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IIIT Manipur JEE Main Cutoff 2025 - JoSAA released the round 1 IIIT Manipur JEE Main 2025 cutoff at, josaa.nic.in. The cutoff of IIIT Manipur is available for each round of JoSAA counselling. IIIT Manipur JEE Main cutoff percentile 2025 is available in the form of opening and closing rank. The authorities considers various factors for determining the JEE Main cutoff for IIIT Manipur like the number of applicants, the difficulty level of the JEE Main 2025 exam, etc. IIIT Manipur cutoff percentile will act as a screening process for admission. Candidates who will qualify the cutoff of JEE Main for IIIT Manipur will have higher chances of admission. However, the candidates must fulfil all other eligibility criteria for admission at the institute along with qualifying the cutoff. Read the full article to know more about IIIT Manipur JEE Mains cutoff 2025.
For JEE Main 2026 session 2, candidates who passed the Class 12 or equivalent examination in 2023 or earlier are not eligible to appear for JEE Main 2026 session 2. The passing year is counted as the year in which a candidate is declared ‘pass’ for the first time in Class 12. Subsequent attempts, subject additions, or improvement exams will not be considered as a fresh passing year for eligibility purposes.
Candidates can check below the IIIT Manipur JEE Main cutoff 2025 for round 1. The data for other rounds will also be updated, once released by the authorities. The table below is sorted for open category, gender-neutral quota.
| Course Name | Opening Rank | Closing Rank |
| Computer Science and Engineering (4 Years, Bachelor of Technology) | 37847 | 48986 |
| Computer Science and Engineering with specialization in Artificial Intelligence and Data Science (4 Years, Bachelor of Technology) | 36550 | 48433 |
| Computer Science and Engineering with specialization in Cyber Security (4 Years, Bachelor of Technology) | 38252 | 52895 |
| Computer Science and Engineering with specialization in Quantum Technologies (4 Years, Bachelor of Technology) | 32256 | 45341 |
| Electronics and Communication Engineering (4 Years, Bachelor of Technology) | 44414 | 53718 |
| Electronics and Communication Engineering with specialization in VLSI and Embedded Systems (4 Years, Bachelor of Technology) | 43009 | 56825 |
Candidates can refer to the given tables to know about the previous year JEE Main cutoff of IIIT Manipur to understand how the cutoffs have changed over the years.
| Academic Program Name | Opening Rank | Closing Rank |
| Computer Science and Engineering (4 Years, Bachelor of Technology) | 30685 | 42798 |
| Computer Science and Engineering with specialization in Artificial Intelligence and Data Science (4 Years, Bachelor of Technology) | 33556 | 41738 |
| Electronics and Communication Engineering (4 Years, Bachelor of Technology) | 32474 | 48388 |
| Electronics and Communication Engineering with specialization in VLSI and Embedded Systems (4 Years, Bachelor of Technology) | 40333 | 50091 |
| Course Name | Gender | Opening rank | Closing rank |
| Computer Science and Engineering | Male | 28532 | 36456 |
Computer Science and Engineering with specialization in Artificial Intelligence and Data Science | Male | 22839 | 34502 |
| Electronics and Communication Engineering | Male | 36464 | 42135 |
Electronics and Communication Engineering with specialization in VLSI and Embedded Systems | Male | 34841 | 41482 |
| Course Name | Category | Male | Female |
| Computer Science and Engineering | OBC | 14622 | 19554 |
| EWS | 6755 | 6755 | |
| General | 43500 | 46086 | |
| SC | 8441 | 9107 | |
| ST | 4221 | 4221 | |
| Computer Science and Engineering with specialization in Artificial Intelligence and Data Science (4 Years, Bachelor of Technology) | OBC | 14940 | 14940 |
| EWS | 6564 | 6564 | |
| General | 42875 | 42875 | |
| SC | 8379 | 8379 | |
| ST | 4049 | 4049 | |
| Electronics and Communication Engineering | OBC | 1620 | 21667 |
| EWS | 8114 | 8927 | |
| General | 50493 | 55551 | |
| SC | 9386 | 10473 | |
| ST | 4375 | 4672 | |
| Electronics and Communication Engineering with specialization in VLSI and Embedded Systems (4 years, Bachelor of Technology) | OBC | 16398 | 16398 |
| EWS | 8202 | 8202 | |
| General | 52341 | 52341 | |
| SC | 9561 | 9561 | |
| ST | 4525 | 4525 |
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| Course Name | Category | Male | Female |
| Computer Science and Engineering | OBC | 9858 | 9858 |
| EWS | 5270 | 5270 | |
| General | 25851 | 25851 | |
| SC | 6353 | 6353 | |
| ST | 3349 | 3349 | |
| Electronics and Communication Engineering | OBC | 11893 | 11893 |
| EWS | 5888 | 5888 | |
| General | 33945 | 33945 | |
| SC | 8163 | 8163 | |
| ST | 3586 | 3586 |
Course Name | Category | Male | Female |
Computer Science and Engineering | OBC | 15500 | 15500 |
EWS | 7026 | 7026 | |
General | 41632 | 41632 | |
SC | 9951 | 9951 | |
ST | 5231 | 5231 | |
Electronics and Communication Engineering | OBC | 16565 | 16565 |
EWS | 7916 | 7916 | |
General | 45452 | 45452 | |
SC | 10709 | 10709 | |
ST | 6065 | 6065 |
Course Name | Category | Gender Neutral | Female Only |
Computer Science and Engineering | OBC | 13798 | - |
EWS | 5211 | - | |
General | 38509 | - | |
SC | 8468 | - | |
ST | 4505 | - | |
Electronics and Communication Engineering | OBC | 14167 | - |
EWS | 5498 | - | |
General | 42303 | - | |
SC | 8726 | - | |
ST | 4860 | - |
Course Name | Categories | Opening Rank | Closing Rank |
Computer Science and Engineering | General | 24389 | 40151 |
OBC-NCL | 7350 | 14370 | |
SC | 5841 | 9546 | |
ST | 3043 | 5476 | |
Electronics and Communication Engineering | General | 27976 | 44229 |
OBC-NCL | 9431 | 15500 | |
SC | 5996 | 10286 | |
ST | 3193 | 6092 |
Courses | Categories | Opening Rank | Closing Rank |
Computer Science and Engineering | General | 19509 | 37474 |
OBC-NCL | 8605 | 13578 | |
SC | 4260 | 8473 | |
ST | 3404 | 6704 | |
Electronics and Communication Engineering | General | 25306 | 39507 |
OBC-NCL | 9424 | 13970 | |
SC | 5563 | 8608 | |
ST | 3164 | 7232 |
Courses | Categories | Closing Rank |
Computer Science and Engineering | General | 36738 |
OBC-NCL | 12397 | |
SC | 7420 | |
ST | 4082 | |
Electronics and Communication Engineering | General | 40661 |
OBC-NCL | 13101 | |
SC | 7694 | |
ST | 3936 |
The authorities consider various factors to determine the cutoff of IIIT Manipur JEE Jain cutoff 2025 -
Difficulty level of JEE Main 2025.
Number of candidates who appeared for the entrance exam.
Number of seats at IIIT Manipur.
Previous Year JEE Main cutoff Trends.
The Indian Institute of Information Technology, Manipur offers two B.Tech programmes for the students- Computer Science and Engineering (CSE) and Electronics and Communication Engineering (ECE). The institute has 24 seats in each program.
Frequently Asked Questions (FAQs)
IIIT Manipur JEE main cutoff is available online.
Engineering admissions at IIIT Manipur are based on the cutoff score released by JoSAA, thus, it is important to know the ranks for which admissions are served.
Yes, there is a separate cutoff for female candidates.
Two B. tech specializations are offered at IIIT Manipur, i.e. Computer Science Engineering and Electronics & Communications Engineering.
The Joint seat allocation authority releases the JEE Main cutoff on its website.
On Question asked by student community
JEE Main 28 Jan shift 2 exam will end at 6 PM. The complete analysis and memory-based questions will solution will be updated in the below article. Keep checking the page-
Math and Chemistry was difficult and Physics was moderate. The complete analysis is available here- https://engineering.careers360.com/articles/jee-main-2026-january-28-shift-1-question-paper-with-solutions-pdf
You can also check the memory-based questions and detailed solutions for JEE Main Jan 28 shift 1 paper.
Hi Lucky,
Please refer to this link and you can download the free pdf.
HI Manisha Maharana
You can download the JEE Mains 10 Free Mock Test with Detailed solutions. Its a feely downloadable pdf.
https://engineering.careers360.com/download/sample-papers/jee-main-10-full-mock-test-and-explanations-pdf
Also, you can check ad attemp the online mock test on our platform.
https://learn.careers360.com/test-series-jee-main-free-mock-test/
A general equation of a circle is
$
x^2+y^2+2 g x+2 f y+c=0
$
Since it passes through $(0,0)$,
$
c=0
$
So the equation becomes
$
x^2+y^2+2 g x+2 f y=0
$
It cuts the x -axis at ( $a, 0$ ).
Substituting:
$
a^2+2 g a=0
$
g=-a/2
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