Sir I got 96.0365%tile in jee paper 2 for B.arch and I am from OBC category. can I get admission in nit
Hello,
Your percentile is 96.035 in JEE paper 2 . So, you can calculate your rank by using the formula given below :
(100 - your total score) X 112679 /100
According to your percentile , your rank is between 4000-4500.
The last year cutoff for B Arch for 2019-2020 for NIT's was :
National Institute of Technology, Calicut - 693
National Institute of Technology, Hamirpur - 24979
National Institute of Technology , Patna - 3900
National Institute of Technology, Bhopal - 3739
For detailed cutoff , you can visit the below link :
https://engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b-planning
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