Joint Entrance Examination (Main)
Hello,
You have not mentioned your category. Cut off percentile is different for each category. NTA has published the JEE Main 2020 official cut off. The cut off is as follows
So, check the cut off. If you
Same problem bro... I am also not able to login..its saying invalid application number.
Hello student ,
Here it will be difficult to tell you the exact about your admission possibilities but as per the p revious year cutoff with your category the chance seems less to get a seat in NITs, I IITs & GFTIs however you can try for state counseling like
Dear Aspirant,
Eve though your rank is 40000 and you are in EWS category getting a seat in NIT would be a little challenging. Even if you get a seat it would not be a desirable branch.So I would suggest you to try in any other private college for any
Hi
Yes you are eligible for JEE advanced examination.
This is because,
You belong to sc category and qualifying cutoff of JEE mains to be eligible for JEE advanced for sc category is 50.1760245 percentile and you have got more than that i.e 81 percentile, so you are definitely eligible
National Testing Agency (NTA) will soon release the JEE Paper 2 result 2020 at jeemain.nta.nic.in. JEE Main paper 2 result date 2020 is not announced by NTA yet. Students can however soon expect JEE Main paper 2 result 2020 September as paper 1 result has been announced. The dates
Hello!
It's not possible to get a seat in any of the NITs with this rank. You can refer this link for last year's cutoff.
https://engineering.careers360.com/articles/jee-main-cutoff-for-top-nits
But you should try for other state counselings like M.P, Maharashtra, Punjab, etc.
Good luck!
Sorry for the late response.But the link for registration of JEE Advanced registration this year is from 12th September and the correction window opens on 20th September.Hope your query has been resolved.Best of luck.
To choose from the best colleges you can also follow the below link-
https://engineering.careers360.com/jee-main-college-predictor?icn=QnA&ici=qna_answer
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