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IIIT Bangalore JEE Main Cutoff 2025- IIITB has released the JEE Main 2025 round 3 cutoff on the official website, iiitb.ac.in. Candidates can check the IIIT Bangalore JEE Main cutoff 2025 on this page. The JEE Main cutoff is the minimum rank required for admission into the institute. IIIT Bangalore offers BTech programs in Computer Science and Engineering (CSE), Data Science and Artificial Intelligence (DSAI), and Electronics and Communication Engineering (ECE). Only those candidates who qualify the JEE Main exam will be considered for admission. Read the complete article for details on the IIIT Bangalore JEE Mains cutoff 2025.
To be eligible for JEE Advanced 2026, candidates must secure a place among the top 2.5 lakh students in JEE Main. For the General category, this usually corresponds to a score of around 90–115+ marks, or roughly the 93–95 percentile. In contrast, the qualifying cutoffs for reserved categories are lower, generally falling in the range of about 60–82 percentiles.
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The authority has released the JEE Main cutoff for IIIT Bangalore for three online rounds. The following tables mentions the IIIT Bangalore JEE Main cutoff percentile as well as the IIIT-B Internal Rank (IIR ranks). Candidates can refer to the information below for the IIIT Bangalore cutoff JEE Mains for CSE and other disciplines as well. Students can refer to the IIIT Bangalore JEE Main 2025 cutoff in the table below.
Programme | JEE Main CRL/JEE Advanced CRL |
|---|---|
BT-CSE | 5286 |
BT-AI&DS | 6517 |
BT-ECE | 7054 |
IMT-CSE | 6470 |
IMT-ECE | 7546 |
Programme | JEE Main CRL/JEE Advanced CRL |
|---|---|
BT-CSE | 4683 |
BT-AI&DS | 5425 |
BT-ECE | 5761 |
IMT-CSE | 5552 |
IMT-ECE | 6282 |
| Programme | JEE Main CRL/JEE Advanced CRL |
|---|---|
| BT-CSE | 3677 |
| BT-AI&DS | 4709 |
| BT-ECE | 4840 |
| IMT-CSE | 4679 |
| IMT-ECE | 5171 |
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Programme | JEE Rank cut-off | IIR Cut-off |
2106 | 156 | |
2833 | 245 | |
3142 | 292 | |
IM.Tech. CSE | 3015 | 269 |
IM.Tech. ECE | 4079 | 402 |
Note that the authority does not provide IIIT Bangalore Cutoff JEE Mains OBC and other reservation categories.
Programme | JEE Rank cut off | IIR Cut-off |
B.Tech. CSE | 3173 | 296 |
B.Tech. DSAI | 4172 | 414 |
B.Tech. ECE | 4431 | 448 |
IM.Tech. CSE | 4204 | 418 |
IM.Tech. ECE | 5008 | 528 |
Programme | Round 3 opening rank (JEE Main CRL) | Round 3 closing rank (JEE Main CRL) |
B.Tech. CSE | 4149 | 4149 |
B.Tech. DSAI | 5039 | 5002 |
B.Tech. ECE | 5259 | 5149 |
IM.Tech. CSE | 5075 | 5057 |
IM.Tech. ECE | 5628 | 5606 |
Programme | Round 3 opening rank (IIR Rank) | Round 3 closing rank (IIR Rank) |
B.Tech. CSE | 411 | 411 |
B.Tech. DSAI | 533 | 527 |
B.Tech. ECE | 571 | 549 |
IM.Tech. CSE | 539 | 535 |
IM.Tech. ECE | 624 | 622 |
Programme | JEE Main CRL Closing Rank (Campus round) | JEE Main CRL Closing Rank (Extended campus round) |
B.Tech. CSE | 7260 | 8706 |
B.Tech. DSAI | 9486 | 11669 |
B.Tech. ECE | 10300 | 12315 |
IM.Tech. CSE | 7987 | 9323 |
IM.Tech. ECE | 11037 | 13176 |
IIIT Cutoffs-
The list of factors considered by the institute to determine the cutoff of IIIT Bangalore 2025 is given below.
Each candidate’s JEE Mains NTA CRL Rank is taken into consideration. That is, applicants with better rank in JEE Main have higher chances of admission.
The number of seats available per course at the time of selection is also important.
Course preference as stated by the candidate while filling the application form.
Frequently Asked Questions (FAQs)
Yes, the IIIT Bangalore cut off 2025 has been released on the official website, iiitb.ac.in.
The IIR cut off is the minimum internal rank required to be eligible for a particular course at IIIT B. The institute assigns each applicant with an internal rank (IIR). It is the position of the candidate among all those who have applied to IIIT B for B.Tech. and Integrated M.Tech. programmes.
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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