Post:- NTA CSIR UGC NET June Online Application Form Reopen 2020
Eligibility, Salary, Admit Card, Online Form, Application Fee, Vacancy Detail, and Full Notification of NTA CSIR NET JRF Online Form 2020 – Reopen.
NTA CSIR NET JRF Online Form 2020 – Reopen Notification
- Application Begin: 16/03/2020
- Last Date For Apply Online: 30/06/2020
- Last Date for Fee Payment: 30/06/2020
- Re-open Form: 22/08/2020 to 10/o9/2020
- Correction Last Date: 15/07/2020
- Answer Key Available: After Exam
NTA CSIR NET JRF Online Form 2020 – Reopen Application Fee
- Gen- 1000/-
- OBC- 500/-
- SC / ST/PH:250/–
- Pay the Examination Fee Through E Challan Deposit Fee at Any Branches of Indian Bank.
NTA CSIR NET JRF Online Form 2020 – Reopen Age Limit
- Minimum Age: 18Years
- Maximum Age Male: 28years
- Maximum Age Female: 28Years
NTA CSIR NET JRF Online Form 2020 – Reopen Admit Card
- Exam Date: Notified Soon
- Admit Card Available: Notified Soon
NTA CSIR NET JRF Eligibility/Qualification
- M.Sc / Equivalent Degree with 55% Marks for General / OBC Candidates.
- For SC / ST and PH Candidates: 50% Marks.
- Integrated Course and B.E/ B.Tech / B.Pharma and MBBS Candidates Are Also Eligible for CSIR NET 2018
NTA CSIR NET JRF Salary
- As per Government Norm.
Apply Online (Re-Open)
Download ReOpen Notice
Download Date Extended Notice
Download Notification
Download Syllabus
Official Website
Tips to Get Job Easily
- Candidate must read the official notification properly from the Official website.
- If this is an Online Vacancy, click on the above link or visit the official website.
- Now register yourself instead of filling all your details like personal details, education, and fee.
- Do not forget to upload the correct photograph and signature.
- If this is an Offline Vacancy, then fill the form and attach all your documents carefully.
- Now Submit your form if the vacancy is Online or if the vacancy is Offline send it to the given address.
- Now keep your eyes on the website for updating the Admit Card or Exam Date.
- If you are still facing any problem kindly comment on the post given below.
