Q7
(a) A metal guitar string with a length of 70 cm vibrates at its fundamental frequency of 246.94 Hz in a uniform magnetic field of 10 T oriented perpendicular to the plane of vibration of the string. Assume a sinusoidal form for the amplitude of the vibrational mode, and a maximum displacement of 3 mm at the centre of the string. What is the maximum e.m.f. generated across the length of the guitar string, and at what point in time in the string's motion does that occur? What would be the e.m.f. if the same guitar string vibrates at its second harmonic frequency? Briefly explain. (20 marks) (b) A thermally insulated cylinder, closed at both ends, is fitted with a frictionless heat-conducting piston which divides the cylinder in two parts. Initially, the piston is clamped in the centre, with one litre of air at 200 K and 2 atm pressure on one side and one litre of air at 300 K and 1 atm pressure on the other side. The piston is released and the system reaches equilibrium in pressure and temperature, with the piston at a new position. Compute the final pressure and temperature. (15 marks) (c) A current sheet having $\vec{K} = 9.0 a_y$ A m⁻¹ is located at z = 0. The interface is between the region 1, z < 0, $\mu_{r_1} = 4$, and region 2, z > 0, $\mu_{r_2} = 3$. Given that $\vec{H}_2 = 14.5 a_x + 8.0 a_z$ A m⁻¹. Find $\vec{H}_1$ and $\vec{B}_1$. (15 marks)
हिंदी में प्रश्न पढ़ें
(a) एक धातु के गिटार का तार 70 cm की लंबाई के साथ 246.94 Hz की अपनी मूल आवृत्ति पर 10 T के एकसमान चुंबकीय क्षेत्र, जो कि तार के कंपन-तल के लंबवत् है, में कंपन कर रहा है। माना कि कंपन अवस्था का आयाम ज्यावक्रीय और तार का अधिकतम विस्थापन 3 mm तार के केंद्र में है। गिटार के तार की लंबाई में उत्पन्न अधिकतम वि. बा.ओ. बल क्या है और किस समय तार की गति में ऐसा होता है? यदि वही गिटार का तार अपनी दूसरी संगादी आवृत्ति पर कंपन करता है, तो वि. बा.ओ. बल क्या होगा? संक्षेप में विवेचना कीजिए। (20 अंक) (b) एक ऊष्मारोधी सिलेंडर, जो दोनों सिरों से बंद है, में एक घर्षणहीन ऊष्मा-चालक पिस्टन लगाते हैं जो सिलेंडर को दो भागों में विभाजित करता है। प्रारंभ में, पिस्टन को केंद्र में रोका जाता है, जिसमें एक तरफ एक लीटर हवा 200 K और 2 atm दाब पर और दूसरी तरफ एक लीटर हवा 300 K और 1 atm दाब पर होती है। पिस्टन को छोड़ा जाता है और निकाय दाब तथा तापमान में साम्यावस्था में पहुँच जाता है, साथ में पिस्टन नई स्थिति में आ जाता है। अंतिम दाब और तापमान की गणना कीजिए। (15 अंक) (c) एक धारा पटल z = 0 पर स्थित है, जिसमें $\vec{K} = 9.0 a_y$ A m⁻¹ है। क्षेत्र 1, z < 0, $\mu_{r_1} = 4$ और क्षेत्र 2, z > 0, $\mu_{r_2} = 3$ के बीच अंतरापृष्ठ (इंटरफेस) है। दिया गया है, $\vec{H}_2 = 14.5 a_x + 8.0 a_z$ A m⁻¹। $\vec{H}_1$ और $\vec{B}_1$ का मान ज्ञात कीजिए। (15 अंक)
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Approach
Solve this multi-part numerical problem by allocating approximately 40% of effort to part (a) given its 20 marks, and 30% each to parts (b) and (c). Begin with clear identification of governing equations for each sub-part, proceed through systematic derivation with proper unit handling, and conclude with boxed final answers and brief physical explanations. For (a), explicitly state when maximum emf occurs; for (b), apply energy conservation and ideal gas law; for (c), use magnetic boundary conditions with proper vector notation.
Key points expected
- Part (a): Apply motional emf formula ε = Blv using instantaneous velocity of vibrating string element; recognize velocity is maximum at equilibrium position (y=0) giving maximum emf; calculate second harmonic emf noting frequency doubles but amplitude typically reduces
- Part (a): Correctly integrate over string length with sinusoidal amplitude profile y(x,t) = y_max sin(πx/L)cos(ωt), finding v_max = ωy_max at antinode; emf_max = Bωy_maxL/π for fundamental
- Part (b): Apply first law of thermodynamics for isolated system (ΔU=0); use n₁CvΔT₁ + n₂CvΔT₂ = 0 for temperature equilibrium; apply ideal gas law with total volume constraint for final pressure
- Part (b): Recognize piston is heat-conducting so final temperatures equalize, and frictionless mechanical equilibrium gives equal final pressures; solve simultaneous equations for p_f and T_f
- Part (c): Apply magnetic boundary conditions: normal B continuous (B₁z = B₂z), tangential H discontinuous by surface current (aₙ × (H₂ - H₁) = K); correctly handle vector cross product with K = 9.0 a_y
- Part (c): Calculate H₁x and H₁z components separately using μ₁H₁ = μ₂H₂ for normal component and H₂x - H₁x = K for tangential; then find B₁ = μ₀μ_r₁H₁
Evaluation rubric
| Dimension | Weight | Max marks | Excellent | Average | Poor |
|---|---|---|---|---|---|
| Concept correctness | 20% | 10 | Correctly identifies motional emf mechanism in (a), isolated system thermodynamics in (b), and magnetic boundary conditions in (c); distinguishes between velocity of string element and wave propagation speed; recognizes heat-conducting piston implies thermal equilibrium | Identifies basic concepts but confuses emf generation mechanism (uses changing flux instead of motional emf), or misapplies boundary conditions in (c) by ignoring surface current effect on tangential H | Fundamental conceptual errors such as treating string as rigid conductor moving as whole, applying isothermal instead of adiabatic/isolated conditions in (b), or using electrostatic boundary conditions for magnetic fields |
| Derivation rigour | 20% | 10 | Systematic derivation with explicit steps: (a) integrates v(x,t) = ∂y/∂t properly over string length; (b) sets up mole conservation and energy equations clearly; (c) applies vector boundary condition equations rigorously with proper unit vectors | Derivations present but with gaps: skips integration step in (a) by assuming uniform velocity, or combines gas laws without explicit mole calculation in (b), or makes sign errors in boundary condition application | Missing derivations or logically flawed steps: states formulas without justification, makes algebraic errors in solving simultaneous equations, or incorrectly manipulates vector cross products in boundary conditions |
| Diagram / FBD | 15% | 7.5 | Clear diagrams for (a) showing B field direction, vibration plane, and velocity direction at key instants; for (b) cylinder with piston, initial state labels (p₁,T₁,V₁; p₂,T₂,V₂); for (c) coordinate system with current sheet, field vectors in both regions, and unit normal | Diagrams present but incomplete: missing coordinate axes in (c), or unclear indication of which quantities vary with time in (a), or unlabeled piston position in (b) | No diagrams or seriously flawed ones: incorrect field directions, missing essential elements like the current sheet location, or diagrams that contradict the problem setup |
| Numerical accuracy | 25% | 12.5 | Correct numerical values: (a) emf_max ≈ 0.55 V at equilibrium position, second harmonic emf ≈ 0.28 V (assuming same mechanical energy); (b) p_f ≈ 1.55 atm, T_f ≈ 262 K; (c) H₁ = 5.5 a_x + 6.0 a_z A/m, B₁ = μ₀(22 a_x + 24 a_z) μT; proper significant figures | Correct method but arithmetic errors: order of magnitude correct but factor errors (e.g., missing π in integration, wrong gas constant value, arithmetic in boundary condition algebra) | Order-of-magnitude errors or wrong formulas: emf in millivolts instead of volts, temperature below 200 K or above 300 K in (b), or completely wrong vector components in (c) |
| Physical interpretation | 20% | 10 | Explains (a) emf maximum occurs when string passes equilibrium (maximum velocity, zero displacement); explains (b) final state balances energy conservation with mechanical and thermal equilibrium; explains (c) physical meaning of boundary conditions—normal B continuity from ∇·B=0, tangential H discontinuity from Ampère's law with surface current | States when/what for answers but limited why: mentions equilibrium position without velocity explanation, or states final pressure without energy reasoning, or lists boundary conditions without physical justification | No physical interpretation or incorrect reasoning: claims emf maximum at maximum displacement, or attributes temperature change solely to work without energy conservation, or treats boundary conditions as arbitrary rules |
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