IIT Bhubaneswar - Indian Institute of Technology

Get insights from 221 questions on IIT Bhubaneswar - Indian Institute of Technology, answered by students, alumni, and experts. You may also ask and answer any question you like about IIT Bhubaneswar - Indian Institute of Technology

Follow Ask Question
221

Questions

0

Discussions

25

Active Users

543

Followers

New answer posted

11 months ago

0 Follower 3 Views

A
Atul Goyal

Contributor-Level 7

IIT Bhubaneswar JAM cutoff 2025 was released for admission to various PG courses. The cutoff was released for different categories belonging to the All India quota. For admission to MSc+PhD in Physics, the cutoff in respect to the General All India category candidates was closed at 566. For the OBC AI category candidates, the cutoff for the same course was closed at 847. For SC AI category candidates, the cutoff was closed at 1819

New answer posted

11 months ago

0 Follower 3 Views

S
Saakshi Garg

Contributor-Level 6

Yes, it is possible to get admission to IIT Bhubaneswar for MSc+PhD in Mathematics with a rank of 250 in IIT JAM for candidates belonging to the General AI category. Round 4 of the cutoff list has been released and the cutoff rank for admission to MSc+PhD in Mathematics was closed at 527 for candidates belonging to the General AI category. For candidates belonging to the OBC AI category, the cutoff rank was closed at 736. For candidates belonging to the SC AI category, the cutoff was closed at 1295. For more categories such as ST AI and EWS AI category, getting admission to the college with this rank is possible. 

 
 

New question posted

a year ago

0 Follower 10 Views

New answer posted

a year ago

0 Follower 6 Views

S
Sanjana Singh

Contributor-Level 6

Yes, it is possible to get admission to IIT Bhubaneswar with a rank of 10,000 in JEE Advanced. The closing ranks as per the IIT Bhubaneswar JEE Advanced cutoff 2025 varied between 3855 and 14467, for the General AI category candidates. A few specialisations in which candidates can easily get admission at 10,000 rank are B.Tech. in Engineering Physics and B.Tech. in Metallurgical and Materials Engineering. Some of the most competitive specialisation among the General AI category turned out to be BTech in Computer Science and Engineering and B.Tech. in Mathematics and Computing, for which it would not be possible to get admissi

...more

New question posted

a year ago

0 Follower 3 Views

New answer posted

a year ago

0 Follower 10 Views

S
Saakshi Garg

Contributor-Level 6

IIT Bhubaneswar JEE Advanced cutoff 2026 was declared for admission to BTech course and its various specialisations. 

For the General AI category, the round 1 cutoff ranks were between 4582 and 14763. Hence, students need to secure at least a rank of 14763 or lower to be eligible for admission. Students can see the table below to know the round 1 cutoff ranks for this category. 

CourseRound 1 (Closing Rank)
B.Tech. in Computer Science and Engineering4582
B.Tech. in Mathematics and Computing4775
B.Tech. in Electronics and Communication Engineering5807
B.Tech. in Electrical Engineering7580
B.Tech. in Mechanical Engineering10367
B.Tech. in Engineering Physics10738
B.Tech. in Metallurgical and Materials Engineering13988
B.Tech. in Civil Engineering14763

Note: Other categories will have different cutoffs.

New answer posted

a year ago

0 Follower 4 Views

N
Nishtha Mishra

Contributor-Level 6

Yes, it is possible to get admission to IIT Bhubaneswar with a 10000 rank in JEE Advanced. According to the IIT Bhubaneswar BTech cutoff 2025 for the General AI category, the cutoff ranged from 3785 in the first round to 16268 in the last round. In conclusion, yes, students who belong to this category can avail admission at this institution.

Refer to the table below to know which specialisations are up for grabs with a rank of 10K for the aforementioned category. 

CourseJEE Advanced cutoff range 2025
B.Tech. in Mechanical Engineering10078 - 11678
B.Tech. in Metallurgical and Materials Engineering13695 - 16268
B.Tech. in Engineering Physics10494 - 12466
B.Tech. in Civil Engineering13957 - 16156

Note: Other categories will have different qualifying cutoffs. 

New question posted

a year ago

0 Follower 2 Views

New answer posted

a year ago

0 Follower 13 Views

S
Shiksha Divya

Contributor-Level 10

Students need to obtain a competitive rank in JEE Main and JEE Advanced so that they can become eligible for admission procedures at IIT Bhubaneswar. Candidates are required to achieve a cut-off rank, as decided by the course. For instance, the cut-off rank needed for BTech Civil Engineering was 14,782, whereas for Computer Science and Engineering (CSE) it was approximately 3,685. Interested candidates can refer to the table below for information on cutoffs from previous years.

Course

Round 1 (Closing Rank)

Last Round (Closing Rank)

B.Tech. in Civil Engineering

13018

14782

B.Tech. in CSE

3404

3685

B.Tech. in Electrical Engineering

6505

7661

B.Tech. in Electronics and Communication Engineering

4798

5157

B.Tech. in Engineering Physics

9974

11307

New answer posted

a year ago

0 Follower 8 Views

P
Parul Jain

Contributor-Level 7

As per the IIT Bhubaneswar JEE Advanced cutoff 2024, to get BTech in Electrical Engineering program, candidates need to score a rank less than or equal to 1762 under SC AI quota. Further, it is mandatory for the candidates to have a valid JEE Advanced rank and candidates must also adhere to the mandatory eligibility before applying for the JoSAA counselling rounds.

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

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 705k Reviews
  • 1850k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.