lf I get 98 percentile which nit will I get.can I get nit suratkal mech ? my catogery is OBC
Hello,
Your expected rank will be around 17380.You can calculate your expected rank using the formula (100-your total percentile)*869010/100
Where 869010 is the total number of candidates appeared in January session.
In April session 1200000 candidates are expected to appear then your expected rank can be calculated by the formula (100-your total percentile)*1200000/100
Your category rank is expected to be in between 4000-5000.
In 2019 a candidate with rank 8502 belonging to OBC category (Home state) National Institute Technology, Surathkal in Mechanical Engineering.A candidate with rank 2844 belonging to OBC category(other state) secured seat in the same.So if you are home state candidate you may get seat considering previous year cut off.You may also get seat in
- NIT, PATNA
- NIT, NAGALAND
- NIT, PUDUCHERRY
- NIT, RAIPUR
- NIT, SIKKIM
However, cut off is subjected to change it may increase or decrease slightly.
You can predict the colleges in which you may get seat by using our JEE main college predictor
Hello aspirant
You have got too good percentile, your expected all India rank with this percentile will be around. 17380.2 . As you are from OBC category you have a decent chance of getting a a seat in the top NIT.
Talking about Surat Kal , mechanic Engineering. It depends on cut off of Surat Kal 2020. Cut off of nit surathkal depends upon total number of candidates appeared for jee main 2020., Total number of questions asked in the new main exam , difficulty level of exam, previous year's cut off, performance of the candidates in JEE main .
2019 cut off if NIT surathkal for mechanical engineering for OBC class was 8502 for Home state and 2844 for other state.
For more information you can refer to following link
https://engineering.careers360.com/articles/jee-main-cutoff-for-nit-surathkal
