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[
{
"slide": 1,
"fragments": [
{
"fragment_index": -1,
"text_description": "Physics Sample Paper Review\nYour roadmap to a perfect 70 in Physics.",
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]
},
{
"slide": 2,
"fragments": [
{
"fragment_index": 1,
"text_description": "Question Distribution & Your Strength",
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},
{
"fragment_index": 2,
"text_description": "Chapter\nPaper Marks\nYour Accuracy\nStrength\nElectrostatics\n8\n90%\nStrong\nCurrent Electricity\n7\n60%\nModerate\nMagnetism & Matter\n5\n70%\nModerate\nEMI & AC\n8\n50%\nWeak\nOptics\n14\n80%\nStrong\nDual Nature, Atoms & Nuclei\n10\n55%\nWeak\nSemiconductor Electronics\n9\n92%\nStrong\nCommunication Systems\n4\n65%\nModerate",
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},
{
"fragment_index": 3,
"text_description": "Source: Self-assessment mock test, April 2024",
"image_description": ""
}
]
},
{
"slide": 3,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nElectric Charges & Fields\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nUsing Gauss’s theorem, derive \\(E(r)\\) at distance \\(r\\) from an infinitely long straight wire of uniform linear charge density \\( \\lambda \\).\nGiven:\nInfinite straight wire (length ≫ \\(r\\)).\nUniform linear charge density \\( \\lambda \\).\nMedium: free space with \\( \\varepsilon_0 \\).\nTo Find:\nElectric field magnitude \\(E(r)\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nGauss’s law (cylindrical surface)\nExploit cylindrical symmetry; flux through curved area only.\nMethod: Algebraic\n2\nDirect integration\nIntegrate Coulomb’s law along wire; calculus-heavy, same result.\nMethod: Calculus\n3\nNumerical simulation\nDiscretise charges and sum fields; useful for teaching only.\nMethod: Computational",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nSymmetry\nField is radial and identical over a circle of radius \\(r\\).\nGaussian surface\nCoaxial cylinder of length \\(L\\), radius \\(r\\).\nFlux calculation\n\\( \\Phi = E(2\\pi r L) \\); end caps give zero flux.\nGauss’s law\nSet \\( \\Phi = \\frac{\\lambda L}{\\varepsilon_0} \\) and solve for \\(E\\).",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nChoose Gaussian surface\nTake a cylinder with radius \\(r\\) and length \\(L\\) coaxial with the wire.\nSurface area curved = \\(2\\pi r L\\)\n2\nCompute electric flux\nField is perpendicular to curved surface; zero on end caps.\n\\( \\Phi = E \\times 2\\pi r L \\)\n3\nApply Gauss’s law\nSet \\( \\Phi = q_{\\text{enc}}/\\varepsilon_0 = \\lambda L/\\varepsilon_0 \\).\n\\( E = \\dfrac{\\lambda}{2\\pi \\varepsilon_0 r} \\)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nLinear symmetry leads to \\(E \\propto 1/r\\), not \\(1/r^2\\).\nFlux through end caps is zero because field lines are parallel.\nAvoid treating \\( \\lambda \\) as surface charge or adding forgotten caps.",
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}
]
},
{
"slide": 4,
"fragments": [
{
"fragment_index": -1,
"text_description": "Moving Charges & Magnetism\nEquilibrium of a charged sphere\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
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{
"fragment_index": 1,
"text_description": "Problem Statement\nA +10 mC sphere rests inside a vertical tube. The tube moves east–west through a\n uniform 2 T horizontal magnetic field. Find the minimum speed and field\n orientation needed so the magnetic force exactly cancels the sphere’s weight.\nGiven:\nCharge \\(q = +10\\,\\text{mC} = 0.01\\,\\text{C}\\)\nMagnetic field \\(B = 2\\,\\text{T}\\) (horizontal, magnitude known)\nGravitational acceleration \\(g = 9.8\\,\\text{m s}^{-2}\\)\nTo Find:\nLeast velocity \\(v\\) and the required field direction for equilibrium.",
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},
{
"fragment_index": 2,
"text_description": "Solution Approaches",
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},
{
"fragment_index": 3,
"text_description": "1\nForce-balance method\nSet \\(F_B = mg\\) with \\(F_B = qvB\\) when \\(\\mathbf{v}\\perp\\mathbf{B}\\).\nComplexity: —",
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},
{
"fragment_index": 4,
"text_description": "2\nRight-hand rule\nChoose \\(\\mathbf{B}\\) so \\(q\\mathbf{v}\\times\\mathbf{B}\\) points upward.\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "3\nMinimum-speed check\nSpeed is minimum when \\(\\mathbf{v}\\perp\\mathbf{B}\\), giving maximum \\(F_B\\).\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Logical Breakdown",
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},
{
"fragment_index": 7,
"text_description": "Magnetic force\n\\(\\displaystyle F_B = qvB\\) when \\(\\mathbf{v}\\perp\\mathbf{B}\\).",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Equilibrium condition\nSet \\(qvB = mg\\) for vertical balance.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Minimum velocity\nAchieved when \\(\\mathbf{v}\\perp\\mathbf{B}\\) so \\(F_B\\) is maximal.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Direction selection\nUse right-hand rule: for +ve charge, \\(\\mathbf{v}\\times\\mathbf{B}\\) upward.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Step-by-Step Solution",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "1\nWrite force balance\nFor equilibrium, magnetic force must equal weight.\n\\(qvB = mg\\)",
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},
{
"fragment_index": 13,
"text_description": "2\nSolve for velocity\nRearrange to obtain minimum speed.\n\\(v_{\\min} = \\dfrac{mg}{qB}\\)",
"image_description": ""
},
{
"fragment_index": 14,
"text_description": "3\nSet field direction\nChoose \\(\\mathbf{B}\\) so that for eastward motion, \\(\\mathbf{v}\\times\\mathbf{B}\\) is upward. Hence \\( \\mathbf{B}\\) must point south.\nRHR → \\(\\mathbf{B}\\) southward",
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},
{
"fragment_index": 15,
"text_description": "Key Insights\nBalancing forces requires explicit use of \\(qvB = mg\\).\nMinimum speed occurs when velocity is perpendicular to the magnetic field.\nRight-hand rule fixes field direction; avoid swapping velocity and field vectors.",
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}
]
},
{
"slide": 5,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nAlternating Current\nDifficulty: Medium\nTime: 15 min",
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},
{
"fragment_index": 1,
"text_description": "Problem Statement\nIn a series LCR circuit at resonance \\(V_R = V_C = V_L = 10\\text{ V}\\). If the capacitor is short-circuited instantly, predict the voltage across the inductor just after the change.\nGiven:\nSeries LCR at resonance (\\(X_L = X_C\\)).\n\\(V_R = V_C = V_L = 10\\text{ V}\\).\nSame supply frequency throughout.\nTo Find:\nInductor voltage \\(V'_L\\) immediately after the capacitor is shorted.",
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},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nPhasor method\nCurrent stays unchanged at the instant; calculate \\(V'_L = I X_L\\).\nComplexity: Conceptual\n2\nImpedance ratio\nRe-evaluate circuit impedance without \\(C\\); find new current, then \\(V_L\\).\nComplexity: Algebraic\n3\nEnergy view\nUse stored magnetic energy continuity to infer same \\(V_L\\).\nComplexity: Qualitative",
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},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nResonance facts\nIn a series LCR circuit, \\(X_L = X_C\\) and current \\(I\\) is maximum.\nInitial phasors\nPhasor magnitudes: \\(V_R = V_C = V_L = 10\\text{ V}\\).\nInstantaneous change\nShorting \\(C\\) removes \\(X_C\\) but \\(I\\) cannot change suddenly (inductor property).\nVoltage prediction\nBecause \\(I\\) and \\(X_L\\) stay same, \\(V'_L = I X_L = 10\\text{ V}\\).",
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},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nWrite resonance condition\nAt resonance \\(X_L = X_C\\) and voltages across \\(L\\) and \\(C\\) cancel in the phasor sum.\n\\(V_R = V_C = V_L = 10\\text{ V}\\)\n2\nDetermine initial current\n\\(I = V_R / R\\). Exact \\(R\\) not needed; ratio suffices.\n\\(I = 10 / R\\)\n3\nShort the capacitor\nCurrent through \\(L\\) cannot change instantly, so \\(I\\) remains \\(10/R\\).\n\\(I_{\\text{new}} = I\\)\n4\nCompute new inductor voltage\n\\(V'_L = I X_L = 10\\text{ V}\\).\n\\(V'_L = 10\\text{ V}\\)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nPhasor reasoning simplifies series LCR voltage questions.\nInductor current continuity lets us predict immediate voltage after circuit modifications.\nLearning outcome achieved: you can now forecast voltage changes when a component is altered.",
"image_description": ""
}
]
},
{
"slide": 6,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nWave Optics – Young’s Double-Slit\nDifficulty: Medium\nTime: 15 min",
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},
{
"fragment_index": 1,
"text_description": "Problem Statement\nA Young’s double-slit (Y-D-S) set-up is illuminated with light of wavelengths \\( \\lambda_1 = 400\\,\\text{nm} \\) and \\( \\lambda_2 = 600\\,\\text{nm} \\) together. Find the least distance from the central maximum where a dark fringe appears for both wavelengths.\nGiven:\nWavelengths: \\( \\lambda_1 = 400\\,\\text{nm} \\), \\( \\lambda_2 = 600\\,\\text{nm} \\).\nY-D-S dark fringe condition: \\( \\Delta = (m+\\tfrac{1}{2})\\lambda \\).\nScreen distance \\( D \\) and slit separation \\( d \\) (answer will include \\( \\tfrac{D}{d} \\)).\nTo Find:\nLeast distance \\( y_{\\min} \\) from the central maximum where both colours give a dark fringe.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nPath-difference LCM\nList odd-half multiples of each wavelength and pick the first common value.\nComplexity: —\n2\nParity Check\nSolve \\( 400(2m_1+1)=600(2m_2+1) \\) for integers to test coincidence.\nComplexity: —\n3\nRatio Analysis\nUse wavelength ratio \\( \\tfrac{3}{2} \\) to scale odd-half sequences and find overlap.\nComplexity: —",
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},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\n1. Dark Fringe Rule\nA Y-D-S dark fringe occurs when \\( \\Delta = (m+\\tfrac12)\\lambda \\).\n2. Path Difference Sets\nGenerate odd-half multiples for both \\( \\lambda_1 \\) and \\( \\lambda_2 \\).\n3. Common Value\nFind the smallest path difference common to both sets (LCM idea).\n4. Convert to y\nUse \\( y = \\frac{\\Delta D}{d} \\) to get distance on the screen.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nWrite Dark Conditions\nFor each colour, \\( \\Delta = (m+\\tfrac12)\\lambda \\).\n\\( \\Delta_1 = (m_1+\\tfrac12)\\lambda_1 \\) \\( \\Delta_2 = (m_2+\\tfrac12)\\lambda_2 \\)\n2\nFind Least Common \\( \\Delta \\)\nListing values shows the first match at \\( \\Delta = 600\\,\\text{nm} \\) ( \\( \\tfrac{3\\lambda_1}{2} = \\lambda_2 \\) ).\n\\( \\Delta_{\\min} = 600 \\times 10^{-9}\\,\\text{m} \\)\n3\nConvert to Screen Distance\nUsing the relation \\( y = \\dfrac{\\Delta D}{d} \\) gives the required distance.\n\\( y_{\\min} = 6.0 \\times 10^{-7}\\,\\dfrac{D}{d}\\,\\text{m} \\)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nA coincident dark fringe needs equal path difference for both beams.\nOdd-half multiples of each wavelength form the dark-fringe series.\nDistance on the screen scales with \\( \\tfrac{D}{d} \\); ratio of wavelengths fixes the position.",
"image_description": ""
}
]
},
{
"slide": 7,
"fragments": [
{
"fragment_index": -1,
"text_description": "Nuclear Binding Energy Insight\nBinding Energy Curve & Nuclear Stability\nDifficulty: Medium\nEst. Time: 5 min\nPrevious Problem\nProblem 7 of 10\nNext Problem",
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},
{
"fragment_index": 1,
"text_description": "Problem Context\nThe graph plots binding energy per nucleon versus mass number. Four nuclei—W, X, Y, Z—are marked for analysis.",
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{
"fragment_index": 2,
"text_description": "",
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"fragment_index": 3,
"text_description": "Question\nUsing the B.E. per nucleon curve, identify which nucleus suits (i) fission and (ii) fusion. Give a brief reason.\na) Fission-favoured nucleus\nb) Fusion-favoured nucleus",
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{
"fragment_index": 4,
"text_description": "Helpful Hints\nHint 1\nHeavy nuclei far right, low B.E./A; splitting makes them more stable.\nHint 2\nLight nuclei far left gain B.E./A when they merge toward the peak.\nHint 3\nCompare how far W and Z lie below the Fe–Ni peak at \\(A\\approx56\\).",
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},
{
"fragment_index": 5,
"text_description": "Things to Consider\nPeak B.E./A at Fe–Ni signifies maximum nuclear stability.\nEnergy released ∝ Δ(B.E./A)×A; larger gap yields more energy.\nAvoid choosing the peak itself for fission or fusion.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Related Concepts\nBinding Energy\nNuclear Fission\nNuclear Fusion",
"image_description": ""
}
]
},
{
"slide": 8,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nElectromagnetic Waves – Maxwell Correction\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nA parallel-plate capacitor (area 0.001 m², gap 0.0001 m) is charged so that its voltage rises at \\(10^{8}\\,\\text{V s}^{-1}\\). Find the displacement current between the plates.\nGiven:\nPlate area \\(A = 0.001\\,\\text{m}^2\\)\nSeparation \\(d = 0.0001\\,\\text{m}\\)\n\\(\\dfrac{dV}{dt}=10^{8}\\,\\text{V s}^{-1}\\), \\(\\varepsilon_0 = 8.85\\times10^{-12}\\,\\text{F m}^{-1}\\)\nTo Find:\nMagnitude of displacement current \\(I_d\\).",
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{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nDirect Maxwell current\nUse \\(I_d=\\varepsilon_0\\frac{A}{d}\\dfrac{dV}{dt}\\).\nComplexity: 1 step\n2\nCapacitance first\nFind \\(C=\\varepsilon_0A/d\\), then \\(I_d=C\\dfrac{dV}{dt}\\).\nComplexity: 2 steps\n3\nElectric flux method\nEmploy \\(I_d=\\varepsilon_0\\dfrac{d\\Phi_E}{dt}\\) for charging plates.\nComplexity: Conceptual",
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},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nMaxwell correction\nIntroduces \\(I_d\\) so Ampere’s law works for charging capacitors.\nCapacitance\nFor parallel plates \\(C=\\varepsilon_0A/d\\).\nVoltage change\nCharging causes \\(\\dfrac{dV}{dt}\\neq0\\) → time-varying electric field.\nNumerical substitution\nInsert values to obtain \\(I_d\\) in amperes.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nCalculate capacitance\n\\(C=\\varepsilon_0\\frac{A}{d}=8.85\\times10^{-12}\\times\\frac{0.001}{0.0001}=8.85\\times10^{-11}\\,\\text{F}\\).\n\\(C = 8.85 \\times 10^{-11}\\,\\text{F}\\)\n2\nApply charging relation\n\\(I_d=C\\dfrac{dV}{dt}\\).\n\\(I_d = 8.85\\times10^{-11}\\times10^{8}\\)\n3\nCompute value\n\\(I_d = 8.85\\times10^{-3}\\,\\text{A} = 8.85\\,\\text{mA}\\).\nAnswer: \\(8.85\\; \\text{mA}\\)",
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{
"fragment_index": 5,
"text_description": "Key Insights\nDisplacement current keeps Ampere’s law valid for time-varying fields.\nFor a capacitor, \\(I_d\\) matches the conduction current in the connecting wire.\nMagnitude depends on area-to-gap ratio and \\(\\dfrac{dV}{dt}\\) during charging.",
"image_description": ""
}
]
},
{
"slide": 9,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nElectrostatic Potential & Capacitance\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nTwo parallel plates are separated by distance \\(d\\). Insert centrally a slab of\n(a) dielectric constant \\(k\\) and thickness \\(t<d\\),\n(b) metal thickness \\(t\\). Compare the resulting capacitances with the empty-plate value \\(C_0\\).\nGiven:\nPlate separation \\(d\\)\nInserted slab thickness \\(t\\;(t<d)\\)\nDielectric constant \\(k\\) (case a)\nTo Find:\nRatios \\(C_{\\text{dielectric}}/C_0\\) and \\(C_{\\text{metal}}/C_0\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nSeries model for dielectric\nTreat dielectric slab and two air gaps as three capacitors in series.\nKey formula: \\(\\frac{1}{C}=\\frac{d- t}{2\\varepsilon_0 A}+\\frac{t}{k\\varepsilon_0 A}\\)\n2\nMetal reduces gap\nField vanishes inside metal; effective separation becomes \\(d-t\\).\nCapacitance: \\(C=\\frac{\\varepsilon_0 A}{d-t}\\)\n3\nCheck consistency\nFor \\(t\\to0\\) both results must return to \\(C_0\\).\nStatus: Verified",
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{
"fragment_index": 3,
"text_description": "Logical Breakdown\nAir gaps\nEach gap thickness \\((d-t)/2\\); permittivity \\(\\varepsilon_0\\).\nDielectric slab\nPermittivity \\(k\\varepsilon_0\\); thickness \\(t\\).\nSeries combination\nAdd inverse capacitances to get \\(C_{\\text{dielectric}}\\).\nMetal insert\nInside metal \\(E=0\\) ⇒ only air gap \\(d-t\\) contributes.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nEquivalent for dielectric\nUse series model: \\( \\frac{1}{C_d}= \\frac{d-t}{2\\varepsilon_0 A}+ \\frac{t}{k\\varepsilon_0 A} \\).\n\\( C_d=\\frac{C_0}{\\displaystyle 1-\\frac{t}{d}+\\frac{t}{kd}} \\)\n2\nSimplify ratio\nDivide by \\(C_0=\\varepsilon_0 A/d\\): \\( \\displaystyle \\frac{C_d}{C_0}= \\frac{1}{1-\\frac{t}{d}+\\frac{t}{kd}} \\).\nCheck: \\(k\\to1\\) gives \\(C_0\\).\n3\nMetal case\nEffective separation \\(d-t\\): \\(C_m=\\frac{\\varepsilon_0 A}{d-t}= \\frac{C_0}{1-\\frac{t}{d}}\\).\n\\( \\displaystyle \\frac{C_m}{C_0}= \\frac{1}{1-\\tfrac{t}{d}} \\)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nPartial dielectric acts like capacitors in series; metal deletes its own thickness.\nCapacitance grows with decreasing effective gap: \\(C\\propto 1/\\text{gap}\\).\nAvoid using \\(k\\) for metal; its permittivity is effectively infinite.",
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}
]
},
{
"slide": 10,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nDual Nature of Radiation & Matter\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nThree photoelectric I–V curves (A, B, C) for the same metal are given. Identify (i) the two curves with equal light intensity and (ii) the two with equal photon frequency.\nGiven:\nSame metal target and apparatus.\nPlateau current = saturation current \\(I_{\\text{sat}}\\).\nVoltage where current becomes zero = stopping potential \\(V_0\\).\nTo Find:\nPairs with equal intensity and pairs with equal frequency.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nCompare plateau heights\nEqual \\(I_{\\text{sat}}\\) ⇒ equal intensity.\nComplexity: O(1)\n2\nCompare stopping potentials\nEqual \\(V_0\\) ⇒ equal frequency (\\(\\nu\\)).\nComplexity: O(1)\n3\nCross-check results\nUse both criteria to avoid misclassification.\nComplexity: O(1)",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nSaturation current\nHigher \\(I_{\\text{sat}}\\) → more emitted electrons → greater intensity.\nStopping potential\n\\(V_0 = \\frac{h\\nu-\\phi}{e}\\); depends only on photon frequency.\nIntensity inference\nCurves with identical plateaus share light intensity.\nFrequency inference\nCurves with identical \\(V_0\\) correspond to equal frequency.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nMeasure \\(I_{\\text{sat}}\\)\nRead the plateau current for A, B, C and note equal values.\nIntensity ∝ \\(I_{\\text{sat}}\\)\n2\nMeasure \\(V_0\\)\nLocate the voltage where current becomes zero for each curve.\n\\(eV_0 = h\\nu - \\phi\\)\n3\nDraw conclusions\nMatch curves with equal \\(I_{\\text{sat}}\\) for intensity and equal \\(V_0\\) for frequency.\nState pairs: ( … )",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\n\\(I_{\\text{sat}}\\) signals intensity; it is frequency-independent.\n\\(V_0\\) depends only on photon energy \\(h\\nu\\).\nSlopes before saturation carry no information for this analysis.",
"image_description": ""
}
]
},
{
"slide": 11,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nSemiconductor Electronics • Bias Judgement\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nA cell, ideal diode (anode to +), bulb and switch are in series. Will the bulb glow (a) when the switch is open, (b) when the switch is closed?\nGiven:\nIdeal diode: zero \\(R_f\\), infinite \\(R_r\\).\nAnode connected to cell’s positive terminal ⇒ forward bias possible.\nBulb glows only when current ≠ 0.\nTo Find:\nGlow status for each switch position.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nCurrent-path check\nTrace continuity with switch open and closed.\n2\nBias analysis\nDetermine forward or reverse bias of the ideal diode.\n3\nEquivalent circuit view\nReplace diode with short (forward) or open (reverse) to predict bulb behaviour.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Logical Breakdown",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Switch state\nOpen ⇒ circuit broken. Closed ⇒ path may exist.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Diode bias\nAnode at higher potential gives forward bias; reverse otherwise.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Current flow\nNeeds closed switch & plus forward-biased diode.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Bulb response\nGlows only when \\(I \\gt 0\\).",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Step-by-Step Solution",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "1\nSwitch open\nCircuit incomplete, current zero, diode bias irrelevant, bulb OFF.\n\\(I = 0\\)",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "2\nSwitch closed\nPath complete. Anode is positive, so diode forward biased ⇒ behaves like short.\n\\(V_D = 0,\\; R_D = 0\\)",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "3\nResult\nCurrent flows, bulb glows. If diode were reverse biased, bulb would remain dark.\nBulb ⇒ ON",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "Key Insights\nIdeal diode conducts only under forward bias; blocks completely in reverse bias.\nAn open switch mimics infinite resistance, stopping current regardless of diode state.\nAlways judge bulb glow by tracing current path and verifying diode bias together.",
"image_description": ""
}
]
},
{
"slide": 12,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nCurrent Electricity\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nA 100 V battery with internal resistance \\(r = 1\\,\\Omega\\) sends 10 A through a heater at \\(20^{\\circ}\\text{C}\\). When the heater reaches \\(320^{\\circ}\\text{C}\\) the current becomes steady. Calculate the power dissipated inside the battery at the higher temperature. Take temperature coefficient of resistance \\( \\alpha = 4.0 \\times 10^{-3}\\,^{\\circ}\\text{C}^{-1} \\).\nGiven:\nBattery emf \\(E = 100\\text{ V}\\)\nInternal resistance \\(r = 1\\,\\Omega\\)\nInitial current \\(I_0 = 10\\text{ A}\\) at \\(20^{\\circ}\\text{C}\\)\nFinal temperature \\(T = 320^{\\circ}\\text{C}\\)\nTemperature coefficient \\( \\alpha = 4.0 \\times 10^{-3}\\,^{\\circ}\\text{C}^{-1}\\)\nTo Find:\nPower lost in internal resistance after heating.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nDirect α-method\nUse \\(R = R_0[1+\\alpha\\Delta T]\\) to update heater resistance, then compute current and battery loss.\nComplexity: Simple math\n2\nGraphical I–R analysis\nPlot \\(I\\) vs \\(R\\) for fixed \\(E,r\\); read current for new \\(R\\).\nComplexity: Visual\n3\nPercent-change shortcut\nEstimate current drop using small-change approximation for \\(R\\) increase.\nComplexity: Approximate",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nInitial heater resistance \\(R_0\\)\nDerived from Ohm’s law using given current and internal resistance.\nTemperature rise \\(\\Delta T\\)\n\\(320^{\\circ}\\text{C} - 20^{\\circ}\\text{C} = 300^{\\circ}\\text{C}\\).\nNew heater resistance \\(R\\)\nApply \\(R = R_0(1+\\alpha\\Delta T)\\) to include temperature coefficient.\nBattery loss \\(P_r\\)\nUse \\(P_r = I^{2} r\\) with updated current to quantify internal resistance power.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nCompute \\(R_0\\)\n\\(R_0 = \\frac{E}{I_0} - r = \\frac{100}{10} - 1 = 9\\,\\Omega\\).\n\\(R_0 = 9\\,\\Omega\\)\n2\nUpdate heater resistance\n\\(R = 9[1 + (4.0\\times10^{-3})(300)] = 9 \\times 2.2 = 19.8\\,\\Omega\\).\n\\(R = 19.8\\,\\Omega\\)\n3\nFind current & battery loss\n\\(I = \\frac{100}{19.8 + 1} \\approx 4.8\\,\\text{A}\\).\n\\(P_r = I^{2} r \\approx (4.8)^{2}(1) \\approx 23\\,\\text{W}\\).\n\\(P_r \\approx 23\\,\\text{W}\\)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nResistance rises linearly with temperature for metal heaters: \\(R \\propto 1+\\alpha\\Delta T\\).\nBattery loss depends on internal resistance power \\(I^{2}r\\); lower current reduces unwanted heating.\nLinking temperature coefficient to circuit analysis meets the learning goal: analyse temperature impact on battery efficiency.",
"image_description": ""
}
]
},
{
"slide": 13,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nElectromagnetic Induction – AC Generator Equation\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nDerive the instantaneous emf \\( \\varepsilon = N A B \\omega \\sin \\omega t \\) for a coil of \\(N\\) turns and area \\(A\\) rotating with angular speed \\( \\omega \\) in a uniform magnetic field \\(B\\).\nGiven:\nUniform magnetic field \\(B\\).\nCoil with \\(N\\) turns and area \\(A\\).\nAngular speed \\( \\omega \\); plane makes angle \\( \\omega t \\) with field.\nTo Find:\nExpression for induced emf as a function of time.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "1\nFaraday derivative\nDifferentiate \\( \\Phi = N A B \\cos \\omega t \\) to get emf.\nEffort: One derivative",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "2\nPhasor shift\nEmf leads flux by \\(90^\\circ\\); cosine becomes sine.\nEffort: Conceptual",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "3\nFlux-cutting view\nSine arises from component of \\(B\\) cut per unit time.\nEffort: Geometric",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Logical Breakdown",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Flux expression\n\\( \\Phi(t)= N A B \\cos \\omega t \\).",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Differentiate\n\\( d\\Phi/dt = - N A B \\omega \\sin \\omega t \\).",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Apply Faraday\n\\( \\varepsilon = - d\\Phi/dt \\).",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Result\n\\( \\varepsilon = N A B \\omega \\sin \\omega t \\).",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Step-by-Step Solution",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "1\nWrite magnetic flux\nFlux through coil changes as cosine of angle.\n\\( \\Phi(t)= N A B \\cos \\omega t \\)",
"image_description": ""
},
{
"fragment_index": 13,
"text_description": "2\nApply Faraday’s law\nInduced emf equals negative time rate of flux change.\n\\( \\varepsilon = -\\dfrac{d\\Phi}{dt} \\)",
"image_description": ""
},
{
"fragment_index": 14,
"text_description": "3\nDifferentiate & simplify\nDerivative converts cosine to sine; negative signs cancel.\n\\( \\varepsilon(t)= N A B \\omega \\sin \\omega t \\)",
"image_description": ""
},
{
"fragment_index": 15,
"text_description": "Key Insights\nEmf leads magnetic flux by \\(90^\\circ\\); hence sine form.\nSign errors or omitting \\(N\\) are common pitfalls.\nGreater \\(N, A, B,\\) or \\( \\omega \\) increases generator output.",
"image_description": ""
}
]
},
{
"slide": 14,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nRay Optics – Right-angled Prism & TIR\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nA ray just grazes face AC of a right-angled prism. (a) Find the prism’s refractive index. (b) When AC is immersed in a liquid of \\(n=\\dfrac{2}{\\sqrt{3}}\\), will the ray undergo total internal reflection or refraction?\nGiven:\nRight-angled prism (\\(45^\\circ\\), \\(45^\\circ\\), \\(90^\\circ\\))\nRay grazes face AC ⇒ angle of refraction \\(=90^\\circ\\)\nInitial surrounding: air \\((n=1)\\); later: liquid \\(\\displaystyle n=\\frac{2}{\\sqrt{3}}\\)\nTo Find:\nPrism refractive index and the ray’s fate (TIR or refraction) in the liquid.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "1\nSnell’s law at grazing\nSet \\(r=90^\\circ\\) at AC; solve \\(n=\\sin i/ \\sin r\\) for the prism.\nComplexity: Easy",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "2\nCritical angle method\nUse \\(\\sin C = \\dfrac{n_{\\text{out}}}{n_{\\text{prism}}}\\) with internal angle \\(45^\\circ\\).\nComplexity: Easy",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "3\nGraphical ray tracing\nSketch rays inside the right-angled prism to verify TIR condition.\nComplexity: Moderate",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Logical Breakdown",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Prism geometry\nRight-angled prism has \\(45^\\circ\\) faces, so internal incidence on AC is \\(45^\\circ\\).",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Snell’s law\nAt grazing emergence, \\(r=90^\\circ\\) ⇒ \\(n_{\\text{prism}}=\\dfrac{1}{\\sin 45^\\circ}\\).",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Critical angle in liquid\nWith liquid outside, \\(\\sin C'=\\dfrac{n_{\\text{liq}}}{n_{\\text{prism}}}\\).",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "TIR condition test\nIf internal angle > critical angle ⇒ TIR; else ray refracts.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Step-by-Step Solution",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "1\nFind \\(n_{\\text{prism}}\\)\nFor grazing emergence, internal incidence = \\(45^\\circ\\).\n\\(n=\\dfrac{1}{\\sin 45^\\circ}= \\sqrt{2}\\)",
"image_description": ""
},
{
"fragment_index": 13,
"text_description": "2\nCritical angle in liquid\n\\(\\sin C'=\\dfrac{n_{\\text{liq}}}{n}=\\dfrac{\\dfrac{2}{\\sqrt{3}}}{\\sqrt{2}}=\\dfrac{2}{\\sqrt{6}}\\).\n\\(C' \\approx 54^\\circ\\)",
"image_description": ""
},
{
"fragment_index": 14,
"text_description": "3\nDecide ray’s fate\nInternal angle \\(45^\\circ < C' (54^\\circ)\\). Condition for TIR unmet.\nRay refracts into liquid; no TIR.",
"image_description": ""
},
{
"fragment_index": 15,
"text_description": "Key Insights\nGrazing emergence fixes the internal angle at the critical angle.\nCritical angle varies with the surrounding medium: \\(\\sin C = n_{\\text{out}}/n_{\\text{in}}\\).\nRight-angled prisms need \\(n \\ge \\sqrt{2}\\) for guaranteed TIR at 45° faces.",
"image_description": ""
}
]
},
{
"slide": 15,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nAtoms – Rutherford Scattering Energy\nDifficulty: Medium\nTime: 15 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nAn \\(\\alpha\\)-particle moves toward a nucleus with speed \\(V\\) and stops momentarily at closest approach \\(d\\). What speed is required for a new closest approach of \\(d/2\\)?\nGiven:\n\\(\\alpha\\)-particle charge \\(+2e\\), mass \\(m\\) unchanged\nNuclear charge \\(+Ze\\)\nInitial case: speed \\(V\\), closest approach \\(d\\)\nTo Find:\nRequired speed \\(v\\) for closest approach \\(d/2\\)",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nEnergy Conservation\nSet initial kinetic energy equal to electrostatic potential energy at minimum distance.\nComplexity: Conceptual\n2\nDimensional Reasoning\nRecognise \\(KE\\propto 1/d\\) for identical charges; scale speed accordingly.\nComplexity: Rapid check\n3\nNumerical Substitution\nInsert values after deriving the general relation to verify \\(v=\\sqrt{2}V\\).\nComplexity: Minimal",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nInitial Kinetic Energy\n\\( KE_1=\\frac12 mV^2 \\)\nElectrostatic PE at \\(d\\)\n\\( U_1=\\frac{1}{4\\pi\\varepsilon_0}\\frac{2Ze^2}{d} \\)\nSecond Scenario PE\n\\( U_2=\\frac{1}{4\\pi\\varepsilon_0}\\frac{2Ze^2}{d/2}=2U_1 \\)\nSpeed Relation\nSet \\(KE_2=U_2\\Rightarrow\\frac12 mv^2=2U_1=2KE_1\\Rightarrow v=\\sqrt{2}V\\).",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nEquate Energies (Case 1)\nFor closest approach \\(d\\): \\( \\frac12 mV^2 = \\frac{1}{4\\pi\\varepsilon_0}\\frac{2Ze^2}{d} \\).\n\\( KE_1 = U_1 \\)\n2\nSet Up Case 2\nFor closest approach \\(d/2\\): \\( \\frac12 mv^2 = \\frac{1}{4\\pi\\varepsilon_0}\\frac{2Ze^2}{d/2} = 2U_1 \\).\n\\( KE_2 = 2U_1 \\)\n3\nTake Ratio & Solve\n\\( \\frac{v^2}{V^2} = \\frac{KE_2}{KE_1}=2 \\Rightarrow v = \\sqrt{2}\\,V \\).\nAnswer: \\( v = \\sqrt{2}\\,V \\)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nClosest approach varies inversely with initial kinetic energy in Rutherford scattering.\nHalving distance requires doubling the kinetic energy, hence speed increases by \\(\\sqrt{2}\\).\nAlways compare energies, not forces, to meet the learning outcome.",
"image_description": ""
}
]
},
{
"slide": 16,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Take-aways",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Formula Flash\nPrepare a recap sheet. Memorise \\(E=\\frac{hc}{\\lambda}\\), \\(V=IR\\), \\(v=u+at\\) and similar core relations.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Unit Check\nTop thing to remember: always convert to SI; dimensional checks catch errors instantly.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Vector & Sign\nRemember directions. Apply right-hand rules and sign conventions to avoid negative surprises.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Diagram & Graph\nNeat sketches of circuits, rays or motion graphs communicate logic and save explanation time.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Time Plan\nScan paper quickly, tackle sure shots first, reserve final five minutes for OMR or recheck.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Review & Present\nFinish with a clean review—units, significant figures, and margins tidy—so no marks leak.",
"image_description": ""
}
]
}
]