Can I get SGSITS for CSE branch with 531212 rank in JEE Mains 2018? I am class 12th student & want to know about the cut-off & closing rank.

Students who belong to the General AI category need to secure a rank between 1528 and 2158 to be eligible for BTech CSE branch. On the other hand, the OBC AI category students need a rank between 495 and 3690. Please note that other categories will have different cutoffs accordingly. 

The BTech CSE last round rank for the General AI category in 2025 was 4245. In 2024 and 2023, the cutoff in the last round was 3365 and 4218, respectively. This difference in the ranks suggests the cutoff trend has been fluctuating over the years. Please note that other categories will have differen

To get admission at NIT Rourkela, General AI category students need to attain a rank of 85536 or below to be eligible, according to the last round NIT Rourkela JEE Main cutoff 2025. 

For the students who belong to the General HS category, the required rank needed for admission was at least 12735

NIT Rourkela JEE Mains cutoff 2025 concluded with the release of the final round cutoff. For the General AI category, the BTech CSE overall cutoff ranged from 3122 to 4245. On the other hand, for the General HS category, the overall cutoff varied between 7371 and 7853. Students should know that the

 IIITDM Kurnool BTech cutoff 2025 was released for various All India categories. For the BTech in Artificial Intelligence and Data Science course, admissions were closed at a rank of 46325 for the General AI category. In 2024, and 2023, the same course had a cutoff of 41928 and 44621, respectiv