My son' s percentile in JEE mains 2020 is 99.25. Kindly give the status of the possibility of joining at NIT Rourkela in BTech course provided he has HOME STATE quota as he is from Odisha.

1. NIT,Rorkela - Electrical engg, Mechanical engg, Electrical and telecommunication engg, Computer science engg, ECE)
2. IIIT,Bhubaneshwar ( CSE)
3. NIT,Jalandar(CSE)
4. NIT, Durgapur( CSE)
5. DTU( Mechanical)

Hello,

We can calculate the rank using the below formula

100 - your total percentile score is the percentage  of people ahead of you. 869010 are the number of candidates appeared for Jan exam. So rank formula is (100 - your total score) X 869010 /100. If the number of candidates for April exam increased to 1200000 then your final rank shall be (100 - your total score) X 1200000/100

So the approximate rank will be around 6517. Which is the general rank and will change after april exam.

As you haven't mentioned any category its difficult to tell. Considering the general rank and the last year cutoff then your son might get in NIT Rourkela

You can use the college predictor to know what colleges your son can get. Please click the below link to get the college predictor

https://engineering.careers360.com/jee-main-college-predictor?icn=jee-main_clgpd&ici=exit-intent

Hello! Hope you're doing good.

Congratulations on such amazing performance. The expected rank of your child is 6,517. Your son has very high chance of getting into a good course in NIT Rourkela. It can be considered almost final that he'll get into a number of courses in NIT Rourkela. Do make the right choice of the course.

Refer to the following link for the cutoff of NIT Rourkela: https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-nit-rourkela/amp