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{
"slide": 1,
"fragments": [
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"fragment_index": -1,
"text_description": "Physics Sample Paper Review\nDecode the paper, ace the exam.",
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{
"slide": 2,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question Distribution & Your Strength",
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{
"fragment_index": 1,
"text_description": "Chapter\nMarks (70)\n% of Paper\nYour Proficiency\nElectrostatics\n8\n11 %\nHigh\nCurrent Electricity\n7\n10 %\nMedium\nMagnetism & EMI\n15\n21 %\nLow\nElectromagnetic Waves\n3\n4 %\nHigh\nOptics\n14\n20 %\nMedium\nDual Nature & Matter Waves\n6\n9 %\nLow\nAtoms & Nuclei\n8\n11 %\nHigh\nSemiconductor Devices\n9\n14 %\nMedium",
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{
"fragment_index": 2,
"text_description": "Source: CBSE SQP 2024-25 & Self-assessment log",
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},
{
"slide": 3,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electric Charges & Fields – Challenge\nQ28 (3 marks)\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
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{
"fragment_index": 1,
"text_description": "Problem Statement\nUsing Gauss’s law, derive the magnitude of the electric field \\(E(r)\\) at a radial distance \\(r\\) from an infinitely long straight wire of linear charge density \\(\\lambda\\).\nGiven:\nInfinite straight wire (high symmetry).\nUniform linear charge density \\(\\lambda\\).\nVacuum permittivity \\(\\varepsilon_{0}\\).\nTo Find:\nExpression for \\(E(r)\\) using Gauss’s law and symmetry.",
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{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nGauss law + Cylindrical Symmetry\nChoose coaxial cylinder, use flux only through curved surface, get \\(E=\\frac{\\lambda}{2\\pi\\varepsilon_{0}r}\\).\nComplexity: Direct\n2\nCoulomb Integral\nIntegrate contributions of charge elements along the wire; longer and prone to error.\nComplexity: Tedious\n3\nNumerical Simulation\nDiscretise line charge and compute field numerically; unnecessary for this symmetric case.\nComplexity: High",
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{
"fragment_index": 3,
"text_description": "Logical Breakdown\n1 Symmetry Check\nField is radial and uniform over any circle around the wire.\n2 Gaussian Surface\nChoose cylinder of radius \\(r\\) and length \\(\\ell\\) coaxial with wire.\n3 Flux Calculation\nFlux through ends is zero; curved area flux \\(=E(2\\pi r\\ell)\\).\n4 Apply Gauss Law\nSet flux equal to enclosed charge \\(\\lambda\\ell/\\varepsilon_{0}\\) and solve for \\(E\\).",
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{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nSelect Gaussian cylinder\nRadius \\(r\\), length \\(\\ell\\), axis along wire.\n\\(\\text{Area}_{\\text{curved}} = 2\\pi r\\ell\\)\n2\nCompute electric flux\nEnds contribute zero; curved surface flux \\( \\Phi = E\\,2\\pi r\\ell \\).\n\\(\\Phi = E(2\\pi r\\ell)\\)\n3\nApply Gauss’s law & solve\n\\(\\Phi =\\lambda\\ell/\\varepsilon_{0}\\) ⇒ \\(E=\\lambda/(2\\pi\\varepsilon_{0}r)\\).\n\\(E(r)=\\dfrac{\\lambda}{2\\pi\\varepsilon_{0}r}\\)",
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"fragment_index": 5,
"text_description": "Key Insights\nSymmetry lets us ignore flux through cylinder ends.\nUsing only the curved area prevents the common area mistake.\nAlways keep \\(\\varepsilon_{0}\\) in the final expression.",
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},
{
"slide": 4,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nMoving Charges & Magnetism – Q26\nDifficulty: Medium\nTime: 15 min",
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{
"fragment_index": 1,
"text_description": "Problem Statement\nA +10 mC, 10 g sphere moves east → west inside a vertical insulating tube through a 2 T uniform magnetic field. What minimum speed keeps it level, and which way must the field point?\nGiven:\nCharge \\(q = +10\\,\\text{mC} = 1.0\\times10^{-2}\\,\\text{C}\\)\nMass \\(m = 10\\,\\text{g} = 0.01\\,\\text{kg}\\); \\(g = 9.8\\,\\text{m s}^{-2}\\)\nMagnetic field \\(B = 2\\,\\text{T}\\) (uniform)\nTo Find:\nMinimum speed \\(v_{\\text{min}}\\) and direction of \\(\\vec B\\) that gives upward magnetic force balancing weight (Lorentz force balance).",
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"fragment_index": 2,
"text_description": "Solution Approaches\n1\nForce-balance equation\nSet \\(qvB = mg\\) and solve for \\(v\\).\nComplexity: Algebraic\n2\nVector cross-product check\nUse \\(\\vec F = q(\\vec v \\times \\vec B)\\) to ensure force is upward.\nComplexity: Conceptual\n3\nFleming’s left-hand rule\nQuickly picks the correct field orientation.\nComplexity: Heuristic",
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"fragment_index": 3,
"text_description": "Logical Breakdown\nForces present\nWeight \\(mg\\) down, magnetic force \\(qvB\\) up.\nLorentz expression\n\\(\\vec F = q(\\vec v \\times \\vec B)\\) gives magnitude \\(qvB\\).\nBalance condition\nSet magnitudes equal: \\(qvB = mg\\).\nDirection check\n\\(\\vec v\\) west ⇒ \\(\\vec B\\) south gives \\(\\vec F\\) upward.",
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{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nWrite force balance\nFor equilibrium, upward magnetic force equals weight.\n\\(mg = qvB\\)\n2\nSolve for speed\nInsert values and compute.\n\\(v = \\frac{mg}{qB} = \\frac{0.01 \\times 9.8}{0.01 \\times 2} = 4.9\\,\\text{m s}^{-1}\\)\n3\nFind field direction\nUsing right-hand rule, \\(\\vec B\\) must point south.\nRequired \\(\\vec B\\): Southward",
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{
"fragment_index": 5,
"text_description": "Key Insights\nMagnetic force can exactly counter weight when \\(qvB = mg\\).\nRequired speed falls as charge or field strength rises.\nDirection comes from the \\(\\vec v \\times \\vec B\\) rule, not from gravity.",
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},
{
"slide": 5,
"fragments": [
{
"fragment_index": 1,
"text_description": "Alternating Current – Quick Test\nQ6 (1 mark)\nDifficulty: Medium\nTime: 15 min",
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{
"fragment_index": 2,
"text_description": "Problem Statement\nIn a series \\(LCR\\) circuit the voltages across \\(R\\), \\(L\\) and \\(C\\) are each 10 V (r.m.s.). The capacitor is short-circuited. Calculate the new voltage across the inductor.\nGiven:\nSeries \\(LCR\\) at resonance, so \\(X_L = X_C\\).\n\\(V_R = V_L = V_C = 10\\text{ V}\\).\nSource voltage \\(V_s = 10\\text{ V}\\) (equals \\(V_R\\)).\nTo Find:\nNew inductor voltage \\(V_L'\\) after removing the capacitor.",
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{
"fragment_index": 3,
"text_description": "Solution Approaches\n1\nPhasor diagram\nShow \\(V_R\\) and \\(V_L\\) at right angles after \\(C\\) is removed; apply Pythagoras.\nComplexity: –\n2\nImpedance method\nUse \\(Z=\\sqrt{R^{2}+X_L^{2}}\\), find current \\(I\\), then \\(V_L'=I X_L\\).\nComplexity: –\n3\nQuick check\nResult must be <10 V because current drops when \\(C\\) is gone.\nComplexity: –",
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"fragment_index": 4,
"text_description": "Logical Breakdown\nResonance fact\nAt resonance \\(X_L = X_C\\) and reactive voltages cancel.\nSupply equals \\(V_R\\)\nBecause only \\(V_R\\) is in phase with current.\nFind \\(R\\)\n\\(V_L = I X_L = I R\\) ⇒ \\(R = X_L\\).\nAfter shorting \\(C\\)\nCircuit becomes \\(L\\!-\\!R\\); phasors no longer cancel.",
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{
"fragment_index": 5,
"text_description": "Step-by-Step Solution\n1\nRelate \\(R\\) and \\(X_L\\)\nFrom \\(V_R = V_L\\) and \\(I = V_R/R\\), obtain \\(R = X_L\\).\n\\(R = X_L\\)\n2\nCompute current\nWith \\(C\\) removed, \\(Z = \\sqrt{R^{2}+X_L^{2}} = R\\sqrt{2}\\).\n\\(I' = \\dfrac{V_s}{R\\sqrt{2}} = \\dfrac{I}{\\sqrt{2}}\\)\n3\nFind new \\(V_L\\)\n\\(V_L' = I' X_L = \\dfrac{V_R}{\\sqrt{2}} = \\dfrac{10}{\\sqrt{2}} ≈ 7.1\\text{ V}\\).\n\\(V_L' ≈ 7.1\\text{ V}\\)",
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{
"fragment_index": 6,
"text_description": "Key Insights\nAt resonance, supply voltage equals \\(V_R\\) even when reactive voltages are larger.\nRemoving \\(C\\) converts the series \\(LCR\\) to an \\(L\\!-\\!R\\) circuit; use phasor addition, not arithmetic.\nVoltage division with impedance quickly yields the new component voltage.",
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{
"slide": 6,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nWave Optics – Interference Puzzle | CBSE Grade 12\nDifficulty: Medium\nTime: 15 min",
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{
"fragment_index": 1,
"text_description": "Problem Statement\nIn a Young’s double-slit experiment, coherent light of wavelengths 400 nm and 600 nm falls on the slits. Find the smallest distance from the central bright fringe where a dark fringe for\nboth\nwavelengths would occur, if possible.\nGiven:\n\\(\\lambda_1 = 400\\,\\text{nm}\\)\n\\(\\lambda_2 = 600\\,\\text{nm}\\)\nDouble-slit separation \\(d\\) and screen distance \\(D\\)\nTo Find:\nLeast non-zero fringe distance \\(y\\) where both colours are simultaneously dark, i.e. common beat dark fringe.",
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{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nIntegral-matching path difference\nSolve \\((m_1+\\tfrac12)\\lambda_1 = (m_2+\\tfrac12)\\lambda_2\\) for integers \\(m_1,m_2\\).\nComplexity: O(1)\n2\nLCM of half-wavelengths\nFind the least common multiple of \\(\\tfrac{\\lambda_1}{2}\\) and \\(\\tfrac{\\lambda_2}{2}\\) to predict a candidate \\(\\Delta\\).\nComplexity: O(1)\n3\nBeat dark-fringe reasoning\nAnalyse whether parity permits any common dark; else conclude “no solution”.\nComplexity: Conceptual",
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{
"fragment_index": 3,
"text_description": "Logical Breakdown\n1 | Dark condition\nFor each colour: \\(\\Delta=(m+\\tfrac12)\\lambda\\).\n2 | Common path\nSet \\((2m_1+1)\\lambda_1 = (2m_2+1)\\lambda_2\\).\n3 | Parity test\n\\((2m+1)\\) is odd. Equation demands even = odd ⇒ impossible.\n4 | Result\nNo simultaneous dark fringe forms; beat minima never reach zero.",
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{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nWrite dark conditions\nFor 400 nm and 600 nm: \\( \\Delta_1=(m_1+\\tfrac12)\\lambda_1,\\; \\Delta_2=(m_2+\\tfrac12)\\lambda_2\\).\n\\( \\Delta_1 = (2m_1+1)\\dfrac{\\lambda_1}{2} \\)\n2\nSeek common \\(\\Delta\\)\nEquate: \\((2m_1+1)\\lambda_1 = (2m_2+1)\\lambda_2\\).\n\\( (2m_1+1)400 = (2m_2+1)600 \\)\n3\nParity check & conclusion\nAfter dividing by 200, equation becomes \\(2p = 3q\\) with \\(p,q\\) odd → even = odd, contradiction.\nNo integer solution ⇒ \\(y\\) does not exist.",
"image_description": ""
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{
"fragment_index": 5,
"text_description": "Key Insights\nIn double-slit interference, simultaneous dark needs odd-multiple parity to match.\nFor 400 nm and 600 nm, parity mismatch kills the common beat dark fringe.\nLearning outcome met: we proved no fringe distance satisfies both wavelengths simultaneously.",
"image_description": ""
}
]
},
{
"slide": 7,
"fragments": [
{
"fragment_index": -1,
"text_description": "Visual Image Problem\nNuclear Binding Energy Curve\nDifficulty: Moderate\nEst. Time: 3 min\nPrevious Problem\nProblem 7 of 10\nNext Problem",
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{
"fragment_index": 1,
"text_description": "Problem Context\nThe graph plots binding energy per nucleon \\( \\text{(BE/A)} \\) versus mass number \\( A \\). Points W, X, Y and Z mark specific nuclei.",
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"fragment_index": 2,
"text_description": "",
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"fragment_index": 3,
"text_description": "Question\nUsing the curve, identify the nucleus that favours (a) fission and (b) fusion. Support each choice with the BE/A trend.",
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"fragment_index": 4,
"text_description": "a) Suitable for fission :\nb) Suitable for fusion :",
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"fragment_index": 5,
"text_description": "Helpful Hints\nHint 1\nHeavy nuclei (A ≫ 56) gain BE/A when they split.\nHint 2\nLight nuclei (A ≲ 30) gain BE/A when they merge.\nHint 3\nPeak of curve is near A ≈ 56 (Fe). Reactions moving toward this peak release energy.",
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"fragment_index": 6,
"text_description": "Things to Consider\nCompare BE/A before and after the nuclear reaction.\nEnergy released = increase in total binding energy.\nFission and fusion are opposite nuclear reactions aiming for the same peak.",
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{
"fragment_index": 7,
"text_description": "Related Concepts\nBinding Energy/A\nNuclear Fission\nNuclear Fusion",
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},
{
"slide": 8,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nElectrostatic Potential & Capacitance – Q23\nDifficulty: Medium\nTime: 15 min",
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{
"fragment_index": 1,
"text_description": "Problem Statement\nA parallel-plate capacitor (area \\(A\\), gap \\(d\\)) receives an inserted slab of thickness \\(t<d\\). (i) Dielectric with relative permittivity \\(\\varepsilon_r\\). (ii) Perfect conductor. Derive the new capacitances \\(C_d\\) and \\(C_c\\) and state which is larger.\nGiven:\nParallel-plate capacitor, plates area \\(A\\), separation \\(d\\).\nSlab thickness \\(t<d\\) inserted centrally.\nCase (i) dielectric \\(\\varepsilon_r\\); case (ii) perfect conductor.\nTo Find:\nExpressions for \\(C_d\\) and \\(C_c\\); decide which >.",
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{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nDielectric as Series Combination\nTreat air gap \\((d-t)\\) and dielectric \\((t)\\) as two series capacitors: \\(C_d=\\frac{\\varepsilon_0A}{d-t+t/\\varepsilon_r}\\).\nComplexity: n/a\n2\nConductor as Reduced Gap\nConductor enforces zero field inside; effective separation becomes \\(d-t\\): \\(C_c=\\frac{\\varepsilon_0A}{d-t}\\).\nComplexity: n/a\n3\nComparison\nSince \\(\\varepsilon_r>1\\) ⇒ \\(d-t+t/\\varepsilon_r>d-t\\), hence \\(C_c>C_d\\).\nComplexity: n/a",
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"fragment_index": 3,
"text_description": "Logical Breakdown\nGeometry Split\nCapacitor now has two regions: air \\((d-t)\\) and slab \\((t)\\).\nBasic Formula\nFor any region: \\(C=\\frac{\\varepsilon A}{\\ell}\\).\nSeries Rule\nSeries capacitors: \\(1/C=1/C_1+1/C_2\\).\nConductor Effect\nZero field inside conductor removes \\(t\\) from effective gap.",
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{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nModel Arrangement\nIdentify two dielectrics in series: air \\((\\varepsilon_0)\\) and slab \\((\\varepsilon_0\\varepsilon_r)\\).\n\\(C_1=\\frac{\\varepsilon_0A}{d-t},\\; C_2=\\frac{\\varepsilon_0\\varepsilon_r A}{t}\\)\n2\nCompute \\(C_d\\)\nUse series rule.\n\\(C_d=\\frac{\\varepsilon_0A}{d-t+t/\\varepsilon_r}\\)\n3\nCompute \\(C_c\\) & Compare\nFor conductor, distance is reduced.\n\\(C_c=\\frac{\\varepsilon_0A}{d-t}\\;\\Rightarrow\\;C_c>C_d\\)",
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{
"fragment_index": 5,
"text_description": "Key Insights\nDielectric raises capacitance by lowering effective gap; formula depends on \\(t\\) and \\(\\varepsilon_r\\).\nConducting insert maximises capacitance for the same thickness.\nEffective separation concept simplifies multi-medium capacitor problems.",
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},
{
"slide": 9,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nElectromagnetic Waves – Displacement Current | Q5 (1 mark)\nDifficulty: Easy\nTime: 3 min\nSolution Approaches",
"image_description": ""
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{
"fragment_index": 1,
"text_description": "Problem Statement\nA parallel-plate capacitor (\\(A = 0.001\\,\\text{m}^2,\\;d = 1.0\\times10^{-4}\\,\\text{m}\\)) is charged so that \\(\\frac{dV}{dt}=1.0\\times10^{8}\\,\\text{V s}^{-1}\\). Calculate the displacement current between its plates.\nGiven:\n\\(A = 0.001\\,\\text{m}^2\\)\n\\(d = 1.0 \\times 10^{-4}\\,\\text{m}\\)\n\\(\\frac{dV}{dt}=1.0\\times10^{8}\\,\\text{V s}^{-1}\\)\n\\(\\varepsilon_0 = 8.85\\times10^{-12}\\,\\text{F m}^{-1}\\)\nTo Find:\nDisplacement current \\(I_d\\).",
"image_description": ""
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{
"fragment_index": 2,
"text_description": "1\nMaxwell correction formula\nApply \\(I_d = \\varepsilon_0 \\frac{A}{d}\\frac{dV}{dt}\\).\nComplexity: Direct calc",
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},
{
"fragment_index": 3,
"text_description": "2\nField-rate method\nFind \\(\\frac{dE}{dt}\\) then use \\(I_d = \\varepsilon_0 A \\frac{dE}{dt}\\).\nComplexity: Direct calc",
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},
{
"fragment_index": 4,
"text_description": "3\nFlux-change viewpoint\nRelate to change in electric flux in Ampere-Maxwell law.\nComplexity: Conceptual",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Logical Breakdown\nMaxwell correction\nAdds displacement current term to Ampère’s law.\nLink E & V\n\\(E = \\frac{V}{d}\\) for uniform field between plates.\nRate of change\n\\(\\frac{dE}{dt} = \\frac{1}{d}\\frac{dV}{dt}\\).\nNumeric evaluation\nInsert values to obtain \\(I_d\\).",
"image_description": ""
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{
"fragment_index": 6,
"text_description": "Step-by-Step Solution\n1\nWrite governing formula\nUsing Maxwell correction, displacement current is proportional to changing electric flux.\n\\(I_d = \\varepsilon_0 \\frac{A}{d}\\frac{dV}{dt}\\)\n2\nInsert numerical values\nSubstitute all known quantities.\n\\(I_d = 8.85\\times10^{-12}\\times\\frac{0.001}{1\\times10^{-4}}\\times1\\times10^{8}\\)\n3\nEvaluate\nCalculate the product to obtain the current.\n\\(I_d = 8.85\\times10^{-3}\\,\\text{A} = 8.85\\,\\text{mA}\\)",
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{
"fragment_index": 7,
"text_description": "Key Insights\nMaxwell’s displacement current term ensures Ampère’s law works inside a capacitor gap.\nMagnitude depends on \\(\\varepsilon_0\\), plate geometry, and the rate of voltage change.\nHere \\(I_d = 8.85\\,\\text{mA}\\) equals the conduction current in the external circuit.",
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]
},
{
"slide": 10,
"fragments": [
{
"fragment_index": -1,
"text_description": "Dual Nature – Photoelectron Wavelength\nQ17 (2 marks)\nDifficulty: Medium\nTime: 15 min",
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{
"fragment_index": 1,
"text_description": "Problem Statement\nLight of frequency \\(1.6\\times10^{15}\\,\\text{Hz}\\) falls on platinum \\((\\phi = 5.63\\,\\text{eV})\\). Calculate the minimum de Broglie wavelength of the emitted photoelectrons.\nGiven:\nWork function \\( \\phi = 5.63\\,\\text{eV}\\)\nIncident frequency \\( \\nu = 1.6\\times10^{15}\\,\\text{Hz}\\)\nConstants: \\(h = 6.626\\times10^{-34}\\,\\text{J·s}\\), \\(m_e = 9.11\\times10^{-31}\\,\\text{kg}\\)\nTo Find:\nMinimum de Broglie wavelength \\(\\lambda_{\\text{min}}\\) of photoelectrons.",
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"fragment_index": 2,
"text_description": "Solution Approaches",
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{
"fragment_index": 3,
"text_description": "1\nEinstein Equation\nPhotoelectric effect gives \\(K_{\\text{max}} = h\\nu - \\phi\\).\nMath: Basic algebra",
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},
{
"fragment_index": 4,
"text_description": "2\nElectron Momentum\nUse \\(p = \\sqrt{2m_e K_{\\text{max}}}\\) to link energy to momentum.\nMath: Square-root",
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{
"fragment_index": 5,
"text_description": "3\nde Broglie Relation\nFinally, \\(\\lambda = \\frac{h}{p}\\) gives wavelength.\nMath: Ratio",
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{
"fragment_index": 6,
"text_description": "Logical Breakdown\nPhoton Energy\nCompare \\(h\\nu\\) with work function to initiate photoelectric effect.\nUnit Conversion\nConvert \\(\\phi\\) from eV to joules to keep units consistent.\nElectron Momentum\nRelate \\(K_{\\text{max}}\\) to \\(p\\) using energy–momentum link.\nWave Nature\nApply de Broglie formula to express momentum as wavelength.",
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{
"fragment_index": 7,
"text_description": "Step-by-Step Solution",
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{
"fragment_index": 8,
"text_description": "1\nFind \\(K_{\\text{max}}\\)\n\\(K_{\\text{max}} = h\\nu - \\phi = 1.58\\times10^{-19}\\,\\text{J}\\).\n\\(K_{\\text{max}} = 6.626\\times10^{-34}\\times1.6\\times10^{15}-5.63\\,\\text{eV}\\)",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "2\nCompute Momentum\n\\(p = \\sqrt{2m_eK_{\\text{max}}}=5.4\\times10^{-25}\\,\\text{kg·m s^{-1}}\\).\n\\(p=\\sqrt{2\\times9.11\\times10^{-31}\\times1.58\\times10^{-19}}\\)",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "3\nDetermine Wavelength\n\\(\\lambda_{\\text{min}}=\\frac{h}{p}\\approx1.2\\,\\text{nm}\\).\n\\(\\lambda = \\frac{6.626\\times10^{-34}}{5.4\\times10^{-25}}\\)",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Key Insights\nEinstein equation links photon energy to electron kinetic energy in the photoelectric effect.\nElectron momentum bridges kinetic energy and de Broglie wavelength.\nProper unit conversion avoids major numerical errors.",
"image_description": ""
}
]
},
{
"slide": 11,
"fragments": [
{
"fragment_index": -1,
"text_description": "Semiconductors – Power Supply Blocks\nQ22 (3 marks)\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
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{
"fragment_index": 1,
"text_description": "Problem Statement\nA power-supply diagram shows two unlabeled blocks X and Y in series. Name each block and sketch its output waveform.\nGiven:\nInput: 50 Hz sinusoidal AC.\nBlock X feeds block Y directly.\nOscilloscope available for waveform sketch.\nTo Find:\nLabels of X and Y, plus their respective output waveforms.",
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{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nIdentify Full-Wave Rectifier\nLook for diode bridge/centre-tap giving same-polarity halves.\nConceptual\n2\nSpot Filter Circuit\nSearch for RC/LC components smoothing rectifier output.\nConceptual\n3\nCombine for Low-Ripple DC\nSequence rectification then filtering to obtain usable DC.\nConceptual",
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{
"fragment_index": 3,
"text_description": "Logical Breakdown\nAC Input\nSupplies alternating voltage of both polarities.\nFull-Wave Rectifier (X)\nFlips negative half cycles, giving double-frequency pulses.\nFilter Circuit (Y)\nCapacitor/inductor smooths pulses, reducing ripple.\nPulsating DC Output\nVoltage hovers near steady level, suitable for biasing circuits.",
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{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nLabel X\nX is a full-wave rectifier using diode bridge or centre-tap transformer.\n\\(V_{\\text{out}} = |V_{\\text{in}}|\\)\n2\nLabel Y\nY is an RC or LC filter reducing ripple by charging and slow discharge.\nRipple \\(\\approx \\frac{V_{\\text{p}}}{2fRC}\\)\n3\nDraw Waveforms\nRectifier: double-frequency pulses touching zero. Filtered: nearly horizontal line with small saw-tooth ripple.\nSketch in answer sheet.",
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{
"fragment_index": 5,
"text_description": "Key Insights\nFull-wave rectification doubles output frequency and polarity.\nFilters cannot create pure DC; they only reduce ripple amplitude.\nCorrect waveform sketches secure the final mark.",
"image_description": ""
}
]
},
{
"slide": 12,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nCurrent Electricity – Temperature Effects\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nA \\(100\\text{ V}\\) battery with internal resistance \\(r=1\\:\\Omega\\) drives a heater that draws \\(10\\text{ A}\\) at \\(20^{\\circ}\\text{C}\\). The heater’s temperature rises to \\(320^{\\circ}\\text{C}\\); its temperature coefficient is \\(\\alpha = 3.7\\times10^{-4}\\,^{\\circ}\\text{C}^{-1}\\). Calculate the power wasted inside the battery after the rise.\nGiven:\n\\(E = 100\\text{ V}\\), \\(r = 1\\:\\Omega\\)\nInitial current \\(I = 10\\text{ A}\\) at \\(20^{\\circ}\\text{C}\\)\n\\(\\alpha = 3.7\\times10^{-4}\\,^{\\circ}\\text{C}^{-1}\\), \\(\\Delta T = 300^{\\circ}\\text{C}\\)\nTo Find:\nPower loss \\(P_b\\) inside the battery at \\(320^{\\circ}\\text{C}\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nDirect Calculation\nUse \\(\\alpha\\) to update heater resistance, then find new current and \\(P_b\\).\nComplexity: O(1)\n2\nRatio Method\nCompare total resistance before and after heating to scale current quickly.\nComplexity: O(1)\n3\nGraphical Insight\nPlot I-V curve including \\(r\\) to visualise how temperature shift lowers current.\nComplexity: O(1)",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nTemperature Coefficient\n\\(\\Delta R = R_0\\alpha\\Delta T\\) links temperature rise to resistance change.\nInitial Heater Resistance\n\\(R_0 = \\frac{E}{I} - r\\) uses internal resistance \\(r\\).\nNew Current\n\\(I' = \\frac{E}{R + r}\\) once heater warms.\nBattery Power Loss\n\\(P_b = I'^2 r\\) quantifies wastage inside the battery.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nFind \\(R_0\\)\n\\(R_0 = \\frac{100}{10} - 1 = 9\\:\\Omega\\).\n\\(R_0 = \\frac{E}{I} - r\\)\n2\nUpdate Heater Resistance & Current\n\\(R = 9\\bigl(1 + 3.7\\times10^{-4}\\times300\\bigr) \\approx 10\\:\\Omega\\).\n\\(I' = \\frac{100}{10+1} \\approx 9.09\\:\\text{A}\\).\n\\(R = R_0(1+\\alpha\\Delta T)\\)\n3\nCompute Battery Loss\n\\(P_b = (9.09)^2 \\times 1 \\approx 82.6\\text{ W}\\).\n\\(P_b = I'^2 r\\)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nHeater resistance rises linearly with temperature via the temperature coefficient.\nInternal resistance must be included to get the correct new current.\nLearning outcome achieved: you can now compute current change and battery power loss after a temperature shift.",
"image_description": ""
}
]
},
{
"slide": 13,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nElectromagnetic Induction – AC Generator (Q32 (I))\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nAn \\(N\\)-turn coil of area \\(A\\) rotates with angular speed \\(\\omega\\) in a uniform magnetic field \\(B\\). Explain the working of an AC generator. Using Faraday’s law, derive \\(e = N A B \\omega \\sin \\omega t\\). State the mechanical source of energy converted to electrical form.\nGiven:\nNumber of turns \\(N\\)\nCoil area \\(A\\)\nUniform field \\(B\\) and angular speed \\(\\omega\\)\nTo Find:\nExpression for induced emf and source of energy (mechanical → electrical).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "1\nFlux Method\nRelate changing magnetic flux to emf using Faraday’s law.\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "2\nPhasor View\nTreat coil normal as a rotating vector to show sinusoidal variation.\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "3\nEnergy Perspective\nLink mechanical torque from turbine/engine to electrical output.\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Logical Breakdown",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Magnetic Flux\n\\(\\Phi = B A \\cos \\theta\\) varies as coil rotates.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Faraday’s Law\n\\(e = -N \\frac{d\\Phi}{dt}\\) gives induced emf.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Sinusoidal Form\nWith \\(\\theta = \\omega t\\), emf becomes \\(NAB\\omega \\sin \\omega t\\).",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Energy Conversion\nMechanical work from turbine/engine is converted to electrical energy.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Step-by-Step Solution",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "1\nWrite Flux\nFor angle \\(\\theta\\) between field and coil normal, \\(\\Phi = B A \\cos \\theta\\).\n\\( \\Phi(t) = B A \\cos (\\omega t) \\)",
"image_description": ""
},
{
"fragment_index": 13,
"text_description": "2\nApply Faraday’s Law\nDifferentiate flux to find emf.\n\\( e = -N \\frac{d\\Phi}{dt} = N A B \\omega \\sin (\\omega t) \\)",
"image_description": ""
},
{
"fragment_index": 14,
"text_description": "3\nIdentify Energy Source\nMechanical torque from a turbine or engine maintains rotation, supplying the electrical output power.\nMechanical → Electrical",
"image_description": ""
},
{
"fragment_index": 15,
"text_description": "Key Insights\nFaraday’s law links changing flux to induced emf.\nUniform rotation leads to a pure sinusoidal emf.\nGenerators convert mechanical work into electrical energy—energy is not created.",
"image_description": ""
}
]
},
{
"slide": 14,
"fragments": [
{
"fragment_index": 0,
"text_description": "Analytical Problem Solver\nRay Optics – Telescope Computation | Q33 (I)\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nA telescope has \\(f_o = 15\\,\\text{m}\\) and \\(f_e = 1\\,\\text{cm}\\). \n (a) Find its angular magnification in normal adjustment. \n (b) Compute the diameter of the Moon’s image at the objective. \n Given: Moon’s diameter \\(3.48\\times10^{6}\\,\\text{m}\\); distance \\(3.8\\times10^{8}\\,\\text{m}\\).\nGiven:\n\\(f_o = 15\\,\\text{m}\\)\n\\(f_e = 0.01\\,\\text{m}\\)\n\\(D_\\text{Moon}=3.48\\times10^{6}\\,\\text{m}\\)\n\\(R_\\text{Moon}=3.8\\times10^{8}\\,\\text{m}\\)\nNormal adjustment (final image at ∞)\nTo Find:\n1. Angular magnification \\(M\\)\n2. Image diameter at objective \\(d_i\\)",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "1\nFocal-length ratio\nUse \\(M=\\frac{f_o}{f_e}\\) and small-angle formula \\(d_i=f_o\\theta\\).\nEffort: 1 min",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "2\nAngular definition\nFind object angle \\(\\theta\\), then image angle, verify \\(M=\\theta_i/\\theta\\).\nEffort: 2 min",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "3\nLens-formula cross-check\nApply lens equation for the objective to get same \\(d_i\\).\nEffort: 3 min",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Logical Breakdown",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Angular magnification\nFor normal adjustment \\(M=\\frac{f_o}{f_e}\\).",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Object angle\n\\(\\theta\\approx\\frac{D}{R}\\) (small-angle approximation).",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Image formation\nLinear size: \\(d_i=f_o\\theta\\).",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Unit vigilance\nKeep \\(\\theta\\) in radians; ignore eyepiece in image-size step.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Step-by-Step Solution",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "1\nMagnification\n\\(M=\\frac{15}{0.01}=1500\\).\n\\(M = 1.5\\times10^{3}\\)",
"image_description": ""
},
{
"fragment_index": 13,
"text_description": "2\nAngular size of Moon\n\\(\\theta=\\frac{3.48\\times10^{6}}{3.8\\times10^{8}}\\approx9.2\\times10^{-3}\\,\\text{rad}\\).\n\\(\\theta \\approx 0.0092\\)",
"image_description": ""
},
{
"fragment_index": 14,
"text_description": "3\nImage diameter\n\\(d_i=f_o\\theta=15\\times0.0092\\approx0.14\\,\\text{m}\\approx14\\,\\text{cm}\\).\n\\(d_i \\approx 1.4\\times10^{-1}\\,\\text{m}\\)",
"image_description": ""
},
{
"fragment_index": 15,
"text_description": "Key Insights",
"image_description": ""
},
{
"fragment_index": 16,
"text_description": "Magnification of a refracting telescope equals \\(f_o/f_e\\) when adjusted for infinity.",
"image_description": ""
},
{
"fragment_index": 17,
"text_description": "Image size at the objective depends on object angle and objective focal length only.",
"image_description": ""
},
{
"fragment_index": 18,
"text_description": "Always express small angles in radians to avoid conversion errors.",
"image_description": ""
}
]
},
{
"slide": 15,
"fragments": [
{
"fragment_index": 1,
"text_description": "Atoms – Spectral Line Mystery\nQ14 (Assertion-Reason)\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Problem Statement\nHydrogen holds one electron, yet its discharge tube shows many spectral lines. Explain this using the Bohr model.\nGiven:\n≈10\n23\nhydrogen atoms in the lamp.\nCollisions excite electrons to Bohr orbits n = 2, 3, 4…\nPhoton energy equals \\(E_i-E_f\\) for each electronic transition.\nTo Find:\nReason multiple discrete emission lines appear in the hydrogen spectrum.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Solution Approaches\n1\nBohr energy-level view\nMany atoms occupy n = 2, 3, 4…; each level can later drop.\nNature: Conceptual\n2\nCount transition pairs\nFor highest level n, possible lines \\(N=\\frac{n(n-1)}{2}\\).\nNature: Algebraic\n3\nSeries identification\nGroup transitions ending at same level: Lyman, Balmer, Paschen…\nNature: Organisational",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Logical Breakdown\nMultiple atoms\nEach atom behaves independently.\nExcited populations\nCollisions raise electrons to high n states.\nElectronic transitions\nElectrons drop to lower orbits, emitting photons.\nDistinct energies\nDifferent \\(E_i-E_f\\) values create unique wavelengths.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Step-by-Step Solution\n1\nRecall Bohr levels\nEnergy \\(E_n=-13.6/n^2\\) eV gives discrete orbit values.\n\\(E_n=-\\frac{13.6}{n^2}\\,{\\rm eV}\\)\n2\nIdentify transitions\nFor highest occupied n, ordered pairs (n\ni\n, n\nf\n) define photons.\n\\(N=\\frac{n(n-1)}{2}\\)\n3\nRelate to spectrum\nPhotons from countless atoms overlap, producing many bright lines grouped into series.\nSeries: Lyman (n\nf\n=1), Balmer (2), Paschen (3)…",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Key Insights\nOne-electron atom still owns infinite Bohr energy levels.\nMany atoms + many downward electronic transitions = many spectral lines.\nObserved wavelengths precisely match Bohr model predictions, validating the theory.",
"image_description": ""
}
]
},
{
"slide": 16,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Take-aways",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Formula Recap\nKeep core equations handy: \\(v = u + at\\), \\(F = qvB\\), \\(E = \\frac{hc}{\\lambda}\\), \\(P = \\frac{V^2}{R}\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Units Matter\nThings to remember: write SI units, convert early, and verify dimensional consistency.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Solve Smart\nSketch free-body or circuit, spot symmetry, apply conservation laws to trim algebra & save time.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Time Management\nAim for 1.5 min per mark; if stuck beyond 2 min, flag, move on, return later.",
"image_description": ""
}
]
}
]