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JEE Main Exam Centres in Odisha 2026- NTA has published the list of JEE Main 2026 exam centres in Odisha at jeemain.nta.nic.in. The authority has released the JEE Main Odisha exam centres 2026 list online. A new JEE Main exam centre, Jharsuguda (OR22) has been included in the list. Candidates had to check the exam centres list carefully during the JEE Main registration. The list of exam centres in Odisha lets the candidates know where the JEE Main exam will be held in Odisha. The allotted exam centre address in Odisha has been intimated to the candidate through the JEE Main admit card. This year, there are 20 exam cities for the JEE Main exam in Odisha.
As per JEE Main 2026 exam trends, the expected JEE Main cutoff for the general category is 93- 95 percentile, for OBC-NCL it is 80-82 percentile, and for SC and ST it is 60-49 percentile.
The address of the allotted exam centre is mentioned on the JEE Main admit card. Candidates can check the JEE Main 2026 exam centres in Odisha from the table below.
| Exam City in Odisha | City Code |
|---|---|
| Balasore (Baleswar) | OR02 |
| Berhampur / Ganjam | OR03 |
| Bhubaneswar | OR04 |
| Cuttack | OR05 |
| Dhenkanal | OR06 |
| Rourkela | OR08 |
| Sambalpur | OR09 |
| Angul | OR10 |
| Bhadrak | OR11 |
| Baripada/Mayurbanj | OR12 |
| Jajpur | OR13 |
| Kendrapara | OR14 |
| Kendujhar (Keonjhar) | OR15 |
| Puri | OR16 |
| Jagatsinghpur | OR17 |
| Jeypore(Odisha) | OR19 |
| Balangir | OR20 |
| Baragarh | OR21 |
| Rayagada | OR26 |
| Jharsuguda | OR22 |
| Exam Centre | Centre Address |
|---|---|
ION Digital Zone iDZ and ABA college | |
Balasore College of Engineering & Technology | |
iDZ Khallikote College Area | |
Roland Institute of Technology | Surya Vihar, on the NH, Golanthara, Brahmapur, Odisha 761008 |
Vignan Institute of Technology And Management(VITAM), Brahmapur | NearBhairabi Temple, PO: Mantridi, Via: Golanthara, Berhampur, Odisha 761008 |
TCS iON DIGITAL ZONE iDZ 2 | |
iON Digital Zone (iDZ) 1 | |
Eastern Academy of Science & Technology | At:, Prachi Vihar Anantapur, Bubaneswar, Phulnakhara, Odisha 75400 |
ION Digital Zone iDZ Banamaliprasad | |
Raga - the taste of health | |
Ion Digital Zone, Baraipali, Sambalpur | Green Solution & Services, Shiva Complex, Baraipali Chowk, Khetrajpur, Sambalpur, Odisha 768150 |
Ion Digital Zone,Majhipali,Sambalpur | |
Ion Digital Zone,Majhipali,Sambalpur |
Frequently Asked Questions (FAQs)
There are a total of 20 JEE Main exam centres in Odisha.
The JEE Main 2026 exam dates are January 21 to 29, 2026 and April 2 to 9, 2026.
Candidates can check the JEE Main exam centre pdf to know the names of test cities. Detailed information on the allotted exam centre will be mentioned in the admit card of JEE Main 2026.
Candidates can check the JEE Main city allotment on the website, jeemain.nta.nic.in.
JEE Main exam is held at dedicated exam centres.
There are 308 JEE centres in India in 2026.
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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