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[
{
"slide": 1,
"fragments": [
{
"fragment_index": -1,
"text_description": "Physics Sample Paper Deep-Dive\nCrack the code to 70/70 with physics precision.",
"image_description": ""
}
]
},
{
"slide": 2,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question Distribution & Your Strength\nSource: Sample Question Paper – Physics (XII) 2024-25 & Self-Assessment Data",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Chapter\nPaper Weightage (%)\nYour Score (%)\nStrength\nMotion\n15\n85\nStrong\nForce & Laws of Motion\n12\n70\nModerate\nGravitation\n10\n55\nNeeds Revision\nWork & Energy\n12\n80\nStrong\nSound\n8\n60\nModerate",
"image_description": ""
}
]
},
{
"slide": 3,
"fragments": [
{
"fragment_index": 1,
"text_description": "Analytical Problem Solver\nGauss’s Law – Field of an Infinite Wire\nDifficulty: Hard\nTime: 5 min",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Problem Statement\nUsing Gauss’s theorem, derive the electric-field magnitude \\(E(r)\\) at a distance \\(r\\) from a very long straight wire carrying uniform linear charge density \\(\\lambda\\).\nGiven:\nLinear charge density \\(=\\lambda\\)\nCylindrical symmetry about wire\nObservation point at radius \\(r\\)\nTo Find:\nExpression for \\(E(r)\\)",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Solution Approaches\n1\nCylindrical Gaussian surface\nUse symmetry so flux crosses only the curved surface of a coaxial cylinder.\nComplexity: medium\n2\n—\n—\n—\n3\n—\n—\n—",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Logical Breakdown\nChoose surface length \\(L\\)\nCreates closed cylinder coaxial with wire.\nEnclosed charge \\(Q_{\\text{enc}}\\)\n\\(Q_{\\text{enc}}=\\lambda L\\)\nFlux through surface\n\\(\\Phi =E(2\\pi r L)\\)\nApply Gauss’s law\n\\(\\Phi=Q_{\\text{enc}}/\\varepsilon_0\\)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Step-by-Step Solution\n1\nCompute flux\nOnly the curved surface contributes.\n\\(\\Phi =E(2\\pi r L)\\)\n2\nEnclosed charge\nCharge inside cylinder equals linear density times length.\n\\(Q_{\\text{enc}}=\\lambda L\\)\n3\nEquate and solve\nSet \\(\\Phi =Q_{\\text{enc}}/\\varepsilon_0\\) and isolate \\(E\\).\n\\(E(r)=\\frac{\\lambda}{2\\pi\\varepsilon_0 r}\\)",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Key Insights\n\\(E\\propto 1/r\\); the field weakens linearly with distance.\nDirection is radially outward for positive \\(\\lambda\\) and inward for negative \\(\\lambda\\).\nResult is independent of chosen cylinder length \\(L\\).",
"image_description": ""
}
]
},
{
"slide": 4,
"fragments": [
{
"fragment_index": 1,
"text_description": "Electron in Uniform Magnetic Field\nMoving Charges & Magnetism\nDifficulty: hard\nEst. Time: 4 min",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Problem Context\nAn electron travels with speed \\(v_0\\) in a circular path of radius \\(r_0\\) under a uniform magnetic field \\(B\\).",
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{
"fragment_index": 3,
"text_description": "",
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{
"fragment_index": 4,
"text_description": "Question\nPredict how (a) the radius and (b) the time period change when the electron’s speed doubles to \\(2v_0\\).",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Helpful Hints\nHint 1\nUse the centripetal balance: \\(qvB = \\dfrac{mv^{2}}{r}\\).\nHint 2\nTime period \\(T = \\dfrac{2\\pi m}{qB}\\) does not involve the speed.\nHint 3\nCyclotron frequency \\(f = \\dfrac{qB}{2\\pi m}\\) stays constant for a given particle and field.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Things to Consider\nMass \\(m\\) and charge \\(q\\) of the electron are constant.\nMagnetic field \\(B\\) is uniform and steady.\nThe field does no work; only the path geometry changes with speed.",
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},
{
"fragment_index": 7,
"text_description": "Related Concepts\nLorentz force\nCircular motion\nCyclotron frequency",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Previous Problem\nProblem 3 of 10\nNext Problem",
"image_description": ""
}
]
},
{
"slide": 5,
"fragments": [
{
"fragment_index": 1,
"text_description": "Analytical Problem Solver\nVoltage Across L After C is Shorted\nDifficulty: Medium\nTime: 3 min",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Problem Statement\nA series LCR circuit is at resonance with \\(V_R = V_L = V_C = 10\\,\\text{V(rms)}\\). After the capacitor is short-circuited, find the new rms voltage across the inductor.\nGiven:\nR, L and C connected in series\nInitial resonance so \\(X_L = X_C\\)\nTo Find:\nNew \\(V_L\\) after the capacitor is removed",
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},
{
"fragment_index": 3,
"text_description": "Solution Approaches",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "1\nImpedance Method\nCompare total impedance before and after shorting and redraw phasor diagram.\nComplexity: Medium",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "2\n—\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "3\n—\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Logical Breakdown\nResonance Condition\nAt resonance \\(X_L = X_C\\) and \\(Z = R\\).\nAfter Shorting C\nImpedance becomes \\(Z' = \\sqrt{R^{2}+X_L^{2}}\\).\n—\n—",
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},
{
"fragment_index": 8,
"text_description": "Step-by-Step Solution",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "1\nDetermine R\nAt resonance \\(I_0 = \\frac{V_R}{R} = \\frac{10}{R}\\).\n\\(R = \\frac{10}{I_0}\\)",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "2\nCompute New Current\nWith \\(X_L = R\\), \\(Z' = R\\sqrt{2}\\) so \\(I' = \\frac{10}{R\\sqrt{2}} = \\frac{I_0}{\\sqrt{2}}\\).\n\\(I' = \\frac{10}{\\sqrt{R^{2}+X_L^{2}}}\\)",
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},
{
"fragment_index": 11,
"text_description": "3\nFind \\(V_L'\\)\n\\(V_L' = I' X_L = \\frac{I_0}{\\sqrt{2}}\\times R = 10\\sqrt{2}\\,\\text{V(rms)}\\).\n\\(V_L' = 10\\sqrt{2}\\,\\text{V}\\)",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "Key Insights\nShorting the capacitor destroys resonance and raises impedance to \\(R\\sqrt{2}\\).\nCircuit current falls by a factor \\(1/\\sqrt{2}\\).\nInductor voltage rises to \\(10\\sqrt{2}\\,\\text{V}\\) despite lower current because reactance is unchanged.",
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}
]
},
{
"slide": 6,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nFirst Common Dark Fringe (400 nm & 600 nm)\nDifficulty: Medium\nTime: 4 min",
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},
{
"fragment_index": 1,
"text_description": "Problem Statement\nTwo wavelengths — 400 nm and 600 nm — shine on a Young’s double-slit. Find the nearest distance from the central maximum where both produce a dark fringe.\nGiven:\n\\( \\lambda_1 = 400 \\text{ nm} \\)\n\\( \\lambda_2 = 600 \\text{ nm} \\)\nTo Find:\nSmallest \\(y\\) where minima coincide.",
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},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nMatch order numbers\nSet \\(m\\lambda_1=(m+\\tfrac12)\\lambda_2\\) and search smallest integer \\(m\\).\nComplexity: Hard\n2\n—\n—\n—\n3\n—\n—\n—",
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},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nFind integer \\(m\\)\nSolve the equality for the smallest positive \\(m\\).\nDistance formula\nUse \\(y=m\\lambda_1\\frac{L}{d}\\) once \\(m\\) is known.\n—\n—\n—\n—",
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},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nSet equality\nCoincidence condition: \\(m\\lambda_1=(m+\\tfrac12)\\lambda_2\\).\n\\(m\\lambda_1=(m+1/2)\\lambda_2\\)\n2\nSolve for \\(m\\)\nSmallest integer solution is \\(m=3\\).\n\\(m=3\\)\n3\nDistance from centre\n\\(y=3\\lambda_1\\frac{L}{d}=3(400\\text{ nm})\\frac{L}{d}\\).\n\\(y=3\\lambda_1L/d\\)",
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},
{
"fragment_index": 5,
"text_description": "Key Insights\nChoose the lowest integer orders that satisfy both wavelengths.\nUsing ratios eliminates slit separation and screen distance during calculation.\nResult: common dark at \\(y=3\\lambda_1L/d\\).",
"image_description": ""
}
]
},
{
"slide": 7,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nFission vs Fusion on B.E./A Curve\nDifficulty: Medium\nTime: 3 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nFour nuclei W(190), X(90), Y(60) and Z(30) lie on the binding-energy-per-nucleon curve. Identify which one is expected to undergo (a) fission and (b) fusion, and justify your choice.\nGiven:\nRelative \\( \\text{BE}/A \\) values from the curve\nMass numbers: 190, 90, 60, 30\nTo Find:\nMost probable fission and fusion candidates",
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},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nEnergy-gain check\nA reaction is favored if the products shift toward higher \\( \\text{BE}/A \\); compare with the peak at \\(A\\approx56\\).\nComplexity: Easy\n2\nAlternative not required\nFirst method fully resolves the task.\nComplexity: N/A\n3\n—\nNo further approach needed.\nComplexity: N/A",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nHeavy nuclei → fission\nSplitting a very heavy nucleus raises its \\( \\text{BE}/A \\).\nLight nuclei → fusion\nCombining light nuclei moves them toward the peak.\nPeak at \\(A\\approx56\\)\nIron region has the maximum \\( \\text{BE}/A \\).\nEnergy release = BE gain\nGreater \\( \\text{BE}/A \\) means lower mass, releasing energy.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nRank mass numbers\nW 190 > X 90 > Y 60 > Z 30.\nW is heaviest; Z lightest.\n2\nRead slope of curve\n\\( \\text{BE}/A \\) rises to the peak at \\(A\\approx56\\) then decreases for heavier nuclei.\nPeak \\( \\text{BE}/A \\) ≈ 8.8 MeV\n3\nApply energy-gain rule\nSplitting W moves fragments toward the peak ⇒ fission. Merging two Z nuclei moves left side nuclei toward the peak ⇒ fusion.\nFission: W → X + Y Fusion: Z + Z → Y",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nW (A = 190) sits on the downward slope; fission raises its \\( \\text{BE}/A \\).\nZ (A = 30) lies left of the peak; fusion propels it upward on the curve.\nEnergy released equals the gain in total binding energy per nucleon.",
"image_description": ""
}
]
},
{
"slide": 8,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nCapacitance with Slabs Inside\nDifficulty: Hard\nTime: 5 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nDerive expressions for the capacitance of a parallel-plate capacitor\n (plate area \\(A\\), separation \\(d\\)) when\n(i) a dielectric slab of thickness \\(t\\) and relative permittivity \\(k\\),\n(ii) a perfect-conductor slab of thickness \\(t\\;(t<d)\\) are inserted between the plates.\nGiven:\nDielectric constant \\(k\\)\nMetal behaves as an equipotential conductor\nTo Find:\nCapacitances \\(C_{\\text{dielectric}}\\) and \\(C_{\\text{metal}}\\)",
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},
{
"fragment_index": 2,
"text_description": "Solution Approaches",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "1\nSeries combination of gaps\nTreat the slab and remaining air gap as two capacitors in series.\nComplexity: Conceptual",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "2\n—\n—\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "3\n—\n—\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Logical Breakdown",
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},
{
"fragment_index": 7,
"text_description": "Air gap\nThickness \\(d-t\\) behaves as capacitor with permittivity \\(\\varepsilon_0\\).",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Inserted slab\nThickness \\(t\\); either dielectric (\\(k\\varepsilon_0\\)) or perfect conductor.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Series model\nCapacitances add as \\(\\frac{1}{C}=\\frac{1}{C_1}+\\frac{1}{C_2}\\).",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Effective separation\nMetal slab removes field, reducing gap to \\(d-t\\).",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Step-by-Step Solution",
"image_description": ""
},
{
"fragment_index": 12,
"text_description": "1\nDielectric slab\nCapacitances in series: \\(C_1=\\dfrac{\\varepsilon_0 A}{d-t}\\), \\(C_2=\\dfrac{k\\varepsilon_0 A}{t}\\).\n\\(C_{\\text{dielectric}}=\\dfrac{\\varepsilon_0 k A}{k(d-t)+t}\\)",
"image_description": ""
},
{
"fragment_index": 13,
"text_description": "2\nMetal slab\nField exists only in air gap of thickness \\(d-t\\).\n\\(C_{\\text{metal}}=\\dfrac{\\varepsilon_0 A}{d-t}\\)",
"image_description": ""
},
{
"fragment_index": 14,
"text_description": "3\nComparison\nBecause \\(d-t<d\\) and no series addition, the metal slab gives the higher capacitance.\n\\(C_{\\text{metal}} > C_{\\text{dielectric}} > C_0\\)",
"image_description": ""
},
{
"fragment_index": 15,
"text_description": "Key Insights\nUse series-capacitor formula when different media share the field path.\nA perfect conductor removes electric field within it, shortening the effective plate gap.\nReducing effective separation increases capacitance more than adding a dielectric alone.",
"image_description": ""
}
]
},
{
"slide": 9,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nDisplacement Current Through Capacitor\nDifficulty: easy\nTime: 2 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nA \\(0.001\\;\\text{m}^2\\) parallel-plate capacitor with \\(1\\times10^{-4}\\,\\text{m}\\) gap experiences a voltage rise of \\(1\\times10^{8}\\,\\text{V s}^{-1}\\). Find the displacement current \\(I_d\\).\nGiven:\n\\( \\varepsilon_0 = 8.85\\times10^{-12}\\;\\text{C}^2\\text{N}^{-1}\\text{m}^{-2} \\)\nArea \\(A = 1.0\\times10^{-3}\\;\\text{m}^2\\)\n\\(\\dfrac{dV}{dt}=1.0\\times10^{8}\\;\\text{V s}^{-1}\\)\nTo Find:\nDisplacement current \\(I_d\\)",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nMaxwell relation\nUse \\(I_d = \\varepsilon_0 A \\dfrac{dV}{dt}\\).\nComplexity: easy\n2\n—\nNot required.\nComplexity: —\n3\n—\nNot required.\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nMultiply constants\nCompute \\( \\varepsilon_0 A \\) first.\nUnit check\nConfirm result in amperes.\n—\n—\n—\n—",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nInsert numbers\nSubstitute given values into the formula.\n\\( I_d = 8.85\\times10^{-12}\\times1.0\\times10^{-3}\\times1.0\\times10^{8} \\)\n2\nCalculate\nMultiply to get the current.\n\\( I_d = 8.85\\times10^{-7}\\,\\text{A} \\)\n3\nState answer\nDisplacement current ≈ \\(0.885\\,\\mu\\text{A}\\).\n\\( I_d \\approx 0.885\\;\\mu\\text{A} \\)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nChanging electric field produces current; no charges move across the gap.\nDisplacement current equals conduction current in the external leads.\nEnsures continuity in the Ampère–Maxwell law.",
"image_description": ""
}
]
},
{
"slide": 10,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nElectron de Broglie Wavelength from Platinum\nDifficulty: Hard\nTime: 4 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nPlatinum with work function \\( \\phi = 5.63\\,\\text{eV} \\) is hit by light of frequency \\( 1.6\\times10^{15}\\,\\text{Hz} \\). Find the minimum de Broglie wavelength of the emitted electrons.\nGiven:\n\\(h = 6.63\\times10^{-34}\\,\\text{J s}\\)\n\\(\\phi = 5.63\\,\\text{eV}\\)\n\\(f = 1.6\\times10^{15}\\,\\text{Hz}\\)\nTo Find:\nMinimum de Broglie wavelength \\( \\lambda_{min} \\)",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nPhotoelectric equation\nUse \\( hf = \\phi + K_{max} \\) to obtain electron kinetic energy.\nComplexity: Medium\n2\n—\nNo alternate method required.\nComplexity: —\n3\n—\n—\nComplexity: —",
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},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nCompute photon energy\nFind \\(E = hf\\) in joules and electron-volts.\nConvert to kinetic energy\nSubtract work function to get \\(K_{max}\\).\nUse \\( \\lambda = h/\\sqrt{2mK} \\)\nRelate momentum to de Broglie wavelength.\nInterpret result\nQuote \\( \\lambda_{min} \\) and check plausibility.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nPhoton Energy\nCalculate energy carried by one photon.\n\\(\n E = (6.63\\times10^{-34})(1.6\\times10^{15}) \\approx 1.06\\times10^{-18}\\,\\text{J} \\approx 10.6\\,\\text{eV}\n \\)\n2\nMaximum Kinetic Energy\nSubtract work function of platinum.\n\\(\n K_{max}=10.6-5.63\\approx4.97\\,\\text{eV}=7.96\\times10^{-19}\\,\\text{J}\n \\)\n3\nMinimum de Broglie Wavelength\nApply de Broglie relation to fastest electrons.\n\\(\n \\lambda_{min}= \\frac{h}{\\sqrt{2m_e K_{max}}}\\approx 5.5\\times10^{-10}\\,\\text{m}\n \\)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nHigher photon frequency increases \\(K_{max}\\) and shortens \\( \\lambda_{min} \\).\nWork function is the energy barrier; surplus energy becomes kinetic.\nde Broglie relation bridges wave–particle duality for electrons.",
"image_description": ""
}
]
},
{
"slide": 11,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nIdentify Blocks in Full-Wave Rectifier\nDifficulty: Medium\nTime: 4 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nIn the AC-to-DC circuit, name blocks X and Y, draw their output waveforms, and state how the waveform changes if the centre-tap shifts toward diode D₁.\nGiven:\nInput: sinusoidal AC\nTo Find:\nNames of X and Y and their output wave shapes",
"image_description": ""
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{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nBlock naming method\nIdentify transformer and rectifier stages; match expected outputs.\nComplexity: easy\n2\n—\n—\nComplexity: —\n3\n—\n—\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nTransformer action\nStep-down coil with centre-tap provides two equal but opposite half-cycles.\nDiode conduction\nD₁ conducts during positive half, D₂ during negative half, flipping polarity.\n—\n—\n—\n—",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nOutput of X\nLower-voltage sine appears across the secondary.\n\\(V_s(t)=V_m\\sin\\omega t\\)\n2\nOutput of Y\nBoth half-cycles are rectified, giving pulsating DC.\n\\(V_o(t)=|V_s(t)|\\)\n3\nShifted centre-tap\nPositive pulses (D₁) grow, negative pulses (D₂) shrink; ripple and DC level rise.\nUnequal amplitude halves",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nCentre-tap position controls symmetry of the full-wave output.\nAdding a capacitor filter converts pulsating DC into smoother DC.\nUnequal half-cycles increase ripple and DC offset.",
"image_description": ""
}
]
},
{
"slide": 12,
"fragments": [
{
"fragment_index": 1,
"text_description": "Analytical Problem Solver\nCurrent Through Arm AD in Circular Bridge\nDifficulty: Hard\nTime: 5 min",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Problem Statement\nA uniform 12 Ω wire is bent into a complete circle. The ends of one diameter, C and D, are connected by a 10 Ω resistor. A battery of 8 V (negligible internal resistance) is connected across another pair of opposite points, A and B. Find the current flowing through the arc AD.\nGiven:\nWire resistance uniformly distributed\nBattery internal resistance ≈ 0\nTo Find:\nCurrent in arm AD, \\(I_{AD}\\)",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Solution Approaches\n1\nSymmetry split\nReplace each arc by resistance proportional to its angle; treat the circle as an unbalanced Wheatstone bridge.\nComplexity: Conceptual\n2\n—\n—\n—\n3\n—\n—\n—",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Logical Breakdown\nArc AB = 60°\nResistance \\(R_{AB}= \\frac{12}{360}\\times60 = 2\\;\\Omega\\).\nBridge structure\nCircular arcs form four arms; the 10 Ω resistor is the bridge branch.\nUnequal arms\nArms adjoining AB have 2 Ω each; opposite arms are 4 Ω each.\nKirchhoff equations\nApply loop and junction laws to find branch currents.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Step-by-Step Solution\n1\nConvert arcs to resistances\nEach 60° arc equals \\(2\\;\\Omega\\); each 120° arc equals \\(4\\;\\Omega\\).\n\\(R = \\frac{12}{360}\\theta\\)\n2\nRedraw as bridge\nTwo 2 Ω arms (AB) opposite two 4 Ω arms, with 10 Ω across CD.\n2 Ω – 4 Ω\n\\| 10 Ω \\|\n3\nApply Kirchhoff & solve\nSolving simultaneous equations gives \\(I_{AD}\\approx0.67\\;\\text{A}\\).\n\\(I_{AD}= \\frac{8}{12} = 0.67\\,\\text{A}\\)",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Key Insights\nUniform wire lets resistance scale directly with angle.\nUnbalanced Wheatstone bridges require Kirchhoff analysis.\nSymmetry arguments can simplify complex circular networks.",
"image_description": ""
}
]
},
{
"slide": 13,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nEmf Expression for AC Generator\nDifficulty: Medium\nTime: 4 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nA coil of \\(N\\) turns and area \\(A\\) rotates with angular speed \\( \\omega \\) in a uniform field \\( B \\). Derive its instantaneous emf.\nGiven:\nMagnetic flux \\( \\Phi = N B A \\cos \\omega t \\)\n—\n—\nTo Find:\n\\( \\varepsilon(t) \\)",
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},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nFaraday’s Law\nApply \\( \\varepsilon = -\\dfrac{d\\Phi}{dt} \\)\nComplexity: Easy\n2\n—\nSingle approach sufficient.\nComplexity: —\n3\n—\n—\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nDifferentiate cosine\nConvert \\( \\cos \\) to \\( \\sin \\) with factor \\( \\omega \\).\nPeak emf \\( \\varepsilon_0 \\)\n\\( \\varepsilon_0 = N B A \\omega \\)\n—\n—\n—\n—",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nDifferentiate flux\nUsing Faraday’s law, take time derivative of \\( \\Phi \\).\n\\( \\varepsilon = N B A \\omega \\sin \\omega t \\)\n2\nAccount for sign\nNegative sign obeys Lenz’s law, opposing change in flux.\n\\( \\varepsilon = -\\dfrac{d\\Phi}{dt} \\)\n3\nEnergy source\nMechanical work from the turbine is converted to electrical energy.\nInput (rotational) → Output (AC emf)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nFrequency of emf equals rotational frequency \\( \\omega/2\\pi \\).\nPeak emf rises with turns, area, field strength, and speed.\nAverage emf over one full cycle is zero; power comes from mechanical input.",
"image_description": ""
}
]
},
{
"slide": 14,
"fragments": [
{
"fragment_index": 1,
"text_description": "Analytical Problem Solver\nRight-Angled Prism with Liquid\nDifficulty: Hard\nTime: 5 min",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Problem Statement\nA ray grazes face AC of a right-angled prism (∠C = 45°). (a) Find the prism’s refractive index. (b) Decide the ray’s fate when face AC is immersed in a liquid of index \\(n = \\frac{2}{\\sqrt{3}}\\).\nGiven:\nIncidence equals grazing angle 45° (critical condition).\nRay initially undergoes total internal reflection.\nTo Find:\n1. Prism index \\(n_{\\text{prism}}\\). 2. Whether the ray reflects or refracts after immersion.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Solution Approaches\n1\nCritical-angle method\nUse grazing condition to equate 45° with critical angle and compute \\(n\\).\nComplexity: medium\n2\nSnell comparison\nCompare incident angle with new critical angle when liquid surrounds the face.\nComplexity: easy\n3\nRay-tracing check\nVisualise ray path to confirm decision on TIR vs refraction.\nComplexity: conceptual",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Logical Breakdown\nSnell law at grazing\nGrazing ray ⇒ incident angle equals critical angle.\nIndex comparison\nCompute new critical angle using \\(n_{\\text{liq}}\\) and compare with 45°.\n—\n—",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Step-by-Step Solution\n1\nFind prism index\nFor grazing incidence, 45° = critical angle \\(c\\).\n\\(n_{\\text{prism}} = \\frac{1}{\\sin 45^\\circ} = \\sqrt{2}\\)\n2\nCompute new critical angle\nLiquid now surrounds the interface.\n\\( \\sin c' = \\frac{n_{\\text{liq}}}{n_{\\text{prism}}} = \\frac{2/\\sqrt{3}}{\\sqrt{2}} = \\sqrt{\\tfrac{2}{3}} \\)\n3\nDecide ray’s fate\n45° < \\(c' \\approx 53^\\circ\\) → no TIR.\nRay refracts into the liquid.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Key Insights\nImmersion raises surrounding index, increasing critical angle.\nWhen incident angle < new critical angle, TIR disappears and refraction occurs.\nGrazing incidence is a quick test for determining refractive index via critical angle.",
"image_description": ""
}
]
},
{
"slide": 15,
"fragments": [
{
"fragment_index": -1,
"text_description": "Analytical Problem Solver\nMany Lines from One‐Electron Atom\nDifficulty: Easy\nTime: 2 min",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Problem Statement\nWhy does a hydrogen atom, which has only one electron, still produce many distinct spectral lines in its emission spectrum?\nGiven:\nSample contains a very large number of hydrogen atoms.\nTo Find:\nReason for multiplicity of lines in hydrogen spectrum.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Solution Approaches\n1\nEnsemble view\nDifferent atoms are excited to different energy levels; each emits its own transition photon.\nComplexity: Easy\n2\n—\n—\nComplexity: —\n3\n—\n—\nComplexity: —",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Logical Breakdown\nExcitation causes varied n→n′ transitions\nEnergy input lifts electrons to many quantum numbers \\(n>1\\).\nEach transition gives unique photon\nPhoton energy \\(E=h\\nu\\) equals the difference between two levels.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Step-by-Step Solution\n1\nPopulation of levels\nExcitation distributes electrons among levels \\(n=2,3,4,\\dots\\).\nn > 1\n2\nReturn paths\nAtoms relax by one or more jumps; each drop emits a photon.\n\\(\\Delta E = E_n - E_{n'} = h\\nu\\)\n3",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Key Insights\nEach allowed \\(n \\rightarrow n'\\) transition produces a unique wavelength.\nLarge ensembles contain atoms in many excited states at once.\nOne atom emits one photon, but millions together reveal the full spectral series.",
"image_description": ""
}
]
},
{
"slide": 16,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Take-aways",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Formula Flash\nMemorise core laws: \\(v = f\\lambda\\), \\(F = ma\\), \\(P = VI\\). Write them first in exam.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Diagram & Graph Snaps\nQuickly sketch ray diagrams or motion graphs to visualise and avoid wordy descriptions.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Unit Check\nAlways cross-verify units; convert cm→m early to stop last-minute errors.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "60-40 Rule\nSpend 60% time on sure-shot questions, 40% on tricky ones to maximise marks.",
"image_description": ""
}
]
}
]