CSIR UGC-NET June 2020 Notification Out

National Testing Agency (NTA) has published job posting for CSIR UGC NET. The advertisement for the job post was published on 17 Mar 2020. The minimum educational qualification for the post of CSIR UGC NET is As per subject. The job application may be submitted to National Testing Agency (NTA) latest by 15 Apr 2020

17 Mar 2020

CSIR UGC NET

As per subject

National Testing Agency (NTA)

15 Apr 2020

National Testing Agency (NTA) has released the notification regarding the registration for CSIR UG NET-June 2020 exam. The examinations will be conducted in different cities at the All India Level. Joint CSIRUGC NET is a test being conducted to determine the eligibility of Indian nationals ‘for Junior Research Fellowship (JRF) and for Lectureship (LS) /Assistant Professor in Indian universities and colleges subject to fulfilling the eligibility criteria laid down by UGC. Interested Candidates may apply online from 16 March to 15 April 2020.

CSIR UGC NET-June 2020 Exam Notification Out

The examination will be held in the subjects mentioned below:

  • Chemical Science
  • Earth, Atmospheric, Ocean and Planetary Science
  • Life Sciences
  • Mathematical Sciences
  • Physical Sciences

Educational Qualification

(a) 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.

(b) B.Sc (Hons) or equivalent degree holders or students enrolled in Integrated MS-PhD program with at least 55% marks for General (UR)/General-EWS and OBC candidates; 50% marks for SC/ST, Persons with Disability (PwD) candidates are also eligible to apply.

(c) Candidates with Bachelor’s degree will be eligible for CSIR fellowship only after getting registered/enrolled for Ph.D./Integrated Ph.D. program within the validity period of two years.

(d) Candidates possessing only Bachelor’s degree are eligible to apply only for Junior Research Fellowship (JRF) and not for Lectureship (LS)/ Associate Professorship.

Age Limit

  • Junior Research Fellow (JRF): Max. 28 years as on 01 Jan 2020
  • Lectureship (LS)/Assistant Professor: No upper age limit

JRF Stipend

The stipend of a JRF selected through CSIR- National Eligibility Test (NET) will be Rs.31, 000/- p.m for the first two years. In addition, an annual contingent grant of Rs.20, 000/- per fellow will be provided to the University / Institution.

On completion of two years as JRF and if the Fellow is registered for Ph.D., the Fellowship will be upgraded to SRF (NET) and the stipend will be increased to Rs.35,000/- p.m for the 3rd and subsequent years, on the basis of an assessment of Fellows’ research progress/ achievements through the interview by an Expert Committee consisting of the Guide, Head of the Department and External Member from outside the University/institution who is an expert in the relevant field, not below the rank of Professor/ Associate Professor.

Examination Pattern

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 Science

Part A

Part B

Part C

Total

Total Questions

20

40

60

120

Max. No. of Questions

15

35

25

75

Marks for each correct answer

02

02

04

200

Earth Science

Part A

Part B

Part C

Total

Total Questions

20

50

80

150

Max. No. of Questions

15

35

25

75

Marks for each correct answer

02

02

04

200

Life Science

Part A

Part B

Part C

Total

Total Questions

20

50

75

145

Max. No. of Questions

15

35

25

75

Marks for each correct answer

02

02

04

200

Mathematical Science

Part A

Part B

Part C

Total

Total Questions

20

40

60

120

Max. No. of Questions

15

25

20

60

Marks for each correct answer

02

03

04.75

200

Physical Science

Part A

Part B

Part C

Total

Total Questions

20

25

30

75

Max. No. of Questions

15

20

20

55

Marks for each correct answer

02

03.5

05

200

Examination Schedule

CSIR UGC NET 2020 examination will be going to held on 21st June 2020. The examination schedule is given below:

Date of Examination

21 June 2020

First Shift

Second Shift

Timing of Examination

09:30 AM – 12:30 AM

02:30 PM – 05:30 PM

Duration of Examination

03 Hours without break

Entry into the Examination Centre

07:30 AM – 08:30 AM

12:30 PM – 01:30 PM

Entry in the Examination Hall

08:45 AM – 09:00 AM

01:45 PM – 02:00 PM

Checking of Admit Cards by the invigilator

09:00 AM – 09:15 AM

02:00 PM – 02:15 PM

Sitting on the seats

09:15 AM

02:15 PM

Instruction by the invigilator

09:15 AM – 09:25 AM

02:15 PM – 02:25 PM

Test Commences

09:30 AM

02:30 PM

Test Concludes

12:30 PM

05:30 PM

Application Fees

  • Rs.1000/- for General/EWS Candidates
  • Rs.500/- for OBC-NCL Candidates
  • Rs.250/- for SC/ST Candidates
  • NIL for PwD Candidates

Important Dates

Starting Date of Online Application: 16 March 2020

Last Date of Online Application: 15 April 2020

Downloading of Admit Cards: 15 May 2020

Date of Examination: 21 June 2020

Important Links

Download Official Notification: Click Here

Apply Online: Click Here

CSIR UGC NET-June 2020 Exam Notification Out

CSIR UGC NET Uttarakhand Govt - 2024 Job Vacancy Summary

Job Posted on : 2020-03-17 00:00:00
Last Date : 2020-04-15 23:59:59
Employment Type : FULL_TIME
Qualification : As per subject
Hiring Organization : National Testing Agency (NTA)
Job Location : New Delhi, New Delhi, India 248001
Salary : 0.0 per month (approx)
Posted in : B Tech Jobs, BE Jobs, BSc Jobs, Employment News 2024, Jobs for Graduates, Jobs for Post Graduates, M Sc Jobs, MBBS Jobs, PhD Jobs, Professor Vacancy

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