The question paper and preliminary answer key2019 for the CSIR NET exam has been released by the National Testing Agency (NTA). The exam was held on 15th and 27th December 2019. Candidates can go through the question paper and answer key from the official website of NTA.
The objections can be raised against the released answer key till 3rd of January till 11:50 pm and the payment of the answer key will be accepted till January 4, 2020. A fee of Rs 1000 need to be paid for raising objections.
The result of 2019 exam will be declared based on the final answer key. Earlier it was announced by NTA to declare the result for CSIR NET on December 31 whereas later it was decided to be released on 14th January 2020.
Candidates can check the answer key and get more details on the exam at csirnet.nta.nic.in or www.nta.ac.in .
NTA CSIR UGC NET answer key 2019: How to raise objections?
- Visit the official website of NTA CSIR UGC NET. Click on the ‘view question paper and to challenge answer key’ button.
- Click on the tab you wish to log-in through.
- Log-in using credentials.
- Check the Answer key present in the dashboard.
- After log-in, click on the ‘challenges/s regarding answer key’.
- Click on the ‘option IDs’ (where currently correct option is marked) under the ‘question id’ you think is wrong.
- Upload supporting documents and Click on ‘save your claim finally’.
- Check and Review the preview of your challenge and click on ‘pay here’.
- Finally make payment and submit.
In CSIR NET exam, of the 2,82,117 candidates who registered, as many as 2,25,889 candidates appeared for the exam.
Stay tuned to NTA official website for more updates and timely information on NTA CSIR-UGC NET 2019 examination.
Also read, CSIR UGC NET December 2019.
