I got 97.7655 in jee paper 2 2020 January exam. Could i get into NIT Calicut with this score? I belong to OBC. Could anyone help me with my rank calculation?

Hello!!!!

Hope you are doing great!!!

Your All India Rank would be 19418 approximately using the formula 100-97.7655*869010/100 where 97.7655 is your percentile and 869010 is the number of candidates who appear for january session.As per the previous year scenario,With this rank you can not get seat in NIT Calicut in any of the branches of engineering.you may have to appear for July session .For checking the probabilities of colleges or NIT's,you can use our college predictor.The link is provided below:

https://engineering.careers360.com/jee-main-college-predictor?ici=cp-listing_engineering&icn=JEE-Main-College-Predictor Hope it helps!!!!!

All the best!!!!!!

Hello,

You can calculate your expected rank by the formula (100-your total percentile)*112679/100 where 112679 is the total number of candidates appeared in JEE Main Paper 2 in January session.By this your rank is expected to be around 2518 and you rank based on your category is expected to be around 500.In 2019 the cut off NIT, calicut for OBC (home state) was 693 and OBC (other state was 804).Considering this you may have chances of getting a seat in NIT, Calicut.However, cut off increase or decrease depending on number of seats, number of candidates appeared, difficulty of exam etc.

You may try our JEE main college predictor to know the other colleges in which you may have chanes to get in

JEE Main College Predictor 2020 - Predict Colleges based on your JEE Main Rank

JEE Main College Predictor – Know your College Admission Chances Based on your Rank/Percentile, Home State, Caste, Gender, Exam Result & Get ...