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{
"slide": 1,
"fragments": [
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"fragment_index": -1,
"text_description": "CBSE Class XII Physics Sample Paper Review\nMaster every mark while beating the three-hour clock.",
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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 Weightage\nYour Proficiency\nElectrostatics\n08\nMedium\nCurrent Electricity\n07\nHigh\nMagnetism & M.E.C.\n08\nLow\nEM Induction & AC\n08\nMedium\nOptics\n14\nHigh\nDual Nature & Modern Physics\n12\nLow\nCommunication Systems\n03\nMedium",
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{
"fragment_index": 2,
"text_description": "Source: CBSE Sample Paper & your latest mock results",
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}
]
},
{
"slide": 3,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nElectric Charges & Fields – Tough Spot\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
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{
"fragment_index": 1,
"text_description": "Problem Statement\nUsing Gauss’s theorem, derive the electric field \\(E(r)\\) at a distance \\(r\\) from an infinitely long straight wire with uniform linear charge density \\( \\lambda \\).\nGiven:\nInfinite straight wire → cylindrical symmetry\nUniform linear charge density \\( \\lambda \\)\nObservation point at radial distance \\( r \\)\nTo Find:\nMagnitude and direction of \\( E(r) \\) using Gauss theorem.",
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{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nCylindrical Gaussian Surface\nExploit symmetry; field constant on curved surface, zero through caps.\nComplexity: Conceptual\n2\nLine-element Integration\nIntegrate Coulomb’s law along wire; confirms \\(E \\propto 1/r\\).\nComplexity: Higher algebra\n3\nDimensional Check\nVerify units and \\(1/r\\) dependence for linear charge distributions.\nComplexity: Minimal",
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},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nSymmetry\nInfinite length ⇒ field radial, depends only on \\(r\\).\nGaussian Surface\nChoose cylinder of radius \\(r\\) and length \\(L\\).\nFlux Calculation\nFlux \\( \\Phi = E(2\\pi rL) \\); caps contribute zero.\nGauss Theorem\nSet \\( \\Phi = Q_{\\text{enc}}/ \\varepsilon_0 = \\lambda L / \\varepsilon_0\\).",
"image_description": ""
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{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nConstruct Gaussian Cylinder\nRadius \\(r\\), length \\(L\\), axis along wire.\nSurface area: \\( A_{\\text{curved}} = 2\\pi r L \\)\n2\nEvaluate Electric Flux\nCaps perpendicular to field ⇒ no flux. Curved surface gives \\( \\Phi = E\\,2\\pi rL \\).\n\\( \\Phi = E\\,(2\\pi r L) \\)\n3\nApply Gauss Theorem & Solve\nSet \\( E\\,2\\pi rL = \\lambda L/\\varepsilon_0 \\) ⇒ \\( E(r) = \\dfrac{\\lambda}{2\\pi \\varepsilon_0 r} \\) radially outward.\n\\( E(r) = \\frac{\\lambda}{2\\pi \\varepsilon_0 r} \\)",
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{
"fragment_index": 5,
"text_description": "Key Insights\nGauss theorem turns a difficult integral into one line when symmetry exists.\nFor linear charge distributions, \\(E \\propto 1/r\\), not \\(1/r^{2}\\).\nExclude end-caps in flux; including them doubles the result—common error.",
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},
{
"slide": 4,
"fragments": [
{
"fragment_index": 1,
"text_description": "Moving Charges & Magnetism – Radius Change\nCBSE Grade 12 • Physics\nDifficulty: Medium\nEst. Time: 2 min",
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{
"fragment_index": 2,
"text_description": "Problem Context\nAn electron moves in a uniform magnetic field \\( \\mathbf{B} \\). It describes a circle of radius \\( r_0 \\) with period \\( T_0 \\). Its speed is suddenly doubled to \\( 2v_0 \\).",
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"fragment_index": 3,
"text_description": "",
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"fragment_index": 4,
"text_description": "Question\nAfter the speed doubles, calculate the new orbit radius \\( r \\) and time period \\( T \\).\na) Radius \\( r=? \\)\nb) Period \\( T=? \\)",
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{
"fragment_index": 5,
"text_description": "Helpful Hints",
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"fragment_index": 6,
"text_description": "Hint 1\n\\( r=\\frac{mv}{qB} \\); radius is proportional to speed.",
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{
"fragment_index": 7,
"text_description": "Hint 2\nCyclotron period \\( T=\\frac{2\\pi m}{qB} \\) is independent of speed.",
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{
"fragment_index": 8,
"text_description": "Hint 3\nDoubling \\( v \\) ⇒ \\( r=2r_0 \\) while \\( T \\) remains \\( T_0 \\).",
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{
"fragment_index": 9,
"text_description": "Things to Consider\nLorentz force \\( q\\mathbf{v}\\times\\mathbf{B} \\) only changes direction, not speed.\nCyclotron period depends only on \\( m, q, B \\).\nAvoid assuming period varies with velocity.",
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{
"fragment_index": 10,
"text_description": "Related Concepts\nLorentz force\nCyclotron motion\nUniform circular motion",
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},
{
"fragment_index": -1,
"text_description": "Previous Problem\nProblem 4 of 10\nNext Problem",
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}
]
},
{
"slide": 5,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nAlternating Current – LCR Surprise\nDifficulty: Medium\nTime: 15 min",
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{
"fragment_index": 1,
"text_description": "Problem Statement\nA series LCR circuit has \\(V_R = V_L = V_C = 10\\text{ V}\\) at the supply frequency. If the capacitor is short-circuited, determine the new voltage across the inductor.\nGiven:\nSeries LCR at resonance ⇒ \\(V_L = V_C\\) and supply \\(V = V_R\\).\n\\(V_R = V_L = 10\\text{ V}\\).\nFrequency unchanged; hence \\(X_L\\) stays same after change.\nTo Find:\nVoltage \\(V_L'\\) across the inductor after the capacitor is shorted.",
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{
"fragment_index": 2,
"text_description": "Solution Approaches",
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{
"fragment_index": 3,
"text_description": "1\nPhasor Method\nUse phasor triangle: find new current with \\(Z_{RL}\\) then compute \\(V_L' = I'X_L\\).",
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{
"fragment_index": 4,
"text_description": "2\nImpedance Calculation\nTreat circuit as RL series: \\(Z = \\sqrt{R^{2}+X_L^{2}}\\); derive \\(I'\\) from Ohm’s law.",
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{
"fragment_index": 5,
"text_description": "3\nQualitative Reasoning\nRecognise that removing \\(X_C\\) raises impedance by \\(\\sqrt{2}\\); hence current and \\(V_L\\) fall by \\(\\sqrt{2}\\).",
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{
"fragment_index": 6,
"text_description": "Logical Breakdown",
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{
"fragment_index": 7,
"text_description": "Resonant Baseline\nAt resonance: \\(I = V/R\\) and \\(X_L = R\\).",
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{
"fragment_index": 8,
"text_description": "Reactance Relationships\nWith \\(X_C\\) removed, only \\(X_L\\) opposes \\(R\\).",
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{
"fragment_index": 9,
"text_description": "New Impedance\n\\(Z_{RL} = \\sqrt{R^{2}+X_L^{2}} = R\\sqrt{2}\\).",
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{
"fragment_index": 10,
"text_description": "Current Adjustment\n\\(I' = V/Z_{RL} = I/\\sqrt{2}\\).",
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{
"fragment_index": 11,
"text_description": "Step-by-Step Solution",
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{
"fragment_index": 12,
"text_description": "1\nFind \\(R\\) and \\(X_L\\)\nEqual voltages give \\(R = X_L\\).\n\\(I = \\frac{V_R}{R} ,\\quad X_L = R\\)",
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{
"fragment_index": 13,
"text_description": "2\nCompute New Current\nAfter shorting \\(C\\): \\(Z_{RL}=R\\sqrt{2}\\).\n\\(I' = \\frac{10}{R\\sqrt{2}} = \\frac{I}{\\sqrt{2}}\\)",
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{
"fragment_index": 14,
"text_description": "3\nVoltage Across \\(L\\)\n\\(V_L' = I'X_L = \\dfrac{10}{\\sqrt{2}}\\text{ V} \\approx 7.1\\text{ V}\\).\n\\(V_L' = 7.07\\text{ V}\\)",
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{
"fragment_index": 15,
"text_description": "Key Insights\nAt resonance, reactive voltages cancel but each can exceed supply voltage.\nRemoving the capacitor raises impedance by \\(\\sqrt{2}\\) in a series RL network.\nPhasor relations let us quickly recalculate voltages after circuit changes.",
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]
},
{
"slide": 6,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nWave Optics – First Common Dark Fringe\nDifficulty: Medium\nTime: 15 min",
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{
"fragment_index": 1,
"text_description": "Problem Statement\nIn Young’s double-slit experiment, two monochromatic beams of wavelengths \\( \\lambda_1 = 400\\,\\text{nm}\\) and \\( \\lambda_2 = 600\\,\\text{nm}\\) illuminate the same slits. Find the smallest distance \\(y\\) from the central maximum where both wavelengths produce a dark fringe simultaneously.\nGiven:\nYoung’s double-slit set-up (YDS)\n\\( \\lambda_1 = 400\\,\\text{nm},\\; \\lambda_2 = 600\\,\\text{nm}\\)\nSlit separation \\(d\\) and screen distance \\(D\\)\nTo Find:\nNearest common dark-fringe position \\(y\\) (express in \\(D\\) and \\(d\\)).",
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"fragment_index": 2,
"text_description": "Solution Approaches\n1\nPath-difference Equality\nSet a single \\( \\delta\\) satisfying dark-fringe conditions for both wavelengths and solve for \\(y\\).\nComplexity: O(1)\n2\nFringe-width LCM\nFind the lowest common multiple of individual dark-fringe spacings.\nComplexity: O(1)\n3\nGraphical Intersection\nPlot minima positions for each colour and read the first overlap.\nComplexity: O(n)",
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"fragment_index": 3,
"text_description": "Logical Breakdown\nYDS Geometry\nPath difference \\( \\delta = \\dfrac{d\\,y}{D}\\).\nDark-Fringe Rule\nMinima when \\( \\delta = (n+\\tfrac12)\\lambda\\).\nCommon Condition\nSet \\( (n_1+\\tfrac12)\\lambda_1 = (n_2+\\tfrac12)\\lambda_2\\).\nConvert to \\(y\\)\nUse \\( y = \\delta D / d \\) or \\( y = \\dfrac{\\delta}{\\lambda}\\beta\\).",
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{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nSet Equality\nLet integer \\(m\\) satisfy \\( m\\lambda_2 = (m+\\tfrac12)\\lambda_1\\).\n\\( m(\\lambda_2-\\lambda_1)=\\tfrac12\\lambda_1 \\)\n2\nSolve for \\(m\\)\nSubstitute values to get \\( m = 1\\).\n\\( m = \\dfrac{0.5\\times400}{600-400}=1 \\)\n3\nFind \\(y\\)\nPath difference \\( \\delta = m\\lambda_2 = 600\\,\\text{nm}\\). Hence \\( y = \\dfrac{600\\,\\text{nm}\\,D}{d}\\).\n\\( y = \\frac{\\delta D}{d} \\)",
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{
"fragment_index": 5,
"text_description": "Key Insights\nCommon dark fringes satisfy one shared path difference across both wavelengths.\nThe condition converts to a simple integer equation in \\(m\\).\nResult: \\( y = \\dfrac{600\\,\\text{nm}\\,D}{d}\\); independent of chosen integer once minimal.",
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},
{
"slide": 7,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nNuclei – Fission vs Fusion\nDifficulty: Medium\nTime: 15 min",
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{
"fragment_index": 1,
"text_description": "Problem Statement\nUsing the B.E./A curve, predict which of the nuclei W (190), X (90), Y (60) and Z (30) are likely to undergo fission or fusion. Give qualitative reasons.\nGiven:\nB.E./A peaks near mass number A ≈ 56.\nHigher binding energy per nucleon implies greater nuclear stability.\nEnergy is released when total binding energy increases.\nTo Find:\nProcess (fission/fusion) favoured by each nucleus based on nuclear stability.",
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"fragment_index": 2,
"text_description": "Solution Approaches\n1\nRead B.E./A Trend\nLocate each mass number relative to the 56-peak.\nComplexity: Conceptual\n2\nApply Stability Rule\nFor A > 56, fission raises B.E./A; for A < 56, fusion does.\nComplexity: Conceptual\n3\nCompute Energy Gain\nEstimate \\( \\Delta E \\) from change in total binding energy.\nComplexity: Conceptual",
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{
"fragment_index": 3,
"text_description": "Logical Breakdown\nBinding-Energy Peak\nMaximum stability at A ≈ 56 (Fe, Ni).\nHeavy Nuclei (A > 56)\nLower B.E./A; splitting raises stability → fission.\nLight Nuclei (A < 56)\nLower B.E./A; combining raises stability → fusion.\nEnergy Criterion\n\\( \\Delta E \\propto \\Delta(\\text{B.E.}/A)\\,A \\); positive ΔE means feasible.",
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{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nLocate on Curve\nW 190, X 90 → far right of peak; Y 60 → near peak; Z 30 → left of peak.\nA > 56 or A < 56?\n2\nApply Rule\nA > 56 → fission favoured; A < 56 → fusion favoured; near-peak gives negligible gain.\nDecision logic\n3\nState Outcome\n• W 190 & X 90: fission increases binding energy.\n• Z 30: fusion increases binding energy.\n• Y 60: almost no change, so neither strongly favoured.\n\\( \\Delta E \\) positive for W, X (fission) and Z (fusion)",
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{
"fragment_index": 5,
"text_description": "Key Insights\nBinding-energy per nucleon curve predicts nuclear processes at a glance.\nRaising B.E./A drives reactions toward greater nuclear stability.\nInterpretation skill meets the learning outcome: decide fission vs fusion qualitatively.",
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}
]
},
{
"slide": 8,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electrostatic Potential & Capacitance – Inserting Slabs\nGrade 12 Physics\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 a central slab of thickness \\(t<d\\). (i) The slab is dielectric of constant \\(\\kappa\\). (ii) The slab is metallic. Derive the new capacitance in each case and state which configuration gives the larger value.\nGiven:\nPlate area \\(A\\)\nPlate separation \\(d\\)\nInserted slab thickness \\(t\\) (\\(t<d\\))\nDielectric constant \\(\\kappa\\) (case i)\nTo Find:\nCapacitance \\(C_d\\) (dielectric), \\(C_m\\) (metal) and compare.",
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"fragment_index": 2,
"text_description": "Solution Approaches\n1\nSeries-capacitor model\nTreat air gaps and slab as capacitors in series.\nFast & direct\n2\nEnergy comparison\nMinimise stored energy for given charge.\nConceptual check\n3\nField mapping\nAnalyse uniform fields in each region.\nVisual insight",
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{
"fragment_index": 3,
"text_description": "Logical Breakdown\n1. Series capacitors\nAir gaps and slab share charge; voltages add.\n2. Dielectric effect\nField weakens inside slab by factor \\(\\kappa\\).\n3. Metallic slab\nActs as equipotential; eliminates \\(t\\) from gap.\n4. Comparison\nSmaller effective separation ⇒ larger capacitance.",
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{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nDerive \\(C_d\\) (dielectric)\nTwo air gaps of \\((d-t)/2\\) each and one dielectric slab form three capacitors in series.\n\\(C_d=\\dfrac{\\varepsilon_0 A}{d-t+\\dfrac{t}{\\kappa}}\\)\n2\nDerive \\(C_m\\) (metal)\nMetallic slab becomes part of the plates; effective separation reduces to \\(d-t\\).\n\\(C_m=\\dfrac{\\varepsilon_0 A}{d-t}\\)\n3\nCompare\nSince \\(\\kappa>1\\), \\(d-t<d-t+\\dfrac{t}{\\kappa}\\) ⇒ \\(C_m>C_d>C_0\\).\nHighest \\(C\\): metallic slab.",
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{
"fragment_index": 5,
"text_description": "Key Insights\nCapacitance rises when effective plate gap falls.\nDielectric insertion requires series-capacitor treatment, not parallel.\nMetal slab gives greater increase than any dielectric with finite \\(\\kappa\\).",
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},
{
"slide": 9,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nElectromagnetic Waves – Displacement Current\nDifficulty: Medium\nTime: 15 min",
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{
"fragment_index": 1,
"text_description": "Problem Statement\nA \\(0.001\\ \\text{m}^2\\) parallel-plate capacitor, plate gap \\(d = 10^{-4}\\ \\text{m}\\), is charged so that \\(\\frac{dV}{dt}=1\\times10^{8}\\ \\text{V s}^{-1}\\). Find the displacement current between the plates.\nGiven:\nArea \\(A = 0.001\\ \\text{m}^2\\)\nSeparation \\(d = 10^{-4}\\ \\text{m}\\)\n\\(\\frac{dV}{dt}=1\\times10^{8}\\ \\text{V s}^{-1}\\)\n\\(\\varepsilon_0 = 8.85\\times10^{-12}\\ \\text{F m}^{-1}\\)\nTo Find:\nDisplacement current \\(I_d\\) using Ampere-Maxwell law.",
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{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nDirect formula\nUse \\(I_d=\\varepsilon_0 A\\,\\frac{dE}{dt}\\) with \\(E=V/d\\).\nComplexity: Simple algebra\n2\nAmpere-Maxwell loop\nApply \\(\\oint \\mathbf{B}\\cdot d\\mathbf{l}=\\mu_0(I_c+I_d)\\) with \\(I_c=0\\).\nComplexity: Conceptual\n3\nEnergy method\nRelate rate of change of stored energy to equivalent current.\nComplexity: Higher",
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{
"fragment_index": 3,
"text_description": "Logical Breakdown\nDisplacement current idea\nTime-varying electric field inside a capacitor acts like a current in Ampere-Maxwell law.\nField–voltage link\n\\(E=\\frac{V}{d}\\) for uniform field between plates.\nRate of change\n\\(\\frac{dE}{dt}=\\frac{1}{d}\\frac{dV}{dt}\\).\nCompute \\(I_d\\)\nInsert values in \\(I_d=\\varepsilon_0 A\\,\\frac{dE}{dt}\\).",
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"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nRelate \\(E\\) and \\(V\\)\n\\(E=\\frac{V}{d}\\) ⇒ \\(\\frac{dE}{dt}=\\frac{1}{d}\\frac{dV}{dt}\\).\n\\(\\frac{dE}{dt}=\\frac{1}{10^{-4}}\\times1\\times10^{8}=1\\times10^{12}\\ \\text{V m}^{-1}\\text{s}^{-1}\\)\n2\nApply displacement current formula\n\\(I_d=\\varepsilon_0 A\\,\\frac{dE}{dt}\\).\n\\(I_d = 8.85\\times10^{-12} \\times 0.001 \\times 1\\times10^{12}\\)\n3\nNumerical value\nCompute the product.\n\\(I_d = 8.85\\ \\text{A}\\)",
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{
"fragment_index": 5,
"text_description": "Key Insights\nDisplacement current replaces missing conduction current in capacitors.\nAmpere-Maxwell law unifies magnetic effects of both conduction and displacement currents.\nMastering this link lets you apply Maxwell’s correction inside any time-varying dielectric region.",
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},
{
"slide": 10,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nDual Nature – Photoelectron Wavelength\nDifficulty: Medium\nTime: 15 min",
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{
"fragment_index": 1,
"text_description": "Problem Statement\nA platinum surface (\\( \\phi = 5.63\\,\\text{eV} \\)) is illuminated with light of frequency \\( \\nu = 1.6 \\times 10^{15}\\,\\text{Hz} \\). Determine the minimum de Broglie wavelength of the emitted photoelectrons.\nGiven:\nPlanck constant \\( h = 6.63 \\times 10^{-34}\\,\\text{J s} \\)\nElectron mass \\( m_e = 9.11 \\times 10^{-31}\\,\\text{kg} \\)\n\\( 1\\,\\text{eV} = 1.6 \\times 10^{-19}\\,\\text{J} \\)\nTo Find:\nShortest de Broglie wavelength \\( \\lambda_{\\min} \\) of photoelectrons",
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{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nPhotoelectric Equation\nApply \\( h\\nu = \\phi + K_{\\max} \\) to find electron kinetic energy.\nFormula\n2\nConsistent Units\nConvert \\(\\phi\\) and \\(K_{\\max}\\) from eV to joule.\nStep\n3\nMatter-Wave Relation\nUse \\( \\lambda = \\frac{h}{\\sqrt{2m_e K_{\\max}}} \\) to compute \\( \\lambda_{\\min} \\).\nStep",
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{
"fragment_index": 3,
"text_description": "Logical Breakdown\nPhoton Energy\nCalculate \\( E = h\\nu \\).\nWork Function\nEnergy needed to free an electron: \\( \\phi \\).\nKinetic Energy\n\\( K_{\\max} = E - \\phi \\).\nde Broglie Wavelength\n\\( \\lambda = \\frac{h}{\\sqrt{2m_e K_{\\max}}} \\).",
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{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nPhoton energy\n\\( E = h\\nu = 6.63 \\times 10^{-34}\\,\\text{J s} \\times 1.6 \\times 10^{15}\\,\\text{Hz} = 1.06 \\times 10^{-18}\\,\\text{J} \\)\n\\( E = 1.06 \\times 10^{-18}\\,\\text{J} \\)\n2\nMaximum kinetic energy\n\\( \\phi = 5.63\\,\\text{eV} = 9.01 \\times 10^{-19}\\,\\text{J} \\)\n\\( K_{\\max} = E - \\phi = 1.06 \\times 10^{-18} - 9.01 \\times 10^{-19} = 1.59 \\times 10^{-19}\\,\\text{J} \\)\n\\( K_{\\max} = 1.59 \\times 10^{-19}\\,\\text{J} \\)\n3\nde Broglie wavelength\n\\( \\lambda = \\frac{h}{\\sqrt{2m_e K_{\\max}}} = \\frac{6.63 \\times 10^{-34}}{\\sqrt{2 (9.11 \\times 10^{-31})(1.59 \\times 10^{-19})}} \\approx 1.2 \\times 10^{-9}\\,\\text{m} \\)\n\\( \\lambda_{\\min} \\approx 1.2\\,\\text{nm} \\)",
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{
"fragment_index": 5,
"text_description": "Key Insights\nPhoton energy beyond the work function converts to electron kinetic energy.\nKinetic energy sets momentum, linking photoelectric effect to matter waves.\nUnderstanding this link satisfies the learning outcome: relate photon energy to de Broglie wavelength.",
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}
]
},
{
"slide": 11,
"fragments": [
{
"fragment_index": 1,
"text_description": "Semiconductor Electronics – Rectifier & Filter\nIdentify stages that convert AC to smooth DC.\nDifficulty: Medium\nTime: 15 min",
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{
"fragment_index": 2,
"text_description": "Problem Statement\nIn the chain AC → X → Y → load, name X and Y, and illustrate their output waveforms.\nGiven:\nInput: sinusoidal AC mains.\nX contains p–n junction diodes (rectification).\nY uses RC/LC network (filter circuit).\nTo Find:\nStage names and corresponding waveforms at outputs of X and Y.",
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"fragment_index": 3,
"text_description": "Solution Approaches\n1\nHalf-wave rectifier + RC filter\nIdentify X as single-diode rectifier; Y as capacitor across load.\nRipple: high → moderate\n2\nBridge rectifier + π-filter\nUse four-diode bridge for full-wave pulses; π (C-L-C) smooths further.\nRipple: low\n3\nCompare ripple levels\nEvaluate reduction from X to Y to validate correct identification.\nRipple: quantified by \\(r = \\frac{V_{\\text{ac}}}{V_{\\text{dc}}}\\)",
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{
"fragment_index": 4,
"text_description": "Logical Breakdown\nDiode Rectification\nDiode permits current one way, producing pulsating DC from AC.\nHalf vs Full Wave\nFull-wave doubles pulse frequency, reducing initial ripple.\nFilter Circuits\nCapacitor stores charge; inductor resists change, smoothing voltage.\nRipple Reduction\nProper RC/LC choice lowers \\(r\\) without altering frequency.",
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{
"fragment_index": 5,
"text_description": "Step-by-Step Solution\n1\nIdentify X\nX is the rectifier: half-wave (one diode) or full-wave (bridge/center-tap).\nOutput: positive pulses only.\n2\nIdentify Y\nY is the filter circuit: RC, LC or π-section that smooths the pulsating DC.\nOutput: nearly steady DC with small ripple.\n3\nSketch Waveforms\nDraw half/full-wave pulses at X; add capacitor-charged peaks to show smooth line at Y.\nRipple \\(r \\approx \\frac{1}{2fRC}\\) for RC filter.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Key Insights\nRectifier (X) turns AC into unidirectional pulses using diodes.\nFilter (Y) employs capacitors/inductors to cut ripple, not frequency.\nAvoid calling the filter a regulator; regulation is a separate stage.",
"image_description": ""
}
]
},
{
"slide": 12,
"fragments": [
{
"fragment_index": 1,
"text_description": "Analytical Problem Solver\nCurrent Electricity – Heating Element Analysis\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Problem Statement\nA heater connected to a 100 V battery with internal resistance 1 Ω draws 10 A at 20 °C. After heating to 320 °C the current stabilises lower. Find the power lost inside the battery at the final temperature.\nGiven:\nSupply voltage \\(V = 100 \\text{ V}\\)\nInternal resistance \\(r = 1 \\,\\Omega\\)\nInitial current \\(I_0 = 10 \\text{ A at }20^{\\circ}\\text{C}\\)\nFinal temperature \\(T = 320^{\\circ}\\text{C}\\)\nTemperature coefficient \\(\\alpha = 3.7\\times10^{-4}\\,^{\\circ}\\text{C}^{-1}\\)\nTo Find:\nPower dissipated in the battery’s internal resistance at 320 °C.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Solution Approaches",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "1\nDirect α-based Calculation\nFind heater resistance at 20 °C, scale it with \\(R_T = R_0(1+\\alpha\\Delta T)\\), then compute current and \\(P = I^2 r\\).\nSteps : 3",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "2\nRatio Method\nUse \\(I_T/I_0 = 1/\\big(1+\\alpha\\Delta T\\big)\\) to get current directly, then evaluate \\(P = I_T^{2} r\\).\nSteps : 2",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "3\nGraphical Insight\nPlot \\(R\\) versus \\(T\\) to visualise how the temperature coefficient influences current and power loss.\nSteps : —",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Logical Breakdown",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Initial Heater Resistance\n\\(R_0 = \\frac{V}{I_0} - r\\)",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Temperature Coefficient Role\nResistance rises linearly with \\(\\alpha\\Delta T\\).",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Current at 320 °C\nUse \\(I_T = V /(R_T + r)\\).",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Power Loss Inside Battery\nCalculate \\(P = I_T^{2} r\\).",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "Step-by-Step Solution",
"image_description": ""
},
{
"fragment_index": 13,
"text_description": "1\nFind \\(R_0\\)\n\\(R_0 = \\frac{100}{10} - 1 = 9\\,\\Omega\\).\n\\(R_0 = (V/I_0) - r\\)",
"image_description": ""
},
{
"fragment_index": 14,
"text_description": "2\nCompute \\(R_T\\)\n\\(\\Delta T = 300^{\\circ}\\text{C}\\); \\(R_T = 9[1 + (3.7\\times10^{-4})(300)] \\approx 10\\,\\Omega\\).\n\\(R_T = R_0(1+\\alpha\\Delta T)\\)",
"image_description": ""
},
{
"fragment_index": 15,
"text_description": "3\nEvaluate \\(I_T\\) and \\(P_{batt}\\)\nTotal resistance \\(= 10 + 1 = 11\\,\\Omega\\); \\(I_T = 100/11 \\approx 9.09 \\text{ A}\\). Power loss \\(P_{batt} = (9.09)^2 \\times 1 \\approx 83 \\text{ W}\\).\n\\(P_{batt} = I_T^{2} r\\)",
"image_description": ""
},
{
"fragment_index": 16,
"text_description": "Key Insights\nTemperature coefficient raises heater resistance, lowering current.\nInternal power loss is governed by \\(I^{2} r\\), not by battery voltage directly.\nAccounting for temperature-dependent resistance prevents under-estimating power loss in circuits.",
"image_description": ""
}
]
},
{
"slide": 13,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nElectromagnetic Induction – AC Generator Emf\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nDerive the instantaneous emf \\( \\varepsilon(t) \\) of an \\( N \\)-turn coil of area \\( A \\) rotating with angular speed \\( \\omega \\) in a uniform magnetic field \\( B \\).\nGiven:\nGenerator principle: rotating coil cuts magnetic flux.\nUniform field \\( B \\); coil area \\( A \\); turns \\( N \\).\nAngular speed \\( \\omega \\); initial angle \\( \\theta_0 = 0 \\).\nTo Find:\nExpression \\( \\varepsilon(t) = \\varepsilon_0 \\sin \\omega t \\) where \\( \\varepsilon_0 = N B A \\omega \\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nFlux–Derivative Method\nWrite magnetic flux, differentiate using Faraday’s law.\nComplexity: Conceptual\n2\nPhasor View\nTreat flux as rotating vector; emf is its vertical component.\nComplexity: Visualization\n3\nEnergy Perspective\nRelate mechanical power \\( \\tau \\omega \\) to electrical power \\( \\varepsilon I \\).\nComplexity: Qualitative",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nGenerator Principle\nRotation converts mechanical work into emf via changing flux.\nMagnetic Flux\n\\(\\Phi = N B A \\cos \\theta\\).\nRotational Emf\nTime-varying angle \\( \\theta = \\omega t \\) gives sinusoidal flux.\nSign & Phase\nNegative sign sets phase; dropping it shifts waveform by \\( \\pi \\).",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nWrite Flux\nAngle between \\( \\mathbf{B} \\) and normal is \\( \\theta = \\omega t \\).\n\\( \\Phi(t) = N B A \\cos \\omega t \\)\n2\nApply Faraday’s Law\nInduced emf equals negative rate of change of flux.\n\\( \\varepsilon(t) = -\\frac{d\\Phi}{dt} \\)\n3\nDifferentiate & Simplify\nDerivative of cosine gives sine; amplitude is \\( N B A \\omega \\).\n\\( \\varepsilon(t) = N B A \\omega \\sin \\omega t \\)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nSinusoidal emf emerges directly from rotational motion plus Faraday’s law.\nAmplitude \\( \\varepsilon_0 = N B A \\omega \\) scales with turns, field, area, and speed.\nCorrect sign ensures correct phase; dropping it shifts the waveform by 180°.",
"image_description": ""
}
]
},
{
"slide": 14,
"fragments": [
{
"fragment_index": -1,
"text_description": "Ray Optics – Cassegrain Telescope\nCBSE Grade 12 Physics\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nDraw a neat ray diagram of a Cassegrain reflecting telescope and state any two reasons it outperforms a similar-sized refracting telescope.\nGiven:\nPrimary concave parabolic mirror with central hole\nSecondary small convex mirror facing the primary\nEyepiece placed behind the primary mirror\nTo Find:\nComplete ray path and two performance advantages (compact length, zero chromatic aberration).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nTrace Rays\nShow two reflections and final focus at eyepiece.\nComplexity: N/A\n2\nIdentify Components\nLabel mirrors, focal points and central hole.\nComplexity: N/A\n3\nCompare Designs\nList reflector advantages over refractor.\nComplexity: N/A",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nOptical Path\nParallel light → primary mirror → convex secondary → through central hole → eyepiece.\nTube Length\nFolded path gives long focal length inside a short, manageable tube.\nAberrations\nAll-mirror system removes chromatic dispersion found in lenses.\nCommon Errors\nEyepiece must be behind primary; include hole in primary mirror.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nSketch Mirrors\nDraw large concave primary (with hole) and small convex secondary near its focus.\nRay diagram\n2\nTrace Incident Rays\nShow parallel rays reflecting at primary, converging to secondary, then passing through hole to focus.\nRay paths\n3\nNote Advantages\nWrite: (i) No chromatic aberration (ii) Large aperture with compact length.\nAdvantages list",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nReflection is wavelength-independent → no chromatic aberration.\nLarge parabolic mirror gathers more light than an equal-diameter lens and is lighter.\nSecondary mirror folds path, giving long focal length in a short tube.",
"image_description": ""
}
]
},
{
"slide": 15,
"fragments": [
{
"fragment_index": 1,
"text_description": "Analytical Problem Solver\nAtoms – Many Lines, One Electron\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Problem Statement\nHydrogen has only one electron, yet its emission spectrum shows many lines. Explain why.\nGiven:\nSample contains about \\(10^{23}\\) hydrogen atoms.\nCollisions excite electrons to various Bohr energy levels.\nBohr model allows discrete electron transitions.\nTo Find:\nReason for multiple spectral lines from a one-electron atom.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Solution Approaches\n1\nMany Atoms, Many States\nLarge ensemble ensures electrons occupy several excited levels simultaneously.\nComplexity: —\n2\nCascade Transitions\nEach electron can drop through several energy gaps, emitting photons of different energies.\nComplexity: —\n3\nBohr Quantisation\nEnergy difference \\(E_{n_i}-E_{n_f}=h\\nu\\) sets a unique wavelength for every allowed pair \\((n_i,n_f)\\).\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Logical Breakdown\n1 . Ensemble Size\nDifferent atoms act independently, adding their emissions.\n2 . Excitation\nCollisions lift electrons to \\(n>1\\) states.\n3 . Multiple Drops\nEach excited electron can follow many downward paths.\n4 . Distinct Energies\nEnergy gaps differ, so photons have different wavelengths.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Step-by-Step Solution\n1\nQuantised Energies\nBohr gives \\(E_n=-13.6\\,\\text{eV}/n^2\\).\n\\(E_n=-\\dfrac{13.6}{n^2}\\,\\text{eV}\\)\n2\nPopulation of Levels\nThermal or electrical collisions move electrons to higher \\(n\\).\n\\(n_i = 2,3,4,\\dots\\)\n3\nEmission of Photons\nWhen an electron drops from \\(n_i\\) to \\(n_f\\), photon energy \\(h\\nu = E_{n_i}-E_{n_f}\\).\n\\(\\lambda = \\dfrac{hc}{E_{n_i}-E_{n_f}}\\)",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Key Insights\nMultiple atoms and levels create a rich emission spectrum.\nEach Bohr transition yields one precise wavelength.\nMistakes: blaming extra electrons or forgetting ensemble effects.",
"image_description": ""
}
]
},
{
"slide": 16,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Take-aways",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Formula Vault\nRecap core equations: \\(v=u+at\\), \\(F=q(E+v\\times B)\\), lens & capacitor rules for instant recall.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Units & Constants\nWrite \\(c, h, e\\) values on rough sheet. Convert units before substitution to avoid negative marking.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Strategic Scan\nRead twice, underline data, decide concept, then pick a formula. This strategy cuts re-work.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Time Tactics\n60 min theory, 30 min numericals, 10 min review. Skip tough items and return—classic time-saving tip.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Neat Work\nBox answers, show steps, draw labelled diagrams. Presentation secures method marks.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Final Review\nThings to remember: tally sub-parts, check sig-figs, use elimination for MCQs in the last sweep.",
"image_description": ""
}
]
}
]