The result of the examination is
generally declared in the month of April and October respectively. In
conjunction with the result, the notification inviting online application form
for the next exam is also made available.
The procedure and criteria for
declaration of CBSE UGC NET June 2015 examination will comprise of following
steps:
Step I: Minimum marks
to be obtained in NET for considering a candidate for the award of JRF and
eligibility for Assistant Professor:
The
candidates are required to obtain minimum marks separately in Paper-I, Paper-II
and Paper –III as given below:
|
Category
|
Minimum
Marks (%) to be obtained
|
||
|
|
Paper – I
|
Paper – II
|
Paper - III
|
|
General
|
40 (40%)
|
40(40%)
|
75 (50%)
|
|
OBC (Non-creamy layer) PWD/SC/ST
|
35 (35%)
|
35 (35%)
|
60 (40%)
|
Step II: Amongst those
candidates who have cleared step I, a merit list will be prepare subject-wise
and category-wise using the aggregate marks of all the three papers secured by
such candidates.
Step III: Top
15% candidates (for each subject and category), from the merit list mentioned
under step II, will be declared NET qualified for eligibility for Assistant
Professor only.
Step IV: A separate
merit list for the award of JRF will be prepared from amongst the NET qualified
candidates figuring in the merit list prepared under step III.
It may be noted that the above
qualifying criteria decided by UGC/CBSE is final and binding.
