Hello,

Your son really got good percentile. You haven't mentioned category and home state. This year total 112679 students attempted jee paper2. You can calculate his approximate rank using :(100-percentile)*112679/100. That is 2,242. Yes he will get an NIT for sure. You can see last year cut off of Rourkela using the below link:

https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b-planning/amp

You can also predict the colleges he will het with this link below.

Hope this is useful.

Dip, you can calculate your son's expected JEE Mains paper 2 rank for the January session using the following formula:

(100-your total percentile) x 112679/100

So with 98.01 percentile, his rank would be around 2242.

Also, if the number of students increases to 1.5 lakh in the April session, the formula for rank calculation would change to:

(100-your total percentile) x 150000/100

Hence the expected rank in this case would be around 2985.

The cutoff rank for NIT Rourkela in 2019 was 2903 for Home State Quota and 1949 for Other state students. Now the cutoff tends to vary every year but your son might get a seat if your home state is Odisha (for Rourkela). Else it might be difficult to get a seat in BArch in NIT Rourkela.

Some other colleges in which he can get admission in BArch (based on previous years' cutoff data) are:

• NIT Jaipur
• NIT Bhopal
• NIT Raipur
• NIT Hamirpur
• NIT Nagpur

You can go to the below link to find the cutoff rank for admission in BArch in NITs or CFTIs for the years 2016 to 2019: https://engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b-planning/amp

You can also go through our college predictor given below to check which colleges your son is eligible for:

All the best.