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[
{
"slide": 1,
"fragments": [
{
"fragment_index": -1,
"text_description": "CBSE Class 12 Physics – Sample Paper Review\nCrack the pattern, conquer the paper.",
"image_description": ""
}
]
},
{
"slide": 2,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question Distribution & Your Strength",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Chapter distribution vs. your proficiency map",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Chapter\n# Questions\nYour Strength\nElectric Charges & Fields\n3\nLow\nMoving Charges & Magnetism\n3\nLow\nAlternating Current\n3\nLow\nWave Optics\n3\nLow\nNuclei\n2\nLow\nElectrostatic Potential & Capacitance\n3\nMedium\nMagnetism & Matter\n2\nMedium\nElectromagnetic Waves\n2\nMedium\nDual Nature of Radiation & Matter\n2\nMedium\nSemiconductor Electronics\n3\nMedium\nCurrent Electricity\n4\nHigh\nElectromagnetic Induction\n3\nHigh\nRay Optics & Optical Instruments\n3\nHigh\nAtoms\n2\nHigh",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Goal: see how the paper maps to your syllabus strengths.",
"image_description": ""
}
]
},
{
"slide": 3,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nClass XII Physics – Electric Charges & Fields\nDifficulty: Challenging\nSelf-Discovery\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nA non-conducting spherical shell of radius \\(R\\) carries surface charge density \\(\\sigma(\\theta)=\\sigma_0\\cos\\theta\\). Using Gauss’s law, find the electric field (magnitude & direction) (i) inside the shell \\((r<R)\\) and (ii) outside the shell \\((r>R)\\).\nSpherical shell with \\(\\sigma=\\sigma_0\\cos\\theta\\)",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nGauss’s law: \\( \\Phi_E = Q_{\\text{encl}}/\\varepsilon_0 \\).\nOnly enclosed charge influences flux through a closed surface.\n\\(\\int \\sigma_0\\cos\\theta\\,dA=0\\); the shell’s net charge is zero, resembling a dipole.\nGauss law\nAngular charge density",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nChoose a spherical Gaussian surface of radius \\(r\\) concentric with the shell.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nEvaluate \\(Q_{\\text{encl}}=\\int \\sigma_0\\cos\\theta\\,dA\\) using \\(dA=R^2\\sin\\theta\\,d\\theta\\,d\\phi\\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nThe integral is zero for any \\(r\\). Hence flux is zero, so \\(E=0\\) inside. Outside, treat the shell as a dipole with moment \\(p=\\frac{4\\pi R^{3}\\sigma_0}{3}\\) and use the standard dipole field.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Break It Down\nFind enclosed charge first; then relate \\( \\Phi_E \\) to \\(E\\).",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Connect to Prior Knowledge\nCompare with a uniformly charged shell and with a point dipole.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize It\nSketch positive and negative hemispheres to see the dipole pattern.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nReplace \\(\\cos\\theta\\) with a constant to check your method gives Coulomb’s law.",
"image_description": ""
}
]
},
{
"slide": 4,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nCBSE Class 12 • Physics\nDifficulty: Challenging\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 proton enters a uniform 0.20 T magnetic field with speed \\(2\\times10^{6}\\,\\text{m s}^{-1}\\) at \\(60^{\\circ}\\) to the field. Find (a) the radius of its circular path and (b) the pitch of the helix it traces.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nLorentz force: \\( \\mathbf{F}=q(\\mathbf{v}\\times\\mathbf{B}) \\).\nOnly \\(v_{\\perp}\\) causes circular motion; \\(v_{\\parallel}\\) stays constant.\nPitch = distance moved along field in one revolution.\nLorentz force\nPitch of helix",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nResolve the velocity into \\(v_{\\perp}\\) and \\(v_{\\parallel}\\) using sine and cosine of \\(60^{\\circ}\\).",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nThe Lorentz force bends only \\(v_{\\perp}\\). Use \\( r=\\frac{m v_{\\perp}}{qB} \\) for the circle.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nCalculate \\( T=\\frac{2\\pi m}{qB} \\). Then pitch = \\( v_{\\parallel} T \\). Take \\( m_p=1.67\\times10^{-27}\\,\\text{kg} \\), \\( q_p=1.6\\times10^{-19}\\,\\text{C} \\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Break It Down\nTreat perpendicular and parallel motions separately.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Connect to Prior Knowledge\nRecall \\(r=mv/qB\\) from uniform circular motion in a magnetic field.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Visualize It\nDraw the helical path around straight field lines.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Test a Simpler Case\nImagine \\(60^{\\circ}=90^{\\circ}\\) to check understanding of pure circular motion.",
"image_description": ""
}
]
},
{
"slide": 5,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nClass 12 Physics – Alternating Current\nDifficulty: Hard\nSelf-Discovery\nProgressive Hints\n(Reveal only when needed)\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nAn ideal series L-C-R circuit has L = 50 mH, C = 80 µF, R = 40 Ω and is driven by a 200 V, 50 Hz source. Find (a) the current amplitude and (b) the phase angle between current and source voltage.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nInductive reactance: \\(X_L = 2\\pi f L\\).\nCapacitive reactance: \\(X_C = \\frac{1}{2\\pi f C}\\).\nImpedance: \\(Z = \\sqrt{R^{2} + (X_L - X_C)^{2}}\\); phase angle \\(\\tan\\phi = \\frac{X_L - X_C}{R}\\).\nImpedance\nPhase Angle",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nStart by comparing \\(X_L\\) and \\(X_C\\) at 50 Hz. Which one is larger?",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nNumeric check: \\(X_L \\approx 15.7 Ω\\), \\(X_C \\approx 39.8 Ω\\). The circuit is capacitive.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nCompute \\(Z\\) with the above values, use \\(I_0 = V_0 / Z\\). Find \\(\\tan\\phi\\); negative sign means current leads the voltage.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Thinking Strategies\nBreak It Down\nTreat reactances, resistance and voltage separately before combining.\nConnect to Prior Knowledge\nRecall DC Ohm’s law—impedance plays the same role in AC.\nVisualize It\nSketch the impedance triangle to see how \\(R\\) and \\((X_L-X_C)\\) combine.\nTest a Simpler Case\nSet C very large; notice how the circuit approaches a pure inductor–resistor pair.",
"image_description": ""
}
]
},
{
"slide": 6,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nWave Optics – CBSE Class 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:\nIn a YDS set-up (λ = 600 nm, d = 0.5 mm, D = 1.5 m) a mica sheet (μ = 1.5) covers one slit so the central maximum shifts to the 3rd bright fringe. Find the sheet thickness t.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nYDS fringe width \\( \\beta = \\frac{\\lambda D}{d} \\).\nSheet adds optical path \\( \\Delta = (\\mu - 1)t \\).\nCentral fringe shifts by \\( m \\) when \\( \\Delta = m\\lambda \\).\nYDS Experiment\nOptical Path",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nWhich path difference makes the central bright coincide with the 3rd bright?",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nCompute \\( \\Delta = 3\\lambda \\) and link it to the sheet by \\( (\\mu - 1)t \\).",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nSet \\( (\\mu - 1)t = 3\\lambda \\). With \\( \\lambda = 600\\,\\text{nm} \\) and \\( \\mu = 1.5 \\), you will get \\( t = 3.6\\,\\mu\\text{m} \\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Break It Down\nIsolate fringe shift first, then apply sheet optics.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Connect to Prior Knowledge\nRecall how glass slabs create phase difference in interference.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Visualize It\nSketch fringe pattern before and after inserting mica.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Test a Simpler Case\nSolve for \\( m = 1 \\) to check your method, then scale.",
"image_description": ""
}
]
},
{
"slide": 7,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nNuclei – Fusion\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:\nTwo deuterons fuse: \\(^{2}\\text{H}+^{2}\\text{H}\\rightarrow{}^{3}\\text{He}+n\\). Given \\(BE(^{2}\\text{H}) = 1.113\\ \\text{MeV}\\) and \\(BE(^{3}\\text{He}) = 7.718\\ \\text{MeV}\\), compute the released energy \\(Q\\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nMass defect converts to binding energy: \\( \\Delta m c^{2}=BE \\).\n\\(Q\\)-value equals change in total binding energy.\n1 MeV ≈ \\(1.602\\times10^{-13}\\) J for energy conversion.\nMass defect\nEnergy conversion",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nAdd the binding energies of both reactant deuterons first.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nThe neutron is free, so its binding energy is zero; only \\(^{3}\\text{He}\\) contributes for products.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nUse \\(Q = BE_{\\text{products}} - BE_{\\text{reactants}}\\). A positive \\(Q\\) means energy is released.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Break It Down\nCompute reactant and product totals separately, then compare.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Connect to Prior Knowledge\nRecall \\(E=\\Delta m c^{2}\\): mass defect converts directly to energy.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize It\nSketch energy levels; the deeper (more negative) level is more stable.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nTry fusing one deuteron with a neutron to verify your calculation steps.",
"image_description": ""
}
]
},
{
"slide": 8,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nElectrostatic Potential & Capacitance\nDifficulty: Challenging\nSelf-Discovery\nProgressive Hints\n(Reveal only when needed)\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nThree identical parallel-plate capacitors, each of capacitance \\(C\\), are connected in series to a 300 V source. While the circuit is live, the middle capacitor is completely filled with a dielectric of constant \\(\\kappa = 4\\). Determine (i) the charge on every capacitor and (ii) the total energy stored in the combination.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nSeries rule: \\(1/C_{\\text{eq}} = 1/C_1 + 1/C_2 + 1/C_3\\).\nDielectric effect: \\(C' = \\kappa C\\).\nEnergy stored: \\(U = \\tfrac12 C V^2\\).\nseries capacitance\nenergy storage",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nTreat the middle capacitor as \\(C' = 4C\\) once the dielectric is inserted.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nFind \\(C_{\\text{eq}}\\) of the series set, then compute common charge \\(Q = C_{\\text{eq}} \\times 300\\text{ V}\\).",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nVoltage drops: \\(V_1 = Q/C\\), \\(V_2 = Q/C'\\), \\(V_3 = Q/C\\). Check they sum to 300 V, then add energies: \\(U = \\tfrac12 \\sum C_i V_i^2\\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Thinking Strategies\nBreak It Down\nAnalyze each capacitor, then combine their effects.\nConnect to Prior Knowledge\nRemember: charge is identical across capacitors in series.\nVisualize It\nDraw potential drops across the three plates.\nTest a Simpler Case\nSet \\(\\kappa = 1\\) first; compare the result to see the dielectric’s impact.",
"image_description": ""
}
]
},
{
"slide": 9,
"fragments": [
{
"fragment_index": 1,
"text_description": "Question with Hints and Nudges\nCBSE Class 12 Physics\nDifficulty: Hard\nSelf-Discovery",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "The Question:\nA bar magnet oscillates freely in Earth’s horizontal field \\(B_H = 3.6 \\times 10^{-5}\\,\\text{T}\\). Its time period is \\(2\\,\\text{s}\\).\n(a) Find its magnetic moment \\(M\\).\n(b) How far from the centre on the axial line will the net field be zero?\nDiagram (if needed)",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Helpful Concepts\nTorsional oscillation of a magnetic dipole \\(T = 2\\pi\\sqrt{\\frac{I}{M B_H}}\\)\nEarth’s horizontal field acts like a uniform reference field\nAxial field of a dipole \\(B = \\frac{\\mu_0}{4\\pi}\\frac{2M}{r^3}\\)\nOscillation method\nNull points",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Gentle Nudge\nStart with the oscillation relation linking \\(T\\), \\(I\\), \\(M\\) and \\(B_H\\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Direction Pointer\nRearranged: \\(M = \\frac{4\\pi^2 I}{B_H T^2}\\). For a slender rod, \\(I \\approx m\\ell^2/12\\).",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Guiding Framework\nFor part (b) set the dipole’s axial field opposite to Earth’s: \\(\\frac{\\mu_0}{4\\pi}\\frac{2M}{r^3}=B_H\\). Solve for \\(r\\).",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Thinking Strategies\nBreak It Down\nTackle part (a) to get \\(M\\); use it immediately in part (b).\nConnect to Prior Knowledge\nRemember tangent law uses the same \\(M/B_H\\) ratio—consistent ideas help.\nVisualize It\nSketch field lines; mark where the magnet’s axial field cancels Earth’s.\nTest a Simpler Case\nAssume \\(M = 1\\,\\text{A·m}^2\\) to verify your algebra before inserting numbers.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Struggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
}
]
},
{
"slide": 10,
"fragments": [
{
"fragment_index": -1,
"text_description": "Electromagnetic Waves – Toughest Question\nPhysics – Class XII\nDifficulty: Hard\nSelf-Discovery\nProgressive Hints\n(Reveal only when needed)\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nAn EM wave with peak electric field \\(E_0 = 6\\;\\text{V\\,m}^{-1}\\) travels along +x. (i) Write expressions for \\(\\mathbf{E}(x,t)\\) and \\(\\mathbf{B}(x,t)\\). (ii) Find (a) the magnetic field amplitude and (b) the average energy flux.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nMaxwell link: \\(c = 1/\\sqrt{\\mu_0\\varepsilon_0}\\)\n\\(\\mathbf{S} = \\mathbf{E}\\times\\mathbf{B}/\\mu_0\\) → energy flux\nEnergy density \\(u = \\varepsilon_0 E^2\\)\nMaxwell Equations\nEnergy Density",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nStart with \\(B_0 = E_0/c\\) from Maxwell’s equations.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nChoose \\(\\mathbf{E}\\parallel\\hat{j}\\), \\(\\mathbf{B}\\parallel\\hat{k}\\) so both are ⟂ to +x.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nWrite sine wave forms, compute \\(B_0\\), then use \\(S_{\\text{avg}} = E_0B_0/(2\\mu_0)\\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Break It Down\nTreat amplitude, orientation, and energy flux as separate mini-tasks.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Connect to Prior Knowledge\nRecall that \\(c = 3\\times10^{8}\\,\\text{m s}^{-1}\\) links E and B fields.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Visualize It\nSketch mutually perpendicular \\(\\mathbf{E}\\), \\(\\mathbf{B}\\) and propagation axes.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Test a Simpler Case\nCheck with \\(E_0=1\\;\\text{V m}^{-1}\\) to avoid confusion between rms and peak values.",
"image_description": ""
}
]
},
{
"slide": 11,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nDual Nature of Radiation & Matter\nDifficulty: High\nSelf-Discovery\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nLight of wavelength 248 nm falls on a metal with work function 2 eV.\n(a) Calculate the maximum kinetic energy of the emitted electrons.\n(b) What accelerating potential must be applied so their de Broglie wavelength becomes 0.50 Å?",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nPhotoelectric equation \\(K_{\\max}=E_{\\gamma}-\\phi\\) (includes work function).\nde Broglie wavelength \\( \\lambda=\\frac{h}{\\sqrt{2mK}} \\).\nTotal kinetic energy after acceleration: \\(K=K_{\\max}+eV_{\\text{acc}}\\).\nWork Function\nKinetic Energy",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)\nGentle Nudge\nFirst find photon energy from its wavelength, then subtract the work function.\nDirection Pointer\nConvert \\(E_{\\gamma}=\\frac{hc}{\\lambda}\\) to eV. Use \\(K_{\\max}=E_{\\gamma}-2\\text{ eV}\\). For part (b), add \\(eV_{\\text{acc}}\\) to \\(K_{\\max}\\) before using the de Broglie formula.\nGuiding Framework\nStep 1: Compute \\(K_{\\max}\\).\nStep 2: Set \\( \\lambda =0.50\\; \\text{Å}=5.0\\times10^{-11}\\,\\text{m}\\).\nStep 3: Solve \\( \\lambda =\\frac{h}{\\sqrt{2m(K_{\\max}+eV_{\\text{acc}})}} \\) for \\(V_{\\text{acc}}\\).",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Thinking Strategies\nBreak It Down\nTackle part (a) completely before using its result for part (b).\nConnect to Prior Knowledge\nRecall how work function limits electron energy in photoelectric effect.\nVisualize It\nSketch an energy bar showing photon energy, work function, \\(K_{\\max}\\), and extra energy from acceleration.\nTest a Simpler Case\nTry λ = 1 Å first to see how potential affects wavelength before calculating the exact value.",
"image_description": ""
}
]
},
{
"slide": 12,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nSemiconductor Electronics – CE Amplifier\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:\nFor a silicon npn transistor in common-emitter mode, \\(R_C = 2\\,\\text{k}\\Omega\\), \\(R_E = 1\\,\\text{k}\\Omega\\) and \\(\\beta = 120\\). A \\(1\\,\\text{mV rms}\\) input is fed through a \\(1\\,\\text{k}\\Omega\\) base resistor.\n(a) Find the small-signal voltage gain \\(A_v\\).\n(b) Discuss the bias stability of this amplifier.\nNo additional diagram required",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nSmall-signal emitter resistance \\(r_e \\approx \\frac{25\\,\\text{mV}}{I_C}\\).\nVoltage gain \\(A_v \\approx -\\beta R_C / \\bigl(r_e + (\\beta+1)R_E\\bigr)\\).\nVoltage-divider bias improves thermal stability of \\(I_C\\).\nCE amplifier\nBias stability",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nCalculate \\(I_b = \\frac{1\\,\\text{mV}}{1\\,\\text{k}\\Omega}\\) and then find \\(I_C = \\beta I_b\\).",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nUse \\(r_e = 25\\,\\text{mV}/I_C\\); expect \\(r_e\\) to be a few hundred ohms.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nInsert \\(r_e\\) into \\(A_v\\). Note \\((\\beta + 1)R_E\\) dominates the denominator, so \\(A_v \\approx -2\\). Large \\(R_E\\) lowers gain but locks \\(I_C\\), giving excellent bias stability.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Break It Down\nSeparate dc bias calculation from ac gain evaluation.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Connect to Prior Knowledge\nRecall that emitter degeneration reduces gain but improves stability.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Visualize It\nSketch the small-signal model to see resistances in the signal path.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Test a Simpler Case\nIgnore \\(R_E\\) briefly to observe how gain would rise to \\(-\\beta R_C/r_e\\).",
"image_description": ""
}
]
},
{
"slide": 13,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nCBSE Class 12 Physics – Current Electricity\nDifficulty: Hard\nSelf-Discovery\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nA uniform wire of resistance \\(2\\,\\Omega\\) is bent into a square. Determine the resistance between opposite corners. The wire is then stretched so one side becomes twice its original length, volume constant. Find the new resistance between the same opposite corners.\nSketch the square and label resistances.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\n\\(R = \\rho \\dfrac{l}{A}\\); stretch keeps volume \\(lA\\) constant.\nBalanced Wheatstone bridge ⇒ no current through diagonal.\nSeries – parallel reduction simplifies symmetric networks.\nWheatstone bridge\nMeter bridge",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nFirst, compute resistance between corners when each side is \\(0.5\\,\\Omega\\).",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nTreat the square as a balanced Wheatstone bridge; ignore the diagonal branch.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nStretching doubles one side’s length. Remaining three sides shorten equally. Use \\(R \\propto l^2\\) then re-apply the bridge reduction.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Break It Down\nAnalyse pre-stretch and post-stretch networks separately.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Connect to Prior Knowledge\nRecall meter-bridge balance ⇒ equal potential nodes cancel internal branch.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize It\nSketch potentials on the square; mark equipotential corners.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nImagine stretching until three sides vanish; check if trend matches answer.",
"image_description": ""
}
]
},
{
"slide": 14,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nClass XII Physics – Electromagnetic Induction\nDifficulty: Hard\nSelf-Discovery\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nA circular coil of 200 turns and area 0.02 m² rotates at 300 rpm about its diameter in a uniform 0.04 T field. (a) Derive the expression for the instantaneous induced emf. (b) Calculate the peak emf produced.\nVisualise the rotating coil of an AC generator.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nFaraday’s law: \\( \\varepsilon = -N \\frac{d\\Phi}{dt} \\).\nFlux for rotating loop: \\( \\Phi = BA\\cos\\omega t \\).\nPeak emf of an AC generator: \\( \\varepsilon_{0}=N B A \\omega \\).\nInduced emf\nAC generator",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)\nGentle Nudge\nStart by writing the magnetic flux through one turn as a time-dependent cosine.\nDirection Pointer\nDifferentiate \\( \\Phi \\) with respect to \\( t \\) to get \\( \\varepsilon \\); then multiply by 200 turns.\nGuiding Framework\nConvert 300 rpm to \\( \\omega = 2\\pi n \\). Use \\( \\varepsilon_{0}=N B A \\omega \\) to find \\( \\approx 5.0\\; \\text{V} \\).",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Thinking Strategies\nBreak It Down\nFirst derive the formula, then plug numbers.\nConnect to Prior Knowledge\nRecall how generators convert rotation to alternating voltage.\nVisualize It\nSketch coil positions for \\( \\cos \\) and \\( \\sin \\) angles.\nTest a Simpler Case\nAnalyze a single-turn loop, then scale by 200.",
"image_description": ""
}
]
},
{
"slide": 15,
"fragments": [
{
"fragment_index": -1,
"text_description": "Question with Hints and Nudges\nRay Optics & Optical Instruments\nDifficulty: Hard\nSelf-Discovery\nStruggle is normal! Try to solve on your own before checking the hints.\nPrevious\nNext",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "The Question:\nA refracting telescope has an objective of focal length 80 cm and an eyepiece of focal length 4 cm. The final image is at infinity.\n(a) Find the lens separation.\n(b) Determine the angular magnification.\n(c) What objective diameter gives 200× the light-gathering power of a 6 mm pupil?",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nNormal adjustment: \\(d \\approx f_0 + f_e\\).\nAngular magnification \\(M = \\frac{f_0}{f_e}\\).\nLight-gathering power \\(\\propto D^2\\).\nTelescope\nAngular Magnification",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Progressive Hints\n(Reveal only when needed)",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Gentle Nudge\nIn normal adjustment, the objective image lies at the eyepiece focal plane.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Direction Pointer\nUse \\(d = f_0 + f_e\\) for separation and \\(M = f_0 / f_e\\) for magnification.",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Guiding Framework\nCalculate \\(M = 80/4\\). For light power, set \\(\\left(\\frac{D_0}{6\\text{ mm}}\\right)^2 = 200\\) and solve for \\(D_0\\).",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Thinking Strategies",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Break It Down\nSolve parts (a), (b), and (c) one after another.",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Connect to Prior Knowledge\nLink magnification to focal-length ratios you learned for simple lenses.",
"image_description": ""
},
{
"fragment_index": 10,
"text_description": "Visualize It\nDraw a ray diagram of the telescope in normal adjustment.",
"image_description": ""
},
{
"fragment_index": 11,
"text_description": "Test a Simpler Case\nTry equal focal lengths to check if your formulas give \\(M = 1\\).",
"image_description": ""
}
]
},
{
"slide": 16,
"fragments": [
{
"fragment_index": -1,
"text_description": "Atoms – Toughest Question\nAtoms • Hydrogen Spectrum\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:\nUsing Bohr postulates, derive the wavelength expression for the Balmer series (transitions to \\(n_1 = 2\\)) and calculate its shortest-wavelength limit.",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Helpful Concepts\nBohr energy levels: \\(E_n = -13.6\\,\\text{eV}/n^2\\).\nPhoton energy–wavelength link: \\( \\Delta E = hc/\\lambda \\).\nRydberg formula: \\(1/\\lambda = R\\,(1/n_1^2 - 1/n_2^2)\\).\nEnergy Levels\nRydberg Formula",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Gentle Nudge\nWrite \\( \\Delta E = 13.6\\,\\text{eV}\\,(1/n_1^2 - 1/n_2^2) \\) for the photon released.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Direction Pointer\nSubstitute \\(n_1 = 2\\). Use \\( \\Delta E = hc/\\lambda \\) to obtain \\(1/\\lambda\\).",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Guiding Framework\nLet \\(n_2 \\to \\infty\\). Then \\(1/\\lambda_{\\min} = R/4\\), so \\( \\lambda_{\\min} \\approx 364.6\\,\\text{nm} \\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Break It Down\nIdentify initial and final levels, compute \\( \\Delta E \\), then convert to \\( \\lambda \\).",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Connect to Prior Knowledge\nRemember Balmer lines lie in the visible region; check if your answer fits.",
"image_description": ""
},
{
"fragment_index": 8,
"text_description": "Visualize It\nSketch energy levels and draw arrows for transitions to \\( n=2 \\).",
"image_description": ""
},
{
"fragment_index": 9,
"text_description": "Test a Simpler Case\nCalculate \\( \\lambda \\) for \\( n_2 = 3 \\) first to verify your method.",
"image_description": ""
}
]
},
{
"slide": 17,
"fragments": [
{
"fragment_index": -1,
"text_description": "Key Take-aways",
"image_description": ""
},
{
"fragment_index": 1,
"text_description": "Check Symmetry First\nQuick recap—verify symmetry, then apply \\( \\oint \\vec{E}\\!\\cdot\\!d\\vec{A}=q_{\\text{enc}}/\\varepsilon_0 \\).",
"image_description": ""
},
{
"fragment_index": 2,
"text_description": "Split Velocity Components\nResolve \\( \\vec{v} \\) into parts ⟂ and ∥ \\( \\vec{B} \\) for instant path insight.",
"image_description": ""
},
{
"fragment_index": 3,
"text_description": "Compare \\(X_L\\) & \\(X_C\\)\nJudge lead/lag and impedance size before any AC number-crunching.",
"image_description": ""
},
{
"fragment_index": 4,
"text_description": "Measure Path Difference\nUse \\( \\Delta = d\\sin\\theta \\) to locate fringes swiftly in interference.",
"image_description": ""
},
{
"fragment_index": 5,
"text_description": "Find Δm First\nCompute mass or binding-energy change, then apply \\(E=\\Delta m c^{2}\\).",
"image_description": ""
},
{
"fragment_index": 6,
"text_description": "Include \\( r_e \\) in Model\nAdd emitter resistance in BJT small-signal diagrams for accurate gain.",
"image_description": ""
},
{
"fragment_index": 7,
"text_description": "Mind Your Units\nLast-minute things to remember: write rad, J, T clearly to avoid slips.",
"image_description": ""
}
]
}
]