My jee main rank is 76220 in general

- To get admission into the undergraduate engineering program offered by any of the National Institute of Technology, candidates are not only required to qualify JEE Main exam, but also have to meet the cutoff specified by the respective NIT. There are a total of 31 NITs in India and each institute releases its own JEE Main 2020 cutoff to offer seats in the B.Tech courses. The minimum rank required by the candidates to be eligible for admission into the institute is made available in the cutoff of JEE Main. Candidates will be able to check their obtained rank in the JEE Main result 2020. NTA will release the rank list of JEE Main 2020 on the official website. Each NIT will release separate cutoffs category as well as programme wise in the form of the opening and closing rank. Since the authorities have not yet released the ranks required in JEE Main 2020 to get admission into NITs, candidates can check the previous years opening and closing ranks for reference purposes. At the time of preparing the cutoff, several parameters such as difficulty level of the exam, total number of candidates, previous years trends, etc. are taken into consideration. please use this link- shiksha predictor thanks and best of luck!!

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