CSIR NET 2025: The National Testing Agency (NTA) will close the extended registration-cum-application window for the Council of Scientific and Industrial Research- University Grants Commission National Eligibility Test (CSIR UGC NET 2025) today at csirnet.nta.ac.in.

Previously, the registration deadline was June 23, but it was extended to June 26.
As per the revised schedule, the window to pay the exam fee will close on June 27. The application form correction window will open on June 28 and close on June 29.
CSIR NET 2025 will be held from July 26 to 28. The test will be bilingual (English and Hindi). Candidates will have to attempt the paper as per the language option exercised by them in the application forms.
If there is any ambiguity in the question papers, the English version will be treated as final, NTA said.
{{/usCountry}}If there is any ambiguity in the question papers, the English version will be treated as final, NTA said.
{{/usCountry}}The exam will be held in the computer-based test (CBT) mode. The test will have three parts with objective-type, multiple-choice questions (MCQs). There will be no break between the question papers.
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CSIR NET 2025: How to apply for the exam
Go to the official website, csirnet.nta.ac.in.
Click on the registration link for the CSIR UGC NET June exam 2025.
Enter the requested information and register.
Now, log in to your account.
Fill out the CSIR NET application form.
Upload the required documents and pay the application fee.
Submit the form. Take a printout of the confirmation page.
NTA said a candidate can submit one application only, and it will take strict action against candidates who submit more than one application form.
CSIR-UGC NET is a national-level test to determine the eligibility of Indian nationals for
- Award of Junior Research Fellowship and appointment as Assistant Professor
- Appointment as Assistant Professor and admission to PhD and
- Admission to Ph.D. only in Indian universities and colleges.
Candidates can visit the official website for more details.