Hello Abhinav

congrats for scoring 97.56 percentile in JEE mains paper 2.

You have  performed extremely well  and from your percentile we can predict your rank.

There is a pedicted formula from our team to predict rank

 Total No of students = 145386

Formula

Probable Rank= ( 100 - your NTA percentile ) x 145386 / 100

So your predicted rank  =3547 approx

Our Team predicted that the qualify cutt off would be above 85 percentile .But it just a prediction .So pleae wait till april.

Your scores are really outstanding  so according to me you will get admission in following NIT's...

• National Institute of Technology, Tiruchirapalli, Tamilnadu
• National Institute of Technology, Warangal, Andhra Pradesh
• National Institute of Technology, Surathkal, Karnataka
• Moulana Azad National Institute of Technology, Bhopal, M.P
• Motilal Nehru National Institute of Technology, Allahabad, U.P
• National Institute of Technology, Calicut, Kozhikode, Kerala
• National Institute of Technology, Durgapur, West Bengal
• National Institute of Technology, Hamirpur, Himachal Pradesh
• Malaviya National Institute of Technology, Jaipur, Rajasthan
• Dr. B R Ambedkar National Institute of Technology, Jalandhar, Punjab

I hope you get my point.

Good Luck!!!

Dear parent

It's outstanding. The rank based on the score will be nearly (100-97.56) *145386/100=3547.And your son has chances with one of the NIT based on previous year trends.To know previous year cutoff, go to

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