CSIR UGC NET June 2020 – Notification, Application Form, Eligibility & Syllabus PDF. The Council of Scientific and Industrial Research (CSIR) has entrusted the responsibility of conducting CSIR-UGC NET in CBT mode to NTA. CSIR-UGC NET is a test being conducted to determine the eligibility ‘for Junior Research Fellowship (JRF) and for Lectureship (LS) /Assistant Professorship’ in Indian universities and colleges subject to fulfilling the eligibility criteria laid down by UGC. Online applications for JRF/NET are invited twice a year on all India basis through Press Notification of CSIR-UGC NET for JRF and LS/AP.
CSIR and UGC provide Research Fellowships for training in methods of research under expert guidance of faculty members/scientists working in University Department/ National Laboratories and Institutions in various fields of Science. CSIR-UGC NET Fellowships are tenable in Universities/IITs/Post Graduate Colleges/Govt. Research Establishments including those of the CSIR, Research & Development establishments of recognized public or private sector industrial firms and other recognized institutions. Only bonafide Indian Citizens are eligible for the test. CSIR-UGC NET Fellowship is tenable in India. The programme is aimed at National Science & Technology Human Resource Development.
A large number of JRFs are awarded each year by CSIR to candidates holding BS-4 years program/ BE/B.Tech./B. Pharma/MBBS/ Integrated BS-MS/M.Sc. or Equivalent degree/B.Sc. (Hons) or equivalent degree holders or students enrolled in integrated MS-Ph.D. program with at least 55% marks for General/General-EWS & OBC (50% for SC/ST candidates and Persons with Disability) after qualifying the Test conducted by CSIR twice a year in June and December.
This Test also determines the eligibility of candidates (i.e it is eligibility criteria only) for Lectureship/Assistant Professorship in Indian University/Colleges. Those who qualify for JRF are eligible for Lectureship/Assistant Professorship also, subject to fulfilling the eligibility criteria laid down by UGC. Some aspirants are declared successful in eligibility for Lectureship/Assistant Professorship only based on their performance in the test.
The award of CSIR-UGC NET Fellowship is for a fixed tenure and does not imply any assurance or guarantee for subsequent employment by CSIR-UGC NET to the beneficiary.
CSIR-UGC NET Examination, June 2020
Joint CSIR-UGC NET Examination would be conducted by NTA on Sunday, 21st June, 2020for determining the eligibility of Indian Nationals for the award of Junior Research Fellowships (JRF) and for determining eligibility for Lectureship (LS)/Assistant Professorship in certain subject areas falling under the faculty of Science & Technology.
Upper Age limit:
(a) JRF: Maximum 28 years as on 01.07.2019 {upper age limit may be relaxable up to 5 years in case of SC/ST/Persons with Disability(PwD)/ female applicants and 03 years in case of OBC (Non Creamy Layer) applicants}.
(b) Lectureship (LS)/ Assistant Professorship: No upper age limit.
Eligibility Criteria:
M.Sc. or equivalent degree/ Integrated BS-MS/BS-4 years/BE/B. Tech/B. Pharma/MBBS with at least 55% marks for General (UR)/General-EWS and OBC candidates and 50% for SC/ST, Persons with Disability (PwD) candidates. Candidates enrolled for M.Sc or having completed 10+2+3 years of the above qualifying examination as on the closing date of online submission of Application Form, are also eligible to apply in the above subject under the Result Awaited (RA) category on the condition that they complete the qualifying degree with requisite percentage of marks within the validity period of two years to avail the fellowship.
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Mode of Examination:
The Examination shall be conducted as Computer Based Test (CBT) only
Pattern of Examination:
The Test will consist of three parts. All the parts will consist of objective type, multiple choice questions. There will be no break between papers. The subject-wise scheme of examination is as per details below:
| CHEMICAL SCIENCES | PART A | PART B | PART C | TOTAL |
| Total questions | 20 | 40 | 60 | 120 |
| Max No of Questions to attempt | 15 | 35 | 25 | 75 |
| Marks for each correct answer | 2 | 2 | 4 | 200 |
| Marks for each incorrect answer (Negative marking) | 0.5 | 0.5 | 1 |
| EARTH, ATMOSPHERIC, OCEAN AND PLANETARY SCIENCES | PART A | PART B | PART C | TOTAL |
| Total questions | 20 | 50 | 80 | 150 |
| Max No of Questions to attempt | 15 | 35 | 25 | 75 |
| Marks for each correct answer | 2 | 2 | 4 | 200 |
| Marks for each incorrect answer (Negative marking) | 0.5 | 0.5 | 1.32 |
| LIFE SCIENCES | PART A | PART B | PART C | TOTAL |
| Total questions | 20 | 40 | 75 | 145 |
| Max No of Questions to attempt | 15 | 35 | 25 | 75 |
| Marks for each correct answer | 2 | 2 | 4 | 200 |
| Marks for each incorrect answer (Negative marking) | 0.5 | 0.5 | 1 |
| MATHEMATICAL SCIENCES | PART A | PART B | PART C | TOTAL |
| Total questions | 20 | 40 | 60 | 120 |
| Max No of Questions to attempt | 15 | 25 | 20 | 60 |
| Marks for each correct answer | 2 | 2 | 4.75 | 200 |
| Marks for each incorrect answer (Negative marking) | 0.5 | 0.75 | 0 |
| PHYSICAL SCIENCES | PART A | PART B | PART C | TOTAL |
| Total questions | 20 | 25 | 30 | 120 |
| Max No of Questions to attempt | 15 | 20 | 20 | 55 |
| Marks for each correct answer | 2 | 3.5 | 5 | 200 |
| Marks for each incorrect answer (Negative marking) | 0.5 | 0.875 | 1.25 |
Chemical Sciences Syllabus : The candidate is required to answer a maximum of 15, 35 and 25 questions from Part-A, Part-B and Part-C respectively. If more than required number of questions are answered, only first 15, 35 and 25 questions in Part A, Part B and Part C respectively will be taken up for evaluation. Below each question in Part A, Part B and Part C, four alternatives or responses are given. Only one of these alternatives is the “correct” option to the question. The candidate has to find, for each question, the correct or the best answer.
Earth Sciences: The candidate is required to answer a maximum of 15, 35 and 25 questions from Part-A, Part-B and Part-C respectively. If more than required number of questions are answered, only first 15, 35 and 25 questions in Part A, Part B and Part C respectively will be taken up for evaluation. Below each question in Part A, Part B and Part C, four alternatives or responses are given. Only one of these alternatives is the “correct” option to the question. The candidate has to find, for each question, the correct or the best answer.
Life Sciences Syllabus : The candidate is required to answer a maximum of 15, 35 and 25 questions from Part-A, Part-B and Part-C respectively. If more than required number of questions are answered, only first 15, 35 and 25 questions in Part A, Part B and Part C respectively will be taken up for evaluation. Below each question in Part A, Part B and Part C, four alternatives or responses are given. Only one of these alternatives is the “correct” option to the question. The candidate has to find, for each question, the correct or the best answer.
Mathematical Sciences: The candidate is required to answer a maximum of 15, 25 and 20 questions from Part-A, Part-B and Part-C respectively. If more than required number of questions are answered, only first 15, 25 and 20 questions in Part A, Part B and Part C respectively will be taken up for evaluation. Below each question in Part A, Part B, four alternatives or responses are given. Only one of these alternatives is the “correct” option to the question. The candidate has to find, for each question, the correct or the best answer. In Part C each question may have “ONE” or “MORE” correct options. Credit in a question shall be given only on identification of “ALL” the correct options in Part C.
Physical Sciences: The candidate is required to answer a maximum of 15, 20 and 20 questions from Part-A, Part-B and Part-C respectively. If more than required number of questions are answered, only first 15, 20 and 20 questions in Part A, Part B and Part C respectively will be taken up for evaluation. Below each question in Part A, Part B and Part C, four alternatives or responses are given. Only one of these alternatives is the “correct” option to the question. The candidate has to find, for each question, the correct or the best answer.
Examination Centres -Cities:
The City of Examination Centres where the test shall be conducted are given at Annexure-I. It is mandatory for candidates to select four cities of their choice while filling Online Application Form of CSIR-UGC NET December 2019. Effort will be made to allot Centre of examination to the candidates in order of the City choice opted by them in their Application Form. However, due to administrative reasons, a different city of nearby area may be allotted.
Syllabus of the Test:
Syllabus & Scheme of Examination of the MCQ Paper may be seen at CSIR HRDG website: www.csirhrdg.res.in.
CSIR-UGC (NET) EXAM FOR AWARD OF JUNIOR RESEARCH FELLOWSHIP AND ELIGIBILITY FOR LECTURERSHIP
EXAM SCHEME FOR SINGLE PAPER CSIR-UGC NET Exam
CSIR-UGC NET Exam for Science stream is conducted by CSIR in the following areas: –
The pattern for the Single Paper MCQ test shall be as given below:-
- The MCQ test paper of each subject shall carry a maximum of 200 marks.
- The exam shall be for duration of three hours.
- The question paper shall be divided in three parts
Part ‘A’ shall be common to all subjects. This part shall be a test containing a maximum of 20 questions of General Aptitude. The candidates shall be required to answer any 15 questions of two marks each. The total marks allocated to this section shall be 30 out of 200
Part ‘B’ shall contain subject-related conventional MCQs. The total marks allocated to this section shall be 70 out of 200. The maximum number of questions to be attempted shall be in the range of 20-35.
Part ‘C’ shall contain higher value questions that may test the candidate’s knowledge of scientific concepts and/or application of the scientific concepts. The questions shall be of analytical nature where a candidate is expected to apply the scientific knowledge to arrive at the solution to the given scientific problem. The total marks allocated to this section shall be 100 out of 200.
Negative marking for wrong answers.
Important Dates
| Examination | CSIR-UGC NET Examination(CSIR-UGC NET)-June 2020 |
| Mode of Examination | Computer Based Examination |
| Registration Dates | 16thMarch to 15 th April, 2020 |
| Downloading of Admit Cards | 15th May, 2020 |
| Dates of Examinations | 21st June, 2020 |
| Date of Announcing Result | 5 th July, 2020 |
Instructions for filling Online Application Form:
❖ Log on to NTA website csirnet.nta.nic.in and click on the “Apply Online” button under CSIR-UGC NET December 2019.
❖ Download Information Bulletin and Replica of Application Form. Read these carefully to ensure your eligibility and acquaint yourself with the requirements for submission of Online Application Form.
Step-1: Apply for Online Registration using unique Email Id and Mobile No.
Step-2: Complete Online Application Form and note down the system generated Application No.
Step-3: Upload scanned images of Candidate’s Photograph and Candidate’s Signature in JPG/JPEG format.
Step-4: Make payment of fee through SBI/Syndicate/ HDFC/ ICICI/ Paytm Payment Gateway (Debit/Credit Card or Net Banking/UPI) and keep proof of fee paid.
- Download save and print copy of Confirmation Page after successful remittance of fee for your record and future reference.
❖ For submission of Online Application Form, Steps 1 to 4 must be completed. All Steps are mandatory. ❖ The final submission of Online Application Form will remain incomplete if step 2 and step 3 are not followed through. Such Online Application Forms would stand rejected and no correspondence on this account will be entertained.
❖ Candidate will get Application Number and One Time Password (OTP) message through e-mail and SMS for submission of Online Application Form. (Future Login will be allowed only with the Application Number and Password, so these must be preserved)
❖ Download, save and print four copies of the Confirmation Page for your record and future reference.
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