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[
{
"slide": 1,
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"text_description": "CBSE Class 12 Physics – Sample Paper Review\nCrack the pattern, ace the physics paper!",
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]
},
{
"slide": 2,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question Distribution & Your Strength\nChapter distribution and student proficiency map—see how the paper maps to your syllabus strengths.",
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"fragment_index": 1,
"text_description": "Chapter\nQuestions\nStrength\nElectric Charges and Fields\n4\nLow\nMoving Charges and Magnetism\n3\nLow\nAlternating Current\n4\nLow\nWave Optics\n3\nLow\nNuclei\n3\nLow\nElectrostatic Potential and Capacitance\n4\nMedium\nMagnetism and Matter\n2\nMedium\nElectromagnetic Waves\n2\nMedium\nDual Nature of Radiation and Matter\n3\nMedium\nSemiconductor Electronics\n4\nMedium\nCurrent Electricity\n5\nHigh\nElectromagnetic Induction\n3\nHigh\nRay Optics and Optical Instruments\n4\nHigh\nAtoms\n2\nHigh",
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{
"slide": 3,
"fragments": [
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"fragment_index": -1,
"text_description": "Electric Charges & Fields – Toughest Question (Board-style Application)",
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"fragment_index": 2,
"text_description": "Locate the Zero-Force Point\nTwo +3 μC charges are fixed 0.5 m apart. Where should a −2 μC charge experience zero net electrostatic force?\nApply superposition and the inverse-square law to locate the point where forces cancel.",
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"text_description": "Key Points:\nSketch charges on a straight line; choose origin at left +3 μC.\nLet −2 μC be \\(x\\) metres from the left charge, outside the 0.5 m pair.\nSet magnitudes equal: \\( \\frac{1}{x^{2}} = \\frac{1}{(0.5 + x)^{2}} \\).\nSolve: \\( x = 0.25\\ \\text{m} \\) left of the first + charge.\nBetween the + charges, forces add; zero force cannot occur there.",
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{
"slide": 4,
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"fragment_index": -1,
"text_description": "Moving Charges & Magnetism – Toughest Question\nHelical motion analysis\nA proton enters a 0.3 T field at 60° with speed \\(2\\times10^{6}\\,\\text{m s}^{-1}\\). Find the pitch of its helical path.\nLorentz magnetic force bends the charged particle into a circle for \\(v_{\\perp}\\) while \\(v_{\\parallel}\\) stays uniform along the field.",
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"text_description": "Key Points:\nResolve velocity: \\(v_{\\parallel}=v\\cos60^{\\circ}\\), \\(v_{\\perp}=v\\sin60^{\\circ}\\).\nMagnetic force on \\(v_{\\perp}\\) gives circular motion; none acts on \\(v_{\\parallel}\\).\nTime period \\(T=\\frac{2\\pi m}{qB}\\) for a proton.\nPitch \\(p=v_{\\parallel}T\\approx0.22\\,\\text{m}\\).\nUse \\(v_{\\parallel}\\), not total \\(v\\), when computing pitch.",
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},
{
"slide": 5,
"fragments": [
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"fragment_index": -1,
"text_description": "Alternating Current – Toughest Question",
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"fragment_index": 2,
"text_description": "Transformer & losses\nIdeal transformer: \\(N_p = 500\\), \\(V_p = 220\\,\\text{V}\\). Refrigerator needs \\(V_s = 110\\,\\text{V}\\), \\(I_s = 4\\,\\text{A}\\). Find \\(N_s\\), \\(I_p\\) and efficiency with 5 % copper loss.\nAC-circuit relations: \\( \\frac{V_p}{V_s} = \\frac{N_p}{N_s} \\). Power balance: \\( V_p I_p = V_s I_s \\). Efficiency: \\( \\eta = \\frac{P_{\\text{out}}}{P_{\\text{in}}} \\).\nKey Points:\nTurns ratio gives \\(N_s = 250\\) turns.\nPower conservation ⇒ \\(I_p = 2.0\\,\\text{A}\\).\n5 % copper loss ⇒ \\(\\eta = 95\\%\\). Avoid mixing primary & secondary.",
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]
},
{
"slide": 6,
"fragments": [
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"fragment_index": -1,
"text_description": "Wave Optics – Toughest Question",
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{
"fragment_index": 1,
"text_description": "YDSE with medium change",
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{
"fragment_index": 2,
"text_description": "Two-slit interference: \\(d = 0.25\\,\\text{mm}\\), \\(\\lambda = 600\\,\\text{nm}\\), measured fringe width \\(\\beta = 2.4\\,\\text{mm}\\).",
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{
"fragment_index": 3,
"text_description": "Use \\( \\beta = \\frac{\\lambda D}{d} \\) to find screen distance \\(D\\). Replace \\(\\lambda\\) by \\( \\lambda' = \\lambda/\\mu \\) to predict new fringe width in water.",
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"fragment_index": 4,
"text_description": "Key Points:\nScreen distance: \\(D = \\frac{\\beta d}{\\lambda} = 1.0\\,\\text{m}\\).\nIn water, new wavelength \\( \\lambda' \\approx 451\\,\\text{nm}\\).\nNew fringe width: \\(\\beta' = \\beta/\\mu \\approx 1.8\\,\\text{mm}\\) (reduced by factor 1/1.33).\nPredict fringe shift by comparing original and altered interference patterns.",
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},
{
"slide": 7,
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{
"fragment_index": -1,
"text_description": "Nuclei – Toughest Question\nSecular equilibrium\nActivity ratio in secular equilibrium\nRadioactivity chains reach secular equilibrium when the parent’s half-life is far longer than the daughter’s.\nUsing the given half-lives, show that the activity of \\(^{234}\\text{Th}\\) tracks \\(^{238}\\text{U}\\) for \\(4.5\\times10^{9}\\,\\text{y}\\).",
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"text_description": "Key Points:\nEquilibrium condition: \\(A_{\\text{U}} = A_{\\text{Th}}\\).\nDecay constant: \\(\\lambda = \\ln 2 / T_{1/2}\\).\nAfter \\(4.5\\times10^{9}\\,\\text{y}\\): \\(e^{-\\lambda_{\\text{Th}}t}\\!\\approx\\!0\\), \\(e^{-\\lambda_{\\text{U}}t}\\!\\approx\\!0.5\\).\nThus \\(A_{\\text{Th}}/A_{\\text{U}} = 1\\); equilibrium persists.\nPitfalls: treating 24.1 d as long, or confusing activity with mass.",
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},
{
"slide": 8,
"fragments": [
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"fragment_index": -1,
"text_description": "Electrostatic Potential & Capacitance – Toughest Question\nConcentric Shells: Potential Profile\nIsolated shells: \\(r_1=10\\,\\text{cm}, Q_1=+8\\,\\text{nC}\\); \\(r_2=20\\,\\text{cm}, Q_2=-4\\,\\text{nC}\\).\nUse Gauss law to find \\(E\\). Then apply \\(V(r)=kQ/r\\) for each shell, superpose, sketch \\(V(r)\\) and obtain \\(\\Delta V\\).",
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"fragment_index": 2,
"text_description": "Key Points:\nInside any conductor \\(E=0\\) ⇒ flat potential regions.\nEvaluate three zones: \\(r<10\\text{ cm}\\), \\(10<r<20\\text{ cm}\\), \\(r>20\\text{ cm}\\).\nFor \\(r>20\\text{ cm}\\), \\(V(r)=k\\left(\\frac{Q_1}{r}+\\frac{Q_2}{r}\\right)\\).\nPotential difference: \\(\\Delta V=V_{10}-V_{20}=360\\,\\text{V}\\) (inner shell higher).",
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},
{
"slide": 9,
"fragments": [
{
"fragment_index": -1,
"text_description": "Magnetism & Matter – Toughest Question\nAtomic moment estimation",
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{
"fragment_index": 1,
"text_description": "A toroid with \\( \\mu_r = 2000 \\), mean radius \\( r = 0.15\\,\\text{m} \\) and 600 turns carries a current of 3 A.",
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{
"fragment_index": 2,
"text_description": "Find (a) the magnetic field \\( B \\) inside the core and (b) the magnetic moment per iron atom when the atomic density is \\( 8.5\\times10^{28}\\,\\text{m}^{-3} \\).",
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{
"fragment_index": 3,
"text_description": "Key Points:\n\\( B = \\mu_0 \\mu_r \\dfrac{NI}{2\\pi r} \\) — include relative permeability.\n\\( H = \\frac{NI}{2\\pi r} \\), \\( M = \\frac{B}{\\mu_0}-H \\) connects field and magnetisation.\nAtomic moment \\( m = \\frac{M}{n} \\approx 2.1\\,\\mu_B \\).\nCore relation \\( B = \\mu_0(H+M) \\) links macroscopic \\( B \\) to microscopic \\( m \\).\nPitfalls: ignoring permeability or the \\( H \\) term overestimates magnetisation.",
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},
{
"slide": 10,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electromagnetic Waves – Toughest Question\nDisplacement Current via Maxwell’s Correction\nA parallel-plate capacitor (area 50 cm², gap 2 mm) is driven by \\(V = 200\\sin(1.5\\times10^{5}t)\\,\\text{V}\\). Find the displacement current between its plates.\nSolution showcases Maxwell’s corrective term in the Ampère–Maxwell law.",
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"text_description": "Key Points:\n\\(A = 50\\,\\text{cm}^{2} = 5.0\\times10^{-3}\\,\\text{m}^{2}\\).\nCapacitance \\(C = \\varepsilon_{0}A/d\\).\nElectric flux \\( \\Phi_E = C V / \\varepsilon_{0}\\).\nDisplacement current \\(I_d = \\varepsilon_{0}\\,d\\Phi_E/dt = C\\,dV/dt\\) (Ampère–Maxwell).\n\\(dV/dt = 200(1.5\\times10^{5})\\cos(1.5\\times10^{5}t)\\).\nUse SI units; do not confuse with conduction current.",
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]
},
{
"slide": 11,
"fragments": [
{
"fragment_index": 1,
"text_description": "Dual Nature – Toughest Question",
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{
"fragment_index": 2,
"text_description": "",
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{
"fragment_index": 3,
"text_description": "Work function deduction",
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{
"fragment_index": 4,
"text_description": "Zn photocell stops current at \\(V_s = 1.3\\,\\text{V}\\) for \\(\\lambda = 250\\,\\text{nm}\\). Determine the work function \\(\\phi\\) and predict \\(V_s\\) for \\(\\lambda = 300\\,\\text{nm}\\).",
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{
"fragment_index": 5,
"text_description": "Key Points:\nUse \\(eV_s = h\\nu - \\phi\\) – Einstein photoelectric equation.\nConvert wavelength to frequency: \\(\\nu = \\frac{c}{\\lambda}\\).\nFirst data ➜ \\(\\phi = h\\nu - eV_s\\).\nInsert \\(\\phi\\) into second wavelength to obtain new \\(V_s\\).\nKeep energies in eV; avoid mixing Joules and eV.",
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]
},
{
"slide": 12,
"fragments": [
{
"fragment_index": -1,
"text_description": "Semiconductor Electronics – Toughest Question\nPlot load line & set Q-point",
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{
"fragment_index": 1,
"text_description": "Question: For a Si BJT in CE mode with \\(V_{CC}=12\\,\\text{V}\\), \\(R_C=2\\,\\text{k}\\Omega\\), \\(R_B=200\\,\\text{k}\\Omega\\) and \\(\\beta =120\\), find the Q-point and comment on thermal runaway.",
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"fragment_index": 2,
"text_description": "Solution: \\(I_B=\\frac{V_{CC}-V_{BE}}{R_B}= \\frac{12-0.7}{200\\,\\text{k}\\Omega}=56\\,\\mu\\text{A}\\). Hence \\(I_C=\\beta I_B \\approx 6.8\\,\\text{mA}\\), but the load-line limit is \\(I_{C\\,sat}= \\frac{V_{CC}}{R_C}=6\\,\\text{mA}\\). The transistor therefore saturates; Q-point is near \\((I_C \\approx 6\\,\\text{mA},\\, V_{CE} \\approx 0\\,\\text{V})\\).",
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"fragment_index": 3,
"text_description": "Key Points:\nSubtract \\(V_{BE}\\) when computing base bias current.\nDraw the \\(I_C\\!-\\!V_{CE}\\) load line to visualise active, cut-off and saturation regions.\nSelect a Q-point in the active region for bias stability; extremes cause distortion.\nNear saturation, \\(P = V_{CE} I_C\\) is low, so thermal runaway is unlikely.\nBias stabilisation safeguards BJT amplifiers against temperature rise.",
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},
{
"slide": 13,
"fragments": [
{
"fragment_index": -1,
"text_description": "Current Electricity – Toughest Question\nPotentiometer calibration",
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"text_description": "A 1 m potentiometer wire (10 Ω) is fed by a 2.5 V DC source with internal resistance 0.5 Ω.\nNull deflection occurs at 65 cm for the unknown DC cell.\nUsing the null-deflection method we obtain \\(E_x \\approx 1.55\\,\\text{V}\\).",
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"fragment_index": 3,
"text_description": "Key Points:\n\\(k=\\frac{E_D}{R+r}\\times\\frac{R}{L}\\approx2.38\\,\\text{V m}^{-1}\\); therefore \\(E_x=k\\times0.65=1.55\\,\\text{V}\\).\nSensitivity rises when the wire is longer or the current is reduced with a series resistance.\nUse driver cell emf \\(E_D\\), not terminal voltage; always include internal \\(r\\).\nNull deflection keeps the test cell in an open circuit, avoiding loading error in DC circuits.",
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},
{
"slide": 14,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electromagnetic Induction – Toughest Question\nMoving Loop Problem\nA square loop of side 10 cm (R = 1 Ω) is pulled out of a 0.4 T region at 3 m s⁻¹, leaving the field in 0.2 s.\nDetermine \\(I(t)\\) and the mechanical work done. Relate the changing magnetic flux to induced emf and work.",
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"text_description": "Key Points:\nFaraday law: \\(\\varepsilon=-\\frac{d\\Phi}{dt}=-B\\,\\frac{dA}{dt}\\).\nUniform pull ⇒ \\(dA/dt=-l v\\); motional emf constant: \\(\\varepsilon=Bvl=0.12\\;\\text{V}\\).\nInduced current: \\(I(t)=\\varepsilon/R=0.12\\;\\text{A}\\;(0\\le t\\le0.2\\text{ s})\\), zero afterwards.\nWork–energy: \\(W=\\int_0^{0.2}I^{2}R\\,dt\\approx2.9\\times10^{-3}\\,\\text{J}\\).\nAvoid using constant flux; keep Lenz’s-law sign for direction.",
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},
{
"slide": 15,
"fragments": [
{
"fragment_index": -1,
"text_description": "Ray Optics – Toughest Question\nCompound microscope",
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"fragment_index": 2,
"text_description": "Objective \\(f_o = 0.8\\;\\text{cm}\\), eyepiece \\(f_e = 2.5\\;\\text{cm}\\), tube length 15 cm. Object is 1 cm from objective. Calculate total magnifying power for near-point viewing.",
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"fragment_index": 3,
"text_description": "Goal: design the required magnification by chaining lens formulae and the microscope relation.",
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{
"fragment_index": 4,
"text_description": "Key Points:\nUse \\( \\frac{1}{f_o} = \\frac{1}{v_o} - \\frac{1}{u_o}\\) with \\(u_o = -1\\;\\text{cm}\\).\nMatch instrument tube: \\(v_o + v_e = 15\\;\\text{cm}\\).\n\\(\\frac{1}{f_e} = \\frac{1}{v_e} - \\frac{1}{u_e}\\) gives eyepiece object distance.\nTotal magnification \\(M = \\frac{L}{f_o}\\left(1 + \\frac{D}{f_e}\\right)\\) with \\(L = 15\\;\\text{cm},\\; D = 25\\;\\text{cm}\\).\nComputed value: \\(M \\approx 150\\).\nAvoid sign errors; never drop the \\((1 + D/f_e)\\) factor.",
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]
},
{
"slide": 16,
"fragments": [
{
"fragment_index": -1,
"text_description": "Atoms – Toughest Question\nBohr model limits\nUse Bohr theory to find the electron speed in hydrogen for \\(n = 2\\).\nCompute the photon wavelength when the electron drops from \\(n = 2\\) to \\(n = 1\\).\nDiscuss how these results fit the observed spectra and where the model falls short.",
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"text_description": "Key Points:\nSpeed: \\(v_2=\\frac{e^{2}}{2\\varepsilon_0 h}\\frac{1}{2}\\).\nEnergy gap: \\(\\Delta E = 13.6\\,\\text{eV}\\left(1-\\frac{1}{4}\\right)=10.2\\,\\text{eV}\\).\nWavelength: \\(\\lambda = \\frac{hc}{\\Delta E}\\approx 1.22\\times10^{-7}\\,\\text{m}\\) (Lyman-α line).\nLimitations: no fine-structure, fails for multi-electron atoms.\nCommon errors: ignore J–eV conversion, apply non-relativistic \\(v\\) at high \\(n\\).",
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},
{
"slide": 17,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Take-aways",
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{
"fragment_index": 1,
"text_description": "Start with Laws\nRecall the governing law, list knowns, and set the formula—your quick recap foundation.",
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{
"fragment_index": 2,
"text_description": "Draw & Visualise\nNeat diagrams reveal field directions and cut algebra mistakes.",
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},
{
"fragment_index": 3,
"text_description": "Units Audit\nCheck dimensions; unit mismatches flag most hidden errors.",
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},
{
"fragment_index": 4,
"text_description": "Isolate First\nSolve symbolically, then plug numbers—saves time and reduces calculator slips.",
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},
{
"fragment_index": 5,
"text_description": "Link to Concept\nTie each answer back to the concept; memory sticks to meaning—things to remember!",
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]
}
]