What is the lowest rank to get into IIT Kanpur?
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1 Answer
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IIT Kanpur accepts JEE Advanced ranks for admission to the BTech and other courses. JEE Advanced cutoff 2025 has been released for admission to IIT Kanpur. The round 1 cutoff for the General AI category candidates is ranged between 270 and 9662. the lowest rank one can achieve is, 270 for B.Tech. in Computer Science Engineering for the General All India category candidates. In addition to that, CSE was the most competitive course among all the courses. Similarly, for the OBC AI category candidates, the CSE still having the lowest rank of 142. To know more cutoffs, candidates can click here.
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Overall, 50-55% seats are reserved in IITs. The break-up of reservation is - 27% for OBC-NCL, 15% for SC, 7.5% for ST, 10% for Gen-EWS and 5% horizontal reservation for PwD candidates.
Physics is the most difficult of all three subjects asked in the JEE Advanced exam. Many students also finds Maths as the toughest. It largely depends upon the understanding of candidates and how well the students are equipped with basic knowledge of any subject.
It is mandatory to appear in both papers of JEE Advanced exam. The total marks are calculated by adding the sum of scores in both papers.
JEE Advanced question paper will be available online on the website jeeadv.ac.in. IIT Roorkee will release both paper 1 and 2 on the same day or the next day of exam.
JEE Advanced question paper 2026 will be released by IIT Roorke after the exam. The question paper is usually released on the same day or the next day of the exam. Candidates will be provided the download link ofthe paper on this page once released.
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