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[
{
"slide": 1,
"fragments": [
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"fragment_index": -1,
"text_description": "Gauss’s Law Refresh\nGauss’s Theorem\nThe net electric flux through any closed surface equals the enclosed charge divided by \\( \\varepsilon_0 \\). Use this law to spot symmetries and predict electric-field patterns.\nKey Characteristics:\nExample: Infinite Line Charge\nWrap a coaxial cylinder (radius \\( r \\), length \\( \\ell \\)) around the wire. Flux \\( = E(2\\pi r\\ell) \\). Enclosed charge \\( = \\lambda \\ell \\). Hence \\( E = \\dfrac{\\lambda}{2\\pi \\varepsilon_0 r} \\), radially outward.",
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{
"fragment_index": 1,
"text_description": "Mathematically \\( \\oint \\vec E\\!\\cdot\\! d\\vec A = \\frac{q_{\\text{encl}}}{\\varepsilon_0} \\).",
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},
{
"fragment_index": 2,
"text_description": "Choose a Gaussian surface that matches the charge symmetry.",
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},
{
"fragment_index": 3,
"text_description": "Flux depends only on enclosed charge, not surface shape.",
"image_description": ""
}
]
},
{
"slide": 2,
"fragments": [
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"fragment_index": -1,
"text_description": "Example: Field of Line Charge\nGoal: derive \\(E(r)\\) for an infinite line charge using Gauss’s law — essential derivation practice for Q28.\n1\nIdentify symmetry\nInfinite wire has cylindrical symmetry; choose a coaxial cylinder of radius \\(r\\) and length \\(l\\).\n2\nEnclosed charge\nLinear density \\( \\lambda \\) gives \\( q_{\\text{enc}} = \\lambda l \\) inside the surface.\n3\nElectric flux\nField is radial and uniform on curved surface, zero on ends → \\( \\Phi = E(2\\pi r l) \\).\n4\nApply Gauss’s law\nSet \\( \\Phi = q_{\\text{enc}}/\\varepsilon_0 \\) so \\( E(2\\pi r l) = \\lambda l / \\varepsilon_0 \\).\n5\nField expression\n\\[ E(r) = \\frac{\\lambda}{2\\pi \\varepsilon_0 r} \\] directed radially outward (inward if \\( \\lambda < 0 \\)).\nPro Tip:\nRemember \\(E \\propto 1/r\\); the same five-step logic solves any cylindrical symmetry problem swiftly.",
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}
]
},
{
"slide": 3,
"fragments": [
{
"fragment_index": 1,
"text_description": "Drift Velocity & Current",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Drift Velocity \\(v_d\\)",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Within the Drude model, free electrons undergo random collisions yet experience a net force \\(eE\\). The average time between collisions, called relaxation time \\( \\tau\\), gives a steady drift speed \\(v_d = \\dfrac{eE \\tau}{m}\\). Because \\(J = n e v_d\\), a longer \\( \\tau\\) directly increases both drift velocity and current.",
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},
{
"fragment_index": 4,
"text_description": "In a wire of area \\(A\\): \\(I = n e A v_d\\). The \\(v_d\\!-\\!\\tau\\) plot is a straight line through the origin.",
"image_description": ""
},
{
"fragment_index": -1,
"text_description": "Source: Drude free-electron model",
"image_description": ""
}
]
},
{
"slide": 4,
"fragments": [
{
"fragment_index": -1,
"text_description": "Walkthrough: Drift Question\nGoal: predict average electron drift velocity for AC current (Q21 II).",
"image_description": ""
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{
"fragment_index": 1,
"text_description": "1\nRelate \\(v_d\\) and \\(I\\)\nIn a conductor \\(v_d \\propto I\\); precisely \\(v_d = I\\!/\\!(n e A)\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "2\nWrite time function\nGiven \\(I = I_0 \\sin 2\\pi\\nu t\\), so \\(v_d(t) = v_0 \\sin 2\\pi\\nu t\\) where \\(v_0 = I_0\\!/\\!(n e A)\\).",
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},
{
"fragment_index": 3,
"text_description": "3\nAverage over a cycle\n\\(\\langle v_d \\rangle = \\frac{1}{T}\\!\\int_{0}^{T}\\! v_0 \\sin 2\\pi\\nu t\\,dt = 0\\). Electrons merely oscillate; net drift is zero.",
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},
{
"fragment_index": 4,
"text_description": "Pro Tip:\nThe integral of any sine or cosine over its full period is always zero.",
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}
]
},
{
"slide": 5,
"fragments": [
{
"fragment_index": -1,
"text_description": "Charged Particle in B-Field\nLorentz Force\nA charge q moving with velocity\nv\nin a field\nB\nexperiences \\( \\mathbf F = q\\,\\mathbf v \\times \\mathbf B \\), always perpendicular to\nv\n; speed stays unchanged while direction bends.\nKey Characteristics:\nExample:\nAn electron entering a 0.1 T field at 3 × 10\n6\nm s⁻¹ (perpendicular) travels in a circle of radius ≈ 0.17 mm and period ≈ 3.6 ns.",
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{
"fragment_index": 1,
"text_description": "If \\( \\mathbf v \\perp \\mathbf B \\) → uniform circle; radius \\( r = \\dfrac{m v}{q B} \\).",
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},
{
"fragment_index": 2,
"text_description": "With component \\( v_\\parallel \\) along\nB\n→ helical path; pitch \\( p = v_\\parallel T \\).",
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},
{
"fragment_index": 3,
"text_description": "Period of revolution \\( T = \\dfrac{2\\pi m}{q B} \\); independent of speed and radius.",
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},
{
"fragment_index": 4,
"text_description": "Magnetic force does no work; kinetic energy and speed remain constant.",
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}
]
},
{
"slide": 6,
"fragments": [
{
"fragment_index": -1,
"text_description": "Walkthrough: Motion in B\nGoal – find how radius & time change when electron speed doubles so you can answer parts (i) & (ii) of Case Q29.\nPro Tip:\nFor Q29 pick (i) 2 r₀ and (ii) T₀ — radius doubles, time stays the same.",
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},
{
"fragment_index": 1,
"text_description": "1\nKey Relations\nFor circular motion in \\(B\\): \\(r=\\frac{mv}{qB}\\Rightarrow r\\propto v\\). Period \\(T=\\frac{2\\pi m}{qB}\\); it is speed-independent.",
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},
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"fragment_index": 2,
"text_description": "2\nRadius Change\nSpeed doubles: \\(v' = 2v_0\\). Hence \\(r' = 2r_0\\). Radius doubles.",
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{
"fragment_index": 3,
"text_description": "3\nPeriod Change\nSince \\(T\\) is independent of \\(v\\), doubling speed leaves \\(T\\) unchanged: \\(T' = T_0\\).",
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}
]
},
{
"slide": 7,
"fragments": [
{
"fragment_index": -1,
"text_description": "Interference Basics\nFringe Spacing \\( \\beta \\)\nIn Young’s double-slit set-up, fringe spacing is \\( \\beta = \\lambda D / d \\). A dark fringe appears when the path difference is \\((m + ½)\\lambda\\). Using two wavelengths overlaps their patterns, forming slow ‘beat’ fringes useful in mixed-colour analysis.\nQuiz: What happens to \\( \\beta \\) if slit separation \\( d \\) is increased?\nSource: CBSE Sample Question Paper 2024-25",
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]
},
{
"slide": 8,
"fragments": [
{
"fragment_index": -1,
"text_description": "Walkthrough: Two Wavelengths\nGoal → locate the nearest dark fringe when 400 nm and 600 nm beams interfere (Q18 I).",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "1\nSet the Darkness Condition\nSimultaneous dark bands need \\(m_1\\lambda_1=(m_2+\\tfrac12)\\lambda_2\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "2\nChoose Smallest Orders\nInsert \\( \\lambda_1=400\\text{ nm}, \\lambda_2=600\\text{ nm}\\). Smallest integers giving half-wavelength shift: \\(m_1=3, m_2=2\\).",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "3\nCompute Fringe Position\nDistance from central maximum: \\(y=m_1\\beta_1=3\\beta_1\\) where \\(\\beta_1=D\\lambda_1/d\\). Thus the least dark fringe is three 400 nm spacings away.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Pro Tip:\nUse the LCM mindset—search lowest integer pair satisfying odd-even rule before plugging into \\(y=m\\beta\\).",
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}
]
},
{
"slide": 9,
"fragments": [
{
"fragment_index": -1,
"text_description": "Photoelectric Snapshot\nEinstein’s Equation\nA photon of energy \\(h\\nu\\) liberates an electron by spending the metal’s work function \\(\\Phi\\); the rest appears as maximum kinetic energy \\(K_{\\max}\\), giving \\(K_{\\max}=h\\nu-\\Phi\\).\nKey Characteristics:",
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},
{
"fragment_index": 1,
"text_description": "Threshold frequency \\(\\nu_{0}=\\Phi/h\\); no photoelectrons for \\(\\nu<\\nu_{0}\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "For \\(\\nu>\\nu_{0}\\), \\(K_{\\max}\\) rises linearly with incident frequency.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Stopping potential \\(V_{s}=K_{\\max}/e\\) converts this energy into a measurable voltage.",
"image_description": ""
}
]
},
{
"slide": 10,
"fragments": [
{
"fragment_index": -1,
"text_description": "Walkthrough: Sodium Work Φ\nGoal: compute sodium’s work function so you can solve Q30 (III).\nPro Tip:\nShorter threshold wavelength means higher work function—keep units consistent.",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "1\nSet up\nGiven threshold wavelength λ₀ = 500 nm; work function formula \\( \\Phi = \\dfrac{hc}{\\lambda_0} \\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "2\nSubstitute\nUse \\( h = 6.63\\times10^{-34}\\,\\text{J·s} \\), \\( c = 3.0\\times10^{8}\\,\\text{m s}^{-1} \\), \\( \\lambda_0 = 5\\times10^{-7}\\,\\text{m} \\).",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "3\nCompute\n\\( \\Phi = 3.98\\times10^{-19}\\,\\text{J} \\approx 4\\times10^{-19}\\,\\text{J} \\).",
"image_description": ""
}
]
},
{
"slide": 11,
"fragments": [
{
"fragment_index": 1,
"text_description": "PN Junction & Rectifier",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Full-Wave Centre-Tapped Rectifier",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "It converts AC to DC using two PN-junction diodes and a centre-tapped transformer. In alternate half-cycles, D₁ or D₂ conducts, giving consecutive positive pulses across load RL. A shunt capacitor smooths ripple.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Now, sketch the full-wave output and label D₁, D₂ and RL on the circuit.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Source: CBSE Physics SQP 2024-25",
"image_description": ""
}
]
},
{
"slide": 12,
"fragments": [
{
"fragment_index": -1,
"text_description": "Walkthrough: Rectifier Output\nGoal: solve Q22—predict waveform when centre tap moves toward D₁.\nPro Tip:\nAverage output remains positive; ripple increases because peaks are unequal.",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "1\nRecall ideal output\nWith equal halves each diode conducts alternately, producing equal-height positive pulses—full-wave DC.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "2\nUnequal halves effect\nMoving the tap toward D₁ shortens its secondary segment. Voltage feeding D₁ drops while that for D₂ rises.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "3\nSketch the waveform\nDraw alternate pulses: lower peaks for D₁, higher peaks for D₂. The train stays unipolar but now asymmetric.",
"image_description": ""
}
]
},
{
"slide": 13,
"fragments": [
{
"fragment_index": -1,
"text_description": "Binding Energy Curve\nMost Stable Nuclei\nOn the BE/A–A curve, binding energy per nucleon climbs rapidly, peaks near A≈60 and then declines slowly. Mid-mass nuclei (Fe, Ni) sit at this peak and are most stable. Fusion of light nuclei and fission of heavy nuclei push products toward the peak, increasing BE/A and liberating energy.",
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}
]
},
{
"slide": 14,
"fragments": [
{
"fragment_index": -1,
"text_description": "Walkthrough: Fission or Fusion\nQ20 asks: From nuclei W(190), X(90), Y(60), Z(30), decide which one fissions and which one fuses.\nPro Tip:\nTo justify choices in Q20, always compare each nucleus’s position to the 60-amu peak of the binding-energy curve.",
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},
{
"fragment_index": 1,
"text_description": "1\nRead the Binding-Energy Curve\nUsing curve reasoning: BE / A rises to a peak near A≈60, then declines. Energy is gained by splitting heavy nuclei or fusing very light ones.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "2\nChoose the Fission Candidate\nW has A = 190, far right of the peak, so splitting it raises BE / A and releases energy. W is most likely to undergo nuclear fission.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "3\nChoose the Fusion Candidate\nZ has A = 30, well left of the peak. Fusing two such light nuclei moves toward A≈60, increasing BE / A. Hence Z is most likely to undergo nuclear fusion. X and Y lie near the peak, so little energy gain either way.",
"image_description": ""
}
]
},
{
"slide": 15,
"fragments": [
{
"fragment_index": -1,
"text_description": "Things to Remember\nRecall key formulas and avoid common pitfalls before the test.",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Line Charge Field\nUse symmetry: \\(E = \\lambda / 2\\pi\\varepsilon_0 r\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "AC Drift Velocity\nAcross one pure AC cycle, average electron drift equals zero.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Charge in B-Field\nRadius \\(r \\propto v\\); period \\(T = 2\\pi m / qB\\) stays constant.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Fringe Overlap\nFirst common dark fringe appears at path difference equal to LCM of wavelengths.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Photoelectric Formula\n\\(K_{\\text{max}} = h\\nu - \\Phi;\\; V_s = K_{\\text{max}}/e\\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Full-Wave Rectifier\nCentre tap must sit exactly midway to produce equal positive pulses.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Binding Energy Curve\nHeavy nuclei favor fission, light nuclei favor fusion—energy gain follows the curve.",
"image_description": ""
}
]
}
]