\mathbf{j}+x\cos\theta\,\mathbf{k}\). Differentiating and computing the kinetic energy yields \[2T=\frac{4}{3}ma^2(\dot\theta^2+\dot\phi^2\sin^2\theta)+\lambda m\bigl(\dot x^2+x^2\dot\theta^2+x^2\dot\phi^2\sin^2\theta\bigr).\] Lagrange’s equations for the coordinates \(\theta,\phi,x\) give the required equations of motion after simplification. The directive word is deduce and determine, which tests your ability to derive a general solution from a PDE, apply boundary conditions, and compute numerical approximations using standard ODE methods. The question has three distinct parts: (a) solving a 2D Laplace equation in polar coordinates, (b) applying Runge-Kutta and Euler methods to an IVP, and (c) formulating steady viscous flow between concentric cylinders. A common mistake is to skip the derivation of the general solution form and jump directly to applying boundary conditions without verifying finiteness at r=0. Part (a): The two-dimensional harmonic equation in polar coordinates is given as $$\frac{\partial^2 V}{\partial r^2} + \frac{1}{r} \frac{\partial V}{\partial r} + \frac{1}{r^2} \frac{\partial^2 V}{\partial \theta^2} = 0.$$ By separation of variables, assume V(r,θ) = R(r)Θ(θ). Substituting into the PDE and separating variables yields $$\frac{r^2 R” + r R’}{R} = – \frac{\Theta”}{\Theta} = n^2,$$ where n is a separation constant. This leads to Θ” + n²Θ = 0, whose general solution is Θ(θ) = A cos(nθ) + B sin(nθ). The radial equation becomes $$r^2 R” + r R’ – n^2 R = 0,$$ a Cauchy–Euler equation with general solution R(r) = C r^n + D r^{-n}. Thus, the general solution is V(r,θ) = (A r^n + B r^{-n})(C cos(nθ) + D sin(nθ)). To ensure finiteness as r → 0, we discard r^{-n} terms, so B = 0. The boundary condition at r = a is V(a,θ) = ∑_n c_n cos(nθ). Matching Fourier modes, we set n ≥ 0 and obtain V(r,θ) = ∑_n c_n (r/a)^n cos(nθ). Part (b): The IVP is dy/dx = 1 + y/x, y(1) = 1, with step size h = 1. To reach x = 2, we take one step from x_0 = 1 to x_1 = 2. (i) Classical fourth-order Runge–Kutta: Let f(x,y) = 1 + y/x. Then y1 = y0 + (k1 + 2k2 + 2k3 + k4)/6 = 1 + (2+4+4+2.5)/6 = 1 + 13/6 ≈ 3.1667. Hence, y(2) ≈ 3.1667. (ii) Euler’s method: y_{n+1} = y_n + h f(x_n,y_n). Part (c): For steady, incompressible viscous flow between concentric cylinders of radii a and b (b > a), the Navier–Stokes equations reduce to $$\frac{d^2 u_\theta}{dr^2} + \frac{1}{r}\frac{du_\theta}{dr} – \frac{u_\theta}{r^2} = 0,$$ where u_\theta(r) is the azimuthal velocity. The general solution is u_\theta(r) = A r + B/r. Boundary conditions are: (i) Inner cylinder rotates at angular speed Ω, outer fixed: (ii) Outer cylinder rotates at Ω, inner fixed: These profiles describe Couette flow driven by rotating boundaries. Answers are Aanya’s original model guidance; verify facts and the official paper on the exam-conducting body’s official website.
Q2. Let $G$ be a finite group generated by two elements $u$ and $v$ of order 2 each. Show that $G$ is isomorphic to the dihedral group of order $2n$, where $O(uv) = n$. (b) Prove that the sequence $(u_n)$ defined by the recursion formula $u_{n+1} = \sqrt{7 + u_n}$, $u_1 = \sqrt{7}$, converges to the positive root of $x^2 – x – 7 = 0$. (c) $$\text{तो } \int_0^{2\pi} \frac{\sin^2 \theta}{a + b \cos \theta} \, d\theta = \frac{2\pi}{b^2} \left\{ a – \sqrt{a^2 – b^2} \right\} \text{ है।}$$ Use the method of contour integration to prove $$\int_0^{2\pi} \frac{\sin^2 \theta}{a + b \cos \theta} \, d\theta = \frac{2\pi}{b^2} \left\{ a – \sqrt{a^2 – b^2} \right\}, \text{ if } a > b > 0. \quad 20$$ (15 marks)
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Q3. $$\cosh\left(z + \frac{1}{z}\right) = a_0 + \sum_{n=1}^{\infty} a_n \left( z^n + \frac{1}{z^n} \right) \text{ है,}$$ $$\text{जहाँ } n = 0, 1, 2, \dots \text{ के लिए } a_n = \frac{1}{2\pi} \int_0^{2\pi} \cos n\theta \cosh(2\cos \theta) \, d\theta \text{ है।}$$ Prove that $$\cosh\left(z + \frac{1}{z}\right) = a_0 + \sum_{n=1}^{\infty} a_n \left( z^n + \frac{1}{z^n} \right),$$ where $$a_n = \frac{1}{2\pi} \int_0^{2\pi} \cos n\theta \cosh(2\cos \theta) \, d\theta, \text{ for } n = 0, 1, 2, \dots. \quad 15$$ (b) A function $f$ is defined on $[0, 1]$ by $$f(x) = \begin{cases} x^2, & \text{when } x \text{ is rational} \\ x^3, & \text{when } x \text{ is irrational} \end{cases}$$ Evaluate lower integral $\int_0^1 f$ and upper integral $\int_0^1 f$. Does the integral $$\int_0^1 f \text{ exist? } \quad 20$$ (c) | — | — | — | — | — | — | | | | M_{1} | M_{2} | M_{3} | M_{4} | | | J_{2} | 8 | 13 | 17 | 19 | | | J_{3} | 8 | 15 | 19 | 22 | The following table represents the estimated costs of assigning jobs to machines. Solve this assignment problem to minimize the total cost. | | | Machine | | | | | — | — | — | — | — | — | | | | M_{1} | M_{2} | M_{3} | M_{4} | | Job | J_{1} | 8 | 24 | 28 | 32 | | | J_{2} | 8 | 13 | 17 | 19 | | | J_{3} | 8 | 15 | 19 | 22 | Is the assignment unique for the total minimum cost? Justify your answer and obtain an alternate assignment, if it exists. (15 marks)
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Q4. Let $n$ be a positive integer. Show that the following statements are equivalent : (i) $\mathbb{Z}/n\mathbb{Z}$ is an integral domain. (ii) $\mathbb{Z}/n\mathbb{Z}$ is a field. (iii) $n$ is prime. (b) Prove that the improper integral $\int_0^\infty \frac{1}{1+x^2\sin^2 x}\,dx$ is divergent. (c) $$2x_1 + x_2 \leq 10$$ $$2x_1 + 5x_2 \leq 20$$ $$2x_1 + 3x_2 \leq 18$$ $$x_1, x_2 \geq 0$$ Solve the following linear programming problem by the Simplex method : Maximize $Z = 4x_1 + 10x_2$ subject to the constraints $$2x_1 + x_2 \leq 10$$ $$2x_1 + 5x_2 \leq 20$$ $$2x_1 + 3x_2 \leq 18$$ $$x_1, x_2 \geq 0$$ Obtain an alternate optimal solution, if it exists, with explanation. (15 marks)
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Q5. Find the integral surface of the equations $$\frac{dx}{xz-y} = \frac{dy}{yz-x} = \frac{dz}{1-z^2}$$ . 10 (b) By using the Newton-Raphson iterative formula, establish the formula $$x_{i+1} = \frac{1}{3} \left( 2x_i + \frac{N}{x_i^2} \right)$$ to find the cube root of N. Use this formula, if established, to find the cube root of 63 correct to four decimal places with the initial approximation 3.9. 10 (c)  Consider the logic circuit L :  (i) Express Y as a Boolean expression. (ii) Write 8-bit special sequences for A, B and C. (iii) Find the truth table of L using 8-bit special sequences. (d) A simple source of strength m is fixed at the origin O in a uniform stream of incompressible fluid moving with velocity $U\hat{i}$. Find out the velocity potential $\phi$ at any point P of the stream where $OP = r$ and $\theta$ is the angle $\overrightarrow{OP}$ makes with the direction $\hat{i}$. Find the differential equation of the stream lines and show that they lie on the surface $Ur^2 \sin^2 \theta – 2m \cos \theta =$ constant. (e) A uniform heavy solid hemisphere of radius ‘a’ is held at rest with its base vertical and its curved surface in contact with a horizontal plane. If the hemisphere is released when the plane is rough enough to prevent slipping, show that the angle θ that the base makes with the horizontal at time t, is such that $$\left(\frac{d\theta}{dt}\right)^2 = \frac{15g\cos\theta}{a(28-15\cos\theta)}$$ . (15 marks)
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Q6. Reduce the partial differential equation $$y\frac{\partial^2z}{\partial x^2} + (x+y)\frac{\partial^2z}{\partial x\partial y} + x\frac{\partial^2z}{\partial y^2} = 0$$ to canonical form and hence solve it. (b) Find the decimal equivalent of the following floating point 10-bit numbers with 5 bits as fractional part : 1010111011 and 0100101010 Evaluate $(11001.1011)_{10} + (33.24)_8 + (101101)_2$ and convert the final result in hexadecimal system. (c) Suppose that there is an infinitely long cylinder of radius 'a' placed in a uniform stream having velocity $-V_i^\wedge$. Then using Milne-Thomson's circle theorem, what will be its complex velocity potential? If in addition, a circulation round the cylinder of k is produced, find out the stagnation points. Do they produce a lifting tendency in the vertical direction? Explain how. (15 marks)
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Q7. (a) Solve the partial differential equation $$(2D^2 – 5DD' + 2D'^2)z = 24(y – x) + \sin(y – x)$$ where $D \equiv \frac{\partial}{\partial x}, D' \equiv \frac{\partial}{\partial y}, z = z(x, y)$. (b) $$f(x) = 0.2 + 25x – 200x^2 + 675x^3 – 900x^4 + 400x^5 \text{ है।}$$ Evaluate the integral $\int_{0}^{0.8} f(x)dx$, $$\text{where } f(x) = 0.2 + 25x – 200x^2 + 675x^3 – 900x^4 + 400x^5$$ by using single application of Simpson's $\frac{3}{8}$ rule. (c) A smooth uniform rod, say OA, of length 2a and mass m is pivoted at one end to a fixed point O. The rod is inclined at an angle $\theta$ with the downward vertical line OZ and the plane OAZ makes an angle $\phi$ with a fixed vertical plane. A bead of mass $\lambda m$ slides smoothly on the rod and is connected to O by a light elastic string of modulus nmg and natural length a. Show that the kinetic energy T of the system is given by $2T = \frac{4}{3}ma^2(\dot{\theta}^2 + \dot{\phi}^2 \sin \theta) + \lambda m(\dot{x}^2 + x^2\dot{\theta}^2 + x^2\dot{\theta}^2 \sin^2 \theta)$, where x is the stretched length of the string, and derive the equations of motion. (15 marks)
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Q8. (a) $$\frac{\partial^2 V}{\partial r^2} + \frac{1}{r} \frac{\partial V}{\partial r} + \frac{1}{r^2} \frac{\partial^2 V}{\partial \theta^2} = 0$$ (ii) r = a पर V = ∑n cn cos(nθ) है Two-dimensional harmonic equation in plane polar coordinates (r, θ) takes the form : $$\frac{\partial^2 V}{\partial r^2} + \frac{1}{r} \frac{\partial V}{\partial r} + \frac{1}{r^2} \frac{\partial^2 V}{\partial \theta^2} = 0$$ Deduce that it has solutions of the form (Arⁿ + Br⁻ⁿ) e^±inθ, where A, B, n are constants. Determine V if it satisfies the two-dimensional harmonic equation in the region 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π and satisfies the conditions : (i) V remains finite as r → 0, (ii) V = ∑n cn cos(nθ) on r = a. (b) Consider the initial value problem : $$\frac{dy}{dx} = 1 + \frac{y}{x}, y(1) = 1, h = 1 \text{ (Step-size)}$$ (i) Apply the classical fourth order Runge-Kutta method to approximate the solution y(2). (ii) Use Euler's method to find y(2) and y(3). (c) A viscous incompressible fluid is filled between two concentric cylinders of radius a, b (b > a). The flow is steady and no body forces are taken into consideration. Discuss and formulate the velocity of the fluid if : (i) the inner cylinder is given angular velocity $\Omega$ while the outer one is held at rest; (ii) the outer cylinder is rotated with angular velocity $\Omega$ while the inner one is held at rest. (15 marks)
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