Hey Rashmi,

This year's cut-off are yet to release, but according to 2019 cut-off for CS branch, general category in female and home state quota, the closing cut-off rank was 10,746. To get this rank, you must have 98-99 percentile score. The results have already been released, so you can check your percentile and then calculate your rank using the formula:

(100- percentile)869010/100.

Here, 869010 is the number of candidates appeared in January session of JEE mains.
Kindly check the link below to see the previous year cut-offs:
https://engineering.careers360.com/articles/jee-main-cutoff-for-nit-rourkela/amp

Also, you can use our College predictor for knowing the possibilities:




Hope this helps!

NIT Rourkela

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