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[
{
"slide": 1,
"fragments": [
{
"fragment_index": -1,
"text_description": "Physics Sample Paper Review\nCrack the 70-mark physics paper with confidence.",
"image_description": ""
}
]
},
{
"slide": 2,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question Distribution & Your Strength\nSee how the paper maps to your syllabus strengths.",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Chapter\nMarks\nWeightage (%)\nYour Proficiency\nElectrostatics\n8\n11\nHigh\nCurrent Electricity\n7\n10\nMedium\nMagnetism\n6\n8\nLow\nEM Induction & AC\n7\n10\nMedium\nEM Waves\n3\n4\nHigh\nOptics\n14\n19\nMedium\nDual Nature & Matter\n4\n6\nLow\nAtoms & Nuclei\n5\n7\nHigh\nSemiconductor Electronics\n6\n8\nMedium",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Source: CBSE Sample Question Paper 2024-25 & your mock test analytics",
"image_description": ""
}
]
},
{
"slide": 3,
"fragments": [
{
"fragment_index": 1,
"text_description": "Question with Hints and Nudges\nElectric Charges & Fields – Challenge\nDifficulty: hard\nSelf-Discovery",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "The Question:\nState Gauss’s law and derive the electric field around an infinitely long, uniformly charged straight wire of linear charge density \\( \\lambda \\).\nNo diagram provided",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Helpful Concepts\nGauss’s surface symmetry\nFlux–field relation\nCylindrical Gaussian surface\nGauss law\nLinear charge",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Gentle Nudge\nChoose a coaxial cylindrical Gaussian surface of radius \\( r \\) and length \\( L \\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Direction Pointer\nOnly the curved surface contributes: \\( \\Phi_E = E(2\\pi r L) \\).",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Guiding Framework\nSet \\( \\Phi_E = \\lambda L/\\varepsilon_0 \\) and solve: \\( E = \\dfrac{\\lambda}{2\\pi \\varepsilon_0 r} \\).",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Break It Down\nUse symmetry: field is radial and uniform over the curved surface.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Connect to Prior Knowledge\nVerify \\( E \\) has units of N C\\(^{-1}\\).",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Visualize It\nSketch the wire and surrounding cylinder to see why only the curved surface matters.",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "Test a Simpler Case\nIf \\( r \\) doubles, the expression shows \\( E \\) halves. Does that make sense?",
"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": 4,
"fragments": [
{
"fragment_index": 1,
"text_description": "Question with Hints and Nudges\nPhysics | CBSE 12\nDifficulty: Hard\nSelf-Discovery",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "The Question:\nAn electron moves in a uniform magnetic field describing a circle of radius \\(r_0\\) with period \\(T_0\\). What happen to the radius and time period if its speed becomes \\(2v_0\\)?",
"image_description": "https://asset.sparkl.ac/pb/sparkl-vector-images/img_ncert/HwekryFoMsjimFCRQedENIGhC1iSyErTe4mDuLqm.png"
},
{
"fragment_index": 3,
"text_description": "Helpful Concepts\nLorentz force: \\(F = qvB\\)\nCentripetal need: \\(F = \\frac{mv^{2}}{r}\\)\nCyclotron frequency: \\(\\omega = \\frac{qB}{m}\\)\ncyclotron motion\nradius-period relation",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Relate Forces\nStart with \\(qvB = \\frac{mv^{2}}{r}\\) to express \\(r\\) in terms of \\(v\\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Radius Ratio\nCompute \\(r_{new}/r_0\\) after substituting \\(2v_0\\); it doubles.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Period Expression\nUse \\(T = \\frac{2\\pi m}{qB}\\). Since \\(v\\) cancels, \\(T_{new} = T_0\\).",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Parameter Isolation\nHold \\(q\\) and \\(B\\) fixed; see how changing \\(v\\) alone affects each formula.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Common Mistake\nDo not plug \\(2v_0\\) into \\(T\\); velocity is absent from the period expression.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Visualize It\nDraw two circles; observe the radius doubles while rotation rate stays constant.",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "Test a Simpler Case\nImagine halving \\(v\\); predict \\(r\\) and \\(T\\) to check your reasoning pattern.",
"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 – Concept Test\nPhysics | CBSE 12\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_C = V_L = 10\\,\\text{V}\\). If the capacitor is short-circuited, what is the new voltage across the inductor?\nNo diagram required.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nPhasor addition of AC voltages\nResonance: \\(X_L = X_C\\)\nImpedance of a series RL circuit\nAlternating Current\nImpedance",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nEqual voltages signal resonance, so \\(X_L = X_C\\).",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nShorting the capacitor leaves an RL series circuit with \\(Z=\\sqrt{R^2+X_L^2}\\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nFind \\(I\\) from \\(V_R = I R\\) (source voltage unchanged). Then compute \\(V_L = I X_L\\) using the same frequency.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Break It Down\nContrast the circuit at resonance with the RL circuit after the short.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Connect to Prior Knowledge\nAt resonance \\(Z=R\\); remember \\(I = V/R\\).",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize It\nRedraw the phasor triangle before and after removing \\(C\\).",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nImagine \\(R=0\\) to see how \\(V_L\\) responds purely to \\(X_L\\).",
"image_description": ""
}
]
},
{
"slide": 6,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nWave Optics – Multi-wavelength YDSE | Physics | CBSE 12\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:\nIn a Young’s double-slit experiment, light of wavelengths 400 nm and 600 nm illuminate the same slits. Find the minimum distance \\(y\\) from the central bright fringe where both colours produce a dark fringe simultaneously. Express \\(y\\) in terms of slit separation \\(d\\) and screen distance \\(D\\).\nNo diagram provided",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nDark fringe condition: \\( \\Delta = (m+\\tfrac{1}{2})\\lambda \\).\nUse LCM to match path differences for two wavelengths.\nRelation to screen: \\( y = \\frac{\\Delta D}{d} \\).\nWave Optics\nInterference",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)\nDark Fringe Rule\nFor each colour, dark fringes satisfy \\( \\Delta = (m+\\tfrac{1}{2})\\lambda \\).\nCommon Position\nFind the smallest path difference that is a half-integer multiple of both wavelengths.\nTranslate to Screen\nOnce \\(\\Delta_{\\text{min}}\\) is known, use \\( y = \\frac{\\Delta_{\\text{min}}D}{d} \\) to express the required distance.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Thinking Strategies\nBreak It Down\nWrite separate dark-fringe equations before combining them.\nConnect to Prior Knowledge\nLCM of numbers aligns repeating patterns—apply that to path differences.\nVisualize It\nSketch the two fringe sets and mark overlaps for insight.\nTest a Simpler Case\nImagine \\(\\lambda_2 = 2\\lambda_1\\) to see how common darks appear.",
"image_description": ""
}
]
},
{
"slide": 7,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nPhysics | CBSE 12\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:\nUsing the B.E./A versus mass-number curve, decide which labelled nuclei (W, X, Y, Z) are most suitable for (i) fission and (ii) fusion, giving reasons.\nNo diagram required.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nPeak at \\(A \\approx 60\\) gives maximum B.E./A.\nMoving toward the peak releases energy.\nHeavy nuclei split; light nuclei fuse to approach the peak.\nNuclear energy\nFusion vs fission",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nCheck each label’s mass number relative to the \\(A\\approx60\\) peak on the curve.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nFor fission, pick the heaviest labelled nucleus that still has a noticeably lower B.E./A than the peak.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nFor fusion, choose the lightest labelled nucleus far left of the peak; products should shift significantly toward \\(A\\approx60\\).",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Energy Slope\nSteeper rise toward the peak implies a larger possible energy gain.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Common Slip\nAvoid picking nuclei already at or near the peak; they cannot release much extra energy.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize It\nDraw arrows from the chosen nuclei toward the peak to see the direction of energy release.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nRecall how splitting \\(^{235}\\text{U}\\) or fusing deuterons releases energy; apply the same idea here.",
"image_description": ""
}
]
},
{
"slide": 8,
"fragments": [
{
"fragment_index": 1,
"text_description": "Electrostatic Potential & Capacitance – Dielectrics\nPhysics | CBSE 12\nDifficulty: Hard\nSelf-Discovery",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "The Question:\nDerive expressions for the capacitance of a parallel-plate capacitor of area \\(A\\) and plate separation \\(d\\) when (i) a dielectric slab of thickness \\(t\\) and permittivity \\(\\varepsilon\\) and (ii) a metal slab of the same thickness (\\(t<d\\)) are inserted between the plates.\nNo figure provided",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Helpful Concepts\nSeries capacitance model for layered media\nElectric field is zero inside a perfect conductor\nCapacitance is inversely proportional to effective plate gap\ncapacitor modification\ndielectric insertion",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Gentle Nudge\nSplit the capacitor gap into two parts: the slab (\\(t\\)) and the remaining air (\\(d-t\\)).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Direction Pointer\nFor each region use \\(C=\\varepsilon A/\\Delta x\\). Take \\(\\varepsilon=\\varepsilon_0\\) for air, \\(\\varepsilon=\\kappa\\varepsilon_0\\) for the dielectric.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Guiding Framework\nCombine the two capacitances in series: \\(1/C=1/C_1+1/C_2\\). For a metal slab, the potential drop occurs only across the air gap \\((d-t)\\); effectively \\(C=\\varepsilon_0 A/(d-t)\\).",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Compare Results\nMetal removal of gap boosts capacitance more than any finite-κ dielectric.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Avoid Error\nA conductor does not merely raise κ; it sets \\(E=0\\) inside, cancelling that part of the distance.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Check Limits\nVerify your formula for \\(t\\to0\\) and \\(t\\to d\\) to detect algebra slips.",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "Visualize It\nSketch field lines and potential drop across each region to clarify the series model.",
"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": 9,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nElectromagnetic Waves – Displacement 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:\nA parallel-plate capacitor of area 0.001 m² and plate separation 0.0001 m is connected to a source whose voltage rises at \\(1\\times10^{8}\\,\\text{V s}^{-1}\\). Find the displacement current through the capacitor.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\n\\(I_d=\\varepsilon_0 A\\,\\dfrac{dE}{dt}\\)\n\\(E=\\dfrac{V}{d}\\)\n\\(\\varepsilon_0=8.85\\times10^{-12}\\,\\text{F m}^{-1}\\)\nMaxwell correction\nCapacitor charging",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)\nGentle Nudge\nUse \\(E=V/d\\); hence \\(\\dfrac{dE}{dt}=\\dfrac{1}{d}\\,\\dfrac{dV}{dt}\\).\nDirection Pointer\nInsert this \\(dE/dt\\) into \\(I_d=\\varepsilon_0 A dE/dt\\) and substitute the numerical values.\nGuiding Framework\nKeep track of powers of ten; the final answer should be a few mA.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Thinking Strategies\nConstant Mistake\nDo not drop the area \\(A\\); forgetting it makes the answer 10⁴ times smaller.\nConcept Link\nDisplacement current ensures Ampere–Maxwell law holds when the capacitor is charging.\nBreak It Down\nList given values, write the formula, then substitute step by step.\nVisualize It\nSketch field lines between plates to see how a changing field creates \\(I_d\\).",
"image_description": ""
}
]
},
{
"slide": 10,
"fragments": [
{
"fragment_index": 1,
"text_description": "Dual Nature – Photoelectron Wavelength\nPhysics | CBSE 12\nDifficulty: Medium\nSelf-Discovery",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "The Question:\nPlatinum (work function 5.63 eV) is illuminated by \\(1.6\\times10^{15}\\,\\text{Hz}\\) light. Find the minimum de Broglie wavelength of the emitted electrons.\nNo diagram provided.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Helpful Concepts\n\\(K_{\\text{max}} = h\\nu - \\phi\\)\n\\(\\lambda = \\frac{h}{p}\\)\nWork function \\(\\phi\\) sets the threshold\nphotoelectric effect\nmatter waves",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Progressive Hints\n(Reveal only when needed)\nGentle Nudge\nUse \\(E = h\\nu\\). Convert photon energy from joules to electron-volts.\nDirection Pointer\nCompute \\(K_{\\text{max}} = E - \\phi\\). Keep everything in SI units for accuracy.\nGuiding Framework\nFind \\(p = \\sqrt{2m_e K_{\\text{max}}}\\). Finally, \\(\\lambda = \\frac{h}{p}\\) gives the minimum wavelength.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Thinking Strategies\nUnit Consistency\nWork in SI until the last step to avoid conversion errors.\nCommon Slip\nDo not use \\(c = \\lambda \\nu\\) for electrons—only photons follow that relation.\nVisualize It\nDraw an energy bar: photon energy, work function, and leftover kinetic energy.\nTest a Simpler Case\nCheck what happens when \\(\\nu\\) is just at the threshold to verify your method.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Struggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
}
]
},
{
"slide": 11,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nSemiconductor Electronics – Diode Logic\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:\nAn ideal diode, a cell, switch S, and a bulb are in series. Will the bulb glow (a) when S is open, (b) when S is closed? Justify each answer using band-diagram reasoning.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nForward vs reverse bias\nCircuit continuity\nBand-diagram interpretation\npn-junction\nIdeal diode",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nWhich diode terminal is linked to the positive cell terminal?",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nWith S open, ask: does any closed path exist for charge to reach and return from the bulb?",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nWhen S closes, check the diode's bias. If forward-biased, model it as a short; if reverse-biased, as an open. Then trace current through the bulb.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Break It Down\nAnalyse the open-switch and closed-switch cases separately.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Connect to Prior Knowledge\nRemember: forward bias ≈ short circuit, reverse bias ≈ open circuit for an ideal diode.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize It\nSketch energy bands for forward and reverse bias to see carrier movement.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nAssume zero diode drop and verify if current can complete the loop.",
"image_description": ""
}
]
},
{
"slide": 12,
"fragments": [
{
"fragment_index": 1,
"text_description": "Question with Hints and Nudges\nPhysics | CBSE 12\nDifficulty: Hard\nSelf-Discovery",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "The Question:\nA 100 V battery (internal resistance 1 Ω) drives 10 A through a heater at 20 °C. After the heater reaches 320 °C, calculate the power dissipated inside the battery. Use \\( \\alpha = 3.7\\times10^{-4}\\,^{\\circ}\\mathrm{C}^{-1} \\).",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Helpful Concepts\n\\( R_T = R_0(1+\\alpha \\Delta T) \\)\n\\( P_{\\text{loss}} = I^{2} r \\)\n\\( P = I^{2} R \\) for resistive heating\ntemperature coefficient\ninternal resistance",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Progressive Hints\n(Reveal only when needed)\nGentle Nudge\nFirst, find the heater’s hot resistance using \\( R_T = R_0(1+\\alpha \\Delta T) \\).\nDirection Pointer\nWith the new \\( R \\), current is \\( I = \\frac{V}{R + r} \\).\nGuiding Framework\nFinally, compute \\( P_{\\text{battery}} = I^{2} r \\). Finish the arithmetic yourself.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Thinking Strategies\nSeparate Resistances\nKeep the heater’s \\( R \\) and battery’s \\( r \\) distinct when forming equations.\nQuick Estimate\nCurrent should drop only slightly; a huge change signals an error.\nVisualize It\nDraw a simple circuit diagram with updated resistance values.\nTest a Simpler Case\nSet \\( \\alpha = 0 \\) to verify your equations before using actual data.",
"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": 13,
"fragments": [
{
"fragment_index": 1,
"text_description": "Electromagnetic Induction – AC Generator\nPhysics | CBSE 12\nDifficulty: medium\nSelf-Discovery",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "The Question:\nA coil with \\(N\\) turns and area \\(A\\) rotates at angular speed \\( \\omega \\) in a uniform field \\(B\\). Derive the instantaneous emf induced in the coil and name the energy source in an a.c. generator.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Helpful Concepts\nFaraday’s law of electromagnetic induction\nMagnetic flux \\( \\Phi = BA\\cos\\theta \\)\nLenz’s law for sign convention\ngenerator principle\nFaraday law",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Progressive Hints\n(Reveal only when needed)\nGentle Nudge\nExpress magnetic flux through the rotating coil first.\nDirection Pointer\nUse \\( \\Phi = N B A \\cos(\\omega t) \\).\nGuiding Framework\nDifferentiate flux to get \\( e = -\\frac{d\\Phi}{dt} = N B A \\omega \\sin(\\omega t) \\). Mechanical work from the prime mover is the energy source.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Thinking Strategies\nSign Convention\nApply Lenz’s law to decide the negative sign in \\(e\\).\nMistake Watch\nDon’t forget the factor \\(N\\) while writing the emf expression.\nVisualize It\nSketch coil positions at \\( \\omega t = 0^\\circ \\) and \\(90^\\circ\\) to see flux change.\nTest a Simpler Case\nCheck your formula for \\( \\omega t = 0 \\) and \\( \\omega t = \\pi/2 \\).",
"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": 14,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nRay Optics – Telescope Magnification | Physics CBSE 12\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 refracting telescope has an objective of focal length 15 m and an eyepiece of 1 cm. Calculate (i) its angular magnification in normal adjustment, and (ii) the diameter of the moon’s image formed at the objective.\nNo diagram required.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\n\\(M = \\frac{f_o}{f_e}\\)\n\\(y = \\theta\\,f_o\\)\n\\(\\theta \\approx \\frac{D}{R}\\) for small angles\nastronomical telescope\nangular size",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nPut both focal lengths in metres before taking their ratio for magnification.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nUse the small-angle formula \\( \\theta \\approx \\frac{3.48\\times10^{6}}{3.8\\times10^{8}} \\) rad to estimate the moon’s angular size.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nAfter finding \\( \\theta \\), multiply it by the 15 m focal length to obtain the image diameter in metres, then convert to millimetres.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Break It Down\nSolve part (i) first; reuse its result to sanity-check part (ii).",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Connect to Prior Knowledge\nRecall that normal adjustment places the final image at infinity, so magnification is just the focal-length ratio.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Visualize It\nSketch parallel rays from the moon hitting the objective to see where the image forms.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Test a Simpler Case\nTry using \\( \\theta = 0.01 \\) rad to estimate image size, then refine with the exact angle.",
"image_description": ""
}
]
},
{
"slide": 15,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nPhysics | CBSE 12 – Atoms\nDifficulty: easy\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:\nWhy does a sample of hydrogen, with only one electron per atom, still show many distinct spectral lines in its emission spectrum?",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nMany atoms → many excited states at once\nPhoton wavelength set by energy-level gap\nBohr’s quantised orbits in hydrogen\nhydrogen spectrum\nenergy levels",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)\nPopulation Variety\nDifferent atoms can occupy different excited states at the same moment.\nTransition Diversity\nEach allowed drop, such as n = 4→2 or 3→1, releases its own unique wavelength.\nGuiding Framework\nCombine ideas: a large ensemble of identical one-electron atoms + all possible downward jumps predicted by Bohr → many discrete spectral lines.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Thinking Strategies\nCommon Mistake\nOne electron cannot jump multiple gaps at once; each photon comes from a single transition.\nLink Back\nRecall Bohr’s postulate: energy depends only on quantum numbers n\ni\nand n\nf\n.\nVisualize It\nSketch the energy-level diagram and draw every downward arrow you can.\nTest a Simpler Case\nImagine a single atom: one line per jump. Add many atoms, lines accumulate.",
"image_description": ""
}
]
},
{
"slide": 16,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Take-aways",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Formulas First\nQuickly recap and write core formulas; remember them and avoid late corrections.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Units & SI\nAlways state answers with correct SI units—simple thing to remember for full marks.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Diagram Strategy\nDraw neat, labeled diagrams; they earn instant points and clarify concepts.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Numerical Flow\nSubstitute symbols before numbers; compute step-wise to avoid arithmetic slips.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Time Management\nFinish sure-shot questions first; return to lengthy ones with remaining time.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Presentation\nBox the final result; examiner sees it instantly and awards marks quickly.",
"image_description": ""
}
]
}
]