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{
"slide": 1,
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"text_description": "Question Distribution & Your Strength\nUnit / Chapter\nMarks\nWeight %\nYour Strength\nElectrostatics (Ch 1-2)\n11\n16 %\nStrong\nCurrent Electricity (Ch 3)\n7\n10 %\nModerate\nMagnetism (Ch 4-5)\n10\n14 %\nNeeds Work\nEMI & AC (Ch 6-7)\n11\n16 %\nModerate\nEM Waves (Ch 8)\n3\n4 %\nStrong\nOptics (Ch 9-10)\n12\n17 %\nStrong\nDual Nature + Atoms + Nuclei (Ch 11-12)\n9\n13 %\nNeeds Work\nSemiconductors & Comms (Ch 13-14)\n7\n10 %\nModerate\nSource: CBSE Physics Sample Question Paper 2024-25 & Self-assessment survey",
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{
"slide": 2,
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"fragment_index": -1,
"text_description": "Electric Charges and Fields – Toughest Question\nLearning outcome: Apply Gauss’s law confidently to symmetric charge distributions.",
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"fragment_index": 1,
"text_description": "1\nQuestion Stem\nUsing Gauss’s law, derive the electric field near a uniformly charged infinite plane sheet of surface charge density \\( \\sigma \\).",
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"fragment_index": 2,
"text_description": "2\nSolution Strategy\nPlace a symmetrical “pillbox” cylinder of face area \\( A \\) cutting the sheet. Field is normal and equal on both sides, so flux passes only through the two flat faces.",
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"fragment_index": 3,
"text_description": "3\nDerive Field\nGauss’s law: \\( 2EA = \\dfrac{\\sigma A}{\\varepsilon_0} \\) ⇒ \\( E = \\dfrac{\\sigma}{2\\varepsilon_0} \\). Direction: outward, opposite on the two faces.",
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{
"fragment_index": 4,
"text_description": "Pro Tip:\nDo not forget both faces of the pillbox. Omitting one doubles the answer to \\( \\sigma/\\varepsilon_0 \\).",
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}
]
},
{
"slide": 3,
"fragments": [
{
"fragment_index": -1,
"text_description": "Moving Charges & Magnetism – Toughest Question\n1\n2\n3",
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{
"fragment_index": 1,
"text_description": "Question Stem\nA 10 g sphere carrying +10 mC moves freely inside a vertical pipe. The pipe is dragged westward through a 2 T uniform field. Find the minimum speed \\(v\\) so that magnetic force balances the weight, and state the direction of \\(\\mathbf B\\).",
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"fragment_index": 2,
"text_description": "Solution Strategy\nBalance forces: \\(qvB = mg\\) (90° between \\( \\mathbf v \\) and \\( \\mathbf B \\)).\nInsert values: \\(v = \\dfrac{0.01 \\times 9.8}{0.01 \\times 2} = 4.9\\ \\text{m s}^{-1}\\).\nUse Fleming’s left-hand rule: force needed upward, velocity West → \\( \\mathbf B \\) must point North.",
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},
{
"fragment_index": 3,
"text_description": "Common Errors\n• Forgetting the \\( \\sin\\theta =1 \\) term when \\( \\mathbf v \\perp \\mathbf B \\).\n• Using grams or millicoulombs directly—always convert to kg and C.\n• Choosing wrong field direction by applying the hand rule for a negative charge.",
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{
"fragment_index": 4,
"text_description": "Pro Tip:\nWhen magnetic force equals weight, you can quickly check units: N = C × m s⁻¹ × T, ensuring your substitutions are consistent.",
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]
},
{
"slide": 4,
"fragments": [
{
"fragment_index": -1,
"text_description": "Alternating Current – Toughest Question\nGoal: visualise how the inductor phasor changes when the capacitor is removed.\n1\nQuestion Stem\nA series LCR circuit has \\(V_R = V_C = V_L = 10\\text{ V}\\). If the capacitor is short-circuited, find the new voltage across the inductor.\n2\nSolution Strategy\nAt resonance, \\(V_s = V_R = 10\\text{ V}\\) and \\(X_L = R\\).\nAfter shorting C, impedance \\(Z = \\sqrt{R^2 + X_L^2}=R\\sqrt{2}\\).\nCurrent \\(I' = V_s/Z = 10/(R\\sqrt{2})\\).\nHence \\(V_L' = I'X_L = 10/\\sqrt{2}\\text{ V} \\approx 7.1\\text{ V}\\).\n3\nCommon Errors\n• Adding 10 V + 10 V algebraically and getting 20 V.\n• Assuming current stays 10 V / R and predicting \\(10\\sqrt{2}\\) V.\nAlways use phasors and recalculate current after altering the circuit.\nPro Tip:\nRotate the phasor diagram: removing the capacitor drops \\(V_L\\) by \\(1/\\sqrt{2}\\) and shifts it to 90° ahead of \\(V_R\\).",
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},
{
"slide": 5,
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{
"fragment_index": -1,
"text_description": "Wave Optics – Toughest Question\nPro Tip:\nPredict fringe positions by writing one clear equation for path difference—this technique works for any multi-wavelength interference problem.",
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{
"fragment_index": 1,
"text_description": "1\nQuestion Stem\nIn Young’s double-slit set-up, wavelengths \\(400\\,\\text{nm}\\) and \\(600\\,\\text{nm}\\) interfere. Find the nearest distance from the central maximum where a dark fringe appears.",
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"fragment_index": 2,
"text_description": "2\nSolution Strategy\nCommon dark fringe occurs when \\(m\\lambda_{600}= \\bigl(m+\\tfrac12\\bigr)\\lambda_{400}\\). Smallest integer \\(m=1\\) gives path difference \\(=600\\,\\text{nm}\\). Hence distance from centre \\(y=\\dfrac{600D}{d}= \\dfrac{3}{2}\\beta\\) where \\(\\beta=\\dfrac{400D}{d}\\). Predict fringe position quickly by comparing path differences, not fringe orders.",
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{
"fragment_index": 3,
"text_description": "3\nCommon Errors\nStudents often equate fringe numbers instead of path differences, mix up bright-dark conditions, or forget that \\(D\\) and \\(d\\) cancel, leaving the answer in terms of \\(\\beta\\).",
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},
{
"slide": 6,
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{
"fragment_index": -1,
"text_description": "Nuclei – Toughest Question\nGoal: Use binding-energy trends to predict whether a given nucleus prefers fission or fusion.\n1\n2\n3\nPro Tip:\nEnergy is released whenever a reaction moves a nucleus toward the iron-56 peak on the BE/A curve.",
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"fragment_index": 1,
"text_description": "Question Stem\nFrom the binding-energy-per-nucleon graph, decide which of W, X, Y, Z is likely to undergo (i) fission and (ii) fusion, with reasoning.",
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"fragment_index": 2,
"text_description": "Solution Strategy\nLow-BE/A heavy nuclei (left of iron-56) gain energy by splitting; low-BE/A light nuclei (far right) gain energy by merging. Therefore, W (A≈190) favours fission, while Z (A≈30) favours fusion.",
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"fragment_index": 3,
"text_description": "Common Errors\nPicking the nucleus with the highest BE/A (X or Y) – these are already near the peak, so neither fission nor fusion releases extra energy. Also, confusing the heavy and light sides of the curve.",
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},
{
"slide": 7,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electrostatic Potential & Capacitance – Toughest Question\nGoal: Analyse capacitors with a partial dielectric or conductor and predict which raises capacitance more.\n1\nQuestion Stem\nFor a parallel-plate capacitor \\(A, d\\): derive \\(C\\) when (a) a dielectric slab of thickness \\(t\\) and constant \\(k\\), (b) a metal slab of thickness \\(t<d\\) is inserted. Which yields larger \\(C\\)?\n2\nSolution Strategy\nModel two regions in series. Dielectric: \\(C=\\varepsilon_0 A /(d-t+t/k)\\). Metal: gap becomes \\(d-t\\), so \\(C=\\varepsilon_0 A /(d-t)\\). Because \\(d-t < d-t+t/k\\), the metal slab gives the higher capacitance.\n3\nCommon Errors\nTreating the two regions as parallel, skipping the \\(t/k\\) term, or assuming a dielectric always boosts \\(C\\) more than a conductor.\nPro Tip:\nCapacitance rises fastest when you shorten the effective gap; a metal insert beats increasing permittivity.",
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{
"slide": 8,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electromagnetic Waves – Toughest Question\nGoal: compute displacement current in a capacitor when voltage varies with time.\n1\n2\n3\nPro Tip:\nUse the on-screen slider for \\(dV/dt\\); watch \\(I_d\\) grow linearly to cement the concept.",
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{
"fragment_index": 1,
"text_description": "Question Stem\nParallel-plate capacitor: \\(A = 1.0\\times10^{-3}\\,\\text{m}^2\\), \\(d = 1.0\\times10^{-4}\\,\\text{m}\\). Voltage rises at \\(dV/dt = 1.0\\times10^{8}\\,\\text{V\\,s}^{-1}\\). Find the displacement current \\(I_d\\).",
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{
"fragment_index": 2,
"text_description": "Solution Strategy\nUse \\(I_d = \\varepsilon_0 A \\frac{1}{d}\\frac{dV}{dt}\\). Substitute \\(\\varepsilon_0 = 8.85\\times10^{-12}\\,\\text{F m}^{-1}\\). Hence \\(I_d = 8.85\\times10^{-4}\\,\\text{A}\\) (0.885 mA).",
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{
"fragment_index": 3,
"text_description": "Common Errors\n• Omitting the \\(1/d\\) factor. • Mixing SI units and prefixes. • Using \\(I_d = C\\,dV/dt\\) without first finding \\(C\\), causing power-of-ten slips.",
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},
{
"slide": 9,
"fragments": [
{
"fragment_index": -1,
"text_description": "Dual Nature of Radiation & Matter – Toughest Question\nGoal: Interpret photoelectric I-V graphs to connect saturation current with intensity and stopping potential with frequency.",
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{
"fragment_index": 1,
"text_description": "1\nQuestion Stem\nThree photoelectric I–V curves A, B and C are shown. Identify which curves share equal light intensity and which share equal photon frequency.",
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},
{
"fragment_index": 2,
"text_description": "2\nSolution Strategy\nCheck saturation current: identical heights ⇒ equal intensity. Check stopping potential \\(V_s\\): identical intercepts ⇒ equal frequency. Result: A & B same intensity; B & C same frequency.",
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},
{
"fragment_index": 3,
"text_description": "3\nCommon Errors\nDo not use slope or initial current to judge intensity, and never link frequency to curve gradient; only saturation current and \\(V_s\\) matter.",
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},
{
"fragment_index": 4,
"text_description": "Pro Tip:\nLabel the axes first; it prevents mixing up intensity (current axis) with energy (voltage axis).",
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}
]
},
{
"slide": 10,
"fragments": [
{
"fragment_index": -1,
"text_description": "Semiconductor Electronics – Toughest Question\nGoal: predict how centre-tap shift alters a full-wave rectifier output.\nPro Tip:\nFull-wave rectifiers always double the input frequency; centre-tap shift only skews amplitudes and ripple, not frequency.",
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"fragment_index": 1,
"text_description": "1\nQuestion Stem\nIdentify blocks X & Y in a centre-tapped full-wave rectifier, sketch load waveforms, and predict change when the centre tap shifts toward D1.",
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"fragment_index": 2,
"text_description": "2\nSolution Strategy\nX = two diodes (D1 & D2). Y = load R\nL\n(plus filter C). Ideal output: equal positive half-cycles, frequency 2f. Moving tap toward D1 makes secondary voltages unequal—pulses via D1 peak higher, via D2 lower—raising ripple and lowering average DC.",
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{
"fragment_index": 3,
"text_description": "3\nCommon Errors\nDrawing negative half-cycles—output is always positive.\nThinking frequency reverts to f; it stays at 2f.\nIgnoring drop in average DC with asymmetry.",
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}
]
},
{
"slide": 11,
"fragments": [
{
"fragment_index": -1,
"text_description": "Current Electricity – Toughest Question\nPro Tip:\nCurrent changes when resistance changes. Keep the source voltage and internal resistance in every recalculation.",
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{
"fragment_index": 1,
"text_description": "1\nRead the Stem\nA \\(100\\text{ V}\\) battery with \\(r = 1\\,\\Omega\\) sends \\(10\\text{ A}\\) through a heater at \\(20^{\\circ}\\text{C}\\). At \\(320^{\\circ}\\text{C}\\) find power lost inside the battery. Given \\(\\alpha = 3.7\\times10^{-4}\\,^{\\circ}\\text{C}^{-1}\\).",
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"fragment_index": 2,
"text_description": "2\nUpdate Heater Resistance\nInitial \\(R_0 = V/I = 10\\,\\Omega\\). Temperature rise \\(\\Delta T = 300^{\\circ}\\text{C}\\). New \\(R = R_0\\bigl(1+\\alpha\\Delta T\\bigr)=10\\!\\left(1+3.7\\times10^{-4}\\times300\\right)=11.11\\,\\Omega\\).",
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"fragment_index": 3,
"text_description": "3\nFind Circuit Current\nTotal resistance \\(R_{\\text{tot}} = R + r = 11.11 + 1 = 12.11\\,\\Omega\\). Current \\(I = \\dfrac{100}{12.11} \\approx 8.26\\text{ A}\\).",
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"fragment_index": 4,
"text_description": "4\nBattery Power Loss\nInternal dissipation \\(P = I^{2}r = (8.26)^2 \\times 1 \\approx 68\\text{ W}\\). Learning outcome achieved: temperature change & internal resistance combined.",
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}
]
},
{
"slide": 12,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electromagnetic Induction – Toughest Question",
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{
"fragment_index": 1,
"text_description": "1\nQuestion Stem\nAC generator: N-turn coil (area A) spins with angular speed ω in field B. Derive \\( \\varepsilon(t) \\) and cite the energy source.",
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"fragment_index": 2,
"text_description": "2\nSolution Strategy\nFlux: \\( \\Phi = N B A \\cos \\omega t \\).\nDifferentiate to get \\( \\varepsilon = -\\dfrac{d\\Phi}{dt} = N B A \\omega \\sin \\omega t \\).\nPeak emf \\( \\varepsilon_0 = N B A \\omega \\).\nMechanical work of the prime mover becomes electrical energy.",
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{
"fragment_index": 3,
"text_description": "3\nCommon Errors\nOmitting the minus sign or using cos instead of sin in \\( \\varepsilon(t) \\) breaks the correct phase relation.",
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},
{
"fragment_index": 4,
"text_description": "Pro Tip:\nRotational change in magnetic flux produces a sinusoidal emf—remember this link whenever you analyse generators.",
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}
]
},
{
"slide": 13,
"fragments": [
{
"fragment_index": -1,
"text_description": "Ray Optics & Optical Instruments – Toughest Question\n1\nQuestion Stem\nTelescope with \\(f_o = 15\\,\\text{m}\\) and \\(f_e = 1\\,\\text{cm}\\). (a) Find angular magnification in normal adjustment. (b) Determine the Moon’s image diameter formed by the objective.\n2\nSolution Strategy\nUse \\(M = f_o / f_e\\). So \\(M = 15\\,\\text{m} / 0.01\\,\\text{m} = 1500\\). Moon’s angular size \\( \\theta = D/R \\approx 9.16\\times10^{-3}\\,\\text{rad}\\). Image diameter on focal plane: \\(y = f_o \\theta = 15 \\times 9.16\\times10^{-3} \\approx 0.137\\,\\text{m}\\) (≈ 13.7 cm). Large \\(f_o\\) links big angular magnification to a sizeable real image.\n3\nCommon Errors\nAvoid using linear magnification. Always link image size to \\(y = f_o \\theta\\); the eyepiece only alters viewing angle, not the physical image on the focal plane.\nPro Tip:\nBigger objective focal length boosts both angular magnification and the actual image, making celestial details easier to study.",
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},
{
"slide": 14,
"fragments": [
{
"fragment_index": -1,
"text_description": "Atoms – Toughest Question\n1\nQuestion Stem\nHydrogen has only one electron, yet its emission spectrum shows many lines. Explain this observation.\n2\nSolution Strategy\nA discharge lamp holds billions of hydrogen atoms. Each atom’s electron may occupy a different excited level. When it falls to a lower level, one photon of that gap’s wavelength is emitted. All gaps are sampled across the ensemble, so the combined light displays every allowed wavelength—hence many spectral lines.\n3\nCommon Errors\nAssuming one atom emits all lines simultaneously. In reality, a single hydrogen atom can release only one photon per transition; spectral richness arises from many atoms making different transitions.\nPro Tip:\nOne electron → one photon. Multiple photons require many atoms—key to understanding spectral line multiplicity.",
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},
{
"slide": 15,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Take-aways",
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},
{
"fragment_index": 1,
"text_description": "Formula & Constants Recap\nMemorise \\(c, h, e, \\varepsilon_0\\) and pivotal equations; write them quickly for ready reference.",
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},
{
"fragment_index": 2,
"text_description": "Section Strategy\nTarget all 1-mark MCQs in 20 min; postpone lengthy derivations to secure early marks.",
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},
{
"fragment_index": 3,
"text_description": "Smart Working\nState givens, pick one formula, substitute, box answer; this saves time and earns method marks.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Units & Sig-figs\nConvert everything to SI units and keep correct significant digits to avoid rework.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Derivation Template\nBegin with law statement, draw diagram, derive; ideal for Gauss, Faraday, Ohm proofs.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "No-Calculator Tips\nPractise mental powers-of-ten, square-root and ratio tricks; indispensable under calculator ban.",
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}
]
}
]