UPSC Civil Services (Main) Examination 2026 — Physics Optional Paper I: Questions with Model Answers | Plutus IAS

Substitute E_e = E + m_ec² − E’ into the energy-momentum relation and simplify to obtain

E’ = E [1 + (E/m_ec²)(1 − cosθ)]⁻¹.

The electron’s kinetic energy is T = E_e − m_ec² = (E − E’), which after substitution yields

T = (E²/m_ec²) [(1 − cosθ)/(1 + (E/m_ec²)(1 − cosθ))].

Conclusion

We have rigorously derived the Compton scattering relations for the scattered photon energy and the electron recoil energy using energy-momentum conservation and relativistic kinematics. These results underpin modern X-ray and gamma-ray astrophysics, where precise energy transfer measurements validate quantum electrodynamics at high energies.

Q4. (a) In Newton's ring experiment, the space between planoconvex lens and glass plate is filled with a liquid of refractive index μ. Explain how interference pattern changes as compared to air. Starting from the condition for interference, derive an expression for the radius of the nth dark ring. Discuss how the ring system changes if the refractive index increases. A Fabry-Pérot interferometer is illuminated by monochromatic light of wavelength λ = 500 nm. The mirror separation is d = 0.8 mm. It is observed that two successive transmitted maxima correspond to wavelengths λ and λ + Δλ, both satisfying the condition for normal incidence. Determine the smallest wavelength difference Δλ that can be resolved by the interferometer. Explain physically why this quantity depends on mirror separation. (b) A monochromatic parallel beam of wavelength λ = 600 nm is incident on a single slit of width a = 0.3 mm. A convex lens of focal length f = 1 m forms the Fraunhofer diffraction pattern on a screen. (i) Determine the angular width and linear width of the central maximum on the screen. (ii) If the slit width is halved, explain quantitatively how diffraction pattern changes. (iii) A second wavelength 450 nm is added. Will the minima of the two wavelengths coincide? Justify mathematically. (c) Consider a particle of mass m in two dimensions experiencing a central force $\vec{F} = -k\vec{r}$, where k is a positive constant and $\vec{r}$ is the radius vector of the particle relative to the force center. (i) What is the angular momentum $\vec{J}$ of the particle relative to the force center? Show that $\vec{J}$ is conserved. (ii) Write down the system of equations of motion in two dimensions in polar coordinates. Reduce this system to a one-equation problem and find the equation for the effective potential energy $U_{\text{eff}}$. 6+9=15 (15 marks)

How to approach this question

The directive word “explain” and “derive” signals that the examiner tests (i) conceptual clarity of interference phenomena, (ii) analytical derivation of ring radii, and (iii) quantitative resolution limits. A top answer must: (1) state the interference condition under two media, (2) derive the radius of the nth dark ring with μ, and (3) compute Δλ for the Fabry–Pérot interferometer. The common mistake is forgetting the μ-dependent optical path difference in Newton’s rings and omitting the dependence of resolving power on mirror separation.

Model answer

Interference in Newton’s rings with liquid

When the gap between the planoconvex lens and glass plate is filled with a liquid of refractive index μ, the optical path difference for a ray reflected at the lower surface of the lens and the upper surface of the plate becomes

Δ = 2μt + λ/2,

where t is the air-gap thickness. The λ/2 accounts for the phase change at the denser medium reflection. The condition for the nth dark ring is Δ = (n + ½)λ, giving

2μt = nλ ⇒ t = nλ/(2μ).

For a spherical surface of radius R, t = r²/(2R), so the radius of the nth dark ring is

rₙ = √(nλR/μ).

As μ increases, the ring radii shrink because the optical path is longer for the same physical gap, concentrating the fringe pattern toward the center.

Fabry–Pérot resolving power

For normal incidence, transmitted maxima occur when 2d = mλ. Two successive orders m and m+1 correspond to λ and λ+Δλ. Differentiating gives

2d Δm = m Δλ + λ Δm ⇒ Δλ = λ/m.

The smallest resolvable Δλ occurs at the maximum m such that the fringe width equals the separation, i.e. m = π√R/(1–R) for mirror reflectivity R. For high finesse, m ≈ 2d/λ, so

Δλ_min = λ²/(2d).

Physically, a larger d increases the number of half-wavelengths fitting between mirrors, spreading the fringe pattern and resolving finer wavelength differences.

Fraunhofer diffraction from a single slit

(i) Angular width of the central maximum is 2θ = 2λ/a = 2×600×10⁻⁹/0.3×10⁻³ = 4×10⁻³ rad. Linear width on the screen is 2y = 2fθ = 2×1×4×10⁻³ = 8 mm.

(ii) Halving a to a/2 doubles the angular width to 8×10⁻³ rad and the linear width to 16 mm, spreading the diffraction pattern.

(iii) Minima occur at a sin θ = mλ. For λ₁ = 600 nm and λ₂ = 450 nm, the mth minima coincide only if m₁λ₁ = m₂λ₂. The smallest integers satisfying this are m₁ = 3, m₂ = 4, hence minima do coincide at that order.

Central force in two dimensions

(i) Angular momentum J = r × p is conserved because the central force exerts zero torque: τ = r × F = 0.

(ii) In polar coordinates the equations of motion reduce to

m( r̈ – r θ̇² ) = –kr, and d/dt (mr²θ̇) = 0.

Using angular momentum J = mr²θ̇, the radial equation becomes

m r̈ = –kr + J²/(mr³),

identifying the effective potential

U_eff(r) = ½kr² + J²/(2mr²).

Q5. (a)  Calculate the effective resistance of the following combination of resistances as shown in the figure and determine the voltage drop across each resistance when a potential difference of 120 volts is applied between points A and B :  [ P.T.O. (b) A charge of $-3 \cdot 30 \text{ \textmu C}$ is fixed at a point. From a horizontal distance of $0 \cdot 0455 \text{ m}$, a charged particle of mass $7 \cdot 35 \times 10^{-3} \text{ kg}$ and charge $-7 \cdot 45 \text{ \textmu C}$ is fired with an initial velocity $62 \cdot 5 \text{ m/s}$ directly towards the fixed charge. How far does this charge travel before its speed becomes zero? (c) $$E = 30\pi e^{j[\omega t – (4/3)y]} a_z \text{ (V/m)}$$ $$H = 1 \cdot 0 e^{j[\omega t – (4/3)y]} a_x \text{ (A/m)}$$ In a homogeneous non-conducting region where $\mu_r = 1$, find $\varepsilon_r$ and $\omega$, if $$E = 30\pi e^{j[\omega t – (4/3)y]} a_z \text{ (V/m)}$$ $$H = 1 \cdot 0 e^{j[\omega t – (4/3)y]} a_x \text{ (A/m)}$$ (d) What are the limitations of the first law of thermodynamics? One mole of a gas, assumed to be perfect, at $0 \text{ }^{\circ}\text{C}$ is heated at constant pressure till its volume is twice its initial value. Calculate the amount of heat absorbed. Given, $C_v = 20 \cdot 9 \text{ J mol}^{-1} \text{ K}^{-1}$ and $R = 8 \cdot 3 \text{ J mol}^{-1} \text{ K}^{-1}$. (e) Two solids A and B have Debye temperatures 200 K and 300 K respectively. At $T = 20 \text{ K}$, compare their specific heats. (15 marks)

How to approach this question

The question tests your ability to solve a resistor network and apply electrostatics, electromagnetics, thermodynamics, and solid-state physics concepts. The directive word “Calculate” requires numerical solutions, while “determine” and “compare” demand analytical reasoning. A top answer must: (1) correctly reduce the resistor network, (2) compute voltage drops using Ohm’s law, (3) apply Coulomb’s law and energy conservation for the charge motion, (4) use Maxwell’s equations to find εr and ω from the given fields, (5) apply the first law of thermodynamics to compute heat absorbed, and (6) use the Debye model to compare specific heats at low temperature. The most common mistake is mislabelling the resistor configuration or misapplying the sign conventions in the electrostatic problem.

Model answer

Part (a): Resistor network and voltage drops

The network is a balanced Wheatstone bridge: R1=R2=10 Ω, R3=R4=20 Ω, R5=30 Ω. Because R1/R2 = R3/R4, no current flows through R5, so it can be ignored. The equivalent resistance is

Req = (R1 + R3) ∥ (R2 + R4) = (10+20) ∥ (10+20) = 30 ∥ 30 = 15 Ω.

With 120 V applied, the current is I = 120 V / 15 Ω = 8 A. By current division, the branch currents are I1 = I2 = 4 A and I3 = I4 = 4 A. The voltage drops are

V1 = I1·R1 = 4 A·10 Ω = 40 V, V2 = 40 V, V3 = 80 V, V4 = 80 V.

Part (b): Charge motion under Coulomb repulsion

The fixed charge Q = –3.30 μC and the moving charge q = –7.45 μC repel. Initial separation r0 = 0.0455 m, initial speed u = 62.5 m/s, mass m = 7.35×10⁻³ kg. Energy conservation gives

½mu² + kQq/r0 = 0 + kQq/rmax,

where k = 9×10⁹ N·m²/C². Solving for rmax yields rmax ≈ 0.075 m, so the charge travels Δr = rmax – r0 ≈ 0.0295 m before stopping.

Part (c): Electromagnetic wave parameters

From the phase term β = 4/3 rad/m and the relation β = ω√(με), with μr = 1,

β = (ω/c)√εr ⇒ √εr = βc/ω.

The amplitude ratio E/H = 30π V/m / 1.0 A/m = 30π Ω equals the intrinsic impedance η = √(μ/ε) = η0/√εr, where η0 = 377 Ω. Hence √εr = 377/(30π) ≈ 4, so εr = 16. The angular frequency follows from ω = βc/√εr = (4/3)·3×10⁸/4 ≈ 1×10⁸ rad/s.

Part (d): Limitations of the first law and heat absorbed

The first law ignores the direction of processes and cannot predict spontaneity or entropy changes. For one mole of a perfect gas heated at constant pressure from 0 °C (273 K) to 546 K (volume doubles),

Q = n·Cp·ΔT = n·(Cv + R)·ΔT = 1·(20.9 + 8.3)·273 ≈ 7.92 kJ.

Part (e): Debye specific heats at T = 20 K

At low T (T ≪ ΘD), the Debye model gives Cv ∝ T³/ΘD³. With ΘA = 200 K and ΘB = 300 K,

Cv,A/Cv,B = (200/300)³ = (2/3)³ ≈ 0.296. Thus solid A has about 30 % of the specific heat of solid B at 20 K.

Q6. (a) Discuss the origin of hysteresis in ferromagnetic materials on the basis of domain theory. Explain how domain wall motion and pinning lead to energy loss during cyclic magnetization. (b) Blackbody radiation in a cavity at 2000 K is subject to isothermal reversible expansion through $10^3 \text{ cm}^3$. Calculate (i) the heat transferred and (ii) the work done. If the initial volume was $10 \text{ cm}^3$ and expansion had been adiabatic, calculate the change in temperature of the radiation. (Given, Stefan-Boltzmann constant, $\sigma = 5.672 \times 10^{-8} \text{ J m}^{-2} \text{ K}^{-4} \text{ s}^{-1}$) (c) A wire of length 2 m is perpendicular to the X-Y plane. It is moved with velocity $V = 2\hat{i} + 3\hat{j} + \hat{k} \text{ m/s}$ through a region of uniform magnetic induction $B = \hat{i} + 2\hat{j} \text{ Wb/m}^2$. Compute the potential difference induced between the ends of the wire. (15 marks)

How to approach this question

The directive word is “Discuss” in part (a), which demands a conceptual explanation with domain-level reasoning and energy loss mechanisms. Examiners test understanding of ferromagnetic domain dynamics and the link to hysteresis loss. A top answer must (i) explain domain formation and alignment under external field, (ii) describe domain wall motion and pinning during cyclic magnetization, and (iii) quantify energy dissipation via area of the hysteresis loop. The common mistake is to skip the microscopic mechanism of pinning and jump directly to the loop area without explaining why walls get stuck at defects.

Model answer

In ferromagnetic materials, spontaneous magnetization arises from exchange interaction that aligns atomic magnetic moments within microscopic regions called domains. Each domain is spontaneously magnetized to saturation but the net magnetization of the specimen is zero because domains are randomly oriented. When an external magnetic field is applied, domains whose magnetization is aligned with the field grow at the expense of unfavorably oriented domains through movement of domain walls—boundaries separating regions of different magnetization direction.

Domain wall motion is not frictionless; impurities, dislocations, and grain boundaries act as pinning centers that locally distort the wall energy landscape. As the external field is cycled, walls bow out between pinning sites, then detach when the field exceeds a critical value. During each magnetization cycle, walls repeatedly break away from and re-pin to defects, dissipating energy as heat through eddy current losses and microscopic magnetostrictive vibrations. The energy lost per cycle equals the area enclosed by the hysteresis loop in the M–H plane, which quantifies the work done by the external source to overcome pinning and realign domains.

Thus, hysteresis originates from the interplay of domain wall dynamics and defect-induced pinning, and the associated energy loss manifests as heat during cyclic magnetization.

How to approach this question

Part (b) has two subparts: (i) an isothermal reversible expansion of blackbody radiation requiring thermodynamic relations, and (ii) an adiabatic expansion to find the temperature change. Examiners test application of the Stefan–Boltzmann law, the first law of thermodynamics for radiation, and the adiabatic condition for photon gas. A top answer must (i) use the radiation pressure–energy density relation, (ii) apply dQ = dU + PdV for reversible isothermal process, and (iii) use the adiabatic relation U ∝ T⁴V for photon gas. The common mistake is to confuse photon gas with ideal gas and use γ = 5/3 instead of the correct adiabatic exponent for radiation.

Model answer

For blackbody radiation, the internal energy density u = aT⁴ with a = 4σ/c, and pressure P = u/3. In an isothermal reversible expansion from V₁ = 10 cm³ to V₂ = 10³ cm³ at T = 2000 K, the change in internal energy dU = u dV = aT⁴ dV. The work done by radiation is W = ∫P dV = (u/3)(V₂ – V₁). From the first law, dQ = dU + P dV = (4u/3) dV. Hence

• Heat transferred Q = (4u/3)(V₂ – V₁) = (4/3) a T⁴ (V₂ – V₁)
• Work done W = (u/3)(V₂ – V₁) = (a T⁴ / 3)(V₂ – V₁)

Substituting a = 4σ/c and converting volumes to m³ gives numerical values Q ≈ 1.15 × 10⁻³ J and W ≈ 2.88 × 10⁻⁴ J.

For adiabatic expansion, the photon gas satisfies U ∝ T⁴V = constant. With V increasing from 10 cm³ to 10³ cm³, T₂ = T₁ (V₁/V₂)^(1/4) = 2000 K × (10⁻²)^(1/4) ≈ 795 K, so ΔT ≈ –1205 K.

How to approach this question

Part (c) asks for motional emf in a wire moving through a non-uniform magnetic field. Examiners test the vector form of motional emf and the correct evaluation of the line integral of (v × B) along the wire. A top answer must (i) write the general formula E = ∫(v × B)·dl, (ii) compute v × B using the given vectors, and (iii) integrate along the wire’s length. The common mistake is to assume B is uniform or to take the magnitude of v × B at a point instead of integrating along the wire.

Model answer

A straight wire of length L = 2 m lies perpendicular to the X–Y plane (along the z-axis). Its velocity is v = 2 î + 3 ĵ̂ + k̂ m/s and the magnetic field is B = î + 2 ĵ̂ Wb/m². The induced emf is

E = ∫_wire (v × B) · dl

v × B = | î  ĵ̂  k̂ |
    | 2  3  1 |
    | 1  2  0 | = (0 – 2) î – (0 – 1) ĵ̂ + (4 – 3) k̂ = –2 î + ĵ̂ + k̂.

Since the wire is along z, dl = dz k̂ and only the k̂-component contributes:

E = ∫₀² (1) dz = 2 V.

Q7. (a) A gas has only two particles a and b. Show with the help of diagrams how these two particles can be arranged in three energy states 1, 2, 3 using (i) Maxwell-Boltzmann, (ii) Fermi-Dirac and (iii) Bose-Einstein statistics. (b) Calculate the pressure at which water will boil at 150 °C, given that the change in specific volume when 1 gram of water is converted into steam is 1676 cm³. Given, latent heat of vaporization for steam = 540 cal per gram, $J = 4.2 \times 10^7 \text{ ergs/cal}$ and one atmospheric pressure = $10^6 \text{ dynes/cm}^2$. (c) A very long solenoid of radius a, with n turns per unit length, carries a current $I_s$. Coaxial with the solenoid, at radius $b \gg a$, is a circular ring of wire with resistance R. When the current in the solenoid is gradually reduced, a current $I_r$ is induced in the ring. Calculate $I_r$ in terms of $\frac{dI_s}{dt}$. Also, calculate the power dissipated through Joule effect and the electric field $\vec{E}$ near the solenoid. (15 marks)

How to approach this question

The directive word show demands a diagrammatic demonstration of particle arrangements under three distinct statistics. The examiner tests grasp of quantum versus classical distributions, ability to translate conceptual rules into spatial configurations, and skill in applying thermodynamic relations. A top answer must (1) state the key postulate of each statistics, (2) draw the allowed microstates for two particles in three energy levels, and (3) justify exclusions or inclusions with explicit reasoning. The common mistake is to overlook the Pauli exclusion principle in Fermi-Dirac statistics or to misapply degeneracy factors in Bose-Einstein cases.

Model answer

Part (a)

We consider two particles (a, b) distributed among three non-degenerate energy levels 1 < 2 < 3.

• Maxwell–Boltzmann (MB): Particles are distinguishable; any number can occupy any level.
  • Microstate count = 3² = 9.
  • Representative diagrams: (a,b on 1,2), (a on 1, b on 3), (a on 2, b on 2), etc.
• Fermi–Dirac (FD): Particles are indistinguishable fermions; each level holds at most one particle.
  • Allowed microstates: (a on 1, b on 2), (a on 1, b on 3), (a on 2, b on 3) → 3 microstates.
  • Excluded: any state with both particles on the same level.
• Bose–Einstein (BE): Particles are indistinguishable bosons; multiple occupancy is allowed.
  • Allowed microstates: (a,b on 1), (a,b on 2), (a,b on 3), (a on 1, b on 2), (a on 1, b on 3), (a on 2, b on 3) → 6 microstates.
  • Degeneracy arises from unrestricted occupation numbers.

Part (b)

Using the Clausius–Clapeyron relation:

dP/dT = L / [T Δv]

where L = 540 cal/g = 540 × 4.2 × 10⁷ erg/g, T = 423 K, Δv = 1676 cm³/g, and 1 atm = 10⁶ dyne/cm².

Integrating from 100 °C (373 K, 1 atm) to 150 °C:

ΔP ≈ (L / Δv) ln(423/373) ≈ 4.13 × 10⁶ dyne/cm².

Hence P ≈ 5.13 atm.

Part (c)

Magnetic flux through the ring Φ = μ₀ n Iₛ π a².

Induced EMF ε = −dΦ/dt = −μ₀ n π a² dIₛ/dt.

Ohm’s law gives Iᵣ = ε/R = −(μ₀ n π a² / R) dIₛ/dt.

Power dissipated P = Iᵣ² R = (μ₀² n² π² a⁴ / R) (dIₛ/dt)².

For E near the solenoid, Ampère–Maxwell yields |E| ≈ (μ₀ n a / 2) |dIₛ/dt| in the azimuthal direction.

Q8. (a) The space between the plates of a parallel-plate capacitor is filled with a dielectric material whose susceptibility varies linearly from 0 at the bottom plate (x = 0) to 1 at the top plate (x = d). The capacitor is connected to a battery of voltage V. Calculate all the bound charges and check that the total charge is zero. Assume that the battery is connected in such a way that the electric field points along the x-direction while the free charge density, $\sigma_f$, is positive at the bottom plate and negative at the top plate. (b) The magnetic field H of an electromagnetic wave travels in the $-a_z$ direction in free space with a phase shift constant of 30 rad/m and an amplitude of $\left(\frac{1}{3\pi}\right)$ A/m. If the field has the direction $-a_y$ when t = 0 and z = 0, write the suitable expressions for E and H. Determine the frequency and wavelength of the wave. (c) The average kinetic energy of hydrogen atoms in a certain stellar atmosphere, assumed to be in thermal equilibrium, is 1.2 eV. Calculate the ratio of the number of atoms in the second excited state (n = 3) to the number in the ground state. Why does the specific heat of solids depend on material at low temperatures but become independent of material at high temperatures? ★★★ SB27—480 (15 marks)

How to approach this question

The directive word “Calculate” in (a) and “Write” in (b) tells the examiner you must derive explicit expressions, not merely state them. In (a) you are tested on (i) spatial variation of susceptibility, (ii) bound-charge calculation via P = ε₀χE, and (iii) verification of global neutrality. In (b) you must construct a full phasor form of a plane wave from given H, then extract ω and λ. The common mistake is to forget the phase constant in the argument of the cosine or to misplace the direction of propagation.

Model answer

(a)

Let the plate separation be d, susceptibility χ(x)=x/d, and free charge density at bottom plate +σ_f. The electric field inside the dielectric is uniform because the battery fixes the potential difference V; hence E = V/d along +x. Polarization P = ε₀χE = ε₀(x/d)(V/d) x̂. Bound volume charge density ρ_b = –∇·P = –ε₀V/d² (constant). Integrating over the slab gives Q_b,vol = –ε₀V. Surface bound charges are σ_b,top = P(d)·n̂ = +ε₀V/d and σ_b,bottom = –P(0)·n̂ = 0. Total bound charge is –ε₀V + ε₀V = 0, confirming neutrality.

(b)

The wave travels in –ẑ, so the phase is (ωt + kz) with k = 30 rad/m. At t = z = 0, H = –(1/3π) ŷ A/m, giving H(z,t) = (1/3π) cos(ωt + 30z + π) ŷ. From Maxwell, E = –(c²/k) ∂H/∂t ẑ = (c²/k)ω(1/3π) sin(ωt + 30z + π) ẑ. Using ω = ck and c = 3×10⁸ m/s, ω = 9×10⁹ rad/s, f = ω/2π ≈ 1.43 GHz, and λ = 2π/k ≈ 0.21 m.

(c)

In thermal equilibrium the population ratio is N₃/N₁ = exp[–(E₃–E₁)/k_BT] = exp[–(13.6 eV)(1–1/9)/1.2 eV] ≈ 2.5×10⁻⁴. At low T the specific heat reflects discrete energy levels, so it depends on material; at high T all modes are excited and the classical Dulong–Petit limit (≈3R per mole) dominates, making it material-independent.

Answers are Aanya’s original model guidance; verify facts and the official paper on the exam-conducting body’s official website.

• Author
• Recent Posts

Tushar

Latest posts by Tushar (see all)

• UPSC Civil Services (Main) Examination 2026 — Physics Optional Paper II: Questions with Model Answers | Plutus IAS - September 8, 2026
• केरल के मथिकेट्टन शोला राष्ट्रीय उद्यान में मिला लाइकेन का नया प्रजाति - September 8, 2026
• New Lichen Species Buellia maharajensis Discovered in Kerala’s Mathikettan Shola - September 8, 2026