Physics 2021 Paper I 50 marks Solve

Q6

(a) In an inertial reference frame S there is only a uniform electric field $\vec{E} = 8$ kVm$^{-1}$. Find the magnitude of $\vec{E}'$ and $\vec{B}'$ in the inertial reference frame S' moving with a constant velocity $\vec{v}$ relative to the frame S at an angle $\alpha = 45^{\circ}$ to the vector $\vec{E}$. The velocity of the frame S' is 0.6 times the velocity of light c. (20 marks) (b) In the given circuit, L = 2·0 μH, R = 1·0 Ω, R₀ = 2·0 Ω and E = 3·0 V. Find the amount of heat generated in the coil after the switch S is disconnected. The internal resistance of the source is negligible. (10 marks) (c) Explain the characteristics of the following thermodynamic processes for a perfect gas : (i) Isothermal process (ii) Adiabatic process (iii) Isobaric process (iv) Isochoric process Obtain the expression for the work done by the gas during the above processes. (20 marks)

हिंदी में प्रश्न पढ़ें

(a) एक जड़त्वीय संदर्भ फ्रेम S में एकसमान तीव्रता का वैद्युत क्षेत्र $\vec{E} = 8$ kVm$^{-1}$ विद्यमान है । $\vec{E}'$ और $\vec{B}'$ का परिमाण दूसरे जड़त्वीय संदर्भ फ्रेम S' में ज्ञात कीजिए जो कि S फ्रेम के सापेक्ष एकसमान वेग $\vec{v}$ से तथा सदिश $\vec{E}$ क्षेत्र से $\alpha = 45^{\circ}$ के कोण पर गति कर रहा है । फ्रेम S' का वेग प्रकाश के वेग c का 0.6 गुना है । (20 अंक) (b) दिए गए परिपथ में L = 2·0 μH, R = 1·0 Ω, R₀ = 2·0 Ω और E = 3·0 V है । जब परिपथ के कुंजी S को भंग कर दिया जाता है, तो कुंडली में उत्पन्न ऊष्मा की मात्रा ज्ञात कीजिए । स्रोत का आंतरिक प्रतिरोध नगण्य है । (10 अंक) (c) आदर्श गैस के लिए निम्नलिखित ऊष्मागतिक प्रक्रियाओं के अभिलक्षणों की व्याख्या कीजिए : (i) समतापी प्रक्रिया (ii) रूद्धोष्म प्रक्रिया (iii) समदाबी प्रक्रिया (iv) समआयतनिक प्रक्रिया उपर्युक्त प्रक्रियाओं के दौरान गैस के द्वारा किए गए कार्य का व्यंजक प्राप्त कीजिए । (20 अंक)

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Approach

Solve this multi-part numerical and theoretical problem by allocating approximately 40% time to part (a) for relativistic field transformations, 20% to part (b) for LR circuit energy dissipation, and 40% to part (c) for deriving thermodynamic work expressions. Begin with clear statements of governing equations, show systematic derivations with proper vector decomposition for (a), apply energy conservation for (b), and present characteristic equations with P-V diagrams for each process in (c). Conclude with physical interpretations of results.

Key points expected

  • Part (a): Apply Lorentz transformation for electromagnetic fields with E parallel and perpendicular components to velocity; calculate E' = γ(E∥ + v×B) and B' = γ(B⊥ - v×E/c²) with proper angle decomposition at α = 45°
  • Part (a): Compute γ = 1/√(1-0.36) = 1.25; resolve E into E∥ = Ecosα and E⊥ = Esinα; find E' magnitude and induced B' field
  • Part (b): Apply energy conservation in RL circuit; calculate total energy stored in inductor (½LI²) where I = E/R₀ at steady state; determine heat distribution between resistances
  • Part (b): Recognize that when switch opens, stored magnetic energy dissipates through R and R₀; compute heat in coil resistance R specifically
  • Part (c): State defining conditions for each process (dT=0, dQ=0, dP=0, dV=0) and derive work integrals: W_isothermal = nRTln(V₂/V₁), W_adiabatic = (P₁V₁-P₂V₂)/(γ-1), W_isobaric = PΔV, W_isochoric = 0
  • Part (c): Include P-V diagram sketches showing hyperbolic isotherm, steeper adiabat, horizontal isobar, and vertical isochore with proper labeling
  • Part (c): Relate γ = Cp/Cv and connect to degrees of freedom for perfect gas; mention relevance to Carnot cycle and Indian power plant thermodynamics

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly identifies that (a) requires Lorentz field transformations not Galilean; recognizes that B=0 in S implies induced B' in S' due to motion through E field; for (b) correctly applies that inductor current cannot change instantaneously and energy dissipates through resistances; for (c) accurately states all four process conditions and constraintsUses correct general formulas but makes errors in transformation interpretation (e.g., treats E as invariant) or confuses process conditions (e.g., adiabatic with isothermal); partial understanding of energy conservation in (b)Fundamental misconceptions: uses classical velocity addition, ignores relativity of fields entirely, treats inductor as open circuit, or states incorrect work expressions (e.g., W=0 for isobaric)
Derivation rigour20%10For (a): Explicitly decomposes E into parallel/perpendicular components, writes full Lorentz transformation matrix, shows γ calculation; For (b): Derives I = E/R₀, computes U = ½LI², shows heat ratio R/(R+R₀); For (c): Integrates PdV with proper substitution of process equation, shows intermediate steps for each work expressionSkips key algebraic steps or assumes results without derivation; correct final formulas but missing justification for component resolution in (a) or integration limits in (c); arithmetic errors in intermediate stepsNo derivations shown—only final formulas stated; or major errors in derivation logic (e.g., integrates without process equation, confuses initial/final states)
Diagram / FBD15%7.5For (a): Clear vector diagram showing S and S' frames, E and v directions with 45° angle labeled, resulting E' and B' directions; For (b): Circuit diagram with switch position indicated, current direction marked; For (c): Four distinct P-V diagrams with proper curves (hyperbola, steep curve, horizontal, vertical), labeled axes, arrows indicating process direction, and shaded work areasSome diagrams present but incomplete: missing frame labels in (a), or only 2-3 process diagrams in (c) with sketchy curves; axes labeled but no arrows or work indicationNo diagrams despite explicit need for P-V representations; or completely incorrect diagrams (e.g., straight lines for isotherms, inverted axes)
Numerical accuracy25%12.5Precise calculations: γ = 1.25 exact; E' = √[(γEcosα)² + (Esinα)²] = 8.72 kV/m; B' = γ(vEsinα/c²) = 14.5 μT; For (b): I = 1.5 A, U = 2.25 μJ, heat in coil = U×R/(R+R₀) = 0.75 μJ; All values with proper units and significant figuresCorrect method but arithmetic slips (e.g., γ = 1.2, E' ≈ 8.5 kV/m); unit conversion errors (mT instead of μT); correct final formulas with substituted values showing calculation trailOrder-of-magnitude errors; incorrect γ calculation; ignores angle in (a) giving E' = E; wrong resistance combination in (b); no units or inconsistent units throughout
Physical interpretation20%10For (a): Explains that pure E in S appears as E'-B' pair in S'—manifestation of electromagnetic field unity; discusses limit v→c; For (b): Interprets heat as magnetic energy dissipation, notes current decay time constant; For (c): Connects process slopes on P-V diagram to compressibility, discusses practical examples (isothermal: Carnot cycle; adiabatic: sound propagation, diesel engine; isobaric: boiling water; isochoric: constant volume heating), relevance to Indian thermal power plantsBrief mention of physical meaning without elaboration; generic statements about 'energy conservation' or 'thermodynamic efficiency' without specific process connection; no real-world applications citedPurely mathematical treatment with zero physical insight; no interpretation of why fields transform, what heat represents, or how processes differ physically; misses that isochoric work is zero because dV=0

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