What is the opening and closing rank in NIT Surathkal for open category and OBC?

I want to know the ranks for 2015.

274 Views|Posted 2016-03-04 15:30:14
2 Answers
Deepanwita Sahu
7 years ago
NIT Surathkal: (general) For home state it is around- 300-3000 GMR For other state- 200-1200 GMR

Thumbs Up IconUpvote Thumbs Down Icon
rohan n
2016-03-15 14:55:45
Hi Raghavendra, In order to get past NIT Surathkal , one must secure a rank in top 20800 in JEE. An engineering aspirant's category and the quota he or she might be eligible for, plays an important role in the admission to institutes that accept JEE Main scores including the NITs (National Instit

...Read more

Thumbs Up IconUpvote Thumbs Down Icon

Similar Questions for you

Yes, it is easy to apply for various courses at the National Institute of Technology Surathkal. Students just have to go to the institute's official website to apply. The admission is through many entrance tests for specific courses. 

National Institute of Technology or NIT Surathkal is in Karnataka and was established in 1960. It has 13 departments, 1 School, and 12 supporting centres.  NIT Surathkal gives UG, PG and PhD courses in the areas of Engineering, Management, and Science. 

NIT Surathkal BTech cutoff 2025 was 3934 in round 1 for the General AI category, which later increased to 5316 in last round for the BTech ECE course. Students can refer to the table below to know the overall cutoff range for all other All India categories. 

All India CategoryJEE Main Cutoff Range 2025
General3934 - 5316
OBC1687 - 7933
SC808 - 888
ST459 - 459
EWS599 - 6196

Please note that other quotas, catego

...Read more

NIT Surathkal CSE cutoff 2025 for the General AI category female students was 3531 in the round 1, which later increased to 4892 in the last round. For the OBC AI category female candidates, the overall cutoff ranged from 1723 to 1792. Students should know that other categories and quotas will

...Read more

Students who belong to the General AI category need to secure a rank between 1528 and 2158 to be eligible for BTech CSE branch. On the other hand, the OBC AI category students need a rank between 495 and 3690. Please note that other categories will have different cutoffs accordingly. 

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.9M
Answers

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Have a question related to your career & education?

or

See what others like you are asking & answering