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🎯 NIT Rourkela JEE Main 2025 Cut-off
Program | 🟢 Opening Rank | 🔴 Closing Rank |
|---|---|---|
Computer Science & Engineering (B.Tech) | 580 | 3,000 |
Artificial Intelligence (B.Tech) | 3,100 | 4,050 |
Electronics & Communication Engineering | 3,900 | 5,900 |
Electrical Engineering | 6,500 | 10,750 |
Electronics & Instrumentation Engineering | 7,400 | 11,050 |
Mechanical Engineering | 12,400 | 15,100 |
Chemical Engineering | 15,300 | 19,200 |
Civil Engineering | 17,800 | 25,900 |
Biomedical Engineering | 28,000 | 38,300 |
Metallurgical & Materials Engineering | 25,200 | 36,400 |
Ceramic Engineering | 30,000 | 36,500 |
Food Process Engineering | 34,500 | 42,100 |
Mining Engineering | 29,150 | 32,350 |
📌 Factors Influencing Cut-offs
Exam Difficulty: A more challenging JEE Main paper can lead to lower cut-off ranks.
Competition Level: An increase in the number of applicants can result in higher closing ranks.
Seat Matrix: Programs with fewer seats typically have higher closing ranks due to increased competition.
Historical Trends: Past years' cut-off data help in predicting current year trends.
Reservation Policies: Cut-offs vary across different categories (SC/ST/OBC-NCL/EWS/PwD) due to reservation norms.
📋 Admission Process at NIT Rourkela
Qualify JEE Main: Secure a valid All India Rank (AIR) in JEE Main.
Meet 12th Criteria: Obtain at least 75% in Class XII (or top 20 percentile); 65% for SC/ST/PwD candidates.
Participate in JoSAA Counselling: Engage in choice filling, seat allocation, and confirmation processes.
Complete Admission Formalities: Submit required documents and pay the admission fee to finalize enrollment.