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{
"slide": 1,
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"fragment_index": -1,
"text_description": "Question Distribution & Your Strength\nChapter\nQuestions / Marks\nYour Score (%)\nPriority Action\nElectrostatics\n4 / 10\n85\nRevise formulas\nMagnetism\n5 / 12\n78\nMore MCQs\nEMI & AC\n6 / 14\n60\nSolve numericals\nEM Waves\n2 / 4\n90\nQuick revision\nOptics\n8 / 16\n55\nConcept videos\nDual Nature\n4 / 8\n70\nPast papers\nAtoms & Nuclei\n2 / 5\n88\nMaintain pace\nSemiconductors\n2 / 5\n65\nClarify basics\nSource: CBSE Physics Sample Paper 2024-25 & Your Mock Test Analytics",
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{
"slide": 2,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electric Charges & Fields – Tough Spot\nClass 12 Physics • CBSE\nQuestion\nUsing Gauss’s law, derive the electric field \\(E\\) at a distance \\(r\\) from an infinitely long straight wire carrying uniform line charge density \\(\\lambda\\).\nProgressive Hints\nTry on your own first. Reveal hints one by one if stuck.\nThings to Consider\nFlux is only through the curved cylindrical surface.\nDo not cancel \\(2\\pi\\); keep the full \\(2\\pi rL\\) area.\nCharge enclosed is \\(\\lambda L\\), matching cylinder length.\nFinal field varies as \\(1/r\\); check your r-dependence.\nConcept Notes\nGauss’s Law\nNet electric flux through a closed surface equals enclosed charge divided by \\(\\varepsilon_0\\). Powerful for high-symmetry cases.\nCylindrical Symmetry\nFor an infinite line charge, field magnitude depends only on radial distance; direction is radial outward (or inward for negative \\(\\lambda\\)).\nResulting Field:\n\\(E=\\dfrac{\\lambda}{2\\pi\\varepsilon_0 r}\\)\nPrevious Question\nCheck Solution",
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},
{
"fragment_index": 1,
"text_description": "Hint 1: Gaussian surface\nChoose a cylindrical Gaussian surface of radius \\(r\\) and length \\(L\\) coaxial with the wire.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Hint 2: Symmetry check\nBy cylindrical symmetry, \\(E\\) is radial and constant on the curved surface; flux through flat end-caps is zero.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Hint 3: Apply Gauss’s law\n\\(E(2\\pi rL)=\\lambda L/\\varepsilon_0\\) ⇒ \\(E=\\dfrac{\\lambda}{2\\pi\\varepsilon_0 r}\\).",
"image_description": ""
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},
{
"slide": 3,
"fragments": [
{
"fragment_index": -1,
"text_description": "Moving Charges & Magnetism – Challenge\nCBSE Class 12 Physics\nQuestion\nAn electron describes a circle of radius \\(r_0\\) with speed \\(v_0\\) and period \\(T_0\\) in a uniform magnetic field \\(B\\). Predict qualitatively how the radius \\(r\\) and period \\(T\\) change if its speed is doubled to \\(2v_0\\).\nProgressive Hints\nTry first, then reveal hints one at a time.\nThings to Consider\nCompare magnetic force with required centripetal force for uniform circular motion.\nDecide which variables are fixed by the field \\((B,q,m)\\) and which vary with speed.\nPeriod depends on angular frequency, not on linear speed.\nAvoid mixing up how radius and period respond to speed change.\nConcept Notes\nMagnetic Force\nA charge moving with velocity \\( \\mathbf{v}\\) in a field \\( \\mathbf{B}\\) experiences \\( \\mathbf{F}=q\\,\\mathbf{v}\\times\\mathbf{B}\\) perpendicular to motion.\nUniform Circular Motion\nWhen \\( \\mathbf{F}\\) is always perpendicular to \\( \\mathbf{v}\\), speed stays constant and the particle follows a circle with constant angular frequency.\nKey Formulas:\n\\( qvB = \\dfrac{mv^{2}}{r}\\) | \\( T = \\dfrac{2\\pi m}{qB}\\)\nPrevious Question\nCheck Solution",
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},
{
"fragment_index": 1,
"text_description": "Hint 1: Identify the forces\nMagnetic (Lorentz) force supplies centripetal force: \\(qvB = \\dfrac{mv^{2}}{r}\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Hint 2: Relate radius to speed\nFrom the above, \\(r = \\dfrac{mv}{qB}\\). For fixed \\(B,\\,q,\\,m\\), radius is directly proportional to speed.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Hint 3: Check the period\nAngular frequency \\( \\omega = \\dfrac{qB}{m}\\) is speed-independent, so \\(T = 2\\pi/\\omega\\) remains \\(T_0\\). Radius doubles.",
"image_description": ""
}
]
},
{
"slide": 4,
"fragments": [
{
"fragment_index": -1,
"text_description": "Alternating Current – Quick Probe\nSeries LCR • Impedance & Reactance\nQuestion\nIn a series LCR circuit, the RMS voltages across R, C and L are each 10 V. The capacitor is suddenly short-circuited. Estimate the new RMS voltage across the inductor.\nProgressive Hints\nTry to solve first. Reveal hints as needed.\nThings to Consider\nCurrent must be recalculated after C is removed.\n\\(X_L\\) stays unchanged; frequency and L are constant.\nAt resonance, \\(R = V/I\\); keep this value for later steps.\nVoltage redistribution follows impedance, not supply change.\nConcept Notes\nResonant Series LCR\nAt resonance \\(X_L = X_C\\); impedance is purely resistive and minimum, equal to \\(R\\).\nImpedance After Change\nRemoving a reactive element adds its reactance vectorially: \\(Z = \\sqrt{R^{2}+X^{2}}\\). Current and element voltages adjust accordingly.\nImportant Formulas\n\\(X_L = \\omega L\\) • \\(Z = \\sqrt{R^{2} + (X_L - X_C)^{2}}\\)\nPrevious Question\nCheck Solution",
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},
{
"fragment_index": 1,
"text_description": "Hint 1: Spot the condition\nEqual voltages on R, L and C imply resonance, so \\(X_L = X_C\\) and circuit current is \\(I = V/R\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Hint 2: New impedance\nShort-circuiting \\(C\\) removes \\(X_C\\). Net impedance becomes \\(Z = \\sqrt{R^{2}+X_L^{2}}\\).",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Hint 3: Re-compute voltage\nCompare currents: \\(I_{\\text{before}} = V/R\\), \\(I_{\\text{after}} = V/Z\\). New inductor voltage is \\(V_L' = I_{\\text{after}}\\,X_L\\).",
"image_description": ""
}
]
},
{
"slide": 5,
"fragments": [
{
"fragment_index": -1,
"text_description": "Wave Optics – Interference Twist\nCBSE Class 12 • Physics\nQuestion\nIn a Young’s double-slit experiment, light of wavelengths 400 nm and 600 nm is used together. Find the smallest distance from the central bright fringe where both colours give a dark (minimum-intensity) fringe simultaneously.\nProgressive Hints\nReveal hints one at a time only if you need them.\nThings to Consider\nUse the destructive-interference condition, not the bright-fringe rule.\nIgnore the trivial \\(m=n=0\\) case; it is the central maximum.\nFind the lowest common multiple of the two wavelengths.\nTranslate fringe order to distance using \\(\\beta = \\lambda D/d\\).\nConcept Notes\nPath Difference & Minima\nIn a YDS experiment, dark fringes occur when \\( \\Delta = m\\lambda \\) with m = 1, 2, 3… .\nSimultaneous Minima\nTwo wavelengths give a common dark fringe when integer orders satisfy \\( m\\lambda_{1}=n\\lambda_{2} \\).\nImportant Formula:\n\\( \\displaystyle \\beta = \\frac{\\lambda D}{d} \\) – fringe width (distance between successive minima).\nPrevious Question\nCheck Solution",
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},
{
"fragment_index": 1,
"text_description": "Hint 1: Start with the condition\nFor a dark fringe in YDS, the path difference must equal an integer multiple of the wavelength: \\( \\Delta = m\\lambda \\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Hint 2: Simultaneous minima\nFind integers \\(m,n\\) such that \\(m\\lambda_{1}=n\\lambda_{2}\\) with \\(\\lambda_{1}=400\\text{ nm},\\,\\lambda_{2}=600\\text{ nm}\\).",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Hint 3: Lowest orders\nThe least non-zero solution is \\(3 \\times 400 = 2 \\times 600 = 1200\\text{ nm}\\). Hence \\(m=3, n=2\\). The required distance = \\(m\\beta_{1}=n\\beta_{2}\\), where \\(\\beta=\\lambda D/d\\).",
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]
},
{
"slide": 6,
"fragments": [
{
"fragment_index": -1,
"text_description": "Nuclei – Energy Curve Insight\nBinding Energy per Nucleon & Nuclear Stability\nQuestion\nFrom the binding-energy curve, nuclei W (190), X (90), Y (60) and Z (30) are marked. \n Which one favours (i) fission and (ii) fusion? State why.\nBinding-energy curve (schematic)\nProgressive Hints\nSolve first, then reveal hints in order.\nThings to Consider\nCompare each mass number with the 56-region peak.\nEnergy is released when BE/A increases after a reaction.\nHeavy >140 ➜ fission; light <10 ➜ fusion trends.\nAvoid picking the already most stable nucleus for fission.\nConcept Notes\nBinding Energy per Nucleon\nHigher BE/A means stronger binding and greater nuclear stability. The maximum lies near iron (A≈56).\nPredicting Reaction Type\nA nucleus will fission or fusion if products sit higher on the BE/A curve, releasing the difference as energy.\nImportant Rule:\nEnergy released ≈ A × Δ(BE/A)\nPrevious Question\nCheck Solution",
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},
{
"fragment_index": 1,
"text_description": "Hint 1: Peak Position\nThe BE/A curve peaks near A ≈ 56 with about 8.8 MeV per nucleon.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Hint 2: Heavy Side\nNuclei with A > 140 gain stability by splitting into fragments nearer the peak.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Hint 3: Light Side\nVery light nuclei (A < 10) release energy when they fuse toward A ≈ 56.",
"image_description": ""
}
]
},
{
"slide": 7,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electrostatic Potential & Capacitance – Composite Dielectric\nClass 12 Physics • CBSE\nQuestion\nA parallel-plate capacitor has plate area \\(A\\) and separation \\(d\\). Derive its capacitance when (i) a dielectric slab of relative permittivity \\(\\kappa\\) and thickness \\(t\\), and (ii) a metallic slab of thickness \\(t\\,(<d)\\) are fully inserted between the plates. Which insertion yields the larger capacitance? Give reason.\nProgressive Hints\nReveal one hint at a time.\nThings to Consider\nSeries combination: \\(1/C=1/C_1+1/C_2\\).\nDielectric constant \\(\\kappa\\) scales \\(\\varepsilon_0\\) inside the slab.\nFor conductors, \\(\\kappa\\to\\infty\\) ⇒ zero potential drop inside.\nAvoid mixing parallel and series models.\nConcept Notes\nComposite Capacitors in Series\nWhen different media fill the space between plates along the field, treat each region as a separate capacitor connected in series.\nRole of Dielectric Constant\nThe dielectric constant \\(\\kappa\\) multiplies permittivity, reducing electric field and increasing capacitance by a factor \\(\\kappa\\) for the same geometry.\nImportant Formulae:\n\\(C_{\\text{dielectric}}=\\dfrac{\\varepsilon_0 A}{d - t + t/\\kappa}\\)\n\\(C_{\\text{metal}}=\\dfrac{\\varepsilon_0 A}{d - t}\\)\nPrevious Question\nCheck Solution",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Hint 1: Identify the model\nTreat the system as two capacitors in\nseries\n: the slab region and the remaining air gap.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Hint 2: Write individual capacitances\n\\(C_{\\text{slab}}=\\kappa \\varepsilon_0 A / t\\), \\(C_{\\text{air}}=\\varepsilon_0 A /(d-t)\\).\nUse \\(1/C=1/C_{\\text{slab}}+1/C_{\\text{air}}\\).",
"image_description": ""
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{
"fragment_index": 3,
"text_description": "Hint 3: Compare with metal insert\nFor a metal, \\(\\kappa\\to\\infty\\) ⇒ effective gap \\(d-t\\).\nDielectric: \\(C=\\varepsilon_0A /(d-t+t/\\kappa)\\).\nSince \\(d-t+t/\\kappa > d-t\\), the\nmetal insert\ngives higher capacitance.",
"image_description": ""
}
]
},
{
"slide": 8,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electromagnetic Waves – Displacement Current\nCBSE Class 12 Physics\nQuestion\nA parallel-plate capacitor of area 0.001 m² and plate gap 0.0001 m is being charged so that the voltage across it rises at \\(10^{8}\\,\\text{V s}^{-1}\\). Estimate the displacement current flowing through the gap.\nProgressive Hints\nReveal hints step-by-step only if needed.\nCommon Pitfalls\nApplying conduction current \\(I = C\\,dV/dt\\) instead of displacement formula.\nForgetting to divide by plate separation \\(d\\).\nMixing electric field and voltage rate units.\nDropping powers of ten when multiplying constants.\nConcept Notes\nMaxwell’s Correction\nIn free space \\( \\nabla \\times \\mathbf{B} = \\mu_0\\left(\\mathbf{J} + \\varepsilon_0 \\dfrac{\\partial \\mathbf{E}}{\\partial t}\\right) \\). The extra term represents displacement current, ensuring continuity in time-varying fields.\nCharging Capacitor\nBetween plates, no real charges move. A changing electric field substitutes with \\(I_d\\), making Ampère’s law valid for the loop linking the capacitor.\nImportant Formula:\n\\(I_d = \\varepsilon_0\\,A\\,\\dfrac{dE}{dt} = \\varepsilon_0\\,A\\,\\dfrac{1}{d}\\dfrac{dV}{dt}\\)\nPrevious Question\nCheck Solution",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Hint 1: Key Formula\n\\(I_d = \\varepsilon_0 A\\,\\dfrac{dV/dt}{d}\\)",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Hint 2: Substitute Values\nUse \\(\\varepsilon_0 = 8.85\\times10^{-12}\\,\\text{F m}^{-1}\\), \\(A = 0.001\\,\\text{m}^2\\), \\(d = 1\\times10^{-4}\\,\\text{m}\\), \\(dV/dt = 10^{8}\\,\\text{V s}^{-1}\\).",
"image_description": ""
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{
"fragment_index": 3,
"text_description": "Hint 3: Final Evaluation\nCalculate: \\(I_d \\approx 8.85\\times10^{-4}\\,\\text{A}\\). Choose the option closest to this value.",
"image_description": ""
}
]
},
{
"slide": 9,
"fragments": [
{
"fragment_index": -1,
"text_description": "Dual Nature – Reading Photoelectric Graphs\nPhotoelectric Effect · Einstein Equation\nQuestion\nPhotoelectric I–V curves A, B and C show different saturation currents and stopping potentials. Relate these differences to changes in (a) light intensity and (b) light frequency for the same metal surface.\nProgressive Hints\nReveal hints one by one only when needed.\nThings to Consider\nSaturation current ∝ intensity, not frequency.\nStopping potential reveals maximum electron energy and depends only on \\( \\nu \\).\nWork function \\( \\phi \\) and threshold frequency remain fixed for the metal.\nVertical shift of I–V curve = intensity change; horizontal shift = frequency change.\nConcept Notes\nEinstein’s Photoelectric Equation\nEnergy of one photon \\(h\\nu\\) is partitioned into work function \\( \\phi \\) and kinetic energy \\( eV_0 \\). Thus \\(V_0\\) tracks frequency.\nShape of I–V Curve\nCurrent rises quickly, saturates when every emitted electron is collected, then reverses sign past stopping potential.\nImportant Formula:\n\\( eV_0 = h\\nu - \\phi \\)\nPrevious Question\nCheck Solution",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Hint 1: Getting Started\nSaturation current depends on how many photoelectrons are emitted per second, i.e. on photon flux (intensity).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Hint 2: Key Approach\nEinstein’s equation \\( eV_0 = h\\nu - \\phi \\) shows stopping potential \\(V_0\\) varies with frequency, not with intensity, for a given work function \\( \\phi \\).",
"image_description": ""
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{
"fragment_index": 3,
"text_description": "Hint 3: Almost There\nCurves with equal \\(V_0\\) but larger saturation current were produced by the same frequency at higher intensity. Curves with higher \\(V_0\\) came from higher-frequency light.",
"image_description": ""
}
]
},
{
"slide": 10,
"fragments": [
{
"fragment_index": -1,
"text_description": "Semiconductors – Band Diagram & Bulb Test\nCBSE Class 12 Physics\nQuestion\nSketch the energy-band diagram of a P-type semiconductor at (a) 0 K and (b) room temperature. For the circuit shown, will the bulb glow when the switch is (i) open and (ii) closed? Justify.\nCircuit with PN diode, switch and bulb\nProgressive Hints\nTry on your own first. Reveal hints one by one if needed.\nThings to Consider\nLocate the Fermi level for a P-type crystal at different temperatures.\nRemember: holes appear when acceptor levels donate electrons.\nForward bias ⇒ P-side positive, depletion region narrows, current flows.\nBulb needs a complete circuit; open switch breaks the path.\nConcept Notes\nEnergy Bands in P-Type\nAcceptor level lies just above valence band. Ionisation creates holes, shifting Fermi level slightly above the valence band as temperature rises.\nPN Diode Biasing\nUnder forward bias, majority carriers cross the junction and current flows; reverse bias widens depletion region and blocks current.\nImportant Rule:\nForward conduction when \\(V_{\\text{anode}}-V_{\\text{cathode}} > V_B\\), where \\(V_B≈0.7\\text{ V (Si)}\\).\nPrevious Question\nCheck Solution",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Hint 1 • Getting Started\nAt 0 K the valence band is full and no holes exist; mark the Fermi level just above it.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Hint 2 • Key Approach\nAt room temperature acceptor atoms ionise, create holes, and pull the Fermi level nearer the valence band.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Hint 3 • Almost There\nThe diode conducts only when its P-side (anode) is at higher potential. Check polarity and confirm that current flows—and the bulb glows—only with switch closed.",
"image_description": ""
}
]
},
{
"slide": 11,
"fragments": [
{
"fragment_index": -1,
"text_description": "Current Electricity – Heating Element Puzzle\nClass 12 • CBSE Physics\nQuestion\nA heater draws 10 A from a 100 V battery whose internal resistance is 1 Ω at 20 °C. When the heater reaches 320 °C, calculate the power dissipated inside the battery. Take the temperature coefficient of resistance \\( \\alpha = 3.7 \\times 10^{-4}\\,\\text{°C}^{-1} \\).\nProgressive Hints\nSolve first, then reveal hints one by one.\nThings to Consider\nOhm’s law assumes constant resistance; temperature breaks this limit.\nThe battery’s internal resistance \\( r \\) is always in series.\nUse \\( \\alpha \\) with \\( \\Delta T \\) in °C for consistency.\nRequired power is \\( I^{2} r \\), not the heater’s power.\nConcept Notes\nTemperature Coefficient of Resistance\nFor small ranges, \\(R = R_0(1 + \\alpha \\Delta T)\\); resistance rises with temperature for metals.\nInternal Resistance & Joule Heating\nA real battery behaves as \\(E\\) in series with \\(r\\); heat inside it equals \\(I^{2} r\\), reducing useful output.\nKey Formula:\n\\(R_0 = \\dfrac{V}{I} - r\\)\nPrevious Question\nCheck Solution",
"image_description": ""
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{
"fragment_index": 1,
"text_description": "Hint 1: Cold resistance\nInclude the source resistance: \\(R_0 = \\dfrac{V}{I} - r\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Hint 2: Hot resistance\nUse temperature rise \\( \\Delta T = 300\\,^\\circ\\text{C} \\): \\(R_T = R_0[1 + \\alpha \\Delta T]\\).",
"image_description": ""
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{
"fragment_index": 3,
"text_description": "Hint 3: Current & power\nFind \\(I = \\dfrac{V}{R_T + r}\\), then battery loss \\(P = I^{2} r\\).",
"image_description": ""
}
]
},
{
"slide": 12,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electromagnetic Induction – Generator Math\nCBSE Class 12 Physics\nQuestion\nIn an AC generator, a coil with N turns and area A spins at angular speed ω in a uniform magnetic field B. Derive the instantaneous emf and name the energy source that keeps the generator running.\n[No diagram provided]\nProgressive Hints\nReveal hints one at a time as you need them.\nThings to Consider\nConcept Notes\nFaraday’s Law\nInduced emf is proportional to the negative time rate of change of magnetic flux through a circuit.\nRotational emf\nA coil rotating in a uniform field produces a sinusoidal emf because the projected area facing the field varies as \\(\\cos \\omega t\\).\nImportant Formula:\n\\( \\varepsilon(t)=\\varepsilon_0 \\sin \\omega t,\\; \\varepsilon_0=N B A \\omega \\)\nPrevious Question\nCheck Solution",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Hint 1: Magnetic Flux\nWrite flux through the rotating coil: \\( \\Phi(t)=N B A \\cos \\omega t \\).\nKeep the negative sign in Faraday’s law when differentiating.",
"image_description": ""
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{
"fragment_index": 2,
"text_description": "Hint 2: Apply Faraday’s Law\nUse \\( \\varepsilon(t)= -\\dfrac{d\\Phi}{dt} \\) to differentiate the flux with respect to time.\nEmf varies sinusoidally; it is not constant.",
"image_description": ""
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{
"fragment_index": 3,
"text_description": "Hint 3: Final Expression & Source\nAfter differentiating, \\( \\varepsilon(t)=N B A \\omega \\sin \\omega t\\). Peak emf \\( \\varepsilon_0=N B A \\omega\\). The coil is driven by mechanical energy from a turbine or engine.\nAxis of rotation must be perpendicular to B for maximum flux change.",
"image_description": ""
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{
"fragment_index": 4,
"text_description": "Mechanical work converts to electrical energy—verify energy conservation.",
"image_description": ""
}
]
},
{
"slide": 13,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Concept Notes\nRay Optics – Telescope Calculations\nQuestion\nA refracting telescope has an objective of focal length 15 m and an eyepiece of 1 cm.\n(a) Find its angular magnification in normal adjustment.\n(b) Calculate the diameter of the Moon’s image formed by the objective lens.\nNo diagram provided\nProgressive Hints\nTry to solve the problem on your own first. If you need help, reveal the hints one by one.\nThings to Consider\nUse small-angle approximation: \\( \\sin\\theta \\approx \\theta\\) (radians).\nOnly the objective decides image size; tube length is not used.\nKeep focal lengths in the same unit system throughout.\nExpress answers with proper significant figures and units.\nConcept Notes\nAngular Magnification\nFor an astronomical telescope in normal adjustment, \\(M = -\\dfrac{f_o}{f_e}\\). The sign indicates an inverted image; magnitude is used.\nImage Formation\nA distant object subtends angle \\( \\theta \\). The objective forms a real image of height \\(y = f_o \\theta\\) at its focal plane.\nImportant Formula:\n\\( \\theta_{\\text{Moon}} \\approx \\dfrac{D_{\\text{Moon}}}{R_{\\text{orbit}}} \\)\nPrevious Question\nCheck Solution",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Hint 1: Getting Started\nConvert \\(f_e = 1\\,\\text{cm}\\) to metres and remember the final image is at infinity in normal adjustment.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Hint 2: Key Approach\nUse \\(M = f_o / f_e\\). With \\(f_o = 15\\,\\text{m}\\) and \\(f_e = 0.01\\,\\text{m}\\), find \\(M\\).",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Hint 3: Almost There\nThe Moon’s angular diameter is \\(\\theta \\approx D/R = 3.48\\times10^{6} / 3.8\\times10^{8}\\,\\text{rad}\\). Image diameter: \\(y = f_o \\theta\\).",
"image_description": ""
}
]
},
{
"slide": 14,
"fragments": [
{
"fragment_index": -1,
"text_description": "Atoms – Spectral Lines Riddle\nBohr Model & Atomic Spectra\nQuestion\nWhy does hydrogen, with only one electron, exhibit many spectral lines in its emission spectrum?\nProgressive Hints\nTry to solve the problem on your own first. If you need help, reveal the hints one by one.\nThings to Consider\nOne photon line = one electron transition.\nSample contains billions of atoms, not a single atom.\nBohr energies follow \\(E_n\\propto -1/n^2\\).\nSeries (Lyman, Balmer, Paschen) arise from different lower levels.\nConcept Notes\nBohr Energy Levels\nAllowed orbits have fixed energies; electrons jump between them by absorbing or emitting photons.\nTransition Rules\nDownward jumps produce emission lines; energy difference sets photon frequency \\( \\nu = \\Delta E/h \\).\nImportant Formula/Rule:\n\\(E_n = -\\dfrac{13.6\\ \\text{eV}}{n^2}\\;\\;\\;\\Delta E = 13.6\\Bigl(\\dfrac{1}{n_1^2}-\\dfrac{1}{n_2^2}\\Bigr)\\)\nPrevious Question\nCheck Solution",
"image_description": ""
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{
"fragment_index": 1,
"text_description": "Hint 1: Getting Started\nA discharge tube holds countless hydrogen atoms, each with its own electron.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Hint 2: Key Approach\nDifferent atoms get excited to different Bohr levels \\(n = 2,3,4,\\dots\\).",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Hint 3: Almost There\nEach electron can drop between two levels, releasing photons of energy \\(E = 13.6\\,(1/n_1^2-1/n_2^2)\\:\\text{eV}\\), creating many lines.",
"image_description": ""
}
]
},
{
"slide": 15,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Take-aways",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Constants Cheat-Sheet\nWrite c, h, e, μ\n0\n, ε\n0\nand N\nA\non the answer sheet margin for instant recall.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Formula First, Numbers Next\nAlways state the equation before substitution to secure partial marks.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Dimensional Cross-Check\nVerify unit consistency to spot errors quickly, especially in MCQs.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Smart Time Split\n45 min (A+B), 60 min (C), 35 min (D), 40 min (E); keep last 10 min for review.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Attempt Low-Risk First\nStart with sure-shot one-markers to build momentum and confidence.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Mental Math Matters\nPractice quick powers-of-ten estimates; calculators are not allowed.",
"image_description": ""
}
]
}
]