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[
{
"slide": 1,
"fragments": [
{
"fragment_index": -1,
"text_description": "CBSE Class 12 Physics – Sample Paper Review\nDecode the exam blueprint to power your 70-mark success.",
"image_description": ""
}
]
},
{
"slide": 2,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question Distribution & Your Strength\nMatch chapter-wise question load with your proficiency to target revision smartly.",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Chapter (Syllabus)\nQuestions in Paper\nYour Strength\nElectric Charges & Fields\n3\nLow\nMoving Charges & Magnetism\n4\nLow\nAlternating Current\n2\nLow\nWave Optics\n3\nLow\nNuclei\n2\nLow\nElectrostatic Potential & Capacitance\n2\nMed\nMagnetism & Matter\n0\nMed\nElectromagnetic Waves\n1\nMed\nDual Nature of Radiation & Matter\n2\nMed\nSemiconductor Electronics\n3\nMed\nCurrent Electricity\n4\nHigh\nElectromagnetic Induction\n1\nHigh\nRay Optics & Optical Instruments\n3\nHigh\nAtoms\n1\nHigh",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Source: CBSE Grade 12 Self-assessment, 2024 Paper",
"image_description": ""
}
]
},
{
"slide": 3,
"fragments": [
{
"fragment_index": 1,
"text_description": "Question with Hints and Nudges\nElectric Charges & Fields – Gauss’s Law Application\nDifficulty: Hard\nSelf-Discovery",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "The Question:\nUsing Gauss’s law, derive the electric field \\(E(r)\\) at distance \\(r\\) from an infinitely long straight wire carrying linear charge density \\( \\lambda \\).\nImagine a coaxial cylindrical Gaussian surface.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Helpful Concepts\nGauss’s law \\( \\Phi = \\dfrac{Q_{\\text{encl}}}{\\varepsilon_0} \\)\nCylindrical symmetry → radial & uniform field\nCurved area of cylinder: \\( 2\\pi r L \\)\nDerivation practice\nGauss’s theorem",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Gentle Nudge\nThink of a Gaussian surface that shares the wire’s cylindrical symmetry.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Direction Pointer\nEnclose length \\(L\\) of wire inside a coaxial cylinder of radius \\(r\\); ignore the flat caps.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Guiding Framework\nApply \\( \\Phi = E(2\\pi r L) = \\lambda L / \\varepsilon_0 \\). Solve to get \\( E = \\dfrac{\\lambda}{2\\pi \\varepsilon_0 r} \\). Do not count end-caps.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Break It Down\nIdentify symmetry, choose surface, compute flux, then isolate \\(E\\).",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Connect to Prior Knowledge\nRecall the Gauss’s-law derivation for an infinite plane to spot parallels.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Visualize It\nSketch the wire and a dashed cylindrical surface to see where flux exits.",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "Test a Simpler Case\nCheck if \\(E \\propto 1/r\\); doubling \\(r\\) should halve your answer.",
"image_description": ""
},
{
"fragment_index": -1,
"text_description": "Struggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
}
]
},
{
"slide": 4,
"fragments": [
{
"fragment_index": 1,
"text_description": "Question with Hints and Nudges\nGrade 12 Physics – Moving Charges & Magnetism\nDifficulty: High\nSelf-Discovery",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "The Question:\nA \\(+10\\;\\text{mC}\\), \\(10\\;\\text{g}\\) sphere inside a vertical insulating tube stays at rest while the tube slides horizontally\neast → west\nthrough a uniform \\(2\\;\\text{T}\\) magnetic field. Find (i) the minimum tube speed and (ii) the field direction that keeps the sphere suspended.\nNo diagram required",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Helpful Concepts\nMagnetic force on a moving charge: \\(F_B = qvB\\) (perpendicular case).\nEquilibrium condition: \\(qvB = mg\\).\nRight-hand rule: direction of \\(q\\,\\mathbf{v}\\times\\mathbf{B}\\).\nLorentz force\nGravity balance",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Gentle Nudge\nIdentify all forces and set their vector sum to zero.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Direction Pointer\nUse \\(qvB\\) upward to cancel \\(mg\\) downward; thus \\(qvB = mg\\).",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Guiding Framework\nSolve \\(v = \\frac{mg}{qB} = \\frac{0.01 \\times 9.8}{0.01 \\times 2} = 4.9\\;\\text{m s}^{-1}\\). For a westward \\( \\mathbf{v} \\) and positive \\(q\\), \\( \\mathbf{B} \\) must point south so that \\( \\mathbf{v}\\times\\mathbf{B} \\) is upward.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Break It Down\nList forces: gravity (down) and magnetic (unknown). No tension or normal force acts.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Connect to Prior Knowledge\nRecall the Lorentz force expression and its dependence on speed and field.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Visualize It\nSketch \\( \\mathbf{v} \\) west, choose \\( \\mathbf{B} \\) south, verify \\( \\mathbf{v}\\times\\mathbf{B} \\) points up.",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "Test a Simpler Case\nSet \\(q = 1\\;\\text{C}\\) mentally to see how speed scales with charge.",
"image_description": ""
},
{
"fragment_index": 13,
"text_description": "Struggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
}
]
},
{
"slide": 5,
"fragments": [
{
"fragment_index": 1,
"text_description": "Alternating Current – Toughest Question\nLCR Circuit Surprise\nDifficulty: Hard\nSelf-Discovery",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "The Question:\nIn a series LCR circuit \\(V_R = V_C = V_L = 10\\,\\text{V}\\). The capacitor is short-circuited. What is the new rms voltage across the inductor?",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Helpful Concepts\nSeries resonance: \\(X_L = X_C\\); net reactance zero.\nAt resonance, supply voltage equals \\(V_R\\).\nShorting \\(C\\) leaves an \\(L\\!-\\!R\\) series path with the same \\(X_L\\).\nresonance\nreactance",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Gentle Nudge\nEqual voltages on \\(R, L, C\\) indicate the circuit is at resonance.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Direction Pointer\nAt resonance, \\(V_{\\text{supply}} = V_R = 10\\text{ V}\\). So \\(I = 10 / R\\) and \\(X_L = R\\).",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Guiding Framework\nWith \\(C\\) shorted, \\(Z = \\sqrt{R^{2}+X_L^{2}} = R\\sqrt{2}\\). New \\(I = 10/(R\\sqrt{2})\\). Hence \\(V_L = I X_L = 10/\\sqrt{2} \\approx 7.1\\text{ V}\\).",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Break It Down\nFirst decide what resonance tells you about \\(X_L\\) and \\(X_C\\).",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Connect to Prior Knowledge\nRecall impedance triangles to combine \\(R\\) and \\(X_L\\).",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Visualize It\nSketch phasors for \\(V_R, V_L, V_C\\) before and after removing \\(C\\).",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "Test a Simpler Case\nTry \\(R = 10\\,\\Omega, X_L = 10\\,\\Omega\\) numerically to verify the formula.",
"image_description": ""
},
{
"fragment_index": -1,
"text_description": "Struggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
}
]
},
{
"slide": 6,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nWave Optics – Young’s Double-Slit\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:\nLight of 400 nm and 600 nm passes through a double-slit. How far from the central maximum will the first common dark fringe appear?\nDiagram not required for this calculation.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nDark fringe: path difference \\( \\delta =(m+\\tfrac12)\\lambda \\).\nCommon darkness when \\( (m+\\tfrac12)\\lambda_{400}=m\\lambda_{600} \\).\nScreen position: \\( y=\\delta \\,D/d \\) for the interference pattern.\nInterference pattern\nMixed λ fringe",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nWrite the dark-band condition for each colour, then set the path differences equal.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nAssume \\( (m+\\tfrac12)\\lambda_{400}=m\\lambda_{600} \\). Solve for the smallest positive integer \\( m \\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nYou get \\( m=1 \\), so \\( \\delta =600\\text{ nm} \\). Insert this into \\( y=\\delta D/d \\) for the required distance.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Break It Down\nHandle each wavelength separately, then combine results.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Connect to Prior Knowledge\nRemember bright-overlap rules; here add the half-order shift for darkness.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize It\nSketch both fringe sets; mark where the first overlap vanishes.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nTry λ ratio 1 : 2 to see the pattern before tackling 400 : 600 nm.",
"image_description": ""
}
]
},
{
"slide": 7,
"fragments": [
{
"fragment_index": -1,
"text_description": "Nuclei – Toughest Question\nFission vs Fusion Decision\nDifficulty: Challenging\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:\nOn the binding-energy per nucleon graph, points W(190), X(90), Y(60) and Z(30) are marked. Which nucleus can release energy by fission and which by fusion? Give a brief reason.\nBinding energy per nucleon vs mass number (schematic)",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nBinding energy per nucleon (B.E./A) gauges stability.\nCurve peaks near A ≈ 56–60 (iron–nickel region).\nSystems move toward the peak to release nuclear energy.\nnuclear energy\nB.E./A curve",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nFirst, see how far each A value lies from the peak around 60.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nHeavy nuclei to the right split; very light nuclei to the left combine—both moves aim for A ≈ 60.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nW(190) can increase B.E./A by splitting into fragments near 90 → fission. Z(30) can gain B.E./A by joining another light nucleus to reach ~60 → fusion.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Break It Down\nOrder the four A values, then compare each with 60.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Connect to Prior Knowledge\nRecall that iron-nickel are most stable due to highest B.E./A.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize It\nSketch the B.E./A curve and plot W, X, Y, Z.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nThink of uranium fission and hydrogen fusion as extreme examples.",
"image_description": ""
}
]
},
{
"slide": 8,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electrostatic Potential & Capacitance – Toughest Question\nCBSE Grade 12 Physics\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:\nParallel-plate capacitor: plates area \\(A\\), air gap \\(d\\). Case 1 – dielectric slab thickness \\(t\\), permittivity \\(\\varepsilon_r\\). Case 2 – metal slab thickness \\(t<d\\). Derive capacitances \\(C_1, C_2\\) and decide which is larger.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\n\\(C=\\frac{\\varepsilon_0 A}{d}\\) for a parallel-plate capacitor.\nSeries dielectrics: \\(C=\\frac{\\varepsilon_0 A}{\\sum d_i/\\varepsilon_{ri}}\\).\nInside a conductor \\(E=0\\); it simply shortens the effective gap.\nElectrostatics\nCapacitors",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nTreat the slab and remaining air as two capacitors in series.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nWrite \\(V=E_1 t + E_2(d-t)\\) and substitute \\(E=\\sigma/\\varepsilon\\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nObtain \\(C_1=\\frac{\\varepsilon_0 A}{d - t + t/\\varepsilon_r}\\). For the metal slab \\(E=0\\) inside, so \\(C_2=\\frac{\\varepsilon_0 A}{d-t}\\). Compare denominators.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Break It Down\nSeparate the problem into slab region and air region.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Connect to Prior Knowledge\nRemember electric field behaviour inside conductors and dielectrics.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize It\nDraw a side view labelling \\(t\\), \\(d-t\\), and field directions.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nCheck results for \\(t=0\\) or \\(\\varepsilon_r=1\\) to build confidence.",
"image_description": ""
}
]
},
{
"slide": 9,
"fragments": [
{
"fragment_index": -1,
"text_description": "Magnetism & Matter – Representative Question\nField Inside a Current-Carrying Wire\nDifficulty: Moderate\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:\nFor a long solid wire of radius \\(a\\) carrying uniform current \\(I\\), find the ratio \\(B_{\\text{above}}:B_{\\text{below}}\\) where “above” is \\(\\frac{a}{2}\\) outside the surface and “below” is \\(\\frac{a}{2}\\) inside the surface.\nNo diagram required",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nAmpère’s law: \\( \\oint \\mathbf{B}\\cdot d\\mathbf{l}= \\mu_{0} I_{\\text{enc}}\\).\nInside (\\(r<a\\)): \\(B(r)=\\frac{\\mu_{0} I r}{2\\pi a^{2}}\\) — linear magnetic field distribution.\nOutside (\\(r>a\\)): \\(B(r)=\\frac{\\mu_{0} I}{2\\pi r}\\) — inverse magnetic field distribution.\nAmpère’s law\nMagnetic field distribution",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nIdentify whether each point lies inside or outside the wire by calculating its distance from the axis.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nUse \\(B(r)=\\frac{\\mu_{0} I r}{2\\pi a^{2}}\\) for \\(r=\\frac{a}{2}\\) and \\(B(r)=\\frac{\\mu_{0} I}{2\\pi r}\\) for \\(r=\\frac{3a}{2}\\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nCompute \\(B_{\\text{below}}=\\frac{\\mu_{0}I}{2\\pi a^{2}}\\cdot\\frac{a}{2}\\) and \\(B_{\\text{above}}=\\frac{\\mu_{0}I}{2\\pi}\\cdot\\frac{2}{3a}\\). Then form the ratio.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Break It Down\nDraw the wire cross-section, label \\(r=\\frac{a}{2}\\) and \\(r=\\frac{3a}{2}\\).",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Connect to Prior Knowledge\nRecall that inside a uniformly current-filled region \\(B\\propto r\\); outside \\(B\\propto 1/r\\).",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize It\nSketch concentric circular field lines with varying spacing inside and outside the wire.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nEvaluate \\(B\\) at the centre and at the surface to verify you are using the correct formula.",
"image_description": ""
}
]
},
{
"slide": 10,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nEM Spectrum (Grade 12 Physics)\nDifficulty: Tough\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:\nMatch each electromagnetic wave with its usual production mechanism.\n1. Radio waves a) Rapid oscillation of conduction electrons in an aerial\n2. Microwaves b) Klystron / magnetron valve\n3. Infra-red c) Vibrations of atoms & molecules in matter\n4. Visible light d) Electronic transitions within atoms\nWrite the correct 1-a, 2-b … pairs.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nEM spectrum orders waves by increasing frequency.\nDifferent frequency bands arise from different physical processes.\nRecall antenna currents, vacuum tubes, molecular heat, and atomic jumps.\nWave Spectrum\nSources",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nStart with the extremes: radio waves come from antenna currents.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nThink of the device that powers radar and microwave ovens—the klystron/magnetron.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nMap now: conduction electrons → radio, vacuum tube → microwave, molecular vibration → IR, atomic transition → visible.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Break It Down\nList the four mechanisms, then match one wave at a time.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Connect to Prior Knowledge\nRecall how your radio and microwave oven generate waves.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Visualize It\nSketch the EM spectrum and label each source.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Test a Simpler Case\nIf you only knew radio waves, what mechanism might the next higher band use?",
"image_description": ""
}
]
},
{
"slide": 11,
"fragments": [
{
"fragment_index": -1,
"text_description": "Dual Nature – Toughest Question\nDecoding Photoelectric Curves\nDifficulty: Challenging\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:\nThree photoelectric I–V curves A, B and C are recorded with different monochromatic lights on the same metal. For every pair (A,B), (A,C) and (B,C), state whether the two lights share intensity, frequency or neither, and justify from the curves.\nCurve sketch goes here if provided",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nSaturation current ∝ photon flux → beam intensity.\nStopping potential \\(V_{0}= \\frac{h\\nu - \\phi}{e}\\) tracks photon frequency.\nPlateau height & horizontal intercept reveal intensity and frequency on I–V graphs.\nPhotoelectric Equation\nDual Nature",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nMatch curves whose plateaus reach the same current; those beams have identical intensity.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nEqual stopping potentials mean equal photon energy, hence same frequency.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nIdentify pairs with matching plateau → same intensity; matching intercept → same frequency. If neither matches, both differ.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Thinking Strategies\nBreak It Down\nExamine current plateau and cutoff voltage separately before combining clues.\nConnect to Prior Knowledge\nRecall Einstein’s explanation linking photon energy to electron kinetic energy.\nVisualize It\nSketch vertical lines from each plateau to see overlaps quickly.\nTest a Simpler Case\nImagine two identical lamps with different filters to predict graph changes.",
"image_description": ""
}
]
},
{
"slide": 12,
"fragments": [
{
"fragment_index": 1,
"text_description": "Semiconductor Electronics – Toughest Question\nGrade 12 Physics | Rectification\nDifficulty: Hard\nSelf-Discovery",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "The Question:\nIn the AC → X → Y → DC block, name X and Y. Sketch the voltage after each stage. How will the DC waveform change if the transformer’s centre-tap moves closer to \\(D_1\\)?",
"image_description": "https://asset.sparkl.ac/pb/sparkl-vector-images/img_ncert/ckkcmoZnMj9gdTY2voD6eNUP314n6f9aMI2aZIj9.png"
},
{
"fragment_index": 3,
"text_description": "Helpful Concepts\nFull-wave bridge and centre-tap rectifiers\nTransformer: step-down & isolation\nCapacitive filter to cut ripple\nPower Supplies\nRectification",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Gentle Nudge\nFirst, identify what lowers AC voltage before diodes take over.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Direction Pointer\nX = step-down transformer. Y = full-wave rectifier followed by a smoothing capacitor.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Guiding Framework\nDraw sine after X. After Y, both half-cycles become positive (pulsating DC). Moving the tap toward \\(D_1\\) enlarges one half-cycle and shifts average DC upward, adding ripple.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Break It Down\nMatch each block to its role: transform, rectify, filter.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Connect to Prior Knowledge\nRecall lab work on bridge and centre-tap rectifiers.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Visualize It\nSketch voltage across the load after each stage.",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "Test a Simpler Case\nAssume ideal diodes and equal halves before considering tap shift.",
"image_description": ""
}
]
},
{
"slide": 13,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nCurrent Electricity – Network Analysis\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 wire of total resistance 12 Ω is bent into a circle. A 10 Ω resistor joins the opposite points C and D. A battery of emf 8 V is connected across another pair of opposite points A and B. Determine the current through the arc A → D.\nSketch the equivalent Wheatstone bridge on your notepad.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nKirchhoff’s current and loop laws for network analysis.\nWheatstone bridge equivalence and balance condition.\nSeries–parallel reduction of resistors to simplify circuits.\nnetwork analysis\nsymmetry",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nThink of the circle as two equal 6 Ω semicircular arcs between A–C–B and A–D–B.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nRedraw the arrangement as a Wheatstone bridge where the 10 Ω resistor is the bridge arm CD.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nApply Kirchhoff’s loop law to two independent loops or test for bridge balance, then solve for branch currents without yet computing the final numeric answer.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies\nBreak It Down\nConvert the circular layout into a four-resistor Wheatstone bridge diagram.\nConnect to Prior Knowledge\nRecall that equal opposite arms can balance a bridge, potentially nullifying current through the 10 Ω path.\nVisualize It\nLabel currents in each branch and mark assumed directions before writing \\( \\sum V = 0 \\) equations.\nTest a Simpler Case\nImagine removing the 10 Ω resistor; predict current split, then re-introduce it to see the change.",
"image_description": ""
}
]
},
{
"slide": 14,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electromagnetic Induction – Toughest Question\nGrade 12 Physics – CBSE\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:\nA coil with \\(N\\) turns and area \\(A\\) rotates at angular speed \\( \\omega \\) in a uniform magnetic field \\(B\\). Using Faraday’s law, derive the instantaneous emf \\( \\varepsilon = N A B \\omega \\sin \\omega t \\). Also state the practical source of this electrical energy.\nNo diagram provided",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nFaraday’s law: \\( \\varepsilon = -\\dfrac{d\\Phi}{dt} \\)\nMagnetic flux: \\( \\Phi = B A \\cos \\theta \\)\nLenz’s law (negative sign)\nEM Induction\nAC Generator",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nExpress the magnetic flux as \\( \\Phi = B A \\cos (\\omega t) \\).",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nDifferentiate \\( \\Phi \\) with respect to time and apply Faraday’s law: \\( \\varepsilon = -\\dfrac{d\\Phi}{dt} \\).",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nYou should arrive at \\( \\varepsilon = N A B \\omega \\sin \\omega t \\) with peak value \\( \\varepsilon_0 = N A B \\omega \\). In real generators, mechanical work from turbines supplies the energy.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Break It Down\nFirst write flux, then differentiate—two clear steps.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Connect to Prior Knowledge\nRecall Faraday’s and Lenz’s laws determine magnitude and sign.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Visualize It\nPicture the coil’s plane sweeping through the magnetic field lines.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Test a Simpler Case\nTry one loop (\\(N=1\\)) to confirm the sinusoidal nature of \\( \\varepsilon \\).",
"image_description": ""
}
]
},
{
"slide": 15,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nRay Optics – Prism\nDifficulty: Hard\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 right-angled prism (\\(C = 45^{\\circ}\\)) receives a ray as shown. Side AC is immersed in a liquid of refractive index \\(n = \\frac{2}{\\sqrt{3}}\\). Will the emerging ray graze, refract, or undergo total internal reflection? Draw its path.\n[Diagram as given in paper]",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nCritical angle \\(C\\) : \\(\\sin C = \\dfrac{n_{\\text{medium}}}{n_{\\text{prism}}}\\)\nTotal internal reflection when incident angle > critical angle.\nSnell’s law links angles and refractive indices.\nSnell’s Law\nPrisms",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nRecall the prism’s refractive index you found earlier when the ray just grazed in air.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nCompute new critical angle at AC: \\(\\sin C = \\dfrac{n_{\\text{liquid}}}{n_{\\text{prism}}}\\).",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nCompare \\(45^{\\circ}\\) with the critical angle. If \\(45^{\\circ}\\) < \\(C\\), TIR; if equal, grazing; if >, refraction into liquid.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Break It Down\nCalculate critical angle first, then decide the ray’s fate.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Connect to Prior Knowledge\nYou earlier assumed air outside—update that step for liquid.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Visualize It\nDraw normals at AC and mark incident \\(45^{\\circ}\\) angle.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Test a Simpler Case\nImagine liquid is air to see why outcome changes.",
"image_description": ""
}
]
},
{
"slide": 16,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nGrade 12 Physics – Bohr Model\nDifficulty: Challenging\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:\nHydrogen has only one electron. Why does its emission spectrum still show many discrete lines?\nEnergy-level diagram could be sketched here",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nBohr levels: \\(E_n=-13.6\\,\\text{eV}/n^2\\)\nPhoton energy: \\(h\\nu = E_i-E_f\\) during a downward jump\nA discharge tube contains billions of atoms acting independently\nAtomic Spectra\nEnergy Quantization",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nConsider how many hydrogen atoms are present in a glowing sample.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nDifferent atoms can have their lone electron in different excited states at the same time.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nEach occupied excited level \\(n_i\\) can decay through several lower levels \\(n_f\\). Every allowed transition emits a photon with a unique energy, producing multiple spectral lines.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Thinking Strategies\nBreak It Down\nList energy levels, then count possible downward jumps.\nConnect to Prior Knowledge\nRecall the Balmer lines and their level diagram.\nVisualize It\nSketch Bohr levels and draw arrows for transitions.\nTest a Simpler Case\nImagine only two levels and observe how one line appears.",
"image_description": ""
}
]
},
{
"slide": 17,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Take-aways",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Symmetry Saves Time\nQuick recap: use symmetry with Gauss’s law and Ampère’s rule to write fields instantly.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Spot Resonance\nRemember: at resonance current peaks as \\(X_L = X_C\\); reactances switch sign beyond this point.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Interference Pits\nDark fringes shift for two colours; recall \\(d\\sin\\theta=(m+1/2)\\lambda\\) separately.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Binding Energy Peak\nRemember: nuclei are most stable at \\(A \\approx 60\\); fusion below and fission above release energy.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Rectifier Chain\nRecap power-supply sequence: transformer → rectifier → filter conditions voltage for circuits.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Sketch Before Math\nThing to remember: sketch and label angles before equations; clarity prevents sign errors.",
"image_description": ""
}
]
}
]