I've got 93.9 percentile in jee mains january , can i get Nit Jamshedpur under home state quota for obcncl girl student from Jharkhand ? If yes whuch branches can i get ?
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Your rank can be predicted by using the formula, (100-your total percentile)*869010/100. In your case, your approximate rank would be 53009. Let's say the number of candidates for April exam increased to 12,00,000 then your final rank shall be (100 - your total score) X 12,00,000/100.
In 2019, the cut offs for NIT Jamshedpur, for female OBC Home State Candidate have been as follows:
Civil Engineering: 16612
CSE: 8148
Electrical: 12083
ECE: 11607
ME: 14011
Metallurgical: 16769
Production: 15011
Now these are 2019 cut offs and cut offs change every year depending on various factors such as Total no. of seats available, difficulty of the exam, previous years trends, no. of Candidates appearing for the exam and no. of Candidates applying for a course. So anything less than 95 percentile would be difficult for a seat in Home State NIT. But if you want to check the closing ranks of other NITs, please visit our page: https://engineering.careers360.com/articles/jee-main-cutoff
Now to predict colleges, please use our college predictor:
Hi Shalu,
Its difficult to predict actual rank or JEE advanced qualify marks. you may calculate your expected rank by this formula
To Calculate the rank for B. Tech:
So rank formula is (100 - your total score) X 869010 /100
But Your predicted JEE Main AIR rank for 2020 is
61474-64710.
At this Score or Rank your Chances To get NIT Jamshedpur are very less.
I hope you will increase your nta score in April and take admissions in nit or iit
Hope this helps! All The Best!!
Feel free to ask if you have any more queries
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