Hi

The cutoff varies depending on whether or not you belong to the home state. Please go through the following link to check for your case

https://engineering.careers360.com/articles/jee-main-cutoff-for-nit-rourkela

In this link, ranks are provided. To go from your percentile p to rank r, use the following

(100-p)*9.34 lakh/100 = r, 9.34 lakh is the number of students that appeared for JEE Mains Paper-1 this year

Hi

The cutoff would depend on whether or not you belong to the home state.Please refer the following link to see the cutoff ranks corresponding to your situation.

https://engineering.careers360.com/articles/jee-main-cutoff-for-nit-rourkela

In this link, the ranks are mentioned. To go from rank to percentile, simply use the following

percentile= 100-((rank/no. of students who appeared for jee mains)*100)

This year, around 9.34 lakh candidates had given the exam,so you can take the denominator accordingly. Hope this helps!

Dear Aspirant ,

Since you mentioned any branch I am considering food processing engineering as this branch has the least cutoff .  57k was closing rank for admission in spot round 2 privious year and for this you must have percentile more than 95 .

But percentile , ranks and cutoff can change every year based on number of students and difficulty level of question papers . But if you want to have a secure seat in this NIT you must score more than 96 percentile to be in safe zone .

You can predict your rank here :-