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[
{
"slide": 1,
"fragments": [
{
"fragment_index": -1,
"text_description": "Physics Sample Paper Review\nDecode the blueprint for acing the 2024–25 exam.",
"image_description": ""
}
]
},
{
"slide": 2,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question Distribution & Your Strength",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Chapter\nMarks in Paper\nYour Proficiency\nElectrostatics\n8\nStrong\nCurrent Electricity\n7\nStrong\nMagnetic Effects of Current\n6\nAverage\nE.M. Induction & AC\n6\nAverage\nOptics\n14\nStrong\nDual Nature of Radiation\n7\nNeeds Work\nAtoms & Nuclei\n8\nAverage\nSemiconductor Electronics\n10\nNeeds Work\nCommunication Systems\n4\nNeeds Work",
"image_description": ""
}
]
},
{
"slide": 3,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nElectric Fields\nDifficulty: hard\nSelf-Discovery\nProgressive Hints\n(Reveal only when needed)\nThinking Strategies\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nDerive the electric field \\(E\\) at a distance \\(r\\) from an infinitely long straight wire carrying uniform linear charge density \\(\\lambda\\) using Gauss’s law.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nCylindrical Gaussian surface\nRadial symmetry of field lines\n\\(\\Phi=\\frac{q_{\\text{enc}}}{\\varepsilon_0}\\) relation\nGauss’s Law\nCylindrical Symmetry",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Visualise Surface\nWrap a coaxial cylinder of length \\(L\\) and radius \\(r\\) around the wire. Charge enclosed \\(= \\lambda L\\).",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Flux Segments\nField is radial. The flat end-caps are perpendicular to \\(E\\), so their flux is zero; only the curved surface contributes.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Apply Gauss\n\\(\\Phi = E(2\\pi r L) = \\frac{\\lambda L}{\\varepsilon_0}\\). Therefore \\(E = \\frac{\\lambda}{2\\pi \\varepsilon_0 r}\\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Common Mistake\nAdding flux from end caps doubles the area and yields an incorrect field.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Fix\nCheck field direction. Since \\(E\\) is radial, \\(E \\cdot dA = 0\\) on the caps.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Visualize It\nDraw outward field lines; they should be evenly spaced around the wire.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Test a Simpler Case\nDouble \\(r\\): your formula should halve \\(E\\). Quick sanity check.",
"image_description": ""
}
]
},
{
"slide": 4,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nMoving Charges & Magnetism\nDifficulty: hard\nSelf-Discovery\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nTwo identical particles of charge\n+q\nmove parallel to the\nY-axis\nwith speed\n\\(2.4 \\times 10^{5}\\, \\text{m s}^{-1}\\)\n. Each starts\n0.5 m\nfrom the axis but on opposite sides, heading toward it. Determine the direction and magnitude of a uniform magnetic field that makes the particles collide head-on at the axis.\nNo diagram provided",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\n\\( \\vec{F}=q\\vec{v}\\times\\vec{B} \\)\nOpposite sides ⇒ forces must point inward simultaneously\nUniform circular motion: \\( r=\\dfrac{m v}{q B} \\)\nLorentz Force\nCircular Motion",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nEach particle must experience a magnetic force directed toward the Y-axis.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nUse the right-hand rule on the left and right particles. Which choice, \\(+\\hat{z}\\) or \\(-\\hat{z}\\), bends both paths inward?",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nSet the required curvature: \\( r = 0.5\\,\\text{m} \\). Apply \\( r=\\dfrac{m v}{q B} \\) to find \\( B \\). Keep the sign that gave inward forces.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Break It Down\nSketch each trajectory and mark its centre of curvature midway on the axis.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Connect to Prior Knowledge\nRecall \\( \\vec{F}=q\\vec{v}\\times\\vec{B} \\) points perpendicular to both velocity and field.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize It\nDraw a top view showing the Y-axis and initial particle positions to test field directions quickly.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nConsider only one particle first; choose \\( \\vec{B} \\) so its path curves toward the axis.",
"image_description": ""
}
]
},
{
"slide": 5,
"fragments": [
{
"fragment_index": -1,
"text_description": "Series LCR – Capacitance Shorted\nAlternating Current\nDifficulty: medium\nSelf-Discovery\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nIn a series LCR circuit \\(V_R = V_L = V_C = 10\\text{ V}\\). The capacitor is suddenly short-circuited. What is the new value of \\(V_L\\)?",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nPhasor triangles for series circuits\nAt resonance, \\(X_L = X_C\\)\nImpedance after removing \\(C\\): \\(Z=\\sqrt{R^{2}+(\\omega L)^{2}}\\)\nLCR circuits\nCBSE Grade 12",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Before Change\nEqual voltages across R, L, C mean resonance. Therefore \\(X_L = X_C\\) and \\(Z=R\\).",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "After Short\nWith \\(C\\) shorted, only \\(R\\) and \\(L\\) remain: \\(Z'=\\sqrt{R^{2}+X_L^{2}}\\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Voltage Division\nCurrent rises from \\(I=\\frac{10}{R}\\) to \\(I'=\\frac{10}{\\sqrt{R^{2}+X_L^{2}}}\\). New \\(V_L = I'X_L\\).",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Pitfall\nDo not assume \\(V_L\\) stays 10 V; the current changes when \\(C\\) is removed.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Check\nFirst find new current with updated impedance, then compute \\(V_L\\).",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Break It Down\nTreat the problem in two phases: before and after shorting the capacitor.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Visualize It\nDraw phasor diagrams to see how current and voltage phasors rotate.",
"image_description": ""
}
]
},
{
"slide": 6,
"fragments": [
{
"fragment_index": -1,
"text_description": "Two-Colour YDSE Dark Fringe\nWave Optics\nDifficulty: hard\nSelf-Discovery\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nTwo coherent sources in Young’s double-slit experiment emit light of wavelengths 400 nm and 600 nm. How far from the central bright fringe will the first common dark fringe appear on the screen?\nNo diagram provided",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nCommon dark fringe: \\(m_1\\lambda_1 = m_2\\lambda_2\\)\nLeast path difference = LCM of wavelengths\nFringe position: \\(y = \\frac{\\lambda D}{d} \\times m\\)\nInterference\nYoung’s Experiment",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nWrite \\(m_1\\lambda_1 = m_2\\lambda_2\\) for minima and look for the smallest non-zero integer pair.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nThe least common path difference equals the LCM of 400 nm and 600 nm, i.e. 1200 nm.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nConvert \\(Δ = 1200\\text{ nm}\\) to position: \\(y = ΔD/d\\). Since \\(β_{400} = \\lambda_1 D/d\\), we get \\(y = 3β_{400}\\).",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies\nCommon Mistake\nStudents often average the two wavelengths. This never gives the correct common minimum.\nFix the Error\nUse integer multiples so each wavelength independently satisfies the dark-fringe condition.\nBreak It Down\nStep 1: find least common path difference. Step 2: translate that distance to screen position.\nVisualise It\nSketch two fringe patterns; mark where dark bands coincide to see the pattern repeat every 3 bright spacings.",
"image_description": ""
}
]
},
{
"slide": 7,
"fragments": [
{
"fragment_index": 1,
"text_description": "Binding-Energy Curve Analysis\nGrade 12 Physics – Nuclei\nDifficulty: hard\nEst. Time: 5 min",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Problem Context\nRefer to the binding-energy-per-nucleon (B.E./A) versus mass number A graph labelled W, X, Y and Z.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Binding energy per nucleon curve",
"image_description": "https://asset.sparkl.ac/pb/sparkl-vector-images/img_ncert/LAyoBcF0IxM2tDJRivHdWZTSGiUmuyddZhzra3HQ.png"
},
{
"fragment_index": 4,
"text_description": "Question\nIdentify which labelled nucleus is most suitable for (a) fission and (b) fusion. Justify using the features of the curve.\na) Likely fission nucleus: ______\nb) Likely fusion nucleus: ______",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Helpful Hints\nHint 1\nFission favours heavy nuclei with lower B.E./A.\nHint 2\nFusion is profitable when two light nuclei move up the curve.\nHint 3\nCheck slopes near A≈60 (peak) and A≈200 (valley).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Things to Consider\nEnergy released ≈ Δ(B.E./A) × A.\nPeak stability occurs near iron (A≈56).\nHigher B.E./A means a more tightly bound nucleus.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Related Concepts\nMass-energy equivalence\nNuclear stability",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Previous Problem\nProblem 7 of 12\nNext Problem",
"image_description": ""
}
]
},
{
"slide": 8,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nElectrostatic Potential & Capacitance\nDifficulty: medium\nSelf-Discovery\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nBetween parallel plates separated by distance \\(d\\) lies area \\(A\\). A slab of thickness \\(t<d\\) is inserted. (i) When the slab is a dielectric of relative permittivity \\(\\varepsilon_{r}\\), derive the new capacitance \\(C_{d}\\). (ii) Repeat when the slab is a conductor and obtain \\(C_{m}\\). Express answers using \\(A,d,t,\\varepsilon_{0},\\varepsilon_{r}\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nCapacitors in series share common charge.\nDielectric introduces factor \\(\\varepsilon_{r}\\) in field.\nA conductor nullifies electric field inside it.\nSeries Combination\nEffective Separation",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nFor the dielectric, treat the system as two capacitors: one filled with dielectric of thickness \\(t\\), the other air gap \\(d-t\\).",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nWrite total potential as \\(V=E_{1}t+E_{2}(d-t)\\) with \\(E_{1}= \\frac{\\sigma}{\\varepsilon_{0}\\varepsilon_{r}}\\) and \\(E_{2}= \\frac{\\sigma}{\\varepsilon_{0}}\\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nFor the metal slab, the field inside it is zero, so effective plate spacing is \\(d-t\\). Hence \\(C_{m}= \\frac{\\varepsilon_{0}A}{d-t}\\), which is larger than \\(C_{d}\\).",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies\nSlip-up Alert\nDo not sum plate separations directly for a dielectric; treat potential drops separately.\nQuick Fix\nWrite the electric field in each region, add the drops, then relate \\(Q=CV\\).\nVisualize It\nSketch the slab between plates and label regions to see series arrangement.\nCheck Extremes\nLet \\(t=0\\) and \\(t=d\\) to verify if your expressions reduce to known limits.",
"image_description": ""
}
]
},
{
"slide": 9,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nElectromagnetic Waves\nDifficulty: Medium\nSelf-Discovery\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nMatch each band with its usual production mechanism:\n(a) Infra-red (b) Radio (c) Visible light (d) Microwave\nwith\n(i) molecular vibrations (ii) oscillating aerial electrons (iii) atomic electron transitions (iv) klystron/maser cavities.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nSpectrum runs Radio → Microwave → IR → Visible → … (frequency rises).\nSources depend on energy scale: bulk charges vs atomic events.\nWave sources\nEnergy & frequency",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)\nRecall Sources\nMolecular vibrations emit IR; atom electron jumps give visible light.\nMicrowave\nMicrowaves are produced in resonant cavities such as klystrons or masers.\nRadio\nRadio waves arise from rapid oscillation of free electrons in an antenna.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Thinking Strategies\nCommon Mistake\nDo not swap IR (low energy) with microwave sources.\nEnergy Check\nHigher frequency bands require higher-energy processes; compare with frequency order.\nVisualize It\nSketch the EM spectrum and label typical sources beside each region.\nTest a Simpler Case\nAsk: which source could you feel as heat? That guides IR placement.",
"image_description": ""
}
]
},
{
"slide": 10,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nDual Nature (Photoelectric Effect)\nDifficulty: medium\nSelf-Discovery\nProgressive Hints\n(Reveal only when needed)\nThinking Strategies\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nThree I–V curves A, B and C are obtained for the same metal. Rank the incident beams by (i) intensity and (ii) frequency. Explain your reasoning.\nCurves not shown – focus on interpreting plateau and cut-off regions.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nSaturation current indicates photon flux → intensity\nStopping potential relates to photon energy → frequency\nEinstein equation \\( eV_0 = h\\nu - \\phi \\)\nPhotoelectric Graphs\nCBSE XII",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nFirst compare the plateau heights: a taller plateau means more emitted electrons.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nNext, check which curve cuts off at the most negative voltage; that beam has the highest photon energy.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nIdentify any two curves with equal \\( I_{\\text{sat}} \\): they share intensity. Any two with equal \\( V_0 \\) share frequency. Rank the odd one out accordingly.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Common Error\nDo not judge intensity from the stopping potential; they are unrelated.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Correct Approach\nTreat current as a measure of how many electrons leave and voltage as how much energy each carries.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Visualize It\nSketch each curve, marking the plateau and zero-current points to see differences clearly.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Test a Simpler Pair\nFirst compare just A and B. Once that pattern is clear, slot C into the order.",
"image_description": ""
}
]
},
{
"slide": 11,
"fragments": [
{
"fragment_index": -1,
"text_description": "Full-Wave Rectifier Output\nSemiconductor Electronics\nDifficulty: hard\nSelf-Discovery\nProgressive Hints\n(Reveal only when needed)\nThinking Strategies\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nIdentify blocks X and Y in an AC-DC converter and sketch their output waveforms. Predict what happens to the DC output when the transformer centre-tap is shifted toward diode \\(D_{1}\\).\nNo external diagram provided",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nDifference between rectifier and filter stages\nEach diode conducts during half of the input cycle\nAsymmetric secondary voltages change output amplitude and ripple\nAC-DC Conversion\nRectifier Circuit",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Block Roles\nX contains the diodes that rectify; Y smooths the pulsating DC using a capacitor or inductor.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Ideal Output\nFirst draw the full-wave pulsating DC from the bridge or centre-tap pair, then overlay the filtered, near-steady DC.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Tap Shift\nIf the centre-tap moves toward \\(D_{1}\\), one half-cycle gains voltage while the other loses it, giving a DC output with unequal peaks and a superimposed ripple bias.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Common Mistake\nDo not confuse amplitude change with frequency change; shifting the tap only alters voltage levels.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Fix\nSketch conduction intervals for each diode to visualise how unequal secondary voltages affect output.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Visualise It\nPlot the original sine waves, then mark the rectified halves to see symmetry or lack of it.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Test a Simpler Case\nAssume a small intentional offset in tap position and calculate resulting peak voltages before generalising.",
"image_description": ""
}
]
},
{
"slide": 12,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nCurrent Electricity\nDifficulty: Hard\nSelf-Discovery\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nA uniform 12 Ω wire is bent into a circle with points A, B, C and D marked clockwise. \n A 10 Ω resistor joins C and D, while an 8 V battery is connected between A and B. \n Find the current through arm A D.\nDiagram not provided.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nSymmetry in a circular resistor network\nResistance ∝ subtended angle (\\(60^{\\circ}\\Rightarrow2\\,\\Omega\\))\nKirchhoff’s loop & junction rules\nWheatstone Bridge\nNetwork Analysis",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Segment Resistances\nTotal 12 Ω over 360°. Each 60° arc is 2 Ω. Label A B, B C, C D, D A accordingly.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Bridge Equivalent\nThe four 2 Ω arms form a Wheatstone bridge; the 10 Ω resistor is the bridge between C and D.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Apply Kirchhoff\nWrite loop and junction equations for the two outer loops. Solve for branch currents, then read \\(I_{AD}\\).",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Pitfall\nDon’t skip translating the circular geometry into resistances; symmetry only helps after that step.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Fix\nFirst assign 2 Ω to each 60° segment, then reduce the network before writing equations.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize It\nSketch the bridge with four equal arms and the 10 Ω resistor to see symmetry clearly.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nImagine the 10 Ω removed; check if currents split equally—then add it back to spot changes.",
"image_description": ""
}
]
},
{
"slide": 13,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nElectromagnetic Induction\nDifficulty: Medium\nSelf-Discovery\nProgressive Hints\n(Reveal only when needed)\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nA coil with \\(N\\) turns and area \\(A\\) rotates at angular speed \\( \\omega \\) in a uniform magnetic field \\( B \\). Derive an expression for the induced emf and state the physical source of this electrical energy.\nNo diagram provided",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nFaraday’s law: \\( \\varepsilon = -N \\frac{d\\phi}{dt} \\)\nFlux: \\( \\phi = B A \\cos(\\omega t) \\)\nMechanical work converts to electrical energy\nAC Generator\nAlternating EMF",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nBegin by writing magnetic flux through the coil as a time–dependent function.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nUse \\( \\phi = B A \\cos(\\omega t) \\) and apply Faraday’s law by differentiating with respect to time.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nYou should obtain \\( \\varepsilon = N B A \\omega \\sin(\\omega t) \\). Remember the energy comes from the external torque driving the coil.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Thinking Strategies\nBreak It Down\nExpress flux, then differentiate step-by-step; avoid sign errors.\nConnect to Prior Knowledge\nRecall Faraday’s minus sign indicates Lenz’s law.\nVisualize It\nSketch coil at 0° and 90° to see how flux changes.\nTest a Simpler Case\nAnalyse a single-turn loop; then extend to \\(N\\) turns.",
"image_description": ""
}
]
},
{
"slide": 14,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nRay Optics — Total Internal Reflection\nDifficulty: Hard\nSelf-Discovery\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nA 45°–45°–90° prism has face AC immersed in a liquid of refractive index \\( n=\\frac{2}{\\sqrt3} \\). A ray enters normally at AB and strikes AC at \\( 45^{\\circ} \\) inside the glass. Will it graze along AC, undergo total internal reflection, or refract into the liquid? Draw its path.\nDiagram (if needed)",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nSnell’s law \\( n_1 \\sin i = n_2 \\sin r \\)\nCritical angle \\( \\sin \\theta_c = \\dfrac{n_2}{n_1} \\) when \\( n_1 > n_2 \\)\nRelative refractive index \\( \\mu_\\text{rel}= \\dfrac{n_1}{n_2} \\)\nRay Optics\nPrisms",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Find \\( \\mu_\\text{prism} \\)\nFor grazing emergence in air earlier, \\( \\sin 45^{\\circ}=1/ \\mu \\). Hence \\( \\mu_\\text{prism}= \\sqrt2 \\).",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Compute \\( \\mu_\\text{rel} \\)\n\\( \\mu_\\text{rel}= \\dfrac{\\mu_\\text{prism}}{n_l}= \\dfrac{\\sqrt2}{2/\\sqrt3}= \\frac{3}{2}=1.5 \\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Check critical angle\n\\( \\sin \\theta_c = 1/\\mu_\\text{rel}=2/3 \\Rightarrow \\theta_c \\approx 41.8^{\\circ} < 45^{\\circ} \\). The incident angle is larger—TIR occurs.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Common Slip-up\nMixing up which medium is \\( n_1 \\) in Snell’s law.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "How to Fix\nAlways write the incident side first: \\( n_\\text{incident}\\sin i = n_\\text{refracted}\\sin r \\).",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize Angles\nSketch the normal at AC and mark 45° to see if it exceeds \\( \\theta_c \\).",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nTry \\( n_l = 1 \\) first. If TIR happens there, it must also happen for a denser liquid.",
"image_description": ""
}
]
},
{
"slide": 15,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nAtoms\nDifficulty: Medium\nSelf-Discovery\nProgressive Hints\n(Reveal only when needed)\nThinking Strategies\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nExplain why hydrogen, which has only one electron, still exhibits many spectral lines.\nNo diagram provided",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nLarge ensemble of hydrogen atoms\nDiscrete quantum energy levels\nPhoton emission during level transitions\nHydrogen Spectrum\nEmission Lines",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Population\nDifferent atoms can start in different excited states.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Transitions\nOne electron may drop across many possible level gaps; each gap emits its own photon wavelength.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Series\nGroups like Lyman, Balmer, and Paschen share a common lower level, so multiple photons form each series.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Common Misconception\nDo not assume a single electron gives only one line. Consider many atoms and many starting levels.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Clarify\nInfinite atoms × multiple excited states produce a rich emission spectrum.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Visualize It\nSketch the energy level diagram and draw possible downward arrows.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Test a Simpler Case\nImagine only two upper levels; notice you already get several lines.",
"image_description": ""
}
]
},
{
"slide": 16,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Take-aways",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Formula Flash\nKeep key relations ready: \\(v=u+at\\), \\(E=h\\nu\\), \\(Q=mc\\Delta T\\). No time for derivations.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Diagram First\nQuick sketches of rays, fields, or circuits clarify data and earn method marks.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "60-40 Split\nAllocate 60 % time to short answers, 40 % to numericals; avoid last-minute panic.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Units & Sig-figs\nWrite SI units and correct significant figures; lose no easy marks.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Rough Work Code\nUse side margins, label steps; evaluator spots each scoring point easily.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Mindful Review\nUse last 10 min to check signs, powers, and decimals; rescue avoidable errors.",
"image_description": ""
}
]
}
]