Council Of Scientific and Industrial Research CSIR Online Are Invited Online Application Form for the NET / JRF Examination 2018. Those Candidates Are Interested in the Following Vacancy and Completed the All Eligibility Criteria Can Read the Full Notification and Apply Online.
Important Dates
Application Begin | 25/02/2019. |
Last Date for Complete Form | 22/03/2019. |
Date of Admit Card. | June 2019 |
Date of Examination | 16/09/2019. |
Date of Result. | Notified Soon |
Application Fees
General | Rs.1000/-. |
OBC | Rs.500/-. |
SC / ST / PH | Rs.250/-. |
Mode of Payment | Application Fee Can be Submitted through E-Challan Making Payment at Any Branch of Indian Bank. |
CSIR UGC NET / JRF June 2019 Examination Online Form Subject1. Chemical Science. | 2. Earth, Atmospheric, Ocean and Planetary Sciences. | 3. Life Sciences. | 4. Mathematical Sciences. | 5. Physical Sciences. |
CSIR awards Junior Research Fellowship (JRF) to a large number of candidates each year. Educational background of candidates includes BS-4 years program, BE, B.Tech, B.Pharma, MBBS, Integrated BS-MS, MSc or equivalent, B.Sc Hons. |
Educational Qualification for CSIR UGC NET / JRF June 2019 Examination Online Form
UGC NET / JRF June 2019 ExaminationEducational Qualifications | Candidate Must Have Passed M.Sc or Equivalent Degree / BE / B.Tech / B.Pharma / MBBS with at least 55% Marks (50% for SC / ST and PH Candidates) or Integrated Course and B.E/ B.Tech / B.Pharma and MBBS Candidates Are Also Eligible for CSIR NET 2018. |
For More Details Please Check the Detailed Brochure. |
Age Limit Details of CSIR UGC NET / JRF June 2019 Examination Online Form |
As on 01/01/2019Maximum Age for JRF | 28 Years. | Maximum Age for NET | No Age Limit. |
Age Relaxation As Per Govt Rules. |
Selection Process of CSIR UGC NET / JRF June 2019 Examination Online Form
Selection May be Based on.
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CSIR UGC NET / JRF June 2019 Examination Online Form SyllabusJoint CSIR-UGC NET examination syllabus is formed keeping in mind the purpose of this exam. Syllabus of CSIR NET is divided into three parts namely Part A, Part B, Part C.- 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.
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CSIR UGC NET / JRF June 2019 Exam Pattern |
Joint CSIR-UGC NET will be a single paper with multiple choice questions. Candidate will appear in any one subject. It is a 3 hours duration exam of 200 marks.Subjects | - Chemical sciences
- Earth, Atmospheric, Ocean and Planetary Sciences
- Life sciences
- Mathematical Sciences
- Physical sciences
| Marks | 200 | Duration of exam | 3 hours | Mode of exam | Pen and paper based | Languages in which question paper will be available | Hindi, English | Question type | MCQs | Morning session timings | 9 am to 12 noon | Afternoon session timings | from 2 pm to 5 pm |
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How to Apply about CSIR UGC NET / JRF June 2019 Examination Online Form 1. The Candidates can Apply through Online mode. | 2. Click Apply Online Link is given on Our Website in Important link Section. | 3. Candidates are required to Register before Applying for the Post and Already
Register Candidate Not needs to register Again. | 4.After Successful Registration system will Generate a Login ID and Password on your Screen. | 5. Login with the ID and Password then Start Filling Your Application form and Read all the instruction before Applying for the Post. | 6. Upload the Scanned Copy of Documents in Prescribed Size and format. | 7. Take Printout of your Application form for future Reference. | 8. Online Application can be Submitted on or Before 22nd March 2019. |
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Important Links to Apply for CSIR UGC NET / JRF June 2019 Examination Online Form |