CSIR UGC NET 2019 – CSIR National Eligibility Test is a national level exam, conducted to determine eligibility for appointment of Lecturer and to award Junior Research Fellowship. It is a 200 marks exam and is conducted for Life Science, Physical Science, Chemical Science, Mathematical Science, Earth, Atmospheric and Planetary Sciences. You can check CSIR UGC NET 2019 answer key, result, cut off, and more details from this page.
Latest: Additional Result and cut off of CSIR NET 2019 December exam is out; check here.
CSIR UGC NET 2019
CSIR UGC NET is held twice a year, in the month of June and December. NTA has also released dates for CSIR UGC NET June 2020 exam and you can check it at the end of this page. To know all the dates for the December exam, refer to the table below:
| CSIR UGC NET 2019 | Important Dates |
| CSIR NET application form 2019 available from | 09 Sep 2019 |
| The closing of online Application | |
| Last date to pay Application fee | 10 oct 2019 |
| Application Form Correction from | 18 -25 Oct 2019 |
| Application form for J&K candidates only reopen on | 04 – 08 Nov 2019 |
| Issuing of Admit Card | 09 Nov 2019 |
| Exam Date | 15 Dec 2019 |
| Issue of admit card for Assam and Meghalaya State | 23 Dec 2019 |
| Date of exam for Assam and Meghalaya State | 27 Dec 2019 |
| Display of question paper and answer Key by NTA and challenging them | 01-03 Jan 2020 |
| Result (scores) declaration date | 15 Jan 2020 |
| Final result and cut off | 30 Jan 2020 |
Quick Links
CSIR UGC NET Result 2019
After releasing of the final CSIR NET Answer Key 2019, the result is declared at the official website. The result contains the name and roll number of qualified candidates. The CSIR UGC NET result is based on the performance of the candidates in the examination and is in the form pdf. The candidates whose name are in the final result are those who are will finally considered qualified for lectureship or JRF (for which the candidates applied for).
CSIR UGC NET Answer key 2019
The NTA releases the Provisional Answer Key of the questions on the NTA website, csirnet.nta.nic.in t. Candidtes can use the answer key to cross check the answers and to calculate the probable scores. It also provide an opportunity to the candidates to challenge the Provisional Answer Key of CSIR NET 2019. The Answer Keys is displayed only for two to three days. The Candidates should be given an opportunity to make a challenge online against the Provisional Answer Key on the payment of Rs. 1000/- per question challenged as processing charges.
CSIR UGC NET Exam Pattern 2019
Before starting the preparation candidates should know about the exam pattern because this help the candidates to know about the subject name, exam timing, marks, types of questions etc. From below candidates can check the detailed exam pattern of CSIR NET 2019.
- Mode of an exam – Online (Computer Based Test)
- Duration of the exam – 3 hours (180 mins)
- Examination Language – Hindi, EnglishThe selection will be based on the cutoff marks and the marks secured by the candidates in the CSIR UGC NET 2019 Examination.
- Question type – MCQ
- Total No of marks – The test paper carry a maxium of 200 marks for each subject.
- Parts of Exam – The test will consist of three parts (Part A, Part B and Part C)
The subject-wise scheme of examination are as follows:
Pattern for Chemical Science (701):
- Part A: +2 marks for each correct answer and -0.5 for each incorrect answer.
- Part B: +2 marks for each correct answer and -0.5 for each incorrect answer.
- Part C: +4 marks for each correct answer and -1 for each incorrect answer.
| No. of Parts | Total no of questions | Max no of Questions to attempt |
| Part A | 20 | 15 |
| Part B | 40 | 35 |
| Part C | 60 | 25 |
| Total | 120 | 75 |
Pattern for Earth, Atmospheric, Ocean and Planetary Sciences (702):
- Part A: +2 marks for each correct answer and -0.5 for each incorrect answer.
- Part B: +2 marks for each correct answer and -0.5 for each incorrect answer.
- Part C: +4 marks for each correct answer and -1.32 for each incorrect answer.
| No. of Parts | Total no of questions | Max no of Questions to attempt |
| Part A | 20 | 15 |
| Part B | 50 | 35 |
| Part C | 80 | 25 |
| Total | 150 | 75 |
Pattern for Life Sciences (703):
- Part A: +2 marks for each correct answer and -0.5 for each incorrect answer.
- Part B: +2 marks for each correct answer and -0.5 for each incorrect answer.
- Part C: +4 marks for each correct answer and -1.32 for each incorrect answer.
| No. of Parts | Total no of questions | Max no of Questions to attempt |
| Part A | 20 | 15 |
| Part B | 50 | 35 |
| Part C | 75 | 25 |
| Total | 145 | 75 |
Pattern for Mathematical Sciences (704):
- Part A: +2 marks for each correct answer and -0.5 for each incorrect answer.
- Part B: +3 marks for each correct answer and -0.75 for each incorrect answer.
- Part C: +4.75 marks for each correct answer and 0 for each incorrect answer.
| No. of Parts | Total no of questions | Max no of Questions to attempt |
| Part A | 20 | 15 |
| Part B | 40 | 25 |
| Part C | 60 | 20 |
| Total | 120 | 60 |
Pattern for Physical Sciences (705):
- Part A: +2 marks for each correct answer and -0.5 for each incorrect answer.
- Part B: +3.5 marks for each correct answer and -0.875 for each incorrect answer.
- Part C: +5 marks for each correct answer and -1.25 for each incorrect answer.
| No. of Parts | Total no of questions | Max no of Questions to attempt |
| Part A | 20 | 15 |
| Part B | 25 | 20 |
| Part C | 30 | 20 |
| Total | 75 | 55 |
Syllabus Of CSIR UGC NET 2019
From below candidates will get the full detail of CSIR NET Syllabus 2019. Syllabus helps the candidates to know about
- Part A – This is common to all subjects. Questions will be on General Aptitude with emphasis on logical reasoning, graphical analysis, analytical and numerical ability, quantitative comparison, series formation, puzzles.
- Part B – These will be subject-related conventional multiple choice questions.
- Part C – Here there will be higher value questions to test knowledge of scientific concepts and/or application of scientific concepts. To answer such analytical questions, scientific knowledge has to be applied.
Get Here CSIR NET 2019 Complete Syllabus
Re-Evaluation/Re-Checking of the result: There shall be no re-evaluation/re-checking of the CSIR UGC NET 2019 result. No correspondence in this regard shall be entertained.
CSIR UGC NET 2019 Exam Frequently Asked Questions
Ans : Please search the same in your email for CSIR UGC 2019 Admit Card. If you are not able to find the same in the inbox, try a different approach. Write an email to NET helpdesk- [email protected]
Ans : CSIR NET Exam is conducted for Lectureship and Junior Research Fellowship and the full form for CSIR is Council of Scientific and Industrial Research.
Ans : The career opportunities after clearing CSIR are very vast. You can be Research Fellow at CSIR Research Laboratories, a Lecturer or an Assistant Professor at a government University, a doctorate to your current qualification, work in the public sector or Private Corporate Firms.
Ans : 7525 candidates qualified this exam for the year 2018.
Official website: csirnet.nta.nic.in
For more details regarding CSIR UGC NET 2019 (Dec), download the Information Brochure – English | Hindi.
Mistakenly i applied for lectureship onli but i got jrf cut off..will i get jrf?
I’ve scored 59.467 nta score in joint csir ugc net Dec 2019 do I have any chance of qualifying it ?