How Golpo AI Turned IIT JEE Advanced 2026 Physics Paper 1 Into Video Solutions

Use the uploaded JEE Advanced 2026 Paper 1 provisional question paper and its supplied answer key.

For each selected question:
1. Codex isolates a clean image containing the complete question and all answer choices. Exclude the printed answer key from that learner-facing image, but retain the key for QA.
2. Codex sends the question image and a step-by-step lesson request to Golpo.
3. Golpo solves the problem and returns its initial narration and solution script before any video is rendered.
4. Codex collaborates as reviewer: check every reasoning step, calculation, unit, and final result, then map the result to the supplied correct answer choice.
5. If any check fails, Codex edits Golpo's script and runs the script QA again. Authorize rendering only after the revised script passes.
6. Send the approved script and question image to Golpo through the API. Use horizontal 16:9, Auto timing, and a Golpo Canvas Compact style.
7. Keep [IMAGE 1] on screen from the first frame while Golpo reads the complete question and every choice aloud. Do not show the answer key.
8. After [IMAGE 1], Golpo must teach the solution with progressive native illustrations, equations, diagrams, and worked examples. Do not use static solution screenshots.
9. Golpo ends with the complete final result and the correct answer choice, held visibly through the final frame.
10. Codex reviews the actual MP4 for decode, duration, audio/ASR, question-image continuity, visual accuracy, teaching effectiveness, and final-answer alignment. Retry any failure and publish only an approved render.

{
  "prompt": "JEE Advanced 2026 Physics Paper 1 Question 1. Use the exact reviewed narration below, preserving the single IMAGE 1 block. Follow the separate video instructions for progressive native illustrations.\nUse no logo, no watermark, and no pen or hand overlay.\n\n[START][IMAGE 1]Question one. Consider a large disk of radius R and two smaller disks, each of radius r equal to R divided by fifty, lying on its circumference as shown. The smaller disks are initially in contact, with angular separation delta theta between their centers. They roll without slipping in opposite directions with constant angular velocities omega and two omega, while the large disk is stationary. Find the time tau when the smaller disks are again in contact. Use sine delta theta equal to delta theta and ignore gravity. Option A: tau equals fifty-one times the quantity two pi minus four over fifty-one, divided by omega. Option B: tau equals fifty-one times the quantity two pi minus two over fifty-one, divided by the product of three and omega. Option C: tau equals fifty-one times the quantity two pi minus four over fifty-one, divided by the product of three and omega. Option D: tau equals fifty-one times the quantity two pi minus two over fifty-one, divided by omega.[END][IMAGE 1]\n\nBegin with the geometry of the initial contact. Each small-disk center is a distance R plus r from the large-disk center O. Join O to the two small centers and join those centers to each other. Their center-to-center distance is two r. Bisecting this isosceles triangle gives\n\nr equals open parenthesis R plus r close parenthesis times sine of delta theta over two.\n\nBecause r equals R over fifty, R plus r equals fifty-one r. Therefore sine of delta theta over two equals one over fifty-one. The angle is small, so delta theta over two is approximately one over fifty-one, and delta theta is approximately two over fifty-one radians.\n\nNext convert each disk's spin into motion of its center. Rolling without slipping on the fixed large disk means the point of contact is instantaneously at rest. Therefore the center speed equals the small radius times the spin rate. For the first disk, v one equals r omega. For the second, v two equals two r omega.\n\nBoth centers travel on a circle of radius R plus r, which is fifty-one r. Their orbital angular speeds about O are therefore omega over fifty-one and two omega over fifty-one. The centers move in opposite directions, so their relative angular speed is the sum: three omega over fifty-one.\n\nNow track the relative angle. Initially it is delta theta. The disks first move apart. They touch again when the relative angle reaches two pi minus delta theta. Hence the relative angle swept is two pi minus two delta theta, not two pi minus delta theta. Using delta theta equal to two over fifty-one, the required sweep is two pi minus four over fifty-one.\n\nTime equals relative angle divided by relative angular speed. Thus tau equals the quantity two pi minus four over fifty-one, divided by three omega over fifty-one. This simplifies to fifty-one times the quantity two pi minus four over fifty-one, divided by the product of three and omega.\n\nThe factor three comes from adding the opposite orbital rates. The factor four over fifty-one comes from subtracting the initial contact angle at both ends of the relative sweep. Therefore the correct answer is option C.",
  "custom_script": "[START][IMAGE 1]Question one. Consider a large disk of radius R and two smaller disks, each of radius r equal to R divided by fifty, lying on its circumference as shown. The smaller disks are initially in contact, with angular separation delta theta between their centers. They roll without slipping in opposite directions with constant angular velocities omega and two omega, while the large disk is stationary. Find the time tau when the smaller disks are again in contact. Use sine delta theta equal to delta theta and ignore gravity. Option A: tau equals fifty-one times the quantity two pi minus four over fifty-one, divided by omega. Option B: tau equals fifty-one times the quantity two pi minus two over fifty-one, divided by the product of three and omega. Option C: tau equals fifty-one times the quantity two pi minus four over fifty-one, divided by the product of three and omega. Option D: tau equals fifty-one times the quantity two pi minus two over fifty-one, divided by omega.[END][IMAGE 1]\n\nBegin with the geometry of the initial contact. Each small-disk center is a distance R plus r from the large-disk center O. Join O to the two small centers and join those centers to each other. Their center-to-center distance is two r. Bisecting this isosceles triangle gives\n\nr equals open parenthesis R plus r close parenthesis times sine of delta theta over two.\n\nBecause r equals R over fifty, R plus r equals fifty-one r. Therefore sine of delta theta over two equals one over fifty-one. The angle is small, so delta theta over two is approximately one over fifty-one, and delta theta is approximately two over fifty-one radians.\n\nNext convert each disk's spin into motion of its center. Rolling without slipping on the fixed large disk means the point of contact is instantaneously at rest. Therefore the center speed equals the small radius times the spin rate. For the first disk, v one equals r omega. For the second, v two equals two r omega.\n\nBoth centers travel on a circle of radius R plus r, which is fifty-one r. Their orbital angular speeds about O are therefore omega over fifty-one and two omega over fifty-one. The centers move in opposite directions, so their relative angular speed is the sum: three omega over fifty-one.\n\nNow track the relative angle. Initially it is delta theta. The disks first move apart. They touch again when the relative angle reaches two pi minus delta theta. Hence the relative angle swept is two pi minus two delta theta, not two pi minus delta theta. Using delta theta equal to two over fifty-one, the required sweep is two pi minus four over fifty-one.\n\nTime equals relative angle divided by relative angular speed. Thus tau equals the quantity two pi minus four over fifty-one, divided by three omega over fifty-one. This simplifies to fifty-one times the quantity two pi minus four over fifty-one, divided by the product of three and omega.\n\nThe factor three comes from adding the opposite orbital rates. The factor four over fifty-one comes from subtracting the initial contact angle at both ends of the relative sweep. Therefore the correct answer is option C.\n",
  "visual_instructions": "Create a progressive illustrative physics lesson in Horizontal 16:9 using Golpo Canvas Whiteboard Compact, with 4 min selected and private visibility. Use the exact supplied narration. Only one source image is supplied: question.png. Show that exact answer-key-free question-and-choices screenshot from the first frame and keep it fully fitted, sharp, and continuously visible through the entire IMAGE 1 block. Do not reveal the answer during the opening. Finish reading option D before transitioning.\nUse no logo, no watermark, and no pen or hand overlay.\n\nAfter IMAGE 1, use only native drawings and progressive equations. Do not reuse the question screenshot, insert solution screenshots, render static equation pages, or show document-like slides.\n\n1. Initial contact geometry: Draw a large fixed circle centered at O and two equal external rolling disks. Join O to centers C1 and C2. Draw chord C1C2 = 2r, radii OC1 = OC2 = R+r = 51r, and angle C1OC2 = delta theta. Bisect the triangle and reveal r = (R+r) sin(delta theta/2), then delta theta approximately 2/51. Keep all lengths geometrically consistent.\n\n2. No-slip kinematics: Animate each small disk rotating in the source-specified direction and its center moving along the radius-(R+r) orbit. Mark the contact point with the large disk as instantaneously at rest. Reveal v1=r omega and v2=2r omega, followed by orbital rates Omega1=omega/51 and Omega2=2omega/51. Use different arrow colors and show that the orbital motions are opposite.\n\n3. Relative-angle sweep: Use a circular angular track with two center markers. Start them separated by delta theta. Animate them moving apart until their next contact configuration has separation delta theta on the other side. Shade the relative sweep from delta theta to 2pi-delta theta and write Delta phi_rel=2pi-2delta theta=2pi-4/51. Then reveal Omega_rel=3omega/51.\n\n4. Final calculation: Progressively form tau=Delta phi_rel/Omega_rel=51(2pi-4/51)/(3omega). Before the closing narration begins, display the complete exact result and \"Correct answer: C\" beside a compact version of the angular-track drawing. Hold the result and choice through the final words and ending pause.\n",
  "narration_instructions": "Read the exact supplied physics narration clearly and at a measured teaching pace. Read every question detail and all four answer choices while IMAGE 1 remains visible. Preserve fraction grouping, squared quantities, Greek-letter names, subscripts, and the distinction between spin angular speed and orbital angular speed. Never narrate image tags or production directions. Do not ad-lib. Use Female 1, English, normal pacing, and no music.\n",
  "timing": "4",
  "golpo_video_engine": "golpo_canvas",
  "canvas_style_variant": "whiteboard_compact",
  "narration_voice": "female-1",
  "narration_language": "en",
  "onscreen_text_language": "en",
  "video_orientation": "horizontal",
  "visibility": "private",
  "enable_color": true,
  "enable_script_only_mode": false,
  "enable_podcast_engine": false,
  "onscreen_text_from_script_only": false,
  "watermark": false,
  "scene_pacing": "normal",
  "visual_animation_style": "sketch",
  "custom_images": [
    "[SANITIZED_SIGNED_MEDIA_URL]"
  ],
  "custom_images_description": [
    "Original JEE Advanced 2026 Physics Paper 1 Question 1 and all answer choices, with the printed answer key removed. Show this exact image from frame one and continuously through the complete Image 1 narration block only. Do not reuse it during the solution; use native progressive illustrations afterward."
  ],
  "keep_original_images": [
    true
  ],
  "disable_image_animation": [
    true
  ],
  "video_id": "[GENERATED_BY_GOLPO]"
}