Q5 50M Compulsory calculate
(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 : (10 marks)
(b) A charge of –3·30 μC is fixed at a point. From a horizontal distance of 0·0455 m, a charged particle of mass 7·35 × 10⁻³ kg and charge –7·45 μC is fired with an initial velocity 62·5 m/s directly towards the fixed charge. How far does this charge travel before its speed becomes zero? (10 marks)
(c) In a homogeneous non-conducting region where μᵣ = 1, find εᵣ and ω, if
E = 30π e^{j[ωt – (4/3)y]} a_z (V/m)
H = 1·0 e^{j[ωt – (4/3)y]} a_x (A/m) (10 marks)
(d) What are the limitations of the first law of thermodynamics? One mole of a gas, assumed to be perfect, at 0 °C is heated at constant pressure till its volume is twice its initial value. Calculate the amount of heat absorbed. Given, Cᵥ = 20·9 J mol⁻¹ K⁻¹ and R = 8·3 J mol⁻¹ K⁻¹. (10 marks)
(e) Two solids A and B have Debye temperatures 200 K and 300 K respectively. At T = 20 K, compare their specific heats. (10 marks)
हिंदी में पढ़ें
(a) चित्र में दर्शाए अनुसार प्रतिरोधों के निम्नलिखित संयोजन के प्रभावी प्रतिरोध की गणना कीजिए तथा जब बिंदु A और B के बीच 120 वोल्ट का विभवांतर लगाया जाता है, तो प्रत्येक प्रतिरोध पर वोल्टता पात का निर्धारण कीजिए : (10 अंक)
(b) एक बिंदु पर –3·30 μC का एक आवेश स्थिर है। 0·0455 m की क्षैतिज दूरी से 7·35 × 10⁻³ kg द्रव्यमान और –7·45 μC आवेश वाले एक आवेशित कण को 62·5 m/s के प्रारंभिक वेग से सीधे स्थिर आवेश की ओर प्रक्षेपित किया जाता है। यह आवेश अपनी गति शून्य होने से पहले कितनी दूरी तय करता है? (10 अंक)
(c) एक समांगी अचालक क्षेत्र में, जहाँ μᵣ = 1 है, εᵣ और ω का मान ज्ञात कीजिए, यदि
E = 30π e^{j[ωt – (4/3)y]} a_z (V/m)
H = 1·0 e^{j[ωt – (4/3)y]} a_x (A/m) (10 अंक)
(d) उष्मागतिकी के प्रथम नियम की क्या सीमाएँ हैं? 0 °C पर किसी गैस के एक मोल, जिसे आदर्श गैस माना गया है, को स्थिर दाब पर तब तक गर्म किया जाता है जब तक कि उसका आयतन उसके प्रारंभिक मान का दोगुना न हो जाए। अवशोषित उष्मा की मात्रा की गणना कीजिए। दिया गया है, Cᵥ = 20·9 J mol⁻¹ K⁻¹ और R = 8·3 J mol⁻¹ K⁻¹. (10 अंक)
(e) दो ठोस A और B के डेबाई तापमान क्रमशः 200 K और 300 K हैं। T = 20 K पर उनकी विशिष्ट ऊष्माओं की तुलना कीजिए। (10 अंक)
Answer approach & key points
(a) From the printed figure the candidate must (i) reduce the network between A and B to a single effective resistance R_eff and (ii) with 120 V applied across A-B, compute the main and branch currents and hence the voltage drop across EVERY individual resistor. The network is: from A the upper branch is 3 Ω in parallel with 6 Ω, that pair in series with 18 Ω; the lower branch is a single 5 Ω; the two branches are in parallel between A and node X; and 8 Ω runs from X to B. Both parts carry marks; a bare R_eff is a half answer. (b) Using conservation of mechanical energy for two like (both negative, hence mutually repelling) point charges, find the distance of closest approach of the moving charge and hence how far it travels from its starting point before its speed falls to zero. (c) For the given uniform plane wave in a lossless non-conducting medium with μ_r = 1, determine (i) the relative permittivity ε_r from the intrinsic impedance ratio E₀/H₀ and (ii) the angular frequency ω from the given phase constant β = 4/3 rad/m. (d) Two parts: (i) state the limitations of the first law of thermodynamics (it says nothing about direction/feasibility, no concept of entropy or unavailable energy, cannot distinguish reversible from irreversible, gives no efficiency limit), and (ii) numerically compute the heat absorbed when one mole of an ideal gas at 0 °C is heated isobarically until its volume doubles, given C_V = 20.9 and R = 8.3 J mol⁻¹ K⁻¹. (e) Using the Debye T³ law, compare the lattice specific heats of solids A (θ_D = 200 K) and B (θ_D = 300 K) at T = 20 K, first justifying that 20 K lies in the low-temperature regime for both, and obtaining the ratio C_A/C_B.
- Correct series–parallel reduction: 3 ∥ 6 = 2 Ω, upper branch 20 Ω, 20 ∥ 5 = 4 Ω
- R_eff = 12 Ω and I = 10 A
- Drops: 8 Ω 80 V, 5 Ω 40 V, 18 Ω 36 V, 3 Ω and 6 Ω 4 V each
- Like charges repel; energy conservation ½mv² + kQq/r_i = kQq/r_min
- ½mv² = 14.36 J, U_i = 4.86 J, E = 19.21 J
- r_min = 1.15 × 10⁻² m; distance travelled = 3.40 × 10⁻² m
- η = E₀/H₀ = 30π Ω and η = 120π/√εᵣ ⇒ εᵣ = 16
- β = (ω/c)√εᵣ with β = 4/3 rad/m
- ω = 1.0 × 10⁸ rad/s
- At least three limitations: no direction of processes, no feasibility criterion, no limit on heat-to-work conversion/efficiency, no entropy concept
- V ∝ T at constant P ⇒ ΔT = 273 K; C_P = C_V + R = 29.2 J mol⁻¹ K⁻¹
- Q = C_PΔT = 7.97 × 10³ J
- T = 20 K ≪ θ_D for both ⇒ Debye T³ law C_V = (12π⁴/5)R(T/θ_D)³
- At fixed T, C_V ∝ 1/θ_D³
- C_A/C_B = (300/200)³ = 3.375; A has the larger specific heat
(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. (20 marks)
(b) Blackbody radiation in a cavity at 2000 K is subject to isothermal reversible expansion through 10³ cm³. Calculate (i) the heat transferred and (ii) the work done. If the initial volume was 10 cm³ and expansion had been adiabatic, calculate the change in temperature of the radiation. (Given, Stefan-Boltzmann constant, σ = 5·672×10⁻⁸ J m⁻² K⁻⁴ s⁻¹) (15 marks)
(c) A wire of length 2 m is perpendicular to the X-Y plane. It is moved with velocity V = 2î + 3ĵ + k̂ m/s through a region of uniform magnetic induction B = î + 2ĵ Wb/m². Compute the potential difference induced between the ends of the wire. (15 marks)
हिंदी में पढ़ें
(a) प्रक्षेत्र (डोमेन) सिद्धांत के आधार पर लोहचुम्बकीय (फेरोमैग्नेटिक) पदार्थों में शैथिल्य (हिस्टेरेसिस) की उत्पत्ति की विवेचना कीजिए। समझाइए कि चक्रीय चुम्बकन के दौरान डोमेन भित्ति की गति और पिनिंग किस प्रकार ऊर्जा हानि का कारण बनती हैं। (20 अंक)
(b) 2000 K तापमान पर एक गुहा (कैविटी) में कृष्णिका विकिरण का 10³ cm³ तक समतापी उत्क्रमणीय प्रसार होता है। (i) स्थानांतरित ऊष्मा और (ii) किए गए कार्य की गणना कीजिए। यदि आरंभिक आयतन 10 cm³ था, और प्रसार रुद्धोष्म होता, तो विकिरण के तापमान में होने वाले परिवर्तन की गणना कीजिए। (दिया गया है, स्टेफ़ॉन-बोल्ट्ज़मान स्थिरांक, σ = 5·672×10⁻⁸ J m⁻² K⁻⁴ s⁻¹) (15 अंक)
(c) 2 m लम्बाई का एक तार X-Y तल के लम्बवत् है। इसे V = 2î + 3ĵ + k̂ m/s के वेग से एकसमान चुम्बकीय प्रेरण B = î + 2ĵ Wb/m² वाले क्षेत्र से गुजारा जाता है। तार के सिरों के बीच प्रेरित विभवांतर की गणना कीजिए। (15 अंक)
Answer approach & key points
(a) Explain the microscopic origin of the hysteresis loop in ferromagnets using Weiss domain theory, and then explain in detail how reversible and irreversible domain-wall motion, domain-wall pinning at defects, and Barkhausen jumps convert magnetic work into heat, making the loop area the energy loss per cycle per unit volume. (b) For blackbody (photon-gas) radiation: (i) compute the heat transferred and (ii) the work done in a reversible isothermal expansion at 2000 K through ΔV = 10³ cm³, and then (iii) compute the temperature change if instead the same expansion from an initial volume of 10 cm³ were carried out reversibly and adiabatically. (c) A 2 m wire lying along the z-axis (perpendicular to the X-Y plane) moves with velocity V = 2î + 3ĵ + k̂ m/s through a uniform field B = î + 2ĵ Wb/m². Compute the motional emf induced between its ends using ε = ∫(V × B)·dL.
- Weiss domains from exchange interaction; Bloch walls (exchange vs anisotropy)
- Magnetisation by reversible wall motion, irreversible wall jumps, then domain rotation
- Pinning of walls at defects ⇒ Barkhausen jumps ⇒ irreversibility, remanence and coercivity
- Loss per cycle per unit volume W = ∮H dB = loop area, dissipated as heat
- u = aT⁴, a = 4σ/c = 7.57 × 10⁻¹⁶ J m⁻³ K⁻⁴; P = aT⁴/3; S = (4/3)aT³V
- (i) Q = (4/3)aT⁴ΔV = 1.61 × 10⁻⁵ J
- (ii) W = (1/3)aT⁴ΔV = 4.04 × 10⁻⁶ J
- (iii) VT³ = constant ⇒ T₂ ≈ 429 K, ΔT ≈ −1.57 × 10³ K
- Wire along z: L = 2k̂ m
- ε = (V × B)·L
- V × B = −2î + ĵ + k̂ V/m
- ε = 2 V
(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. (20 marks)
(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 = 4·2×10⁷ ergs/cal and one atmospheric pressure = 10⁶ dynes/cm². (15 marks)
(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 >> 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 dI_s/dt. Also, calculate the power dissipated through Joule effect and the electric field E⃗ near the solenoid. (15 marks)
हिंदी में पढ़ें
(a) एक गैस में केवल दो कण a और b हैं। आरेखों की सहायता से दर्शाइए कि इन दो कणों को तीन ऊर्जा अवस्थाओं 1, 2, 3 में (i) मैक्सवेल-बोल्ट्ज़मान, (ii) फर्मी-डिराक और (iii) बोस-आइंस्टीन सांख्यिकी का उपयोग करके किस प्रकार व्यवस्थित किया जा सकता है। (20 अंक)
(b) उस दाब की गणना कीजिए, जिस पर 150 °C पर पानी उबलेगा, यह देखते हुए कि जब 1 ग्राम पानी भाप में परिवर्तित होता है, तो विशिष्ट आयतन में परिवर्तन 1676 cm³ होता है। दिया गया है, भाप के लिए वाष्पीकरण की गुप्त ऊष्मा = 540 cal प्रति ग्राम, J = 4·2×10⁷ ergs/cal और एक वायुमंडलीय दाब = 10⁶ dynes/cm². (15 अंक)
(c) a त्रिज्या वाली बहुत लम्बी परिनालिका में, जिसकी प्रति इकाई लम्बाई में n फेरे हैं, I_s धारा प्रवाहित हो रही है। परिनालिका के साथ समाक्ष रूप से, b >> a त्रिज्या पर, R प्रतिरोध वाला एक वृत्ताकार तार का वलय (रिंग) स्थित है। जब परिनालिका में प्रवाहित धारा को धीरे-धीरे कम किया जाता है, तो वलय में I_r धारा प्रेरित हो जाती है। I_r की गणना dI_s/dt के पदों में कीजिए। साथ ही, जूल प्रभाव के कारण क्षयित शक्ति और परिनालिका के निकट विद्युत्-क्षेत्र E⃗ की गणना कीजिए। (15 अंक)
Answer approach & key points
(a) For a gas of exactly two particles a and b distributed over three energy states 1, 2, 3, enumerate and DRAW all distinct microstates under (i) Maxwell–Boltzmann, (ii) Fermi–Dirac and (iii) Bose–Einstein statistics, obtaining 9, 3 and 6 arrangements respectively, and explain the assumptions (distinguishability, spin, exclusion principle) that produce the different counts. (b) Using the Clausius–Clapeyron equation for the liquid–vapour transition, compute the pressure at which water boils at 150 °C, given L = 540 cal g⁻¹, Δv = 1676 cm³ g⁻¹, J = 4.2 × 10⁷ erg cal⁻¹ and 1 atm = 10⁶ dyne cm⁻². (c) For a long solenoid (radius a, n turns per unit length, current I_s) coaxial with a resistive ring of radius b ≫ a and resistance R, derive (i) the induced ring current I_r in terms of dI_s/dt, (ii) the Joule power dissipated in the ring, and (iii) the induced electric field E⃗ in the region near (outside) the solenoid.
- MB: distinguishable, multiple occupancy allowed ⇒ gⁿ = 9 arrangements, all shown
- FD: indistinguishable, Pauli exclusion ⇒ ³C₂ = 3 arrangements, all shown
- BE: indistinguishable, no exclusion ⇒ ⁴C₂ = 6 arrangements, all shown
- Clausius–Clapeyron dP/dT = L/[T(v₂ − v₁)] with T = 373 K
- L = 2.268 × 10¹⁰ erg g⁻¹ ⇒ dP/dT = 3.63 × 10⁴ dyne cm⁻² K⁻¹
- ΔP = 1.81 × 10⁶ dyne cm⁻² for ΔT = 50 K
- P ≈ 2.81 × 10⁶ dyne cm⁻² ≈ 2.81 atm
- B = μ₀nIₛ inside, ≈ 0 outside ⇒ flux through ring Φ = μ₀nIₛπa² (not πb²)
- Iᵣ = −(μ₀nπa²/R) dIₛ/dt, sense given by Lenz's law
- P = Iᵣ²R = μ₀²n²π²a⁴(dIₛ/dt)²/R
- E = −(μ₀na²/2s)(dIₛ/dt) φ̂ for s > a (azimuthal)
(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, σ_f, is positive at the bottom plate and negative at the top plate. (20 marks)
(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 (1/3π) 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. (15 marks)
(c) (i) 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. (10 marks)
(ii) Why does the specific heat of solids depend on material at low temperatures but become independent of material at high temperatures? (5 marks)
हिंदी में पढ़ें
(a) एक समांतर-प्लेट संधारित्र की प्लेटों के बीच का स्थान एक परावैद्युत पदार्थ से भरा है, जिसकी सुग्राहिता निचली प्लेट (x = 0) पर 0 से लेकर ऊपरी प्लेट (x = d) पर 1 तक रैखिक रूप से परिवर्तित होती है। संधारित्र को V वोल्टेज वाली एक बैटरी से जोड़ा गया है। सभी बद्ध आवेशों की गणना कीजिए और जाँच कीजिए कि कुल आवेश शून्य है। मान लीजिए कि बैटरी को इस प्रकार जोड़ा गया है कि विद्युत्-क्षेत्र x-दिशा में इंगित करता है, जबकि मुक्त आवेश घनत्व, σ_f, निचली प्लेट पर धनात्मक और ऊपरी प्लेट पर ऋणात्मक है। (20 अंक)
(b) एक विद्युत्-चुम्बकीय तरंग का चुम्बकीय क्षेत्र H मुक्त आकाश में −a_z दिशा में 30 rad/m के कला विस्थापन स्थिरांक और (1/3π) A/m के आयाम के साथ गति करता है। यदि t = 0 और z = 0 पर क्षेत्र की दिशा −a_y हो, तो E और H के लिए उपयुक्त व्यंजक लिखिए। तरंग की आवृत्ति और तरंगदैर्ध्य निर्धारित कीजिए। (15 अंक)
(c) (i) किसी तारकीय वायुमंडल में मौजूद हाइड्रोजन परमाणुओं की औसत गतिज ऊर्जा 1·2 eV है, और यह माना गया है कि वे तापीय संतुलन में हैं। दूसरी उत्तेजित अवस्था (n = 3) में परमाणुओं की संख्या और आद्य (निम्नतम) अवस्था में परमाणुओं की संख्या का अनुपात ज्ञात कीजिए। (10 अंक)
(ii) ठोसों की विशिष्ट ऊष्मा कम तापमान पर पदार्थ पर क्यों निर्भर करती है, जबकि उच्च तापमान पर यह पदार्थ से स्वतंत्र हो जाती है? (5 अंक)
Answer approach & key points
(a) For a parallel-plate capacitor filled with a dielectric of linearly graded susceptibility χ_e(x) = x/d (0 at x = 0, 1 at x = d) and held at voltage V, find the free surface charge density σ_f in terms of V, then compute ALL bound charges — the bound volume charge density ρ_b(x) and the bound surface charge densities at x = 0 and x = d — and verify that the total bound charge is zero. (b) For a plane EM wave in free space travelling along −a_z with β = 30 rad/m, |H| = 1/3π A/m and H along −a_y at t = 0, z = 0: write the complete time-domain expressions for H⃗ and E⃗ (amplitude, direction and phase), and determine the frequency and wavelength. (c(i)) From the given mean kinetic energy of 1.2 eV, obtain the temperature of the stellar hydrogen using the equipartition result (3/2)kT, then apply the Boltzmann distribution with the correct statistical weights g_n = 2n² to find the population ratio N₃/N₁ between the second excited state (n = 3) and the ground state. (c(ii)) Explain why the lattice specific heat of solids is material-dependent at low temperature (Debye T³ regime, where C ∝ (T/θ_D)³ and θ_D encodes the material's elastic constants, atomic mass and bonding) but becomes material-independent at high temperature (Dulong–Petit limit, C = 3R per mole).
- D = σ_f uniform; E(x) = σ_f/[ε₀(1 + x/d)]; V = (σ_f d/ε₀) ln 2 ⇒ σ_f = ε₀V/(d ln 2)
- P = σ_f x/(d + x); ρ_b = −dP/dx = −σ_f d/(d + x)²
- σ_b(0) = 0, σ_b(d) = +σ_f/2
- Total bound charge −σ_f/2 + 0 + σ_f/2 = 0
- ω = βc = 9 × 10⁹ rad/s, f = 1.43 GHz, λ = 0.209 m
- E₀ = η₀H₀ = 40 V/m; phase (ωt + 30z) for −a_z travel
- H = −(1/3π) cos(9 × 10⁹ t + 30z) a_y A/m
- E = 40 cos(9 × 10⁹ t + 30z) a_x V/m (from a_E × a_H = a_k)
- (3/2)k_BT = 1.2 eV ⇒ k_BT = 0.80 eV (T ≈ 9.28 × 10³ K)
- ΔE = E₃ − E₁ = 12.09 eV; g₃/g₁ = 18/2 = 9
- N₃/N₁ = 9 e^{−15.11} ≈ 2.46 × 10⁻⁶
- Only modes with ħω ≲ k_BT contribute
- Low T: Debye T³ law with material-dependent θ_D
- High T: all 3N modes excited, equipartition ⇒ C_V = 3R ≈ 25 J mol⁻¹ K⁻¹ for all solids