How many mark in JEE Mains 2019 we have to secure to get a seat in NIT, Rourkela for B.Tech in computer science and engineering?

427 Views|Posted 7 years ago
Asked by Kishan Behera
6 Answers
Srinarayan Prajapati
6 years ago
Hello Please go with this college predictor this will help you to choose best for you https://www.shiksha.com/b-tech/resources/jee-mains-college-predictor Thank you.

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Jyoti  Prajapati
6 years ago
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Yash Gupta
6 years ago
Hi Chinnu Rao, Kindly predict your colleges in this college predictor. JEE Mains rank predictor is an online tool which predicts the candidates rank range and probable colleges based on the difficulty level of the examination, average marks scored in JEE Main and seats available in IITs, NITs, GFTI

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S
7 years ago
Hi, Placements are really good at NIT-R. Average salary for computer Science branch is around INR. 7 lacs - INR. 8 lacs per annum. The highest package offered is lacs 39 lacs p. a. the lowest is lacs 3 lacs p. a. not only technical but management related jobs are also offered at NIT-r. 30 percent

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S
7 years ago
Well, every year, the rank changes and the marks for each institute changes. Definitely, you have to score more than 170 marks and yes, category can play a big role in getting admission specially if its a CSE branch. All the best.

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pradeep kumar
7 years ago
Hi, the opening and closing ranks in B.Tech CSE for NIT, Rourkela for year 2018 is 4209 and 8524 respectively. Hence you have to score more than 150.

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BranchGeneral Category Closing RankOBC/SC/ST Closing RankNotes
Computer Science (CSE)~12,000–15,000Up to ~25,000–30,000Most competitive branch
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Civil Engineering~30,000–35,000~60,000–70,000Lower demand
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