CSIR UGC NET June 2022 Online Form Notification @csirnet.nta.nic.in
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CSIR UGC NET June 2022 Recruitment: CSIR UGC NET Exam 2022 Eligibility, Exam Pattern, Selection Process, CSIR UGC NET Syllabus and Apply Link 2022 @csirnet.nta.nic.in
Council of Scientific & Industrial Research (CSIR) has given a Notification for the Recruitment of CSIR UGC NET Junior Research Fellowship (JRF) & Lectureship (LS) Examination June 2022. Those Candidates are interested in the Following Recruitment complete the Required Eligibility Criteria for CSIR UGC NET Exam 2022, Can read the Full Notification and Apply Online.
Candidates should have possess 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, PwD candidates.
Subjects of the Test
Subject Code
Subject
701
Chemical Sciences
702
Earth, Atmospheric, Ocean and Planetary Sciences
703
Life Sciences
704
Mathematical Sciences
705
Physical Sciences
CSIR UGC NET June 2022 Selection Process
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:
Topic
Part A
Part B
Part C
Total
Chemical Sciences
Total Questions
20
40
60
120
Max No. of Qs. Attempt
15
35
25
75
Marks for Each Answer
2
2
4
200
Negative Marks
0.5
0.5
1
–
Earth, Atmospheric, Ocean and Planetary Sciences
Total Questions
20
50
80
150
Max No. of Qs. Attempt
15
35
25
75
Marks for Each Answer
2
2
4
200
Negative Marks
0.5
0.5
1.35
–
Life Sciences
Total Questions
20
50
75
145
Max No. of Qs. Attempt
15
35
25
75
Marks for Each Answer
2
2
4
200
Negative Marks
0.5
0.5
1
–
Mathematical Sciences
Total Questions
20
40
60
120
Max No. of Qs. Attempt
15
25
20
60
Marks for Each Answer
2
3
4.75
200
Negative Marks
0.5
0.75
0
–
Physical Sciences
Total Questions
20
25
30
75
Max No. of Qs. Attempt
15
20
20
55
Marks for Each Answer
2
3.5
5
200
Negative Marks
0.5
0.875
1.25
–
Interested Candidate Can Read the Full Notification Before Apply Online.