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"slide": 1,
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"text_description": "Class 12 CBSE Physics – Targeted Revision Session",
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},
{
"fragment_index": 2,
"text_description": "Weak-to-Strong Chapter Walk-through",
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},
{
"fragment_index": 3,
"text_description": "Roadmap: decode the 2024-25 sample paper, flag your weak concepts, and practise targeted problems to convert them into scoring strengths.",
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{
"fragment_index": 4,
"text_description": "Sample paper snapshot – 33 questions, 5 sections (A–E), 70 marks, 3 h: our plan mirrors this pattern.",
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{
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"text_description": "Source: CBSE Physics Sample Question Paper 2024-25 (Code 042)",
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"slide": 2,
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"text_description": "Chapter-wise Question Distribution & Your Strength\nRevision Roadmap\nMore questions mean higher exam weightage. Target “low” chapters first, consolidate “med”, then polish “high”.\nChapter\n# Qs\nStrength\nElectric Charges & Fields\n3\nLow\nMoving Charges & Magnetism\n3\nLow\nAlternating Current\n1\nLow\nWave Optics\n3\nLow\nNuclei\n2\nLow\nElectrostatic Potential & Capacitance\n2\nMed\nMagnetism & Matter\n1\nMed\nElectromagnetic Waves\n2\nMed\nDual Nature of Radiation & Matter\n3\nMed\nSemiconductor Electronics\n2\nMed\nCurrent Electricity\n4\nHigh\nElectromagnetic Induction\n2\nHigh\nRay Optics & Optical Instruments\n3\nHigh\nAtoms\n2\nHigh",
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{
"slide": 3,
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"fragment_index": -1,
"text_description": "Electric Charges & Fields – Tough Question\nGauss’s Law • Infinite Line Charge\nGauss’s theorem:\nNet electric flux through a closed surface equals enclosed charge divided by \\( \\varepsilon_0 \\).\nPitfalls: choosing a spherical surface breaks symmetry; remember no flux through end-caps.",
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{
"fragment_index": 1,
"text_description": "1. Gaussian surface —\ncoaxial cylinder of radius \\( r \\) and length \\( L \\).",
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{
"fragment_index": 2,
"text_description": "2. Flux evaluation —\n\\( E \\) is radial & uniform on curved area; ends contribute zero, so \\( \\Phi_E = E(2\\pi rL) \\).",
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{
"fragment_index": 3,
"text_description": "3. Apply law —\n\\( E(2\\pi rL) = \\lambda L/\\varepsilon_0 \\).",
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{
"fragment_index": 4,
"text_description": "Result:\n\\( E(r) = \\dfrac{\\lambda}{2\\pi \\varepsilon_0 r} \\).",
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},
{
"slide": 4,
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"fragment_index": -1,
"text_description": "Moving Charges & Magnetism – Tough Question\nMagnetic Force Balance\nImportant formula: \\(F = qvB = mg\\)",
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{
"fragment_index": 1,
"text_description": "A 10 g ball carrying +10 mC slides freely inside a vertical, frictionless tube. The tube is moved horizontally from east to west through a uniform 2 T magnetic field. Find (i) the minimum tube speed that keeps the ball stationary relative to the tube, and (ii) the direction of the magnetic field.",
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"fragment_index": 2,
"text_description": "Force balance:\nset \\(qvB = mg\\) so magnetic force equals weight.",
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},
{
"fragment_index": 3,
"text_description": "Solve speed:\n\\(v = \\dfrac{mg}{qB} = \\dfrac{0.01\\,\\text{kg}\\times 9.8}{0.01\\,\\text{C}\\times 2\\,\\text{T}} \\approx 4.9\\ \\text{m s}^{-1}\\).",
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},
{
"fragment_index": 4,
"text_description": "Direction:\nRight-hand rule → with \\(v\\) west and \\(F\\) up, \\(B\\) must point south.",
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{
"fragment_index": 5,
"text_description": "Key idea:\n\\(F = qvB\\) for motion ⟂ \\(B\\).",
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},
{
"slide": 5,
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"fragment_index": -1,
"text_description": "Alternating Current – Tough Question\nNew \\(V_L\\) after Short\nQuestion: In a series LCR circuit \\(V_R = V_C = V_L = 10\\,\\text{V}\\) (resonance). The capacitor is short-circuited. Find the new voltage across the inductor.\nHints\n1. Resonance ⇒ \\(X_L = X_C\\) and \\(R = X_L\\).\n2. Without the capacitor, \\(Z = \\sqrt{R^{2}+X_L^{2}} = R\\sqrt{2}\\); current becomes \\(I/\\sqrt{2}\\).\n3. Hence \\(V_L = IX_L/\\sqrt{2} = 10/\\sqrt{2} ≈ 7.1 V\\).\nAnswer: \\(10/\\sqrt{2}\\,\\text{V}\\).\nPitfalls: assuming current unchanged or forgetting \\(V_C = 0\\) after the short.\nKey relation: \\(Z = \\sqrt{R^{2}+(\\omega L)^{2}}\\)",
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},
{
"slide": 6,
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"fragment_index": -1,
"text_description": "Wave Optics – Tough Question\nDual-wavelength YDSE\nQ.\nIn a Young’s double-slit set-up using\nλ\n1\n= 400 nm\nand\nλ\n2\n= 600 nm\n, find the smallest distance\ny\nfrom the central maximum where a common dark fringe appears.\nKey relation\n\\( \\Delta = m\\lambda_1 = \\left(m+\\tfrac{1}{2}\\right)\\lambda_2 \\)\nPitfalls: mix-ups between bright & dark criteria, forgetting to convert nm to metres.",
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{
"fragment_index": 1,
"text_description": "Apply the condition above to obtain the least positive integer \\( m \\).",
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},
{
"fragment_index": 2,
"text_description": "Calculate the corresponding path difference \\( \\Delta \\).",
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},
{
"fragment_index": 3,
"text_description": "Translate to screen position: \\( y = \\dfrac{\\Delta D}{d} \\) (standard YDSE geometry).",
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}
]
},
{
"slide": 7,
"fragments": [
{
"fragment_index": -1,
"text_description": "Nuclei – Tough Question\nBinding-Energy Curve Insight\nQuestion:\nUsing the binding-energy per nucleon curve, decide which nucleus—W(190), X(90), Y(60) or Z(30)—is most likely to (i) undergo fission and (ii) undergo fusion. Give a brief reason.\nEnergy is released whenever a nucleus moves toward the curve’s peak at A≈60. Don’t confuse total binding energy with binding energy per nucleon.",
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"fragment_index": 1,
"text_description": "Hint 1 – Curve Slope:\nFor A > 120 the curve falls; such heavy nuclei release energy by splitting.",
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"fragment_index": 2,
"text_description": "Hint 2 – Light Nuclei:\nVery light nuclei (A < 10) lie low on the curve; they gain energy by combining.",
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"fragment_index": 3,
"text_description": "Hint 3 – Apply Values:\nW (190) is heavy ⇒ fission. Z (30) is light ⇒ fusion.",
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},
{
"slide": 8,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electrostatic Potential & Capacitance – Tough Question\nDielectric vs Metal Slab\nProblem\nA parallel-plate capacitor (plate area \\(A\\), gap \\(d\\)) receives\n(i) a dielectric slab of thickness \\(t\\) and permittivity \\(\\kappa\\),\n(ii) a conducting slab of thickness \\(t\\) ( \\(t<d\\) ).\nDerive \\(C\\) in each case and state which insertion increases capacitance more.\nHint 1\nModel dielectric case as two series sections:\nair gaps \\((d-t)\\) and dielectric slab \\(t/\\kappa\\).\nHint 2\nEquivalent capacitance\n\\(C_{\\text{dielectric}}=\\dfrac{\\varepsilon_0A}{d-t+t/\\kappa}\\).\nHint 3\nIn a metal, \\(E=0\\); effective separation becomes \\(d-t\\), so\n\\(C_{\\text{metal}}=\\dfrac{\\varepsilon_0A}{d-t}\\).\nBecause \\(d-t < d-t+t/\\kappa\\), we get \\(C_{\\text{metal}} > C_{\\text{dielectric}}\\).\n⚠︎ Common slips: adding capacitances in parallel, omitting \\(\\kappa\\).\nCapacitance grows as the effective plate separation shrinks.\nSource: CBSE Physics Sample Question Paper (2024 – 25)",
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{
"slide": 9,
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"fragment_index": -1,
"text_description": "Magnetism & Matter – Conceptual Question\nField Ratio at \\(a/2\\) Below & Above\nA solid wire of radius \\(a\\) carries uniform current \\(I\\).\nFind \\(\\displaystyle \\frac{B_{\\,a/2\\;{\\text{below}}}}{B_{\\,a/2\\;{\\text{above}}}}\\).\nHints\n1. Ampère’s law → \\(B_{\\text{in}}=\\mu_0 I r /(2\\pi a^{2})\\), \\(B_{\\text{out}}=\\mu_0 I /(2\\pi r)\\).\n2. Radii: inside point \\(r=a/2\\); outside point \\(r=1.5a\\).\n3. Ratio \\(= (a/2)/(1.5a)=1/3\\).\nExpected answer : \\(1:3\\).\nSource: CBSE Physics Sample Paper 2024 – Q3",
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{
"fragment_index": 1,
"text_description": "Watch for mixing inside/outside formulas or using surface distances instead of center distances. Remember: inside \\(B\\propto r\\); outside \\(B\\propto 1/r\\).",
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]
},
{
"slide": 10,
"fragments": [
{
"fragment_index": 1,
"text_description": "Electromagnetic Waves – Tough Question",
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},
{
"fragment_index": 2,
"text_description": "Spectrum Production Match",
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},
{
"fragment_index": 3,
"text_description": "Match each electromagnetic wave with its typical production mechanism: Infra-red, Radio, Light and Microwave.",
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},
{
"fragment_index": 4,
"text_description": "Hints\nIR – vibrations of atoms or molecules.\nRadio – oscillating electrons in aerial/LC circuit.\nLight – electronic transitions between atomic levels.\nMicrowave – klystron or maser action.\nMap each wave to one source only; don’t mix IR with microwave.\nRemember: production process follows the wave’s frequency range.",
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},
{
"slide": 11,
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"fragment_index": -1,
"text_description": "Dual Nature – Tough Question\nInterpret the I–V Curves\nTask →\nFrom the three photoelectric I–V plots (A, B, C), rank the incident beams by\nintensity\nand by\nfrequency\n.\nHint 1:\nHigher saturation current ⇒ higher intensity.\nHint 2:\nMore negative stopping potential ⇒ higher photon frequency (energy).\nHint 3:\nCompare both features to classify A, B, C.\nPitfall – Stopping potential is independent of intensity.\nEinstein’s Photoelectric Equation: \\(eV_0 = h\\nu - \\phi\\). Thus \\(K_{\\text{max}} = eV_0\\).",
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{
"slide": 12,
"fragments": [
{
"fragment_index": -1,
"text_description": "Semiconductor Electronics – Tough Question\nP-Type Band Diagram & Circuit\nQuestion :\nDraw energy-band diagrams for a P-type semiconductor at 0 K and room temperature. Decide if the bulb glows when the switch is (a) open, (b) closed in the ideal-diode circuit.\nHints\n• 0 K: Fermi level lies just above valence band, near acceptor level.\n• Room T: holes populate valence band; \\(E_F\\) rises slightly.\n• Switch open — diode reverse-biased; circuit broken; bulb off.\n• Switch closed — diode forward-biased; current flows; bulb on.\nWatch out :\nMisplacing \\(E_F\\) in conduction band or assuming reverse leakage in an ideal diode.\nP-N Junction: Forward bias lowers the depletion barrier, allowing majority carriers to conduct.",
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{
"slide": 13,
"fragments": [
{
"fragment_index": -1,
"text_description": "Current Electricity – Tough Question\nTemperature-Dependent Resistance\nA heating element on a 100 V battery (internal resistance 1 Ω) draws 10 A at 20 °C. After warming to 320 °C, find the power dissipated inside the battery. Temperature coefficient α = 3.7 × 10\n−4\n°C\n−1\n.\nRemember: apply α only to the element, not to the constant internal resistance. Power lost inside a source is always \\(I^{2}r\\).\nKey formula: \\(R_T = R_0(1 + \\alpha\\,\\Delta T)\\)",
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{
"fragment_index": 1,
"text_description": "Hint 1:\nHot resistance \\(R_h = R_{20}[1 + \\alpha\\,\\Delta T]\\)",
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},
{
"fragment_index": 2,
"text_description": "Hint 2:\nSteady current \\(I = \\dfrac{V}{r + R_h}\\)",
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{
"fragment_index": 3,
"text_description": "Hint 3:\nBattery loss \\(P_{int} = I^{2}r\\)",
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},
{
"slide": 14,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electromagnetic Induction – Tough Question\nA.C. Generator: Instantaneous EMF\nQuestion – Derive the instantaneous emf induced in an N-turn coil of area A rotating with angular speed \\( \\omega \\) in a uniform field \\( B \\).\nEnergy comes from the mechanical work done by the prime mover rotating the coil.",
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"fragment_index": 1,
"text_description": "1. Magnetic flux: \\( \\Phi = N B A \\cos \\theta \\), where \\( \\theta = \\omega t \\).",
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{
"fragment_index": 2,
"text_description": "2. Faraday’s law: \\( e = -\\dfrac{d\\Phi}{dt} \\).",
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"fragment_index": 3,
"text_description": "3. Differentiate ⇒ \\( e = N B A \\omega \\sin \\omega t \\).",
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{
"fragment_index": 4,
"text_description": "Peak value: \\( e_{\\text{max}} = N B A \\omega \\).",
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},
{
"fragment_index": 5,
"text_description": "Negative sign shows Lenz’s law; we usually quote magnitude only.",
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},
{
"slide": 15,
"fragments": [
{
"fragment_index": -1,
"text_description": "Ray Optics – Tough Question\nCassegrain Telescope Imaging\nTask: Draw a ray diagram of a Cassegrain telescope and state two advantages over a refractor.\n• Primary concave and secondary convex mirrors fold the optical path.\n• Secondary reflects rays through the central hole in the primary to the final focus.\n• Advantages: no chromatic aberration; long focal length with a short, light tube.\nAvoid: forgetting the hole in the primary mirror or claiming lower spherical aberration without proof.\nSource: CBSE Physics SQP 2024-25 Q24",
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},
{
"slide": 16,
"fragments": [
{
"fragment_index": -1,
"text_description": "Atoms – Tough Question\nHydrogen Emission Line Puzzle\nA discharge lamp contains an enormous number of hydrogen atoms. Each one-electron atom can be excited to different quantum levels. When electrons in different atoms return to lower levels, they emit photons with energies equal to the level gaps, \\(E = 13.6\\,\\text{eV}\\,(1/n_f^{2} - 1/n_i^{2})\\). The many possible \\(n_i \\rightarrow n_f\\) transitions across the ensemble create the full set of discrete spectral lines, even though every atom supplies only one electron.\nCommon mistake: imagining one electron makes all lines at once. In reality, many atoms, each in a single state, together produce the spectrum.",
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{
"slide": 17,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Takeaways\nQuick Recap\n• Exploit symmetry before using Gauss’s or Ampere’s law.\n• Levitation needs \\(qvB = mg\\).\n• At resonance \\(V_L = V_C\\); shorting \\(C\\) changes impedance.\n• Interference minima: path difference = \\((m+\\tfrac12)\\lambda\\).\n• Peak energy release near \\(A \\approx 60\\).\n• Capacitance rises as effective gap shrinks.\n• Wire field: \\(B \\propto r\\) inside, \\(B \\propto 1/r\\) outside.\n• Photon energy shifts stopping potential, not saturation current.\n• Forward bias narrows depletion layer, enabling current.\n• I²R losses heat conductors—monitor current.\n• Generator emf \\(e_{\\text{max}} = NBA\\omega\\); mechanical work → electricity.\n• Reflecting telescopes avoid chromatic aberration and stay compact.\n• Many atoms yield multiple spectral lines, not one electron.\nNext steps: revisit weaker chapters first, then drill derivations and numericals daily.",
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