Physics 2024 Paper I 50 marks Compulsory Derive

Q1

(a) A particle of mass m kg having an initial velocity V₀ is subjected to a retarding force proportional to its instantaneous velocity. Obtain the expression for the velocity and position of the particle as a function of time. 10 marks (b) Show that the kinetic energy of a system of n particles is given by T = ½ MV²_cm + ½ Σⁿᵢ₌₁ mᵢV'²ᵢ where M is the total mass, V_cm is the velocity of the centre of mass, V'ᵢ is the velocity of the particles about the centre of mass and mᵢ is the mass of the ith particle. 10 marks (c) A charged π-meson with rest mass of 273mₑ at rest decays into a neutrino and a μ-meson of rest mass 207mₑ. Find the kinetic energy of the μ-meson and the energy of the neutrino. (mₑ is the rest mass of the electron) 10 marks (d) The intensity at the central maximum observed on a screen in a double-slit experiment is 2×10⁻³ W/m². If the path difference between interfering waves reaching a point on the screen is λ/6, where λ is the wavelength of the light used in the experiment, determine the intensity at that point. 10 marks (e) A telescope has an objective lens of diameter 10 cm. Determine whether this telescope can resolve two stars having an angular separation of 2·4 seconds of arc. (Assume the wavelength of starlight as 550 nm) 10 marks

हिंदी में प्रश्न पढ़ें

(a) आरंभिक वेग V₀ और द्रव्यमान m kg के एक कण पर एक मंदक बल लगाया जाता है, जो उसके तात्क्षणिक वेग के समानुपाती है। समय के फलन के रूप में कण के वेग और उसकी स्थिति के लिए व्यंजक प्राप्त कीजिए। 10 अंक (b) दर्शाइये कि n कणों के एक निकाय की गतिज ऊर्जा को T = ½ MV²_cm + ½ Σⁿᵢ₌₁ mᵢV'²ᵢ से व्यक्त किया जा सकता है, जहाँ M कुल द्रव्यमान है, V_cm द्रव्यमान केंद्र का वेग है, V'ᵢ द्रव्यमान केंद्र के परितः कणों का वेग है और mᵢ, ith कण का द्रव्यमान है। 10 अंक (c) विराम द्रव्यमान 273mₑ का एक आवेशित π-मेसॉन विरामावस्था में एक न्यूट्रिनो में और विराम द्रव्यमान 207mₑ के एक μ-मेसॉन में क्षयित होता है। μ-मेसॉन की गतिज ऊर्जा और न्यूट्रिनो की ऊर्जा ज्ञात कीजिए। (mₑ इलेक्ट्रॉन का विराम द्रव्यमान है) 10 अंक (d) एक द्वि-स्लिट प्रयोग में स्क्रीन पर प्रेक्षित केंद्रीय उच्चिष्ठ पर तीव्रता 2×10⁻³ W/m² है। यदि स्क्रीन के एक बिंदु पर पहुँची हुई व्यतिकरण करती तरंगों के बीच पथांतर λ/6 है, जहाँ λ प्रयोग में प्रयुक्त प्रकाश का तरंगदैर्घ्य है, तो उस बिंदु पर तीव्रता ज्ञात कीजिए। 10 अंक (e) एक दूरदर्शी के अभिदृश्यक लेंस का व्यास 10 cm है। निर्धारित कीजिये कि क्या यह दूरदर्शी 2·4 सेकंड चाप (आर्क) के एक कोणीय पृथक्कन वाले दो तारों का विभेदन कर सकता है। (तारों के प्रकाश का तरंगदैर्ध्य 550 nm मान लीजिये) 10 अंक

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Directive word: Derive

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How this answer will be evaluated

Approach

Begin with a brief conceptual overview, then systematically solve each sub-part: (a) set up and solve the differential equation for damped motion; (b) prove the König theorem using vector decomposition; (c) apply relativistic energy-momentum conservation for decay; (d) use interference intensity formula with phase difference; (e) apply Rayleigh criterion for resolution. Allocate approximately 15-18 minutes per part, showing all steps clearly.

Key points expected

  • Part (a): Correct setup of differential equation m(dv/dt) = -kv, integration to get v(t) = V₀e^(-kt/m), and x(t) = (mV₀/k)[1 - e^(-kt/m)]
  • Part (b): Decomposition of position vectors as rᵢ = R_cm + r'ᵢ, expansion of kinetic energy, and cancellation of cross term using definition of centre of mass
  • Part (c): Application of energy-momentum conservation in relativistic decay, use of E² = p²c² + m²c⁴, and calculation of kinetic energy as E - mc²
  • Part (d): Phase difference φ = (2π/λ)(λ/6) = π/3, application of I = I_max cos²(φ/2) or I = 4I₀cos²(δ/2) with proper substitution
  • Part (e): Application of Rayleigh criterion θ_min = 1.22λ/D, calculation of resolving power, and comparison with given angular separation
  • Correct handling of units throughout (seconds of arc to radians, nm to m, electron mass units)
  • Physical interpretation of results: exponential decay nature, separation of CM and relative motion, mass-energy conversion, interference pattern modulation, and astronomical resolution limits

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly identifies all physical principles: viscous damping for (a), König's theorem for (b), relativistic kinematics for (c), interference conditions for (d), and diffraction limit for (e); no conceptual errors in any sub-partIdentifies most principles correctly but may confuse classical with relativistic mechanics in (c), or misapply interference formula in (d)Fundamental misconceptions such as treating (c) classically, using wrong interference condition, or applying wrong criterion for resolution
Derivation rigour20%10Complete, step-by-step derivations with proper justification for each step: integration constants evaluated, vector algebra explicit in (b), relativistic invariants used correctly in (c), trigonometric identities justifiedDerivations mostly complete but skips key steps like evaluation of integration constants, or assumes results without proof in (b)Missing derivations, jumps to final formulas, or contains algebraic errors that propagate through calculations
Diagram / FBD15%7.5Clear free-body diagram for (a) showing velocity and opposing force; vector diagram for (b) showing CM and relative velocities; decay schematic for (c); wavefront diagram for (d); telescope ray diagram for (e) with Airy disksSome diagrams present but incomplete or poorly labeled, missing key elements like coordinate systems or angle labelsNo diagrams where needed, or diagrams that misrepresent the physical situation (e.g., wrong force direction)
Numerical accuracy25%12.5All numerical calculations correct: final expressions in (a)-(b), kinetic energy ~4.08 mₑc² and neutrino energy ~66 mₑc² in (c), intensity 1.5×10⁻³ W/m² in (d), and correct comparison showing resolvability (θ_min ≈ 1.34" < 2.4") in (e)Correct approach but arithmetic errors, or correct final answers with intermediate steps unclear; unit conversion errorsMajor calculation errors, wrong order of magnitude, or failure to complete numerical parts in (c), (d), or (e)
Physical interpretation20%10Insightful commentary: characteristic time τ = m/k in (a); physical meaning of CM and internal energy separation in (b); significance of mass difference as energy release in (c); visibility and contrast in (d); practical implications for telescope design like Vainu Bappu Observatory in (e)Brief mention of physical meaning without depth, or interpretation only for some partsNo physical interpretation provided, or incorrect interpretation of results (e.g., negative kinetic energy, unphysical conclusions)

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