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i have secured 98.98 percentile in jee main . i belong to general class of odishastate.what are the chances of getting computer engg in nit raurkela.


gyan ranjan 21st Jan, 2020
Answers (2)
Naufal Mohamed 21st Jan, 2020
In the following link you will find the expected cutoff for the 2020 for NIT Rourkela and you'll also find the previous year cutoffs
https://engineering.careers360.com/articles/jee-main-cutoff-for-nit-rourkela

You'll find cutoffs of previous year for all NITs in the following link
https://engineering.careers360.com/articles/jee-main-cutoff
Nihal Kumar Gupta Student Expert 21st Jan, 2020

Hi Gyan Ranjan,

If you have secured 98.98 percentile in JEE Main paper, you rank would be approximately in between 10000 to 12000.  Also, Above is the range of your possible rank. The predicted rank may vary due to JEE Main tie breaker rules for the same score.

Also, you can check your expected rank using the formula:

(100-your total percentile) x 869010/100

So with 98.71 percentile, your rank would be around 10000.

If you have secured this rank you have a chance to get CSE Branch in these colleges:

NIT Silchar

IIEST Shibpur

NIT Hamirpur

NIT Meghalaya

NIT Sikkim

NIT Uttarakhand

NIT Srinagar

These are the college you will get if you have scored these marks. The last year cutoff for CSE branch in NIT rourkela was approx 3000-4000 rank. So, you will not get CSE in NIT Rourkela. If your rank come under 4000 after april session exam, you will get NIT Rourkela with CS branch.

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