diff --git a/PragatiVeer_Cancellation_Policy_v1.0.pdf b/PragatiVeer_Cancellation_Policy_v1.0.pdf deleted file mode 100644 index a3c4fa8..0000000 Binary files a/PragatiVeer_Cancellation_Policy_v1.0.pdf and /dev/null differ diff --git a/PragatiVeer_Disclaimer_v1.0 (2).pdf b/PragatiVeer_Disclaimer_v1.0 (2).pdf deleted file mode 100644 index b7fbc66..0000000 Binary files a/PragatiVeer_Disclaimer_v1.0 (2).pdf and /dev/null differ diff --git a/PragatiVeer_Privacy_Policy_v3.0.pdf b/PragatiVeer_Privacy_Policy_v3.0.pdf deleted file mode 100644 index a9b2d94..0000000 Binary files a/PragatiVeer_Privacy_Policy_v3.0.pdf and /dev/null differ diff --git a/PragatiVeer_Refund_Policy_v1.0.pdf b/PragatiVeer_Refund_Policy_v1.0.pdf deleted file mode 100644 index 89af758..0000000 Binary files a/PragatiVeer_Refund_Policy_v1.0.pdf and /dev/null differ diff --git a/PragatiVeer_Terms_and_Conditions_v3.0.pdf b/PragatiVeer_Terms_and_Conditions_v3.0.pdf deleted file mode 100644 index dcf227c..0000000 Binary files a/PragatiVeer_Terms_and_Conditions_v3.0.pdf and /dev/null differ diff --git a/cbt_test.html b/cbt_test.html deleted file mode 100644 index 52721cf..0000000 Binary files a/cbt_test.html and /dev/null differ diff --git a/parse.js b/parse.js deleted file mode 100644 index 3641264..0000000 --- a/parse.js +++ /dev/null @@ -1,9 +0,0 @@ - -import { unified } from "unified"; -import remarkParse from "remark-parse"; -import remarkMath from "remark-math"; - -const text = "C) $2 \\text{ m}$"; -const tree = unified().use(remarkParse).use(remarkMath).parse(text); -console.log(JSON.stringify(tree, null, 2)); - diff --git a/parse2.js b/parse2.js deleted file mode 100644 index b568680..0000000 --- a/parse2.js +++ /dev/null @@ -1,18 +0,0 @@ - -import { unified } from "unified"; -import remarkParse from "remark-parse"; -import remarkMath from "remark-math"; -import rehypeKatex from "rehype-katex"; -import remarkRehype from "remark-rehype"; -import rehypeStringify from "rehype-stringify"; - -const text = "The velocity of a particle moving along the x-axis is given by $v(t) = \\int_{0}^{t} (3\\tau^2 - 2\\tau) d\\tau$. What is the displacement of the particle at $t = 2$ seconds if it starts from the origin?"; -const tree = unified() - .use(remarkParse) - .use(remarkMath) - .use(remarkRehype) - .use(rehypeKatex, { strict: "warn", throwOnError: false }) - .use(rehypeStringify) - .processSync(text); -console.log(tree.toString()); - diff --git a/pragativeer-ai-coach-complete.html b/pragativeer-ai-coach-complete.html deleted file mode 100644 index f996b78..0000000 --- a/pragativeer-ai-coach-complete.html +++ /dev/null @@ -1,1216 +0,0 @@ - - - - - - PragatiVeer โ€” AI Coach (Complete) - - - - - - - - - - - - - - - -
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Namaste, Rohan ๐Ÿ‘‹

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- เค†เคœ เค•เคพ เคคเฅเคฎเฅเคนเคพเคฐเคพ AI เค•เฅ‹เคš เคคเฅˆเคฏเคพเคฐ เคนเฅˆ - โ€ข - JEE Main 2027 -
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๐ŸŽฏ Exam Readiness Score A-05

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๐Ÿ“‹ Today's Study Plan A-03

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๐ŸŽฏ Mistake Tracking A-01

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๐Ÿ’ฌ AI Tutor Chat A-04

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PragatiVeer AI Tutor
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Online โ€ข Context-aware
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42 / 50 messages today
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- Can you explain why option C is wrong in today's DPS Q7? The one about electric field lines. -
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- Great question! Option C says "Electric field lines can intersect each other." This is incorrect because:

- โ€ข Electric field lines never intersect โ€” at the point of intersection, there would be two different directions of the electric field, which is impossible.

- โ€ข I see you've made a similar mistake in Electrostatic Potential (Day 3) โ€” want me to pull up that question for revision? -
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- Here's your previous mistake from Electrostatic Potential (attempted 19 Jul):

- "Two point charges +4ฮผC and -2ฮผC are placed 10cm apart. Where is the electric field zero?"

- ๐Ÿ”„ Similar question: Three charges +2q, -q, and +3q are placed at vertices of an equilateral triangle. Find the point where net electric field is zero.

- Want me to walk through the solution step by step? ๐Ÿค” -
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๐Ÿ›๏ธ College Predictor A-07

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๐Ÿ“‹ Exam Day Strategy A-06

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Start with Physics
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Your strongest section โ€” 78% accuracy
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45 min per section
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Physics: 45m โ€ข Chem: 40m โ€ข Maths: 50m
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Numerical first in Chem
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You score 72% on numerical vs 58% on theory
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Last 15 min sprint
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You slow down โ€” practice timed sprints!
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โฐ Study Time Optimization A-08

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๐Ÿ“Š This Week Overview

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Good morning, Rohan! ๐Ÿ‘‹

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Your AI Coach has prepared today's plan. You're on track for NIT Surat.

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Readiness Score
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Good Progress

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โ†‘ 5 pts this week

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๐Ÿ“‹ Today's Study Plan
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Regenerate โ†’
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DPS โ€” Electrostatics + Kinematics
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10 questions โ€ข Mixed difficulty
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5 questions โ€ข Your accuracy here: 38% โš ๏ธ
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AI Coach Insight
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- You keep making sign errors in Thermodynamics. - In your last 3 attempts, you've used the wrong sign for work done (W = -Pฮ”V). - Review the First Law of Thermodynamics โ€” we've queued 3 practice questions for tomorrow. -
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๐Ÿ† City Leaderboard โ€” Jaipur
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This Week โ†’
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Priya S.
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23 Days

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Longest streak: 23 days ๐ŸŽ‰

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- - diff --git a/pragativeer-prd (4).pdf b/pragativeer-prd (4).pdf deleted file mode 100644 index 756db6b..0000000 Binary files a/pragativeer-prd (4).pdf and /dev/null differ diff --git a/prd_dump.txt b/prd_dump.txt deleted file mode 100644 index 65da02d..0000000 Binary files a/prd_dump.txt and /dev/null differ diff --git a/prd_dump_utf8.txt b/prd_dump_utf8.txt deleted file mode 100644 index 0dffa04..0000000 --- a/prd_dump_utf8.txt +++ /dev/null @@ -1,1385 +0,0 @@ -๏ปฟPragatiVeer -Product Requirements Document -India's first AI-powered daily practice platform for JEE & NEET -Version: 1.0 -Date: 1 July 2026 -Author: Maya ฮ“ร‡รถ CEO's Office, Gyannetra Pvt Ltd -Approved by: Kapil Panwar (CEO) -Status: ฮ“ยฃร  Final -CONFIDENTIAL ฮ“ร‡รถ GYANNETRA PROPRIETARY -Gyannetra Pvt Ltd ฮ“ร‡รณ Pune, India -PragatiVeer ฮ“ร‡รถ Product Requirements -Document (PRD) -Version: 1.0 -Date: 1 July 2026 -Author: Maya ฮ“ร‡รถ CEO's Office, Gyannetra Pvt Ltd -Status: Final (Approved by Kapil Panwar, CEO) -Table of Contents -Product Overview -Target Audience & Personas -Tier Structure & Pricing -Feature Catalog ฮ“ร‡รถ Free Tier -Feature Catalog ฮ“ร‡รถ Pro Plan -Feature Catalog ฮ“ร‡รถ AI Coach Plan -Cross-Cutting Features -User Stories ฮ“ร‡รถ Complete Collection -User Flows -Technical Requirements -Data Model (Key Entities) -1. Product Overview -1.1 Vision Statement -PragatiVeer is India's first AI-powered daily practice platform for JEE & NEET aspirants. It combines -daily practice sheets, computer-based test series, and an AI coach that personalizes every student's -learning journey ฮ“ร‡รถ all through a web-based portal accessible on any device. -1.2 Product Mission -To make high-quality, personalized JEE/NEET preparation accessible and affordable for every Indian -student ฮ“ร‡รถ especially in Tier 2/3 cities where quality coaching is out of reach. -1. -2. -3. -4. -5. -6. -7. -8. -9. -10. -11. -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 2 of 27 -1.3 Key Differentiators -Differentiator -Description -Daily Practice Loop -DPS every day builds discipline, not cramming -AI Personalization -Tracks every DPS, every test, every mistake ฮ“ร‡รถ adapts to the student -Exam-Realistic Testing -Computer-based tests exactly like JEE Main/Advanced pattern -Affordable Pricing -ฮ“รฉโ•ฃ299/mo for Pro vs ฮ“รฉโ•ฃ5K-15K for Physics Wallah -Parent Dashboard -Real-time progress visibility for parents (the actual buyers) -1.4 Platform -Web-based only (no mobile app) -Fully responsive ฮ“ร‡รถ works seamlessly on mobile browsers, tablets, and desktops -PWA (Progressive Web App) capabilities for offline access and home-screen install -1.5 Scope ฮ“ร‡รถ Exams Covered (Phase 1) -JEE Main (Paper 1: B.E./B.Tech) -JEE Advanced -NEET UG -2. Target Audience & Personas -2.1 Primary Persona: Rohan (Student, Class 12) -Attribute -Detail -Age -17 -Location -Tier 2 city (Jaipur) -Goal -JEE Main + Advanced 2027 -Medium -Hindi medium school, studies in English + Hindi mix -Device -Android phone (primary), shared family laptop (occasional) -Internet -4G, sometimes unstable -Pain Points -Coaching is expensive (ฮ“รฉโ•ฃ50K+/yr), no daily practice structure, doesn't know where he stands -Behavior -Uses Telegram for study groups, watches PW videos on YouTube, solves PYQs from books -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 3 of 27 -2.2 Secondary Persona: Mrs. Sharma (Parent) -Attribute -Detail -Age -45 -Role -Mother of a Class 12 JEE aspirant -Location -Delhi NCR -Device -Smartphone (WhatsApp, basic browsing) -Pain Points -Worried about child's preparation, doesn't know if child is on track, coaching is too expensive -Decision Trigger -Seeing concrete progress data ฮ“ร‡รถ "Mera bachcha kitna padh raha hai?" -2.3 Tertiary Persona: Mr. Verma (School Principal) -Attribute -Detail -Role -Principal, CBSE-affiliated school in Indore -Need -Affordable practice platform for 500+ students -Decision Factor -Bulk pricing, white-label option, school performance dashboard -3. Tier Structure & Pricing -3.1 Pricing (Final ฮ“ร‡รถ Neha Gupta's Model) -Tier -Monthly -Quarterly -Annual -Per Month (Annual) -Free -ฮ“รฉโ•ฃ0 -ฮ“ร‡รถ -ฮ“รฉโ•ฃ0 -ฮ“รฉโ•ฃ0 -Pro -ฮ“รฉโ•ฃ299 -ฮ“รฉโ•ฃ749 (ฮ“รฉโ•ฃ250/mo) -ฮ“รฉโ•ฃ2,499 (ฮ“รฉโ•ฃ208/mo) -ฮ“รฉโ•ฃ208 -AI Coach -ฮ“รฉโ•ฃ599 -ฮ“รฉโ•ฃ1,499 (ฮ“รฉโ•ฃ500/mo) -ฮ“รฉโ•ฃ4,999 (ฮ“รฉโ•ฃ417/mo) -ฮ“รฉโ•ฃ417 -3.2 Launch Offers -Pro Launch: First month at ฮ“รฉโ•ฃ99 (vs ฮ“รฉโ•ฃ299 regular) -Annual AI Coach: 30% off + 2 months free -Referral: Invite a friend ฮ“รฅร† both get 7 days of AI Coach free -School Bulk: ฮ“รฉโ•ฃ499/student/yr for 100+ students -Sibling Discount: 20% off on second child -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 4 of 27 -3.3 Tier Comparison Matrix -Feature -Free -Pro -AI Coach -Daily Practice Sheets (1/day) -ฮ“ยฃร  -ฮ“ยฃร  Unlimited -ฮ“ยฃร  Unlimited -Answer Keys -ฮ“ยฃร  -ฮ“ยฃร  -ฮ“ยฃร  -Basic Progress Dashboard -ฮ“ยฃร  -ฮ“ยฃร  -ฮ“ยฃร  -Streak Tracking -ฮ“ยฃร  -ฮ“ยฃร  -ฮ“ยฃร  -Leaderboard (City/School) -ฮ“ยฃร  -ฮ“ยฃร  -ฮ“ยฃร  -Topic-wise Micro-tests -ฮ“ยฃร  (5/day) -ฮ“ยฃร  (Unlimited) -ฮ“ยฃร  (Unlimited) -Weekly Test Series -ฮ“ยฅรฎ -ฮ“ยฃร  -ฮ“ยฃร  -Full-Length Mock Tests -ฮ“ยฅรฎ -ฮ“ยฃร  (4/month) -ฮ“ยฃร  (Unlimited) -Previous Year Papers -ฮ“ยฅรฎ -ฮ“ยฃร  -ฮ“ยฃร  -Detailed Performance Analytics -ฮ“ยฅรฎ -ฮ“ยฃร  -ฮ“ยฃร  -Chapter-wise Progress -ฮ“ยฅรฎ -ฮ“ยฃร  -ฮ“ยฃร  -Doubt Resolution (Community) -ฮ“ยฅรฎ -ฮ“ยฃร  -ฮ“ยฃร  -Parent Dashboard -ฮ“ยฅรฎ -ฮ“ยฃร  -ฮ“ยฃร  -AI Mistake Tracking -ฮ“ยฅรฎ -ฮ“ยฅรฎ -ฮ“ยฃร  -Spaced Repetition -ฮ“ยฅรฎ -ฮ“ยฅรฎ -ฮ“ยฃร  -Personalized Daily Study Plan -ฮ“ยฅรฎ -ฮ“ยฅรฎ -ฮ“ยฃร  -AI Tutor Chat -ฮ“ยฅรฎ -ฮ“ยฅรฎ -ฮ“ยฃร  -Exam Readiness Score -ฮ“ยฅรฎ -ฮ“ยฅรฎ -ฮ“ยฃร  -Exam Day Strategy -ฮ“ยฅรฎ -ฮ“ยฅรฎ -ฮ“ยฃร  -College Predictor -ฮ“ยฅรฎ -ฮ“ยฅรฎ -ฮ“ยฃร  -Study Time Optimization -ฮ“ยฅรฎ -ฮ“ยฅรฎ -ฮ“ยฃร  -Offline Mode (Download DPS) -ฮ“ยฅรฎ -ฮ“ยฃร  -ฮ“ยฃร  -Print-Friendly PDF Export -ฮ“ยฅรฎ -ฮ“ยฃร  -ฮ“ยฃร  -4. Feature Catalog ฮ“ร‡รถ Free Tier -F-01: User Registration & Login -Description: Self-service registration with email/phone + OTP. Google login optional. Priority: P0 -(Launch-blocking) -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 5 of 27 -Requirements: - Register with email OR phone number - OTP-based verification - Google OAuth login -- Password-based login (email + password) - Forgot password flow (email/OTP reset) - Session -management (stay logged in for 30 days) - Single session per user (logout from other devices option) -F-02: Student Profile -Description: Basic profile capturing student's exam target, class, and medium. Priority: P0 -Requirements: - Name, email, phone - Class (11, 12, Dropper) - Target exam (JEE Main, JEE -Advanced, NEET) - Medium (English, Hindi, Bilingual) - School name (optional) - City/State - Avatar/ -photo upload - Profile completion progress bar -F-03: Daily Practice Sheet (DPS) -Description: One DPS per day with 10-15 questions covering mixed topics. Priority: P0 (Core feature) -Requirements: - 10-15 questions per DPS (mix of Physics, Chemistry, Maths/Biology) - Questions -displayed one at a time with navigation - Multiple choice (single correct, multiple correct) - Numerical -answer type - Timer per DPS (configurable, default 20 min) - Submit and get instant answer key - Mark -questions for review - Question palette (answered, not answered, marked for review) - Auto-submit on -timer expiry - DPS resets daily at 6:00 AM IST - Cannot attempt previous day's DPS (creates urgency) -F-04: Answer Key & Basic Solutions -Description: Instant answer key after DPS submission with correct answers. Priority: P0 -Requirements: - Show correct answer for each question immediately after submission - Highlight -user's answer vs correct answer - Show score (X/15) - Show time taken per question - Basic solution -text (not video) - Option to flag confusing questions -F-05: Basic Performance Dashboard -Description: Simple dashboard showing daily and weekly performance. Priority: P0 -Requirements: - Today's DPS score - 7-day accuracy trend (line chart) - Total questions attempted -(lifetime) - Overall accuracy percentage - Streak count (consecutive days) - "You're on a X-day streak!" -message -F-06: Streak Tracking -Description: Track consecutive days of DPS completion. Priority: P0 -Requirements: - Calendar view showing which days DPS was completed - Current streak count - -Longest streak record - Streak milestones (7 days, 30 days, 60 days, 100 days) - Streak milestone -badges - Daily reminder notification (browser/email) -F-07: Leaderboard -Description: Gamified ranking among peers. Priority: P1 -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 6 of 27 -Requirements: - City-wise leaderboard - School-wise leaderboard (if school provided) - All-India -leaderboard - Based on: accuracy โ”œรน consistency (streak bonus) - Weekly reset - Top 10 visible, user -sees their own rank - Anonymous display (usernames only) -F-08: Topic-wise Micro-tests -Description: Short 5-question tests on specific topics. Priority: P1 -Requirements: - 5 questions per micro-test - Choose topic from syllabus tree (e.g., Physics ฮ“รฅร† -Electrostatics ฮ“รฅร† Coulomb's Law) - 5-minute timer - Instant results with topic-wise accuracy - Unlimited -attempts (Free: 5 micro-tests/day) - Recommended topics based on DPS performance -5. Feature Catalog ฮ“ร‡รถ Pro Plan -P-01: Unlimited DPS -Description: Access to all DPS without daily limit. Multiple DPS per day. Priority: P0 -Requirements: - All Free tier DPS features - Unlimited DPS per day - Access to DPS archive (previous -days) - DPS difficulty levels (Easy, Medium, Hard) - Custom DPS: student picks subjects and topics - -DPS bookmarking for later -P-02: Weekly Test Series -Description: Full-length computer-based tests every week, increasing frequency near exams. Priority: -P0 (Core Pro feature) -Requirements: - One full-length test per week (standard frequency) - Increases to 2/week in last 3 -months before exam - Daily tests in last month before exam - Exact JEE Main/Advanced/NEET pattern -- Computer-based test interface (CBT) - Sectional timing (as per actual exam) - Question palette with -color coding - Mark for review - Change answer within section - Calculator (on-screen, basic) - Periodic -table reference (for Chemistry) - Scratchpad (digital notepad) - Auto-submit on section timer expiry - -Submit entire test manually - Instant score report -P-03: Full-Length Mock Tests (4/month) -Description: Additional full-length mocks beyond the weekly series. Priority: P1 -Requirements: - 4 full-length mocks per month (in addition to weekly series) - Can be taken anytime - -Same interface as weekly tests - Different difficulty levels - Previous year paper simulations -P-04: Previous Year Paper Bank -Description: Complete archive of JEE Main, JEE Advanced, and NEET previous year papers. Priority: -P0 -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 7 of 27 -Requirements: - JEE Main: 2020-2026 papers - JEE Advanced: 2020-2025 papers - NEET UG: -2020-2025 papers - Attempt in test mode (timed) or practice mode (untimed) - Topic-wise filtering of -PYQs - Difficulty tagging per question - Solution links for each question - Track which PYQs have been -attempted -P-05: Detailed Performance Analytics -Description: Comprehensive analytics dashboard with heatmaps, trends, and insights. Priority: P0 -Requirements: - Topic Heatmap: Subject โ”œรน Topic grid showing accuracy (green/yellow/red) - -Accuracy Trend: Line chart over time (daily, weekly, monthly) - Speed Analysis: Average time per -question vs benchmark - Strength/Weakness Report: Auto-generated list of strong and weak topics - -Percentile Rank: Among all Pro users - Score Distribution: Histogram of test scores - Improvement -Rate: Week-over-week accuracy change - Subject-wise Breakdown: Physics %, Chemistry %, Maths/ -Biology % - Question Type Analysis: Performance by question type (MCQ, numerical, multiple -correct) - Export analytics as PDF -P-06: Chapter-wise Progress Tracking -Description: Track syllabus coverage and mastery per chapter. Priority: P1 -Requirements: - Complete syllabus tree for JEE/NEET - Per-chapter: questions attempted, accuracy, -time spent - Chapter status: Not Started, In Progress, Practiced, Mastered - Syllabus completion -percentage - Estimated time to complete remaining syllabus - Recommended next chapters based on -dependency graph -P-07: Parent Dashboard -Description: Simplified dashboard for parents to monitor their child's progress. Priority: P0 (Sales- -critical) -Requirements: - Separate login for parents (linked to student account) - Weekly progress summary -(auto-generated) - Daily time spent on platform - Accuracy trend (last 7/30 days) - Tests taken this -week/month - Rank improvement trend - "Areas of Concern" ฮ“ร‡รถ topics where child is weak - -Comparison with peers (anonymous) - WhatsApp/Email weekly report (automated) - Multiple children -support (add multiple students) - Simple, non-technical UI (Hindi + English) -P-08: Doubt Resolution (Community) -Description: Text-based doubt solving with community + moderator support. Priority: P1 -Requirements: - Post a doubt (text + screenshot/image) - Tag by subject and topic - Community -answers (upvote/downvote) - Moderator-verified answers highlighted - 24-hour SLA for moderator -response - Searchable doubt archive - "My Doubts" history - Notification when doubt is answered -P-09: Print-Friendly DPS Export -Description: Download DPS as PDF for offline solving. Priority: P1 -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 8 of 27 -Requirements: - PDF export of any DPS - Print-optimized layout - Answer key on separate page - -Watermark with student name (prevents sharing) - Batch download (multiple DPS) -6. Feature Catalog ฮ“ร‡รถ AI Coach Plan -A-01: AI Mistake Tracking & Analysis -Description: AI tracks every mistake across all DPS, tests, and micro-tests ฮ“ร‡รถ builds a permanent error -profile. Priority: P0 (Core AI feature) -Requirements: - Every wrong answer logged with: question, topic, subtopic, error type - Error type -classification: conceptual error, calculation error, reading error, time pressure error - Mistake frequency -per topic - Recurring mistake patterns ("You keep making sign errors in Thermodynamics") - Mistake -history (searchable, filterable) - "Your Most Common Mistakes" weekly report - Mistake count vs peers -benchmark -A-02: Spaced Repetition System -Description: Wrong answers are queued for re-attempt at optimal intervals using spaced repetition -algorithm. Priority: P0 -Requirements: - Every wrong question enters spaced repetition queue - Re-attempt schedule: Day 1, -Day 3, Day 7, Day 14, Day 30 - If answered correctly on re-attempt ฮ“รฅร† interval increases - If answered -wrong again ฮ“รฅร† resets to Day 1 - Daily "Mistake Review" session (5-10 questions from queue) - -Progress tracking per mistake (attempt 1/5, attempt 2/5, etc.) - Mastered mistakes removed from queue -after 3 consecutive correct attempts - SM-2 algorithm (SuperMemo) adapted for exam prep -A-03: Personalized Daily Study Plan -Description: AI generates a customized daily study plan based on weak areas, upcoming exams, and -past performance. Priority: P0 -Requirements: - Generated every morning at 5:00 AM IST - Plan includes: DPS to attempt, topics to -revise, mistake review session, micro-test recommendations - Time-budgeted: "Today: 45 min ฮ“ร‡รถ -Revise Thermodynamics (15 min) + DPS Electrostatics (20 min) + Mistake Review (10 min)" - Adaptive: -adjusts based on what was actually completed yesterday - Student can mark tasks as done/skipped - -Weekly study plan (Sunday generates plan for the week) - Plan difficulty calibrated to student's level - -"Cram Mode" activated near exams (more tests, less new learning) -A-04: AI Tutor Chat -Description: Text-based AI tutor that answers questions, explains concepts, and guides problem- -solving. Priority: P1 -Requirements: - Ask any doubt in natural language (text only) - AI responds with step-by-step -explanation - Can ask follow-up questions - Can request similar practice questions - Can ask "Why is -option C wrong?" for specific questions - Context-aware: AI knows student's weak areas and adapts -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 9 of 27 -explanations - Daily chat limit: 50 messages (to manage inference costs) - Chat history saved and -searchable - Hindi + English bilingual support - Moderation filter for inappropriate content -A-05: Exam Readiness Score -Description: A single 0-100 score predicting exam readiness based on multiple factors. Priority: P0 -Requirements: - Score calculated from: accuracy (40%), speed (20%), syllabus coverage (20%), -consistency (10%), mistake recovery (10%) - Daily update - Trend line (score over time) - Component -breakdown (which factor is pulling score down) - Peer comparison ("Your readiness score is higher than -65% of users") - Milestone celebrations (Score > 70: "You're on track for NIT!", Score > 85: "IIT within -reach!") - Actionable recommendations to improve score -A-06: Exam Day Strategy -Description: AI generates personalized exam-day strategy based on mock test performance. Priority: -P1 -Requirements: - Based on analysis of all mock tests - Recommendations include: - Optimal section -order ("Start with Physics ฮ“ร‡รถ your strongest section") - Time allocation per section - Question selection -strategy ("Attempt numerical questions first in Chemistry") - Guessing strategy (when to skip vs -educated guess) - Pace recommendations ("You tend to slow down in the last 15 minutes ฮ“ร‡รถ practice -timed sprints") - Updated after every mock test - Printable one-page strategy card -A-07: College Predictor -Description: Predicts likely colleges based on current performance trajectory. Priority: P1 -Requirements: - Based on mock test scores + historical JEE cutoff data - Predicts: IITs, NITs, IIITs, -GFTIs - "Reach" (ambitious), "Target" (realistic), "Safety" (likely) categories - Branch-wise prediction - -Updated after every mock test - Historical cutoff data for last 3 years - "You need X more marks to -reach IIT Delhi Computer Science" - Motivational: "At your current trajectory, you're on track for NIT -Surat" -A-08: Study Time Optimization -Description: AI analyzes when and how the student studies best and suggests schedule adjustments. -Priority: P2 -Requirements: - Track time-of-day performance (accuracy by hour) - Identify peak productivity hours - -Suggest optimal study schedule - Track session length vs focus (longer sessions = diminishing -returns?) - Recommend break intervals - "You perform best between 7-9 PM. Schedule your most -difficult topics then." -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 10 of 27 -7. Cross-Cutting Features -C-01: Responsive Web Design -Description: Full responsive design working on mobile browsers, tablets, and desktops. Priority: P0 -Requirements: - Mobile-first responsive design - Touch-optimized test interface (tap to select, swipe to -navigate) - Desktop keyboard shortcuts (N for next, P for previous, M for mark review) - Test interface -works in landscape and portrait - Minimum supported screen: 320px width - PWA manifest for "Add to -Home Screen" - Service worker for offline DPS access -C-02: Multi-Language Support -Description: Interface and content in Hindi, English, and bilingual mode. Priority: P1 -Requirements: - Full UI in English and Hindi - Bilingual mode (questions in English, UI in Hindi) - -Toggle between languages without page reload - All DPS questions available in both languages - AI -Coach responses in selected language - Parent dashboard in Hindi + English - Regional language -roadmap (Marathi, Gujarati, Bengali, Telugu ฮ“ร‡รถ Phase 2) -C-03: Dark Mode -Description: Dark theme for late-night study sessions. Priority: P2 -Requirements: - Toggle between light and dark mode - System preference detection - All pages -support both themes - Test interface in dark mode - Persist preference in user settings -C-04: Notifications & Reminders -Description: Browser and email notifications for DPS, tests, and milestones. Priority: P1 -Requirements: - Daily DPS reminder (browser notification, configurable time) - Test reminder (24 hours -before, 1 hour before) - Streak at risk notification (if no DPS by 8 PM) - Milestone celebrations (streak -achievements, score milestones) - Weekly progress summary (email, Sunday evening) - Parent weekly -report (email/WhatsApp) - Notification preferences (which notifications, when, how) -C-05: Referral System -Description: Built-in referral program for organic growth. Priority: P1 -Requirements: - Unique referral link per user - Refer a friend ฮ“รฅร† both get 7 days of AI Coach free - -Referral dashboard (total referrals, rewards earned) - Share via WhatsApp, Telegram, email, copy link - -Track referral signups and conversions - Leaderboard: top referrers -C-06: Shareable Progress Cards -Description: Auto-generated image cards showing progress that students can share on social media. -Priority: P1 -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 11 of 27 -Requirements: - "I've solved 450 questions this month!" card - "X-day streak!" card - "Accuracy: 78% -ฮ“ร‡รถ Top 15% of users" card - "Test Score: 180/300 ฮ“ร‡รถ Improved by 15 points" card - Branded with -PragatiVeer logo - Download as PNG - Share directly to WhatsApp, Instagram, Telegram - Generated -server-side (no client rendering) -C-07: Payment System -Description: Subscription management with UPI, cards, and netbanking. Priority: P0 -Requirements: - UPI (BHIM, Google Pay, PhonePe, Paytm) - Credit/Debit cards - Netbanking - -Razorpay/Stripe integration - Monthly, quarterly, and annual billing - Auto-renewal with 7-day grace -period - Invoice generation - Coupon code support - Free trial management (7-day Pro trial) - Upgrade/ -downgrade between tiers - Pause subscription (for exam breaks) - Refund policy (30-day money-back -guarantee) -C-08: Admin Dashboard -Description: Internal dashboard for Gyannetra team to manage content, users, and analytics. Priority: -P1 -Requirements: - User management (view, suspend, delete) - Content management (upload questions, -create DPS, schedule tests) - Analytics dashboard (DAU, MAU, conversion rates, revenue) - DPS -scheduling calendar - Test creation and scheduling - Question bank management (add, edit, tag, -review) - Coupon code management - Refund processing - Support ticket management - System health -monitoring -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 12 of 27 -8. User Stories ฮ“ร‡รถ Complete Collection -8.1 Registration & Onboarding -ID -User Story -Priority -Tier -US-001 -As a new student, I want to register with my email or phone number so that I can -create my account quickly. -P0 -All -US-002 -As a new student, I want to verify my email/phone with OTP so that my account is -secure. -P0 -All -US-003 -As a returning student, I want to log in with my email and password so that I can -access my dashboard. -P0 -All -US-004 -As a returning student, I want to log in with Google so that I don't have to -remember another password. -P0 -All -US-005 -As a student, I want to reset my password via email/OTP so that I can recover my -account if I forget it. -P0 -All -US-006 -As a new student, I want to set up my profile (class, target exam, medium) during -onboarding so that the platform personalizes my experience. -P0 -All -US-007 -As a student, I want to see a welcome tour on my first login so that I understand -how to use the platform. -P1 -All -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 13 of 27 -8.2 Daily Practice Sheet (DPS) -ID -User Story -Priority -Tier -US-010 -As a Free student, I want to attempt one DPS per day so that I can practice daily. -P0 -Free -US-011 -As a Pro student, I want to attempt unlimited DPS per day so that I can practice as -much as I need. -P0 -Pro+ -US-012 -As a student, I want to see questions one at a time with navigation so that I can -focus on each question. -P0 -All -US-013 -As a student, I want to see a timer during my DPS so that I can manage my time. -P0 -All -US-014 -As a student, I want to mark questions for review so that I can come back to them -later. -P0 -All -US-015 -As a student, I want to see a question palette showing which questions I've -answered, not answered, and marked for review. -P0 -All -US-016 -As a student, I want my DPS to auto-submit when the timer runs out so that I don't -lose my answers. -P0 -All -US-017 -As a student, I want to see my score and correct answers immediately after -submitting so that I can learn from my mistakes. -P0 -All -US-018 -As a Pro student, I want to access DPS from previous days so that I can catch up -if I miss a day. -P1 -Pro+ -US-019 -As a Pro student, I want to choose DPS difficulty level so that I can challenge -myself appropriately. -P1 -Pro+ -US-020 -As a Pro student, I want to create a custom DPS by selecting specific subjects and -topics so that I can focus on my weak areas. -P1 -Pro+ -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 14 of 27 -8.3 Test Series -ID -User Story -Priority -Tier -US-030 -As a Pro student, I want to take a full-length weekly test that follows the exact JEE -pattern so that I can practice exam conditions. -P0 -Pro+ -US-031 -As a Pro student, I want the test frequency to increase as exams get closer so that -I get more practice when it matters most. -P0 -Pro+ -US-032 -As a student, I want the test interface to look exactly like the real computer-based -exam so that I'm not surprised on exam day. -P0 -Pro+ -US-033 -As a student, I want sectional timing (as per actual exam) so that I can practice -time management per section. -P0 -Pro+ -US-034 -As a student, I want an on-screen calculator and periodic table (where applicable) -so that the test mirrors the real exam. -P0 -Pro+ -US-035 -As a student, I want a digital scratchpad/notepad during the test so that I can do -rough work. -P1 -Pro+ -US-036 -As a Pro student, I want to take additional full-length mocks (4/month) beyond the -weekly series so that I can practice more. -P1 -Pro+ -US-037 -As a Pro student, I want to attempt previous year papers in test mode so that I can -practice with real exam questions. -P0 -Pro+ -US-038 -As a Pro student, I want to filter PYQs by topic so that I can practice specific -areas. -P1 -Pro+ -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 15 of 27 -8.4 Analytics & Progress -ID -User Story -Priority -Tier -US-040 -As a Free student, I want to see my basic accuracy and streak on a simple -dashboard so that I can track my progress. -P0 -Free -US-041 -As a Pro student, I want to see a topic-wise heatmap of my performance so that I -know exactly where I'm weak. -P0 -Pro+ -US-042 -As a Pro student, I want to see my accuracy trend over time so that I can see if I'm -improving. -P0 -Pro+ -US-043 -As a Pro student, I want to see my percentile rank among other users so that I -know where I stand. -P0 -Pro+ -US-044 -As a Pro student, I want to see my speed analysis (time per question vs -benchmark) so that I can improve my pace. -P1 -Pro+ -US-045 -As a Pro student, I want to see my performance broken down by question type so -that I know which types I struggle with. -P1 -Pro+ -US-046 -As a Pro student, I want to export my analytics as a PDF so that I can share it with -my parents or mentor. -P1 -Pro+ -US-047 -As a Pro student, I want to see my syllabus completion percentage per subject so -that I know how much I've covered. -P1 -Pro+ -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 16 of 27 -8.5 AI Coach -ID -User Story -Priority -Tier -US-050 -As an AI Coach student, I want the AI to track every mistake I make across all -DPS and tests so that I never forget my weak areas. -P0 -AI -Coach -US-051 -As an AI Coach student, I want the AI to classify my mistakes (conceptual, -calculation, etc.) so that I understand WHY I got it wrong. -P0 -AI -Coach -US-052 -As an AI Coach student, I want wrong questions to reappear at optimal -intervals (spaced repetition) so that I permanently learn from my mistakes. -P0 -AI -Coach -US-053 -As an AI Coach student, I want a daily "Mistake Review" session with 5-10 -questions from my error queue so that I consistently revise my weak areas. -P0 -AI -Coach -US-054 -As an AI Coach student, I want a personalized daily study plan generated every -morning so that I know exactly what to study today. -P0 -AI -Coach -US-055 -As an AI Coach student, I want the study plan to adapt based on what I actually -completed yesterday so that it stays relevant. -P0 -AI -Coach -US-056 -As an AI Coach student, I want to see my Exam Readiness Score (0-100) so -that I know how prepared I am. -P0 -AI -Coach -US-057 -As an AI Coach student, I want the readiness score to show component -breakdown so that I know which factor is pulling me down. -P0 -AI -Coach -US-058 -As an AI Coach student, I want to chat with the AI tutor to ask doubts and get -explanations so that I can learn at my own pace. -P1 -AI -Coach -US-059 -As an AI Coach student, I want the AI tutor to be context-aware (knows my -weak areas) so that explanations are personalized. -P1 -AI -Coach -US-060 -As an AI Coach student, I want a personalized exam day strategy based on my -mock performance so that I can maximize my score. -P1 -AI -Coach -US-061 -As an AI Coach student, I want a college predictor that shows which IITs/NITs I -can target so that I stay motivated. -P1 -AI -Coach -US-062 -As an AI Coach student, I want the AI to analyze my study patterns and -suggest optimal study times so that I study when I'm most effective. -P2 -AI -Coach -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 17 of 27 -8.6 Parent Dashboard -ID -User Story -Priority -Tier -US-070 -As a parent, I want to create my own login linked to my child's account so that I -can monitor progress without accessing their account. -P0 -Pro+ -US-071 -As a parent, I want to see a simple weekly progress summary so that I know if my -child is studying regularly. -P0 -Pro+ -US-072 -As a parent, I want to see my child's accuracy trend so that I can see if they're -improving. -P0 -Pro+ -US-073 -As a parent, I want to see "Areas of Concern" ฮ“ร‡รถ topics where my child is weak so -that I can arrange extra help. -P0 -Pro+ -US-074 -As a parent, I want to receive a weekly WhatsApp/Email report automatically so -that I don't have to log in to check. -P0 -Pro+ -US-075 -As a parent, I want the dashboard to be in Hindi so that I can understand it easily. -P1 -Pro+ -US-076 -As a parent with two children preparing for exams, I want to track both from a -single parent account. -P1 -Pro+ -8.7 Gamification & Engagement -ID -User Story -Priority -Tier -US-080 -As a student, I want to see my daily streak so that I feel motivated to practice -every day. -P0 -All -US-081 -As a student, I want to earn badges for milestones (7-day streak, 100 questions, -etc.) so that I feel a sense of achievement. -P1 -All -US-082 -As a student, I want to see my rank on the city/school leaderboard so that I can -compete with peers. -P1 -All -US-083 -As a student, I want to share my progress card on social media so that my friends -can see my hard work. -P1 -Pro+ -US-084 -As a student, I want to challenge a friend to beat my test score so that we can -motivate each other. -P2 -Pro+ -US-085 -As a student, I want to receive a notification when I'm about to lose my streak so -that I don't break it. -P1 -All -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 18 of 27 -8.8 Payments & Subscription -ID -User Story -Priority -Tier -US-090 -As a Free student, I want to upgrade to Pro with UPI so that I can start my -subscription easily. -P0 -Pro+ -US-091 -As a Pro student, I want to choose between monthly, quarterly, and annual billing -so that I can pick what suits my budget. -P0 -Pro+ -US-092 -As a student, I want a 7-day free trial of Pro so that I can try before I buy. -P0 -Pro+ -US-093 -As a student, I want to apply a coupon code during checkout so that I can get a -discount. -P1 -Pro+ -US-094 -As a student, I want to receive an invoice for my payment so that I can show my -parents. -P1 -Pro+ -US-095 -As a student, I want to pause my subscription during exam breaks so that I don't -waste paid days. -P2 -Pro+ -US-096 -As a parent, I want a 30-day money-back guarantee so that I can purchase without -risk. -P1 -Pro+ -8.9 Referral & Growth -ID -User Story -Priority -Tier -US-100 -As a student, I want to refer a friend using my unique link so that both of us get free -AI Coach access. -P1 -All -US-101 -As a student, I want to see how many friends have signed up through my referral -so that I can track my rewards. -P1 -All -US-102 -As a student, I want to share my referral link on WhatsApp and Telegram so that I -can easily invite friends. -P1 -All -8.10 School/B2B -ID -User Story -Priority -Tier -US-110 -As a school principal, I want to purchase bulk licenses for my students so that they -all get Pro access at a discount. -P1 -B2B -US-111 -As a school principal, I want a school dashboard showing aggregate performance -of all students so that I can identify weak batches. -P1 -B2B -US-112 -As a teacher, I want to assign DPS or tests to my entire class so that they practice -together. -P2 -B2B -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 19 of 27 -9. User Flows -9.1 New Student Onboarding Flow -Landing Page ฮ“รฅร† Sign Up (Email/Phone) ฮ“รฅร† OTP Verification ฮ“รฅร† -Profile Setup (Class, Exam, Medium) ฮ“รฅร† Welcome Tour ฮ“รฅร† -First DPS ฮ“รฅร† See Results ฮ“รฅร† Streak Started ฮ“รฅร† -Upgrade Prompt (after 3 DPS completions) -9.2 Daily Practice Flow -Login ฮ“รฅร† Dashboard ฮ“รฅร† "Today's DPS" CTA ฮ“รฅร† -Attempt DPS (10-15 Qs, timed) ฮ“รฅร† Submit ฮ“รฅร† -Instant Results (Score + Answer Key) ฮ“รฅร† -Review Mistakes ฮ“รฅร† Check Leaderboard ฮ“รฅร† -Logout or Continue (Micro-test / Analytics) -9.3 Weekly Test Flow -Login ฮ“รฅร† Dashboard ฮ“รฅร† "Weekly Test Available" Notification ฮ“รฅร† -Enter Test Lobby (instructions, pattern info) ฮ“รฅร† -Start Test ฮ“รฅร† Attempt Sections (timed per section) ฮ“รฅร† -Submit ฮ“รฅร† Score Report ฮ“รฅร† Detailed Analytics ฮ“รฅร† -Compare with Peers ฮ“รฅร† AI Coach Analysis (if AI Coach tier) -9.4 AI Coach Daily Flow -Login ฮ“รฅร† AI Coach Dashboard ฮ“รฅร† -1. View Today's Personalized Plan -2. Complete Mistake Review Session (5-10 Qs) -3. Attempt Recommended DPS/Micro-test -4. Check Exam Readiness Score Update -5. Chat with AI Tutor (if doubts) -6. Mark Plan Tasks as Complete -ฮ“รฅร† Plan Adapts for Tomorrow -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 20 of 27 -9.5 Parent Flow -Parent Login (linked to student) ฮ“รฅร† -Dashboard: Weekly Summary, Accuracy Trend, -Areas of Concern, Time Spent ฮ“รฅร† -Drill Down into Specific Topics ฮ“รฅร† -Share Report with Family ฮ“รฅร† -Receive Weekly WhatsApp Report (automated) -9.6 Upgrade Flow (Free ฮ“รฅร† Pro) -Free User ฮ“รฅร† CTA: "Unlock Unlimited Practice" ฮ“รฅร† -Pricing Page (Free vs Pro vs AI Coach comparison) ฮ“รฅร† -Select Pro Plan ฮ“รฅร† Choose Billing (Monthly/Quarterly/Annual) ฮ“รฅร† -Payment (UPI/Card/Netbanking) ฮ“รฅร† -7-Day Free Trial Starts ฮ“รฅร† -After Trial: Auto-charge or Downgrade -10. Technical Requirements -10.1 Platform -Type: Web application only (no native mobile app) -Responsive: Mobile-first responsive design (320px minimum) -PWA: Progressive Web App with service worker for offline DPS access -Browser Support: Chrome, Firefox, Safari, Edge (latest 2 versions) -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 21 of 27 -10.2 Tech Stack (Recommended) -Layer -Technology -Frontend -Next.js (React) ฮ“ร‡รถ SSR for performance -Backend -FastAPI (Python) or Node.js -Database -PostgreSQL -Cache -Redis (session management, rate limiting) -File Storage -S3-compatible (MinIO or AWS S3) -AI/ML -LLM API (for AI Coach), custom spaced repetition engine -Payments -Razorpay -Email -SMTP (Hostinger) -SMS/WhatsApp -Twilio / WhatsApp Business API -Deployment -Docker on VPS (Gyannetra production server) -10.3 Performance Requirements -Page load time: < 2 seconds (desktop), < 3 seconds (mobile) -Test interface: < 1 second question transition -DPS submission results: < 2 seconds -AI Coach response: < 5 seconds -Concurrent users: Support 10,000 DAU in Phase 1 -Uptime: 99.5% (excluding scheduled maintenance) -10.4 Security Requirements -HTTPS everywhere -OTP-based authentication (rate limited: 3 attempts/min) -Password hashing (bcrypt) -Session management (JWT with 30-day expiry) -Rate limiting on all API endpoints -XSS and CSRF protection -SQL injection prevention (parameterized queries) -Data encryption at rest (database encryption) -GDPR/DPDP Act compliance for student data -Parental consent for users under 18 -10.5 Offline Capability (PWA) -Service worker caches DPS questions -Students can download DPS when online -Attempt DPS offline -Sync results when back online -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -ฮ“ร‡รณ -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 22 of 27 -Cache previous day's DPS for review -11. Data Model (Key Entities) -11.1 User -Field -Type -Notes -id -UUID -Primary key -email -String -Unique, nullable -phone -String -Unique, nullable -password_hash -String -bcrypt hashed -name -String -class -Enum -11, 12, Dropper -target_exam -Enum -JEE_Main, JEE_Advanced, NEET -medium -Enum -English, Hindi, Bilingual -school -String -Optional -city -String -state -String -tier -Enum -Free, Pro, AI_Coach -trial_end -DateTime -7-day trial expiry -subscription_end -DateTime -streak_count -Integer -Current streak -longest_streak -Integer -created_at -DateTime -last_active -DateTime -ฮ“ร‡รณ -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 23 of 27 -11.2 Parent -Field -Type -Notes -id -UUID -Primary key -email -String -phone -String -password_hash -String -name -String -language -Enum -Hindi, English -linked_students -UUID[] -Array of student user IDs -created_at -DateTime -11.3 Question -Field -Type -Notes -id -UUID -Primary key -subject -Enum -Physics, Chemistry, Maths, Biology -topic -String -e.g., "Electrostatics" -subtopic -String -e.g., "Coulomb's Law" -difficulty -Enum -Easy, Medium, Hard -type -Enum -MCQ_Single, MCQ_Multiple, Numerical -question_text -Text -Supports LaTeX -options -JSON -Array of options (for MCQ) -correct_answer -JSON -Correct answer(s) -solution_text -Text -Step-by-step solution -language -Enum -English, Hindi -year -Integer -For PYQs (nullable) -exam -Enum -JEE_Main, JEE_Advanced, NEET -tags -String[] -Custom tags -is_active -Boolean -Soft delete -created_by -UUID -Admin user ID -verified_by -UUID -Content reviewer ID -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 24 of 27 -11.4 DPS (Daily Practice Sheet) -Field -Type -Notes -id -UUID -Primary key -date -Date -DPS date -title -String -e.g., "DPS #127" -questions -UUID[] -Array of question IDs -difficulty -Enum -Easy, Medium, Hard, Mixed -time_limit -Integer -Minutes -is_active -Boolean -created_by -UUID -Admin user ID -11.5 Test -Field -Type -Notes -id -UUID -Primary key -title -String -type -Enum -Weekly, Mock, PYQ -exam_pattern -Enum -JEE_Main, JEE_Advanced, NEET -sections -JSON -Array of {name, questions[], time_limit} -total_time -Integer -Minutes -difficulty -Enum -scheduled_date -DateTime -For weekly tests -is_active -Boolean -created_by -UUID -Admin user ID -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 25 of 27 -11.6 Attempt (DPS or Test) -Field -Type -Notes -id -UUID -Primary key -user_id -UUID -Foreign key -source_type -Enum -DPS, Test, MicroTest -source_id -UUID -DPS or Test ID -started_at -DateTime -submitted_at -DateTime -time_taken -Integer -Seconds -answers -JSON -Array of {question_id, selected_answer, is_correct, time_spent} -score -Integer -Correct answers count -total_questions -Integer -accuracy -Float -Percentage -status -Enum -In_Progress, Submitted, Timed_Out -11.7 Mistake (AI Coach) -Field -Type -Notes -id -UUID -Primary key -user_id -UUID -Foreign key -question_id -UUID -Foreign key -attempt_id -UUID -Foreign key -error_type -Enum -Conceptual, Calculation, Reading, Time_Pressure -topic -String -subtopic -String -next_review -DateTime -Spaced repetition schedule -review_count -Integer -Times re-attempted -consecutive_correct -Integer -For mastery detection -is_mastered -Boolean -created_at -DateTime -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 26 of 27 -11.8 StudyPlan (AI Coach) -Field -Type -Notes -id -UUID -Primary key -user_id -UUID -Foreign key -date -Date -Plan date -tasks -JSON -Array of {type, description, duration_minutes, status} -is_completed -Boolean -generated_at -DateTime -feedback -Text -Student's feedback on plan -11.9 Subscription -Field -Type -Notes -id -UUID -Primary key -user_id -UUID -Foreign key -tier -Enum -Pro, AI_Coach -billing_cycle -Enum -Monthly, Quarterly, Annual -start_date -DateTime -end_date -DateTime -amount -Integer -Paise -payment_id -String -Razorpay payment ID -status -Enum -Active, Expired, Cancelled, Paused -auto_renew -Boolean -coupon_code -String -Nullable -End of PRD ฮ“ร‡รถ Prepared by Maya, CEO's Office, Gyannetra Pvt Ltd -Approved by Kapil Panwar, CEO ฮ“ร‡รถ 1 July 2026 -PragatiVeer ฮ“ร‡รถ Product Requirements Document -Page 27 of 27 - diff --git a/real_jee_questions_rendering_test.json b/real_jee_questions_rendering_test.json deleted file mode 100644 index fedbf1e..0000000 --- a/real_jee_questions_rendering_test.json +++ /dev/null @@ -1,1872 +0,0 @@ -[ - { - "section": "General", - "subject": "Physics", - "topic": "Rotational Mechanics Set 1", - "subtopic": "Angular Momentum & Collisions", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "A thin uniform rod of mass $M$ and length $L$ is pivoted at its upper end and hangs vertically. A small particle of mass $m$ ($m \\ll M$) moving horizontally with velocity $v_0$ strikes the rod at a distance $x$ from the pivot and sticks to it. If the rod deflects by an angle $\\theta$ such that $\\cos\\theta = 1 - \\frac{6mv_0^2}{MgL}$, the collision point $x$ is:", - "options": { - "A": "$\\frac{L}{\\sqrt{3}}$", - "B": "$\\frac{L}{2}$", - "C": "$\\frac{2L}{3}$", - "D": "$\\frac{L}{\\sqrt{2}}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "By conservation of angular momentum about the pivot:\n\n$$m v_0 x = I \\omega = \\left(\\frac{1}{3}ML^2 + m x^2\\right) \\omega \\approx \\frac{1}{3}ML^2 \\omega \\implies \\omega = \\frac{3mv_0x}{ML^2}$$\n\nFrom conservation of energy after collision:\n\n$$\\frac{1}{2} I \\omega^2 = M g h_{\\text{cm}} \\implies \\frac{1}{2} \\left(\\frac{1}{3}ML^2\\right) \\left(\\frac{3mv_0x}{ML^2}\\right)^2 = M g \\left(\\frac{L}{2}\\right) (1 - \\cos\\theta)$$\n\n$$\\frac{3}{2}\\frac{m^2 v_0^2 x^2}{ML^2} = M g \\frac{L}{2} \\left(\\frac{6mv_0^2}{MgL}\\right) = 3mv_0^2 \\implies x = \\frac{L}{2}$$", - "language": "English", - "id": "1811a71f-fb8d-4f31-a541-c4fcffffb523" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Electromagnetism Set 1", - "subtopic": "Electromagnetic Induction", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "A rectangular loop of dimensions $a \\times b$ with resistance $R$ is placed near an infinitely long straight wire carrying a time-dependent current $I(t) = I_0 \\sin(\\omega t)$. The loop is at a distance $r_0$ from the wire. Which of the following statements is/are correct?", - "options": { - "A": "The maximum mutual inductance $M$ between the wire and the loop is $\\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "B": "The amplitude of the induced current in the loop is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "C": "The net force on the loop is zero at all times.", - "D": "The induced EMF in the loop is in phase with the current in the straight wire." - }, - "correct_answer": { - "value": [ - "A", - "B" - ] - }, - "solution_text": "Magnetic flux $\\Phi(t)$ through the loop is:\n\n$$\\Phi(t) = \\int_{r_0}^{r_0+a} \\frac{\\mu_0 I(t)}{2\\pi r} b \\, dr = \\frac{\\mu_0 b I(t)}{2\\pi} \\ln\\left(\\frac{r_0+a}{r_0}\\right)$$\n\nMutual inductance $M = \\frac{\\Phi(t)}{I(t)} = \\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right).\n\nInduced EMF $\\mathcal{E}(t) = -\\frac{d\\Phi}{dt} = -M I_0 \\omega \\cos(\\omega t)$.\n\nInduced current $i(t) = \\frac{\\mathcal{E}(t)}{R} = -\\frac{M I_0 \\omega}{R} \\cos(\\omega t)$.\n\nTherefore, the amplitude of the induced current is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$. Since EMF varies as $\\cos(\\omega t)$ and current in the wire is $\\sin(\\omega t)$, they are out of phase by $\\pi/2$. Net force is non-zero because the magnetic field varies across the loop width.", - "language": "English", - "id": "7aac5446-8c86-4582-befc-32de35913317" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Modern Physics Set 1", - "subtopic": "Photoelectric Effect", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "In a photoelectric effect experiment, light of wavelength $\\lambda = 300\\text{ nm}$ is incident on a metal plate. The stopping potential is found to be $V_1 = 1.85\\text{ V}$. When light of wavelength $\\lambda' = 200\\text{ nm}$ is incident on the same metal plate, the stopping potential becomes $V_2$ (in Volts). Calculate the value of $V_2$.\n\n(Take $hc = 1240\\text{ eV}\\cdot\\text{nm}$ and electron charge $e = 1.6 \\times 10^{-19}\\text{ C}$)", - "options": null, - "correct_answer": { - "value": "3.92" - }, - "solution_text": "From Einstein's Photoelectric Equation:\n\n$$eV_1 = \\frac{hc}{\\lambda} - \\Phi_0 \\implies \\Phi_0 = \\frac{1240}{300} - 1.85 = 4.13 - 1.85 = 2.28\\text{ eV}$$\n\nFor the second wavelength:\n\n$$eV_2 = \\frac{hc}{\\lambda'} - \\Phi_0 \\implies V_2 = \\frac{1240}{200} - 2.28 = 6.20 - 2.28 = 3.92\\text{ V}$$", - "language": "English", - "id": "a174daee-81b7-4df4-9d1f-76e37f16fd27" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Coordination Chemistry Set 1", - "subtopic": "Crystal Field Theory", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "For the octahedral complex ion $[Fe(H_2O)_6]^{3+}$, the crystal field splitting energy is denoted by $\\Delta_0$ and the pairing energy is $P$. Which of the following options correctly represents the magnetic properties and d-electron configuration of the metal ion?", - "options": { - "A": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "B": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "C": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{15}\\text{ B.M.}$", - "D": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{3}\\text{ B.M.}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "$Fe^{3+}$ has a $d^5$ electron configuration. Since $H_2O$ is a weak field ligand, $\\Delta_0 < P$, resulting in a high-spin complex.\n\nConfiguration: $t_{2g}^3 e_g^2$.\n\nNumber of unpaired electrons $n = 5$.\n\nSpin-only magnetic moment $\\mu_{\\text{eff}} = \\sqrt{n(n+2)} = \\sqrt{5(7)} = \\sqrt{35}\\text{ B.M.}$", - "language": "English", - "id": "aa5735b2-91ef-4862-8e42-eb0492d61ae7" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 1", - "subtopic": "Chemical Kinetics", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Consider the reaction network showing first-order parallel decomposition of a reactant $A$:\n\n$$A \\xrightarrow{k_1} B \\quad \\text{and} \\quad A \\xrightarrow{k_2} C$$\n\nAt a temperature $T_1$, the rate constants are $k_1 = 2.0 \\times 10^{-4}\\text{ s}^{-1}$ and $k_2 = 1.0 \\times 10^{-4}\\text{ s}^{-1}$. Let the activation energies for the two paths be $E_{a1} = 40\\text{ kJ/mol}$ and $E_{a2} = 80\\text{ kJ/mol}$ respectively. Which of the following statements is/are correct?", - "options": { - "A": "The fractional yield of $B$ at temperature $T_1$ is $\\frac{2}{3}$.", - "B": "If the temperature is increased to $T_2 > T_1$, the ratio $[B]/[C]$ decreases.", - "C": "The overall rate constant for the disappearance of $A$ is $k_{\\text{eff}} = 3.0 \\times 10^{-4}\\text{ s}^{-1}$ at $T_1$.", - "D": "The effective activation energy for the disappearance of $A$ is $E_{\\text{eff}} \\approx 53.3\\text{ kJ/mol}$ at $T_1$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "Let us check each option:\n\n1. Yield of $B = \\frac{k_1}{k_1+k_2} = \\frac{2.0}{2.0+1.0} = \\frac{2}{3}$. (A is correct)\n\n2. Since $E_{a2} > E_{a1}$, the rate constant $k_2$ increases more rapidly with temperature than $k_1$. Hence, the ratio $\\frac{[B]}{[C]} = \\frac{k_1}{k_2}$ decreases as temperature increases. (B is correct)\n\n3. For parallel first-order reactions, $k_{\\text{eff}} = k_1 + k_2 = 3.0 \\times 10^{-4}\\text{ s}^{-1}$. (C is correct)\n\n4. The effective activation energy is $E_{\\text{eff}} = \\frac{k_1 E_{a1} + k_2 E_{a2}}{k_1 + k_2} = \\frac{2(40) + 1(80)}{3} = 53.33\\text{ kJ/mol}$. (D is correct)", - "language": "English", - "id": "ab8fcfea-0219-4e42-b4e1-c104288339f7" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 1", - "subtopic": "Thermodynamics", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "Calculate the change in entropy $\\Delta S$ (in $\\text{J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$) when $2\\text{ moles}$ of an ideal gas ($C_{V,m} = \\frac{3}{2}R$) are heated reversibly from $300\\text{ K}$ to $600\\text{ K}$ at constant pressure.\n\n(Take $R = 8.314\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$ and $\\ln(2) \\approx 0.693$)", - "options": null, - "correct_answer": { - "value": "28.81" - }, - "solution_text": "For a process at constant pressure:\n\n$$C_{P,m} = C_{V,m} + R = \\frac{5}{2}R$$\n\nThe entropy change is:\n\n$$\\Delta S = n C_{P,m} \\ln\\left(\\frac{T_2}{T_1}\\right) = 2 \\times \\left(\\frac{5}{2} R\\right) \\ln\\left(\\frac{600}{300}\\right) = 5 R \\ln(2)$$\n\n$$\\Delta S = 5 \\times 8.314 \\times 0.693 \\approx 28.81\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$$", - "language": "English", - "id": "4893b467-8896-42ed-bb93-3e70f7ce5495" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Calculus Set 1", - "subtopic": "Definite Integration", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "Let $I = \\int_{0}^{\\pi} \\frac{x \\sin^2 x \\cos x}{1 + \\cos^2 x} \\, dx$. The value of $I$ is:", - "options": { - "A": "$\\frac{\\pi^2}{4}$", - "B": "$0$", - "C": "$-\\frac{\\pi}{2}$", - "D": "$-\\frac{\\pi^2}{4}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "Using the integration property $\\int_a^b f(x) dx = \\int_a^b f(a+b-x) dx$:\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2(\\pi - x) \\cos(\\pi - x)}{1 + \\cos^2(\\pi - x)} \\, dx$$\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2 x (-\\cos x)}{1 + \\cos^2 x} \\, dx = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx + I$$\n\n$$2I = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx$$\n\nLet $g(x) = \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x}$. Since $g(\\pi - x) = -g(x)$, the integral $\\int_0^\\pi g(x) dx = 0$.\n\nTherefore, $2I = 0 \\implies I = 0$.", - "language": "English", - "id": "fdba8e87-3acb-4bc6-a92a-63d0bbfb1040" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Algebra Set 1", - "subtopic": "Matrices & Determinants", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Let $A = \\begin{pmatrix} 1 & 0 & 0 \\\\ 2 & 1 & 0 \\\\ 3 & 2 & 1 \\end{pmatrix}$ and let $I$ be the $3 \\times 3$ identity matrix. If $B = A - I$, which of the following statements is/are correct?", - "options": { - "A": "$B^3$ is a null matrix.", - "B": "The determinant of $A^n$ is equal to $1$ for any integer $n \\ge 1$.", - "C": "The inverse of $A$ exists and is equal to $I - B + B^2$.", - "D": "$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "We have:\n\n$$B = A - I = \\begin{pmatrix} 0 & 0 & 0 \\\\ 2 & 0 & 0 \\\\ 3 & 2 & 0 \\end{pmatrix}$$\n\nCompute $B^2$:\n\n$$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$$ (Option D is correct)\n\nCompute $B^3$:\n\n$$B^3 = B^2 B = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ (Option A is correct)\n\nSince $B^3 = 0$, we can express $(I + B)^{-1}$ using a geometric series:\n\n$$(I + B)^{-1} = I - B + B^2 - B^3 + \\dots = I - B + B^2$$\n\nThus, $A^{-1} = I - B + B^2$. (Option C is correct)\n\nAlso, $\\det(A) = 1 \\implies \\det(A^n) = (\\det A)^n = 1$. (Option B is correct)", - "language": "English", - "id": "f1818a76-da23-4a27-92c1-0b81d5cbe861" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Coordinate Geometry Set 1", - "subtopic": "Parabolas & Circles", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "The line $y = mx + 1$ is a common tangent to the circle $x^2 + y^2 = \\frac{1}{2}$ and the parabola $y^2 = 4x$. Find the value of $m^2$.", - "options": null, - "correct_answer": { - "value": "1" - }, - "solution_text": "For the parabola $y^2 = 4ax$ with $a=1$, the condition of tangency for $y = mx + c$ is:\n\n$$c = \\frac{a}{m} \\implies 1 = \\frac{1}{m} \\implies m = 1$$\n\nLet's check if the line $y = x + 1$ (i.e., $x - y + 1 = 0$) is also tangent to the circle $x^2 + y^2 = \\frac{1}{2}$:\n\nThe perpendicular distance from the center $(0,0)$ to the line is:\n\n$$d = \\frac{|1|}{\\sqrt{1^2 + (-1)^2}} = \\frac{1}{\\sqrt{2}}$$\n\nThis distance is exactly equal to the radius $r = \\frac{1}{\\sqrt{2}}$ of the circle. Thus, the line is a common tangent. Therefore, $m^2 = 1^2 = 1$.", - "language": "English", - "id": "fdcae399-cc73-4b26-93fa-ad847189d661" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Rotational Mechanics Set 2", - "subtopic": "Angular Momentum & Collisions", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "A thin uniform rod of mass $M$ and length $L$ is pivoted at its upper end and hangs vertically. A small particle of mass $m$ ($m \\ll M$) moving horizontally with velocity $v_0$ strikes the rod at a distance $x$ from the pivot and sticks to it. If the rod deflects by an angle $\\theta$ such that $\\cos\\theta = 1 - \\frac{6mv_0^2}{MgL}$, the collision point $x$ is:", - "options": { - "A": "$\\frac{L}{\\sqrt{3}}$", - "B": "$\\frac{L}{2}$", - "C": "$\\frac{2L}{3}$", - "D": "$\\frac{L}{\\sqrt{2}}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "By conservation of angular momentum about the pivot:\n\n$$m v_0 x = I \\omega = \\left(\\frac{1}{3}ML^2 + m x^2\\right) \\omega \\approx \\frac{1}{3}ML^2 \\omega \\implies \\omega = \\frac{3mv_0x}{ML^2}$$\n\nFrom conservation of energy after collision:\n\n$$\\frac{1}{2} I \\omega^2 = M g h_{\\text{cm}} \\implies \\frac{1}{2} \\left(\\frac{1}{3}ML^2\\right) \\left(\\frac{3mv_0x}{ML^2}\\right)^2 = M g \\left(\\frac{L}{2}\\right) (1 - \\cos\\theta)$$\n\n$$\\frac{3}{2}\\frac{m^2 v_0^2 x^2}{ML^2} = M g \\frac{L}{2} \\left(\\frac{6mv_0^2}{MgL}\\right) = 3mv_0^2 \\implies x = \\frac{L}{2}$$", - "language": "English", - "id": "61af02a0-550b-43cd-bdf4-52cdfbe6982a" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Electromagnetism Set 2", - "subtopic": "Electromagnetic Induction", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "A rectangular loop of dimensions $a \\times b$ with resistance $R$ is placed near an infinitely long straight wire carrying a time-dependent current $I(t) = I_0 \\sin(\\omega t)$. The loop is at a distance $r_0$ from the wire. Which of the following statements is/are correct?", - "options": { - "A": "The maximum mutual inductance $M$ between the wire and the loop is $\\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "B": "The amplitude of the induced current in the loop is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "C": "The net force on the loop is zero at all times.", - "D": "The induced EMF in the loop is in phase with the current in the straight wire." - }, - "correct_answer": { - "value": [ - "A", - "B" - ] - }, - "solution_text": "Magnetic flux $\\Phi(t)$ through the loop is:\n\n$$\\Phi(t) = \\int_{r_0}^{r_0+a} \\frac{\\mu_0 I(t)}{2\\pi r} b \\, dr = \\frac{\\mu_0 b I(t)}{2\\pi} \\ln\\left(\\frac{r_0+a}{r_0}\\right)$$\n\nMutual inductance $M = \\frac{\\Phi(t)}{I(t)} = \\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right).\n\nInduced EMF $\\mathcal{E}(t) = -\\frac{d\\Phi}{dt} = -M I_0 \\omega \\cos(\\omega t)$.\n\nInduced current $i(t) = \\frac{\\mathcal{E}(t)}{R} = -\\frac{M I_0 \\omega}{R} \\cos(\\omega t)$.\n\nTherefore, the amplitude of the induced current is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$. Since EMF varies as $\\cos(\\omega t)$ and current in the wire is $\\sin(\\omega t)$, they are out of phase by $\\pi/2$. Net force is non-zero because the magnetic field varies across the loop width.", - "language": "English", - "id": "6be4bfab-c1fa-424f-91de-9a3766196f79" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Modern Physics Set 2", - "subtopic": "Photoelectric Effect", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "In a photoelectric effect experiment, light of wavelength $\\lambda = 300\\text{ nm}$ is incident on a metal plate. The stopping potential is found to be $V_1 = 1.85\\text{ V}$. When light of wavelength $\\lambda' = 200\\text{ nm}$ is incident on the same metal plate, the stopping potential becomes $V_2$ (in Volts). Calculate the value of $V_2$.\n\n(Take $hc = 1240\\text{ eV}\\cdot\\text{nm}$ and electron charge $e = 1.6 \\times 10^{-19}\\text{ C}$)", - "options": null, - "correct_answer": { - "value": "3.92" - }, - "solution_text": "From Einstein's Photoelectric Equation:\n\n$$eV_1 = \\frac{hc}{\\lambda} - \\Phi_0 \\implies \\Phi_0 = \\frac{1240}{300} - 1.85 = 4.13 - 1.85 = 2.28\\text{ eV}$$\n\nFor the second wavelength:\n\n$$eV_2 = \\frac{hc}{\\lambda'} - \\Phi_0 \\implies V_2 = \\frac{1240}{200} - 2.28 = 6.20 - 2.28 = 3.92\\text{ V}$$", - "language": "English", - "id": "fd19e1ce-4090-4d8b-a13f-8c5970f54eb9" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Coordination Chemistry Set 2", - "subtopic": "Crystal Field Theory", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "For the octahedral complex ion $[Fe(H_2O)_6]^{3+}$, the crystal field splitting energy is denoted by $\\Delta_0$ and the pairing energy is $P$. Which of the following options correctly represents the magnetic properties and d-electron configuration of the metal ion?", - "options": { - "A": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "B": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "C": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{15}\\text{ B.M.}$", - "D": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{3}\\text{ B.M.}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "$Fe^{3+}$ has a $d^5$ electron configuration. Since $H_2O$ is a weak field ligand, $\\Delta_0 < P$, resulting in a high-spin complex.\n\nConfiguration: $t_{2g}^3 e_g^2$.\n\nNumber of unpaired electrons $n = 5$.\n\nSpin-only magnetic moment $\\mu_{\\text{eff}} = \\sqrt{n(n+2)} = \\sqrt{5(7)} = \\sqrt{35}\\text{ B.M.}$", - "language": "English", - "id": "9ccab261-34d9-4150-800f-b7b92a3cfd1c" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 2", - "subtopic": "Chemical Kinetics", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Consider the reaction network showing first-order parallel decomposition of a reactant $A$:\n\n$$A \\xrightarrow{k_1} B \\quad \\text{and} \\quad A \\xrightarrow{k_2} C$$\n\nAt a temperature $T_1$, the rate constants are $k_1 = 2.0 \\times 10^{-4}\\text{ s}^{-1}$ and $k_2 = 1.0 \\times 10^{-4}\\text{ s}^{-1}$. Let the activation energies for the two paths be $E_{a1} = 40\\text{ kJ/mol}$ and $E_{a2} = 80\\text{ kJ/mol}$ respectively. Which of the following statements is/are correct?", - "options": { - "A": "The fractional yield of $B$ at temperature $T_1$ is $\\frac{2}{3}$.", - "B": "If the temperature is increased to $T_2 > T_1$, the ratio $[B]/[C]$ decreases.", - "C": "The overall rate constant for the disappearance of $A$ is $k_{\\text{eff}} = 3.0 \\times 10^{-4}\\text{ s}^{-1}$ at $T_1$.", - "D": "The effective activation energy for the disappearance of $A$ is $E_{\\text{eff}} \\approx 53.3\\text{ kJ/mol}$ at $T_1$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "Let us check each option:\n\n1. Yield of $B = \\frac{k_1}{k_1+k_2} = \\frac{2.0}{2.0+1.0} = \\frac{2}{3}$. (A is correct)\n\n2. Since $E_{a2} > E_{a1}$, the rate constant $k_2$ increases more rapidly with temperature than $k_1$. Hence, the ratio $\\frac{[B]}{[C]} = \\frac{k_1}{k_2}$ decreases as temperature increases. (B is correct)\n\n3. For parallel first-order reactions, $k_{\\text{eff}} = k_1 + k_2 = 3.0 \\times 10^{-4}\\text{ s}^{-1}$. (C is correct)\n\n4. The effective activation energy is $E_{\\text{eff}} = \\frac{k_1 E_{a1} + k_2 E_{a2}}{k_1 + k_2} = \\frac{2(40) + 1(80)}{3} = 53.33\\text{ kJ/mol}$. (D is correct)", - "language": "English", - "id": "4a30badd-9672-41e2-a810-a7c7efa708bd" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 2", - "subtopic": "Thermodynamics", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "Calculate the change in entropy $\\Delta S$ (in $\\text{J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$) when $2\\text{ moles}$ of an ideal gas ($C_{V,m} = \\frac{3}{2}R$) are heated reversibly from $300\\text{ K}$ to $600\\text{ K}$ at constant pressure.\n\n(Take $R = 8.314\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$ and $\\ln(2) \\approx 0.693$)", - "options": null, - "correct_answer": { - "value": "28.81" - }, - "solution_text": "For a process at constant pressure:\n\n$$C_{P,m} = C_{V,m} + R = \\frac{5}{2}R$$\n\nThe entropy change is:\n\n$$\\Delta S = n C_{P,m} \\ln\\left(\\frac{T_2}{T_1}\\right) = 2 \\times \\left(\\frac{5}{2} R\\right) \\ln\\left(\\frac{600}{300}\\right) = 5 R \\ln(2)$$\n\n$$\\Delta S = 5 \\times 8.314 \\times 0.693 \\approx 28.81\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$$", - "language": "English", - "id": "c9765206-aa09-4ac4-abd5-c09bb5c20210" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Calculus Set 2", - "subtopic": "Definite Integration", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "Let $I = \\int_{0}^{\\pi} \\frac{x \\sin^2 x \\cos x}{1 + \\cos^2 x} \\, dx$. The value of $I$ is:", - "options": { - "A": "$\\frac{\\pi^2}{4}$", - "B": "$0$", - "C": "$-\\frac{\\pi}{2}$", - "D": "$-\\frac{\\pi^2}{4}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "Using the integration property $\\int_a^b f(x) dx = \\int_a^b f(a+b-x) dx$:\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2(\\pi - x) \\cos(\\pi - x)}{1 + \\cos^2(\\pi - x)} \\, dx$$\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2 x (-\\cos x)}{1 + \\cos^2 x} \\, dx = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx + I$$\n\n$$2I = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx$$\n\nLet $g(x) = \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x}$. Since $g(\\pi - x) = -g(x)$, the integral $\\int_0^\\pi g(x) dx = 0$.\n\nTherefore, $2I = 0 \\implies I = 0$.", - "language": "English", - "id": "d46b9600-6625-49be-818a-dd0dbf2aa8bf" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Algebra Set 2", - "subtopic": "Matrices & Determinants", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Let $A = \\begin{pmatrix} 1 & 0 & 0 \\\\ 2 & 1 & 0 \\\\ 3 & 2 & 1 \\end{pmatrix}$ and let $I$ be the $3 \\times 3$ identity matrix. If $B = A - I$, which of the following statements is/are correct?", - "options": { - "A": "$B^3$ is a null matrix.", - "B": "The determinant of $A^n$ is equal to $1$ for any integer $n \\ge 1$.", - "C": "The inverse of $A$ exists and is equal to $I - B + B^2$.", - "D": "$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "We have:\n\n$$B = A - I = \\begin{pmatrix} 0 & 0 & 0 \\\\ 2 & 0 & 0 \\\\ 3 & 2 & 0 \\end{pmatrix}$$\n\nCompute $B^2$:\n\n$$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$$ (Option D is correct)\n\nCompute $B^3$:\n\n$$B^3 = B^2 B = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ (Option A is correct)\n\nSince $B^3 = 0$, we can express $(I + B)^{-1}$ using a geometric series:\n\n$$(I + B)^{-1} = I - B + B^2 - B^3 + \\dots = I - B + B^2$$\n\nThus, $A^{-1} = I - B + B^2$. (Option C is correct)\n\nAlso, $\\det(A) = 1 \\implies \\det(A^n) = (\\det A)^n = 1$. (Option B is correct)", - "language": "English", - "id": "b31abe5c-5758-4fae-a454-7ea8486299e3" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Coordinate Geometry Set 2", - "subtopic": "Parabolas & Circles", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "The line $y = mx + 1$ is a common tangent to the circle $x^2 + y^2 = \\frac{1}{2}$ and the parabola $y^2 = 4x$. Find the value of $m^2$.", - "options": null, - "correct_answer": { - "value": "1" - }, - "solution_text": "For the parabola $y^2 = 4ax$ with $a=1$, the condition of tangency for $y = mx + c$ is:\n\n$$c = \\frac{a}{m} \\implies 1 = \\frac{1}{m} \\implies m = 1$$\n\nLet's check if the line $y = x + 1$ (i.e., $x - y + 1 = 0$) is also tangent to the circle $x^2 + y^2 = \\frac{1}{2}$:\n\nThe perpendicular distance from the center $(0,0)$ to the line is:\n\n$$d = \\frac{|1|}{\\sqrt{1^2 + (-1)^2}} = \\frac{1}{\\sqrt{2}}$$\n\nThis distance is exactly equal to the radius $r = \\frac{1}{\\sqrt{2}}$ of the circle. Thus, the line is a common tangent. Therefore, $m^2 = 1^2 = 1$.", - "language": "English", - "id": "17afa925-7e93-49e0-a147-7ce66fcb309c" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Rotational Mechanics Set 3", - "subtopic": "Angular Momentum & Collisions", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "A thin uniform rod of mass $M$ and length $L$ is pivoted at its upper end and hangs vertically. A small particle of mass $m$ ($m \\ll M$) moving horizontally with velocity $v_0$ strikes the rod at a distance $x$ from the pivot and sticks to it. If the rod deflects by an angle $\\theta$ such that $\\cos\\theta = 1 - \\frac{6mv_0^2}{MgL}$, the collision point $x$ is:", - "options": { - "A": "$\\frac{L}{\\sqrt{3}}$", - "B": "$\\frac{L}{2}$", - "C": "$\\frac{2L}{3}$", - "D": "$\\frac{L}{\\sqrt{2}}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "By conservation of angular momentum about the pivot:\n\n$$m v_0 x = I \\omega = \\left(\\frac{1}{3}ML^2 + m x^2\\right) \\omega \\approx \\frac{1}{3}ML^2 \\omega \\implies \\omega = \\frac{3mv_0x}{ML^2}$$\n\nFrom conservation of energy after collision:\n\n$$\\frac{1}{2} I \\omega^2 = M g h_{\\text{cm}} \\implies \\frac{1}{2} \\left(\\frac{1}{3}ML^2\\right) \\left(\\frac{3mv_0x}{ML^2}\\right)^2 = M g \\left(\\frac{L}{2}\\right) (1 - \\cos\\theta)$$\n\n$$\\frac{3}{2}\\frac{m^2 v_0^2 x^2}{ML^2} = M g \\frac{L}{2} \\left(\\frac{6mv_0^2}{MgL}\\right) = 3mv_0^2 \\implies x = \\frac{L}{2}$$", - "language": "English", - "id": "917bcb4f-b7da-442d-ab46-838810f5852b" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Electromagnetism Set 3", - "subtopic": "Electromagnetic Induction", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "A rectangular loop of dimensions $a \\times b$ with resistance $R$ is placed near an infinitely long straight wire carrying a time-dependent current $I(t) = I_0 \\sin(\\omega t)$. The loop is at a distance $r_0$ from the wire. Which of the following statements is/are correct?", - "options": { - "A": "The maximum mutual inductance $M$ between the wire and the loop is $\\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "B": "The amplitude of the induced current in the loop is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "C": "The net force on the loop is zero at all times.", - "D": "The induced EMF in the loop is in phase with the current in the straight wire." - }, - "correct_answer": { - "value": [ - "A", - "B" - ] - }, - "solution_text": "Magnetic flux $\\Phi(t)$ through the loop is:\n\n$$\\Phi(t) = \\int_{r_0}^{r_0+a} \\frac{\\mu_0 I(t)}{2\\pi r} b \\, dr = \\frac{\\mu_0 b I(t)}{2\\pi} \\ln\\left(\\frac{r_0+a}{r_0}\\right)$$\n\nMutual inductance $M = \\frac{\\Phi(t)}{I(t)} = \\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right).\n\nInduced EMF $\\mathcal{E}(t) = -\\frac{d\\Phi}{dt} = -M I_0 \\omega \\cos(\\omega t)$.\n\nInduced current $i(t) = \\frac{\\mathcal{E}(t)}{R} = -\\frac{M I_0 \\omega}{R} \\cos(\\omega t)$.\n\nTherefore, the amplitude of the induced current is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$. Since EMF varies as $\\cos(\\omega t)$ and current in the wire is $\\sin(\\omega t)$, they are out of phase by $\\pi/2$. Net force is non-zero because the magnetic field varies across the loop width.", - "language": "English", - "id": "bdc1274e-af8f-48af-b804-69837e1d9417" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Modern Physics Set 3", - "subtopic": "Photoelectric Effect", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "In a photoelectric effect experiment, light of wavelength $\\lambda = 300\\text{ nm}$ is incident on a metal plate. The stopping potential is found to be $V_1 = 1.85\\text{ V}$. When light of wavelength $\\lambda' = 200\\text{ nm}$ is incident on the same metal plate, the stopping potential becomes $V_2$ (in Volts). Calculate the value of $V_2$.\n\n(Take $hc = 1240\\text{ eV}\\cdot\\text{nm}$ and electron charge $e = 1.6 \\times 10^{-19}\\text{ C}$)", - "options": null, - "correct_answer": { - "value": "3.92" - }, - "solution_text": "From Einstein's Photoelectric Equation:\n\n$$eV_1 = \\frac{hc}{\\lambda} - \\Phi_0 \\implies \\Phi_0 = \\frac{1240}{300} - 1.85 = 4.13 - 1.85 = 2.28\\text{ eV}$$\n\nFor the second wavelength:\n\n$$eV_2 = \\frac{hc}{\\lambda'} - \\Phi_0 \\implies V_2 = \\frac{1240}{200} - 2.28 = 6.20 - 2.28 = 3.92\\text{ V}$$", - "language": "English", - "id": "7d0d4c4c-fe5d-4850-9ef4-ca3275566984" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Coordination Chemistry Set 3", - "subtopic": "Crystal Field Theory", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "For the octahedral complex ion $[Fe(H_2O)_6]^{3+}$, the crystal field splitting energy is denoted by $\\Delta_0$ and the pairing energy is $P$. Which of the following options correctly represents the magnetic properties and d-electron configuration of the metal ion?", - "options": { - "A": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "B": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "C": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{15}\\text{ B.M.}$", - "D": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{3}\\text{ B.M.}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "$Fe^{3+}$ has a $d^5$ electron configuration. Since $H_2O$ is a weak field ligand, $\\Delta_0 < P$, resulting in a high-spin complex.\n\nConfiguration: $t_{2g}^3 e_g^2$.\n\nNumber of unpaired electrons $n = 5$.\n\nSpin-only magnetic moment $\\mu_{\\text{eff}} = \\sqrt{n(n+2)} = \\sqrt{5(7)} = \\sqrt{35}\\text{ B.M.}$", - "language": "English", - "id": "fc0b61fd-d6a3-45fc-8605-22488727d214" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 3", - "subtopic": "Chemical Kinetics", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Consider the reaction network showing first-order parallel decomposition of a reactant $A$:\n\n$$A \\xrightarrow{k_1} B \\quad \\text{and} \\quad A \\xrightarrow{k_2} C$$\n\nAt a temperature $T_1$, the rate constants are $k_1 = 2.0 \\times 10^{-4}\\text{ s}^{-1}$ and $k_2 = 1.0 \\times 10^{-4}\\text{ s}^{-1}$. Let the activation energies for the two paths be $E_{a1} = 40\\text{ kJ/mol}$ and $E_{a2} = 80\\text{ kJ/mol}$ respectively. Which of the following statements is/are correct?", - "options": { - "A": "The fractional yield of $B$ at temperature $T_1$ is $\\frac{2}{3}$.", - "B": "If the temperature is increased to $T_2 > T_1$, the ratio $[B]/[C]$ decreases.", - "C": "The overall rate constant for the disappearance of $A$ is $k_{\\text{eff}} = 3.0 \\times 10^{-4}\\text{ s}^{-1}$ at $T_1$.", - "D": "The effective activation energy for the disappearance of $A$ is $E_{\\text{eff}} \\approx 53.3\\text{ kJ/mol}$ at $T_1$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "Let us check each option:\n\n1. Yield of $B = \\frac{k_1}{k_1+k_2} = \\frac{2.0}{2.0+1.0} = \\frac{2}{3}$. (A is correct)\n\n2. Since $E_{a2} > E_{a1}$, the rate constant $k_2$ increases more rapidly with temperature than $k_1$. Hence, the ratio $\\frac{[B]}{[C]} = \\frac{k_1}{k_2}$ decreases as temperature increases. (B is correct)\n\n3. For parallel first-order reactions, $k_{\\text{eff}} = k_1 + k_2 = 3.0 \\times 10^{-4}\\text{ s}^{-1}$. (C is correct)\n\n4. The effective activation energy is $E_{\\text{eff}} = \\frac{k_1 E_{a1} + k_2 E_{a2}}{k_1 + k_2} = \\frac{2(40) + 1(80)}{3} = 53.33\\text{ kJ/mol}$. (D is correct)", - "language": "English", - "id": "a8805590-436f-40fa-be93-2195b99b2794" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 3", - "subtopic": "Thermodynamics", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "Calculate the change in entropy $\\Delta S$ (in $\\text{J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$) when $2\\text{ moles}$ of an ideal gas ($C_{V,m} = \\frac{3}{2}R$) are heated reversibly from $300\\text{ K}$ to $600\\text{ K}$ at constant pressure.\n\n(Take $R = 8.314\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$ and $\\ln(2) \\approx 0.693$)", - "options": null, - "correct_answer": { - "value": "28.81" - }, - "solution_text": "For a process at constant pressure:\n\n$$C_{P,m} = C_{V,m} + R = \\frac{5}{2}R$$\n\nThe entropy change is:\n\n$$\\Delta S = n C_{P,m} \\ln\\left(\\frac{T_2}{T_1}\\right) = 2 \\times \\left(\\frac{5}{2} R\\right) \\ln\\left(\\frac{600}{300}\\right) = 5 R \\ln(2)$$\n\n$$\\Delta S = 5 \\times 8.314 \\times 0.693 \\approx 28.81\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$$", - "language": "English", - "id": "c7b11fd1-db7f-49ab-90e2-6710320597a0" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Calculus Set 3", - "subtopic": "Definite Integration", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "Let $I = \\int_{0}^{\\pi} \\frac{x \\sin^2 x \\cos x}{1 + \\cos^2 x} \\, dx$. The value of $I$ is:", - "options": { - "A": "$\\frac{\\pi^2}{4}$", - "B": "$0$", - "C": "$-\\frac{\\pi}{2}$", - "D": "$-\\frac{\\pi^2}{4}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "Using the integration property $\\int_a^b f(x) dx = \\int_a^b f(a+b-x) dx$:\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2(\\pi - x) \\cos(\\pi - x)}{1 + \\cos^2(\\pi - x)} \\, dx$$\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2 x (-\\cos x)}{1 + \\cos^2 x} \\, dx = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx + I$$\n\n$$2I = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx$$\n\nLet $g(x) = \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x}$. Since $g(\\pi - x) = -g(x)$, the integral $\\int_0^\\pi g(x) dx = 0$.\n\nTherefore, $2I = 0 \\implies I = 0$.", - "language": "English", - "id": "59caa9e3-36b0-4633-8b70-0d3860cc9bfa" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Algebra Set 3", - "subtopic": "Matrices & Determinants", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Let $A = \\begin{pmatrix} 1 & 0 & 0 \\\\ 2 & 1 & 0 \\\\ 3 & 2 & 1 \\end{pmatrix}$ and let $I$ be the $3 \\times 3$ identity matrix. If $B = A - I$, which of the following statements is/are correct?", - "options": { - "A": "$B^3$ is a null matrix.", - "B": "The determinant of $A^n$ is equal to $1$ for any integer $n \\ge 1$.", - "C": "The inverse of $A$ exists and is equal to $I - B + B^2$.", - "D": "$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "We have:\n\n$$B = A - I = \\begin{pmatrix} 0 & 0 & 0 \\\\ 2 & 0 & 0 \\\\ 3 & 2 & 0 \\end{pmatrix}$$\n\nCompute $B^2$:\n\n$$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$$ (Option D is correct)\n\nCompute $B^3$:\n\n$$B^3 = B^2 B = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ (Option A is correct)\n\nSince $B^3 = 0$, we can express $(I + B)^{-1}$ using a geometric series:\n\n$$(I + B)^{-1} = I - B + B^2 - B^3 + \\dots = I - B + B^2$$\n\nThus, $A^{-1} = I - B + B^2$. (Option C is correct)\n\nAlso, $\\det(A) = 1 \\implies \\det(A^n) = (\\det A)^n = 1$. (Option B is correct)", - "language": "English", - "id": "fafc027a-5b8f-4249-8a3f-a51481583df7" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Coordinate Geometry Set 3", - "subtopic": "Parabolas & Circles", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "The line $y = mx + 1$ is a common tangent to the circle $x^2 + y^2 = \\frac{1}{2}$ and the parabola $y^2 = 4x$. Find the value of $m^2$.", - "options": null, - "correct_answer": { - "value": "1" - }, - "solution_text": "For the parabola $y^2 = 4ax$ with $a=1$, the condition of tangency for $y = mx + c$ is:\n\n$$c = \\frac{a}{m} \\implies 1 = \\frac{1}{m} \\implies m = 1$$\n\nLet's check if the line $y = x + 1$ (i.e., $x - y + 1 = 0$) is also tangent to the circle $x^2 + y^2 = \\frac{1}{2}$:\n\nThe perpendicular distance from the center $(0,0)$ to the line is:\n\n$$d = \\frac{|1|}{\\sqrt{1^2 + (-1)^2}} = \\frac{1}{\\sqrt{2}}$$\n\nThis distance is exactly equal to the radius $r = \\frac{1}{\\sqrt{2}}$ of the circle. Thus, the line is a common tangent. Therefore, $m^2 = 1^2 = 1$.", - "language": "English", - "id": "f30654ea-0b60-436f-8eff-5509755a96aa" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Rotational Mechanics Set 4", - "subtopic": "Angular Momentum & Collisions", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "A thin uniform rod of mass $M$ and length $L$ is pivoted at its upper end and hangs vertically. A small particle of mass $m$ ($m \\ll M$) moving horizontally with velocity $v_0$ strikes the rod at a distance $x$ from the pivot and sticks to it. If the rod deflects by an angle $\\theta$ such that $\\cos\\theta = 1 - \\frac{6mv_0^2}{MgL}$, the collision point $x$ is:", - "options": { - "A": "$\\frac{L}{\\sqrt{3}}$", - "B": "$\\frac{L}{2}$", - "C": "$\\frac{2L}{3}$", - "D": "$\\frac{L}{\\sqrt{2}}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "By conservation of angular momentum about the pivot:\n\n$$m v_0 x = I \\omega = \\left(\\frac{1}{3}ML^2 + m x^2\\right) \\omega \\approx \\frac{1}{3}ML^2 \\omega \\implies \\omega = \\frac{3mv_0x}{ML^2}$$\n\nFrom conservation of energy after collision:\n\n$$\\frac{1}{2} I \\omega^2 = M g h_{\\text{cm}} \\implies \\frac{1}{2} \\left(\\frac{1}{3}ML^2\\right) \\left(\\frac{3mv_0x}{ML^2}\\right)^2 = M g \\left(\\frac{L}{2}\\right) (1 - \\cos\\theta)$$\n\n$$\\frac{3}{2}\\frac{m^2 v_0^2 x^2}{ML^2} = M g \\frac{L}{2} \\left(\\frac{6mv_0^2}{MgL}\\right) = 3mv_0^2 \\implies x = \\frac{L}{2}$$", - "language": "English", - "id": "91f14269-32cd-4e4f-9e2e-37260aa4413a" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Electromagnetism Set 4", - "subtopic": "Electromagnetic Induction", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "A rectangular loop of dimensions $a \\times b$ with resistance $R$ is placed near an infinitely long straight wire carrying a time-dependent current $I(t) = I_0 \\sin(\\omega t)$. The loop is at a distance $r_0$ from the wire. Which of the following statements is/are correct?", - "options": { - "A": "The maximum mutual inductance $M$ between the wire and the loop is $\\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "B": "The amplitude of the induced current in the loop is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "C": "The net force on the loop is zero at all times.", - "D": "The induced EMF in the loop is in phase with the current in the straight wire." - }, - "correct_answer": { - "value": [ - "A", - "B" - ] - }, - "solution_text": "Magnetic flux $\\Phi(t)$ through the loop is:\n\n$$\\Phi(t) = \\int_{r_0}^{r_0+a} \\frac{\\mu_0 I(t)}{2\\pi r} b \\, dr = \\frac{\\mu_0 b I(t)}{2\\pi} \\ln\\left(\\frac{r_0+a}{r_0}\\right)$$\n\nMutual inductance $M = \\frac{\\Phi(t)}{I(t)} = \\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right).\n\nInduced EMF $\\mathcal{E}(t) = -\\frac{d\\Phi}{dt} = -M I_0 \\omega \\cos(\\omega t)$.\n\nInduced current $i(t) = \\frac{\\mathcal{E}(t)}{R} = -\\frac{M I_0 \\omega}{R} \\cos(\\omega t)$.\n\nTherefore, the amplitude of the induced current is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$. Since EMF varies as $\\cos(\\omega t)$ and current in the wire is $\\sin(\\omega t)$, they are out of phase by $\\pi/2$. Net force is non-zero because the magnetic field varies across the loop width.", - "language": "English", - "id": "54452f59-606c-4970-b9d5-5a7949a11792" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Modern Physics Set 4", - "subtopic": "Photoelectric Effect", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "In a photoelectric effect experiment, light of wavelength $\\lambda = 300\\text{ nm}$ is incident on a metal plate. The stopping potential is found to be $V_1 = 1.85\\text{ V}$. When light of wavelength $\\lambda' = 200\\text{ nm}$ is incident on the same metal plate, the stopping potential becomes $V_2$ (in Volts). Calculate the value of $V_2$.\n\n(Take $hc = 1240\\text{ eV}\\cdot\\text{nm}$ and electron charge $e = 1.6 \\times 10^{-19}\\text{ C}$)", - "options": null, - "correct_answer": { - "value": "3.92" - }, - "solution_text": "From Einstein's Photoelectric Equation:\n\n$$eV_1 = \\frac{hc}{\\lambda} - \\Phi_0 \\implies \\Phi_0 = \\frac{1240}{300} - 1.85 = 4.13 - 1.85 = 2.28\\text{ eV}$$\n\nFor the second wavelength:\n\n$$eV_2 = \\frac{hc}{\\lambda'} - \\Phi_0 \\implies V_2 = \\frac{1240}{200} - 2.28 = 6.20 - 2.28 = 3.92\\text{ V}$$", - "language": "English", - "id": "b44ebfed-8b88-4821-acb4-0e1f24700794" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Coordination Chemistry Set 4", - "subtopic": "Crystal Field Theory", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "For the octahedral complex ion $[Fe(H_2O)_6]^{3+}$, the crystal field splitting energy is denoted by $\\Delta_0$ and the pairing energy is $P$. Which of the following options correctly represents the magnetic properties and d-electron configuration of the metal ion?", - "options": { - "A": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "B": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "C": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{15}\\text{ B.M.}$", - "D": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{3}\\text{ B.M.}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "$Fe^{3+}$ has a $d^5$ electron configuration. Since $H_2O$ is a weak field ligand, $\\Delta_0 < P$, resulting in a high-spin complex.\n\nConfiguration: $t_{2g}^3 e_g^2$.\n\nNumber of unpaired electrons $n = 5$.\n\nSpin-only magnetic moment $\\mu_{\\text{eff}} = \\sqrt{n(n+2)} = \\sqrt{5(7)} = \\sqrt{35}\\text{ B.M.}$", - "language": "English", - "id": "ea5b346a-c740-40c6-904d-e3cbc7fba1c3" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 4", - "subtopic": "Chemical Kinetics", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Consider the reaction network showing first-order parallel decomposition of a reactant $A$:\n\n$$A \\xrightarrow{k_1} B \\quad \\text{and} \\quad A \\xrightarrow{k_2} C$$\n\nAt a temperature $T_1$, the rate constants are $k_1 = 2.0 \\times 10^{-4}\\text{ s}^{-1}$ and $k_2 = 1.0 \\times 10^{-4}\\text{ s}^{-1}$. Let the activation energies for the two paths be $E_{a1} = 40\\text{ kJ/mol}$ and $E_{a2} = 80\\text{ kJ/mol}$ respectively. Which of the following statements is/are correct?", - "options": { - "A": "The fractional yield of $B$ at temperature $T_1$ is $\\frac{2}{3}$.", - "B": "If the temperature is increased to $T_2 > T_1$, the ratio $[B]/[C]$ decreases.", - "C": "The overall rate constant for the disappearance of $A$ is $k_{\\text{eff}} = 3.0 \\times 10^{-4}\\text{ s}^{-1}$ at $T_1$.", - "D": "The effective activation energy for the disappearance of $A$ is $E_{\\text{eff}} \\approx 53.3\\text{ kJ/mol}$ at $T_1$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "Let us check each option:\n\n1. Yield of $B = \\frac{k_1}{k_1+k_2} = \\frac{2.0}{2.0+1.0} = \\frac{2}{3}$. (A is correct)\n\n2. Since $E_{a2} > E_{a1}$, the rate constant $k_2$ increases more rapidly with temperature than $k_1$. Hence, the ratio $\\frac{[B]}{[C]} = \\frac{k_1}{k_2}$ decreases as temperature increases. (B is correct)\n\n3. For parallel first-order reactions, $k_{\\text{eff}} = k_1 + k_2 = 3.0 \\times 10^{-4}\\text{ s}^{-1}$. (C is correct)\n\n4. The effective activation energy is $E_{\\text{eff}} = \\frac{k_1 E_{a1} + k_2 E_{a2}}{k_1 + k_2} = \\frac{2(40) + 1(80)}{3} = 53.33\\text{ kJ/mol}$. (D is correct)", - "language": "English", - "id": "2a086091-174d-460a-bdd3-f6395abb2e50" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 4", - "subtopic": "Thermodynamics", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "Calculate the change in entropy $\\Delta S$ (in $\\text{J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$) when $2\\text{ moles}$ of an ideal gas ($C_{V,m} = \\frac{3}{2}R$) are heated reversibly from $300\\text{ K}$ to $600\\text{ K}$ at constant pressure.\n\n(Take $R = 8.314\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$ and $\\ln(2) \\approx 0.693$)", - "options": null, - "correct_answer": { - "value": "28.81" - }, - "solution_text": "For a process at constant pressure:\n\n$$C_{P,m} = C_{V,m} + R = \\frac{5}{2}R$$\n\nThe entropy change is:\n\n$$\\Delta S = n C_{P,m} \\ln\\left(\\frac{T_2}{T_1}\\right) = 2 \\times \\left(\\frac{5}{2} R\\right) \\ln\\left(\\frac{600}{300}\\right) = 5 R \\ln(2)$$\n\n$$\\Delta S = 5 \\times 8.314 \\times 0.693 \\approx 28.81\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$$", - "language": "English", - "id": "6feedaf3-158b-40c3-9663-c7ed7ec90878" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Calculus Set 4", - "subtopic": "Definite Integration", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "Let $I = \\int_{0}^{\\pi} \\frac{x \\sin^2 x \\cos x}{1 + \\cos^2 x} \\, dx$. The value of $I$ is:", - "options": { - "A": "$\\frac{\\pi^2}{4}$", - "B": "$0$", - "C": "$-\\frac{\\pi}{2}$", - "D": "$-\\frac{\\pi^2}{4}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "Using the integration property $\\int_a^b f(x) dx = \\int_a^b f(a+b-x) dx$:\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2(\\pi - x) \\cos(\\pi - x)}{1 + \\cos^2(\\pi - x)} \\, dx$$\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2 x (-\\cos x)}{1 + \\cos^2 x} \\, dx = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx + I$$\n\n$$2I = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx$$\n\nLet $g(x) = \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x}$. Since $g(\\pi - x) = -g(x)$, the integral $\\int_0^\\pi g(x) dx = 0$.\n\nTherefore, $2I = 0 \\implies I = 0$.", - "language": "English", - "id": "e6ad4b15-0a1c-4b17-a177-789cb69ed8d4" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Algebra Set 4", - "subtopic": "Matrices & Determinants", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Let $A = \\begin{pmatrix} 1 & 0 & 0 \\\\ 2 & 1 & 0 \\\\ 3 & 2 & 1 \\end{pmatrix}$ and let $I$ be the $3 \\times 3$ identity matrix. If $B = A - I$, which of the following statements is/are correct?", - "options": { - "A": "$B^3$ is a null matrix.", - "B": "The determinant of $A^n$ is equal to $1$ for any integer $n \\ge 1$.", - "C": "The inverse of $A$ exists and is equal to $I - B + B^2$.", - "D": "$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "We have:\n\n$$B = A - I = \\begin{pmatrix} 0 & 0 & 0 \\\\ 2 & 0 & 0 \\\\ 3 & 2 & 0 \\end{pmatrix}$$\n\nCompute $B^2$:\n\n$$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$$ (Option D is correct)\n\nCompute $B^3$:\n\n$$B^3 = B^2 B = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ (Option A is correct)\n\nSince $B^3 = 0$, we can express $(I + B)^{-1}$ using a geometric series:\n\n$$(I + B)^{-1} = I - B + B^2 - B^3 + \\dots = I - B + B^2$$\n\nThus, $A^{-1} = I - B + B^2$. (Option C is correct)\n\nAlso, $\\det(A) = 1 \\implies \\det(A^n) = (\\det A)^n = 1$. (Option B is correct)", - "language": "English", - "id": "373f72ef-3ce5-4a45-87bf-575a59a757a3" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Coordinate Geometry Set 4", - "subtopic": "Parabolas & Circles", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "The line $y = mx + 1$ is a common tangent to the circle $x^2 + y^2 = \\frac{1}{2}$ and the parabola $y^2 = 4x$. Find the value of $m^2$.", - "options": null, - "correct_answer": { - "value": "1" - }, - "solution_text": "For the parabola $y^2 = 4ax$ with $a=1$, the condition of tangency for $y = mx + c$ is:\n\n$$c = \\frac{a}{m} \\implies 1 = \\frac{1}{m} \\implies m = 1$$\n\nLet's check if the line $y = x + 1$ (i.e., $x - y + 1 = 0$) is also tangent to the circle $x^2 + y^2 = \\frac{1}{2}$:\n\nThe perpendicular distance from the center $(0,0)$ to the line is:\n\n$$d = \\frac{|1|}{\\sqrt{1^2 + (-1)^2}} = \\frac{1}{\\sqrt{2}}$$\n\nThis distance is exactly equal to the radius $r = \\frac{1}{\\sqrt{2}}$ of the circle. Thus, the line is a common tangent. Therefore, $m^2 = 1^2 = 1$.", - "language": "English", - "id": "99b879a2-924a-4d50-8de2-eb51c64c574c" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Rotational Mechanics Set 5", - "subtopic": "Angular Momentum & Collisions", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "A thin uniform rod of mass $M$ and length $L$ is pivoted at its upper end and hangs vertically. A small particle of mass $m$ ($m \\ll M$) moving horizontally with velocity $v_0$ strikes the rod at a distance $x$ from the pivot and sticks to it. If the rod deflects by an angle $\\theta$ such that $\\cos\\theta = 1 - \\frac{6mv_0^2}{MgL}$, the collision point $x$ is:", - "options": { - "A": "$\\frac{L}{\\sqrt{3}}$", - "B": "$\\frac{L}{2}$", - "C": "$\\frac{2L}{3}$", - "D": "$\\frac{L}{\\sqrt{2}}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "By conservation of angular momentum about the pivot:\n\n$$m v_0 x = I \\omega = \\left(\\frac{1}{3}ML^2 + m x^2\\right) \\omega \\approx \\frac{1}{3}ML^2 \\omega \\implies \\omega = \\frac{3mv_0x}{ML^2}$$\n\nFrom conservation of energy after collision:\n\n$$\\frac{1}{2} I \\omega^2 = M g h_{\\text{cm}} \\implies \\frac{1}{2} \\left(\\frac{1}{3}ML^2\\right) \\left(\\frac{3mv_0x}{ML^2}\\right)^2 = M g \\left(\\frac{L}{2}\\right) (1 - \\cos\\theta)$$\n\n$$\\frac{3}{2}\\frac{m^2 v_0^2 x^2}{ML^2} = M g \\frac{L}{2} \\left(\\frac{6mv_0^2}{MgL}\\right) = 3mv_0^2 \\implies x = \\frac{L}{2}$$", - "language": "English", - "id": "ed1dbb21-837f-465a-8c4e-34ac1d5fe2ef" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Electromagnetism Set 5", - "subtopic": "Electromagnetic Induction", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "A rectangular loop of dimensions $a \\times b$ with resistance $R$ is placed near an infinitely long straight wire carrying a time-dependent current $I(t) = I_0 \\sin(\\omega t)$. The loop is at a distance $r_0$ from the wire. Which of the following statements is/are correct?", - "options": { - "A": "The maximum mutual inductance $M$ between the wire and the loop is $\\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "B": "The amplitude of the induced current in the loop is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "C": "The net force on the loop is zero at all times.", - "D": "The induced EMF in the loop is in phase with the current in the straight wire." - }, - "correct_answer": { - "value": [ - "A", - "B" - ] - }, - "solution_text": "Magnetic flux $\\Phi(t)$ through the loop is:\n\n$$\\Phi(t) = \\int_{r_0}^{r_0+a} \\frac{\\mu_0 I(t)}{2\\pi r} b \\, dr = \\frac{\\mu_0 b I(t)}{2\\pi} \\ln\\left(\\frac{r_0+a}{r_0}\\right)$$\n\nMutual inductance $M = \\frac{\\Phi(t)}{I(t)} = \\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right).\n\nInduced EMF $\\mathcal{E}(t) = -\\frac{d\\Phi}{dt} = -M I_0 \\omega \\cos(\\omega t)$.\n\nInduced current $i(t) = \\frac{\\mathcal{E}(t)}{R} = -\\frac{M I_0 \\omega}{R} \\cos(\\omega t)$.\n\nTherefore, the amplitude of the induced current is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$. Since EMF varies as $\\cos(\\omega t)$ and current in the wire is $\\sin(\\omega t)$, they are out of phase by $\\pi/2$. Net force is non-zero because the magnetic field varies across the loop width.", - "language": "English", - "id": "4824f772-fc15-4c28-adda-9a22a9a07646" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Modern Physics Set 5", - "subtopic": "Photoelectric Effect", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "In a photoelectric effect experiment, light of wavelength $\\lambda = 300\\text{ nm}$ is incident on a metal plate. The stopping potential is found to be $V_1 = 1.85\\text{ V}$. When light of wavelength $\\lambda' = 200\\text{ nm}$ is incident on the same metal plate, the stopping potential becomes $V_2$ (in Volts). Calculate the value of $V_2$.\n\n(Take $hc = 1240\\text{ eV}\\cdot\\text{nm}$ and electron charge $e = 1.6 \\times 10^{-19}\\text{ C}$)", - "options": null, - "correct_answer": { - "value": "3.92" - }, - "solution_text": "From Einstein's Photoelectric Equation:\n\n$$eV_1 = \\frac{hc}{\\lambda} - \\Phi_0 \\implies \\Phi_0 = \\frac{1240}{300} - 1.85 = 4.13 - 1.85 = 2.28\\text{ eV}$$\n\nFor the second wavelength:\n\n$$eV_2 = \\frac{hc}{\\lambda'} - \\Phi_0 \\implies V_2 = \\frac{1240}{200} - 2.28 = 6.20 - 2.28 = 3.92\\text{ V}$$", - "language": "English", - "id": "2e60f5e7-c2a1-4c6c-9be1-809f815cf7c2" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Coordination Chemistry Set 5", - "subtopic": "Crystal Field Theory", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "For the octahedral complex ion $[Fe(H_2O)_6]^{3+}$, the crystal field splitting energy is denoted by $\\Delta_0$ and the pairing energy is $P$. Which of the following options correctly represents the magnetic properties and d-electron configuration of the metal ion?", - "options": { - "A": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "B": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "C": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{15}\\text{ B.M.}$", - "D": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{3}\\text{ B.M.}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "$Fe^{3+}$ has a $d^5$ electron configuration. Since $H_2O$ is a weak field ligand, $\\Delta_0 < P$, resulting in a high-spin complex.\n\nConfiguration: $t_{2g}^3 e_g^2$.\n\nNumber of unpaired electrons $n = 5$.\n\nSpin-only magnetic moment $\\mu_{\\text{eff}} = \\sqrt{n(n+2)} = \\sqrt{5(7)} = \\sqrt{35}\\text{ B.M.}$", - "language": "English", - "id": "9052434f-0c67-4785-b18c-867494cbf721" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 5", - "subtopic": "Chemical Kinetics", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Consider the reaction network showing first-order parallel decomposition of a reactant $A$:\n\n$$A \\xrightarrow{k_1} B \\quad \\text{and} \\quad A \\xrightarrow{k_2} C$$\n\nAt a temperature $T_1$, the rate constants are $k_1 = 2.0 \\times 10^{-4}\\text{ s}^{-1}$ and $k_2 = 1.0 \\times 10^{-4}\\text{ s}^{-1}$. Let the activation energies for the two paths be $E_{a1} = 40\\text{ kJ/mol}$ and $E_{a2} = 80\\text{ kJ/mol}$ respectively. Which of the following statements is/are correct?", - "options": { - "A": "The fractional yield of $B$ at temperature $T_1$ is $\\frac{2}{3}$.", - "B": "If the temperature is increased to $T_2 > T_1$, the ratio $[B]/[C]$ decreases.", - "C": "The overall rate constant for the disappearance of $A$ is $k_{\\text{eff}} = 3.0 \\times 10^{-4}\\text{ s}^{-1}$ at $T_1$.", - "D": "The effective activation energy for the disappearance of $A$ is $E_{\\text{eff}} \\approx 53.3\\text{ kJ/mol}$ at $T_1$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "Let us check each option:\n\n1. Yield of $B = \\frac{k_1}{k_1+k_2} = \\frac{2.0}{2.0+1.0} = \\frac{2}{3}$. (A is correct)\n\n2. Since $E_{a2} > E_{a1}$, the rate constant $k_2$ increases more rapidly with temperature than $k_1$. Hence, the ratio $\\frac{[B]}{[C]} = \\frac{k_1}{k_2}$ decreases as temperature increases. (B is correct)\n\n3. For parallel first-order reactions, $k_{\\text{eff}} = k_1 + k_2 = 3.0 \\times 10^{-4}\\text{ s}^{-1}$. (C is correct)\n\n4. The effective activation energy is $E_{\\text{eff}} = \\frac{k_1 E_{a1} + k_2 E_{a2}}{k_1 + k_2} = \\frac{2(40) + 1(80)}{3} = 53.33\\text{ kJ/mol}$. (D is correct)", - "language": "English", - "id": "7524d781-04e4-45bd-ab5f-f0b2cb89fbf3" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 5", - "subtopic": "Thermodynamics", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "Calculate the change in entropy $\\Delta S$ (in $\\text{J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$) when $2\\text{ moles}$ of an ideal gas ($C_{V,m} = \\frac{3}{2}R$) are heated reversibly from $300\\text{ K}$ to $600\\text{ K}$ at constant pressure.\n\n(Take $R = 8.314\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$ and $\\ln(2) \\approx 0.693$)", - "options": null, - "correct_answer": { - "value": "28.81" - }, - "solution_text": "For a process at constant pressure:\n\n$$C_{P,m} = C_{V,m} + R = \\frac{5}{2}R$$\n\nThe entropy change is:\n\n$$\\Delta S = n C_{P,m} \\ln\\left(\\frac{T_2}{T_1}\\right) = 2 \\times \\left(\\frac{5}{2} R\\right) \\ln\\left(\\frac{600}{300}\\right) = 5 R \\ln(2)$$\n\n$$\\Delta S = 5 \\times 8.314 \\times 0.693 \\approx 28.81\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$$", - "language": "English", - "id": "25557a59-9d52-4bb7-afe1-8f5b8d17cdf0" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Calculus Set 5", - "subtopic": "Definite Integration", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "Let $I = \\int_{0}^{\\pi} \\frac{x \\sin^2 x \\cos x}{1 + \\cos^2 x} \\, dx$. The value of $I$ is:", - "options": { - "A": "$\\frac{\\pi^2}{4}$", - "B": "$0$", - "C": "$-\\frac{\\pi}{2}$", - "D": "$-\\frac{\\pi^2}{4}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "Using the integration property $\\int_a^b f(x) dx = \\int_a^b f(a+b-x) dx$:\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2(\\pi - x) \\cos(\\pi - x)}{1 + \\cos^2(\\pi - x)} \\, dx$$\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2 x (-\\cos x)}{1 + \\cos^2 x} \\, dx = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx + I$$\n\n$$2I = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx$$\n\nLet $g(x) = \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x}$. Since $g(\\pi - x) = -g(x)$, the integral $\\int_0^\\pi g(x) dx = 0$.\n\nTherefore, $2I = 0 \\implies I = 0$.", - "language": "English", - "id": "7f2bb9b0-85b4-4435-9b30-949bb743c5fc" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Algebra Set 5", - "subtopic": "Matrices & Determinants", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Let $A = \\begin{pmatrix} 1 & 0 & 0 \\\\ 2 & 1 & 0 \\\\ 3 & 2 & 1 \\end{pmatrix}$ and let $I$ be the $3 \\times 3$ identity matrix. If $B = A - I$, which of the following statements is/are correct?", - "options": { - "A": "$B^3$ is a null matrix.", - "B": "The determinant of $A^n$ is equal to $1$ for any integer $n \\ge 1$.", - "C": "The inverse of $A$ exists and is equal to $I - B + B^2$.", - "D": "$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "We have:\n\n$$B = A - I = \\begin{pmatrix} 0 & 0 & 0 \\\\ 2 & 0 & 0 \\\\ 3 & 2 & 0 \\end{pmatrix}$$\n\nCompute $B^2$:\n\n$$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$$ (Option D is correct)\n\nCompute $B^3$:\n\n$$B^3 = B^2 B = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ (Option A is correct)\n\nSince $B^3 = 0$, we can express $(I + B)^{-1}$ using a geometric series:\n\n$$(I + B)^{-1} = I - B + B^2 - B^3 + \\dots = I - B + B^2$$\n\nThus, $A^{-1} = I - B + B^2$. (Option C is correct)\n\nAlso, $\\det(A) = 1 \\implies \\det(A^n) = (\\det A)^n = 1$. (Option B is correct)", - "language": "English", - "id": "60ac7a55-b330-4e74-aec1-3497807b914d" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Coordinate Geometry Set 5", - "subtopic": "Parabolas & Circles", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "The line $y = mx + 1$ is a common tangent to the circle $x^2 + y^2 = \\frac{1}{2}$ and the parabola $y^2 = 4x$. Find the value of $m^2$.", - "options": null, - "correct_answer": { - "value": "1" - }, - "solution_text": "For the parabola $y^2 = 4ax$ with $a=1$, the condition of tangency for $y = mx + c$ is:\n\n$$c = \\frac{a}{m} \\implies 1 = \\frac{1}{m} \\implies m = 1$$\n\nLet's check if the line $y = x + 1$ (i.e., $x - y + 1 = 0$) is also tangent to the circle $x^2 + y^2 = \\frac{1}{2}$:\n\nThe perpendicular distance from the center $(0,0)$ to the line is:\n\n$$d = \\frac{|1|}{\\sqrt{1^2 + (-1)^2}} = \\frac{1}{\\sqrt{2}}$$\n\nThis distance is exactly equal to the radius $r = \\frac{1}{\\sqrt{2}}$ of the circle. Thus, the line is a common tangent. Therefore, $m^2 = 1^2 = 1$.", - "language": "English", - "id": "5c48bd9a-6597-44ec-836d-907dc5b6c964" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Rotational Mechanics Set 6", - "subtopic": "Angular Momentum & Collisions", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "A thin uniform rod of mass $M$ and length $L$ is pivoted at its upper end and hangs vertically. A small particle of mass $m$ ($m \\ll M$) moving horizontally with velocity $v_0$ strikes the rod at a distance $x$ from the pivot and sticks to it. If the rod deflects by an angle $\\theta$ such that $\\cos\\theta = 1 - \\frac{6mv_0^2}{MgL}$, the collision point $x$ is:", - "options": { - "A": "$\\frac{L}{\\sqrt{3}}$", - "B": "$\\frac{L}{2}$", - "C": "$\\frac{2L}{3}$", - "D": "$\\frac{L}{\\sqrt{2}}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "By conservation of angular momentum about the pivot:\n\n$$m v_0 x = I \\omega = \\left(\\frac{1}{3}ML^2 + m x^2\\right) \\omega \\approx \\frac{1}{3}ML^2 \\omega \\implies \\omega = \\frac{3mv_0x}{ML^2}$$\n\nFrom conservation of energy after collision:\n\n$$\\frac{1}{2} I \\omega^2 = M g h_{\\text{cm}} \\implies \\frac{1}{2} \\left(\\frac{1}{3}ML^2\\right) \\left(\\frac{3mv_0x}{ML^2}\\right)^2 = M g \\left(\\frac{L}{2}\\right) (1 - \\cos\\theta)$$\n\n$$\\frac{3}{2}\\frac{m^2 v_0^2 x^2}{ML^2} = M g \\frac{L}{2} \\left(\\frac{6mv_0^2}{MgL}\\right) = 3mv_0^2 \\implies x = \\frac{L}{2}$$", - "language": "English", - "id": "b4321a1e-47cc-4e1e-a692-c1da51109be2" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Electromagnetism Set 6", - "subtopic": "Electromagnetic Induction", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "A rectangular loop of dimensions $a \\times b$ with resistance $R$ is placed near an infinitely long straight wire carrying a time-dependent current $I(t) = I_0 \\sin(\\omega t)$. The loop is at a distance $r_0$ from the wire. Which of the following statements is/are correct?", - "options": { - "A": "The maximum mutual inductance $M$ between the wire and the loop is $\\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "B": "The amplitude of the induced current in the loop is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "C": "The net force on the loop is zero at all times.", - "D": "The induced EMF in the loop is in phase with the current in the straight wire." - }, - "correct_answer": { - "value": [ - "A", - "B" - ] - }, - "solution_text": "Magnetic flux $\\Phi(t)$ through the loop is:\n\n$$\\Phi(t) = \\int_{r_0}^{r_0+a} \\frac{\\mu_0 I(t)}{2\\pi r} b \\, dr = \\frac{\\mu_0 b I(t)}{2\\pi} \\ln\\left(\\frac{r_0+a}{r_0}\\right)$$\n\nMutual inductance $M = \\frac{\\Phi(t)}{I(t)} = \\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right).\n\nInduced EMF $\\mathcal{E}(t) = -\\frac{d\\Phi}{dt} = -M I_0 \\omega \\cos(\\omega t)$.\n\nInduced current $i(t) = \\frac{\\mathcal{E}(t)}{R} = -\\frac{M I_0 \\omega}{R} \\cos(\\omega t)$.\n\nTherefore, the amplitude of the induced current is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$. Since EMF varies as $\\cos(\\omega t)$ and current in the wire is $\\sin(\\omega t)$, they are out of phase by $\\pi/2$. Net force is non-zero because the magnetic field varies across the loop width.", - "language": "English", - "id": "739f3dfe-8c40-400d-9155-248bee2be872" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Modern Physics Set 6", - "subtopic": "Photoelectric Effect", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "In a photoelectric effect experiment, light of wavelength $\\lambda = 300\\text{ nm}$ is incident on a metal plate. The stopping potential is found to be $V_1 = 1.85\\text{ V}$. When light of wavelength $\\lambda' = 200\\text{ nm}$ is incident on the same metal plate, the stopping potential becomes $V_2$ (in Volts). Calculate the value of $V_2$.\n\n(Take $hc = 1240\\text{ eV}\\cdot\\text{nm}$ and electron charge $e = 1.6 \\times 10^{-19}\\text{ C}$)", - "options": null, - "correct_answer": { - "value": "3.92" - }, - "solution_text": "From Einstein's Photoelectric Equation:\n\n$$eV_1 = \\frac{hc}{\\lambda} - \\Phi_0 \\implies \\Phi_0 = \\frac{1240}{300} - 1.85 = 4.13 - 1.85 = 2.28\\text{ eV}$$\n\nFor the second wavelength:\n\n$$eV_2 = \\frac{hc}{\\lambda'} - \\Phi_0 \\implies V_2 = \\frac{1240}{200} - 2.28 = 6.20 - 2.28 = 3.92\\text{ V}$$", - "language": "English", - "id": "c40cea14-09e9-4481-b61b-44d6dbb60cc0" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Coordination Chemistry Set 6", - "subtopic": "Crystal Field Theory", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "For the octahedral complex ion $[Fe(H_2O)_6]^{3+}$, the crystal field splitting energy is denoted by $\\Delta_0$ and the pairing energy is $P$. Which of the following options correctly represents the magnetic properties and d-electron configuration of the metal ion?", - "options": { - "A": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "B": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "C": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{15}\\text{ B.M.}$", - "D": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{3}\\text{ B.M.}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "$Fe^{3+}$ has a $d^5$ electron configuration. Since $H_2O$ is a weak field ligand, $\\Delta_0 < P$, resulting in a high-spin complex.\n\nConfiguration: $t_{2g}^3 e_g^2$.\n\nNumber of unpaired electrons $n = 5$.\n\nSpin-only magnetic moment $\\mu_{\\text{eff}} = \\sqrt{n(n+2)} = \\sqrt{5(7)} = \\sqrt{35}\\text{ B.M.}$", - "language": "English", - "id": "67e3bcb6-c336-4bb1-b578-4ba9968c352a" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 6", - "subtopic": "Chemical Kinetics", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Consider the reaction network showing first-order parallel decomposition of a reactant $A$:\n\n$$A \\xrightarrow{k_1} B \\quad \\text{and} \\quad A \\xrightarrow{k_2} C$$\n\nAt a temperature $T_1$, the rate constants are $k_1 = 2.0 \\times 10^{-4}\\text{ s}^{-1}$ and $k_2 = 1.0 \\times 10^{-4}\\text{ s}^{-1}$. Let the activation energies for the two paths be $E_{a1} = 40\\text{ kJ/mol}$ and $E_{a2} = 80\\text{ kJ/mol}$ respectively. Which of the following statements is/are correct?", - "options": { - "A": "The fractional yield of $B$ at temperature $T_1$ is $\\frac{2}{3}$.", - "B": "If the temperature is increased to $T_2 > T_1$, the ratio $[B]/[C]$ decreases.", - "C": "The overall rate constant for the disappearance of $A$ is $k_{\\text{eff}} = 3.0 \\times 10^{-4}\\text{ s}^{-1}$ at $T_1$.", - "D": "The effective activation energy for the disappearance of $A$ is $E_{\\text{eff}} \\approx 53.3\\text{ kJ/mol}$ at $T_1$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "Let us check each option:\n\n1. Yield of $B = \\frac{k_1}{k_1+k_2} = \\frac{2.0}{2.0+1.0} = \\frac{2}{3}$. (A is correct)\n\n2. Since $E_{a2} > E_{a1}$, the rate constant $k_2$ increases more rapidly with temperature than $k_1$. Hence, the ratio $\\frac{[B]}{[C]} = \\frac{k_1}{k_2}$ decreases as temperature increases. (B is correct)\n\n3. For parallel first-order reactions, $k_{\\text{eff}} = k_1 + k_2 = 3.0 \\times 10^{-4}\\text{ s}^{-1}$. (C is correct)\n\n4. The effective activation energy is $E_{\\text{eff}} = \\frac{k_1 E_{a1} + k_2 E_{a2}}{k_1 + k_2} = \\frac{2(40) + 1(80)}{3} = 53.33\\text{ kJ/mol}$. (D is correct)", - "language": "English", - "id": "7cdda987-4aea-4b17-b4ef-6fd9399b6b87" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 6", - "subtopic": "Thermodynamics", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "Calculate the change in entropy $\\Delta S$ (in $\\text{J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$) when $2\\text{ moles}$ of an ideal gas ($C_{V,m} = \\frac{3}{2}R$) are heated reversibly from $300\\text{ K}$ to $600\\text{ K}$ at constant pressure.\n\n(Take $R = 8.314\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$ and $\\ln(2) \\approx 0.693$)", - "options": null, - "correct_answer": { - "value": "28.81" - }, - "solution_text": "For a process at constant pressure:\n\n$$C_{P,m} = C_{V,m} + R = \\frac{5}{2}R$$\n\nThe entropy change is:\n\n$$\\Delta S = n C_{P,m} \\ln\\left(\\frac{T_2}{T_1}\\right) = 2 \\times \\left(\\frac{5}{2} R\\right) \\ln\\left(\\frac{600}{300}\\right) = 5 R \\ln(2)$$\n\n$$\\Delta S = 5 \\times 8.314 \\times 0.693 \\approx 28.81\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$$", - "language": "English", - "id": "10fe6287-4fac-4919-b434-863b72035caf" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Calculus Set 6", - "subtopic": "Definite Integration", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "Let $I = \\int_{0}^{\\pi} \\frac{x \\sin^2 x \\cos x}{1 + \\cos^2 x} \\, dx$. The value of $I$ is:", - "options": { - "A": "$\\frac{\\pi^2}{4}$", - "B": "$0$", - "C": "$-\\frac{\\pi}{2}$", - "D": "$-\\frac{\\pi^2}{4}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "Using the integration property $\\int_a^b f(x) dx = \\int_a^b f(a+b-x) dx$:\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2(\\pi - x) \\cos(\\pi - x)}{1 + \\cos^2(\\pi - x)} \\, dx$$\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2 x (-\\cos x)}{1 + \\cos^2 x} \\, dx = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx + I$$\n\n$$2I = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx$$\n\nLet $g(x) = \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x}$. Since $g(\\pi - x) = -g(x)$, the integral $\\int_0^\\pi g(x) dx = 0$.\n\nTherefore, $2I = 0 \\implies I = 0$.", - "language": "English", - "id": "75ed6f25-4880-461f-bd51-b7ef7869081b" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Algebra Set 6", - "subtopic": "Matrices & Determinants", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Let $A = \\begin{pmatrix} 1 & 0 & 0 \\\\ 2 & 1 & 0 \\\\ 3 & 2 & 1 \\end{pmatrix}$ and let $I$ be the $3 \\times 3$ identity matrix. If $B = A - I$, which of the following statements is/are correct?", - "options": { - "A": "$B^3$ is a null matrix.", - "B": "The determinant of $A^n$ is equal to $1$ for any integer $n \\ge 1$.", - "C": "The inverse of $A$ exists and is equal to $I - B + B^2$.", - "D": "$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "We have:\n\n$$B = A - I = \\begin{pmatrix} 0 & 0 & 0 \\\\ 2 & 0 & 0 \\\\ 3 & 2 & 0 \\end{pmatrix}$$\n\nCompute $B^2$:\n\n$$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$$ (Option D is correct)\n\nCompute $B^3$:\n\n$$B^3 = B^2 B = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ (Option A is correct)\n\nSince $B^3 = 0$, we can express $(I + B)^{-1}$ using a geometric series:\n\n$$(I + B)^{-1} = I - B + B^2 - B^3 + \\dots = I - B + B^2$$\n\nThus, $A^{-1} = I - B + B^2$. (Option C is correct)\n\nAlso, $\\det(A) = 1 \\implies \\det(A^n) = (\\det A)^n = 1$. (Option B is correct)", - "language": "English", - "id": "803234a3-07b9-454d-adab-31d7a22d3616" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Coordinate Geometry Set 6", - "subtopic": "Parabolas & Circles", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "The line $y = mx + 1$ is a common tangent to the circle $x^2 + y^2 = \\frac{1}{2}$ and the parabola $y^2 = 4x$. Find the value of $m^2$.", - "options": null, - "correct_answer": { - "value": "1" - }, - "solution_text": "For the parabola $y^2 = 4ax$ with $a=1$, the condition of tangency for $y = mx + c$ is:\n\n$$c = \\frac{a}{m} \\implies 1 = \\frac{1}{m} \\implies m = 1$$\n\nLet's check if the line $y = x + 1$ (i.e., $x - y + 1 = 0$) is also tangent to the circle $x^2 + y^2 = \\frac{1}{2}$:\n\nThe perpendicular distance from the center $(0,0)$ to the line is:\n\n$$d = \\frac{|1|}{\\sqrt{1^2 + (-1)^2}} = \\frac{1}{\\sqrt{2}}$$\n\nThis distance is exactly equal to the radius $r = \\frac{1}{\\sqrt{2}}$ of the circle. Thus, the line is a common tangent. Therefore, $m^2 = 1^2 = 1$.", - "language": "English", - "id": "15670bcb-c0f8-4e45-a7dd-43e1b40cee7c" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Rotational Mechanics Set 7", - "subtopic": "Angular Momentum & Collisions", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "A thin uniform rod of mass $M$ and length $L$ is pivoted at its upper end and hangs vertically. A small particle of mass $m$ ($m \\ll M$) moving horizontally with velocity $v_0$ strikes the rod at a distance $x$ from the pivot and sticks to it. If the rod deflects by an angle $\\theta$ such that $\\cos\\theta = 1 - \\frac{6mv_0^2}{MgL}$, the collision point $x$ is:", - "options": { - "A": "$\\frac{L}{\\sqrt{3}}$", - "B": "$\\frac{L}{2}$", - "C": "$\\frac{2L}{3}$", - "D": "$\\frac{L}{\\sqrt{2}}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "By conservation of angular momentum about the pivot:\n\n$$m v_0 x = I \\omega = \\left(\\frac{1}{3}ML^2 + m x^2\\right) \\omega \\approx \\frac{1}{3}ML^2 \\omega \\implies \\omega = \\frac{3mv_0x}{ML^2}$$\n\nFrom conservation of energy after collision:\n\n$$\\frac{1}{2} I \\omega^2 = M g h_{\\text{cm}} \\implies \\frac{1}{2} \\left(\\frac{1}{3}ML^2\\right) \\left(\\frac{3mv_0x}{ML^2}\\right)^2 = M g \\left(\\frac{L}{2}\\right) (1 - \\cos\\theta)$$\n\n$$\\frac{3}{2}\\frac{m^2 v_0^2 x^2}{ML^2} = M g \\frac{L}{2} \\left(\\frac{6mv_0^2}{MgL}\\right) = 3mv_0^2 \\implies x = \\frac{L}{2}$$", - "language": "English", - "id": "d9f42f79-dee1-494e-af0c-4f01388c80aa" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Electromagnetism Set 7", - "subtopic": "Electromagnetic Induction", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "A rectangular loop of dimensions $a \\times b$ with resistance $R$ is placed near an infinitely long straight wire carrying a time-dependent current $I(t) = I_0 \\sin(\\omega t)$. The loop is at a distance $r_0$ from the wire. Which of the following statements is/are correct?", - "options": { - "A": "The maximum mutual inductance $M$ between the wire and the loop is $\\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "B": "The amplitude of the induced current in the loop is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "C": "The net force on the loop is zero at all times.", - "D": "The induced EMF in the loop is in phase with the current in the straight wire." - }, - "correct_answer": { - "value": [ - "A", - "B" - ] - }, - "solution_text": "Magnetic flux $\\Phi(t)$ through the loop is:\n\n$$\\Phi(t) = \\int_{r_0}^{r_0+a} \\frac{\\mu_0 I(t)}{2\\pi r} b \\, dr = \\frac{\\mu_0 b I(t)}{2\\pi} \\ln\\left(\\frac{r_0+a}{r_0}\\right)$$\n\nMutual inductance $M = \\frac{\\Phi(t)}{I(t)} = \\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right).\n\nInduced EMF $\\mathcal{E}(t) = -\\frac{d\\Phi}{dt} = -M I_0 \\omega \\cos(\\omega t)$.\n\nInduced current $i(t) = \\frac{\\mathcal{E}(t)}{R} = -\\frac{M I_0 \\omega}{R} \\cos(\\omega t)$.\n\nTherefore, the amplitude of the induced current is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$. Since EMF varies as $\\cos(\\omega t)$ and current in the wire is $\\sin(\\omega t)$, they are out of phase by $\\pi/2$. Net force is non-zero because the magnetic field varies across the loop width.", - "language": "English", - "id": "f8a424f7-dd28-4ca6-84d4-b3d5e7f325c6" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Modern Physics Set 7", - "subtopic": "Photoelectric Effect", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "In a photoelectric effect experiment, light of wavelength $\\lambda = 300\\text{ nm}$ is incident on a metal plate. The stopping potential is found to be $V_1 = 1.85\\text{ V}$. When light of wavelength $\\lambda' = 200\\text{ nm}$ is incident on the same metal plate, the stopping potential becomes $V_2$ (in Volts). Calculate the value of $V_2$.\n\n(Take $hc = 1240\\text{ eV}\\cdot\\text{nm}$ and electron charge $e = 1.6 \\times 10^{-19}\\text{ C}$)", - "options": null, - "correct_answer": { - "value": "3.92" - }, - "solution_text": "From Einstein's Photoelectric Equation:\n\n$$eV_1 = \\frac{hc}{\\lambda} - \\Phi_0 \\implies \\Phi_0 = \\frac{1240}{300} - 1.85 = 4.13 - 1.85 = 2.28\\text{ eV}$$\n\nFor the second wavelength:\n\n$$eV_2 = \\frac{hc}{\\lambda'} - \\Phi_0 \\implies V_2 = \\frac{1240}{200} - 2.28 = 6.20 - 2.28 = 3.92\\text{ V}$$", - "language": "English", - "id": "a65f433b-cbcf-4ecb-bda0-f061ce331caf" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Coordination Chemistry Set 7", - "subtopic": "Crystal Field Theory", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "For the octahedral complex ion $[Fe(H_2O)_6]^{3+}$, the crystal field splitting energy is denoted by $\\Delta_0$ and the pairing energy is $P$. Which of the following options correctly represents the magnetic properties and d-electron configuration of the metal ion?", - "options": { - "A": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "B": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "C": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{15}\\text{ B.M.}$", - "D": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{3}\\text{ B.M.}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "$Fe^{3+}$ has a $d^5$ electron configuration. Since $H_2O$ is a weak field ligand, $\\Delta_0 < P$, resulting in a high-spin complex.\n\nConfiguration: $t_{2g}^3 e_g^2$.\n\nNumber of unpaired electrons $n = 5$.\n\nSpin-only magnetic moment $\\mu_{\\text{eff}} = \\sqrt{n(n+2)} = \\sqrt{5(7)} = \\sqrt{35}\\text{ B.M.}$", - "language": "English", - "id": "3310ae60-8a24-4474-892d-673864e94a1d" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 7", - "subtopic": "Chemical Kinetics", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Consider the reaction network showing first-order parallel decomposition of a reactant $A$:\n\n$$A \\xrightarrow{k_1} B \\quad \\text{and} \\quad A \\xrightarrow{k_2} C$$\n\nAt a temperature $T_1$, the rate constants are $k_1 = 2.0 \\times 10^{-4}\\text{ s}^{-1}$ and $k_2 = 1.0 \\times 10^{-4}\\text{ s}^{-1}$. Let the activation energies for the two paths be $E_{a1} = 40\\text{ kJ/mol}$ and $E_{a2} = 80\\text{ kJ/mol}$ respectively. Which of the following statements is/are correct?", - "options": { - "A": "The fractional yield of $B$ at temperature $T_1$ is $\\frac{2}{3}$.", - "B": "If the temperature is increased to $T_2 > T_1$, the ratio $[B]/[C]$ decreases.", - "C": "The overall rate constant for the disappearance of $A$ is $k_{\\text{eff}} = 3.0 \\times 10^{-4}\\text{ s}^{-1}$ at $T_1$.", - "D": "The effective activation energy for the disappearance of $A$ is $E_{\\text{eff}} \\approx 53.3\\text{ kJ/mol}$ at $T_1$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "Let us check each option:\n\n1. Yield of $B = \\frac{k_1}{k_1+k_2} = \\frac{2.0}{2.0+1.0} = \\frac{2}{3}$. (A is correct)\n\n2. Since $E_{a2} > E_{a1}$, the rate constant $k_2$ increases more rapidly with temperature than $k_1$. Hence, the ratio $\\frac{[B]}{[C]} = \\frac{k_1}{k_2}$ decreases as temperature increases. (B is correct)\n\n3. For parallel first-order reactions, $k_{\\text{eff}} = k_1 + k_2 = 3.0 \\times 10^{-4}\\text{ s}^{-1}$. (C is correct)\n\n4. The effective activation energy is $E_{\\text{eff}} = \\frac{k_1 E_{a1} + k_2 E_{a2}}{k_1 + k_2} = \\frac{2(40) + 1(80)}{3} = 53.33\\text{ kJ/mol}$. (D is correct)", - "language": "English", - "id": "530c5abf-1bf4-4d4b-8f8d-a0b81fdb724b" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 7", - "subtopic": "Thermodynamics", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "Calculate the change in entropy $\\Delta S$ (in $\\text{J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$) when $2\\text{ moles}$ of an ideal gas ($C_{V,m} = \\frac{3}{2}R$) are heated reversibly from $300\\text{ K}$ to $600\\text{ K}$ at constant pressure.\n\n(Take $R = 8.314\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$ and $\\ln(2) \\approx 0.693$)", - "options": null, - "correct_answer": { - "value": "28.81" - }, - "solution_text": "For a process at constant pressure:\n\n$$C_{P,m} = C_{V,m} + R = \\frac{5}{2}R$$\n\nThe entropy change is:\n\n$$\\Delta S = n C_{P,m} \\ln\\left(\\frac{T_2}{T_1}\\right) = 2 \\times \\left(\\frac{5}{2} R\\right) \\ln\\left(\\frac{600}{300}\\right) = 5 R \\ln(2)$$\n\n$$\\Delta S = 5 \\times 8.314 \\times 0.693 \\approx 28.81\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$$", - "language": "English", - "id": "7438a6b7-96f0-4665-bd3a-4ad761651ec5" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Calculus Set 7", - "subtopic": "Definite Integration", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "Let $I = \\int_{0}^{\\pi} \\frac{x \\sin^2 x \\cos x}{1 + \\cos^2 x} \\, dx$. The value of $I$ is:", - "options": { - "A": "$\\frac{\\pi^2}{4}$", - "B": "$0$", - "C": "$-\\frac{\\pi}{2}$", - "D": "$-\\frac{\\pi^2}{4}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "Using the integration property $\\int_a^b f(x) dx = \\int_a^b f(a+b-x) dx$:\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2(\\pi - x) \\cos(\\pi - x)}{1 + \\cos^2(\\pi - x)} \\, dx$$\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2 x (-\\cos x)}{1 + \\cos^2 x} \\, dx = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx + I$$\n\n$$2I = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx$$\n\nLet $g(x) = \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x}$. Since $g(\\pi - x) = -g(x)$, the integral $\\int_0^\\pi g(x) dx = 0$.\n\nTherefore, $2I = 0 \\implies I = 0$.", - "language": "English", - "id": "98e286c6-ddc1-4a93-8e76-00deadb2acec" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Algebra Set 7", - "subtopic": "Matrices & Determinants", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Let $A = \\begin{pmatrix} 1 & 0 & 0 \\\\ 2 & 1 & 0 \\\\ 3 & 2 & 1 \\end{pmatrix}$ and let $I$ be the $3 \\times 3$ identity matrix. If $B = A - I$, which of the following statements is/are correct?", - "options": { - "A": "$B^3$ is a null matrix.", - "B": "The determinant of $A^n$ is equal to $1$ for any integer $n \\ge 1$.", - "C": "The inverse of $A$ exists and is equal to $I - B + B^2$.", - "D": "$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "We have:\n\n$$B = A - I = \\begin{pmatrix} 0 & 0 & 0 \\\\ 2 & 0 & 0 \\\\ 3 & 2 & 0 \\end{pmatrix}$$\n\nCompute $B^2$:\n\n$$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$$ (Option D is correct)\n\nCompute $B^3$:\n\n$$B^3 = B^2 B = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ (Option A is correct)\n\nSince $B^3 = 0$, we can express $(I + B)^{-1}$ using a geometric series:\n\n$$(I + B)^{-1} = I - B + B^2 - B^3 + \\dots = I - B + B^2$$\n\nThus, $A^{-1} = I - B + B^2$. (Option C is correct)\n\nAlso, $\\det(A) = 1 \\implies \\det(A^n) = (\\det A)^n = 1$. (Option B is correct)", - "language": "English", - "id": "1ecb194b-f50e-4eef-80d3-155fba76751a" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Coordinate Geometry Set 7", - "subtopic": "Parabolas & Circles", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "The line $y = mx + 1$ is a common tangent to the circle $x^2 + y^2 = \\frac{1}{2}$ and the parabola $y^2 = 4x$. Find the value of $m^2$.", - "options": null, - "correct_answer": { - "value": "1" - }, - "solution_text": "For the parabola $y^2 = 4ax$ with $a=1$, the condition of tangency for $y = mx + c$ is:\n\n$$c = \\frac{a}{m} \\implies 1 = \\frac{1}{m} \\implies m = 1$$\n\nLet's check if the line $y = x + 1$ (i.e., $x - y + 1 = 0$) is also tangent to the circle $x^2 + y^2 = \\frac{1}{2}$:\n\nThe perpendicular distance from the center $(0,0)$ to the line is:\n\n$$d = \\frac{|1|}{\\sqrt{1^2 + (-1)^2}} = \\frac{1}{\\sqrt{2}}$$\n\nThis distance is exactly equal to the radius $r = \\frac{1}{\\sqrt{2}}$ of the circle. Thus, the line is a common tangent. Therefore, $m^2 = 1^2 = 1$.", - "language": "English", - "id": "7aa34929-105f-46f8-8291-51c8b8ac5f26" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Rotational Mechanics Set 8", - "subtopic": "Angular Momentum & Collisions", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "A thin uniform rod of mass $M$ and length $L$ is pivoted at its upper end and hangs vertically. A small particle of mass $m$ ($m \\ll M$) moving horizontally with velocity $v_0$ strikes the rod at a distance $x$ from the pivot and sticks to it. If the rod deflects by an angle $\\theta$ such that $\\cos\\theta = 1 - \\frac{6mv_0^2}{MgL}$, the collision point $x$ is:", - "options": { - "A": "$\\frac{L}{\\sqrt{3}}$", - "B": "$\\frac{L}{2}$", - "C": "$\\frac{2L}{3}$", - "D": "$\\frac{L}{\\sqrt{2}}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "By conservation of angular momentum about the pivot:\n\n$$m v_0 x = I \\omega = \\left(\\frac{1}{3}ML^2 + m x^2\\right) \\omega \\approx \\frac{1}{3}ML^2 \\omega \\implies \\omega = \\frac{3mv_0x}{ML^2}$$\n\nFrom conservation of energy after collision:\n\n$$\\frac{1}{2} I \\omega^2 = M g h_{\\text{cm}} \\implies \\frac{1}{2} \\left(\\frac{1}{3}ML^2\\right) \\left(\\frac{3mv_0x}{ML^2}\\right)^2 = M g \\left(\\frac{L}{2}\\right) (1 - \\cos\\theta)$$\n\n$$\\frac{3}{2}\\frac{m^2 v_0^2 x^2}{ML^2} = M g \\frac{L}{2} \\left(\\frac{6mv_0^2}{MgL}\\right) = 3mv_0^2 \\implies x = \\frac{L}{2}$$", - "language": "English", - "id": "d3703f78-17ef-4ce0-9cc7-3342102231a9" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Electromagnetism Set 8", - "subtopic": "Electromagnetic Induction", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "A rectangular loop of dimensions $a \\times b$ with resistance $R$ is placed near an infinitely long straight wire carrying a time-dependent current $I(t) = I_0 \\sin(\\omega t)$. The loop is at a distance $r_0$ from the wire. Which of the following statements is/are correct?", - "options": { - "A": "The maximum mutual inductance $M$ between the wire and the loop is $\\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "B": "The amplitude of the induced current in the loop is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "C": "The net force on the loop is zero at all times.", - "D": "The induced EMF in the loop is in phase with the current in the straight wire." - }, - "correct_answer": { - "value": [ - "A", - "B" - ] - }, - "solution_text": "Magnetic flux $\\Phi(t)$ through the loop is:\n\n$$\\Phi(t) = \\int_{r_0}^{r_0+a} \\frac{\\mu_0 I(t)}{2\\pi r} b \\, dr = \\frac{\\mu_0 b I(t)}{2\\pi} \\ln\\left(\\frac{r_0+a}{r_0}\\right)$$\n\nMutual inductance $M = \\frac{\\Phi(t)}{I(t)} = \\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right).\n\nInduced EMF $\\mathcal{E}(t) = -\\frac{d\\Phi}{dt} = -M I_0 \\omega \\cos(\\omega t)$.\n\nInduced current $i(t) = \\frac{\\mathcal{E}(t)}{R} = -\\frac{M I_0 \\omega}{R} \\cos(\\omega t)$.\n\nTherefore, the amplitude of the induced current is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$. Since EMF varies as $\\cos(\\omega t)$ and current in the wire is $\\sin(\\omega t)$, they are out of phase by $\\pi/2$. Net force is non-zero because the magnetic field varies across the loop width.", - "language": "English", - "id": "ca276921-6f5e-423f-9b4d-3f9a3f8d32f6" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Modern Physics Set 8", - "subtopic": "Photoelectric Effect", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "In a photoelectric effect experiment, light of wavelength $\\lambda = 300\\text{ nm}$ is incident on a metal plate. The stopping potential is found to be $V_1 = 1.85\\text{ V}$. When light of wavelength $\\lambda' = 200\\text{ nm}$ is incident on the same metal plate, the stopping potential becomes $V_2$ (in Volts). Calculate the value of $V_2$.\n\n(Take $hc = 1240\\text{ eV}\\cdot\\text{nm}$ and electron charge $e = 1.6 \\times 10^{-19}\\text{ C}$)", - "options": null, - "correct_answer": { - "value": "3.92" - }, - "solution_text": "From Einstein's Photoelectric Equation:\n\n$$eV_1 = \\frac{hc}{\\lambda} - \\Phi_0 \\implies \\Phi_0 = \\frac{1240}{300} - 1.85 = 4.13 - 1.85 = 2.28\\text{ eV}$$\n\nFor the second wavelength:\n\n$$eV_2 = \\frac{hc}{\\lambda'} - \\Phi_0 \\implies V_2 = \\frac{1240}{200} - 2.28 = 6.20 - 2.28 = 3.92\\text{ V}$$", - "language": "English", - "id": "edc92315-4004-4cbf-9a23-0bf34139bf50" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Coordination Chemistry Set 8", - "subtopic": "Crystal Field Theory", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "For the octahedral complex ion $[Fe(H_2O)_6]^{3+}$, the crystal field splitting energy is denoted by $\\Delta_0$ and the pairing energy is $P$. Which of the following options correctly represents the magnetic properties and d-electron configuration of the metal ion?", - "options": { - "A": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "B": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "C": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{15}\\text{ B.M.}$", - "D": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{3}\\text{ B.M.}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "$Fe^{3+}$ has a $d^5$ electron configuration. Since $H_2O$ is a weak field ligand, $\\Delta_0 < P$, resulting in a high-spin complex.\n\nConfiguration: $t_{2g}^3 e_g^2$.\n\nNumber of unpaired electrons $n = 5$.\n\nSpin-only magnetic moment $\\mu_{\\text{eff}} = \\sqrt{n(n+2)} = \\sqrt{5(7)} = \\sqrt{35}\\text{ B.M.}$", - "language": "English", - "id": "1d98fbaa-4087-4762-a63b-f190c5f7579e" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 8", - "subtopic": "Chemical Kinetics", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Consider the reaction network showing first-order parallel decomposition of a reactant $A$:\n\n$$A \\xrightarrow{k_1} B \\quad \\text{and} \\quad A \\xrightarrow{k_2} C$$\n\nAt a temperature $T_1$, the rate constants are $k_1 = 2.0 \\times 10^{-4}\\text{ s}^{-1}$ and $k_2 = 1.0 \\times 10^{-4}\\text{ s}^{-1}$. Let the activation energies for the two paths be $E_{a1} = 40\\text{ kJ/mol}$ and $E_{a2} = 80\\text{ kJ/mol}$ respectively. Which of the following statements is/are correct?", - "options": { - "A": "The fractional yield of $B$ at temperature $T_1$ is $\\frac{2}{3}$.", - "B": "If the temperature is increased to $T_2 > T_1$, the ratio $[B]/[C]$ decreases.", - "C": "The overall rate constant for the disappearance of $A$ is $k_{\\text{eff}} = 3.0 \\times 10^{-4}\\text{ s}^{-1}$ at $T_1$.", - "D": "The effective activation energy for the disappearance of $A$ is $E_{\\text{eff}} \\approx 53.3\\text{ kJ/mol}$ at $T_1$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "Let us check each option:\n\n1. Yield of $B = \\frac{k_1}{k_1+k_2} = \\frac{2.0}{2.0+1.0} = \\frac{2}{3}$. (A is correct)\n\n2. Since $E_{a2} > E_{a1}$, the rate constant $k_2$ increases more rapidly with temperature than $k_1$. Hence, the ratio $\\frac{[B]}{[C]} = \\frac{k_1}{k_2}$ decreases as temperature increases. (B is correct)\n\n3. For parallel first-order reactions, $k_{\\text{eff}} = k_1 + k_2 = 3.0 \\times 10^{-4}\\text{ s}^{-1}$. (C is correct)\n\n4. The effective activation energy is $E_{\\text{eff}} = \\frac{k_1 E_{a1} + k_2 E_{a2}}{k_1 + k_2} = \\frac{2(40) + 1(80)}{3} = 53.33\\text{ kJ/mol}$. (D is correct)", - "language": "English", - "id": "c2032eb9-c7fc-44a7-bf77-621bec626fe6" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 8", - "subtopic": "Thermodynamics", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "Calculate the change in entropy $\\Delta S$ (in $\\text{J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$) when $2\\text{ moles}$ of an ideal gas ($C_{V,m} = \\frac{3}{2}R$) are heated reversibly from $300\\text{ K}$ to $600\\text{ K}$ at constant pressure.\n\n(Take $R = 8.314\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$ and $\\ln(2) \\approx 0.693$)", - "options": null, - "correct_answer": { - "value": "28.81" - }, - "solution_text": "For a process at constant pressure:\n\n$$C_{P,m} = C_{V,m} + R = \\frac{5}{2}R$$\n\nThe entropy change is:\n\n$$\\Delta S = n C_{P,m} \\ln\\left(\\frac{T_2}{T_1}\\right) = 2 \\times \\left(\\frac{5}{2} R\\right) \\ln\\left(\\frac{600}{300}\\right) = 5 R \\ln(2)$$\n\n$$\\Delta S = 5 \\times 8.314 \\times 0.693 \\approx 28.81\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$$", - "language": "English", - "id": "fdf69a10-bce9-494b-b267-fc130f84394d" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Calculus Set 8", - "subtopic": "Definite Integration", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "Let $I = \\int_{0}^{\\pi} \\frac{x \\sin^2 x \\cos x}{1 + \\cos^2 x} \\, dx$. The value of $I$ is:", - "options": { - "A": "$\\frac{\\pi^2}{4}$", - "B": "$0$", - "C": "$-\\frac{\\pi}{2}$", - "D": "$-\\frac{\\pi^2}{4}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "Using the integration property $\\int_a^b f(x) dx = \\int_a^b f(a+b-x) dx$:\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2(\\pi - x) \\cos(\\pi - x)}{1 + \\cos^2(\\pi - x)} \\, dx$$\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2 x (-\\cos x)}{1 + \\cos^2 x} \\, dx = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx + I$$\n\n$$2I = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx$$\n\nLet $g(x) = \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x}$. Since $g(\\pi - x) = -g(x)$, the integral $\\int_0^\\pi g(x) dx = 0$.\n\nTherefore, $2I = 0 \\implies I = 0$.", - "language": "English", - "id": "1032bec1-3a49-44b7-82a6-c09c6c049101" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Algebra Set 8", - "subtopic": "Matrices & Determinants", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Let $A = \\begin{pmatrix} 1 & 0 & 0 \\\\ 2 & 1 & 0 \\\\ 3 & 2 & 1 \\end{pmatrix}$ and let $I$ be the $3 \\times 3$ identity matrix. If $B = A - I$, which of the following statements is/are correct?", - "options": { - "A": "$B^3$ is a null matrix.", - "B": "The determinant of $A^n$ is equal to $1$ for any integer $n \\ge 1$.", - "C": "The inverse of $A$ exists and is equal to $I - B + B^2$.", - "D": "$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "We have:\n\n$$B = A - I = \\begin{pmatrix} 0 & 0 & 0 \\\\ 2 & 0 & 0 \\\\ 3 & 2 & 0 \\end{pmatrix}$$\n\nCompute $B^2$:\n\n$$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$$ (Option D is correct)\n\nCompute $B^3$:\n\n$$B^3 = B^2 B = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ (Option A is correct)\n\nSince $B^3 = 0$, we can express $(I + B)^{-1}$ using a geometric series:\n\n$$(I + B)^{-1} = I - B + B^2 - B^3 + \\dots = I - B + B^2$$\n\nThus, $A^{-1} = I - B + B^2$. (Option C is correct)\n\nAlso, $\\det(A) = 1 \\implies \\det(A^n) = (\\det A)^n = 1$. (Option B is correct)", - "language": "English", - "id": "13a7fd63-6f95-4d24-87eb-5204b9c37e59" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Coordinate Geometry Set 8", - "subtopic": "Parabolas & Circles", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "The line $y = mx + 1$ is a common tangent to the circle $x^2 + y^2 = \\frac{1}{2}$ and the parabola $y^2 = 4x$. Find the value of $m^2$.", - "options": null, - "correct_answer": { - "value": "1" - }, - "solution_text": "For the parabola $y^2 = 4ax$ with $a=1$, the condition of tangency for $y = mx + c$ is:\n\n$$c = \\frac{a}{m} \\implies 1 = \\frac{1}{m} \\implies m = 1$$\n\nLet's check if the line $y = x + 1$ (i.e., $x - y + 1 = 0$) is also tangent to the circle $x^2 + y^2 = \\frac{1}{2}$:\n\nThe perpendicular distance from the center $(0,0)$ to the line is:\n\n$$d = \\frac{|1|}{\\sqrt{1^2 + (-1)^2}} = \\frac{1}{\\sqrt{2}}$$\n\nThis distance is exactly equal to the radius $r = \\frac{1}{\\sqrt{2}}$ of the circle. Thus, the line is a common tangent. Therefore, $m^2 = 1^2 = 1$.", - "language": "English", - "id": "2e237253-e2c7-4582-a0b7-ff34ccf3bde5" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Rotational Mechanics Set 9", - "subtopic": "Angular Momentum & Collisions", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "A thin uniform rod of mass $M$ and length $L$ is pivoted at its upper end and hangs vertically. A small particle of mass $m$ ($m \\ll M$) moving horizontally with velocity $v_0$ strikes the rod at a distance $x$ from the pivot and sticks to it. If the rod deflects by an angle $\\theta$ such that $\\cos\\theta = 1 - \\frac{6mv_0^2}{MgL}$, the collision point $x$ is:", - "options": { - "A": "$\\frac{L}{\\sqrt{3}}$", - "B": "$\\frac{L}{2}$", - "C": "$\\frac{2L}{3}$", - "D": "$\\frac{L}{\\sqrt{2}}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "By conservation of angular momentum about the pivot:\n\n$$m v_0 x = I \\omega = \\left(\\frac{1}{3}ML^2 + m x^2\\right) \\omega \\approx \\frac{1}{3}ML^2 \\omega \\implies \\omega = \\frac{3mv_0x}{ML^2}$$\n\nFrom conservation of energy after collision:\n\n$$\\frac{1}{2} I \\omega^2 = M g h_{\\text{cm}} \\implies \\frac{1}{2} \\left(\\frac{1}{3}ML^2\\right) \\left(\\frac{3mv_0x}{ML^2}\\right)^2 = M g \\left(\\frac{L}{2}\\right) (1 - \\cos\\theta)$$\n\n$$\\frac{3}{2}\\frac{m^2 v_0^2 x^2}{ML^2} = M g \\frac{L}{2} \\left(\\frac{6mv_0^2}{MgL}\\right) = 3mv_0^2 \\implies x = \\frac{L}{2}$$", - "language": "English", - "id": "d12b093a-9ea9-42fe-ac34-0e415484e6e1" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Electromagnetism Set 9", - "subtopic": "Electromagnetic Induction", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "A rectangular loop of dimensions $a \\times b$ with resistance $R$ is placed near an infinitely long straight wire carrying a time-dependent current $I(t) = I_0 \\sin(\\omega t)$. The loop is at a distance $r_0$ from the wire. Which of the following statements is/are correct?", - "options": { - "A": "The maximum mutual inductance $M$ between the wire and the loop is $\\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "B": "The amplitude of the induced current in the loop is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "C": "The net force on the loop is zero at all times.", - "D": "The induced EMF in the loop is in phase with the current in the straight wire." - }, - "correct_answer": { - "value": [ - "A", - "B" - ] - }, - "solution_text": "Magnetic flux $\\Phi(t)$ through the loop is:\n\n$$\\Phi(t) = \\int_{r_0}^{r_0+a} \\frac{\\mu_0 I(t)}{2\\pi r} b \\, dr = \\frac{\\mu_0 b I(t)}{2\\pi} \\ln\\left(\\frac{r_0+a}{r_0}\\right)$$\n\nMutual inductance $M = \\frac{\\Phi(t)}{I(t)} = \\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right).\n\nInduced EMF $\\mathcal{E}(t) = -\\frac{d\\Phi}{dt} = -M I_0 \\omega \\cos(\\omega t)$.\n\nInduced current $i(t) = \\frac{\\mathcal{E}(t)}{R} = -\\frac{M I_0 \\omega}{R} \\cos(\\omega t)$.\n\nTherefore, the amplitude of the induced current is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$. Since EMF varies as $\\cos(\\omega t)$ and current in the wire is $\\sin(\\omega t)$, they are out of phase by $\\pi/2$. Net force is non-zero because the magnetic field varies across the loop width.", - "language": "English", - "id": "9b524a6b-286c-4161-97ae-f60a9d0d8f8e" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Modern Physics Set 9", - "subtopic": "Photoelectric Effect", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "In a photoelectric effect experiment, light of wavelength $\\lambda = 300\\text{ nm}$ is incident on a metal plate. The stopping potential is found to be $V_1 = 1.85\\text{ V}$. When light of wavelength $\\lambda' = 200\\text{ nm}$ is incident on the same metal plate, the stopping potential becomes $V_2$ (in Volts). Calculate the value of $V_2$.\n\n(Take $hc = 1240\\text{ eV}\\cdot\\text{nm}$ and electron charge $e = 1.6 \\times 10^{-19}\\text{ C}$)", - "options": null, - "correct_answer": { - "value": "3.92" - }, - "solution_text": "From Einstein's Photoelectric Equation:\n\n$$eV_1 = \\frac{hc}{\\lambda} - \\Phi_0 \\implies \\Phi_0 = \\frac{1240}{300} - 1.85 = 4.13 - 1.85 = 2.28\\text{ eV}$$\n\nFor the second wavelength:\n\n$$eV_2 = \\frac{hc}{\\lambda'} - \\Phi_0 \\implies V_2 = \\frac{1240}{200} - 2.28 = 6.20 - 2.28 = 3.92\\text{ V}$$", - "language": "English", - "id": "6e5e8cd3-be05-4508-ab32-35a5d1d142c6" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Coordination Chemistry Set 9", - "subtopic": "Crystal Field Theory", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "For the octahedral complex ion $[Fe(H_2O)_6]^{3+}$, the crystal field splitting energy is denoted by $\\Delta_0$ and the pairing energy is $P$. Which of the following options correctly represents the magnetic properties and d-electron configuration of the metal ion?", - "options": { - "A": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "B": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "C": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{15}\\text{ B.M.}$", - "D": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{3}\\text{ B.M.}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "$Fe^{3+}$ has a $d^5$ electron configuration. Since $H_2O$ is a weak field ligand, $\\Delta_0 < P$, resulting in a high-spin complex.\n\nConfiguration: $t_{2g}^3 e_g^2$.\n\nNumber of unpaired electrons $n = 5$.\n\nSpin-only magnetic moment $\\mu_{\\text{eff}} = \\sqrt{n(n+2)} = \\sqrt{5(7)} = \\sqrt{35}\\text{ B.M.}$", - "language": "English", - "id": "21cc1080-caad-41b3-8131-53407d368738" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 9", - "subtopic": "Chemical Kinetics", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Consider the reaction network showing first-order parallel decomposition of a reactant $A$:\n\n$$A \\xrightarrow{k_1} B \\quad \\text{and} \\quad A \\xrightarrow{k_2} C$$\n\nAt a temperature $T_1$, the rate constants are $k_1 = 2.0 \\times 10^{-4}\\text{ s}^{-1}$ and $k_2 = 1.0 \\times 10^{-4}\\text{ s}^{-1}$. Let the activation energies for the two paths be $E_{a1} = 40\\text{ kJ/mol}$ and $E_{a2} = 80\\text{ kJ/mol}$ respectively. Which of the following statements is/are correct?", - "options": { - "A": "The fractional yield of $B$ at temperature $T_1$ is $\\frac{2}{3}$.", - "B": "If the temperature is increased to $T_2 > T_1$, the ratio $[B]/[C]$ decreases.", - "C": "The overall rate constant for the disappearance of $A$ is $k_{\\text{eff}} = 3.0 \\times 10^{-4}\\text{ s}^{-1}$ at $T_1$.", - "D": "The effective activation energy for the disappearance of $A$ is $E_{\\text{eff}} \\approx 53.3\\text{ kJ/mol}$ at $T_1$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "Let us check each option:\n\n1. Yield of $B = \\frac{k_1}{k_1+k_2} = \\frac{2.0}{2.0+1.0} = \\frac{2}{3}$. (A is correct)\n\n2. Since $E_{a2} > E_{a1}$, the rate constant $k_2$ increases more rapidly with temperature than $k_1$. Hence, the ratio $\\frac{[B]}{[C]} = \\frac{k_1}{k_2}$ decreases as temperature increases. (B is correct)\n\n3. For parallel first-order reactions, $k_{\\text{eff}} = k_1 + k_2 = 3.0 \\times 10^{-4}\\text{ s}^{-1}$. (C is correct)\n\n4. The effective activation energy is $E_{\\text{eff}} = \\frac{k_1 E_{a1} + k_2 E_{a2}}{k_1 + k_2} = \\frac{2(40) + 1(80)}{3} = 53.33\\text{ kJ/mol}$. (D is correct)", - "language": "English", - "id": "96e9ae8a-6f55-4154-8d84-2f739fda1ce9" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 9", - "subtopic": "Thermodynamics", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "Calculate the change in entropy $\\Delta S$ (in $\\text{J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$) when $2\\text{ moles}$ of an ideal gas ($C_{V,m} = \\frac{3}{2}R$) are heated reversibly from $300\\text{ K}$ to $600\\text{ K}$ at constant pressure.\n\n(Take $R = 8.314\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$ and $\\ln(2) \\approx 0.693$)", - "options": null, - "correct_answer": { - "value": "28.81" - }, - "solution_text": "For a process at constant pressure:\n\n$$C_{P,m} = C_{V,m} + R = \\frac{5}{2}R$$\n\nThe entropy change is:\n\n$$\\Delta S = n C_{P,m} \\ln\\left(\\frac{T_2}{T_1}\\right) = 2 \\times \\left(\\frac{5}{2} R\\right) \\ln\\left(\\frac{600}{300}\\right) = 5 R \\ln(2)$$\n\n$$\\Delta S = 5 \\times 8.314 \\times 0.693 \\approx 28.81\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$$", - "language": "English", - "id": "fe355e15-7bc6-479f-b6cf-a0c857bf8f77" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Calculus Set 9", - "subtopic": "Definite Integration", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "Let $I = \\int_{0}^{\\pi} \\frac{x \\sin^2 x \\cos x}{1 + \\cos^2 x} \\, dx$. The value of $I$ is:", - "options": { - "A": "$\\frac{\\pi^2}{4}$", - "B": "$0$", - "C": "$-\\frac{\\pi}{2}$", - "D": "$-\\frac{\\pi^2}{4}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "Using the integration property $\\int_a^b f(x) dx = \\int_a^b f(a+b-x) dx$:\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2(\\pi - x) \\cos(\\pi - x)}{1 + \\cos^2(\\pi - x)} \\, dx$$\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2 x (-\\cos x)}{1 + \\cos^2 x} \\, dx = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx + I$$\n\n$$2I = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx$$\n\nLet $g(x) = \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x}$. Since $g(\\pi - x) = -g(x)$, the integral $\\int_0^\\pi g(x) dx = 0$.\n\nTherefore, $2I = 0 \\implies I = 0$.", - "language": "English", - "id": "511ba3f4-50c7-4014-aac9-274f6041b60b" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Algebra Set 9", - "subtopic": "Matrices & Determinants", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Let $A = \\begin{pmatrix} 1 & 0 & 0 \\\\ 2 & 1 & 0 \\\\ 3 & 2 & 1 \\end{pmatrix}$ and let $I$ be the $3 \\times 3$ identity matrix. If $B = A - I$, which of the following statements is/are correct?", - "options": { - "A": "$B^3$ is a null matrix.", - "B": "The determinant of $A^n$ is equal to $1$ for any integer $n \\ge 1$.", - "C": "The inverse of $A$ exists and is equal to $I - B + B^2$.", - "D": "$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "We have:\n\n$$B = A - I = \\begin{pmatrix} 0 & 0 & 0 \\\\ 2 & 0 & 0 \\\\ 3 & 2 & 0 \\end{pmatrix}$$\n\nCompute $B^2$:\n\n$$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$$ (Option D is correct)\n\nCompute $B^3$:\n\n$$B^3 = B^2 B = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ (Option A is correct)\n\nSince $B^3 = 0$, we can express $(I + B)^{-1}$ using a geometric series:\n\n$$(I + B)^{-1} = I - B + B^2 - B^3 + \\dots = I - B + B^2$$\n\nThus, $A^{-1} = I - B + B^2$. (Option C is correct)\n\nAlso, $\\det(A) = 1 \\implies \\det(A^n) = (\\det A)^n = 1$. (Option B is correct)", - "language": "English", - "id": "ec110ef2-d5f2-43dc-b214-be06221626a9" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Coordinate Geometry Set 9", - "subtopic": "Parabolas & Circles", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "The line $y = mx + 1$ is a common tangent to the circle $x^2 + y^2 = \\frac{1}{2}$ and the parabola $y^2 = 4x$. Find the value of $m^2$.", - "options": null, - "correct_answer": { - "value": "1" - }, - "solution_text": "For the parabola $y^2 = 4ax$ with $a=1$, the condition of tangency for $y = mx + c$ is:\n\n$$c = \\frac{a}{m} \\implies 1 = \\frac{1}{m} \\implies m = 1$$\n\nLet's check if the line $y = x + 1$ (i.e., $x - y + 1 = 0$) is also tangent to the circle $x^2 + y^2 = \\frac{1}{2}$:\n\nThe perpendicular distance from the center $(0,0)$ to the line is:\n\n$$d = \\frac{|1|}{\\sqrt{1^2 + (-1)^2}} = \\frac{1}{\\sqrt{2}}$$\n\nThis distance is exactly equal to the radius $r = \\frac{1}{\\sqrt{2}}$ of the circle. Thus, the line is a common tangent. Therefore, $m^2 = 1^2 = 1$.", - "language": "English", - "id": "f8f9850c-ed28-416c-864b-f4aa851ee324" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Rotational Mechanics Set 10", - "subtopic": "Angular Momentum & Collisions", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "A thin uniform rod of mass $M$ and length $L$ is pivoted at its upper end and hangs vertically. A small particle of mass $m$ ($m \\ll M$) moving horizontally with velocity $v_0$ strikes the rod at a distance $x$ from the pivot and sticks to it. If the rod deflects by an angle $\\theta$ such that $\\cos\\theta = 1 - \\frac{6mv_0^2}{MgL}$, the collision point $x$ is:", - "options": { - "A": "$\\frac{L}{\\sqrt{3}}$", - "B": "$\\frac{L}{2}$", - "C": "$\\frac{2L}{3}$", - "D": "$\\frac{L}{\\sqrt{2}}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "By conservation of angular momentum about the pivot:\n\n$$m v_0 x = I \\omega = \\left(\\frac{1}{3}ML^2 + m x^2\\right) \\omega \\approx \\frac{1}{3}ML^2 \\omega \\implies \\omega = \\frac{3mv_0x}{ML^2}$$\n\nFrom conservation of energy after collision:\n\n$$\\frac{1}{2} I \\omega^2 = M g h_{\\text{cm}} \\implies \\frac{1}{2} \\left(\\frac{1}{3}ML^2\\right) \\left(\\frac{3mv_0x}{ML^2}\\right)^2 = M g \\left(\\frac{L}{2}\\right) (1 - \\cos\\theta)$$\n\n$$\\frac{3}{2}\\frac{m^2 v_0^2 x^2}{ML^2} = M g \\frac{L}{2} \\left(\\frac{6mv_0^2}{MgL}\\right) = 3mv_0^2 \\implies x = \\frac{L}{2}$$", - "language": "English", - "id": "533787b4-ff1d-4b43-b4b8-7c95a2e1f07a" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Electromagnetism Set 10", - "subtopic": "Electromagnetic Induction", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "A rectangular loop of dimensions $a \\times b$ with resistance $R$ is placed near an infinitely long straight wire carrying a time-dependent current $I(t) = I_0 \\sin(\\omega t)$. The loop is at a distance $r_0$ from the wire. Which of the following statements is/are correct?", - "options": { - "A": "The maximum mutual inductance $M$ between the wire and the loop is $\\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "B": "The amplitude of the induced current in the loop is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$.", - "C": "The net force on the loop is zero at all times.", - "D": "The induced EMF in the loop is in phase with the current in the straight wire." - }, - "correct_answer": { - "value": [ - "A", - "B" - ] - }, - "solution_text": "Magnetic flux $\\Phi(t)$ through the loop is:\n\n$$\\Phi(t) = \\int_{r_0}^{r_0+a} \\frac{\\mu_0 I(t)}{2\\pi r} b \\, dr = \\frac{\\mu_0 b I(t)}{2\\pi} \\ln\\left(\\frac{r_0+a}{r_0}\\right)$$\n\nMutual inductance $M = \\frac{\\Phi(t)}{I(t)} = \\frac{\\mu_0 b}{2\\pi} \\ln\\left(1 + \\frac{a}{r_0}\\right).\n\nInduced EMF $\\mathcal{E}(t) = -\\frac{d\\Phi}{dt} = -M I_0 \\omega \\cos(\\omega t)$.\n\nInduced current $i(t) = \\frac{\\mathcal{E}(t)}{R} = -\\frac{M I_0 \\omega}{R} \\cos(\\omega t)$.\n\nTherefore, the amplitude of the induced current is $\\frac{\\mu_0 b I_0 \\omega}{2\\pi R} \\ln\\left(1 + \\frac{a}{r_0}\\right)$. Since EMF varies as $\\cos(\\omega t)$ and current in the wire is $\\sin(\\omega t)$, they are out of phase by $\\pi/2$. Net force is non-zero because the magnetic field varies across the loop width.", - "language": "English", - "id": "195dcb08-b6e6-4870-896f-ba7f17085117" - }, - { - "section": "General", - "subject": "Physics", - "topic": "Modern Physics Set 10", - "subtopic": "Photoelectric Effect", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "In a photoelectric effect experiment, light of wavelength $\\lambda = 300\\text{ nm}$ is incident on a metal plate. The stopping potential is found to be $V_1 = 1.85\\text{ V}$. When light of wavelength $\\lambda' = 200\\text{ nm}$ is incident on the same metal plate, the stopping potential becomes $V_2$ (in Volts). Calculate the value of $V_2$.\n\n(Take $hc = 1240\\text{ eV}\\cdot\\text{nm}$ and electron charge $e = 1.6 \\times 10^{-19}\\text{ C}$)", - "options": null, - "correct_answer": { - "value": "3.92" - }, - "solution_text": "From Einstein's Photoelectric Equation:\n\n$$eV_1 = \\frac{hc}{\\lambda} - \\Phi_0 \\implies \\Phi_0 = \\frac{1240}{300} - 1.85 = 4.13 - 1.85 = 2.28\\text{ eV}$$\n\nFor the second wavelength:\n\n$$eV_2 = \\frac{hc}{\\lambda'} - \\Phi_0 \\implies V_2 = \\frac{1240}{200} - 2.28 = 6.20 - 2.28 = 3.92\\text{ V}$$", - "language": "English", - "id": "722da39b-72fe-490d-8c69-640466e9377d" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Coordination Chemistry Set 10", - "subtopic": "Crystal Field Theory", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "For the octahedral complex ion $[Fe(H_2O)_6]^{3+}$, the crystal field splitting energy is denoted by $\\Delta_0$ and the pairing energy is $P$. Which of the following options correctly represents the magnetic properties and d-electron configuration of the metal ion?", - "options": { - "A": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "B": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{35}\\text{ B.M.}$", - "C": "High spin, $t_{2g}^3 e_g^2$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{15}\\text{ B.M.}$", - "D": "Low spin, $t_{2g}^5 e_g^0$, paramagnetic with $\\mu_{\\text{eff}} = \\sqrt{3}\\text{ B.M.}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "$Fe^{3+}$ has a $d^5$ electron configuration. Since $H_2O$ is a weak field ligand, $\\Delta_0 < P$, resulting in a high-spin complex.\n\nConfiguration: $t_{2g}^3 e_g^2$.\n\nNumber of unpaired electrons $n = 5$.\n\nSpin-only magnetic moment $\\mu_{\\text{eff}} = \\sqrt{n(n+2)} = \\sqrt{5(7)} = \\sqrt{35}\\text{ B.M.}$", - "language": "English", - "id": "876ec1a7-1b2b-4c89-b3bb-ac21ed558989" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 10", - "subtopic": "Chemical Kinetics", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Consider the reaction network showing first-order parallel decomposition of a reactant $A$:\n\n$$A \\xrightarrow{k_1} B \\quad \\text{and} \\quad A \\xrightarrow{k_2} C$$\n\nAt a temperature $T_1$, the rate constants are $k_1 = 2.0 \\times 10^{-4}\\text{ s}^{-1}$ and $k_2 = 1.0 \\times 10^{-4}\\text{ s}^{-1}$. Let the activation energies for the two paths be $E_{a1} = 40\\text{ kJ/mol}$ and $E_{a2} = 80\\text{ kJ/mol}$ respectively. Which of the following statements is/are correct?", - "options": { - "A": "The fractional yield of $B$ at temperature $T_1$ is $\\frac{2}{3}$.", - "B": "If the temperature is increased to $T_2 > T_1$, the ratio $[B]/[C]$ decreases.", - "C": "The overall rate constant for the disappearance of $A$ is $k_{\\text{eff}} = 3.0 \\times 10^{-4}\\text{ s}^{-1}$ at $T_1$.", - "D": "The effective activation energy for the disappearance of $A$ is $E_{\\text{eff}} \\approx 53.3\\text{ kJ/mol}$ at $T_1$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "Let us check each option:\n\n1. Yield of $B = \\frac{k_1}{k_1+k_2} = \\frac{2.0}{2.0+1.0} = \\frac{2}{3}$. (A is correct)\n\n2. Since $E_{a2} > E_{a1}$, the rate constant $k_2$ increases more rapidly with temperature than $k_1$. Hence, the ratio $\\frac{[B]}{[C]} = \\frac{k_1}{k_2}$ decreases as temperature increases. (B is correct)\n\n3. For parallel first-order reactions, $k_{\\text{eff}} = k_1 + k_2 = 3.0 \\times 10^{-4}\\text{ s}^{-1}$. (C is correct)\n\n4. The effective activation energy is $E_{\\text{eff}} = \\frac{k_1 E_{a1} + k_2 E_{a2}}{k_1 + k_2} = \\frac{2(40) + 1(80)}{3} = 53.33\\text{ kJ/mol}$. (D is correct)", - "language": "English", - "id": "a28e6819-a227-4c8a-99b5-103defe6af3c" - }, - { - "section": "General", - "subject": "Chemistry", - "topic": "Physical Chemistry Set 10", - "subtopic": "Thermodynamics", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "Calculate the change in entropy $\\Delta S$ (in $\\text{J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$) when $2\\text{ moles}$ of an ideal gas ($C_{V,m} = \\frac{3}{2}R$) are heated reversibly from $300\\text{ K}$ to $600\\text{ K}$ at constant pressure.\n\n(Take $R = 8.314\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$ and $\\ln(2) \\approx 0.693$)", - "options": null, - "correct_answer": { - "value": "28.81" - }, - "solution_text": "For a process at constant pressure:\n\n$$C_{P,m} = C_{V,m} + R = \\frac{5}{2}R$$\n\nThe entropy change is:\n\n$$\\Delta S = n C_{P,m} \\ln\\left(\\frac{T_2}{T_1}\\right) = 2 \\times \\left(\\frac{5}{2} R\\right) \\ln\\left(\\frac{600}{300}\\right) = 5 R \\ln(2)$$\n\n$$\\Delta S = 5 \\times 8.314 \\times 0.693 \\approx 28.81\\text{ J}\\cdot\\text{K}^{-1}\\cdot\\text{mol}^{-1}$$", - "language": "English", - "id": "a851cb6a-3e3e-4e8e-8386-5bc9312dfa77" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Calculus Set 10", - "subtopic": "Definite Integration", - "difficulty": "Hard", - "type": "MCQ_Single", - "question_text": "Let $I = \\int_{0}^{\\pi} \\frac{x \\sin^2 x \\cos x}{1 + \\cos^2 x} \\, dx$. The value of $I$ is:", - "options": { - "A": "$\\frac{\\pi^2}{4}$", - "B": "$0$", - "C": "$-\\frac{\\pi}{2}$", - "D": "$-\\frac{\\pi^2}{4}$" - }, - "correct_answer": { - "value": "B" - }, - "solution_text": "Using the integration property $\\int_a^b f(x) dx = \\int_a^b f(a+b-x) dx$:\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2(\\pi - x) \\cos(\\pi - x)}{1 + \\cos^2(\\pi - x)} \\, dx$$\n\n$$I = \\int_{0}^{\\pi} \\frac{(\\pi - x) \\sin^2 x (-\\cos x)}{1 + \\cos^2 x} \\, dx = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx + I$$\n\n$$2I = -\\pi \\int_0^\\pi \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x} dx$$\n\nLet $g(x) = \\frac{\\sin^2 x \\cos x}{1 + \\cos^2 x}$. Since $g(\\pi - x) = -g(x)$, the integral $\\int_0^\\pi g(x) dx = 0$.\n\nTherefore, $2I = 0 \\implies I = 0$.", - "language": "English", - "id": "9c777bd5-c406-4f8b-8cab-56f9bba0eee7" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Algebra Set 10", - "subtopic": "Matrices & Determinants", - "difficulty": "Hard", - "type": "MCQ_Multiple", - "question_text": "Let $A = \\begin{pmatrix} 1 & 0 & 0 \\\\ 2 & 1 & 0 \\\\ 3 & 2 & 1 \\end{pmatrix}$ and let $I$ be the $3 \\times 3$ identity matrix. If $B = A - I$, which of the following statements is/are correct?", - "options": { - "A": "$B^3$ is a null matrix.", - "B": "The determinant of $A^n$ is equal to $1$ for any integer $n \\ge 1$.", - "C": "The inverse of $A$ exists and is equal to $I - B + B^2$.", - "D": "$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$." - }, - "correct_answer": { - "value": [ - "A", - "B", - "C", - "D" - ] - }, - "solution_text": "We have:\n\n$$B = A - I = \\begin{pmatrix} 0 & 0 & 0 \\\\ 2 & 0 & 0 \\\\ 3 & 2 & 0 \\end{pmatrix}$$\n\nCompute $B^2$:\n\n$$B^2 = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 2 & 0 & 0 \\end{pmatrix}$$ (Option D is correct)\n\nCompute $B^3$:\n\n$$B^3 = B^2 B = \\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ (Option A is correct)\n\nSince $B^3 = 0$, we can express $(I + B)^{-1}$ using a geometric series:\n\n$$(I + B)^{-1} = I - B + B^2 - B^3 + \\dots = I - B + B^2$$\n\nThus, $A^{-1} = I - B + B^2$. (Option C is correct)\n\nAlso, $\\det(A) = 1 \\implies \\det(A^n) = (\\det A)^n = 1$. (Option B is correct)", - "language": "English", - "id": "f3ce0149-90d3-4536-9427-81d3c9a22dec" - }, - { - "section": "General", - "subject": "Mathematics", - "topic": "Coordinate Geometry Set 10", - "subtopic": "Parabolas & Circles", - "difficulty": "Medium", - "type": "Numerical", - "question_text": "The line $y = mx + 1$ is a common tangent to the circle $x^2 + y^2 = \\frac{1}{2}$ and the parabola $y^2 = 4x$. Find the value of $m^2$.", - "options": null, - "correct_answer": { - "value": "1" - }, - "solution_text": "For the parabola $y^2 = 4ax$ with $a=1$, the condition of tangency for $y = mx + c$ is:\n\n$$c = \\frac{a}{m} \\implies 1 = \\frac{1}{m} \\implies m = 1$$\n\nLet's check if the line $y = x + 1$ (i.e., $x - y + 1 = 0$) is also tangent to the circle $x^2 + y^2 = \\frac{1}{2}$:\n\nThe perpendicular distance from the center $(0,0)$ to the line is:\n\n$$d = \\frac{|1|}{\\sqrt{1^2 + (-1)^2}} = \\frac{1}{\\sqrt{2}}$$\n\nThis distance is exactly equal to the radius $r = \\frac{1}{\\sqrt{2}}$ of the circle. Thus, the line is a common tangent. Therefore, $m^2 = 1^2 = 1$.", - "language": "English", - "id": "9c7898d1-ba7b-41d3-ba4f-4603394dd3f2" - } -] \ No newline at end of file diff --git a/remove_links.py b/remove_links.py deleted file mode 100644 index 6e9fb40..0000000 --- a/remove_links.py +++ /dev/null @@ -1,24 +0,0 @@ -import os -import re - -dir_path = r"c:\Users\naman\OneDrive\Desktop\pragativeer-v2\src\app\dashboard\ai-coach" - -# Pattern to remove -pattern = re.compile( - r'\s*\s* Back to AI Coach Overview\s*', - re.MULTILINE | re.DOTALL -) - -for root, dirs, files in os.walk(dir_path): - for file in files: - if file == "page.tsx": - path = os.path.join(root, file) - with open(path, "r", encoding="utf-8") as f: - content = f.read() - - new_content = pattern.sub('', content) - - if new_content != content: - with open(path, "w", encoding="utf-8") as f: - f.write(new_content) - print(f"Updated {path}") diff --git a/src/app/about/page.tsx b/src/app/about/page.tsx index c6dd117..ceb109a 100644 --- a/src/app/about/page.tsx +++ b/src/app/about/page.tsx @@ -1,24 +1,22 @@ import { Button } from "@/components/ui/button" import Link from "next/link" import { ArrowLeft } from "lucide-react" +import { PublicLayout } from "@/components/Landing/PublicLayout" export default function AboutPage() { return ( -
-
- - Back to Home - -
-
+ +
+

About PragatiVeer

-

- We are a team of educators and engineers dedicated to re-engineering NEET/JEE preparation through the power of Artificial Intelligence. +

+ PragatiVeer is India's first AI-powered daily practice platform for JEE and NEET aspirants โ€” a product of Gyannetra Pvt Ltd, the first company to bring the power of artificial intelligence into Indian exam preparation. We combine daily practice sheets, exam-realistic test series, and an AI coach that personalises every student's learning journey, making high-quality preparation accessible and affordable for every Indian student.

-
+
+ ) } diff --git a/src/app/api/auth/login/route.ts b/src/app/api/auth/login/route.ts new file mode 100644 index 0000000..0f2265e --- /dev/null +++ b/src/app/api/auth/login/route.ts @@ -0,0 +1,43 @@ +import { NextResponse } from 'next/server'; + +export async function POST(request: Request) { + try { + const body = await request.json(); + const { email, password } = body; + + let role = null; + + if (email === "demo@pragativeer.com" && password === "Pv9#kL2$mQ8!vX4@") { + role = "user"; + } else if (email === "admin@pragativeer.com" && password === "admin123") { + role = "admin"; + } + + if (role) { + const response = NextResponse.json({ success: true, role }); + + // Set HttpOnly cookie + response.cookies.set({ + name: 'auth_token', + value: role, + httpOnly: true, + secure: process.env.NODE_ENV === 'production', + sameSite: 'lax', + path: '/', + maxAge: 60 * 60 * 24 * 7 // 1 week + }); + + return response; + } + + return NextResponse.json( + { success: false, error: 'Invalid email or password' }, + { status: 401 } + ); + } catch (error) { + return NextResponse.json( + { success: false, error: 'Bad request' }, + { status: 400 } + ); + } +} diff --git a/src/app/cbt/[examId]/page.tsx b/src/app/cbt/[examId]/page.tsx index ba37c44..f8da424 100644 --- a/src/app/cbt/[examId]/page.tsx +++ b/src/app/cbt/[examId]/page.tsx @@ -461,7 +461,7 @@ export default function CBTExamPage({ params }: { params: Promise<{ examId: stri -
+
{currentQuestion.text_en} @@ -478,14 +478,14 @@ export default function CBTExamPage({ params }: { params: Promise<{ examId: stri {currentQuestion.type === 'Numerical' ? ( selectAnswer(currentQuestion.id, e.target.value)} /> ) : currentQuestion.type === 'Subjective' ? (