Hi dear,

You have scored 98.5 percentile in JEE main that means your rank should be around 13035. You can also calculate your rank

Rank=(100-your percentile)×869010÷100

However his final rank will depend on number of student appear in April month exam. He has a chance to get cse in NIT Patna and NIT Jamshedpur.

Hope it helps.

Hey there hope you're doing well.
Dear aspirant,according to your percentile score in JEE MAIN examination your expected rank is about 13,035. With this rank you are expected to be in the top 1k or 2k rank in your respective category. Being the residents of  Bihar you can easily expect CSE at NIT Patna as home state closing rank is 4531. Imvattaching a link where you can check cutoffs for CSE for NITs for OBC category.
Link-https://engineering.careers360.com/articles/nits-cutoff-for-top-btech-branches.
Thank you hope this helps you!

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